The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
Well, OK, Professor Frey invited me to give the two lectures this week on first order equations, like that one, first order dy dt.
And the lectures next week will be on second order equation.
So we're looking for, you could say, formulas for the solution.
We'll get as far as we can with formulas, then numerical methods.
Graphical methods take over in more complicated problems.
This is a model problem.
It's linear.
I chose it to have constant coefficient a, and let me check the units.
Always good to see the units in a problem.
So let me think of this y, as the money in a bank, or bank balance, so y as in dollars, and t, time, in years.
So we're looking at the ups and downs of bank balance y.
The rate of change, so the units then are dollars per year.
So every term in the equation has to have the right units.
So y is in dollars, so the interest rate a is percent per year, say 6% a year.
So a could be 6%-- that's dimensionless-- per year, or half a percent per month if we change.
So if we change units, the constant a would change from 6 to a half.
But let's stay with 6.
And then q of t represents deposits and withdrawals, so that's in dollars per year again.
Has to be.
So that's continuous.
We think of the deposits and the interest as being computed continuously as time goes forward.
So if that's a constant-- and I'll take that case first, q equal 1-- that would mean that we're putting in, depositing $1 per year, continuously through the year.
So that's the model that comes from a differential equation.
A difference equation would give us finite time steps.
So I'm looking for the solution.
And with constant coefficients, linear, we're going to get a formula for the solution.
I could actually deal with variable interest rate for this one first order equation, but the formula becomes messy.
But you can still do it.
After that point, for a second order equations like oscillation, or for a system of several equations coupled together, constant coefficients is where you can get formulas.
So let's go with that case.
So how to solve that equation?
Let me take first of all, a constant, constant source.
So I think of q as the source term.
To get one nice formula, let me take this example, ay plus 1, let's say.
How do you find y of t to solve that?
And you start with some initial condition y of 0.
That's the opening deposit that you make at time 0.
How to solve that equation?
Well, we're looking for a solution.
And solutions to linear equations have two parts.
So the same will happen in linear algebra.
One part is a solution to that equation, so we're just looking for one, any one, and we'll call it a particular solution.
And the associated null equation, dy dt equal ay.
So this is an equation with q equals 0.
That's why it's called null.
And it's also called homogeneous.
So more textbooks use that long word homogeneous, but I use the word null because it's shorter and because it's the same word in linear algebra.
So let me call yn the null solution, the general null solution.
And y, I'm looking here for a particular solution yp, and I'm going to-- here's the key for linear equations.
Let me take that off and focus on those two equations.
How does solving the null equation, which is easy to do, help me?
Why can I, as I plan to do, add in yn to yp?
I just add the two equations.
Can I just add those two equations?
I get the derivative of yp plus yn on the left side.
And I have a times yp plus yn.
And that is a critical moment there when we use linearity.
I had a yp a yn, and I could put them together.
If it was y squared, yp squared and yn squared would not be the same as yp plus yn squared.
It's the linearity that comes, and then I add the 1.
So what do I see from this?
I see that yp plus yn also solves my equation.
So the whole family of solutions is 1 yp plus any yn.
And why do I say any yn?
Because when I find one, I find more.
The solutions to this equation are yn could be e to the at, because the derivative of e to the at does bring down a factor a.
But you see, I've left space for any multiple of e to the at.
This is where that long word homogeneous comes from.
It's homogeneous means I can multiply by any constant, and I still solve the equation.
And of course, the key again is linear.
So now I have-- well, you could say I've done half the job.
I've found yn, the general yn.
And now I just have to find one yp, one solution to the equation.
And with this source term, a constant, there's a nice way to find that solution.
Look for a constant solution.
So certain right hand sides, and those are the like the special functions for the special source terms for differential equation, certain right hand sides-- and I'm just going to go down a list of them today.
The next one on the list-- can I tell you what the next one on the list will be?
y prime equal ay.
I use prime for-- well, I'll write dy dt, but often I'll write y prime.
dy dt equal ay plus an exponential.
That'll be number two.
So I'm just preparing the way for number two.
Well, actually number one, this example is the same as that exponential example with exponent s equal 0, right?
If s is 0, then I have a constant.
So this is a special case of that one.
This is the most important source term in the whole subject.
But here we go with a constant 1.
So we've got yn.
And what's yp?
I just looked to see.
Can I think of one?
And with these special functions, you can often find a solution of the same form as the source term.
And in this case, that means a constant.
So if yp is a constant, this will be 0.
So I just want to pick the constant that makes this thing 0.
And of course, their right hand side is 0 when yp is minus 1 over a.
So I've got it.
We've solved that equation, except we didn't match the initial condition yet.
Let me if you take that final step.
So the general y is any multiple, any null solution, plus any one particular solution, that one.
And we want to match it to y of 0 at t equals 0.
So I want to take that solution.
I want to find that constant, here.
That's the only remaining step is find that constant.
You've done it in the homework.
So at t equals 0, y of 0 is-- at t equals 0, this is C. This is the minus 1 over a.
So I learn what the C has to be.
And that's the final step.
C is bring the 1 over a onto that side, so C will be y of 0 e to the at minus 1 over a e to at.
And here we had a minus 1 over a.
Well, it'll be plus 1 over a e to the at.
So now I've just put in the C, y of 0 plus 1 over a. y of 0 plus 1 over a has gone in for C. And now I have to subtract this 1 over a.
Here, I see a 1 over a, so I can do it neatly.
Got a solution.
We can check it, of course.
At t equals 0, this disappears, and this is y of 0.
And it has the form.
It's a multiple of e to the at and a particular solution.
So that's a good one.
Notice that to get the initial condition right, I couldn't take C to be y of 0 to get the initial condition right.
To get the initial condition right, I had to get that, this minus 1 over a in there.
Good for that one?
Let me move to the next one, exponentials.
So again, we know that the null equation with no source has this solution e to the at.
And I'm going to suppose that the a in e to the at in the null solution is different from the s in the source function, which will come up in the particular solution.
So you're going to see either the st in the particular solution and an e to the at in the null solution.
And in the case when s equals a, that's called resonance, the two exponents are the same, and the formula changes a little.
Let's leave that case for later.
How do I solve this?
I'm looking for a particular solution because I know the null solutions.
How am I going to get a particular solution of this equation?
Fundamental observation, the key point is it's going to be a multiple of e to the st.
If an exponential goes in, then that will be an exponential.
Its derivative will be an exponential.
I'll have e to the st's everywhere.
And I can get the number right.
So I'm looking for y try.
So I'll put try, knowing it's going to work, as some number times e to the st.
So this would be like the exponential response.
Response, do you know that word response?
So response is the solution.
The input is q, and the response is Y.
And here, the input is e to the st, and the response is a multiple of e to the st.
So plug it in.
The timed derivative will be Y.
Taking the derivative will bring down a 1. e to the st equals aY.
A aY e to the st plus 1 e to the st. Just what we hoped.
The beauty of exponentials is that when you take their derivatives, you just have more exponential.
That's the key thing.
That's why exponential is the most important function in this course, absolutely the most important function.
So it happened here.
I can cancel e to the st, because every term has one of them.
So I'm seeing that-- what am I getting for Y?
Getting a very important number for Y.
So I bring aY onto this side with sY.
On this side I just have a 1.
Maybe it's worth putting on its own board.
Y is, so Ys aY comes with a minus, and the 1, 1 over-- so Y was multiplied by s minus a.
That's the right quantity to get a particular solution.
And that 1 over s minus a, you see why I wanted s to be different from a. I If s equaled a in that case, in that possibility of resonance when the two exponents are the same, we would have 1 over 0, and we'd have to look somewhere else.
The name for that-- this has to have a name because it shows up all the time.
The exponential response function, you could call it that.
Most people would call it the transfer function.
So any constant coefficient linear equation's going to have a transfer function, easy to find.
Everything easy, that's what I'm emphasizing, here.
Everything's straightforward.
That transfer function tells you what multiplies the exponential.
So the source was here.
And the response is here, the response factor, you could say, the transfer function.
Multiply by 1 over s minus a.
So if s is close to a, if the input is almost at the same exponent as the natural, as the null solution, then we're going to get a big response.
So that's good.
For a constant coefficient problem second order, other problems we can find that response function.
It's the key function.
It's the function if we have, or if we were to look at Laplace transforms, that would be the key.
When you take Laplace transforms, the transfer function shows up.
Then when you take inverse Laplace transforms, you have to find what function has that Laplace transform.
So did we get the-- we got the final answer then.
Let me put it here.
y is e to the st times this factor.
So I divide by s minus a.
A nice solution.
Let me also anticipate something more.
An important case for e to the st is e to the i omega t. e to the st, we think about as exponential growth, exponential decay.
But that's for positive s and negative s. And all important in applications is oscillation.
So coming, let me say, coming is either late today or early Wednesday will be s equal i omega, so where the source term is e to the i omega t. And alternating, so this is electrical engineers would meet it constantly from alternating voltage source, alternating current source, AC, with frequency omega, 60 cycles per second, for example.
Why don't I just deal with this now?
Because it involves complex numbers.
And we've got to take a little step back and prepare for that.
But when we do it, we'll get not only e to the i omega t, which I brought out, but also, it's real part.
You remember the great formula with complex numbers, Euler's formula, that e to the i omega t is a combination of cosine omega t, the real part, and then the imaginary part is sine omega t. So this is looking like a complex problem.
But it actually solves two real problems, cosine and sine.
Cosine and sine will be on our short list of great functions that we can deal with.
But to deal with them neatly, we need a little thought about complex numbers.
So OK if I leave e to the i omega t for the end of the list, here?
So I'm ready for another one, another source term.
And I'm going to pick the step function.
So the next example is going to be dy dt equals ay plus a step.
Well, suppose I put H of t there.
Suppose I put H of t. And I ask you for the solution to that guy.
So that step function, its graph is here.
It's 0 for negative time, and it's 1 for positive time.
So we've already solved that problem, right?
Where did I solve this equation?
This equation is already on that board.
Because why?
Because H of t is for t positive.
That's the only place we're looking.
This whole problem, we're not looking at negative t. We're only looking at t from 0 forward.
And what is H of t from 0 forward?
It's 1.
It's a constant.
So that problem, as it stands, is identical to that problem.
Same thing, we have a 1.
No need to solve that again.
The real example is when this function jumps up at some later time T. Now I have the function is H of t minus T. Do you see that, why the step function that jumps at time T has that formula?
Because for little t before that time, in here, this is-- what's the deal?
If little t is smaller than big T, then t minus T is negative, right?
If t is in here, then t minus capital T is going to be a negative number.
And H of a negative number is 0.
But for t greater than capital T, this is a positive number.
And H of a positive number is 1.
Do you see how if you want to shift a graph, if you want the graph to shift, if you want to move the starting time, then algebraically, the way you do it is to change t to t minus the starting time.
And that's what I want to do.
So physically, what's happening with this equation?
So it starts with y of 0 as before.
Let's think of a bank balance and then other things, too.
If it's a bank balance, we put in a certain amount, y of 0.
We hope.
And that grew.
And then starting at time, capital T, this switch turns on.
Actually, physically, step function is really often describing a switch that's turned on, now.
This source term act begins to act at that time.
And it acts at 1.
So at time capital T we start putting money into our account.
Or taking it out, of course.
If this with a minus sign, I'd be putting money in.
Sorry, I would start with some money in, y of 0.
I would start with money in.
Yeah, actually, tell me what's the solution to this equation that starts from y of 0?
What's the solution up until the switch is turned on?
What's the solution before this switch happens, this solution while this is still 0?
So let's put that part of the answer down.
This is for t smaller than T. What's the answer?
This is all common sense.
It's coming fast, so I'm asking these questions.
And when I asked that question, it's a sort of indication that you can really see the answer.
You don't need to go back to the textbook for that.
What have we got here?
Yeah?
Is it the null solution [INAUDIBLE]?
It'll be this guy.
Yeah, the particular solution will be 0.
Right, the particular solution is 0 before this is on.
I'm sorry, the null solution is 0, and the particular solution, well, the particular solution is a guy that starts right.
I don't know.
Those names were not important.
And then the question is-- so it's just our initial deposit growing.
Now, all I ask, what about after time T?
What about after time T?
For t after time T, and hopefully, equal time T, what do you think y of t will be?
Again, we want to separate in our minds the stuff that's starting from the initial condition from the stuff that's piling up because of the source.
So one part will be that guy.
I haven't given the complete answer.
But this is continuing to grow.
And because it's linear, we're always using this neat fact that our equation is linear.
We can watch things separately, and then just add them together.
So I plan to add this part, which comes from initial condition to a part that-- maybe we can guess it-- that's coming from the source.
And how do we have any chance to guess it?
Only because that particular source, once it's turned on, jumps to a constant 1, and we've solved the equation for a constant 1.
Let me go back here.
I think our answer to this question-- so this is like just first practice with a step function, to get the hang of a step function.
So I'm seeing this same y of 0 e to the at in every case, because that's what happens to the initial deposit.
I'll say grow, assuming the bank's paying a positive interest rate.
And now, where did this term comes from?
What did that term represent?
The money that [INAUDIBLE]
The money that, yeah?
They had each of [INAUDIBLE]
The money that came in and grew.
It came in, and then it grew by itself, grew separately from that these guys.
So the initial condition is growing along.
And the money we put in starts growing.
Now, the point is what?
That over here, it's going to look just like that.
So I'm going to have a 1 over a.
And I'm going to have something like that.
But can you just guess what's going to go in there?
When I write it down, it'll make sense.
So this term is representing what we have at time little t, later on, from the deposits we made, not the initial one, but the source, the continuing deposits.
And let me write it.
It's going to be a 1 over a e to the a something minus 1.
It's going to look just like that guy.
When I say that guy, let me point to it again-- e to the at minus 1.
But it's not quite e to the at minus 1.
What is it?
t minus [INAUDIBLE]
t minus capital T, because it didn't start until that time.
So I'm going to leave that as, like, reasonable, sensible.
Think about a step function that's turned on a capital time T. Then it grows from that time.
Of course, mentally, I never write down a formula like that without checking at t equal to T, because that's the one important point, at t equal capital T. What is this at t equal capital T?
It's 0.
At t equal capital T, this is e to the 0, which is 1 minus 1 altogether 0.
And is that the right answer?
At t equal capital T is 0, should I have nothing here?
Yes?
No?
Give me a head shake.
Should I have nothing at t equal capital T?
I've got nothing.
e to the 0 minus 1, that's nothing?
Yes, yes that's the right thing.
Because at capital T, the source has just turned on, hasn't had time to build up anything, just that was the instant it turned on.
So that's a step function.
A step function is a little bit of a stretch from an ordinary function, but not as much of a stretch as its derivative.
In a way, this is like the highlight for today, coming up, to deal with not only a step function, but a delta function.
I guess every author and every teacher has to think am I going to let this delta function into my course or into the book?
And my answer is yes.
You have to do it.
You should do it.
Delta functions are-- they're not true functions.
As we'll see, no true function can do what a delta function does.
But it's such an intuitive, fantastic model of things happening over a very, very short time.
We just make that short time into 0.
So we're saying with the delta function, we're going to say that something can happen in 0 time.
Something can happen in 0 time.
It's a model of, you know, when a bat hits a ball.
There's a very short time.
Or a golf club hits a golf ball.
There's a very short time interval when they're in contact.
We're modeling that by 0 time, but still, the ball gets an impulse.
Normally, for 0 time, if you're doing things continuously, what you do over 0 time is no importance.
But we're not doing things continuously, at all.
So here we go.
You've seen this guy, I think.
But if you haven't, here's the time to see it.
So the delta function is the derivative of-- so I've written three important functions up here.
Let me start with a continuous one.
That function, the ramp is 0, and then the ramp suddenly ramps up to t. Take its derivative.
So the derivative, the slope of the ramp function is certainly 0 there.
And here, the slope is 1.
So the slope jumped from 0 to 1.
The slope of the ramp function is the step function.
Derivative of ramp equals step.
Why don't I write those words down?
Derivative of ramp equals step.
So there is already the step function.
In pure calculus, the step function has already got a little question mark.
Because at that point, the derivative in a calculus course doesn't exist, strictly doesn't exist, because we get a different answer 0 on the left side from the answer, 1 on the right side.
We just go with that.
I'm not going to worry about what is its value at that point.
It's 0 up for t negative, and it's 1 for t positive.
And often, I'll take it 1 for t equals 0, also.
Usually, I will.
That's the small problem.
Now, the bigger problem is the derivative of the-- so this is now the derivative of the step function.
So what's the derivative of this step function?
Well, the derivative along there is certainly 0.
The derivative along here is certainly 0.
But the derivative, when that jumped, the derivative, the slope was infinite.
That line is vertical.
Its slope is infinite.
So at that one point, you have an affinity, here, delta of 0.
You could say delta of 0 is infinite.
But you haven't said much, there.
Infinite is too vague.
Actually, I wouldn't know if you gave me infinite or 2 times infinite.
I couldn't tell the difference.
So I'll put it in quotes, because it sort of gives us comfort.
But it doesn't mean much.
What does mean much?
Somehow that's important.
Can I tell you how to work with delta functions, how to think about delta functions?
It's the right way to think about delta function.
So here's some comment on delta function.
Giving the values of the function, 0, and infinity, and 0, is not the best.
What you can do with a delta function is you can integrate it.
You can define the function by integrals.
Integrals of things are nice.
Do you think in your mind when you take derivatives, as we did going left to right, we were taking derivatives.
The function was getting crazy.
When we go right to left, take integrals, those are smoothing.
Integrals make functions smoother.
They cancel noise.
They smooth the function out.
So what we can do is to take the integral of the delta function.
We could take it from any negative number to any positive number.
And what answer would we get?
What would be the right, well, the one thing people know about the delta function is-- and actually, it's the key thing-- the integral of the delta function.
Again, I'm integrating the delta function from some negative number up to some positive number.
And it doesn't matter where n is, because the function is 0 there and there.
But what's the answer here?
Put me out of my misery.
Just tell me the number I'm looking for, here, the integral of the delta function.
Or maybe you haven't met it
[INAUDIBLE]
It's?
It's the one good number you could guess.
It's 1.
Now, why is it 1?
Because if the delta function is the derivative of the step function, this should be the step function evaluated between N and P. This should be the step function, , here, minus the step function, there And what is the step function?
You have to keep it straight.
Am I talking about the delta function?
No, right now, I've integrated it to get H of t. So this is H of P at the positive side, minus H of N. That's what integration's about.
And what do I get?
1, because H of P, the step function here, H is 1.
And here, it's 0, so I get 1.
Good, that's the thing that everybody remembers about the delta function.
And now I can make sense out of 2 delta function, 2 delta of t. That could be my source.
So if 2 delta of t was my source, what's the graph of 2 delta of t?
Again, it's 0 infinite 0.
You really can't tell from the infinity what's up, but what would be the integral of 2 delta of t, the integral of 2 delta of t or some other?
Well, let me put in the 2, here?
What's the integral of 2 delta of t, would be 2H of t. Keep going.
What do I get here?
2
It would be 2 of these guys, 2 of these, 2 of these, 2.
All right?
So we made sense out of the strength of the impulse, how hard the bat hit the ball.
But of course, we need units in there.
We have to have units.
And here, the value for that unit was 2.
Now, I'm going to-- because this is really worth doing with delta functions.
I didn't ask at the start have you seen them before.
But they are worth seeing.
And they just take a little practice.
But then in the end, delta functions are way easier to work with than some complicated function that attempts to model this.
We could model that by some Gaussian curve or something.
All the integrations would become impossible right away.
We could model it by a step function up and a step function down.
Then the integrations would be possible.
But still, we have this finite width.
I could let that width go to 0 and let the height go to infinity.
And what would happen?
I'd get the delta function.
So that's one way to create a delta function, if you like.
If you're OK with step functions, then one way to create delta is to take a big step up, step down, and then let the size of the step grow and the width of the steps shrink.
Keep the area 1, because area is integral.
So I keep this, that little width, times this big height equal to 1.
And in the end, I get delta.
Now again, my point is that delta functions, that you really understand them.
What you can legitimately do with them is integrate them.
But now in later problems, we might have not a 1 or a 2, but a function in here, like cosine t, or e to the t, or q of t. Can I practice with those?
Can I put in a function f of t?
I didn't leave enough space to write f of t, so I'm going to put it in here.
f of t delta of t dt.
And I'm going to go for the answer, there.
My question is what does that equal?
You see what the question is?
I got my delta function, which I only just met.
And I'm multiplying it by some ordinary function.
f of t gives us no problems.
Think of cosine t. Think of e to the t. What do you think is the right answer for that?
What do you think is the right answer?
And this tells you what the delta function is when you see this.
What do I need to know about f of t to get an answer, here?
Do I need to know what f is at t equals minus 1?
You could see from the way my voice asked that question that the answer is no.
Why do I not care what f is at minus 1?
Yeah?
Because you're multiplying by [INAUDIBLE]
Because I'm multiplying by somebody that's 0.
And similarly, at f equal minus 1/2, or at f equal plus 1/3, all those f's make no difference, because they're all multiplying 0.
What does make a difference?
What's the key information about f that does come into the answer?
f at?
At just at that one point, f at?
[INAUDIBLE]
0, f at 0 is the action.
The impulse is happening.
The bat's hitting the ball.
So we're modeling rocket launching, here.
We're launching in 0 seconds instead of a finite time.
So in other words, well, I don't know how to put this answer down other than just to write it.
I guess I'm hoping you're with me in seeing that what it should be.
Can I just write it?
All that matters is what f is at t equals 0, because that's where all the action is.
And that f of 0, if f of 0 was the 2 that I had there a little while ago, then the answer will be 2.
If f of 0 is a 1, if the answer is f of 0 times 1-- and I won't write times 1.
That's ridiculous.
Now we can integrate delta functions, not just a single integral of delta, but integral of a function, a nice function times delta.
And we get f of 0.
So can I just, while we're on the subject of delta functions, ask you a few examples?
What is the integral of e to the t delta of t dt?
It's 1
Yeah, say it again?
It's 1
It's 1.
It's 1, right.
Because e to the t, at the only point we care about, t equal 0 is 1.
And what if I change that to sine t?
Suppose I integrate sine t times delta of t?
What do I get now?
I get?
0
0, right.
And actually, that's totally reasonable.
This is a function, which is yeah, it's an odd function.
Anyway, sine, if I switch t to negative t, it goes negative.
0 is the right answer.
Let me ask you this one.
What about delta of t squared?
Because if we're up for a delta function, we might square it.
Now we've got a high-powered function, because squaring this crazy function delta of t gives us something truly crazy.
And what answer would you expect for that?
1
Would you expect 1?
So this is like?
I'm just getting intuition working, here, for delta functions.
What do you think?
I'm looking at the energy when I square something.
OK, so we had a guess of 1.
Is there another guess?
Yeah?
A third?
Sorry?
1/3
1/3, that's our second guess.
I'm open for other guesses before I-- OK, we have a rule here for f of t. And now what is the f of t that I'm asking about in this case?
It's delta of t, right?
If f of t is delta of t, then that would match this.
And therefore, the answer should match.
Do you see what I'm shooting for, yeah?
It'd be infinity?
It'd be infinity.
It would be infinity.
That's delta of t squared is that's an infinite energy function.
You never meet it, actually.
I apologize, so so write it down there.
I could erase it right away because you basically never see it.
It's infinite energy.
Well, I think you'd see it.
I mean, we're really going back to the days of Norbert Wiener.
When I came to the math department, Norbert Wiener was still here, still alive, still walking the hallway by touching the wall and counting offices.
And hard to talk to, because he always had a lot to say.
And you got kind of allowed to listen.
So anyway, Wiener was among the first to really use delta functions, successfully use delta functions.
Anyway, this is the big one.
This is the big one.
Now, so what's all that about?
I guess I was trying to prepare by talking about this function prepare for the equation when that's the source.
So dy equal ay plus a delta function.
Let me bring that delta function in at time T. So how do you interpret that equation?
So like part of this morning's lecture is to get a first handle on an impulse.
So let me write that word impulse, here.
Where am I going to write it?
So delta is an impulse.
That's our ordinary English word for something that happens fast.
And y of t is the impulse response.
And this is the most important.
Well, I said e to the st was the most important.
How can I have two most important examples?
Well, they're a tie, let's say.
e to the st is the most important ordinary function.
It's the key to the whole course.
Delta of t, the impulse, is the important one because if I can solve it for a delta function, I can solve it for anything.
Let's see if we can solve it for a delta function, a delta function, an impulse that starts at time T. Again, I'm just going to start writing down the solution and ask for your help what to write next.
So what do you expect as a first term in the solution?
So I'm starting again from y of 0.
Let's see if we can solve it by common sense.
So how do I start the solution to this?
Everybody sees what this equation is saying.
I have an initial deposit of y of 0 that starts growing.
And then at time capital T I make a deposit.
At that moment, at that instant, I make a deposit of 1.
That's an instant deposit of 1.
Which is, of course, what I do in reality.
I take $1 to the bank.
They've got it now.
At time T, I give them that $1, and it starts earning interest.
So what about y of t?
What do you think?
What's the first term coming from y of 0?
So the term coming from y of 0 will be y of 0 to start with, e to at.
That takes care of the y of 0.
Now, I need something.
It's like this, plus I need something that accounts for what this deposit brings.
So up until time T, what do I put?
So this is for t smaller than T and t bigger than T. So what goes there?
For t smaller than T, what's the benefit from the delta function?
0, didn't happen yet.
For t bigger than T, what's the benefit from the delta function?
[INAUDIBLE]
For t bigger than T, well, that's right.
OK, but now I've made that deposit at time capital T. Whatever's going there is whatever I'm getting from that deposit.
At time capital T, I gave them $1, and they start paying interest on it.
What's going to go there?
So if I gave them $1 at that initial time, so that $1 would have been part of y of 0.
What did I get at a later time?
e to the at.
Now I'm waiting.
I'm giving them the dollar at time capital T, and it starts growing.
So what do I have at a later time, for t later than capital T?
What has that $1 grown into?
e to the a times the-- right, it's critical.
It's the elapsed time.
It's the time since the deposit.
Is that right?
So what do I put here?
t minus capital T?
t minus capital T, good.
Apologies to bug you about this, but the only way to learn this stuff from a lecture is to be part of it.
So I constantly ask you, instead of just writing down a formula.
I think that looks good.
So suddenly, what does this amount to at t equal capital T?
Maybe I should allow t equal capital T. At t equal capital T, what do I have here?
1
1.
That's my $1.
At t equal capital T, we've got $1.
And later it's grown.
So we have now solved.
We have found the impulse response.
We have found the impulse response when the impulse happened at capital T. That was good going.
Now, I've given you my list of examples with the pause on the sine and cosine.
I pause on the sine and cosine because one way to think about sine and cosine is to get into complex numbers.
And that's really for next time.
But apart from that, we've done all the examples, so are we ready?
Oh yeah, I'm going to try for the big thing, the big formula.
So this is the key result of section 1.4, the solution to this equation.
So I'm going back to the original equation.
And just see if we can write down a formula for the answer.
So let me write the equation again.
dy dt is ay plus some source.
I think we can write down a formula that looks right.
And we could then actually plug it in and see, yeah, it is right.
So what's going to go into this formula?
We got enough examples, so now let's go for the whole thing.
So y of t, first of all, comes whatever depends on the initial condition.
So how much do we have at a later time when our initial deposit was y of 0?
So that's the one we've seen in every example.
Every one of these things has this term growing out of y of 0.
So let me put that in again.
So the part that grows out of y of 0 is y of 0 e to the at.
That's OK.
So that's what the initial.
So our money is coming from two sources, this initial deposit, which was easy, and this continuous, over time deposit, q of t. And I have to ask you about that.
That's going to be like the particular solution, the particular solution that comes from the source term.
This is the solution it comes from the initial condition.
So what do you think this thing looks like?
I just think once we see it, we can say, yeah, that makes sense.
So now I'm saying what?
If we've deposited q of t in varying amounts, maybe a constant for a while, maybe a ramp for awhile, maybe whatever, a step, how am I going to think about this?
So at every time t equal to s, so I'm using little t for the time I've reached.
Right?
Here's t starting at 0.
Now, let me use s for a time part way along.
So part way along, I input.
I deposit q of s. I deposit it at time s. And then what does it do?
That money is in the bank with everybody else.
It grows along with everything else.
So what's the growth factor?
What's the growth factor?
This is the amount I deposited at time s. And how much has it grown at time t?
This is the key question, and you can answer it.
It went in a time s. I'm looking at time t. What's the factor?
Is it e to the a t minus s
e to the a t minus s. So that's the contribution to our balance at time t from our input at time s. But now, I've been inputting all the way along.
s is running all the way from here to here.
So finish my formula.
Put me out of my misery.
Or it's not misery, actually.
Its success at this moment.
What do I do now?
I?
Integrate
I integrate, exactly.
I integrate.
I integrate.
So all these deposits went in.
They grew that amount in the remaining time.
And I integrate from 0 up to the current time t. So you see that formula?
Have a look at it.
This is a general formula, and every one of those examples could be found from that formula.
If q of s was 1, that was our very first example.
We could do that integration.
If q of s was e to the-- anyway, we could do every one.
I just want you to see that that formula makes sense.
Again, this is what grew out of the initial condition.
This is what grew out of the deposit at time s. And the whole point of calculus, the whole point of learning [?
1801 ?
], the integral equation part, the integrals part, is integrals just add up.
This term just adds up all the later deposits, times the growth factor in the remaining time.
And as I say, if I took q of s equal 1-- the examples I gave are really the examples where you can do the integral.
If q of s is e to the i omega s, I can do that integral.
Actually, it's not hard to do because e to the at doesn't depend on s. I can bring an e to the at out in this case.
That formula is just worth thinking about.
It's worth understanding.
I didn't, like, derive it.
And the book does, of course.
There's something called an integrating factor.
You can get at this formula systematically.
I'd rather get at it and understand it.
I'm more interested in understanding what the meaning of that formula is than the algebra.
Algebra is just a goal to understand, and that's what I shot for directly.
And as I say, that the book also, early section of the book, uses practice in calculus.
Substitute that in to the equation.
Figure out what is dy dt.
And check that it works.
It's worth actually looking at that end of what you need to know from calculus It's is.
You should be able to plug that in for y and see that solves the equation.
Right, now I have enough time to do cosine omega t. But I don't have enough time to do it the complex way.
So let me do as a final example, the equation.
Let me just think.
I don't know if I have enough space here.
I'm now going to do dy dt-- can I call that y prime to save a little space-- equal ay plus cosine of t. I'll take omega to be 1.
Now, how could we solve that one?
I'm going to solve it without complex numbers, just to see how easy or hard that is.
And you'll see, actually, it's easy.
But complex numbers will tell us more.
So it's easy, but not totally easy.
So what did I do in the earlier example if the right hand side was a 1, a constant?
I look for the solution to be a constant.
If the right hand side was an exponential, I look for the solution to be an exponential.
Now, my right hand side, my source term, is a cosine.
So what form of the solution am I going to look for?
I naturally think, OK, look for a cosine.
We could try y equals some number M cosine t. Now, you have to see what goes wrong and how to fix it.
So if I plug that in, looking for M the same way I look for capital Y earlier, I plug this in, and I get aM cosine t cosine t. But what do I get for y prime?
Sine t. And I can't match.
I can make it work.
I can't make a sine there magic a cosine here.
So what's the solution?
How do I fix it?
I better allow my solution to include some sine plus N sine t. So that's the problem with doing it, keeping things real.
I'll push this through, no problem.
But cosine by itself won't work.
I need to have sines there, because derivatives bring out sines.
So I have a combination of cosine and sine.
I have a combination of cosine and sine.
So the complex method will work in one shot because e to the i omega t is a combination of cosine and sine.
Or another way to say it is when I see cosine here, that's got two exponentials.
That's got e to the it and e to the-- anyway.
Let's go for the real one.
So I'm going to plug that into there.
So I'll get sines and cosines, right?
When I plug this into there, I'll have some sines and some cosines, and I'll just match the two separately.
So I'm going to get two equations.
First of all, let me say what's the cosine equation?
And then what's the sine equation?
So when I match cosine terms, what do I have?
What cosine terms do I get out of y prime, here?
The derivative.
Well, the derivative of cosine is a sine.
That that's not a cosine term.
The derivative of sine is cosine.
I think I get, if I just match cosines, I think I get an N cosine.
N cosine t equal ay.
How many cosines do I have from that term?
ay has an M cosine t. I think I have an aM, and here I've got 1.
That was a natural step, but new to us.
I'm matching the cosines.
I have on the left side, with this form of the solution, the derivative will have an N cosine t. So I had N cosines, aM cosines, and 1 cosine.
Now, what if I match signs?
What happens there?
We're pushing more than an hour, so hang on for another five minutes, and we're there.
Now, what happens if I match sines, sine t?
How do I get sine t in y prime?
So take the derivative of that, and what do you have?
Minus [INAUDIBLE]
Minus M sine t. That tells me how many sine t's are in there.
And on the right hand side, a times y, how many sine t's do I have from that?
You have N t's
N, good thinking.
And what about from this term?
None, no sine there.
So I have two equations by matching the cosines and sines.
Once you see it, you could do it again.
And we can solve those equations, two ordinary, very simple equations for M and N. Let's see if I make space.
Why don't I do it here, so you can see it.
So how do I solve those two equations?
Well, this equation gives me-- easy-- gives me M as minus aN.
So I'll just put that in for N. So I have N equals aM.
But M is minus aN.
I think I've got minus a squared N plus that 1.
All I did was solve the equation, just by common sense.
You could say by linear algebra, but linear algebra's got a little more to it than this.
So now I know M, and now I know N. So now I know the answer.
y is M, so M is minus aN.
Oh, well, I have to figure out what N is, here.
What is N?
This is giving me N, but I better figure it out.
What is N from that first equation?
And then I'll plug in.
And then I'm quit
[INAUDIBLE]
1 over, yeah
1 plus a squared
1 plus a squared, good.
Because that term goes over there, and we have 1 plus a squared.
So now y is M cosine t. So M is minus aN.
So minus aN is 1 over 1 plus a squared cosine t. Is that right?
That was the cosines.
And we had N sine t. But N is just 1-- I think I just add the sine t. Have I got it?
I think so.
Here is the N sine t, and here is the M cos t. It was just algebra.
Typical of these problems, there's a little thinking and then some algebra.
The thinking led us to this.
The thinking led us to the fact we needed cosines in there, as well as cosines.
But then once we did it, then the thinking said, OK, separately match the cosine terms and the sine term.
And then do the algebra.
Now, I just want to do this with complex.
So y prime equals ay plus e to the it.
To get an idea, you see the two.
And then I have to talk about it.
You see, I'm only going to go part way with this and then save it for Wednesday.
But if I see this, what solution do I assume?
This is like an e to the st.
I assume y is some Y e to the it.
See, I don't have cosines and sines anymore.
I have e to the it.
And if I take the derivative of e to the it, I'm still in the e to the it world.
So I do this.
I plug it in.
Uh-huh, let me leave that for Wednesday.
We have to have some excitement for Wednesday.
So we'll get a complex answer, and then we'll take the real part to solve that problem.
So we've got two steps, one way or the other way.
Here, we had two steps because we had to let sines sneak in.
Here, we have two steps because I could solve it, and you could solve that right away.
But then you have to take the real part.
I'll leave that.
Is there questions?
Do you want me to recap quickly what we've done
Yes
I try to leave on the board enough to make a recap possible.
Everything was about that equation.
We have only solved-- I shouldn't say only-- we have solved the constant coefficient, model constant coefficient, first order equation.
Wednesday comes nonlinear equation.
This one today was strictly linear.
So what did we do?
We solved this equation, first of all, for q equal 1; secondly, for q equal e to the st; thirdly, for q equal a step; fourthly for q equal-- where is it?
Where is that delta of t?
Maybe it's here.
Ah, it got erased.
So the fourth guy was y prime equal ay plus delta of t, or delta of t minus capital T. So those were our four examples.
And then what did we finally do?
So if we're recapping, compressing, we're compressing everything into two minutes.
We solved those four examples, and then we solved the general problem.
And when we solved the general problem, that gave us this integral, which my whole goal was that you should understand that this should seem right to you.
This is adding up the value at time t from all the inputs at different times s. So to add them up, we integrate from 0 to t. And finally, we returned to the question of cos t, all important question.
But awkward question, because we needed to let sine t in there too.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation, or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
Monday's lecture was all linear equations.
And I thought I would start today with nonlinear equations, still first order.
And we can't deal with every nonlinear equation.
That's too much to ask.
These are going to be made easier by a property called "separable."
So these will be separable nonlinear equations.
And let me start with a couple of examples and then you'll see the whole idea.
So one example would be the simplest nonlinear equation I can think of, with a y squared.
So how to get there?
Here's the trick.
This is the separable idea.
You're going to see it in one shot.
We can separate, put the Ys on one side and the Ds on the other.
So I write this as dy over y squared equal dt.
I put the dt up and brought the y squared down.
So now they're separated, in a kind of hookie way with infinitesimals.
But I'll makes sense out of that by integrating.
I'll integrate both sides.
I'll integrate time from 0 to t. And I have an initial condition, y of 0, always.
And since this one is about y, when t starts at 0, this guy starts at y of 0, up to, this ends at t, so this ends at y of t. OK. Now the point is, also the problem was nonlinear, we've got two separate ordinary integrals to do.
And we can do them.
We can certainly do the right hand side.
I get t. And on the left hand side, what do I get?
Well, that maybe I better leave a little space to figure out this one.
But the point is we can integrate 1 over y squared.
And I guess we get minus 1 over y.
So I get minus 1 over y between y of 0 and y of t. In other words, I'm getting let's see, so what the right, the derivative of the integral of 1 over y squared is minus 1 over y because I always check the derivative gives me that back.
So now I'm ready to plug-in those limits.
So I'll do the bottom limit first because it comes with a minus sign, canceling that minus, 1 over y of 0 minus 1 over y of t. Got it.
And that equals the other integral, which is just t. So that's the answer as it comes directly from integration.
And we can do more.
You can see that finding the solution when these things are separable has boiled down to two integrals.
And we could have a function of t here, too.
And that would be allowed, a function of t multiplying this guy, because then I would leave the function of t on that side.
And I would have to integrate that.
And I would bring the y.
You see, I've just separated the y.
In general, these equations look like dy, dt is some function of t divided by some function of y.
Maybe the book calls the top one g, I think, and the bottom one f. And everybody in this room sees that I can put the f of y up there.
I can put the dt up there.
And I've separated it.
OK.
So that's sort of the general situation.
This is a kind of nice example, nice example, dy, dt equals y squared.
Can we just play with this a little bit?
Let me take y of 0 to be 1, just to make the numbers easy.
So if y of 0 is 1, then I have, I'll just keep going a little bit.
You do have to keep going a little bit because when you finish the integral right there, you haven't got y equal.
You've got some equation that involves y, but you have to solve for y.
So I have to solve that equation for y.
Let me just do it.
So how would I solve it?
And let me take y of 0 to be 1.
So now, if I just write it below, I'm at 1 minus 1 over y equals t. Good?
So I'm going to put the 1 over y of t on that side and the t on that side.
So if I just continue here, I've got 1 over y of t on this side and, do I have 1 minus t on that side?
Yeah.
Looking good.
So solution starting from y of 0 equal 1 is y of t equal 1 over 1 minus t. You could do that.
You could do that.
And I can always, like, mentally I check the algebra at t equals 0.
That gives me the answer, 1.
But let's step back and look at that answer.
I mean, that's part of differential equations is to do some algebra, if possible, and get to a formula.
But if we don't think about the formula, we haven't learned anything.
Right there, yes.
Good.
So what happens?
I want to compare with the linear case that was like e to the t. This was y prime equal y, right?
And that led to e to the t. Y prime equals y squared leads to that one.
So first observation.
I haven't got exponentials anymore in that solution.
Exponentials are just like perfection for linear equations.
For nonlinear equations, we get other functions.
Professor Fry had a hyperbolic tangent function in his first lecture.
Other things happen.
OK. Now, how do those compare if I graph those?
It's just like, why not try?
So they both started at 1.
And e to the t went up exponentially.
E to the t. And, I don't know, we use exponential.
In our minds, we think, that's pretty fast growth.
!
mean, that's the common expression, grew exponentially.
But here, this guy is going to grow faster because y is going to be bigger than 1.
So y squared is going to be bigger than y.
That one's going to grow faster.
Faster than exponential.
This has the exponential growth.
Pretty fast.
Polynomial, of course, some parabola or something would be hanging way down here, left behind in the dust.
But this 1 over 1 minus t, that's going to grow really fast.
And what's more, it's going to go to infinity.
So that y prime equal y squared, the solution to that doesn't just-- e to the t goes to infinity at time infinity.
At any finite time, we get an answer.
Eventually, at t equal infinity, it's gone above every bound.
But this one, 1 over 1 minus t is what I want to graph now.
I believe that that takes off and at a certain point, capital T, it's going to infinity.
It's blown up.
So it's blow up in finite time.
Blow up in finite time.
And what is that time?
What's the time at which the y prime equal y squared has taken off, gone off the charts?
T equal--
1
--1.
Because when t reaches 1, I have 1 over 0, and I'm dividing by 0, and so that's the blow up.
Finite time blow up.
OK.
So this can happen for some nonlinear equations.
It wouldn't happen for a linear equation.
For a linear equation, exponentials are in control.
OK.
So that's one nice example.
Oh, another nice thing about that example.
Well, I say nice if you're OK with infinite series.
I just want to compare.
The book mentions the infinite series for these guys because that's an old way to solve differential equations is term-by-term in an infinite series.
It's sort of fun to see the two series.
Well, because they're the two most important series in math.
Actually, they're the two series that everybody should know.
The power series, Taylor series-- whatever word you want to give it for those two guys.
So let me do them.
E to the t, I'll put that one first, and 1 over 1 minus t. These are the great series of math.
Shall I just write them down and sort of talk through them?
Because this is not a lecture on infinite series by any means.
But having these two in front of us, coming from these two beautiful equations, y prime equal y squared and y prime equal y, I can't resist seeing what they look like this way.
So e to the t, do you remember e to the t?
It starts at 1.
What's the slope of e to the t?
At t equals 0.
So I'm doing everything-- this series is going to be, both of the series are going to be, around t equals 0.
That's my, like, starting point.
So this e to the t thing has a tangent.
It has a slope there.
And what's the slope of e to the t at t equals 0?
1
1.
It's derivative.
The derivative of e to the t is e to the t. The slope is 1.
So that tangent line has coefficient 1.
That's how it starts.
That's the linear approximation.
That's the heart of calculus, is this.
But we're going to go better.
We're going to get the next term.
So what's the next term?
That gave us the tangent line.
Now I'm going to move to the tangent parabola.
So the parabola has got another is still going to be below the real thing.
Can I squeeze in the words "line" and "parab," for "parabola?"
Parabola has bending.
I'm really explaining the Taylor series in what I hope is a sensible way.
Here is the starting point.
This has the slope.
The next term has the bending.
The bending comes from what derivative?
What derivative tells us about bending?
Second derivative.
Second derivative tells us how much it curves.
Well, the second derivative of e to the t is still e to the t. So the bending is 1.
The bending is also 1.
Now that comes in with a factor of a 1/2.
There is the tangent parabola.
And you will see what these numbers become.
Let me just go to, the third derivative would be responsible for the t cubed term.
And its coefficient would be 1 over 3 factorial.
So 2 is the same as 2 factorial.
3 factorial is 3 times 2 times 16.
So the numbers here go 1, 6, 24, 120, whatever the next one is.
720 or something.
They grow fast.
So that series always gives a finite answer.
It does grow with t. But it doesn't spike with t. Now compare that famous series.
And of course, this is 1 over 1 factorial, everything consistent.
Compare that with the series for 1 over 1 minus t. That's the other famous series that they learned in algebra.
I'll just write it.
That's 1 plus t plus t squared plus t cubed plus 1 so on, with coefficient 1.
This had 1 over n factorials.
Those make the series converge.
These don't have the n factorials.
This is 1, 1, 1, 1.
And, well, I could check that formula.
But do you remember the name for that series?
1 plus t plus t squared plus t cubed plus so on?
Algebra is taught differently in many high schools now.
And maybe that never got a name.
I guess I would call it the Geometric series.
Geometric series.
And you see, it's beautiful.
It's the other important series.
But it's quite different from this one because, what's the difference about this series?
Yeah?
It goes to infinity
It's a--
It goes to infinity
It goes to infinity.
But where?
At what time?
At what value of t is this sum going to fall apart?
Blow up?
At t equal 1.
When have 1 plus 1 plus 1 plus 1, I'm getting infinity.
So this blows up.
And of course, we see that it should because this blows up.
Left side blows up at t equal 1, the right side blows up a t equal 1.
Where the exponential series, which is the heart of ordinary differential equations, never blows up because of these big numbers in the denominator.
OK, I'm good for this first simple example, y prime equals y squared.
It has so much in it, it's worth thinking about.
I'm ready, you OK for a second example?
A second important separable equation.
I'm going to pick one.
So I'm going to pick an equation that starts out with our familiar linear growth.
This could be, you know, last time it was growth of money in a bank.
It could be growth of population.
The number of, to a sum first approximation, the rate of growth of the population comes from, like, births minus deaths.
And with modern medicine, births are a larger number than deaths.
So a is positive, and that grows.
But if we're talking about the, I mean, the United Nations tries to predict, everybody tries to predict, population of the world in future years.
And so this could be called the Population Equation.
But just to leave it as pure exponential is obviously wrong.
The world can't grow forever.
The population can't grow forever.
And the, I guess I hope it doesn't grow like 1 over 1 minus t. So this is at least a little slower.
But somehow competition for space, competition for food, for oil, for water-- which is going to be the big one-- is in here.
Competition here, of people versus people, a reasonable term, a first approximation, is a y squared, with a minus, is a y squared and with some coefficient.
That's a very famous equation.
A first model of population is it grows.
But this is a competition term, y against y.
And so, the same would be true if we were talking about epidemics.
That's a big subject with ordinary differential equations, epidemiology.
Or say, flu.
How does flu spread?
And how does it get cured?
So partly, people are getting over the flu.
But then y against y is telling us how many infected, how many new infections.
So we would like to solve that equation.
And it's separable.
I can do what I did before, dy over ay minus by squared equal dt.
And I can integrate, starting from year of 0.
Well, why don't we start from year 2014, with the population y at now-- the present population?
That would be a model that the UN would consider using.
That other people with very important interest in measuring population and measuring every resource would need equations like this.
And then they would put on more terms, like a term for immigration.
All sorts, many improvements have to go into this equation.
Let me just look at this as it is.
Well, I've got two choices here.
Well, it's this integral that I'm looking at.
That is a doable integral.
It's the type of integral that we saw in the Rocket problem.
The Rocket problem was more constant minus.
This was a drag term, when we were looking at rockets.
And this was a constant, say, gravity.
So it was still a second degree.
Still second degree, but a little different.
This has the linear in second degree terms.
If you look up that integral, you'll find it.
Or there's a systematic way to do it.
That's in 1801, I guess, called partial fractions.
It's not a lot of fun.
I don't plan to do it.
It's in the book.
Has to be because that's the way you can integr-- you can integrate polynomials over polynomials by partial fractions.
That's what they're for, but there's a neat way to do this one.
There's a neat trick that Bernoulli discovered to solve that equation, to turn it into a linear equation.
And of course, if we can turn it into a linear equation, we're on our way.
So the neat trick is let z be 1 over y.
You can put this in the category of lucky accidents, if you like.
So now I want an equation for z.
So I know that dz, dt if I take the derivative of that, that's y to the minus 1.
So it's minus 1 y to the minus 2 dy, dt.
That's the chain rule.
Take the derivative of 1 over 1, you get minus 1 over y squared.
Multiply by the derivative of what's inside.
That's the chain rule.
OK.
So I plan to substitute those in here.
So dy, dt, let's see.
Can you see me?
You can probably do it better than me.
So dy, dt is minus.
I'll bring that up.
Dy, dt I'm going to put-- I hope this'll work all right-- for dy, dt, I'm going to put in dz.
Using this, I'm going to put minus y squared dz, dt.
Did that look right?
I don't think I'm necessarily doing this the most brilliant way.
But dy, dt-- I put this up here and I got that-- equals ay.
So that's a over z. Oh, y Is 1 over z.
So get this, I want all Zs now.
So that's this part.
And ay is over z minus by squared is minus b over z squared.
Would you say OK to that?
I've got Zs now, instead of Ys.
I just took every term and replaced y by 1 over z. Y is 1 over z and dy, dt I can get that way.
OK. Yeah.
Now what?
Now look what happens, if I multiply through by z squared or by minus z squared.
Let me multiply through by minus z squared.
I get dz, dt.
Multiplying by minus z squared gives me a minus az.
And what do I get when I multiply this one by minus z squared?
Plus b
I get plus b.
Look what happened.
By this, like, some magic trick.
You could say, all right.
That was just a one time shot.
But it was a good one.
We ended up with a linear equation for z.
A linear equation for z.
And we solved that equation last time.
So let me squeeze in the solution for z, and then elsewhere.
So what was the solution for z of t?
It was some multiple of, no, yeah.
This is perfect review of last time.
We have a constant times z.
And so that's going to go into the exponential.
This will be the, it's a minus a, notice.
That will be the, what part of the solution is that one called?
That's the null solution.
The null solution, when b is o.
And now I add in a particular solution.
A particular solution.
And one good particular solution is choose the z to be a constant.
Then that'll be 0.
So I want that to be 0.
So what constant z makes that 0?
I think it's b over a, don't you?
I think b over a.
Does that work good?
That's every null solution plus one particular solution.
Let me say now, and I'll say again, looking for solutions which are steady states, b over a-- of this particular solution, that particular solution made this 0.
So it made this 0.
So it's a solution that's not going anywhere.
It's a constant solution.
It's a solution that can live for all time.
OK. B over a.
Let me put that word there, steady state.
OK. And now I would want to match the initial conditions using c. Yeah.
I'd better do that.
OK. And I have to get back to y. I have y is 1 over z.
So I'm going to have to flip this upside down.
I'm going to have to flip this upside down is what will actually happen.
Let me make it easy to flip.
Let me, I'll change c, which is just some constant to some constant d over a.
So then it's a is everywhere down below.
And I just write it here in the middle.
That makes it easier to flip.
So finally I get their solution.
Solution to the population equation.
But that's the famous word for it, the Logistic equation.
This is section 1.7 of the text on the differential equations in linear algebra.
It's a very, very much studied example.
It's a great example.
It fits the growth of human population with some, it's our first level approximation to growth of or other populations or other things.
It's a linear term giving us exponential growth, and a quadratic term of competition slowing it down.
And let's see that slow down.
So now that was a bit of algebra.
Much nicer than partial fractions.
The bit of algebra just came from this idea of going to z.
And now I want to go back to y.
So y is 1 over z.
So it's a over d e to the minus at plus b.
That's our solution.
A and b came out of the equation.
And d is going to be the number that makes the initial value correct.
So at t equals 0, I would have y of 0, whatever the initial population is, is a over d. T Is 0, so that's just 1 plus b.
So that tells me what d is.
D equals something.
It comes from y of 0.
So the answer, let me circle that answer.
That answer has three numbers in it, a, b, and d. a and b come from the equation.
D also involves the initial starting thing, which is exactly what it showed.
So you could say we've solved it.
But if you ever solve an equation like this, you want to graph it.
You want to graph it.
So let me draw its graph.
This is important picture.
So here is time.
Here is population.
Here's, maybe it started there.
This is times 0.
And now I want to graph this.
I want to graph that function.
Really, this is where we're going somewhere.
What happens for a long time?
At t equal infinity, what happens to the population?
Does it grow, like e to the t?
Just remember the examples here.
We had a growth like e to the t. We had a growth faster than e to the t that actually blew up.
What about this guy?
What will happen as t goes to infinity with that population?
It goes to?
A over b
A over b.
A over b.
That's the key number in the whole thing.
It keeps growing, but it never passes a over b.
This is y at infinity.
That's the final population.
So how does it do this?
If I draw this graph-- and what about negative time?
Let's go backwards in time.
What is it at t equals minus infinity?
Then you really see the whole curve.
At t equal minus infinity, what is this doing?
0 infinity
It's 0.
Good.
Good.
Good.
T equal minus infinity, this is enormous.
This is blowing up.
It's in the denominator.
We're dividing by it.
So the whole thing is going to 0.
So here's what the logistic curve looks like.
It creeps up.
And it's beautifully, there's a point of symmetry here.
The growth is increasing here.
And then, as a point of inflection you could say, growth is bending upwards for a while.
At this point, it starts bending downwards.
From that point on, ooh, let's see if I can draw it.
It'll get closer, and exponentially close.
That wasn't a bad picture.
The population here is half way.
Here, the population, the final population, is a over b.
And just by beautiful symmetry, the population here is a 1/2 of a over b.
At this point.
If this was the actual population of the world we live in-- I think we're pretty close to this point.
I believe, well, of course, nobody knows the numbers, unfortunately, because the model isn't perfect.
If the model was perfect, then we could just takes the census and we would know a and b.
But the model isn't that great.
But it's sort of, we're at a very interesting time, close to a very interesting time.
I believe that with reasonable numbers, this a over b might be maybe 12 billion.
And we might be, I think we're a little above six billion.
I think so.
So we're a little bit past it.
This is now.
This is halfway.
That's the halfway point.
It's perfectly symmetric.
It's called an S curve.
And many, many equations in math biology involve S curves.
So math biology often gives rise, with simple models, to a kind of problem we've had here with a quadratic term slowing things down.
Enzymes, all kinds of.
Ordinary differential equations are core ideas in a lot of topics, lot of areas of science.
OK. Do I want to say more about the logistic equation?
I guess I do want to distinguish one thing.
Yeah.
One thing about logistic equations and will of course come back to this.
OK. Let me look at that logistic equation.
Here's my equation.
So I've managed to solve it.
Fine.
Great.
Even graph it.
But let me come back to the question, suppose I just look at.
I can see two constant solutions, two steady states, two solutions where the derivative is 0.
So nothing will happen.
So in other words, I want to set this thing set to 0 equal to 0 to find steady solutions.
Steady means the derivative is 0.
So this side has to be 0.
So what are the two possible steady states where, if y of 0 is there, it'll stay there?
0
0.
Y equals 0 is one.
And the other?
A over b
Is a over b.
So steady equal to 0.
And I get two steady states.
Let me call them capital Y equals 0 because that's certainly 0 of, if we have 0 population, we'll never move.
Or setting this to 0, ay is by squared cancel y's divide by b a over b.
So the two steady states are here.
That's a steady state and that's a steady state.
Those are the only two in this problem.
You see how easy that was to find the steady states?
That's an important thing to do.
And then the other important thing to do is to decide, are those steady states stable?
When the population's near a steady state, does it approach that, does it go toward that steady state or away?
So what's the answer?
For this steady state, that steady state, y is a over b.
Is that stable or unstable?
So I'll write the word stable.
And I'm prepared to put in "un," unstable, if you want me to.
This is a key, key idea.
And with nonlinear equations, you can answer this stability stuff without formulas.
Without formulas.
That's the nice thing.
And then that comes in a later class.
But here's a perfect example.
So do we approach this answer or do we leave it?
We approach it, the solutions.
This is stable, yes.
And here's the other stationary point, capital Y.
The other steady state is that nothing happens.
So now if I'm close to that, if y is a little number, like 2, will that 2 drop to 0, will it approach this steady state, or will it leave it?
Leave it
Leave it.
So this steady state is
Unstable
Unstable.
Unstable.
Right.
Right.
With linear equations, we really only had one steady state, like 0.
Once it started, it took off forever.
Here, it doesn't go infinitely high.
It bends down again to that limit, that carrying capacity is what it's called, a over b. I guess I hope you think a nonlinear equation like got a little more to it.
Little more interesting, but a little more complicated, than linear equations.
Yep.
Yep.
Yep.
And similarly, the rocket equation, we could at the right time soon in the course, ask the same thing, a rocket equation was something like that.
What are the steady states?
Are they stable?
Are they unstable?
Can you find a formula?
Here.
This.
We got a formula.
And there are other nonlinear equations, which we'll see.
OK.
I could create more separable equations, but I guess I hope that you see with separable equations, you just separate them and integrate a y integral and a t integral.
Is that OK any question on this nonlinear separable stuff?
Differential equations courses and the subject tends to be types of equations as can solve.
And then there are a hell of a lot of equations that are not on anybody's list, where you could maybe solve them by an infinite series, but not by functions that we know.
OK.
I'm ready to do the other topic for today.
It's the topic that I left incomplete on Monday.
So I'm staying with first order equations, but actually this topic is essential for second order equations.
So I'm going to topic two for today.
So topic two will involve complex numbers.
So we have to deal with complex numbers.
And the purpose of introducing these complex numbers is to deal with what we met last time when the right hand side, the forcing term, was a cosine.
Typical alternating current, oscillating, rotating, rotation.
All these things produce trig functions.
Maybe rotation is more of a mechanical engineering phenomenon.
Alternating current more of an EE phenomenon.
But they're always there.
And what was the point?
The point was we had some linear equation, and we had some forcing by something like cos omega t. Or it could be A cos omega t and B sine omega t. Either just cosine alone, or maybe these come together.
And then the solution was y equals some combination of those same guys.
In other words, what I'm saying is cosines are nice right hand forcing functions.
Fortunately, because we see them all the time.
But they do lead to cosines and sines.
I emphasized that last time.
If we just have cosines in the forcing function, we can't expect that there's any damping, we can't expect only cosines.
We have to expect some sines.
In other words, we have to deal with combinations of them.
And the question is, how do you understand cos omega t plus 3.
Or let me take a first example.
Example-- cos t plus sine t. That's a perfect example.
So what is omega here in this example that I'm starting with?
1
1.
So I just read off the coefficient of t is 1, 1 hertz here.
But we have got this combination.
And the question is, how do we understand that cosine plus sine?
Two very simple functions, but they're added, unfortunately.
And there's a much better way to write this so you really see it.
You really see this.
That's called a sinusoid.
And the rule that want to focus on now is that everything of that kind, of this kind, of this kind, of a cosine plus a sine, can be compressed into one term.
One term.
Of course, it's got to have two constants to choose because that had an a and a b.
This had an m and an n. This had a 1 and a 1.
But the term I'm looking for is some number R times a pure cosine of omega t, but with a phase shift.
So you see there are two numbers here to choose.
It's really like going from rectangular to polar.
Say in complex numbers, let's just remember the first fact about a complex number.
If the real part is 3, and the imaginary part is, let's say 2, then here's a complex number, 3 plus 2 i.
So this was the real axis.
This was the imaginary axis.
I went along 3, I went up 2, I got to that number.
There it is.
I plotted the number 3 plus 2 i in the complex plane.
And for me, that number 3 plus and so on, really saying something important.
And maybe it's not entirely new.
I'm saying something important about complex numbers, this is their rectangular form.
Something plus something.
That form is nice to add to another complex number.
If I added 3 plus 2 i to 1 plus i, what would I get?
4 plus 3 i
4 plus 3 i.
But if I multiply, multiply, 3 plus 2 i times, let's say I square it.
I multiply 3 plus 2 i by 3 plus 2 i.
What do I get?
If I do it with this rectangular form, I get a mess.
I can't see what's happening.
It's the same over here.
This is like having a 1 and a 1, with an addition.
This is like a polar form where it's one term.
OK.
So let me answer the question here and then let me answer the question there.
And then you've got a good shot at what complex numbers can do, and why we like the polar form for squaring, for multiplying, for dividing.
What's the polar form?
Well, I'm using that word "polar" in the same way we use polar coordinates.
What are the polar coordinates of this point?
They're the radial distance, which is what?
So what's that distance?
That's the R you could say.
It corresponds to this R here.
So I'm just using Pythagoras.
That hypotenuse is what?
Square root of 13
Square root of 13.
Thanks.
9 plus 4, square root of 13.
And what's the other number that's locating this in polar coordinates?
The angle.
And the angle.
What can we say about that angle?
Let's call it phi is-- what's the angle?
Well, it's some number.
It's between 0 and pi over 2, I'm sure of that.
What do I know about that angle?
I know that this is 2 and this is 3.
So that's telling me the angle.
Well, what is that really telling me immediately?
It's telling me the
Tangent
Tangent of the angle.
So the tangent of the angle is 2 over 3.
And the magnitude is square root of 13.
OK.
So those beautiful numbers, 2 and 3, have become a little weirder.
Square root of 13, inverse tangent of 2/3.
You could say, well, that's not so nice.
What was I going to do?
I was going to try squaring that number.
So if I square 3 plus 2 i, or if I take the 10th power of 3 plus 2 i, or the exponential, all these things, then I'm happy with polar coordinates.
Like, what would be the magnitude of the square?
And where will the square of that number, so I want to put in 3 plus 2 i squared, which I can figure out in rectangular, of course-- a 9, and 6 i, or 12 i, or 4 i squared, stuff like that.
It's not pleasant.
What's the magnitude, what's the R for this guy?
What's the size of that number squared?
Yes?
Say that again
13
13.
Right.
I just have to square this square root so I get 13.
And the angle will be, what's the angle for the square there?
I don't want a number.
I guess I'm just doing this.
R e to the i phi squared is R squared.
And what's the angle here?
E to the i phi squared is e to the 2 i phi.
It's the angle doubled.
E to the 2 i phi.
The angle just went from phi to 2 phi.
The lengths went from square root of 13 to 13.
Squaring, multiplying is nice with complex numbers.
Maybe can I before I go on and on about complex numbers, I should ask you, how many know all this already?
Complex numbers are familiar?
Mostly.
Correctly, with a wiggle.
OK.
I won't go more about complex numbers.
Let me come back to my question here.
Let me come back to the application.
So here it is with complex numbers.
Here it is with sinusoids.
And the little beautiful bit of math is that the sinusoid question goes completely parallel to the complex number question.
So you have an idea on those complex numbers.
We'll see them again.
Let me go to this.
So I want this to be the same as this, OK. Maybe I'm going to have to use a new board for this.
Can I start a new board?
So I want cos t plus sine t to be some number R times cosine of t 1.
I can see omega's 1, so I just put to 1 minus some angle.
OK. And I want to choose R and phi to make that right.
You see what I like about it?
This tells me the magnitude of the oscillation.
It tells me how loud the station is.
When I see cos t and, separately, sine t, or I might see 3 cos t and 2 sine t. 3 cos t is a cosine curve.
2 sine t is a sine curve shifted by 90.
I put them together, it bumps, it bumps, bumps.
Not completely clear.
It seems to me just beautiful that if I put together a cosine curve that we know, that starts at 0, with a sine curve that starts at 0, the combination is a cosine curve.
Isn't that nice?
I mean, you know, that sometimes math gets worse and worse whatever you do.
But this is really nice that we can put the two into one.
But you see, it's going to-- well, let's do it.
What would R and phi be here?
So I'll use a trig fact here.
A cosine of a difference of angles, so this is R times cosine t, do you member this?
This was the whole point of going to high school.
Plus sine t sine phi.
So now, how do I get R and phi?
I use the same idea that worked last time.
I match the cosine terms and I match the sine terms.
So the cosine t has a 1.
1 cosine t is R cos phi.
That's what's multiplying cosine t. And the sine t has a 1.
And that has to agree with R sine phi.
So I'm in business if I solve those two equations.
And well, they're not linear equations.
But I can solve them.
Of course, the one fact that you never forget is that sine squared plus cosine squared is 1.
Right?
So if I square that one, and square that one, and add, what will I get?
1 squared and 1 squared will be 2, on the left hand side.
On the right hand side, I'll have R squared cos squared, R squared cos squared, and plus R squared sine squared.
And what's that?
What's R squared cosine squared plus r squared sine squared?
R squared
It's just R squared.
So all that added up to R squared.
In other words, it's just like polar coordinates.
R is the square root of 2.
That's telling us the magnitude of the response.
Square root of 2.
You see, it's just like complex numbers.
It's like the cosine gave us a real part and the sine gave us an imaginary part.
And R was the hypotenuse.
And that's really nice.
So R is the square root of 2.
OK. Now, the angle is never quite as nice.
But how can we get something about an angle out of there?
All we could get in this case here was the tangent of the angle.
And I'll be happy with that again here because it's a totally parallel question.
How am I going to get the tangent of the angle?
What do I have?
From these two equations, I want to eliminate R. So how do I eliminate R?
What do I do?
Divide.
Divide.
I guess if I want tangent sine over cosine, I'll divide this one in the top by this one in the bottom.
So I take the ratio.
That'll cancel the Rs perfectly.
It'll leave me with 10 phi.
And here it happens to be 1.
OK.
So what have I learned?
I've learned that when these two add up together, they equal what?
R square root of 2.
You see how easy it is.
Square root of 2 came from the square root of 1 plus.
It's like Pythagoras.
Pythagoras going in circles, really.
Times the cosine of t minus.
And what is phi?
Its tangent is 1, so what's the angle phi?
Pi over 4
Pi over 4.
Right.
So that's the sinusoidal identity when the numbers are 1 and 1.
But you saw the general rule.
Let me just take it.
Suppose this is the output, and cos omega t plus n sine omega t. What is the gain?
What's the magnitude, the amplitude, the loudness of the volume in this when I'm tuning the radio?
What's the R for this guy?
What's this R?
If we just follow the same idea.
So if we have m times a cosine and n times a sine, what's your guess?
What's your guess for R, the magnitude?
I'm guessing a square root of what?
Yeah?
You got.
What is it?
n squared-- [INTERPOSING VOICES]
Plus n squared.
Way to go.
M squared plus N squared.
And the angle is like the phase shift.
I'm not great at graphing, but let me try to go back to my simple example.
If I tried to add up on the same graph cosine t, which would start from 1 and drop to 0, go like that, right?
Something like that would be cosine.
And now I want to add sine t to that.
So that climbs up to 1 back to 0, down.
And now if I add those two, this formula is telling me that it comes out neat.
Neatly.
That one plus that one is another sinusoid with height square root of 2.
If I had different chalk, I've got at least a little bit different.
But does it start here?
Of course not.
It starts here, I guess.
But it goes up, right?
Because this comes down, but this is going up.
All together, it's up to, where is the peak?
Where is the peak on the sum?
So I'm adding, everybody sees what I'm doing?
I'm adding a cosine curve and a sine curve.
And it goes up.
And where does it peak?
What angle is it going to peek at?
What's the biggest value this gets to?
[INAUDIBLE]
At pi over 4, it'll peak.
At pi over 4, it'll be the cosine of 0, which is 1.
It's height'll be the magnitude, the gain, square root of 2.
So it'll peak at pi over 4, which is probably about there, right?
Peak at pi over 4 and, I don't know if I got it right frankly.
I did my best.
That's the sum.
That right there.
The first key point is it's a perfect cosine.
The second key point is it's a shifted cosine.
The third key point is its magnitude is the square root of 1 squared plus 1 squared, or n squared plus n squared, or a squared plus b squared.
So that's the sinusoidal identity.
A key identity and being able to deal with forcing terms, source terms, that are sinusoids.
OK. Now, I'm going to take one more step since we have just like 10 minutes left, and let the number i get in here properly.
Get a complex number to show up here.
OK. Before I start on this, let me recap.
Let me recap today's lecture.
It started with nonlinear separable equations.
And a great example was the logistic equation up there, with the S curve.
That took half the lecture.
The second half of the lecture has started with things real with sinusoids that are combinations of cosine and sine and has written them in a one term way.
And now I want to get the same one term picture from using complex numbers.
OK. OK. And everything I do would be based on this great fact from Euler that e to the i omega t. The real part is cosine omega t. And the imaginary part is sine omega t. That's a central formula.
Let me draw it rather than proving it.
Let me draw what that means.
I'm in the complex plane again.
Real part is the cosine.
The imaginary part is the sine.
That number there is e to the i omega t because it's got that real part and that imaginary part.
And what's its magnitude?
What's the R, the polar distance for cos omega t plus, for this number, which is for this number?
What's the hypotenuse here?
Everybody knows
1
1.
Hypotenuse is 1.
Cos squared plus sine squared is 1.
So e to the i omega t is on a circle of radius 1.
That's the most important circle in the complex world, the circle of radius 1.
And all these points are on it.
And their angles are omega t. And as t increases, the angle increases, and you go around the circle.
You've seen it.
Physics couldn't live without this model.
OK.
So that's basically what we have to know.
And now, how do we use it?
Well, the idea is to deal with the equation.
Like, the equation I had last time was dy, dt equals y plus cos t. That gave us some trouble because the solution didn't just involve cosines, it also involved sines.
Yeah.
So I want to write that equation differently, in complex form.
And this is the key point here.
So I'm going to look at the equation dz, dt equals z plus e to the i t. Well, I'll make that cos omega t just to have a little more, the units are better, everything's better if I have a frequency there.
Units of this are seconds and the units of this are 1 over seconds.
Now, question.
What's the relation between the solution z to that complex equation and the solution y to that equation?
Of course, they have to be related, otherwise it was stupid to move to this complex one.
My claim is that complex equation is easy to solve.
And it gives us the answer to the real equation.
And what's the connection between y and z?
So y's the real part
Y is, exactly, say it again
The real part--
Of z. Y is the real part of z.
So that gives us an idea.
Solve this equation and take it's real part.
If I can solve this equation without getting into cosine and sine separately and matching, I can stay real.
I solve the equation by totally real methods up to now.
Now I'm going to say, here's another approach.
Look at the complex equation, solve it, and take the real part.
You may prefer one method.
You may like to stay real.
In a way, it's a little more straightforward.
But the complex one is the one that will show us, it brings out this R, it brings out the gain, it brings out the important-- engineering quantities are important, if I do it this way.
Now, I believe that the solution to that is easy.
Actually, it is included in what I did last time.
It's a linear equation with a forcing term that's a pure exponential.
And what kind of solution do I look for?
I'm looking for a particular solution.
If I see an exponential forcing term, I say, great.
The solution will be an exponential.
So the solution will be sum.
Z is sum capital Z e to the i omega t. Plug it in.
What happens if I plug that in to find capital Z, which is just a number?
Right.
This is my method.
This is a linear equation, with one of those cool right hand sides, where the solution has the same form with a constant, and I just have to find that constant.
So I plug it in.
Dz, dt.
Take the derivative of this, z i omega will come down.
E to the i omega t. Z is just this, z to the i omega t. And this is just 1 e to the i omega t. So I plugged it in, hoping things would be good.
And they are because I can cancel e to the i omega t, that's the beauty of exponentials, leaving just a 1 there.
So what's capital Z?
What's capital Z then?
I've got a z here.
I better bring it over here.
And I've got the 1 there.
I think the z is 1 over.
When I bring that z over here, do you see what I'm getting?
It's all multiplying this e the i omega t. It's a number there.
Z times i omega and comes over as a minus z.
What do I have multiplying z here?
I see the i omega.
And what else have I got multiplying the z?
Minus
A negative 1 because it came over with a minus sign.
Done.
Equation solved.
Equation solved.
Complex equation solved.
So the point is, the complex equation was a cinch.
We just assumed the right form, plugged it in, found the number, we're done.
But there's one more step, which is what?
Take the real part.
So I have to take the real part of this.
So the correct answer is y is the real part of that number, 1 over i omega minus 1 times e to the i omega t. I'm tempted to stop there, but just with a little comment.
How am I going to find that real part?
And what form will it have?
What form will that real part have?
Yeah, maybe just to say what form will it have?
The real part, it's going to be a sinusoid.
But I have a complex number multiplying this guy.
The real part is going to be exactly of the form we-- well, of course, it had to be the form because that was another way to solve the equation.
It's going to be some number.
And I'll call it g, for gain, times the real part.
And so the real part will be a cosine.
Yeah, it's just perfect.
A cosine of omega t. And there'll be a phase.
Yeah.
i haven't taken that step fully.
I got to that fully.
And then I said that that, if I use some complex arithmetic, will come out to be this.
And you see the beauty of that answer, which was way better than a sum of sines and cosines.
We see the gain.
We see the amplitude.
And we see the phase shift.
Yeah.
So I don't know, that would be a good exercise in complex numbers.
Find g and find phi, in taking the real part of this thing.
Yeah.
It's a pure exercise in using complex numbers.
I don't feel like doing it today.
If we do it, you just see a lot of formulas.
Here, you see the point.
The point was that the complex equation could be solved in one line.
We just did it.
But that left us the problem of taking the real part.
That was the e to the i omega t there.
Left us the problem of taking the real part.
And that's a practice with complex arithmetic.
So you've got the choice.
Either stay real-- sign plus cosine.
And then use the sinusoidal identity, polar form.
Or get the polar form from here.
Same answer both ways.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free.
To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
This week is my second pair of lectures.
Last week the two lectures were about first order differential equations, and this week second order.
Those are the two big topics in differential equations.
Let me start with most basic second order equation.
We see the second derivative and the function itself, and we don't see yet the first derivative term.
This is the nice case, when I just have y double prime and y.
In general, I-- I'm taking constant coefficients today.
Because if the coefficients depend on time, the problem gets much, much harder now.
So let's stay with constant coefficients, meaning we have a mass, for example, we have a spring.
The stiffness of the spring is k, the mass is m, and the y, the unknown displacement, tells us the movement of the mass.
The classical problem.
You will have seen it before.
Because you have an exam this afternoon, I wanted to start with things that-- they are about second order equations, but they're still close to the exam idea, particularly the idea of exponentials.
With constant coefficients, that's the fundamental message.
Exponentials in, exponentials out.
But it's not quite so clear when we had first order, y prime equal ay, we knew that the exponent was a.
The solution was e to the at.
Now we've got second derivatives coming in, and it won't be so much e to the at type thing.
Either the at was growth for a positive, decay for a negative.
Now we're going to see oscillation.
It's still exponentials, but oscillation.
Things going up and down, things going around.
Harmonic motion, you call it.
Sines and cosines.
And sines and cosines connect to complex exponentials.
So that instead of e to the at-- so now oscillations-- they're going to be coming from e to the i omega t. In other words, instead of an a, we're going to have an i omega.
Or, if we like to stay real, we can stay with cos-- cosine-- and sine.
And actually, I've written two real guys there, so I better have two complex ones.
And it will turn out to be plus or minus.
There are two frequencies there.
Plus i omega, minus i omega, and they turn into cosine and sine.
So in this case, with no damping term, we can stay entirely real without creating any problems.
We can work with cosines and sines.
The first question is, what's omega.
What is the frequency of oscillation.
And of course, another similar picture would be a pendulum, a linear pendulum, swinging side to side, keeping time, because that frequency will stay constant.
Always I'll start with zero on the right hand side.
Just look at these equations.
Constants there.
I'm looking for solutions.
And I'm looking for null solutions, looking for the natural motion of this spring, the natural up and down motion of this spring.
Classical problem.
Won't be brand new, but it's the right starting point for the full second order equation.
It'll get a little complicated on Wednesday, when damping gets in there.
The formula got a little messy, because you've got a mass-- you've got an m-- and a k, still, but you also will have a damping constant.
Then complex numbers really come in.
Here they're optional.
So this is my equation to solve.
Because we don't have a first derivative, a cosine will solve that.
So let me look for-- I could look for exponentials.
Maybe I should do that first, look for an exponential solution.
Yeah, that's a good idea.
And let me not jump ahead to know that the exponent has that i omega form.
Let me discover it.
So I look for no solutions-- because I have that zero there-- no solutions of the form e to the st, some exponent.
Plug it in.
That's the message with constant coefficients.
Look for exponentials, substitute them in, discover what s will be.
So let's just do that.
This is the most basic step.
For null solutions will be exponentials, I substitute into the equation.
I get ms squared from two derivatives.
It will bring s down twice.
This is just ke to the st, and I'm looking for null solutions.
Zero.
No forcing.
So this is undamped, unforced.
Undamped, unforced.
Natural motion.
What do I do now?
Plugged in an exponential, got this equation.
And the beauty is that the exponentials cancel.
An exponential is never zero, so I can safely divide by it.
So I cancel those, and I get ms squared plus k equals zero.
The key equation-- and it's so simple-- it's just we're doing algebra now.
The calculus, the derivative we took when we plugged it in, but now it's an algebra question.
And of course, solving that system is easy.
There's no s term, no damping term.
So the frequency, s, is-- put k on the other side, divide by n. s is-- s squared, let's say-- is k over m-- is minus k over m. Critical point.
That tells me, with that minus sign there, that s is an imaginary number.
A complex number has a real part and an imaginary part.
In this case, all imaginary.
No real part at all.
It's natural to think-- s is the square root of that, so I'm going to write-- everybody writes-- s, the frequency s, is i omega.
So if I plug that in, I have omega squared equal k over m. i squared and the minus 1 deal with each other.
So the frequency omega-- here is the great fact-- square root of k over m. That's-- yes?
What difference does having imaginary parts to answer affect the oscillation?
To-- OK. Oscillation, just pure oscillation-- which is what we would see here with no damping-- is the frequency, e to the-- the solution-- the displacement, I could write-- the displacement up and down, y, will involve e to the i omega t, and e to the minus i omega t. We've got second order equation.
Let me just go back to that key point.
When we have second order equations, we look for two-- we expect and we want and we need two-- solutions.
There will be two.
I didn't put it here, and I should.
s, the frequency, is plus or minus i omega, because in both cases when we square, it comes out right.
So we get two frequencies, and here they are.
So let's see how to answer your question.
The presence of this i is only telling me that, essentially, I've got sines and cosines.
That's really what-- when it's a pure imaginary number-- I would call that a pure imaginary number, there's no real part at all-- then equally cos omega t and sine omega t. I can now, if I want, go real.
I can say, OK, these were the general null solutions.
Let me put this down, then.
The null solution-- I'm looking only right now at null solutions-- is some combination of e to the i omega t and e to the minus i omega t. That's what we got from plugging in e to the st, discovering that s was an imaginary number, and we got these guys.
But equally-- equally-- yn is a combination of cos omega t and sine omega t. And maybe you'll like those better.
I think everybody practically likes those better.
Do you see that these guys are the same as these guys?
The c's are a little different because, well, we know that we can switch from one to the other.
We remember that basic fact that e to the i omega t is cos omega t plus i times sine omega t. You're used to maybe seeing that omega t as theta, e to the i theta is cos theta plus i sine theta.
And e to the minus i omega t, of course, is cos omega t minus i sine omega t. I hope you won't think I'm filling the blackboard with formulas, because I'm really just writing down-- well anyway, they're beautiful formulas.
So if I have these guys, then I have these and vice versa.
If I have cos-- how would you write cos omega t using the exponentials?
I want to just see totally clearly that I can go back and forth between complex imaginary exponentials and cosines and sines.
So how would I, I want to go in the opposite direction and write the cosine and the sine as combinations of these, just to show if I've got combinations of one, I've got combinations of the other.
Combinations of these are the same as combinations of those.
So what is cos omega t in terms of these guys?
[INAUDIBLE] some of them divided by 2?
Exactly.
If I add those two, this part cancels.
I've got two of these, so I have to divide by 2, as you say.
It's a half of the first plus a half of the second.
And how about sine omega t?
Sine omega t is always slightly more annoying, because it's the one-- it's the imaginary part that brings in an i.
What would be the same formula?
How could I produce sine omega t out of that?
Yes?
The difference divided by 2i
Yes.
If I take the difference, that'll cancel the cosines.
So I'm going to take e to the i omega t minus e-- minus e to the minus i omega t. Take the difference.
But then I've got 2i multiplying this sine.
Up here I had a 2, but now I've got-- when I take the subtract, these i's are in there, so I divide by 2i.
So this just tells me that I can go either way.
Next time, we'll see what happens when there is damping and there are complex numbers instead of pure i omegas.
We're golden here.
We've found the great quantity with the right units.
The right units of omega are 1 over time.
Actually the units are radians per second, would be the typical appropriate unit.
Radians per second.
I'll use the word frequency for that, but there's another definition of frequency, cycles per second.
I just want to think about steady motion around a circle.
So this tells me how many radians per second.
And if this is 2pi-- if omega happened to be 2pi-- then I would go once around the circle.
If omega was 2pi, then when t reached one, I would be around the circle.
Let me draw a circle in a minute.
So there's a 2pi here hiding behind the word radians.
And in many cases, you'll want also a definition in cycles per second.
So f is omega divided by the 2pi, and that's in cycles per second.
Full revolutions per second.
And that's hertz.
I think I misspoke last time in confusing these two, so let's get them straight here.
There's no complicated math in here, it's just a factor 2pi, but of course that factor is important.
So a typical frequency in everyday life would be like f, 60 cycles per second, 120pi radians per second.
So I'm going around in a circle.
Now I'm ready to have initial conditions.
This connects, again, to the afternoon exam.
We found the general solution with some constants, like here.
Let's keep that real form.
And now those constants get determined by the initial conditions.
Conditions plural, because we have an initial position, like I stretch it-- maybe I stretch it and let go.
Maybe I stretch the spring and then I let go.
What happens?
By stretching it, I'm giving it an initial displacement.
And I'm giving it zero initial velocity, because I stretched it and just let go.
Another possibility would be to strike it.
If I hit that mass, that would be a different initial condition.
What would be the initial condition if it's sitting there in equilibrium quietly minding its own business and I hit it?
Then I've given it an initial velocity, with initial displacement zero.
So those would be the two extreme possibilities.
Pull it down, let go, or strike it when it's sitting in equilibrium.
Anyway, we've got two initial conditions.
You see why-- y double prime is showing up because essentially we've got Newton's Law.
This is Newton's Law.
Mass times acceleration is equal to minus ky-- that's the, with the minus sign, and the all-important minus sign, that's the acceleration.
That's a force, sorry.
Mass, acceleration, this thing with a minus sign is the force, and the force is pulling back.
If y is stretching, the force is restoring.
Let me just go ahead with what you know.
The initial conditions.
And I want to solve my double prime plus ky equals 0.
So I'm still talking about the unforced with given y of 0 and y prime of 0.
Just think for a moment.
Could you do that?
This is the most basic second order equation.
We know what the solutions look like.
Let's do this one in a box, cosines and sines.
We know what omega is.
Omega had to be square root of k over m. Then the equation was solved.
All I've got left is to get c1 and c2.
All I have left is to match-- choose c1 and c2 to match the two initial conditions.
So let me just do that.
What are c2 and c2?
At time zero, I have to match an initial displacement.
So at time zero, this is a 1, cosine of zero is a one, and that's a zero.
So at t equals 0, I have y of 0.
The displacement matches c1 times cosine of omega t, which is a 1, plus c2 times 0.
I'll put it in there plus c2 times 0.
C1 cos 0 plus c2 sine 0.
I've learned c1.
And also-- what do I do next?
I want to get c2.
And where is c2 coming from?
Now I would like to know what's the coefficient of the-- the initial conditions are supposed to determine that coefficient.
It'll be that initial condition that determines it.
y prime of 0.
The initial velocity should match the derivative.
OK, so what's the derivative?
y prime.
So the derivative of the cosine will be a sine.
And that will disappear at t equals 0.
The derivative of this sine will be a cosine with a factor omega.
So I'll have y prime of 0 will be the c2 omega cos of 0.
Which is what?
That tells me c2.
You could do all this without my pointing the way.
I'm solving this equation.
I have the solution in general form with two constants.
Now I'm determining those constants, and cosine and sine just determine them perfectly because cosine is 1 and sine is 0 at the start.
So we've got the answer.
The solution is y of t is c1 is y of 0 cos omega t, and c2 is-- now you'll notice, c2 is y prime at 0 divided by omega.
y prime of 0 divided by omega sine omega t. There we go.
Finished.
Finished.
Unforced problem solved.
Everybody in this room could get to that point.
Let me make some comments about that.
It's a combination of cosine and sine.
They're both running at the same frequency, omega.
I'm going to give a special name to that frequency, omega, this famous formula, all-important.
Lots of physics in that formula.
I call that the natural frequency, because the next step will be to drive the system by a driving frequency, which would be different from omega.
So we need to-- we've got 2 omegas.
Actually when I first wrote the book I thought, we've got to keep these two separate.
Everybody has to keep them separate.
My first attempt was to use little omega and big omega for the two.
I concluded after looking at it for a while that it was better to be more conventional.
People had figured out a good way to do it.
And the good way is to call this the natural frequency and put a subscript, m. So all the omegas that you see on this board should be omega n. I can change them all, but let me just change it here.
I'll change omega n. So that's omega n. We've only got that one omega right now, because we don't have a driving term yet.
So natural frequency has the advantage, which kind of made me smile, that the n stands for natural, and everybody calls it the natural frequency.
And n also stands for null, and we're talking here about the null solution, because there's no forcing.
So I could have subscripts on all the y's.
Eventually I'll need subscripts on the y to separate what we've done.
This is really yn of t, the null solution.
Good?
Now we could take one more step.
This is a combination of cosine and sine.
And we learned last time that that could be put in a polar form, but I don't plan to do this.
Let me just say I could do it.
This would be some amplitude, some gain-- maybe g for gain-- no, a for amplitude is good-- times-- what is this second optional form, which I'm just going to write here, say that we could do it, remember a little about it, but not make a big deal-- what is it I'm after here?
I'm looking to write this combination of cosine and sine, which is two oscillations, a cosine curve and a sine curve, but with the same frequency.
Then I can combine them into a single cosine, a single cosine of omega nt.
But now what else have I got in this form?
There's a phase shift, minus phi.
Thanks.
So there's an a phi, two constants, or there's y of 0, y prime of 0, two constants.
Let me not write again the formula for a or for phi, I don't plan to do anything with it.
It just could be done.
In other words, what we've done so far is just to see that the single spring oscillates with the frequency omega n. That's really what we've done.
A single spring oscillates with a frequency omega n. Saying that makes me think, let me look ahead to the linear algebra part of the course.
So where is linear algebra going to come in?
It's going to come in for a system of springs.
When I have another spring.
Can I draw another spring and another mass and another spring?
Say six springs, six masses.
Then-- and they could be different k's, different m's, or not.
Then we've got six displacements-- six differential equations-- coupled together, because the whole system is coupled together.
So what happens at that point?
That's the point where linear algebra, where matrices are coming in.
You want to see what's the point of matrices.
It's not a separate course by any means.
It's a most necessary part, because a single spring happens in reality but also systems today are coupled.
Big, actually, there are many, many things.
You have an electric circuit with thousands or tens of thousands of elements.
You have a coupled system with many gears, many oscillations going on.
So we need matrices at that point.
Can I even just add one more word about the language?
When we had-- here we have a frequency of motion for one spring.
What are we going to have for two springs or six springs?
The motion will be a combination of six different frequencies.
And so you'll see that it's a much more interesting, much more not so simple motion.
A combination of six pure frequencies.
And those frequencies are determined from the six eigenvalues of the matrix.
I'm just using that word looking ahead.
We will have a 6x6 matrix to describe the coupled system.
That matrix will have six eigenvalues.
It will tell us six natural frequencies, and our solution will be a combination of all six oscillations.
Here, it's 1.
Here it's 1.
That spring is not there.
So the problem we've solved now is the fundamental, basic problem, and I have to-- next step is forcing.
I now want to add a force that drives the motion.
In general, it could be any function of time.
Calling it f of t. So that's what I'm going to put in now.
But in reality, very, very, very often f of t is also a simple harmonic motion.
It's also a cosine.
But at a different frequency, at a driving frequency.
So I'm going to-- the next equation to solve is to put in cosine-- let's stay real for now-- at another, driving frequency.
At a driving frequency.
And of course, it could have an amplitude.
But let me take that amplitude as 1 to keep things simple.
So now I'm talking about forced motion.
Can we solve it?
How can we solve this equation?
Let me take out the 0 or-- take out the 0-- equals cosine omega t. With a different omega.
If the two omegas were the same, if the driving frequency is the same as the natural frequency, the formulas have to be slightly adjusted.
There's still an answer, but it's a case of resonance and you have to look separately.
But let's say, no.
Let's say omega d is different from omega n. How are you going to solve this?
I have to think myself.
How do I solve that.
Let's start a fresh board.
my double prime plus ky equals cosine of omega dt-- or often, I won't put the d. I don't have to put the d anymore.
Omega will now represent the driving frequency, because I've got omega n, the natural frequency, as the square root of k over m. What am I looking for now?
I found the null solution.
I'm looking for a particular solution.
I'm trying to keep the whole thing systematic.
Null solutions are now dealt with.
Took a little more time than just ce to the at for first order equations, because we've now got a two-dimensional collection of null solutions, but we've got them.
Now I'm taking a forcing term.
So I'm looking for a particular solution.
I'm looking for any solution to this equation.
I'm looking for a particular guy.
What do you suggest?
Again, it's a neat problem because of that particular forcing term, a cosine, an oscillation.
So I'm going to look for yp is some gain times [INAUDIBLE].
This is the next and, fortunately, a highly, highly important case, in which the particular solution has the same form as the forcing term.
It's just a multiple of the forcing term.
That's best possible.
That's best possible, is to have the forcing term reveal to me-- the forcing term immediately reveals a particular solution.
Once I know what I'm looking for, what do I do?
Substitute it in.
So I substitute that particular solution in here.
And notice everything is going to be a cosine, mgy double prime.
So what do I get when I plug this in for that guy?
I want to-- you can do it quickly, but let's stay together and do it together, because we can with this case.
What happens when I plug that in and take its second derivative?
I get the g. And then what's the second derivative?
[INAUDIBLE]
We have a negative, because two derivatives of the cosine bring out a minus omega d, will come out twice.
And I'll keep writing omega d for a moment, but then I'll give up on the d. Cosine of omega dt.
And then k times this, g cosine of omega dt, equals the forcing term, cosine of omega dt.
It worked.
This is one of that small family of nice functions where the solution has the same form as the function.
Actually that list of what you could call best possible forcing functions, where the form of the forcing function tells you the form of the solution.
That's a small family.
But it's fortunately a very important one.
Cosines, sines are included, and we'll see all the other guys that are included.
Most forcing functions we couldn't just assume that the solution had the same form.
It's only these nice ones.
But cosines are nice.
So what do I do now?
Everything is multiplying cosines, so I just look at-- I have minus m omega squared g-- g is going to factor out-- minus m omega squared and a k times g. Let me remove that off for the moment.
I'm canceling cosine omega, so my right hand side is 1.
That's it.
We looked for a solution with that simple format, and we found it.
Now we know g, the gain.
So the solution is-- this is g is 1 over k minus m omega squared times cosine of omega t. And omega is omega d. Omega is omega d now.
Does that look good to you?
This is the periodic solution going at the driving.
This is what the-- this g is the gain, the driving force.
The driving force is 1 times cosine omega d, then that 1 gets multiplied by this number.
This is, you could say, the amplifying factor.
I guess frequency response would be the right word.
Can I bring in that word, response, again?
Response is a word for a solution.
It's what comes out.
When the input is this, a pure frequency, the output, the response, is a pure frequency-- same frequency, of course-- multiplied by that.
That is the frequency response factor.
Notice we could write that a cool way, by remembering that omega squared-- that's wrong as it stands.
What have I forgotten in writing k minus m omega squared in that denominator?
I forgot a subscript, which is n. Which is n. This is n. This is-- is that right?
No.
Is it?
Or is it d?
Maybe I didn't make a mistake.
Is it d?
You're seeing a kind of critical moment.
Which is it?
[INAUDIBLE]
It's d, isn't it?
Yeah.
It's d. Sorry.
It's d. But when I see this and remember what omega n squared is-- omega n squared is k over m-- I can see that I can get an omega-- I can use this in here to make it even more interesting.
So it'll be equals-- let me get this box ready-- cosine of omega dt divided by-- now I just want to rewrite that.
I want to take out an m. I'm going to write this as m times k over m. m times k over m. Safe to do that.
Now I have a factor, m, that I can bring out.
And what is m multiplying?
That's the neat thing.
What is m multiplying?
k over m is-- omega n squared.
And this is minus m omega d squared.
Minus omega d squared.
That's pretty terrific.
The gain is this multiplier, 1 over m, times that.
And we see that the gain is bigger and bigger when the frequency is near the natural frequency.
And of course everybody has seen the pictures of that bridge-- wherever the heck was that bridge?
Somewhere in the Northwest, I think.
You know the bridge I'm talking about?
[INAUDIBLE] Tacoma, Washington
Yeah, I think Tacoma, that's right.
The Tacoma Narrows Bridge.
Right.
Tacoma, Washington.
Where the natural-- when you build a bridge, you've built in a natural frequency.
And then when traffic comes, it's doing a driving frequency.
And if you haven't got those two well-separated, you're in trouble, as this shows.
Or similarly, when an architect designs a skyscraper, there's going to be a frequency of oscillation, a natural frequency, at which that skyscraper swings.
And then there's wind.
Actually I talked yesterday to the-- by chance, the math department is not a very party-going department, but once a year we let it out.
And so we had our party at Endicott house out in the suburbs, and all the usual people-- that's all the professors I know-- came, of course.
But also, there was a really cool person.
He's the key architect for Building 2.
You've noticed that Building 2 is under wraps and we're moved out.
And we move back in January 2016.
So we've been out a year and a quarter and we have another year and a quarter to go.
It's going to be cool.
And you may say, well, who cares.
But the key point is Building 1 is next, and Building 1 is going to have the same cool addition of a fourth floor.
We're putting in a fourth floor, which all the-- Buildings 3, 4, 5, 6 go up to four, but Buildings 2 and 1 stopped at the third floor.
But there's a lot of space up there under the roof.
And they've discovered they could put a fourth floor up there.
Here was one interesting thing, though.
These buildings that we're sitting in are sinking.
You know that MIT was built on marshy land, just the way the Back Bay-- which is like the greatest idea in the history of Boston, the Back Bay and the dam that makes the Charles River beautiful-- was built by bringing in trainloads of earth from Needham.
So whole mountains and hills in Needham have come into Boston and come here.
So anyway, we're sinking.
You may say something like 3/16 of an inch a year is not something to worry about, but now it's been more than 100 years that these buildings have been here.
Anyway, not good to sink faster.
So the weight had to be controlled.
So by putting in a fourth floor, that put in a lot a new weight, and faster sinking, probably by some formula here.
Probably there.
So the weight had to get subtracted out.
It turns out that the ceiling, the roof to Building 1-- Building 2 and no doubt to Building 1-- was more than a foot thick of concrete.
Really heavy.
And some more asbestos probably, which we don't want to think about.
That's much reduced.
A whole lot of weight came out of the roof.
I think they probably did the calculation right, so we won't get rain coming through, but it won't weigh as much and the fourth floor is acceptable.
All this was a big decision by MIT to pay for that, or to raise money and pay for the new fourth floor.
But it's going to be fantastic.
And it'll be fantastic in Building 1 also.
So all that is discussion of that formula.
That's the frequency response, this factor to frequency, omega d, or omega, is this factor.
I guess I should say something about resonance.
What happens when that formula breaks down?
When the driving force equals the natural frequency, then we're dividing by 0, and something is different.
The formula isn't right anymore.
What enters in the formula-- let me just tell you what enters, and then we'll see it in a simple example.
When I have this repeated thing, two things are equal, what tends to happen is a factor, an extra factor, t, appears.
So an extra factor, t, will appear in the case omega n equal omega d. The solution, y, will be some factor, I'll still call it g-- no, I don't want to call it g, let me call it a.
There'll be a factor, t, times cosine of omega t. So in this case, there's really only one frequency.
We're driving it.
So the oscillation grows.
As you know, when you push a child on a swing, the whole point of pushing that child is to push at the natural frequency.
You wait for this swing to swing back naturally and you drive it again with that-- at that-- maintain that frequency.
And of course you see the amplitude-- the child swing higher and higher.
Presumably you stop pushing before disaster for that child.
But that's a case of resonance.
And it's what happened in the Tacoma Narrows Bridge, and there was nothing to-- nobody stopped, traffic just kept coming.
The movie is amazing, because there's one car that shows up after it's already swinging wildly, some crazy person still driving across.
And you might think, OK, that's ancient history.
But you know the bridge in London, the pedestrian bridge, the Millennium Bridge-- it's just a walking bridge across the Thames-- a big feature of modern London, and it had the same problem.
It was swaying.
People could not walk across.
They couldn't keep their balance.
So they had change it.
So it's not trivial to anticipate.
So now we've solved it-- we've solved the the null equation with no force, and we've solved the driving force equal to a cosine.
And of course, we could do a sine.
What other driving force should we do?
I think we should do a delta function.
I think we have to understand the fundamental solution is the case when, if we can solve it-- there's always this general rule, if we can solve with a delta function, that will give us a formula for every driving force, because every function is some combination of delta functions.
So if we could do it with a delta-- really the great right hand sides are-- well, cosines and sines I'll include as great right hand sides.
Those are the exponentials in disguise.
So the great right hand sides are really exponentials at different frequencies and delta functions.
Delta of impulses.
So now I want to find the impulse response.
That's the next-- that's really a job.
At this point, in these last 20 minutes when I solve my double prime plus ky equal a delta function-- well, what I was going to say was I'm now taking you to something that you won't see on the exam this afternoon.
But maybe you will.
Delta function, right hand side.
I haven't seen it yet.
Or I haven't looked recently.
You won't see second derivatives, I guess.
So what is it?
So now this is of the form with an f of t, a very special f of t. And that very special f of t makes that extremely easy to solve.
That's really my point here, is it it's going to be a cinch to solve that, and we practically have done it already to solve that with a delta function.
And the reason is sort of physical.
We have here our spring.
And what am I doing with that force?
I'm hitting the mass.
I'm striking the mass.
Let me say, and I'll write it on the board, the point I want to make about this.
That point is that this equation with a delta function force starting from 0-- say, y of 0 equal y prime of 0 equals 0, let's give it starting from rest-- it starts from rest by hitting it.
And that hit, that impulse, is in no time at all.
It's not stretched out.
It's hit over one second.
So this has the same solution.
This is the beauty.
This is why we can solve it so easily.
Same solution as-- let me write it and see what you think-- as my double prime plus ky equal 0.
We know how to solve those.
With-- it's still, when I hit it-- when I hit it, what happens in that split second?
In that split second, it doesn't have time to move.
It doesn't move.
It still has y of 0 equals 0.
But in that split second, we've given it a velocity.
We've given it a velocity.
And that velocity will be y prime.
The initial velocity is 1-- because here I had a 1-- over an m. We have to have the units right.
So here's a point, and we will stay with it.
We'll come back to this point next time.
Maybe the first thing for you to take in is the fact that it's such a nice thing.
We have this equation with this mysterious delta function, and I'm saying that the solution is the same as this equation with no force, but starting from a mass.
I'm tempted to take an example to make this point.
Let me take an example where the whole thing is a lot simpler.
y double prime equal delta of t. I've taken the spring away, so the k is gone, the mass is 1.
What's the solution to y double prime equal delta of t?
If we concentrate on this example, we're good for today.
So my point is the same solution as-- now, what's the other problem?
I'm just repeating here, but making it simple by taking k equals 0 and m equal 1.
So the same solution as y double prime equals 0, with y of 0 equal what, and y prime of 0 equal what.
I just wanted to repeat here what I've said there, and then we'll solve it and we'll see that it's all true.
If I look for a solution to y double prime equal delta starting from 0-- this was starting from 0-- if I say that's the same as this, what should y of 0 be here?
Zero
Zero, right.
It hasn't had time to move.
It hasn't had time to move.
But in that instant, what happened to y prime?
It jumped to 1.
That's right.
That's right.
Exactly.
Now just solve that equation for me.
Solve this example for me.
Suppose y double prime-- yeah.
Here we go.
What's the solution if y double prime is 0?
What are the solutions to y double prime equals 0?
[INAUDIBLE]
Constant and linear.
a plus bt, right, have second derivative 0.
Now what's the solution that starts from 0 that kills the a and has slope 1?
What's the answer to that question?
[INAUDIBLE]
t. t. The solution to this equation is a ramp.
It's zero everything in this course, is zero up until time 0.
At time 0, in this example, all the action happens.
Everything happens.
And what happens is it gets a velocity of 1, and the solution is y equal to t. y is 0 here, of course.
At that point, that's the key point, t equals 0, right there-- it gets a slope.
We don't have a step function.
There's no jump in y.
The jump is in y prime, the y prime the velocity jumped from 0 to 1.
That's exactly-- I think when I introduced delta functions and drew a picture.
What is the derivative, the first derivative, y prime, for that guy?
Let's just review, because this is what we've seen already.
The first derivative is--
[INAUDIBLE] step
A step.
And the second derivative is delta.
The second derivative of this is the first derivative of a step.
The derivative of a step is 0 everywhere except at the step, at the jump when it jumps to 1.
So that's the solution in this example.
And now to end the lecture, let's solve it in this example.
Again, let me just say-- why do I like this forcing term?
Mathematically, I like it because, if I can solve that guy-- as we're doing, we are solving it-- if I can solve that one, I can solve all forces.
Over here, I could solve when I had a very happy f of t, a perfect f of t, where I could guess the answer and push through.
Now with a delta, I can build everything out of delta functions.
That's why I like it mathematically.
Why do I like it physically?
Because it's a very physical thing to have an impulse.
That happens in real time, in real things.
And by the way, let's just, before I write down any more formula, what would-- I would like to be able to solve it for a step function.
[?
Heavy thud.
?]
I would like to be able to do that one.
I'm going to have to erase something, or I'll write it right above just for the moment.
I would also like to solve my double prime plus ky equal a step function.
So I would call the solution, y, the step response.
And what would be a step function start?
A step function start would be like turning a switch.
Suddenly things happen.
That's forcing by a step, so I'm looking for the step response.
And how do you think these two are related?
I look at the relation at the right hand sides.
What's the relation of this step to the delta?
Yeah?
One's a derivative
One's a derivative of the other.
And we've got linear equations.
So the right hand sides.
The step response, y step, and the delta response, y delta-- I'll use a different letter for this because it's so important.
One is the derivative of the other.
The great thing about linear equations is we have linear equations, differentiation, integration.
Those are linear operations.
The step function is just like a steady-- anyway.
I was going to-- I won't-- is the integral of the delta.
Step function is the integral of the delta, so the step response is the integral of the delta response.
I guess to finish the lecture, why don't we solve this problem, which looks tricky because it's got a delta.
Instead, we'll solve this problem, which doesn't look tricky at all.
It's exactly what we started the lecture with.
Zero forcing and some initial conditions.
So let me just finally make space for the big deal from today's lecture, which would be the fundamental solution with a force by a delta.
I'm just going to write down the answer when you tell me what it is.
What's the answer to that?
What's the solution to this second order constant coefficient unforced equation with those initial conditions?
We probably had it here.
I may just have erased it.
But now let's get it.
So y is y delta.
This is the impulse response.
y of t-- and I'll give it later another name.
So here's a perfect review question.
What's the solution to this problem?
Everybody remembers-- what are the solutions, what's the general form for the solution to the equation?
I'm reviewing today's lecture.
The solution to that equation looks like what?
[INAUDIBLE]
It's a cosine and a sine, right.
And then how much of a cosine do we have and how much of a sine do we have?
The initial condition will tell me how much of a cosine we have.
And what's the answer?
None?
No cosine.
This condition, this initial velocity, will tell me how much of a sine we have, because the sines are the things that have initial velocities.
So it would be a sine of-- the sine of what?
Square root of k over m, right?
omega nt, right?
And what's the number?
What's the number so this has the right-- let me write again what I want.
I want y prime at 0 to be 1 over m. What's the number that I put in there?
I've got something, its derivative, at zero.
This is some number-- I'll call it little a for the moment, but I want to find out what it is.
Are we right?
Yeah?
I think we're right.
Yeah.
The derivative is at zero, so I just plug that into here, take the derivative at zero-- of course that makes it a cosine, which will be 1-- but it also brings out that factor.
So a times-- well, that factor will be 1 over m, and that tells me what a has to be.
Well, this is omega.
So a is-- this is omega a equal m. This is 1 over m omega.
And that's omega.
Sorry I'm erasing stuff which I-- this is the formula I'm after.
Sine omega t over omega.
I think we're good.
Are we?
Yeah?
Yeah.
I'll come back to this in-- Wednesday is my day to move to damping terms.
I've intentionally stayed with undamped equations here, because you're thinking about that level of equation.
Damping is going to bring in new stuff, and that should wait till Wednesday.
Shall I recap today?
I'll just recap today, and then we're done.
Today started with the unforced equation.
We solved it by assuming-- by not thinking ahead, just assume I have an exponential, because the beauty of exponentials is, when I plug it in, the exponential cancels.
And that told me that s was pure imaginary.
It told me that it had this form, e to the i omega t. And there were two s's.
Two possible s's, plus and minus.
I get to make a little comment about this example here.
What was omega?
What's the natural frequency in this problem?
What's the natural frequency here?
I guess this is a case where-- what's the natural frequency?
I guess this is a case where m is 1 and k is 0, is that right?
This does fit into that pattern, but it's a little special.
This is a case where m is 1 and k is 0.
So what's the natural frequency in this?
Zero
Zero.
Zero.
This is a crazy case of resonance.
It's a case in which the natural frequency and the driving frequency, say in this-- I'll have to do it here-- this simplest of all equations is, in a way, special.
It's a case when the natural frequency is zero and the driving frequency is zero and they're equal.
And what happens with resonance?
What's the new formula, the new term that comes in with resonance?
It's t. You saw it happen for this example, and we didn't have to use the word resonance.
We knew that we had a ramp.
We just used the word ramp, not resonance.
But this is a case of resonance.
When omega n is zero and omega d is zero.
And the factor t up here.
Anyway.
Just that small comment there.
And now, just going back to the recap.
The recap was, we tried exponentials.
We learned that they were pure oscillations.
We realized that we could do cosines and sines instead, and we did.
And we took off.
We got the formula.
Then of course the-- so this is section 2.1 of the book.
And it goes through all those steps carefully.
Section 2.2 of the book tells us about complex numbers, and section 2.3 brings damping in.
So that's what's coming next time.
So the recap again.
We found the null solution, we found a particular solution-- oh there's just one comment I want to make, and then I'm done.
Where was our particular solution?
Yeah.
This was our particular solution.
Here's my comment.
Here's my comment.
Suppose I want to solve this basic equation starting from a given y of 0 and a y prime of 0.
I'm going to do it in two parts, I think.
I've got the null solution, and I've got this particular solution.
Now here's my point.
If I want to get y of 0-- how shall I say this.
You can't just put together-- it's an easy mistake to make-- solve the null equation with the initial conditions and then add in the particular solution.
You'd think, I just followed all the rules.
But this particular solution that you added in has a-- at t equals 0, it's not zero.
So you have to change.
So the correct thing, the correct yp plus yn-- let me make that point.
Just a warning.
So in words the warning is, remember that the particular solution has some initial condition-- in that case, g-- and then that is going to affect the right null solution.
So again, y is y null plus y particular-- plus y of particular-- so it's some c1 cos omega nt plus some c2 sine omega nt plus this particular guy, g, cosine of omega dt.
All correct.
All correct.
But now, put in the initial conditions.
y of 0 is given.
And what do I get on the right hand side when I put in t equals 0?
I get c1 here.
What do I get when I put t equals 0 in there?
Nothing.
What do I get when I put t equals 0 in here?
g. So it's not c1 equal y of 0 anymore.
That's the easy mistake that I'm correcting.
When you put in this particular solution, it has an initial value.
That initial value is going to come in here.
So c1, then, the correct c1 is y of 0 minus g. End of story.
Just don't be too quick to just add the two pieces and think you can do them completely separately, because you're putting them together.
And then you have to put them together in the initial condition.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
So morning.
This is my fourth lecture on differential equations, that part of the course.
And I haven't said anything about the textbook.
That's Differential Equations and Linear Algebra.
I wrote it because so many courses like this one want to combine those two topics.
They're the two major topics of undergraduate math, after calculus, the two major directions, and the book connects those two directions.
And I just wanted to give you the website for lots of things connected with the book.
So it's the math website, dela for differential equations and linear algebra.
And so today is differential equations, second order, with a damping term, with a first derivative term.
So that in many engineering problems, those coefficients A, B, C would have the meaning of mass, damping, and stiffness.
Mass, damping, and stiffness.
And physically, we know what the mass comes from.
The stiffness comes from a spring, as I drew before.
And traditionally we describe the source of damping as a dashpot.
I guess in my whole life I've never seen a dashpot, but maybe it's-- think of a piston going up and down within a cylinder of oil or something, with resistance.
OK, so that's the left-hand side.
Linear constant coefficients still.
We don't have formulas, it's not easy to see what's happening when things are varying, when the equations non-linear, so this is the starting place.
OK. With some forcing.
So this is damped forced motion.
This is the ultimate within linear equations.
OK. And now, what's the forcing?
So now, always we solve first for the null solution.
With no force, what are the natural motions?
And we'll find a formula for those.
Then the big one, the special right-hand side is always an exponential.
So this is going to be y is going to be the null solution yn, and we'll get a formula for that.
This, the exponentials.
Always, the response to an exponential is an exponential.
That's like the most important fact in getting solutions.
So the response will be some ye to the st at the same frequency-- well, I say frequency.
If s is a real number that would be a growth or a decay, but very frequently s is an imaginary number, like last time.
And this is comes from a rotation or an oscillation.
And then, the complete picture comes from being able to solve it with an impulse.
So then y is the impulse response, which I write as g of t. Let me introduce just that letter g. That stands for growth factor in first order equations stands for Green's function, so that word Green is getting in here, his name.
And it represents the impulse response.
OK, one, two, three.
And then the point of doing this one is that then we can do it, then we can get a formula for any f of t. So this is the pattern that we followed for first order equations.
We followed it for second order equations that didn't have damping.
And now we're doing the big one with damping.
So what's the damping going to do?
What's going to be effect of damping?
Say if I had a right side of one, just a unit force?
The damping.
So I'll still have oscillation-- you'll see this-- I'll still have oscillation, but its amplitude damps out.
Like that like everything we know, like something swinging back and forth, but friction is it is eventually damping out that motion.
So it'll be oscillating with exponentially decaying amplitude.
Let's find this solution.
OK. How do we find the null solution?
I've got constant coefficients here.
So the nice, the right thing to look at is exponentials.
In first order equations it was e to the at.
Here, we'll have to see what it is.
So I'm going to try a particular-- I'm going to look for the right exponentials.
Certain exponentials will be null solution.
Can I do it?
So I take that equation and put in y equal e to the st.
Remember now, I'm doing 0 here so it's not the same s as the right-hand side.
OK. What happens if I put in-- so I'm looking for the null solution.
I'll try y equal e to the st. Plug it in.
I get m. Two derivatives gives me s squared e to the st, b.
One derivative gives me an s, e to the st, and k gives me k e to the st. And that's supposed to equal to 0 for the null solution.
OK. Have we made a good guess at what will work?
Yes, because I can cancel e to the st, and I come to the most important equation.
The equation that governs everything in this whole lecture is ms squared plus bs plus k equals 0.
That's called the characteristic equation, or it has many names.
It's obviously the big deal.
It's the thing.
It's an equation for s, for the special frequencies that solve the null equation with no force.
OK. And how many values of s do we expect?
Two.
So I expect solutions s1 and s2, and then my null solution is e to the s1t is a null solution.
E to the s2t is another one.
My equation is linear.
I can multiply that by any constant.
I can multiply this by any constant.
And I can superimpose, i can add, I can combine, because I can do linear algebra here.
This is the most important operation in linear algebra, multiply things by constants and add.
That's called a linear combination.
It's the basic operation in linear algebra and it's a basic operation here, because we're doing linear algebra with functions.
Do you see that that's it?
We've got it, except we should really write a formula and draw some pictures to show s1 and s2.
What would be the formula for s1 and s2, to the two solutions?
We remember that from school, it's the quadratic formula.
The two solutions to this.
Everybody remember this one?
Let's see if I do.
Does it start with-- a minus b.
And then there's going to be a denominator that I'll remember, which is 2a.
No.
I said a but I mean m. 2m.
And now this is plus or minus-- what goes into here?
B squared minus 4ac.
The key quantity in this whole business, b squared minus 4ac.
Ah, what is it?
Mk, thank you.
4mk.
Great.
And can I just remark, a little remark about units.
B squared has the same units as 4mk.
It has to or such a formula would be crazy.
So we will see that actually there's something called the damping ratio that involves the ratio of these guys.
OK, that's the formula.
But it's not like-- OK it's got a square root, and what's the point of-- the thing we have to remember with this square root is that if it's the square root of a positive number then we have a plus or minus, ordinary real numbers.
If this thing is negative, then what?
What's up if b squared is smaller than 4mk?
So if b squared is-- if there's not much damping, if it's underdamped, b squared would be smaller than 4mk.
And what then?
We've got a negative number here.
When we take its square root we have an imaginary number.
That's oscillation.
Under-damping is going to show oscillation.
Let me draw this.
Let me draw that curve for different choices of m, b, and k. This is a good picture to see.
So let me draw first of all one that has b equals 0.
OK.
So there's a curve.
What I'm drawing is, up on this curve is this ms squared plus bs plus k. Ms squared and bs and k. And here, this one is one with no damping at all.
This is just s squared plus 0 s plus 1.
That's what that curve is That's the example we saw before.
Y double prime plus y. Y double prime plus y equals 0.
This is coming from Y double prime plus y equals 0.
This is pure oscillation.
OK. Now let me bring in some damping.
Now, as I bring in damping the curve will move.
And it takes a little patience to see where it moves.
Let me have a little damping a little damping, so s squared plus 1 s plus 1.
I think the curve-- so this is s here.
And it's a parabola.
Everything I draw here is a parabola, is just that different parabolas come from different choices of b.
So I think that this choice, I think it goes a little bit like that.
That's the next one.
It goes down a bit.
Now, let me do s squared plus 2s plus 1.
Now, let's see, I really should stop for a moment and solve the equation, find the roots for each of these guys.
And then I'm going to have an s squared plus 3s plus 1.
So we're doing something very straightforward, parabolas.
But it shows us the different possibilities.
And we could give them names.
OK.
So I would call this one undamped.
And what are the roots of that equation?
S squared plus 0 s plus 1 equals 0.
What are the s1 and s2 for that guy.
So the roots of s squared plus 1 equals 0 are?
Stay with me here.
S squared plus 1 equals 0, that's the equation that has roots
I and minus i
I and minus i. I and minus i.
So this is undamped.
S1 is i and s2 is minus i.
That's the pure oscillation.
Pure oscillation, that's the case where b is 0 here, I have a square root of a negative number, and it gives me plus or minus 2i, and then the 2s cancel and I get plus or minus i.
So I can always go back to this but I'll try to choose numbers that come out nicely.
Now, what happens with some damping?
This guy.
What are the roots for this one?
Well, I better use this formula.
Now, I'm always keeping m equal 1 and k equal 1, in all those samples.
But now b has increased to 1.
So if b is 1, what do I have?
I have s is-- the roots are minus 1, plus or minus the square root-- And it's going to be just 2 down below.
What's in the square root, the all important square root.
When m and k and b are all 1.
Just do the calculation with me, so you see it.
It's negative?
Three
Negative 3.
So what's the point there?
The point is, this is going to be square root of 3i or minus i.
We have oscillation at a frequency square root of 3, and we have decay from s minus one-half.
The real part of this is giving us the drop off.
We didn't have any drop off at all in this case.
They were pure imaginary.
Now the s1 and s2 are whatever I have a minus 1, plus or minus square root of 3i over 2.
Those are the roots.
All right, I'm ready for this guy, and it's particularly nice.
It's particularly nice.
What do you see for s squared plus 2s plus 1?
When you see this parabola, now b has moved up to 2.
What's up with that one?
It's going to be-- let's see-- that looks like a perfect square to me.
Right, inside the square root is 0.
Right, exactly.
Inside the square root, b squared is 4, and 4mk is 4, so I have 0 inside this square root.
So now I have s equals minus 1 plus or minus 0 over 2.
That's when this was the case, when b moved up to 2
[INAUDIBLE] minus 2
I'm messing it up?
[INAUDIBLE] equals minus 2, plus or minus 0
Minus 2 plus or minus 0.
Thank you.
So what do I have?
What are the roots of this guy?
Negative 1.
What's the other one?
Negative 1.
A double root.
It's critical damping.
Critical damping-- it's not underdamped.
It's not overdamped.
It's right on the borderline.
And I see that, when you first saw quadratics, before anybody brought up that awful formula, you would have factored this into s plus 1 squared equals 0, and you would have discovered that s was minus 1 twice.
A double root.
So the picture there would be-- there's minus 1.
Yeah.
Everybody recognize that this, we're hitting 0, height 0.
We're hitting 0 twice at s equal minus 1.
So this is now the case.
This was b equal 0, no damping.
This was b equal 1, under-damping.
This is b equal 2, critical damping, just on the border.
And what do you think the s squared plus 3s plus 1 is going to look like.
Again, we can find it.
No, let me do the s squared plus-- let me take b equal 3 and find the roots and draw the picture.
If you're with, me that picture and these formulas tell you the difference between these four cases.
So what do I get?
Minus 3 plus or minus the square root of, 9 minus 4, is 5, over 2.
So I have two negative roots.
I have decay.
I have decay at a fast rate and a slow rate but both are giving decay.
So the curve now is coming down here and back up there, and it hits there, and it's got the two roots.
These two roots are x and x.
So these are s1 and s2.
Let me copy them over here.
S1 and s2 are minus 3 plus or minus the square root of 5 over 2.
Yeah.
Real.
Real roots.
So this is two real roots.
This is a double real root.
This is two complex roots.
And this is two pure imaginary roots.
The four possibilities.
The famous four.
OK. And for me, that picture is a-- so this was the b equal 3 curve, more overdamped.
It's a little interesting that overdamping has this root that's pretty near 0.
So overdamping doesn't mean that you go to 0 real fast.
Actually, the one that goes fastest to 0 is the critical damping.
Then, as b grows, one root gets closer to 0, so it's like slower decay, and another root is going off to fast decay.
I think you have to know those guys, because the physically that's very important, where's the damping.
But we've now found the null solution completely.
The null solution completely is-- let me write it again here-- is anything times e to the s1t and anything times e to the s2t.
And where do those constants, c1 and c2 get decided?
By the?
[INAUDIBLE]
Initial conditions.
Right.
These two conditions determine c1 and c2.
OK. Are we good?
So the null solution has already separated the different cases that depend on how much damping.
I'm ready for number two.
Number two now.
OK.
So the idea is I now have a forcing term, some frequency e to the st. And I will assume that it's not-- s not equal s1 or s2.
That's to make my life easy.
Just as last time, the formulas will break down.
And I'll have to put in something different if there's resonance, if the driving force is at the same frequency as a natural frequency.
In that case, there's a resonance.
And the way you spot a resonance formula is there's an extra factor of t. There's a growth of t. Up here, we're seeing no factors of t. Over there in the exponent, of course.
I mean, down below, there would be a t times e to the st. And the book does that carefully.
I don't see that it has a place in the very first lecture on this topic.
OK.
So this is saying no resonance.
All right.
What's the solution then?
The solution is a multiple of e to the st.
The input was e to the st.
The response is a multiple of e to st, so that's the frequency response.
Capital Y is telling us the response to s. And it's a very, very important function.
It's called the transfer function.
It's just the key to everything.
I probably say that so often, that this is the key to everything.
Well, it's partly because I just have one lecture to do a big part of differential equations, and it's got some key ideas.
And the first key idea is s1 and s2.
And the second key idea is Y.
And let's go for it.
All right, so this is the s1 and s2 picture.
I'll move that up, and now, in the equation, try Y e to the st. We hope it works.
What is this?
I'm trying, this is my solution, and it's a particular solution now.
I got the null solution.
I've moved to a particular And this is when the force is e to the st.
In other words, I'll just say again, if we have an exponential force, try an exponential response.
1803 would call this the exponential response formula.
So you could use the word exponential response, very appropriate.
You could use the word frequency response.
That frequency response is kind of the right word when s is an imaginary number giving an oscillation frequency.
OK, I'm going to plug that in.
So into the differential equation, m. Second derivative of that is s squared Y e to the st, right?
That's m Y double prime.
The beauty of exponentials.
Take every derivative, just brings down an s. The next term.
What's the next term?
Maybe do it with me, do it for me.
What happens when I plug in this as Y into b, so there'll be a b Y prime.
So what's Y prime?
[INAUDIBLE]
s, Y, this constant, e to the st. And then the final term is k times this Y itself, no derivative, so it's just Y e to the st, and that's matching e to the st. That's the force.
This is f. F, this is an exponential force.
Of course, I could and should have a constant here to give me the units of force.
Let me just keep the formula as clean as possible by taking units, so that's one.
OK what do I do?
[INAUDIBLE]
I cancel.
The nice part of 287 is canceling e to the st. That's the most fun you get.
And now, I have Y's on the left.
So Y is-- can I see what Y is?
Y is this 1, divided by the coefficient of Y.
And what's the coefficient of Y?
We know it.
We've seen that coefficient.
Y on the left is multiplied by ms squared plus bs plus k ms squared, again, ms squared, bs, and k. Multiplying Y, I divide by that, so I put it down here.
Ms squared plus bs plus k. And because s is not one of the roots, that's not 0, so we're golden.
That problem took five minutes.
The null solution took half an hour.
The exponential response is clear.
And you can see what it would be.
And let's give it a name.
This is the transfer function.
Widely used name.
Other names could be given but that's the best.
So it's a function of s. It's a function of s. And so, again, when f is e to the st, it sort of transfers the input into the output.
That's the way I think of the transfer function.
Here is the input.
The output is just multiplied by the transfer function.
And the transfer function is just that nice expression.
Just that nice expression.
So we are golden for a frequency, for a linear equation with an exponential forcing function.
What would be another example?
I'm using mass, dashpot, spring here.
If this was in electrical engineering, what three things would I be using instead of mass, dashpot, spring.
The three guys would be?
What would correspond to the dashpot?
So can I just draw here a little-- let me put on a low voltage and put on something that does this, and then something like this.
And then there's something like this.
And give me a break, tell me what these things are.
This guy is?
Inductor
An inductor.
This guy is a?
Resistor
Resistor.
Now that's the one that's like damping.
This resistor here is like damping.
Like the damping term, or maybe 1/b.
I'm not getting its units right because I haven't got any equation here at all.
Resisting is-- there's friction in that resistor.
It burns up heat.
And similarly, the dashpot slows things down.
And then this guy is a--
[INAUDIBLE]
Capacitor, right.
In other words, you can do the mechanical application and the electrical application with exactly the same ideas, just a change of letter, and of course, different units, but same problem.
OK, so that's a comment that you've seen before.
What else do I want to comment on?
Because this example was really so straightforward.
I think what I want to mention, and this is important, is that this is the central starting point for the Laplace transform.
So I can't do Laplace transforms all today by any means, and so Professor Fry will talk about the Laplace transform next week.
But what is the point of the Laplace transform?
The point of the Laplace transform is to get your money's worth out of the simple formula for exponentials.
Having an exponential there turns the whole differential equation problem into an algebra problem.
We just have quadratic equations.
We just have a division by a quadratic.
That's the great thing about the Laplace transform.
It turns the t domain, the time domain, where we have exponentials, into the s domain, the exponent domain, the frequency domain, where we just have quadratics.
And then first order equation's just linear.
And we can even get from second order of the quadratic to linear, because I can factor that guy.
If I factor into s minus s1 times s minus s2, I've two linear pieces.
And that's the first step in the Laplace transform, in the algebra.
So all of the algebra in the Laplace transform is this algebra for e to the st. And then the job of the Laplace transform-- and this is the tricky part.
So let me even take a little board space on that.
So this is like a heads up for next week.
So the Laplace transform.
OK.
So for any forcing function, f of t. That's the thing that we're going to take the Laplace transform of and the response and its response, y of t. OK.
So here's the idea of transforms in general.
I choose some terrific functions like exponentials.
So I want to convert my problem to exponentials.
e to the st for all s, s between let's say 0 and infinity.
So what do I do?
I take my function, and I figure out how much of every exponential is in it.
That's the Laplace transform.
I take my function f of t, and I go to what you'll see next week, its Laplace transform.
Let me call it capital F of s, or it's sometimes written the Laplace transform of F of s. Something like that.
I'm not going to do that.
I'm not going there.
So F of s is like the amount of a particular exponential in my function.
If my function is just the sum of two exponentials, then the Laplace transform just is a big bump on one and a big bump on the other.
But most functions, like some forcing function, has some e to the st is for all for a whole range of s. So I figure out how much of this is.
OK now second step.
Using linearity, I can solve the problem for when the right hand side is just that.
Then solve for right hand side, F of s, e to the st.
Solve for each frequency, each s separately.
And what does that mean?
That means just dividing by this.
So this was the easy step.
So this is step one, is take the Laplace transform.
The Laplace transform tells you how much of each exponential is in it.
Now step two is a cinch.
Step two is a cinch.
I just multiply by the transfer function.
I divide by this, bs plus k. And now, what's step three.
Step three, I have the solution for each separate exponential, but I've got a whole lots of exponentials, so I have to do an inverse Laplace transform, add up, figure out what function has this Laplace transform.
And that's often the place where the algebra gets harder.
In principle, we can do it for any F of t. We can take its Laplace transform, we can solve for each frequency in the transform, we can assemble them all together.
That's the inverse Laplace transform, so this is the inverse Laplace to get the solution y of t. So this is really the Laplace transform of the solution, and we have to get back to the solution itself.
Can I just let you think about those ideas?
I'm not up to describing the algebra here.
The point is the Laplace transform takes this into separate exponentials.
Each of those right hand sides is simple algebra, divide by that.
And then you have the job of okay, what function has this Laplace transform, to go backwards.
And that's usually the hard part.
So people make tables of Laplace transforms.
Everybody remembers the Laplace transforms of a few functions.
You could say that those few functions are the golden functions of differential equations.
The golden functions of differential equations are the ones where you know their Laplace transform and you can go back and forth easily.
And what are those golden functions?
Well, you might guess exponentials, simple polynomials, 1 t, t squared.
You can do Laplace transform for those.
What else do you think would be nice?
Cosine and sine.
And you could multiply those together.
We could deal with t times cosine t. So a little bit different.
But having said that, the reality is I've given you the whole list of nice functions.
Those functions show up in every simple method for solving equations.
There's a method called undetermined coefficients and what does it amount to?
I'm sorry, I don't have time to say it this morning, but it can come later.
Undetermined.
It just means if the right hand side is one of these nice guys-- shall I write down again the golden functions?
e to the st is like the platinum function.
And then some golden functions are like t and t squared and so on.
Of course, when we get that, we're close to cosine omega t, and sine omega t. All these guys, their Laplace transforms are nice.
We can deal with them completely.
Or multiply any of those together.
And when the right hand side is one of these golden functions, you can write down the answer.
We've focused on this one because it's the platinum one.
And we did these two too, because they come from s equal i omega.
And then these guys are a little bit of juice.
But that's it.
I'm sorry the list isn't longer.
It'd be nice to have--and of course, people for centuries have worked with the next hardest functions.
You know, the silver functions.
Famous functions have names like Bessel's function or Legendre's function.
Others where you can get pretty far.
Those are the best.
Then you have the famous ones where you get pretty far, in the web has the Laplace transforms, and then you get the general function, F of t. Oh, could you get anywhere with delta of t?
Oh, yes.
Does it belong on the list of golden functions?
Yes, it does.
I almost forgot it, and it's like-- I'll call it-- Yeah, delta of t. Yeah, that's a beauty.
That's a beauty.
The Laplace transform of delta of t happens to come out 1 over s. You can't ask for more than that.
Or maybe it's one.
Yeah, it's probably one.
Yeah, the Laplace transform of that is one.
Yeah.
It's got all exponentials in sort of, you could say, equal amounts.
OK.
So that's some thoughts about that about Laplace transforms, just sort of the big picture that takes the differential equation, turns it into an algebra problem, and then at the end, you have to get back, and that's the part that's not always doable.
OK.
So what's left for today is this guy.
Now this one now.
Have I got to that point?
So this will be the final ideas in this course, this four unit core elective.
What is the impulse response when there's damping?
What's the impulse response when there's damping?
OK. OK.
So that means that I would really like to solve m y double prime, b y prime, k Y equals an impulse, delta of t. Because if I can solve this, then I can solve everything.
And I can solve this.
It turns out to be easy.
Now why is it easy?
You might think, my gosh, we've got second order equations here, we've got a delta function there, where do we go?
And so my advice is go this way.
The solution to that is the same.
This with y of 0 equals 0 and y prime of 0 equal 0.
So you have a spring, so again we have a spring with a mass.
And that spring is in a damper.
So can I just, without knowing what I'm doing, draw a damper around it.
So the idea is I'm striking that mass at time 0.
Striking that mass at time 0.
What happens?
What happens immediately?
And then I don't touch it again.
I strike it and that's it.
I've set off, and what have I-- so my point is this has the same solution as m y double prime plus b y prime.
And Ky equals 0.
Nothing happens beyond time 0.
But what are y of 0-- Well, let me give this a letter g, just to emphasize it's special and deserves its own name.
So now, what is the starting position and the starting velocity for this picture?
So I'm saying that the y here is the same as the g there.
And really if you see it physically then you see it best.
So again, at the instant t equals 0, I'm hitting that mass with a unit impulse.
So what is the position of that mass, instantly after I've hit it?
0
Still 0.
Good for you.
Good for you.
And what is the velocity of that mass instantly after I've hit it?
1
1 is essentially the right answer.
1 is the right answer.
But the way I've set it up is, there is an m there.
If I put an m there, which would be nicer, then the answer would be 1.
But the way I've written it here, I have an m there, and I haven't fixed the units.
So that turns out to give me a 1/m there.
I could explain why it's a 1/m, but let me just for the moment say it's my fault.
It's because I didn't get any units right that we have 1/m.
No big deal.
But now, what's good here?
What's good?
This was a problem with a mysterious delta function.
This is a problem with 0.
And the only price we're paying is the impulse gave the mass a little velocity.
And you can imagine that the velocity gives it is 1/m because the strike didn't tell us about that.
So what I'm saying is we can solve that equation for g. We can find g, and in fact, we have.
So this really brings the lecture full circle.
What do I have here?
I have a null solution.
So this g here is a null solution.
So what form does it have?
What can you tell me about g of t?
And it's the same as y of t. So y is the same as g, and what's the form for it?
Yes?
Tell me?
I'm taking it back to the very beginning of the lecture where I solved it with a 0
[INAUDIBLE]
C1, thanks
[INAUDIBLE]
e to the?
C1, e to the s1, t, the two s's, s1 and s2, the special two s's, the special roots of the key equation.
That equation gives us s1 and s2, by this formula, or in a homework or an exam problem, we hope that it would come out easily.
So this is that part of it.
What's the rest of it?
C2 e to the s2 t. The impulse response is the particular null solution that starts with a shot, starts with an impulse, starts with a strike.
OK.
So I just have to find c1 and c2 here.
All right?
We've come back to the basic problem in differential equations.
We've got the solution.
We've got two constants.
We've got two equations.
We just plug that function into that equation.
It gives us one fact about c1, c2.
This gives us the second fact.
We solve them.
Why don't I just write down the answer?
It turns out to be e to the s1 t, and the other guy will come in with an opposite sign e to the s2t over s1 minus s2.
I think that gives us the c's.
Oh, there'd be an m. There'd be an m times this, because of my messing with units.
So can we just check?
At t equals 0, what do I get out of this?
0
0.
What I'm supposed to.
What's the derivative at t equals 0?
The derivative at t equals 0, this derivative is going to bring down an s1.
The derivative here will bring down an s2.
At t equals 0 the exponentials will all be one so I'll just have the s1 minus the s2.
It'll cancel that and it'll be 1 over m. So this is the neat formula for the impulse response.
That's the neat formula for the impulse response.
And then why-- can I use this little corner of the board?
Why do I want that impulse response?
What can I use it for?
It gives me the answer, not just for the impulse but for everything.
The particular solution is for any forces, force, I multiply by whatever the-- let me write the formula and I'll show you what it says.
Yes, there is the formula.
I'm sorry it's squeezed.
But really, the goal here was simply to get a handle on what is the response to any f. And again, I look at that this way.
F of s is the input at time s. G is the growth factor over the remaining time up until time t. So Y at time t, I take all the inputs up to time t, And each input gets multiplied by its growth factor.
It was e to the a, t minus s in the first order equation.
Now we've got two exponentials.
But that's the solution of the general problem.
So we have now in one lecture completed a solution to the second order constant coefficient differential equation.
Right.
Yeah.
By finding the impulse response.
Yes?
[INAUDIBLE] would we still be able to [INAUDIBLE] if s1 is equal to s2?
Ah, if s1 equals s2.
That's the case where formulas need a patch.
They need a patch.
If s1 equals s2, what do you think happens?
If s1 equal s2, everybody sees, I have 0/0.
And so this is like a technical question that I wasn't going to ask myself.
You asked it.
You're responsible.
What do we do for 0/0?
What did you learn in calculus?
Who's the crazy guy who figured out how to deal with 0/0?
In a way, calculus is all about 0/0, right?
Delta y over delta x, they're both headed to 0.
And suppose you have-- let me take the most famous example of 0/0.
It's like sine x over x, as x goes to 0.
So x going to 0.
Sine x goes to 0, x goes to 0.
What's the answer, by the way?
What happens to that ratio as x goes to 0?
This is maybe the most famous example.
Sine x over x, when x gets very small, is close to?
1
1, thanks.
And now just help me out with the name of the guy.
It's a crazy spelling name, and do you remember?
L'Hopital.
L'Hopital.
Everybody despises him.
Probably hi friends despised him.
But anyway, L'Hopital says in a situation like that, when you're going to 0/0, you're allowed to do something a little strange.
You're allowed to take the derivative of the top, so it has the same limit.
Instead of looking at this 0/0, you can take the derivative of the top, cos x, divided by the derivative of the bottom, 1.
And now you can let x go to 0, and you get the right answer.
So this gave a 0/0.
Unclear.
Fuzzy.
This gives-- what's the right answer then?
Just tell me again.
When x goes to 0 this becomes?
1
1.
Right.
So that's L'Hopital.
So that's what I would have to do here.
I would take the derivative of this, and the derivative of this-- the only sort of tricky part is it's the s derivative.
It's s1 going to s2.
Let me just tell you the result.
Since you asked.
A factor t comes out.
It's t e to the s1, or s1 is the same as s2, divided by the m. It actually looks simpler.
There's only one.
This is in the case s1 equal s2.
So s1 is the same as s2.
I just chose s1.
Yeah.
L'Hopital gives a simpler answer.
And it's got this suspicious and recognizable factor t. That came from L'Hopital.
OK, I won't do that stiff.
So let me say again, we've now done the second order constant coefficient equation I do just have 10 minutes of something to make it better.
And that is that the famous quadratic formula for s, for s1 and s2 is not beautiful.
It's correct.
It's correct.
But it's a little bit of a mess.
You've got three things, b and m and k playing around.
And we saw in this picture, we saw all the differences.
I guess in this example I kept m1 and I kept k1, and I increased b. I could do other examples where I increase the k, I make it stiffer and stiffer.
All these examples.
And engineers have worked for 100 years to see, out of this formula what are the important parameters, what are the important numbers, and hopefully, where possible dimensionless.
So I just want to- the final minutes would be-- back to high school-- playing with this formula, to get better numbers in there.
May I do that?
I just think, because then you'll see-- I learned this, actually, so this is like something math professors have no reason to do.
Look at that.
That's the formula.
End of story.
But the Web, 1803 website, has a class in which Professor Miller from the math department was teaching this subject, doing these, exactly these, but also Professor Vandiver from Engineering was putting in his suggestion of what are the good parameters?
What are the parameters that engineers look at?
So that would be my final comment, and I won't do it as well as Professor Vandiver did.
But can I just take that-- let me erase these two special examples, and look at this question.
Again, the book will do it.
So one nice-- b/2m is a pretty natural parameter to use.
Let me introduce that as one of them.
I'm going to, by taking ratios like b/2m, let me call that p. Let me call b/2m.
So that's a ratio of damping to mass.
And then this has got to come out simpler.
What does that come out?
If you'll allow me, I'm going to open the book so I don't write the wrong thing here.
This is in the book, on page 99.
The title is Better Formulas for s1 and s2.
Better Formulas for s1 and s2.
And here's my first better formula.
You can see that I get a minus b plus or minus the square root of something.
And that something will turn out to be p squared.
And then it'll be a minus something.
And that something will turn out to be the natural frequency squared.
Isn't that nice?
So what's the natural frequency?
Somehow, the natural frequency's coming in from this and this?
And just remind me what that second parameter is, omega n squared, the natural frequency of oscillation with no damping.
Tell me again what that is, because that was the fundamental ratio from last time.
It's central to all of engineering.
It's?
k/m
k/m.
Thanks, k/m.
So I believe-- and maybe Professor Fry could make this assignment a homework question, which is just algebra question.
Everybody sees that I have a minus p here.
And with a little care you get p squared, which is quite nice.
Which is quite nice.
And so we see that the decision between overdamping-- remember now?
Overdamping is when you damped so much that this became negative and you got an imaginary number in there.
Underdamping, it's still positive.
Overdamping, it's negative.
And so really that separation between overdamping and underdamping is the ratio of p to omega n. P to omega n is the damping ratio.
I think.
There may be a factor too.
Let me try to-- everybody sees that's the battle between these, if you accept that formula, and if it's in the book it's got to be right.
OK. And so the damping ratio is just that.
That's the damping ratio.
Now that's called zeta.
The Greek letter zeta.
I'm not Greek and not good writing zeta.
So I have unilaterally decided to use a capital zeta, which is a Z. Zeta is the Greek letter for Z. I could try to write it but you wouldn't be impressed.
So it's that damping ratio.
So now what does this mean?
Z smaller than 1, z equal to 1, c greater than 1?
Tell me what those-- obviously, when it's smaller than 1, p is smaller than omega n. Yeah, so what's going on here?
Which one is underdamping, which one is critical damping, which one is overdamping?
Because there's no difficult stuff here.
We're coasting in the last minutes here by just choosing words and notation that have turned out over a century to be more revealing than b squared minus 4mk, and this Z.
So z less than 1 will be what?
Z less than 1 will be p smaller than omega n. p smaller than this.
What's the story on that case then?
[INAUDIBLE]
That is underdamping, I guess.
p small has to go.
p is like b, the damping.
And small damping is underdamping.
So this is underdamp.
Underdamp.
And that's the case, in which we're going to have some imaginary stuff.
We're going to have some oscillation with the decay coming from there.
Now, what about z equal to 1?
z equal to 1 means p equals omega m, so that equals that so it's a big 0 in there.
What case is that?
[INAUDIBLE]
Critical damping.
It's this case in that picture.
It's that case with a double 0, equal s's.
Formulas that have to take account of that, this is critical.
And then finally, z greater than 1 is what?
Overdamped.
Overdamping.
Z bigger than 1 means p is big.
P big means b is big, damping is big, it's overdamped.
Overdamped.
So we've got it down to one parameter, the damping ratio, to tell us these things.
Rather than previously we had to say is b squared smaller than 4mk?
Is it equal to 4mk?
Is it bigger than 4mk?
Now we've got those words down to a single number z.
And let me just write next to us here that the z turns out to be the ratio of the damping to-- I think it's right-- it's the damping divided by the square root of 4mk, I think.
Can I just put a question mark there.
You couldn't mess around with the letters p and z.
But to get some variation from some other, but the point is, you see how much cleaner that is compared to this?
You're directly comparing that number to that number.
And that ratio is that number.
Yeah.
So all the formulas come out nicely.
Yeah, the formulas come out nicely.
And I guess what we see here-- final comment-- what we see here is what is the frequency of underdamped oscillations.
So I want to be in this underdamping case where there is oscillation.
There is an imaginary number coming out of that.
But there's also a real number.
Is the frequency of underdamped oscillation the same as omega m, the natural frequency?
No.
The frequency of that number-- so final comment, let me put it just here.
I would like this whole thing to be i, to give me oscillation, times omega d, the damped frequency.
And let me just say what that is.
So omega d, the damped frequency, squared, is this omega natural frequency squared minus the p squared.
If I had longer and we didn't have blackboards already full of formulas, I could-- it's the thing whose square root we're taking here.
So this is minus p plus or minus i, omega damped.
Omega damped is this square root.
There, we succeeded to fit in the better ratios, the good quantities to look at.
So again, the good quantities to look at are p, z, the damping ratio, omega d the damped frequency.
I think in a first lecture you could say, well, we already had correct formulas, we should just leave it there, and that's absolutely true, but anyway, this is what-- these are the letters people have introduced to make the formulas easier to understand in an engineering problem.
OK.
I'm all done except for questions.
Yes?
Don't ask me about resonance again.
Yes, OK.
Yes?
In the case of where we have the delta function, what is the velocity [INAUDIBLE]?
What is the what?
Why is the velocity equal to [INAUDIBLE]?
A-ha.
Okay.
You're right on the ball.
The question is where did this come from.
Where did that come from?
Can I tell you?
So if I integrate everything here, if I take the integral of everything, between 0, a little bit left of 0-- can I call that 0 minus?
Just a little bit left of 0.
This is crazy.
No math professor or whatever should ever do this.
To a little bit right of 0, just a real short time.
So what am I going to call a little bit right of 0?
0 plus.
OK now what is that integral?
Between a little left of 0 and a little right of 0, you know what the integral of the delta function is.
It is?
1
1.
Good.
Now, what are these ridiculous things?
Well, y is not changing in this tiny, tiny time.
So this is something, it's not getting big.
I'm integrating it over this tiny little time.
It's nothing.
Forget it.
Similarly here, y prime, the velocity is not climbing to infinity.
There's no-- and I'm just integrating over this infinitesimal little time.
Nothing here.
So this term has to be responsible for the 1.
And now you can tell me the integral of m y double prime.
What's the integral of m y prime?
m y prime
m y prime.
So m y prime plus, at 0 plus, minus m y prime at 0 minus.
But at 0 minus, it's 0.
You see what's happening?
And on the right side I'm getting a 1.
This is-- no person who had any skill with a blackboard would allow this to happen.
But that happened.
OK.
So these lower order terms are typical of math.
Lower order terms in the limit, forget them.
This is the top term, and it has to have something there, because it has to balance the 1.
And what it has is the jump in y prime.
So this is the instant jump in y prime in velocity, is times m gives 1.
So the instant jump jumped us from 0 to 1 over m. That's where I came from.
Well, that was a good question, and a kind of crazy answer but there it is.
OK, so we've got a mention of the Laplace transform as the algebra tool that works when you're staying with exponentials and nice functions.
And you'll see more of that.
So it's a frequently used tool to turn problems into algebra.
My name's Lorna Gibson.
I'm the professor for 3.054, it's a course on cellular solids.
And I've been working on cellular solids since I was a graduate student, since I did my Ph.D. And cellular solids are materials that are made up of an interconnected network of struts or plates.
And there's examples like engineering honeycombs and foams, and there's lots of examples in nature.
Things like wood and cork and there's a type of porous bone.
And there's lots of examples in medicine too.
Tissue engineering scaffolds, for example.
So my background is in civil engineering, and in civil engineering we study structures.
And typically people think of large structures like bridges or buildings.
But in fact when we analyze the cellular solids, we use the same kind of mechanics.
It's just the scale is very much smaller.
So we're looking at structures where the scale might be hundreds of microns or millimeters, things like that, but the same sort of mechanical principles apply to that.
OK, so I grew up in Niagara Falls, in Ontario.
And people always think of Niagara Falls as being the waterfall and all the tourist stuff, there's a casino there now.
But in fact, there's loads and loads of big civil engineering works in Niagara Falls, mostly associated with the hydroelectric power station.
So when they make hydroelectric power in Niagara Falls, the power station is actually about a mile downstream from the Falls.
And what they do is they have a big hydraulic gate that goes into the river and it diverts water from the river above the Falls into a whole series of canals and tunnels and there's a big reservoir where they store water.
And then the water from this reservoir goes into the penstocks, the tubes that go down to the turbines and then make the electricity.
Niagara Falls is not a big town, but if you drive around Niagara Falls, you see these canals, you see the reservoir, you see the big power station.
And so there's these really huge, impressive civil engineering works.
And my father worked for an engineering company in Niagara Falls and they specialized in the design of hydroelectric power stations, and I think that's how I got interested in engineering.
So I've been interested in bird watching for some time.
Mostly just because birds are beautiful and there's all sorts of interesting behaviors you can see with birds.
But since I started doing research on cellular solids and, in particular, teaching this course, I realize there's lots of examples of things about birds that have to do with cellular materials.
So for instance, some people had once told me that woodpeckers avoid head injury and brain injury by having a special cellular material in between their brain and their skull.
And that this acted kind of like a foam in a bicycle helmet.
That it would absorb the energy of the impact.
And I thought oh, well, I like bird watching and I study cellular materials, I should find out about this.
So I started looking into it and people had looked at the anatomy of the woodpecker skull and brain.
And, in fact, there is no special cellular material.
But by that point, I was kind of hooked.
And I actually did a project at one point looking at why it was that woodpeckers don't get brain injury.
And it's largely a scaling law.
It has to do with the fact that their brains are very small.
Another aspect of birds that has to do with cellular solids is how birds make themselves very light.
And here we have an owl skull.
This owl, unfortunately, had an accident with a car.
But somebody picked up its body and took it to Mass Audubon, and I got this from somebody at the Massachusetts Audubon Society.
And if you look at the skull-- I don't know if you can do a close-up here-- if you look at the skull, you can see there's a dense layer of bone on the outside and there's another dense layer bone on the inside, and there's a sort of foamy layer bone in between.
And that's called a sandwich structure.
And this foamy type of bone is called trabecular bone.
And that's one of the things that I study.
And it turns out that particular structure gives you a very stiff, strong, lightweight structure.
So you can see an example of how cellular materials are used in engineering but here sort of manifested in the owl's skull in making the skull very light.
Another part of the course is that I also have a project in the course.
And students do the projects in pairs.
Partly, because I just think it's nice for them to have somebody to work with.
And they've done a huge range of projects.
I let them do whatever project they want.
It has to have something to do with cellular solids.
And some of them have done, sort of, analytical things where they've done numerical finite element analysis of some cellular structure.
Some of them done very experimental things.
And some have been a lot of fun.
So, for instance, almost every year there are students who want to work on food foams.
And one year, for example, there was a group made bread.
And they looked at sort of the processing of the bread.
And they looked at how if you used more yeast, or you let the yeast rise, or the bread rise for longer, how that affected the cellular structures.
So they sort of changed the chemistry of how they made the bread.
And then they looked at the micro structure of the bread that they got.
I think they even did some mechanical tests of the bread.
So that was kind of fun.
And I think, probably, the most interesting project any student has done was on elephants skulls.
So you could imagine an elephant skull is huge, and it's bony.
So it would be very, very heavy, if it didn't have some pores in it.
And elephant skulls, it turns out, have some very large pores in them.
I think partly to reduce the weight of the skull, and the head, and the bone.
The neck has to, kind of, carry it all.
So these two students came to me.
And they said they had heard somewhere or they'd read somewhere that these pores in the elephant skull had an effect on how the elephants perceived sound and the acoustic transmission of sound waves through the skull.
And they wanted to do a project on elephant skulls.
So I was kind of intrigued by this.
I love all natural history kinds of things.
And I've worked before with people at Harvard's Museum of Comparative Zoology, where they have bones.
They have stuffed bodies.
They have all kinds of animals over there.
And I called up a colleague over there, and it turned out they had elephant skulls over there.
So I went with one of the students.
And it was an attic of the building, this kind of dingy place.
And there was this huge room.
And it was full of elephant bones.
And they had several skulls, which are like the size of this table.
They're huge.
They're this big.
And some of the skulls were cracked.
And you could see these big pores in the skulls.
And then the students found out that University of Texas at Austin has CT scans, Computed Tomography scans of all sorts of bones.
And sure enough, they had elephant skulls scanned.
And so they got a three-dimensional representation of the elephant skull through this University of Texas at Austin program.
And then they used that as input to a 3D printing set up.
So they 3D printed a sort of mimic of the elephant skull with some ceramic powder.
And they made a skull was about this big.
And then they wanted to look at the acoustic properties of it.
So what they did was they suspended the skull from a string.
And they had a speaker, and the speaker had a sound.
And they put an accelerometer on the skull.
And they measured the vibration of the skull.
And then to compare it with the skull that didn't have these pores.
And they got the CT image from Austin.
And they 3D printed this dolphin skull.
And they did the same thing.
They suspended the skull from as a thread.
And they measured the vibration.
And they could show-- and I've forgotten the details of their results-- but they could show, basically, that the two skulls had a different frequency response to the vibrations.
And they thought maybe part of it was because the pores.
So these two sections here are two sections of their 3D printed elephant skull.
I don't have the entire thing.
But you can see this is the orbit, where the eyes would have gone in here.
And if I turn it over and you look inside, you can see these pores in here.
And also if you look at this section lower down on the skull, you can see this whole porous structure here, as well.
And if we flip it over there's a little bit more over there.
So that was probably the most intriguing project that the students did as part of this course.
That was quite something.
So typically in the course we start off talking about the structure of these cellular materials.
We do some modeling of honeycomb and foam type materials, and that takes not quite half the term, but most of the first part of the term anyway.
And I give them more problem sets at the beginning of the term, and they don't really start on the projects at the beginning.
So I give them some background information to get them going.
And I assign the project at the beginning of the course, and I think there's three kind of deadlines.
One is they have to give me a proposal.
I think that's typically about a month into the course, and that can be fairly brief, but I want to at least know they've got a team.
They've got an idea.
They've got some idea of how they're going to carry out their project.
Then in about another month they have to send me a sort of update on the project, and by that point, I would have expected them, if they're doing a literature review, to start reading some papers and be able to tell me something about the background to their project.
If they're doing experiments, I would have expected them to at least gone into the lab and maybe made some materials, or bought some materials, and done some preliminary tests.
And I give them feedback at that point.
And at the end of the term, they hand in the final project.
And I see them twice a week in class, and I don't really have a formal recitation, but what I do do is every week that a problem set is due I have office hours.
And in fact, I don't actually have it in my office.
I book a room, and we do a little tutorial.
So there's times throughout the term they can see me and come and ask me questions about the project as well.
So the question is what kind of feedback do I give the students and how do they interact with me on the projects, and typically there's about 20 students that take the course.
So if they do it in pairs, there's roughly 10 projects.
Some of the projects students just do literature searches, and I don't really get that involved with those.
They're perfectly capable of going to the library and doing a literature search.
And some of the students who are taking it are graduate students, and they often do finite element numerical calculations.
And again, I give them some advice about how to set it up, but often they have experience doing this, and it's sort of applying what they already know to something new, but they kind of know what to do.
The way I get the most involved is if students do experimental projects.
So for instance, in this elephant skull project, I had this connection at the Museum of Comparatives Zoology, and I took the student up there.
They had this idea about 3D printing it, and we have a 3D printer in the department.
And one of the technical staff, Mike Tarkanian, is very, very good with the students and very helpful in getting them set up on the 3D printer, so he helps with that.
And I give them sort of general advice.
When they said now we want to measure some sort of acoustical response, I suggested maybe you could suspend it from a wire, or thread, or something and then put an accelerometer and measure the vibrations.
So I try to give them some general advice like that, but they then carry the experiment out themselves.
They have to figure out how to actually put it into practice and how to do it.
And I think they get a big kick out of that.
MIT students enjoy that kind of thing so that's kind of fun, and obviously I was very delighted too with this elephant skull project.
So there's different things I try to help with-- mostly giving general advice about how they can do their experimental projects.
So what are some of the challenges that students encounter in their projects.
So I think one thing is students sometimes are little over ambitious in what's possible to accomplish, because typically we have to cover some material before they can even start the project.
So typically they don't start the project until a month or six weeks into the term, and the term is only three months long more or less.
And so they really have a fairly limited amount of time-- maybe six weeks or eight weeks, something like that-- to actually do the project.
So if they want to make materials, if they want to do some sort of processing to make a foam, for example, they don't really have a lot of time, because often it takes some trial and error to be able to do that.
So the projects have to be fairly focused so that they can actually get something interesting out of it in a fairly short amount of time.
And sometimes students might want to do something where they would have to order materials, and it may take a few weeks for the materials to come in.
So again, I try to discourage them from doing projects where there's going to be a long lead time on getting some critical thing that they need for the project.
So there are some limitations, partly because of just the time we have for them to actually do it.
And they're not just taking my course.
They're taking other courses, so there's sort of a limited amount of time they can spend on it, too.
So really the way I set up the lectures is I write out notes for myself of what I want to cover, and the notes are pretty detailed.
And in the past I always just had the notes for me.
And even though they were reasonably neat and I could read them, I didn't hand them out to the students.
But because I've turned my fall course into an MITx course and I want to make the lecture notes available for that, when I was doing that course for MITx I made really nice notes.
And a friend of mine who lectures, I was asking her how she did it, and she said she actually goes and measures the chalkboard, and she measures the aspect ratio, how tall to wide the chalkboard is.
And then she sets up her notes so that they're the same aspect ratio, and she plans out exactly where she's going to put everything on the board on her notes.
And so I started doing that.
And when I'm doing the lecture I put it all on the board, and I find a lot of students-- even though I hand the notes out now-- they like to write their own notes.
And I think it helps them pay attention in class and helps them kind of focus on the material.
So another thing I do in the lectures, when I first started lecturing and for a long time I just focused on the engineering, you know, on the equations, this is the derivation, this is an example, this is how you use this.
And it was all just about kind of the engineering of whatever I was working on.
And I found over the last few years, actually in both courses, in the fall one on mechanical behavior and in the spring course on cellular solids, I look for more interesting examples and stories, kind of stories about the people who discovered some of the principles that we talk about.
I tell them stories about engineering situations that came up and there was some interesting thing happened.
And the students love it.
I mean, they really like having the kind of hard core mechanics broken up with some sort of stories.
So I do that a lot more now than I used to do.
So probably most lecturers have some kind of interesting example or historical thing that I talk about.
So for instance, when I teach the fall course, the mechanical behavior materials, one of the first things we talk about is stress.
So stress is a force per unit area.
If I take this piece a wood and I pull on it like this I'm pulling on it with a force that goes out like this.
Stress is just that force divided by that area.
And the unit of stress in the SI system is called the Pascal.
And it's named after Blaise Pascal who's a French mathematician.
And a couple of years ago I was in France for a conference and I was in a little town called Clermont-Ferrand, which is in the middle of France.
It's a pretty little town.
And I'm just walking around one day kind of seeing the square and the cathedral and all this stuff, and I see there is sign, there's like Pascal something or another.
And this was apparently the site of Blaise Pascal's house.
And right next door to it is Cafe Pascal.
So of course, I have to take a photograph of Cafe Pascal.
And in this other course, when we get to the bit about stress and I tell them about the Pascal I show them the picture of the Cafe Pascal.
And the other amusing thing that they kind of get a kick out of because of Boston, you know how in Boston there's the Freedom Trail and there's a red line goes around all these historical sites all this colonial and revolutionary stuff around Boston.
It's kind of cool, all the tourists to it.
Well in Clermont-Ferrand there's a Pascal Trail.
And there's little metal medallions of the portrait of Blaise Pascal put in the sidewalk and you can walk around Clermont-Ferrand doing the Pascal Trail.
So I kind of keep an eye out for stuff like that and I put that into the course now, and I never used to do that kind of stuff.
I have a picture from the Library of Congress, which I went to just for fun a while ago.
And one of the main buildings is this old, beautiful historical building for the Library of Congress.
And they have a marble staircase that goes up the middle.
And the staircase has all these little cherubs.
And there's an agriculture cherub and he's holding a sheaf of wheat.
And there's a wine cherub and he's holding a little thing of grapes.
Well, it turns out there's a mechanics cherub and he's holding a gear.
So at the end of the first lecture I show them the mechanic's cherub with the gear.
So there's these cute little things.
So I just keep an eye out for these cute little things.
And last fall, this past fall, one of the students came up at the end of the first lecture and he said, I really like art.
And he says, is there going to be more art in the class?
And I'm like, not usually.
This has kind of exhausted my art in mechanics.
And I said, but I'll keep an eye out.
And through the term there actually were different things that I saw that had to do with mechanics and art I showed the class.
So one of them was at the Peabody Essex Museum this fall there's been and display of the mobile sculptures of Alexander Calder.
And they also made these very large, I think they're called stabile sculptures, as well.
The big sail at MIT is one of his sculptures.
And these mobiles are actually a really nice example of free body diagrams in mechanics and balancing the forces.
So anyway, I got a picture of one of his mobiles and I went up to the exhibit.
And when I was at the exhibit I found that he did a degree in mechanical engineering.
He actually was a mechanical engineer.
And so the students were very, very tickled by the sculpture, the fact that he studied engineering.
So anyway, I look out for stuff like that.
So there's a lot of images in the class, partly because we're studying these materials and you can see just with the ones sitting in front of me, they have this porous, cellular structure.
So I show lots of images of materials.
We also look at how the images deform under load.
And I think, perhaps, that's something that might be a little unexpected.
So for instance, we have a stage in the electron microscope where we can actually deform things in the microscope.
And the stage is set up that it has a load cell on it.
And we could also measure how much it deforms, the materials.
So we can actually watch the materials as they're deforming.
So even though the cells may be 100 micron size, you can watch how they deform and how they fail, and you're going to learn a lot about the mechanics from these sorts of observations.
So we have both video of these deformations and also still photography at different time points that we show.
Another interesting little video clip that I show in the class is we look at the interactions between biological cells and tissue engineering scaffolds.
So for example, people who've looked at trying to heal, say, burns in skin where there's a large area of skin missing, one of the ways that you can do that is by using a collagen-based tissue engineering scaffold.
And when you have a burn in your skin and it just heals the normal way and you get all the scar tissue forming, that scar tissue is thought to form in conjunction with a process of what's called wound contraction.
So cells will actually migrate into the wound bed and actually mechanically pull the edges of the wound together, and that partly closes the wound and then scar forms as well.
And those two processes are thought to be related to each other.
So I've done a collaboration with Professor Ioannis Yannas here at MIT who developed one of these scaffolds for burn patients.
And he and I have been interested in this wound contraction problem.
And it turns out if you just put fibroblast skin cells into a dish of culture medium with one of these tissue engineering scaffolds, they will contract the scaffold.
And you can use an optical microscope to actually watch the cells do this.
So you can focus on an individual cell and you can see the cell elongating.
You can see the scaffold itself contracting.
And you can see this whole process.
So this is one of the little video clips that we show in class.
And the students always find that fascinating.
So I think one of the special things about this course and about my kind of research on cellular materials is that I spend a fair amount of time talking about materials in nature.
So I talk about wood, for instance, and what it is about the cellular structure that gives rise to the density dependence of wood properties and the anisotropy in wood properties.
I talk about trabecular bone, and I talk about the structure and properties of the bone, but also we do a little bit of modeling on how you might look at bone loss and osteoporosis.
So if you lose a certain fraction of the bone density, what residual strength would you expect the bone to have.
We have a project on bamboo right now.
We talk about the structure of bamboo.
Bamboo is actually a grass.
And this is a Chinese species of bamboo called moso bamboo, and you can see how big this one is.
They even get bigger, maybe six or eight inches across.
And what we're interested in doing is making something called structural bamboo products.
And this is an example of a bamboo oriented strand board.
So the same way people take wood, and they chop wood up into strands and make oriented strand boards for housing construction, you could, in principal, do the same thing with bamboo.
And we have a project that's in collaboration with some colleagues at the University of British Columbia.
They're the ones who are actually making the bamboo oriented strand board.
And with some architects in England, in Cambridge, England, we're looking at things like how you might modify wood building codes that talk about wood structural products, how you would modify that for bamboo structural products.
And what we're doing here at MIT is we're looking at the structure of the bamboo and doing some modeling of the mechanical properties of the bamboo itself.
So that's one example.
Here's another example.
This is a bamboo laminate.
So this is a little bit like a glue laminated wood member, but in fact, this one is made out of bamboo instead of out of wood.
So it's the same kind of idea.
Let's see.
We've also had a project in the past on cork.
So this is a cork from a wine bottle, and we've looked at that.
Cork has an unusual mechanical property.
You know, if you take a rubber band and you pull on it, if you can make it longer this way, it gets narrower that way.
Well, if you load cork in one direction, if you, say, pull on it or compress it, it doesn't get any wider or narrower in the other direction.
It just stays the same kind of size.
And you can show that that's related to the structure of the cells.
The cells are like little bellows, or like a little concertina.
So you can imagine if you have a little concertina, and you push on it this way, it doesn't really get any bigger that way or smaller that way.
It just stays the same dimension.
And the cork cells look a little bit like that, and that's why they do that.
So we have all these natural materials.
We talk a little about the hierarchical structure in plants.
The cell walls are fiber composites, and then there's a cellular structure.
And plant materials have a sort of hierarchical structure, with several different levels of hierarchy, and we talk about that in the class as well.
And we talk a little bit about why that makes these materials mechanically efficient and how you might look at designing engineering materials based on that.
So there's a little bit of biomimicking in the class as well.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
OK, so what we're going to talk about in this course are materials that have a cellular structure.
So they're all very porous.
And typically they have low volume fractions of solids, like less than 30% solid.
And we're going to talk about different types of cellular solids.
So one type are honeycombs.
And this would be kind of your standard hexagonal honeycomb, something like that.
We're going to talk about foams.
And I'm sure you've all seen polymer foams.
Here's an aluminum foam.
We'll talk about other sorts of foams as well.
We're going to talk about some medical materials.
And I brought in some bone samples, some trabecular bone samples here.
I brought in a little tissue engineering sample here.
And then, we're going to talk about materials in nature that have a cellular structure too.
So we're going to talk a little bit about wood.
And I brought in a piece of balsa wood.
I'll pass these around in a minute so you can play with them.
This is the lightest wood.
And I brought in lignum vitae.
This is the densest wood.
Lignum vitae is so dense that it sinks in water.
And I have a couple of projects right now on natural materials.
So I thought I'd talk a little bit about those too.
We have one on bamboo and making structural products out of bamboo.
So this is like a beam made out of bamboo.
And this is a piece of oriented strand board made out of bamboo.
So the same way you have like wood oriented strand board, you can do the same thing with bamboo.
And we have a project that involves this, so I thought talk a little bit about that later on in the course.
So we'll talk a little bit about the processing, how you make these materials, the structure.
We'll talk a little bit about how we do the modeling.
So we're going to start with the honeycombs, partly because they have a nice, simple unit cell that you can repeat and you can analyze fairly exactly.
And then we're going to talk about modeling the foams.
And then once we've got the modeling background, so we're going to get equations describing the mechanical properties, the stiffness and the strength, once we've got that, then we can apply it to lots of things.
So we can apply it to understanding the trabecular bone, the tissue engineering scaffolds, or how cells interact with scaffolds.
We're going to look at energy absorption in foams because they're good for absorbing energy.
We're going to look at a lightweight sandwich panels as well.
So we're going to look at all these different things.
And I have some slides I'm going to go over today that have more pictures of all of this.
And I'll pass this around to a sec, but let me go through the logistics first.
So this first hand out has a few of the kind of details.
There's two books that I've coauthored with colleagues.
The one on cellular solids, if you wanted to buy a book, is probably the most relevant one to buy.
If you don't want to buy it, you certainly don't have to.
I'm sure the library has multiple copies of it.
And the other more recent one is called Cellular Materials in Nature and Medicine.
And that's a little more specialized.
And again the library has that.
So you don't need to run out and buy that.
You can just get it there, but let me just mention there's that reference and it might be helpful.
OK, so let me talk a little bit about the projects.
What we do in this class is everybody does a project.
And I like for you to do it in pairs, just so that you have somebody to work with.
I think it's nice to have somebody to do it with.
And the project has to be something on cellular materials, but I really leave it pretty open ended what the project is.
It's really up to you to decide what you'd like to do.
And to give you some idea what people have done, people have done all kinds of stuff in the past.
So people have worked on negative Poisson's ratio honeycombs.
I didn't bring any of those in with me, but you can design these honeycombs so that they have this property that when you push it this way-- instead of if you make the strain smaller in this direction, it gets wider in this direction normally, but with negative Poisson's ratio materials, it will contract in that direction.
So I've had people do projects on that.
I've had people work on osteoporosis.
I had a group once who worked on elephant skulls and I brought part of their project with me.
So it turns out elephant skulls have a sort of porous layer to them.
And this is-- they have these large pores in the top of the skull.
I don't have a whole skull, but this is part of it.
And what they did was they had heard that elephant skulls have these pores and that the pores somehow affect sound transmission and how the elephant hears.
And so they wanted to do a project on that.
And that was really all they knew to start with, elephants have pores and we want to do a project on it because it's cool.
So I helped them put together this project.
We went up to Harvard.
We went up to the Museum of Comparative Zoology where they have elephant skulls, which are like about this big around, huge skulls.
And some of them had the outer part of the bone was broken and you could see these big pores.
So they got some nice pictures of elephant skulls.
And then they found that the University of Texas at Austin has computer tomography images of all sorts of bones of different animals.
And sure enough, they had elephant skulls.
So they got the file for the micro CT image, which gives you the sort of 3-D picture of the skull.
And then they use that to 3-D print small versions of the skull.
So that's what this is, they printed smaller versions.
And these were just a couple of slices that I got from them at the end project.
And then what they did with one of the skulls, they suspended it from a wire, and they took speakers, and they had sound that vibrated the skull, and then they put an accelerometer on the skull and they measured the vibration response of the skull.
And they also made, I think, it was a dolphin skull, which did not have these pores, and they kind of compared the vibration response from the sound for these two skulls with two different structures.
So that was a project for this class.
People have worked on tissue engineering scaffolds.
Often there's people who work on food foams.
So one year, people worked on bread.
They did bread processing.
They made bread by having different amounts of yeast, different rise times, different ingredients.
And they made bread.
And I have a little-- I like these historical things.
And I have a little thing here I wanted to show you.
So the people in 3032 last year, last fall, will know that I went to England in November.
And I went to the Royal Society for an editorial board meeting.
But I also went and looked at their archives.
And one of the things they showed me was this article, which is in the sort of an original, sort of archives of the Royal Society from 1600s.
And it's by a guy called John Evelyn.
He's famous for writing a book called Silva about trees and wood.
But he's written this article here.
And I love the title.
It's "The Several Manors of Making Bread in France, Where by General Consent, the Best Bread in the World is Eaten," by Mr. Evelyn.
So here we have in the Official Royal Society bread making science.
And in fact, the article was several pages long.
There was quite a lot on bread and how you make it in France, where the best bread is eaten.
So if you want to do something on food foams, people have done meringue before, various things.
They look at typically changing something about the recipe, or the composition, or the processing, or the baking.
And they look at the structure.
Sometimes they look at mechanical properties too.
So anyway, there's a whole list of things there.
You can think of other things.
If you look through the books, you might get some ideas as well for projects.
So what I'd thought I'd do next is I wanted to just kind of give an overview of what we're going to talk about.
And I've got a bunch of slides.
Let's see I forgot some things, because I do that.
Books, yeah, OK, let me pass some of these things around so you get to play with them too.
So here's some honeycombs.
I don't know if we can pass all these things around because it's a little unwieldy.
So this is an aluminum honeycomb.
I like bringing toys in.
These are little rubber honeycombs.
This is a little ceramic honeycomb.
This is a little paper honeycomb.
And let's see, we have some foams here.
So here's a metal foam and a ceramic foam.
And this is a sort of, not quite a foam, it's made of hollow spheres by centering hollow spheres together.
So you can play with that.
And what else do we have?
We have little lattice things here.
So here's a sort of 3-D lattice material.
So this has sort of a cellular structure, but it's very, very regular.
It's not like a foam.
So that's called a 3-D lattice material or sometimes a 3-D truss.
And what else should we pass around?
We need to do the bones.
Here's the wood.
You can feel how different the densities of the two woods are.
I would like these all at the end because I show these around for different classes.
Here's some bone, you can see of the bone looks like the foam.
And this is a tissue engineering scaffold for generating skin.
All right, so while those are getting passed around, I'll talk a little bit about what we're going to cover in the class with some slides.
OK, let's see, should I dim the lights?
Would that be a good thing, Craig, if I dimmed the lights?
They're kind of preset, you can try maybe two
OK.
Doesn't seem to-- there we go.
How about that?
That OK?
All right, so I like these historical things.
So this is a picture of Robert Hooke's drawing of cork from his book Micrographia.
And he was the first person who used the word cell to describe a biological cell.
And it comes from the Latin cella, which means a small compartment.
So you can think of the cells as small compartments.
That kind of makes sense.
And he kind of very modestly says, "I no sooner to discerned these, which were indeed the first microscopical pours I have ever saw, but me thought I had with the discovery of them perfectly hinted to me the true and intelligible reason for all the phenomena of cork."
So he's saying by looking at the structure of cork, he thinks he understands everything about the properties of cork-- very modest.
But in fact, this is kind of the foundation of material science.
So material science is all about looking at the structure of materials and trying to say something about the properties of the behavior of the materials.
And that sentence kind of sums that up.
So that's why I like that sentence.
So what we're going to do is look at different kinds of cellular materials.
We'll look at engineering ones.
And these we typically refer to as honeycombs or foams.
Honeycombs have two dimensional prismatic cells, while foams have three dimensional polyhedral cells.
And we'll look at applications for the honeycombs and foams in things like lightweight sandwich panels, in energy absorption devices, and things like thermal insulation.
We'll talk a little bit about the thermal properties.
We're also going to talk about cellular materials in medicine, so trabecular bone.
We'll talk about osteoporosis and how loss of bone reduces the strength, and how you might estimate that.
We'll talk about tissue engineering scaffolds, something about their mechanical properties.
You may think the mechanical properties aren't probably the most important thing.
But in fact, the mechanical properties do have some effect on how the cells interact with the scaffold.
So we'll talk a little bit about cell scaffold mechanics.
And then we're going to talk about cellular materials in nature at the end of the course.
So we'll talk a little bit about honeycomb like materials, like wood and cork, and foam like materials, like the trabecular bone.
There's also a type of tissue in plants called parenchyma that looks just like a foam.
And there's some sponges that have some interesting features.
And often, in nature the cellular material appears in combination with some solid material.
And so it's sort of a structural component.
And you can see sandwich structures in nature, leaves and skulls.
You can see materials that have density gradients, palm stems and bamboo are examples of that.
And you can see materials that have cylindrical shells with compliant cores, and things like plant stems and animal quills are like that.
So that's kind of the range of materials that we're going to talk about.
And one of the interesting things about cellular materials is that you can make cellular materials out of almost anything now.
And this is really a huge range of materials and lots of different applications for this.
So one of the fundamental things we're going to do is look at the mechanisms by which the materials deform and how they fail.
And we'll use a structural analysis to obtain the bulk mechanical properties, so things like the stiffness, the moduli, the strength, the fracture, toughness.
We'll look at how you can control the design of the microstructure to get the properties that you might want, and also how you might select for the best material for a given engineering application in engineering design.
So we're going to get those three things.
So let me start by showing you some more examples of engineering cellular solids, and in particular, some micrographs.
So that you can see what it looks like on a small scale too.
So these are the honeycombs.
These are the sorts of things that I'm passing around now.
The aluminum one, paper resin, and the ceramic ones.
The aluminum and the paper resin ones are typically used in the cores of sandwich panels.
And the ceramic ones are used in the catalytic converter in your car.
So what they do is they block off every other cell at one end and then the opposite cells at the other end and exhaust gas from your car is forced to go through those channels in the honeycomb.
And the walls, if you look at that triangular one, those triangular walls themselves are porous, and they're coated with the catalyst, which is platinum.
And that forces the exhaust gas through the wall in contact with the platinum, and then comes out the other end.
And they're ceramic, obviously, because the gas is hot and they need something that has a high thermal resistance.
So these are some examples of honeycombs.
These are some examples of engineering foams.
When I tell people I work on foams, they always think of polymer foams, like polystyrene or something.
And there's lots of polymer foams.
But you can actually foam any materials now.
There's metal foams.
There's ceramic foams, and glass foams, carbon foams, all sorts of foams.
So those are some examples.
You can see when you look at these images here that the foams have a low volume fraction of solids, like if you look at say this polyethylene one here.
Say we look at this guy up here, then you can see there's not much solid, there's a lot of gas.
So the volume fraction of solids is fairly low on that foam there.
So one of the things we're going to talk about is how the volume fraction of solids affects the properties.
You can also see on the top left and the top right, the top left one has what we call open cells.
There's just edges along the polyhedra, there's no faces over the membranes.
And the right hand one is a closed cell foam.
So there's like membranes that cover the faces of the cells.
So we're going to talk a little bit about the differences and the behavior of open cell and closed cell foams too.
These are food foams.
So I've already said you might want to do a project on food foams.
And these are just some examples of different kinds of foods that are in fact foams.
And it turns out the food industry spends quite a lot of time and effort thinking about the mechanical properties of food.
And it turns out if the texture of the food isn't right, then people don't like the way it feels in their mouth.
There's something they actually called mouth feel.
So it turns out if your cereals too soggy, it's icky.
If it's too crunchy, it's icky.
So it's sort of a happy medium.
And food companies spend quite a lot of time and money worrying about the mechanical properties of food.
This is an example of showing that the cells could be antisotropic, the cells could be elongated in one direction.
For instance, in the top one on the polyurethane foam.
And if they're elongated in one direction, it's not too surprising, you might have different properties in that direction from the plane perpendicular to it.
And then the bottom images is of pumice.
Pumice is a volcanic rock.
And you can see how the pores are kind of flattened out there.
And they're flattened out because that was once molten lava.
And the molten lava was flowing down a mountain side of a volcano.
And as it flowed, it got sheared.
And the shape of those pours reflects the shearing as the molten lava flow down the volcano.
And so this kind of sort of stretched out cell shape is going to give you antisotropic properties, different properties in different directions.
This is the 3-D truss that I'm passing around.
I don't know if it's exactly the same one, but it's a similar one.
And these trusses are triangulated structures.
And we'll talk a little bit about their properties too.
And then we also are going to talk about some applications.
So obviously, these materials are mostly air.
And that gives them a low weight.
And that means they're often used in structural sandwich panels as the core of the panel.
And these panels have stiff faces separated by a lightweight core.
And the idea is to make it a little bit like an I-beam.
So the way you have the flanges on the I-beam, the faces are like the flanges, and the porous core is like the web of the I-beam.
They can also undergo large deformations at relatively low stress.
And that means they can absorb a lot of energy.
So if you think of the energy as the area under the stress strain curve, if there's big strains and big deformations, then there's going to be a large area.
And that sort of energy absorption occurs at a fairly low stress.
So typically, when you want to absorb energy, it's not just how much energy you want to absorb.
You have to do it without actually breaking the thing you're trying to protect.
So you don't want to generate high stresses as you go along, and foams are good at this.
Foams are also good at being thermal insulators.
They have a low thermal conductivity.
And that's because they're largely made of gas and the gas has a lower conductivity than the solids.
So that gives them a lower conductivity.
And they have a large surface area.
And the smaller the pore size, the bigger the surface area per unit volume.
And that makes them good for things like carriers for catalysts.
And that's why they're used for these catalytic converters too.
OK, so here's some examples of cellular materials in medicine.
So here's some examples of trabecular bone.
Trabecular bone exists at the ends of your long bones.
So say in your hip or in your knee.
It also exists in your vertebrae in the middle of the spine in the vertebrae there.
And it also exists in your skull.
And you can see it's a porous type of bone.
It looks very similar to the foams and the sorts of mechanical models we make for foams can be applied to the bone as well.
And so that's one of the things we're going to do later in the course.
These are two slides showing what happens when people get osteoporosis.
The left hand slide is from a 55-year-old female to the same bone, the same slice.
And the right hand one is from an 86-year-old female.
This thing here, row star over row S, that's the relative density, the density of the bone divided by the solid that it's made from.
That's the same as the volume fraction of solids.
And so on the normal bone on the left it's about 17% solid.
And on the osteoporotic bone on the right it's, about 7%.
So you can kind of see the bone density has gotten lower, partly by thinning of the struts, but partly by resorption of the struts, as well.
And obviously the one on the right is going to have a lot lower strength than the one on the left.
These are micro CT images of bone.
And again, you can see how the structure looks different at different relative densities.
The one on the left is sort of in the middle at around 11% dense.
The one in the middle is the most dense 25%.
And the one on the right is 6% dense.
So it's not too surprising that the one on the right would have a much lower strength from the other ones.
And we'll look at how we can model that.
This is just showing some deformation in bone.
I have a colleague, Ralph Mueller who's got a micro CT machine, which allows you to do compression tests in the micro CT.
So he can make these sort of images where he scans it at zero strain.
He compresses it a little bit.
He scans it again.
He compresses it a bit more.
He scans it again.
And he these are stills from his images, but he makes animations from this.
And if you look at the top right up here, you see these struts here.
They're pretty straight in this one.
They're a little bent and starting to buckle here.
And then if you look at that one strut there, you can see how it's buckled right over.
So you can look at the deformation mechanisms by looking at the CT scans and things like that.
People are starting to think about using metal foams for coatings of orthopedic implants.
So one of the issues with implants is that say you have a hip implant or a knee implant, you remove the bone that's preexisting, and then you replace it with some sort of implant.
Typically, the implant has a stem that fits into the hollow part of the bone and then has a sort of joint piece to it that fits into the joint.
And you can get some loosening of the stem in the remaining bone.
And one idea is that you use porous coatings to minimize that.
And right now, typically what they do is center beads, metal beads on to the stem.
And another idea is maybe you could use a metal foam.
And these are some different types of metal foams that people are looking at.
Another type of cellular material in medicine is a tissue engineering scaffold.
And this just shows some different examples made by different processes.
And we'll talk more about these later on in the course.
This one here at the top left is a collagen based one made by a freeze drying process.
And I don't know if you saw MIT's website yesterday and today, Ioannis Yannas was the one who developed this.
And he's just being inducted into the National Inventors Hall of Fame.
And this is what he really was inducted for is he's invented a skin-- well, he calls it a dermal regeneration template for regenerating skin, mostly in people with serious burns.
Then these are some other sorts of scaffolds that are made by different processes.
This is by a sort of rapid prototyping process here.
The bottom two, these are kind of interesting.
These are actually the extracellular matrix in the body.
And they've had all the cells removed from it.
So these tissue engineering scaffolds are really designed to mimic the extracellular scaffold in your body or extracellular matrix in your body.
And you can see how when you remove the cells, the structure of those two things looks a lot like a closed cell foam.
So that's the kind of structure you're trying to replicate.
We'll look a bit at cell mechanics.
This is a cell contraction of a scaffold.
So here these sort of very thin transparent bits of the collagen based scaffold, and this is a fiber blast on it right here.
And I had a student, Toby [?
Fryman, ?]
who worked with me and Ioannis Yannas on this.
And you can see from the video the cell is actually contract in the scaffolding making it deform.
And you can calculate what forces the cell must be imposing on the scaffold by knowing something about the geometry of the struts and how the cells attached.
And then it's going a little bit more.
So we'll talk more about that.
And then there's a final picture down here, where you can see these two, the two points up here and down here have now been brought pretty much right together.
So we'll talk about that in more detail.
We've also done studies on cell attachment and how that attachment rate or the amount of cells that attach is related to the surface area, the surface area per unit volume.
So these are just some tests from that done by [?
Fergal ?]
O'Brien, a post-doc that worked with me.
We've also done some studies on cell migration.
And Brendan Harley was the student who did this.
And he stained the cells with one stain and he stained the scaffold with something else.
So red are the scaffold and green are the cells.
And then he used a confocal microscope to track the cells.
And he tracked the cells and where they moved versus time.
And if he has the location at different times he can get the velocity.
And one of the things he did was he changed the stiffness of the scaffold and he found that the migration speed depended on how stiff the scaffold was.
So he was looking at sort of interactions between mechanical properties of the scaffold and behaviors like cell migration.
And then we're going to look at materials in nature.
So here is wood.
So you can see the cellular structure of wood.
It's a lot like the honeycombs.
It has sort of a prismatic structure.
That one happens to be cedar, but other woods look similar.
Now, this is balsa wood.
And this is showing just how the balsa deforms.
I think this was loaded from top to bottom.
And this is at zero load.
And then this is more load, and more load, and more load.
And if you look at that cell there with this little kind of tear in it here, that's the same as the cell down here, and that's the tear there.
So you can see how the cell walls bend and how they deform.
And you can model that using the honeycomb models.
This is just another image showing actual failure of wood, buckling of the cell walls.
This is cork.
So these are modern scanning electron micrograph of cork.
And one of the interesting things is the cork cells have these little corrugations.
You see how they're not flat, they have little kind of wrinkles in them.
And that gives rise to sort of an interesting property of cork.
If you take cork and you load it.
So here we were pulling it in the direction of these arrows, pulling it like along this direction here.
And again, this is the same set of cells.
That tear there is the same as that tear there.
And you see all these little corrugations here, they've all straightened out when we're pulling on it.
And the Poisson's ratio of the cork is zero.
It's kind of like a bellows.
Like if you had an old camera, or you have an accordion bellows.
If you pull the bellows in and out, it doesn't get any wider this way or the other way.
You're just sort of opening the bellows and closing the bellows.
And the cork cells are doing kind of the same thing.
And it gives them this Poisson's ratio of zero.
Which it happens is one of the things that makes it easy to get the cork into your wine bottle, because as you're pushing on it, it's not pushing out in all directions.
This is only for this one direction, but it's not pushing out in all directions.
These are parenchyma cells in plants, in carrots and potatoes.
All those little blobs in the potato, those are starch blobs.
This is called a Venus flower basket sponge, and Joanna Aizenberg, at Harvard, has studied this quite a lot.
This has a hierarchical structure.
If you look at the overall sponge and then you look at each of the sort of struts that make up the lattice, that too has a hierarchical structure.
And she's looked at the optical properties of this glass sponge.
It's kind of a beautiful thing.
And then there's some cellular structures in nature as well.
There's sandwich structures.
There's density gradients.
And there's tubes with a cellular core.
So here's some examples of that.
Here's the iris leaf, you know the iris plant has these long kind of leaves that stand up like this.
And it's just like a sandwich panel.
The parenchyma are kind of like a foam in the middle here.
And there's very dense fibers called sclerenchyma that run up and down the length of the leaf.
And they're like fibers in a fiber composite.
And here's a bull rush or a cat tail leaf.
And they're like little I-beams almost.
It's like a whole series of little I-beams.
And again that's sort mechanically efficient.
These are examples of sandwich structures in bird skulls.
Some of you know I'm a birder.
So I sort of sneak in bird stuff from time to time.
But you can see how these birds skulls are all sandwich panels, and obviously birds want to minimize their weight for flight.
And this is one of the ways that they do that.
This is a horseshoe crab, sort of similar kind of thing.
This is from Mark Myers in San Diego.
He did a study on the crab in its shell as a sandwich.
And this is the ever so handsome cuttlefish.
And the cuttlefish has something called a cuttlefish bone.
This is the bone here and the bone is made up of these kind of sandwich type structures.
The cuttlefish is related to octopus and squid and things like that.
And it's hard to see in this picture here, but these little things here are actually like little tentacles.
There's several tentacles that it eats stuff with.
The cuttlefish is actually a mollusk.
All those things are mollusks.
It's called a fish, but it's not really a fish.
And here's an example of a natural material that has a radial density gradient.
Have you ever noticed if you look at a palm, like you see those pictures of Hollywood in LA, and the palm trees, you know, they line the street.
But they're all about the same diameter from the bottom to the top.
And when you think of like an oak tree, it's not like that.
It's big diameter at the bottom, skinny diameter at the top.
So when wood grows, the wood has more or less the same density in the bottom and at the top.
So as it's growing, the density is more or less the same.
And it resists the bigger loads from getting taller by adding circumference.
So it gets wider and wider as it gets older and older.
But palms don't do that.
They come out of the ground a certain diameter.
And most palms just grow that same diameter.
As they get taller and taller, you can imagine there's wind forces, and different kinds of forces are on it, the stresses get bigger and bigger.
And the way they resist those is that the cell walls get thicker, and they preferentially get thicker on the outside.
And if you think of moment of inertia, remember moment of inertia is increased more with the material on the outside of a beam.
And that's kind of what the palm is doing.
So if you look at young cell walls and old cell walls, here's some SCM pictures of young ones and it's sort of skinny.
And SCM pictures of older ones, and the cell wall has gotten thicker.
So we're going to talk a little bit about that.
And it turns out, this is an incredibly efficient way to deal with getting taller and needing to resist bigger loads.
Another material that has a radial density gradient is bamboo.
So this again shows these sort of dense sclerenchyma fibers.
Do you see these kind of dense parts here?
And you can see there's not very many of them here.
And there's more and more as you get towards the outside there.
So there's a density gradient there.
So we'll talk about that.
And some plant materials have a cylindrical shell with a compliant core.
Plant stems or commonly like.
This is a milkweed stem.
And you can see it's got these sort of fibers that are almost completely dense.
And then a sort of lower density core, cellular core, here, and a void in the very middle.
And you can show that that core helps prevent local buckling.
So if you take a drinking straw and you bend it, you get that kinking kind of failure.
You can show that having this sort of foam like core in the middle helps to resist that.
Imagine you have a drinking straw and now you put foam in the middle.
It's going to be harder to get it to kink like that.
So that's what the plants are doing.
Animal quills do the same thing.
That's a porcupine and a hedgehog quill.
And all of this stuff is in these two books.
So it doesn't really matter to me if you go out and buy the book.
I don't make very many much money on these books.
So this is not an income producing thing.
But those are the books that all these pictures have been taken from.
All right, so I think that's my sort of introduction to the class.
Are there any questions about how we're organized or what we're going to be doing?
Are we good?
It's OK?
OK.
So then I think what I'm going to do for the rest of the time is start the next section of the course which is on processing of cellular materials.
Now, I have another little hand out here.
So I don't know if I'll remember to do this for every lecture, but I like to have a little outline, that partly makes me be organized.
So it's just a little outline for the lecture
[INAUDIBLE]
You asked me to do what?
Put the room light back up
Put the what up?
Lights
Oh, the lights.
I'm going have another set of slides though.
So let me get out of the intro slides.
I know I have another set of slides.
So I'm going to just leave the screen up.
And kind of put stuff on the board and talk about the slides
That's fine
You're good?
OK. OK.
So I wanted to talk a little bit about processing of cellular solids and then, next time we'll start talking about the structures.
It seemed good to talk about the processing before we got to the structure.
So I'm going to talk a little bit about honeycombs and how they make honeycombs, and then foams, and then lattice materials.
Yeah?
The slide you're showing with the the shell with the foam inside it, are there techniques for analysis of it?
Well, I don't think we're going to get into all the gory details, but I can certainly give you references.
That's something that one of my students did at one point.
And in fact, I've been collaborating with Jennifer Lewis up at Harvard and she has a student who's making sort of cylindrical shells with foams out of ceramic foam.
So he's 3-D printing sort of with coaxial nozzles a cylindrical shell that's pretty solid.
And then a ceramic foam on the inside.
And that's one of the things he's playing around with.
So he's looking at ways you might make engineering versions of that.
So I wanted to start with looking at honeycombs and how they make honeycombs.
And I thought what I'd do is I've got some slides.
And I'm going to talk about the slides.
And then I'll put some notes on the board to kind of describe what we're doing, OK?
So this is the first sort of slide on the honeycombs.
And the two main techniques that they're made by, especially those aluminum honeycombs and paper honeycombs that I passed around, one technique is an expansion process.
So what they do is they take flat sheets of some metal, say aluminum, and they put little stripes of glue on it in different places.
So these little kind of specialty things are where the glue goes.
And then they stack those guys up in a sort of particular arrangement.
And then what they do is they pull it all apart, kind like a paper doll thing.
They pull it all apart and when they pull it apart, they get the hexagonal shape.
So let me just show on the board how you do the gluing and how that works.
So they would start with some sheets.
Say we start with two sheets like that.
They'd put some glue down, say there.
And then there's a gap.
And then they put glue on the opposite side over there.
And then there's another gap.
And then they put glue there.
And then they do the same positions but the opposite sides on the next sheet.
So they do that.
And then if you glue those together-- well, let me do another one.
Maybe do a couple more.
So then if I do one like that, it's glued there, and there, and there.
That guy gets there, and there and there.
And when glue that-- when you push that together and then take it apart, you've got something that looks like this.
So say I call that one, two, three, and four.
Then this would be 2 and 3.
OK, so this thing here is that.
Where it's not glued, you get them doing that.
And then it's glued again down here.
And so you get this kind of pattern.
And one of the things about these honeycombs that are made by the expansion process is these inclined walls have a thickness t. And because there's two sheets up here, the vertical walls have a thickness 2t.
So that's typically what you see in the commercial honeycombs that are made by this way.
And this process is used for aluminum honeycombs, for paper resin honeycombs, for Kevlar honeycombs.
And I'll just say note that the inclined walls have a thickness t, and the vertical walls are 2t.
So that's the expansion process.
And the process that's commonly used for honeycombs is called a corrugation process.
And for the corrugation process, it's just like the lower schematic here shows.
You take a flat sheet.
So you've got a roll of a flat sheet.
And you've got some rollers that have the right shape to give you the corrugated profile that you want.
You pass the sheet through the rolls and you get individual sheets out.
And each sheet is kind of a half hexagon.
And then you put the sheets together and that forms the whole hexagon.
So you have a flat sheet that's fed through a shaped wheel to form half hexagonal sheets, which you then bond together.
And it's the same kind of thing that the inclined walls have thickness t and the vertical walls have thickness 2t.
And this corrugation process, you can only really use it in materials that you can deform a fairly large amount to get the corrugations.
So typically, this is for metals.
And aluminum is probably the most common metal that this is used for
How are the corrugated sheets attached to each other?
I think they're just bonded with epoxy.
Yeah, so obviously if you wanted to use it for high temperature performance-- you know, all of these things are bonded with some sort of epoxy or some sort of resin.
So there's an issue if you wanted to use it higher temperatures.
So another process that's used to make ceramic honeycombs is an extrusion process.
And you just take a ceramic slurry and you pass it through a die.
And you can make a ceramic honeycomb by doing that.
And I believe that's how they make the ceramic honeycombs I passed around, the catalytic converter ones.
Other techniques involve rapid prototyping.
You can 3-D print honeycombs.
And Jennifer Lewis has a project on 3-D printing of honeycombs up at Harvard.
And one of the interesting things they're looking at is not just printing with an ink, but printing with a fiber reinforced ink.
So they're making cell walls of the honeycomb that are fiber reinforced.
And one of the tricks is trying to orient the fibers in the way that you want them to be oriented.
So there's rapid prototyping techniques as well.
You can use also selective laser centering.
Let's call it selective laser scanning.
So you can have a photosensitive polymer and use a laser to cure that and build up a honeycomb type structure.
And you can also cast honeycomb structures.
So those rubber honeycombs that I pass around, those are made by casting.
You take a liquid silicone rubber and you add a hardener and you pour it into a mold.
Another kind of interesting way that people have made-- well here's another example of the honeycombs that are 3-D printed.
And this is an example of-- or a couple of examples of looking at a bio carbon template.
So what that means is that these materials are based on the wood, but none of them are actually wood.
So what they do is they take wood, like they take pine, or they take beech or something.
They take some kind of wood and they carbonize it.
So that they do the same processes as used for making carbon fibers.
So you put the wood in an inert atmosphere.
And you pyrolyze it.
You heat it up to I think 800 degrees C. And all you're left with is the carbon.
And it preserves the structure.
And you replicate the structure.
You just get the same structure.
There's some shrinkage.
The shrinkage is about 30%.
But you get the same structure as the wood.
So this material up here is actually all carbon.
It's just replicating the wood that was used, the pine that was used.
An then what people have been doing is using that carbon structure and then infiltrating that with gaseous silicon.
And they form silicon carbide.
So these structures down here are all silicon carbide replicas of wood.
And they're thinking about using that for things like filters for high temperatures or for catalyst carriers.
And one of the attractive features is wood has fairly small cells.
The cells around 50 microns across.
And so you get a large surface area.
And this is a similar thing here.
These two are the carbon template.
And here they've used silicon and they've actually filled in the voids.
And so they've got silicon carbide where the cell walls used to be.
And they've got silicon where the void used to be.
So people are playing around with this is another way of making a honeycomb type of structure.
And they use other kinds of plants besides wood as well.
But that's the kind of general idea.
So the idea is that wood has a honeycomb like structure.
And the cells are fairly small.
the cells are in the order of 50 microns sort of in diameter and maybe a few millimeters long.
And this bio carbon template replicates the wood structure.
So the wood is pyrolized at 800 degrees C in an inert atmosphere.
So say an inert gas.
And that gives you the bio carbon template.
And you maintain the structure, although there's some shrinkage.
structures And then this carbon structure can then be further processed.
So for example, you can infiltrate it with a gaseous silicon.
And you end up with a silicon carbide wood replica.
So possible applications are things like high temperature filters, or catalyst carriers.
I think that's it on the honeycombs.
Are we good with all sorts of methods?
And my little talk here on processing is certainly not comprehensive.
I'm sure there's other ways people have developed.
These are some main ways.
All right then, I want to talk about foams as well.
People have developed different types of processes for different types of solids, so polymers, and metals, and ceramics.
So I just go through each class of solid and talk about that.
So the idea with polymer foams is that you want to introduce gas bubbles into either a liquid monomer or a hot polymer.
And then you want the bubbles to grow.
And then you want to stabilize them and solidify it by other cross linking or by cooling the hot polymer.
So there's a variety of ways of doing that, but let me just put that down.
So there's a few ways to get the bubbles in there in the first place.
One, is just by mechanical stirring.
So if you've ever made meringue, you know what that is, you just take a whisk and you beat egg whites and bubbles.
The air will get enveloped in the egg whites.
They also do that with polymers.
Or you can use a blowing agent.
And there's several varieties of blowing agents.
So the blowing agents are divided into physical and chemical blowing agents.
And the physical ones, they force the gas into solution under high pressure, and then you reduce the pressure, and the gas bundles expand.
So you can use physical blowing agents.
Or you can introduce liquids that, if you're using a hot polymer, that at the temperature of the hot polymer, they form a gas.
So that the liquid just turns into a gas.
And that would form vapor bubbles.
And then the chemical blowing agents.
There's a couple of different ways that those work.
You can either use chemical blowing agents where you have two parts that react together to form a gas.
And so that gas then blows the foam.
Or you can have a chemical blowing agent that reacts with the polymer to form a gas and that blows the foam.
So either way.
And you can have them decompose on heating.
So the same kind of thing.
Evolved the gas when it gets into the hot polymer.
So there's these different ways of blowing the foams.
And there's many, many different types of these blowing agents.
But, these are kind of the general techniques.
And whether or not the foam forms an open cell or a closed cell structure depends on the rheology of the polymers, so the viscosity of it, and also the surface tension.
Another way to make a foam is to make something called a syntactic foam.
A syntactic foam is made by taking thin walled hollow spheres and then using, say a resin, like an epoxy resin, to bond them together.
So you end up with something that's porous.
And you've got the void from the hollow sphere, but you don't foam it in the same way that you blow bubbles through it in some way.
One other thing about polymer foams is they sometimes have a skin on the surface.
So when you blow them, say you've got a mold, there will be a skin that forms against the mold, and sometimes the process is designed in a way that the skin is thick enough that it acts like a skin in a sandwich panel.
So they control the mold in a way and the blowing process so that they get a foam in the middle and thicker skins on the top and the bottom surface.
And that forms a sandwich panel.
Those are called structural foams.
Let's see.
So I think what I'm going to do next is the next section's on metal foams.
And I've got a few slides on that.
So I think I'm going to run through the schematics and just talk about it.
But, I'll put the notes on the board next time.
And there is one thing I forgot to do at the beginning.
I like to tell you a little about me.
And I want to hear about you.
So I wanted to leave a few minutes for that.
So let me just wait until people are finished writing stuff down.
And I'll go through these in a few minutes, and then we'll through it in more detail next time.
And I'll write notes down.
OK?
Are we good?
So there's a whole variety of ways of making foamed metals.
And most of them have been focused on aluminum.
But you could in theory do them with other types of metals.
So this was one of the first processes and it just involved taking a molten aluminum, so here's the aluminum down here in a crucible.
They added silicon carbide powder to it and then they just used a stirring paddle, like they just stirred it up and mixed gas in that way.
And they found that they got bubbles that rose up.
And then they used conveyor belts to just kind of pull the foam off.
And the thing about the silicon carbide was that if you didn't have that, then the bubbles wouldn't be stable enough that you could do this.
The bubbles would collapse before you got to be able to pull them up.
But the silicon carbide I think makes the aluminum melt more viscous and it helps prevent sort of drainage and collapse of the bubbles.
And so that's one way.
And there's a type of foam called Cymat, and this is an example of the foam that's made with that process.
Maybe I'll bring it next time and we can pass it around.
Another method is to use a metal powder and titanium hydride powder.
Then you can consolidate that.
So here's-- it's hard to see the writing, but this is a aluminum powder.
This would be the titanium hydride powder.
You mix them together and then you compact them.
You press them together.
And the titanium hydride decomposes and forms the hydrogen gas at a temperature at which the aluminum is not really quite molten but it's kind of viscous-y, kind of softening.
And so when the aluminum is soft like that and the titanium hydride decomposes and forms the hydrogen gas, you get a foam from that process.
And I think, somewhere, yeah, this foam here I think was made by that process.
Then in a similar thing, you can just put titanium hydride powder into molten aluminum, and again the titanium hydride powder evolves the hydrogen gas and you get foamed aluminum from that.
And I think this foam here was made with that process.
They all look kind of similar.
Another method is by replicating an open cell polymer foam.
So I think I passed an open cell polymer foam around.
And that's made by taking an open celled-- an open cell aluminum foam-- it's made by taking an open cell polymer foam, you fill up all the voids with sand.
You then burn off the polymer, but now you've got sand in all the sort of places where there were voids.
And then you infiltrate molten metal into that.
And then you get rid of the sand.
And then you're left with an aluminum foam that replicates the polymer that you started off with.
So this replication technique.
There's a vapor deposition technique.
And this was developed by Inco to make nickel foams.
So they take again an open cell polymer foam, that's kind of this thing here is, and they infiltrate into it nickel CO4.
The only teeny detail that's a problem with this process is that happens to be highly toxic.
So they put this gas through here.
And then they get nickel depositing on the polymer and they burn the polymer out and they center it.
So it is possible to do this.
They have done this.
But it's not that practical because of the toxicity of the gas.
Now another method is something called entrapped gas expansion.
And here, what you do is you take a can, like a metal can.
This one's titanium, a titanium alloy.
And then you put a titanium powder in here.
You evacuate the can.
So the can has a little valve on it, so you can evacuate it.
And then you backfill it with argon gas and you pressurize the argon gas.
So you've got a powder with sort of pressurized gas inside of a can.
And then you hot isostatically press it.
So you heat it up and you press it uniformly in all three directions.
And you compact it.
And then, if you want you can roll it.
Sometimes people roll it because they want to make sandwich panels, and they want to have a certain thickness, and they want to have faces on the panel.
But then you heat that up.
And as you heat it up, the gas evolves again.
And the thing expands and you get a foam that way.
So that's another method.
Another method is by making hollow spheres and then bonding the spheres together.
And in this process-- this was developed at Georgia Tech.
They used the titanium hydride again.
They made a slurry of the titanium hydride in an organic binder in a solvent.
And then they had a little kind of needle that they injected gas.
And so they had this slurry and they were blowing gas through this needle and they got hollow spheres of the titanium hydride.
And then they heated it up, and again evolved the hydrogen gas off.
But now they're just left with titanium spheres.
And then they bonded the spheres together.
And these aren't titanium.
This is an iron chromium, like it's not quite a stainless steel.
But this is the same thing.
I can pass that around next time too.
Those are the little beads that they make.
And then there's also fugitive phase technique.
So you can take say salt particles, put them in a mold and pour a liquid metal into that, and then, leach the salt away.
And I think that's it for the metal foams.
So I think I'm going to stop there for today in terms of the lecture.
I'll go over that again next time.
And I'll write stuff on the board.
But I wanted to tell you a bit about me.
So the people in 3032, they already know me because they had me in the fall.
But I see some unfamiliar faces.
So I thought I would tell you a little about me.
So I grew up in Niagara Falls in Canada, big power station.
Lots of big civil engineering works in Niagara Falls.
And my father worked at an engineering company that specialized in the design of hydroelectric power stations.
It was founded by the guy who designed the Niagara Falls power station.
And then I went to university in Toronto.
And I did a degree in civil engineering in Toronto.
And then when I finished my degree in civil engineering, I wasn't really sure what I wanted to do next.
And I applied for some jobs, and I applied to graduate school.
I applied to MIT.
And I didn't get in-- ouch.
But I did OK, it turns out.
And I ended up-- I had an advisor when I was in Toronto who had taken a sabbatical in Cambridge, England.
And he said he thought I might enjoy Cambridge, England.
And I ended up going to Cambridge, England to doing my PhD there.
And I worked on the cellular solids for my PhD.
And it was a nice combination because I was interested in material behavior and mechanics, but I had a background in me in civil engineering.
And these are just like civil engineering structures, but they're on a little teeny weeny scale, not like big buildings or bridges or something, like little teeny things.
And really all of this has come out of doing that PhD in Cambridge.
But when I was there, I never even thought about being an academic.
And I never applied for any academic jobs.
I didn't think I wanted to be an academic.
And I went and got a job in Calgary in the oil business.
And I was working at a consulting firm that did work for the oil business.
I hated it.
I just hated it.
It was like I had a boss.
I hated having this boss.
And, you know, the projects were too short term.
The winter in Calgary-- if you think this is bad, you've seen nothing.
Like less snow, but cold.
I mean like 30, 40 below, everyday, cold.
Real cold.
So I stayed there one winter.
And somewhere along the way there, I decided maybe the academic job thing would be good.
And I just sent my CV out to a bunch of Canadian universities.
And I ended up getting a job at the University of British Columbia in Vancouver.
And I lived in Vancouver for two years.
And I probably would have stayed there, except there was a gigantic recession, and it was all very depressing, and there was no money.
And that the universities in Canada are almost all run by the provincial governments.
And the government had no money.
It was all, you know, frustrating.
And I sort of thought, oh, I'll look around and see what else I can get.
And I answered a little ad in Civil Engineering Magazine for a job at MIT.
And I got the job at MIT.
And I was in the civil engineering department for about 12 years.
And then I moved over to the materials department.
Because my work started off on sort of sandwich panels and structural things.
And then it kind of became more biomedical stuff and had less and less to do with civil.
And I've been in the materials department since then.
And this is kind of what I do, this kind of work.
So that's kind of my little five minute story.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
I think, last time, we got as far as talking about processing of foams and we talked a bit about processing of polymer foams.
And today, I want to pick up where we left off.
And I was going to talk about processing metal foams.
We'll talk a little bit about carbon foams, ceramic foams, and glass foams.
We'll finish this section on processing today and then we'll start talking about the structure of cellular materials.
And hopefully-- we won't finish that today, but we'll finish it the start of tomorrow's lecture.
And then, we'll start doing mechanics of honeycombs tomorrow.
OK?
So that's the scheme.
I thought what I'd do-- I have a whole series of slides that show, schematically, a variety of different processes for making metal foams.
So I thought I would just go through the slides- and I think I did this really quickly last time-- but I would write down a little bit of notes on each of the slides so that you've got some notes on it, too.
This is the first method here.
And many of these methods were developed for aluminum foams.
But you could, in principle, use them for other types of foams, as well.
This first method here-- let me see if I can get my little pointer.
[INAUDIBLE] The idea here is that you take molten aluminum.
Down in this bath here, you've got molten aluminum.
And they put, into the aluminum, silicon carbide particles.
And the silicon carbide particles adjust the viscosity of the melt.
They make it more viscous.
And then they just have a-- I've got this thing here-- they've got a tube here that they blow gas in with.
And if you go to the bottom of the tube, there's an impeller, or a little paddle, that stirs the gas up.
And so the gas just forms in the molten aluminum here.
And then they have conveyors which pull off the molt-- or the metal foam, and then it cools.
The idea is that if you just had the aluminum, you couldn't really do this because the bubbles would collapse before they cooled down and became a solid.
But adding the silicon carbide particles increases the viscosity of the melt.
It helps prevent drainage of the foam.
Normally, if you have a liquid foam, just from gravity, you're going to have some drainage.
It's going to tend to-- some of the liquid is going to tend to sink, just from gravity.
By putting the silicon carbide particles in, you increase the viscosity.
It helps prevent the drainage.
And you can get the foam bubbles to be stable.
Let me just write one little note here.
The first method here involves just bubbling gas into the molten aluminum.
And that molten aluminum is stabilized by silicon carbide particles.
Sometimes people use aluminum particles and the particles increase the viscosity of the melt.
And that reduces the drainage and it stabilizes the foam.
That process was developed at Alcan in Canada and at Norsk Hydro in Norway.
And this foam here is an example of the foam that they've made.
And you can kind of see-- there's a density gradient in this foam.
I'll pass it around.
You can see it.
But the bubbles are smaller down here and there's fewer of them than there are up here.
And that's partly from the drainage.
The molten aluminum is draining down and you get more liquid at the bottom and then you get more solid at the bottom.
So that's the Alcan foam.
OK. Another-- there's half a dozen of these processes for metal foams.
Another version of the process involves taking metal powder and combining it with titanium hydride powder.
And then you consolidate it and you heat it.
So if I can show through the schematics here.
In the first step, you take the two powders, you mix them up.
So the second thing down here is mixing the powders up.
And then you press them together.
There's a dye here that you press it, and then you get pieces.
And then you can heat that up.
And the way that this works is that the titanium hydride decomposes at a temperature of about 300 degrees C. And if the other powder is something like aluminum-- aluminum melts at something like 660 degrees C-- the aluminum has become soft at 300 degrees C, but it's not molten.
And the titanium hydride-- when it decomposes-- forms hydrogen gas.
So the hydrogen gas forms the bubbles that you need to make the foam.
Are you riding your bicycle?
You are tough.
Very tough.
I ride my bike, but I gave up a couple weeks ago.
Well, three weeks ago, I guess.
When it started snowing, I gave up.
The idea with this process is you can use titanium hydride powder with aluminum and the aluminum becomes soft at the temperature that the titanium hydro decomposes.
And when it forms the hydrogen gas, that gives you the bubbles, and so you get the foam made by that process.
And that was developed by a place called Fraunhofer.
And I have one of their little foams here.
This is an example of their foam made by that powder metallurgy process there.
OK. Let me just write a little note about that.
So you can mix up titanium hydride powder with a metal powder and then heat that up.
When you do this, you need to have a metal powder that's going to be deforming by, say, high temperature creep at the temperature that the decomposition of the titanium hydride happens.
And for aluminum, that works.
So you need to have the metal material be able to deform fairly easily, in order to get these bubbles to form a nice foam.
But that's another way you can do it, is by consolidating these two powders.
Here's another way to do it.
You can also make use of this property of the titanium hydride-- that it'll decompose and form hydrogen gas-- by just putting it into molten aluminum.
And in this example here, you've got an aluminum melt in here.
And this time, they've added 2% calcium to it.
Again, to adjust the viscosity.
And then they add the titanium hydride in this step here and they've got a little mixer thing here that's spinning around and will mix it.
And then they'll put a lid on it to control the pressure and the titanium hydride will decompose, the hydrogen gas will evolve, and you'll get the foam made by that method, too.
So you can stir titanium hide right into a molten metal, as well.
OK.
So that's another method.
And I think that was made by something called the Alporas process.
And this is an example of one of their foams.
I'm pretty sure that's an Alporas foam there.
Yep
[INAUDIBLE]
They can't really control it perfectly.
In that first example that's going around-- maybe you might have missed it-- there's this drainage.
The first one's made by a molten process and you get drainage.
And you get different sized bubbles.
And you get different-- you get a density gradient in the thing, as well.
So you can't control these things perfectly.
OK.
Here's another method for the metal foams.
This one here involves replication.
In this method here, what you do is you start out with an open-cell polymer foam.
In this step up here, there's an open-cell polymer foam.
You fill that with sand.
So you fill up all the open parts of the cells with sand.
Then you burn off the polymer.
And so you've got a little channels where the polymer used to be.
And then you infiltrate those channels with the molten metal that you want to use.
So you replicate the polymer foam structure.
This is the infiltration process here.
This little thing here is your furnace.
And then you get rid of the sand and then you're left with a metal replica of the-- a replica of the original polymer foam.
And this example here is one of these things that's made by this replication process.
So that's an open-celled aluminum.
I think the-- if you look at the density of these things, they're fairly low.
And so there's quite a large volume of pores and they're all interconnected.
So it's not that hard to get the sand out.
This method would involve replication of the open-cell foam-- the polymer foam-- by casting.
So you fill the open-cell polymer foam with sand.
You burn off the polymer.
And then you infiltrate the metal into that.
And then you remove the sand.
OK?
Then, another process involves just using vapor depositions.
So you take an open-cell polymer foam again.
Here-- let me use my little arrow.
Here's the open-cell polymer foam up here.
And you have a furnace here with a vapor deposition system.
And they use a nickel CO4 system to do this.
You then burn out the polymer.
You're left with a metal foam with hollow cell walls, where the polymer used to be.
And then usually, what they do is they center it.
They heat it up again to try to densify the walls.
The only teeny weeny problem with this process is that the gas they use this incredibly toxic.
And so it's not cheap and it's got health hazards, as well, associated with it.
But you can do it.
That gives you a nickel foam when you're finished with that.
And you could also use an electrodeposition technique that's similar.
OK. That's another method.
Another method is shown here.
This is the entrapped gas expansion method.
And what they do in this method is they have a can and the can has a metal powder in it.
It's whatever metal you want to make the foam out of.
In this example here-- it's probably too small for you to read in the seats-- but it's titanium.
They've taken a titanium alloy.
They've got a powder of titanium alloy.
They then evacuate the can.
They take all the air out of it.
And then they back fill it with argon gas.
They put in an inert gas in there.
And then what they do is they pressurize and heat the thing up and so the gas is internally pressurized by doing that.
And then, sometimes when they do this, they want to have a skin on the two faces.
So in this next little image here, it's done where they roll the can and produce a panel that's got solid faces.
And then when you heat it up, the gas expands and you end up with a sandwich panel by doing this.
So this bottom figure down here-- I'm not having much luck with the pointer.
This bottom figure down here, they've heated it up.
The gas expands and you've got this solid skin on the thing, which is from where the can use to be.
So that's trapping of a gas.
OK?
There's a couple more methods for the metal foams.
One involves centering hollow metal spheres together.
And the trick there is to make the hollow spheres.
And the way that can be done is by taking titanium hydride again, if you want to make titanium spheres.
You put it in an organic binder-- in a solvent-- so you've got a slurry here.
And you've got a tube that you blow gas through.
And as you do that, you get hollow titanium hydride bubbles.
And then you can do the same thing where you heat that up.
The hydrogen gas evolves off, and you're left with titanium.
You're left with titanium spheres down at the bottom here.
And then you can pack those together and press those and form a cellular material that way.
You can also center hollow metal spheres.
And the last method I've got for the metal foams is that you can use a fugitive phase.
With the fugitive phase, you would take some material that you could get rid of at the end of the process.
Say, something like salt, that you could leach out.
Here, we have our bed filled with salt.
And then you would infiltrate that with a liquid metal, typically under pressure.
And then, after the metals cooled, you leech out the salt.
You get rid of that.
You can pressure infiltrate a leachable bed of particles, and then leach the particles out.
OK. We have a whole variety of methods that have been developed to make metal foams.
And most of these have been developed, probably, in the last 20 years.
Something like that.
Some of them are a little bit older than that.
But there's been a lot of interest in this recently.
Those are all methods to make metal foams.
I wanted to talk just a little bit about a few other types of foams.
People make carbon foams.
And they use the same kind of method as they do to make those bio carbon templates I told you about.
When you take wood and you heat it up in an inert atmosphere and it turns into a carbon template, you can do the same thing where you take a polymer foam, you heat it up in an inert atmosphere, and everything except the carbon is driven off.
And you're left with a carbon foam.
It's the same process they used to make carbon fibers.
There's carbon foams.
There's also ceramic foams that you can make.
I brought the little sample of ceramic foaming again.
You can pass that around.
And those are typically made by taking an open-cell polymer foam and passing a ceramic slurry through the polymer foam so that you coat the cell walls.
And then you fire it so that you bond the ceramic together and you burn the polymer off.
And you're left with a foam that's got hollow cell walls.
You can also make ceramic foams by doing a CVD process on the carbon foam that you could make by the previous process.
And people also make glass foams.
And to make glass foams, they use some of the same kinds of processes as people use for polymer foams.
I'll just say similar processes to polymer foams.
OK. That covers making the foams, and I think we talked about the honeycombs last time.
I wanted to talk a little bit about making what are called 3D lattice materials, or 3D truss materials, as well.
Let me [?
strip ?]
that up there.
Yeah?
[INAUDIBLE]
Chemical vapor deposition
[INAUDIBLE] use this for metal foams?
Well, people were quite interested in using them for sandwich panels-- the cores of sandwich panels-- lightweight panels.
There was interest in using them for energy absorption, say, car bumpers.
The automotive industry was quite interested in this, in terms of trying to make components with sandwich structures that would be lighter weight, or energy absorption for bumpers.
Or filling up-- if you take-- say you take a metal tube.
If you think of a car chassis and it's made of tubes, if you fill those tubes with a foam-- especially if you fill them with a metal foam-- you can increase the energy absorption quite substantially.
So when you have a tube in a chassis, if it's loaded axially, it will fold up and you get all these wavelengths of buckling.
And if you've put a foam in there, it changes the buckling wavelength and it increases the amount of energy you can absorb.
So not only is the energy absorbed by the foam itself, it actually changes the buckling of the tube so you can absorb more energy.
So there was a lot of interest in that.
There was an interest in using them for cooling devices for, say, electronic components.
The idea was you would take, say, an aluminium open-cell foam and you would flow air through that.
And say you have your device that's generating heat, you'd have a foam underneath it.
And the aluminum conducts the heat fairly well.
And then you would blow air through it to try to cool it off.
So there were a bunch of different applications people have had in mind for them
What about glass?
Glass foams, I think, are largely used for insulation in buildings, believe it or not.
I think, actually, one of the dorms at MIT-- maybe the Simmons dorm-- has a glass foam insulation
[INAUDIBLE]
Well, I think because the foam-- because the cells are closed, the gas is trapped in the cell.
Whereas with a fiberglass, gas could move through the fibers more easily.
So I think that's partly how it works.
OK. Well, let me talk a little bit about the lattice materials, too, and how we make those.
We're going to start talking about the modeling of honeycombs and foams.
And when we do that, we're going to see that if we have a structure that deforms by bending, the properties vary with the amount of material, in a certain way.
But if we have materials where the deformation is controlled by axial deformation, the stiffness and the strength are going to be higher at the same density.
People made these lattice-type materials to try to get something with a more regular structure, and especially a triangulated structure.
You see how these things are like little trusses?
Triangulated?
Triangulated structures, when you load them-- say I load this like this-- there's just axial components-- axial forces in each of the members.
And so, theoretically, this would be higher stiffness and strength for a given weight than, say, a foam would be.
So people were interested in these lattice material.
This one here is made of aluminum.
And I wanted to talk a little bit about how you can make these things.
One way you can do it is by injection molding.
And this here is just the centerpiece of something that would look like this.
So there would be a-- I didn't bring it, but there's a top face and a bottom face that fit onto this.
And they would be injection molded as three different pieces, and then assembled together.
So you can make a mold in this complicated geometry, and you can make a lattice material by injection molding.
We'll start with polymer lattices first.
One way is injection molding.
Another way to do it is by 3-D printing.
You can generate a structure like that by 3-D printing.
You can also make trusses in 2D and you can make them so that you can snap fit those together.
So you can make little 2D trusses.
Here's a little truss here and here's a little piece of a truss here.
And you can make a little snap fit joints.
Do you see how these ones have little divots in them?
And you can make it so that these things will fit together.
I think these guys-- can I do it?
No.
You'd have to take-- oops.
Wait a minute.
No, it's not that way.
There we go.
So you can snap them together like that.
OK.
I can't get it to-- there we go.
So you can do that.
And you do that over and over again.
And if you do it over and over again, you get something that looks like that.
OK?
You can make a snap fit thing.
Let me pass all these guys around and you can play with those.
That's the injection molding one.
This is the snap-fit one.
Got that?
Let's see.
I think I have a little picture here.
This is an example of the snap fit truss here.
It's the thing that's getting passed around.
And another clever way that was developed was by taking a monomer that's sensitive to light.
You take a photo sensitive monomer.
And you put a mask on top of it and the mask has holes in it.
And then you shine collimated light on it.
You shine, say, a laser on it.
And the light goes through the holes in the mask.
And it starts to polymerize the polymer because it's photosensitive.
And then, as the polymer-- as it polymerizes and becomes solid, it then acts as a waveguide and draws the light down deeper into the monomer.
And so the polymer acts as a wave guide.
It brings the light down.
And this is a schematic over here.
This is a schematic showing the set up.
And these are some examples of some 3D trusses that they've made using this technique.
And one of the nice things about this is you can get a very small size cell size.
So this is-- I think that bar-- it says 1,500 microns.
That's what?
One and a half millimeters.
So you can get a nice, small cell size if you want that.
Let me write that down.
You take a photosensitive monomer and then you have it in a mold beneath a mask.
And then you shine collimated light on it.
And as the light shines on it, it polymerizes the monomer.
So then it solidifies and then it guides the light deeper into the monomer.
OK. That's that.
And then finally, there's metal lattices, as well.
And so this is, obviously, a metal lattice here.
It's an aluminum alloy.
And the metal lattice is made by taking that polymer lattice that was made by the injection molding technique.
You coat that with a ceramic slurry.
You burn off the polymer, and then you infiltrate the metal where the polymer used to be.
OK. That's the section on processing of the honeycombs and the foams and the lattices.
So there's a variety of different techniques that people have developed for making these kinds of materials.
And I thought it'd be useful to just describe some of the techniques.
As I think I said last time, this isn't comprehensive.
This doesn't cover every technique.
But it gives you a flavor of what techniques people have developed for making these kinds of materials.
OK?
Are we good?
We're good.
OK.
The next part, I want to do on the structure of cellular materials.
And I have a little overview.
I don't think we'll finish this today, but we'll finish it tomorrow.
Yeah?
[INAUDIBLE]
Down here?
What happens to the ceramic?
They get rid of the ceramic.
Typically, the ceramic is not very strong and it's just fired enough so they can infiltrate it with the metal.
And then they-- yeah, I think with mechanical smushing around, you can get rid of the ceramic.
And the ceramic's brittle, so if you break the ceramic, you're not going to break the metal
I'm wondering if you could make a type of metal lattice [INAUDIBLE] with reducing [?
the ?]
oxides?
I guess you could, if you could-- but you'd have to then make the oxide in that shape, too.
You've always got to make something in that shape
[INAUDIBLE]
Yeah, maybe you could make a foam.
But to make these lattices, you need this really regular kind of structure and be able to control the structure.
OK. Let me scoot out of this set of slides and get the next set up.
OK. We want to talk about the structure of cellular solids.
And we classify cellular materials into two main groups.
One's called honeycombs.
This thing down here is a honeycomb.
And honeycombs have polygonal cells that fill a plane and then they're prismatic in the third direction.
So you can think of them as just being a prismatic-- and they can be hexagons, they can be squares, they can be triangles-- but you can think of them as prismatic cells.
And the cells are just in a 2D plane.
And then we also have foams.
All of these ones over here are foamed materials.
And they're made up of polyhedral cells.
The cells themselves are three-dimensional polyhedra.
And this slide here shows a number of different types of foams.
These ones are polymers up here.
These are two metals.
These are two ceramics.
This is a glass foam down here.
And this is another polymer foam down here.
OK?
[INAUDIBLE]
No.
I just know that

I took those pictures so I know that.
No, you can't tell by looking at them.
In fact, that's one of the things about how we model the cellular materials.
The fact that their structure is so similar is what gives them similar properties.
And they behave in similar ways because they've got similar structures.
OK. We've got 2D honeycombs, where we have polygonal cells that pack to fill a plane.
And then they're prismatic in the third direction.
And then we have what we call 3D cellular materials, which are foams, which have polyhedral cells.
And then they pack to fill space.
The properties of all of these materials depend, essentially, on three things.
They're going to depend on the solid that you make it from.
If you make the material from a rubber or from an aluminum, you're going to get different properties.
So they depend on the properties of the solid.
And some of the properties that we're going to use that are important for this type of modeling are a density of the solid-- which I'm going to call rho s-- a Young's modulus of the solid-- which I'm going to call es-- and some sort of strength of the solid-- which I'm going to call sigma ys for now.
And you could think of other things.
There could be a fractured toughness of the solid.
There could be other kinds of things.
One thing that the properties of the cellular material depend on is the properties of the solid.
Another is the relative density of the cellular material.
And that's the density of the cellular thing divided by the density of the solid.
And that's equivalent to the volume fraction of solids.
So it makes sense that the more solid you've got, the stiffer and stronger the material's going to be.
So it's going to depend on how much material you've got.
And it also depends-- the properties also depend on the cell geometry.
The cell shape can control things like whether or not the honeycomb or the foam is isotropic or anisotropic.
You can imagine, if you have a foam, and you've got equiaxed cells, you might expect to have the same properties in all directions.
But if you had cells that were elongated in some way, you might expect you'd have different properties in the direction that they're elongated relative to the other plane.
So cell shape can lead to anisotropy.
For the foams, you can also have what we call open-cell and closed-cell foams.
If you look at this slide here, and we look at this top right images-- these two up here-- the one on the left in the top is an open-celled foam.
There's just material in the edges.
There's no faces.
And so a gas can flow between one cell and another.
And then if you look at the one on the right, this is a closed-celled foam.
There's faces.
If you think of the polyhedra, you've got solid faces covering the faces of the polyhedra.
For an open-cell foam, you've only got solid in the edges of the polyhedra.
And the voids are continuous, so they're connected together.
And for a closed-cell foam, you've got solid in the edges and the faces.
And then the voids are separated off from each other.
So we'll say, the cells are closed off from one another.
Another feature of the cell geometry is the cell size.
And the cell size can be important for things like the thermal properties of foams.
It's important for things like the surface area per unit volume.
But typically, for the mechanical properties, it's not that important.
And we'll see why that is when we do the modeling.
OK.
Yes?
For the closed-cell foams-- because we can't really see it without cutting it, is it that all of the faces are closed?
Or is it like some fraction of the faces are closed?
If you look at this one on the top right here, they're pretty much all closed.
But the reason we have this little picture down here is some of them are closed and some of them are open.
So you can get ones that are in between.
But typically-- this is kind of unusual.
Usually, they're either all open or all closed.
If we look at the mechanical properties of cellular materials, typically the cell geometry doesn't have that much of an effect.
The relative density is much more important.
The relative density, we define as the density of the cellular solid.
And when I use a parameter like rho or e or something, if it's got a star, it's for the cellular thing and if it's got an s, it's for the solid.
So rho star is going to be the density of the cellular material.
And rho s is going to be the density of the solid it's made from.
And so the relative density is just rho star over rho s. And I just wanted to show you how this is the volume fraction of solids.
So rho star is going to be the mass of solid over the total volume.
Imagine you've got a honeycomb or a foam and you've got, say, a unit cube of it, the sum total volume of the whole thing-- the density of the cellular material is going to the mass of the solid over the whole volume.
The density of the solid is going to be the mass of the solid over the volume of the solid.
This is really just equivalent to the volume fraction of solids, how much solids you've got.
And that's also n equal to 1 minus the porosity.
Typical values for cellular materials-- I think last time I passed around one of those collagen scaffolds-- those tissue enineering scaffolds.
It was in a little plastic bag.
That collagen scaffold has a relative density of 0.005, so its 0.5% solid and 99.5% air.
And if we look at typical polymer foams, the relative density is typically between about 2% and 20%.
And if we look at something like softwoods-- wood is a cellular material.
And we look at softwoods, the relative density is usually between about 15% and about 40%.
Something like that.
OK?
As the relative density increases, you get more material on the cell edges, and if it's closed-cell foam, on the cell faces.
And the pore volume decreases.
And you can think of some limit.
If you keep increasing the relative density more and more and more, eventually you've got-- it's not really a cellular material anymore.
It's more like a solid with little isolated pores in it.
And so there's two bounds.
And the density has to be less than a certain amount for you to consider it a cellular material in the models that we're going to derive to be valid.
And if the relevant density is more than a certain amount, people model it as a solid with isolated holes.
If I have a unit square of material, if it's a cellular material, you might expect that you've got pores that would look like this.
And you've got relatively thin cell walls, relative to the length of the material.
And for a cellular material, typically, the relative density is less than about 0.3.
And when we come to the modeling for the honeycombs and the foams, we're going to see that the cell walls deform, in many cases, by bending.
And that you can model the deformation by modeling the bending.
And that the bending dominates the behavior if the density is less than about that.
And at the other extreme, you can imagine if you had just little teeny pores.
I have a little pore here and one here and one there and one there.
That's not really a cellular material.
It's just got a teeny weeny little bit of pores.
And that could be modeled as isolated pores in a solid.
Each one is acting independently.
And people have found that that is appropriate if the relative density is greater than about 0.8.
And then, in between, there's a transition in behavior between the cellular solid and the isolated pores in the solid.
OK?
Are we OK?
The next thing I wanted to talk about was unit cells.
Especially for honeycombs, people often use unit cells.
A hexagonal cell is an obvious one to use to model this kind of behavior.
For honeycomb materials, you can have unit cells and you can have different ones.
On the left here, we've got triangles, in the middle, I've got squares.
On the right-hand side, I've got hexagons.
And you can see, even if you have a certain unit cell, there's also different ways to stack it.
So the number of edges that meet at a vertex is different for, say, this example on the top left and this example on the bottom left.
Here, we've got six members coming into each vertex, and here, we've got four.
And again, this stacking for the two square cells is also different.
So you can have different numbers of edges per vertex.
Another thing to note that's kind of interesting-- if you look at the honeycomb cells here, this one on the top left-- this equilateral triangle one with the stacking-- and this one on the top right-- the regular hexagonal cells-- those two are isotropic for linear elastic behavior, whereas all the other ones are not.
So we have 2D honeycomb unit cells.
We can have triangles, squares, and hexagons.
They can be stacked in more than one way.
And that gives different numbers of edges per vertex.
And in that figure, a and e are isotropic, for linear elasticity.
OK.
When we come to modeling the honeycombs, we're going to focus on the hexagonal cells.
We'll talk a little bit about the square and triangular cells, as well.
And then, for foams, when you look at the structure of a foam, it's obviously not a unit cell that repeats over and over again.
But people started off trying to model the mechanical behavior of foams by looking at periodic repeating polyhedral cells.
And there's three cells here that are prismatic.
We're not really going to talk about those beyond this.
So they're not really physically realistic or interesting.
But people would use these two cells here in initial attempts to model foams.
And this one here is called the rhombic dodecahedra.
Rhombic because each of the faces has four sides and dodecahedra because each polyhedra has 12 faces.
I forget if I've bored you with my Latin already.
Hedron means face in-- oh, this is Greek, I think.
Hedron means face.
Do is two, deca is 10.
So dodeca is two plus 10.
It's got 12 faces.
OK?
So that's the rhombic dodecahedra over here.
And then this bottom one down here is a tetrakaidecahedra.
It's a similar thing.
Tetra's four, kai mean and.
Four and 10-- tetra kai deca-- it's got 14 faces.
OK?
And those two pack to fill space.
I think those are the only uniform polyhedra that pack to fill space.
Here is the 3D foams.
We have the rhombic dodecahedron and the tetrakaidecahedron.
And the tetrakaidecahedron packs in a bcc packing.
Initial models for foams-- they took these two unit cells.
And what they would do is have an infinite array of them to make up the whole material.
And then they would isolate a unit cell.
And they would apply loads-- some say compressive stress, for example.
And then they would figure out what the load, or force, was in every single member, and how much that member deformed.
And they would figure out the component of the deformation in the same direction that they were putting the load on.
And they would figure out things like a Young's modulus, or they would figure out when there was some failure of one of these struts, and they would figure out a strength for the foam.
But you can kind of imagine, geometrically, not that easy to keep straight.
A little bit complicated.
So one way to model foams is by using these unit cells.
But we're going to talk about a different way to do it, as well.
OK.
So those are unit cells.
When they make foams, as we just talked about, one way to make a foam is by blowing a gas into a liquid.
And if you blow a gas into a liquid, then the surface tension is going to have an effect on the cell geometry and on the shape of the cells.
And if the surface tension is isotropic-- if it's the same in all directions-- then the structure that you get is one that minimizes the surface area per unit volume.
And so people were interested in what sort of cell shape minimizes the surface area per unit volume.
And Lord Kelvin, in the 1800s, was the person who worked this out.
And this is called the Kelvin tetrakaidecahedron.
And it's not just a straight tetrakaidecahedron.
There's a slight curvature to the cells here, to the faces.
And you can kind of see it in some of the edges here.
Like if we-- let me get my little pointer.
If you look at that edge, it's not straight.
This edge here is not straight.
It's got a little bit of a curvature to it.
But this minimizes the surface area per unit volume.
And then more recently, in the 1990's, there were two people-- Dennis Weaire and Robert Phelan-- discovered that this structure here-- which isn't a single polyhedron, but it's made up of a few polyhedra.
That has a slightly smaller surface area per unit volume.
Smaller by 0.3%.
So, a tiny bit smaller.
OK. Let's see.
What I'll say here is that foams are often made by blowing a gas into a liquid.
And if the surface tension controls and it's isotropic, then the structure will minimize the surface area per unit volume.
OK. That's relevant if the foam is made by blowing a gas into a liquid and surface tension is the controlling factor.
Sometimes foams are made by supersaturating a liquid with a gas, and then you nucleate bubbles, and then the bubbles grow.
So there's a nucleation and growth process.
So that's a little bit different.
And if you have a nucleation and growth process, you get a structure that is similar to something called a Voronoi structure.
In an idealized case, imagine that you have random points that are nucleation points and that you start to grow bubbles at those nucleation points.
So you start off with these random points.
And the bubbles all start to grow at the same time and they all grow at the same linear rate.
If you have that situation, then you end up with this Voronoi kind of structure.
And I've shown a 2D version of it here just because it's easier to see in 2D, but you can imagine a 3D system.
And in order to make one of these Voronoi honeycombs, you can imagine-- if you have random points-- here, say that little point there is one of the nucleation points, and here's another point here-- you form the structure by drawing the perpendicular bisectors between each pair of points.
This is the bisector between these two points.
Here's a bisector between those two points.
And then you form the envelope of those lines around each nucleation point.
And that, then, gives you that structure.
And you can see, this structure here is kind of angular.
It doesn't look that representative of something like a foam.
But if you have an exclusion distance, where you say that you're not going to allow the nucleation points to be closer than some given distance-- your exclusion distance-- then you get this structure here.
And this starts to look a lot more like a foamy kind of structure.
So these Voronoi structures are representative of structures that are related to nucleation and growth of the bubbles, or nucleation and growth processes.
Let me write down something about Voronoi things.
And these Voronoi structures were first developed to look at grain growth in metals.
They weren't developed for cellular materials.
But you can use them to model cellular materials, as well, as long as it's a nucleation and growth process.
We'll say that foams are sometimes made by supersaturating a liquid with a gas, and then reducing the pressure so that the bubbles nucleate and grow.
So initially, the bubbles are going to form spheres.
But as the spheres grow, they start to intersect with each other and form polyhedral cells.
And you get the Voronoi structure by thinking about an idealized case in which you randomly nucleate the-- you have nucleation points at a randomly distributed space.
They start to grow at the same time and they grow at the same linear rate.
OK.
The Voronoi honeycomb, or the foam-- you can form that by drawing the perpendicular bisectors between the random points.
So each cell contains the points that are closer to the nucleation point than any other point-- or any other nucleation point.
And if we just do this process as I've described here, you end up with a Voronoi structure, where the cells appear kind of angular.
And if you specify an exclusion distance, where you say the nucleation points can't be closer than a certain distance, then the cells become less angular, and of more similar size.
OK.
So are we good with the Voronoi honeycomb nucleation and growth idea?
Alrighty.
All right.
If we think about cell shape-- if we start with honeycombs and we just think about it hexagonal honeycombs, if I have a regular hexagonal honeycomb so that all the edges are the same length and this angle here is 30 degrees, then that is an isotropic material in the plane in the linear elastic regime.
One of the things we're going to do is calculate-- if I loan it this way on, what's the Young's modulus?
If I load it that way on, what's the Young's modulus?
And we're going to find they're the same, in fact, no matter which way on I loaded it.
It would be the same.
But if I now have my honeycomb, and imagine that I stretched it out-- and I'm kind of exaggerating how much we might stretch it out.
But if we did something like that, it wouldn't be too surprising to think that the properties are going to be different if I loaded it this way on and that way on.
And in terms of the cell geometry, I'm going to call that vertical cell edge length h. And I'm going to call this one-- the inclined one-- of length l. I'm going to say that angle is the angle theta.
And the cell shape can be defined by the ratio of h over l and that angle theta.
OK.
When we derive equations for the mechanical properties of the honeycombs, we're going to find that they depend on some solid properties.
Say, the modulus of the honeycombs can depend on the modulus of the solid.
It's going to depend on the relative density raised to some power.
And we're going to figure out what that is.
And then it's going to depend, also, on some function of h over l and theta.
And that function really represents the contribution of the cell geometry to the mechanical properties.
OK. That's the honeycombs.
It's fairly straightforward to characterize the shell shape for the honeycombs.
It's a little more involved to do it for the foams.
And the technique that's used is called the mean intercept length.
At least, that's one technique that's used.
Let me wait until you've finished writing because I want you to see the picture as I talk about it.
OK?
OK.
Here's-- whoops.
My pointer keeps disappearing.
This top left picture here shows an SEM image of a foam.
And you can see, you've got some big cells and little cells and there's no obvious way to characterize the cell shape.
And what people do to calculate this mean intercept length is they would take an image.
They would then sketch out just the cell edges that touch a plane's surface.
So all these black lines are just the-- if you took your-- if you put ink on your foam and you just put it on a pad and put it on a piece of paper, you would get this outline of the edges of the cells, where they intersect that plane.
And then what people do is they draw test circles.
Here's the test circle here.
And they draw parallel equiaxed, or equidistant lines.
So the lines here are parallel.
They're all at, say, zero degrees.
And they're all the same distance apart.
And then they count the number of intercepts.
They count-- say we went out here.
The cell wall intercepts here.
There's one that intercepts here.
And then, it'd go up here.
Here's another intercept.
Here's another intercept.
So they count the number of intercepts of the cell wall with the lines.
And then they get a mean intercept length, which is characteristic of the cell dimension.
And then what they do-- because this is just in one orientation-- you would then rotate those parallel lines by, say, 5 degrees and do it all over again.
And get another length at 5 degrees and one at 10 degrees one at 15.
And so you get different lengths for the intercepts as you rotate your parallel lines around.
And then you make a polar plot, and that's what the thing is down at the bottom here.
And so you plot your mean intercept length as a function of the angle that you measured it at.
And you can fit it to an ellipse.
And if you do it in three dimensions, you fit it to an ellipsoid.
And the major and minor axes of that ellipse, or ellipsoid, are characteristic of how elongated the cell is in the different directions.
And the orientation of that ellipsoid is characteristic of the orientation of the cells.
Those of you who took 303, too, you remember Mohr's circles?
Is this beginning to look familiar?
It's the same kind of idea as Mohr's circles.
Same way we have principal stresses and orientation of principal stresses, now we have principal dimensions and the orientation of the principal dimensions.
So it's the same kind of idea.
OK?
Let's see.
I feel like I'm getting to the end here.
Maybe I'll stop there for today.
But next time, I'll write down the whole technique about how we get these mean intercepts and get this ellipsoid.
And I'm going to write the mean intercepts down as a matrix.
But you could also write it as a tensor.
And there's something called the fabric tensor, which characterizes the shape of the cells.
And as you might imagine, the same is with the honeycomb.
If you have equiaxed cells in the foams, you might expect you would get isotropic properties.
If you have cells that are stretched out in some way-- so you've got different principal dimensions for them-- then you've got anisotropic material.
And you can relate how much anisotropy to the shape of the cells.
OK.
I'm going to stop there for today.
I'll see you tomorrow.
Seems very sudden.
I'll see you tomorrow.
I'll pick up and I'll finish this section on the structure.
We've got a bit more to do.
And then we'll start looking at honeycombs and modeling honeycombs.
The honeycombs are simpler to model just because they have this nice simple unit cell.
So we'll start with that, and then we'll move from there to the foams.
OK?
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
All right, so I guess I should start.
So I think last time we were talking about cell structure and cell geometry.
And I got as far as putting this image here up.
And I talked a little bit about how this works.
And I was going to go over it again and then write the notes down today.
So we'll start from there.
So we're going to do a little bit more on cell structure today.
We're going to talk about some topological laws for cellular materials for polyhedral cells.
And then we'll start talking about modeling honeycomb materials, and talk about how we look at the mechanical properties of honeycombs.
So I think where we left off last time was we started talking about mean intercept length.
So the idea is that with-- hello, hello-- if you have a honeycomb, it's fairly easy to define the cell shape in terms of the ratio of the cell edge lengths, h over l, and the angle that the inclined cell wall is to the vertical one.
But for a foam it's a little more difficult.
And what people do is they measure this mean intercept length.
So last time I talked about it briefly.
So the idea is you take an image of the structure you're interested in.
You sort of draw out the outline of a planar surface.
And then you put on that a test circle.
And you superimpose on the test circle equidistant parallel lines.
So say we start off at theta equals 0.
We then count how many times the cell walls intercept those lines.
And we get a number of cells per unit length.
And that gives us an intercept length for that orientation.
Then we rotate it around a little bit, say 5 degrees or something.
And we measure a new intercept length for that orientation.
And we keep doing it all the way around, 180 degrees.
And then you get-- then what you do is you plot those points-- if I can find my little pointer-- we plot those points.
And it will form an ellipse.
And the major and minor axes of the ellipse correspond to the principal dimensions of the cells on that plane.
And then you can do the same for perpendicular planes and form an ellipsoid.
And you can get the three principal dimensions.
You can also get the orientation of the cells by getting the orientation of this ellipse, or the ellipsoid in three dimensions.
So let me just write down some notes that kind of summarize all of that.
So I'm going to say that for foams, we characterize the cell shape and orientation by using these mean intercepts-- mean intercept lengths.
So we consider a circular test area on a plane section.
So you want to use a circle and not say, a square or a rectangle.
Because if you had a square, then you'd have different total lengths of line, depending on what orientation you took it.
So you want to use a circular test section.
And then you draw equidistant parallel lines.
So for example you might start at theta equals to 0.
And then you count the number of intercepts of the cell walls with the lines.
So if we say that Nc is the number of cells per unit length of line, then we can get the intercept length for that orientation.
Say for theta equals to 0, it works out to 1.5/Nc.
So it's not just 1/Nc, because you may not be cutting the cell-- you know, you're cutting the cell-- say you've got a three dimensional cell like this.
You're cutting it at different places along here.
So people have worked out the stereology of that.
And there's a constant that's 1.5 that fits into there.
So then you just repeat that process for different increments of theta.
So you increment theta by some amount, something like say, 5 degrees.
And then you repeat that whole process.
And then you plot a polar diagram of the intercept lengths versus theta.
And you then fit an ellipse to the points.
And if you did it in 3D, you'd have an ellipsoid.
And then the principal axes of the ellipsoid are the principal dimensions of the cells.
And the orientation of the ellipsoid is the orientation of the cells.
Hello.
And you can write the equation of the ellipsoid.
So it would be something like Ax1 squared plus B x2 squared plus Cx3 squared plus 2Dx1x2 plus 2Ex1x3 plus 2Fx2x3.
And that would all equal 1.
And you can write those coefficients as a matrix.
And if you do that, the first three, A, B, C, are the diagonals, and D, E, F are the off diagonals.
And you can also represent this is a tensor.
And if you do, it's called the fabric tensor.
So the fabric-- I think they use that term for other types of materials-- it says something about the orientation of the material.
And if you have this matrix here, if D, E, and F are all of 0, then A, B, and C are the principal dimensions of the cell.
So if all non-diagonal elements are 0, then A, B, and C are the principal dimensions.
Are we good with how this works?
So it's kind of a crank and churn kind of thing.
But it gives you a way of characterizing the cell shape and the orientation of the cells in the foams.
So the next thing I wanted to talk about was the connectivity of the cells.
So imagine we have an array of cells.
And if we have an array we have vertices that are connected by edges.
And edges surround faces.
And the faces enclose the cells.
And there's something called the edge connectivity, which is usually given the symbol Ze.
And that's the number of edges that meet at a vertex.
And there's a face connectivity, and that's the number of faces that meet at an edge.
And it's very common for honeycombs to be three connected.
So Ze is 3.
And it's common for foams to be four connected.
And there's some topological laws that are kind of interesting.
And so we'll get into those after this bit on connectivity.
So for the connectivity we have vertices, which are connected by edges, which surround faces, which enclose cells.
So we're going to talk about the vertices, the edges, the faces, and the cells.
So the edge connectivity, Ze, is the number of edges that meet at a vertex.
And for a honeycomb, say a hexagonal honeycomb, Ze is 3.
So I have a little honeycomb here.
If we look at the number of edges, there's three edges, connect at a vertex.
So Ze is 3.
And for a foam, Ze is typically four.
So I'll just say typically here.
And if I go back, if you look at the sketches here of the rhombic dodecahedra and the tetrakaidecadedra, if you look at the tetrakaidecadedra, you can see the number of edges.
So here's one edge.
Here's another one here.
There's another one here.
And there's another one here.
So that's the four edges that are meeting at that vertex there.
So if you look at arrays of cells, you can convince yourself that Ze is typically 4 for a foam.
And then we also have the face connectivity, Zf.
And that's the number of faces that meet at an edge.
And that's typically three for foams.
And so again if you look at these pictures you can sort of see how the face connectivity is three.
So if you look at this bottom one here, there's faces here.
And then there's another face-- well, let's see if we can-- it's kind of hard to show.
If you look at this one here, there's this cell.
There's that face there.
And then there's another one coming out of the page, I think.
OK.
So that's connectivity.
And the next thing are some of these topological laws.
So the first one I'm going to talk about is called Euler's law.
If you remember Euler from buckling, same guy.
So Euler's law relates the number of vertices and faces and cells and edges for a large array of cells.
So we can relate the total number of edges, so I'm going to call that E, vertices V, faces F, and cells C. That total number is related by Euler's law for a large aggregate of cells.
So in 2D Euler's law is that the total number of faces minus the number of edges plus the number of vertices is equal to 1.
And in 3D, you just add a minus the number of cells in front.
So minus the number of cells, plus the number of faces, minus the number of edges, plus the number of vertices is equal to 1.
So there's some interesting things you can figure out using Euler's law.
And one of the things I wanted to look at was if we had an irregular honeycomb that was three connected.
So say we have a honeycomb that's not just a regular hexagonal honeycomb, or not even one that's got repeating hexagonal cells.
But say we had a honeycomb where some of the cells had 5 sides to them, some of them had 7, some of them had 6.
You can ask, what's the average number of sides per face?
And you can use Euler's law to figure that out.
So we're going to look at an irregular three connected honeycomb.
Let's see, I can start that up here.
So when I say irregular, I mean we've got cells with different numbers of sides.
So we have an irregular-- oops.
You OK, Greg?
--that is three connected.
And what is the average number of sides per face?
And I'm going to call that n bar.
So we're told that it's three connected.
So that's saying that the edge connectivity is 3.
So there's three edges coming to each vertex.
So imagine that you had just a regular hexagonal honeycomb, like this.
Here's our vertex here.
And there's our kind of three edges that come into it.
And each vertex is going to have half of each edge, right?
Because the next vertex-- each edge is shared between two vertices.
So if Ze is equal to 3, then the number of edges per vertex is 3/2.
Because each edge is shared between two vertices.
And I'm also going to define something I'm going to call Fn.
And Fn is going to be the number of faces with n sides.
So I could have some number of the cells have six sides, some number have five sides, some number have seven sides.
So these are-- Fn is the number of faces with some particular number of sides, n. So if we have Fn is the number of faces with n sides, then if I sum up n times Fn and divide it by 2, that's the number of edges.
So imagine I have all these faces.
I take n times Fn.
So say we had four five-sided faces.
That would be four of those.
There's be 20 edges associated with that, 20 sides.
Then I have some number of six-sided, some number of seven-sided.
I add them all up.
And I have to divide by 2 to get the number of edges, because each edge separates two faces.
OK. And then I can use Euler's law.
So I can say F minus E plus V is going to equal 1.
And here I've got an F minus E plus-- I can put V here.
V's going to be 2/3 of E. So I've got that.
So that's the same as F minus 1/3 of E is equal to 1, like that.
And then for E I can substitute this thing here in.
So that gives me F minus 1/3 of the sum of n times Fn over 2 is equal to 1.
And then what I'm going to do is multiply everything by 6 so I can get rid of my denominator here.
So 6F minus sum of nFn is going to be 6.
And then I'm going to divide that through by F. So that's 6 minus sum of nFn over F is equal to 6/F.
And then if I let F go to a large number-- so say I've got a large aggregate of cell.
I've got lots of faces.
I'm going to let F become large.
So if F becomes large, then 6/F is going to tend to 0.
Let me get the other rubber
Professor?
Mhm?
That's-- F is like the total number of faces, not the total number of faces per cell
F is the total number of faces.
And Fn is the number of faces with n sides.
We've got some number with 5 sides, some number with 6, some number with 7.
So we're almost at the end here.
So 6/F goes to 0.
And so that says that the sum of n times Fn over F is equal to 6.
And that just is n bar.
That is the average number of sides per face.
This is your-- kind of your total number of sides in the whole thing.
And you're dividing by the total number of faces.
So that is the average number of sides per face.
So what this is saying is if you have a three connected honeycomb, the average number of sides per face is always 6.
So that if you introduce a cell with 5 sides, somewhere you have to introduce a cell with 7 sides, so that they compensate and you come back out to 6.
And I have a little soap bubble picture here that kind of illustrates that.
So here we have a soap honeycomb.
So you can make these just by putting two glass sheets close together with a soap bubble froth in between them.
And you can kind of see in this picture here, they've numbered how many sides each cell or each face has.
So here's one with 5, here's one with 7.
And I'm not going to add these up all together.
But you can kind of see that the 5's and the 7's kind of compensate for each other, and that the average works out to about 6.
And this is true for a large aggregate of cells.
So if you only have a few, it's not going to work out perfectly.
You need to have a large aggregate to have it work.
Yeah?
And that 5, 7 balance doesn't have to be touching, right?
No, no, no, overall.
Yeah, overall.
And, you know, you could have one with 4 sides, and you'd need two 7's or one 8 or something.
But you'd need to have that balance out and match up.
OK.
So that's the Euler law.
Now there's a couple more of these kinds of things.
There's something called the Aboav-Weaire law.
And so the Aboav-Weaire is sort of related to the Euler because it's looking at this idea that if you have a three connected honeycomb, if you introduce a 5-sided cell, you have to introduce a 7-sided cell to compensate.
And what Aboav noticed was that generally cells with more sides than average have neighbors with fewer sides than average.
Let's see.
So we'll say the introduction of a 5-sided cell requires the introduction of a 7-sided cell.
And I'll just put-- this is all for a 3 connected net, a 3 connected honeycomb.
So that generally, cells with more sides than average have neighbors with fewer sides than average.
And in 3D, you could say the same thing about the faces.
The cells that have more faces than average have neighbors with fewer faces than average.
So Aboav made observations of this.
And it was Dennis Weaire who made a proof of it.
And the equation that they came up with relates the average number of sides in the cell surrounding a candidate cell.
Let's see-- are we going to be able to put it in here?
Maybe.
So say you have a candidate cell and say it has n sides.
So you look at one particular cell and you say, count up, it's got n sides around it.
And then you count up the number of sides of all the cells surrounding it.
It has n neighbors because it has n sides, so then the average number of sides of the n neighbors is called m bar.
It has n sides.
So then the average number of sides of its n neighbors is m bar.
And m bar is equal to 5 plus 6 over n for a 2D honeycomb kind of cell structure.
OK, so that's the Aboav-Weaire law.
And then there's one more of these things.
The last one is called Lewis' rule.
And Lewis looked at biological cells and 2D cell patterns, and he found that the area of the cell varied linearly with the number of the sides.
And he found just an empirical relationship.
It looks like this.
We can say what everything is in a minute.
So he found that the area of a cell with n sides, that's A n, was linearly related to the number of sides, so this n over here, and naught's just a constant and A n bar is the area of the cell with the average number of sides.
So here, A n is the area of cells with n sides and A n bar.
And then n naught is just a constant.
And he found that for 2D cells n naught was equal to 2.
And if you look at voronoi honeycombs-- remember last time we talked about those voronoi honeycombs-- you can show that this holds for voronoi honeycombs.
And then you could write a 3D version of this as well.
And in 3D, it's the volume and faces instead of the areas and the number of sides.
So you can say that the volume of the cells with f faces relative to the volumes of the cells with the average number of faces, they vary linearly with the number of faces.
So there's sort of an exactly analogous expression here.
And here, this f naught is another constant.
And in 3D it's about equal to 3.
So these are all just kind of interesting topological rules that are nice to know about.
OK. Yeah?
Basic question.
In two dimensions, what is a face?
Like what does that refer to?
So-- and let me put my little slides again.
So say we have some honeycombs like this, then say we look at this guy.
So that's a vertex there, that's an edge, and this thing in the middle here is the face.
So in 2D, the face and the cell is kind of the same thing.
All right?
And then--
[INAUDIBLE] sides, what's different with the edge?
It's not quite.
Like if I count up the number of edges, I say one, two, three, four, five, and I count those up like that.
But if I say sides of the face-- see that's a face or it's a cell-- I would say that had six sides.
Right?
So typically when people say it has-- a cell or a face has so many sides, they count up how many all around it.
But because each side is-- or each edge is-- shared between two faces, the number of edges is actually half that.
Right?
Because there's one edge, but if I count up the number of sides for this face and I count up the number of sides for that face, I'm counting that twice.
So it's a little bit-- so I try to see edges when I'm talking about adding them all up and I try to say sides when I'm talking about, here's a face, how many sides does it have.
OK?
It's a little bit confusing.
Anybody else?
[INAUDIBLE] what is n bar naught?
n bar naught, did I put an n bar naught?
Oh, that's where I didn't erase this enough.
There you go.
It's just n naught.
OK.
So we've been talking about the structure of the honeycombs in the foams.
And what we ultimately want to do is be able to model the mechanical behavior or the thermal behavior, some sort of behavior of the cellular material.
And for looking at mechanical behavior, there's three main approaches that people take.
So you have to model the structure somehow.
So there's three main approaches.
And the first one is to use the unit cell.
So say, for example, for the honeycombs, if you have a hexagonal honeycomb you'd just use that unit hexagonal cell.
And that's what we're going to do probably starting later on today.
So for the hexagonal honeycomb, it's kind of obvious.
You would use a unit cell, the thing's periodic, you figure out how that unit cell behaves, you're all set.
For a foam, it was not so obvious a unit cell to use.
But the thing-- people use different things, but one of the common ones they use is a tetrakaidecahedron.
So the nice thing about the tetrakaidecahedron is that it packs to fill space.
So it's a repeating single cell.
Packs to fill space.
But, in fact, real foams aren't tetrakaidecahedrons, so this is a bit of an idealization.
So we'll just say foam cells are not tetrakaidecahedrons.
OK, so that's one approach.
And then a second approach is to use something called dimensional analysis.
So in dimensional analysis, what you want to do here, with this technique, is model the mechanisms of deformation and failure in the structure.
But you don't necessarily represent the cell geometry exactly.
And so what you do is, you say one thing is proportional to another and there's some constant of proportionality, and you just wrap all of those constants of proportionality up at the end.
So for instance, when people look at foams, the geometry of the foams is kind of complicated.
There's cells with different number of faces, there's different sizes of cells, it's kind of a mess.
So we could just say the geometry is complex and it's difficult to model.
And with dimensional analysis, instead what we do is we model the deformation mechanisms and failure mechanisms.
And you can get quite a bit out of just modeling those.
So when we look at modeling foams, we're going to do this, and you'll see how it works.
And then the third method is to use finite element analysis.
So this is a numerical technique and it's a very standard numerical technique.
And one of the nice things about this is you can apply it to random structures.
So for example, those voronoi structures we saw, if you want to try to see-- say you had a certain amount of material and you had a honeycomb that was a regular hexagonal honeycomb.
You had the same amount of material and you put it in a voronoi honeycomb, and you want to know how does having a random material affect the properties relative to the uniform material?
You could figure that out using finite element analysis.
So you can apply it to random structures.
And another thing you can do with finite element analysis-- and people in the orthopedics end of the world often do this-- if you're interested in trabecular bone, that porous type of bone I showed you the first day, people can take micro-computed tomography images of trabecular bone and they get a file which basically says, every voxel, says if it's solid or it's void.
And you can use those files as input to a finite element analysis.
And so you can analyze exactly how a piece of a trabecular bone would deform.
So it's nice for that.
So one of the things we're going to talk about later is we'll look at the structure of trabecular bone and the sort of mechanics of trabecular bone, and I'll show you some results that a former student of mine did, where we look at-- say you have a sort of intact structure of a certain density and then you reduce the density by fitting the cell walls or you reduce the density by removing cell walls-- because in osteoporosis sometimes the walls resorb all together-- and you can see what residual strength you would have for some given amount of density loss.
So it's good for looking at that kind of a thing.
You can also use it for looking at local effects.
So you can look at defects, for instance.
So if you think of the trabecular bone as having missing trabeculae, that would be a defect in the structure.
You can look at that.
You can look at size effects.
So when we were studying some of those metal foams, some of them have very large cells, and if you have a small sample relative to the cell size, you may only have four or five cells across the dimension.
Where you've cut the cells, you've got edges that are less constrained than in the bulk of the material because at the outer edge you've cut them.
They're not connected to anything else.
And so you can look at edge effects that relate to the size effects in foams as well.
So these are basically the three approaches that people use for modeling cellular materials.
And what we're going to do is we're going to start with looking at honeycombs and we're going to look at these hexagonal kind of honeycombs like this.
And one reason to start with them is that they have this unit cell structure.
And if you can analyze how that unit cell deforms and fails, you can say something about the whole structure.
And it turns out that the honeycombs, they deform and fail by the same mechanisms as the foams.
So if you can understand through this simple structure, it gives you a lot of insight into how the foams behave.
So if I deform this a little bit, the cell walls bend.
And you can show that if you deform this guy a little bit, the cell walls bend.
And so you can learn a lot by looking at the honeycombs and then sort of applying that to the foams.
So we're going to start off-- so that's the end of the section on sort of the structure of the cellular materials-- and now we're going to look at modeling the mechanical properties and we're going to start with the honeycombs and then we're going to do foams.
So when we talk about the honeycombs, we've got this hexagonal structure here and we're going to call properties in this plane the in-plane property.
So if I load it this way on or that way on, those are in-plane, and if I load it that way on, those are out of plane.
So think of the cells are in the plane and the prismatic direction is the out of plane.
And clearly, the honeycombs are going to have similar properties this way and that way, but they are going to have very different properties that way.
So we're going to start with the in-plane behavior.
So the honeycombs have these prismatic cells and they're widely available in different materials, polymers, metals, ceramics.
And they're used in a variety of applications.
So one of the most common is to use them in sandwich panels.
So I brought a couple of sandwich panels with me today.
So here's a couple of sandwich panels that have honeycomb cores.
This one's an aircraft flooring panel.
It's got carbon fiber faces and a Nomex core, honeycomb core.
And this is an aluminum honeycomb.
It's got aluminum honeycomb core and then aluminum faces, and it's kind of amazing how stiff that little panel is.
And each of the pieces is really not that stiff at all, but the thing put together is quite stiff, and we'll talk more about that when we get to sandwich panels.
So it's used in sandwich panels.
They're used for energy absorption.
So sometimes you'll see there's some natural disaster area and they fly in helicopters and they drop big crates of supplies, and they will have it like a pallet with a big crate thing.
Often they have a honeycomb, like a metal honeycomb, that is kind of oriented this way.
So the pallet would be like this and the honeycomb's like that.
And the idea is that when they drop it-- they sort of bring it down as close as they can-- and then the honeycomb absorbs some of the energy from the impact.
And they're also used, I think, sometimes in car bumpers.
So they're used for energy absorption.
And they're used as carriers for catalysts.
So the catalytic converter in your car looks like this, this is kind of the material that's used in the catalytic converter in your car.
And the way that works is, the cell walls here are actually porous and they're coated in the platinum, which is the catalyst, and half of the cells are blocked off on this end.
So every other cell on this end is blocked off and then every other cell over here, the opposite ones, are blocked off.
And so the gas is forced down a channel but then through the wall and then out the next channel.
And that's where the reaction actually occurs, is in the cell wall.
So they're used as carriers there.
And some natural materials also have a cellular structure, have a honeycomb structure.
So for example, things like woods and cork have a honeycomb structure, too.
So I said the mechanisms of deformation and failure in the hexagonal honeycombs parallel those in foams.
So we can learn a lot about foams by understanding the honeycombs.
And then, similarly, the mechanisms of deformation and failure in triangulated honeycombs parallel those in the lattice materials.
Remember I brought those lattice materials in?
The sort of trust type materials.
The triangular honeycombs have a behavior similar to those lattice materials.
OK.
So let me scoot over here.
OK.
So let me scoot out of this one.
OK.
So here is our kind of hexagonal geometry.
This is kind of an idealized geometry here.
And, as we talked about in this section on the structure, we're going to call these vertical walls, we're going to say they have a length h, the inclined walls have a length l, the wall thickness is t, and there's going to be an angle between the horizontal and that inclined wall of theta.
And I'm going to define three axes here, an x1, an x2, and an x3 axis.
So the x1, x2 plane is the in-plane and x3 is out of plane.
OK?
So if we load our honeycomb up, we get stress strain curves that look like this.
So the ones on the left here, over here these three are all in compression, and the ones on the right, these three are all in tension.
OK, so let's talk about the compression ones first.
And let's start at the top.
This is in elastomeric material, so like one of these rubber honeycombs.
This would be a material that yields plastically, so like an aluminum honeycomb.
And this would be a honeycomb that fails in a brittle manner, like one of those ceramic honeycombs.
OK?
So if we go up to the elastomeric one, up here, if I compress it I get a linear elastic part first, and then at some point, I get a stress plateau where the stress is almost constant for strains that are quite large.
And then finally, the stress starts to rise quite sharply at the end here.
So the strain here goes from 0 to 1.
So that's a strain of 100%.
You've completely flattened the thing there.
So that's a large strain.
So initially, when we're loading it up to smaller strains like this, we've got linear elastic behavior and these cell walls bend.
And you can relate the modulus here-- let's see where'd my little arrow go-- you can relate this Young's modulus here to the bending of those cell walls.
And we're going to-- I don't know if we'll finished that today-- but we'll start that today.
Then-- oops, lost my arrow, where'd my arrow go-- then at the stress plateau here, that plateau is related to collapse of the cells.
So if I have my little rubber honeycomb and I load it, at some point, the cell walls buckle.
So you see how they've buckled there?
And that stress plateau, I can smush that to quite large strains that are roughly constant stress.
OK?
So what's happening here is that we're buckling the cells.
And as we go along here, the buckling deformation gets bigger and bigger.
And then if I smush it and I've got it kind of like that, at some point it becomes much harder to press it together again because the cell walls are now touching each other and they're pressing against themselves, and to get a certain amount of strain, it gets much more difficult to do that.
And the stress required to do that gets much bigger.
And that's what leads to this last piece of the stress strain curve here, which is called densification because you've almost eliminated the pores.
It's hard to eliminate them entirely but the porosity has gone way down by the time you're up there.
So this type of stress strain curve where you've got linear elasticity and then a stress plateau and then the densification is classic for compressive behavior of cellular materials.
So elastomeric ones will have a plateau that's related to elastic buckling.
I lost my arrow again.
There we go.
If we had a metal honeycomb, we'd again have linear elasticity related to cell wall bending.
This plateau here would be related to yielding of the cell walls.
So say we had aluminum honeycomb, the aluminum could yield, and that would again cause this stress plateau.
And then we get densification.
If I had a brittle honeycomb like the ceramic one, we'd have initial linear elasticity.
Then you've got a stress plateau that's kind of a lot of up and down here.
And the serrated nature is related to the fracture of individual cell walls.
So it goes up and down because when you break a cell wall, the stress drops off, and then the other cell walls will try to pick the stress up again.
And so each one of those little up and downs corresponds to breaking a cell wall.
But if you kind of took an average of that, you can see there's a stress plateau and then there's the densification region.
So in compression, the shape of the curves is very similar and the mechanism of the plateau varies a little bit.
So let's see if I can write some of that down.
So in compression, we can say we have three regimes of behavior.
So we've got the linear elastic regime initially and we're going to see that's related to cell wall bending.
Oh, thanks.
And then we've got a stress plateau.
And for elastomeric materials, that's caused by buckling of the walls.
For metals, it would be caused by yielding.
And for ceramics, it would be caused by a brittle crushing.
And then we've got densification.
And that's related to the cell walls touching.
And if we increase the ratio of the thickness of the cell walls relative to their length, we're going to increase the stiffness, so the Young's modulus of the honeycomb.
The stress plateau I'm going to call sigma star.
And we decrease the strain at which that densification occurs, which I'm going to call epsilon D. So you see over on the right hand side of these plots here, that strain there is the densification strain, epsilon D. So that's in compression.
And these materials are very often loaded in compression.
In tension, we still get linear elasticity initially.
And that's going to, again, be related to the bending of the cell walls.
But if we look at the stress plateau, if you look at these three curves here on the right, the stress plateau only exists if the material has a yield point and you get some plastic yielding there.
If you have an elastomer, if I pull on this-- you don't get buckling in tension, so you're not going to get a stress plateau in tension.
And for a brittle material like one of those brittle honeycombs, if you pull that in tension, you would just get a crack propagate and the thing would break into two pieces and so you don't get a stress plateau there.
So the stress plateau in tension only exists if the material yields.
So you don't get any buckling in tension.
And for a brittle honeycomb, you would just get fracture.
OK?
These inclined walls are going to bend.
We're going to see that in more gory detail in one minute.
If you can wait one minute.
OK.
So are we good with a stress strain curves yet?
More of an abstract question, but is it possible [INAUDIBLE] to get collapse this way before the plateau?
I haven't seen that.
But maybe it it's possible, I don't know.
I don't think so, but maybe.
Anybody else?
OK. OK, so here's some photographs of honeycombs.
So these ones are white but these ones are just the same.
So most of them are just a regular hexagonal honeycomb and one of them is this funny shaped honeycomb here.
These two guys at the top, these two here, are unloaded.
And then the rest of them have some loading.
So if you look at this one down here-- so here I'm taking the honeycomb and I'm loading it.
This is the x1 direction, that way on.
And if you look very carefully, you can see what happens is these vertical walls just move sideways and that is going to zoot, with the sound effects.
And then the-- I can't help making sound effects-- and then these guys here bend.
OK, so this guy here-- it's maybe a little hard to see it on the image but it's actually a little bit bent there.
So that's loading it this way on.
And then similarly, if I take it and I load it that way on, that's the x2 direction.
And now these guys here, the vertical guys, are just going to compress a little.
But these guys here, the incline guys, are still going to bend a little bit.
And I've got some schematics that's going to show that a little bit better.
And then if I shear it, if I took it like this and I-- I can't do it because my hands aren't glued to the rubber-- but if I could sort of shear it this way on, then you would get this kind of deformation here and that also involves bending.
You can imagine these guys here would bend if I do that to them.
And then the buckling deformation looks like this.
If I scooch that like that, it should look kind of like that picture there.
OK?
So that's kind of what the deformation looks like for these sorts of honeycombs.
So these were elastomer rubbers.
These are just some images from an aluminum honeycomb.
So here's, the top one's the undeformed honeycomb, the middle one's loading it in the one direction from left to right, and the bottom one's loading it in the two direction from top and bottom like that.
OK?
So you can imagine that you've got little cell walls.
If you load it up high enough, those walls are going to yield and we're going to see that the yielding in the walls causes the formation of something called plastic hinges and the walls can rotate and they can produce these kind of shapes.
OK.
So then this is sort of a schematic stress strain curve here.
And this is showing what happens if you increase the thickness of the walls relative to the length, or you increase the relative density, you increase the volume per action of solids.
And this kind of shows the different regimes.
So over here-- well let's, first of all, start with the different relative density.
So this one here is the lowest relative density, then this is higher, and higher, and higher.
And not too surprisingly, the more you increase the density, the more material you've got, the stiffer it is, the stronger it's going to be.
And the more material you've got, the sooner it's going to densify.
If you've got more solid in here, you're going to reach that densification strain at a smaller number.
OK, so the shape of the curves looks like this.
And you can define these three kind of regimes.
So everything in here is linear elastic, everything in this big sort of envelope here is the plateau region, and everything up here is this densification region.
So this is just a bigger picture kind of plot.
OK, so let me skip through all of that.
All right.
OK, where are we?
Chalk?
Here we go.
OK, so I think I mentioned this before, but let me just go over it again.
So there's three types of things that affect the properties, the mechanical properties, of the honeycombs.
And probably the most important thing is the relative density of the honeycomb.
So remember, we said this was the same as the volume fraction of solids, and for a hexagonal honeycomb, you can show that's equal to the thickness to length ratio times h over l plus 2 divided by 2 cos theta times h over l plus sin theta.
And if you have regular hexagons, so h over l is equal to 1, all the sides are of equal length, theta is equal to 30 degrees, the relevant density is 2 over 3 times t over l. So it just goes linearly with t over l. The thickness to length ratio of the walls.
So it depends on how much solid you've got.
Depends on the solid properties.
So the Young's modulus of the solid, yield strength if it's a metal, some sort of fracture strength if it's brittle.
It also depends on the cell geometry, which we can describe with h over l and theta.
So if we think of a cell here-- that's our edge length h, that's our edge length l, that's our angle theta, here's the cell wall thickness t, and then we've got some set of solid properties here.
OK?
So that's kind of the set up.
And we're going to define x1, x2 axes like that.
And we're going to make a few assumptions just to make life a little bit simpler.
So we're going to assume that t over l is small, so that also means the relative density is small.
And what that means is that we're going to be able to neglect axial and shear deformations.
So you can imagine, if I have a thin wall and I'm applying loads that produce moments and produce bending, if the wall is very thin then the axial and the shear deformations are going to be small.
I'm also going to assume the deformations are small.
And what that means is that I'm going to neglect any changes in the geometry of the cell during the deformation.
And I'm going to assume that the cell wall is linear, elastic, and isotropic.
And we're going to start off with looking at in-plane behavior, and we're going to start with the elastic moduli.
And if we look at the elastic moduli, we're going to be talking about Hooke's law.
And Hooke's law and the elastic behavior of the material can be described by a set of elastic constants.
And if you recall, the number of independent elastic constants, how many constants you need to describe the material, depends on its symmetry.
And these materials are orthotropic.
So the regular hexagonal honeycomb is actually transversely isotropic, but imagine that h was not equal to l, then it would be orthotropic.
So remember, orthotropic means that you can rotate the structure 180 degrees about three mutually perpendicular axes and the structure looks the same.
So if I take this and I do that, it looks the same, and if I do that, it looks the same, if I-- no matter how I rotate this, about three mutually perpendicular axes, the structure remains unchanged.
So it's orthotropic.
So I'm going to write down Hooke's law for our orthotropic material, and then we'll talk about the constants that we're going to work out.
OK, so this is Hooke's law for our orthotropic material.
And let me just remind you what our notation is here.
So epsilon 1 is epsilon 1 1.
Epsilon 2 is epsilon 2 2.
Epsilon 3 is epsilon 3 3.
Epsilon 4 is gamma 2 3.
Epsilon 5 is gamma 1 3.
And epsilon 6 is gamma 1 2.
So these are the normal strains here, epsilon 1, 2, and 3, and these are the shear strains here, epsilon 4, 5, and 6.
And you remember this convention where the subscripts add up to 9.
So 4 plus 2 plus 3 is 9, 5 plus 1 plus 3 is 9, 6 plus 1 plus 2 is 9.
And then the stresses are a similar thing.
So sigma 1, sigma 2, and sigma 3 are the normal stresses.
And then sigma 4, sigma 5, and sigma 6 are the shear stresses.
And for the in-plane moduli, so we're dealing with the x1, x2 plane, there's four independent elastic constants.
So we could think of it as E1, E2, a Poisson's ratio 1 2 and a shear modulus in the 1 2 plane.
OK?
And the compliance matrix is symmetric, so there's the reciprocal relationship between the moduli and the Poisson's ratios.
And then the notation I'm going to use for Poisson's ratio, I'm going to say that mu i j is minus the ratio of the strain in the j direction divided by the strain in the i direction.
OK.
So what we're going to do next is we're going to calculate some of the elastic moduli.
I'm going to show you the derivation for E1 star and mu 1 2 star, and you can get the other two in a similar way.
So I'm not going to do all of them but next time we'll do the derivations for the Young's modulus in the Poisson's ratio.
OK?
And then we're going to talk about the out of plane direction later and we'll get the moduli for the out of plane direction as well.
OK?
So I think I'm going to stop there for today.
And then we'll start doing the derivations next time.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
OK, so it's five after.
We should probably start.
So last time we were talking about honeycombs, and I just wanted to quickly kind of review what we had talked about, and then today I'm going to start deriving equations for the mechanical properties of the honeycombs, OK?
So this is a slide of our honeycomb setup here.
These are the hexagonal cells we're going to look at.
We talked about the stress-strain behavior.
The curves on the left-hand side are for compression and the ones on the right-hand side are for tension.
And so what we're going to be doing today is we're going to start out by calculating a Young's modulus, this slope here.
We're going to calculate the stress plateaus for failure by elastic buckling in elastomeric honeycombs, by failure from plastic yielding in, say, a metal honeycomb, and by failure by a brittle crushing in, say, a ceramic honeycomb.
And if we have time, we'll get to the tension stuff.
I don't know if we'll get to that today or next time.
So we're going to start calculating those properties today.
And these were the deformation mechanisms.
Remember, we said the linear elastic behavior was related to bending of the cell walls, and then the plateau was related to buckling if it was an elastomer.
And the plateau was related to yielding if it was, say, a metal that had a yield point.
And then this was sort of an overview of the stress-strain curve showing those different regions, OK?
So what I'm going to talk about today to start is the linear elastic behavior.
And we're going to be starting with the in-plane behavior.
So in-plane means in the plane of the hexagonal cells.
And then next time we'll do the out-of-plane behavior, this way on.
So if I had to form my little honeycomb like this, what initially happens is the inclined cell walls bend.
So if you can see over here, we've kind of exaggerated it on this sketch.
So this wall here is bent.
This one here just kind of moves along, goes for the ride.
And this guy here is bent.
So this is for loading in the what we're calling the x1 direction, sigma 1.
And the same kind of thing happens when we load in the other direction, in the sigma 2 direction, these guys still bend.
Now the honeycomb gets wider that way.
It gets shorter this way, wider that way.
And we can calculate the Young's modulus if we can relate the load on the beam in the moments to this deflection here, right?
So the Young's modulus is going to be related to the stiffness and the stiffness is going to be related to how much deformation you get for a certain amount of load that you put on the beam.
So I'm going to calculate the modulus for the x1 direction, the thing on the left there.
And you can do the same thing for the x2 direction on the right, but I won't calculate that because it's exactly the same kind of process.
OK, so let me start here.
Get my chalk.
So I'm going to draw a one-unit cell here.
So here's my unit cell there, like that.
And this member here is of length h. That member there is of length l. That angle there is theta.
I'm going to say all the walls have equal thickness, and I'm going to call it t. And I'm going to define an x1 and x2 axis like this.
So the horizontal is x1 and the vertical axis is x2.
And I'm going to say that I apply a sort of global stress to it, sigma 1.
So there's a stress in the one direction there, sigma 1, OK?
And I'm going to say my honeycomb has a depth b into the page, but the depth s-- is because the honeycomb's prismatic, the b's are always going to cancel out of all the equations that we're going to get, because everything's uniform in that direction.
And we can think about a unit cell here, and in the x1 direction, we could say the length of our unit cell is 2l cos theta.
So in the x1 direction, that's our unit cell there, and that's 2l cos of theta.
And in the x2 direction, you might think that you go from this vertex up here down to that vertex there, but if you did that, then on the next layer of cells, you wouldn't have the same distance.
So the unit length in the x2 direction is actually from here to here, and then you can see the next cell, you would get the same thing.
You get this bit here from the inclined member, and then you would get h down here from the next member.
So this bit here is equal to h plus l times sine of theta.
So I can say in the x2 direction, the length of the unit cell is h plus l sine theta, OK?
So that's kind of the setup.
And then what we want to look at is that inclined member that bends, we want to look at how this guy bends under the load.
And if we can relate the forces on it to the deflection, and we need the component of the deflection in the one direction, then we're going to be able to get the modulus.
So I'm going to draw that inclined member again over here, and it's going to see some loads that I'm going to call p, and that's going to cause this thing to bend.
So I've kind of exaggerated it there, but there's the bending.
And there's some end deflection there, delta.
And there's moments at either end of the beam, or either end of that member, as well.
And this member here has a length.
That length is l. OK, are we good?
So it's just kind of the setup.
And I'm going to draw the deflection delta bigger over here.
So say that's delta.
That's the same parallel as this guy here.
What I'm going to want is the deflection in the x1 direction, and when I come to calculate the Poisson's ratio, I'm going to want the deflection in the x2 direction.
And if this angle here is theta, between the horizontal and the inclined member, then this angle up here is also theta, and so this bit here is delta sine theta.
And this bit here is delta cos theta.
Ba-doop-ba-doop-ba-doop.
So the Young's modulus is going to be the stress in the one direction divided by the strain in the one direction.
So I need to get the stress and the strain in the one direction.
So here the stress in the one direction.
If I'm applying my load, like this, sigma 1, the stress in the one direction is going to be this load p-- so this load, say, p on this member here, divided by this length here, the unit cell length, and then divided by b into the board, the width end of the board.
So sigma 1 is going to be p divided by h plus l sine theta times b.
And epsilon 1 is going to be the strain in the one direction, is going to be the deformation in the one direction divided by the unit cell length in the one direction.
So that's going to be delta sine theta divided by l cos theta.
So here, even though I've said this unit cell is 2l cos theta, there'd be two of the members here that would be twice that deflection in the one direction.
So it's, for one beam, it's delta sine theta over l cos theta.
Are we OK so far?
So far so good?
So for the hexagons, because we're going to figure out the equations more or less exactly, we're going to keep track of all the geometrical factors.
When we come to the foams, we're not going to keep track of all the geometrical factors.
So one of the things that makes us look a little kind of hairy is just the fact that we're keeping track of all these sines and cosines and all the dimensions and things.
All right, whoop.
So I need to be able to relate my load p to my deformation delta to get a stiffness out of this, to get a modulus, OK?
So the way I do that is, remember in 3032, we did those bending moment diagrams and we did the deflection of the beams?
This is where this comes in handy.
So I'm going to draw my beam a little bit differently now.
I'm going to turn it on its side.
So this is still my length l, but I'm going to turn on my side just so that you can see it the same kind of way we did the bending moment diagrams.
So this is still my length l across here.
And there's end moments, M and M here.
And P sin theta is just the perpendicular component of the load p. So p sin theta is just the component perpendicular to my beam.
So I could draw a shear diagram here and I could draw a bending moment diagram here.
And, if you remember, the shear diagram, if I have no concentrated load along here, and I have no distributed load along here, if this is zero, down here, it's just going to go up my P sin theta, and then be horizontal, and then come down by P sin theta, OK?
So that's the shear diagram.
And then the bending moment diagram, I'm going to draw down here.
So I've got some moment at the end here, and this would tend to bend like that.
So this would be a negative moment.
Remember, bending moments were negative if there was tension on the top, and they were positive if there was tension on the bottom.
So over here we'd have tension on the top, so that would give us a negative bending moment.
And then, if you also remember, the moment at a particular point is equal to the integral from, say, A to B-- well maybe I should write this another way.
And B minus Ma is the integral of the shear diagram between the two points.
A little sloppy.
OK.
So if I know how I have some moment here minus M, if I integrate this shear diagram up, then this is just going to be linear here, and then I'm going to be at plus M over there.
So if you look at this shear and bending moment diagram, it's really just the same as the shear and bending moment diagram for two cantilevers that are attached to each other.
So let me just draw over here what the cantilever looks like.
Let's see.
So imagine I just had a cantilever like this, and I have some force F on it like that.
And I call this distance here capital L, and I'm going to call that deflection capital delta, like that.
If I drew the shear diagram for that, there'd be a reaction here, F, there would be a moment here, FL.
Doot, doot, yup.
So this would look-- whoops-- it's a little too long.
Shear diagram here would look like this.
That would be zero.
This would be FL.
And the moment diagram would look like this.
Whoops.
A little too long again.
And that would be minus FL.
And that would be zero.
So do you see how the shear and the bending moment diagram here are really just like two cantilevers, OK?
So I know that the deflection for a cantilever, delta, is equal to F capital L cubed over 3EI.
It's kind of a standard result.
And so I can take this and apply that to this beam here.
So instead of working everything out from first principles, I'm just going to say that my beam here is like two cantilevers, and instead of F, I've got P sin theta.
And instead of capital L here as the length, I've got l/2 because l/2 would be the length of one of the cantilevers.
OK, so for the honeycomb, I've got two cantilevers of length l/2.
So delta for the inclined member on the honeycomb is going to be 2-- because I've got two cantilevers-- the force, instead of having F, I'm going to have P sin theta.
And instead of having capital L, I'm going to have l/2, all cubed.
So this is like F capital L cubed over 3.
And here, the modulus that I want is the modulus of the solid cell wall material, so I'm going to call that ES, and over the moment of inertia.
So you see how I've done it?
Is that OK?
So then I can just kind of simplify this thing here.
I've got P sin theta l cubed.
1/2 cubed is going to be 1/8.
So this is going to be 2, if that's 1/8, times 3 is 24.
So 2/24 is 12.
So I've got delta for my honeycomb member is P sin theta l cubed over 12 EsI.
And here I is the moment of inertia of that inclined member of the honeycomb.
And that's BT cubed over 12, OK?
So B is the depth into the board, and T is the thickness.
We'll cube that and divide by 12.
It's a rectangular section.
Yeah?
What was ES again?
ES is the Young's modulus of the solid that it's made from.
So clearly, if my honeycomb is made up of these members, whatever material the members are made of is going to affect the stiffness of the whole thing.
Are we good?
Because once we have this part, then we just combine these equations for the stress and the strain in the one direction.
And we have this equation relating delta and P, and we're going to be able to get our Young's modulus, OK?
We're happy?
OK. All right.
So I'm going to call the Young's modulus in the one direction E star 1.
So everything with a star refers to a cellular solid property, and 1 because it's in one direction.
So that's going to be sigma 1 over epsilon 1.
So if I go back up there, I can say sigma 1 is equal to P divided by h plus l sin theta b.
And epsilon 1 is equal to delta sine theta in the denominator over l cos theta.
And now instead of having delta here, I can substitute this thing here in for delta.
And then I'm going to able to cancel the P's out.
So delta was equal to P sine theta l cubed over 12 Es, and there was an I, a moment of inertia, and I was equal to bt cubed over 12.
And let's see here.
So that's delta.
And there's another sine theta here so I'm just going to square that sine theta there.
So now the P's cancel out.
The b's are going to cancel out.
The 12s are going to cancel out.
And I'm going to rearrange this a little bit.
So I'm going to write Young's modulus of the solid out in the front.
Then I've got a term here of t cubed and I'm going to multiply that by 1/l squared, and then everything else-- well, let's see.
We can take this l cubed here.
I can take that.
Put it underneath that, so that's going to give me t/l cubed.
And then I've got an h plus l sine theta here, and I've got an l there, so I'm going to take that to be h/l plus sine theta.
Boop-boop-da-doop.
So I've got this term with [?
h/l's ?]
in the thetas.
There's a cos theta from the numerator here.
This term here turns into h/l plus sine theta, and then I've got my sine squared thetas down there.
And that's my result for the Young's modulus in the one direction.
OK?
Let's make sure that seems right.
It seems good.
OK.
So one of the things to notice here is there's three types of parameters that are important.
So one is the solid properties.
So the Young's modulus of the solid comes into this.
So the stiffness of the whole thing depends on the stiffness of whatever it's made from.
There's this factor of t/l cubed-- that's directly related to the relative density or the volume fraction of solids.
So what this is saying is the relative density goes as t/l, so the Young's modulus depends on the cube of the relative density.
So it's very sensitive to the relative density.
And then this factor here really is just a factor that depends on the cell geometry.
Remember when we talked about the structure of the honeycomb, we said we could define the cell geometry by the ratio of h/l and theta, OK?
And since we often deal with regular hexagonal honeycombs, I'm just going to write down what this works out to be for regular hexagonal honeycombs.
So for a regular hexagonal honeycomb, h/l is 1.
All the members have the same length.
And theta's 3, and the modulus works out to 4 over root 3 times Es times t/l cubed, OK?
So do you see how we do these things?
So all the other properties work in a similar kind of way.
You have to say something about what the sort of bulk stress is on the whole thing and relate that to the loads on the members.
You have to say something about how the loads are related to deflections, or when we look at the strengths, we're going to look at moments and how the moments are related to failure moments of one sort or another.
But it's all just like a little structural analysis, OK?
Are we good?
You good, Teddy?
I thought you were going to put your hand up?
No?
You're OK?
OK. OK, so the next property we're going to look at is Poisson's ratio.
And I'm going to look at it for loading in the one direction.
So Poisson's 1 2, say we load uniaxially in the one direction, we want to know what the strain is in the two direction, it's minus epsilon 2 over epsilon 1.
And again, if I look at my inclined member, and I say that member's going to bend something like that, and that's my deflection delta there, and, say, got the same x1 and x2 axes.
And again, if I look at delta here, it's the same little sketch I had before.
That's delta sine theta.
And this is delta cos theta.
I'm going to need those components to get the two strains in the different directions.
So epsilon 1 is going to be delta sine theta over l cos theta.
And if I'm compressing it, that would get shorter.
And we get-- and epsilon 2 is going to be delta cos theta divided by h plus l sine theta.
And that would get longer.
So these two have opposite signs, and so the minus sign is going to disappear here.
Doodle-doodle-doot.
So then I can get my Poisson's ratio by just taking the ratio of those two guys.
So I could put a minus sign there and say that's the opposite sign to epsilon 2.
Then this would be delta cos theta divided by h plus l sine theta.
And epsilon 1 would be delta sine theta over l cos theta.
And the thing that's convenient here is that the two deltas just cancel out.
So the Poisson's ratio is the ratio of two strains.
Each one of the strains is going to be proportional to delta, and so the two deltas are just going to cancel out.
And so I can rewrite this thing here as cos squared theta divided by h/l plus sine theta times sine theta.
And so one of the interesting things to notice that the Poisson's ratio only depends on the cell geometry.
It doesn't depend on what solid the material is made from.
It doesn't depend on the relative density.
It only depends on the cell geometry.
Oops.
OK, and then we can also work out what the value is for a regular hexagonal cell.
And if we plug-in h is equal to l and theta's equal to 30, you get that it's equal to 1.
So one is kind of an unusual number for a Poisson's ratio.
When we think of most materials, it's around 0.3, so it's kind of unusual that it's that large.
The other thing that's interesting is that it can be negative.
So if theta is less than 0, then you can get a negative value.
If the cos squared is going to be a positive value, but you've got a sine theta down here, then that's going to give you a negative value.
So you can get negative values.
So let me just plug in an example.
So say h/l is equal to 2, and theta is equal to minus 30 degrees, then this turns out to be 3/4.
So cos of 30 is root 3/2, so square of that is 3/4.
h/l is 2, sine theta is 1/2, but it's minus 1/2.
So 2 minus 1/2 is 1 and 1/2.
And then the sine theta is minus 1/2.
And so it works out to be minus 1 for that particular combination.
And I brought my little honeycomb that has a negative Poisson's ratio in.
So this guy here-- let's see, I don't think there's an overhead here.
No overhead?
Guess not.
I'll just pass it around.
So if you take it, put your hands on the flat side and load it like this, and don't smoosh it like that.
Just load it a little bit, because you [?
want to be ?]
linear elastic.
If you load it just a little bit, you can see that as you push it this way, it contracts in sideways that way.
So don't smash it.
Just load it a little bit and you can kind of see it with your hands.
And if you put on a piece of lined paper, it's easier to see it.
OK, so that's kind of interesting.
So are we good with getting the Young's modulus in the one direction and the Poisson's ratio for loading in one direction?
OK, so you can do the same sort of thing to get the Young's modulus in the two direction and the Poisson's ratio for loading in the two direction.
And you get slightly different formulas, but it's the same idea.
And you can also get a shear modulus this way, and in-plane shear modulus.
It's a little bit-- the geometry of it's a little bit more complicated.
So all of those things are derived in the book, in the cellular solids book.
So if you wanted to figure those out, look at that, you could look at the book.
So let me just comment on that.
All right, so those are the in-plane linear elastic moduli, and remember we said that four of them describe the in-plane properties for an anisotropic honeycomb.
And you can use that reciprocal relationship to relate the two Young's moduli and the two Poisson's ratios.
All right, so the next thing I wanted to talk about was the compressive strength.
So let me just back up here a second.
So if we go back to here, remember we had for an elastomeric honeycomb, this stress plateau was related to elastic buckling.
So we're going to look at that buckling stress first.
And this plateau here is related to yielding.
And then we'll look at the yielding stress next.
And then this plateau here is related to a brittle sort of crushing, and we'll do that one third.
So we're going to go through each of those next.
And this is kind of a schematic for the elastic buckling.
So when you look at the elastic buckling, one of the things to note is that when you load the honeycomb this way on, if you load it in the one direction, you don't get buckling.
It just sort of continues to-- whoops, if I can keep it in plane.
It all just kind of folds up, so you just get larger and larger bending deflections.
You don't really get buckling.
But when you load it this way on, these vertical members here, the ones of length h, they're going to buckle.
So, see if I do that, my honeycomb looks like those cells up on the schematic there, OK?
So, whoops.
So we're going to look at the compressive stress or strength next.
That's sometimes called the plateau stress.
So we can get cell collapse by elastic buckling, if, for instance, the honeycomb is made of a polymer.
And then the stress-strain curve looks something like that.
And what happens is you get buckling of those vertical struts throughout the honeycomb And then you could also get a stress plateau by plastic yielding.
And what happens when you get plastic yielding is you get localization of the deformation.
So one band of cells will begin to yield initially, and then as the deformation proceeds, that deformation ban will propagate and get bigger and bigger, and you get a wider and wider band of cells yielding and failing.
So you get localization of yield, and then as deformation progresses, the deformation band widens throughout the material.
So if I go back and look-- if I look at this one here, when you look at this middle picture here, you can see how one band of cells has started to collapse and started to fail.
And as you continue to compress that in the one direction, this way on, then more and more neighboring cells are going to collapse and the whole thing will get wider until the whole thing has collapsed.
And that's kind of characteristic of the plastic failure.
And then the third possibility is brittle crushing.
And then you get these kind of serrated plateau.
And the peaks and valleys correspond to fractures of individual cell walls.
OK, so we're going to start off with the elastic buckling failure.
And I'm going to call these plateau stresses sigma star, for the sort of compressive strength.
And el means it's by elastic buckling.
And as I mentioned, you don't get it in the one direction.
The cells just fold up.
You only get it for loading in the two direction, so it's going to be sigma star el 2.
Oops, need a different piece of chalk.
So you get this elastic buckling for loading in the x2 direction, and the cell walls of length h buckle.
And you don't get it for loading in the one direction, the cells just fold up.
So again, let me draw a little kind of unit cell here.
And here is our stress sigma 2, like that.
And here's our little wall of length h that's going to buckle.
So if I load it up, initially it'll be linear elastic.
And then eventually, at some stress, it will get large enough that this wall here will buckle.
And we can relate to that plateau stress, or that compressive stress, to some Euler buckling load.
So you remember, if we have a pin-ended column, so just a single column, pins on either end, the Euler buckling load says you get buckling when the critical load is equal to some end constraint factor, n squared.
So n squared pi squared E, and here it's E of the solid, I over the length of the column, and in this case, the column length is h-- so h squared.
OK, so that's just the Euler formula.
And here, n is an end constraint factor.
And if you remember for a pin column, so if our column is pinned at both ends like that, and just buckles out like that, then n is equal to 1.
And if the column is fixed at both ends, something like that, then the column looks like that and then it's equal to 2, OK?
So if I know what the end condition is, I know what n is and I can use my Euler formula here.
So the trick to this is that it's not so obvious what n is.
Yes?
So, when you're loading in the x2 direction here, the first thing you're going to get is the incline members deforming?
Yeah
And then at some point, you hit a P critical that will cause the vertical members to buckle?
Exactly

Exactly.
That's exactly right.
Hello.
So the trick here is that we don't really know what this n is, initially.
They're not really pinned, pinned; they're not fixed, fixed.
And if you think about the setup with the honeycomb here, the constraint on that vertical member depends on how stiff the adjacent members are.
So you can kind of imagine, if I'm looking at one of these vertical members here, if these two adjacent inclined members were big honking thick things, it would be more constrained.
And if they were little thin, kind of teeny little membranes, it would be less constrained.
And you can think of it in terms of a rotational stiffness, that when the honeycomb buckles, you kind of see the member h goes from being horizontal to sort of it buckles over like this.
But that whole end joint, see the end joint at the top here or the end joint at the bottom, that whole joint rotates a little bit.
And so there's some rotational stiffness of that joint.
And that rotational stiffness depends on how stiff the member h is and how stiff those inclined members are.
So there's a thing called the elastic line analysis that you can use to calculate what n is.
And basically what that does is it matches the rotational stiffness of the column h with the rotational stiffness of those inclined members.
So we're not going to get into that.
I'm just going to tell you what the answer is.
But if you want to go through it, it's in an appendix in the book.
So you can look at it, if you want.
So here I'm just going to say that the constraint n depends on the stiffness of the adjacent inclined members.
And we can find that by something called the elastic line analysis.
And if you have the book, you can look in the appendix and see how that works.
But essentially what it does is it matches the rotational stiffness of the column h with the rotational stiffness of the inclined members.
So what you find is that n depends on the ratio of h/l.
And I'm just going to give you a table with a few values.
So for h/l equal to 1, then n is equal to 0.686.
For h equal to 1.5, it's equal to 0.76.
And for h/l equal to 2, it's equal to 0.806.
OK, so now if we have values for n, we can just substitute in to get the critical buckling load.
And if I take that load and divide it by the area of the unit cell, I'm going to get my buckling stress.
So it's pretty straightforward from this.
So my buckling stress is going to be that critical load divided by my unit cell area.
So it's divided by the unit cell length in the x1 direction to l cos theta times the depth b into the page.
So it's equal to n squared pi squared Es times I.
And I is bt cubed over 12.
Divided by the length of the column, h squared, and then divided by the area of the unit cell, 2l cos theta b, OK?
And I can rearrange that somewhat to put it in terms of dimensionless groups.
So if I pull all the constants out, it's n squared pi squared over 24 times the modulus of the solid, t/l cubed in the numerator divided by h/l squared times cos theta in the denominator.
So again, you can see that the buckling stress, the compressive sort of elastic collapse stress, depends on the solid property.
So here is the modulus of the cell wall in here.
Depends on the relative density through t/l cubed.
And then it depends on the cell geometry through h/l cos theta, and n depends on h/l as well, OK?
And then we can do the same thing where we figure out what it is for regular hexagonal cells.
And it's 0.22 Es times t/l cubed.
And then we can also notice that since E in the 2 direction, for a regular hexagonal cell, E is the same in the 2 direction and the 1 direction.
It's isotropic.
So E2 is also equal to 4 over root 3 Es times t/l cubed.
That's equal to-- whoops-- it's equals to 2.31 Es t/l cubed.
And we can say that the strain at which that buckling happens is just equal to a constant.
And for regular hexagonal honeycombs, it works out to a strain of 10%.
Are we good?
So we have a buckling load.
We divide by the area.
The only complicated thing is finding n. And you can find it by this elastic line analysis thing.
So each of these calculations is like a little structural analysis, only on a little teeny weeny scale of the cells.
So you see where my background in civil engineering comes in handy.
Yup.
OK.
So the honeycombs involve the most sort of complicated equations.
When we come to do the foams, we're going to use a dimensional analysis and all the equations are going to be much simpler.
So this is the most kind of tedious part of the whole thing.
So the next property I want to look at is the plastic collapse stress.
Say we had a metal honeycomb and we wanted to calculate the stress plateau for a metal honeycomb.
So we have this little schematic here, and say we load it in the one direction again.
So we're loading it here.
And we've got some load P, like that.
And if we have our honeycomb, we load it this way on, initially, the cell walls bend.
And you have linear elasticity and you have some Young's modulus.
But if you have a metal, if you continue to deform it and you continue to load it more and more, eventually you're going to hit the yield stress and the cell wall.
So the stresses in the cell wall are going to hit the yield stress.
And initially, the stresses are just going to be-- remember, if you have a beam, the stresses are maximum at the top and the bottom of the beam.
So initially you're going to hit the yield stress at the top and the bottom of the beam first.
But as you continue to load it, you're going to end up yielding the cross-section through the entire section.
So the entire section is going to be yielded.
And once the entire section yields, it forms what's called a plastic hinge.
Once the whole thing's yielded, then you can add more force and the thing just rotates.
And because it rotates, it's called a plastic hinge.
You know, if you take a coat hanger, and you bend it back and forth and bend it back and forth.
If you bend it enough, you form a plastic hinge because it just can bend easily.
So these little schematics here, if you look at the, say, one of these inclined members, the moments are maximum at the end.
So Remember when we had the linear elastic deformation and I looked at the little bending moment diagram?
The moments are maximum at the ends, and you're going to form those plastic hinges initially at the ends.
And so these little ellipsey things here, all the ends, those kind of show where the plastic hinges are.
So those plastic hinges are forming.
So here's for loading in the x1 direction, and here's for loading in the x2 direction, there.
So the thing we want to calculate is what stress does it take to form those plastic hinges and get this kind of plastic plateau stress?
OK, so we can say we get failure by yielding in the cell walls.
And I'm going to say the yield strength of the cell wall is sigma ys.
So sigma y for yield and s for the solid.
And the plastic hinge forms when the cross-section has fully yielded.
So let's look at the stress distribution through the cross-section when its first linear elastic.
So say that's the thickness t of the member.
And if the beam was linear elastic, the stress would just [?
vary ?]
linearly, like that, right?
And this would be the neutral axis, here, where there is no normal stress.
So that's what happens if it's linear elastic, and I'm hoping you remember something vaguely like that.
Sounds good?
But as we increase the load on it, and we increase the sort of external stress, this stress in the member is going to get bigger and bigger, and eventually, that's going to reach the yield stress, OK?
And once that reaches the yield stress, if we continue to load it, what happens is the yielding propagates down through the thickness of the thing here.
So we get yielding through the whole cross-section.
So let me scoot over here
Professor?
Yup?
When it starts to yield, does this curve change?
Yes.
I'm going to draw it for you
Oh,
That's the next step.
That would be the next thing.
OK, so once the stress at the outer fiber is the yield strength of the solid, then the yielding begins and it progresses through the section as the load increases.
So the stress distribution starts to look something like this once it yields.
OK, so that's sigma y of the solid.
Actually, let me rub that out because then I can show you something else.
So in 3D, this would be through the thickness of the beam.
That would be the thickness of the beam there.
And boop, boop.
It would look something like that.
OK?
And then this is still our neutral axis here.
And then eventually, as you load it more and more, the whole cross-section is going to yield.
Whoops.
And I'm assuming that the material is elastic, perfectly plastic.
So the stress-strain curve from the solid I'm idealizing as-- whoops.
That's not quite right.
I'm idealizing as that, OK?
So when you get to this point here, the entire cross-section has yielded, and that means you form the plastic hinge.
The idea here is that the section then just rotates like a pin.
All right.
So we can figure out the plateau stress that corresponds to this by looking at the moment that's associated with the plastic hinge formation.
So there's some internal moment associated with that.
And then equating that to the applied moment from the applied stress.
So doodle-loodle-oot.
Let me see me, maybe back up here.
So there's some-- if I have the stress distribution here, I could say this whole kind of stress block is equivalent to some force acting out like that and some force acting out like that.
It would be sigma ys times b comes t/2 would be f. And I can say there's some plastic moment.
If I think of the force here and the force there, they act as a couple and they have some moment, and that's called the plastic moment.
So that's like an internal moment when the plastic hinge forms.
So I'll say the internal moment at the formation of a plastic hinge.
I'm going to call that Mp, for plastic moment.
And we can work out Mp by looking at that stress distribution when the entire cross section has yielded.
The force F is going to be sigma ys times b comes t/2.
It's the stress times that area.
And then the moment arm between the two forces is also t/2.
And so that plastic moment is just sigma ys bt squared over 4, OK?
Are we good?
Sonya?
What's the second [INAUDIBLE]?
OK, so this is the force.
This thing here is the force F. And I have to-- if I'm getting a moment, I'm saying that that force, if I doot-doot-doot-- the distance between those two forces there is t/2.
So each force acts through the middle of the block, and so the distance between [?
it is ?]
t/2.
And I'm going to equate that moment to the applied moment from the sort of applied stress.
And then if I go back to my inclined member-- whoops, let's see.
Let me get a little more inclined.
That's my inclined member, there, of length l. I've got modes p that are applied at the end from sigma 1.
And I've got moments that are induced at the ends.
And that angle there would be theta.
This length here is l, like that.
And if I just use static equilibrium on that, I can say that I've got 2 times the moment, so I've got one at each end-- they're both the same sign-- minus P. And then the distance between these two P's, say I take moments about here, I've got M applied plus M applied, I've got minus P times l sine theta.
That's equal to 0.
So the applied moment there is just Pl sine theta over 2.
So now what I'm going to do is I'm going to equate this applied moment with this plastic moment, and I'm going to relate P to my applied stress sigma 1.
And then I'm going to get a strength in terms of the yield strength of the solid, there's going to be a t/l factor and there's going to be some geometrical factor.
So that's just the last step.
Boop-ba-doop-ba-doop.
So we get plastic collapse of the honeycomb.
And the stress I'm going to call sigma star plastic with a 1, because I'm going to look at the one direction.
And that happens when that internal plastic moment equals the applied moment.
So let's see.
I've got that.
Let me also write down over here, I've also got this sigma 1 is equal to P over h plus l sine theta times b.
So here I can write P in terms of sigma 1, in this thing.
And then write that, get the applied moment in terms of that, and then equate it to that.
So this term on the left-hand side corresponds to this expression for the applied moment where I've plugged in.
For P, I've plugged in sigma 1 times h plus l sine theta times b.
And that's my plastic moment on the right-hand side.
So if I just rearrange this, I can then solve for this plastic collapse stress.
So it's equal to the yield strength of the solid times t/l squared, and then times another geometrical factor.
2 times h/l plus sine theta times sine theta.
Doop-doop-doop.
So the same kind of thing, there's a solid property, a t/l, a relative density term, in then a cell geometry term.
And we can calculate with this for regular hexagonal cells.
And we can do a similar kind of calculation for loading in the other direction.
And you can get a shear strength if you want to do that, too
If you're going in the other direction, only the E or the M apply changes, right?
[?
Or ?]
like that section
Yeah, this thing here is the same
That stays
Right.
And this is-- there's a different geometry to it.
Because now you're loading it this way on.
OK, so we've calculated an elastic buckling plateau stress and a sort of plastic collapse plateau stress.
And if you have thin enough walled, say, even aluminum honeycombs, then the elastic buckling could precede the plastic collapse.
And so I'm just going to work out what the criterion would be for that to happen.
So the two stresses can be equated.
And then that's going to give us some criterion.
So the two are equal, I'm just going to write down the equations that we had.
So the buckling stress was n squared pi squared over 24 times E of the solid times t/l cubed divided by h/l squared times cos of theta.
And the plastic collapse stress for the 2 direction was sigma ys times t/l squared divided by 2 cos squared theta.
So I can write this-- because this has a t/l cubed term, and that has a t/l squared term, I can write this in terms of a t/l critical.
So if I leave it t/l here and I put everything else on the other side, I've got 12 over n squared pi squared, then h/l squared over cos theta times sigma ys over Es.
So if t/l is less than that, I'm going to get elastic buckling first.
And if it's more than that, I'm going to get plastic yielding first.
And we can work out an exact number for regular hexagonal honeycombs, so I'm going to do that.
So if I have a particular geometry, I can figure out what n is.
So for regular hexagonal honeycombs, t/l critical just works out to 3 times the yield strength of the solid over the Young's modulus of the solid.
So if we know that ratio of the yield strength of the modulus of the solid, we can get some idea of what that critical t/l would be.
So we'll do that next
And you said if t/l is less than that critical, then you're going to get the yielding first
No, if it's less, you get the buckling first.
If it's really skinny, it tends to buckle first.
So, for example, for metals, the yield strength over the modulus is roughly 0.002, like the 0.2% yield strength.
And so that means that t/l, the sort of transition or the critical value is at 0.6%.
So most metal honeycombs are denser than that.
That's a pretty low density.
But if we look at polymers, you can get polymers with a yield strength relative to the modulus of about 3% to 5%, and then that critical t/l is equal to about 10%, 15%.
So low-density polymers with yield points may buckle before they yield.
So we have one more of these compressive plateau stresses, and that's for the brittle honeycomb.
So I don't think I'm going to finish this today, but let me set it up and then we'll finish it next time.
So the idea here is that if you have a ceramic honeycomb-- remember I showed you some of those ceramic honeycombs-- that if you compress them, they can fail by a brittle crushing mode.
So ceramic honeycombs can fail in a brittle manner.
And again, initially there would be some cell wall bending, but at some point, you're going to reach the bending strength of the material.
And bending strengths are called modulus of rupture.
So you reach the modulus of rupture of the cell wall.
So I'm not going to write the equations down today because we're not going to get very far, so I'll do that next time.
But we're going to set this up exactly the same as we did for the last one, for the plastic yielding.
But instead of getting that sort of blocky, fully yielded cross-section stress distribution, we're just going to have the linear elastic stress distribution, and when the maximum stress reaches that modulus [?
rupture, ?]
the thing's going to fail.
So the form of the equations is going to be very similar to what we had for the plastic collapse stress, but there's a slightly different geometrical factor-- that's all.
So we'll do that next time.
And then next time we're also going to talk about the tensile behavior of honeycombs in-plane.
We'll work out a fracture toughness and then we'll start talking about the out-of-plane properties, as well.
So on Wednesday, we'll do the out-of-plane properties, OK?
So hopefully we'll finish the out-of-plane properties Wednesday.
And then next week, I was going to talk about some natural materials that have honeycomb-like structures, so things like wood and cork, OK?
All right, so this is the kind of most equationy lecture in the whole course.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
All right.
I should probably start.
Last time, we were talking about the honeycombs and doing some modeling of the mechanical behavior and we started off talking about the in plane behavior.
We're talking about loading it in this direction or that direction there.
And we talked about the elastic modulus.
I think I derived a Young's modulus for the one direction, a Poisson's ratio for loading in the one direction.
And then we started talking about the stress plateau and we went over the elastic buckling stress, for one of these elastomeric honeycombs like this.
And we went through the plastic collapse stress, for, say, a metal honeycomb that would yield.
And I think I started talking about a brittle honeycomb and brittle crushing.
The idea with a brittle honeycomb-- like a ceramic honeycomb-- is it could fail in a brittle manner.
And the failure is going to be controlled by the cell wall in bending.
And when that bending stress reaches the modulus of rupture, or the bending strength of the material, then you get wall fracture.
I think that's where we left it last time, right?
I had written down something about cell wall fracture.
Now, I wanted to do the little derivation.
Here's our little schematic up here.
Here's the honeycomb.
You've loaded it with sigma 1 here to such an extent that one of these cell walls has reached the modulus of rupture and has broken.
And this is the little free body diagram that corresponds.
I'm going to go through sigma 1 for loading in the one direction.
This is the same thing for loading in the two direction.
And the result for that's in the book.
OK.
If I have loading in the one direction, I can relate that horizontal force p to the stress in the one direction.
The little p is equal to sigma 1 times h plus sin theta times b.
And remember, b's the depth into the page.
And I'm going to define sigma fs as the modulus of rupture of the cell wall material.
It's the bending strength of the cell wall material.
And we're going to say that we get fracture of that bent wall when the applied moment is equal to the fracture moment.
From the plastic collapse stress from last time, we had the applied moment was equal to p times l sin theta over 2.
That was just using static equilibrium, looking at that free body diagram of the beam.
And if I write p, in terms of sigma 1 up here, I can just write that like this sigma 1 times h plus l sin theta times b.
And then I've got this other term of l sin theta and we divide that whole thing by 2.
That's the applied moment.
And we're going to get fracture when we reach the fracture moment.
I'm going to call that mf-- the moment at fracture.
Last time, we figured out a plastic moment to form a plastic hinge.
And this is an analogous thing.
But in this case, remember, if we have a beam and we have the stress profile through the cross section of the beam, it's going to look something like that.
So for our beam, that's going to be the thickness of the beam there.
So if it's linear elastic, we get the maximum stress at the top and the bottom.
And the neutral axis is here in the middle.
There's no stress there.
This is the normal stress distribution here.
And as we increase the stress for a brittle material that's going to be linear elastic till fracture, this is going to stay linear like this until we reach this modulus of rupture stress here.
When we reach that stress, then we're going to get fracture of the beam.
And we can say that there's some moment associated with that.
I could say that this stress block here is equivalent to some concentrated force and this stress block down here is also equivalent to the-- it's going to the same magnitude, but the opposite direction force.
And I can get the fracture moment by figuring out how big those forces are and multiplying by this moment arm between the two forces.
OK?
The magnitude of those forces is just going to be the volume, essentially, of this stress block here.
Imagine there's stresses there.
It's a triangle, so the area of it's going to be a half times t over 2 times of sigma fs.
And it's going to go b into the page.
So if you think of the force-- if this was the stress-- if that stress was constant, it would just be sigma fs times b times t. But it's not constant.
It's a linear relationship.
So I'm taking the area of that triangle.
That's the force.
And then I want to multiply that times the moment arm.
And the moment arm between those two forces-- each of these forces acts through the centroid of the area.
The centroid of the area is not in the middle for a triangle, and that total distance is 2/3 of the thickness, t. OK?
That's the moment arm that you get by figuring out where the centroid of these areas are.
I multiply that times 2/3 t and one of the 2's is going to cancel.
I can rewrite that and sigma fs times b times t squared over 6.
OK?
I think, last time when we talked about the plastic moment, we did a similar calculation and it worked out to sigma y bt squared over 4.
So now this is sigma fs, the modulus of rupture, times bt squred over 6.
The 6 is just slightly different because we've got a triangle here instead of a square shape like we had before.
And now I can get the brittle crushing strength and compression by just equating that applied moment to this fracture moment.
And if you do that, the result you get is this plateau stress for brittle crushing and compression.
In the one direction, it's sigma fs-- the modulus of rupture of the cell wall-- times t over l squared.
And then divided by a geometrical factor.
And her regular hexagons, it works out to 4/9 of the modulus of rupture times t over l squared.
OK?
Are we good?
We've got the in plane compressive properties now.
We've got the elastic moduli and we've got the three plateau stresses that correspond to the three mechanisms-- to the elastic buckling failure mechanism, the plastic yielding mechanism, and then the fracture mechanism for brittle crushing.
OK?
If you think of the stress-strain curve of these materials in compression, the stress strain curves all look something like that.
And now we've figured out equations that give us the modulus here and our collapse stress there.
OK?
So we can describe that stress-strain curve.
All right.
That's compression.
And the next thing I wanted to talk about is tension.
And if we think about the tensile behavior, the elastic moduli are going to be just the same.
So the moduli are the same in tension and compression.
And then, if we think about the stress plateau, we don't really have a stress plateau for an elastomeric material because there's no elastic buckling.
If you pull it in tension, you're not going to get buckling in tension.
You only get buckling if it's in compression.
If you have a material that yields like a metal, you can get a plastic collapse stress and a plastic plateau.
And that's very similar in tension and compression.
There's a very small geometrical difference, but you can, essentially, ignore it.
If you're loading the material in compression-- and imagine this was a metal-- if you load it in compression, the cell walls are getting a little further apart when I compress it.
And if you're loading in tension, like this, the cell walls are getting a little closer together.
So there's a small geometrical difference.
But if we ignore that, we can say that the plateau stress for plastic behavior is about the same in tension and compression.
And so really, the only property that's left is to look at a brittle honeycomb.
And for a brittle honeycomb, you can have fast fracture and we can calculate a fracture toughness.
So this next slide describes the fracture toughness calculation that we're going to do.
Here's our honeycomb.
I'm going to load it in the sigma 1 direction here.
I've turned the honeycomb 90 degrees, so this is still sigma 1.
And imagine now that we've got a crack here.
And I'm going to consider a situation where the crack is very large, relative to the cell size.
So it's not a crack in the cell walls.
It's a crack that goes through multiple cells.
I'm going to assume the crack is large, relative to the cell size.
I'm going to assume that the bending is the main deformation mode.
And what I'm going to do is look at-- if I have my crack tip here, I'm going to look at this cell wall a just ahead of the crack tip.
And I'm going to say, that cell wall is bent.
And I'm going to figure out something about the stress in that cell wall and look at when that fails.
And I'm going to assume that the cell wall has a constant modulus of rupture.
So the cell wall has a constant strength.
You can imagine the cell wall could have little tiny cracks in it, too.
And if a cell wall has a bigger crack, it's going to fail at a lower stress.
But let's imagine that the cell walls are all the same strength and they all have a constant modulus of rupture.
Let me write some of this down.
In tension, the elastic moduli are going to the same as in compression.
There's no elastic buckling in tension, so that's not going to happen.
The plastic plateau stress in tension is going to be very similar to that in compression.
As I mentioned, there's a small geometrical difference, but we're going to ignore that.
And then, if we had a brittle honeycomb, like one of those ceramic honeycombs I showed you, then we can have fast fracture.
What we want to calculate is the fracture toughness.
And I'm going to make a few assumptions here.
I'm going to assume that the crack length is large compared to the cell size.
And if I do that, I can say that I'm going to use the continuum assumption.
Hello.
We'll come back to that.
I'm going to say that axial forces can be neglected.
We're just going to look at bending forces.
And I'm also going to assume that the modulus of rupture is constant for the cell wall.
First, let's just think again about the continuum.
Imagine we just had a solid and we have a plate of the solid and it's loaded in tension with some remote stress-- some far away stress-- sigma 1.
And the plate has a crack of length 2c perpendicular to that normal stress.
And we're going to look at the stress-- local stress at the crack tip-- some distance r ahead of the crack tip there.
In fracture mechanics, it's been worked out what that local stress field is.
And it depends on the crack length, and then how far ahead of the crack tip you are.
So you can say that if you've got a crack length of 2c in a linear elastic solid, and the crack is normal to a remote tensile stress-- which I'm going to call sigma 1-- then that crack is going to create a local stress field at the crack tip.
And we're going to use this equation for the local stress field.
The local stress field is equal to the far away field divided by-- or multiplied by the square root of pi c and divided by the square root of 2 pi r. So there's a stress singularity at the crack tip.
And then the local stress decays as you move away from the crack tip
And what is r?
r is the distance from the crack tip.
So if that's the tip of my crack there, then r is my distance out.
OK.
In the honeycomb wall, if we look at the crack here, and then we look at that cell wall a that's just ahead of the crack tip, that cell wall is bent.
So in the honeycomb, we're going to be looking at the bent cell wall.
And that wall is going to fail when the applied moment equals the fracture moment.
If we look at wall a, we could say that the applied moment is going to be proportional to p times l. Getting ahead of myself there.
I'm going to do this-- because it's hard to say exactly where the crack tip is because there's a void there.
I'm going to use that argument here where I make everything proportional.
The moment's going to be proportional to p times l on wall a.
And the fracture stress is going to be proportional to sigma fs times bt squared.
Last time, we said it was sigma fs bt squared over 6.
It's the same thing.
I'm just dropping the 6 out.
And then I can also say that this applied moment, if it goes as pl-- p is just going to be my local stress times lb.
And then I multiply times l. So if you think of just thinking about-- if you got a load p on this member here, l, there's going to be some local stress there.
And p is just going to be that local stress times the cell wall length times the width into the page.
And then, that local stress, sigma l, I can replace with that equation over there.
So that local stress is going to go as sigma 1 times the root of c over the root of r. And I'm going to say the distance ahead of the crack tip goes as l. Instead of having r, I'm going to say it's l. It's not necessarily exactly l, but it's going to be some fraction of l. That's my local stress there.
And then I've got an l squared times b.
And if I set that equal to the fracture moment, that's going to be proportional to sigma fs bt squared.
Are we good here?
You have to think of the crack tip.
And there's some local stress field ahead of the crack tip.
And we're saying that the load p is equal to that local stress times a cell length times the depth into the board.
And then multiply it times l to get the moment.
And then I replace that local stress with this standard equation for the remote stress and the crack length and the distance ahead of the crank tip.
So here, the b's are going to cancel out.
And now I can solve for a fracture stress in the one direction.
And that's going to-- well, let me get proportional-- that's going to be proportional to sigma fs.
Then there's going to be a t over l squared.
And then, this is going to go as the square root of l over c like that.
And now, if I want to get a fracture toughness, the fracture toughness is just the strength times the square root of pi times the half crack length c. So here, my fracture toughness is just that strength times the root of pi c. So I can say that's equal to a constant times sigma fs times t over l squared.
And now, times the square root of l. These root of c's have canceled out.
So that's my equation for the fracture toughness.
And one of the interesting things here is that the fracture toughness depends on the cell size.
This is the first property that we've derived an equation for where it depends on the cell size.
OK. And here, c's just going to be a constant.
All right.
Now we've got a set of equations that describe the in plane properties.
We've got equations that describe the linear elastic moduli in the plane.
We've got three equations that describe the compressive stress for elastic buckling failure, for plastic yielding failure, and for a brittle crushing failure.
And we've got an equation for the fracture toughness, as well.
OK?
We've got a description of the in plane behavior of these hexagonal honeycombs.
The next thing I wanted to do was talk a little bit about in plane behavior, but for a different cell shape-- for triangular honeycombs.
Because they deform by a different mechanism.
And they can be used to represent the lattice materials that we looked at earlier, too.
If we have a triangular honeycomb with triangular cells, triangulated cells behave like a truss.
And you can analyze trusses by just saying that the joints are pin jointed.
There's no moments at the end of the joints or end of the members.
And the forces are all axial and so the behavior's a little bit different.
I wanted to show you how these triangular honeycombs work, too.
I can scoot this up.
Imagine that you've got a honeycomb that's an array of triangular cells like this.
And say we're applying some bulk stress sigma to it, like that.
And say it's got a depth b into the page.
When we have a triangulated structure like this, it behaves like a truss.
And we can analyze it as being a pin-jointed structure.
There's no moments at the joint.
And if it's pin-jointed and there's no moments at the nodes, then we just get axial forces along the members.
And even if the nodes were fixed-- as they are in these ceramic honeycombs-- you can show that if it's triangulated, even if you accounted for any bending, it really is a very tiny contribution to the deformation in the forces.
It's less than a couple of percent.
I'll say even if the ends are fixed-- I'll just say the bending contributes less than 2% to the forces in the deformation.
If I have a triangular cell like that, and say I pick a unit cell like this, and I say that the bulk stress produces a load of p on the top and p over 2 at each of the bottom nodes there, then the force in each member is going to be proportional to p. And for a given geometry of triangle, you can figure out exactly what the force distribution would be in each of the members.
But I'm going to use one of these proportional arguments again, just to get a general result.
Because I don't really care that much about the details of the geometry.
OK.
If I have a little set up like this, I can say that the overall stress is going to be proportional to p over lb.
And the stress in each member is going to be proportional to p over l times the thickness-- or b times the thickness.
This is the overall stress.
The overall strain is going to be proportional to some deflection of the triangle divided by the length.
So if I said, say, this length here was a length l. And then the deflection of each member is going to be proportional to p times l over es times the cross sectional area of the member, and that's just b times t. OK?
So this is the stress on the whole thing, the strain on the whole thing, and relating the delta to the p. And then, the modulus of the whole honeycomb is going to go as the stress over the strain.
So that's p over lb divided by delta over l. These l's here cancel.
And delta here is pl over es bt, and so the b's cancel and the p's cancel.
And the modulus I get for the honeycomb is just some constant related to the cell geometry times the modulus of the solid times t over l. And if you did an exact calculation for equilateral triangles, you'd find that that constant's 1.15.
The interesting thing to note here is that the modulus for these triangular honeycombs goes as t over l, not as t over l cubed.
For the hexagonal honeycomb, it went as t over l cubed.
And here, because the deformations are axial-- not bending-- it's much stiffer.
And it's much stiffer to have one of these triangulated structures.
I'll just say, here, that the modulus goes as t over l cubed for the hexagonal honeycombs due to the bending.
One of the reasons that people are interested in those lattice materials is that they, too, have moduli that go as t over l. That basically go with the relative density, rather than with the relative density cubed.
So they're much stiffer than, say, a hexagonal honeycomb.
OK?
Are we good with the triangulated honeycombs?
Yes?
What is c?
c's just a constant related to the cell geometry.
For equilateral triangles, it's 1.15.
You could work it out, but it just makes the whole thing a little more complicated to do that.
OK. That's the in-plane behavior.
And next, I wanted to talk about the out-of-plane behavior.
Remember, we said the hexagonal honeycombs are orthotropic and the orthotropic materials have nine elastic constants.
And we've figured out four so far.
We've figured out the four in-plane elastic constants.
There's five out-of-plane elastic constants to describe the elastic behavior completely.
And so we want to talk about these other elastic constants.
The honeycombs are also-- I should just back up a little bit.
The honeycombs are used in sandwich panels.
And when they're used in sandwich panels-- I brought a little panel in with carbon fiber faces and a nomex core.
If you bend that panel like that, you're going to get shear stresses in the core.
And the shear stresses are going to be going this way and this way on, and that way, that way on.
And so those shear stresses are out-of-plane.
They're in the x1, x3, or x2, x3 planes.
And so you need the out-of-plane properties for the shear properties in the sandwich panels.
Honeycombs are also sometimes used as energy absorption devices.
Not these rubber ones, but imagine there was a metal one.
And when they're used for energy absorption devices, they're typically loaded this way on.
Again, that's the out-of-plane direction and you need the out-of-plane properties.
And for the out-of-plane properties, the cell walls don't tend to bend.
Instead, they just extend or contract.
And you get stiffer and stronger properties.
Let me just write something down and then we'll start to derive some of those properties.
The cell walls contract or expand instead of bending, and that gives stiffer and stronger properties.
OK. OK.
There's five elastic constants in the out-of-plane directions.
We'll start with the Young's modulus.
And if I take my honeycomb and I load it in the x3 direction-- just taking this thing here and just loading it like that-- the cell walls just axially contract and the stiffness just depends on how much cell wall I've got.
So the modulus in the three direction is just equal to the area fraction times the modulus of the solid.
That's just the same as the volume fraction, or the relative density.
So it's quite straightforward.
The cell walls contract or extend axially.
e3 is just es times the relative density.
And that's just es times t over l. And then there's a geometrical factor here.
h over l plus 2 over 2. h over la plus sin theta times cos theta.
Again, a little bit like those triangular honeycombs.
The thing to notice here is that in the three direction, the modulus goes linearly with t over l, whereas in the in-plane directions, it goes with t over l cubed.
So there's a huge anisotropy in the honeycombs because of this difference.
Imagine a honeycomb might be 10% dense.
t over l might be something like a tenth-- 0.1.
So e star 3 is going to 0.1 of es, roughly, and in the other direction, it's going to be 1/1000th.
So there's a huge anisotropy because of this.
Let me just-- square honeycombs.
This just shows looking at the out-of-plane directions and the different stresses and properties that we're going to look at here.
The next one we're going to look at is Poisson's ratio.
And first, we're going to look at loading in the x3 direction.
And if we load it in the x3 direction, the cell wall's just strain by whatever the Poisson's ratio is for the solid times the strain in the three direction in the other two directions.
We'll say for loading in the x3 direction, the cell wall's strain by nu of the solid times whatever the strain is in the three direction in the other two directions.
If we load it in the x3 directions and everything contracts by that much in the other two directions, that just means that the Poisson's ratios-- nu 3 1 and nu 3 2 are going to be the same.
And they're just going to be equal to the Poisson's ratio of the solid.
So if each wall is going to contract by that amount, the whole thing's going to contract by that amount.
And that's going to give you that Poisson's ratio.
Let me just say, here, also-- and remember that I'm defining nu ij as minus epsilon j over epsilon i.
We're loading in the three direction here.
And then you can get the other two Poisson's ratios using those reciprocal relationships.
So nu 1 3 and nu 2 3 can be found from the reciprocal relations.
And remember, those relationships come from saying that the compliance tensor, or the stiffness tensor, is symmetric.
We can write, for instance, that nu 1 3 over e1 is equal to nu 3 1 over e3.
So I can write that like that.
And then I can say nu 1 3-- that is going to equal to nu 3 1 times e1 over e3.
And we just saw that nu 3 1 was equal to nu s. And we see, from before, the e1 is equal to some constant.
Let me just call it c1 times es times t over l cubed.
And e3 is going to be some other constant times es time t over l. The es's are going to go.
And if t over l is small-- even if it's say, a 10th-- and this is going as t over l cubed and that's going as t over l, then I can say this thing is about equal to 0.
It's going to be small.
It's not to be exactly [INAUDIBLE] 0, but it's going to be small so we're going to say it's 0.
So I'll just say for small t over l. And then, similarly, nu 2 3 is going to be close to 0, as well.
So there's the Poisson's ratios.
We've got the Young's modulus, the Poisson's ratios, and next we want to get the Shear moduli.
And the shear moduli is little more complicated.
The cell walls are loaded in shear but the neighboring cell walls constrain them and they produce some non-uniform strain.
I'm talking about shearing it this way on.
You can see on this figure here, we're talking about shearing it, like tau 2 3 or tau 1 3, this way.
And so each wall is going to shear, but the walls are attached to each other so they can't just do it independently.
They have to be constrained by each other.
And the exact solution is a little bit complicated.
And I'm just going to give you an estimate of what that modulus is.
And we're going to see that it depends linearly on t over l, as well.
I'll just say the cell walls are loaded in shear.
An estimate is g star 1 3 is equal to g of the solid times t over l times a geometric function.
It's cos theta over h over l plus sin theta.
And for regular hexagonal honeycombs, it's 1 over root 3 times gs times t over l. Again, just note the linear dependence of the modulus on t over l. And in the book, there's a method using upper and lower bounds that gives an estimate for g 2 3.
I'm not going to go into it.
I just want you to notice that the shear moduli go as t over l, just like the Young's modulus does.
And Sardar, who's sitting in the back there, has done even more involved calculations and analysis of the shear moduli of the honeycombs in this direction.
So I'm not going to go into all the gory details on that.
OK. That gives us the moduli now.
So now we've got all nine elastic moduli.
OK?
And the next thing to do is, then, to figure out the compressive strength.
So we're going to look at compression again, and then we'll look at tension.
If we look at compressive strengths, again, we've got different modes of failure.
And if I have an elastomeric honeycomb like this one here-- if these cell walls were a little longer, I might be able to actually do it.
If you compress this enough, you produce buckling in the cell walls.
And this is a schematic of this buckling pattern here.
And you can see there's a diamond pattern where it alternates up and down in the different cell walls.
We're going to do some approximate calculations, but you can see the idea of how the material behaves in this direction, just from these approximate calculations.
OK. Say we have our honeycomb like this, and here's the prism axis this way.
And now, we're going to load it up with some stress in the three direction.
I'm going to call this sigma star elastic 3 when it buckles.
And what we're going to do is just look at a single plate.
And look at the buckling of a plate.
We're going to analyze it just looking at a single plate and then adding up how many plates we have per unit cell.
It's actually more complicated than this because, obviously, the plates are attached together and there's some constraint by attaching the plates.
But we're not going to worry about that.
If you have a column-- just a, say, circular cross-section column-- and you apply a compressive load to it, it buckles at the Euler load.
And similarly, there's an Euler load for plates.
And that equation is usually written as a p critical is equal to some end constraint factor.
For plates, it's usually called k instead of n. So this is an end constraint factor.
It depends on the modulus of the plate.
It goes as t cubed.
Then, there's a factor of 1 minus mu of the solid squared and the length of the plate.
Say this plate here-- actually the width of the plate there is h and the length here is b.
And this thickness here is t like that.
Here, k is an end constraint factor.
And it's going to depend on the stiffness of the adjacent cell walls.
If I had a honeycomb, and say it was-- these walls here-- the adjacent walls-- were thicker, then you can imagine those thicker walls-- it'd be harder to get them to deform.
And the end constraint for the plate is going to depend on those thicker walls.
So that the end constraint, k, depends on these-- say I'm looking at this wall here of width h here, then how stiff these other two walls are is going to affect that end constraint factor.
What we're going to do is just do something very approximate.
We're going to say if these vertical edges here-- if this edge here and that one there-- if they were simply supported-- if they're just pinned to the next column, the next member-- then k has some value.
And if they're fixed, it has some other value.
And we're going to pick a value in between.
So we're going to do something very approximate.
I'll say if those vertical edges are simply supported-- that means they're free to rotate-- then k is equal to 2.0.
And this is if b is bigger than three times the length.
So this is h here, or we could say l. Either way.
It's really the-- it's the length when we look at the honeycomb this way on, but it's the width in that picture there.
And if the vertical edges are clamped, or fixed, then k is equal to 6.2.
These are values you can look up in tables of plate buckling.
And we're just going to approximate it by saying k is equal to 4.
We're just picking a value that's in between those two.
And then, the p total is going to be the sum of the p criticals for the columns that make up a unit cell.
For the unit cell, I have one wall of length h and two of length l. And if you just take that total load and divide by the area of the cell, you get that this compressive strength for elastic buckling is approximately equal to es over 1 minus nu s squared times t over l cubed.
And then there's a geometrical factor here.
And if you had regular hexagonal cells, this buckling stress works out to 5.2 times es times t over l cubed.
If you remember, for the loading in the two direction-- in the in-plane direction-- it has the same form.
And goes as es times t over l cubed, but it's much smaller.
This number here was, I think, 0.2.
It was much smaller.
So it has the same form, but it's a lot bigger.
OK?
Are we good with that?
The idea is we just use the standard equations for plate buckling.
We make some estimate of what that end constraint factor is.
And we just have an approximate calculation here.
OK. That's the elastic buckling.
If I had a metal honeycomb, then it might not fail by elastic buckling like that.
Instead, we'd probably get yielding.
If it was dense enough, we could just get axial yielding that-- if you just loaded it, you'd have axial forces.
And at some point, you'd reach the yield stress.
And so you can get failure by just uniaxial yield.
That's one option.
And if you get that, then it just depends on how much solid you've got again.
So it's just the yield strength of the solid times the relative density.
But usually, the honeycomb is thinner walled than that.
And usually, you get plastic buckling proceeding that.
In plastic buckling, you can think of it as-- say if you have a tube-- this is just shown for an individual tube here.
You can see how the tube folds up.
And you can get that same kind of thing with the honeycomb.
Here's a single tube.
It's been loaded along the prism axis of the tube.
And you can see, you get these folds, and the more you load it, the more number of folds you get.
And the more the folds concertina up.
To do an exact analysis for the honeycombs, you would have to take into account not just one tube, but the constraint of the neighboring tubes again.
And again, that gets to be a complicated, messy thing.
So again, we're going to do a more approximate thing.
What we're going to do is just say that we have members that are folding up like that.
So the same geometry.
But we're just going to look at a single cell wall and see what the single cell wall does.
And someone else has done the more exact calculation.
We'll just compare our approximate calculation to the exact one.
OK. We're going to consider an approximate calculation.
What we're going do is look at our isolated cell wall.
And if you look at the figure here, the wall is going to be vertical, initially.
And as we load it, eventually it's going to buckle and we're going to form one of those plastic hinges in the middle here.
And then, the thing is then going to rotate about that plastic hinge and just fold up.
So at the bottom here, it's completely folded up.
OK?
And we're going to do a little work calculations.
We're going to look at the internal work done and we're going to look at the external work done.
The external work is just going to be this load p times that deflection delta that the p moves through.
And if we say this is half of a wavelength-- if you think of this thing going through multiple wavelengths, just consider when it folds up like that, that's a half of a wavelength.
It would go two of those to get a full wavelength.
That's lambda over 2.
And so to go from this stage to that stage over here, the external work done is going to be approximately p times lambda over 2.
Say that it's thin and that 2t is small compared to lambda.
So it's going to be about p times lambda over 2.
And then, we're also going to look at the work done by the plastic moment.
And when we form the plastic hinge here, there's a plastic moment.
And that moment is going to rotate through an angle of pi.
So we start out straight here, we end up folded up like that, and we've gone from straight to that.
We had to go through 180 degrees to get there.
So it goes through an angle of pi.
And if you have a moment going through a rotation, the work done is the moment times the rotation.
We're going to equate those two works done.
We're going to look at the rotation of the cell wall by an angle of pi at the plastic hinge.
Our plastic moment-- it's going to be the yield strength of the solid again times t squared over 4, the same as when we were talking about the plastic moment before for the other loading direction.
But now, instead of multiplying this times b, we're multiplying it times 2l plus h. That's the length of the cell wall that's associated with one cell.
And now, it's not b because now we've turned the thing the other way on.
We're loading it the other way on.
And this plastic hinge-- if I think of-- if this was b before.
And now that b is l plus 2h-- or 2l plus h, rather.
That's the dimension of the-- let me draw a little hexagon so maybe you can see.
OK. Now we're forming a plastic hinge halfway down the board.
Imagine that this has some length b that way and we're halfway down the board.
And now, the plastic hinge has to form all the way around these members here for one cell.
Or you could think about it as this guy plus these guys is one cell.
You can think about the unit cell different ways, but it's one h plus two l's.
OK?
Are we OK with that?
OK. Then the internal plastic work is that plastic moment times the rotation pho-- or pi, rather.
Sorry.
Are we OK with this?
That the work done is m times our angle?
Imagine-- let me get rid of my honeycomb here.
Imagine you have a point here and you have some force over here.
Let's call that f. And say, the force is at distance r from f. And say that it moves through some distance.
The moment here would be r times f. And if that rotates, say, through some angle-- let's call it alpha-- and here is f here, then this distance here that the force moves through is just r times alpha.
So the work done is going to be r times alpha times f, or just the moment times alpha.
OK?
So that's all that we're doing.
OK. That's the internal plastic work.
And now we have to look at the external work done.
And that's equal to p times lambda over 2.
Here, lambda is the half wavelength of the buckling.
I'm going to say for these tubular kinds of things, it's in the order of l. So if you look at that last slide here-- oops.
Rats.
How'd that happen?
Let me scoot back down here.
There.
If we look at that guy again, the magnitude of the buckling wavelength is on the order of l. And here, below p, can be related to the stress in the three direction.
We'll just multiply it times the area of the unit cell.
And so if I equate the internal work and the external work, I can say p times lambda over 2 is equal to pi times my plastic moment.
And then, for p, I can write sigma 3 h plus l sin theta times 2l cos theta.
And then, lambda is l divided by 2 is equal to pi.
And then I've got my plastic moment over there.
And then if I solve for sigma 3, that's my compressor strength.
I've got pi by 4, the strength of the solid, sigma ys, times t over l squared.
Then h over l plus 2 divided by h over l plus sin theta times cos theta.
And for the regular hexagons, this works out to about 2 sigma ys times to over l squared.
And the exact calculation for regular honeycombs is equal to 5.6 times sigma ys times t over l to the 5/3 power.
This power here-- 5/3-- is a little less than 2.
And that's because the additional constraint of the neighboring cell walls.
But the main thing we're interested in, in these sorts of calculations, is the power dependence on the density and this simple calculation.
Obviously, it's not exact, but it gets you close.
OK.
I'm just going to wait for people to catch up a little.
OK.
The next property I'm going to look at is out-of-plane brittle fractures.
Say we loaded in tension, and if we had no cracks in the walls, we'd just see uniaxial tension and the strength would just be the strength of the solid times the relative density times the amount of solid.
We'll just say if defect free, the walls see uniaxial tension.
And then the fracture stress in the three direction is just equal to the relative density times the fracture strength of the solid.
If the cell walls are cracked, and if the crack length is very much bigger than the cell length, then the crack propagates normal to x3.
Then we can say the toughness gc-- or the critical strain energy release rate-- is just equal to the volume fraction of solid times gc for the solid.
And then the fracture toughness, k1c, is equal to the square root of the Young's modulus times gc.
And that's just equal to the relative density times the modulus of the solid.
And then the relative density times the toughness of the solid.
So it's just equal to the relative density times the fracture toughness of the solid.
It's just straightforward there.
Then we've got one last out-of-plane property.
And that's brittle crushing and compression.
And if we have some compressive strength of the cell wall-- say I call it cs-- then it's just the relative density times that strength.
And for brittle materials, that crushing strength is typically around 12 times the modulus of rupture, or fracture strength.
We could say that's about equal to 12 times the relative density times sigma fs, a fracture strength.
OK. That's the modeling of the honeycombs.
I know there's been a lot of equations and derivations, but that's the basis of a lot of the things we're going to do in the rest of the course.
The modeling we're going to do on the foams is based on this and the mathematics is just easier because we're going to use these dimensional arguments.
We're not going to figure out all these geometrical parameters.
Before we get to the foams, I wanted to talk a little bit about honeycombs in nature.
And we've only got a couple minutes left, so I won't really get that far.
But I wanted to talk a little bit about honeycomb materials in nature.
And the two examples we're going to talk about are wood and cork.
I'm going to talk a little bit about the structure of wood next time.
Then, we'll see how we can apply these models to understanding how wood behaves.
And we'll see how you can use these models to predict the density dependence of wood properties and also the anisotropy in wood properties.
And I guess we'll probably, maybe, start it Wednesday next week.
We'll talk about cork, as well.
Those of you who took 3032 know that I like cork because of Robert Hooke and his drawing of cork.
And I made a new video that I'm going to show you.
Remember in 3032, I showed you the video from the Bodleian Library, where they had the first edition of Hooke's Micrographia.
Well, it turns out Harvard has a first edition.
Harvard has three first editions.
Yeah.
Exactly.
MIT has zero first editions.
Gee, why does that surprise me?
And I have a friend who's a librarian at Harvard and she arranged for me to go and make a little video with the first edition of Micrographia.
So I can-- I don't if we'll play the whole thing, but I'll show you the first little bit of it.
And you can watch it at your leisure.
And Sardar came.
You came and saw it with me.
You came and saw the first edition with me, right?
Yes
Yeah.
Yeah.
It's very beautiful and you'll see some of the nice drawings.
And I talk about the cellular structure of some of the drawings.
So we'll talk about wood and cork next time.
But I think I'm going to stop there because that seems like enough equations for now.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
So I should probably get started.
So I just wanted to mention this Friday the libraries are having Furry Friday.
So they have therapy dogs come, and if you like dogs, it's kind of fun to go get cuddled by the dog.
The other thing I wanted to mention, last term, there was a student taking 3032 who was interested in art.
And I kept trying to find art pictures for him, and he's not here.
But I thought everybody else can like the art too.
So I belong to the Peabody Essex Museum in Salem, Mass.
And they have an exhibit right now on wood and on sort of using wood as a sculptural material.
And this is one of their posters to advertise it.
So this thing was carved out of a single piece of wood, I think.
And they've got lots of other sort of sculptural wood.
So I thought you might like to see that.
If you wanted to go to Salem, there's a couple of options, if you don't have a car.
You can take a commuter rail to Salem.
You can also take the ferry.
And if you take the ferry to Salem, it's like a five minute walk to where the ferry lets you off to get to the Peabody Essex.
And it's a kind of neat museum.
It's not too big.
It's kind of small.
But it's a beautiful building, and they have neat stuff there.
So you could go to the Peabody Essex Museum.
Hmm?
It's very nice
Yeah, you've been there?
Yeah, it's really nice.
So I was going to talk about honeycomb-like materials in nature today.
And I'm going to talk about wood today, and I might finish this today.
I might not.
And then I'm going to talk about cork for a little bit on Wednesday, and then we'll start talking about foams after that.
So I have a couple of sort of cute little language historically things.
And you know how I like that stuff too.
So I have two things about words that are related to wood.
So the word "materials--" you know where the word materials comes from?
It comes from the Latin.
So there's a Latin "materies materia."
And materies materia means "wood" or "the trunk of a tree."
So if you think of studying materials, in olden times that was like studying wood.
And another cute thing that I found was that in old Irish the names of the first few letters of the alphabet are named after trees.
So the letter A, that's called alem in old Irish, and alem is the word for elm.
And B is-- I don't know if I'm saying these right.
It's called beith, and that's the word for birch.
And C is call.
That's the word for hazel.
And D is dair, and that's the word for oak.
And so they sort of named the letters of the alphabet after different kinds of trees, different kinds of woods.
So I just thought those were kind of interesting historical things.
So I wanted to start by talking about wood structure.
And then we're going to look at how would deforms and fails, and talk about the data that people have measured for the wood properties, things like stiffness and strength.
And then we'll talk a little bit about how the honeycomb models can be applied to understanding the mechanical properties of woods.
So this is kind of a generic trunk of the tree here.
And we're defining three axes.
The radial axis comes radially out of the tree.
There's the tangential axis, so that's the x1 and the x2 axes.
And then there's the longitudinal or the axial axis, x3.
So if you think of the wood as being, in a very, very simple way, just like the honeycomb, the radial would be this way on.
The tangent would be that way on.
And that axial would be that way on.
So it's like that.
And if you neglect the growth rings, you can say that other woods orthotropic, and that's typically what people do.
They neglect the growth rings, and they say that it's orthotropic.
And the density of the woods, the relative density ranges from about 5% for balsa wood to about 80% for lignum vitae.
So I brought in some pieces of wood.
So this is balsa wood.
You're probably familiar, making different kinds of models with balsa.
So balsa's very light.
It grows in Ecuador.
And it's the lightest wood.
And this is lignum vitae.
You're probably not so familiar with that.
This actually grows in Florida, and it's the densest wood.
It has a relative density of 0.8.
And it's so dense, that if you put it in water, it sinks.
So it's a very dense wood.
And the way the wood cells grow is that if you look at the sort of structure here of a tree, there's the bark on the outside here, and then there's the kind of wood cells inside the bark.
And there's a layer of cells in between the bark and the wood called cambial layer.
And that's really the layer of cells that are alive and are dividing.
So if you think of the wood cells, they're living when they're in that little cambial layer there.
And they're dividing.
And that cambial layer, the cells have a plasma membrane and a protoplast.
And then they sort of exude the plant cell wall.
So a little like bone cells, like if you think of the bone in your body, there's osteoblasts and osteoclasis, different kinds of bone cells.
But the bone cells secrete the sort of collagen and the calcium phosphate that are the sort of hard mineral part of the bone that you think about in the bone.
And that's not a living thing.
The cells are the living thing.
That's like an extracellular matrix.
In the trees, it's a little bit the same.
So there's the living cells that are just under the bark, and they have this plasma membrane and the protoplasm.
And over a few weeks they excrete the plant cell wall, and then they die.
So the living cells die, and you're left with the plant cell walls.
And then as the tree grows, you're always having a layer of these cambial cells, and it forms bark on the outside and wood on the inside.
So there's sort of a layer of cells that are differentiated, such that on the outer layer they form the bark, and on the inner layer they form the wood.
And as the tree grows, that cambial layer is kind of expanding out radially.
So let me write some of these things down.
Let's see.
So let me just write down those two little word things because I think they're cute.
So the word materials is from the Latin materies materia.
And that means "wood" or the "trunk of a tree."
And here's the little old Irish thing.
It's not like I think-- I'm not going to put this on the test or something.
I just thought it was cute.
So the letter A is alem, which is elm.
And letter B is beith, which is birch.
Letter C is call.
That's hazel.
And D was dair, and that's oak.
So that's just for general interest.
So then the wood structure we can think of it as orthotropic, if we ignored the growth rings.
And if you have a sort of large diameter tree and you take a piece of wood not from the very center, but from somewhere near the outside, then that's not a bad approximation.
Now, the relative density of the woods ranges from about 0.05 for balsa to about 0.8 for lignum vitae.
Any Latin scholars here?
I took one year of Latin in high school.
Anybody take Latin?
No, no Latin.
So I think lignum vitae I think is "tree of life."
"Vitae" is the sort of life.
And when it has this ending A-E it means "of life."
So I think that's the "tree of life" is lignum vitae.
So trees have cambial layer beneath the bark.
And the cell division occurs in that cambial layer.
So the new cells on the outer part turn into bark, and the new cells on the inner part turn into wood.
And then we have the living plant cells that have the plasma membrane and the protoplast.
And those cells then secret the plant cell wall, which sort of surrounds them.
So in trees, the living cells lay down the plant cell wall over a period of a few weeks.
And then the living cells die.
Oops.
Back.
Here.
Now you always retain a layer of those cambial cells.
So you may have heard if you have a tree and you cut a ring around the tree through the bark, if you go into those cambial cells and you destroy them, you kill the tree, because you're killing that layer of living cells.
So then we want to look at the cellular structure of the woods as well.
And I've got a couple of slides here.
This one is of softwoods.
And softwoods have two types of cells.
That have tracheids, which are the bulk of the cells here, and the tracheids provide structural support.
And the tracheids also have little holes along the length of them at their ends called pits, and those pits allow fluid transport up and down the tree.
And then the softwood also has these ray cells here.
So those are examples of ray cells.
So this is a transverse section.
This is a longitudinal section here.
And the rays are parenchyma cells which store sugars.
So softwoods have tracheids and rays.
And then hardwoods, here's an example of a hardwood oak.
They have three types of cells.
There's cells called fibers, so these guys all in here would be fibers.
They provide the structural support.
They have vessels, these really large cells that provide fluid transport up and down the tree.
And they also have rays.
So here are some rays here.
And again, those rays are parenchyma cells that store sugars in the tree.
So let me just write down what all these cells are.
So in softwoods most of the cells are these tracheids, so they make up the bulk of the tree, something like 90% of the tree.
And they provide structural support.
They have holes in the cell wall for fluid transport, and those are called pits.
And to give you some idea of what size they are, they're are a few millimeters long, so something like 2 and 1/2 to seven millimeters long.
And then they're tens of microns in the other two directions, so they're something like 20 to 80 microns across.
And the cell wall thickness, t, is usually a few microns, so something between about two and seven microns.
So typically, the denser the wood, the thicker the cell wall's going to be.
Whoops, let's see if I can get the rays down here.
Put it on the same board.
So the rays are parenchyma cells that store sugar.
And then the hardwoods have three types of cells.
They have the fibers that provide the structural support.
And the amount of cells that are fibers varies, depending on the species, but it's usually somewhere around 35% to 70% of the cells.
And then they the vessels, which are the sap channels.
That provides for the conduction of fluids.
And that's between about 6% and 55% of the cells.
And then, again, there's rays that store sugars, and they usually make a boat 10% to 30% of the cells.
So there's the structure of this sort of cellular structure, at this kind of length scale of tens of microns.
And then there's also a structure within the cell wall itself.
And the cell wall itself is made up of cellulose fibrils in a matrix of lignin and something called hemicellulose.
So if you look at the cellulose structure, the cellulose has a regular structure, a sort of periodic lattice.
And it's crystalline for most of the length of the fibrils.
So this is the structure of the cellulose here, and this is showing it at a slightly larger length scale.
It might have a crystalline region here and then a non-crystalline region here.
And these macro fibrils, which are made up of bundles of micro fibrils, are about 10 to 25 nanometers.
And each one of the micro fibrils might be three to four nanometers across.
So you have these cellulose fibers.
And then the cell wall is made up of different layers.
So there's what's called the primary wall here, which has a random arrangement of the cellulose fibrils.
Then there's an outer layer here.
These are all called secondary layers.
This is S-- I think that's S1.
Yeah, it's S1.
And it has this arrangement of the fibrils.
Then there's a layer called S2, and it's generally the thickest layer in the cell wall.
And the cellulose fibrils are aligned not perfectly vertical, but a little off the vertical.
And the angle between the vertical and the orientation of the cellulose fibers is called the microfibril angle.
And then there's a third layer here, S3, with, again, a different winding of the fibers.
So because S2 layer is the thickest layer and because the fibrils are closest to the vertical axis, the S2 layer actually contributes the most to the longitudinal modulus and stiffness and strength of the cell wall.
So that's kind of the arrangement of the cell wall.
And then so that one cell would have that.
Another cell would have that.
And in between the two, there's a layer called the middle lamella that kind of glues them together.
So that's the arrangement of the cells.
Let me scoot over here.
So the cells are often modeled as a fiber reinforced composite that has four layers to it.
And in each layer there's different volume fraction of the fibers and different orientation of the fibers.
So the cell wall has this fiber-reinforced structure.
Here's the cellulose fibers in a matrix of lignin and hemicellulose.
And there's four layers, each with the fibers in a different orientation.
And then there's the middle lemella between the two cells.
So in doing the modeling of a material like wood, you need to know what the properties of the cell wall material are, because, obviously, the properties of the wood would depend on the cell wall properties.
And it turns out that they're similar.
They're not exactly the same, but they're similar in different species of wood, so we're going to call them more or less the same.
So the density of the solid is 1,500 kilograms per cubic meter.
The modulus of the solid in the axial direction is 35 gigapascals.
The modulus in the tangential direction or transverse direction is 10 gigapascals.
And the strength of the solid in the axial direction is 350 megapascals.
And the strength of the transverse direction is about 135.
So here A means Axial direction, and T is transverse.
And just for comparison, if you just look at cellulose, cellulose has some pretty amazing properties.
The modulus of cellulose is about 140 gigapascals, which is very high for a polymer.
And the strength of cellulose fibers run between about 700 and 900 megapascals.
So the cellulose fibers have very impressive properties.
And that's one of the things that gives wood very good properties.
So the next thing is I want to show you some stress-strain curves for wood.
And you'll see how similar they are to the honeycombs that we looked at before.
And then we'll look at how the cells are deforming as they're getting loaded.
And from that, we're going to do some modeling.
So let me just wait till people get caught up.
Are we caught up?
More or less?
OK.
So these are all compression curves, so I'm just going to talk about compression.
So these are curves for different types of woos.
And on the left, the wood is loaded in the tangential direction.
So in terms of the sort of honeycomb model, it's loading at kind of this way on, like that.
And on the right, are a set of curves for wood load it in axial compression.
So in axial compression loading, we're loading it that way on.
And we've got different species here.
So the lowest density is balsa, around about 100 kilograms per cubic meter.
The densest species on this plot is beech, which is around 700 kilograms per cubic meter.
And then there's pine and willow, some other species in between here.
So you can see the shapes of the curve look just like the curves that we had for the honeycombs.
So here there's a linear elastic bit.
There's a stress plateau.
And there's a densification bit.
And then, as the density goes up, it gets stiffer, and the strain at which the densification occurs gets smaller, and the strength gets higher.
And if we look at the axial properties, the shape of the curve is similar.
We get linear elastic stress plateau densification.
But if you look at the scale here, this scale goes from 0 to 100, whereas that scale went from 0 to 20.
And so the stiffness and the strength along the grain are much higher than they are across the grain.
And you probably already know that.
Wood is stronger and stiffer along the grain than across the grain.
So that's what the stress-strain curves looked like.
And the fact that we're getting the curves that look like that makes us think maybe the mechanisms of deformation and failure are similar to the honeycomb, too.
So here's a set of curves for balsa all plotted on the same scale.
And, again, you can see for loading across the grain, either in the radial or the tangential direction, the stiffness and the strength is a lot less than if you load it in the axial direction.
So a number of years ago, we had a project on balsa.
And the thing we were interested in doing was looking at how the cells deformed and failed.
And because balsa's a low-density wood, it was easier to see the deformation in the cells, because the cells were thin.
So that's why we chose balsa.
I actually have a project on balsa right now.
And [?
Sardar ?
], my postdoc, is doing more detailed kind of finite [INAUDIBLE] modeling, trying to represent the structure of balsa.
And I think I mentioned, the reason we're interested in it is the balsa's used as a core in sandwich panels in wind turbine blades.
It's actually the best material that they can find, it's better than any engineering material.
So that's comparing the three curves for balsa.
And then if you look at a specimen that's loaded in the [?
SCM ?
], with a loading stage, you can measure the stress-strain curve, and you can take photographs of what the cellular structure looks like at different stages of loading.
So here, this picture one, is unloaded.
And these four images here are looking at the same section of cells, the same area of cells.
And you can see there's a big vessel here, and that's the same vessel there.
So here, this image two, is at this point on the stress-strain curve.
Here's three, at that point.
And four is at that point.
So if you look at this carefully-- and I've got another higher mag picture I'll show you in a second-- you can see that what's happening is the cell walls are bending.
So it's kind of like taking my honeycomb like this, and I'm doing that to it, and the cell walls are bending, so just the same as the honeycomb.
And then eventually, if I load it enough, you get to this sort of densified stage.
And you're doing this, and the stress-strain curve increases sharply.
So here you can see how the cells have densified over here.
It kind of looks a lot like my honeycomb when I-- maybe I do it this way-- when I smush it up like that.
It looks kind of similar.
So if we look at the higher mag picture, again, these four images are the same area of the cells.
And if you look at that a little bit of crud, it's the same on all four of them there.
So this is the unloaded one.
And these were loaded from top to bottom.
And this is loaded to some extent/ that's loaded more, and that's loaded more.
So if you look at this cell here, it's got this little tear on it.
So you can sort of find it again.
If you look at that cell there, that's what it looks like unloaded.
And here you can see-- see that wall there?
You can see how it's bent up.
So it's bent like the honeycomb walls.
And here it's bent even more.
And eventually, it has this sort of a shape here, and it's deformed permanently.
It's formed one of these plastic hinges.
So it's like the aluminum honeycombs almost, that it's filled like that.
So in the balsa wood, when we load it in the tangential direction, we're getting bending of the cell walls and then yielding and plastic hinges forming, just the same as we would in an aluminum honeycomb.
Are we good with that?
Well, I'll go through all three directions.
And then I'll write down the notes.
So this is loading the balsa in the radial direction.
And these things here are the rays.
So we're loading it in that direction.
And here you see this also bending occurs, but the rays act a little bit like fiber reinforcement.
So the rays are a little bit stiffer, and they sort of reinforce the thing a bit.
And this is the loading platten here, and you can kind of see that the failure starts at the loading platten, and as you sort of load up more, it progresses in from the loading platten.
So we're going to look at the modeling of the balsa in the radial direction, and we're going to count for the rays, at least in a crude way.
And then when you load them balsa in the axial direction, initially you don't really see much happening.
So if one is unloaded, one's down here.
And two is at this peak stress up here.
And really, if you look between one and two, you just don't see an awful lot of difference.
And that's because what you're doing is you're taking the wood and you're loading it this way on.
And it's so stiff, you just don't see much deformation.
So there's not really much to see.
But then eventually, something starts to fail.
And in this case, what fails are the end caps.
So the balsa wood has these long cells here.
But then at the end of the cells, there's little caps on the ends.
And the cells kind of fit together like that.
So that eventually, if you keep smushing it, those end caps start to fail.
Here you can see how bright it gets, and the cells are starting to crush together and kind of fail those end caps.
And in fact, each one of these serrations here, if you look at, say, from that peak up to that peak, that corresponds to a length of about the length of the cell, or the length of the cell between the end caps.
So in axial deformation you're just actually deforming the cells until you break those end caps.
If you look at denser words, they fail in slightly different ways.
This is a Douglas fir, which is much denser.
This particular specimen, the whole thing is kind of buckled over.
So it's not really so representative of the structure itself.
This is Douglas fir in radial compression.
You can see this picture, it looks just like what I showed you for the balsa wood, that sort of propagation of the failure.
These long things here are the rays.
And this is a Norway spruce in axial compression.
And this is fairly common in denser words.
You get this buckling formation.
And what happens is, I think, you get some yielding of the cell walls initially, but that leads to buckling, like a plastic buckling.
And you can see on this higher mag picture down here, you get these really small wavelength buckles in the cell wall.
And the two-- you get a plane that kind of shears over itself.
And you can see in the top image, this top half has shared over relative to the bottom half.
And all the deformation is in this little band here.
So this stuff here is all going on in that band up there.
So let me write down some notes about how these things to form and fail.
And then we'll get to the modeling in little bit.
So we can say the stress-strain curves resemble those for honeycombs.
And I'll say the mechanisms of deformation and failure are most easily identified in low density balsa wood.
So for balsa, if we look at the tangential loading, we see bending of the cell walls and then eventually plastic yielding.
And for radial loading, the rays act as reinforcing.
And for axial loading, you get axial deformation and then the failure of the end caps.
And I'll just say failure by plastic buckling is also observed, say, in the denser woods.
[INAUDIBLE] So then we can look at some data for the properties of woods.
And these charts plot relative Young's modulus and relative strength against relative density.
So here the modulus of the wood is divided by the modulus of the solid cell wall material.
And here we've normalized everything by the modulus of the solid cell wall material in the axial direction, because the cell wall itself is anisotropic.
And so here's the relative modulus, and here's the relative density.
These are log-log plots.
And we see that when we load the wood in the axial direction, the moduli is just linearly related to the density.
And when we load it across the grain, it varies with the cube of the relative density.
So do you remember our little honeycomb models?
If I took the honeycomb and I loaded it this way, it went as the cube of T over L. And that's because the bending.
And so the wood doesn't lie perfectly on that cube line, but it's fairly close.
And then similarly, if we took the honeycomb and we loaded it this way on, it deformed axially.
The modulus depended linearly on the density.
So you get the same kind of relationships there.
And then if you look at the strength, the strength along the grain goes linearly, and the strength across the grain goes with the square.
And we'll see when we get to the modeling in a minute, that if we loaded, say, an aluminum honeycomb this way on, the strength would go linearly with the density, if we're just yielding the cell walls.
And if we loaded it this way on, it went as the square of T over L. So these things kind of correspond.
And you can see the structure of the wood is a lot more complicated than just a simple honeycomb.
And so these models are sort of first order, and they're fairly crude.
They don't try to capture every detail of the wood structure.
But they can give you a sense of where the wood properties are coming from.
So let me just write down some of these observations.
So the data for the wood-- the modulus along the grain goes linearly with density.
It goes more or less as the cube for loading in the tangential direction.
And the radial direction is somewhat stiffer different than that.
The strength in the axial direction goes linearly with the density.
And the strength across the grain goes with the square of the density.
And then there's data for the Poisson's ratios too.
So let me just write them down.
So the modeling based on the honeycomb is sort of a simplified model that gives you kind of a first-order description of the behavior.
And it doesn't really attempt to capture all the details of the softwood and hardwood structure.
And in the equations, I'm going to take the cell wall properties along the grain, or along the axial direction.
And we're going to have a bunch of constants that describe the cell geometry, and those constants are also going to reflect the cell wall anisotropy.
So we can model the wood structure as something that's a bit more of a simplified thing, just like this.
And we say we've got cells that are roughly hexagonal, and then we've got some cells that are more or less rectangular that are the ray cells.
And if you look at lots of micrographs, you can get some idea what the dimensions of the cells are.
And these dimensions were measured for a particular density of balsa wood.
So if we look at the linear elastic moduli, we can start off with a tangential loading.
And if we have the tangential loading, we can model it as a honeycomb loaded in the plane, and we get cell wall bending.
And from the cell wall bending in the honeycomb model, you would get that the tangential modulus varies with the relative density cubed.
And the structure's not quite that simple.
There's ray cells.
There's end caps.
And they act to stiffen it a little bit.
And the data lie a little bit above this line.
Then if we look at the radial loading, the rays kind of line up with the radial direction, and the rays act as reinforcing plates.
And so you can just use kind of an upper-bound composites idea to get the modulus.
And the rays tend to be a bit denser than the fibers.
So if I say a Vr is the volume fraction of rays, and R is the ratio of the relative density of the rays compared to the fibers, so it's rho over rho S for the rays divided by rho over rho S for the fibers.
And that varies a little bit from what one species to another, one specimen to another.
But it's something a little over 1, something between 1 and 2.
Then I can say the modulus in the radial direction is the volume fraction of the rays times R cubed times the tangential modulus plus 1 minus the volume fraction of rays times the tangential modulus.
And that works out to be about 1.5 times the tangential modulus.
I wanted to work this out in terms of the tangential modululs, so I've put this in terms of the tangential modules in the first term there.
So we get that the radial modulus is slightly larger than the tangential, but also goes roughly as the cube of the density.
And then for the axial loading, we just have axial deformation in the cell wall.
And the Young's modulus just varies linearly with the density.
So these are kind of simple models, but they kind of explain to first order the density dependence of the wood moduli and the anisotropy.
So it's kind of nice because they're fairly simple models, and it gives you kind of a big picture.
So if you wanted to know the modulus of a particular piece of wood, this probably isn't the best way to figure it out.
But if you wanted to kind of compare how do woods behave in general and how does the density affect the properties and why are they anisotropic, this is a pretty good way to do it.
We could also look at the Poisson's ratios.
And just because I didn't want to write them down again, I've just left on what the data were down here.
But let me just write what the model would give us for nu RT and nu TR, the model would give us one if we had regular hexagonal cells.
And these are the values we get here.
This might be 0.6, 0.7 would be a typical value, somewhere around 0.4 in there, so they're not quite one, but they're close to it.
And I think the reason they're a little less is because the rays in the end caps provide some constraint.
If you have the honeycomb, if I just had these cells, and I squeeze it like this, these guys can move out.
If it's a regular hexagonal honeycomb, the strain that I'm applying here is equal to the strain going out that way.
But if I have rays this way that sort of constrain it or end caps, it means that the Poisson's ratio is going to be a little bit less.
So I'll just say constraining effect of the end caps and rays-- constraining.
Then for nu RA and nu TA, the model says the value we would get would be zero.
And these are pretty close to zero.
They're not quite zero, but pretty close.
And then the last pair nu AR and nu AT, the model says that we would get nu of the solid.
And the data's close to 0.4, which we might expect would be about the nu of the solid.
So, again, there's some variation in the Poisson's ratios.
They're not all just one number.
But you can see these ones here are about zero, and that's roughly what the model says.
These ones here are closer to 1.
And then these ones here are closer to what you might expect for a solid material.
So it gives you the kind of general idea.
Are we good?
We're good, yeah?
So we can do a similar thing for the compressive strength.
So for tangential loading, we get plastic hinges forming and the bent cell walls, just like in an aluminum honeycomb.
Then we get that the strength over the cell wall strength goes as the relative density squared, so just like the honeycomb.
the radial loading, we can do the composites thing again.
So we can say the strengths in the radial direction is about equal to the volume fraction of rays times R squared times the tangential strength plus 1 minus the volume fraction of rays times the tangential strength.
And for balsa, I have some values here.
VR is about equal to 0.14.
R is about equal to 2.
And so the radial strength is about equal to 1.4 times the tangential.
And in higher density woods, the value of R is a little bit smaller, and in general, the radial strength is a bit larger than the tangential, and both depend on the density squared And then for axial loading, if the failure's initiated by yielding in the cell walls, then the axial strength's just going to depend linearly on the density.
So the idea with these models isn't that they kind of describe a particular piece of wood exactly.
It's more that it gives you a general picture of how the cells are deforming and failing, and how the properties scale with density and why the wood's anisotropic.
Are we good?
Yeah?
Caught up.
So there are a couple more sort of interesting things we can do with looking at the wood properties.
So we've been talking about how to model the cellular structure.
But people have also looked at how to model the cell wall as a fiber composite.
And this plot and the next one kind of show you how you can combine all of that together.
So remember, I said the modulus of the cellulose was around 140 gigapascals.
So here's the modulus of the cellulose, at least the crystalline part of the cellulose plotted in that little envelope there.
The lignin and the hemicellulose have a modulus around 2 or 3 gigapascals, so it's down there.
And if you made composites with cellulose fibers in lignin and hemicellulose matrix, those composites would have a modulus that fell in this envelope here.
They've got to be in between those two limits, right?
The modulus have to be between those two limits.
The density have to be within the densities of the constituents.
And if you look at the modulus of the wood cell wall, it lies in this envelope here.
Along the grain it'd be here, and then across the grain is further down here.
So the cell wall modulus is in here.
And then if you take that cell wall and you make it into the honeycomb-type material that wood is, if you load it along the grain, you're going to get this linear dependence of modulus on density.
And if you load it across the grain in the radial or the tangential direction, you're going to get this cubed dependence here.
So here's a set of data for different woods of different densities.
And that envelope kind of encompasses all of them.
But if you look at the slope of that data, it's roughly equal to a slope of 1.
And so it corresponds to that equation there.
And similarly, here's a set of data for different species of woods of different densities loaded perpendicular to the grain.
And they lie on a line that has more or less a slope of 3.
And this set of data here along the grain intersects the wood cell wall towards the top of that envelope, and this set of data here intersects closer to the bottom of that envelope for the cell wall material.
So this gives you a way of sort of putting everything together on one plot-- the cell wall as well as the cellular structure.
So that plot does it for the modulus.
And you can do the same kind of thing for the strength.
Here's the cellulose up here.
Here's the lignin down there.
Here's the wood cell wall, the composite made from those two.
And then here's data for different kinds of woods loaded along the grain and for load across the grain.
So it gives you a way of putting all this modeling into one set of plots.
So let me just write a couple of little things about that.
So we could say you could model the cell wall as a fiber composite.
And you can use the composition upper and lower bounds to give an envelope.
And then you can also show the cellular solids models on the same plot.
So overall, it shows you how the hierarchical structure fits together and can be modeled.
Now there's some more cute things we can see.
So another thing I want to talk about is material selection, because it turns out wood is very good compared to other materials in certain applications.
So we're going to look at, say, having a beam of a given stiffness at a given span, and say it's just a square cross-section beam of edge length T. And the question is, what material would minimize the mass of the beam?
So say we have some span we have to have.
It's got to have some rectangular cross-section, some given stiffness.
And the question is, what's the material that minimizes the mass?
So there's a little short calculation we can do to figure that out.
And then I've got another plot, and you can compare different materials on this other plot.
Then you'll see how good wood is compared to other materials.
So from beam of a given stiffness and given span, and say it's a square cross-section, then the question is, what material minimizes the mass of the beam?
So the mass is just going to be the density times t squared times l And if it's a beam, say it's got some central load on it, a concentrated load, the deflection's going to go as pl cubed divided by some constant and divided by the Young's modulus and the moment of inertia I.
So the stiffness, if I just rearrange this, the stiffness, p over delta, that's going to go as p over delta CEI, and I's going to go as t to the fourth over l cubed.
And then I can solve that for t squared.
And I want t squared because I'm going to plug it back into the equation for the mass.
So if I solve this for t squared, I've got my stiffness p over delta.
I've got l cubed divided by CE.
And then I take that whole thing to the 1/2 power like that.
And then I plug the t squared back into the little equation for the mass.
So I've got density minus p over delta times l cubed over CE.
And we'll take that whole thing to the 1/2 power-- [INAUDIBLE] another l. And so to minimize the mass, you want to look at the material properties.
And here, the material properties are the density and the Young's modulus.
And to minimize the mass, you want to minimize rho over E the 1/2 power.
Or conversely, you want to maximize E to the 1/2 over rho.
So if you just had a bar that you were just pulling on, you would just want to maximize E over rho.
But if it's a beam and bending, it works out that you want to maximize E to the 1/2 over rho.
And if we look at the next slide, this next slide then plots on a log-log scale, it plots the modulus on this axis and the density on that axis.
And here this plotted data for lots of different materials.
So there's engineering alloys.
Metals are up here.
Engineering ceramics are here.
Composites are here.
Polymers are down here.
Elastomer is way down here.
Foamy things, down here.
And this envelope here is woods.
And notice log scale here.
The lowest stiffness polymer foams here are 0.1 gigapascal, and diamond is up here at 1,000 gigapascal.
So there's like five orders of magnitude difference in the modulus here.
So then, if you look at the bottom right corner here, there's a bunch of old dashed lines.
And this red one here is E to the 1/2 over rho.
So if it's log-log plot, E to the 1/2 over rho's going to show up as a straight line.
And every point on that line has the same value of E to the 1/2 over rho.
And the material that would be the best for a beam of a given stiffness would be the one that has the biggest value of E to the 1/2 over rho.
And if I move the line up to the top left here, I'm increasing E. I'm decreasing rho.
It's got the biggest value of E to the 1/2 over rho.
So the materials that are on this line here, they all have the same value of E to the 1/2 over rho.
And they've got the biggest value-- well, virtually the biggest value.
So let's look at what those materials are.
There's things like engineering ceramics, like diamond that maybe are not the most convenient thing to make our beam out of, and tend to be brittle and might break.
So we have some issues.
There's engineering composite, so things like carbon fiber reinforced plastics.
And at this sort of tip of the composites, there'd be things like unidirectional fiber composites.
And then here's other woods down here.
So the woods have the same performance index, this is called, the same value of E to the 1/2 over rho as the best engineering composites.
And so they have very good properties for their weight.
And one of the interesting things is if you look at this performance index of E to the 1/2 over rho, this is the performance index for the wood.
This is for the solid cell wall material that the wood's made from, so E to the 1/2 of the solid over rho for the solid.
And from the modeling of the wood, just looking at the axial modulus, this thing here is equal to that times rho S over rho.
So if you look at this, this is the performance index for the wood.
This is the solid it's made from.
This number here is bigger than 1, right?
Because the density of the solid is bigger than the density of the wood, and so this is saying the wood is more efficient than the thing that it's made from, than the solid that it's made from.
And so that's the sort of plot for the stiffness.
And there's a similar plot for the strength.
That if you do the same little kind of calculation, you find that the performance index for the strength is some failure strength raised to the 2/3 power over a rho.
And again, here we're plotting strength versus density on a log-log plot.
And this red line here is the strength of the 2/3 over rho.
And again, if we scoot over here so we have a parallel line, every point on that line has the same value of the strength to the 2/3 power over the density.
And these are the materials that have the highest values.
And again, here's engineering composites.
These are ceramics.
But the ceramics, they have a high compressive strength, but they tend to be brittle.
So it's not really a practical strength.
These are metals in here.
And here's the woods down here.
So it's kind interesting just to see that the wood has such a good property.
Yes?
So I realize why this is valuable setting up the problem this way.
But if you're actually trying to design something, why would you want to fix your cross-section?
You could change your material and change your cross-section
So this is the starter version of this problem.
And there's another part two of the problem is to change the shape.
And you could look at what shape's efficient.
There's something called a shape factor that gives you the efficient shapes.
So you could take the material and turn it into a different shape and have a more efficient thing because it was a different shape.
So you can account for that
So then if you varied, like let's say you made your cross-section smaller, like even if it was still square, you could just still make it smaller
Yeah.
I'm saying we've got a given stiffness.
So if we're given a certain stiffness and a certain span, we would need a certain cross-section to get to that stiffness.
Are we happy?
OK.
So that's one thing.
Let's see here.
So let me just write a few more notes about the material selection, and then there's one more thing I wanted to show you about the woods.
Hmm?
C is just a constant.
So it's just a number.
So if you had a beam in three-point bending, then C would be three.
If you had a beam that was simply supported with a central load, C would be 48.
C is just a number
One more question.
So follow your line there, and the choice is really just about cost
No, it's not about cost.
There's nothing on cost here.
It's all really about the properties.
What's the best combination of properties to minimize the mass, and then which material has that combination of properties.
You can do charts like this that include cost.
You can make these charts with whatever property
[INAUDIBLE].
I guess maybe there's a difference off two or so of strength
Between pine and balsa?
Yeah, maybe more than that.
I think-- I can't quite see where-- pine's close to 100, and balsa's, I don't know, 20 or something
[INAUDIBLE]
Yeah, and it's not-- the point of this isn't so much looking at the absolute value of a strength.
It's looking at the value of this performance index.
And what you want to do is maximize that index to get the material that's going to minimize the mass.
So let me just read a couple notes about this.
So we have these-- these are called material selection charts.
So you plot the log of one property versus the log of some other property.
And then we have a line of constant E to the 1/2 over rho.
I'll just say it's shown in red because you're going have the same plots.
And the materials with the largest values are in the upper left.
So the woods have similar values to engineering composites.
And you can do a similar thing for strength.
So I have a few more minutes.
So I have a few minutes, and I want to talk about a couple of uses of woods.
So one is in old ships.
So I don't know if you know Professor Lechtman has this course Materials and the Human Experience, and they talk about sort of ancient uses of materials.
And I did a section, a module, on woods and the use of woods in old colonial ships, like The Constitution that's in Boston Harbor.
So this is kind of a schematic of an old ship.
And the thing that was interesting and the thing I talked about in this module was that people chose particular species for particular parts of the boat.
And they would choose a particular species depending on its properties.
And a lot of the hull was made of oak.
So oak's a very dense wood.
But they would get something they called straight oak, and they would get something they called compass oak.
And you can see this little thing down here, this little kind of schematic here, this little sketch.
This is straight oak, just a straight trunk.
And this thing here would be the compass oak.
And what they would do is they would use the straight oak for straight parts of the boat, so something like this, these pieces here.
And then they would actually look for trees that had the curve of the branches to match some part of the boat that they were looking for.
So, for instance, if you have the hull out here and the deck here and they had their cannons here, there's something called a knee, which is sort of a bracing piece that goes between the deck and the hull.
And that bracing piece is curved.
And they would actually look for trees in which the branches curved at the same kind of curvature as they were looking for in that piece.
And then they would use it for that piece.
And the advantage of this is they basically had the grain running along the curve, and so they got the best properties out of the wood by doing that.
So they had this straight oak and compass oak, and that was one cute thing.
And often they used white oaks.
And I brought a piece of white oak in.
You can see how dense it is.
And the US Navy often used something called live oak.
Live Oak grows in the South.
Anybody from the South?
You see these big trees with huge sort of spreading branches.
Those are the live oaks.
And apparently, the US Navy, I read somewhere, still has a forest somewhere with live oak for doing things like repairing The Constitution.
So let me just pass those guys around.
So those are a couple of oaks they would use for the hull.
Then they would use white pine for the masts.
And the reason they used white pine for the mast is because the white pine grows very, very tall and very straight.
And white pine was actually like a strategic resource in the 1600s, the 1700s.
And it turns out that when the British Royal Navy was doing all that colonial stuff in the 1600 and 1700s, Britain actually ran out of trees for masts for boats, and they would actually import masts from New England.
And there were these people called surveyors who would go around and they would mark certain trees that were supposed to be saved for these masts for the British Royal Navy.
And the thing was that the size of the boat and how many cannons you could put on the boat depended on how big the mast was.
So the size of the boat depended on the mast, because the mast height controlled how much sail area you could get.
So the taller the mast, the more sail/ the more sail, the bigger the ship.
The bigger the ship, the more cannons.
And so having these tall Eastern white pines was a sort of a strategic resource.
And I have a piece of white pine.
Unfortunately, my dog got to this one.
And be careful.
It's a bit splintery.
But you can see it's a lighter kind of wood.
And if you go around New England, if you go to the arboretum, you can see white oak.
You can see Eastern white pine.
The other wood they used is lignum vitae, that first dense one that I passed around.
And if you notice that lignum vitae has kind of a waxy feel to it.
And they used that in the block and tackle, so like pulleys and stuff like that.
And it was thought to be self-lubricating because of that kind of waxy layer on it.
And because it's very dense, if you think of like a block and tackle and you've got like a rope going over a pulley, you've got a pressure from everything sort of fitting together and the bits bearing against each other.
And the fact that was very dense made it very good for the block and tackle.
And so they used the lignum vitae for that.
And there's one other cute story about lignum vitae.
I don't know if any of you've ever read Dava Sobel's Longitude.
Anybody read that?
I'm a sucker for those history of science books.
So her book Longitude is about the development of an instrument to measure longitude.
Originally, they could get the latitude from the stars, but they were really bad at getting the longitude.
And so boats would go off, and they wouldn't really be able to figure out where they were, until they had a method to measure longitude.
And there was some British board of something or another.
They put forward a prize for somebody who could produce a way of measuring longitude accurately.
And there was a guy called John Harrison, and he built a clock.
He built a very accurate chronometer.
And if you knew when sunrise was and sunset was, and you knew the time and where you left, you could figure out where you-- it's kind of like time zones.
You could figure out where you are today.
And he built a chronometer, and one version of his chronometer used a lignum vitae for the same reason, because it was very dense, and it was very stable.
And the clock that he eventually won the prize with was in the 1700s, 1759.
I think they went on some trip with it.
It was 81 days at sea, and it lost five seconds over 81 days.
So that's pretty impressive for 250 years ago.
So that was the lignum vitae in the clock.
I have one more picture, and then I can finish up the thing on wood.
And we'll start the cork next time.
So this is another example of using wood.
And this is sort of a more modern use.
So this bridge here is made with a glue-laminated wood.
So this big beam here, the big arch, is made up of sections of wood which are glued together.
And you can glue the sections in a curved shape if you want.
They sort of have molds to do that.
And when they make this glue-laminated wood, they cut the defects out.
So they cut knots out, and they control the pieces of each laminate that they use to get the best quality.
And the glue-laminated wood actually has better properties than just two-by-fours or whatever you would cut down, lumber that you would cut from a tree.
So glue-laminated wood is kind of a nice kind of wood structure that's used now.
And you see it all the time in things like ice rink arenas, like large spans.
It's kind of beautiful.
You can see the wood grain in the curve in the wood grain when they make these things.
So that's the wood lecture.
I'm going to stop there.
So next time I'll talk about cork.
I just have a little bit about cork.
And then we'll start talking about foams.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
I wanted to talk a little bit about cork partly because cork is kind of interesting.
Cork has a structure that's a little bit like one of those honeycomb things.
What I'm going to do is I'm just going to talk and go through the slides.
I'm not going to write the notes on the board.
There's only a few pages of notes, and it's in the Stellar site.
I'm not going to write notes for this because it's just really for fun.
This first slide starts off with historical uses of cork.
Cork was used by the Romans.
They used it for the soles of sandals, the same as we do.
And they used it for stopping bottles of wine, the same as we do.
But they didn't realize that you could just use the cork.
They would take the cork, put it in the wine bottle, and then they would use pitch which is a tarry stuff from a tree.
They would use the pitch and seal the bottle with the pitch.
In the 1600s, there were some Benedictine monks that realized that you could just use the cork and not use the pitch.
They were the ones who really perfected the use of corks in wine bottles for sealing it.
So what is cork?
Cork is the bark of a tree called Quercus suber, the cork oak.
Here's a piece of the cork bark.
All trees have a layer of cork in them.
But the thing that's different about Quercus suber is that it's very thick and the corks are obtained from this thick layer.
Quercus suber is a Mediterranean type of tree.
It grows, Portugal is the main place that exports cork, but also places like Algeria, Spain.
You can grow it in California.
The cork is kind of unusual because-- let me scoot onto the next slide-- it's kind of unusual.
So here's a little picture.
I went to Portugal when I was a graduate student doing this project.
Here's the cork tree here.
Here's the little mini we rented.
Here's the cork being harvested here.
Cork's unusual because you can remove the bark from a cork oak tree and it regrows.
So most trees if you did this, you would kill the tree.
But cork doesn't get killed by doing this.
What happens is they plant trees.
You have to wait something like 10 or 15 years for the tree to get big enough to harvest the cork.
And then the first harvest is poor quality and they don't use that.
And then you have to wait another 10 years or so before you can actually harvest the cork.
So you can imagine if a cork orchard or forest gets chopped down to build a skyscraper, or apartment buildings, or something, it's not an economically feasible thing to plant cork trees these days.
So when the trees are cut down, they don't tend to get replanted at this point.
And there's a number of artificial substitutes for corks.
You've probably seen wine bundles with foam plastic corks in them too.
OK.
The reason it's called Quercus suber is that the cell walls in this particular type of cork oak are covered with a waxy substance called "suberine."
That's where the Quercus suber-- "Quercus" is "oak."
Every oak is Quercus something.
"Quercus alba" is white oak.
Quercus just means oak.
If we look at the structure of the cork, we can see that it's got these different views of the cork that are seen here.
This is a drawing by Robert Hooke in the 1600s from his book Micrographia.
He was the first person to really draw cork like this.
You can see he drew sort of boxy cells on this side.
And then, the other perpendicular plane, the cells have this structure here, sort of more rounded.
And here's a little sketch he's got of the cork tree.
Over here are SEM pictures.
These two planes here correspond to this plane in Hooke's drawing.
This plane here corresponds to this plane here.
This over here is Hooke's actual microscope.
I think the Royal Society still has that microscope that Hooke used in the 1600s.
Some of you know that Hooke wrote this book called Micrographia.
He got one of the first microscopes.
He looked at a lot of different materials, and he drew these beautiful drawings.
He wrote a page or two about each of the drawings.
Harvard has a first edition of Micrographia, and I made a little video on it.
So I thought I'd show you the video because the drawings are beautiful.
But the url's on the slides, and so you can watch it yourself.
There's more on this and on how he came to be so good at making scientific apparatus, how he came to do the Micrographia book.
At the end of it, there's a comparison of a number of his drawings with modern SEM images of the same thing.
He had this very famous picture where he draws a flea.
Don Galler, the person who runs the SEM for me, I had him put a flea in.
And he has essentially the same kind of image.
The thing that's spectacular is you can see how much of the detail that Hooke was able to capture in his drawings.
There's some really beautiful drawings.
OK.
So let's go back to cork.
Hooke was the first person to use the word "cell" to describe biological cells, and he described the cell in cork.
That's the structure looking at the SEM micrographs and his optical micrographs.
These are just some more higher resolution, higher magnification, images.
One of the things that you can see is that the cork has these little corrugations on the cell walls.
See those little wrinkles?
All the cell walls have those little wrinkles.
This plane here is the perpendicular plane.
If you look down into the cells, you can see those blurry things.
Those are the corrugations in the cell walls.
That's the structure.
Here's a schematic.
Here's the cork tree.
The cork is the layer just beneath the bark.
This is a picture of how the cells are oriented relative to their radial, and tangential, and axial directions.
You can think of them as roughly hexagonal.
They've got these little corrugations.
This is a schematic of an individual cell.
We measured the dimensions of the cells, and these are some average dimensions here.
Typically, the cells are tens of microns long, and the cell wall is about a micron thick.
Something like that.
OK?
One of the things we're going to see is that corrugated structure gives rise to some of the interesting properties of cork.
If we load cork and just do mechanical tests on it-- this is just a uniaxial stress-strain curve.
You can see the stress-strain curve looks like all these other curves we've seen for honeycombs.
There's a linear elastic part here.
There's a stress plateau here.
And then there's a densification part here.
Typically, the relative density of cork is around 0.15.
Something like that.
It densifies at a strain of something less than 0.85.
It's a stress-strain curve.
And when we did our little project on cork, we measured the properties in the three directions because in one direction, it's roughly hexagonal cells.
That plane is isotropic.
The e1,e2 plane is isotropic.
This compares the measured values of the properties versus what we calculated from the honeycomb model.
And really, we just used the sorts of equations that we talked about in class over the last couple of lectures, apart from loading in the x3 direction because in the x3 direction, you've got those corrugated walls.
And you have to take those corrugations into account.
So there's another complication that I'm not going to go into.
But there's a sort of modification you can do to account for that.
But you can see there's actually pretty good agreement here between the elastic moduli that we measured and what we calculated.
The compressive strengths down here are not quite so good.
They're off by a factor of two, more or less.
But they're in the right ballpark for the cork.
So those are some of the structures, some of the properties.
One of the interesting things about cork is what it's used for.
The uses of cork really exploit the mechanical properties.
Obviously, it's used for stoppers for bottles.
I brought a champagne cork along with me, and I brought a couple of other little pieces of cork here.
I'll pass those around in a minute.
One of the things to look at is just the still wine cork, which is the one on the right, and the champagne cork, which is the one on the left.
If you notice the still wine cork is just made of one piece of cork that's cored out from the bark.
And if you notice these little channels.
These little channels here are called "lenticels."
On the still wine cork, they go this way.
And on the champagne cork, they go that way.
They're oriented perpendicular.
It turns out that they are normal to the isotropic plane in the cork.
If you look at the champagne cork, this plane here is the isotropic plane.
If you think of that being put into a bottle, I think part of the reason they orient it this way is because it gives you a uniform compression against the neck of the bottle and gives a nice seal.
So that's one of the things about corks.
Another thing that's interesting is that cork has a Poisson's ratio equal to zero if you load it in that direction.
Let's see.
Did I not bring that picture?
Maybe I didn't bring that picture.
Hang on a sec.
Nope.
I guess I didn't.
Sorry.
I thought I brought a picture where I had the deformation of the cells when you load it in that direction.
When you load it-- so say the cells are corrugated this way on.
When you load it that way on, it's like having a bellows and folding up a bellows, or unfolding a bellows.
So when you load it that way on, there's no expansion or contraction this way on.
And so you get zero Poisson's ratio.
And if you think of trying to get the cork into the bottle, it's rather convenient to have zero Poisson's ratio because you don't get as much expansion.
As you're pushing it into the bottle, you don't get as much expansion that way out, pressing against it.
In fact, if you compare wine corks with rubber stoppers, this is a kind of typical rubber stopper.
Wine corks are always just cylinders.
In fact, even the champagne corks are cylinders when they start off.
When they put it into the bottle, it's just a straight cylinder.
It gets deformed like that from being in the bottle for some time.
They have straight sides and you can just squeeze them.
There's these funnel things that wine makers have for putting the cork in the bottle.
You can just squeeze them into the bottle top.
You can do that because the Poisson's ratio is zero and because the Young's modulus and the bulk modulus are both small.
But if you look at a rubber stopper, rubber stoppers always have these tapered sides to them.
And that's because the Poisson's ratio of the rubber is 0.5.
As you squeeze it in, it's trying to move out this way.
You couldn't get the stopper in unless it had those tapered sides.
So that's sort of an interesting thing about cork.
Let's see.
Another application of cork is it's used for gaskets for the same sorts of reasons.
It's relatively compliant.
It takes up any slack between two pieces that you want to press together.
It's often used for musical instruments that come in pieces.
Things like clarinets, there'll be a piece of cork-- you a clarinet player?
Yeah?
Yeah.
One of the interesting things about the clarinet is that, if you can see here at the ends, there's a piece of cork there.
And I think there's a piece of cork down there.
And the other pieces mate up with that.
The cork provides a seal.
And the way the cork is cut, it's cut in such a way that those lenticels go radially out like this, which means that the plane of isotropy and the direction that's got the zero Poisson's ratio is that radial direction.
When you're squeezing, say, the second part onto it, the cork does not expand circumferentially.
So as you're squeezing it this way, it doesn't expand that way.
It doesn't wrinkle or anything on your other part.
They use the cork in a particular orientation for that reason, I think.
So it's used for gaskets It's also used because it's got a good friction property.
It's got a property that is taken advantage of in things like flooring and shoes.
Cork has a high friction even if it gets wet.
Some sources of friction are from adhesion, from a surface effect.
Then if, say, the floor gets wet, then you break that, and it could be slippery.
But the source of friction in cork is from an energy loss and dissipation as you're deforming it.
Imagine you have a wheel here.
The wheel is rotating on this cork floor.
And here, a piece of cork is getting deformed as the wheel rolls over it.
As it gets deformed, there's some histeresis loop.
Cork has quite a lot of damping in it.
There's quite a lot of energy lost in that histeresis loop.
What that means is that's characteristic of the cork itself.
It's not a surface effect.
That means that if you use it for floors or for shoes, it doesn't lose that friction and damping when it gets wet.
Here's some measurements we did of the coefficient of friction for cork versus doing it dry and doing it with a liquid, soapy surface.
You can see the soap doesn't make any difference.
Cork is seen as a very attractive material for things like flooring.
It's actually not a cheap material to make your floors out of, but it's an attractive material for flooring.
Part of the reason it's used for floors and for the soles of shoes is because of these friction properties.
Another feature of cork is that it's got very small cells compared to a lot of engineering polymers.
The cells are on the order of tens of microns, whereas many polymer foams, the cells are hundreds of microns or millimeters.
We'll get into this later, but this plot here is really saying that the thermal conductivity of a cellular material depends, in part, on the cell size.
The cell size for foam plastics is in here.
And that for cork is down here.
Because it has a smaller cell size, it has a lower thermal conductivity.
Cork was at one point used to some extent for thermal insulation.
If you go to Portugal, where cork comes from, there's hermit caves.
There were these old hermit, religious people who had holed up in a cave.
And they would line the caves with cork to try to make it a little more insulated, a little more warm.
The other place you see this is if you look at cigarettes, you know cigarettes have that little brown tip on the part that touches your lips?
That's meant to look like cork.
And if you look, it has little dots on it.
The little dots are the little lenticels.
Apparently, they used cork originally in cigarettes as a sort of thermal insulation between the cigarette and your mouth.
So it was used for that too.
And then, one final thing.
Cork's also used, obviously, for bulletin boards.
If you push a pin into cork, then you get this local deformation here.
Here's our pin.
And here's cells locally deformed.
When you pull the pin out, you'll recover some of that deformation because the deformation is elastic.
So the hole will partly close.
So that's my little spiel on cork.
And that's just because it's interesting.
There's no test on cork or anything like that.
OK.
So are we good with cork?
Any questions?
We're OK?
OK. Let me scoot out of there.
Then the next part of the course, I wanted to talk about foams.
Let me just park the cork thing.
Let me pass these corks around.
So you can play with those too.
Oops.
There's little bits of cork.
There you go.
There's the champagne cork, the rubber cork, some little cork layers.
OK.
So the next part of the course, I wanted to talk about foams.
And I want to talk about how we model the mechanical behavior of foams.
If we look at the stress-strain curve for foams, these are some examples for foams made out of different materials with different characteristics.
The polyurethane and the polyethylene here are examples of elastomeric foams, really.
This one here is an open-celled foam.
This one's a closed-cell foam.
Polymethacrylamide is a polymer that has a yield point.
Mullite is a ceramic.
You can see the shapes of these curves resemble the shapes that we saw for the honeycombs.
Right?
They look exactly the same, in fact.
And the mechanisms of deformation in the foams are very similar to the honeycombs.
Even though the foams have a much more complicated geometry, we can use some of the ideas from the honeycombs to understand how the foams behave.
So that was part of the reason for doing the honeycomb analysis.
Let me back up.
These curves here were all in compression.
These curves here were all in tension.
So again, these ones in tension also look like the curves for the honeycombs.
Remember, in tension, we don't get any elastic buckling.
So if the foam was made of an elastomer, we don't see any stress plateau.
If the foam is made of material with a yield stress, then we get a very short yield plateau because of a slight geometrical difference between pulling and compressing the foam.
And if it's a brittle material, then we just get fracturing.
There's going to be some fracture toughness that's going to characterize it.
We can look at the deformation and the failure in these foams and look at the mechanisms.
And what we're going to do is model the mechanisms and not worry too much about the cell geometry.
So we're going to use dimensional arguments here.
Here's a foam in compression.
It was compressed from the top to the bottom.
And you can see this strut that's circled in red.
This is unloaded.
And then, this is after some load.
You can see this has bent somewhat.
And then, you can see this vertical strut here.
As the load gets larger, you can see that strut's buckled.
In an elastomeric foam, you get bending and buckling just the same as we did in the honeycomb.
Then here's a metal foam.
You form plastic hinges in the metal foam.
So here's a cell wall here.
And it's a little bent to start with at zero load.
But you can see it becomes more bent in this image over here.
And here's a cell wall that's more or less vertical.
And you can see that wall buckles.
It's a plastic buckling in this case.
There's a permanent deformation there.
Here's a brittle foam.
And you can see cell walls in this foam fracture.
So this region here is equivalent to this region here.
That little glitch there is the same as that little glitch there.
You can see there's a couple of cell walls here that are fractured.
So we get fracture.
The idea is is that the mechanisms of deformation in the foams parallel those in the honeycombs.
OK?
All right.
So let me write some of this stuff on the board.
We're going to start off by talking about open-celled foams, so foams where there's just solid in the edges, but not in the faces of the polyhedral cells.
Then we'll talk about close-cell foams where there's solid in the faces, as well.
But the open-celled ones are easier.
So we'll start with that.
In compression, we see the same three regimes as we did before for the honeycombs.
There's a linear elastic regime that corresponds to bending of the cell walls.
There's a stress plateau.
And for elastomeric foams, that corresponds to buckling.
For metal foams, that corresponds to the formation of plastic hinges.
And then, for ceramic or brittle foams, that corresponds to brittle crushing, so fracturing of the cell walls.
Then, if you load the foam up to higher strains and higher stresses, eventually you get to densification.
And in tension, just like the honeycombs, for the elastomeric materials there is no buckling.
We can get a stress plateau from plastic hinges if there's, say, a metal foam.
And for a brittle foam, we would get a fracture toughness and brittle fracture in tension.
So the idea is the mechanisms of deformation and failure just parallel what we've seen in the honeycombs.
So we'll start off with the linear elastic behavior.
And we'll start with open-cell foams.
The initial linear elasticity is due to bending of the cell walls.
And if the thickness of the cell edges relative to the length is small, the bending dominates the deformation.
But as the thickness to length ratio increases, then axial deformations can become important too.
What we're going to do is we're going to consider dimensional arguments.
We're going to set the dimensional arguments up so that we replicate the mechanisms of deformation and failure.
But we don't worry too much about the cell geometry.
What I'm going to start with is considering a cubic cell.
And I've arranged the cubic cell so that the cell edges are staggered.
That's going to give me the bending.
The edge length is going to be L. I'm going to say we have a square cross-section, t squared.
Here's our idealized model here with a cubic cell.
All the members have a length, l. All of them have a square cross-section, t squared.
That's an open-cell model.
We've got just solid on the edges and nothing on the faces.
The idea is that if I bend that, or if I load that in compression, so I apply, say, a stress out here that puts forces on those members there, because I've staggered these cell walls with these ones here, we're going to get bending in this cell edge here.
That bending is going to be what we model.
I'm going to set this up so that one thing is proportional to something else.
These relationships are going to be true regardless of the cell geometry.
So I could have picked a tetrakaidecahedra if I wanted to, and I would have had these same relationships.
I'm just picking a cubic thing because it's easier to think about.
So first of all, we look at the relative density.
Remember, the relative density is the volume fraction of solid.
So it's the volume of the solid over the total volume.
And that's going to go as t squared l over l cubed, or just t over l all squared.
You remember for the honeycomb, the relative density went linearly with t over l. For the open-celled foam, it goes as t over l squared.
The moment of inertia in this case is going to go as t to the fourth.
Remember, we have a square section, t squared.
So if it's bh cubed over 12, b is t, h is t. It's going as t to the fourth.
Then what I'm going to say is the stress is going to go as F over an area length squared.
OK?
So if I look at my little square thing here, I look at having my force here.
Here we have a force f. And it's acting over an area that's somehow related to l squared.
Right?
I don't know exactly what that constant is, and I'm going to not try to calculate that.
But it goes as F over l squared.
Similarly, I can write that the strain is going to go as delta over l. So the strain is going to go as this bending deflection here, that delta divided by the height of the cell.
And that's also l. Then I also know from structural mechanics that delta is going to go as Fl cubed over E of the solid and I.
Then I'm just going to put all these things together to get the modulus.
The modulus of the foam is going to go as the stress over the strain.
If I plug in what I have for the stress, it's F over l squared.
If I plug in what I have for the strain, it's delta over l. So this is F over l and delta.
I'm going to replace delta by Fl cubed over Es.
I'm going to use, instead of I, I'm going to use t to the fourth.
Then the F's are going to cancel out.
I've got that the modulus goes as the modulus of the solid times t over l to the fourth power.
Then I can put that in terms of the relative density.
It's going to go as the relative density squared.
So I can summarize all of this by saying that the Young's modulus of the foam is some constant C1, I'm going to call it, times the modulus of the solid times the relative density squared.
OK.
So this has the same kind of form as those equations we had for the honeycombs.
Right?
There's a solid-cell wall property.
The solid module's here.
For the honeycombs, I put it in terms of t over l. But the t over l was related to the relative density.
How much solid you've got is reflected in the relative density.
And then, this constant C1 wraps up all the geometrical constants that I've said, one thing's proportional to another, and something else is proportional to another.
C1 just wraps up all of those.
OK?
I'm just going to say here C1 includes all the geometrical constants.
We have to get C1 by looking at data.
If we look at data for the Young's modulus, we find that C1 is just about equal to one.
People have also done more sophisticated analyses than this.
There's a group of people who looked at doing a full-scale, structural analysis of an open-celled tetrakaidecahedral cell.
Remember, I said they pack to fill space.
So you can look at a unit cell.
They also made their cells such that the thickness along the length of the cell was not constant.
The thickness varied as something called a "plateau border."
If you have a foam that's made by surface tension, the edges will tend to have these plateau borders.
And the thickness will vary along the length of the edge.
So when they did all this whole, complicated thing, they could calculate a value for C1.
They calculated 0.98.
So it's very close to 1.
I'll say analysis of open-cell tetrakaidecahedron cells with these plateau borders give C1 equal to 0.98.
OK.
So that's the Young's modulus.
We can also look at the shear modulus.
The shear modulus is also going to be controlled by bending of the cell walls.
And so the shear modulus is just going to be some other constant times Es times the relative density squared, so a similar kind of relationship.
It's just a different constant.
And if the foam's isotropic, and if the Poisson's ratio is a third, then C2 is equal to 3/8.
Remember, if we have isotropy, then the shear modulus is equal to E over 2 1 plus nu.
And so you can get the C2 from that if you say nu is a third.
Then we can also get Poisson's ratio for the foam.
If the foam is isotropic, so we'll say for an isotropic foam here, nu is equal to E over 2G minus 1.
That's just rearranging this expression here.
And because E and G both depend on the relative density squared, they both depend on Es squared, that's all going to cancel out.
So this is going to be equal to C1 over 2 C2 minus 1.
So that's going to equal to a constant.
That constant's going to be independent of whatever material the foam is made from and the relative density.
The constant just depends on the cell geometry.
Remember, in honeycombs we found the same thing.
The Poisson's ratio for the honeycombs only depended on the cell geometry.
It didn't depend on the solid properties.
It didn't depend on the relative density.
So this is an exactly parallel thing here for the foams.
Yeah?
I have a silly kind of question.
What is the difference between foam and the honeycombs?
Oh.
The honeycombs have cells in a plane, and they're prismatic in the third direction.
And the foams have polyhedral cells.
You know what a tetrakaidecahedron, a 3D, polyhedral cell.
OK?
The honeycombs are prismatic.
And the foams have polyhedral cells.
OK?
Are we good?
OK.
So there's a couple more interesting things about Poisson's ratio.
The same way we can make honeycombs with negative Poisson's ratios, we can also make foams with negative Poisson's ratios.
They do it the same way as for the honeycombs, really.
The honeycombs had negative Poisson's ratios if the cell walls looked like this, this sort of arrangement.
So that sort of a thing.
We said that theta was negative for the honeycombs.
And if you invert the cell walls on a foam, you also get negative Poisson's ratios.
And the way they do that is they take a thermoplastic foam, and then, they load it hydrostatically.
So they compress it in all three directions.
And they smush the cells in on each other.
And then, they heat it up to a temperature above the glass transition temperature while it's still smushed.
And then, they cool it down.
So they end up with that structure frozen in.
And if they do that, they get a negative Poisson's ratio.
I'll just say they invert the cell angles analogous to the honeycomb.
They load the foam hydrostatically and heat to a temperature above Tg.
And then, they cool and release it.
I have a photograph here of a foam with a negative Poisson's ratio.
You can see how the cells have been smushed in.
It's the equivalent of the way it's done for the honeycomb.
OK?
Are we good?
OK. That's the linear elastic moduli for open-celled foams.
The next thing I wanted to do was closed-cell foams.
If we look at a closed-cell foam, we can idealize it in this kind of a way here.
I've set it up so that the edge thickness is different from the face thickness.
That's really representing the fact that in foams, many foams are made using a liquid, and the foaming is controlled by surface tension.
Often, the surface tension draws material into the edges and away from the faces.
So the faces tend to be thinner than the edges.
When we have deformation of the closed-cell foam, we've got bending of the edges the same as we did for the open-celled foam.
But the faces can stretch.
So they can have an axial stretching.
You can think of that as a cell membrane stretching here.
So imagine if I either pull on the foam or I compress it, there's going to be some axial load in the faces.
So when we analyze them, we have to account for both bending of the edges and axial deformation in the faces.
I'll just say we have edge bending as in open-cell foams.
And we also get a face stretching.
Another thing that can happen in the closed-cell foams is we can get compression of the gas.
In an open-cell foam, the gas can move from one cell to the next.
But in a closed-cell foam, the gas is trapped.
And as the volume of the cell changes, the gas gets compressed.
So we have another effect here.
So we'll say for polymer foams, surface tension tends to draw material to the edges during processing.
We define two thicknesses, one for the edge and one for the face.
And then we apply a force to this cubic structure.
And we can do an analysis a little bit like what we did for the open-cell foam.
Let me rub all this off.
OK.
I'm going to set this up a little bit differently than for the open-celled foam.
We're going to do a work argument.
We're going to look at the external work done by the force F going through a deformation, delta.
And that's going to have to equal the internal work done by the edges bending and by the faces stretching.
So let me set that up.
We're going to say the external work done, that's going to be proportional to F times delta.
So delta is how much the whole thing is going to deform.
Then, I've got internal work from bending of the edges.
That internal work is going to be proportional to F over delta times delta squared.
I'm going to end up with an expression where everything's in delta squared.
So I want to keep the delta squared there for now.
And F over delta is the stiffness.
That's going to go as E of the solid times I over l cubed.
I is going to go is Te to the fourth here because it's I of the edges.
I've also got internal work from stretching of the faces.
That internal work is going to be-- I'm going to run into my other equations here.
Let me put it down a little.
That's going to go as the stress on the face times the strain in the face times the volume of the face.
Or I could write that, instead of stress of the face, I can put it in terms of Hooke's law and say it's E of the solid times the strain in the face squared times the volume of the face.
And I can replace the strain in the face by delta over l. So it's delta over l squared.
And then, the volume of the face is going to be t of the face times l squared.
Then I want to balance the internal work and the external work.
I can say F times delta is going to equal some constant I'm going to call "alpha" times E of the solid times t of the edge to the fourth, that's this guy up here, over l cubed times delta squared plus some other constant I'm going to call "beta" times E of the solid times delta over l squared tf l squared.
So far, I've got this in terms of the force I'm applying.
But I want to get a modulus of the foam out of this.
I want to relate this force here to the modulus of the foam.
I can say the modulus of the foam is going to be related to F over l squared, that's the stress, divided by the strain, delta over l. I can write the force is proportional to the modulus times the deflection, delta, and times the member length, l. And then, I'm going to plug that guy into this expression up here.
And I'm going to get a delta squared on the left.
And I'm going to get delta squareds in each of the right-hand terms.
And so I'm going to cancel out the deltas.
If I put all that together, I have the modulus in the foam times delta squared times l. There's a delta here, and there's a delta there.
That gives me delta squared.
And that's going to equal alpha Es Te to the fourth over l cubed times delta squared plus beta times Es times delta squared tf, if I cancel out one of those l squareds.
So here I can get rid of the delta squareds in all these terms.
And if I just divide by l, I'm going to have the modulus of the foam.
I get a term here that it goes as alpha E of the solid times t of the edge over l to the fourth power plus beta times E of the solid times tf over l. Are we good?
OK. And what I'd like to do is instead of putting it in terms of te and tf, I'd like to put it in terms of the relative density.
I'm going to look at two limits.
I'm going to say if we just had open cells, if there were no faces on the membranes, if we just had open cells, and we had a uniform t, then the relative density would go as t over l squared.
If I just had closed cells, and I had a uniform t, then the relative density just goes linearly with t over l. The relative density is the volume fraction of solids.
It's the volume of the solid over the total volume.
For a closed-cell foam with a uniform t, the volume of the solid is going to be t times l squared.
And then, the volume total is l cubed.
So it's t over l. Now, I'm going to define one more thing.
If I say that phi is equal to the volume fraction of the solid in the edges, then I can say te over l is some constant times phi to the 1/2 times the relative density to the 1/2.
And I can say tf over l is equal to some other constant, c prime, I'll call it, times 1 minus phi, that's how much is in the faces, times the relative density.
Then I could combine all of this.
Can I put it in here?
Maybe I'll just stick it down here.
Hang on.
OK.
Put it up here.
This is my final expression here.
And this term here arises from the edge bending.
This term here arises from the face stretching.
So I think I should wait a bit for you to catch up.
The idea is the edges are bending.
The faces are stretching.
We're looking at the work done by deforming the edges and the faces and equating that to the work done by the externally applied load, f. OK?
That gives us two terms, one that accounts for the edge bending, and one that accounts for the face stretching.
To be comprehensive, we want to take into account the compression of the gas, as well.
So there's one more term I'm going to add on to that.
Typically the gas effect is only significant for elastomeric foams.
If you had a metal foam or a ceramic foam that was closed-cell, it wouldn't really contribute much.
But just to be complete, we'll go through this.
The idea here is we say we've got a cubic element of the foam.
Initially, it has a volume, v0.
If we apply a stress, a uniaxial stress, there's some change in the volume of the foam.
So there's some volume, v, after we compress it by some strain, epsilon.
If we can figure out what the volume is relative to the initial volume, we can figure out how the change of the amount of gas goes from the initial state to the compressed state.
Then we can use Boyle's law.
The idea is P1V1 equals to P2V2.
Then we can figure out, using Boyle's law, what the pressure must be.
And then, that pressure is what's contributing to the modulus.
When I do this, I'm going to write down some equations that just have the main results.
There's a whole bunch of algebra just to get from one to the other.
And I'm not going to write them all down.
When I write the equations, don't panic if it's not obvious how you get from one to the other.
In the notes that I'm going to put online, there's all the details of how you get from one to the other.
The idea is that we start with a cubic element of foam.
Initially, before it's loaded, it has a volume V0.
Then we apply a uniaxial stress.
And we say the axial strain in the direction of the stress, I'm going to call epsilon.
Just from the geometry, you can calculate the deformed volume.
So after you load it, the volume is V. And that volume on loading, V, divided by the initial volume, V0, you can show.
It's fairly straightforward.
It's 1 minus epsilon times 1 minus 2 times the Poisson's ratio of the foam.
Here, I'm taking compressive strain as positive.
And if you do this whole volume thing, you'll get terms in epsilon squared and epsilon cubed.
But because if it's linear elastic, epsilon is going to be small.
I'm going to ignore the epsilon squared and the epsilon cubed terms.
That's the total volume of the foam after and before the compression.
And then, what I want is the volume of the gas.
And the volume of the gas is just going to be the volume minus the volume of the solid.
And I can get that by just subtracting off the relative density.
So Vg over Vg0.
Again, Vg0 is the volume of the gas initially.
Vg is the volume of the gas when I'm compressing it.
Remember, the relative density is the volume fraction of solids.
So I'm essentially subtracting out the amount of solid to get the amount of gas.
Then we can use Boyle's law.
Here, p is going to be the pressure after the strain and p0 is going to be the initial pressure.
The building seems rather making unhappy noises for some reason.
I'm not sure what that's all about.
There'll be some initial pressure in the gas even before the strain, or before the stress is applied.
That's p0.
So the pressure we need to overcome is p minus p0.
Again, I'm missing out a bunch of steps.
But using that expression and this expression and this expression, you could find that p prime is equal to p0 times epsilon times 1 minus 2 nu divided by 1 minus epsilon 1 minus 2 nu minus the relative density.
Then the contribution of the gas to the modulus you can get by just taking the derivative of that pressure with respect to the strain.
So remember, modulus is stress over strain.
It's the same kind of idea.
So I'm going to call it E star g for the contribution from the gas.
So that's going to dp prime d epsilon.
And that's going to equal p0 times 1 minus 2 nu over 1 minus the relative density.
OK?
As I said, I'll scan the pages that have the details and put it online.
The final expression we get combines all of these things.
The modulus of the foam relative to the solid is phi squared rho over rho s squared plus 1 minus phi, that's the amount of material in the faces, times the relative density and then plus this gas compression term.
Like I said, for most foams, the gas compression is negligible.
So if p0 is equal to the atmospheric pressure, so about 0.1 megapascal, then the gas term's negligible, except for elastomeric foams.
So that's the Young's modulus.
Then we can do a similar thing for the shear modulus.
And the thing to notice in shear is that when you shear something, there's no volume change.
So if you shear it and there's no volume change, there's no gas compression.
There's no pressure built up because there's no change in the volume.
So for the shear modulus, you just have the first two terms.
And if the foam is isotropic and you use a third, then this constant here is going to be 3/8.
And then, Poisson's ratio is just going to be a function of the cell geometry, again.
And roughly, we could say it's going to be around a third.
OK. Are we good?
Let me wait a little bit.
So the idea is we're looking at the deformation of the bending in the cell edges, and the stretching in the cell faces, and the gas compression.
And we're accounting for those different terms.
The next thing is to compare these model equations with some data.
And here's data for Young's modulus.
Here, we're plotting the relative Young's modulus, so the modulus of the foam divided by the solid against the relative density.
Again, these are log-log scales.
On this plot, everything with open symbols is an open-cell foam, and everything with filled symbols is a closed-cell foam.
Here's our equation for the open-celled foams, the simplest thing that goes with density squared.
And that's that thick line there.
And you can see these are all open-cell foams.
There's some more up here.
So that gives a reasonable description of that data.
And then, these are two lines for closed-cell foams for different values of phi, so different values of the amount of solid in the edges.
And you can see all the filled symbols are the closed-cell foams.
And they're in between this line here and this line here, basically.
So that gives a fairly good description for the modulus of the foams considering how crude this model is.
We're not trying to account for the cell geometry.
We're just modeling the mechanisms of deformation and failure.
And then, here's a similar plot for the shear modulus.
Here's for open-cells, 3/8 times the relative density squared.
These are open-celled foams down here.
These are closed-cell foams up here.
If the amount of material in the faces is small, then you would just get the shear modulus varying with the relative density squared.
Then here's data for Poisson's ratio.
We don't really know what the constant is because we don't know what all these geometrical parameters are.
But here's a value of a third.
And you can see there's a lot of scatter here.
This is like less than 0.2.
This is more than 0.5.
And the scatter really represents differences in the cell geometry.
If the foams were, say, anisotropic, and the cells were stretched in one direction, then you would get different values of the Poisson's ratios.
So that's the Poisson's ratios there.
I'm thinking I'm going to stop there because that seems like enough equations for today.
And then, next time we'll start doing the stress plateau and we'll figure out the elastic collapse stress from buckling and a plastic collapse stress from yielding, and so on, and so on.
We'll finish doing the modeling next time.
And we'll probably start doing a little bit of other stuff on foams next time.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
All right.
And I really wanted to show you my little hook video and I downloaded it so I thought we'd start just by watching that and then I'll pick up about modeling phones.
So this takes like nine or 10 minutes, but I just thought it was cute.
And I made it and I want you to see it.
So let's do that to start.
[VIDEO PLAYBACK] We're here at the Harvard University Botany Library, looking at a first edition of Robert Hooke's Micrographia, published-- How do I get rid of the bar, Greg?
Oh, there it is.
show the microscopic structure of materials.
And it has a number of remarkable drawings in it.
Here we see drawings of silk.
These are two different silks.
On the top here, we have a fine-waled silk.
And in this more details drawing down here, you can see the patterned weaving of the silk.
The bottom image here is a drawing of watered silk.
And over here, there's another higher magnification image.
And you can see the pattern here is more sharply angled.
And it appears that this sharper angle here gives the different texture to the surface finish of the silk.
So here we see a drawing of charred wood.
And one of the things I find interesting about this drawing is how similar it is to modern electron micrographs which we've seen before.
And in this drawing, we can see two of the main features.
We see these small cells, which are fibers that provide structural support to the tree.
And we see these larger cells here, which are vessels which allow fluids to go up and down the tree.
And here we see a drawing of the surface of a rosemary leaf, with the unexpected, tiny, little bars.
And this is something that you can only see with the microscope.
You wouldn't expect to see those when you just feel the surface of the rosemary leaf.
So it's kind of interesting that with the microscope, you can see these features that are invisible to the naked eye.
One of the main themes of material science is that the property of materials are related to their structure.
And so being able to see the structure at a microscopic scale is very helpful.
And today, we can even see the structure at the atomic scale.
Robert Hooke understood this idea.
And in the description of the cork, Hooke states, "I no sooner discerned these-- which were the first microscopical pores I ever saw-- but methought that I had with the discovery of them, perfectly hinted to me the true and intelligible reason for all of the phenomena of cork."
So what he's saying here is that by looking at the structure and looking at the cells here in the drawing, he thinks he can understand the properties of cork or the phenomena of cork.
What was it about Robert Hooke that allowed him to make this book?
Why was it him and not somebody else?
Well, Robert Hooke had kind of an interesting history.
He grew up on the Isle of Wight.
And as a boy, he loved making drawings.
And he got quite skilled at making drawings.
The other thing was, he loved making models of things.
He made models of ships.
He made a wooden clock that was a working clock when he was a kid.
And as a teenager, he moved to London.
And he became an apprentice to Sir Peter Lilley, who was a famous painter of the time.
So his drawing was good enough that he would be working with a very well-known painter.
After he did that, he went to the Westminister School.
And he studied classics.
He studied mathematics.
But he also learned to use a lathe.
And this was also very helpful in him making various sorts of apparatus.
And as a student at Oxford, he worked in the lab of Robert Boyle.
And his job in that lab was to develop scientific apparatus.
And he did things like he built pumps that allowed Robert Boyle to do the experiments that led to Boyle's Law.
When he returned to London after Oxford, he became the Curator of Experiments at the Royal Society.
And one of the things he did was he got a microscope.
He improved that microscope, increasing their magnification, which was what allowed him to make the beautiful drawings that we see today.
And here in the preface of the book, we see that he even made a drawing of his microscope.
So this thing down here-- this is Robert Hooke's microscope.
The development of new microscopes with higher and higher magnifications continues to this day.
Scanning electron microscopes were invented in the 1960s.
And today, we have transmission electron microscopes and atomic force microscopes with even higher magnifications.
At these higher magnifications, we can see details that Hooke was unable to see because of the limitations of the microscope that he had-- the optical microscope.
But it's interesting to see today the images we see in a scanning electron microscope at a similar magnification to those that Hooke saw in his optical microscope.
And it's remarkable to see how many of the features that we see in these much more fancy microscopes that he was able to capture in his drawings with his simple optical microscope.
So here we have a picture of cork.
We have Hooke's drawings showing two perpendicular planes.
We also have this nice, little drawing of a cork branch here.
Cork is the bark from the cork oak tree.
And in Hooke's drawings of the microstructure, we can see these cells here are roughly box-like.
They're more or less rectangular.
And these cells here look more or less circular.
So there's these two different perpendicular planes in the cork.
And when we look at these scanning electron micrographs, we can see very similar structure.
There are some cells that are roughly boxlike, and others that are more or less hexagonal or roughly rounded.
One feature that Hooke was not able to see, though, that you do see on the scanning electron micrographs, is the waviness in the cell walls.
And that was because the resolution of his microscope was insufficient to see that level of detail.
And here in this illustration on the bottom here is a drawing of sponge.
And when we look at the scanning electron micrograph, we see that the structure is remarkably similar to what Hooke has drawn.
So here we have Hooke's drawing of feathers.
And we can see he's made several drawings at different length scales.
And if we look at this one here, we see the barbule.
And you can see these little hooked regions there.
And those hooks lock into the little feathers over on this side over here of the adjacent barbule.
And in the higher magnification picture, you can see on one barbule, there's hooks on one side but not on the other.
And it's this hooking of the two sections together that allows the feathers to maintain a smooth surface for the wing when the bird is flying.
And you can see the same sort of thing when you look at the electron micrograph.
So you can see the little hooks on one side of the barbules.
And you can see how they interconnect together with the next barb.
One of the most reproduced images from Hooke's book is that of the flea-- this image we see here.
And you can see why.
It's a gorgeous image.
And it shows details that people had never seen before.
People were amazed to see that the little flea that they might have found on their dog or something was actually made up of this compound body, with all these little plates and little hairs here.
And you can see these little tiny claws on the legs, and the legs have all these hairs.
Nobody had any idea that this is what a flea actually looked like.
And so it was an amazing drawing.
And it was something that people were just stunned by when Hooke's book came out.
And if we look at a modern electron micrograph, we can see it's remarkably similar if we look at the same magnification.
So Hooke showed many of the same details, showed some of the same hairs on the legs, showed the same sorts of plates, showed the claws at the ends of the legs.
And our modern image is probably from a different species of flea.
We don't know what species of flea that Hooke actually looked at.
But you can see there's a tremendous similarity between the two images.
And it's remarkable how many of the features that Hooke was able to capture in his drawing.
And here we have the compound eye of the fly.
And this, again, was astonishing to people in Hooke's day.
And even today, people look at this image, and they're pretty amazed at the detail in this drawing.
And again, we can compare this with a modern electron micrograph.
And again, you can see the similarities between what Hooke saw and what we see in a modern scanning electron microscope at a similar magnification.
In the 1980s, atomic force microscopes were invented, which have a resolution down to tens of nanometers.
And today, there's transmission electron microscopes, which allow you to see the atomic structure.
So for instance in a crystal lattice, you can see the individual atoms and the regular crystal structure.
Today, most experimental studies of materials include photographs of the microscopic structure of the material taken through some sort of microscope.
And the remarkable thing is that all of these studies really trace back to this book here that we're looking at today-- to Robert Hooke's Micrographia.
[END PLAYBACK] There you have it.
So I just thought that was kind of cute.
You might enjoy that.
So that was that.
All right, let's get out of there.
Stop.
So let's go back to the foams.
So I think last time, we got as far as talking about the linear elastic behavior of foams and modeling that.
But we didn't quite get to looking at the compressive strength of the foam.
So I think we got as far as comparing the models with these equations here, and these plots of the data.
And what I wanted to pick up with today was looking at the compressive strength.
And we'll look at the fracture toughness as well in tension.
So we're going to start with nonlinear elasticity and the elastic collapse stress.
So if we have an open-cell foam, the derivation for the elastic collapse stress is really pretty straightforward.
We say the elastic collapse occurs when the cell walls buckles.
So in this schematic here, you can see the vertical cell walls have buckled.
And so there's going to be some Euler load that's related to that buckling.
And that's just the usual Euler load-- n squared, the n constraint factor, pi squared E. This is going to be E of the solid, I over l squared-- the length of the member.
And then the stress that corresponds to that is just going to be proportional to that buckling load over the area of the cell, which is just l squared, so just P critical over l squared.
So that just goes as Es.
I is going to go as t to the fourth, because we have that square sectioned member.
And now this is going to be l squared.
And that's an l squared.
So that's l to the fourth.
And so if I combine all of that together, I can say that the elastic buckling stress is going to be some constant-- and I think we're up to C4 now-- times the Young's modulus of the solid times the relative density of the foam squared.
So that's our equation for the elastic buckling stress.
And if you compare this with data, you can make an estimate of what C4 is.
And we find that C4 is about 0.05.
And you can also say that 0.05 really corresponds to the strain at which the buckling occurs.
Because the Young's modulus goes as the constants 1 times Es times the relative density squared.
So the strain's just going to be the stress over the modulus.
So that does correspond to the strain.
So that's saying that buckling compressive stress occurs at a strain of about 5%.
So that's open cells.
And then if we look at closed cells, if you recall when we looked at the moduli we looked at a couple of extra terms.
One was associated with face stretching for the modulus.
And the other was associated with the compression of the gas.
For the buckling, the faces don't really contribute that much, because typically the faces are very thin relative to the struts.
And because they're so thin, they buckle at a much lower load, and they don't contribute too much.
So we're not going to worry about that contribution.
So I'm just going to say that the thickness of the face is often small compared to the edges.
And that really is from the surface tension in processing that draws material away from the face and into the edges.
There can be some contribution from the internal pressure.
So if the internal pressure is greater than atmospheric pressure, then the cell walls are pre-tensioned, and you'd have to account for that.
So the buckling would have to overcome that pressure as well.
So then you would have the buckling stress would just be what we have up there-- C4 times Es times the relative density squared.
And then we just add on that factor P0 minus P atmospheric.
The thing with the gas which tends to affect more than the buckling stress, though, is the post-collapse behavior.
So let me just show you a couple of things here.
So here's some data for the elastic collapse stress.
And you can see on the y-axis, we've got the stress normalized by the Young's modulus of the solid.
And on the x-axis, we've got the relative density.
And that solid line there-- sort of solid, dark line-- is that equation there, which is the same as this one up here.
And you can see the data lie fairly close to that.
But what's interesting is if you look at the-- why is this not working?
Maybe my batteries finally died.
If we look at the post-collapse behavior, you can see if these are the stress-strain curves, they're not flat here.
They have some rise to them.
And this is a closed-cell foam.
And you can imagine as you're compressing the closed-cell foam, you're reducing the volume of the cell.
And as you doing that, you're increasing the pressure inside the cell from the gas.
And you can calculate what that is.
And I'll do that in a second.
And if you subtract off that gas pressure contribution, that works out to this line here.
Then these lines will be more flat, like this.
And we already really pretty much worked out that gas contribution.
So I'll just say for the post-collapse behavior, the stress rises due to the gas compression.
And that's as long as the faces don't rupture.
So if you have an elastomeric foam, typically they don't rupture.
And what we had worked out before was that that pressure-- we called it P prime-- it was P0 minus P atmospheric-- that was equal to P0 times the amount of strain, epsilon, times 1 minus 2 times the Poisson's ratio divided by 1 minus epsilon times 1 minus 2 nu minus the relative density.
And once you get to the buckling stress, then the Poisson's ratio becomes 0.
So if you take a foam-- so I brought a little foam in so you can play around with this one-- so if you take a foam like this and you compress it, once you've buckled it like this, it's not getting any wider this way.
And part of the reason for that is you've got all these pores in here.
And the cells just collapse into the pores.
They don't really need to move out sideways.
So you can smush that yourself, and try to convince yourself that the Poisson's ratio is just 0.
Yes, Matt
[INAUDIBLE] I guess I want to measure [INAUDIBLE] the gas contribution?
Yes, so there is a strain rate effect with these things.
But I wasn't going to get into that here.
If you look in the book, it's described in the book.
So I think there's two things.
One is that the solid itself can have a rate dependency.
And then there could be something connected
[INAUDIBLE]
Yeah, I mean, I'm not going to go into that here.
But one could look at that.
So let me just write down one more thing here, because if we let nu be 0, then this thing here becomes simpler.
So we could say the stress post collapse as a function of strain would be our buckling stress and then plus this factor here.
So that curve on the bottom over here-- if this is the stress-strain curve-- this little dashed line here-- that's the gas contribution.
And that is this term here.
So you can kind of see how the shape of the curves reflects that gas contribution.
And when you subtract it out, you get pretty much a horizontal plateau over here.
Are we happy?
Yeah?
[INAUDIBLE]?
This is for the closed cell.
Because the closed cell are the ones that are going to have the gas pressure.
If it's open cells, the gas can just move out of the cells
[INAUDIBLE]?
Oh, sorry, that was to show you that the Poisson's ratio was 0.
And that's true for both of them.
So then we can look at the plastic collapse stress.
Say we had a metal foam.
And we do a calculation a little bit like the one we did for the honeycombs, too.
So we say the failure occurs when the applied moment equals the plastic moment.
And the applied moment is proportional to the applied stress times the length cubed.
So I'm going to call that applied stress-- our strength sigma star plastic times the length cubed.
So if you think of, say, the little schematic up here, the force is going to go with stress times the length squared.
And the moment's going to force times the length.
So it's the stress times the length cubed.
And then the plastic moment goes as the yield strength times the thickness cubed.
And then if I just combine those, I get that the plastic collapse stress in compression is another constant-- I'm going to call it C5-- times the yield strength times the relative density to the 3/2 power.
And if we look at data, we find that the constant is about equal to 0.3.
And if I go to the next slide, here's a plot of the yield strength or the plastic collapse strength of the foam divided by the yield strength of the solid, plotted against the relative density.
And that dark, bold line is this equation here.
And you can see the data lie pretty well on that line.
There's one data set that's a little bit above the line.
But you can see the slope of that data set is still about 3/2.
OK, and the same as in the honeycombs, we could say that we can get elastic collapse before the plastic collapse if we were at a low density.
You can get the same thing in the foams.
And you calculate out what the critical relative density is for that the same kind of way.
So we can say we can get elastic collapse precedes the plastic collapse if the elastic buckling stress is less than the plastic collapse stress.
So all we do is make those two things equal to figure out the critical relative density where you get the transition from one to the other.
So the relative density has to be less than 36 times the yield strength of the solid over the Young's modulus of the solid squared in order to get buckling before yielding.
And let's see, where can I put that?
So for rigid polymers, that ratio of the strength of the solid over the modulus of the solid is about one over 30.
And so the critical relative density for the transition is about 0.04.
So you'd have to have a pretty low-density foam, but it's possible.
And for metals, that ratio is about 1/1,000.
And then the critical transition density is less than 10 to the minus 5.
So essentially, it never happens for the metal foams.
And then for the closed-cell foams, we could include the terms for face stretching and for the gas.
But in practice, the faces don't really contribute very much.
And typically for foams like say metal foams or a rigid polymer that had a yield point, the faces rupture.
And then if the faces rupture, then you don't get the gas compression term, either.
So I'll just write the full thing down.
But typically, you don't need to use it.
So the first term would be from the edges bending.
And the second term would be from the faces stretching.
And this would be from the gas.
But in practice, the first term is really the only one that is significant.
So for closed-cell foam, this equation works pretty well, too-- the same one as for the open-cell foams.
OK, so if we had, say, a ceramic foam that was brittle, there'd be a brittle crushing strength.
And then we get failure when the applied moment M is equal to the fracture moment Mf.
And this works very similar to the plastic yield strength.
So we find the applied moment goes as the global stress times the length cubed.
And the fracture moment goes into the cell wall strength times the cell wall thickness cubed.
So the brittle crushing strength goes as another constant-- let's call it C6-- times the wall strength times the relative density to the 3/2 again.
And C6 is about equal to 0.2.
And typically, ceramic foams have open cells.
So I'm just going to leave it at the open-celled formula there.
So there's one last thing for the compressive behavior, and that's the densification strain.
And we just have an empirical relationship for the densification strain.
So if you compress the foam and you get to very large strains, then the cell walls start to touch, and the stress starts to rise steeply.
And there's some strain at which that occurs.
And we call that the densification strain.
And in the limit, the modulus at that point would go to the modulus of the solid.
If you could completely squeeze all the pores out, the stiffness of that would go to the modulus of the solid.
And you might expect that that densification strain is just 1 minus the relative density, but it actually occurs at a slightly smaller strain.
So in a large compressive stress, or strain, I guess we could say, cell walls touch, and we start to get this densification.
So the modulus in the limit would go to the modulus of the solid.
And you might expect that the densification strain was just equal to 1 minus the relative density.
But it occurs at a little bit less than that.
So empirically, we find that it's just 1 minus 1.4 times the relative density.
And then I have this plot here, which is really just fitting a line to that data for densification strain.
So those equations describe the compressive stress or the compressive behavior of the foam.
So we've got the moduli, we've got the three compressive strengths, and we've got the densification strain.
So what we're going to do later on in the course is we'll use those models to look at how we can use foams and things like sandwich panels and looking at energy absorption.
And we'll also look at these equations in terms of some biomedical materials-- looking at trabecular bone, and looking at tissue engineering scaffolds.
So there's one last property I wanted to go over, and that's the fracture toughness.
So if we were pulling the foam in tension, and we had a crack in the foam, we'd want to know what the fracture toughness would be for a brittle foam.
And this follows the same sort of argument as we had for the honeycomb.
So all of these equations really are just following the same kinds of arguments.
But you can kind of see how having the honeycomb calculations makes it easier to do the foam ones.
So we'll do the fracture toughness calculation, and then I want to talk a little bit about material selection and selection charts for foams.
So that's less equation-y.
OK, and we're just going to look at open cells here.
So imagine we have a crack of length 2a.
And we have some remote stress applied, so remote tensile stress, so I'm going to call that sigma infinity-- the far-away stress.
And then we have a local stress on the cell walls.
I'm going to call that signal local.
So I have a little schematic that kind of shows what we're doing.
So we're pulling on it.
There's some crack.
The crack length is large compared to the cell size.
And we want to know what the fracture toughness is.
So we can say from fracture mechanics the local stress is going to be equal to some constant times the faraway stress times the square root of pi a over the square root of 2 pi r. And that's at a distance r from the head of the crack tip.
And if we look at our little schematic here, we could say it's hard to say exactly where the crack tip is, but it would be somewhere in here.
And we'd say this next unbroken cell wall is a distance r ahead of the crack tip.
And that r is going to be related to l. It's going to be some function of l. So I can say the next unbroken wall ahead of the crack tip at some distance r is going to be related to l. And that's subject to a force, which is going to be the local stress times l squared.
So that force is going to go as local stress times l squared.
And the local stress-- I can substitute this thing here in-- that's going to be proportional to the faraway stress.
And I'm going to get rid of the pi's.
And I'm going to substitute for r. I'm going to put in l. So it's going to be proportional to the faraway stress times the root of a over l and times l squared there.
And then we're going to say, again, the edges are going to fail when the applied moment equals the fracture moment.
And the fracture moment is going to go as the modulus of rupture of the cell walls times t cubed.
And the applied moment is going to go as f times l. And I've got f from up there, so that goes as the faraway stress, sigma infinite, times the root of a over l. And now I've got l cubed, because there's an l squared there and there's an l down here.
And then if I just equate those, then this is going to go as sigma fs times t cubed, like that.
So then I can say the fracture strength is equal to my faraway stress.
That's going to go as my modulus of rupture times the root of l for a times t over l cubed.
And then my fracture toughness is going to be this tensile stress times the root of pi a.
So there's going to be some other constant here, which I'm going to call that C8.
We've got the modulus of rupture of the solid.
I've got the square root of l, and I'm going to multiply it by pi so it's like other fraction mechanics kinds of equations.
And then we multiply that times the relative density to the 3/2 power.
And here, if we look at data, we find that that constant is about equal to 0.65.
And here's another one of these plots.
So here I've normalized the fracture toughness of the foams by the modulus of rupture of the cell walls times the root of pi l. So I've taken the cell size into account here, and I've plotted against the relative density.
And that equation there is the same as this equation I've got down on the board.
And this is the only one of the properties that we've looked at that depends on the cell size.
There's a cell size dependence here.
All right, so I think that's all the modeling of the foams.
Are we good?
I gave you a lot of equations.
We're good?
All right.
So I want to talk about how we might design foams to improve their properties.
And then I want to talk about how we might select foams for certain applications and look at selection charts.
So when we've been talking about the foams, especially the open-cell foams, we've been saying their deformation is largely by bending of the cell edges.
And if we could do something to increase the stiffness of the edges or the strength of the edges, then that would increase the overall properties of the foam.
And there's a couple of ways to think about doing that.
So the foam properties-- if the foam is controlled by bending of the edges, and the edges have some flexural rigidity, EI, if we could increase that EI of the edges, we would increase the properties of the foam.
And one way to do that is by making the edges hollow.
So if we had hollow edges, and you had a tube, then that would increase the EI.
And we can work out how much it's going to increase them.
And I have a little example here of-- a natural example of hollow foam struts.
So this is a grass.
I don't know what kind of grass it is.
I just saw this grass.
And we picked some different grasses, and we took some SEM pictures.
And it has a really kind of common structure for grasses.
It's very common for grass stems to have sort of a solid outer part and then a foam-like inner part.
It's so common that botanists have a name for it.
They call it the core-rind structure.
And if you take one of these grass stems, and you look at the sort of foamy bit in the middle, and you do a SEM picture of that, you can see that the little cell walls are actually little hollow tubes.
So one of these things-- it's a little hollow tube.
So what I wanted to do is work out how much the modulus of the foam would increase if you could make all the edges into little hollow tubes.
So we're going to start by saying the foam behavior is dominated by cell bending, so edge bending.
And the foam properties can be increased by increasing the EI of the cell wall.
So there's a couple of ways we could do that.
So the first one is looking at hollow walls.
So imagine I have a thin-walled tube-- just a circular, thin-walled tube.
There's my little wall there.
It has some radius little r, and a wall thickness t. And then imagine I have the same amount of mass, but now I have a solid circular section.
And I'm going to say the radius of that is big R. So for our thin-walled tube, the moment of inertia is pi r cubed times the thickness, t, if it's thin.
And for our solid circular section, I is going to be pi big R to the 4th over 4.
And if I say I want to set this up so that the masses are equal, then the areas of the cross-sections have to be equal-- say it's from the same material.
So the masses are going to be equal if pi R squared is equal to 2 pi r t. So I'm going to solve here for R. So the pi's are going to cancel out.
So the masses are equal if R is equal to the square root of 2 times r times t. And then what we're going to do is see how the big is the moment of inertia of the tube relative to the solid.
And the tube is pi little r cubed t. And the solid was pi R to the 4th, divided by 4.
And I'm going to get rid of the R here, and get rid of the pi's there.
So R to the 4th is going to be 4r squared t squared.
So the 4s are going to go.
And this boils down to r over t. So if I had a thin-walled tube, the moment of inertia is going to be r over t bigger than if I had the same mass in a solid circular section.
So you can see for the little plant here, by making a thin-walled tube, you're increasing the stiffness of the foam with the same amount of material.
That's the idea.
And you can do a similar kind of analysis for other properties.
So that's if we have hollow tubes.
So another option is we could have cell walls that are sandwich structures.
So imagine if the cell walls themselves were little, tiny sandwich structures.
So when you have a sandwich beam, what you have is too stiff, strong faces that are separated by some sort of porous core, like a honeycomb or a foam or balsa wood.
And the idea with the sandwich structure-- if I draw a little sketch of the sandwich, here's my faces.
So imagine those are solid.
So they might be aluminum sheets, or they might be fiber reinforced composites.
And then we have some sort of cellular thing here as the core.
And the idea is, that's analogous to an I-beam.
So in the sandwich beam, we have two, stiff, strong faces separated by a lightweight core.
So the core is typically a honeycomb, or a foam, or balsa wood.
And the idea is, you increase the moment of inertia of the cross-section with little increase in weight.
And if you think of an I-beam, an I-beam has a large moment of inertia, because you're separating the flanges by the web.
And the sandwich beam works in the same way.
You're separating the faces by the core.
But the core doesn't weigh very much, because it's a cellular thing.
So the faces of the sandwich are like the flanges in the I-beam.
And then the core is like the web.
So the idea is to make something called a micro-sandwich foam.
So what you want to do is make the cell walls into sandwiches.
And one way to do that is to disperse a large volume fraction of thin-walled spheres into the foam.
And you have to get the geometry right to make it work.
So let me draw a little kind of sketch here of how it works.
So here's our thin-walled spheres.
And then you're going to distribute those in a foam.
Here's another sphere over here.
The spheres are not perfect.
Let's say there's another one in here.
And then the idea is this stuff in here would be the foam.
So these guys are hollow spheres.
And say the spheres have a diameter D. And say they have a wall thickness here of t. And say that the separation of the spheres I'm going to call c. You can see that there.
And then the cell size of the foam I'm going to call e. So there's a bunch of parameters you have to kind of play with to get this to work.
So you have to have thin-walled spheres so the faces are thin.
The sandwich panels work best when the faces are thin.
So you need the thickness of the sphere to be much less than D. You need the faces to be stiff relative to the foam.
So you need the modulus of the sphere material to be greater than the modulus of the foam.
And you need the volume fraction of the spheres to be relatively high to get the spheres close enough together for this to work.
So you want that volume fraction to be something like 50% to 60%.
And for the foam, you need to have the foam cell size less than the separation between the spheres.
You need to have a number of-- you can't just have one pore in here.
That's not really like a foam.
It won't behave like a foam as a continuum.
So you need to have a number of different cell sizes in between each sphere.
And so you need the cell size of the foam to be a lot less than the separation of the spheres there, c. But if you can control this geometry, you can get the sandwich effect.
And you can get improved properties by doing that.
So there's ways you can play around with the structure of the foams to improve their properties.
So that was one thing I wanted to say.
Another way to improve the properties of a foam-like material is to use one of those lattice materials.
So we've been talking about ways to improve the bending stiffness.
But if you could get rid of the bending altogether and have axial deformation in the cell walls, that would be much stiffer.
And you can get axial deformation by having those 3D truss kind of materials.
So I have a picture of this.
There we go, so there's one of those 3D truss materials.
So another alternative is to sort of get rid of the bending altogether, and to try to make a truss-type material.
So there's various ways to make these.
I think that we talked about a few of them earlier on.
And you can analyze them as truss-type structures.
And I can just run through a sort of little dimensional argument to get the modulus.
So the modulus is going to go as the stress over the strain.
The stress is going to go as a force over a length squared.
The strain's going to go as a deformation over l. So this is just like what we had before for the foams.
But in this case, the deformation is going to go with the force times the length over the area of the cross-section divided by Es, because we're pulling it or pushing it axially.
So that goes as Fl over t squared Es.
And if I just put that back in the equation here for the modulus, I get that we've got F over l. And I've got delta here, so that's F l t squared Es.
And you just get the modulus goes as the modulus of the solid times t over l squared.
And that goes as the modulus of the solid times the relative density.
So for the open-celled foams, the modulus went as the relative density squared.
So if it was 10% solid, the modulus would be 0.01.
And this is saying if it's 10% solid, the modulus is 0.1.
So it's much bigger.
So this is all sort of well and good.
The only difficulty is that when you look at the modulus, you can do reasonably well.
But when you look at the strength, some of the members are going to be inevitably in compression.
When you have these truss materials, some members are going to be in tension.
Some members are going to be in compression.
And the compression members tend to buckle.
And once the compression members buckle, then you're back to the same kind of strength relationship that you have for the foam.
So that's one of the difficulties of this.
So let me say that the strength-- so if the strength was controlled by uni-axial yield, it would go linearly with relative density.
But if it goes with buckling, it goes as the square.
So I'll just say the compression members can buckle.
And say you had a metal lattice.
Then there's some interaction between the plastic behavior and the buckling.
And you use what's called the tangent modulus instead of just the Young's modulus.
And the tangent modulus is lower.
And there's also what's called knock-down factors that can be large, too.
So the knock-down factor can be like 50%.
So the measured strength can be half of what you thought it was going to be.
This should be a squared over here.
Sorry.
So even though the stiffness of these 3D trusses can be quite good, the strength often isn't quite as good as one might hope.
So that's one of the issues with them.
All right.
So do you see the idea, though, with all these different micro structures, is that you can control the structure in a way to try to increase the bending stiffness or get rid of the bending stiffness and increase the axial stiffness?
So there's things you can do to play around with that.
And I wanted to talk a bit today about material selection charts for foams.
So when we talked about woods, we started talking about this.
Remember, I derived a little performance index.
We said if we had a material and we wanted to have a given stiffness, and we wanted to minimize the mass, we had that performance index that was E to the 1/2 over rho.
And we had a chart of modulus versus density.
And we saw that wood was really good.
You can do that for other sorts of properties, not just modulus.
So you can make-- depending on what the mechanical requirement is, you can work out different performance indices.
So I want to go into that in a little bit more detail.
So the question is, how do we select the best material for some mechanical requirement?
So in the wood section, we looked at the minimum mass of a beam of a given stiffness.
And we saw that the performance index was E to the 1/2 over rho.
So let me do another one of these little examples, and then I'll show you some more of them.
So another example would be what material-- minimize the mass of a beam of a given strength or a given failure load.
So we'll call the failure load Pf.
And we can see the maximum stress in the beam is going to be the moment in the beam times the distance from the neutral axis y, and divided by the moment of inertia.
So here, M is the maximum moment in the beam.
And y is the maximum distance from the neutral axis.
And I is the moment of inertia.
And I'm going to say i goes as t to the 4.
And I'm going to define a failure stress of the material sigma f. So sigma max is going to go as my failure load times the length.
That would be the moment.
The distance from the neutral axis is going to go as t. And the moment of inertia is going to go as t to the 4th.
And that's going to be the failure strength there.
So I can solve this for t. And then I'm going to write the mass in terms of t, and put that in there.
So here t goes as Pf l divided by sigma f. And that's going to be to the 1/3 power.
I guess I can scoot over here.
Then we can say that the mass M goes as the density of times t squared times l. So the mass M is going to go as rho times l times t squared.
So that whole thing goes to the 2/3 power.
So if we look at the material properties, the mass goes as the density times the failure stress raised to the 2/3 power.
So if we want to minimize the mass, we want to minimize rho over sigma f to the 2/3, or we want to maximize sigma F to the 2/3 over rho.
So that's the performance index for that case.
So we can obtain these performance indices for different loading configurations and different mechanical requirements.
And I don't want to go through a whole lot of them, but I'm going to put this up with the notes.
So this is from Mike Ashby's book on Material Selection in Mechanical Design.
And this is a whole series of these performance indices for different situations, for things loaded in torsion, for columns and buckling, for panels and bending.
So these ones are all for stiffness.
And they all involve a modulus raised to some power divided by a density.
So a tie in tension, c over rho, the beam in bending is E to the 1/2 over rho.
A plate in bending is E to 1/3 over rho.
So you don't need to memorize those.
But you can see you can derive these for different situations.
And here's another one for strength-limited design.
So the shaft is, depending on what the specifications are, it's the strength raised to the 2/3 power over rho.
The beam loaded in bending-- the top one there-- sigma f to the 2/3 over rho.
That's what we just did.
So there's all these different kind of performance indices.
So depending on what your situation is, you would pick one of these indices.
And then what you can do is use these material selection charts, which plot one property against another on log-log scales.
And because all of these performance indices involve a power, they always end up being a straight line on your log-log plot.
And here this one, I think, is the same as what I showed you for the wood.
This one's the modulus here plotted against density.
So foams are down here.
And other engineering materials are over here.
And these guidelines here are the different performance indices.
So this one's E over rho.
This one's E to the 1/2 over rho.
This one's E to the 1/3 over rho.
And for this case here, as you move the lines up to the top left-hand corner, E is getting bigger.
Rho is getting smaller.
And so the actual value of the performance index is getting bigger.
So you can use this to select a material.
So we've made these charts for foams as well.
So here's a couple of charts for foams.
And I think what I'm going to do is just go through them quickly.
And there aren't really that many notes, so I'll just put the notes on the website.
And you can come and write all the notes down.
And then we can finish this today.
So this one here is the Young's modulus versus density.
And these are all sorts of different foams.
So the low modulus ones tend to be flexible.
The higher modulus ones tend to be more rigid.
And you could use this to select foams, if you wanted.
You can also see what the range of values is.
So the values of the modulus here goes from a little less than a 100-- because this is two orders of magnitude here, I think, each one of these-- down to about 10 to the minus 4 or a little less than that.
So there's a huge range.
There's almost a range of a factor of a million in those moduli.
And the same with the strengths here.
The strengths go from 10 to the minus 3 mega-pascals up to about maybe 30 mega-pascals, something like that.
And you can see for the modulus and the strength, things like the metal foams are good.
The balsa's good.
Here's the balsa up here.
Metal foam's up there.
So you can kind of see the range of properties that you could get.
And then you could also-- need a drink, hang on.
You can also plot the specific property.
So here's the compressive strength divided by the density plotted against the Young's modulus divided by the density.
And here you want to be up at this end.
So you would have a high strength and a high stiffness.
So the balsa and the metal foams are good up here.
This next plot-- this is the compressive stress at 25% strain.
And this is the densification strain.
And if you think of having your stress-strain curve looks like this, something like that, so you could say that's a strain of 0.25 and that's the stress that corresponds to that.
So that stress times the densification strain, which is out here someplace, is an estimate of the energy underneath the stress-strain curve.
So you can think of this right-hand plot here-- those dashed lines-- these lines like this and this and this-- each one of those corresponds to how much energy you would absorb under the stress-strain curve.
So points that lie on here would have an energy of 0.001 megajoules per cubic meter.
And over here, we're at 10 joules per cubic meter.
So again, the balsa and the metal foams are good over here.
So you can use these plots to try to identify foams for particular applications.
And I think there's a couple more.
It doesn't have to be mechanical properties.
Here is thermal conductivity versus compressive strength.
So you can imagine if you wanted some insulation, you wanted to have a certain thermal conductivity value, you probably also need at least some minimal compressive strength.
You could also have something like a maximum service temperature, that maybe the foam is going to melt at some temperature.
You can't go beyond that.
So there's some property there.
And I think there's one more here.
You can look at things like the density in terms of the buoyancy of a foam, if you have some buoyancy application.
And you can look at cell size on this one here.
And cell size can be important for things like filtration and catalysis.
So the amount of surface area goes as 1 over the cell size-- the surface area per unit volume.
And so the cell size can be important for those sorts of applications.
So the idea is, you can make these material selection charts for foam.
And you can put data on there.
And you can compare foams.
And you can use these performance indices.
So I'm going to leave it at that.
There is a little bit more notes.
But I'll just put them on the website, and you can get them from there.
So I think we're good for today.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at OCW.MIT.edu
So what I wanted to do today was talk about thermal properties of foams.
And foams are often used for thermal insulation.
And that's always closed cell foams that are used for thermal insulation.
And we'll see why.
And the foams tend to have a low thermal conductivity.
And that's largely because gases have lower conductivity than solids.
And if you have mostly gas, you're going to have a lower conductivity.
So they have a low conductivity because they have a high volume fraction of gas.
And they've got a low volume fraction of the solid.
They also have cells.
And the heat is transferred partly by radiation and convection.
And if you have small cells, you reduce the amount of convection and radiation.
And we'll see that.
So that, by having a cellular structure, and in particular, by having small cells, you can decrease the heat transfer.
OK, so let me write some of the stuff down.
So closed cell foams are widely used for thermal insulation.
And the only materials with lower thermal conductivity than closed cell foams are aerogels gels.
And I'll I talk a little bit more about aerogels later on today.
But the difficulty with aerogels is that they tend to be very weak and brittle, like ridiculously weak and brittle.
So we had a project on aerogels a couple of years ago.
And the students who I was collaborating with would make aerogels.
And they'd bring it up to my office.
And I would pick it up and like-- I would pick it up like this, and it would break.
So they have very low thermal conductivity, but they're very brittle.
And I brought a few of our samples of aerogels, just so you can see what they look like.
And I'll pass them around in that little tube, so you can kind of play with them.
OK, so we're going to focus on foams.
And whoops-- And we can say the low thermal conductivity of foam arises mostly from the high volume fraction of gas and that the gas has a low lambda, a low thermal conductivity.
So lambda is thermal conductivity, so I'm just going to put lambda there.
Then it has a small volume fraction of solid, which has a higher thermal conductivity.
And then the foams have a relatively small cell size.
So one of the things we're going to look at is how does the cell size effect the thermal transfer, the thermal conductivity.
OK, so there's lots of applications for foams.
And I guess one of the main ones is in buildings, insulating buildings-- also insulating refrigerated vehicles, things like LNG tankers.
So there's lots of the applications for using foams for thermal insulation.
Foams, in addition to having a low thermal conductivity, they also have good thermal shock resistance.
So thermal shock is if you have a material, you heat it up, and then you suddenly cool the surface of it, for example.
So say you takes something and you quench it in water or quench it in some fluid, then the surface, it wants to shrink because its temperature drops, but it's connected to everything underneath it and it can't really shrink.
And so it's constrained and you can get cracking and spalling.
And so, it turns out foams have a good resistance to that thermal shock kind of loading.
And we'll see why that is, too.
Roughly, you can see if the thermal expansion strain is the thermal expansion coefficient times the change in temperature.
And the stress that you might generate is just going to be related to the modulus times alpha times delta-t. And because we're going to see that alpha for the foam is the same as alpha for the solid, but the E foam is going to be a lot less that E of a solid would be.
So because the modulus is smaller, you would get a better thermal shock resistance.
OK, so I wanted to go over a couple of sort of laws of heat conduction, so we can talk about what thermal conductivity is and how we define it.
So the first one here-- --first one here is for steady state conduction.
So when we say steady state conduction, what we mean is that the temperature is constant with time, the temperature doesn't change with time.
So time's not going to come into the equation here.
And heat transfer for steady state conduction, where there is no change in the temperature with time, described by Fourier's Law.
And that says that the heat flux q is equal to minus lambda times the gradient in temperature.
And if you want to think about just a one diversion of that, it's equal to minus lambda times dt by dx.
So here, q is our heat flux.
So that would have units of joules per meter squared per second.
So how much heat transfer per unit area per unit time.
Lambda is the thermal conductivity.
And it has units of watts per meter k, so degrees kelvin.
And then delta or-- and then this is our temperature gradient.
OK, so that's Fourier's Law, and we're going to use that later on when we talk about the foams.
And then, just so that we have things a little more complete, if you have a non-steady heat conduction, if the temperature varies with time, then there's a difference equation that involves the thermal diffusivity.
So if we have non-steady heat conduction-- so t varies with time.
I'm going to call a time tau.
Then a partial differentiation, the partial derivative of temperature with respect to time is equal to the diffusivity, that's given the symbol a, times the second derivative of temperature with respect of distance, so with respect x squared.
So here a is the thermal diffusivity.
And it is equal to the thermal conductivity divided by the density and divided by the specific heat.
So here, rho is the density and cp is the specific heat.
The specific heat is the heat required to raise the temperature of a unit mass by the unit temperature.
And so, the density times cp is the volumetric heat capacity.
It's how much energy you would need to raise a certain volume by, say, 1 degree k instead of a certain mass.
OK, so on the table here, on the screen, we have different materials.
And we have the thermal conductivity lambda.
And we have the thermal diffusivity, a.
And I guess I should also say a has units of meters squared per second.
So this table is arranged in order of decreasing thermal conductivity.
So here's copper at the top, 384, watts per meter k. Here's, you know, different metals.
You've got aluminum.
Here's a couple of ceramics.
They're about a factor of 10 less than the metals.
Here's the polymers, another factor of 10 less than that.
And here's some gases.
Air is about 0.025.
Carbon dioxide is less than that.
Triclorofluoromethane, which used to be used as a gas in foams because it's got such low thermal conductivity, is 0.008.
But it's no longer used because it's a-- what you call it?
A fluorocarbon.
Anyway, it decreases our ozone layer.
So they don't use that anymore.
Now here's some wood.
So that's one sort of cellular solid.
And they're around 0.04-- something like that.
And here's a group of polymer foams.
And they're a little over 0.025.
So if you think of-- if you had the gas, air-- if the air was the gas inside the foam, 0.025 is lambda for the gas.
So you're not going to get lower than that.
And you have to use a low conductivity gas to get these values, like 0.025 here, 0.020, 0.017.
And then, hear some other sorts of sort of mineral fibers, glass foams, glass wools.
OK, so that's just a table so that you have some data there.
All right?
Yes?
[INAUDIBLE] --foams, if they are closed cell, with a different gas rate.
Because if they're open cell--
--Right.
The gas is going to--
--it would just always be air
It's going to go.
And in fact, one of the difficulties with using the lower conductivity of any gases is there's a phenomenon called aging, that if, you know, you've got your gas inside your foam, it's going to diffuse out into the air.
And air's going to diffuse in.
So over time, the thermal conductivity tends to increase because you're getting air coming and the local-- conductivity gas going out.
But I think, typically, that process takes a number of years.
It doesn't happen in a week.
But if you're designing a building and want the building to be there for 50 years, it occurs faster than that.
So it's not ideal from that point of view.
All right.
So let me talk a little bit more about thermal diffusivity.
Let me scoot over here.
So materials with a high value of that thermal diffusivity, a.
They rapidly adjust their temperature to their surroundings.
So if they have a high value of a, what it really means they've got a, say, a high value of lamda-- so high thermal conductivity.
And, say, a low value of this volumetric heat capacity.
So it doesn't take much energy to change their temperature.
And they also conduct heat well.
So they tend to adjust their temperature to their surroundings quickly.
OK, so then, let's talk about the thermal conductivity of a foam.
So I'm going to call that lambda star.
So the star is the foam.
And then we'll talk about-- lambda s will be the lambda for the solid that it's made from.
So if you think of the thermal conductivity of the foam, there's contributions from different types of heat transfer.
So you could have conduction through the solid.
I'm going to call that lambda s. You could have conduction through the gas.
You could have convection within the cell.
So convection has to do with having, say, within the cell, it might be a different temperature on one side of the cell to the other side of the cell.
And the warmer side of the cell, the gas is going to tend to rise to the warmer side and fall to the cooler side.
And you get a convection current set up.
So you can get heat transfer from that.
And you can also get heat transfer by radiation.
So radiation can cause heat transfer, as well.
So we're going to have contributions from conduction through the solid.
So the amount of conduction in the foam from the solid-- I'm going to call lambda star s. So lambda s would be the conductivity of the solid.
And lambda star s is the thermal conductivity contribution from the solid in the foam.
So we get kind of-- through the solid.
We have conductivity through the gas.
So it's lambda star g for gas.
And then we could have convection within the cells.
We'll call that lambda star c. And then we could get radiation through the cell walls and across the voids.
We'll call that lambda star r. And so, the thermal conductivity of the foam is just the sum of those four contributions.
So we're just going to go through each of those contributions, in turn, and work out how much thermal conductivity you get from each of them.
And it turns out most of the thermal conductivity comes through the gas.
So if we first look at just conduction through the solid, we've got that contribution to the conductivity of the foam from the solid, it's just equal to some efficiency factor times the thermal conductivity of the solid times the volume fraction of the solid or the relative density.
And here, eta is an efficiency factor.
And it accounts for that tortuosity in the foam.
So if you think of the solid in the foam, it's not like we have little fibers that just go from one side to the other like this and the heat just moves along those fibers.
You know, the foam cells have some complicated geometry and the heat has to kind of run along that complicated geometry.
And people have made estimates of what this is.
And it's roughly a factor of 2/3.
So I guess it would depend on exactly the foam cell geometry.
But typically it's around 2/3.
So that's conduction through a solid.
That's straight forward.
Conduction through gas is similarly straightforward.
It's just the conductivity of the gas times the amount of the gas.
And the volume fraction of the gas is just 1 minus the volume fraction of the solid.
So it's just 1 minus the relative density.
So the conduction through the gas is just lambda g times 1 minus the relative density.
So we can do a little example here.
And you can see how much of the conduction comes from the solid in the gas.
So for example, if we look at a foam that's 2.5% dense and say it's a closed cell poly-- what are we doing-- polystyrene.
So the total thermal conductivity of the foam is about 0.04 watts per meter k. And the thermal conductivity of polystyrene is 0.15 watts per meter k. And the thermal conductivity of air is 0.025.
So let's assume it's just blown with air.
And then if I just add up, what's the contribution of conduction through the solid and conduction through the gas-- so I just use those two little equations-- conduction through the solid-- it's going to be 2/3 of this value of lambda s times the amount of the solids-- that's 0.025 and then plus lambda g, which is 0.025 times the amount of the gas, which is 0.975.
And if I work those two things out, this is 0.003 and this is 0.024.
So that total is 0.027 watts per meter k. So you can see if the total is 0.04, most of it's come from the gas.
A little bit's come from the solid.
And the rest is going to be from convection and radiation.
And that's typical.
And that's the reason that they sometimes use low thermal conductivity gases to blow foams for thermal insulation because the gas makes up such a big fraction of the total conductivity.
If you can reduce that, you reduce the overall conductivity.
So, we'll say foams for insulation are blown with low conductivity gases.
But as I mentioned, you have this problem with aging that, over time, that gas is going to diffuse out and air is going to diffuse in.
Then the overall thermal conductivity of the foam is going to increase.
So that's the conduction.
And then the next contribution is from convection.
So imagine we have one of our little cells here.
And it's hotter on that side than it is on that side.
And hot air is going to rise.
Cold air is going to fall.
So you get a convection current set up.
And because of the density changes, you get a buoyancy force in the air.
So that's kind of driving the convection.
But you also have a viscous drag.
So the air is moving past the wall of the foam.
And there's going to be some viscous drag associated.
And how much convection you can get depends on the balance between this buoyancy force and the viscous drag.
So we'll say the gas rises and falls due to density changes with temperature.
And the density changes give rise to buoyancy forces.
But we also have these viscous forces from the drag of the air against the walls of the cell.
So air moving past the walls-- this is kind of a fluid mechanics thing-- so that air is a fluid.
And in fluid mechanics, they often use dimensionless numbers.
And there's a dimensionless number called the Rayleigh number.
And the Rayleigh number, you can think of it-- it's not quite the balance of the buoyancy force against the viscous forces.
But it involves those forces.
And convection is important if this Raleigh number's over 1,000.
And here's what the Rayleigh number is.
It's the density of the fluid times the acceleration of gravity times beta.
Beta's the volume expansion coefficient for the gas-- times the temperature change.
And we're going to look at a temperature change across a cell.
And then, times the length.
That's going to be the cell size.
And we divide that by the fluid viscosity and the thermal diffusivity.
So let me write down what all these things are.
So rho is the density of the gas.
So the g's gravitational acceleration.
Beta is the volume expansion of the gas.
And for a constant pressure that's equal to 1 over the temperature.
Then delta tc is the temperature difference across a cell.
And l is the cell size.
Mu is the dynamics viscosity the fluid.
And a is our thermal diffusivity.
So what I'm going to do is just work out, for a typical example, how big of a cell size do you need to get this Rayleigh number to be 1,000.
And we're going to see that, typically, that cell size is big.
It's like 20 millimeters.
So in most foams, the convection really isn't very important at all.
So it's typically-- people don't worry about convection.
And let me just show you how that works.
So for our Rayleigh number, which is ra-- for the Rayleigh number to be 1,000-- say we had air in the cells.
And say the temperature was room temperature.
Then the volume coefficient of expansion is just 1 over t. So it's 1 over 300, say.
degrees k to the minus 1.
Let's say our change in temperature across one cell was 1 degree k. Bless you.
The viscosity of air is 2 times 10 to the minus 5, pascal seconds.
The density of air is 1.2 kilograms per cubic meter.
And the thermal diffusivity for air is 2 times 10 to the minus 5 meters squared per second.
And if you plug all of these into that equation for the Rayleigh number and you solve for the cell size, you find that the cell size, l, is 20 millimeters.
So that says convection is only important if the cell size is bigger than that.
And so most foams have cells much smaller than that.
And convection is negligible.
So I have enclosed cells and the heat's not transferred so easily from one cell to another by the gas moving.
And by having small cells the convection drops out.
So you don't have to worry about that.
So the last contribution to heat transfer is from radiation.
And there's something called Stefan's law that describes the heat flux for radiated heat transfer from a surface at one temperature to another surface at a different temperature across a vacuum.
So we can say we have a heat flux qr not from a surface of one temperature.
So I'm going to call that t1-- to one at a lower temperature.
I'm going to call tnot-- with a vacuum in between them.
So this is [?
Stefan's ?]
law so this is the radiative heat flux is equal to the emissivity of the surfaces, which is beta 1 times a constant called Stefan's constant-- sigma times the fourth power of temperatures.
I'm taking the difference of the temperatures so here are the Stefan's Constant-- is sigma.
And that's equal to 5.67 times the 10 to the minus 8.
And that's in watts per meter squared per k to the fourth.
And beta is a constant describing the emissivity of the surfaces.
So it gives the radiant heat flux per unit area of the sample relative to a black body.
And that's a characteristic of the emissivity.
All right, so then, so if we-- yes?
[INAUDIBLE]
Now-- so right now, forget the foam.
We have no foam.
We just have two surfaces with a vacuum between them.
And now I'm going to stick a foam between the surfaces.
And we're going to see how that changes the heat flux, OK?
So the next step is we put the foam between those two surfaces.
And the heat flux is going to be reduced because the radiation is going to be absorbed by the solid and reflected by the cell walls.
And so we're going to characterize how much it's reduced.
So there's another law called Beer's Law, which characterizes the reduction in the heat flux.
Piece of chalk's getting to small OK, so Beer's Law gives us the attenuation, so the sort of reduction in the heat flow.
So qr is equal to qr not.
That would be the heat flux, if we just had the vacuum.
And then there's an exponential law.
And it's the exponential of minus k star t star.
And here, k stars in an extinction coefficient for the foam.
Talk a little bit more about that in a minute.
and t star is just the thickness of the foam.
And then this thing is called Beer's Law.
So we have very thin walls and struts.
And we're just going to consider optically thin walls and struts to make life easy.
Then we can say that, if they're optically thin, they're transparent to radiation.
They're optically thin if they're less than about 10 microns.
Then this extinction coefficient is just the amount of solid times the extension coefficient for the solids.
So it's just the relative density times the extinction coefficient for the solid.
OK, and then I can say, the heat flux by radiation.
I can use two equations to write that down now.
And then I'm going to let them be equal to get the thermal conductivity.
I can say qr is going equal to lambda r times dt by dx.
So that's the Fourier's Law that we started out with.
And then I've also got the qr that I'm going to get by combining the Stefan's Law with the Beer's Law up there.
So if I do that, I get that qr is beta 1 times sigma times t1 to the fourth minus t not the fourth.
So that's the qr not up there from down there.
And then I've got an exponential for the attenuation.
And instead of k star, I'm going to put the relative density of times ks.
and then I've got the thickness of the foam, t star, as well.
OK, so that's qr, but that has to equal lambda times dt by dx.
So I'm going to use some approximations.
Here and I'm going to end up with an expression for the contribution from radiation to heat transfer in the foam.
Yeah?
So when you say optically thin walls, where t is less than 10 microns, you mean like the walls of the foam?
Yeah, yea
So it's different t than the--
t star is the thickness of the whole thing, yeah.
So imagine we had our two surfaces.
And they might be like 100 millimeters apart or something.
t star is the sort of thickness of the foam in between the two surfaces.
And the optically thin is the cell walls, which are microns kind of thickness.
OK.
So I'm going to make some approximations here.
And that's going to allow me to solve for t star.
So I'm going to say that dt by dx x is approximately equal to just t1 minus t not over the thickness of the foam or I'll call that delta t over t star.
And then, the other approximation I'm going to use is that t1 to the 4th minus t not to the fourth is equal to 4 times delta t times the average temperature cubed.
So here t bar is the average temperature, t1 plus t not over 2.
So then, if I use those two approximations, I can write that qr, our heat flux from radiative transfer.
I got the beta 1.
I've got the sigma.
And instead of the difference of the fourth power, I'm going to write 4 delta t t bar cubed.
And then I've got my exponential.
Blah, blah, blah, blah, blah.
So then, here's the relative density times ks times t star, the overall thickness.
That's going to equal the radiative contribution to the thermal conductivity of the foam.
And instead of dt by dx, I'm going to have delta t over t star here.
So part of the reason for doing these approximations I end up with a delta t term on both sides.
Now I can cancel that out.
And if I just take this mess here and multiply it by t star, then I've got lambda r star.
That's our thermal conductivity contribution from radiation.
So one of the things to notice here is that, as the relative density goes down, then the contribution from radiation to the thermal conductivity of the foam goes up.
OK, so this chart here shows thermal conductivity as a function of relative density.
And it breaks down the contributions from the gas, g, the solid, s, and the radiation, r. And you kind of see the gas contribution doesn't change that much.
These are relative densities between a little over 2 and a little less than 5%.
So the amount of gas-- it's mostly gas in all of these things.
The solid contribution increases as the relative density increases.
So you'd expect that.
And then, as I just said, as the relative density goes down, the amount of radiation contribution goes up.
And so you can kind of see how that all fits together.
Another plot that shows the thermal conductivity versus the relative density.
These are for a few different types of foams.
You can see for this plot here, you reach a minimum in the thermal conductivity.
And that's because you've got this trade off between the contribution from the solid and the contribution from the radiation.
And those two kind of trade off and you get to a minimum.
So let me write some of this down.
So I'll just say that-- hang on.
Write this over again.
This is looking at the overall thermal conductivity.
And we can see the relative contributions of lambda, solid, lambda, gas, lambda, radiation.
I'll just say this shown in the figure.
I'm going to say the next figure shows a minimum in the thermal conductivity.
Then I'll just say there's a trade off between the conduction through the solid and I can direction from the radiation.
And then we also have a plot here that shows the conductivity versus the cell size.
And you can see that the conductivity increases with cell size.
And the reason for that is the bigger the cells get, the radiation is reflected less often.
And one thing I wanted to mention with the cell size is that if you look at aerogels, the way aerogels shells work is that they have a very small cell size, a very small pore size.
So typically, it's less than 100 nanometers.
And the mean free path of air is 68 nanometers.
So the mean free path is the average distance the molecules move before they collide with another molecule.
And if your pore size is less than the mean free path, then that reduces the thermal conductivity.
It reduces the ability of the atoms to pass the heat along between one another.
So the way the aerogels work is they have a very small pore size.
And what's important is how big the pores are relative to the mean free path of air.
OK, so that's the thermal conductivity.
I wanted to talk about a few other thermal properties of foams, as well, today.
So one is the specific heat.
And since the specific heat is the energy required to raise the temperature by a unit mass, then the mass is the same-- you know, if you have a certain mass of foam or a certain mass of solid-- the specific heat from the foam is the same as the solid.
So the specific heat for the foam is the same as the specific heat for the solid.
So that would have units of joules per kilogram per degree k. And the next property is the thermal expansion coefficient.
And it's a similar thing.
The thermal expansion coefficient for the foam is equal to the thermal coefficient of expansion for the solid.
So imagine you have-- say you had something like a honeycomb.
If you heat it up a certain amount, every member is going to expand by alpha.
And if every member expands by alpha, the whole thing expands by alpha.
And this is the same.
And it's the same idea with the foam.
So if every member just gets longer by alpha, then the whole thing gets bigger by alpha.
OK, so the last topic I wanted to talk about was the thermal shock resistance.
And thermal shock is the idea is that if you have something that's hot, and say you quench it in a liquid-- so you put it suddenly in a liquid-- the surface is going to cool down faster than the bulk of it.
And because the surface is trying to contract because it's cooling down, but it's attached to the bulk of it and it's constrained, it can't really cool down, then you generate stresses.
And if the stresses are big enough, you can cause fracture and have the thing crack and spall.
So we'll say if the materials is subjected to a sudden change in the surface temperature, that induces thermal stresses at the surface and can induce spalling and cracking.
So we're going to think about a material at one temperature that's dropped into, say, a liquid at a different temperature.
So the surface temperature is going to drop to the cooler liquid temperature and it's going to contract the surface layers.
And the fact that they're bound to the layers underneath that are not contracting as quickly, it means that you generate a thermal strain.
So the thermal strain is going to be the coefficient of thermal expansion times the change in temperature.
So you're going to constrain the surface to the original dimensions.
And then you're going to induce the stress.
So if it's a plane or thing, it's e alpha delta t. And then, there's a factor of 1 minus nu, just because it's a plane, in a plane.
And then you'll get cracking or spalling when that stress equals some failure, stress.
So I can rearrange this and solve for the critical delta t that you can withstand without getting cracking.
So I just rearranged this and say sigma's equal to sigma f. That would be sigma f times 1 minus nu over e and over alpha.
So that's the critical change in temperature to just cause cracking.
So now what I can do is I can substitute in there for what you would have for the foam.
And I'm going to do it just for the open cells just because it's easier to write the equations.
So for the foam, I would have some sort of fracture strength.
So when we did the modeling of the foams, we said that was equal to about 0.2 times the modulus of rupture times the relative density to the 3/2's power and 1 minus nu.
And if I divide by the modulus of the foam, that's es times the relative density squared.
And then we just had alpha for the foam was the same as alpha s. So then, I can rearrange this slightly and say it's equal to 0.2 over the relative density to the 1/2 power.
So I'm canceling out these relative densities here.
And then I can combine all the solid properties together.
And I'm going to say that nu for the solid is about equal to the same as nu for the foam.
So what I can do here is I can group all the solid properties together.
And this just is delta t critical for the solid, right?
So this is saying that the critical temperature range before you get cracking in the foam is equal to the range for the solid, but multiplied by this factor of 0.2 and divided by the square root of the relative density.
So if the square of-- the relative density is going to less than 1.
So this number here is going to be bigger than 1.
So it's saying that the temperature range that will give you spalling in the foam is going to be bigger than the temperature range in the solid.
So the foam's going to be better than the solid, OK?
And that uses our little models from before.
So I think I'm going to stop there, probably cause my throat is starting to get too sore.
There's a little case study in the notes.
And I'll just put that on the notes on the Stellar site.
It's like one page and it's really straightforward.
You can just read that, OK?
So this is the end of the bit on thermal conductivity.
That's just this one lecture.
And this is really the end of the whole section on modeling of the honey combs and the foams.
So that's kind of the first half of the term is modeling the honey combs and the foams.
And the second half of the term, we're kind of applying those models to different situations.
So next week, we'll have the review on Monday, have a test on Wednesday, week after that is Spring break.
I can't believe we're at Spring break already.
And then after that we'll start we'll do the trabecular bone for a week.
We'll do tissue engineering scaffolds and cell mechanics for two or three lectures.
We'll look at some other applications to engineering design, look at energy absorption and sandwich panels.
And then, I'm going to talk about plants a little bit at the very end, OK?
So we've already covered a lot of the kind of deriving equations part of the course.
The rest of the course is more applying the equations to lots of different situations, OK?
So I'm going to stop there just because my throat is giving out.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
So I was just going to be here to answer questions
Just clarifying, What was the material that we were covering?
In the test?
Yeah
So the test covers everything up to the end of the part on modeling foams, but not the bit on the performance indices, and the material selection charts for foams.
So I think up to the end of the fractured toughness of foams
I see.
OK.
So not covering past [INAUDIBLE]
Not covering thermal properties, no.
It doesn't cover thermal properties.
Here you go.
So you know I got this MacVicar award, and we had a lunch on Friday at the Catalyst restaurant.
And I had to get up and speak.
And just as I got up and spoke, there was a red-tailed hawk swooped by the window.
It was perfect.
It was perfect.
Yeah.
Hi.
So I'm just here to answer questions.
So come on.
Somebody must have questions.
It's all perfectly clear?
You want me to do--
Talk through test one from last year a little bit
You would have to give me test one from last year.
I didn't bring it with me.
I just brought the problem sets
I have it on my computer, and I could read you the problems or just hand you the laptop.
Whichever you prefer
Why don't you hand me the laptop and I'll try to do it.
Is that OK?
OK.
So the question is about the test for 2014.
OK.
So the first question was, describe four processes for making honeycombs, and comment on the type of material usually used for each process.
So I did post the solutions, right?
Did you look at that?
Yeah, I looked at them.
I guess I just feel like I don't fully understand why things are there.
But I can look at it some more
Well, I can go over it, if you want
I'll just look at it some more
No, I can go over it.
So four processes.
Let's see.
So there's the expansion process, where you take sheets and you glue the sheets together, and then you pull them apart.
So you can only really use that process for materials that are going to have large plastic deformations.
So you could use it for metals, some polymers.
But you couldn't really use it for ceramics.
You couldn't use it for glass because as soon as you yanked on it, you'd break the sheets, right?
So--
Is it rigid polymers that you can use it for?
Well, something like nylon you can use it for.
Something that's got a little yield, that will have some sort of yield point.
Not like if you had epoxy, you couldn't use it for epoxy.
So, OK.
So that's one process.
Another process is a corrugation process where you have a wheel that has little gear knobs on it.
And you run your flat sheet through that and it comes out with the half hexagonal profile, and you glue those together.
So again, you need something that's going to yield.
So that would typically be a metal that you would use that with.
Let's see.
Another processor making honeycombs is 3D printing.
You can 3D print honeycombs.
And there's different ways to do it.
One way is by having an ink.
So if you want to print in ink, typically that's some sort of polymer that you're printing.
I suppose physically it's possible to print glass or to print a metal, but you'd have to have some very high temperature setup to do that.
So typically a resin of some sort.
Let's see.
Other ways to make honeycombs.
You can extrude honeycombs.
So the ceramic honeycombs we saw were made by extruding a ceramic slurry.
And typically, you would do that with a ceramic that's a slurry and a powder.
You wouldn't necessarily do that with a metal.
I don't think I've seen any metal honeycombs that are extruded like that.
OK?
Are we good with number one?
Yeah, that makes sense
We now need your password
Oh, sorry.
So is there an actual difference between 3D printing and the extrusion process?
Yeah.
So the extrusion you have a die, and you squeeze the material through the die, right?
So extrusion's kind of like the toothpastey thing.
And 3D printing, you can have an ink or you can do 3D printing where you have, let's say, a powder.
And then you print the binder.
And then you heat it up some way to get the binder to cure.
And then you get rid of the powder that's not bound.
That's another way to do the 3D printing.
OK?
Can you also pour it into a mold?
Yeah.
Yeah, those silicon rubber honeycombs that I showed you, those are all made by pouring a liquid into a mold and then curing it.
Yeah.
Yeah, there's other ways, it just asked for four, so I randomly thought of four.
OK. OK, are we good?
So the next one is a hexagonal titanium alloy honeycomb has h/l is two, theta is 45, and t/l is 0.05.
It says, the end constraint factor for elastic buckling is n equals 0.806.
The titanium has a modulus of 110 gigapascals, and yield strength of 880.
And then you have to calculate some properties.
So would you like me to do that?
Yeah
Yeah, you would?
Does anybody else want that?
I don't see a lot of other people wanting anything else, so I might as well do that.
Let's see.
I would need a piece of chalk.
Here we go.
OK.
So it's a titanium alloy honeycomb, and we're told h/l is 2, theta's 45, and t/l is 0.05.
And we're told that n-- why don't I put the n over here-- n is 0.806.
And we're told the modulus of the solid is 110 gigapascals, and the yield strength of the solid is 880 mega pascals.
OK.
So it says calculate the value of and describe the mechanism of deformation failure for-- and the first part is the Young's modulus in the two direction, e star 2.
OK.
So I don't remember these formulas either, so I need to look at my notes.
And oh, I don't have the formula for e 2 in my notes.
Let me see.
Is it in any of the problems?
I have the formula sheet
You have the formula sheet?
I didn't bring the formula sheet with me.
You have it?
Yeah, I think it's there.
I'm pretty sure it's there.
OK, so this is just like substituting, and there's nothing complicated about this.
So it's equal to Es times t/l cubed times h/l plus sine theta divided by cos cubed theta.
So then you just plug everything in.
So this is 110 gigapascals.
And I'm going to put 110,000 mega pascals, because it's probably going to be less than a gigapascal.
Then t/l is 0.05.
So that's 0.05 cubed.
And then h/l is 2 plus sign of 45 is 0.707.
And then we divide by cos theta cubed, 0.707 cubed.
And then I'm not going to work it out, but that's OK.
I assume that's what's in the solution.
Are we good?
So it's just substituting.
That's all it is.
In the second part-- OK. You need to change the time on this, because it just keeps timing out
Sorry, I don't know how to change it, but I'll try
Oh, is that what this is?
OK, this is the test.
Oh, this is from 2013.
Oh, the next-- if we keep going.
Here we go.
He's got it, so you take your computer
That's probably better
That's perfect for everybody.
OK.
The second part is the plateau stress for loading in the x2 direction.
So that's that.
For loading in the x2 direction, it could either be an elastic buckling collapse stress, or a plastic buckling collapse stress.
So I'm going to calculate both, and then whichever one is lower, that's the one it would be.
So let's see here.
So that's going to be-- I'm missing my formulas again.
Here we are.
So here's the buckling one.
It's n squared pi squared over 24 times t cubed over lh squared times 1 over cos theta.
So you put 0.806 squared in here.
Pi squared 24.
So here you go.
This is t over l cubed times h over l squared.
So that would be 0.05 cubed.
And that would be 1 over 2 squared, and then 1 over 0.707, and then whatever that equals.
OK?
Are we good with that?
And then you'd want to calculate sigma star to plastic.
And that's equal to sigma ys, t over l squared, and 1 over 2 cos squared theta.
All right?
So then sigma ys was 880 mega pascals.
And t over l was our 0.05.
And that's 1 over 2 times 0.707 squared, and then whatever that equals.
OK?
And then whichever one of those would be less is the plateau stress.
So I don't-- yeah?
So I know-- I think I made this mistake in the problem set, but here, because it's titanium, we don't consider it brittle
Right.
Right.
I mean, you could.
But you don't need to
That was something with ceramics?
Yeah, or if it was a glass, or ceramic, or maybe an epoxy.
Something that was brittle.
And besides which, if you look at the question, I only give you a yield strength and a solid modulus.
So to get the brittle thing, I would have to give you a fracture strength
That's true with all the problems
I usually give you what you need.
Especially on the test, I'm going to give you what you need.
You're not going to be looking things up.
OK?
So we're good so far?
OK, let me go back to the question.
So then the third one is the out of plane Young's modulus in the x3 direction.
And that's just going to be Es times the relative density.
And the relative density-- let's see.
So it's Es times rho star over rho s. So that's 110,000 mega pascals.
And then there's also the very first equation on this is the relative density.
So it's t/l times h/l plus 2 divided by 2 cos theta h/l plus sine theta.
So I'm not going to substitute everything in, OK?
So that was it.
It was very plug and chug.
Are you good?
Can I ask a question about a specific question
I'm going to go through the rest of them.
She wants me to do the whole test.
You want me to do the whole test, right?
That would be great, but if other people have other questions it's fine
Why don't I do the whole test.
And then there's going to be time, I think
[INAUDIBLE]
Let me do the test from last unit.
OK, so that's number one.
Or that's one and two.
And then three is, a closed cell elastomeric polyethylene foam has a relative density of 0.05 and a volume fraction of solid in the edges of 0.6.
They give you the Young's modulus of the solid is 0.2 gigapascals.
The pressure within the cell walls is atmospheric, 0.1 mega pascals.
And the Poisson's ratio of the foam is 0.3.
And you're asked to get the Young's modulus of the foam, the compressive plateau stress.
And then there's a question about why does the Young's modulus depend on the solid modulus and relative density, while the Poisson's ratio does not.
So let me go through, then.
So the first one is, what's e star.
And we're told relative density is 0.05.
The volume fraction in the edges is 0.6.
So remember, that was what we called phi.
So phi's 0.6.
The Young's modulus of the solid is 0.2 gigapascals.
The initial pressure within the cells is 0.1 mega pascal.
and Poisson's ratio for the foam is 0.3.
And it's a closed cell.
So if you remember-- oh, let's see.
I think back here, was I supposed to-- I was supposed to say something about the mechanism of deformation and failure for the first one.
So the mechanism of deformation in the modulus in the 2 direction is bending.
The mechanism of failure here is buckling.
The mechanism of failure here is yielding.
And the mechanism of deformation here was axial deformation, OK?
So I forgot to say that.
OK, let me go back here.
So for this one, if it's a closed cell foam, remember there were three terms to the modulus.
There was one from bending of the edges, one from stretching of the faces, and one from the gas contribution if you've got gas inside the cells.
So again, I'm going to have to peek at the equation.
Foams.
Here we go, foams.
OK. And then this gas one.
Get rid of that.
OK, so this one is just the same kind of thing.
It's just plug and chug.
Do I need to put all the numbers in?
No.
I guess I'm a little bit confused why you need the pressure term.
Because you talked about faces bursting
Ah, so if it's the modulus, remember the modulus-- so the question is, why do you need to worry about the pressure because I talked about the faces bursting.
And remember, the stress strain curve looks something like that.
Maybe the slope of the curve is a little bit higher over here.
But this is the modulus down here, right?
The modulus is related to the initial stress strain relationship.
And initially, they're not going to burst.
You'd have to load it up to some amount of stress before the faces burst, right?
So when you're down here, the faces certainly down at the beginning, they're not burst, right?
You have to get some stress before they're going to burst.
And in some materials, when you get up around here, around the plateau stress-- let's just call that sigma star-- then they might burst.
And then the pressure term would disappear and the face term would disappear.
So if I don't tell you to ignore them, if I don't say they're going to burst, or I don't say they're negligible, I would calculate them.
And then if they're small, then you say, well, they're negligible.
OK?
Are we good with that?
You're good?
You're good?
Sardar, you don't need to be here.
But you can stay if you want, but you don't need to be here.
OK, are you good?
Everybody else good?
OK. That was A.
B, what's the compressive plateau stress of the foam.
So here, they want to know what sigma star is.
You're told it's elastomeric.
So if it's elastomeric, it's like a rubber.
It's rubbery.
So if it's rubbery, it's going to buckle.
It's not going to yield.
It's not going to be brittle.
So you can just calculate the elastic stress here.
And if we flip over to our handy dandy list of equations-- blah, blah, blah, blah.
Oh, pooh.
I'm realizing-- yeah, so I don't have the term here for the faces or for the gas, so here we could assume it's going to rupture.
So let's assume it's going to rupture.
So if we assume that it's going to rupture, it would be just like the open celled foams.
And then you can just use that.
And that's been found to work fairly well for the open celled and the closed cell foams
And we assume that faces rupture because--
Well, to be honest, I can't remember if last year I got to this point in the test and somebody said, we don't have the equation, and I gave them the equation with the other terms, or if we just assumed that the faces ruptured.
I can't remember.
To be honest, I think probably for elastomeric foam, you could assume that they'd probably don't rupture unless they're very, very thin
So what kind of foams do they typically rupture in?
So certainly if you had a metal foam, they'd probably rupture.
If you had, say, a polymer foam that was more rigid, like a rigid polyurethane.
So polyurethanes can be flexible, which means they're made out of an elastomer, or they can be rigid.
And the foams that are typically used for insulation, thermal insulation, are typically closed celled polyurethane foams.
And those typically have very thin faces, and they would rupture.
Yeah?
Are you looking for the Young's modulus in this problem?
The Young's modulus was the first part, right?
So part A was the Young's modulus
In B--
In B is the collapse stress, the compressor strength
So I think in my notes, I think it has this.
If the--
Right, if p0 is bigger than-- so what she's showing me in her notes is I had a little note in the class, in the lecture, that if the initial pressure in the cells is greater than atmospheric, then the cell walls are pre-stressed and you have to overcome that in the buckling
Is that atmospheric?
That is atmospheric pressure
So you don't need the [INAUDIBLE]
No.
Yeah.
OK?
OK. Yeah, you don't know that that's atmospheric.
So do you ever do things in PSI?
No, you don't.
Because when I was a student a long time ago, the thing I remember learning was atmospheric pressure is 14.7 PSI, more or less.
And the conversion between mega pascals and PSI is there's more or less 145 PSI to a mega pascal, so the atmospheric pressure is about 0.1 mega pascals.
Yeah?
I don't know if this is a silly question, but for part B, how do we get from the Young's modulus of the foam--
Oh, sorry, sorry, sorry, sorry.
I put the wrong thing down here.
Sorry, my mistake.
OK, now you happy?
Sorry.
OK, shall I move on to the next part?
Another question?
Well, I had a question about number two, but maybe we can come back to that--
OK, let me finish this, and then we'll go back to number two.
So this one here, the part C is why does the Young's modulus foam depend on the solid modulus and the relative density while the Poisson's ratio does not.
So when I write the equation for the Young's modulus, the solid modulus comes into it, the relative density comes into it.
And remember when we had the Poisson's ratio, it's just a constant that depends on the cell geometry.
So here's C, nu star just as a constant.
And that constant just depends on the cell geometry.
So if you think of the Poisson's ratio, it's the ratio of two strains, right?
So say I have my foam here.
So say that's just a block of foam.
Little cells in it here.
Little cells.
And say I press on it this way here.
And let's call this the one direction and the two direction.
If I press it in the two direction, the Poisson's ratio is then just nu would be-- let's see, this would be 2, 1.
It'd be the strain in the one direction over the strain in the two direction.
So it's the ratio of two strains, right?
And if you think of our model for the elastic behavior of the foam, each of those strains is going to be related to some bending deformation in the cell walls or the cell struts.
And this strain here is going to be-- let me make this proportional.
It's going to be proportional to a delta over l. And this one here is going to be proportional to the delta over l. So this might be delta in the one direction, and this will be delta in the two direction.
But those two things are both going to be related to the bending deflection of the beams, right?
And since both of those deltas are related to the bending deflection of the beams, we could write them-- if you want, I could write that as f l cubed over E of the solid times t to the fourth.
And then that's times 1 over l. And then this thing here is also f l cubed over E of the solid t to the fourth 1 over l. So everything cancels out except the geometrical constant.
And if you remember when we did the honeycombs, it looked exactly the same.
When we looked at the Poisson's ratio of the honeycombs, we had the strain in one direction over the strain in another direction.
And each of those strains was related to some component of delta.
There was a delta 1 and a delta 2.
But the delta 1 might be delta sine theta and delta 2 was delta cos theta.
So if the deltas are the same, then it all just cancels out.
And all you're left with is a geometrical constant.
OK?
Do you get physically why that is?
So I have a question about this.
Because we're given most of the equations that we need, is it only in conceptual questions that we should know how we actually derived that version?
I'm not going to ask you to derive that equation for a closed cell foam
I meant this last part
Well yeah, you should be able to explain that.
But I mean just at this level.
Nothing very mathematically involved
OK, cool
OK, are we good?
So that was the end of the test for the undergraduates, OK?
And then for the graduate students, just like the problem sets, I just have an extra question.
And that's what I did this year, too.
So the graduate students have one extra question.
So you and you.
Is anybody else a graduate student?
I think it's just the two of you.
You're post post-graduate.
OK. OK.
So let's see.
So this one says, the performance-- so this is on the performance indices which I told you you didn't need to know for this test, partly because, remember, we missed two lectures.
We're not exactly on the same spot as we were last year.
But I can do it if you want.
Do you want me to do it, or should we do other questions?
What do the grad students think?
Sure.
OK.
So the question is, the performance index to minimize the mass of a beam of a given bending stiffness, length and square cross-section is e to the one half over rho.
So you remember, we derived that e to the one half over rho in class.
In the section on wood, we saw that this performance index for wood is higher than that for the solid cell wall material in wood.
Do you remember that?
The e to the one half over rho for the wood was, I think, rho s over rho star to the one half times Es to the one half over rho s. So explain why wood has a higher value of e to the one half over rho than the solid cell wall material.
And then part B is, suggest a design for an engineering material based on wood that has high values of e to the one half over rho.
So one way to explain it is to say that if you're looking at e to the one half over rho, you can say for wood, E over Es is equal to rho star over rho s for loading in the axial direction.
So this will be for loading actually along the grain.
And that's what we were looking at.
So that's what I'm talking about here.
So I think-- let me just see if this is right.
Yeah.
So this equation here is exactly the same as that equation there, right?
And this is basically saying that this is the performance index for the wood.
This is the performance index for the solid.
And this factor here is bigger than 1, because the solid density is higher than the wood density.
OK?
So really, all you have to do is say that for the wood, the modulus in the longitudinal or the axial direction along with grain varies linearly with the relative density.
And it probably would be a good idea to say that this is a result of the cell walls deforming axially.
So when you take the cells, if you think of the wood cells as being something like that and you're loading it this way on, the cells just actually shorten, and the modulus depends on the-- it just is the volume fraction of solid times the modulus of the solid.
And that's where this comes from.
And once you have this, that basically gives you that.
OK?
Are we good with that?
So that's why it's higher.
Another way to look at it as sort of more of a hand-wavy argument is that if you have a certain amount of solid-- so say you have a certain mass of solid.
If it's solid, it takes up a certain cross-sectional area.
So say that your beam's a certain length, that's going to have a certain cross-sectional area.
And if you have wood, if you have a cellular material, if you have the same mass, you're essentially making the dimensions of that piece bigger.
So you're moving the material further away.
And as you're making it bigger, you're increasing the moment of inertia.
And so you're increasing the bending resistance of it.
That's another, more hand-waving way to talk about it.
And then the second part is to suggest a design for an engineering material based on wood that would have high values of e to the one half over rho.
So remember when we looked at those material performance charts, we said that wood was similar to engineering fiber composites.
But those data for fiber composites are assuming that it's solid, the fiber composite's a solid.
So if you could take fiber composites and make little tubes of fiber composites and assemble the tubes together so that it was like wood, would get something that would be even higher.
So if you could make, say, a fiber composite honeycomb material, and you'd want to have the fibers aligned along the prism axis of the honeycomb, then you would get higher values.
It would be the same-- it'd be this sort of argument again, but now with a fiber composite.
So you'd want-- if this was your fiber composite like this, you'd want the fibers-- well, in wood, they're at a little bit of an angle.
But say they were lined up like that.
You'd want them something like that, then loading it that way on, right?
And if one way to think about those charts is if you-- say we had a plot.
And say this was log of the modulus and that was log of the density.
And I think I'll just draw the envelope.
So foams were somewhere down here, metals were somewhere over here, and with elastomers we're somewhere in here.
I think ceramics were up here.
And then I can't remember exactly where composites were, but composites were around about here.
I'll just say FRC for fiber reinforced composites.
And I think woods were kind of in here.
Something like that.
And then we had our performance index, right?
So remember, there was a performance index, something like that.
And that slope of that was e to the one half over rho.
So every point on that line had the same value of e to the one half over rho.
And essentially, if you had the fiber composite and you made a honeycomb out of it, you would be taking the data from here and shifting them out that way.
You'd be pushing them out over here, so you'd get a higher value of that performance index.
OK?
So that's the test from last year.
That's the end-- yeah?
So for along right here or different it would be cubed
Yeah, so the thing about the honeycombs is--
The opposite
Right
[INAUDIBLE]
It'd be worse, that's true.
So the thing about the honeycombs is they're very stiff in the axial direction, but you pay for that in the other directions.
And it's the same for wood.
So the wood is very good when you load it along the grain, but you pay for it the other way.
But if you think of from the tree's point of view-- if you're a tree.
So here's my little tree.
So here, say we have a tree trunk, and we have some branch.
Branch over here, branches, tree.
So the grain is lined up this way.
And then when there's a branch, the grain turns around and goes that way, right?
So if you think of the tree as a whole, the whole tree blows in the wind like this.
So it's like a column like this, and everything's lined up that way.
And you're loading it this way.
So that is the stiff direction, right?
And if you're a branch, the branches are more loaded by gravity.
So they're loaded that way.
And then because the fibers, the grain turns around, they're also oriented in the good direction.
So from the tree's point of view, it's optimized things.
Then you remember when I talked about the old wooden sailing ships, when they made the old wooden sailing ships, if this was the deck here and that was the haul there, they would get pieces of wood to fit in here that were called the knees.
That was the knee.
And they would try to get a piece that was from a branch like this.
And they would try to match the curve of that joint with the branch with the curve that they needed in here so that the grain followed the pattern of what they needed for the boat.
OK. Other questions?
I'm not quite sure what the difference between tangential versus ray here
Oh, OK.
So in the wood, you mean?
Yes
OK.
So can I rub this stuff off?
We're happy?
Let's see.
Say again?
So tangential and radial.
OK.
So say the wood cells look something like this.
So these would be the fiber cells or the tracheids And then the rays typically are more rectangular cells.
So they might look something like that.
And then they would be some more fibers or tracheids, depending on if it was a soft wood or a hardwood.
So these would be either fibers or tracheids in a hardwood or a soft wood.
And then these would be the ray cells in here.
So they have a different structure.
They just look different.
They're different shape
This is the top?
From the top?
Yeah, this is looking from the top down.
And then if you think of the tree-- so the tree's going to have growth rings, right?
So the growth rings are going to look-- obviously I'm not making perfect circles, but you get the idea-- something like that.
And then the rays go this way.
They go radially.
OK?
So this would be the radial direction, and then those are the rays.
Are we good?
So which way's the tangential?
So the tangential would be this way on, OK?
So if I loaded this way like that, that would be loading it in the tangential direction.
And if I loaded it this way, that would be loading it in the radial direction.
The length of the rays runs in the radial direction.
The length this way on.
So this thing here corresponds to one of these lines I've drawn here.
And then these guys here are the stuff in between here
How do you know what the tangential, the Young's modulus and stiffness is?
So say we were loading it tangentially, we're loading it like that.
Then--
If you have a tree, how do you apply a tangential load on it?
Oh, well it's not the whole tree, right?
So say we have a piece of wood that we cut out like this.
So say I have that.
And if I loaded it this way on, I'd be loading it tangentially.
The tree's big, right?
So I'm not talking about loading the whole tree, I'm talking about taking a piece of wood out of the tree and loading it
OK, so you can't really load tangentially for the entire trunk
No, I'm talking about taking a piece out and loading that piece
How about a ray here?
Do you take the--
So the same thing.
You'd take-- say this was the piece of wood that you were looking at.
Now you would just load it this way on.
OK?
I think-- I brought my thing because I have the slides.
Let me see if I can find-- I think there was a slide that showed this.
OK. That was Furry Fridays.
That was the wood sculptor.
Here we go.
OK, so imagine that that cube is your piece of wood that you're loading, right?
So imagine this is the-- you cut a little piece out.
Then you're loading it tangent.
Can you see, then?
You can load it-- you're not loading the whole tree
Yeah, I was thinking about loading the entire tree and then applying the tangential load on it
Yeah, because then-- I see the problem
It's going to give us shears
Yeah.
Yeah, because I could say, well, if I was trying to load the whole thing.
Say I was loading it from here to there.
Well if you look at it one way, it looks tangential.
If you look at it the other way, it looks radial.
So think of cutting a piece out, because that is what you're going to do.
You're going to cut a piece out.
OK?
All right.
Are there other questions?
Yes?
With that formula sheet, do you only give that formula for the honeycombs, or also for the foams?
I'm going to give you-- if you look at, I think, problem set 2, I gave you a sheet that had three pages of equations.
And it looked exactly like this.
So there was one saying, properties of two dimensional cellular solids-- honeycombs.
There was a whole thing of in plane properties and out of plane properties.
That was one page.
The next page was properties of regular hexagonal honeycombs.
And then the next page was properties of three dimensional cellular solids foams
OK, excellent.
Thank you
OK?
Like I just said, I think there's maybe one or two equations missing from this.
But if it was something you needed, I would give it to you.
I would give it to you.
OK?
So I should have scrolled down for it?
What?
You should have scrolled?
So you're like me.
This happens to me all the time.
I have some website.
I'm looking at it.
I'm like, OK.
I got it.
I think I've got everything.
And then I realized I'm supposed to-- I missed something because I was supposed to scroll down
Problem set two was only the honeycombs.
That's why
Well, I think that was probably all we covered was the honeycombs on that problem set
So that's why I only got that part
OK. All right, yeah.
So I'm going to give you, this will be attached to the test.
OK?
So I think on the test that I posted it was attached, wasn't it?
Yeah.
[INAUDIBLE], did you have a question?
For 2 part D, I don't think we went over that.
I was confused as to how-- is it just that equilateral triangular cells always have-- is always truss behavior?
Let's see
Oh, sorry.
It's the 2014 test
2014-- oh, sorry.
I missed a part.
Sorry.
Yeah, so it says, the same titanium alloy is used to make a honeycomb with equilateral triangular cells.
And what is the in plane Young's modulus for loading in the x 2 direction of the triangular honeycomb?
So this is-- say you have cells that look like that now.
And that's x 1.
That's x 2.
OK.
So the Young's modulus for this-- so I happen to remember the formula.
So I guess I'm thinking you might have put this on your cheat sheets.
It's 1.15 times Es times that, times the relative density.
So I'm trying to remember.
Do I have that on here?
I don't have it on here
Are we just supposed to know that?
Are we just supposed to--
Well, I guess what I would hope that you would know is maybe not the constant, but that it should go as Es and linearly with the relative density.
Because it's a truss and because it deforms axially.
I don't really expect that you would remember the 1.15.
Yeah?
So does that mean for all the foams, of which there were like 10 constants, we don't need to write all down, like what C1 equals--
I think that's what this thing gives you.
Let's see.
It gives you all the Cs.
All right, then I'll make sure I give you the Cs.
I'll make sure I give you the Cs.
But who's got a pen so I can write that down to make sure that I do that?
Does somebody got a piece of paper.
Or I could write it down here.
Oh, here we go.
I can write it down this little sticky thing here.
OK.
So I'll stick that on there so I remember to do that
Will you also be writing what Cs?
Because you have, in your equation, C1, C2--
Well, I think I would say-- say I asked you for, I don't know, the yield stress for a foam in compression that yields plastically.
I would say, the constant for that is 0.3.
I wouldn't give you C2.
I wouldn't do it by numbers, I would tell you what the number was for the thing you needed, because I mean the way the numbers are, the only reason they're numbered is because that's the number they are in the book.
They're just ordered sequentially in the book.
But I don't expect you to remember which-- it's C6, or C5 or something.
So anyone else?
Can you explain the difference between uniaxial yield and plastic buckling?
Oh, OK.
So if you have something and it fails by uniaxial yield-- so say you have a honeycomb like this, and you're loading it this way on, OK?
So if you're loading it that way on, these walls of the honeycomb are just axially deforming, initially.
Right?
So the elastic behaviors, they just axially deform.
So it works out that if these cell walls are very thick, then you can reach a yield stress before any buckling occurs.
And then the strength would just be that yield stress of the cell wall material times the relative density.
OK?
But the cell walls have to be thick for that to happen.
So then imagine that the cell walls aren't thick.
Imagine that the cell walls are thin.
So say I have the same honeycomb like this.
If the walls are thin, and say the solid material itself-- so this is for the solid-- it has some stress strain curve.
And it may have a linear elastic part, and then a yield thing like that.
So say this is the yield strength here.
Say we compress that this way on the same thing in the three direction.
Then if a material's got a yield point, there can be an interaction between plastic yielding and elastic buckling.
And you can get plastic buckling.
And the plastic buckling, you're going to get the wrinkles that go along the length of it this way.
remember I showed you that tube that kind of collapsed and folded up kind of thing?
That's plastic buckling.
OK?
And typically, people use what's called the tangential modulus to calculate the buckling stress for plastic buckling.
And the tangential modulus would be something related to the tangent over there.
I don't expect you to be able to derive plastic buckling equations.
But the plastic buckling-- you know what elastic buckling is, right?
Yeah.
Yeah.
So one way to think about plastic buckling is, if you have-- and I'm trying to remember.
This is called the slenderness ratio.
And I'm trying to remember, is that l over r?
Imagine you had just a circular cross section and you had a length, l. So you have a column here, and it's got a length, l, and it's got a radius, r. Like that, OK?
So the longer it gets, the more slender it is, the higher the slenderness ratio is.
And this, I think, is some sort of stress.
If the slenderness ratio of just a single column is short-- if it's stubby, if you had a column that looked kind of like that, It's not going to buckle.
It's going to yield.
And so if you compress that, it would just yield.
And it's just going to yield at the yield stress, right?
It's just going to yield at sigma y of the solid, whatever the solid is.
If you have a long column, it would buckle elastically by an Euler buckling.
And Euler buckling-- let's see.
I'm going to run out of room here.
If you think of it in terms of a stress instead of a load, it's going to be in squared pi squared Es i.
Let's say i goes as r to the fourth.
And this is going to be l squared r squared.
Right?
This is going to be a pi in here.
There's going to be a pie in here.
I might have lost a factor of 4, but it's going to be-- let me make this proportional, OK?
So the slenderness ratio, there's going to be an l over r squared term here.
So I could cancel out the four there and put a squared.
So sigma Euler is going to go as Es times r over l squared.
Like that.
And so this is the Euler buckling stress here, OK?
So this would be elastic.
And right here at this little corner, it turns out life isn't quite that mathematically exact.
If you're near that corner, it's not like here it's buckling elastically, and here, it's buckling plastically.
What happens is, if you looked at data, data might do something like that.
So the sum interaction between the elastic and the plastic.
And that's kind of what's going on with this thing here.
Does that makes sense?
Plastic buckling can-- OK, so if you unload plastic buckling, you get some of the elastic part back?
You're not going to get much back
You're not?
No, because once-- to get the plastic buckling, you're very close to this.
By the time you get that deformation, you've got locally, it's yielded.
It's not all elastic everywhere.
It's going to yield in places.
And once it starts yielding, it's-- if you think of these buckles forming, it's not like you're at one spot on this curve throughout the whole thing.
Some of it's more deformed, and some of it's less deformed.
Let me pull up those plastically buckled columns, those tubes.
Get rid of that one.
Let me try and remind myself where they were.
I think-- honeycombs, I want honeycombs.
Out of plane, that's what I want.
It was this thing here.
So you see when you have-- that's just one tube, but the whole honeycomb would-- imagine that you have groups of tubes put together.
They would have to fail in some compatible way.
But the deformation and the stresses are not going to be uniform through this whole thing, right?
One part of it's going to be at one stress, and something else is going to be at another stress.
So parts of it are going to yield plastically, and you're not going to recover that.
OK?
So in fact, they use these sorts of things for energy absorption devices, like in cars and things like that.
To absorb the energy from the impact.
More questions?
Let him have a turn
Sorry, I have a question about I think 2013
2013,
The last question.
It's about the plastic
2013.
Let me rub some of this stuff off.
OK.
Here we go.
OK. Oh, we haven't covered this at all.
So the last question, this one here?
Right
Yeah, so this question's on energy absorption.
We haven't got there yet
How about--
And the third question's on sandwich structure.
So when I taught the course in 2013, I did the topics in a different order.
So I did honeycombs, and I did foams.
And then I think I did sandwich panels, and I did energy absorption.
And I left the stuff on the wood and the cork to the end.
So we haven't done that.
So don't panic if you haven't-- if you can't do that.
OK?
You should have known.
Come on, you should have known that if it talked about things we haven't covered yet, I'm not going to give it on the test.
OK, what else?
Can you explain more plastic hinges?
Plastic hinge, OK.
So let's just say we have a beam in bending, OK?
And say it just has a load p in the middle, all right?
Are we good?
So this load in the middle, then this reaction is p over 2, and that reaction is p over 2.
And if I drew the sheer force diagram, it'd go p over 2 up, we go over, go p over 2 down, like 2 p over 2 down.
Over and back up.
OK?
And then if I drew the bending moment diagram, it would go up and down like that.
And that would be zero.
And that would be zero.
And this would be PL over 4.
OK?
Are we OK with that?
We haven't got to the end of the answer, but--
I have a question.
We're not expected to--
No, no, you don't need to do this.
I'm just trying to explain it now.
You don't need to retain that information, for heaven's sakes.
No.
Come on, I'm so disappointed
For the moment, is it positive for the counterclockwise turns?
Oh, so you remember for the beam bending, there's a different convention.
It's positive if it's tension on the bottom.
Are you in mechanical engineering?
I though you were in mechanical engineering
No, I take physics
Oh, you do physics.
All I really want to say is the moment's maximum in the middle, OK?
So let me just say the moment's maximum in the middle, OK?
So then let's look at the cross section.
So say I look at a cross section here.
Let's just make it rectangular to make it easy for me to draw.
So it has width, b, and a height, h. OK?
So this would be h on this picture over here.
And remember, the neutral axis goes through the middle on the cross-section here.
And so one half of the beam is in tension, and the other half of the beam is in compression.
So for this situation here, this half of the bean is going to see compression, and that half of the beam is going to see tension, OK?
Are we happy with that?
We're happy with that.
OK. Now let me draw the stress distribution.
So if it's linear elastic and it hasn't yielded yet, the stress distribution is going to look like this.
So this is h again.
That's the height.
And now b is into the board.
And I'm plotting the stress this way.
So this thing here is my neutral axis.
It has no stress.
Remember, there was one plane that has no stress, and for a rectangular cross-section, it goes through the middle of the cross-section.
It goes through the centroid.
Is this ringing a bell?
I'm hoping this is ringing a bell.
Come on.
I know we did this in 302, too.
I know we did.
OK. OK, so this is all linear elastic, right?
So at some point-- so I'm going to get to the plastic hinge.
At some point, If you keep loading it and p gets bigger and bigger, the moment gets bigger and bigger, the stress gets bigger and bigger remember, the equation here for the stress is equal to My over i.
This moment, the maximum moment's going to be this moment here.
The maximum y is going to be h over 2, the distance from the neutral axis.
And i is going to be bh cubed over 12 for the rectangular section.
So if I keep loading it up, at some point, the maximum stress is going to equal the yield stress.
right?
And in our cellular things, is going to equal the yield stress of the solid.
So our beam is one of our edges in the foam, or struts in the honeycomb.
So at some point, is going to equal the yield strength of the solid.
So let me draw the stress distribution again, where we start to have plasticity.
So here, the stress is equal the yield stress.
And it's going to equal the yield stress at the bottom, too, because it's all symmetric, right?
And the neutral axis is still going to be in the middle here.
So it's going to 0 down there.
So initially, when it's just barely reached the yield stress at the outer part of the beam, then this stress distribution would still be linear in between.
But once you start to load it more than that, then the plastic region starts to seep in from the outside inwards.
And what we do here is we assume that the solid is elastic perfectly plastic.
And if you remember, when we said things were perfectly plastic, or if they were elastic perfectly plastic, they look like that.
The stress strain curve for the solid, I'm assuming, looks like that.
So I'm assuming the yield stress in the solid is just a constant.
That if I strain it more, there's no work hardening.
I'm neglecting work hardening
You said inward that the--
In board?
Inward, you said something like--
Inward.
So this is-- let's see.
I didn't bring a bean with me today.
No beam.
Do we have anything beam-like?
Ah, here we have a beam-like thing.
OK.
So say this is my beam, and I'm loading it this way on, OK?
And this is b, and that's h. So this picture here is looking at it that way on, OK?
And this picture here, I've drawn h, but now I'm just looking at the stress distribution across h. And b is into the board.
Is that OK?
Does that answer your question?
No, I mean for the plastic
For this part?
OK.
I'm working up.
I haven't finished it yet.
So this is the same kind of view as here.
I drew it a little bigger.
It shouldn't have draw bigger, I should have drawn it the same height.
But it's the same thing.
OK?
So you'll buy that at some point, we reach the yield strength here.
And if I keep loading it up and I assume that the solid is perfectly plastic, that there's no work hardening, then the stress distribution would look like this.
OK?
And then if it yields more, then it's going to look like that.
And if it yields more, eventually I'm going to get to the stage here, where it's-- let me redraw this.
That would be-- you get the idea, OK?
This will go over here, and down here, and like that.
OK?
OK.
So are we happy with this stress distribution across the cross-section?
Yeah, OK.
So that's when it forms the plastic hinge.
So when it forms a plastic hinge, the stress distribution looks like this.
So these are supposed to be the same size.
They're not quite.
Let's see.
So one of the things I talked about was the plastic moment that kind of characterized that plastic hinge.
And the plastic moment is just the internal amount of moment that the beam can withstand when it's yielded completely across the whole cross-section.
So when we're at this point here.
So you calculate that by saying, that stress there is equivalent to a force.
That stress there is equivalent to a force.
And you get the moment by multiplying those forces times that distance there.
OK?
Because you think of those two forces as being a couple, and the moments the force times the distance between them.
So the plastic moment was sigma ys.
And say we're talking about our honeycomb or foam or something.
That was our cell wall thickness t. So let me call it t instead of h for the foams and the honeycombs.
So this force here is going to be the stress time the area over which it acts.
And let's say we look at it for a honeycomb.
Then I've got the stress is acting over this distance here, and then times the depth into the page, right?
And if it's the honeycomb, that depth into the page is just b, OK?
And then this moment arm here is just t over 2 as well, because that's t over 4, and that's t over 4.
So it's t over 2 again.
So it's sigma ys bt squared over 4 for the honeycomb.
And say we have an open cell foam.
The edges aren't of thickness b, they're of thickness t. So then it's m p is just sigma ys t cubed over 4.
Are we happy?
And really, physically what that is is it means that the beam can't hold any more force.
You can't apply any more force to it.
It's just going to rotate like this once you've gotten to that plastic moment.
That's why it's called a hinge, because it just can rotate like hinge rotates.
Like a door hinge, OK?
How does it rotate?
Well, where's my original picture.
So if this was the beam, when you form the plastic hinge, your beam would just look like that.
And this would be your hinge point.
I'm a civil engineer originally.
We try to avoid this.
So that's why, in the foam and in the honeycomb, that's when it fails is when you get that plastic hinge forming.
OK?
All right.
We have a few more minutes
That kind of looks like plastic buckling
Well yeah, it's not buckling, but it's plastic, yeah.
It's permanent.
Anyone else?
OK. No other questions?
Should we call it a day?
Is that helpful?
All right, then.
It's what I do.
Come on.
It's what I do.
All right.
So I'll see you Wednesday.
And my plan is to grade the tests before spring break, so I shall have it back to you.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
So we're going to start talking about trabecular bone, and we're going to do a bone today and on Wednesday.
And I'm hoping we can more or less finish it on Wednesday.
So hello.
Hold on a sec.
So these are some images of trabecular bone, and you can see that it has a foam-like structure.
And trabecular bone exists in certain places in the body.
There's three main places that it exists.
So it exists at the end of the long bones.
And over here this is a femur.
This is the very top of the femur, and you can see this is all trabecular bone in here.
This is a tibia, and this is the top of your knee there.
And you can see how the bone get more bulbous at the ends, and it's filled with a trabecular bone.
This is a vertebrae.
So vertebrae are actually mostly trabecular bone, and they have a really thin shell of what's called cortical bone, the dense bone, on top of it.
So trabecular bone exists at the ends of the long bones, it exists in the core of the vertebrae, and it also exists in sort of shell or plate-like bones.
So in your skull, for example, there's a layer of trabecular bone in between two layers of the compact dense bone.
And in your pelvis, it's the same thing.
So those of you who took 3032, you remember when I passed around the bird skulls, there was that very porous kind of trabecular bone.
So trabecular bone is of interest medically in three main kind of medical situations.
So the first one is osteoporosis.
So I want to talk a little bit about osteoporosis now, and then we'll talk about it in more detail later on.
I guess, we'll probably start today.
Another medical issue is osteoarthritis, and the properties of the trabecular bone are important in arthritis, and the third issue is in joint replacements.
And so we're going to talk a little bit about osteoporosis, osteoarthritis, and then joint replacements.
And then I'll talk a little bit more about modeling bone like a foam and how it deforms and how it fails.
And then we'll talk a little bit how we can model osteoporosis.
So let me write down some of these things.
I guess we'll start here.
So trabecular bone has a foam-like structure, and what we're going to see is that we can use the models for foams to describe the mechanical behavior of the bone.
It exists at the ends of the long bones.
And at the ends of the long bones, the bones become more bulbous.
And really what that's for is to increase the surface area so that there's cartilage between the ends of the two bones.
So there would be a bone here and a bone there, and there's cartilage in between them that sort of lubricates that joint and makes low friction at the joint.
And the bone gets larger to decrease the stresses on the cartilage.
So if you have the same force and you have a larger area, you're going to have a smaller stress.
So that's why the bone gets bulbous like that.
And then by having the trabecular bone, because it's so porous and lightweight, you're not having a big, dense hunk of bone at the end of the long bones.
So it exists at the ends of the long bones.
And I'll just say the ends have a larger area than the shafts.
And that's to distribute the loads on the cartilage or to reduce the stress on the cartilage.
And then the trabecular bone reduces the weight.
So it also exists in the core of the vertebrae, and in fact, it makes up most of the vertebrae and then in things like the skull and the pelvic bones in shell and plate-like bones.
And so it's the core of a sandwich structure there.
So it's of interest in osteoporosis, in osteoarthritis, and in joint replacements.
So if we start by thinking about osteoporosis, you probably know that osteoporosis is a disease where the bone mass becomes reduced and there's a greater risk of fracture, so there's especially a greater risk of hip fractures and vertebral fractures.
And it turns out in both of those sites, if you just look at these bones here, typically if you have a hip fracture, what happens is the neck of the femur breaks.
So this is called the neck here.
This is called the head, this spherical bit there.
So the neck has a fracture, and you can see most of the bone there is trabecular bone, so it's really carrying most of the load and the same with the vertebrae.
This sort of cylindrical part of the vertebrae here carries most of the load.
It has a shell of really thin cortical bone, but it's mostly trabecular bone.
And when the loads are vertical like this, that's really the trabecular bone that's carrying most of the load.
And people get sometimes what are called wedge fractures where instead of having a sort of a cylinder with parallel faces like this, the trabecular bone fails, and the bone ends up like that so that there's-- yeah, I know.
You make that wincing expression.
It's like ouchy.
And in fact, it's very ouchy for people who get that.
And when you see little old ladies who are all hunched over, that's why.
The bone has actually failed.
It's actually been crushed into these wedge fractures, and there's no way they can straighten it out, and it's quite painful.
So people who look at osteoporosis are quite interested in the mechanical properties of trabecular bone for this sort of reason.
The hip fractures are particularly serious because people become immobilized and then sometimes because they're immobilized, they get pneumonia, and in elderly people they sometimes die.
So something like 40% of elderly patients who are over 65 die within a year of having hip fracture.
So it's not that the hip fracture kills them, it's that they become so immobile, and they can't move, and they can't walk around, and they end up getting pneumonia.
So it's quite a serious thing.
And there's something like 300,000 hip fractures a year in the US, and the cost of treating these hip fractures is something like $19 billion.
So yeah, it's a huge problem.
And as the population is aging, as baby boomers like me get older and older, there's going more people having hip fractures.
So it's a huge deal, osteoporosis.
So we'll say bone mass decreases with age, and osteoporosis is extreme bone loss.
And a little later today I'll show you some pictures of what it looks like when people have osteoporosis.
So the most common fractures are of the hip and the vertebrae, and at both sites, most of the load is carried by the trabecular bone.
And the hip fractures you are the most serious.
40% of elderly patients pass away within a year.
So that's sort of a little introduction to osteoporosis.
The next issue that people are interested in is osteoarthritis.
And in osteoarthritis, there's a degradation of cartilage at the joints, and the stress on the cartilage is affected by the modulus of the bone that presses against the cartilage.
You can kind of magic if you have a fiber compass, for instance, most of the stresses, if you're loading it along the fibers, is carried by the fibers because they're stiffer.
So if you have like say trabecular bone that has varying density, the denser bits are going to have higher moduli, and it's there's going more stress associated with that.
And so the modulus of the trabecular bone can affect how the loads are distributed in the cartilage, and that can affect the damage in the cartilage.
And the shell, as I mentioned before, this sort of shell of cortical bone or the dense bone at the joints can be quite thin.
It can be less than a millimeter.
So I brought my little bones along with me again.
So this is the head of a femur here, and this is a piece of a knee joint here from a tibia.
And you can see just looking at these how thin the cortical shell is.
So you can get an idea of how thin that is.
So osteoarthritis involves a degradation of the cartilage at the joints.
And the stress on the cartilage is affected by the moduli of the underlying bone, and the cortical shell, the totally dense bone, can be quite thin.
So the mechanical properties of the trabecular bone can affect the stress distribution on the cartilage.
And if osteoarthritis gets particularly bad, then sometimes people have joint replacements.
So when it gets really bad, the cartilage is degraded completely, and the bone is rubbing on bone, and that's quite painful.
And when it gets to that point, people generally have a joint replacement.
And so the way the joint replacements are done is say somebody who is going to have a hip replacement, what they do is they chop off the top of the femur.
So they would chop the femur off somewhere around here, and then they have a metal implant that has a spherical ball.
That's like the head of the femur.
And then it has a sort of stem and a shaft here that goes into the hollow part of the long part of the shaft of the femur.
And so they use a number of different metals for this, titanium and stainless steel, and there's a cobalt-chromium alloy are also used.
So you need metals that are biocompatible, aren't going to corrode, aren't going to have degradation products.
And then the bone grows around that implant, and the bone grows in response to mechanical loads.
So the density of the bone depends on the magnitude of the load, and the orientation of the trabeculae depends on the orientation of the principle stresses that are applied.
So let me write that down.
So they cut off the end of the bone, and they insert the implant into the hollow shaft of the remaining bone.
And the metals they use are titanium, stainless steel, and a chromium-cobalt alloy.
And then the bone grows into that implant.
And the bone grows in response to mechanical loads.
So the density of the bone depends on the magnitude of the stresses, and the orientation of the bone depends on the principle stresses.
So one of the issues that comes up in joint replacements is that there's a mismatch in the moduli between the metal and the bone.
So if you think the metal, like something like stainless steel, has a modulus of around 200, 210 gigapascals.
And the cortical bone has a modulus of about 18 gigapascals, and the trabecular bone has a modulus between about 0.01 and 2 gigapascals, depending on its density.
So you're taking the bone out, and you're replacing it with something that's much, much stiffer, and that changes the stress distribution around the remaining bone.
And one of the things that can happen is you can get a loosening of the implant.
So the bone can grow in initially, but over time, you get a different stress field in the bone.
And if you have a different stress field, then the bone can resorb away from the implant and cause what's called loosening.
So if the implant becomes loose, that's clearly not a good thing.
It's a bad thing.
And often orthopedic surgeons don't like to do these joint replacements in young people partly because they don't always loosen, but occasionally they do.
And if they loosen they can go back and do a revision.
But you can kind of imagine after they've chopped the head of the femur off and they put one implant in, it's not that easy to go back in and replace that with another one.
You would need one with a longer stem, and the whole thing becomes a little bit more complicated.
So this issue of stress shielding is what it's called when you have something much stiffer that's shielding the stresses in the bone.
The issue of stress shielding means that they don't like to do the replacements on younger patients unless you can get stress shielding.
And if we just compare-- if we look at the cobalt and chromium alloy, the modulus of that in gigapascals is about 210.
If we look at the titanium alloys that are used, the modulus is about 110.
If we look at the stainless steel-- it's 316 stainless steel-- it has a modulus of around 210.
And then if we look at the bone, the cortical bone has a modulus of about 18, and the trabecular bone has a modulus 0.01 to 2 gigapascals depending on the density.
So after the joint replacement happens, the remodeling of the bone is affected.
So the idea is that the stiffer metal carries more of the load, and then the bone carries less load, and then it resorbs.
And that can lead to this thing called loosening, which is not desirable.
Now, this typically doesn't happen till about 15 years after you've had the implant, so it's not something that would happen right away, but it can happen later on.
So these are all sort of medical reasons why people are interested in trabecular bone because of osteoporosis, osteoarthritis, and joint replacements.
So I wanted to start by talking about the structure of trabecular bone.
And then we'll talk about what the stress-strain curves look like in compression and tension, what are the mechanisms of deformation and failure, and how we can apply our models for foams to the trabecular bone.
So the idea is that the structure of the bone resembles a foam, and here's some SCM images of trabecular bone.
And you can see that the bone has a varying structure.
If it's relatively low density, this is a bone that's almost like an open-cell foam if I didn't tell you that was a bone, you might actually think it was an open-cell foam.
And here's a denser piece of bone, and you can see there's still interconnections between all the openings, so it's not exactly like a closed-cell foam, but it's much denser, and it's almost like there's perforated plates in the structure.
And then as I said the bone can grow in response to loads.
So if you have loads that are more or less vertical, the trabeculae tend to line up and be more or less vertical with some sort of horizontal bracing.
So this is a piece of bone from a knee, the condyle is sort of towards the top of the knee.
And you can see these are sort of plate-like pieces of bone.
They're almost parallel, and not too surprisingly in your knee, the loads are typically vertical, and then there's a little bracing bits that go horizontally here.
So you can get different structures depending on the loading on the bone, and the density of the bone corresponds to the magnitude of the load, and the orientation of the trabeculae corresponds to the orientation of the load
[INAUDIBLE]
Resorb.
So when the bone density goes down, when you lose bone mass, that's called resorption.
So the idea is that the trabecular bone resembles a foam.
And in fact, the word trabecular comes from Latin, and in Latin, it means little beam.
So the foams to form by bending.
They act like little beams, and so the trabeculae are like little beams, even in Latin.
There's a range of relative densities, and you can see in that image up there, you can see that there's a range.
And they range typically between about a 5% in dense and 50% in dense.
So something like 0.1 or 0.2 might be typical.
And the low-density bone resembles an open-cell foam.
And the higher density, it becomes more like perforated plates.
And the structure can be highly anisotropic depending on the stress field.
And then I've got another image here of the trabecular bone.
These images are using what's called micro computed tomography.
So you've probably heard of computed tomography.
Say somebody has cancer, they get put in a CT machine, and they do a scan.
The micro CT is more of a research tool.
It's the same kind of technology, but it's got a much finer resolution, and typically, you put a small specimen into a machine to do this.
So the specimen might be half an inch in diameter and an inch tall, something like that.
So these are done by a colleague, Ralph Muller, who's in Zurich, and this is one of his bread and butter things that he has these images, and he looks at osteoporosis.
And you can see here the difference in the structure for the different densities.
So here's a 26% dense piece of bone in the femoral head.
It looks pretty sturdy and substantial.
Here's an 11% dense piece from the lumbar spine, and here's a 6% dense piece.
And you can kind of see when you go from 26 to 11, the struts get a little bit thinner.
And when you go from 11 to 6, the struts get very thin, and in fact, if they get too thin, the struts resorb altogether, and some of their struts can just disappear.
So when people get osteoporosis, what happens is they first lose bone mass by thinning the struts, but then at some point, the struts just resorb altogether.
And if you think of the struts as a biological material, they have bone cells in them.
So there's little osteoclasts and osteoblasts and osteocytes that live in the bone, the mineral thing, the bony thing.
And those cells have dimensions of 10s of microns, so maybe 20, 30 microns, something like that.
So the struts can't get any thinner than that.
If they get thinner than that, then the cells can't live, and the thing just disappears altogether.
And you can think of from a mechanical point of view, if you lose density by thinning the struts, you can use our sort of foam equations.
And say the density went from 0.2 to 0.1, you could make some estimate of how the modulus and how the strength would vary depending on our foam models.
But if you lose density by resorbing the struts, the struts just disappear altogether, then it's as if you had a steel scaffold or a steel structure of a building.
And now you're starting to remove columns and remove beams.
Yes, I know.
That's not good, not good.
And so we'll talk a little bit more about that when we talk more about osteoporosis, and you can see the consequences of that.
But this image here kind of gives you a little bit of a picture of what the bone structure looks like as it gets less dense.
So I want to talk a little bit more about the bone growing in response to load.
Let me rub off the board.
So you're probably already a little bit familiar with this idea.
So when astronauts go up into space, they often do exercises where they have a treadmill, and they've got springs, and they're pulling on the springs to try to exercise themselves.
And the reason they do that is when they're in microgravity, if they were doing some kind of exercise, they would lose bone mass.
And they will get back to Earth where we have Earth gravity, and they would have a problem.
So you see it in astronauts, in microgravity.
The other place you see this just in everyday life is in professional tennis players.
People have done like x-rays of the bones of professional tennis players, and obviously, they have one arm that they hit the ball with their racquet.
The bones in that arm actually get bigger because they're loading that bone over and over again pretty much every day when they're playing tennis, and they're not loading the other arm.
So their two arms are not symmetrical because of this loading from hitting the racquet over and over.
And the people in 3032 have already seen this, but I couldn't resist bringing up the Guinea fowl experiments again.
So obviously, you can only do x-rays on human.
You can't sacrifice the humans and look at their bones, but you can with Guinea fowl.
And so people have done experiments where they run Guinea fowl on treadmills, and they have one set of Guinea fowl that they run on the treadmill that's horizontal.
They have another set of Guinea fowl that they run on a treadmill that's inclined to 20 degrees, so one would think they might have more stress on their bones from that, and then they have a control group that they don't run on the treadmill at all.
And then what they do is they have a forced plate on the treadmill so as the Guinea fowl is running, they measure the maximum force in they're taking high-speed video.
And then they measure the angle of the knee at that point at which the force is maximum.
And they can see there's a change in the angle of the knee when they put them on the inclined treadmill, not too surprisingly.
And then these are juvenile Guinea fowl that haven't completely matured their bones.
And then after about six weeks of this, they sacrifice the Guinea fowl, and they do scans on the bone, and they look at the orientation of the bone, and they measure what's called the orientation of the peak trabecular density, which is a way of characterizing the orientation of the bone.
And they find that the angle of the knee when the Guinea fowl are running changes by about 14 degrees.
And it turns out the angle of the bone, the orientation of the bone also changes by about 14 degrees.
So the bone has remodeled to match that change in the forces that are applied as the Guinea fowl are running on a treadmill.
So this is all a demonstration just to show that bone grows in response to load.
So let me write down some of this stuff.
So I will say astronauts-- did you see Michael Collins is going to come to the talk at MIT?
When I was a kid in '60s, he was one of the Apollo astronauts.
He was like one of the first NASA astronauts.
Anyway astronauts, so in microgravity, they would lose bone if they don't exercise.
And tennis players, the bones get larger in the arm that they hold the racket with.
And then I'll just write a little bit of notes about the Guinea fowl experiments.
So this was done by-- it's in a paper, Ponzer et al 2006.
So they have one set of Guinea fowl that run on a level treadmill, they have another set that run on a inclined treadmill, and it's inclined at 20 degrees.
And then they have a control group that doesn't run on the treadmill.
And then they measure the angle at the knee at the moment of peak force on the treadmill.
And after six weeks, they sacrificed the Guinea fowl, and they measured the orientation of the peak trabecular density.
And they find that the knee flexion angle changed by 13.7 degrees.
And if you compared the inclined versus the level treadmill, and they found the orientation of the peak trabecular density, which they called OPDD, also changed by 13.6 degrees.
So the idea is that the orientation of the trabeculae changed to match the orientation of the loading.
Then I have a little video here.
Do you like video?
So I have a colleague who's at Harvard who studies animal locomotion, and they didn't do this set of experiments, but they do do experiments on Guinea fowl running on treadmills, and thought you might find this amusing.
So let me see if I can make this work.
[VIDEO PLAYBACK] -Sometimes you walk into a lab and you just think this is what science is all about.
-I just put the Guinea fowl on the treadmill, and this is something that we commonly do.
-Welcome to the Concord Field Station, a defunct Nike missile base turned scientific menagerie.
It's owned by Harvard, and biologist Andy Biewener is the director here.
So think of it as-- -A research lab facility for doing comparative biomechanics and physiology of largely animal movement.
-And the birds are just the tip of the iceberg.
-So do you want to see the baby goat and the emu?
-Obviously.
-OK. We keep it because it's sort of like a mascot.
There used to be a lizard colony.
You can hear the African greys.
Then the jerboas are housed in this room here.
This is where we originally did our pigeon flight studies.
So usually the ones with claws and sharp teeth and aggressive behaviors, you want to watch out for them.
-As you might expect.
But did you know that Guinea fowl-- -They're really lovely to work with.
-Sometimes.
Or that-- -Rats are not very good on treadmills.
-Yes.
That's what a rat treadmill looks like.
And this?
-And this was historically a treadmill of note, the treadmill that they first taught kangaroos on and showed that kangaroos stored energy in their tendons enough that they don't actually increase their metabolic rate when they hop at faster speeds.
-These are the kind of discoveries made here with the use of high-speed video and x-ray machines and semi cooperative animals.
But beyond the basic biology, Biewener says engineers are using this research to build better robots, and it can help improve medical treatment for people with movement disorders.
Today the big excitement at the lab is happening here.
Ivo Ros is studying how heart rate changes when cockatiels fly at different speeds.
So this is a way to look at how much energy it takes to fly, and that cord is measuring heart rate.
But instead of the birds flying faster, the wind changes speed.
-I'm going to turn it on then.
-It's hard to fly fast, and it's hard to fly slow, Ros says.
So the expectation is that the heart rate should be shaped like a U.
But so far Ros is finding that it's a flat line.
It's like the bird goes into a stress reaction when it takes off.
Is that just because of the wind tunnel?
What Ros wants to know is-- - --whether or not they need to be stressed to fly in the first place.
-That's something that Ros is looking into, but today is mostly about training.
-Keep going.
Come on.
-Imagine you're a cockatiel.
A wind tunnel is kind of a strange experience.
-A bird in a wind tunnel has to confront the fact that the world is not moving past, which defies its normal sensory cues.
-Which pretty well sums up the Concord Field Station generally.
For Science Friday, I'm Flora Lichtman.
[END PLAYBACK] And if read The New York Times, Flora Lichtman used to work for NPR and would make these Science Friday videos for them.
But now she has a gig doing things for The New York Times, and she does science videos still.
I don't know if you quite call them videos, but what they do is they have these little paper puppets, and the paper puppets are animated and re-enact different episodes in science.
And it's kind of amazing how they do these little science videos.
So if you Google Flora Lichtman, you'll see more Science Videos with all sorts of things.
I guess the other interesting anecdote is I went to the Concord Field Station once.
And I had done a study on quills and animals that have quills because the quills have a foamy structure in the middle.
So they're carrot, and they have sort of like carrot-like structures.
And they have an outer shell that's dense, and then they have a foamy thing in the middle.
Anyway I did this paper on quills and how they work mechanically.
And Technology reviewed a little article about it, and they said they wanted to take a picture of me with a hedgehog.
The hedgehogs are little European-- they're like little small, cute things.
And I said, well, if you can find a hedgehog, I'm happy to have my photograph taken.
And they had a hedgehog at the Concord Field Station.
So we went out there, and we had these big leather gloves and took a picture.
And I don't if it was Andy or I don't know who it was, but I said one of the people there, what did you do with a hedgehog?
And he said, well, we tried to do the treadmill study.
But hedgehogs are like porcupines.
When they get scared, they curl up into a little ball.
And so they said they would put the hedgehog down under the treadmill, and they would start it up, and it would make a noise, and it would get scared it.
And it would just go into a little ball and kind of slide along to the end, and then it would kind of get flopped off.
So that was the end of the hedgehog experiments.
But they did have wallabies there the day I went.
So they have all sorts of animals that they put on to treadmills, birds that they fly, so it's kind of interesting to go there.
But the main idea here is that there was this set of experiments with Guinea fowl that showed just how precisely the orientation of the bone matches the orientation of the loads
Was there a difference between the control groups?
Ah, so I have slides.
I have slides.
Hang on a sec.
I got distracted by my video.
Sorry.
So here's the sort of schematic of Guinea fowl on treadmill.
And where's the little doo-da here?
So on the level, the knee flexion angle was whatever this is, 76.3, and here was the 62.6, so the difference is 13.7.
And then this kind of table here summarizes these results.
So this is the maximum trabecular density, and this is the angle.
And here we have the incline.
Let's see, the control was the yellow, and the level was the blue, and they've got the values for that peak trabecular density orientation.
So they've got that for the level.
And then they looked at the difference between the incline and the level in the knee angle.
That's what this thing here is.
And then between the level and the control, there wasn't really any difference in the knee angle because the control ones, they were just walking around.
So that's the sort of slide that has the actual data on it.
All right.
And then I showed you the video.
All right.
So we need to do a couple more things before we get to the modeling.
Let me get a drink.
So if we want to use the models for foams to try to describe the trabecular bone, we need to know something about the properties of the solid.
Remember we used the properties of the solid in the models.
So we want to get the properties of the solid in the trabeculae, and there's a couple of ways you can do this.
To get the moduli, you can use an ultrasonic wave propagation method, and you can measure a modulus that way.
And if they do that, they measure a modulus between about 15 and 18 gigapascals.
Another way to do it is to take a piece of bone, do a compression test on it, measure the modulus.
Before you do the test, you put it in the micro CT machine, and you get a picture of the structure, and then you use that as the input to a finite element analysis.
So the finite element analysis is a computer numerical analysis to do mechanical calculations.
And if you know what the modulus of the structure is, you can back out what the modulus of the solid must have been from the fine element thing.
And those sorts of experiments also showed that the modulus was around 18.
And it turns out that moduli is about the same as cortical bone, and the properties of the solid trabeculae are very similar to the solid cortical bone.
So let me scoot over here.
So if you use an ultrasonic wave propagation, people have measured a modulus for the solid in trabecular bone of 18 gigapascals, or you can do a finite element calculation based on micro CT data for the structure.
And then you measure the overall modulus for the trabecular bone, and then you back out the modulus of the solid.
And people who've done that have gotten values of around 18 gigapascals too, and that's very similar to what the cortical bone is.
And so we're going to use the following properties for the solid in the trabecular bone.
We're going to say the density is 1,800 kilograms per cubic meter.
The Young's modulus is 18 gigapascals.
The yield strength has different values in tension and compression.
It's about 182 megapascals in compression.
And it's about 115 megapascals in tension.
So those are the solid properties.
So then if we look at the compressive stress-strain curves, they have the shape shown on the screen there.
And you can see how similar the stress-strain curves are for those for a foam.
So there's the same three regimes that we see for the foam.
There is a linear elastic regime over here, there's a stress plateau here, and there's some densification regime here.
These are three curves for three different relative densities.
As the relative density goes up, the stiffness goes up, the plateau stress goes up, and the densification strain goes down.
So is the same as the foams that we've looked at before.
And if we look at the mechanisms of deformation and failure, people have looked at this.
These are on a whale vertebrae.
So these are tests that are done in a micron CT machine, again, by Ralph Muller's group.
And here the specimen is unloaded, and here's the same specimen loaded.
So you can see this platen has come down a little bit.
And if you look at this column here, this trabecular here, you can see it's bent out and bowed out more.
And people have found that usually the linear elastic behavior is controlled by bending of that trabeculae, and the plateau stress is usually controlled by some sort of buckling.
But it's not elastic buckling.
You don't recover it.
If you take a piece of bone and you compress it, it's going to have a permanent deformation.
So it's inelastic buckling.
And I think we have some more pictures.
This is another example from whale bone from Ralph Muller's group.
So here's the bone unloaded.
Here it's loaded to 4% strain, here it's to 8%.
And you can start seeing right in this area here if you compare with up there, it's starting to form.
And if you go up here to 12% strain, you see that strut right there.
That was this guy up here, and you can see that it's buckled right over.
So people have made measurements like this in observations, and you can actually see the buckling.
And people have also done finite element modeling.
They can take a micro CT scan and input that to the fine element model.
And then if they do the compression and they input the properties of the solid, they can see that they get a buckling kind of failure.
If you have trabeculae that are very aligned-- we have more.
Here's one more in the buckling.
So this is one of Ralph's little movies.
So when it unloads, it looks like it recovers, but this is all just an animation.
He takes several stills and puts them together, and it doesn't actually recover.
It's just the way that it shows.
But again, these are two different specimens of different densities.
You can see how the struts deform.
They bend and then they buckle.
Let me stop there, and I'll put some stuff on the board.
So we'll say the compressive stress-strain curve has the characteristic shape of cellular solids.
And the mechanisms of deformation and failure, usually there is bending followed by, usually, inelastic or plastic buckling.
And sometimes if the trabeculae are aligned like that knee that I showed you, or if the trabecular are aligned or if they're very dense, then the actual deformation is important.
And I'll just say people have found this by making observations using micro computer tomography or by finite element calculations.
And this is a stress-strain curve and tension here, a tension you get failure at small strains, then you get micro cracks in the bone.
And these next plots just show some data for the bone.
So we're plotting the Young's modulus here.
So this is a relative Young's modulus, the modulus of the bone divided by the solid cell wall material.
Here's the relative density.
Here's data for lots of different specimens of bone.
So some of this data is for human bones, some is for bovine bone.
Sometimes the data is taken where the orientation of the trabeculae doesn't line up with the direction of the loading.
So you might have trabeculae that are oriented this way, but you're loading it this way.
There's sometimes different strain rates.
Let's see.
There's different groups, and so there's a huge scatter in the range of the data.
But you can see if you look at it broadly and you look at that whole cluster of data, the data lie close to a line of a slope of 2.
And if you think of the open-celled foam model and you had bending of the cell walls, you'd expect that the modulus would vary [?
to the ?]
density squared.
So that's kind of the limit of how we do the modeling.
We're really just interested in seeing how the properties vary with density.
If you had a particular piece of bone, I don't think you could use the models to exactly predict what the modulus of that bone would be.
And here's the compressive strength here.
So this is the relative compressive strength.
Here we've normalized it with the yield stress of the solid bone, and here's the relative density.
And you can see, again, this line is of slope 2, so that kind of speaks to the buckling-type failure mode.
And I think I have another one here.
This is the tensile strength.
So if you pull the bone in tension, you wouldn't expect to get buckling, you'd expect to get plastic yielding.
And if you got yielding and you use the open-cell foam model, you'd expect a slope of 3/2, so this line has a slope of 3/2.
And this line is sort of towards the upper bound of that set of data.
You can imagine a line that went through it a little bit lower but the same slope.
And so these open-celled foam models, they don't predict the properties of a particular piece of bone because the bone can have some anisotropy to it.
The orientation of these things may not be perfectly lined up with the loading.
But overall the models give you a sense of how the bone is deforming and failing.
So let me write some of this down.
So that's data for the modulus, the compressive strength, and the tensile strength.
And those have been on those plots.
Those values are normalized by data for cortical bone.
I thought somebody was talking.
It's just the chair squeaking.
And as I said the spread in the data is large, and that's due to anisotropy in the bone and misalignment between the bone orientation and the loading direction.
So when people first started doing tests on trabecular bone, they typically were orthopedics labs.
And the orthopedics labs tended to initially cut the bone specimens on anatomical axes.
So they would do you know the superior-inferior, or the medial-lateral, or the posterior-anterior.
But the bone orientation didn't line up with those directions.
So the bone might have been this way, but they were loading it this way, and so that gave this misalignment.
And there could be some variation in the solid properties too.
So you could imagine some solid might have more micro cracks than another.
So if you took say human bone of different ages, you might expect the older bone to have more micro cracks in it.
So these plots put a lot of data together, and then the lines are based on models for open-cell foams.
So the relative modulus goes roughly as a relative density squared, and the cell walls are bending.
And the compressive strength goes roughly as the modulus squared, and that's related to this plastic buckling.
And then the tensile stress or tensile strength depends on the formation of plastic hinges, and it goes roughly as the density to the 3/2 power.
And one observation that people have made is that in compression, if the modulus and the strength both go as a density squared, then the ratio of the strength to the modulus is just a constant, and that, in fact, is just the strain at failure, or the strain for that say, the plateau.
And that's a strain of about 0.7%, and that's pretty consistent in trabecular bone.
Let's see.
And we said sometimes the bone was relatively aligned.
So here's that picture of the femoral condyle again in the knee, and you can see the bones lined up.
If you have bone that's lined up like that and you load it along the direction of alignment, then you can get axial deformation in the trabeculae.
And then you would expect the moduli would go linearly with the density.
And here's some data for the Young's modulus and the compressive strength of bone that was fairly aligned.
So this was selected to be aligned.
So here's the modulus here.
And the square data points are the longitudinal direction, and the little diamond, these little stars, are transverse.
So here's a line of slope 1, and again, they don't all lie perfectly on that line, but roughly the slope is about 1 there.
And then similarly here, this is the compressive strength.
Now the little squares are the longitudinal data, and they're not exactly on a slope of 1, but they're more or less on a slope of 1.
So I'll just say in some regions, the bone may be aligned.
And then axial deformation is important.
And then you would expect the modulus to go linearly with the density and the strength to go linearly with the density in the longitudinal direction.
Then finally I wanted to finish up the bit on the modeling by making one of these plots a little bit like we did for wood.
So here's the Young's modulus of bone plotted against the density.
The trabecular bone is down here.
It's sort the lowest density.
And then this is the collagen that's in the solid part of the bone, and this is hydroxyapatite, the mineral.
So the modulus of hydroxyapatite is around 120 gigapascals, and the modulus of collagen is somewhere around 5.
And if you make composites of collagen and hydroxyapatite, their moduli are going to be in this envelope here, and compact bone, the modulus fits in around here.
Remember I said it was around 18 gigapascals.
So then if you take a compact bone and you turn it into trabecular bone, you'd expect the modulus would go down along a slope of 2.
So here's our little slope of 2, and more or less that's what you see with the trabecular bone.
So the idea is that the models give you kind of a general idea of how the bone is behaving, but it's not really meant to predict a particular piece of bone.
Because a particular piece is going to have a particular geometry.
Typically they're not equi ax and isotropic.
All right.
So are we good with the general overview?
Are we good with how fewer equations there are now that we got past the first part of the course?
So I'm going to talk a bit more about osteoporosis, and I'm going to talk about some modeling that my group did to look at the consequences of osteoporosis.
And then later on we're going to talk a little bit about using metal foams as a possible replacement material for a trabecular bone as well.
And I have a little bit of a talk on using trabecular bone in evolutionary studies to see whether or not a species was bipedal or quadrupedal.
So I think I talked about this a little bit in 3032, but I have more slides and more stuff I'm going to talk about.
So let me get myself organized.
So osteoporosis comes from the Latin, and it actually means porous bones.
So osteo means bone, and not too surprisingly porosis means porous.
So this next slide gives you some idea what osteoporotic bone looks like.
So the top slide is normal bone in a 55-year-old woman.
These are sections from the lumbar spine.
And that bone up here is 17% dense, so the relative density is point 0.17.
And this is a section from the same area of bone in an 86-year-old woman, and it's 7% dense.
So you can see there's a huge difference in the density, and you can start to see what happens when you lose bone mass.
So if you look at this bone up here, it's all well connected.
Each little trabeculae is connected to its neighboring friends.
And you can see down here, I mean, you look at this and you kind of go ouch just looking at it.
Because this piece of bone here is just kind of dangling off, not connected to anything.
And you can see the struts have gotten thinner, so they've lost bone mass by thinning.
And then as I said when the thickness gets less than they are roughly equal to the size of the cells, then the cells can't live anymore, and the bone strut just disappears altogether.
So it's not too surprising that if you lose this much bone mass, there's mechanical consequences, and there's a greater risk of fracture.
And as I said the two most common types of fractures are hip fractures and vertebral fractures.
So let's see here.
So as people age, everybody loses bone mass.
And happily for you and not so happily for me, the bone mass peaks at about 25 years old.
So you're probably either not at the peak or just barely at the peak.
And then it decreases after that every year.
I'm considerably older than you.
And in women, when you go through menopause, the cessation of estrogen production increases the bone loss.
And so typically, osteoporosis is most common in post menopausal women.
And osteoporosis is defined as a bone mass 2.5 standard deviations or more below that of a young, normal mean.
So it's not like you fall and break your hip and they say you have osteoporosis.
It's based on the bone mass.
And as I said, the trabeculae thin and then they resorb completely.
So anybody here take Latin?
Yes.
I did Latin for one year in high school.
So trabeculae, with an E on the end here, trabeculae, I suppose, is the plural.
Trabecula with an A is singular.
And that comes from Latin.
So you don't say trabeculas.
That's a no-no.
All right.
Let me get rid of these.
So if we saw that the strength of the bone varies as the density squared, you can begin to see how sensitive the strength is going to be to this bone mass loss.
So say you went from a density of 0.2 to a density of 0.1, then the densities changed by a factor of 2 so that the densities gone down by 1/2, but the strength is going to go down by a factor of 4.
You're going to have the strength to be a 1/4.
And so you're going to have a big change in the strength.
And you can imagine is the trabeculae thins, this buckling gets easier to happen.
And then once the trabeculae begin to resorb as they disappear altogether, it's like I said.
it's like having a building's framework.
Now, you're removing beams and columns, and the strength is going to go down even more dramatically.
And the way we model the osteoporotic bone is we use finite element analysis.
So before we talked about using the unit cell for the honeycomb.
But to use the unit cell, you have to have repeating unit cells, and obviously, you don't have that.
You've got local variations in what the structure looks like.
And we also used a dimensional analysis.
And the dimensional analysis relies on the geometry being similar from one specimen to another, and you can't really rely on that either for the osteoporotic bone.
And so what we've done is-- and this is what other people do as well, is use finite element modeling to represent the bone.
So initially what we did was we used a 2D Voronoi model.
So remember we talked about Voronoi honeycombs and Voronoi foams.
So I like to start out with simple things, so we started out with 2D Voronoi model for a honeycomb.
Then we did a 2D representation of vertebral bone.
And then we had a 3D Voronoi.
And I had a couple of students who did this.
Matt Silver was the one who did the first two, and Sereca Vagilla was the one who did the last one.
So let's see.
I think that's probably a good place to stop there for today.
So next time I'll talk about the modeling of osteoporotic bone, and we might talk a little bit about metal bones as a substitute for trabecular-- metal foams as a substitute.
I don't think we'll get to the evolution stuff.
We probably won't quite finish next time.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
So last time we were talking about trabecular bone and that it's this porous kind of foam-like type of bone.
And we talked a little bit about the modeling.
And I think I got as far as starting to talk about osteoporosis.
And I wanted to talk today about how we can model osteoporosis using those voronoi honeycombs that we talked about a while ago when we were talking about the structure.
So Bruno has a project on trabecular bone for the class.
And he needed bone samples.
And so we talked about different kind of bone samples we could get.
And the thing is if you get human bone-- well, there's all sorts of issues about just handling human bone and permissions, and it's complicated.
So that was too complicated.
We've used bovine bone before.
You just go to the slaughterhouse and get bovine bone.
But one of the things with trabecular bone is because it grows in response to loads, the geometry of it can be different, sort of architecture can vary from one spot in the bone to another.
And I have a colleague who started doing tests on whale bones, just because it's a way of getting nice, uniform bone.
So I was on a Ph.D. Committee a few years ago for a student who was in the Woods Hole program.
She was at MIT, but was doing Woods Hole thing.
And I don't know if you've heard of the Atlantic right whales, the North Atlantic right whales.
They're endangered.
There's about 500 of them left in the world.
And they migrate between like typically the Bay of Fundy and off the Florida coast.
So they go up and down the coast.
And they sometimes get hit by ships.
And then bones break and that kills them.
And her study was on ship impacts on right whales.
And so I got to know people at Woods Hole who worked on whales.
So Bruno, I called up my friend at Woods Hole.
And the Woods Hole guy didn't have any bones.
But he put me in touch with somebody at the Mass Fish and Wildlife Department.
And he had a couple of whale vertebrae that he was willing to give up.
So I got one of them for you.
So here it is.
So I went out to the Mass Fish and Wildlife yesterday.
And they produced this bone for me.
I could either pass it around.
It's not too heavy.
Or you could come up and look afterwards.
Shall I pass it around?
Or do you want came up afterwards?
Maybe come up-- pass it?
OK.
So one of the things is our like vertebrae, there's these things that stick up like this.
There's one that's missing off of this bone.
There should have been one down here too.
But this is sort of what's called the body of the vertebrae.
And in human vertebrae it's about that big.
But it's the same kind of general structure.
There's these kind of bony plates that come off.
And the body is almost all trabecular bone.
So you can see on this side, this is a growth plate here.
But on this side, there normally would have been a thin shell of the cortical bone.
And when I pass it around, if you look up at this point here, you can see there's just a little bit of that left.
But it's kind of gotten worn out.
And the rest of this, if you look, you can see it's the trabecular bone.
And you can see it's pretty uniform, which is why I thought this might be good for your tests.
So I thought you should get in touch with Mike Tarkanian.
And I talked to him a little bit about cutting it with a water jet cutter.
So I think we could use the water jet cutter.
And I think I emailed you.
If you could cut it in half, I'd like to have some pictures of it cut in half.
And he's got a diamond corer, cylindrical corer.
So you could make little cylindrical specimens.
And I know if you want to do compression tests or beams
Well, I was planning to do compression tests
Yeah, so you could probably use some kind of a bands or something to cut them up.
So, and this is where the spinal column goes through in the whale.
So there you have it.
Anybody want to ask me anything about the whale bone?
OK, so let me pass that around.
And I guess I have to tell you a couple of other stories.
So while I was there, they have a brand new building.
And it's all got solar panels and geothermal heat.
And it's all very groovy.
And the guy who I was talking to about the bone, he wanted to give me a little tour of the building.
And he said, oh, you've got to see your freezer.
OK, so the freezer.
So he opens the freezer door.
The freezer's like a room.
And there's like a bear.
I'm serious, like a bear on the floor, like a dead bear on the floor of the freezer.
And he said it was like a two-year-old bear that had, I guess, just come out of hibernation a couple weeks ago.
I think it had gotten hit by a car or something.
And somebody must have called them up.
And they have it.
So they had this bear.
They had like several deer.
They had a coyote.
They had like boxes full of all kinds of animals.
So anyway it was kind interesting to see all these animals there.
And you know what he said about the deer?
He said, normally deer in Massachusetts don't starve.
Like if you live in the suburbs, you actually have a problem with deer eating your vegetable garden because there's deer all over the place.
But he said, this winter, you know how much snow we've had and how cold it's been?
He said, deer have been starving this winter.
And I think a couple of the deer they'd had had actually starved to death.
And people had called up.
And they had come and kind of collected the carcasses.
So anyway that was my little trip out to Mass Fish and Wildlife yesterday.
OK, so let's go back to talking about the bone.
And I think last time we kind of left off more or less here.
So this was a slide of what osteoporosis looks like.
And you can see the bone loss is a combination of thinning of the struts and resorption of the struts.
And we wanted to try to model this, sort of an engineering sense.
And the way we did that is we used our voronoi honeycombs.
So the bone has an irregular structure.
And we wanted to look and see if we could model something that had an irregular structure.
So we used the voronoi honeycomb.
And if you remember when we talked about the structure of cellular materials earlier on in the course, we said that these are generated by putting down random seed points and then drawing the perpendicular bisectors.
And if you have a constraint that the seed points can't be closer than some exclusion distance, then you get structure where you have cells that are not all exactly the same size, but roughly the same size.
So that's what we did here.
We got this structure here.
And I had a graduate student, Matt Silva, who did a lot of these studies.
And then he used this and analyzed it using finite element analysis.
So the first thing he did was he calculated the elastic moduli of the structure.
So he applied loads.
We calculated deformations.
Figured out elastic moduli.
And so this plot here shows a comparison between the analytical equations that we derived at the first part of the course and what he calculated for these voronoi honeycombs for the final element analysis.
So here's Young's modulus down here, for example.
In the closed form, the line, that's just the analytical equation we had originally.
And the little dashed line is his finite element.
Here's the Shear modulus down here.
And here's the possum's ratio up here.
So you can see, there's a pretty good agreement between these two things here.
So let me write some of this on the board.
And then I can keep going after that.
So for the 2D voronoi honeycomb, we have the random seed points.
And we used the perpendicular bisectors.
And we used a minimum separation distance between the points.
So we generated the structure.
And then we did a finite element analysis.
And from that, we calculated the modulus.
And what we found was the finite element analysis results were pretty close to our closed form analytical model.
And if we think about the modulus, the modulus is the average stiffness over the whole honeycomb.
And when we look at the strengths next, the strength is going to be related to the weakest few struts.
And we're going to find that the strength doesn't work quite the same way.
So first, we got the modulus.
That's sort of the simplest thing to calculate.
And then after that, we wanted to calculate the compressive strength.
So we did a similar thing.
We set up the voronoi honeycomb in the finite element analysis.
We had a few more elements along the length of each strut.
And then we modeled the elastic buckling and the plastic failure behavior.
And we looked at honeycombs that had different densities.
And the lowest densities failed by buckling.
And the higher densities failed by a sort of plastic yielding.
And we assumed that the cell walls were elastic, perfectly plastic.
Remember we said if we have a material where-- this is for the solids-- so say this is the stress in the solid and that's the strain in the solid.
If it behaves like that, we say that's elastic, perfectly plastic.
So we modeled the walls as elastic, perfectly plastic.
And we made the ratio of the solid modulus to the yield strength similar to what it would be for bone, which is about 0.01.
And for that particular value, the transition between the elastic buckling and the plastic yielding failure was at a relative density of about 0.035.
So then what we did was we analyzed structures that were a little lower than that were equal to that and then a little higher than that.
So we would try and see what happened with these different failure modes.
And then we found that if we look at the compressive stress strain curve, this model here had a relative density of 15%.
And the transition occurred at a relative density about 3.5%.
So this one here failed by a plastic failure.
You can see, if we unload it, there's some plastic deformation.
We get a little strange softening here, which is kind of characteristic of the plastic failure.
When we look at the overall deformation of the honeycomb, we saw local kind of failure, like we do in aluminum honeycombs.
So that was the kind of stress strain curve for a relatively dense honeycomb that failed by yielding.
And then this is one for a much lower density.
This is now point 1.5%.
And this one fails by elastic buckling.
And if we load it up and unload it, we recover most of the deformation.
Barry, do you think you could make that stop?
Yeah
There's a chance it could be--
Below.
Just because we're recording it.
It just doesn't seem very good.
OK, so what we did was we made five different voronoi honeycomb.
So we had five sets of seed points.
And we had five slightly different geometries.
And then we averaged the results of those.
So if we make that calculation, this is the strength of the voronoi honeycomb here divided by the strength of the periodic regular hexagonal honeycomb.
And this was plotted against relative density.
And you can see the strengths for the voronoi structure are a little bit lower than for the regular periodic hexagonal honeycomb.
And they reach a minimum here.
And the minimum's around about 0.05 relative density.
So this was the 1.5.
That was a 3.5.
That's 5% and 15%
Why is there like a wide gap between 0.05 and--
Well, I think because we felt-- I think this one failed by some combination of-- there was a limit to how many of these we were going to do.
And this is what we chose to do.
We wanted one that we knew was going to fail by plastic yielding.
And that was this one.
And we wanted one that we knew it was going to fail by elastic buckling.
And we wanted a couple in between.
So we didn't do-- I guess we were lazy is the real reason there's not another point in the middle there
It just seems like there might be a variable--
You think it might go [CAREENING NOISE] like that.
Except there's a physical reason why this happens.
And I'm going to get to that in a minute.
So let me I'll finish explaining this, and then I'll put the notes on the board.
So first of all, the strength is less than the regular hexagonal honeycomb.
And then there's also this minimum here around about 5% density.
And I think the reason that the strength is not the same as in the voronoi in the periodic honeycomb is because if you look at the distribution of strains-- or you can think of the distribution of stresses-- so these were the normal strains at the nodes in the honeycombs.
And the distribution here is for the voronoi honeycomb.
So the voronoi honeycomb, you have lots of members of different lengths.
There are different orientations.
And so there's some distribution of stresses and strains in each member.
And that gives you the distribution.
In the regular hexagonal honeycomb, if you look at just the nodes, there's really just the vertical member, which has a certain strain.
And all the vertical members are going to have the same strain at the nodes.
And then the obliques members, if you just look at the nodes, there's just going to be a maximum tension and a maximum compression at the nodes.
Because there's a unit cell and it repeats, all the oblique ones are going to have the same maximum and the same minimum.
So the dashed lines here are for the regular hexagonal honeycomb.
So the thing to observe here is that the voronoi has some strains and correspondingly some stresses that are outside the range of the regular periodic honeycomb.
And so if there's parts of it that are seeing higher strains and higher stresses, it's going to fail at a lower load.
So I think it's this distribution because you've got this random structure, and you've got different lengths and different orientations of the members.
So that's one reason why these strengths are less than the periodic structure.
I think there's a minimum here, because-- I think before the test I mentioned there's some interaction between elastic buckling and plastic yielding.
And when you get that interaction, that also reduces the strength.
And so there's a minimum near where the crossover is between the elastic buckling and the plastic yielding.
So let me write some notes of the compressive strength.
So we'll say the cell wall-- so the cell wall elastic, perfectly plastic.
And the yield strength relative to the modulus for the solid was 0.1, which is pretty much what it is for bone.
And we assumed that the possum's ratio of the solid was 0.3.
And for this value of sigma Ys over Es, the transition between elastic buckling and plastic yielding is at about 3.5% relative density.
So then we made models with densities that were a little bit less than that and a little bit more than that.
And then the strengths we got from the voronoi were between about 0.6 and 0.8 times what we got from the periodic.
So then what we looked at were these maximum strains of the nodes.
And we found that because the voronoi honeycomb had a much broader distribution of those strains, that led to the lower strengths.
And then we found the minimum strength was at a density of about 5%.
And if you think just about a pin ended column, and you make a plot of the strength-- so I'm just going to say the strength of that columm-- against l/r, the slenderness ratio.
So say it's a cylinder, l would be the length.
r would be the radius.
If you just had Euler buckling, you get a curve that looked like that.
The Euler buckling load is pi squared EI over l. squared.
So I goes as r to the 4th.
And if we get the stress, then it's going to be that divided by the area of the column.
So it's pi squared E. Moment of inertia is pi r to the 4th over 4 l squared.
And the area is pi r squared.
So it goes is as r over l squared, or 1 over l over r squared.
So this would be the Euler buckling.
So that's elastic buckling.
But at some point, the column is going to yield.
If I make it really, really short, then it's going to yield before it buckles.
And at some point, this would be the real stress here.
And this would be failure by the plastic yielding.
And in practice, there's not like a sharp corner here.
You know, if you made columns of progressively longer length, and you tested them, the little short ones would yield.
So they'd be along here.
But they don't kind of yield and the next one buckles.
In fact, you get something like that.
So that when you're near that transition, when you're near this point here, the actual failure stress is a little bit less than that.
And I think that's partly what's going on over there.
It's the minimum.
OK, so those two studies, we just looked at the modeling and the strength of honeycombs.
And we were kind of looking at how does the random structure change the properties.
So the randomness of the structure didn't really change the modulus much at all.
But it did affect the strength.
And then the next thing we did was we looked at putting defects into the bone.
So we knew that the bone, partly the density is reduced by thinning of the struts and partly by resorption when there's a lot of density loss, a lot of bone loss.
So we wanted to look at putting defects where we actually removed some of the struts.
So we did another series of voronoi models.
So we did another series of tests where we looked at it if we have a certain density that we start with and then we look at losing the same bone mass or relative density in our honeycomb, and we look at what happens if we thin the struts versus if we remove struts.
So these plots here show that.
And Matt Silva also did this.
So this is the residual modulus plotted against the reduction in relative density.
So residual modulus is the modulus after we've reduced the density by some amount relative to the initial modulus.
And if we have an intact honeycomb where we just thin the struts, we don't remove any of the struts, if we have an intact honeycomb, as you reduce the density, the modulus just goes down like that.
So that's really just the same as those analytical formulas that we had.
But if you start removing struts, not too surprisingly, the modulus goes down substantially more.
And at this value here, I think it was 30% or 35% density reduction, you reach what's called the percolation threshold.
At the percolation threshold, you have two separate pieces of material.
So obviously the mechanical properties are going to go down to zero when you reach that percolation threshold.
And then this plot over here is the same sort of thing, but now for strength.
So it's the strength of the bone, or the strength of the honeycomb, after you've either thinned or resorbed the wall, divided by the strength of the intact honeycomb.
And again, you're reducing the density here.
And this is for the intact model where you're just thinning the struts.
And this is for the models where you're removing the struts.
So you can see that if you think of-- this is kind of a simple model, but if you think of the bone, if you first thin the struts, you're going to lose a little bit of strength.
But then if you start removing the struts, you're going to lose a lot of strength.
So that's really where people run into-- there's much higher risk of fracture once you get to the point where you might be resorbing struts.
OK, so let me see, what's next thing?
OK, let me just finish the slides, and then I'll put the notes up.
So this is the same sort of thing, just plotting the strength and the modulus on the same plot.
So you can see the shape of the curves is very similar.
The modulus is a little bit more sensitive than the strength.
And here we are thinning, and here we are removing the struts.
And then the next step was that we made a model that was more similar to bones.
So let me write down the notes for this.
And then we'll do that one that's more similar to bone.
Thought it didn't makes sense.
OK, so then we were interested in trying to model something that was a little closer to the structure of bone.
And so we set up this model here.
So we started with just a square voronoi.
So you just force the points into a square, voronoi, or a square pattern, you get a square voronoi.
And then what we did is we just perturbed the points a little bit.
And we got this perturbed voronoi array.
And so we made this model here.
And we took a piece of vertebral bone.
And we measured the angle of orientation of a lot of the struts.
And we matched our voronoi model to that distribution in the bone.
So our model looked something like this.
So you can kind of see how it's more or less vertical and horizontal, but not exactly.
And here's a sort of comparison with a slice of vertebral bone.
And again, because the loads are more or less vertical in the bone, the trabeculae tend to orient that way.
And then have some horizontal struts.
So here you can see on the left, we've got a voronoi model that's more or less representative of the bone.
And we've removed some of the struts.
So we're going to the same thing with this model.
We thin the struts.
And we remove the struts.
And we try to see what the residual strength is.
And you can see there's for of at least a 2d model this isn't a bad representation of the vertebral trabecular bone.
And this was the stress strain curve for both the specimen of the vertebral bone that was tested and then the honeycomb model that we made to kind of match it.
So a similar kind of behavior.
This is how our model failed, this sort of a local band of struts that fail.
Let's call it local deformation band.
And then what we did was we thought about changing the density.
And what we did was we removed either horizontal or vertical struts, or we thinned either the vertical or the horizontal struts.
So these are the same kinds of plots as I showed before.
This is the residual modulus.
This is the residual strength.
This is the density reduction.
And here we're reducing the density by making struts thinner.
So it's still intact.
We haven't removed any struts.
But each of the struts gets thinner.
And this top line is for thinning the longitudinal or the vertical struts.
And this was for thinning the transverse or horizontal struts.
And then this is for removing struts here.
So we're removing a bigger number.
So the more we remove, the more the density changes.
And then this plot here is for the strength.
So again, this is thinning.
So we're moving either the horizontal or the vertical ones-- I'm sorry, thinning either the horizontal ones or the vertical ones.
And then this is for removing struts.
And again removing more to reduce the density more.
So you can kind of see we're kind of working our way to more complex models here.
So this one here was looking at the bone.
Let me write some notes for this.
And then we did a 3D voronoi model.
So I'll do that one next.
Oop, over here.
So I'll just say the model was adapted to reflect the trabecular bones study in the vertebrae more.
So we perturb a square with array of seed points to get a structure that was more like the bone.
And then we looked at the reduction in the thickness and the number of strides in the longitudinal and transverse directions.
OK, and then the next model we did was the 3D version.
So we made a same kind of thing.
But with the 3D model, we had fewer cells in that model, because once it goes to 3D, you've got more struts in each cell.
But this was the same idea.
We uniformly thinned the struts, or we removed the struts.
And again, you can see removing the struts is a much bigger effect.
And also the other thing to look at here is for 3D the percolation threshold is 50%.
So it kind of makes sense that if it's in 3D, and you've got struts in all directions, you're going to have to remove more of them.
You're going to reduce the density more to break it into two separate pieces at the percolation threshold.
So that was the 3D model there.
And then this is just a comparison of the 3D with the 2D for the modulus.
We just did the modulus for the 3D structure because it sort of gets computationally more involved.
So the 3D, these two lines here corresponded to removing the struts and the change in the modulus.
And the little triangles were the 3D voronoi calculation that we did.
The little crosses here were the same sort of calculation done for tetratridecahedron.
One of my former students had loaded that.
And then at the bottom here are the lines for the 2D structures, for either a regular hexagonal cell or for a 2D voronoi cell.
And you can see, there's not a huge difference whether or not you take a regular structure or a voronoi random structure.
But there's a fairly significant difference between the 2D and the 3D.
So the 3D, just you have to remove more material before you get the same reduction in modulus.
So let me write some notes for that.
So it's the same kind of analysis, but just with a 3D model.
So do you see the idea with these models?
It was an attempt to look at a way that you could computationally estimate how much modulus loss or strength loss you get by either thinning the struts or removing the struts.
So I gave you a way to model the osteoperotic bone.
Yes
Can you clarify the percolation threshold?
So the percolation threshold is say you have some network and you start removing things.
If your remove enough, you have two separate pieces of things.
If you remove enough struts, you have two separate pieces.
That's called the percolation threshold.
So I think it's-- you want to know why it's called that?
So I think that originated because it wasn't used in this kind of context.
I think instead it was used in a context where imagine you have 2D plate.
So you put 2D holes in it, little circular holes.
And they we're looking at flow of a fluid through the plate.
So you can imagine, if you put enough holes, eventually they line up or they-- line up is not the right term-- but there's enough of them that they connect.
And you end up with two separate.
You have a path through for the fluid.
That's called the percolation threshold.
But in mechanics it's sort of two separate pieces.
OK, does that makes sense?
So you know, at the percolation threshold, the stiffness is zero because you have two separate things and the strength is 0
So the two doesn't have to be like separated by--
It could be-- yeah, yeah, it doesn't have to be a straight line
Does it make a threshold between whether everything is interconnected?
Yes.
Or there's some path that separates them.
Yeah, that's what it is.
OK, so that's the end of the bit on osteoporosis.
I had a couple more things I wanted to talk about on bone.
So the next part is on the idea of using metal foams as a bone substitute materials.
So when they make hip implants-- so they're typically metals.
They're titanium or stainless steel or something.
Often what they do is they coat the outside with some sort of porous coating.
And the idea is the porous coating allows the bone to grow in.
So you can kind of imagine, like especially on the stem and around the head of a femur, you'd like the bone to grow into that to attach.
And having a porous coating helps.
And there's a couple of ways they do it currently.
They use porous sintered beads.
So they put little beads of metal on the outside.
And the idea is that the bone grows into the little gaps between the beads.
Or sometimes if they have-- not so much for hip implants but sometimes when people have say car accidents and their face get sort of smashed up, and they have to have, say, a plate put in their face, and they need like a flatter plate, they use like a wire mesh.
And they have sort of a wire mesh that goes on the outside.
And it's the same thing.
It's the idea to try to get the bone to grow into that plate.
So some people are thinking instead of using the porous sintered beads, or instead of using one of these wire mesh plates, that you could use metal foams.
And so there's been some interest in using metal foams in coatings of implants.
And longer term, there's been some interest in trying to make sort of a vertebral body that would involve using a metal foam from the vertebral body.
So you know that whale bone that we just passed around, that vetebral body, that cylindrical part, it's almost all trabecular bone.
So there's some interest in trying to use metal foams for spinal surgeries.
Maybe not to replace the whole body.
But to use in part of the surgery.
So I have a little slide here which shows a bunch of metal foams that people have made with the idea in mind that perhaps some of this could be used in orthopedic surgery.
So these are some different kinds of metal foams.
And these ones are made from titanium or tantalum.
So typically, the metals that they use in orthopedic implants are the cobalt chromium alloys.
Titanium, they sometimes use tantalum or stainless steel.
And they use those because they're biocompatible.
And they're very corrosion resistant.
So these ones here are mostly titanium.
And this is one that's a tantalum.
So let me just go over how they make them.
And then I've got another slide that goes over it in more detail.
And I'll write a few notes down.
So this guy on the top left up here, that's made by taking an open cell polyurethane foam, like a seating foam, like a cushion.
And then what they do is they heat that up in an inert atmosphere, so that they are left just with the carbon.
So it's a sort of vitreous carbon.
And then what they do is they coat that carbon by a CVD process with tantalum.
And so they end up with a tantalum foam with a very thin layer of carbon at the core.
The carbon makes up something like 1% of the final composition and the tantalum is the 99%.
So that's sort of a replica process there.
This one here is made by another replica process.
They take an open cell polyurethane foam.
They infiltrate it with a slurry, which has the titanium hydride particles in it.
Remember when we talk about processing of the foams at the beginning, I said, if you heat up the titanium hydride, eventually the hydrogen would be driven off, and you'd be left with the titanium.
So they do that, and then they sinter it, and they get a titanium foam.
This one is made by a fugitive phase process.
So the idea with a fugitive phase is you burn off some phase.
So you could mix powders, consolidate the powders, then you burn off one of the powders.
And you're left with the other one.
And then you need to sinter that together to make it have some reasonable mechanical properties.
This one here is made by using a foaming agent.
This one's made by expansion of argon gas.
I think when we talked about the metal foams, we talked about the idea of packing, say, titanium powder in a can.
And then you evaporate the can.
And you then pressurize the can with argon gas.
And then you heat treat it.
So you heat it up.
And as you heat it up, the argon gas expands.
And you're left with the pores.
This one's made by a freeze casting process.
I have a slide.
And I'll talk about that in a minute.
This one's made by a selective laser sensory.
So it's like a 3D printing.
But instead of printing in ink, this time you have a bed of a powder.
And you've got a laser that selectively sinters.
So you turn the laser on and off.
And where it's on, it's going to bind the material.
When it's off, it's not going to bind the material.
And then you raise the thing.
You make a little bit more powder.
You do it all over again.
And you can get different patterns.
And then this is made by a sort of process in which they take powders and press them and ignite them.
But I think that's not very commonly used.
So this is some more details about how they might do it.
So this is the fugitive phase process here.
You could take a titanium and a filler powder, pack them together.
You know, you'd mix them up, pack them together.
You would raise it to a certain temperature to decompose the filler.
So typically the filler decomposes at a lower temperature than what you sinter the powder, the metal powder that's left.
So you decompose the filler.
You drive that off.
And then you sinter the metal powder.
And you're left with porous titanium.
This one's the expansion of the foaming agent.
So you take your titanium powder.
You might have a binder and then a foaming agent.
Mix those all together.
They heat them up until typically the binder becomes a liquid.
And the foam foams up the liquid binder.
Then they drive off the binder.
And then they sinter at a higher temperature the titanium.
So you you've got a porous titanium left.
This is the freeze casting or the freeze drying process.
So here they would take titanium powder and put it in agar.
And the agar's in water.
So the agar is like a jelly, like a gel.
But it's mostly water.
And then if you freeze it, what happens is the water freezes.
And it drives off the titanium agar into the interstitial zones between the frozen ice crystals.
So these little dots here are the ice crystals.
The ice crystals are growing as it gets colder.
And as the ice crystals get bigger and bigger, you're left with the titanium and the agar in between those ice crystals.
And then if you sublimate the ice off, you're left with a porous thing.
And you can sinter the titanium if you want to make it a little more dense.
And this is a rapid prototyping thing here.
So you spread the powder.
You could either print a binder or you could use a laser to sort of almost weld the particles together.
Then you would drop the piston, put more powder down, have the binder go again until you made your product.
And then you would get rid of all the unbound material.
And you'd have your final product.
OK, so these are some of the methods they can use for making metal foams.
I don't know, should I write notes?
Or are you good if I just put the notes on the website?
I'll put the notes on the website.
And then this is a stress strain curve for a titanium foam.
So it looks like all these other kinds of foams and bone and wood and everything else that we've looked at.
And this is some data for titanium foams that are made by different processes.
And we've just taken different data from the literature and put it together.
So this is the modulus here.
This is a slope of 2.
You can see some of the data is sort of near that slope, but below it.
And there is obviously a large spread in the data too.
But if you go back and look at the different types of structures, then not all these have this kind of typical foam-like structures.
The structures aren't all quite like a foam either.
Yeah?
So is there any objective in this to make the foam similar in structure to the bone that will be growing into it?
Or does it just need to be scaffolding?
I think for this, they just want to make a porous thing.
And they're thinking about coatings.
So the coatings aren't necessarily similar to the bone.
When we finish the section on bone, we're going to start talking, I think, about tissue engineering.
And when we talk about tissue engineering, people have made scaffolds to try to make them the same shape as the anatomical part that they're trying to mimic.
And then they make a cellular kind of core.
So they have made some more of an attempt to do that.
I could give you a sneak preview.
Would you like a preview?
So these are some scaffolds that are generated from a making different kinds of tissue.
These aren't all from bone.
But this one here, for example, is for regenerating bone.
And they've printed it in this exact geometry, because that's going to replace some anatomical piece.
And they want it in that geometry.
So I'll talk more about that.
But so for the scaffolds, they sometimes do that.
But not so much for these bone coatings.
Let's see.
Se we did that.
We did that.
So these are the data.
And then over here, there's the compressive strength.
And again, you know, this is our line with a slope of 3/2.
Again, there's a lot of scatter, because there's a lot of variation in the structure of these things.
But it's in the ballpark.
So are we good with metal foams?
And there's just like a page and a half of little notes.
Should I just put that on the website?
You're good?
OK. OK, so the last topic I wanted to talk about on bone has to do with how people look at the structure of bone in evolution and in evolutionary studies.
So the idea here is, in particular in looking at sort of pre-human species, sort of hominid species, people are interested in when primates went from being quadruplets to bipeds.
So obviously, bipedalism, walking on two legs, is characteristic of us.
And people would like to know if they find some fossil-- you can't just tell from the fossil directly is it a biped or a quadraped.
You can't see the species moving because it doesn't exist anymore.
So you'd like to have some way of making some estimate of whether or not it was a biped or a quadruped.
Let me get my notes together here.
So I wanted to kind of look at the big picture a little bit and look at the evolution of different species.
And this is a phylogenetic sort of chart.
And this is kind of-- have you heard of the tree of life?
This is like the tree of life.
So this piece of it is-- metazoa is for animals.
So not plants, animals.
And this goes back about 1.2 billion years.
So these are millions of years ago.
And then these are different sort of eras and ages that are defined.
And when we have a branch here, this is a common ancestor.
And then this is a branch in one direction, and that's a branch in another direction.
So these points here, like 1 and 2 and so on, the implication there is that there was a common ancestor to everything that traces back to there.
So this point here, 1, is 1.2 billion years ago.
So this was sort of very early kind of species.
And the very first things were, well, multicellular things.
I mean, there were little amoeba type things.
But the more sort of sophisticated animals were sponges.
And there's three kind of classes of the sponges.
There's calcarea.
And they're called calcarea because they're mineralized.
And they're mineralized with a calcium carbonate.
And then there's another branch of them called-- I don't know if I'm going to say this right, but hexactinellida.
And those have glass.
Those are called glass sponges because they have SiO2 is the mineral in those.
And then there's these guys here the demospongiae.
I think some of those have calcium carbonate and some of them don't.
Oh, no, some of them have silica.
And some of them don't.
So these things here are sort of very early multicellular structures.
And the mineral in them is either calcium carbonate or silica.
And then if we move up, I've sort of highlighted a few of them.
The cnideria-- I know there's a C, but that's actually-- when I say s-nideria, my biologist friends laugh at me.
And they say no, no, no, it's nideria.
The cnideria are the corals and the jellyfish.
And corals are also mineralized with calcium carbonate.
And you can see they branched off something like a billion years ago.
Then we get up here.
These are the mollusks.
So the mollusks are things like bivalves, like if you like to eat claims, things like that.
So bivalves, snails, and things like octopi, octopus.
So those are all molluscs.
And molluscs, when they're mineralized, also are calcium carbonate.
So those are the calcium carbonate kind of shell.
So we haven't got up to anything bony yet.
Bone is collagen plus a calcium phosphate.
So we haven't gotten to anything that's a calcium phosphate yet.
Then another large class is arthropoda.
That's insects and spiders and crustaceans.
Those all have a chiton skeleton.
So if you think of like the exoskeleton of an insect or a spider, those are chiton.
And crustaceans, things like lobsters, those also have a chiton shell.
And in crustaceans, it might be mineralized.
But again, the mineral is a calcium carbonate.
So all the way up here, most of these things, if there is any mineral, it's calcium carbonate.
And if we get up finally to the vertebraes, the vertebraes have the calcium phosphate and have a bone, like what we think of as real bone.
So the vertebrates obviously involve things like mammals, birds, snakes, and fish.
So this is kind of the big picture going back.
And sort of one of the interesting things is that bone doesn't come along until you get somewhere over here.
And I have one more little, nice slide here.
So when I talked about the sponges, they were these guys here with the glass sponges.
Joanna Aizenberg at Harvard did a nice study on glass sponges.
And she looked at this one here.
It's called the Venus flower basket is kind of the common name.
And it has this hierarchical structure.
And it's remarkably stiff and tough.
And what she did in her paper was look at optical and mechanical properties.
But they looked at the structure at different length scales.
So there's kind of a cellular structure at this length scale.
It's kind of a tube.
This was just a picture I took in a natural history museum.
But if you look at higher magnification, each little strut is made up of sort of fibers and has a hierarchical structure itself.
So that's just one of the sponges.
And here's a similar chart for the vertebrates.
So this point here is where the other chart kind of branched off.
And if we start with the earliest things again, the earliest ones with the most common ancestor back here is something called cyclostomata.
And those are things like jawless fish, so things like lamprey and hagfish.
Do you know what a hagfish is?
It's this kind of eely thing.
And I have the video for you if you don't know what a hagfish is.
So let me get out of here because it's just an amazing thing, the hagfish.
OK, so let's see, I got my sound on.
[VIDEO PLAYBACK]
Here, at the University of Guelph, about an hour outside Toronto, materials scientist Atsuko Negishi and biologist Julia Herr think that these lovely creatures, called hagfish, may revolutionize how we make strong materials
These are specific hagfish.
They're well known for their unique defense mechanism
So if I wanted to see this, what would we do?
Like could we poke at it with a stick?
I think the best way to do it is to reach in there and grab one
Oh, my gosh, look at that disgusting-- oh, no, I've been slimed.
I feel like an outtake from Ghostbusters.
Look at the quantities of this stuff.
[END PLAYBACK]
He used to do this for The New York Times.
And I think he's got his own going on, but he used to make these little videos.
And he had a show on PBS a year or two ago all about materials.
And there were like four different episodes.
And he talked about different kinds of materials.
And he went to different labs.
But he's quite a character.
But anyway the hagfish have this defense mechanism where they make the slime.
And I don't know if you know Professor McKinley over in mechanical engineering here at MIT, but he collaborates with those people of Guelph.
And he studies what he calls non-Newtonian fluids.
Well, a lot of people call non-Newtonian fluids.
A Newtonian fluid is a thing like water, where the viscosity is a constant no matter what sort of strain rate you shear it at.
And a non-Newtonian fluid does not have that property.
The viscosity changes.
And some of them have a strength, like in the hagfish one has a strength.
So Gareth's studies things like this kind of hagfish slime.
I don't know if he still has them.
He used to have hagfish in his lab over him building, whatever it is, 1 or 3 or something over there.
OK, so that's what the hagfish are, just because that's amusing.
And they don't have bone.
So they and the jawless fish don't have bone, even though they're called vertebrates.
Then the next sort of most recent thing is the chondrocytes.
Those are the cartilaginous fish, so things like shark.
So sharks don't actually have bones.
They have cartilage.
And they have a little bit of mineralization in the cartilage, but they don't actually have bone.
And the first thing that actually has a bone is the ray finned fish, which are these guys here.
And that occurred about 450 million years ago.
And then everything in the vertebrate since then is bony.
So there's coelacanth-- I don't know if you've ever see those.
Every now and then they find one of these things in Florida or something-- lung fish.
There's these guys here, which are the frogs and the toads and the salamanders.
So it's finally getting to be spring after the winter from hell.
And the salamanders are going to come out into the vernal pools and mate.
And it's cute.
So anyway, they come out this time of year.
And then there's the amniota, things that have eggs of one sort or another.
So that includes us, the mammals, the birds, the snakes, and the turtles.
And so those would have branched off about there.
So the last thing I wanted to point out here is that there's this huge kind of diversity of animals that have evolved over millions and millions of years.
And that the first ones that were mineralized used the calcium carbonate and that the bony type materials didn't really evolve or didn't appear until about 450 million years ago.
And then these are the vertebrates that have these kind of bony things.
So as I've said many times now, the bone grows in response to loads.
And the bone structure reflects the mechanical loads and the function.
And evolutionary studies have looked at both cortical bone and trabecular bone architecture to try to say something about the locomotion of the animal or of the species.
So there's ones that look at cortical bone.
But I'm just going to talk about one that deals with trabecular bone.
So this study was done by a group of peoples, the first author was Rook.
And what they studied was the ileum.
So this is a pelvis.
And there's the ileum is one of the bones in the pelvis.
And they found fossils of a hominid species that was about 8 million years old, called Oreopithecus bambolii.
And bambolii refers to the place in Italy where these fossils were found.
And so they found two-- or at least somebody found two pieces of an ilium.
And they took x-rays.
And they made a digital reconstruction so that they would get one ilium-- it turned out that the two pieces were two different parts-- so they could make a whole one out of the two pieces.
And then they looked at the structure of the trabecular bone.
And they compared that structure to the structure of the trabecular bone in humans and other primates.
And they wanted to see is it more like the humans, which are obviously biped, or is it more like some of the primates, which are quadrupeds.
So this just shows for a human and a non-human primate what the structure of the ilium is.
And these little black boxes with the letters are what are called anatomical landmarks.
So they're sort of comparable spots on the bone of different species.
And what they were doing was looking at the trabecular architecture at these different spots.
So you can kind of see how they're more or less analogous in the two different species.
And this is the digitally reconstructed ilium that they put together from their fossil species.
And again, these little letters refer to these anatomical landmarks.
And then what they did was they compared the Oreopithicus, the fossil, with the human and then three non-human primates.
So these four images are all from the fossil.
These are the human.
These are champi, siamang and baboon.
And this square here corresponds to that one there.
This is B, corresponds to that one.
And C and D are those two there.
So they had two fossils, they made the digital reconstruction.
They looked at certain areas.
And then they looked at the same or analogous areas in these other species.
And they tried to look for similarities and differences in the bone structure.
So let's look at this first box A.
You can see there's a very white bit here.
And the white corresponds to more dense.
So there's a white bit that's more dense in their fossil.
And in the human bone, you see is a similar sort of band right there.
And if you look at the other non-human primates, that band is missing.
So that says to them, OK, this feature, this one feature at least, is more similar to the-- sorry, in the fossil here is more similar to the human than it is to the non-human primates.
And then they had three other landmarks that they looked at like that.
So this one here again is a sort of dense regions.
So you can see that white dense region.
There's some there.
There's some here.
So those are the fossil and the human.
But there's a teeny bit here and a teeny bit there.
But it's not as pronounced in the non-human primates.
Yes
From I get at least so far, the portions of the bone that are dense versus not dense seem less relevant to the direction of loading than the orientation of the foamy parts of the trabecular bone
So the density reflects more or less the magnitude of the stresses.
So if the stresses are higher, it's going to be denser.
And the orientation of the bone, like whether or not which way the struts are oriented, that reflects the sort of ratio of the principal stresses.
So if the principal stresses go in a certain direction, the bone's going to tend to line up in that direction.
That's what that Guinea fowl study was kind of about.
OK?
Are we good?
OK. And then these other two, so in B-- and you can't really see it from here, but they looked at the sort of architecture of the trabecular bone.
And they said that it was more similar in the fossil in the human than it was in these other three primates.
And this region here, this looks very similar to that bit there to me.
But I think there was some other feature about this region that they were looking at.
And again, it was more similar to the human than it was to the non-human primates.
So by just looking at the pattern of the bone and the density of the bone and comparing it to these other species, they said that the-- and you know the hip, because we're standing like this, you would kind of expect to see differences in the hip.
That's why they wanted to look at the ilium.
So the conclusion they made was that these observations suggested that the species was bipedal, or at least spent some of its time as a bipedal animal.
And I think that might be it.
Do I have more I have one little summary here.
So just to summarize what we've done on bone this week.
We talked about the structure of the bone.
We talked about mechanical properties in the foam models.
We talked about osteoporosis in the voronoi models, how you can try to represent the loss of bone and the loss of bone strength using those models.
We talked just a little bit about the idea of using metal foams as bone substitutes or coatings as implants, and then this little bit on trabecular architecture and evolutionary studies.
I have some notes, but I think we've got just a couple minutes left.
So maybe I'll just scan those and put them on the website?
Are we good with that?
I have a question about what we just looked at about the different species.
We always consider on the density changes.
Can there always be changes in the solids?
So it changes a little bit from one species to another.
So the question is, does the solid properties of the bone, the solid bone itself change?
So if you look at cortical bone in different species, it changes a little bit, but like 10%, not a huge amount.
So the two most common things people have compared are bovine bone and human bone.
And there is a slight difference between them.
But it's not a huge difference.
One of the other things people have looked at in osteoporosis that I didn't really talk about was there's another whole set of things that can go on that reflect what you're talking about.
So they talk about bone quality as well.
And when they talk about bone quality, they're talking about are there micro cracks in the solid.
So you might imagine as you get older, it's not just that you lose the struts or that the struts get thinner, but the solid bit itself has more cracks in it.
So you can imagine if the solid bone itself had little micro cracks, then it too would be weaker.
And then you think what I put up was bad.
It gets even worse if you put that in as well.
So, yes, people do look at bone quality, which is sort of looking at with age.
And typically it's fairly elderly people that the bone quality is an issue.
I guess there are certain diseases where it's an issue.
But in osteoporosis, it's sort of elderly people.
Any thing else?
Should I stop there for today?
So there were hardly any equations in this.
Did you know that?
So we got to the part where there's lots of equations.
So next week I'm going to talk about tissue engineering.
I think I'm going to talk a little bit about different kinds of scaffolds, how they make scaffolds, how the scaffolds fit sort of into an anatomical things, what is that supposed to represent, how they use them clinically a little bit.
And we had a research project on osteochondrol scaffold, so scaffolds for replacing bone and cartilage.
And I'm going to talk a little bit about that as sort of a case study in tissue engineering scaffolds.
And I have some stuff on cell mechanics and how biological cells interact with scaffold.
I don't if we're going to get to that next week or not, but somewhere close to that.
So the next bit is on tissue engineering scaffolds, osteochondrol scaffolds, cell mechanics.
And there's not that many equations in that part either.
So OK.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
So last time, we finished talking about trabecular bone.
And what I wanted to talk about this week was tissue engineering scaffolds.
So the idea with tissue engineering is you want to be able to repair damaged or diseased tissue.
And typically, that's done by regenerating the tissue in some way.
So in your body, many types of cells, like maybe not blood cells, but most types of cells for sort of structural tissues are attached to an extracellular matrix.
And this is sort of a schematic of the extracellular matrix here.
And the composition depends on the type of tissue that's involved.
For example, in skin, it would be collagen and something called glycosaminoglycans and elastin.
In bone, it would be collagen and a mineral-- a calcium phosphate mineral, hydroxyapatite.
So the composition varies.
But the idea with the tissue engineering scaffold is that you want to make a material that essentially substitutes for the extracellular matrix.
And it does so on a sort of temporary basis.
So the idea is you put something in the body.
The cells attach to that, whatever scaffold you put in.
And the scaffold has to be made in such a way that the material-- that the cells, they can migrate through it.
They can attach to it.
They can differentiate.
They can proliferate.
So the cells can do all their normal function.
And the idea is that as the cells are doing their normal function, they then secrete the natural extracellular matrix.
And the engineered thing that you put in is resorbed.
So there has to be a balance between the rate at which the scaffold you've made resorbs and the rate at which the cells are depositing the new native extracellular matrix.
So that's one of the key things about this.
So this is just an example of sort of a schematic of an extracellular matrix.
In this case here, there's collagen fibers.
So these guys here collagen fibers.
And these kind of hairy-looking things are proteoglycans.
So they have a core of protein with sugars kind of hanging off of them.
And there's different kinds of GAGs, they're called, that hang off of them.
So one's chondroitin sulfate.
The CS here stands for Chondroitin Sulfate.
There's one called dermatan sulfate.
And there's one called heparin sulfate.
So there's different of these glycosaminoglycans.
So let me write down some of this, and then we'll kind of get into this.
So what I wanted to do today was show you some examples of tissue engineering scaffolds.
Show you some of the sort of design requirements.
So you have to have, obviously, a material that's porous so the cells can get in there.
And that's where the cellular solids comes in.
So we'll talk about some scaffolds, some of the sort of design requirements for them.
And also, we'll talk a little bit about processing of the scaffolds and mechanical properties of the scaffolds.
So I'm hoping I can finish most of that today.
And then next time, I have a little case study on osteochondral scaffolds.
So osteo means bone.
And the chondral means cartilage.
So this was sort of a two-layer scaffold that we developed in collaboration with some other people at MIT and some people at Cambridge University.
And it went from a research thing to a startup thing and being used clinically.
So I was going to talk about that Wednesday.
OK.
So let me just get started here.
So the goal of tissue engineering is to regenerate diseased or damaged tissues.
And in the body, the cells attach to the extracellular matrix.
And that's sometimes called the ECM.
Sometimes, the scaffolds are called scaffolds.
And sometimes, they're called matrix because the extracellular matrix is called matrix.
So then, the composition of the ECM depends on the tissue, but it usually involves some sort of structural protein.
So something like collagen or elastin.
It also typically involves some sort adhesive proteins.
So something like fibronectin or laminin.
And it involves these proteoglycans, which are the core of protein with a sugar hanging off them.
Whoops.
And the sugars are typically glycosaminoglycans.
And for short, people call them GAGs, just because it's easier to say.
So some examples of the GAGs are chondroitin sulfate.
And we're going to talk more about that a little later because that's one that we've used in making collagen-based scaffolds with [INAUDIBLE].
So for example, if you look at the composition of extracellular matrix in something like cartilage, it has a collagen component and a hyaluronic acid component and some GAGs.
And this hyaluronic acid is a proteoglycan.
If you look at bone, extracellular matrix in bone, it's made up mostly of collagen and hydroxyapatite.
And if you look at skin, skin is made up of collagen, elastin, and proteoglycans.
And the idea is that the cells have to be attached to this extracellular matrix in order to function.
Or, they have to be attached to this or to other cells in most cases.
So next week, I'm going to talk a bit about cell mechanics.
And we have some video where we watch a cell deforming a scaffold.
And I don't have videos to show you, but I had a student who took videos of cells migrating along with scaffolds.
We'll look at how the stiffness of the scaffold affects how the cells migrate along the scaffold.
So that's kind of the native ECM in the body.
And the idea with tissue engineering is that you want to make a scaffold that's porous that mimics the extracellular matrix in the body.
So, say, there's some tissue that's damaged.
Say there's damaged cartilage.
You want to provide sort of an extracellular matrix that's a sort of synthetic thing that's going to provide the same function as the ECM in the native tissue.
So people have been working on scaffolds for regenerating all sorts of different tissues.
And probably, the most successful one has been used to regenerate skin.
And there's been scaffolds available for regenerating skin for probably almost 20 years now.
And one of the first ones was developed by Professor Yannas in mechanical engineering here at MIT.
And it's actually still sold by a company called Integra.
But this research to develop scaffolds for lots of different tissues, orthopedic tissues, things like bone and cartilage, cardiovascular tissues, nerve-- like Professor Yannas works on peripheral nerve these days.
People have looked at trying to make scaffolds for gastrointestinal tissues.
So all sorts of different tissues.
And at MIT, there's quite a lot of interest in this.
There's a lot of people working on it.
So Bob Langer works on it.
Linda Griffith.
Sangeeta Bhatia.
There's really quite a number of people at MIT who work on it.
Those are just some of the people at MIT.
So the idea is in the body, the cells are constantly resorbing the extracellular matrix and depositing more.
So if you think about bone, for example.
Remember we said bone grows in response to load?
Even in healthy bone, the bone is constantly being resorbed and deposited.
And in normal bone, the rate of resorption and the rate of deposition is roughly the same.
When people get osteoporosis, the thing that happens is that that balance gets out of whack.
And so it's not being deposited at the same rate it's being resorbed.
And the idea with the tissue engineering scaffolds is that they degrade over time.
And that the cells that were attached to them are forming their own extracellular matrix.
So there's kind of a balancing act between the cells depositing the native ECM and the tissue engineering scaffold that was provided, say, by the clinician being resorbed.
So the scaffolds are actually designed to degrade.
And controlling that degradation rate is one of the design parameters of the scaffolds.
OK.
So that's kind of the overall, kind of big picture.
And what I wanted to talk about next was some design requirements for the scaffolds.
So if you think about it, there's different sort of ways you can think about what the requirements are.
So you have to make the scaffold out of some solid.
And there's some requirements for the solid.
So obviously, you want a solid that's biocompatible.
That's one kind of main requirement.
Another requirement is that not only the solid has to be biocompatible, but when the solid decomposes, if it decomposes into other components, they have to be biocompatible too.
So you don't want the solid to degrade during this resorption process into toxic components.
That would be a bad idea.
And then, the other thing is the solid itself has to promote cell attachment, and cell proliferation, and cell migration, all these kinds of things.
So we're going to talk about some different materials for the scaffolds.
And some of them are sort of native proteins.
So some of them are things like collagen.
Collagen already has binding sites for cells to attach to it.
Obviously, it's one of the proteins in the native ECM.
There's also a number of synthetic polymers you can use.
And with the synthetic polymers, they don't have natural binding sites for the cells.
And so you have to coat them with something else.
So you have to coat them with, say, adhesive proteins so that the cells will attach to them.
So we're going to talk about the requirements for the solid.
Then, you make the solid into some sort of porous, foamy thing.
I think I have some slides here.
Here's an example of a collagen GAG scaffold.
And this is one of the ones that is made in Yannas' lab.
And you can see it looks a lot like a foam.
It's very, very porous.
And there's some requirements for the sort of cellular structure, the foamy structure of the scaffold as well.
So typically, you want interconnected pores so it's easy for the cells to migrate in.
Typically, you want pores to be within a certain range.
It turns out if the pores are too small, it makes it difficult for the cells to get in.
Sometimes, they can by eating away at the material.
But typically, the pores-- you want them to be bigger than a certain size.
You also want them to be smaller than a certain size because how much specific surface area, how much surface area per unit volume you've got, depends on the pore size.
The smaller the pores, the more surface area per unit volume you have.
And then, the number of binding sites you have for cells to attach to it depends on that specific surface area.
I'm going to write all this down, so I'll do that.
So there's requirements for sort of the pore structure.
And then, there's also some requirements for the whole scaffold itself.
So for instance, it's got to have some minimal mechanical integrity.
So there's sort of some requirements for that.
So there's requirements for the solid.
There's requirements for the sort of cellular structure, the porous structure.
And there's requirements for the overall scaffold.
So let me write some of these things down.
So there's some requirements for the solid.
So it must be biocompatible.
It must also promote cell attachment and proliferation in the cell functions.
And then it must degrade into nontoxic components.
So there's some requirements for the cellular structure, too.
And what you want to have is a large volume fraction of interconnected pores.
And so you want that to facilitate the cell migration.
And also, the transport of nutrients into the cells.
So you also want the pore size to be within a critical range.
So you need the pores bigger than a lower limit so the cells can migrate through easily, can kind of get in there.
And you want the pore size to be less than an upper limit to have enough surface area to have enough binding sites to actually attach cells.
And for different tissues, there's different critical ranges of the pore size.
So for example, for skin they found that you want to have a pore size between about 20 microns and about 150 microns.
And for bone, the pore sizes that people tend to use are between about 100 and 500 microns.
So there's the pore size.
And one other feature is that the pore geometry should be conducive for the cell type.
So lots of cells are somewhat equiaxed.
Maybe a little elongated.
But if you look at something like nerve cells, like peripheral nerve, they're incredibly elongated.
So you want to have pores in the scaffold that are also very elongated.
And then for the overall scaffold, it needs to have some mechanical integrity.
You guys OK?
Yeah?
We're good.
Oh, achy.
Yeah.
I know.
So you want to have some overall mechanical integrity.
I mean, the thing has to be put into the body in surgery.
And people are going to be pushing and poking at it.
And so it has to have some just overall mechanical integrity.
Also, it turns out that if you put stem cells into scaffolds, the types of cells they differentiate into depends in part on the stiffness of the scaffold.
So you want to be able to control the stiffness of the scaffold
How do they do this research?
Do they use animals?
Or they do it in vitro?
Yeah.
So the question is, how do they do the research?
So they do a sort of series of different things.
So at one level, you could have the scaffold and you'd put cells on it.
Say you're making a bone scaffold.
You'd put osteocytes onto it.
So one level, you just put cells onto it.
And you want to see, are the cells attaching?
Are they dying?
Are they proliferating?
So sometimes, what people will do is put the-- they'll seed a certain number of cells at a certain time.
Say, time 0.
Then, they'll look at how many cells are attached at 24 hours or 48 hours.
And you kind of see the cell attachment.
You can measure relatively easily.
Another thing people do is animal studies.
So for instance, Yannas does research on peripheral nerves and scaffolds for peripheral nerves.
And they cut a piece out of the sciatic nerve of rats.
So obviously, they have a surgeon and do a surgery thing with it.
You can't just kind of do this in the lab.
You've got to get permissions and stuff to do it.
And then, they put in the scaffold.
And the scaffold's actually in a tube.
And so they put the two stumps of the nerve end at either end of the tube, and then the tube's filled with a sort of porous scaffold that we're talking about here.
And then, they wait some period of time.
And they take video of rat running, things like that.
They then sacrifice the rat and they do histology.
And they look at the sort of cross-sections and see what it looks like.
And so I'm going to talk next time about this osteochondral scaffold we worked on.
So we did cell studies.
We did goat studies.
We put into goat knees.
There was a longer term sheep study.
And then, the student who's in Cambridge, England, started up a company and he ended up getting approval in Europe to start clinical trials.
And then he worked with an orthopedic surgeon who started putting it in people.
But typically, they're looking at cells, looking at animals before you get to the people stage.
And one of the things people do when they're making these scaffolds is you want to use materials that already have some sort of regulatory approval.
So say FDA approval or approval in Europe.
So typically, people don't start with a brand new material from scratch.
Because to get approval for that would just take a very long time.
So typically, people start with-- the solid material is already approved for some other sort of use.
OK.
So one requirement for the overall scaffold is it has to have sufficient mechanical integrity.
Sufficient.
And then also, as I mentioned, the stiffness of the scaffold can affect differentiation of cells.
And the other thing that is really a factor for the overall scaffold is you want to control the rate of degradation of the overall scaffold.
So you want that rate to be matched to the rate at which the new tissue is forming.
So it has to degrade at a controllable rate.
OK.
So I want to talk about the materials that people use.
And you can kind of break them down into a few classes.
So one class is natural polymers.
So things like collage.
So you can get collagen.
And that's an example up there of a scaffold that's made with collagen.
Another class of materials is synthetic biomaterials.
And if you've had stitches or surgery, you may know that some of the sutures they use are resorbable.
So some of those polymers that they use for resorbable sutures are also used for tissue engineering scaffold.
And then, there's also hydrogels that people use as well.
So those are probably the three main groups are sort of natural polymers, synthetic biopolymers, and I guess the hydrogels are sort of a subset of the sympathetic biopolymers.
So collagen is probably the most common kind of natural polymer that's used.
They also use GAGs.
And this scaffold up here is made by making a coprecipitative collagen with a GAG chondroitin sulfate.
People also use alginate.
I think one of the project groups-- you guys are going to make some sort of alginate scaffold, right?
No?
[INAUDIBLE] foamy thing?
Yeah.
And people also use something called chitosan, which is a derivative of chiton, which is what's in the exoskeleton of insects and things like lobsters.
So those would be all examples of natural polymers that can be used and people have tried.
I'm going to talk a bit more about collagen, just because it's the most common one.
So collagen is a major component in the natural extracellular matrix.
And not surprisingly, it has binding sites for cells to attach to it.
So if you use that, that kind of takes care of that issue.
Let me put it down here.
So collagen exists in many types of tissues.
Exists in skin.
Exists in bone, cartilage, ligament, tendon-- cartilage.
So it's very common.
It has surface binding sites for cells.
It has a relatively low Young's modulus.
So the Young's modulus is a little less than a gigapascal.
But you can increase the modulus by either cross-linking or by using it in conjunction with some synthetic polymers.
And I'm going to talk a little bit about how you make these scaffolds up here.
And the first step in making those scaffolds is you put the collagen in acetic acid, and then you add the glycosaminoglycan and it forms a coprecipitate.
And the fact that it forms a coprecipitate with the glycosaminoglycan, the GAG, means that you can use a freeze-drying process.
And that's how that's made.
Collagen is one option.
So then synthetic biopolymers is another option.
And as I said, typically they use the materials that are used for resorbable sutures.
So there's several of those.
There's something called PGA.
That's polyglycolic acid.
And something called PLA.
That's polylactic acid.
And then you can combine those two and make something called PLGA polylactic co-glycolic acid.
And you can control the degradation rate of these things by controlling the molecular weight.
And in this case, you can also control it by controlling how much of each of those things you put in.
And there's another one called polycaprolactone.
So those are several synthetic biopolymers that people use.
There's lots of different materials, but these are just some typical ones.
And then another class are hydrogels, which are produced by cross-linking water soluble polymers to form an insoluble network.
And those are typically used for soft tissues.
Sometimes, they're used for things like cartilage.
And again, there's a few different materials that are commonly used.
One's PEG, Polyethylene Glycol.
One's PVA, Polyvinyl Alcohol.
And another one's PAA, Polyacrylic Acid.
So for these synthetic polymers, there's many different processing techniques available.
And I'll talk a little bit about some of the processing techniques.
But one of the limitations is they don't have natural binding sites.
And you have to coat them with some sort of binding agent, like an adhesive protein, to get the cells to attach to them.
And then, as I mentioned before, you have to make sure that whatever material you choose, if it's a synthetic material that when it degrades, it's not toxic to the cells.
Because you don't want to have some sort of toxic reaction or inflammation.
OK.
So there's a couple more things about materials.
So those are all polymer-based materials.
When people are trying to make scaffolds for bone tissue engineering, they also include a calcium phosphate mineral.
And there's different versions of the calcium phosphate.
So they can include-- you can buy, for instance, hydroxy powders now.
There's another calcium phosphate called octacalcium phosphate, which will, with water, turn into hydroxyapatite.
So typically, there is this mineral, some sort of calcium phosphate, is combined with either collagen or with one of these synthetic biopolymers.
And one other option is something called an acellular scaffold.
And what that is they take some natural tissue and they remove all the cell material from it.
And so when they remove all the cell material, what they're left with is the native ECM.
And that's called an acellular scaffold.
So it's a native ECM with all the cell matter removed.
And they remove the cells by-- they can use sort of a physical agitation or chemical, or using enzymatic methods.
Using something like trypsin to get rid of the cell.
So there's ways that they can get rid of the cells.
OK.
So are we good so far?
So there are some requirements for what materials we kind of use, what the cell structure should be, and these are some examples of typical materials.
So I wanted to talk a little bit about the processing of the materials.
Let me wait until people catch up a little.
Oh, and I have some scaffolds I was going to pass around.
So this big sheet is a piece of the collagen GAG scaffold that I showed a minute ago.
And then this little piece is a mineralized version of that.
So this has the collagen plus calcium phosphate plus hydroxyapatite in it.
OK.
So this slide shows some examples of different scaffolds that people have made.
And I was going to talk a little bit about some of these methods.
And why don't I talk about them, and then I'll write some notes on the board.
So this top one here on the top left, that's the collagen GAG scaffold that's made in Yannas' group.
And that's made by a freeze-drying process.
So you put the collagen in acetic acid, then you put in the GAG.
The GAG and the collagen form a coprecipitate.
And then you can freeze that.
And if you freeze it, what happens is-- it's just like if you freeze saltwater.
The water freezes.
The pure water freezes.
And you've got increasingly higher brine content in the bit in between the water grains, or in between the ice.
So you get the sort of solid ice forming.
And the collagen and the GAG are kind of squeezed into the interstitial bits between the ice crystals.
And then if you sublimate the ice off, you're left with this porous kind of structure that looks like a foam.
So that's made by a freeze-drying process.
And I'll go over it in a bit more detail when I write the notes on the board.
You could also foam some of these polymers.
So just blowing a gas.
The same way you can blow a gas through engineering foam.
You do the same thing with some of these polymers.
This one's made by foaming.
You can have a fugitive phase process.
So this is made by salt leaching, the second row on the left there.
So you could imagine you could take a polymer powder.
You could mix it with salt.
You can sort of mix them up, combine them together.
You heat it up to get the polymer to melt and to sort of form a connected mass.
And then you leech out the salt.
And then you get pores where the salt was.
This one here is made by an electrospinning process.
So you have a nozzle.
You feed the polymer through the nozzle.
Then you have plates that are charged and you get fibers forming and kind of scattered in different directions by the electrospinning process.
Then, this one here represents scaffolds that are made by things like 3D printing, selective laser centering.
You can have laser-sensitive polymer.
And you can produce scaffolds that way.
I think the geometry of this one matches some part in the body.
I think it was a knee or something like that.
And then, these two examples down here are the acellular scaffolds.
That's what I was talking about at the very end there.
So those are from porcine pork heart tissue.
You know, pig heart tissue.
And those are mostly elastin.
And they've had all the cell matter removed from them.
OK.
So you can kind of see that these synthetic scaffolds here have a structure that's not so different from these native ECM scaffolds down here.
OK.
So let me write some of the things about the processing on the board.
Let me rub this off.
Start over here.
So these freeze-dried scaffolds are used for skin regeneration.
And I think I have some more slides here.
So it's kind of a two-step process.
In the first step, you make what they call a slurry.
So you make the slurry by taking the collagen.
And for skin, I think you want type 1 collagen.
Yeah, skin is type 1 collagen.
There's different types.
You put it in acetic acid, and then you add the GAG.
And we use chondroitin 6-sulfate is just the particular GAG that we use.
And one of the things that the acid does is that it swells the collagen.
And collagen has a sort of periodic structure in it, sort of periodic banding.
And the acid destroys that periodic banding structure.
And that helps increase the resistance to having some host immune response.
So that you remove the immunological markers and it makes it less likely that the scaffold's going to get rejected by the body.
So then when you put the GAG in, you form a coprecipitate.
So this next step just shows kind of mixing the whole thing up.
And then you've got kind of a little slurry that you can store.
And then, this is the freeze-drying step here.
So you put the slurry, the suspension, into a pan.
Kind of just like a cookie sheet, really.
And then you freeze it.
So if you think of this phase diagram here, where you have temperature and pressure.
So here we have liquid, solid, and vapor.
So if you start off at this point here, you freeze it.
So you've reduced the temperature.
So that forms the ice.
And the ice is surrounded by the collagen and the GAG fibers.
And then if you do the sublimation step, you reduce the pressure and increase the temperature a bit.
And then you get over to the vapor end of the world.
And then you're left with this porous scaffold.
And then, let me see.
Let me do one more step here.
And then you can control the size of the pores by controlling the freezing temperature.
So the size of the pores is exactly related to the size of the ice grains that are forming.
And the faster it freezes, the smaller the grains are going to be.
And then, the smaller your pore size are going to be.
So you can control that.
The type 1 collagen is mixed with acetic acid.
And it then swells the collagen and disrupts periodic banding.
And it removes immunological markers.
And then you add the GAG, the chondroitin 6-sulfate.
And then that cross links with the collagen and forms a coprecipitate.
And then you can freeze dry that to get the porous scaffold.
And typically, the relative densities of these scaffolds is very low.
So typically, they're 0.5% dense.
The relative density is 0.005.
So they're 99.5% air and the rest of it's the collagen and the GAG.
And the pore sizes are typically between about 100 and 150 microns.
And Yannas uses the same scaffolds for the nerve regeneration.
And he uses a directional cooling.
And that then elongates the pores, so that-- the idea is that they elongate so the nerves kind of grow along that length.
So that's one way.
Another way is leaching a fugitive phase.
So let's see.
I think-- yep.
Here we go.
Back to there.
So if you look at the one on the second row on the left, that's done by using salt as the fugitive phase.
People use other things.
You can use wax.
Paraffin wax works as well.
So it doesn't have to be salt.
There's different things you can use.
So you combine a powder of the polymer with your fugitive phase Say, salt.
Then, you heat it up to get the polymer to bind.
And then you leach out the salt.
So you can control the porosity by the volume fraction of the fugitive phase, and then the pore size by the size of whatever the fugitive phase is.
Another technique is electrospinning.
The idea is you produce fibers from a polymer solution that you extrude through a nozzle.
And then you apply a voltage across some plates to spin the fibers.
And then you get a network of these fibers.
And typically, their micron-scale diameter.
And the last method I'm going to talk about is rapid prototyping.
So you can think of using 3D printing.
Or you could use selective laser centering or stereo lithography using a photo-sensitive polymer.
So the idea is-- you know how this works.
You just build up layers of solid, one layer at a time.
And then you can make complex geometries with that.
So that's one of the advantages.
If you wanted to make a part to fit a particular place in the tissue, then it's convenient that you can control the geometry of the whole part.
OK.
So that just kind of summarizes very briefly, kind of how the tissue engineering scaffolds are meant to work, what kinds of materials people make them from, and a few of the processes.
And there's many, many processes.
These are just sort of a few common processes.
I wanted to talk a little bit about the mechanical behavior because that's kind of what I do.
And so this next plot just shows a stress strain curve in compression for a collagen GAG scaffold.
And I'm hoping that by now you're getting the idea all of these cellular materials have this kind of shape of a curve.
So there's the same kind of linear elastic regime, and then a collapse plateau.
These collagen things, as you can imagine just pressing them in your hand, they fail by buckling, by inelastic buckling.
And then there's a densification regime.
So they look like all the other kinds of curves that we've got.
One of the things we've done in the modeling is typically in the model, we want to be able to calculate or measure the modulus of the solid or the strength of the solid from which the thing's made.
So [?
Brendan ?]
Harley was one of my PhD students.
And he took a little microscope, cut a little strut.
The struts are very small.
The pore size is 100 microns.
So the struts are on that order.
He glued one end of the strut to a glass slide, and then he used an AFM probe to do a little bending test on that little strut.
If he could measure the deflection, he knew the length.
He knew the geometry of the strut.
He could figure out what the modulus was for the solid.
So he backs out what the modulus is.
So he did these measurements on a dry strut.
And then by comparing the overall modulus of a dry scaffold with a wet scaffold, he estimated what the modulus of the wet scaffold or the wet strut would be.
So the modulus of the dry collagen GAG was 672 megapascals.
A little less than a gigapascal.
And wet it was about 5.
So there's a huge difference between the wet and the dry.
OK.
So let me just write a few notes about that.
So in compression, there's the three regimes that we see for all these cellular materials.
And so you can estimate the modulus by using the model we have for the foam for the modulus.
The modulus of the foam divided by the modulus of the solid goes as the relative density squared.
And that's related to bending in the cell wall.
And then, the collapse plateau is related to elastic buckling.
And so that's equal to some constant times E of the solid times the relative density squared.
And that's related to elastic buckling.
And then we measured E of the solid doing this little AFM beam bending test.
OK. And for one of these low-density scaffolds, we measured the modulus.
We measured the buckling strength.
And we got pretty good agreement by using these equations here.
And the good agreement was if this constant was 0.2.
That was the strain, that buckling.
Yeah
So what does it mean to be wet in here?
And why is it so much lower?
Well, we make the scaffolds by this freeze-drying thing.
And like this thing I passed around, that was dry.
We just immerse it in water.
We just put it in water.
And then, he does the test.
So he does the test on the whole scaffold dry, and then he does the test on the whole scaffold wet.
And I assume there's some sort of bonding that gets disrupted by having the water.
I don't know the details of how that works, but there must be some change in the bonding to make that happen.
So let's see.
I'm trying to see.
What else should we do today?
Oh, yeah.
So we get pretty good agreement with this sort of simple foamy model.
One of the things we did find was that when we tested higher-density scaffolds, the agreement wasn't so good.
And I think this was because when you get higher density, it's just hard to get the collagen GAG mixture to mix in with the acetic acid.
And so we ended up getting inhomogeneous scaffolds with sort of large voids in them.
So in order for the modeling to work, you have to have a scaffold that's relatively homogeneous.
You don't have kind of big defects in it.
I think that's all I'm going to say about that.
I have one more slide here.
So those are sort of three-dimensional scaffolds we've been talking about.
Sort of foam-like scaffolds.
People have also made honeycomb-like scaffolds as well.
And this just shows some examples of some honeycomb-type scaffolds.
So this one here, I think was made in Sangeeta Bhatia's lab.
You can sort of think of it as a hexagon, but it's also kind of triangulated as well.
These two here were made by George Engelmayr.
He worked with Bob Langer at one point.
And these two scaffolds here, they're both sort of rectangular cells.
And they were designed to look at how the cell geometry or the pore geometry affected the sort of morphology of the cells that attached.
So if you have different, say, [?
porous, ?]
you get different morphology in the cells that you're trying to attach.
And then, George Engelmayr also made these scaffolds here.
And those were designed to be anisotropic and have different mechanical properties in different directions.
And I think what they had done was try to match the anisotropy in the mechanical properties to anisotropy in heart tissue, in cardiac tissue.
So these are some examples of honeycomb-type scaffolds.
So let me just write down a few things about those.
So I think these were more-- obviously, the scaffolds are used-- the ultimate goal is to use them clinically.
But sometimes, people make scaffolds just to study cell behavior.
And some of these, I think, were made just to study how the cells would behave on them.
So they're sort of idealized to do that.
So that triangulated.
It kind of looks like a hexagon, but there's also sort of triangles in there.
I think the thing they were looking at with that was transport of nutrients to cells.
And from a mechanical point of view, if it's triangulated you'd expect, say, the modulus of that to go linearly with how much solid there is there.
So the rectangular honeycomb and the diamond shaped pores, they were used to study the effect of pore geometry on the cell orientation.
They used fibroblasts.
And I'm going to call it the accordion-like honeycomb.
That's mechanically anisotropic.
And the mechanical anisotropy is matched to the cardiac tissue.
OK.
So I think I'm going to stop there for today.
That's sort of the end of this part.
And next time, I'm going to talk about the osteochondral scaffolds.
I don't think it really makes sense to start it for two or three minutes.
So I'll start that tomorrow.
Or Wednesday.
And we should be able to finish that on Wednesday.
So one of the things that these honeycomb-type scaffolds kind of suggests is that the scaffolds are used both to try to regenerate tissue in the body in clinical applications, but they're also used as sort of an environment for cells, in order to study cell behavior.
So next time, I'm going to talk about an osteochondral scaffold.
And the idea with that was to try to use it clinically.
But next week, I'm going to talk about cell mechanics a bit.
And when people study cell mechanics, or look at the mechanics of biological cells, not the cellular structures.
So when people look at trying to study how cells behave, they need some environment to put them on.
Typically, people started by just using flat 2D substrates.
But the flat 2D substrates are kind of easy to study, but they don't really represent the tissue in the body.
And so people are now using tissue engineering scaffold as an environment that they can control to study how cells behave.
So they study cell attachment, cell proliferation, cell migration, and cell differentiation all by using the scaffolds as kind of a controlled environment.
So we'll talk about that, probably not-- I don't know if we'll get to it on Wednesday.
But we might start it on Wednesday and finish it next week.
OK?
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
What I thought I would do today is two things.
I wanted to give this talk about osteochondral scaffolds, and this is just slides.
I'm not going to write anything down.
So there aren't going to be any notes.
And I'll put the slides on the Stellar site, I don't know, this afternoon or tonight or something.
And this isn't going to take the whole time we have today, so the other thing I thought I'd do is I wanted to walk you through this little booklet that I handed out about how to write a paper.
So this is by Mike Ashby, who you know, I've written books with.
He was my PhD advisor and we wrote the books on cellular solids together.
He has written many books.
I looked him up on Amazon this morning, and even though I know he'd written many books, I was shocked how many books.
They had 58 listings for just books with him as a co-author and some are second editions and third editions.
He's written a lot of books.
He's written a lot of materials, papers.
And he's, in the materials community, he's seen as a very clear and lucid writer.
So he's written this thing for his students and I thought we could just walk through it and I can talk to you a little bit about writing.
Because I know you're not quite there with your projects, but later on in the term you're going to want to write up your project, and it's not too early to start thinking about writing.
And there's some really good advice and it's short and it's to the point.
It's really helpful, that little brochure.
So we won't read every single thing, but I wanted to kind of walk through some of the main points in it, and that pretty much should take the hour.
So last time we talked about tissue engineering scaffolds, and today, I wanted to do a case study, and this was a project we had here at MIT and in collaboration with some people in the materials department at Cambridge University.
And we made what we called an osteochondral scaffold.
So osteo means there was a part for regenerating bone and chondral means there was a part for regenerating cartilage.
And really the point of the scaffold was to try to repair small defects in cartilage.
And I'll explain why we did the bone thing as well.
So this is just a little outline.
I'm going to talk a little bit about what the structure of cartilage is.
We have a little schematic of that.
Then talk about how small defects in cartilage are currently treated.
So these are things that, say you're an athlete, you might tear some cartilage, that kind of thing, not like you have osteoarthritis and you need a new knee.
It's not going to do that.
So I'll talk a little bit about cartilage and the current treatments.
I'll talk about some of the things we thought about in making an osteochondral scaffold, like what parameters were important.
And we based it on that collagen-GAG scaffold that I talked about last time.
So the collagen glycoseaminoglycan scaffold that we talked about last time.
And what we did was we had a layer that was a collagen based scaffold for the cartilage and we had another layer that was a mineralized version of that.
And one of the main things we did in this project was figure out how to mineralize that scaffold.
And then we made this two-layer osteochondral scaffold.
So I'll talk about that, OK?
So are we good?
So this is a schematic of articular cartilage.
Articular just means it's in a joint, so between like, say, two of your long bones, and there's several regions.
So this shows both the cartilage and the bone underneath it.
And so this top layer is called the superficial layer, and then there's a transition zone, and then there's this deep cartilage here.
And the little white lines represent how the collagen is oriented in the cartilage.
So the collagen is oriented, more or less, vertically here and then it becomes more kind of woven and horizontal towards the top.
And so those three different zones, the collagen is oriented differently.
Then there's a region down here called the tidemark.
Everything above here is just cartilage and everything below is calcified to some extent.
So this next layer down here it's more cartilage-like, but it's got some calcification in it as well.
And then here's the compact bone.
They call it subchondral bone.
It's the bone below the cartilage.
So chondral is cartilage.
And then here's the trabecular bone.
We talked about trabecular bone a couple of weeks ago, OK?
So one of the things is there's different types of collagen and the different types may have slightly different fiber structures or slightly different compositions.
They're all related, but they're slightly different types.
And bone has what's called type I collagen and cartilage has type II collagen.
So when we made the scaffold, we wanted the bony layer to be made with type I and the cartilage layer to be made with type II.
So I think with the 3032 people have probably seen it.
I think I did a version of this for you guys, didn't I at the end of term, something?
Yeah, but there's other people, so it's really for them.
So if you think of articular cartilage, it has difficulty repairing itself and one of the reasons that it's difficult for the cartilage to repair itself is that there's no blood supply in it and another reason is that there's not very many cells.
So chondrocytes are the cells in cartilage and if there's a low volume fraction of the cells, then it's not so easy for that small number of cells to actually produce the extracellular matrix, which is kind of what you think of as the cartilage.
And it can be damaged, either as I said, from sports injuries, typically that's what young people get, they tear their cartilage in some sporting injury accident, or from osteoarthritis.
So these scaffolds we're talking about, they're not really meant to repair large amounts of cartilage that are damaged.
And as I said, the cartilage has a poor capacity for self-repair.
So there's several treatments, there's three treatments that are given currently and the most common one is called marrow stimulation.
And some of these orthopedic treatments are fairly crude when you look at it.
So this marrow stimulation, what's involved is they take a drill.
So here's the drill.
And they basically drill through the cartilage.
So this would be the cartilage layer here.
And they drill down into the bone and they want to get down into the trabecular bone.
So they want to go below the cortical or the subchondral bone and they want to go down into the trabecular bone.
And the reason they want to do that is the trabecular bone has bone marrow in it and the bone marrow has mesenchymal stem cells, and they want those mesenchymal stem cells to move up into the cartilage layer.
So that's what this red glob is here.
The idea is that you've made a hole and now that hole is going to fill up with a blood clot, which will have these mesenchymal stem cells, and that those will form cartilage.
So it does work to some extent, but it's not really a great result.
There's something like 75,000 of these done a year.
As I said, this is the most common kind of repair.
The next most sophisticated type of repair, what they do is they take plugs of bone and cartilage from another place.
So say this is where the defect is and they want to repair that.
They take bone and cartilage from this spot up here.
So they sort of drill out little cylindrical cores and then they plug them into this bit here.
And that's called an osteochondral autograft.
Sometimes it's called mosaicplasty because they build up a little mosaic from all those little pieces.
And then they just leave these donor sites, where they took the plugs from, they just leave those empty.
So they try to take the bone and cartilage from regions where the loads are lower, but they end up leaving holes, which is not so desirable.
And again, those holes may fill a little bit by the previous mechanism.
Then the most fancy method that they use now is called autologous chondrocyte implantation.
So what they do is they harvest cartilage cells from the patient.
They then take them to a lab and they culture them for two or three weeks and they get them to multiply and proliferate and grow and then they re-implant the cells.
So this works fairly well, but the difficulty is it involves two surgeries, so one to get the cells out and one to put them back in, and there's the cost of the cell culture.
So this is a much more expensive procedure because of the two surgeries and the cost of doing the cell culture, but I think it's the method that works the best.
And when I talked about this in 3032, I mentioned Dara Torres.
Dara Torres is an Olympic swimmer.
She's 47.
She began swimming in the Olympics in 1984, more than 30 years ago.
And she has swum in the Olympics, not every one, but up until 2008.
And even 2008, she won three silvers that year.
She was, I think, the oldest person who's ever won a medal in the Olympics.
And she had this surgery done at the Brigham a few years ago.
And it used to be, they've stopped running these commercials but the Brigham for a time was running these commercials, that featured Dara Torres and saying basically that she had this surgery done there.
I'm assuming she was happy with it.
OK, so that's what they do currently and what we were thinking of and what other groups have thought about is could you repair damage in the cartilage by using a tissue engineering scaffold?
So what we were thinking about when we tried to do this project was we wanted to use a healthy articular cartilage joint as a model for our scaffold.
We wanted to have a layer that would go down into the bone so that you would have access to those mesenchymal stem cells.
And that's why we wanted an osteochondral scaffold, so that we would have a layer for the cartilage but also a layer that would go down into the bone.
We wanted to be able to control scaffold parameters, things like the mineral content and the pore size.
Remember, I said these tissue engineering scaffolds, the pore size is one of the parameters, that you want to have the pore size in a certain range.
And we wanted to use materials that would be appropriate for approval from things like the FDA.
So typically, when people make tissue engineering scaffolds, they don't start with some material that's never been approved before.
They start with something that already has approval for something else and that's what we wanted to do too.
So this was our idea of what we wanted the scaffold to look like.
We wanted an unmineralized type II collagen scaffold up here that would be for the cartilage.
We wanted a mineralized type I collagen scaffold down there for the bone.
And we wanted some region that had some gradient in mineralization because it would be like that layer of cartilage that was slightly mineralized.
So we wanted to duplicate that whole structure.
So that was our picture of what we wanted to do and we had a pretty good idea of how we were going to make this.
We were just going to use that same process that [INAUDIBLE] developed for the skin scaffolds.
He was involved with this project but just used type II collagen instead of type I.
The challenge was really figuring out how to make the mineralized collagen scaffold and then how to get this gradient in the mineralization between the two layers.
So this, I think, I showed you last time.
So this is just the method we used to make the collagen-GAG scaffold.
So we take type II collagen.
We put it in acetic acid.
Remember, that destroys the periodic banding of the collagen and it improves the immunological response.
We add the chondroitin 6-sulfate, the GAG, to crosslink it.
We then just make a slurry out of that.
So we mix that all together.
We keep it as a slurry.
And then the second stage is you put the slurry or the suspension into a pan, and then you do the freeze drying process.
So you go through this process where you start at this room temperature and atmospheric pressure.
You cool it down to freeze it, and then you sublimate it, and you're left with a very porous collagen-GAG scaffold.
So these are pictures.
I think this was with a type I collagen, but it looks the same with the type II collagen.
So that's what the structure looks like.
I think I showed you this last time.
We can control the pore size by controlling the freezing temperature.
So the way the freeze dryer works is there is shelves that you put these pans on, and these cooling elements go through the shelves, and you can set the temperature of those shelves.
So we would set the temperature of those shelves to different values and we got different pore sizes.
So the colder the shelf temperature was, the faster the freezing, and the smaller the size of the ice grains, and then the smaller the size of the pores.
And this is just the mechanical response again.
So we did mechanical tests on it.
These are some of the numbers for the properties.
So we tested it dry and wet, and we measured a modulus and a buckling collapse stress for it.
So those are just some values there.
And then it came to making the mineralized collagen-GAG scaffold, and one of the students in Cambridge, England was really the main person who did this, Andrew Lynn.
And he worked with Brendan Harley, who was our student here, and after they graduated, I had another student, Biraja Kanungo, who worked on this too.
So Andrew Lynn really developed this technique.
And this involved taking the collagen and the GAG, so the same as for the other scaffold, but this time, if we want to make a mineralized scaffold, we somehow have to get calcium phosphate into it.
So it's got to have sort of a hydroxyapatite-ish type of component to it.
So this time we used phosphoric acid instead of the acetic acid and we put some calcium salts into the mixture as well.
And Andrew was really the one who figured out how to do this and what salts to use.
And then the process was very similar.
So we have this slurry.
We mix the slurry up.
We did a freeze drying process, and then we crosslinked it with a chemical crosslinker called EDAC.
So it's just a chemical that you put into this and it crosslinks it all.
So that was how we made the mineralized scaffold.
The mineral we got was something called brushite, which is a calcium phosphate, but it's not exactly the same as hydroxyapatite.
And we could control the amount of brushite by different weight fractions or volume fractions by controlling how much of the calcium salts we put in and what the molarity of the phosphoric acid was.
If you take brushite and you put it in water, it then converts to octacalcium phosphate and then to apatite by a hydrolytic conversion.
So the apatite is related to the hydroxyapatite in bone.
And this was the structure of the mineralized scaffold that we got.
So you can see it looks a little different from the collagen-GAG scaffold.
It's much denser.
Typically, the densities were like 5% or 10% dense.
And remember, the collagen scaffold was 0.5% dense.
So it's a lot denser.
But if you notice, there's a few things to notice here.
So one is the pore size, that's a 500 micron bar.
And we could make pores between about 50 and 1,000 microns, depending on the freezing conditions.
And the range of pore sizes we were shooting for was somewhere between 100 and 500, so we could get in the right ballpark with that.
And you can see, just looking at that picture, if that's 500 microns, those pores are somewhere of that order.
Another thing to look at is this image here, and this just shows, the white little dots are the calcium phosphate mineral, and it just shows that the mineral is uniformly distributed throughout the thickness of the scaffold.
So some people had tried to make scaffolds for regenerating bone where they take, say, one of those polymers for resorbable sutures or they take collagen and they coat it with hydroxyapatite, but the shortcoming of that is that the scaffold is going to resorb over time.
The cells are going to secrete enzymes, which are going to eat away at the scaffold.
And if the scaffold resorbs, you're eating away at the hydroxyapatite first and then you're left with whatever polymer is underneath that.
Whereas this gives you a more uniform composition throughout the thickness of the scaffold, and you've got calcium phosphate everywhere throughout the thickness of the struts in the scaffold.
So that's the structure of that.
We wanted to make sure that the mineral was uniformly distributed throughout the scaffold, and we did some micro-CT imaging, some micro computer tomography.
And this is our sample here, and this red line just says that is the plane at which this image was taken, and the black is the mineral.
And then here's a lower plane and here's another image.
And you can see the calcium phosphate's pretty uniformly distributed throughout that specimen there.
And this was just another way of looking at the same thing using EDX in an SEM to look at where the calcium was and where the phosphate was.
So again, this is uniform distribution of the mineral.
We did mechanical tests on these scaffold as well.
So we measured moduli and collapse stresses.
We get the same kind of stress strain curve as all these other cellular solids.
This was done by- Biraja Kanungo was the student who did this bit.
One of the things we found was that with these mineralized scaffolds you could manually compress them.
If you pushed them down and you hydrated them, they would recover all the deformation, but we were increased in improving the mechanical properties of the mineralized scaffold for improved handling during surgery.
And there's another reason that I'll get to as well.
So the second reason is that, if you look at how bone itself forms, it forms from a collagen precursor, so bone in your body.
And there's something called osteoid, which is this collagen-based precursor to bone, and it has a modulus of about 25 to 40 kilopascals and Angler showed that if you have mesenchymal stem cells and you put them on substrates of different stiffnesses, they differentiate into different kinds of cells, depending on the stiffness of the substrate.
So the substrate stiffness can affect what kind of cells you get.
And the idea here was we thought, well, if we could get a stiffness that was close to this osteoid, what the natural bone formation has, then that might help the mesenchymal stem cells differentiate into the bony cells that we want them to.
So we wanted to try to reach a stiffness of this in the wet state.
And if I back up here, you can see the stiffness we had was around four in the wet state, four kilopascals and we want to get to 30 or 40, something like that.
So these are our equations for the modeling of the mineralized scaffold, the open celled foam models for the modulus and for the collapse strength.
So we could change different things.
We could change the solid properties or we could change the relative density.
The geometry of the thing is probably not going to help us too much.
So basically, that's what we did.
We first started off trying to increase the mineral content.
We thought if there was more mineral content that would make it stiffer.
So Biraja made these more highly mineralized scaffolds and these are just some SEM images of those.
But the thing he found was that the properties actually got worse when he had the more highly mineralized scaffold.
The modulus went down and the strength went down, so that wasn't very helpful.
And when he looked into it more detail, he found that he had more voids in the cell walls and he had more disconnected walls.
This shows some of the micrographs, so this isn't really very quantitative, but you can see there's holes here.
There's a few holes here, but there tended to be a bigger volume fraction of holes in the more highly mineralized scaffolds and more walls that were disconnected.
So we realized increasing the mineral content wasn't going to work very well.
And then the second thing he tried was increasing the relative density.
So we started off at this density of 4 1/2% dense and he developed a method of increasing the relative density up to about almost 20% here.
He did this by, first of all, he tried to just mix more of the constituents into the slurry, that's kind of the most obvious thing.
But as you add more constituents, it gets harder and harder to mix the thing up and have it homogeneously distributed.
So in the end, that's not how he made these things.
He started with the starting mixture and then he had a vacuum system for sucking water out of it.
So he would reduce the amount of water, which essentially increased the amount of solids that was in there.
And you can see in this last one here, the most dense scaffold, he was sucking the water in one direction, and he sucked it so much that he was starting to get the cells collapsing.
So this one here, these cells that have this sort of elongated orientation, that's because the cells are starting to collapse because of the vacuum that he was applying.
But he could get a pretty good difference.
This was almost 5%.
That's almost 20%, so roughly a factor of four.
There are, yeah, four difference between them.
So then he did mechanical tests too, and he measured the relative density.
Here's the moduli wet and then the strength dry and the strength wet.
So you can see here, if you look at the dry moduli, for instance, when you go from this relative density to that relative density, from about 14% to 19%, the modulus actually drops down.
And if you look at the structure, I think, it's because you've got this flattened structure here.
You've collapsed the cells a bit already.
So there's a maximum density that you might want to go to, probably somewhere around 14% or 15%.
But the wet modulus we've got here is around 35 kilopascals, so that's close to the target that we had.
It's close to what we wanted to have.
So one way you could get the right or the appropriate modulus is by increasing the density to that value.
Another way is by playing around with the crosslinking.
So those values were for non-crosslinked scaffolds.
So here, this is the 14% dense scaffold wet, non-crosslinked it was around 35 kilopascals.
This is a dehydrothermal treatment, just basically heating it up.
It gives you a higher modulus.
And then this is a chemical crosslinking technique that increases the modulus again.
So if you put all of this together, you can show these results for the modulus on one table.
So these are all the wet modulus.
So it's pretty clear by playing around with the relative density and the crosslinking, that you can get a scaffold in that range.
And the idea is that that would help get the mesenchymal stem cells to differentiate into these osteoblast-like cells.
So then we wanted to also see if our models for the cellular solids could be applied to this scaffold, and we needed the solid properties.
So Kristyn Van Vliet helped us with this, and we isolated a single strut, bonded it to a glass slide, and, I think, I mentioned this last time, we used an AFM tip to then do a little bending test.
So I think the one I mentioned last time was for the collagen-GAG scaffold.
Then we also did the same thing for the mineralized scaffold.
And here we measured a modulus of about seven gigapascals, so that's a dry modulus.
And just for comparison, the modulus of the solid and trabecular bone is something around 18.
So it's lower, but it's in the same ballpark.
And by nanoindentation, we measured a strength of about 200, and that's similar to what you would get in trabecular bone.
So the solid in the struts themselves is not exactly like trabecular bone in mechanical properties, but not too far off.
And then here's a plot of the scaffold modulus divided by the solid modulus against the relative density.
And this line here, the curve, is a squared relationship.
So that's what we'd expect for the foam models.
And this is the strength.
These things fail by a plastic or a brittle failure, and this curve is a three halves power with relative density.
And again, that's what you'd expect from the cellular solids model.
So that gives you a reasonable description of the behavior of the mineralized scaffold.
So again, these were the considerations in trying to make the osteochondral scaffold.
So now we have a collagen scaffold and we have a mineralized scaffolding, and we want to put the two of them together.
So we wanted to use the joint as a model, and we wanted to have some intermediate layer that had some gradation in the mineralization.
So that was the next step.
And the way we did that was we used what we fancily called liquid-phase co-synthesis.
This just meant that we took the mineralized collagen-GAG slurry, we poured that into a mold, and then we poured the non-mineralized slurry into the same mold.
And then we just allowed the two slurries to interdiffuse for some time period.
And then we did the freeze drying step.
So the idea was, that if you poured the one on top of the other, the mineralized one is denser and then you put the less dense one on top, but over some period of time, there will be some diffusion between the two.
And then we just did the freeze drying step.
So this is the scaffold we ended up with.
This is a micro computer tomography image.
So here's the collagen-GAG scaffold for the cartilage on top, and here's the mineralized collagen-GAG calcium phosphate scaffold on the bottom for the bone.
And that just kind of shows what it looked like.
The porosities and the pore sizes we got, the collagen-GAG was about 98% porous and had a pore size of around 650 microns.
The mineralized scaffold was 95.5% porous and had a pore size of 400, roughly, microns.
And then this is what the structure looked like in the EDX.
You can see, this is the collagen-GAG layer, this is the mineralized layer, and there's some zone in between that's a little bit mineralized, not as much as the bony layer but more than the cartilage layer, and the same with the phosphorus.
So we have this collagen-GAG slightly mineralized layer, and then a more mineralized layer And then finally, there was a student, Scott Vickers, who worked with Myron Spector at the Brigham.
And oops, oops.
No, I don't want the weekly updates, thank you.
Sorry.
Well, let me just get rid of this.
Oop, where's my little mousy mouse?
There we go.
OK, so Scott Vickers worked with Myron Spector, and Myron had a surgeon who could do animal studies.
So we did some animal studies on goats.
I think there were six goats at the Brigham.
And they took a plug out of the knee of the goats, and they put our scaffold in.
So this is one of the surgeries as they're about to poke the scaffold in.
And then, Scott waited, I think it was four months, and then sacrificed the goats, and then did the histology.
And this is one of the images from his PhD.
So this staining shows that you've got tissue growing in.
The scaffold was where these little black dotted lines are.
So you had bony tissue in here, and there was a cartilage-like tissue formed at the top.
It wasn't perfect articular cartilage, but it was something similar to that.
And really was as far as we took the project with the funding that we had.
We had a-- I don't know if you remember that-- well, it's beyond before your time.
But Cambridge and MIT had a big research collaboration and the student exchange was part of that.
And this was done through that research collaboration between Cambridge and MIT.
So this was as far as we took it with that research funding.
Andrew Lynn, who was the student in Cambridge, who developed the mineralized scaffold, he then started up a company called Orthomimetics, and he had longer term animal studies done, and he took it a little further.
In Europe, there's something called CE Mark approval, and CE Mark approval means you can start doing clinical trials.
So he never got FDA approval for it, but he got approval to have clinical trials in Europe.
And the first clinical use was in February of 2009.
And they started off using it for the donor sites for the mosaicplasties.
Remember the second method I talked about?
They take plugs out of one region and put them into the region with the damage.
So what they were doing was using our scaffold to fill up these donor sites here and they found that worked quite well.
And then eventually, they started using it for the primary sites as well.
And as of about April a few years ago, April 2012, they had treated about 200 people with this
Wait, so why not just directly start putting it in the-- why start with the--
I think because these were supposed to be sites that were less loaded, weren't as highly stressed, and they thought that was a more, not as critical a place to put them in.
Yeah?
So what is different about this scaffold?
Is it the lack of the dense bone that allows all of the marrow cells to migrate up to the cartilage?
Well, I think, there's not that many people that have made osteochondral scaffolds, so it's good that we've got these two layers.
And the idea that you try to get the stem cells up into the cartilage.
The stem cells will differentiate into the bone or the cartilage, I think, partly depending on the stiffness of the surrounding tissue
But they don't do that in normal bone because of the dense layer?
No, I think they would.
But the thing is, I mean, in some ways, that very first technique where you just drill holes in, I mean, in some ways, that's what it's counting on, right, is that the marrow mesenchymal stem cells are going to differentiate either into the bone or into the cartilage.
But it's just got a hole to differentiate into.
So this gives the cell something to attach to and I think, gives a better result.
Yeah?
So is part of the scaffold and this bone, are they removing the original bone?
Yeah, they remove-- yeah.
Let me back up a step.
So when they do this thing here, so this is in a goat, but they would do the same thing in a person.
So when they drill the hole to put that in, they go through the cartilage, they go through the compact bone, the dense bone, and they go into the trabecular bone, because you don't really get into the marrow until you're in the pores of the trabecular bone.
OK?
Are we good?
How do they attach it?
How do they attach it?
Yes
I think it's just a press fit.
I think they just drill a hole and stick it in.
And these are all in joints, right?
So there's always another bone pressing against it.
I mean, in the surgery, they kind of peel things apart so they can do the surgery but there's another bone pressing down on it.
So I don't think there was any glue or anything.
So there's that.
So this is just a summary.
So we were able to make this two-layer scaffold with a gradient interface, and we tried to make it so that it mimicked the osteochondral tissues.
And this freeze drawing process allowed us to control things like the mineral content, the porosity, the pore sizes.
And then we used materials that had already been approved for medical devices.
And this was funded by a number of places.
So the Cambridge MIT Institute was that collaboration I mentioned.
I have a chair and I used some money for that.
Brendan Harley was one of the students involved and he got a fellowship from MIT.
And Andrew Lynn got a fellowship through the Cambridge Commonwealth Trust and through St. John's College in Cambridge, that's his college there.
So he had had funding through that.
So I think that's the end of that talk.
So are we good with scaffolds?
Yeah, you're good?
So obviously, this probably whole course is on tissue engineering.
This is just scratching the surface and giving you an introduction to it, but I wanted to show you how a lot of these scaffolds look a lot like the foamy materials that I work on, and that you can use the same models for trying to understand the mechanical properties of the scaffold.
And even though the scaffolds are acting in a biological way, there's actually a connection between the mechanical behavior of the scaffolds and the biological response.
So I'm going to talk on Monday about cell/scaffold interactions, so how the environment biological cells are in can affect how they behave.
So we're going to talk about things like cell adhesion and cell migration and cell contraction, contractile behavior.
So I've got another talk a little bit like this and that means I'll have a few notes I'll put on the board on Monday about how the environment that the cells are in affects how they behave, OK?
So this is leading up to that.
It's sort of a similar thing.
So I thought for the rest of the time today, because I knew this was going to end early, I thought what I would do is just switch gears and talk about writing.
So normally when I give a class, I don't talk about writing all that much.
I talk about the technical stuff.
But it turns out writing is actually a huge part of what scientists do and engineers do and whether or not you end up with an academic job or a job in industry or working for government, no matter where you are, you are going to have to write things.
And the better you write, the better off you're going to be.
So I made copies of this brochure and I have a few little slides.
And you're going to have to write up your project report for me, and I thought it might be helpful to just go through this little brochure.
So I'm not going to write stuff on the board.
Everything is pretty much in this brochure, but I thought I'd just walk you through some of the main parts of it.
Oh, did you get one?
Here, have one.
So you might want to think about this when you're writing up your project report for this class, but this really is, it's general.
It's really for any time you have to write something.
This is helpful.
So Mike Ashby put this together.
And as I mentioned, he's done a lot of scientific writing and especially in material science and engineering.
He also makes paintings.
This is one of Mike's little paintings of him writing a paper.
And what I was going to do is just walk through it.
So there's just a few figures, and mostly it's text, but let me go through some of the figures.
So I'm going to just turn to what's page three, OK?
So one of the things that he says to start, and I think this makes a lot of sense, is that you can think of writing a paper the same way you can think about designing something in engineering.
So when you think about design, there's different stages of design, right?
There's a conceptual design, where you just decide roughly what the thing's going to be.
There's embodiment, where you work out a lot of the more details.
You design one version of it.
And then there's the detailed design, where you do all the fine-tuning.
You do all the final design things.
And before you really start your design, if you were going to design an engineering thing, you would first of all think about what's the market.
And if you're writing, the market is your readers.
And so when you think about what you're going to write, you have to think about who's going to read it.
And it's the same thing when I give a talk.
When I give a talk, before I make a single PowerPoint slide, the first thing I think about is who am I talking to and what do they already know?
What's the audience?
What are they looking to get out of the talk?
What am I trying to convey in the talk?
And the writing is the same thing.
You have to think about your audience.
So for instance, I have technical talks.
Obviously, I go give technical talks at meetings, but I do other sorts of talks too.
I've got the woodpecker talk.
I go give the woodpecker talk at the Mass Audubon.
And people come to listen to that talk who are interested in birds but they're not engineers.
So when I do that talk, I have to make it so somebody could understand it who's intelligent but they're not necessarily an engineer.
So I give a different kind of talk when I do that than when I give an engineering talk.
And I got invited to a student dinner.
I'm going out to somewhere with some students on Friday.
I'm going to give that how I became a professor talk, and when I do the how I became a professor talk, it's more general, and so it's a different audience that I'm thinking about.
And in some ways, when I do these talks, it can be the same group of people, but depending on what I'm talking about, the way they look at it is different, OK?
So the market thing is something to think about.
So if you're writing a thesis, your market is your PhD committee, who's going to read the thesis and examine you on the thesis.
If you're writing a paper, you've got to think about the market as being other people in your research field, some of who are going to be the reviewers, who are going to be reading it and criticizing it.
If you're writing a general popular science book, it's a different kind of audience.
If you're writing a research proposal, the audience is going to be the funding agency.
Are they interested in what you're talking about, but also reviewers, who are going to be deciding whether or not to give you the money and what the criticisms are.
So you have to think about who the market is.
So that's one thing.
And one thing that think about in doing the writing-- you know Mike and I have written several books together, and when I tell my neighbors I've written books, they somehow think-- their first idea is that we start on page one and we start writing the book from page one and then we work our way through to page 500.
And that's not how we do it at all and that's not how I write papers.
That's not how most people write papers.
You've got to think about the big picture and think about, roughly, what goes where.
And then maybe think about a draft that gets the scientific facts right.
You put the information down, and then you try to figure out about how do you make the style really nice?
How to make it read well?
How do you make it easy to understand?
How does one paragraph lead into the next paragraph?
So it's an iterative thing.
It's not like you start at page one or line one and you just start writing.
So it's an iterative process, the same as engineering design is an iterative process.
If you were going to design, I don't know, a skateboard or something, you wouldn't just think you were going to make one and that would be it.
That's how writing is, and often, students don't see it as this iterative thing.
OK, so that's that page there.
Let's see, I already talked about market.
I think I have another little slide about-- here's another slide here about markets.
So this is what I just said.
Who are the readers?
How are they going to use it?
So you've got to think about who you're writing for or if you're giving a talk, who's going to look at that.
OK, now the next phase is to make what Mike calls a concept sheet.
I would just think of this as an outline.
And I always make some sort of outline before I try to write anything.
And there's different ways you can do that.
The thing that Mike's got here and there's a nice little figure on page six, which is what I've got up here, is he takes a big piece of paper.
In Europe, it would be known as the A3.
Here it would be known as the eight and a half by 17.
You take a big piece of paper.
This stationery company actually makes paper with big blank space in the middle and little note space on the outside or you could just use the back of it.
So you take a big piece of paper and you may think this is douffy, but it actually helps.
So you take your big piece of paper, and you just make boxes about each topic.
You're going to have an introduction, you're going to have methods and materials, you're going to have a result section.
And you think about what should go into each of those boxes.
And what you're trying to do here is think about the whole paper and how it all fits together and what goes where and what are you going to include and what are you not going to include.
So people think about writing as sitting at the keyboard and typing, but that's just the-- how when you get a problem on a problem set, you turn and crank, that's the turning and cranking part.
The thoughtful part is figuring out what to put in, what to leave out, what figures you want, how you organize the whole thing, how you put it all together, and this helps you do that.
And you can make this-- it's just for you.
It doesn't have to look great.
You can make this messy if you want.
It doesn't really matter.
But it's good to have some sort of an overview of how you want to put the thing together, and that's what this stage helps you do.
So Mike's put other little things here.
So see papers by so-and-so and so-and-so.
So you've got an introduction.
You want to talk about something.
You know there should be some references go here.
Maybe you think there's some extra references you haven't got.
Maybe something in the method, there's some analogy you can use here.
Maybe you need a figure, needs a good figure.
You don't have to actually have the figure.
You just say I need a figure and you have a vague idea what the figure looks like.
Here there's some discussion point.
Discuss this with collaborators, Ed, all this stuff.
So you just put down roughly what needs to go where and you think about the whole thing and how it's all going to fit together.
And then as you work through it, it looks more like this.
So this would be on page seven.
And this is filled in, and if fact, this particular one, what he's written down here is the overview he made for making this booklet, OK?
So the things refer to this book.
So here's the introduction.
Here's the little chart we went through and through, the concept design, embodiment, and detail, blah, blah, blah.
Here's the need, the market need.
We just talked about that.
Here's the concept thing that we we're just talking about now.
Then we're going to talk about each of these different phases.
And his initials are MFA, Michael F. Ashby.
Think out, that means think about this more.
I haven't figured this out yet.
I need an example here, all these kinds of things.
So this is the way he does it.
There's a couple of other ways you can do it.
One way is to just make a bullet outline, that's typically what I do.
You know you're going to have an introduction, these different headings.
And you might put in the bullet outline these same sorts of things, like I need this figure or we need to get one more set of data or I need to look up this reference.
So it doesn't have to be finished, but it's a thing that tells you what have you got, what do you need to do to make it all come together.
So that's one way to do it.
Another way to do it is by thinking about what figures you want in the paper.
So some people, before they write any words, they say, well, I know I want to have these figures, and then they build the words around the figures.
And they can even sketch out what the figures are.
Some people even start before they start the project, they say what kind of figures do I want at the end of the project?
And they don't know if the data is going to do this or that.
They don't know which way the data is going to go, but they say I want to have a plot of one thing versus another thing.
Oh, you're looking like this is not OK?
No
No?
OK, yeah, some people do this.
So even when you're ready to write, you could say to yourself, well, in the methods, do I need a figure that's a schematic of some apparatus?
In the results, how am I going to present the results in the discussion?
Do we need some other kind of figure?
So if you have a set of figures, you can work the text around those figures.
So that's another way to do this.
So those are just three options, but all of them have in common that you think about the whole paper and how it all fits together and what goes where and what do you have already and what else do you need to get, OK?
So that's that.
And then what I was going to do-- I think that's probably the last figure that's-- yeah, OK.
So there's just this little thing here.
So what I was going to do is just talk about some of the other things in this little booklet that work through each of these stages of the writing.
So that's the concept, and then the next phase would be what's called the embodiment, if you think of the design language, and that would be the first draft.
And I think most people find the most difficult thing is to write the first draft.
I mean, I find that the hardest.
Once you've got something, editing it is relatively straightforward.
You go, oh, this piece should go over there or there's something missing.
But getting the first draft down is the hardest thing.
And one of the things-- it's like when I talked about writing the book, you don't write the draft sequentially either.
I typically tell students to start with the materials and methods because that's the most straightforward and it's the easiest to write.
There's not a whole lot of mystery to how you actually did something you've already done.
So I tell people don't write the introduction first.
That's a bad idea because it's actually often not so easy to write the introduction.
So I tell people to write the materials and methods first.
Write the result section because you've got your results and you know what the results are going to be.
So those are the two easiest things.
And often, I think, what people find is if they find it hard to write it's because they don't know what they want to write or they haven't thought things through.
They haven't got their thoughts together.
And if you've got your thoughts, if you know what you want to say, then the writing becomes easier.
So one of the pieces of advice Mike gave me, and it's in this booklet too, is in the first draft what you should try to do is just get the facts down.
Just get the information down and don't worry about if it doesn't sound quite right or if this sentence doesn't lead into that sentence.
Don't worry about the style of it at all.
Just try to get the facts down.
Just try to get the information down.
And once you've got the information and you've got some kind of framework-- I'm not saying that you don't want to make the style good, you do, but you don't need to do that the first thing.
The first thing is just to get the facts down, and then you can edit it later on.
It's more of this iterative process.
OK, so the first draft, the most important thing is just to get the facts down.
Then, I'm not going to read all of these things because you can just read them yourself, but I'll just comment on a few things.
So one thing he's got-- I'm on page eight now, is the abstract.
So the abstract should be concise.
It should be fairly short and you want to tell people why you're doing what you're doing, what you did, what the key result is, and what the conclusions are.
So it can be pretty short.
There's a section on the introduction.
You can just read that.
Let's see here.
Yeah, I think, you can read yourself these other things.
I don't want to go through the whole thing.
All right, so that's that.
OK, so you get to the end of the first draft, there's a little section on figures here.
Let me just make some comments on that.
People who are busy and may not want to read every word that you've written will look at the figures, and they'll look at the figures to make some judgment about does this look useful?
Does this look like you've got something I want to spend more time on?
So it's important to make the figures easily understandable and to be fairly self-contained.
So you want to make the figures clear.
You don't want to have too much information on it so that people can't follow what's going on, and you want them to illustrate the points that you're trying to make.
So the figures are very important.
And then when you've got the first draft finished, the next step, which I find tremendously useful and students often don't get that this is a step.
The next step is you put it somewhere and you don't do anything with it for a while.
And this is why it's important not to write it up at the last second.
Because there's something about, you work on something, you put it to one side, you don't do it.
You just leave it and you go do something else.
And then when you come back to it, all of a sudden, you see things that you didn't see the first time through.
So the next step is you just put it to one side for a couple days.
But you can only do that step for a class if you started before the night before it's due, OK?
So that's why you have to start earlier.
That's why I'm telling you this now, OK?
So you put it aside and then you go have a cup of coffee.
You see there's several cups of coffee that Mike has happily drawn for us here.
OK, then the next part of this is all about grammar and sentence structure.
I'm not going to go over that.
You can read all that yourself.
But there is one amusing thing I want to tell you because there's a cute story.
I'm on page 16 now.
There's a thing about spelling.
Mike is a terrible speller.
When I was his student, he used to-- there were no word processors.
And we used to write things and he would write things and I would say, I can't read your writing.
And he says, ah, well, I make up for my bad spelling with my bad writing.
So I couldn't tell if he had spelled something right.
The other thing we found was that, we wrote the cellular solid book, the first edition was in the 1980s.
And I grew up in Canada, then I lived in England, then I moved here.
He grew up in Australia, then he moved to England, then he moved to the States then he moved back to England.
And words that end, that we spell I-Z-E, like normalized, like normalized variables.
They're I-Z-E here.
In England, they're all I-S-E. OK, and then some of the words, it turns out, Canadians spell with an I-Z-E and some with I-S-E.
So anyway, we wrote the entire fricking book.
We were almost ready to send it off, and we have all those graphs where the axes are normalized, and we must have used the word normalized like 100 times, and we realized half the time we had spelled it I-S-E and half the time we had spelled it I-Z-E. And then we had to go back and find them all and fix it all.
So anyway, that's just our little story about spelling.
OK, there's a thing about punctuation.
You can read that.
That's kind of boring.
I wanted to move on to the bit about style, because style is important too.
And people remember papers that are well written.
Obviously, papers that have valuable scientific information are memorable too, but if it's well-written, it's more memorable, and the style really is important.
And it's the same with giving a talk.
You can convey the same information, but if you don't have it presented in a clear way, people aren't going to remember it.
So the first rule-- so I'm on page 20.
The first rule, now, is to be clear.
And being clear, it doesn't mean having long-winded sentences with lots of jargon.
It really just means make it short, make it concise, make it clear what you want to say.
So Mike has several examples here of headlines from the newspapers, which were not clear.
So I'll just read these out because I find them amusing.
"Red Tape Holds Up New Bridge."
So, OK, maybe the red tape temporally held it up, but not spatially held it up.
OK, and then here's another one.
"Something Went Wrong in Jet Crash, Expert Says."
In fact, you hear this all the time on the news now.
Obviously, something went wrong.
You don't have to be an engineer to figure out something went wrong in a jet crash.
Then, "Chef Throws Heart In To Help Feed the Hungry."
OK, well, maybe that's a little too much.
This is my favorite one, "Prostitutes Appeal to Pope."
OK, there's different ways you can appeal.
You don't have to appeal to the Pope in that way.
And then, "Panda Mating Fails, Vet Takes Over."
I don't think the vet really took over with the mating, but you know what they mean.
OK, so one thing is clarity.
Another point is don't waffle.
You want things to be concise and get to the point.
So he's got this example here from what he cites as a well-known but anonymous materials text.
"The selection of the proper material is a key step in the design process, because it is the crucial decision that links computer calculations and the lines on an engineering drawing with a real or working design."
But what does it say?
It says material selection is important.
So don't say something in some long-winded way that you could say in a short way.
OK, let's see.
So let me move on to the next page.
The next step that I-- let me emphasize one more time, is 8.5, revise and rewrite.
So writing really is iterative.
Those books that you've seen that Mike and I have written, I cannot tell you how many drafts of each chapter we went through over and over and over again.
So to make it good, you have to do it over and over again.
OK, there's that.
Let me see.
Is there anything else I have here?
Ah, here's a couple things that are not so obvious.
So one is, for each paragraph you should have a good first sentence.
You should tell the reader something they don't already know.
So there's some examples here, and you can read through them.
But it is really helpful for each paragraph to have a good opening sentence.
Another thing is at the end of the paragraph it's good if there's a sentence that links it to the next one.
So it hints at what the next paragraph's going to be about, so it leads the reader through it, and it is a logical progression.
So an opening sentence for the paragraph and the final sentence, you might want to look at in a bit more gory detail.
OK, and then there are some references at the very back, at page 25.
And I brought a couple of books in that I have found helpful.
So you may have seen these.
One is called The Elements of Style and this is by Strunk and White, so William Strunk, Jr. and EB White.
You guys know EB White, Charlotte's Web, same guy.
He's written this small, small, not too long to have a look at book about style.
So there's some rules of usage, which is grammar stuff, but there's principles of composition and there's a chapter called "An Approach to Style."
And so here's some of the headings.
Place yourself in the background.
Write in a way that comes naturally.
Work from a suitable design.
These people are not engineers, but they're saying work from a suitable design.
It's like that concept thing.
Revise and rewrite, did I mention that, revise and rewrite?
Do not overwrite.
Don't make it too complicated.
Avoid the use of qualifiers.
If you say something's very stiff, what does that mean?
You can just say it's stiff.
You don't have to say it's very stiff.
Very compared to what?
Let's see.
Use orthodox spelling, blah, blah, blah, blah, blah.
Avoid fancy words.
OK, be clear, all this kind of stuff.
So Strunk and White is old but good.
And then, there's this book here by Bill Bryson.
Bryson's Dictionary of Troublesome Words.
So this is, it is like a dictionary.
It goes by letter and has different words, but it talks about how people sometimes use a word thinking it means one thing when it doesn't, it actually means something else.
And so it's just words that people use commonly that they sometimes confuse with what they really mean.
So it's a very handy thing too.
Do you know Bill Bryson?
Very funny travel writer.
If you ever go somewhere that Bill Bryson has been you should get the book he's written about it because he's very funny.
OK, so I think that's it.
Oh, and then, I think, the very end of this booklet has some examples of good writing and bad writing.
So you can have a look at that too.
All right, so are we good on how to write a paper?
Do you have questions on how to write?
Writing is important.
So I have one more writing story.
So when I first got this job at MIT, I was living in Arlington, and my mom comes to visit me from Toronto.
And I'm work, work, work, work, working, working, and my mom says to me, she says, my, you spend a lot of your life writing.
I'm like, Mom, that's what I get paid to do.
I get paid to write.
Like she always thought I was should be spending maybe 30 hours a week lecturing or something like that.
Like she, well, you're a teacher.
You only teach three hours a week, what is this?
I'm like, I write, Mom.
That's what I do.
So anyways, so writing is important and giving presentations is important.
And no matter what you end up doing, you're going to end up having to do those two things.
And if you do it well, it's going to make your life better, and if you do it badly, you're not going to be so good.
So that's my message.
So I'm going to stop there because I haven't got the next lecture ready, and I can start that on Monday.
So Monday we'll do the cell mechanic stuff.
And then let me just-- because we're getting ridiculously close to the end of term.
Like four weeks from today is the last test.
I know, it's shocking.
So there's one lecture on cell mechanics, and then there's some more engineering applications of foam.
I'm going to talk about energy absorption in foams and foams in sandwich panels, that kind of stuff.
And then, we're going to talk about some natural materials again.
We're going to talk about natural structures, so like natural sandwich panels and natural cylindrical shells with foamy cores.
So I've got lots of nature things at the last part of the course, and you know how I like those nature things.
So there you have it.
Bob's your uncle, as my mother would say.
You know what Bob's your uncle is?
Bob's your uncle is an English expression.
It just means there you have it.
There you have it.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
All right.
So last time we were talking about tissue engineering scaffolds.
And what we're going to talk about today still has to do with tissue engineering scaffolds, but we're going to look at it from a different perspective.
So last time we were looking more at sort of a clinical perspective, and looking at those osteochondral scaffolds for repairing small defects in cartilage.
And today what we're going to talk about are how cells-- how biological cells, interact with the scaffolds.
And there's various kinds of interactions.
So we're going to go through a bunch of these.
So the first one I'm going to talk about is degradation of the scaffolds.
Then we'll talk about the cell attachment.
Cell morphology-- so the shape of the pores in the scaffold can affect the way the biological cells-- what shape they have.
Biological cells could also contract the scaffold and apply mechanical forces.
So we're going to talk about that.
The stiffness of the scaffold and the pore size can affect the speed of cell migration.
And the stiffness of the scaffold can affect the differentiation of cells, so from one cell type to another.
So I thought today I'd talk about that.
This probably won't take the whole hour.
The next topic is on energy absorption in foams.
And so we'll probably start that towards the end of the lecture.
OK.
So the idea here is that we're looking at how scaffolds are being used, really, to provide a 3D environment to characterize the behavior of cells.
And in particular, how the cells interact with their environment.
So let's write that down.
So how the cell behavior is affected by the substrate it's on.
OK.
So the first thing we're going to talk about is scaffold degradation.
And if you think of the native extracellular matrix, the cells secrete enzymes which resorb that matrix and then they also deposit new matrix.
So it was kind of like what we were talking about with the bone.
The bone is always being resorbed and deposited.
And if there's a balance between that those two, then the density of the bone stays the same.
And if one of the rates gets out of whack, then you get osteoporosis and you lose bone mass.
So the idea is that in just the native extracellular matrix, the cells are producing enzymes that degrade the scaffold.
And those enzymes are also going to degrade the tissue engineering scaffolds as well.
And you want to be able to control the rate of degradation, versus the rate at which the native extracellular matrix gets deposited.
Excuse me, sorry.
So you can kind of imagine if the tissue engineering scaffold did not resorb quickly enough, you'd have scaffold there.
And the cells would be trying to put down their own extracellular matrix, and there wouldn't be a place to put it.
And if it resorbs too quickly, then the cells don't have something to attach to.
So there has to be a balance between the rate at which the enzymes are resorbing the tissue engineering scaffold, versus the rate at which the cells are depositing their own extracellular matrix.
So in the native extracellular matrix, the enzymes produced by the cells are resorbing the extracellular matrix.
And then the cells are also synthesizing so they synthesize ECM to replace it.
So the cells are also going to degrade the tissue engineering scaffold that you put in.
And the length of time that the scaffold is insoluble, or so that it remains in the body as a solid, is called the residence time.
And so then we require the scaffold degradation to occur over a time that balances with the new ECM synthesis.
And so the scaffold residence time must be about equal to the time required to make new native extracellular matrix.
So the degradation rate depends on the composition of the scaffold, on how much cross linking there is, and on the relative density.
Obviously, the more scaffold there is, the longer it's going to take to degrade it.
And with synthetic polymers you can vary the molecular weight of the polymer.
And sometimes if you have copolymers, one may degrade faster than the other.
And you can control the balance of how much of each copolymer you have.
And for natural proteins, like collagen, you can control the amount of cross linking.
So you can do the cross linking by various techniques.
That's what's called physical methods.
There's something called dehydrothermal treatment, where you heat the collagen up to 105 degrees C in a vacuum, in a dry environment.
And that eliminates water and causes more cross linking.
There's a UV treatment, ultraviolet light treatment, you can use.
And there's also chemical cross linkers you can use.
So there's different chemical methods you can also use to cross link the collagen.
OK.
So the next thing I wanted to talk about was cell adhesion.
And let's just wait a minute for people to catch up.
Are we just about there?
So this next slide shows a sort of schematic of how a cell would adhere to a substrate.
So down at the bottom here, all these little squiggly lines are representing the extracellular matrix in the native tissue.
Or you can think of it as a, say, a collagen scaffold.
But here we have the ECM.
And this little blob here is our cell.
This is the nucleus of the cell here, the little green blob in the middle.
And the cell is attached to the ECM through something called focal adhesion points.
And this schematic here is a blow up of that focal adhesion.
And at the focal adhesion there's proteins called integrins.
And integrins pass across the cell membrane.
So the idea is the integrins attach to ligands on the extracellular matrix.
And then they also attached to the sub membrane plaque within the cell.
And then that plaque attaches to the side of skeleton.
Things like actin filaments within the cell.
So this is what attaches the cell as a whole to the extracellular matrix, is these focal adhesion sites here.
And different kinds of cell behaviors-- obviously, things like cell attachment, but also things like cell migration, are affected by those focal adhesions there.
So we have that the cells attach to the ECM at focal adhesions.
And sometimes you see those referred to just as FA.
And at the adhesion point the cell has integrins.
And the integrins are transmembrane proteins, so they go across the membrane.
And they bind to like ends on the ECM.
And then the other end of the integrin is attached to the submembrane plaque within the cell.
And then that connects to the cytoskeleton.
And then different kinds of cell behaviors-- so for example, things like adhesion, and proliferation, and migration.
And that cell contraction, we're going to talk more about that in a minute.
They all depend in part on this adhesion between the cells and the extracellular matrix.
And the biological activity depends on how many binding sites there are.
So if you think of the extracellular matrix, it's got these ligands and it depends on the density of binding sites, how much interaction you can get.
So things like how much cell attachment you can get, depends in part on just how many of these binding sites you've got for the cells to attach to.
And that density of the binding sites or the density of the ligands depends on the composition of the scaffold.
But also on the surface area per unit volume of the scaffold.
So if you think of first, just the composition, if you have native proteins, like collagen, they have binding sites themselves.
They have native binding sites.
But if you think of synthetic polymers, like the resorbable sutured type of polymers that we talked about, they don't have binding sites and you have to coat the scaffold with some sort of adhesive protein.
And then the surface area per unit volume of the scaffold is related to the pore size and the relative density.
Let's call the specific surface area, surface area per unit volume.
And if you think of having some scaffold that's like an open celled foam, you can roughly calculate what the surface area per unit volume is.
So say each strut was a cylinder, then the surface area of each cylinder is going to be 2 pi rl.
If each one has a radius r and a length l. Say we had n of them, that would be your surface area.
And the volume of the whole scaffold, or one cell, would go as l cubed, the length of each strut cubed.
So if we just forget about all the constants here.
Forget about n. This just goes as r over l times 1 over l, and that goes as the relative density to the 1/2 power times 1 over the pore size.
So the specific surface area depends on the relative density and on the pore size.
And if you have a tetrakaidecahedron cell, you can work out exactly what that relationship is.
It's sort of a model.
And that gives you the relationship there.
And in this particular case, I think the relative density was 0.5%.
And so it's a constant over the cell size.
So one of the things we did in my group was look at how cell attachment varied with this specific surface area.
So we seeded cells onto scaffolds of different pore sizes.
We kept the relative density constant and we changed the pore sizes.
Remember I said, when we make these scaffolds by freeze drying we can control the pore size, by controlling the freezing temperature.
And we see that it's just a linear relationship between how many cells attach, or the percentage of the cells that were seeded that attach, and the specific surface area.
In here we used MC 3T3 cells.
It's sort of a standard cell one that you can get.
So Fergal O'Brien was the post-doc in my group who did that.
So I'll just say we find cell attachment is proportional to the specific surface area.
OK.
So that's the cell attachment.
So you can see how the scaffold design is going to affect how the cells attach.
So there's some relationship between them there.
Another thing people have looked at is cell morphology.
And so if you change, the sort of, orientation of the pores, how does that change the orientation of the cells?
So this was a study done in another group.
So here we have randomly oriented fibers that make up the scaffold.
And here they're not perfectly oriented this way, but more or less.
And then these are cells that have been seeded onto them, so that the green staining is the cells.
And you can see if the scaffold is random, the cells themselves line up with that fiber structure and become more or less random.
And if the scaffold has fibers that are aligned, then the cells, they also line up and be aligned.
So the morphology of the cells can be affected by the orientation of the scaffold pores.
Also the cell morphology can be affected by the stiffness of the cells.
Or the stiffness of the substrate.
So this is a substrate.
Here this was a PEG-fibrinogen hydrogel.
And they varied the cross linking of this hydrogel.
So they got different modularly for the hydrogel.
So these numbers here, are all the stiffness of the four different hydrogels.
And you can see the cell morphology changes from being a spread out thing on the least stiff substrate, to being just a little spherical or circular blob on the most stiff substrate.
So the cells respond to the substrate.
And so how the cells behave, depends in part on their environment.
So I wanted to also talk about womb contraction.
And talk about how cells contract scaffolds as well.
So one of the things people have found when they look at say, skin and regeneration of skin-- so say you had somebody with a burn and the surgeons will clean the burnt out.
And then what will happen as it heals, is scar tissue will form.
And the scar tissue forms in conjunction with the wound contracting.
So cells will actually migrate into the wound bed and they'll pull the edges of the wound together to try to close the wound.
And they won't close it completely, but they'll partially close it.
And that's called wound contraction.
And that is thought to be associated with the formation of scar tissue.
So the cells can actually apply mechanical loads.
And they can contract the wound.
And one of the things that Professor Yannas found was that if you use one of his collagen and gag scaffolds, you can inhibit that wound contraction.
And if you can prevent the wound contraction from occurring, you also prevent the formation of the scar tissue.
And that allows normal dermis to form.
So you get normal skin.
So this photograph here is of somebody who had burns over their entire torso.
And they put this tissue injury scaffold on this part at the bottom, but not on that part at the top.
And you can see these lines here are contracture lines from the scar formation.
And you can see this skin down here is relatively normal.
And in fact, when people look at the histology of the skin the forms using these scaffolds, they find that it is pretty much the same as normal dermis.
It doesn't have sweat glands and it doesn't have hair follicles.
So you can't sweat from that skin and you don't grow hair.
But apart from that, it's more or less normal dermis.
So this observation that if you can inhibit the womb contraction, you can prevent scar formation and you can get normal dermis to form.
That's led to some interest in just seeing how is it that the cells do this contract l process.
I think hitting the thing and my battery is dead.
So one of the things people have done, is they've just taken what's called, free floating scaffold.
They've just taken little disks of scaffold and put it in a cell culture medium in a Petri dish.
And they find that if you put, say fiberblast on it, the fiberblast will contract that scaffold.
And people have measured how much the diameter of the scaffold changes.
And so they've kind of measured this contraction just by-- it's almost like measuring a strain.
And what we wanted to do is we wanted to try to measure the forces that were involved.
So we first developed something called a cell force monitor, and I'll show you that.
And then we tried to calculate how much an individual cell could apply in terms of the force.
So we used this scaffold here.
This is the same collagen GAG scaffold I showed you before.
And here's the cell force monitor.
So that's just a schematic of holding a piece of the scaffold between two clamps.
So here it is in elevation view.
And then I'll just build the whole thing up, so you can see how it works.
So it's on a base plate.
It's attached to a horizontal stage that's adjustable.
Then there's a very thin beam here.
So this is another adjustable stage here, and this very thin beam here.
And that's attached to one end of this clamp.
And here's the matrix.
And this is attached to this other adjustable stage here.
And then when we have a proximity sensor-- so what's going to happen is, this is fixed over here.
The scaffold is going to contract with the cells applying these contract l forces.
This beam here is going to bend and the proximity sensor is going to tell us how much it's bent.
So we can measure how much that's bent.
If we know how much that's bent, and we calibrate the beam, we can figure out the force in the beam.
OK.
So we can figure out how much is the total force that the cells are contracting with.
And then this just is a little silicone well with some culture medium.
So that's the whole setup there.
Toby Fryman was a student who did that, who's married to Professor Van Vliet.
And I have a very big soft spot for both of them.
So anyway, that's the set up.
And the thing that Toby measured was the force, by measuring how much that beam deflected.
And he measured the force over time.
And he found that if he put say, a certain number of fiberblasts onto the scaffold, the force would increase and then reach an asymptotic point.
And you could describe these curves by this equation here.
Here's the asymptotic force.
And it's a 1 minus exponential of minus time over a time constant tao.
And then this number here is the number of fiberblast that were attached at 22 hours.
So he ran these tests for 22 hours.
And when he was finished, he could count the number of cells that were attached in the scaffolds.
So you would just wash off any cells that weren't attached and you can do accounting of how many cells are left.
And one of the things that he found was that if you plot that asymptotic force-- if you plot through this force over here, against the number of cells that were attached, you just get a linear relationship.
And the slope of that is roughly the force per cell.
And that's about one nano neutron.
Now this is a little deceptive because not all the cells are contracting.
And not all the cells are lined up in one direction.
So there are cells in different orientations.
But just as an order of magnitude the cells are applying something like one minute per cell.
So that's the effect of the cell number.
Another thing he did was he looked at what happens if you change the stiffness of that beam if.
You make that beam in the device different stiffnesses, how do the cells react.
And so the stiffness here are the stiffness of the system.
So there's 0.7 newtons per meter up to ten, so it's a factor of a little over ten difference.
And you can see the displacement per cell changes.
The stiffer the system is the less the cells can displace it.
But if you then plot the force per cell, you find that the force per cell is about the same.
So you develop about the same force.
So that suggests the cells are capable of applying a certain amount of force, and not any more force.
No larger force.
So he did that.
Then we were interested in what was the mechanism of this.
How were the cells applying this force?
Because I was kind of surprised to find out the cells even could apply forces.
So we were interested in understanding the mechanism of this.
And one of the things we knew that we didn't quite figure out how this all worked together was, we knew that the cells elongated.
If you just take a substrate, like even just a 2d substrate, and you put cells on it they'll be rounded to start out with.
And over time, over a few hours, they'll spread.
And that's pretty standard.
Many types of cells will do that.
So we knew the cells were starting off as rounded and they were spreading.
So the cells are getting longer, but our whole scaffolds getting shorter.
And so it wasn't obvious how was the cells going longer, but the scaffold's getting shorter.
And so the next thing we thought we would do is just watch the cells and see what they did.
And so we measured the aspect ratio of the cells at different time points.
And we did this by just impregnating the scaffold in the cells at different time points with a resin, and then using a stain, and then using digital image analysis.
So what we found was that the fiber of the fiberglass morphology looked like this.
So the long thready things of the scaffold, and these little blobs here are the fiberblast of the cells.
So here at time 0 you can see-- like I said, the cells are pretty rounded they're not very spread out.
Here at eight hours you can see-- here's a cell that's gotten longer.
Here's another one.
This guy here is still rounded, it's not doing much.
22 hours, again, some of the cells are quite elongated.
Some of them are still not that elongated.
So they don't all become active.
But one of the things we noticed, if you look at this image here, you can see these cells are attached at one end, and at the other end.
But they're not attached in the middle.
There's sort of a gap between the cell and the strut.
And this is another example here.
Here's a cell here, and this is the collagen GAG strut that it's attached to.
And you can see it's attached to the two ends, but not in the middle.
And this starts to explain how it is that the cells are elongating but the scaffolds getting shorter.
It's that the cells are just attached at two ends.
And the cells are moving along a strut and they're attached to the two ends.
And if you think of the cells attached through those focal adhesion points, they're applying tension to the cell.
And the actin filaments in the cell are in tension.
Obviously, filaments can't be in compression.
They're only going to be in tension.
And what happens is that puts the stress into compression.
And if the struts in compression, at some point it's going to buckle.
And you can see this strut here has basically buckled under that cell.
And so if the cells are getting longer, and they're buckling the struts, then that's going to shorten the struts and the whole scaffold is going to get shorter.
And so then Toby plotted the aspect ratio of the cell, so that is a measure of their elongation against the time.
And again, he found one of these curves with the same kind of form as the curve for the forced development.
And he found the time constant here for the change in the aspect ratio was about five hours.
And for the development of the force it was about 5.7 hours.
So the time constant for the elongation of the cells, more or less matches up with a time constant for developing the force.
So that's what that says.
And that suggests there's a link between the elongation of the cell population and the macroscopic contraction of the population.
So then we wanted to take it one step further.
And we wanted to look at what the cells were doing live.
Like as they were doing it.
So Toby devised this little schematic thing here.
So he had just an optical microscope.
He had a microscope slide with a fairly thick well in it, so that we could put culture medium in the well.
We put a cell seeded matrix in here.
And he had a heated stage here.
And then he took little videos of what the cells were doing.
And this required some patience because as you could see not all the cells did anything.
Some of them just sat there and did nothing.
So he would set this up for a day, and watch a cell, and it would do nothing.
And then he would have to find another cell.
But he did find some cells that were responsible for the contraction.
And that was it was kind of neat.
So here's the scaffold again.
All these little bits here are the scaffold.
This is a strut of the scaffold.
And this is a fiberblast parked on the scaffold.
And this has a little video here.
And you can see what's happening is the strut here is starting to buckle.
And you can see these two sides here, those two things are coming closer together.
So they originally were this piece here, and that piece there.
And now they're at that point there.
And then if I let it go a little bit longer, it continues to do that process.
And then the final thing-- this kind of smushed up mess here is these two things having me brought completely together.
And this strut here is some strut down over here.
So you can see how the cells are elongating and causing contraction of the scaffold.
Here's a series of stills taken from another video that he did.
So this sort of square thing is the scaffold.
So b is the scaffold.
And a, this little blob here, is the fiberblast.
And you can see, even from this image to this one, you can see that the fiberblast has spread a little.
Do you see how it's kind of oozed out along the scaffold there.
And eventually it attaches over here.
And you can see that it's buckled this strut underneath it.
And here it's a little bit more deformed.
It then grabs on down here somewhere and deforms it even more.
So you can see that's more deformed.
And then Toby put alcohol on the whole thing, which kills the cells and the cell let's go.
And you can see you recover some of the deformation.
You don't recover all of it, but you recover some of it.
This was another example.
And this was kind of interesting.
Here there was a scaffold junction where there were three struts that came together, a little bit like a strut.
And there was a little cell right there.
And you can see the cell elongates.
You see how this elongated and its grabbing on up here somewhere.
But the amount of force the cell was kind of pulling with must have been less than the-- or rather must been more than the force of the focal adhesion.
Because what happens was eventually the focal adhesion let go.
And the cell kind of bounces back and ends up over here.
So the cell was kind of snapped back on to the other focal adhesion over here.
And here it's rounded again.
And here it elongates again.
And then this focal adhesion lets go and now it's moved back over to there.
So these struts here are so stiff.
They're much stiffer, I think, partly because they're triangulated.
And it looks like they're just shorter and a lot thicker.
The cell isn't being able to deform those.
But it's elongating and then focal adhesion was letting go.
So this is a little schematic of what we thinks going on.
So the cell starts out-- it's some elongation here.
It's attached at that point.
It's attached at that point there.
And the cell is getting longer.
And if you think about it as the cell's getting longer-- if you think about the Euler Buckling formula, the buckling load goes as 1 over l squared.
So the longer the length of this piece of the strut of the scaffold underneath the cell is, the smaller the load it takes to actually cause it to buckle.
So at some point it buckles like this.
And this is just a little force diagram.
So the actin fibers are in tension and the matrix strut is in compression.
Sometimes we saw some bending.
So you could see if a cell was spanning between two struts, you could get the cell bending the struts as well.
That was another possibility.
And so we think that the cell elongation was related to the contraction.
The time constants for the two things were almost the same.
And as the cell elongates there's a gap between the cell and the matrix on the central portion.
And then the cell is adhered at the periphery of the adhesion points.
And then the tensile forces in these act.
And filaments inside the cell induce compression in the strut, and that causes buckling.
And then Toby graduated.
And then I got another student, Brendan.
And Brendan saw what Toby did and he wanted to do a little more with that.
Brandon was also involved that osteochondral project that I talked about last time.
And Brendan this other thing as well for his project.
So he wanted to measure the force of an individual cell.
So when we had that cell force monitor, that was the total force of all the cells in that one direction.
But Brendan wanted to know if he could measure the force of a single cell.
And now that we knew that the contractal process was related to buckling, We thought, well, we could just use Euler's formula.
If we knew what the modulus of the solid was, and we knew what the dimensions of the struts were.
So that would allow us to calculate the contractile force of a single fiberblast.
So I think I've shown you this thing here.
So Brendan was the one who did these experiments.
He cut a single strut out of the scaffold.
And the single strut is about 100 microns long.
He used a microscope to do this.
He then glued it onto a glass slide and he used the atomic force microscope probe to bend the strut like a cantilever beam.
And he measured this displacement here.
And from that he could back out what the modulus of the solid was.
He did these tests in the dry state.
But we could extrapolate to the wet state from looking at the behavior of the whole scaffold.
So he had a modulus for the wet scaffold solid.
And then this is our formula for Euler buckling here.
So that's just the standard formula.
I had a student from civil engineering, who looked at hydrostatic loading of a tetrakaidecahedral cell and he looked at buckling.
If you had a tetrakaidecahedral cell and you load it in all three directions.
He looked at the buckling.
And he had calculated that the n constraint factor-- the n squared was point 0.34.
So we have some idea of what that n squared value should be.
Although it's somewhat of an estimate.
I had a UROP student who took Toby's images and measured the dimensions of the struts.
So he measured the diameter and the thickness of the struts.
And from that, we just plugged everything into the Euler formula.
And we found that the average single cell force is somewhere between about 11 and 41 nano neutron.
It was something like 26 nano neutrons.
So it would make sense that it's more than the one nano neutron per cell because not all of those cells were active and they weren't all going in the same direction.
So Brendan Harley and Matt Wong did that part of the project.
OK.
So that's the contraction.
Are we good with contraction?
So it's kind of interesting that cells will contract and we can measure some forces.
So the next type of interaction between the cells and the scaffolds that I wanted to talk about is cell migration.
And these are some studies from the literature.
These are two different studies.
But the top one here, they've measured migration rate as a function of the cross linking treatment of a scaffold.
And the decreasing stiffness goes this way.
And so they're seeing that the speed of migration-- this is in millimeters per day.
Cells don't move too quickly.
They go millimeters per day.
But you can see that the migration speed, the speed at which the cells can move, depends on the stiffness of the scaffold that they're attached to.
And in this study on the bottom here, what they did was they had just a flat 2d substrate.
Just a flat polymer.
And what they did was they cross linked one part of the polymer more than the other part of polymer.
So over here, this was the less cross linked.
That was the soft part.
And this was the more highly cross linked.
This was the stiffer part.
And they found that if they put a cell on the soft part it would migrate onto the stiff part.
But if they put a cell on the stiff part, it would start going this way towards the soft part.
But when it got to the interface it would just spread out along the interface.
And it wouldn't go into the soft part.
So the cells were somehow sensing the stiffness of the substrate.
And for some reason, I don't know what, but for some reason these particular cells seem to prefer being on the stiff substrate.
So this is just really showing that there's some interaction between the substrate stiffness and the way the cells are behaving and migrating.
And then Brendan also wanted to study this.
And he got some of the collagen GAG scaffold.
He made some of the scaffold.
And he stained that with a stain that made it turn red.
So these lines here are all red struts in the scaffold.
And then he put fiberblasts on to the scaffold and stained them green.
So all these little blobs here that are green are the cells.
And then he used confocal microscopy.
And the confocal microscopy allowed him to look at a certain volume of the scaffold.
And he had some software that would track the centroid of each cell as it moved through the scaffold.
And so he had a thing he called spot tracking.
So each of these little spheres here corresponds to a cell.
And the white box is the volume of material that you could see in the scaffold.
And this color scale here really corresponds to time.
So I've forgotten which round.
I think blue is the original time 0, and then red is maybe five seconds, and yellow was 10 seconds.
The different colors correspond to different times.
So he could track the path of each cell and also what the position was at different time points.
So he knew what the position was at different time points.
And obviously from that, he could get the speed of the scaffold.
And he did these experiments on scaffolds of different stiffnesses, as well as, different pore size.
And here you can see the cell speed.
He's measuring it in microns per hour now.
The cell speed increases at first and then decreases with the strut stiffness.
So we don't know exactly why this is.
But there is an effect between the stiffness of the scaffold and the migration speed.
And another thing he did was he looked at how the cell speed varies with the pore size.
And as the pore size gets smaller, the speed goes up.
And we're not entirely sure why that is.
But I think that might be related to this binding site thing too.
As the pore size goes down, the number of binding sites is going to go up.
And if you think of the cells migrating by having these adhesion sites, and the adhesion sites are just at the ends of the cells, and the cells kind of putting out a little extension, and then looking for somewhere else it can bind.
The more binding sites there are, the faster it's going to find a binding site.
And the faster, I think, it's going to move on.
So I think that the cell speed depends on pore size, at least in part because of the increase in the binding sites with smaller pore sizes.
So pore size and the migration.
And then the last thing I wanted to talk about was cell differentiation.
And this is a study study by Engler.
And one of the things he found was he put mesenchymal stem cells on 2d substrates.
Just flat 2d substrates of different stiffnesses.
And again, he could control the stiffness by cross linking.
And what he's showing up here in the first bit is that he's looking at the stiffness of tissues of different kinds.
So here's brain type tissue.
Something like one kilo pascal.
Muscle might be something like 10 kilo pascal.
And collagenous bone-- this is sort of the osteoid that is the precursor of bone, not the bone itself.
Is about 100 kilo pascals.
And what he did was he put these mesenchymal stem cells-- so here's his cell onto his substrate.
And he varied the stiffness of the substrate.
And then he looked at the shape of the cells.
So here's the least stiff substrate, so between point 1 and 1 kilo pascals.
And here's 4 hours, 24 hours, 96 hours.
And these cells formed long processes extending beyond the cell body.
And they looked kind of like neurons.
So they he called those neuron like.
Then there's an intermediate stiffness of substrate here.
And these cells became even more elongated.
And became something like a muscle cell, myoblast like.
And then cells that were put onto a substrate that was between about 25 and 40 kilo pascals, they developed a shape that was something like an osteoblast, like a bone cell.
So one of the things he was looking at here, was how the stiffness of the substrate affected how a stem cell might differentiate into different cell types.
And another thing that he did was he looked at different cell markers.
And he found that the cells were expressing markers that were corresponding to the types of tissue.
So I couldn't tell you the names of all these things and what they are.
But I think the red here is expressing more of a particular marker.
And I think these wounds were related to nerve tissue.
These wounds here, were related more to muscle tissue.
And these wounds here were related more to bone tissue.
So the things the cells were expressing also seemed to correspond to the different types of tissue that they were corresponding to.
So I'm just going to end this part by going through a little summary here.
So what I've tried to show you today is different types of cell behavior that are affected by the scaffold.
And they're affected by things like the number of binding sites, by the pore size, by the stiffness of the scaffold.
So we started with a cell attachment.
We saw that the cell attachment increases linearly with a specific surface area.
We saw that the cell morphology depends on the orientation of the pores.
And that kind of makes sense, they got to line up with the pores.
We talked about the contraction behaviors.
So the cells bind at the periphery, the cells elongate, and that causes this buckling.
And you can calculate the buckling forces.
It's around 10 to 40 nano neutrons.
We looked at the cell migration speed.
That increases with the stiffness of 1D fibers.
And we looked at cell migration in the collagen gag scaffolds.
So that depends on the stiffness of the pore size.
And then there was this final study on the cell differentiation.
So I wasn't going to write any notes on this because the slides I think pretty much explain it.
So I was just going to put the slides on the website at the end after today's lecture.
So are we good with how cells and the scaffolds of the environments interact?
Because I think it's not so obvious that this actual mechanical environment makes a difference.
People think of the chemical, the biochemical environment.
That obviously affects the cells.
But people don't think at first that something like the sort of structure of the pores, the pore size, or the orientation of the pores, or the mechanical properties are going to affect how the cells behave.
But in fact, they do.
So that's it.
And this is all various people who worked with me on these projects.
So it was a lot of fun.
OK.
So hang on a sec here.
What's this all about?
I'm going to get rid of that.
Go away.
Here we go.
OK.
So are we good with cells and substrates?
Yeah?
OK.
So let's just take a little moment and I'll rub the board off.
And then we can start the next bit.
OK. OK.
So that's the end of the medical material stuff.
So we talked about the bone.
We talked about the tissue engineering scaffolds.
And then we talked about the cell scaffold interactions.
So now we're going to go back to more engineering topics.
And the next thing I wanted to talk about was energy absorption in foams.
So foams are very widely used for energy absorption applications, things like bicycle helmets, different kinds of helmets.
You buy a new computer, it comes in foam packaging.
And the reason foams are used so much is they're extremely good at absorbing energy from impact.
And in fact, they're better than the solid that they're made from.
So let's just look at this curve here for a minute.
So here's a stress strain curve in compression for the foam.
And the material that it's made from would have the stiffness something like this.
It would be much, much stiffer than the foam.
And if you think about how much energy you can absorb, the energy you can absorb is just the area under the stress/strain curve.
That's the energy you can absorb in a given volume of foam.
And so when you're thinking about these energy absorption problems, it's not just that you need to absorb a certain energy.
You need to absorb it without exceeding a certain peak stress.
So whatever it is you're trying to protect, at some point it's going to break.
This is what you want to avoid.
You want to avoid it breaking.
So you don't want to have a stress bigger than the stress that's going to break whatever it is, your computer, or your head, or whatever.
So say you have a given peak stress that you can tolerate here.
And we've normalized things by the solid modules.
But just say that's a peak stress here.
The foam is going to absorb this amount of energy up here, this whole little shaded region.
And the solid is going to absorb that little, teeny weeny bit in there.
So what you want to do is absorb the energy without exceeding a certain peak stress.
And the foam is always going to be better than the solid that it's made from.
There's a couple other things that make the foams good because they're more or less isotropic, maybe not perfectly.
But roughly, they have the same properties in all directions.
Sometimes you don't know what direction the impact's going to come from.
And so if you've got the same properties in all directions or roughly the same, that's a good thing.
You also want the protective thing to be light.
If you're paying for shipping for your computer or whatever, the fact that the packaging is light makes the shipping easier.
If you have a helmet for your head, you don't want some big heavy thing.
You want something fairly light.
And foams are cheap.
So the fact that they're roughly isotropic, they're light, they're cheap, this all helps as well.
But from a mechanical point of view, foams are very good at absorbing energy.
And so what we're going to do in the next-- the rest of this lecture and on Wednesday-- we're going to see how we can convert these stress/strain curves into what are called energy absorption diagrams.
We're going to look at some energy absorption diagrams that we just measure from the stress/strain curves.
And we're going to look at how we can predict the energy absorption diagrams as well.
OK.
So the main idea here is that the impact protection has to absorb the energy from the impact but without exceeding a certain peak stress.
So the direction of loading may not be predictable.
And foams are good because they're roughly Isotropic.
And they would have the same energy absorption capacity from any direction.
And foams are also light and cheap.
We can say for a given peak stress the foam is always going to absorb more energy than the solid it's made from.
So other things that make foams good are that they have a capacity to undergo large deformations.
And they do that at roughly constant stress.
So that if you look at the stress strain curve for the foam, you're going to be able to absorb all this energy under here.
And these strains that the foam might go to might be 0.08 to 0.09, so huge strains on an engineering scale.
And then this is your energy-- would absorb is that area under the stress/strain curve.
So I wanted to say something about strain rates too.
So typically we're going to be talking about problems of impact.
And in impact, the strain rates are typically on the order of 10 to 100 per second, something like that.
We're not going to talk about things like blast.
If you have a blast loading, then you have to take inertial effects into account.
And blasts involves strain rates, which are 1,000 to 10,000 per second, much, much higher.
So we're going to talk about strain rates that are about 10 to 100 per second, maybe a bit more than that.
And for instance, you can roughly estimate what one of these impact rates would be.
So you had something that you dropped from a height of 1 meter.
Then the velocity on impact is just if you just equate the potential energy with a kinetic energy.
The velocity and impact is just the square root of 2gh.
So g's the gravity acceleration.
And h is the height.
So that's the square root of 2 plus 9.81 meters per second times 1 meter.
And that comes out to 4.4 meters per second.
And say you had some foam packaging that was 100 millimeters thick.
Then you could say roughly that the strain rate would be approximately equal to that velocity over the thickness, so 4.4 per second over 0.1 meters.
That' would be 44 per second.
So it's somewhere in that range.
Obviously, the thickness could be a little bit smaller, it could be bigger.
But it's in that ballpark.
And if you do tests on servo controlled instrons or you do a drop hammer test, you can get strain rates in that ballpark.
OK.
So we're talking about impact and not blast.
OK.
So most of the energy that's absorbed is really absorbed in that stress plateau.
So if you think of the stress/strain curve, most of the area under the stress/strain curve comes from the area from underneath the stress plateau.
So the mechanisms of absorbing the energy are going to be mechanisms that are associated with a plateau stress.
So for elastomeric foams, we've got elastic buckling of the cells.
And one of the advantages or disadvantages-- depending on what you want-- of this is that the deformation is recoverable and you got to have rebounds.
So if you have an object and you drop it onto elastomeric foam, it's going to bounce around like that.
So the elastic deformation is going to be recovered, and you're going to get rebound.
If you have a foam that has a plastic yield point or is brittle, then the deformation is going to be largely from dissipating plastic work or work of fracture.
And in that case, there's no rebound.
But once you've loaded it, you've crushed the thing, and you've permanently deformed it, and you can't use it again.
So sometimes if you ride your bicycle like I do, if you have a helmet, you should wear your bicycle helmet.
If you have a problem, if you have an accident, and your helmet get smooshed, that's it.
You have to throw your helmet away.
You can't use it again.
And this is why.
[INAUDIBLE], even if it doesn't get smooshed, if you hit your head at all, [INAUDIBLE].
Exactly.
[INAUDIBLE] Yeah.
You need a new helmet.
Yeah.
Go ahead.
Talk about helmets because I'm on a helmet conversion thing.
Yes.
You've got to wear your helmet.
And you should change it every now and then.
Anything else you'd like to add about bicycle helmet safety?
No, absolutely.
You've got to wear your helmet.
So I know several people who would have had their head smooshed had they not been wearing their helmet.
So you have to wear your helmet.
Let's see.
OK.
If you think about natural cellular materials, things like wood, they often have cell walls that are fiber compensates.
And you can dissipate energy by mechanisms related to the fiber nature, so by things like fiber pull out fracture.
And then you can also have open cell foams with fluids.
You can have fluid within the cells.
And if the cells are open cells, the fluid effect is really only going to be important if the cells are extremely small or the fluid is particularly viscous, or the strain rates are very high.
So in most cases, the fluid effects aren't important in open cell foams.
But, for example, you could try to make an open cell foam that had more energy absorption by putting a fluid into it.
So you could put glycerin into the fluid, and that would increase how much energy it would absorb.
Or, you can put this honey into it.
That would make it more energy absorption.
And enclosed cell foams, you may have an effect of the gas within the cells.
But it's really only going to be significant if you have elastimeric foams where the cell faces don't rupture.
The cell faces rupture, then the gas is just going to flow out of them, and that's not going to do much.
So the next step is I want to go from having the stress/strain curve that we've become very familiar with, and make something with that that is a little easier to see graphically that shows how much energy we can absorb.
Remember, what I said what we're really interested in is absorbing a certain amount of energy without exceeding a certain peak stress.
So what I'm going to do is plot another plot that's based on that.
It's going to be the energy absorbed.
So w is going to be energy absorbed per unit volume.
And I'm going to plot that against the peak stress.
OK.
So we're going to look at three different regimes here.
We're going to look at what happens in the linear elastic part, what happens in the stress plateau, and then what happens in the densification part.
So let's think about the elastic regime first.
And if I moved up-- say I moved up to some point right there where the little x is on the stress/strain curve.
Then the amount of energy I absorbed would just be equal to this little bit here.
And if I moved up, and then the peak stress would be this peak stress there.
We'll call that sigma p1 and w1.
And if I moved up over here, I'd be at w2.
And that would be sigma p2, right?
And if I know the modulus, I know what that relationship is.
And I get a relationship.
And these are going to be-- I'm going to do this on log scales here.
There's going to be log, and that's going to be log.
I'm going to get in that linear elastic regime.
The energy is going to go as the peak stress squared over 2 times the modulus of the foam.
Remember, energy is a half stress times strain.
And I can say strain is sigma p over e. So it's 1/2 sigma p squared over e. So on my log1 plot here, this is just going to be a straight line like that.
And then I'm going to get to this value here.
I'm going to get to my collapse stress here.
So let's call that single star.
And at that point, the more I go along here, every point I go along, like that, I'm going to absorb more and more energy.
But the stress isn't going to go up at all.
So then this thing here is going to go like that because I'm absorbing more and more energy.
But the stress just stays the same.
So this is good news if we want to absorb energy.
And then once I get to the densification point, then it's going to do the opposite thing.
As I go along here, at each increment I'm not absorbing that much more energy.
But the stress is going up.
So at some point it turns and starts to look like that.
So this part here corresponds to linear elasticity.
This bit here corresponds to the stress plateau.
And this bit here corresponds to densification.
And the point where I would like to be is right here, because here I'm going to absorb the most energy possible through the peak stress.
So you can think of that as sort of an optimal point.
And I'm going to refer to that as a shoulder because it's the shoulder between where the curve bends over again.
So I've only got a couple minutes left.
But let me just show you one thing and then we'll talk about this more next time.
So I've just done this for one relative density.
But if you look at the screen, you can imagine I would have stress/strain curves for lots of different relative densities.
And let's say these are all at the same temperature and all at the same strain rate.
And I could draw a curve that looks like that for each stress/strain curve.
And if I did that, I'd get a family of them.
So this is our energy absorbed here.
I've normalized it by dividing by the solid modulus.
This is our peak stress here.
And I've normalized that by dividing by a solid modulus.
And I've got a sort of family of these things, right?
They all have the same shape.
But they shift depending on the relative density.
And then the thing that makes life good is that these shoulder points you can connect with a line.
And you can mark off the relative density for those shoulder points on each line.
And then the last step you can do is you can just plot these lines.
And you can repeat this for different strain rates.
So this would be a family of these guys here.
There's a family of those lines at different strain rates.
And then you would join up the points that correspond to each relative density.
So you can make a drawing that looks like this that summarizes the most energy you can absorb for a certain peak stress for foams of different relative densities tested at different strain rates.
You could do it for different temperatures if you wanted to.
So next time, we'll talk about that.
But I'm going to stop there for today.
OK?
Are we good?
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation, or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
All right, so last time we started talking about energy absorption, and I wanted to try to finish that up today.
And then next week we would talk about sandwich panels and using honeycombs and foams in sandwich panels.
So I think we got as far as the idea of introducing what these energy absorption diagrams are.
So let me run through this little sequence again, and then I'll put the notes up on the board.
So the idea is that you have your compressive stress strain curve-- so here's a series of curves for different densities of a foam.
And we would do these all at the same strain rate and temperature, so that those aren't variables.
And then what we would do is we could turn those into energy absorption diagrams.
So notice here, this is a log plot.
So here's that energy absorption here.
Here's the peak stress here-- so that's the peak stress up to some level of energy that you've absorbed.
And we've normalized both of those by the solid modulus.
So say we look at one density here, say we look at the lowest density-- 0.01.
So here's our curve here for the test.
If I go up to some stress level that's still in the linear elastic region, then that's going to translate into somewhere along here on the energy absorption curve.
And one of the things we're going to do today-- I'm going to show you how you draw these, and how you can set them up.
And you can either get them from experiments-- so this is, say the top curve were experiments, doing it from experiments-- or you can use the models for the foams to do it, as well.
So we'll see how you can do from both ways.
So this would be the linear elastic bit here.
This vertical part-- where the energy is increasing, but the peak stress isn't increasing-- that corresponds to the plateau here.
And then this part here, where the energy is not increasing very much, but the stress increases a lot-- that corresponds to the densification part over there.
So what you would do is you would do tests on foams of different densities, and from the test data, you could draw an energy absorption curve for each density.
So you plot-- there's four different densities here, so it forms a family of these curves.
And then what you do is you say-- and I think we talked with this last time-- that the place you want to be is at this shoulder point.
You want to be absorbing as much energy as possible at that plateau stress.
So you want to be at the point just before it turns around to the densification regime.
So you can mark that little shoulder point for each density.
So here's like 0.01, here's 0.03, and so on.
And then those points can be connected by a line-- so that heavy line then connects those shoulder points, or those optimum points.
And then what you can do is then repeat this whole process for different strain rates.
So this family of lines here is at different strain rates-- so this set that goes this way, corresponds to this line here.
And if you do them at different strain rates, where the shoulder appears at a slightly different point-- and so if you mark those shorter points, you can draw these lines here that connect up for one relative density.
So this line on the left hand side here, these are all relative densities of 0.01, these ones are all relative densities of 0.03.
So this diagram down here, doesn't really look like this at all-- it doesn't look like the basic energy absorption diagram.
But it actually has information for what the optimum would be for a range of densities and a range of strain rates.
So typically, these foams are viscoelastic and they have some strain rate sensitivity.
So you'd like to get the strain rate sensitivity into it.
And it's not shown here, but you could do the same thing at a constant strain rate and varying the temperature, too.
So if you had things at different temperatures, you could do the same kind of idea.
OK-- so are we good with how this works?
Because now I'm going to write some notes on the board so you have it in your notes.
So the idea is that you turn your stress strain curve to look something like that, into an energy absorption diagram.
And you can plot on log log scales, the energy versus the peak stress.
And you get something like that.
And this point here, I'm going to call the shoulder point.
And that would use the material in the most efficient way-- you get the most energy absorption without getting higher than that plateau stress.
So we can say at this stress plateau, the energy increases without much increase in the peak stress.
And then as the foam densifies, then you get an increase in that peak stress, with little increase in the energy absorbed.
And so ideally, you want to be at that shoulder point.
So to construct these energy diagrams, you can test the series of foams at different relative densities and constant strain rate and temperature.
And then you make that plot of the energy absorbed normalized by the solid modulus, versus the peak stress normalized by the solid modulus for each curve.
And typically what people do is they would take the solid modulus at a constant strain rate and temperature, so you don't have to introduce that, as well.
Then you would mark the density for each of those shoulder points, and then you would connect them.
And then you could repeat this whole process for different strain rates.
And then you would draw the final diagram at the bottom, where you have this family of lines that describes the shoulder points for different densities and different strain rates.
And you could treat different temperatures in the same way, if you wanted to.
You would hold the strain rate constant, and vary the temperature.
so it's kind of a nice way of just putting a lot of information in one diagram.
So one of the things about this is that, because we're normalizing by Es, and if you think of elastomeric foams, elastomeric foams, both the Young's modulus depends on Es, and the plateau stress depends Es.
So the Young's modulus depends on the stiffness of the solid, and also because the plateau stress is related to elastic buckling, it also depends on the modulus of the solid.
So for elastomeric foams-- it's because you normalized it with respect to Es-- one of these diagrams will represent all elastomeric foams.
So that's rather a nice thing-- so you can have different elastomeric foams, but one of those diagrams is going to represent all of them.
So we can say, elastomeric foams can all be plotted on one plot, or one curve, since both the modulus and the plateau stress are related to Es.
So if we look at this next figure here-- maybe I'll just wait a minute for people to stop writing.
Some people write faster than others.
So if we look at this next plot here, here's a compressive stress and strain.
These tests are done for one density, but at different strain rates-- so it's kind of the other version of this.
But here's the stress strain curves here, and here's the energy absorption diagram that's derived from those.
And then here's a summary diagram that has the different strain rates, and that would have different densities here.
And the idea is this diagram here could represent all elastomeric foams.
So this has been put together for polyurethane, but it should be able to represent other sorts of flexible elastomeric foams.
Sorry?
In those diagrams, then, the intersection of this strain rate line and your relative density line should be the shoulder position?
Yeah, exactly.
And so this line here is for 0.01, and that one's for 0.03.
So 0.02 is going to be-- you'd have to interpolate somewhere in between there.
So we've just put on certain values, because we're not going to put on a million values.
We just put on certain ones, and then you can estimate where other densities would appear on there.
OK?
So here's another example here-- these are curves for two different foams.
So here's a polyurethane a polyethylene, so these are both elastomers.
And one is the dashed line, and one is the solid line.
And you can kind of see how the lines mesh up.
So here's a density of 0.01 for the polyurethane.
Here's 0.05 for the polyurethane.
And here's 0.06 for the polyethylene.
And you can see how the 0.06 and 0.05-- they're not quite on top of each other, but they're pretty close to coming on top of each other.
And then 0.1, 0.12-- and so you get a family of them for the different densities.
And you can also do this for materials that have a yield point, so polymethacrylimid has a yield point, so you do exactly the same kind of thing.
So here's the energy absorption diagram that's been developed from the stress strain curves.
But now this curve here, or this set of curves here, is really just valid for one ratio of sigma [?
ys ?]
to Es-- so the solid yield strength of the solid modulus.
So it's valid for whatever it was for that particular type of foam-- the polymethacrylimid.
So I'll just say, if we have foams that are made from a material with a yield point-- and so they have a plastic collapse stress-- then the curve will be valid for foams with the same ratio of [?
sigma ys ?]
to Es.
So in that case there, for the polymethacrylimid, that ratio is about equal to 1 over 30.
So that plot would probably give not a bad description of other foams, with the same value of sigma ys over Es.
So the idea here is we can generate these diagrams either from data-- the way those ones have been done-- or from the models.
So another way to generate these is to think about the models that we have for the foams, and the foam behavior.
So we have an equation that describes the Young's modulus, we have equations that describe the plateau stresses, we have an empirical equation for the densification strain.
And we can use those to generate these diagrams.
And they're kind of useful, because they show you what's going on a sort of mechanistic basis.
So this is a diagram here that's been generated for open cell elastomeric foams.
Here's our energy absorbed, here's our peak stress down here.
This little inset is sort of a schematic of the idealized foam behavior.
So here's the Young's modulus, here's the elastic [?
collapse ?]
stress, here's the densification over here.
So obviously it's kind of a very idealized set-up.
But we can generate this diagram-- and I'm going to go through the equations that will let us do that.
So one thing to note is here's the curves for each density.
Here is this line that connects the shoulder points.
There's a couple of other lines on here that I just wanted to mention something about.
If you think about the densities-- like that's 0.01, this is 0.03, that's 0.1-- if you had a fully dense solid that was made of the same elastomer, you could plot the curve for that, and that is going to show up over here.
So this is kind of an upper bound.
It can't get any more from that.
And we've also got a dotted line here, which takes into account fluid flow within the cells.
So there can be some fluid flow, and because we haven't talked about that, we're not going to go into that.
So we can just ignore that dotted line for now.
So let me go through how we can do the modeling.
So we're going to divide the stress strain curve up into bits, and I'm going to write equations for the energy absorbed for each bit.
So we're going to start with a linear elastic behavior.
And I'm going to-- let's see-- yeah, so let me just note here that this is the densification strain out here.
And this strain, epsilon naught, corresponds to the strain at which we first reach the stress plateau.
So here, for the linear elastic part, I'm going to say that the strain is less than that.
So the strain is less than absolute naught.
And then I can say, the energy absorbed-- so if you just remember from Hooke's law on linear elasticity, energy under the stress strain curve for the linear elastic part is 1/2 of sigma squared over E. So I'm going to call sigma-- whatever the stress is going to be, the peak stress that we get to.
And now we're going to divide by E of the foam-- because these were our foams here.
And I can use our model here to say-- and now I've got to divide that by Es, because I've normalized here.
So because I know from my modeling that the Young's modulus of the foam is just equal to Es times the relative density squared for the open celled foam, that means I've now got an Es squared in the denominator, so I've got a sigma p over Es squared.
And then I've got a 1 over the relative density squared term.
So that factor there, that equation there, gives you these first set of lines here.
Gives you that bit, and this bit, and that bit.
So it gives you those first parts of the energy absorption diagram.
And then if we look at the stress plateau-- so here we're going to say that epsilon naught is less than epsilon is less than the densification strain-- so we're on the plateau somewhere.
And now the energy absorbed is just going to be our plateau stress times the amount of strain we've got.
So that's epsilon minus epsilon naught.
And if I normalized with respect the solid modulus, I can write down that my plateau stress is 0.05 times the relative density squared, and then multiply that times epsilon minus epsilon naught.
So that equation then corresponds to these vertical parts-- so this part here, that part there, this part here.
It corresponds to those vertical lines on the figure.
Vertical lines on the diagram.
And then the plateau stress is going to end at the densification strain.
And at that point, the energy diagram is just going to become vertical.
OK, and then the last part-- I'll try to rub this off a little better.
And then the last part is when we're at the end of this stress plateau, and the strain is equal to that densification strain.
And so the amount of energy we absorb here is really going to be the maximum.
So this is the energy that's going to correspond to that shoulder point that I've been talking about.
So I'm going to call that W max, and normalize that with respect to Es.
And that's then going to be our plateau stress times the densification strain.
And the densification strain was just 1 minus 1.4 times the relative density.
So here I'm assuming that the densification strain is very much bigger than the strain at which the buckling first occurs, and I'm neglecting that linear elastic part.
So then we could say that the optimum foam is at that shoulder point.
And I can say that the peak stress at that point is just equal to the plateau stress.
And what I want to do is get an equation for that solid line up here that connects all those shoulder points.
So I want an equation, in terms of the energy, and the peak stress, instead of in terms of the density.
So what I'm going to do a solve this for the density, and then plug that back into there.
So I get that the relative density is, then, 20 times the peak stress over the solid modulus, and I take the square root of that.
And now I can substitute this up here for the relative density.
I think I need another board.
So then I've got-- this is my peak stress, or my plateau stress, over Es, and this is all just the densification strain.
And that's just 1 minus 1.4 times the relative density.
But now I'm putting the relative density in terms of the peak stress, or the plateau stress.
If I just simplify that slightly with the constant, it's 1 minus 6.26 times sigma p over Es to the 1/2 power.
So that equation there describes the-- oops, no more updates.
No updates.
Go away.
Ah, so that equation there describes this line here that connects those shoulder points.
And that's the line you're the most interested in, because each of those shoulder points is a point where the foam is being used in the most efficient way, or the optimum way.
And then, let's see-- is that going to fit?
No-- let me try the other board.
And then the last thing we can do is calculate that line that corresponds to the dense solid.
Yeah?
On those ones over there, it says it corresponds to the vertical lines on the diagram.
And then later it says then it becomes vertical.
Is that referring to two different diagrams?
So this stress plateau equation here corresponds to these vertical lines here.
So for relative density of 0.01, it corresponds to that part.
For 0.03 it's this part.
And your 0.01 it's that part
But then some of the [INAUDIBLE] it says, then w [?
versus ?]
sigma becomes vertical.
Should that--
So, well the plateau stress ends at the densification strain.
So the plateau stress ends here.
Oh, let's see-- and then it becomes-- should be horizontal.
Sorry.
OK, sorry.
Happy?
And now I'm going to rub that all off.
OK, did everybody get this?
I can rub it off?
OK, so then the last part is what happens when the foam is densified.
And if it was fully dense-- and you never really can get to this point-- but if it was fully dense, you would get rid of all the pores, and it would just be a solid.
And then the energy absorption curve would be the curve for the solid.
So I'll just say, when fully densified, I'm going to say a curve approaches that for the solid.
And for the solid, you would just have that W over Es is equal to 1/2 the peak stress squared over Es.
So this model curve, the curves have the same shape as when you get the diagrams from the experiments.
And you can see how the different mechanisms of deformation and failure contribute to the diagram, where the diagram comes from.
And I guess one other point is you can see that the foams are always going to be a lot better than the solid.
And remember this is a log log curve, so that a foam that has a density of 3% here, there's a huge difference in the peak stress.
So say you wanted to absorb this amount of energy up here, for a foam that was 0.03 dense, the peak stress would be a little less than 10 to the minus 4, normalized by the modulus.
And for the solid, it would be 10 to the minus 2-- so it's orders of magnitude better to have the foam rather than the solid.
All right, now let's see what else we have.
So we could do a similar thing for closed-cell foams, and you get diagrams that look like this.
One of the differences with the closed-cell foams, if you assume that the faces don't rupture, the plateau stresses and horizontals-- remember we had that gas contribution, and you can take that into account?
So I'm not going to go over the details of that.
The next one I wanted to talk about was looking at foams that have a yield point.
And again, you can generate a similar kind of diagram, but now, instead of having an elastic failure here, you've got a plastic failure-- you form plastic hinges.
And then again, this diagram is less general than the one for elastomeric foams, so this diagram would be valid for whatever ratio of sigma ys over Es you've the calculation for.
So this one here is for 0.01.
So let me just run through the same kind of calculation for the plastic foams.
Can I rub this off, and then I can use this board to start here?
OK, so the linear elastic part is just the same as for the elastomeric foams.
So you get W over Es is 1/2.
Sigma p over Es squared times 1 over the relative density squared, so it's just the same thing.
And the stress plateau-- you get w over Es is just the plastic collapse strength times the strain range that you go up to.
So if you remember the plastic collapse strength was 0.3 sigma ys times the relative density to the 3/2 power, and then times the strain range.
And then at the end of the stress plateau, you've got the maximum energy absorbed.
So normalize that by Es.
And that's going to be your peak stress over Es times the densification strain again.
So this bit here is the densification strain.
And then you can do a similar thing to figure out the equation of that line that joins the shoulder points.
So the first step is to solve for the relative density there.
So if this part here-- 0.3 sigma ys times the relative density to the 3/2 power is the plastic collapse stress, then at the densification point, the relative density, you just rewrite that and it comes out to the 2/3 power, because you turn the power around.
And then you just substitute this up in here.
So you get that the maximum energy absorbed, normalized by the solid modulus, is your peak stress, times 1 minus 1.4 times this thing in brackets to the 2/3 power-- the 3.3 times the peak stress over the yield strength of the solid.
And then if I just rearrange that, and get the constants, it's 1 minus 3.1 times our ratio of the stresses there.
So maybe I'll just put over here-- the curves are less general than for elastomeric foams.
So each family of curves would be for a particular ratio of the solid yield strength to the solid modulus.
So you get the idea?
It's fairly straightforward.
So I wanted to finish up this topic by giving you a few examples of how you can use these curves.
So the next thing is to look at the selection of foams for impact protection.
And typically, you're given some information about the objects-- so typically you want to protect some object.
It could be a computer, it could be your head, some part of your body.
So typically you know something about the object you want to protect.
So you might know its mass, you might know the area of contact, you might say, well, if it's my computer, I want to make sure it doesn't break if I drop it from a height of a meter.
Or you could say whatever, [?
maybe it's ?]
2 meters.
But you pick something.
And so there's a certain amount of energy you know that you need to absorb.
And you may know that whatever the component is, or the body part, or whatever-- there's some maximum acceleration you can tolerate.
So you might say, well, I want to make sure my computer doesn't break under acceleration of so much.
So you're given the acceleration.
And so if you know the mass and the acceleration, and you know the area of contact, you can figure out a force over an area, and that gives you the peak stress.
So typically, you know those things in the problem.
And typically the problem involves choosing a material, or choosing-- like choosing what kind of material do you want to make the foam out of, and what density of foam do you want to use, what thickness of foam do you want to use?
So I've got a couple of examples just to show you how this works.
So let me just write down a couple notes, and then the rest of it I think I'm just going to take from the slides.
So typically, you know what it is you want to protect, and you know something about it.
So if you know what it is, typically you know what the mass is.
You might know what the contact area would be.
Say a maximum drop height, maximum tolerable acceleration.
So say if you're worried about brain injury-- and making a helmet-- you might know what the maximum tolerable acceleration would be.
So typically, you know what the peak allowable stress is, just from whatever the object itself is.
And the variables that you have to play around with-- variables-- are things like the foam material, the foam density, and the foam thickness.
So I have a couple of examples that have different setups here.
And we'll just see how the thing works out here.
So the first example, we're told the mass of the packaged object is 1/2 kilogram.
We're told the area of contact between the foam and the object is going to be point.
0.01 of 1 meter squared.
And we're told that it's supposed to be designed to withstand a drop of 1 meter-- so if the drop height is 1 meter, then the velocity is just the square root of 2gh.
So g is gravity, so you can work out the velocity on impact would be 4.5 meters per second.
And if you have the velocity on impact and the mass, you can figure out the energy to be absorbed.
You can say mv squared screwed over 2, or you could say it's mgh-- either way.
So here it's going to work out to 5 joules.
And in this case, we're told that the maximum deceleration is 10g-- so then if that's 10g, the maximum force is the mass times the acceleration.
That works out to 50 Newtons.
And then that gives us a peak stress, the maximum allowable peak stress of the force over the area it's 5 kiloNewtons per meter squared.
And in this case, we're told that the foam is going to be a flexible polyurethane, and it has a solid modulus of 50 megapascals.
And so we can calculate this normalized peak stress.
So the normalized peak stress here is 10 to the minus 4.
So in this problem here, we need to figure out what's the foam density, and what's the thickness of the foam to protect the object?
And so that last slide is just summarized here.
And I realized that when I was going to put this up, the font is going to be kind of small, so I blew it up a little.
So we have figured out that we're at sigma p over Es-- the peak stress normalized by the solid modulus is 10 to the minus 4.
And we know that we're going to use a flexible elastomeric polyurethane, so we pull out our diagram for elastomeric foams.
And this sort of hashed band here corresponds to all the different strain rates, and all the different densities.
So I haven't plotted each individual line, we've just got this band that represents the whole thing.
So we know that our normalized peak stress is going to be somewhere along this line here.
That's from everything that's given, and what we can calculate.
And we want to know what density of foam to use, and what thickness.
So the way you approach this is the thickness is going to affect the strain rate.
So the strain rate's just going to be the velocity on impact, divided by the thickness-- or it's an approximation for the strain rate.
So we don't know if we're at this point down here, or if we're at this point up there, because we don't know where we are in the strain rate end of things.
So the way you solve this is you just guess a thickness.
And if you guess a thickness, you can calculate a strain rate.
Then if you calculate the strain rate, you know where in this band you are.
You can figure out a value of W over Es.
And you can use an iterative process.
And the way this is set up is we've chosen two very different initial thicknesses, and the point of doing that is to show you that it converges very quickly.
So the first iteration on this side here, we've chosen the thickness of a meter, which is probably unlikely that we need a meter of foam.
And on the side here, we've chosen a millimeter-- 0.001 meters.
So probably, we need more than that.
So we probably need to be somewhere between those two bounds.
So if the thickness was a meter, then the strain rate turns out to be 4.5 per second.
And then we know where we are in this diagram, and we can read off a value of W over Es.
So we know we're on this line here, and for a particular strain rate, we can read off the W over Es.
So here's the value-- 5.25 times 10 to the minus 5.
And if we know Es, which we do, we can then calculate the actual energy absorbed per unit volume.
We get W-- so W is [?
2620 ?]
joules per cubic meter.
And we can use that value-- because that's an energy per unit volume-- we know the area of contact, and we can use that to get another value of the thickness.
So we use that to calculate the next iteration of the thickness.
So U is the total energy in joules, and W is the energy per unit volume.
So U, the energy in joules, is going to equal the energy unit volume times the area times the thickness.
So we know what the area is, too.
We can then calculate a new thickness.
So now our new thickness is 0.19 meters.
We can use that to get a new strain rate-- that's 24 per second, and go through the whole thing again.
And we end up with a revised energy of 3,300 joules per cubic meter.
Now if we started at the other end, if we started with the first guess was a millimeter, then the strain rate is 4.5 times 10 to the minus 3 per second.
That gives us a different value that we read off here for W over Es, and a different value for W. And then we use this value here for W to make another guess for the thickness.
So that value is 0.14.
And then we go through the whole thing again-- we get a revised strain rate a revised W over Es, and a revised W. And you can see after just two iterations, these two things are almost exactly the same.
And then on the third iteration, they both would give you a thickness of point 0.15 meters.
So even though you can pick wildly wrong first iterations, it converges very quickly, and it's a fairly simple calculation to do.
So you know the thickness that you want is in here, and then you can get the optimum density here.
So you know that your sigma p over Es is along this line of 10 to the minus 4--we're somewhere in here.
These two strain rates here-- one was 24, one was 32-- so the final value is going to be somewhere around 30.
And if you look on this thing here, you can see there's a line that corresponds to 10, there's a line that corresponds to 100.
We're going to be right around in there.
So the relative density is going to be right around 0.01.
So you can use the diagram to get the thickness and to get the density.
OK, are we good?
We're good?
So I've written some notes, and I'll just scan those and I'll put them in the Stellar site.
So here's another example here-- and in this example, it's set up a little bit differently.
So in this example here, we're not told the material, but we're told the thickness of the foam.
So this time we want to get the material, and we want to get the density of the foam.
So here we've got the specification.
We're told the mass is 2 and 1/2 kilograms.
The area of contact is 0.025 meters squared.
We're told the thickness-- here thickness is 20 millimeters.
We've got a drop height of 1 meter again.
And the velocity of impact is then going to be 4.5 meters per second again.
And since we know T, we know that the strain rate is going to be around 225 per second.
We can calculate the energy absorbed-- [?
MGH-- ?]
25 joules.
We can calculate the energy absorbed per unit volume, because we've got the area and the thickness-- so I'm just going to divide that by the area and the thickness.
Now we have 5 times 10 to the 4th joules per cubic meter.
And we're told that we're supposed to design it so the package can withstand a deceleration of 100g-- and so that gives you a maximum force.
And here we've got a maximum allowable peak stress of 10 of the 5 Newtons per meter squared.
So here we have our curve for the elastomeric foams again-- let's assume it's going to be an elastomeric foam, but we don't know what kind of foam.
So the way you solve this problem is that you make a guess for what Es is.
So remember, these diagrams are all normalized by Es.
So to plot a point on there, we need to know what Es is.
So here, we're going to make a guess, and we're going to assume that Es is 100 megapascals to start.
And if we had that value of Es, we know W, up here, and we know sigma p there.
So we just divide those values of W and sigma p by Es, and we get these two values here.
And we plot those two values on the thing here.
So here's our point A-- and that corresponds to that first guess of 100 mega Pascals for the modulus of the solid.
So that's not necessarily the final answer, that's not the right answer-- that's just somewhere to start.
So the thing to notice is, if we have that point there, the strain rate there is probably not quite right.
This upper bound here is-- let's see.
Got to get closer.
Oh, let's see-- that's 10 to the minus 2, that's 10 to the 2.
So the strain rate we want to be is closer up to here, it's not quite down there.
So we're not at the right strain rate.
But the thing to notice is that if we draw a line of slope 1-- so there's this dash line of slope 1-- when we move up and down that line, we're just changing Es.
Because everything is normalized with respect to Es-- if that line has a slope of 1, then we're just moving up and down with respect to Es.
So what we do is we scoot up the line to get to the point that's on the right strain rate.
So if this was a strain rate of 100, or around 200, we want to be at point B.
And then from point B, we can read off what's the value of W over Es, and sigma p over Es.
And from that, we can back out what the solid modulus we want is.
So these are the values that we read off the chart.
This gives us an Es of 28 mega Pascals.
And again, you can go to this more detailed diagram here, and if you read off the sigma p over Es, and the W over Es, the density you want is about 0.1.
So it tells you the modulus of the solid, and from that you can pick a solid, and it tells you the density.
Are we good?
OK, so I have one more.
So there's a slightly different way you can do it, too.
So now I want to talk about bicycle helmets-- you're my bicycle helmet person.
So this is another little case study that involves a slightly different way to do this.
And this involves a slightly different diagram.
So the idea here is to choose a material for a bicycle helmet.
And as you probably all know, the bicycle helmets have a hard shell, and they have some sort of foamy liner.
And the foam's usually around 20 millimeters thick.
And you want something that's light, because you don't want your helmet to be too heavy-- but you want something that will absorb the energy from the impact.
So I've set this up here-- so we assume the mass of the head is about 3 kilograms.
And we assume that it can withstand a deceleration of something like 300g-- and so then you can get a force mass times the acceleration, because you have the force.
And I've assumed an area of contact of something like 0.01 meters squared, so that gives you a peak stress.
So this method here is based on the idea that you have these material selection charts for foams.
Remember we talked about that earlier.
And this is the compressive stress at 25% strain.
So the idea is that axis there is meant to represent the plateau stress, and this axis here represents the densification strain.
And these dashed lines here correspond to basically a value of W-- and energy absorbed per unit volume.
So this is an energy absorbed per unit volume of 0.001 megajoules per cubic meter, 0.01, and so on.
So if you know the peak stress that you can tolerate, for the numbers I gave you it works out to be 0.9 mega Pascals-- so here it's just a little less than 1.
So that's the peak stress that you can tolerate there.
And you can see that the material that's going to absorb the most energy-- so you're absorbing more energy as you move over this way on the curve on the plot-- so the material that's going to do the best is something like an expanded polystyrene that's 5% dense.
So I've highlighted that in red-- expanded polystyrene that's 5% dense.
And then this is the densification strain here.
And you know that the lines of the energy absorption are just the stress times the strain.
So you can basically just read off from here that expanded polystyrene that was 5% dense would be a good choice for a bicycle helmet foam.
Would you like to add anything about bicycle helmet foams?
I think that is exactly what they use, yes.
So this is just another way to do this kind of thing.
And I did one more little calculation here-- if you know the thickness of the foam is, say, 20 millimeters, and we've estimated the area of the contact, you can figure out what the energy absorbed is per unit volume, and from that you can back out the energy in terms of joules, and from that you can back out the velocity that's the maximum speed that one would want to get dinged at on your bicycle.
And the maximum speed works out about 22 miles an hour.
So that's just another example.
Are we good?
OK-- yeah, exactly, your head would be hitting the ground at 22 miles an hour which, would be ouchy.
Which is why you want to wear your helmet, because your skull will not be happy if that happens.
And your brain will not be happy.
And you will not be happy.
And your mom and dad will be really, really, really unhappy.
So you don't want that to happen.
so I have a few minutes left-- and I know I've done this for the people in 3032, but I was going to buy woodpecker talk, because it's about energy absorption.
So if you don't want to watch it, if you've already seen it, you can go.
But there are some people-- you guys haven't seen it, and you haven't seen it.
There's a few people that haven't seen.
And it's cute, involves birds.
And it involves energy absorption-- so I thought I would do this woodpecker talk.
So Barry, this is all just slides.
I don't know if you want to do the lights differently.
I'm not going to write anything on the board, I'm just going to talk
Well, if it will help them see the slides better then certainly
Yeah, I think maybe we could turn the lights down a little, please
Let's try this one
Oh, there we go.
That's good.
Yeah, now I don't feel like I'm looking in the spotlight so much.
That's good.
OK, so you guys know that I like to watch birds.
And you know that I work on foams.
And if you look at bird books, sometimes the bird books say that woodpeckers can withstand the impact from pecking because they have a special material between their skulls and their brains.
And I thought, oh, well I study foams, and I'm interested in woodpeckers-- I should find out what this special material is.
So I started looking into this.
And it turns out there is no special material.
People have looked at the anatomy of woodpecker brains and skulls, and there is no special material.
But it turns out there were also a group of neurologists in the late 1970s who got interested in why woodpeckers don't get brain injury, and they took high speed video of a woodpecker pecking.
And it was kind of amazing what they found out.
So it turns out the woodpeckers can withstand incredibly high decelerations-- much higher than our brains could withstand.
And so I kind of got interested in this, and I decided to try to figure out how it works.
So here's an acorn woodpecker.
And let's see-- I got a little video here.
Wait a minute-- there we go.
There we go.
So here's a little acorn woodpecker.
These live in California.
Anybody from California?
Yeah-- have you seen them?
No, OK. Well you have not been looking carefully.
So here's our little acorn-- and you can see they peck-- oh?
Like around San Francisco
[INAUDIBLE].
[LAUGHS]
So they're pecking, and when they're pecking-- they don't just go bonk.
They do this repeatedly-- they can peck at 10 or 20 times per second.
So the question is, why don't they get brain injury?
So, first of all, why do they peck?
So as we in that little video, that woodpecker was foraging-- it was trying to get little things out of the bark.
And woodpeckers eat insects, and so the bird books say that they can actually hear insects scurrying around under the bark, and they'll peck at the bark and forage to try to get insects.
And there's one other anatomical feature of woodpeckers, which is kind of amazing-- and you can see it in this picture here.
So this is the woodpecker tongue, and the tongue is connected to something called the hyoid process.
And the hyoid process wraps around their eyeballs.
And then when they peck-- I mean, the idea is they're making a hole in the tree, and they've got to get their tongue into the hole to get the bug.
And the end of their tongue has little barbs on it.
And when they contract this thing here, their tongue scoots out and gets the little bugs.
So they pick partly to forage, but they also build something called cavity nests.
So they'll find a tree that's started to rot, and they'll drill a sort of horizontal hole, and then they'll drill a cup underneath that, and they lay the eggs and have their nest at the bottom of the cup.
And then they also-- especially at this time of year, in fact, today I heard woodpeckers drumming-- so it's one of these mating things, one of these courtship things.
So woodpeckers will peck on a hollow branch or a hollow tree, just to make a big loud noise to say, here I am, looking for sex, I'm ready, this is my territory.
And so in the spring-- so this time of year you hear woodpeckers drumming.
And I think naturally they do this on hollow trees to try to make a big sound, but they have adapted to metal downspouts, which are also very effective for making this loud noise.
And people who've-- I wrote a paper on this-- people who have read my paper sometimes email me and say, oh, I heard you know about woodpecker pecking.
How can I get them to stop drumming on my downspout?
Because it's kind of annoying, if you're the human inside the house.
So anyway, they peck for these reasons.
And then acorn woodpeckers are special-- acorn woodpeckers are what I think of as the champions of pecking.
And they do one more behavior-- so here's from David Sibley's beautiful Guide to Birds, here's the acorn woodpecker.
They store acorns in what's called a granary, and they look at old tree trunks that are beginning to rot, and they pick holes in the tree trunk and they store the acorns in the holes.
So see all those little dark dots on that trunk?
Those are all holes that the acorn woodpecker has pecked.
And they'll do this-- they live in social groups, so there might be like 10 or 20 of them living together, and they'll peck like 10,000 holes into their granary.
There will be thousands and thousands of these holes.
And here we have the acorn woodpecker with the acorn in its beak, and you can see some of these holes have acorns, and some of them are empty.
So here is the acorn woodpecker in action.
And here's the little video again-- I won't play the video.
So I often give this talk for non-engineers, so I'm going to explain some things-- there's going to be some writing on the slides that you guys already know.
So the impact force depends on the deceleration-- how quickly the brain stops when the beak hits the tree.
And I should mention, these videos are from the Cornell Lab of Ornithology-- they have an amazing collection of bird audio, like bird calls, bird songs, and bird photographs and videos.
It's incredible, the collection that they've got.
And then I explain what acceleration is-- so I'm going to put acceleration in terms of gravity.
And when the beak hits the tree, there's going to be a deceleration on impact.
And just as a comparison, human brain injury occurs roughly at about 100g.
And so the question is, how much deceleration can the woodpecker brain take?
And that's where the neurologists in the Bay Area come into the picture.
They found out that there was a park ranger-- I think maybe at Point Reyes, just north of San Francisco.
And he had an acorn woodpecker that, I don't, had an injured wing or something.
Anyway, he had this acorn woodpecker that he kept.
And they were able to use this acorn woodpecker, and they took high-speed video of the woodpecker pecking.
So from the high-speed video, the video that they took went at something like 2,000 frames a second.
So they have a picture of where the head is every 2,000th of a second.
So if you know the position at these times, you can get the velocity, and you can get the deceleration.
So they measured the deceleration, and they measured some amazing things.
So they measured that on impact, the woodpecker's bill was going something like 15 miles an hour.
And the decelerations were up to 1,500g-- so many times more than what we can withstand.
And they also measured the stopping time-- and they thought it was between about 1/500th to 1/1,000th of a second.
And that's going to be important later on.
So one of the interesting things about this was how they got this whole thing set up.
So I don't know how many of you do UROPs-- but you know, part of the thing in doing UROPs, and doing experiments in the lab is just how do you do the experiments?
So you know, the park rangers got the acorn woodpecker, that's all very good.
But they have to get the woodpecker to peck in front of a camera, and they have to get the camera to turn on as it's pecking-- so there's some experimental challenges.
So the important thing, the critical thing, is the date of the paper.
This was written in 1979, which meant they probably did the experiments in 1978.
And I was a graduate student in 1978.
And I can report there were no Apple computers in 1978-- there were no laptops.
You couldn't just kind of do your little PowerPoint slides.
And most offices had something called an IBM Selectric typewriter-- I don't know if you've ever seen the old IBM typewriters.
But the typewriters, when you typed on the typewriters, they made these noises, and it sounded kind of like a woodpecker pecking.
And the ranger had one of these typewriters in his office, and he had discovered that if he typed on the typewriter, the woodpecker thought, oh, there's another woodpecker.
I'll start pecking.
And so they used the typewriter as a way to get the woodpeckers to peck.
And I think they had some old stump, and I don't know if they put nuts or peanut butter or something into the stump to get the woodpecker to peck at it-- because they needed to peck at a particular spot, so they have the camera all set up.
So anyway, they had this whole arrangement to do this high-speed video of the woodpecker pecking.
OK, so it goes that up to 1,500g, which is kind of amazing.
And then, remember also, they do this repeatedly-- they do it at like 10 or 20 times a second.
Then this is just explaining what stress is-- but you already know what stress is, so I'm going to skip over that.
And so the way you can think about this is to think about it in terms of a scaling argument.
So imagine there's the brain and there's the skull, and when the head hits the tree, the brain's going to accelerate, and the brain and the skull are both going to decelerate.
And you can think of the stress as the force over the area.
So the force is the mass times the deceleration over the area.
So that value for the woodpecker, you can say, is roughly equal to the same values, but for the human.
So what that argument relies on is the idea that the stress to cause damage in the woodpecker brain is similar to the stress to cause damage in the human brain.
And that's not totally unreasonable.
So if you look at bone, for example, like when you measure the strength of the whale bones, it's not going to be that different from what you would measure from human bones.
So when you look at a particular type of tissue, and you look at the strength of it in different species, the properties aren't that different from one species to another.
So let's say the brain tissue gets damaged at the same stress in the two species.
So we can write this sort of equation down.
And the mass is going to depend on the density of the brain tissue times the volume-- and the volume goes as the radius cubed.
And the density is going to be the same for the woodpecker, or for the human-- so assume the brain tissue has the same kind of basic stuff.
So the mass goes as the radius cubed, and the area of contact is going to go as the radius squared.
And so there's a radius term.
So you can say the radius times the deceleration in the woodpecker should be equal to the radius times the deceleration of the human.
And obviously, the woodpecker radius is going to be a lot smaller, so the woodpecker deceleration is going to be a lot bigger.
So part of the thing is the brain is just a lot smaller.
Then there's another factor that comes into it-- these are photographs of an acorn woodpecker skull and a human skull that I got from the Museum of Comparative Zoology at Harvard.
So this is looking down on the top, and this is an elevation view.
And if you think of the brain as roughly a hemisphere-- I mean, obviously it's not a perfect hemisphere, but let's say it's roughly a hemisphere-- the orientation of the brain in the skull is slightly different in the birds and in the human.
So if you think of it as being this way on, and in the woodpecker it's turned roughly this way on, and the contact area-- think of this as the contact area.
The projected area would be a full circle, and in the human, the brain is more this way on.
And then the projected contact area is a semicircle.
So there's a factor of 2 difference, just because of the orientation of the brain.
And so I've put the factor of 2 that accounts for that.
So then I could say the deceleration that the woodpecker can take is twice the ratio of the radii of the human and the woodpecker brain, times the deceleration the human can take.
And this was around 100g, remember.
So then I wanted to know what the ratio of the sizes was, and I thought, oh geez-- I'm going to have to mess around with skulls, and try to make some sort of measurements, and this is going to be a drag.
And then I found this paper-- I couldn't believe it-- I found this paper called "Brain Size in Birds."
And this guy had table after table after table of the mass of the brain in different birds-- and he had the acorn woodpecker, lucky for me.
So if you just assume that the brain is roughly a hemisphere, if you have the mass, you can work out roughly what the radius is.
So it turns out there's a factor of 8 difference between the radii.
And so the human brain is about eight times the size of the woodpecker.
And then if you take into account this factor of 2, that means the woodpecker should be able to tolerate a deceleration of 16 times what the human can.
So remember, I said the human can take about 100-- so this would get the woodpecker up to 1,600.
But they measured 1,500.
And I'm a civil engineer, originally, I like big factors of safety-- that's a little too close for comfort for me.
And so it turns out there's one more factor that matters.
People have studied human brain injury pretty extensively, and one of the things they've looked at is how much acceleration you can tolerate without injury, relative to the duration of the impact.
So this is from a car crash conference-- so here's the tolerable acceleration, and here's the duration of the impact.
And typically, for human head impacts, the deceleration occurs over a few milliseconds-- like over 3 to 10 milliseconds.
So if this is the duration of the typical head impact for a human, here's the range of the tolerable accelerations between about 80g and 160g-- so you know, I said around 100.
So we can take this curve, and now we have this factor of 16, and we can just scale it up by our factor of 16 for the woodpecker.
So if I scale it up by the factor of 16, we're there.
But the duration of the impacts was more around 1/2 a millisecond, to a millisecond, so I've extrapolated this a little bit.
And the duration of them is up in here.
So this is saying the woodpecker can take these sorts of decelerations here, and that adds on another factor of 4.
And these were the measured decelerations-- 1,500 was about the biggest, and I think it went to about a few hundred g. So there's really three factors-- one is the small brain size, so there's this sort of scaling factor.
One is the orientation of the brain, that was another factor of 2.
And then there's this duration of impact, which is a factor of 4.
And so that's how you can get this huge decelerations that they've measured in the woodpecker when they're pecking.
So now I have just a couple more slides.
So woodpeckers have various adaptations to pecking, too.
So one of the things is they have amazingly stiff tail feathers-- you see the tails here?
You can't quite see it because this didn't reproduce quite properly, but the woodpecker is perched on a tree here, and the tail is pressed up against the tree.
And if you think about pecking, like imagine if its tail was not pressed up against the tree-- it would be kind of grabbing on with its feet, and it would be trying to peck, and it would be hard to push.
You know, you need something to push against.
And so it's got these stiff tail feathers, which make it easier to push against the tree.
And so here's the stiff tail feathers here.
It's also got kind of unusual feet for birds-- it's got two toes forward and two toes back.
And again, if it's grabbing with its feet, that helps it get some purchase to push against.
That's called zygodactyl in the bird world.
So it's got these adaptations, the pecking.
And then finally, I just like this slide here from The New Yorker, because it's a woodpecker pecking out a woodpecker, so it seemed kind of cute.
And several people helped me with this project.
Trey Crisco studies head injury at Brown.
He actually does work for the NFL on football brain injuries, and looks at football helmets.
Sharon Swartz is a friend of mine at Brown, who's a biologist.
She studies bat flight.
And she just thought this was kind of a cool project.
So, in fact, I was walking the dog last week, a few days ago, and I saw bats flying around overhead at night.
And it was so great to see the little bats.
So I had to immediately email Sharon, and say, bats-- there's bats in my neighborhood.
Andy Biewener runs the Concord Field Station, and studies animal locomotion, and I talked to him about it.
Jeremy Trimble was the one who gave me the skull pictures, and Matt Dawson was a student helped me do some of the images.
So that's my woodpecker talk.
So that's the end of energy absorption, I just thought that was kind of amusing.
So next time, we'll start talking about sandwich panels.
So, let's see-- Monday's a holiday.
I will be in Toronto on Monday, seeing my family.
In fact, I'm going to see one of my old professors-- I'm going to my old fluid mechanics professor on Monday and have lunch.
I'll see my family on the weekend.
And so we'll meet Wednesday next week, and I'll start the bit on sandwich panels.
So I think there's two lectures on engineering sandwich panels, and then there's a lecture on natural sandwich panels-- so bird skulls, for example, are natural sandwich panels.
So I'll talk about that.
And then I think the last lecture-- there's only a handful of lectures left-- the last one I think I talk about natural materials, because you know, I like that.
So I just do that for fun.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free.
To make a donation, or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
OK, so we should probably start.
So last time we finished up talking about energy absorption in foamy cellular materials.
And today I wanted to start a new topic.
We're going to talk about sandwich panels.
So sandwich panels have two stiff, strong skins that are separated by some sort of lightweight core.
So the skins are typically, say, a metal like aluminum, or some sort of fiber composite.
And the core is usually some sort of cellular material.
Sometimes it's an engineering honeycomb.
Sometimes it's a foam.
Sometimes it's balsa wood.
And the idea is that what you're doing with the core is you're using a light material to separate the faces, and if you think about an I-beam-- so if you remember when we talk about bending and we talk about I-beams, the whole idea is that in bending, you want to increase the moment of inertia.
So you want to make as much material as far away from the middle of the beam as possible to increase the moment of inertia.
So if you think about an I-beam, you put the flanges far apart with the web, and that increases the moment of inertia.
And the sandwich panels and the sandwich beams essentially do the same thing, but they're using a lightweight core instead of a web.
And so the idea is you use a lightweight core.
It separates the faces.
It increases the moment of inertia.
But you don't add a whole lot of weight because you've got this lightweight core in the middle.
So I brought some examples that I'll pass around and we can play with.
So these are some examples up on the screen, and some of those I have down here.
So for instance, the top-- turn my little gizmo on-- the top left here, this is a helicopter rotor blade, and that has a honeycomb core in it.
This is an aircraft flooring panel that has a honeycomb core and has carbon fiber faces.
So that's this thing here.
I'll pass that around in a minute.
This is a downhill ski.
This has aluminum faces and a polyurethane foam core.
And that's the ski here.
And it's quite common in skis now to have these sandwich panels.
This is a little piece of a small sailing boat.
It had, I think, glass fiber faces and a balsa wood core.
And I don't know if any of you sail, but MIT has new sailing boats.
Do you sail?
I do not much these days
OK, but those little tech dinghies that you see out in the river, those have sandwich panel holes to them.
So those are little sandwich panels.
This is an example from a building panel.
This has a dry wall face and a plywood face and a foam core, and the idea with panels for buildings is that usually they use a foam core because the foam has some thermal insulation.
So as well as sort of separating the faces and having a structural role, it has a role in thermally insulating the building.
The foams are a little less efficient than using a honeycomb core.
So for the same weight, you get a stiffer structure with a honeycomb core than a foam core.
But if you want thermal insulation as well as a structural requirement, then the foam cores are good.
And these are a couple examples of sandwiches in nature.
This is the human skull.
And your skull is a sandwich of two dense layers of the compact bone, and you can see there's a little thin layer of the trabecular bone in between.
So your head is like a sandwich, your skull is like a sandwich.
And I don't know if I'll get to it next time, but in the next couple of lectures, I'm going to talk a little bit about sandwich panels in nature, sandwich shells in nature.
You see this all the time.
And this is a bird wing, here.
And so you can see there's got the dense bone on the top and the bottom, and it's got this kind of almost trust-like structure in the middle.
And obviously birds want to reduce their weight because they want to fly, so reducing the weight's very important.
And so this is one of the ways that birds reduce their weight, is by having a sandwich kind of structure.
So I have a couple of things here.
These are the two panels at the top there.
This is the ski, and you can yank those around.
I also have a few panels that people at MIT have made.
And I have the pieces that they're made from.
So you can see how effective the sandwich thing is.
So this was made by a guy called Dirk Moore.
He was a graduate student in ocean engineering.
And it has aluminum faces and a little thin aluminum core.
So you can see, if you try and bend that with your fingers, you really can't bend it any noticeable amount.
And this panel here is roughly the same thickness of the face on that.
And you can see how easy it is for me to bend that-- very easy.
And this is the same kind of thing as the core.
It's thicker than that core, but you can see how easy it is for me to bend this, too.
So each of the pieces is not very stiff at all.
But when you put them all together, it's very stiff.
So that's really the beauty of this.
You can have lightweight components, and by putting them together in the right way, they're quite stiff.
So here's another example here.
This is a panel that one of my students, Kevin Chang made.
And this has actually already been broken a little bit, so it's not quite as stiff as it used to be.
And you can kind of hear, it squeaks.
But you can feel that and see how stiff that is.
And this is the face panel here.
And you can see, I can bend that quite easily with my hands.
Doodle-doot.
And then this is the core piece here.
And again, this is very flexible.
So it's really about putting all those pieces together.
So you get this sandwich construction and you get that effect, OK?
[?
Oop-loo.
?]
All right, so what we're going to do is first of all look at the stiffness of these panels, calculate their deflection.
We're going to look at the minimum weight design of them.
So we're going to look at how for, say, given materials in a given span, how do we minimize the weight of the beam for a given stiffness?
And then we're going to look at the stresses in the sandwich beams.
So there's going to be one set of stresses in the faces, and a different kind of stress distribution in the core.
So we'll look at the stress distribution.
And then we'll talk about failure modes, how these things can fail, and then how to figure out which failure mode is dominant, which one occurs at the lowest load.
And then we'll look at optimizing the design, minimizing the design for a certain strength and stiffness.
So we're not going to get all that way through everything today, but we'll kind of make a start on that.
OK.
So let me start.
So the idea here is we have two stiff, strong skins, or faces, separated by a lightweight core.
And the idea is that by separating the faces, you increase the moment of inertia with little increase in weight.
So these are particularly good if you want to resist bending, or if you want to resist buckling.
Because both of those involve the moment of inertia.
And they work like an I-beam.
So the faces of the sandwich are like the flanges of the I-beam, and the core is like the web.
And the faces are typically made of either fiber reinforced composites or metals.
So typically, something like aluminum, usually you're trying to reduce the weight if you use these things, so a lightweight metal like aluminum is sometimes used.
And the cores are usually honeycombs, or foams, or balsa.
And when they use balsa wood, what they do is-- I brought a piece of balsa here-- what they do is they would take a block like this and chop it into pieces around here.
And then they would lay those pieces on a cloth mat.
So typically the pieces are maybe two inches by two inches.
They lay them on a cloth mat, and because they're not one monolithic piece, they can then shape that mat to curved shapes.
So it doesn't have to be just a flat panel.
They can curve it around a curved surface if they want.
So we'll say the honeycombs are lighter than the foams for a given stiffness or strength.
But the foams provide thermal insulation as well as a mechanical support.
And the overall mechanical properties of the honeycomb depend on the properties of each of the two parts, of the faces and the core, and also the geometry of the whole thing-- how thick's the core, how thick's the face, how dense is the core?
That kind of thing.
And typically, the panel has to have some required stiffness or strength.
And often what you want to do is minimize the weight for that required stiffness or strength.
So often these panels are used in some sort of vehicle, like we talked about the sailboat, or like a helicopter, or like an airplane.
They're also used in like refrigerated trucks-- they would have a foam core because they'd want the thermal insulation.
So if you were going to use it in some sort of a vehicle, you want to reduce the mass of the vehicle and you want to have the lightest panel that you can.
Yup?
So if you saw the base material and you'd have the [INAUDIBLE] sandwich panel, that piece [INAUDIBLE] the sandwich panel with something [?
solid ?]
in the middle?
Well--
So, as we're getting [INAUDIBLE] aluminum piece that's was as thick as a sandwich panel
Yeah
[INAUDIBLE]
Oh, well, if you have the solid aluminum piece that was as thick as the sandwich, it's going to be stiffer, but it's going to be a lot, lot heavier.
So the stiffness per unit weight would not be as good.
OK?
So we're going to calculate the stiffness in just one minute.
And then we're going to look at how we minimize the weight, OK?
OK.
So what I'm going to do is set this up as kind of a general thing.
We're just going to look at sandwich beams rather than plates, just because it's simpler.
But the plates, everything we say for the beams basically applies to the plates.
The equation's just a little bit more complicated.
So we're going to start with analyzing beams.
And I'm just going to start with a beam, say, in three-point bending.
So there's my faces there.
Boop.
And I've made it kind of more stumpy than it would be in real life, just because it makes it easier to draw it.
And then if I look at it the other way on, it would look something like that.
So say there's some load P here.
Say the span of the beam is l. Say the load's in the middle, so each of the supports just sees a load of P/2.
And then let me just define some geometrical parameters here.
I'm going to say the width of the beam is b.
And I'm going to say the face thicknesses are each t. So the thickness of each face is t. And the thickness of the core is c, OK?
So that's just sort of definitions.
And I'm going to say the face has a set of properties, the core has a set of properties, and then the solid from which the core is made has another set of properties.
So the face properties that we're going to use are a density of the face.
We'll call that row f. The modulus of the face, Ef, and some sort of strength of the face, let's imagine it's aluminum and it yields, that would be sigma y of the face.
And then the core similarly is going to have a density, rho star c. It's going to have a modulus, E star c. And it's going to have some strength, I'm going to call sigma star c. And then the solid from which the core is made is going to have a density row s, a modulus Es, and some strength, sigma ys, OK?
So the core is going to be some kind of cellular material, a honeycomb, or a foam, or balsa.
And typically, the modulus of the core is going to be a lot less than the modulus of the face.
So I'm just going to say here that the E star c is typically much greater than Ef.
And we're going to use that later on.
So we're going to derive some equations, for example, for an equivalent flexural rigidity for the section, an Ei equivalent.
And that has several terms.
But if we can say the core stiffness is much less than the face thickness, and also if we can say the core-- the stiffness is less and also the thickness of the core is much greater than the thickness of the face, a lot of the expressions we're going to use simplify.
So we're going to make those assumptions.
So let me just draw the shear diagram here.
So V is shear, so that's the shear diagram.
We have some load P/2 at the support.
There's no other load applied until we get to here.
Than the shear diagram goes down by P, so we're at minus P/2.
Then there's no load here, so this just stays constant, and then we go back up to 0.
And then let me just draw bending moment diagram.
The bending moment diagram for this is just going to look like a triangle.
Remember, if we integrate the shear diagram, we get the bending moment diagram.
And that maximum moment there is going to Pl over 4.
OK.
So initially, I'm going to calculate the deflections.
And I don't really need those diagrams for that, but then I'm going to calculate the stresses, and I'm going to need those diagrams for the stresses.
So just kind of keep those in mind for now.
So to calculate the deflections, sandwich panels are a little bit different from homogeneous beams.
In a sandwich panel, the core is not very stiff compared to the faces.
And we've got some shear stresses acting on the thing.
And the shear stresses are largely carried by the core.
So the core is actually going to shear, and there's going to be a significant deflection of the core and shear as well as the overall bending of the whole panel.
So you have to count for that.
So we're going to have a bending term and a shear term-- that's what those two terms are there.
So we're going to say there's a bending deflection and a shear deflection.
And that shearing deflection arises from the core being sheared and the fact that the core, say, Young's modulus or also the shear modulus, is quite a bit less than the face modulus.
So if you think of the core as being much more compliant than the face, then the core is going to have some deflection from that shear stress.
OK, so we're going to start out with this term here, the bending term.
And if I just had a homogeneous beam in three-point bending, the central deflections-- so these are all the central deflections I'm calculating here-- with Pl cubed over it turns out to be 48 is the number, and divided by EI.
And because we don't have a homogeneous beam here, I'm going to call that equivalent EI.
And to make it a little bit more general, instead of putting 48, that number, I'm just going to put a constant B1.
And that B1 constant is just going to depend on the loading geometry.
So any time I have a concentrated load on a beam, the deflection's always Pl cubed over EI, and then the sum number in the denominator and that number just depends on the loading configuration.
So for three-point bending, it's 48.
For the flexion of a cantilever, B1 would be 3.
So think of that as just a number that you can work out for the particular loading configuration.
So here we'll say B1 is just a constant that depends on the loading configuration.
And I'll say, for example, for three-point bending, B1 is 48.
For a cantilever end deflection, then B1 would be 3.
So it's just a number.
So the next thing we have to figure out is what's the EI equivalent.
So if this was just a homogeneous beam, and it was rectangular, E would just be E of the material and I would be the width B times the height H cubed divided by 12.
So here we don't quite have that because we have two different materials.
So here we have to use something called the parallel axis theorem, which I'm hoping you may have seen somewhere in calculus, maybe?
But, yeah, somebody is nodding yes.
OK, so what we do, what we want to do is get the equivalent EI-- I'm going to put it back up, don't panic-- of this thing here, right?
So I want-- this is the neutral axis here, and I want the EI about that neutral axis there.
So, OK, you happy?
There.
OK, so I've got a term for the core.
OK, the core, that is the middle of the core, right?
So for the core, it's just going to be E of the core times bc cubed over 12.
Remember, for a rectangular section, it's bh cubed over 12 is the moment of inertia.
And here our height for the core is just c, OK?
And then if I took the moment of inertia for, say, one face about its own centroidal axis, I would get E of the face now times bt cubed over 12.
So that's taking the moment of inertia of one face about the middle of the face.
And I have two of those, right?
Because I have two faces.
And the parallel axis theorem tells you what the moment of inertia is going to be if you move it, not to the-- you don't use the centroid of the area, but you use some other parallel axis.
And what that tells you to do is take the area that you're interested in-- so the area of the face is bt, and you multiply by the square of the distance between the two axes that you're interested in.
Oop, yeah.
Let me change my little brackets.
Boop.
So, oop-a-doop-a-doop.
Maybe I'll stick this, make a little sketch over here again.
OK, all right.
So this term here, Ef bt cubed over 12, that would be the moment of inertia of this piece here, about the axis that goes through the middle of that, right?
Its own centroidal axis.
But what I want to do is I want to know what the moment of inertia of this piece is about this axis here.
This is the neutral axis.
So let's call this the centroidal axis.
And the parallel axis theorem tells me what I do is I take the area of this little thing here, so that's the b times t, and I multiply by the square of the distance between those two axes.
So the distance between those axes is just c plus t over 2, and I square it.
And then I multiply that whole thing by 2 because I've got two faces.
Are we good?
[INAUDIBLE]
Yeah?
The center [INAUDIBLE] and the [INAUDIBLE], are those Ed's or Ef's?
These are Ef's because this is the face now, right?
So this term here is for the core.
So here the core is E star c. And these two Ef's are for the face up there, OK?
Because you have to account for the modulus of the material of the bit that you're getting the moment of inertia for.
Are we good?
OK.
So now I'm just going to simplify these guys a little bit.
Doodle-doodle-doodle-do-doot.
OK?
So I've just multiplied the twos, and maybe I'll just write down here this is the parallel axis theorem.
Doot-doot-doot.
Yes, sorry?
So for the term that comes from the parallel axis theorem, why do we only consider Ef and not [INAUDIBLE]
Because I'm taking-- what I'm looking at-- so the very first term, this guy, here--
Yeah, [INAUDIBLE]
Accounts for this, right?
And these two terms both account for the face
Oh, OK, so the face acting--
Yeah, about this axis.
So the parallel axis theorem says you take the moment of inertia of your area about its own centroidal axis, and then you add this term here.
But it's really referring to that face, OK?
Let me scoot that down and then scoot over here.
And this is where we get to say the modulus of the face is much greater than the modulus of the core.
And also, typically c, the core thickness, is much greater than t, the face thickness.
So if that's true, then it turns out this term is small compared to that one.
And also this term is small compared to this one.
And also this term, instead of having c plus t squared, if c is big compared to t, then I can just call it c squared, OK?
So you can see here, if Ec is small, then this is going to be small compared to these.
If t is small, then this guy is going to be small.
So even though it looks ugly, many times we can make this simpler approximation.
OK, so we can just approximate it as Ef times btc squared over 2.
So then this bending term here, we've got everything we need now to get that bit there.
So the next bit we want to get is the shearing deflection.
So what's the shearing deflection equal to?
So say we just thought about the core, and all we're interested in here is what's the deflection of the core and shear?
And so say that's P/2, that's P/2, that's l/2.
We'll say that's-- oops.
That's our shearing deflection there.
We can say the shear stress in the core is going to equal the shear modulus times the shear strain, so we can say P over the area of the core is going to be proportional to the Young's modulus times delta s over l. And let's not worry about the constant just yet.
So delta s is going to be proportional to-- well, let me [?
make it ?]
proportional at this point.
Delta s is going to equal Pl divided by some other constant that I'm going to call B2, and divided by the shear modulus of the core, and essentially the area of the core.
And here B2 is another constant.
So again, B2 just depends on the loading configuration.
Yeah, this is a little bit of an approximation here, but I'm just going to leave it at that.
OK, so then we have these two terms and we just add them up to get the final thing.
Start another board.
OK.
So that would give us an equation for the deflection.
And one thing to note here is that this shear modulus of the core, if the core is a foam, then we have an equation for that.
We also could use an equation if it's a honeycomb.
But I'm just going to write for foam cores.
Whoops.
This is for-- that will be for open-cell foam cores.
Oops, don't want to-- and get rid of that.
We won't update just now, thank you.
OK, so the next thing I want to think about is how we would minimize the weight for a given stiffness.
So say if we're given a stiffness, we're given P over delta, so I could take out the two P's here.
If I divide it through by P, delta over P would be the compliance, P over delta would be the stiffness.
So imagine that you're given the face and core materials, and you're told how long the span has to be, you're told how wide the beam is going to be, and you're told the loading configuration.
So you know if it's three-point bending, or four-point bending, or a cantilever-- whatever it is.
And you might be asked to find the core thickness, the face thickness, and the core density that would minimize the weight.
So I have a little schematic here.
I don't know if you're going be able to read it.
So I'm going to walk through it and then I'll write things on the board.
Whoops, hit the wrong button.
OK, so we start with the weight equation here.
The weight's obviously the sum of the weight of the faces, the weight of the core, so those two terms there.
So I'll write that down in a minute.
And then we have the stiffness constraint here.
So this equation here is just this equation that I have down here on the board, OK?
Then what you do is you solve that stiffness constraint for the density of the core.
So this equation here just solves-- we're solving this equation here in terms of the density, and we get the density by substituting in this equation here for the shear modulus of the core.
So you substitute that there.
It's kind of a messy thing, but you solve that in terms of the density.
Then you put that version of the density here in terms of this weight equation up here.
So then you've eliminated the density out of the weight equation, now you've just got it in terms of the other variables.
And then you take the partial derivative of the weight with respect to the core thickness c, set that equal to 0, and you take the partial derivative of the weight with respect to the face thickness, t, and you set that equal to 0.
And that then gives you two equations and two unknowns.
You've got the core thickness and the face thickness are the two unknowns.
And you've got the two equations, so then you solve those.
So the value you get for the core thickness is then the optimum, so it's going to be some function of the stiffness, the material properties you started with in the beam geometry.
And similarly, you get some equation for the optimum face thickness, t. And again, it's a function of the stiffness and the material properties in the beam geometry.
Then you take those two values for c and t, those two optimum values, and plug it back into this equation here, and get the optimum value of the core density.
And so what you end up are three equations for the optimal values of the core thickness, the face thickness, and the core density in terms of the required stiffness, the material properties, and then the loading geometry.
So I'm going to write down some more notes, because I'll put this on the Stellar site.
But it's hard to read just here.
So let me write it down and I'll also write out the equations so that you have the equations for calculating those optimum values.
So before I do that, though, one of the interesting things though is if you figure out the optimal values of the core thickness and the face thickness and the core density, and you substitute it back into the weight, and you calculate this is the weight of the face relative to the weight of the core, no matter what the geometry is, and what the loading configuration is, the weight of the face is always a quarter of the weight of the core.
So the ratio of how much material is in the core and the face is constant, regardless of the core-- of the loading configuration.
And this is the bending deflection relative to the total deflection.
It's always 1/3.
And the shearing deflection relative to the total deflection is always 2/3.
So regardless of how you set things up, the ratio of what weight the face is relative to the core and the amount of shearing and bending deflections is always a constant at the optimum.
OK, so let's say we're given the face and the core materials.
So that means we're given their material property, too.
And say we're given the beam length and width and the loading configuration.
So that means we're given those constants, B1 and B2.
If I told you it was three-point bending, you would know what B1 and B2 are.
So then what you need to do is find the core thickness, c, the face thickness, t, and the core density, rho c, to minimize the weight of the beam.
So there's two faces, so the weight of the face is 2 rho f g times btl.
And then the weight of the core is rho c g times bcl.
So I'm going to write down the steps and then I'll write down the solution.
So you solve.
So you put this equation for the shear modulus of the core into here, and then you rearrange this equation in terms of the density of the core here.
So you have an equation for the core density in terms of that stiffness, and then you solve the partial derivatives of the weight equation with respect to the core thickness, c, and put that equal to 0.
And then the partial of dw [?
over ?]
dt and set that to 0.
And if you do that, you can then solve for the optimal values of the face and core thicknesses.
Yes?
[INAUDIBLE] for weight, what is g?
Gravity

Just density is mass, mass times gravity-- weight.
That's all it is.
And then you've got a version of this that's in terms of the core density.
You can substitute those values of the optimum face and core thicknesses into that equation and get the optimum core density.
And then in the final equations, you get, when you do all that, and I'm going to make them all dimensionless, so this is the core thickness normalized by the span of the beam is equal to this thing, here.
So you can see each of these parameters here, the design parameters that we're calculating the optimum of.
I've grouped the constants B1 and B2 together that describe the loading configuration so you'd be given those.
C2 is this constant-- oop, which I just rubbed off-- that relates the shear modulus of the foam core.
So you'd be given that.
These are the material properties of the-- you know, say, it's a polyurethane foam core, this would be the density of the polyurethane.
Say it's aluminum faces, that would be the density of the aluminum.
so you'd be given that.
You'd be given the stiffnesses of the two materials, the solid from which the core is made and the face material.
And then this is the stiffness here that you're given, just divided by the width of the beam, B.
So the stiffness, you'd be given the width B.
So you're given all those things, then you could calculate what that optimum design would be.
So the next slide here just shows some experiments.
And these were done on sandwiches with aluminum faces and a rigid polyurethane foam core.
And here we knew what the relationship was for the shear modulus.
We measured that.
And what we did here was we designed the beams to all have the same stiffness, and they all had the same span in the width, B, then we kept one parameter at the optimum value and we varied the other ones.
So here, on this beam, this set of beams here, the density was at the optimum.
And we varied the core thickness, and we varied the face thickness, and the solid line was our model or our sort of optimization.
And the little X's were the experiments.
So you can see there's pretty good agreement there.
Then the second set here, we kept the face thickness at the optimum value and we varied the core thickness, we varied the core density.
So the same thing, the solid line is the sort of theory and the X's are the experiments.
And here we had the core thickness of the optimum value, and we varied the face thickness and the core density.
So you can kind of see how you can see this here.
And over here, just because I forgot to say it, this is the stiffness per unit weight, over here, OK?
So these are the optimum designs here, all right?
So there was pretty good agreement between these calculations and what we measured on some beams.
Do I need to write anything down?
Do you think you've got that?
Yeah?
I was just going to ask, for the optimum design column that you have there, do those numbers like fall out of these equations if you do the math?
They do, yeah.
I mean, it's-- yeah, exactly.
So if you remember the equation we had for the weight, so the weight is equal to 2 rho f gbtl plus the density of the core, bcl, so if you plug these things into there, then-- so this is the way to the face, that's the way to the core, then it drops out to be a quarter.
So it's kind of magical.
I mean, you have this big, long, complicated gory thing, and then, poof, everything disappears except a factor of 1/4.
And the same for the bending deflection.
So we had those two terms, so there was the bending and the shear.
If you just calculate each of those terms and take the ratio of 1 over the total, or the one over the other, everything drops out except that number.
So that's why I pointed it out, because it seemed kind of amazing that everything would drop out except for that one thing.
OK, so then the next thing-- so that's the stiffness in optimizing the stiffness.
Are we happy-ish?
Yeah?
OK.
So the next thing-- oh, well, let's see.
I don't think I need to write any.
I think if you have that graph, I don't really need to write much down.
So the next thing then is the strength of the sandwich beams.
So let me get rid of that.
You guys OK?
Yeah?
Yeah
Yeah, but you're shaking your head like this is very, very helpful for me
[INAUDIBLE]
Oh OK, that's
[INAUDIBLE]
That's OK, you can do that.
I don't mind.
But as long as you don't have questions for me.
OK, and so the first step in trying to figure out about this strength is we need to figure out the stresses in the beams.
So we need to find out about the stresses.
And we're going to have normal stresses and we're going to have shear stresses.
So I'm going to do the normal stresses first and then we'll do the shear stresses.
So you do this in a way that's just analogous to how you figure out the stresses in a homogeneous beam.
So we'll say the stresses in the face-- normally it would be My over I. M is the moment, y is the distance from the neutral axis, I is the moment of inertia.
So this time, instead of having a moment of inertia, we have this equivalent moment of inertia.
And we multiply by E of the face.
So you can think of this as being the strain essentially.
And then you multiply by E of the face to get the stress.
The maximum distance from the neutral axis, we can call c/2.
So that's y.
Then EI equivalent we had Ef btc squared over 2.
And then I have a term of Ef here.
c squared.
So one of the c's goes, the 2's go, the Ef's go.
Then you just get that the normal stress in the face is the moment at that section divided by the width, b, the face thickness, t, the core thickness, c. And I can do the same kind of thing for the stress in the core, except now I multiply by the core modulus.
So if I go through the same kind of thing, it's the same factor of M over btc, but now I multiply times E of core over E of the face.
And since E of the core is a lot smaller than E of the face, typically these normal stresses in the core are much smaller than the normal stresses in the face.
So the faces carry almost all of the normal stresses.
And if you look at an I-beam, the flanges of the I-beam carry almost the normal stresses.
So I want to do one more thing here.
I want to relate the moment to some concentrated load.
So let's say we have a beam with a concentrated load, P. So for example, something in three-point bending, typically we're interested in the maximum stresses, so we want the maximum moment.
So M max is going to be P times l over some number.
And this B3 is another constant that depends on the loading configuration.
So if it was three-point bending, B3 would be 4.
If it was a cantilever, B3 would be 1.
So if I put those things together, the normal stress in the face is Pl B3 divided by btc.
OK, so that's the normal stresses.
And then the next thing is the shear stresses, and the shear stresses are going to be carried largely by the core.
And if you do all the exact calculations, they vary parabolically through the core.
But if we make those same approximations that the face is stiff compared to the core, and that the face is thin compared to the core, then you can say that the shear stress is just constant through the core.
So we'll say the shear stresses vary parabolically through the core.
But if the face is much stiffer than the core and the core is much thicker than the face, then you can say that the shear stress in the core is just equal to the sheer force over the area of the core, bc.
So here, V is the shear force of the cross-section you're interested in.
And bc is just the area of the core.
And we could say the maximum shear force is just going to be V over-- actually, let's make it P, P over yet another constant.
And B4 also depends on the loading configuration.
So if I was giving you a problem, I would give you all these B1, B2, B3, B4's and everything.
So the maximum shear stress in the core is in just the applied load, P, divided by this B4 and divided by the area of the core.
OK, so this next figure up here just shows those stress distributions.
So here's a piece of the cross-section here.
So there's the face thickness and the core thickness.
You can think of that as a piece along the length, if you want.
This is the normal stress distribution, here.
So this is all really from saying plane sections remain plane.
These are the stresses, the normal stresses in the core.
And you can see they're a lot smaller in this schematic than the ones in the face.
And then this is the parabolic stress in the core.
And similarly, there'd be a different parabola in the face.
And these are the approximations.
Typically these approximations are made so the normal stress in the face is just taken as a constant.
The normal stress in the core is often neglected.
And here the shear stress in the core is just a constant here.
So the two things you need to worry about are the normal stress for the face and the shear stress for the core.
Are we good?
We're good?
Yeah, good-ish.
OK, so if we want to talk about the strength of the beam, we now have to talk about different failure modes.
And the next slide just shows some schematics of the failure modes.
So there's different ways the beam can fail.
Say it's in three-point bending just for the sake of convenience.
One way it can fail is, say it had aluminum faces.
This face here would be in tension, and the face could just yield.
So you could just get yielding of the aluminum.
That would be one way.
It could be a composite face and you could have some sort of composite failure mode.
You can get more complicated failure modes for composites, but there could be some sort of failure mode.
This face up here is in compression, and if you compress that face, you can get something called face wrinkling.
You get sort of a local buckling mode.
So imagine you have the face, that you're pressing on it, but the core is kind of acting like an elastic foundation underneath it.
And you can get this kind of local buckling, and that's called wrinkling.
That's another mode of failure.
You can also get the core failing in shear.
So here's these two little cracks, denoting shear failure in the core.
And there's a couple of other modes you can get, but we're going to not pay much attention to those.
The whole thing can delaminate, and, as you might guess, if the whole thing delaminates, you're in deep doo-doo.
Because, remember when I passed those samples around, how flexible the face was by itself and how flexible the core is by itself.
If the whole thing delaminates, you lose that whole sandwich effect and the whole thing kind of falls apart.
We're going to assume we have a perfect bond and that we don't have to worry about that.
The other sort of failure mode you can get is called indentation.
So imagine that you apply this load here over a very small area.
The load can just transfer straight through the face and just kind of indent the core underneath it.
We're going to assume that you distribute this load over a big enough area here, that you don't indent the core.
So we're going to worry about these three failure modes here-- the face yielding, the face wrinkling, and the core failing and shear, OK?
So let me just write that down.
And then you also can have debonding or delamination, and we're going to assume perfect bond.
And then you can have indentation, and we're going to assume the loads are applied over a large enough area that you don't get-- So you can have different modes of failure, and the question becomes which mode is going to be dominant?
So whichever one occurs at the lowest load is going to be the dominant failure mode.
So you'd like to know what that lowest failure mode is.
So we want to write equations for each of these failure modes and then figure out which one occurs first.
So we'll look at the face yielding here.
And face yielding is going to occur just when the normal stress in the face is equal to the yield stress of the face.
So this is fairly straightforward.
So this was our equation for the stress in the face.
And when that's equal to the face yield strength, then you'll get failure.
And the face wrinkling occurs when the normal compressive stress in the face equals a local buckling stress.
And people have worked that out by looking at what's called buckling on an elastic foundation.
So the core acts as elastic support.
You can think that as the face is trying to buckle into the core, the core is pushing back on the face.
And so the core is acting like a spring that pushes back, and that's called an elastic foundation.
So people have calculated this local buckling stress, and they found that's equal to 0.57 times the modulus of the face to the 1/3 power times the modulus of the core to the 2/3 power.
And here, if we use our model for open cell foams, we can say the core modulus goes as the relative density squared times the solid modulus.
And so you can plug that in there.
So then the wrinkling occurs when the stress in the face, the Pl over the B3 btc is equal to this thing here.
OK, so one more failure mode that's the core shear, and that's going to occur when the shear stress in the core is just equal to the sheer strength of the core.
So the shear stress is P over B4 times bc, and the shear strength is some constant, I think it's C11, times the relative density of the core to the three halves power times the yield strength of the solid.
And here, this constant is about equal to 0.15, something like that.
So now we have a set of equations for the different failure modes, and we could solve each of them, not in terms of a stress, but in terms of a load P. The load P is what's applied to the beam, right?
So we could solve each of these in terms of the load, P. And then we can see which one occurs at the lowest load, P. And that's going to be the dominant failure mode.
So one way to do it would be to, for every time you wanted to do this, to work out all these three equations and figure out which one's the lowest load.
But there's actually something called a failure mode map, which we're going to talk about.
So let me just show you it and we'll start now.
I don't know if we'll get finished this.
But there's a way that you can manipulate these equations and plot the results as this failure mode map.
And you'll end up plotting the core density on this plot, on this axis here, and the face thickness to span ratio here, and so this will kind of tell you, for different configurations of the beam, different designs, for these ones here, the face is going to wrinkle, for those ones there, the face is going to yield, and for these ones here, the core is going to shear.
So I'm going to work through these equations, but I don't think we're going to finish it today.
So this is just kind of where we're headed is to getting this map.
So we'll say the dominant failure mode is the one that occurs at the lowest load.
So the question we're going to answer is how does the failure mode depend on the beam design?
And we're going to do this by looking at the transition from one failure mode to another.
So at the transition from one mode to another, the two modes occur at the same load.
So I'm going to take those equations I had for each of the failure modes, and instead of writing this in terms of, say, the stress in the face, I'm going to write it in terms of the load, P. So using that first one over there, the load for face yielding, I'm just rearranging that.
It's B3 times bc times t/l times the yield strength of the face.
And similarly for face wrinkling, I can take this equation down here and solve it for this P here, OK?
And then I can take that equation at the top and solve that for P2 for the core shear, and that's equal to C11 times B4 times bc times sigma ys times-- oops, wrong thing-- times the relative density to the 2/3 power.
OK?
And then the next step is to equate these guys.
So you get a transition from one mode to the other when two of these guys are equal to each other, right?
So there's going to be a transition from face yielding to face wrinkling when these guys are equal.
And I'm not going to start that because we're going to run out of time.
But let me just say that I can pair these two up and say there's a transition between those two.
And that transition is going to correspond to this line here, OK?
So at this line here, that means you get face yielding and face wrinkling at the same load, OK?
And then if I paired up-- let's see here.
If I paired up face wrinkling and core shear, these two guys here, I'm going to get this equation here on that plot.
And then if I paired up these two guys here, the face shielding and the core shear, I would get that line there, OK?
So once I have those lines, that tells me, you know, anything with a lower density core and a smaller face thickness is going to fail by face wrinkling.
Anything with a bigger density is going to fail by face yielding.
And anything with a larger face thickness and a larger density is going to fail by core shearing.
And so you can start to see that it-- I'll work out the equations next time, but you can start to see that it kind physically makes sense.
Intuitively, this face wrinkling, it depends on the normal stress in the face, in compression.
So obviously the thinner the face gets, the more likely that's going to be to happen.
So it's going to happen at this end of the diagram.
And it also depends on that elastic foundation, on how much spring support the foundation has, right?
So the lower the core density, the more likely that is to happen.
Then if you, say you have small t, so the face is going to fail before the core, as you increase the core density, you're making that elastic foundation stiffer and stiffer, and you're making it harder for the buckling to occur.
It can't buckle into the elastic foundation, so then you're going to push it up to the yielding.
And then as you make the face thickness bigger, as t gets bigger, then the face isn't going to fail and the core is going to fail.
So you can kind of see just looking at the relative position of those things, they all kind of make physical sense.
So I'm going to stop there for today and I'll finish the equations for that next time.
And we'll also talk about how to optimize for strength next time.
And we'll talk about a few other things on sandwich panels.
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
All right, then, I guess we may as well start.
So what I wanted to talk about today was natural sandwich panels and sandwich beams.
So there's lots of examples of sandwich structures in nature, and we've been looking at the engineering sandwich structures.
And we've seen that you can get a lightweight structure by having this sandwich construction.
And so there are several examples I was going to talk about today.
And I think because this isn't really on the test, I'm not going to write a lot on the board.
So there's some notes.
I'll just put them on the website, and you can look at that if you want.
Because we have kind of a shorter time today.
I'll just try and talk and explain what's what.
Hey, Bruno.
How are you?
So this is the first example.
So many leaves of Monocotyledon plants have a sandwich structure.
And this is an iris plant and iris leaves.
And for those of you in 3032, I think you know that these are glass flowers.
So the Harvard Museum of Natural History has a glass flower collection that was made in the 1800s.
And there was a botany professor there who made these as sort of a lecture demonstration vehicle.
And so he would bring then to class, and he would show different things about the plants with the glass flowers.
But now they're just in the museum, and they're very realistic.
So I just wanted to show you those.
So let's see, it's not working.
Turn it on.
There we go.
So if we look at a cross-section of an iris leaf, it looks like the diagram on the left.
So here's the iris.
And you can see there's these kind of solid fibers, and those solid fibers are called schlerchyma.
And they only exist at the top and the bottom of the leaf.
So I went out this morning.
And if you look outside of the Stata building, there's that little kind of river-y thing, and there's some iris leaves growing there.
So I went and got some iris leaves.
And you can tell we had a horrible winter because usually when I give this lecture in the spring, the leaves are like twice as big.
But this year, they're just little, short, wimpy ones.
But I'm going to pass it around.
And if you just like move your thumb over the top, you can feel little ridges, little bumps.
And those little ridges that you can feel are these little schlerchyma fibers.
So you kind of see they kind of stick up a little.
And so when you move your thumb over it, you can feel that.
And then you can see that the middle of the iris leaf has this kind of foamy-type structure here, and that's called parenchyma cells.
So you can think of the leaf as very much like one of the sandwiches.
This is like a fiber-reinforced composite at the top and at the bottom.
And then this is kind of like a foam core in between, separating the fiber-reinforced faces.
And so the iris leaf behaves mechanically like a sandwich beam.
So I'm going to talk a little bit about how we can actually demonstrate that using the equations that we developed in class.
This is another example.
This is, I guess, what Americans call the cat tail, but Canadians and English people call it a bull rush.
And you can see this is a slightly different construction, but there's the same sort of idea.
So instead of having a foamy core as in the iris leaf, you've got these kind of webs here that go in between the top and the bottom, and that forms like a series of I-beams almost.
And you can think of that also like a sandwich panel or a sandwich beam.
So you've got two stiff top and bottom pieces, and then you've got these kind of webs that separate them, kind of like a honeycomb core would be.
So that's another example of a leaf that has the sandwich-type structure.
And this is very common in these Monocotyledon leaves.
So if you think of a cat tail or you think of an iris, they tend to be kind of narrow at the base, maybe an inch or two wide at the base, and they can be quite tall.
The iris leaves can get two or three feet tall.
The cat tails can get five or six feet tall.
And they stand up more or less straight.
They bend over a little, but they stand up more or less straight.
And this sandwich structure is one of the things that lets them stand up straight at a fairly low weight.
And from the plant's point of view, there's a sort of metabolic cost associated with making more material.
So it we can minimize the amount of material, it's a better thing for the plant.
These are some other examples of grasses that are sandwich-type constructions.
This is from some papers by Julian Vincent.
And the black little circles here are the schlerchyma, are those sort of dense fibers.
Then you can see in both of these cases, the dense fibers are on the outside, and the parenchyma cells, which is the white, are on the inside.
And so this is sort of another set of micrographs of the iris.
So this is just showing the outside, and these are the ribs viewed from the outside.
And this is the core, just sort of viewed along the length of it.
And so you can idealize the structure as being like a sandwich that's got sort of fibers on the top and on the bottom.
So the top and the bottom are like a fiber composite.
And the middle part, with the parenchyma cells, is kind of like a foam.
And so we did a little project on iris leaves, and we wanted to see if you could show that they behave mechanically, like a sandwich beam.
So you remember that we had that equation for the deflection of the sandwich beam.
There were two terms.
There was a bending term, and then there was a shearing term.
And so we took some little sandwich beams.
We cut little kind of rectangular beams.
We hung little weights.
We measured how much they deflected, and we wanted to see if we could use this equation to predict their stiffness and how much they deflected.
So to do that, we needed to know a bunch of things.
We needed to know some of the geometrical parameters.
So we needed to know what volume fraction of the face is those solid ribs, how thick's the core, how thick's the face?
And so we measured a bunch of these geometrical parameters.
We tested it like a cantilever so we knew what B1 and B2 were for the cantilever.
We knew how long the beam was, so we know what l is.
We knew what loads we applied, so we knew what P was.
But we needed to make some estimate of what the face modulus was and what the core shear modulus was, too.
And so we made some estimates of that.
So this table here just shows some of the dimensions of the leaf.
The leaf tapers, and this is at the thin end, so here's the face thickness.
Here's the sort of length of this.
Some square cells in the face.
This is the core thickness here.
This is the dimensions of the core cells.
This is the diameter of the ribs, the spacing of the ribs, the volume fraction of solids in the ribs.
And we did that at different lengths along the different positions along the length of the rib, or length of the leaf.
So we had the geometrical parameters, but we needed to get this E of the face and G of the core.
And to do that, we looked at the literature.
And people had done tests on the fiber parts of leaves.
They'd done little tensile tests, and they'd measured modulii between about two and 20 gigapascals.
And then we did some tension tests on the iris leaf.
And in tension, those ribs are going to take most of the stress.
And if you know the volume fraction of the ribs, you can back out what the stiffness of the ribs must have been.
If you know the stiffness of the ribs, you can figure out the stiffness of the face.
So we calculated that, and then we looked at the literature.
And people have done tests on parenchyma cells and different types of tissue on things like apples and potatoes and carrots.
And these are the values for the Young's modulus they get.
They're between about 1, and the highest one was 14 megapascals.
But most of these values for the Young's modulus are around about four.
And the shear modulus is roughly about half of the Young's modulus.
So we said the shear modulus was around two.
So we have these values we could plug it in and then calculate what the stiffness would be for the iris leaf.
And so this was a little analysis we did.
So this was the measured beam stiffness up here.
We had four beams, and they were different stiffnesses.
They all had the same length.
They all the same face thickness.
The core thickness varied.
They all had the same width.
We cut them to have the same width so we could calculate a flexural rigidity.
That's the EI equivalent.
We could calculate the bending deflection term, the shear deflection term.
And this is the calculated beam stiffness.
And then this is the ratio of the calculated over the measured.
So it's not exactly right.
Obviously, there's some difference here.
But it's in the same order of magnitude.
It's in the same ballpark.
And one of the complications that we didn't really try to take into account was that the leaf isn't a nice rectangular structure.
The leaf has this kind of curved cross-section to it.
And we made a bit of an approximation to that, but it wasn't that close, really.
We could have probably done better on that.
But I think the idea that the iris behaves like a sandwich is a reasonable one.
So that was the iris leaf.
And then I wanted to show you some other structures in nature that are sandwiches.
So this is a seakelp, help like a seaweed thing, in New Zealand.
This is the largest intertidal seaweed.
The fronds, the sort of long pieces of it, are up to 12 meters long.
So that's almost 40 feet.
So 40 feet is probably like from one side of this room to the other side of the room.
It's quite long.
And you can see, if you look at this section here, this is all like a honeycomb-type section here.
And the honeycomb is like a honeycomb in a sandwich, and the top and the bottom faces are like the face of the sandwich.
So this would be like the face here.
That would be the honeycomb core.
And that would be the other face on the other side over there.
And those honeycomb-like cores, apparently, have some gas-filled pockets that then provide buoyancy to keep the whole thing floating.
So it photosynthesizes.
So one of the things about these leaves is that they have multiple functions.
It's not just that they have to have a certain stiffness so they don't fall over.
The plant wants to photosynthesize, so you want to maximize the surface area as well, and you want to have exposure to the sunlight.
So there's a number of things that the plant's trying to do in having this structure.
So that seakelp is one example.
These are skulls from birds.
And so this is a pigeon here.
This is a magpie.
If you come from the West you see magpies out West.
You see them in Europe as well.
And this is a long-eared owl.
This long-eared owl's around here.
And I brought in a couple of bird skulls as well.
And you can see that all of those birds skulls are sandwich structures.
The one for the pigeon has sort of a foam-like core here.
And you can see that the two faces aren't sort of concentric for the pigeon skull.
They sort of not following each other.
But here, this would be, say, on the top shell of the magpie, where the two, the inner and outer face, are sort of concentric.
Then you get these kind of little ribs of trabecular bone in between them, and then the same with a long-eared owl.
You get these little ribs in between them.
And so you can see that there's a sandwich structure there.
And obviously, birds want to be light.
They have to be light to fly, to take off, and so they want to be light.
So I've got two skulls here.
And I'll pass them around.
Please be careful because they're kind of delicate.
This one is from a screech owl, and you see screech owls around here.
This was a screech owl that had an intersection with a car.
Yeah, so the skull fractured, but you can see the sandwich right there.
You see the two little bits?
So you can see the inner plate and the outer plate and the foam, the trabecular bone.
So that's the screech owl.
And this is a red tail hawk.
So you can't really see the shell and the sandwich structure here.
But I want to pass it around just so you can see how light it is.
So it's amazingly light.
So a red tail hawk is probably about this big, something like that.
And this is one of the things that makes them very light.
So those are the bird skulls.
Oh, yes, so now I have to tell you about the owl.
So I think the people in 3032 have heard this before.
But the other people haven't.
So one of the things about the owl is if you look at the whole skull, if you look at this picture here, one of the things is that this bone here is not symmetrical with that bone there.
Normally, when you think of a body, you think of the bones being symmetrical.
But those bones are not symmetrical, and those bones are near where the ear is.
And it turns out on owls, at least on some owls, the ears are at different heights on their heads.
And people think that one of the things that allows the owls to do is it allows their hearing to sort of pinpoint where something is.
And owls can catch little creatures at night, but they can also catch little creatures underneath the snow.
So they can catch things that they can't even see.
And they have a number of adaptations to improve their hearing, but this is one of them.
So here's a little owl Allison Curtis is a Canadian friend who lives in northern Ontario, and this is looking out of her living room window.
And that's a barred owl.
And you can see the barred owl has caught this little vole here.
And you can see in the background it's winter in Canada.
and there's snow all over the place.
So this owl has probably caught that little vole underneath the snow.
And then it's come to eat it.
And this is another picture of-- you can see this is where an owl landed in the snow.
It's wings hit the snow, trying to catch something underneath.
And this is another kind of beautiful print of the owl's wings hitting the snow in the winter time.
So did I show you the fox video?
Should I show you the fox video?
You saw it, right?
I think I showed it last time in 3032.
But you guys haven't seen it.
Let me show you the fox video because foxes do the same kind of thing.
Their ears are the same as ours.
They're in the same position.
But they have this-- let me see.
Where's the sound thing?
We don't really need the sound for this, but there's BBC sound.
So we get this music, even though the fox can't hear the music.
Here we go, fox no drive.
Check this out.
Is it going to come up?
Is that going to play?
OK [VIDEO PLAYBACK] -It listens for the tiny sounds of its prey moving about below
So you see how it cocks its head, and it does this with its head?
It's putting its ears at different heights when it does that.
So check this out.
And look carefully, you can see the little animal it's got in its mouth when it comes out.
There's a little tail.
So part of the reason dogs and foxes and coyotes do that thing, I think, is because they put their ears at different heights, and it helps them pinpoint where something is.
[END PLAYBACK] You know I love these Nature videos, right?
So that's the fox video.
Let me see if I can stop that.
So that's one of the interesting things about owls.
Let me go back to my little PowerPoints.
So here's another example of a creature that has a sandwich-type structures.
So here's the sandwich here.
Here is the ever so charming looking cuttlefish.
And the cuttlefish is not actually a fish.
It's a mollusk.
So it's related to things like octopus, things like that, and squids.
It's a cephalopod.
And you can't see it so well in this picture, but I'm going to show you something else and you see it.
It's got like little tentacles.
These things here are actually separate little tentacles.
And because it's not a fish, it doesn't have like fins that can kind of swim with.
And it's got this thing called the cuttlefish bone.
And this is a cuttlefish bone here.
And that bone has the sandwich structure here.
And it's not actually a bone.
It's really a shell.
It's a calcium carbonate thing, not a calcium phosphate thing.
But the cuttlefish can control how much air goes into those little pockets.
And it can control its buoyancy by controlling how much air goes into those little pockets.
And I brought with me a cuttlefish bone.
Have you ever owned like, I don't know, like a parrot or a pet bird?
Apparently, pet birds love to sharpen their beaks on this cuttlefish bone.
So if you go to a pet store, you can buy this stuff.
So you won't be able to see the little sandwich structure because it's a very small length scale.
But you can kind of see there's a sort of different material on the inside than there is on the outside of that.
So do people know the other thing that cuttlefish are famous for, besides the bone?
Change colors.
Can I show you a video of cuttlefish changing color?
Yeah, of course.
So let me get rid of this again.
Go back to this.
Let's see, somewhere-- where's the cuttlefish?
Here we go.
Did I do it?
Is it thinking?
Here we go.
Where's the cuttlefish?
So this is another one of these Science Friday videos from National Public Radio with Flora Lichtman.
[VIDEO PLAYBACK] -OK, let's play a game.
[GAME SHOW MUSIC PLAYING] [APPLAUSE]
See it?
-Biologist Sarah Zielinski took these shots.
And if you needed a helping hand to find the cuttlefish, don't feel bad.
-I've certainly taken photos in the past then come back to look at them and gone, I'm sure there was a cuttlefish in there somewhere!
-These cephalopods are master camouflagers.
But while they're hiding their body, they're revealing something about their mind, or at least their visual system.
-In very simple terms, they can tell us what they can see by the body patterns they produce on their skin.
-They produce these body patterns by expanding or contracting chromatophores, these little ink sacks on their skin.
And they use different displays for different reasons, like for male-to-male combat.
-Two males will turn into each other and pass these kind of waves of dark chromatophores over a really bright sort of iridescent stripey body pattern and somehow solve these combats.
Eventually, one male gives up and goes away.
-And then there's this unsolved mystery.
It changes color when it grabs a snack.
-That doesn't make perfect sense because it seems to make it very conspicuous.
So one theory is that it's just a happy signal of how excited it is to have caught something, some response that it doesn't have any control over.
-But most of the time they seem to be using their chromatophores more intentionally, primarily to blend in.
-Because otherwise they're more likely to be eaten, so it's very important they don't make mistakes about ambiguous visual information.
-And ambiguous visual information is specifically what Zielinski's interested in.
So here's the experimental setup.
Print out laminated patterns, like this checkerboard, and stick them in a tank.
-And we place the animals in the tank.
And we record the body patterns that they produce.
-You're seeing them on squares, but they do the same thing on top of circles.
They produce-- - --the disruptive pattern, where you get these blocky components of high-contrast components.
-But when you put a cuttlefish over squiggles, it produces-- - --a sort of mottley pattern, where you get these little groups of dark spots showing across the body.
-So what happens when you put a cuttlefish on something in between, when you put them on incomplete circles?
When we see something like this, our visual system likes to fill in the blanks, something we do constantly, Zielinski says.
-The reason why cartoons and sketches work is because we can recognize objects based on their edges alone.
-And we can identify objects even if they're broken up or-- - --have an object that is occluded by another object.
That's no problem for us.
We can still work out what the object is most of the time.
And I was interested to know whether cuttlefish can solve similar problems.
-And Zielinski and colleagues report this week that cuttlefish do seem to-- - --fill in those gaps and interpret those little segments as a whole circle.
-Or anyway, the broken circles prompted the same camo pattern as full circles.
So if you're wondering, uh, I see these as circles, too.
What's the big deal?
The weird thing here is that there's no reason why cuttlefish, which are-- - --invertebrates, and they're in the same group as slugs and snails.
- --should see the world the way we do.
-Yes, it's like they're alien, but we also seem to have so much in common with them.
-So the next step?
-Because we can't share the perceptive experience of a cuttlefish, it's hard to know exactly what it is that they're doing to fill in that missing information.
And I want to try to get a better grasp on that and also see whether they actually respond to true illusory contours.
-So you're going to show optical illusions to cuttlefish?
-(LAUGHING) That's what I'm hoping to do, yes.
[END PLAYBACK]
So let's go back to sandwiches.
I think I have-- do I have one more?
There we go.
So horseshoe crab shells, so different sorts of arthropods, the shells are sandwiched too.
This is from Mark Myers' work.
So we're looking at the cross-section of a horseshoe crab shell.
So again, it's the same idea-- the animal wants to minimize the amount of material or minimize the weight, and this is a way of doing that.
And I went to the Galapagos about a year ago.
And there was a place where they had these giant Galapagos tortoise shells.
And one of them was broken, and you could see there was a sandwich structure in the Galapagos tortoise shells.
These Galapagos tortoises, their shell is like this big.
They're gigantic.
They're huge.
So those are my examples of sandwich panels and beams and shells and whatnot in nature.
So the idea is that nature too wants to minimize weight and minimize the amount of material, and the sandwich structure is a way of doing that.
So I have one more thing I wanted to talk about today.
So this isn't quite sandwich structures, but it's looking at another kind of natural structure that is designed to reduce the weight of plant stems, in this case, palm stems.
And there's a couple of interesting things about this.
So when you look at palms, like let's pretend we're not in Boston.
We're in California, where they have palms.
And we're in LA, and they don't have winter.
And if you look at the palms growing, when the palm's short, it's about this big in diameter.
And as it gets taller and taller, the diameter doesn't really change.
It gets taller and taller and taller, but the diameter doesn't change, at least in some species.
Whereas if you think of a tree, a tree starts out with a little skinny diameter.
And as the tree gets taller, the diameter gets bigger.
And it sort of tapers and does that whole thing.
So palms don't do that.
And palms are not trees.
They're a botanically different thing from trees.
So here's a coconut palm.
And so the question is, as the stem gets taller and taller, how does it resist the bending loads that get bigger and bigger?
So probably, the main load on these sorts of things is from the wind.
And often these plants are in areas where they have hurricanes.
And you see them in hurricanes, you see the pictures of the palm stem blowing way over.
And so how do they resist the larger internal stresses as they get taller and taller, if the diameter doesn't get bigger and bigger?
And the way they do that is that they deposit additional layers of cell wall as the plant ages.
So if you think of a tree, when a tree grows, it just deposits more and more cells.
And the cells have roughly the same thickness.
So there's ones that are deposited in the spring have thinner walls.
Then the summer and the fall have thicker walls.
But more or less, it's similar.
Whereas the palm, it deposit cells, and then as the trunk of the palm gets taller, as the stem gets taller, it deposits more layers on the cell wall.
So this is an example in an SCM.
You can see here this is a young cell, and it's got-- this one that's not marked is a primary cell wall, and then this is the first layer of the secondary cell wall.
And then this is an older palm.
And you can see here it's got more layers, and so the cell wall itself has gotten thicker.
So that means that the density of the tissue changes as the palm ages.
And it does so in a very kind of clever way.
If you think of the palm as being like a cantilever that's vertical and it's bending in the wind, when we have a cantilever beam or any kind of beam, the stresses are going to be biggest on the periphery, right?
They're going to be biggest on the outside.
And if you think of the palm as having a circular cross-section, that outer periphery is going to see the biggest stresses.
So it would make the most sense if that was the densest tissue.
And that's exactly what the palm does.
So there was a nice study done by Paul Rich quite a number of years ago.
And he studied palms in Central America and looked at the density and measured the mechanical properties.
And I'm going to talk about his stuff today.
So the white is the low density.
The gray's the medium, and the black's the high.
So you can see the low density's on the middle of the young stem, and just at the very base and then the periphery is the dense tissue.
But as the stem gets taller and gets older, then stuff that was low density is now high density.
And only the very middle here is the low density.
And that some stuff that was low density has turned to middle density.
And some stuff that was low density has turned to high density.
So it's done this by adding more and more layers to the cell wall, making the cell wall thicker and making the cells themselves denser.
So this is looking just at a single palm.
So each one of these lines is a single palm.
And this is looking at how the density changes from the periphery to the center of the palm.
So if you cut the palm down and, say, we take a little sample radially from the middle to the outside or from the outside to the middle, he then measured the density.
And it's probably easiest to think about the dry ones because that's kind of what you would compare wood to.
So the dry densities varied from about one gram per CC, that's about 1,000 kilograms per cubic meter, down to almost zero in this particular species here, probably like 50 or something like that.
And if you compare this with woods, this little arrow here is the density of most common woods.
So if you looked at pine and spruce an oak and maple and ash and hickory, they would all be in that little range there.
So a single palm stem can have a bigger range of densities than many different species of wood.
So it has this kind of profile of the density.
And the thing I was interested in is seeing how mechanically efficient that was to put the denser material at the outside.
So I looked at the stiffness of the palm, and I also looked at the strength.
So I just replotted that data on this slightly different axes here.
So this is the radial position relative to the outer radius, and this is the density.
And I subtracted off the minimum and then took the range.
And for this species here, the minimum density was almost zero.
So this expression simplifies to something like that.
And just because it's mathematically simpler, that's what we're going to look at.
So the density goes roughly as the radius squared.
And Paul Rich also did a lot of mechanical tests on the palm, and he took out little beams of different densities.
And he measured the stiffness and the strength of the beams.
So he measured the modulus of elasticity here versus density.
And he measured the modulus of rupture here.
And these are all along the grain.
And he found that the Young's modulus varied with the density to the 2.5 power, and the strength varied as the density squared.
And if the-- [BUZZING SOUND] Oh, hello.
[LAUGHTER] So these were just sorts of empirical findings that he made.
If you have prismatic cells and you deform them axially, and the cell wall was the same in the different specimens, then the solid modulus would be a constant.
And you would expect that the modulus of the beam would go just linearly with the density, sort of like a honeycomb loaded [?
at a ?]
plane.
But what he measured was that the modulus and the strength varied with some power of the density.
And the reason for that really was that the cell walls of the denser material had more layers.
And in the additional layers, the cellulose microfibular angle was probably different, so that the different layers had different stiffnesses.
And if you have layers of differences, then you're going to get this power relationship.
So what I then did was I took his data, and I tried to see how efficient that would be in bending.
So he had found that the density varied with the radius raised to some power.
This power n was 2, but I wanted to do it just for a general case, so I said I was just n. And he said that he found that the modulus varied with the density raised to some other power m. And for him, m was 2 and 1/2.
And so I could write just another equation saying that the modulus goes as the radius to the mn power.
And then you could do a little calculation where you work out with the equivalent flexural rigidity is.
So you have to integrate up.
You kind of say you have a little band at a certain radius.
That radius has a certain modulus.
And you can figure out the moment of inertia that goes with that particular radius.
And then if you integrate it up over the whole thing, you can say that the flexural rigidity for the gradient density is some constant times pi times the outer radius to the fourth power divided by those two powers mn plus 4.
So m was the power here for the modulus.
And n was the power there for the density.
And then you could compare that with having the same mass just uniformly distributed over the whole cross-section.
And then if you take the ratio of the flexural rigidity for the density gradient versus the flexural rigidity for the uniform density, you can show that it's this equation here.
And then if you plug in these measured values for those exponents for n and m, you find that the flexural rigidity with the gradient density relative to the uniform density is a factor of 2 and 1/2.
So the stem is 2 and 1/2 times stiffer by having that density profile.
So there's a huge sort of mechanical advantage to doing that.
And just sort of physically, if you know the stresses our biggest on the outside, it would make sense to put the denser material on the outside.
And then the other thing I looked at was the strength of the palm.
So imagine this is our very schematic palm here, and then there's a circular cross section.
So I wanted to compare the bending stress distribution with the bending strength distribution.
So the stress goes as the modulus times the strain, just Hooke's law.
And here we're assuming that plane sections remain plane, like that's the standard assumption of bending.
So if you assume plane sections remain plane, then the strain goes with the curvature times the distance y from the neutral axis, the distance from the middle.
So this distance here would be the dis-- [?
same ?]
at loaded with a loaded p here.
That distance would be y there.
And then I can plug in some things here.
So instead of E, I'm going to plug-in my relationship with the radius to that mn power.
And here's my curvature, and instead of y, if I say that some radius, I'm going to say y is our r cos theta.
And so I'm going to say that the stress goes-- [SNEEZE] Bless you.
Goes as radius raised to some power mn plus 1.
And again, for the species I know what n and m are, so the stress goes as the radius to the sixth power.
And then I can also compare with what Paul Rich had found for the strength.
He found that the strength-- so sigma star is the strength-- was proportional to the density raised to some power q, and that power was 2 in the measurements that he made.
And so I can say that the strength goes as the radius to this power nq, so to the fourth power.
And then if I plot the stress distribution and the strength distribution-- so imagine, this is through the cross-section here.
So this is the diameter of the stem.
And this is the neutral axis here in the middle.
The strength goes as that solid line there.
It goes as the fourth power.
And the stress goes as that dashed line there, as the sixth power.
So they're not exactly on top of each other, but they're very close to being on top of each other.
So basically what the palm has done is it's arranged the material in such a way that the strength matches the stresses that are applied to it.
So if I just had a constant density, my stress profile would look like that.
And if I had a constant density, the strength profile would kind of like that.
So the strength here would be a constant, and this would be the stress here.
So the stuff in the middle, it's much stronger than it needs to be.
Whereas the palm has arranged things so that it's got just the right amount of strength for the stress, as a function of the radial position.
So it's kind of a clever thing.
So that's kind of a beautiful thing.
And I think that is it.
I think that's-- yeah, that's the end of it.
So all these images came from this other book that we wrote.
And if you wanted to get the sources, you could get them from there.
So that all I wanted to talk about today was some examples of sort of efficient mechanical design in nature and the sandwich panel structures as one, and these radial density gradients is another.
We have a project on bamboo right now, and the bamboo also has a radial density gradient, and it's the same thing.
The densest material's on the outside, and the least dense is on the inside.
So I think I'm going to stop there for today.
So what I was going to do on Monday is talk a little bit about bio-mimicking.
And that won't take the whole class at all.
And I thought we could spend the rest of the class on Monday just doing a review.
So the test's on Wednesday.
So if you want to bring questions, that would be a beautiful thing.
I can't really can I review the whole last six weeks or something in an hour and a half or something.
So if you want to bring questions, I'll be here and we can just go over questions.
Does that sound good?
The following content is provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free.
To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu
All right, well, I guess I may as well start.
I don't know if anybody else is going to come.
So I wanted to finish up by talking a little bit about the biomimicking.
And some of these examples, you've seen before.
But I just thought I'd put them all together, and we could look at them as one thing.
So if you remember, when we talked about wood, one of the things that I showed you was that people have taken wood and pyrolized it.
So they get a carbon template of the wood cells.
And then they infiltrate that with silicon carbide-- or, with a silicon vapor infiltration.
And they make a silicon carbide ceramic.
So they can get a replica of the structure.
So sometimes when people say "biomimicking," some people think of it as replicating something.
But it doesn't have to be just replicating something.
It can be also just some design inspired by the biological material.
But this thing really is a replica.
And this was another version where there was the-- they took the silicon carbide material and then infiltrated that with liquid silicon to get a fiber-reinforced material, really.
So there was the wood composites we talked about.
Jennifer Lewis's group up at Harvard is doing 3-D printing of honeycombs.
And one of the things they've been interested in is not just printing of the pure resin, but having a fiber-reinforced resin.
And the first thing they did was they made these honeycombs like this.
And they had small silicon carbon fibers-- or, were they silicon carbide?
Maybe it was carbon fibers-- in the ink.
And so they would just-- if the ink was being laid down, the fibers would just line up in the direction of the ink.
And so the fibers tended to be in the plane of the honeycomb.
And if you think of things like wood, you want the fibers to normal to that plane.
And more recently, they've-- hello.
Oh, hello.
Oh, look.
Oh, look, almost everybody is here.
So more recently, they've got a technology now where they're rotating the nozzle as they print the honeycomb.
And as they rotate the nozzle, they get some change in the orientation of the fiber.
So they're beginning to be able to make honeycombs that are fiber-reinforced.
And they can get the fibers aligned with the prism axis of the honeycomb, which is more or less what the wood does.
So I wasn't going to write anything on the board today.
I was just going to go over some of the slides and do the review.
So they're beginning to make honeycombs that have the same sort of structure on the cell wall level, or at least a similar structure, as to what the wood composites have.
So that's another example there.
This is just another close-up of their fiber-reinforced walls in honeycomb specimens.
We talked about trabecular bone, and we talked about the fact that people are starting to look at using foamed metals as coatings on orthopedic implants.
And there's been some interest in looking at using foamed metals for more permanent parts of the body, more permanent bone parts, things like vertebral cages, stuff like that.
And so this is just an example here, with the trabecular bone on the left and the tantalum foam that's made by replicating an open-celled polyurethane foam on the right.
And you can see the similarity in the structures of those two things there.
Then we talked about tissue engineering scaffolds.
And if you remember, these two scaffolds here on the bottom were made from pig heart tissue.
And they're made by just removing all the cells.
So that actually is the natural extracellular matrix.
And then, these other structures up here, these were all engineered tissue engineering scaffolds that are made in a synthetic way.
And the idea is to try to mimic the extracellular matrix in the body.
So you can see the similarities there.
And then more recently, we were talking about sandwich panels.
So this is the example from the helicopter rotor blade.
That's from an aircraft flooring panel.
And this was the Irish leaf, and these were the bird skulls.
So the same sort of idea, that there's these engineering lightweight structures, and there's also similar things in nature.
And then, I think, last time, we also were talking about palms.
And the palm stems had density gradients in them.
And one of the things we showed was that by having that density gradient, the stress distribution across, say, the radius of the palm was almost matched by the strength distribution.
So it was a very efficient way to use the material.
And at MIT, that was a student in architecture who was looking at doing this with concretes with aerated or foam concretes and making a radial density distribution.
So he made beams.
He made columns.
He made different kinds of things.
With the concrete, there's a little bit of a limitation, because the concrete is much stronger in compression than it is in tension.
So if you had a beam loaded in tension or a concrete column that might buckle, you'd still have to have some reinforcing bars in there to take the tensile loads.
And one thing I think I didn't really talk about was animal quills and other sorts of plants stems.
And many of these have a structure that's made up of a cylindrical shell with a foam or a honeycomb core.
So here are some examples in nature.
If you look at grass stems, there's a dense layer on the outside, and then a foamy layer on the inside, and then just a hollow layer in the middle.
And if you look at porcupine quills-- this is a porcupine quill-- these are made of keratin.
They're like modified hairs.
So it has a dense layer on the outside, and then this foamy stuff on the inside.
This is a hedgehog spine here.
And again, you can see there's a dense layer on the outside and these ribs on the inside.
And this is the toucan beak-- you know, the toucans that live in Central America.
And the beak has a foam core.
Again, they're keratin structures, and yet the outside is solid.
But the inside has a foamy structure.
And Mark Myers did a paper on this a while ago.
So we got interested in these structures that have a solid shell on the outside but a foamy thing on the inside.
And we wondered if there was a mechanical reason for that.
And you can show that, at least in some of them, if, say, you have a grass stem-- it's really common in plant stems.
Do I have some more plant stems?
Yeah, here we go.
Here's a milkweed stem.
So it's got these dense fibers with this foamy core here.
And blue jay feathers-- feathers have this as well.
So they have an outside layer on the quill that's solid, and then an inside layer that's foamy.
If you look at things like plant stems, they blow in the wind.
And you can look at the buckling resistance of the stem.
And because there's this shell with the foam-like core, it's not just the overall buckling of the whole thing.
You can get that local face-wrinkling mode again.
So the outer shell can wrinkle.
And you can show that having a foam-like core helps prevent that wrinkling from happening, the same as with the sandwich panel.
You remember we talked about the face wrinkling on the sandwich panel?
Well, on these stems, you can get wrinkling of the outer shell.
And the foam helps prevent that from happening.
And you can show that you can actually-- for the same buckling load, you can reduce the weight of the plant stem, or the bird feather quill, or whatever by having that foamy core.
So people have looked at this too.
And there's a group in Germany who had looked at the idea of mimicking the horsetail stem.
So this is a plant stem here, the horsetail plant.
And they made something they called a technical plant stem, where they made this structure here.
And they made it out of fiber-reinforced composites.
And you can see, the little holes here represent those holes there, in the plant stem.
So the idea was to try to get something that was good at resisting the buckling, but at a lower weight.
And they were doing that with this thing here.
And there was a group in Japan that did a similar thing.
They used-- I think they took a copper tube, and then they took copper and aluminum wires and filled the copper tube with the wires.
They extruded that.
And then they melted out the aluminum, which, I think, also helped to soften and make the copper bond together.
And they got these structures here.
And you can see, that's similar to some of the plant stems as well.
So these are all examples of cellular structures that have-- mechanically efficient structures.
They're lightweight, and they're strong, and they're stiff.
And these natural structures have been mimicked in engineering applications.
So that's really all I wanted to talk about today.
But I think we wanted to use the rest of the class as a review.
So I haven't made a one-hour summary of the last six weeks, because that's not really possible.
So I thought I would just answer questions.
If you have questions, I'll try and answer them.
So for the test-- so, the test is on Wednesday.
You can bring one 8 and 1/2 by 11 sheet.
I wasn't going to give you all those honeycomb and foam equations, partly because-- the only thing I would really want you know is open-cell foams.
And I was hoping, by now, that you might have registered those equations somewhere in your brain.
So I'm not going to ask you for-- some equations are obscure.
I might ask you-- expect you to know what the Young's modulus of an open-celled foam is by now, or the axial modulus of a honeycomb or something.
But I don't think I'm going to ask you anything like, calculate air pressure contributions to the modulus of the closed-cell foam.
I don't think we're going to do that on this test.
So I think you should know what the modulus of a-- the Young's modulus of an open-celled foam is, the shear modulus of a foam, because you need that for the sandwich panels.
But you don't need reams and reams of those equations, so I wasn't going to give you those this time.
OK?
So Jenny, did you have-- so I finished the biomimicking thing.
We're just going to do a review.
I don't know if you want to stay or if you want to go.
You want to stay?
Well, whatever.
So do you have questions, Jenny?
I do.
I just wanted to have question 4 from the last pset reexplained.
Because I know that you explained it to in office hours, and I know that the solutions are online, and I looked at them.
I'm still confused
OK.
So, you're going to have to remind me what problem 4 is, because I don't remember
Question 4 says, polymethacrylate foam at solid strength of 3.0-- or, solid Young's modulus, rather, of 3.0-- gigapascals is being considered for the energy absorption layer in a bicycle helmet
OK.
I'll tell you what, I think I have it on my little disk here.
Maybe that's easier, because then I can read it
[INAUDIBLE]
Yeah, let me just see if that's going to come up.
There it is.
OK.
So this one here about the foam, about the--
The one with the graphs
--energy absorption?
I'm just a little bit confused by the graphs
OK. Let me see if I can make this bigger.
Hang on a sec, my computer's thinking.
OK. OK. And I think I gave you-- right, I gave you this graph here.
Right?
Yes
OK.
So can you read what I've got here, or is that-- should I make it bigger?
I can't really read it, [INAUDIBLE]
Does that help?
OK.
So I have to admit, when I put this together, somebody-- I can't remember who it was-- told me you thought it was overconstrained.
And it turned out it was overconstrained.
And then I said, forget the thickness.
Forget that I've given you the thickness, right?
So disregard the thickness
No, the velocity
Oh, speed-- the speed?
The speed-- all right, OK, the speed.
No, I don't want to register.
OK.
So from what I gave you, you can figure out the normalized peak stress, right?
So you can get-- so if I do this, is this good?
You can see what I'm pointing at?
So you can get, the peak stress is just the mass times the acceleration over the area-- are we good with that-- divided by Es, which I gave you.
So I think most people probably got the peak stress here.
And then, because I had given you the velocity and the thickness, I was thinking you could calculate the strain rate.
But if I don't give you the velocity, say we're not calculating the strain rate at this point here, OK?
So here, we're-- did I put this-- I don't know if I have the graph on this solution.
There we go.
So this point here is the 2.5 times 10 to the minus 4 for the peak stress.
And if you're not given the velocity, I think what I thought you would do then would be just assume a velocity.
And then you can check it at the end.
So since I had given you this velocity of 12, let's just say that's what we assumed.
OK?
So then you get a strain rate of 480 per second.
So let's just say we assumed that velocity.
Then we could scoot back over here.
So we know we're on this line here for the sigma p over Es.
And we want to be up towards the top of these different strain rates.
So if you look at the strain rates, see, the very last one at the top is 1,000 per second, and the next one's 100 per second.
So we're halfway in between those.
And they're so close together, you can't really read the difference.
But we're up here somewhere.
So then we read off a w over Es for that.
OK?
Are we good?
Then, if we know the w over Es, this is the number, here, that I read off.
You know what the Es is, so you can get w. If you can get w, w is in joules per cubic meter.
It's in energy per unit volume.
But you know the area, and you know the thickness.
So you can get the energy in joules.
And then-- oh, did I-- I must have rubbed that off.
Didn't have it on this version.
The version in my notebook, I think, calculated what the velocity would be that corresponds to this.
And I think it turned out to be 8 meters per second or something.
So I had assumed 12, and I think it worked out to 8.
And on those log log graphs, whether or not it's a strain rate of 480, or a little bit less, or a little bit more, you can't read the difference on these things.
OK?
Can you explain how this graph relates to the other graphs from lecture that were simpler?
Because we had ones that were all density, and ones that were all the same strain rate.
And I guess I'm just confused why
Oh, OK.
Hang on.
So let me see if I can-- hang on.
No, I think I know what you mean.
Let me see.
I want to pull up the lecture notes.
I think I'm finding the right thing.
Here we are.
So in the lecture notes, there was a thing that looked like this.
Is that what you're talking about?
Yeah.
So the top set are the stress strain curves.
So those are OK. You do a compression test, you measure that.
Then, the middle set, you take, say, for one density, for one stress strain curve-- say that's your curve, the middle one there, 0.03.
Say you loaded it up to some point, or say you looked at some point here, on the curve.
You would figure out, for that stress, what's the area under the curve up to that stress.
So you'd have a stress and an area under the curve up to that stress.
And you could then-- say we know what this foam is.
It's-- I don't know-- a polyurethane or something.
So say we know Es.
Then we could divide those two numbers by Es.
And we would plot that one thing.
Let me just walk over here.
So this is the 0.03.
So if it's in the linear elastic reading, it would be somewhere in here.
Then I would scoot along, say, to here someplace.
I would do a whole bunch of points all the way along there.
And at every point, I would say, what's the stress, and what's the energy absorbed up to that stress.
OK?
And then I would plot-- doot, doot, doot, doot, doot-- up here, I would plot all those points.
OK?
And then when I got to this part here, that corresponds to that part over there.
OK?
Are we all good with that?
OK.
So then we repeat-- so we get one curve on the middle chart.
Then, for the different densities and the different stress strain curves, we plot a different curve for each of the different densities doing the same process.
And the thing we notice is that these points here are really the optimum point.
Because at that point there, you absorb as much energy as you possibly can for that stress.
OK?
And we notice, happily, that those points lie on a line, basically, on a straight line.
And we tick of what the different densities are-- so 0.01, 0.03, 0.1, 0.3.
OK?
And that line there, and all of these stress strain curves, and all these lines here, curves here, they all correspond to one strain rate.
All right?
So now I could take that line there for that first strain rate, and it would be one of these thinner lines here for a particular strain rate.
And I would mark off-- just the same as I've got here, 0.01, 0.03-- I'd go 0.01, 0.03, 0.1, 0.3.
All right?
And then I would repeat this whole process again for a different strain rate.
So I'd go back to doing some mechanical tests at, now, a new strain rate.
And typically, the new strain rate is going to be bigger or smaller by a factor of 10 or so, because you're not going to see much difference in the behavior unless you get big changes in the strain rate.
So you're going to change the strain rate significantly.
You get a new series of these stress strain curves.
Then you get a similar-- very similar, but not quite the same-- series of these curves here.
And what you'd find is you'd have another line here that would be offset a little from the first one.
And the positions of where these densities were would also be offset a little from the first one.
And then you'd draw the second one.
So the second strain rate line would go here.
And the little density positions would be offset a little bit, and you would mark them off.
And you basically repeat it for different strain rates, and then you build this thing up.
And then the density lines connect too.
Is that OK?
So that's how you generate that.
So the idea is, this bottom diagram is really summarizing all of those shoulder points.
The bottom diagram is really summarizing all of these points where the stress starts to scoot up at the densification machine.
Yeah?
So in question 3, we were constructing the graph in the middle for this particular one.
What confused me is that on the graph, the variable on the x-axis is stress, whereas in the question, we were given the stress.
So I drew something that looked as it should have looked.
And it turned out to be right, but I'm very confused as to why
OK. Let me try and get rid of that.
And I'm going to have to remind myself what the question is again.
OK.
So we have an open filled aluminum foam.
I asked you to write the equations for each of the different regimes.
And those were straight out of the notes, I think.
Right?
Yeah, I think the three inputs were different relative densities
Yeah, and then you had to construct the energy absorption curve based on those equations.
And I give you three relative densities.
So this was my solution.
So all this stuff here, I think, was straight out of the notes.
So there was the-- oops, let me back up so we start at the beginning.
So there was the linear elastic part.
There was the stress plateau.
That was just as it starts to densify.
And then, there's that line joining up the points.
So you were OK with all of that?
Yeah.
No, the only part that confused me is because once I simplified-- I inputted all of what we were given such that I had equations in terms of the relative density so I could just apply it to the different ones given.
But then I wasn't very sure how-- because I got constants, I wasn't really sure how-- the slopes should look like, because they were constants
So you got-- I'm not sure what you mean about you got constants.
So what I did was I said, well, I know that the diagram has to have this basic shape, right?
In the first part here is where it's linear elastic.
And this part here is the stress plateau.
And that part there is the densification.
OK?
And I said, well, if I can find these two points that correspond to the change between the linear elastic part and the stress plateau and the point that corresponds to the stress plateau and the densification, then I've got the diagram.
Right?
OK?
Yeah, I think that--
Oh, can I stop talking now?
I mean, if you wanted to--
Well, it's for you.
So you're OK?
I think
You
So what confused me about this question was that in order to find the two points, you'd need to know the strain at which that [INAUDIBLE].
So I wasn't sure how you would find that strain
Yeah, I think I didn't tell you.
Let's see.
Well, let's see.
Do you have to have the strain?
You have that.
I think if you have-- I think you don't have to have it in terms of the strain.
I suppose you could put it in terms of the strain.
I mean, that would be a different way to do it
But isn't the equation for the stress plateau already in terms of the strain?
Yeah, but this assumes that the stress plateau is just perfectly flat, right?
So this thing here would be useful, the strain at which the plateau starts.
But I think that's the same as saying-- that's the same point as saying you're at that point there, because that's where the plateau starts.
OK?
So I think there's different ways you could try to approach this.
But I think-- let's see, I'm just trying to remember what I did here.
Yeah, so what I did here-- so to get point a, I said, well, this is the equation for the linear elastic bit, right?
And this is the equation for the stress plateau.
Where I'm at point a here, this stress is the stress plateau.
So that's why I've put sigma star plastic there.
And then I said, this is-- the sigma star plastic is-- and this works out to some number here, some number times the relative density to the 3/2 power.
And then I just made this little table here, where I said, OK, these are the densities I need to get the curve for.
This is going to be my sigma star plastic for those densities.
And from that, then I can get this column here.
It's basically just that equation with this substituted in.
And do you see that that corresponds to the point a?
So I mean, you could do it by figuring out what that strain is and then figuring out all the points along that line up to that strain.
But you don't have to do it that way.
Do you know what I'm saying?
So you're saying-- if I had-- so say I have my idealized stress strain curve like that, right?
You're saying, well, I had to know that strain there so that I know where this stops, right?
And I'm saying that corres-- and you could do it this way if you wanted to.
You could say that this is the energy absorbed up to that point.
And if you know that the modulus goes as the relative density squared times Es, and you know this, if you know those two things, you can figure out what that strain is.
So you could do it that way if you wanted to, but I just did it a slightly different way
Yeah, so I don't understand how you just were able to get rid of the epsilon minus epsilon on that curve
Oh, let's see.
So I think, in the first part, where it's linear elastic, I just go up to epsilon 0, right?
So this part here doesn't really involve the epsilon, because I've gotten rid of it by taking the square of the stress.
And then here, the other point I'm looking at is this point here.
And I've assumed that epsilon 0 is much smaller than epsilon d, and I've ignored it.
And that's, I think, how I did it in the notes in the class.
OK?
So when you get to that point, this strain here is usually a few percent at most.
And this strain here is typically 80% or 90%.
So it's pretty common to do that.
OK?
Other questions?
So don't forget, the test covers everything from the thermal conductivity of the foams, it covers the stuff on trabecular bone, sandwich panels, and then, the energy absorption stuff.
Yeah?
Are you a little exhausted already?
Yeah.
So I'm guessing that this week and next week, you have everything due.
You have papers, and projects, and tests.
Do you have very many exams left, like final exams?
Yeah, you've got finals?
[INAUDIBLE]
Oh.
All right.
Anybody else have any questions?
Because I think we can just go and do other things if nobody has any other questions.
So the test is on-- so let's just review where we're at for the rest of the term.
So the test's on Wednesday.
And you can bring one cheat sheet, but I am not giving you all those equations with honey combs and foams.
So I think, on your cheat sheet, you would want to put Young's modulus of an open-celled foam, shear modulus of an open-celled foam, compressive strength of an open-celled foam, shear strength of an open-celled foam.
But I don't think there's going to be anything more complicated than that.
And Monday, I was going to do the how I became a professor talk.
So if you've seen it and you don't want to see it again, you're welcome to not come.
But for you guys, I just talk about how I got here.
And it's about my life.
It's not about cellular solids or anything.
And I don't know, you guys liked it.
You like it, don't you?
Yeah
Yeah.
Yeah.
So if you want to come, I'll do that.
And then Wednesday, I thought, I just need to collect the projects.
I wasn't going to do anything on Wednesday.
And so I've been thinking about the how I became a professor talk.
And I think-- so I have this idea that students would like to hear more of these talks from other faculty.
Would that be true?
Yes
Ah.
Because I've been in touch with Cindy Barnhart, and we're going to try and organize something for the fall.
So I'm going to approach some other professors-- not just in our department, across all of MIT-- and see if I can get other people to do the how I became a professor talk.
So you would like that?
OK, yeah.
So we'll see if we can make that happen.
The following content is provided by MIT OpenCourseWare under a Creative Commons license.
Additional information about our license and MIT OpenCourseWare in general is available at ocw.mit.edu
--not of the vindictive sort.
You skip class, you've skipped a lot of important stuff.
But I'll get you on the quiz, that's all.
Kidding aside, there were a number of things that I passed out.
I think nobody needs a set of problem set number 13.
That was the one that had something on symmetry constraints and working with second-rank tensors.
If anybody m