Is there any question on what we did in the portions which we covered in the last class? No. I
hope now all of you know how to deal with isoparametric elements, but we have to continue
a bit because we have not completed isoparametric elements. Quite a bit of things have to be
done before we close this chapter; may be next two to three classes we have to do and
this is very important. As I have been telling you that this is a very important element;
isoparametric element is an important element, because this is a concept. Isoparametric element
is a concept. It can be used for a variety of elements. In fact, after this, may be next
two classes I will derive an element for heat transfer which I gave as a problem in the
test; this will also be derived. In other words, we will look at those problems as we
go along; it will be solved as we go along.
But, what I want to emphasize is the fact that it is very important that we understand
isoparametric elements. We will look at, when we even interpret the results, then also we
would be interested in the theory of isoparametric elements. In other words from further on,
from now on, the course is going to go something like this. I am going to teach isoparametric
element and finish them, next three classes. Then I will deal with beam elements. I am
not going to go into the details of plates and shells because it requires lot more theory
and we do not have the time to cover in a first course. Nevertheless, I am just going
to indicate the type of the shell element and axisymmetric shell, what it means, what
are the types of, say, degrees of freedom and so on. I am going to just talk about that
for a short time.
I will not be in a position to derive them, because that requires lot of background and
also time. Once we complete that, then we will take a break to look at how to interpret
results. I am not going to cover in this course solution techniques. Finite element is a very
vast field. In other words, kd is equal to r or f; that is only one part of it. Then,
I have to solve, ……, apply boundary conditions and solve them. All of you have been solving
this boundary, this kind of problems, small things which I had given you by using Gauss
elimination. That is the technique many people still use, but there are lot more advanced
techniques which make things faster, what are called as preconditioned conjugate gradient
methods, relaxation techniques and so on. Many things are there, but we are not going
to cover all these things in this course. Again that requires a big background in linear
algebra and so on, so we are not going to cover that. But nevertheless, if you know
Gauss elimination which you would have studied in eighth standard that is enough to understand
that you have to solve this problem and enough to go further, because if at all you are going
to use a package, you are not going to solve this. But, if at all you want to write a package
and then with this knowledge you can always pickup as to how to solve problem. They are
all readily available today.
It is a specialized area. Again, I am not going to cover that, but what am I going to
do, after looking at the niceties of results and niceties of interpretation and so on,
we are going to go to an introduction to nonlinear finite element analysis. How exactly nonlinear
finite element analysis is done that is what I am going to talk or in other words algorithms
that are used. We are going to talk specifically about what are called as increments, iteration
and so on. In that respect, I am going to bring in the formulations that are used for
solving manufacturing problems. There are number of formulations that are used. It is
again not possible to cover all these formulations. We need three courses like this, if you want
to cover all those things. Again, it requires lot more basic understanding of what we call
as continuum mechanics.
Hence, again we are going to look at one of the formulations which is very important,
which is extensively used, for example, to solve problems in a metal forming, what we
will call as fluid formulations or flow formulation rather. This flow formulation is what I am
going to emphasize as a next step. So, after beam elements some results and then flow formulations.
If you have time further, we will talk about contact formulations as well. Many of the
principles that are used in metal forming simulations I will indicate it and we will
look at it with some practical examples. Again applications of finite element to say metal
forming, it is a huge area, so, we are not going to talk about again sheet metal, because
that requires knowledge of shells and so on. But, we are going to talk about application
of finite element to bulk-forming processes and where many of the techniques that we have
used will be of use.
Why I am saying this is because, this topic is extremely important. Once you understand
this, the application of this topic to many other fields or many other fields of application
of finite element becomes rather routine. Tomorrow you want to go and do heat transfer
problem or determine temperatures, the issues are very similar to what we have done now.
Hence, you please follow whatever I am saying.
We stopped this problem at this place. We said that one of the major, I would say, issues
here is to write down the stiffness matrix of a triangular element. We also gave a reason,
why triangular elements are important till recently and tetrahedron elements its big
brother, its three dimensional brother is very important even today. Basically, because
most of the softwares, most of the softwares, generate triangular element and now, yes,
there has been some change, the technology has moved. Now, they are capable of generating
a four sided quadrilateral element. But even today, most software, 99% of the softwares,
unless you make efforts, lot of efforts, when you say automatic machine that means you do
not make efforts, the software does only tetrahedron elements. Unless you make an effort by what
are called as mapped meshing, which owes its origin to Professor Zinkevich (Pl check the
spelling) in the early 70’s, then only do you get an element which we call as a hexahedron
element, which has 6 sides or else you still have the big brother of triangular elements
that is the tetrahedron elements.
In other words, what I mean to say is, whatever we are talking about for triangular elements
can be, many of the properties can be, extended to this kind of tetrahedron elements. So,
once you understand the basis of these elements, it is very easy to extend them. There are
two ways in which we can develop the stiffness matrix of a triangular element; there are
two ways. One is to collapse a node which is probably the easiest to understand and
many softwares do it in this way, as well as in another way which I am going to discuss
later.
One way is to collapse. For example, look at this element. This, I have given specific
dimensions in order that we can work out a problem and you get a feel of what is happening.
I do not want to be vague and hence I have given some dimensions to it. Say for example,
if I collapse this 1 onto 2 and get an element, collapse means make this coincide or make
x1 is equal to y1 and x2, sorry x1 is equal to x2 and y1 is equal to y2, when you make
like that then the resulting element becomes a triangular element. Is that clear?
Now, what I am going to do is to proceed in this fashion and workout a B matrix to it.
I want to show some peculiarity with the B matrix, what happens to the B matrix? The
procedure can be written as a code and so on or you can follow a second technique for
the generation of stiffness matrix of a triangular element which people call as area method.
We will keep that, we will just reserve this area method for the next step and let us now
look at this, because you get some nice results with this, so your understanding will also
go up looking at this. I need your help in working out. I am sure all of you are familiar
with this problem, because last time also I left it at that stage. Let us see how to
work out this problem. These are isoparametric elements and hence I can write down say x
by means of our well known formula Ni xi.
By the way what is N1 in this case? I think there has been mistake in this as well. Yes,
look at the way I have numbered. I have deliberately done it. Look at the way I have numbered these
things. See where is 1, where is 2 and so on. It is important that you understand that
you cannot blindly put N1 N2 N3 and N4 and hence I say N1 is equal to 1 by 4 into 1 plus
psi into 1 plus eta. Is that clear?
What is N2 now? 1 by 4 into 1 minus psi into 1 plus eta; N2’s position is 2 here and
three, N3 is equal to 1 by 4 into1 minus psi into 1 minus eta and N4 is equal to 1 by 4
into 1 plus psi into 1 minus eta. It is very simple to just check this up, substitute them in these
equations, I mean, sorry, in this triangular coordinates you will see that familiar properties
are satisfied. Is that clear?
My first step is to substitute this here and as well as y is equal to Ni yi. Please note
what am I doing by writing it like this? I am invoking my isoparametric principle, the
basic principles of natural coordinate system. Is that clear? My first step is to substitute
these equations into this equation and then substitute or put x2 is equal to x1 as well,
so that now I can write down. What do I do ultimately? I am reducing the coordinates
so that I can write down, for example, x in terms of x2 x3 and x4. Let us see how I write
it. x is equal to so one by This becomes I will just write it here, so that you can just
follow this easily.
Now, x1 and x2 becomes the same. What happens to x? 1 by 4 into what? 1 plus eta into, no,
no; first let us write it down in terms of x2 x3 and so on into 2, x2, so that that and
this goes off and so 2 into x2 plus N3 x3 is present for x4.
Similarly, you can write down y as well. Is
that clear? My next step is let me substitute; let me substitute, the coordinates x2 x3 and
x4, from this figure. Let me substitute it from this figure onto my equations, so that
I can write down now x and y.
x is equal to, can someone help me? Take a minute to look at that and then substitute
this. What happens to x2? x2 is zero, x3 is zero. What happens to x4? 2; x4 is here, 2,
so, 1 plus psi into 1 minus eta by 2. Is that clear? How many of you did this problem? How
many of you did this problem? Not even one? How many came up to this step? Very good;
let us see. Just think with me, let us see whether you are able to do it. But, none completed
it. But then, you will be surprised when I do it that this is the one of the simplest
problem, because I am just going to follow exactly what I did for the previous case;
nothing more than that.
What is y? Substitute now for y2 and y3 and so on. Yeah, yeah! 1 plus eta into, no, when
you, whether you write it in terms of x1 and a x2, you can you are saying that you will
write it in terms of x1 x3 and x4. No; x1 is equal to x2; x1 is equal to zero. No, no,
I do not follow your question. You can, if you want you can, eliminate x2 and write it
in terms of x1 x3 x4. x1 is equal to x2, so what does it matter? Is that clear? So, what
is y? 1 plus eta.
Now, why I did this is because, I can keep it as it is. I can keep it in terms of x1
x2 x3 and continue, but then my algebra becomes very big. I am solving a particular problem
that is why I substitute it or else I need not substitute it. I can keep it as it is
so that I will get a much more general picture, but then since we are only illustrating one
problem, I have just substituted. Is that clear? I need not have substituted, if that
is your doubt? I need not have substituted it. But then I will again, I mean fill up
the board with lot numbers, then it becomes very confusing and you may miss the central
result of this problem and that is the reason why I wrote that. Is that clear?
What is my next step? What are we trying to do? I am trying to solve this isoparametric
problem. I mean, I want you to listen, because none of you have done this problem. It is
very important that you understand it. What is that I am trying to do? What do you think
is the next ultimate step, not the immediate step? What do you think is the next ultimate
step? Jacobian; right. Why not I calculate the Jacobian? Is that clear, any question?
Now, calculate the Jacobian.
I will give you 2 minutes, just do that. Just look at these things, look at this and then
calculate and write down Jacobian. What is it? dow x by dow psi dow y by dow psi dow
x by dow eta dow y by dow eta. So, substitute them and check what you get? Who is going to give me the answer? 1 minus eta zero
minus of 1 plus psi 2; beautiful, simple; dow x by dow psi and so on. That is one part
of the story. Now, you have not yet answered, this is the second step Jacobian; you have
not yet answered what is it that I am interested in? What is the problem? We are going to look
at B matrix. We are going to look at B matrix and see how B matrix is going to look like?
My whole idea is to go to B matrix. Is that clear? What is it that I should do in order
that I go to B matrix? DN, J inverse; yes, of course, I have to calculate J inverse.
What is J inverse? Let us do that straight away, because J is available here.
What is J inverse? Just calculate that and let me know. Yes, no. 2 upon 1 minus eta zero
1 plus psi 1 minus eta, beautiful, 1; lovely. See, the only way to learn this subject quite
thoroughly is to do these things, so that you will get familiar with each step. Now,
we have J inverse. Now, what is my next step? What is the other principle that I am going
to invoke? What is the other principle that I am going to invoke? Let me remove this.
I think you know all these N’s and you have it in your note books. What is the other one?
I am going to now write, what is it that I am going to write? u in terms of Ni ui and
v, Ni vi.
Sigma is implied because, ui I am just repeating and sorry u, v is equal to Ni vi. What am
I going to do here, when I collapse? When I collapse, what is that I have done? u1 is
equal to u2. That is all; u1 is equal to u2. So, my degrees of freedom become u2 u3 u4,
v2 v3 v4, period. Get that, so that you can now write it in terms of say N1 or u1.
I will not write that 1 plus; the same thing what we wrote in the previous case, half into
1 plus, half into 1 plus eta into u2 plus N3u3 plus N4u4 and that is equal to half into
1 plus eta into v2 plus N3v3 plus N4v4. We have invoked two principles which we used
for isoparametric elements. What is my next step? What is that I should calculate? DN
or in other words I should calculate dow u by dow psi. Why, because ultimately, what
is that I want? dow u by dow x; very good. So, I want dow u by dow x dow u by dow y and
so on, because in order to calculate B, I need them. Remember, I had that 1’s and
zeros matrix multiplied by our u comma x and so on. My next step is to calculate dow u
by dow psi. What is that equation, if you remember, just kept that in your mind?
That is dow x by say dow u by dow x and say dow u by dow y and so on. What is it? What
does it involve? It involves J inverse. We will go to that step as a next step, but before
that what is that I should calculate?
DN or DN consists of, I have to get dow u by dow psi. That is what my next step is.
I have to calculate dow u by dow psi dow v by dow psi dow u by dow eta and dow v by dow
eta. In other words, you can split this up into DN into, what is that you will get? DN
into d. What is d in this case?
You can write down, say in other words, just to preempt what we are going to do, in other
words B is equal to those zeros and 1 matrix which you have, refer to your book, multiplied
by what? What did we write it as? Yeah, gamma; gamma s we wrote remember? Zero gamma s into
d matrix, small d; small d. What are they? They are the degrees of freedom. Sorry, DN
into d; remove this B. That is actually, no, no; sorry, sorry, sorry and so this whole
thing becomes B. Is that clear? How many of you had come to this step? There is nothing that we have done;
same procedure, whatever we have done.
Has anyone come to this step in the exam? What exactly was the problem? You know, let
me just go back and try why you have not done this? Is there any difficulty in understanding
this? What is the confusion? Whatever step I have written is known to you. In fact B
matrix, I have told you so many times that this is B matrix. The only difference now
is this d is now a compressed d. So, 1 2 3 4, we have only 2 3 4; that is it. Now, I
will give you two minutes, I want you to calculate, say DN. Just calculate DN and tell me what
DN is? Just do that; since you have not done it before, I would like you to do that and
tell me what DN is?
Any question? Because you have not done, is there any confusion? Anyone has not understood
the steps? Yes; now, it is clear, everything is clear. It always happens that after the
exam everything is clear and in the exam you do not work out usually and that is a problem
with all of us. Even when I was a student it was the same problem. Let us see now. Let
us see how many of you get this problem correctly. Let us just stick to only two terms, so that
we will not get all the things. I mean, because it is a very big matrix, we will just stick
to only …… In other words, let me make this problem simpler. Just calculate dow u
by dow x and dow u by dow y, we will not do all the things.
We will just stick to the first one, dow u by dow x, like only one variable; dow u by
dow x and dow u by dow y. Similarly you have to do dow v by dow x and dow v by dow y; just
dow u by dow x and dow v by dow y; they will have, it will just be J inverse, straight
away. J inverse and then we will have another DN term followed by small d’s, followed
by u1 and so on, u2 and so on; just these terms. Yeah; correct, correct. That is what
I want you to workout; that is what I want you to workout. Correct, substitute; that
is what I am asking you. What is it? What is this value? You can directly get it. You
have, what is that you have here? Dow u by, can you fill this equation? Dow u by dow psi
and here dow u by dow eta. What I am trying to ask you is, what is this? What is this?
This can be written as Niui; same way, here also Niui. Is that clear? How do I write this? How do I
write this thing? Dow Ni by dow psi; dow N1 by dow psi dow N2 by dow psi or dow N3. There
are three terms u1 or u1 is equal to u2; so u1 u3 u4. Clear?
You will have some term u1 u3 u4, just for illustration. Then, you can write the same
thing for dow v by dow x dow v by dow y, combine them together; that is all, appropriately
place it. What will be the first term here? Yes, what will be the first term? No, no.
What will be the first term? Dow N1 by dow psi. You remember that, whatever we had written.
It is not actually N1 but a combination; half of 1 plus eta. That would be the first one
because that multiplied by u1. Is that clear? Because, N1 and N2 we have combined them together.
What is that? How did I write? u is equal to half into 1 plus eta into u1 plus N2 into
u2 plus N3 into u3. That is how I wrote. The first term here would be that dow of dow psi
into half of 1 plus eta. Is that clear? What, yes or no? So, first term is zero. What had
happened to my second term?
Why is that it is zero? No; look at the way I have written it. To just make it simple
I have just written only in terms of u’s. If you want, you can write it also in terms
of u as well as v. Just to make things simple right now, I have written it in a simple fashion.
So, minus 1 by 4 into 1 minus eta; minus 1 by four into 1 minus eta plus 1 by four into
1 minus eta; is that clear? What would be the other things? Here it is dow u by or that
one half of 1 plus eta, that has to be differentiated with respect to eta. So, what will be the
first term here? Half, very good; see, so simple nothing there. Then you have to differentiate
the second term. Second term, when you differentiate it minus 1 by 4 into 1 minus psi; very good.
What will be the third term there? Is it minus 1 by 4 or plus 1 by 4? 1 minus, sorry, 1 plus
eta, very good. Now, I want also dow v by dow x dow v by dow y. I can write the same
thing in a similar fashion. Sorry, same I say; whether u1 or u2, u1 is equal to u2,
so, u2 u3 u4. Is that clear? What is my question? My question is, how is this B going to appear?
That is my whole question. How does that B matrix appear? I can look at it even here
to tell how B is going to appear?
What is that I am going to do? Substitute for J inverse; I already have it, multiply
it with this equation or with this matrix and tell me what is that you will get, because
that will get repeated. I have to only readjust, nothing great in it. In other words what I
want you to do is to substitute for J inverse, multiply here, multiply it and tell me how
dow u by dow x and dow u by dow y will appear.
I will just write it here itself; dow u by dow x and dow u by dow y. How it will appear
to me? You had J inverse; what is J inverse? Yeah, 2 by 1 minus eta zero 1 plus psi into
or by 1 minus eta 1; period, that is it. So, substitute it there and multiply it, simple
multiplications. That is the reason why I just stopped with that and just tell me how
it will appear? Now you know how to calculate DN. Our regular DN is what? Just expand this;
expand this to u2v2 u3v3 u4 v4. Similarly, put zero here, zero here, keep on expanding
that. Is that clear? Very simple, no problem; that is done. So, now multiply J inverse and
this and tell me what is that you get out of it and this result is very important.
If you want for your … what is …. J inverse is 2 by 1 minus eta zero 1 plus psi into 1
minus eta and 1. Am I right? Yeah; just for your reference and so just substitute that
here, multiply it and tell me what is that you are going to get?
First one is half; wait, wait, wait; slowly tell me. Let everyone complete it. So, first
term is zero minus half minus half. Next, next one half minus half zero. What does this
tell you? So, my B matrix can be, once I know this structure, I know how B matrix is going
to look like. Is it clear or not? Is there any confusion on this? I have taken a part
of B matrix and worked out because I do not want to write, again fill the board; that
is why. In other words, I can write down dow v by dow x dow v by dow y; it will have the
same form, but will be multiplied by what? v2 v3 v4; period. I can now combine them together
to write my big B matrix. What does this tell you? It is constant throughout the triangle.
B is a constant, the whole of the triangle. You go back and look at what we did with the
other element, our quads; quads were not like this.
What does this give you from using a triangular element? Suppose there is a place where strain
changes. Then obviously, the strain inside a triangular element is going to be a constant;
strain inside the triangular element is going to be a constant and hence I will be hard
put to save myself if I use a triangular element in that area, a big triangular element. But,
there is one big but. If I use many number of elements, then the problem is obviously
to a great extent alleviated, because I can fit any curve by means of, when I take more
and more elements by means of, small steps.
There are other peculiarities to this element. I am not going to answer that now. I want
you to keep this in mind that the constant strain state, I will answer it in the next
class when we look at, what is it that we are going to get out of finite elements? So,
lesson number 1, major lesson, if you want to use triangular element or as I told you,
as I called it as a big brother, that tetrahedron element, then you have to be careful in meshing.
Because especially when there is a strain change, for example, in bending strain changes
linearly, then if you use this kind of elements you will be in trouble and that is what I
showed you in one of the exercises, which we did in one of the classes. Is that clear?
This exercise was not very difficult. I am sure now, you all agree that it is not very
difficult; quite straight forward, but it cannot be that easily extended to higher order
elements or in other words you may look at it and say that it is a simple thing, but
can I have a much more general technique of doing a triangular element in the same fashion
as you did our quadrilateral element? That is our question. Is it possible to have an
element? Why is it that I am looking for a much more general issue? I will come to that
in a minute. Is it clear? Can I go to the next issue? Fine. What we are looking at so
far, are what are called as only a linear element. Next class, I will illustrate that
we can move over to a more quadratic elements.
For example, I can say that a triangular element can have more than 3 nodes, like I have, say,
quadrilateral element can have more than 4 nodes. Say for example, in a triangular element
apart from these three chaps, I can put nodes at other places. Why am I doing this is because
I am stung by the criticism that that B is a constant. Why did B become constant, because
of the simple fact which I illustrated about 4 or 5 classes back, that u is written in
terms of x and y; a0 plus a1x plus a2y. But all those things are camouflaged in whatever
I did. That is why I did that much more elaborately. But if I can use not a linear interpolation,
but a quadratic interpolation, I can to great extent solve this criticism.
In fact, today for many of the very large deformation cases, elements which have been
developed, which are triangular elements with 6 nodes and perform extremely well under much
more difficult conditions. The major advantage is with good mesh generators, triangular meshes
are very easy to generate and I can use those generators. So, I have a need to develop a
much more general technique for a triangular element, which can accommodate not only just
a linear triangle or constant strain triangle, as it is called, but higher order triangular
elements like this which has, say, 6 nodes. Once we do that, we will get back to this
element which has apart from these 4 guys here, 4 nodes there, I will have 4 more at
these places. Is that clear? In order to develop this isoparametric or this element from an
isoparametric fashion, I have to, I have to, now generate what is called as a natural coordinate
system for a triangle. I will leave it with this small indication and then we will continue
it in the next class, because it is a very lovely idea. What is that we did in the last
instant?
I had, say, a point here, a point say P. We expressed it in terms of psi eta x y and so
on. In other words, we had a, we mapped; we mapped or got psi eta in a fashion such that
we can get psi and eta, by joining. How did we get it? By joining midpoints on either
side we got psi and eta, mapped it. Here we are going to have another approach, say for
example, if this is the point P, then I can join this P with the coordinates say 1 2 and
3; 1 2 and 3 and call this area, this area here as A1 and this area as A2 and this area
as A3 and use A1 by A, A2 by A and A3 by A as a natural coordinate system. We will see
that or we will expand this in the next class. How we are going to do it, we will see in
the next class.