In the last class, we saw the potential energy theorem. We actually derived it by using variation
of principles. We also in that process saw the principal of virtual work and we said
that both of them are the same, in the sense that, we can get solution of a problem either
by going through this potential energy theorem or through the virtual work principal. I hope,
you all remember. On one hand, we derived potential energy theorem from the virtual
work principle. On the other hand, we also said that if I have to go through this route,
what is the assumption that we made? We said that strain energy should be defined or in
other words, as far as potential energy theorem is concerned, it is valid when its functional
exists; in a more mathematical term, the functional exist.
In this case, what is the functional? Pip is the functional. In other words, the potential
energy is called as the functional, because it is the function of the functions. This
functional exists, only when strain energy density function exists or in other words
for an elastic problem, we say that we can go through the route of the potential energy.
On the other hand if you look at elastoplastic problems, then we cannot derive the constitutive
equation through the strain energy density. Then we have to resort to the virtual work
principal or in other words, our starting point will not be the potential energy theorem,
but will be one step before it. That is the virtual work principle. We will see that later,
because we are going to derive somethings now. So, we will do that first and then come
back to this problem of defining or determining the stiffness matrix and in that process how
do we go about either from potential energy principal or from the virtual work principle.
We have to wait for, I mean that, that has to wait may be, till next class.
But, before we go ahead, let us look at the problem that we did or that we posed in the
last class. We said that we have to derive or we have to determine what this problem
is. I hope all of you remember this problem. This is one of our earliest problems that
we did. What is that we determined? We determined u2 and u3; all of you remember this problem?
Right. Now, my goal is to do the same problem by using potential energy principle.
You know what potential energy is. It is the strain energy minus the potential lost due
to external loads; due to the work done by the external loads. u minus say omega; let
me write that as u minus omega. Omega is the potential or minus omega is the potential
lost or in other words, in simple terms, it is just P into d. P is the load that is acting,
d is the displacement for that load. Can you help me now to determine u? u is equal to,
please note that I have already put for your sake 1, 2 and 3. There are two ways in which
we proceed; just a word of caution, two ways in which we can proceed. One is to look at
the boundary condition at this stage itself. We look at it like that and start writing
down, but for a bigger problem this may cause a bit of confusion. There is nothing wrong,
but it may cause a bit of confusion. Why, because actually the problem has it that there
is a reaction that acts at 1; there is a reaction that acts at that point or in other words,
the potential is also lost due to reaction. But, it so happens that the displacement at
that place is equal to zero and hence P into d term would b equal to zero. Is that clear?
But, nevertheless in order to complete the picture, in order to complete the picture,
it is a good practice to write it down completely along with the reaction. Help me in writing
that. So, that is equal to half into K1, one of you can say that; K1 into say u2 minus
u1 whole squared; u2 minus u1 whole squared plus half K2 into u3 minus u2 whole squared
minus P into u3 minus R into u1, where R is the reaction that happens to be present at
1. Is this clear? What is the next step? What are my degrees of freedom? u1, u2 and u3;
so, pip is the function of u1 u2 and u3.
What is that we do as a next step? Differentiate this with respect to ui’s. So dow pip by
dow ui is equal to zero. Is that clear? So, what happens to dow pip by dow u1? Yeah, work
it out, because it is a good practice. Instead of just following what I write, it is a good
practice; just you write it down. That will be half into K1 into, so, u1 squared plus
u2 squared minus 2u1 u2. The first term here will be 2u1; 2u1 minus, minus 2u1 u2, so minus
2u2. Is that clear? So, nothing will be present here and then we will have minus R; minus
R and that is equal to zero. Is it clear? Straight forward.
Now, write down what is dow pip by dow u2? dow pip by dow u2. What is it? Yeah, you do
that. You look at this here; look at this thing here and then differentiate it with
respect to u2. May be, I think it is better to write it here, so that you can see it here.
What will happen to the first term? Here also, u2 is there. So, differentiate it and then,
do the same thing with respect to u3 as well. Both of them are equal to zero. I will give
you 2 minutes. Just do that and look at the result and ultimately fill this up for me.
So, just do that you fill it up for me. So, what will you get? We will get K1, there minus
K1, zero; here minus K1 K1 plus K2, sorry minus K2 minus K2 and K2. Look at that expression
now. Is it clear? What is this? Go back and of course that is zero; go back and have a
look at what we did in your earlier classes? Is it exactly the same? In other words, what
is that you have done? You have got the same stiffness matrix, first from equilibrium equation;
remember that we got that from equilibrium equation. Now, you have got the same stiffness
matrix from potential energy theorem. What is here by the way?
R and minus P, because the directions are important; minus P, R and minus P, so on and
so you can work it out. P, R and P, rather R zero P; R zero and P and so the direction
of R automatically be determined from this. I am sure you can or in other words you can
leave the first row, calculate the u2 and u3 from the second and third row and you can
get this thing going. After this, I think I will leave this problem. I just wanted to
show you that the same problem that you did earlier with the equilibrium equation can
also be determined by means of this potential energy theorem. I would like to do another
problem. The other problem is slightly more complicated and I am going to introduce Rayleigh-Ritz
procedure today. Yeah, any question? Is that clear? Let us now look at another problem
and we will introduce what is called as a Rayleigh-Ritz procedure, before we go into
actually the finite element analysis.
Look at the problem, q is equal to Cx. C is equal to force per L squared. Determine the
displacement. Determine the displacement and the stress and the strain. That is the problem;
q is equal to Cx and C is equal to some force per L squared. Determine the displacement.
Now, you have to first write down pip. What is this displacement? Displacement is now
function of x. So, u is some function of x. Is it clear? Now, write down pip first, as
a first step write down pip. It consists of two terms; strain energy term as well as the
potential that is lost due to external loads. What is the strain energy in this case? You
know that, half sigma epsilon. What is sigma? E into epsilon.
What is epsilon? du by dx; now, this is integrated. This is per unit volume, so this is integrated
throughout the bar. I can ultimately write the first term of pip as half zero to L E
into u comma x whole squared into Adx. That is all; Adx. Is that clear, first term? What
is the second term? Minus integral zero to L u into what is now the force?
Cx into dx; Is that clear? So, force is u into Cx into dx. My whole job becomes slightly
more complicated here, because I want to find out now u as a function of x. Now, we will
follow what is called as Rayleigh-Ritz procedure, which actually is a precursor or a pre runner
to finite element analysis. What Rayleigh-Ritz did was to convert an accurate problem into
an approximate. In other words, he found out approximate solutions to this problem. What
he did was that, he introduced an approximate function for u, introduced an approximate
function for u.
For example a polynomial can be introduced for u. u is equal to say a0 plus a1x and a2x
squared and so on. It was found that this polynomial should satisfy the boundary condition.
In other words, u in this case happens to be a1x plus a2x squared plus so on. Is that
clear? So, he said that I will approximate u in terms of some polynomial function which
we say a1x plus a2x squared plus a3x cube and so on. That is the first step. Now, what
he did was to substitute this expression for u into the function pip. What happens when
I substitute like that? This is an approximate solution; remember that it is an approximate
solution. We will comment about the solution later in the class, but right now, let us
say that what we want to do is to approximate u in terms of some a1 a2 and so on.
What happens when I substitute this into this expression, into pip? pip will now become
a function of a1’s, sorry, a’s; that is a1 a2 a3 and so on. In other words, in the
previous expression or previous problem, I had piP in terms of u1 u2 and so on. Now,
I will have it in terms of a1 a2 and so on. In other words, a1 a2 a3 can be looked at
as a degree a freedom. That is correct. So, a1 a2 a3 can be looked at as degrees of freedom.
Is it clear? Yes, what is this degree of freedom? I know I can understand that you immediately
do not catch. There is no physical meaning to this degree of freedom. On the other hand
in my previous problem, you did not have any doubts, because the degree of freedom was
very clearly written as u2 u3 and so on, for which physical meaning was very clear to you.
On the other hand, here I am writing it in terms of a1 a2 and a3 and hence you may have
a doubt as to what this means and that is one of the reasons why, finite element method
modified this Rayleigh-Ritz procedure and introduced this concept of nodes and degree
of freedom, corresponding to the displacements of the nodes. Is it clear? So, as a first
step, let us see how to do the problem with, say, taking only one, u is equal to a1x; only
one term. We will do that with two terms as well, but before that let us look at the final
result. Just write it down at the corner of your note book, so that when you do this problem
with single or a1 alone or one term and then with two terms compare that with the final
result which I am going to give.
The final result can be written as u is equal to 6, sorry C divided by 6 AE into 3L squared
x minus x cube; 3L squared x minus x cube. So, that is the final solution or accurate
solution or analytical solution, by solving this problem analytically. Now, my goal is
to get an approximate solution. In other words, whatever I am talking about, equilibrium is
approximately satisfied; is not accurately satisfied at every point, but on an average
it is satisfied. So, the first thing is take u is equal to a1x, substitute it into this
expression and see what you get.
Let us see how to do that. Yeah, any questions, any doubts? Yeah; half into sigma epsilon.
What is sigma? E epsilon; so half E epsilon squared. What is epsilon? du by dx or u comma
x; What is du by dx is what I write as u comma x. So, E epsilon, sorry, E epsilon squared
becomes E u comma x squared. Is it clear? Yeah. Now, u is equal to a1x. That is what
I want you to work on, so that du by dx becomes a1. Is that clear? Now, you have to help me.
What is pip?
Half zero to L, E into a1 squared into dx minus, Adx minus the other terms. Now, you
fill it up. You know, I will just wait for a minute. I am not going to do this problem.
I am going to give you some five, ten minutes, let us see. What I want you to do is, I want
you to do the first problem with this a1 alone and then I want you to do the second problem
with a1 plus a 2x. Let us get the results for these two problems and then, we will see,
compare it with the final result. Before we proceed, just a minute; so, you will write
it in terms of pip. The next solution is very simple. If you want, I can write that also.
Zero to L, u is what? a1x into Cx dx.
It is a1x into Cx dx. So, pip is now function of a1. Clear? Any question? No; a1 is not
a constant. It is a degree of freedom. What it means is that, u is equal to a function
of a1 plus a2x and I am trying to find out what this distribution is? Please note this
carefully. It is not a constant, because whatever be the loading, I will write it in terms of
a1 plus a2x. This is the polynomial, irrespective of the loading condition. Is that clear? It
is the polynomial irrespective of the loading condition; a1 plus a2x and so on. Now, I will
find out that a1 and that a2 which will satisfy the potential energy theorem.
That is dpip by dai; dow pip the dow ai.
What is that? That is what I am trying to find out. So a1 a2 is something like a variable.
It is not that you can, it is a concept, whereas on the other hand, this form of the equation
a1 plus a2x or a0 plus a1x plus a2x squared plus a3x cubed and so on, that form is the
same. Usually polynomials are used; may be, sometimes sine and cos functions can also
be used but 99% of the case, it is the polynomial that is used. So, a1 a2 a3 and so on becomes
the degree of freedom. They are like u’s; u1 u2 u3 and so on. Is that clear? Now what
you do is, differentiate this. Then, calculate dow pip by dow ai or dow a1, in this case.
There is only one a; so, a1. dpip by da1 and calculate a1. That is the first step. Just
a minute; you know, before you proceed, let us look at the other one.
The next problem or next step in this problem is to compare this result with two a’s.
What happens if there are two a’s? That is a1 plus a2. u is equal to a1x plus a2x
squared, so that du by dx is equal to a1 plus 2 a2x. Substitute it here. Find out what the
results are and let us now compare the results with respect to one degree of freedom and
two degrees of freedom or in other words, a linear term and the quadratic term. We will
stop for 10 minutes and see whether you are able to work out this solution and then later,
one of you can come and work out this and then we will get some very interesting results
to compare with actual solution or analytical solutions. So, you had some 10 minutes time
to work out. I hope you would have done the problems, but let us look at the first problem.
For your sake let me work out or let me look at the first problem or the first part of
the problem. The first part we had said that u is given by means of a1x. Substituting it
here in this expression so that du by dx becomes a1 and hence epsilon x whole squared becomes
just a1 squared, I can write down pip in this format. Then, integrating it, so, this term
becomes, now this tern becomes just L and that term becomes, what is it? L cube by 3,
so that, ultimately a1 becomes CL squared by 3AE and note that dpip by da1 is equal
to zero and that is our initial condition for the potential energy theorem. Is that
clear?
Now let us look at the second part of the problem. The second part of the problem has
u to be defined in terms of a1x plus a2x squared or in other words what is that we have done?
We have taken more terms in the polynomial and all of you know that as I take more and
more terms in a polynomial I can fit, whatever be the order of u I can fit or whatever be
the curve, I can fit it closer and closer. Any curve can be represented by means of polynomials.
I go closer to the curve, as I take many more number of terms. That is the whole idea. You
are right now clear that, when I take more terms to u then, I will go closer to the solution.
We are going to see that in just a minute.
At this point of time compare u with, what you have got as the final solution or the
accurate solution or rather solution that has been arrived at from analytical means
and then compare this also, the stress and strain that you arrive at also, with respect
to the other thing or in other words, immediately you can realize, how do you calculate stress,
by the way?
So, this is equal to, u is equal to a1x. So, du by dx will give me epsilon x and that is
equal to a1 and that is equal to this expression. So, CL squared by 3AE gives me the, what is
that it gives? Strain multiplied by E gives me the stress. So, what does it mean? It means
that, now the stress, if I have taken, when I have taken the one term, is constant throughout
the bar, which is definitely rather an approximation.
Let us look at two term solution. Let us see how better this two term solution is and let
us see whether one of the students can come and workout a two term solution. Can you come?
We can work out the two term solution. You can bring your paper, whatever you have worked
out and look at that. Write boldly, yeah. You work it out; work it out as you have done;
the second one. Yeah; so, you can point out if there is any mistake. Is it correct? a1
squared; what happens to this term actually u comma x; u comma x is what? dow u by dow
x. What is that here? a1 plus a2x whole squared. What? Is he writing correctly? So, 4; so,
this is the second one. a1 a2x, so, the first term is right. What is it? 2, yeah, 2a2x;
not x2, yeah, that is it. Just x squared, yeah, I did not notice it, sorry. 2a2x squared,
so, second term is right. a1x plus a2x squared into C into xdx.
Now integrate this. First term becomes a1 squared into x. When I substitute for L, that
becomes a1L. Yeah, yeah; you left out a1 squared. Yeah; is that the second step? What is the
third step? Please note here that pip is now in terms of a1 a2 only, because we have already
integrated. So, you can clearly see that pip, like in the previous problem or previous discussion,
is a function of D1 and D2 and so on. Now, it is a function of a1 a2 and so on. So, obviously
what we have to do is to differentiate pip with respect to a. That is a1 and a2; dow
a1.
By 3; algebra can be quite complicated, you cannot leave out any term. Is that right?
2a1 L plus, this term comes from the 4a1 a2 L squared divided by 2; from that term you
get that. That is for a1. Similarly, you write down dow pip by dow a2; dow pip by dow a2.
Yeah, you can take this side dow pip by dow a2. Is that okay? Any one has any objections
to it? No algebra mistake; you can rub this off. So, we get now two equations in a1 and
a2. Just I think you can write down the final value of a1 and a2, you can solve for it.
Just get the final values of a1 and a2. You can write down the a1 a2 final values.
CL squared by 12AE; CL by 4AE, a1 is CL by, yeah by 4AE; CL by four AE and a1 is 7CL squared
by 12AE. How does this solution look like? Let us just draw approximately how this solution
looks like and see where we stand with respect to the actual solution? Let us see what
What is the actual solution? How many terms are there or what is the, I think, I had written
down already, I think I have written in a piece of paper. What is the solution? It is
cubic, in terms of x cube. First let us plot u versus x.
Let us say that that is the correct solution or in other words that is the solution. It
has been obtained from analytical means. What is the solution when we took only one term?
It is a linear solution and hence that is what if I plot it, that is what I would have
got and the second term solution is actually very, very close to the actual solution and
looks something like that. In other words, as I take more and more terms, as I take more
and more terms, what is that I do? I approach the actual solution.
Now, let us see how the stress plot looks like? Now, what would be the stress plot for
the first case that is for the actual solution? Will it be a straight line? What is the stress
plot for the actual case? It is in terms of x cube. So, it will be a quadratic. The solution
would look something like this, sigma versus x. Now, how would the first thing look like
or the first solution or first part of the solution? It would be a constant. So, it is
like this and the second one, what about the second one? It would be linear. That is something
like this. That is the case. Is it clear? Can someone Just look at it for a minute.
Can you comment on this graph? There are very interesting things there; there are very interesting
things.
Number one is that, yes, there is a constant; there is a linear variation, that is very
clear. But apart from that, there are other things as well. Can you look at that graph
and tell me? Just take a minute to look at it and tell me. Beautiful; so in other words
in other words That is a very good answer. In other words, what it means is that there
are points where the solution is very accurate. There are points in this particular bar, where
the solution happens to be very accurate. Later on you will see that in finite element
analysis also, there are points where the solutions are very accurate and these points
are called as super convergent points. That is a very interesting result. So, these two
are very interesting, but the point you may ask is how many terms do I take? Do I take
only one term, two terms or three terms and number two whether I can take only x cube
terms, from x cube terms onwards? What will happen and so on? These are some of the questions,
you may ask. So, there are some rules for writing down the displacement functions.
Number 1, the displacement function should be complete, should be complete. What does
it mean? It means that if I take a higher order term, x squared term, then I should
also take the linear term and so on. Why, because for example, if I had started this
whole exercise of writing u in terms of x power 4, if I had taken this in terms of x
power 4, what would have happened? Beautiful; so, we would not have got the solution at
all. It is important that we have a complete series. If I take x power 4, I should also
have x terminate, x squared terminate, x cube terminate and then only x power 4 terms; so,
it should be complete in that sense.
Yes, I can leave out, after a1 and a2 I can leave out other terms, it does not matter.
But I cannot go to the end or n? terms like, after cube or terms with higher order terms
and then leave out lower order terms. Number 2, the u should be continuous. Whatever I
choose, the u should be continuous and should be single valued, or in other words then only
it will give you compatible results. It should be a single valued function; that is important.
We are going to expand these conditions later, when we go to the second degree and so on.
But right now, it is important to realize that this is the one these are the two important
conditions. Now, let us again look back this curve, look at this curve. Now, look at this
curve.
There is one more thing that you would notice from this curve. What is that you notice?
That the displacement that we have arrived at is lower than the actual displacement;
the displacement that we have arrived at is lower than the actual displacement. In other
words, what does it mean, physically? No, what it means physically? Physically it means,
yeah, that is it. Physically it means that the stiffness is more. The stiffness is more,
when? When is that? When I restrict the degree of freedom; see, obviously in a one term solution,
I had only that a1 fellow. On the other hand, when I make it two terms, a1 and a2, I had
two chaps to really tell me, how displacement takes place? So, lesser the degree of freedom,
more stiff it becomes and hence the displacement terms are smaller or displacements are smaller.
Another thing you should notice very carefully between the two graphs. What is that, between
this graph and this graph? What is that you notice between the two?
Please notice that displacements are much more accurate than the strains and the stresses.
Notice that displacements are much more accurate than strain and stress. This is the same as
stress or strain ….. the same; only thing is I have multiplied by E. But, in other words
u is much more accurate than its derivative and notice one more thing carefully. These
observations are very important because we are going to extend this same observations
into finite element analysis, which is nothing but a small extension of the Rayleigh-Ritz
principle.
One more thing can you notice from the stress graph. You see that at the end, stress is
not zero, as is shown in other cases. In other words, stress boundary conditions will not
be exactly satisfied but will also not be accurately satisfied, but also be approximately
satisfied. You can look at Rayleigh-Ritz procedure, what we have done, in another sense; in the
sense that we do Rayleigh-Ritz procedure by assuming as if the whole body is one element.
Now, finite element can be looked at as a Rayleigh-Ritz procedure with the difference
that we split up the body into a finite number of elements and start writing down this kind
of equations, for every element.
There is quite a unity between finite element and this Rayleigh-Ritz procedure, because
both of them are approximate, both of them try to write down the displacement conditions
in terms of certain other factors like a1 or u1 and so on and both of them rely on,
say, for example potential energy minimization or potential energy principle. Is that clear?
We will start the derivation for finite element analysis from potential energy principle in
the next class. But, meanwhile if there is any question, I can answer. Is this procedure
clear? What we are going to do is to extend this procedure and apply it to a number of
elements. I know that the criticism, as someone said that, a1 a2, what is a1 a2 a3? That is
not very clear. That is exactly what we are going to address, when we go to finite element
or in other words what we are going to do is to replace the degrees of freedom a1 a2
a3, which does not seem to have any physical meaning by means of degrees of freedom of
the nodes, which seems to have much more or which looks like it has a much more physical
sense rather than a’s. How are we going to derive this? We will see that in the next
class.