In the last class, we saw the Rayleigh-Ritz procedure. As I told you Rayleigh-Ritz procedure is a
forerunner tool in finite element analysis and I hope you understand. If there is any
question, I can answer that. Yeah; one of the questions that was asked is how is that
a0 became zero? Because, if we remember we started with u is equal to a0 plus a1x plus
a2x squared and so on. Why did you make a0 is equal to zero? In that procedure, it should
be such that the displacement boundary condition that has been given, which is an essential
boundary condition, should be satisfied by that particular polynomial. In other words,
at x is equal to zero, u has to become zero, which means that a0 becomes zero.
Now, we move over to the finite element analysis. What is the difference between finite element
analysis and whatever we did? The difference is very simple. In the Rayleigh-Ritz procedure
we considered the complete body. If you remember what we had put down, we had just put down
this and then we said that we will consider it as a complete body.
In Rayleigh-Ritz procedure, we had the body to be taken as if it is one element. But,
what we are going to do in finite element analysis is to split this body into number
of elements. In other words, we will proceed as if you are going to apply this procedure
to more number of elements. Is it clear? That is what we are going to do, in this case.
Before we proceed, let us look at how to write down the potential energy for a body. The
potential energy term is written down completely like this. You will see lot of terms here.
Have a look at this whole thing together and I will explain each of them very carefully;
each of them a very carefully. Now, look at those terms.
You have
strain energy term. Please note you are used to writing half sigma epsilon. Now, what I
have written? Essentially, for a multi axial case, I have written it in terms of a matrix
notation, epsilon transpose D epsilon. Our old E now takes the value or takes the shape
of D. Is it clear? So, the first term indicates in this particular expression, the popular
strain energy term you all know. Look at the second term. You have now, a new epsilon zero,
a new term called epsilon zero. In this case, what we have essentially done is to consider
a body with an initial strain and an initial stress.
Let us see what happens to a body when there is an initial strain and an initial stress?
Epsilon zero is an initial strain and sigma0, which comes to the next term, is the initial
stress. Is that clear or other words, what does it mean? It means that the body has some
strain due to, say, assembly or fitting or whatever it is, some initial strain. It can
be due to, say thermal stresses, this kind of thing can be present. So, sigma0 and epsilon
zero indicate the state of the body, when you start loading them. Or in other words,
they are something external; due to something external that has happened to the body.
Now, look at this term sigma transpose epsilon zero. Look at the second term sigma transpose
epsilon zero and look at this third term epsilon zero transpose sigma or these two terms are
sorry, I think it should be sigma0 transpose or epsilon transpose sigma0; sorry.
Yeah; because epsilon zero and sigma0, I have just repeated that. Look at these two terms.
What does it mean? It means that when I start loading the body, because of the pre existence
of the strains and the stress, they start doing work and hence these two, there are
minus signs. These two terms come from initial strain and the initial stress. Is that clear?
What is the next term? What is the next term? Due to body forces? Please note these two
are different from body forces. In the previous problem itself, you would have noticed that
what we were specifying as, what did we specify, Cx? C is force per length squared and so on,
was nothing but a sort of a body force that we were specifying. We are specifying throughout
the volume or throughout the length of the body.
Now, what is this term, here? What is this term here? That term is due to surface forces.
It is the work done due to the surface forces. So, you would see all those minuses there.
What does it mean? It means that the potential for these external loads to do work is now
getting lost. Is it clear and look at that last term. Last term, I have just added it
now, specifically; may be you would not have seen it. I have just added it here, because
we are going to also meet concentrated loads in a finite element situation. There are going
to be nodes and you may put a concentrated load at a node. That is why, I have put that.
The last term, I have added it. The last term depicts the work done by concentrated load.
This completes the definition of potential energy for a body, when it has initial strain
and initial stress, surface forces, body forces, concentrated loads and apart from an elastic
type of constitutive equation. What is my goal? My goal is to now develop or reduce
this into an algebraic set of equations. How am I going to do it? Let us see how we are
going to do that? Please note that in my previous derivation on Rayleigh-Ritz procedure, what
is that I did? I had put down a degree of freedom as a1 and a2 and so on. Now, what
I am going to do right away, is to swap this a1 and a2 with u1 and u2 or ui and uj, to
be more general. Is that clear? How am I going to do that?
The first thing that I am going to do is to take an element, say ij. Let us take that
element ij. Let us say that the length of this element is equal to L. Now, I am going
to put a local coordinate system to it. Let me put a local coordinate system, such that
it is depicted by s. What does it mean? s is equal to zero at i and s is equal to L
at j. Is that clear?
Let me now write down an expression for u. How do I write an expression for u? u is equal
to a1 plus a2s. Now, what we are going to do is to replace a1 and a2 in terms of ui
and uj; ui and uj, within quotes, known values or in other words, what we are going to solve?
ui and uj are the ones which we are going to solve. You already know a procedure for
solving them and so on. So, I am going to replace a1 and a2 by means of u1 and u2. Till
now the procedure is the same. Please note here that I am not going to put a1 is equal
to zero, because there is no boundary condition here. We are treating a separate element,
we are going to put the boundary condition later only. Is that clear? So, I am going
to put first u is equal to a1 plus a2. Can you now substitute ui and uj and give me an
expression for u? I will give you one minute, so, let us see whether you do it fast?
Let us see how you are going to express u in terms of ui and uj? What is that you are
going to do? You are going to put s is equal to zero, u is equal to ui. s is equal to L,
s is equal to L, what happens? u is equal to uj. So, you can express this as 1 minus
s by L into ui plus s by L into uj. Is that clear? Let me write that as N ui uj. What
is N? N is nothing but 1 minus s by L; s by L.
So, if you want, this N is what is called as the shape function. Note this carefully
that N is called as the shape function. People call this also as interpolation function,
basis function and so on. Essentially they are, either I mean, They are called by one
of these names, but essentially they are the most important input for finite element analysis.
Is it clear? Have a look at that and see whether you can straight away get one or two properties
from it; 1 minus s L and s by L. Straight away, you can see one or two things. What
are they?
They are nothing but an interpolation function, because it interpolates u by or getting the
values of ui and uj. That is why they are called as interpolation functions. Is that
clear and one more thing you can straight away notice there. What is it? That sigma
of N is equal to 1. If I call this as N1 and N2, then sigma of N that is N1 plus N2 is
equal to 1 and what is the other thing that you can see from this? What happens when s
is equal to zero? N is equal to N1, N2 is equal to zero. So, 1 minus s by L and s by
L is such that N1 becomes, sorry, N1 becomes 1 at s equal to zero and N2 becomes, sorry,
N1 is equal to 1 at s is equal to zero and N2 becomes zero at s is equal to zero and
vice versa.
When s is equal to L, the first fellow goes off and second fellow remains. Is it clear,
any questions on this? Now let me write down, u as Nd. I can write down the same procedure
by means of a much more complicated matrix notation and so on. Because I am just introducing
this topic, I have written down like that. This kind of moving from a basis to this N
basis can be done for much more complex case as well. Is that clear? Have a look at that
first expression, u is equal to a1 plus a2s; u is equal to a1 plus a2s.
Yesterday, if you remember, that we had put a2s plus say a3s squared. Suppose, I put a1
plus a2s plus a3s squared, what am I going to do now? Yes, pardon? Third node, in other
words, I cannot swap now with two nodes. So, I have to put one more node. Now I have, u
is equal to ui, uj and uk or u1 u2 and u3, however you want to call it. You can now have
three nodes and three coefficients, which can be swapped. What does it mean? It means
that as you increase the degrees of the polynomial, I will also have to increase the number of
nodes which go to define the element. Is it clear and these elements, where I have used
higher order or higher order polynomials, are called as higher order elements. If I
now have to do this, like replace a1 a2 and a3 by means of u1 u2 and u3, if I keep on
doing this, this procedure becomes bit difficult, because I have to sort out or solve these
equations like this. So, we are going to work out a better procedure to do it.
We are going to look at some well known interpolation functions later in the day or may be in the
next class, we are going to look at what are called as Lagrange interpolation functions.
Is that clear? Probably you would have already heard about Lagrange interpolation functions
and Hermite interpolation functions and so on. We will look at that Lagrange interpolation,
later in the class. Right now, let us say that we can write down u in terms of N and
d. This is one dimensional problem.
But if I happen to have a 2D problem with, say for example, an element like this, then
how many degrees of freedom I am going to have? u and p, in which case I have to write
to down u as like that which depicts u and v and what is going to happen now to d? How
many entries did I have here? I had only entries, but when I now shift to a two dimensional
case, how many d’s I will have? Is it 4? Look at that in 1 2 3 4 and each of this nodes
will have 2, correct. So, I will have 8 entries.
N will correspondingly change. So, in general I can write down u is equal to N d. Is it
clear, any question? In general I can What is this N? Do not worry about it now. What
is this N for this 2D case? Let us not worry about it. I can get you that very easily.
It is not difficult, but right now, accept that u can be written down in terms of Nd
or in other words accept that shape functions can be determined for different types of elements.
When things become very complicated, we will move away from this very simple thing. We
are going to move away, we are going to talk about what are called as iso parametric element
and so on. Let us not worry about it right now. But right now, we will accept that u
is equal to Nd can be written.
What is my next step? Just think for a minute and tell me, what is my next step? What do
you think I should do as the next step? In order to do that, have a look at this expression
here. Have a look at this big expression here and tell me what is it that I should get from
u is equal to Nd?
Epsilon, right, beautiful! So, you are very logically looking at things; epsilon. Fantastic,
that is what I wanted; so, epsilon can be written in terms of my displacements. So,
I am going to get strain displacement relationship. Remember, how did we write it? In terms of
u, previously we had written it epsilon11, epsilon22, epsilon12 and so on or gamma12;
we had all those things set up already. Just a second, we will write that term. Let us
follow a notation and say that u, by the way you have to tell me which is that you are
going to differentiate? Nd is there, because you have to apply differentiation.Yeah, that
is what I was waiting for. Is that d? What is d? What does d contain?
Yeah, they are constants, they are in terms of degrees of freedom. Say for example, I
am going to differentiate this element with respect to s, which is the local coordinate
system, which is something like x. Epsilon I have to differentiate with respect to x,
so it will be in terms of N.
Let me call this as epsilon is equal to some dow term Nd. What is this dow matrix, can
you think about it? What is this epsilon, I mean, sorry, I think I have to put like
that. For 2D, for example, epsilon is epsilon11 epsilon22 gamma12 and so on. You can keep
expanding epsilon for 3D as well. What is this matrix, dow matrix? No; that is beautiful.
So, that is it. So, that matrix is dow by dow x1 zero zero and so on. Can you fill it
up? The next line will be zero dow by dow x2, then dow by dow x2 and dow by dow x1.
That is now multiplied or this operator, you can say that, dow is an operator. Now, it
operates on N; it operates on N. Is that clear? Now, I am going to ask you one more question.
Just think about it. Even though you do not worry about what is inside N, you have to
tell me what is the size of dow N and d? You have to tell me what is the size of dow N
and d for a 2D case, let us say 2D case. Look at this.
You have to tell me, what are the sizes of that? What is the size of dow N and d? What
do you think? Dow-2 by 2; 2 by 2 or what? 3 by 2 and then N? N 2 by, common, 2 by 3,
2 by 4, 2 by 8; I get all types of answers. See, it is very simple, it is very simple.
Look at d. d is what? What is d? 8 by 1; common, that is it. 8 by 1. So, what is N? 2 by 8,
that is it; 2 by 8. What is that ultimately you will get? How do you get it? How do you
see it? 3 by 1, so, that is what is epsilon?
Yeah, just have a look at that. You should not get confused. This is very, very important
to understand what the sizes of this matrix are, because when we assemble, when we put
them together, it is the sizes which are going to confuse you. That is why I am right now
emphasizing what should be the sizes of these matrices? Having understood that, having understood
that let me call dow N as a B matrix. Let me call this dow N as a B matrix, so that
epsilon can be written as B into d. This is the most important part of the whole finite
element, so, I want you to concentrate. If you have any doubts, just ask me. Yeah, so,
I can write epsilon is equal to Bd.
What is my next logical step? My friend, yeah, he said what is my next logical step? What
is that I should get? What is the next logical step? What is that I should look for? Stress
strain matrix, beautiful. We have already derived that. That is why I did all those
things initially. We know everything now. I already know what this D matrix is. So,
I can write down; that I have already written down, sigma is equal to, I am going to repeat
that sigma is equal to, D epsilon. In our earlier classes we have spent lot of time
looking at D. So, I will just write down sigma is equal to D epsilon.
So, I can say that sigma is equal to DB into D and so on. Any question? Now, having got
all these things, what am I going to do? I am going to substitute them into this expression.
I am going to substitute all of them into this expression.
I would like you to do it, because it is straightforward and gives you an exercise. I will give you
one minute to write this term, just attempt it. What is that you should know that a b
transpose is equal to? What is ab transpose? b transpose a transpose, that is all you should
know. Write it down, I will give you a minute. Have a look at this expression, this expression
and then write it down and after a minute let me hear what you have done. This will
give you a good exercise to remember each of these things, just put that down. Let us
see, have a look at this. Look at each of the terms and replace. All the while think
about the size, for example, for a 2D problem. By the way notice that piP is a scalar quantity;
piP is a scalar quantity.
Do not look at what I am going to write now, because I would like you to write it. I am
going to write the final result, but have a look at this and you keep writing it so
that you will get a feel of what is happening. So, all of you have written the first term?
Is that ok? Minus, epsilon transpose, what is epsilon, sigma transpose epsilon zero;
okay what is sigma transpose? Yeah, epsilon transpose d transpose. So, what is epsilon?
I want you to substitute till the end. Yeah, d transpose; see, all of them are matrices.
I am not putting everything as a matrix, you know they are matrices. So, d transpose, B
transpose, capital D transpose. Is it D transpose? B transpose D transpose yeah, so, what is
the next term? Minus, where is that? Yeah, still you are writing? d transpose B transpose
sigma0 and then next term, minus v d transpose N transpose F dV that is the next term. That is for what? For
body forces. Then for surface forces, d transpose N transpose phi sorry ds; yeah any questions?
Lastly, I have that D transpose P; minus D transpose, minus D transpose P. Just to make
it simplified, I am not going to enter or I am not going to carry on those two terms
that epsilon zero and sigma0 terms. It is very easy to include them, because I have
to write down big expressions. You can very easily take it down, so, I am going to just
remove that for the time being. If you want, you can always add it to make it much simpler.
Now, I am going to define a stiffness matrix as integral B transpose DB dV. Let me define
K, stiffness matrix of an element to be integral B transpose DB. Is that clear and this is
for a general body. But, in the sense that is for an element as well or for the whole
body; you can look at it either way. But, since I have written down this last term here,
it is for the whole body which has been already discretized.
Is that clear? Now, I am going to do one more thing. This small d’s here are actually,
there is a small problem; the small d’s here are for the element degree of freedom.
I am going to expand them actually. That is why I am very careful in writing the last
term. They do not go with this, so, there is always a trouble, as to understand that.
What I am going to do is to expand that small d to the whole of the matrix. In other words,
if I have an element which run say from 1 and 2 nodes, I am going to say I will write
as if it is 1 2 and other things being zero up to say 20 elements, if the problem has
20 elements.
I am going to expand them. Why is that I am going to expand? If I expand like that, it
is easy to add them, right now. But, computer implementations are not like that. They are
quite different. So, let us not worry about that at this moment and so, let us see how
we can write down this particular case or in other words what is that we are doing?
We are expanding or in other words, we are substituting for small d by capital D; substituting
small d by capital D. Capital D indicates a vector which covers all the nodes in the
body, which covers all the nodes in the body. Is that clear?
Write down the expression. Let us see, how you can write down that expression? Define
ke, ke, as integral V B transpose DB dV and when I expand it to D, the total piP of the
whole system can be written in terms of capital D. So, I can write that D transpose D as a
contribution of this element strain energy. Of course, half is there, half; as a contribution
of strain energy of one element to the total system. Is that clear? No? Let me repeat that.
Have a look at this again. What is this small d’s? What is this small d’s? They are
for element. In other words, what it means is each element now contributes for piP. In
strict terms, this piP, what I have written is valid only for an element. But, I have
just added this term. This is not right to do it in the last term. Please note this is
not the right way of doing it, but I just wanted to remind you that there is the last
term there. There is one more contribution from concentrated loads.
I would say that I have just added it more symbolically there. For the time being you
forget it. If there is confusion, forget it for the time being. We will add it in the
next step. I said that I have added it very symbolically. Each of the element contributes
to strain energy and each of the element contributes to potential lost. What is that I should do,
if I want to find out the complete piP of the system? I have to add, add because each
of the elements I have already divided the whole body into number of elements. I have
to add the contribution of each of the elements as far the strain energy is concerned, as
far the potential lost due to internal loads or in other words sorry, due to body loads
or body forces and so on.
In order to do that, in order to add, I cannot add straight away because my d, small d is
there, sitting there. They are different. The first element may have 1 and 2 nodes,
the second element may have 2 and 3 nodes, third element 3 and 4 nodes and so on. So,
how do I now add? Tell me a procedure how to add it? The best thing is to expand that
small d to the whole of capital D, so that others will be zero. 3916I will just give
you a minute to understand this. What I am going to do is to put down my old problem,
you know the old problem and write it down in terms of the entire degree of freedom.
What is that old problem?
Say for example, let me say that I have 4 nodes. I have 4 nodes; 1 2 3 and 4 nodes.
I have an element 1, which connects 1 and 2, element 2 which connects 2 and 3 and element
3 which connects 3 and 4. I have these three guys sitting there. Element 1 has k1, say
for example, k1 minus k1 minus k1 and k1 and the corresponding displacements are u1 and
u2. I am saying u1 u2, make it u1 u2 u3 u4. Make it u1 u2; so, how will the k change now?
That is what I want to ask you. Just a minute, write it down. Try it. It is very, very simple.
I mean, very obvious question but I know that it is not very easy to follow. You please
try it and write it down. I will write it down in a minute what exactly I mean by that?
Have a look at that and then have a look at this and then write it down in terms of, let
us say that, this is the element 1. It is very straight forward question. What I am
saying is that the k matrix will have now 1 2 3 4; u1 u2 u3 and u4.
What will happen to the k matrix? I am expanding it. What will happen to the k matrix? Zero,
that is it. You know that is what I have been looking for. It took you quite a while to
find out. So k1 minus k1 zero zero, minus k1 k1 zero zero and so on. What is the advantage
now? The advantage is that if I write the second element also in the same sense, how
do I write it? Zero zero zero zero, first row zero zero zero zero. Then, second row
becomes zero k2 minus k2 zero and so on. Then these are, straight away they can be added.
They are matrix of the same size and not only that, this u’s that are sitting there, they
are the same u’s. So, u can be taken out and I can add them. Is that clear? No, still
not clear?
What I have done is to facilitate adding of k. What I have done is to facilitate the adding
of k or else I cannot add k straight away, because this k1 corresponds to this 1 and
2; k1’s or k’s corresponds to 1 and 2, the k2 corresponds to 2 and 3. So, I cannot
add k1 and k2 straight away, unless and until I expand it like this. Though it is not an
efficient way of doing things in a computer, let me make it very clear, this is not the
way commercial code is written. Right now, as a student it is easy to understand what
is happening. There are different ways of doing it. We are not looking at computer implementation
straight away. We are not looking at computer implementation. We are only looking at how
to do things, how things are done. Hence this is a nice way of looking at this. Now, get
back to this, get back to this question.
I have already defined ke and u. This is not ke, that is the u, the contribution of u of
an element. Now, complete this thing. I am making you write because it is easier to follow
when you start thinking and writing them. I will give you one minute, just write down
or convert it and write down. I want something like this. I want ultimately K into d is equal
to R. This is the expression I want, KD is equal to R. How are you going to write it
down? That is the question. Just try it for a minute. I will give you all the clues that
are necessary. One is, ke can be summed up or k can be sum of ke’s, capital K. Let
us call that as capital K. Capital K is equal to sum of ke’s. Then you look at also re’s.
What is re? What is re? re is the other term, say for example, D; D transpose, take it out.
So, re is equal to minus N transpose F dV and minus N transpose phi ds.
Is that clear? In other words, what is that you are going to get ultimately? Yeah, write
it down, you have another 2 minutes, just write it down. I will do that. I will write
it down, but just see how you are going to write down. I can write it down as half capital
D transpose KD minus D transpose re. What is the next term I will have or R rather minus
D transpose P will be my piP.
That is the total piP is equal to half D transpose KD. This K is the sum of sigma of ke’s minus
D transpose R. This R is sigma of re minus D transpose P. That will be piP. K is equal
to sigma of ke. This is the matrix addition, because I have already expanded K in terms
of complete D and R is equal to sigma of re’s. Is that clear? Is that clear and the last
term now I have brought in because, my piP now should include the contribution due to
concentrated loads. Please remember what re is. re is nothing but N transpose FdV N transpose
phi ds that is re. In other words, this will come out and that will come out.
What is my next step, what is my next step? Yes, that is it. So, dpiP by d D should be
equal to zero, independently. In other words, dow piP dow Di2 or i, where i is equal to
1 2 and so on, should be equal to zero. What is it that I am going to get? A matrix, correct;
but, what is that? Half D transpose KD, when you differentiate it with respect to D, becomes
KD. You can work it by hand, KD.
The first term now, after I differentiate this becomes KD minus R minus P and that is
equal to zero. So, KD is equal to R plus P. That is my final expression. KD is equal to
R plus P. K comes from small ke, which is, what is it? Integral v transpose DB dV.
Yeah, 1. I think only small thing is that, see usually, I think there is a small error
here. That sigma0 is usually replaced by means of, I think that is the only mistake you did
when you expanded, by D into epsilon zero. That is how you expand.
I think, you said d transpose. That is what I was surprised actually you write sigma0
as D epsilon zero. Did you get it correctly? D epsilon zero and this is equal to sigma
transpose epsilon zero. Anyway, I mean, this is very straight forward. If you understand
sigma transpose epsilon zero and if you put it as sigma0 D epsilon zero, I will explain
this in the next class. What it really means is any strain, initial strain, you can talk
in terms of initial stress or vice versa. We will discuss this in a later class. I hope
you understand this. Any questions, any questions on it? Please go through it, if there are
any questions, I will answer it in the next class.