In the last class we saw a number of applications
of finite element analysis. We saw really the breadth of applications. We saw a number of case studies from different
fields and I am sure that you are all convinced that this is a very good technique which can
be applied to lot of practical problems. From now on, we will study some of the basis
or theoretical basis of finite element analysis. You may have to have background in various
fields in order to understand exactly how this technique works. But as far as possible, in this course, we
will see to it that all these things are covered and the course is completely self contained. We stopped with the definition of stiffness. We said that or we said that it is easy to
understand what stiffness is as far as the spring is concerned, but how do you define
stiffness for other components? We considered that there is force P that is
acting on the spring and if the displacement at this point is u, then we said that it is
possible to define a term called stiffness K such that P is equal to Ku. Now the question is how am I going to extend
this concept to other components or other bodies which are not necessarily nice springs
like this? As a first case let us consider a simple bar
and let us say that I apply a force P. Can I talk something as stiffness for this particular
bar? Is it possible that we can define it? Yes; how do we define it? May be all of you are familiar with it. We know what is the relationship between stress
and strain? What is In a very elementary level you would
have defined stress as P by A where A is the area of this particular bar. So, P by A is equal to E into change in length. What is change in length in this case, say
for example? No; as far as the strain is concerned what
is change in length by original length; that is how you would have defined. Let me call this change in length as u, in
a similar fashion as we say that the displacement of the spring is u divided by L. It is very
simple to now say that P is equal to AE divided by L into u; AE divided by L into u. Now compare this and this. It is quite clear that I can call AE by L
as the stiffness of the bar. Stiffness straight away is not a concept which
is restricted only to a spring but can be extended to say for example a bar. We are going to extend this for all other
types of components as well. Now let us look at another problem. Let us see whether we can extend this kind
of concept to a problem, where a simple case a very simple problem, where I have a P, force
that is acting on a bar whose, for example, cross sections are say A1 and A2. This poses slightly more difficulty when compared
to this problem or it is slightly more difficult. You can, looking at it immediately say that
“Sir it is made up of two bars”? It is made up of two bars. So, I can say that why not you consider it
as two bars? Once I start considering it as two bars you
come into the realm of discretizing this particular piece; you have
discretized it into two bars. Now on we take over or we come into the realm
of finite elements. Yes, that is nice answer, but there is a small
problem. What is the problem? Here again, like this case, we had stiffness
to be only one number AE by L. Because one end was fixed it was nice to me to say that
the displacement is considered only at the bottom of the bar and we arrived at this point. On other hand look at this; it is not like
that. I have now two bars. If you look at these two bars which are joined
at this point, this point starts moving, this point also starts moving. One end is not fixed; easy to handle, but
slightly more complex than this. You can look at another problem where there
are more number of bars. There is a technique to handle this. In order to use that technique first let us
define what is called as a bar element? What is a bar element? Let us take an element. I now started calling this as an element. Let me take that element and put it out. Immediately you see that there are two guys
sitting on either side which are called as nodes. So, let me call those nodes as say i and j;
a very generic term say i and j. Now let me define or determine the stiffness
matrix. Note this, we are now going to define stiffness
matrix of this particular bar which are defined by these nodes i and j. What is also important is the area. Let me call this as A, area of this bar as
well as the length, this length which I would call as L and the Young’s modulus which
I will call as E. Let us start with what we know. In this bar, let me apply a displacement which
I call as say ui and fix j. Now let me call the force that is required
to cause this displacement as Fi. From my previous result I know what Fi is. What is Fi? AE by L into ui. When I fix this j, this j is going to take
a reaction or in another words there is also a force that is acting at j. Let me call this as Fj. From elementary strength of materials I know
that Fj is a reaction and that it is, what is that? Opposite to Fi and of equal magnitude. That’s right; so, Fj is equal to minus Fi;
that’s the first step. Now let me again repeat this with j free or
applying a displacement of uj and fixing i. What is that I am going to do? I am now going to release this ….. and say
that let j move from the original position by an amount uj and let me fix this position. In this situation what happens to Fj? That is straight forward and that happens
to be AE by L into uj and now what happens to my Fi? That is minus of Fj or in other words that
is minus AE by L into uj. Is that clear? Now coming back to my problem here, this particular
bar, this particular bar now has two nodes and is in the position where both the nodes
would start displacing or ui and uj exist for this element. Since I have derived in a very simple fashion
what is Fi and Fj when one or the other are fixed, it is possible for me to straight away
get to a situation where both ui and uj exists. It is nothing but a superposition of these
two results. So, when both ui and uj exist, I can say that
Fi is equal to AE by L into ui minus uj. Is that clear? What happens to Fj? Fj is equal to AE by L into minus ui plus
uj. Is that clear? This can be expressed in a matrix form. What do I mean by that? I can express this as AE by L into 1, minus
1, minus 1 and 1 multiplied by ui uj is equal to Fi Fj. Look at this particular equation quite closely. You can immediately recognize what is that
you can recognize? You can recognize that there is a force on
one side and there is a displacement on the other side. In other words this is a relationship between
force and displacement and that relationship has been brought about by this particular
matrix. Is that clear? Comparing that equation with my first equation
here where it is P is equal to Ku for a simple spring where I called K to be stiffness and
similarly there is a relationship between force and displacement, I can call that matrix
which I derived now to be what is called as the stiffness matrix. That is the stiffness matrix. Let us understand what this stiffness matrix
means. Can I extend the concept of stiffness, the
physical behavior or the physical understanding of stiffness what I have for a spring, to
the stiffness matrix which we derived? If I had asked you to define stiffness and
not write down an equation, you would have said that stiffness is equal to force per
unit displacement. There was no problem because there is only
one place where force is applied and one place where the displacement happens to be, there. What about in this situation? How do you now define here or how do you physically
get a picture of what the stiffness matrix is? That is the question. How do you define it? Can someone tell that or I would call this
as a K11 K12 K21 and K22, along with AE by L. The same matrix can be written for us to interpret
as K11 K12 K21 and K22 that multiplied by say u1 and u2 where i and j’s I have replaced
by u1 and u2. If you want you can keep this as i and j itself. If you are getting confused you can keep it
as i and j itself. My question is what does K11 indicate? What is the physical meaning of K11? What does K12 indicate and so on because here
I said K indicates force per unit displacement. In order to understand that, in order to give
you a clue let me say that I apply a unit displacement. Here it is force per unit displacement. Keeping that spirit let me say that ui is
equal to 1 and let me fix the other side uj is equal to zero and then look at the complete
equation now. What is that in the other side? Fi and Fj. What does the first equation indicate? First equation; K11 is equal to Fi and what
does K21 indicate? K21 is what? Fj. Is that clear? Can someone give me an interpretation of what
is K11 and what is K21? One of you; yes, so it is nothing but, yes,
in other words just to repeat that what it means is K11 is the force that happens to
be present at i or the force to be applied at i in order to have, note that, unit displacement
at i and zero displacement at j and K21 is the force that is required to be applied at
j in order to have a unit displacement at 1 and zero displacement at j. The same thing, same logic can be repeated
for j as well; you can put this as zero and this as 1. You can again interpret this Kij. I can have a physical interpretation of Kij’s
which are entrées here. i is equal to 1, j is equal to 1; it becomes
K11. i is equal to 2, j is equal to 1, it becomes K21 and so on. So, I can have physical interpretation for
Kij. What is the physical interpretation? What does Kij indicate? It is the force that has to be applied at
j in order to keep ….. Now look at this and tell me. See, K21 means, what is the displacement I
have applied for i? It is unit displacement. So, if I have to apply a unit displacement
at 1, and a zero displacement at j then K21 is obtained from Fj. If there is a unit displacement at j, then
Kij indicates the force that has to sit at i. It is the force that is required to make or
to give a unit displacement at j and the force being measured at i, the force that is required
at i for a unit displacement of j. So, that is a simple physical interpretation. Yes, zero displacement at i; force that is
required to be present at i for a unit displacement at j, zero displacement at i. Now let us look at this matrix closely. What are the other things that you see in
this matrix? What are the other things that you see in
this matrix? Symmetric, very good; so, this matrix is a
symmetric matrix. It is not necessary that every matrix that
we are going to come across is symmetric matrix; but this particular matrix because we are
just dealing with a linear elastic problem, this matrix happens to be a symmetric matrix. Very rarely do we come across matrices which
are not symmetric and in this course most of the problems except when we go to contact
we will only be dealing with symmetric matrices. Some of the large deformation problems have
non-symmetric matrix, let us not worry about it. But as you correctly said, the next thing
that you observe is that this matrix is a symmetric matrix. It is very simple now to understand that I
can have say two bars which are here or two sections in a bar to have different A’s,
different E’s and different L’s. The only thing is that A, E and L would change. It will become A1, E1, L1; A2, E2, L2 and
so on. Now let us see how we can determine a composite
or an assembled matrix of a bar which looks like this. This particular bar which we considered as
i and j have what are called as how many degrees of freedom? Two degrees of freedom or i and j whose displacements
are ui and uj give the freedom for this bar to be displaced in the X direction. So, there are two degrees of freedom or if
you consider each node there is one degree of freedom. What do I mean by degree of freedom? It means that this particular node i can be
displaced only in the X direction and j can be displaced in the X direction. As an element this has two degrees of freedom,
one degree of freedom per node. Now let us see how this concept can be applied
to a composite bar and how assembly can be done so that I get a composite stiffness of
this bar. For that the first thing I have to do is to
determine the stiffness of individual bar. But before that I have already done one job. What is it that I have done? I have already discretized this particular
component into two elements. I did it by looking at it and seeing that
there is a change in A, change in E and so on, if it happens to be there, and so I have
used that knowledge in order to discretize it. But there are lots of other considerations
when I discretize it, but that we will see as we go on. But right now it is important to understand
that discretization is very important and can depend upon a number of factors, one factor
being a nice physical picture or an engineering sense and that is what has helped me to discretize
it. Let me now call this as element 1 and let
me call this as element 2 and let me right now worry about fixing this particular node. Let me call that node as 1, sorry 1, 2 and
3. Is that clear? Let me call A1, This is say L1; let me give
some lengths to it. Let me say that it is L1 and this length to
be say, L2. Just to generalize this problem, let me call
the Young’s modulus to be E1 of this bar and this to be E2. Since AE by L comes up as a unit, let me define
K1 to be A1E1 by L1 and K2 to be A2E2 by L2. 2640K1 is equal to A1E1 by L1 and K2 be A2E2
by L2. First what I can do is to write down the stiffnesses
of individual elements and look at this kind of assembly for individual elements or how
does this equation look like? How does it look like for the first element? I can change that here itself. How does it look like? It will be, for the first element, K1, minus
K1; K2 does not come here, we are only looking at the first element minus K1 and again K1. Now what happens to i and j? Now what happens to i and j? u1 and u2. What happens to Fi and Fj and that happens
to be F1 and F2. I am not worried about whatever the values
of F1, how it looks like and so on; just I am repeating it for this particular element. Any questions? i and j was generic indices, now I have replaced
it with the actual index. Let us now look at the second element. The second element can also be written in
the same fashion. I will do it in the same matrix or maybe we
will write it separately so that things become clear. What do I get? Instead of K1, K2; so, K2, minus K2, minus
K2 and K2. Instead of now 1 and 2 what are our i’s
and j’s? u2 and u3; what happens to this guy here? F2 and F3; F2 and F3. Now I have to assemble them. Please bear in mind what are the steps that
we are doing? We discretized it first step. I will write it down at the end of this exercise
but nevertheless as we go along please bear in mind what we are trying to do. We discretized it in terms of what we called
as elements. In this case, the behavior is bar. We will come to what is behavior in a minute,
but in this case we call this as bar element, so, we discretized it. The next step is we wrote down the stiffnesses. That is the next step; we wrote down the stiffnesses
of individual element. The third step is I have to assemble them. How am I going to assemble or what does this
term assemble mean? In order to assemble them let us see or let
us imagine that we are building this bar. Let us not worry about the actual manufacturing
process; it may be turning …… let us not worry about that. We will assemble them as if first I will put
this and later put that. In fact, it may well turn out to be a welding
process because E1 and E2 are different; like we may imagine that they are welded together. So, first I take this piece, the first element,
put it there and identify these guys, these nodes there. Let us imagine that that is what happens. Since I have already identified a node at
3, it is imperative that it also has a displacement u3. Let us say that it has a displacement u3,
though more importantly this element does not support any force there. If I apply a force here, the force happens
to be in the air and this particular element does not support a force. In other words there is no stiffness. See, when can you apply a force? Only when there is a resistance to it or in
other words there is stiffness to it; stiffness to the body on which you are applying. Since it is in vacuum, let us imagine that;
let us not worry about anything else. So, I cannot put a force at that place or
even if I put a force, there is no resistance due to this bar. If I have to now expand u1 as u1 u2 and u3,
how will my stiffness matrix look like from our physical understanding of K? Let us see how it would like? So, I will have K1, no doubt; I will have
minus K1. These two guys are for u1 and u2 and what
will happen to my third term there? Very good, because you may imagine or remember
our Kij definition; so, here I will have minus K1 and then followed by K1 0 0 0 0; that is
very good. Let me go and put the other element here,
weld them together or in other words weld them together. As soon as I do that, this force what we have
mentioned here will be supported. Let us not worry about that right now, but
the process is quite clear and hence some of the zeroes are now going to be replaced
by the stiffness we are giving, by introducing the next element which means that I will have
K2 coming to the aid of the first element; so, K1 plus K2, minus K2, minus K2. Please note that this is the last term in
the matrix that is in the second row and 2, 3 and K2. One of the things you had not noticed in the
last stiffness; when I said that you just said symmetric. I will look at it but you have not noticed
one very important thing in this stiffness matrix which I will come to in a minute but
what is important here is that we have assembled the matrix. First thing is that look at the matrix size. Now there are three nodes and there is a 3
by 3 stiffness matrix and there are u1 u2 u3 that are not known and what is the right
hand side? What is the right hand side? Correct. It is F1 F2 and F3; remember that they are
nothing but the forces that can be applied. It is not necessary that they should exist
but that can be applied to 1, 2 and 3. So, this is the complete equation which I
am interested in after assembling. What is my next step? I have to apply what is called as boundary
condition. I have to apply what is called as boundary
condition. But before we go to boundary condition let
us look at this matrix closely and observe one more point. Again physically there is lot of mathematical
implications to it. At this stage let us not worry about the mathematical
implications but physically let us look at it and see whether we know why it happens
to be in this fashion. Look at that matrix. You will see that there are some positive
terms and negative terms. K1 is positive, K2 is positive; so, there
are some positive terms and there are some negative terms. See that there is no negative term in the
diagonals. All the diagonal terms happens to be positives. Apart from that you also see one more thing. What is that you see? Zeroes; so, there are positive terms, negative
terms as well as some zeroes and it is very clear that the diagonal terms are positive,
off diagonal terms, there are negative terms as well as zero terms. In this particular problem there are negative
terms as well as there are zero terms. The question is, is it that these terms that
are sitting in the diagonals are positive only for this problem it happens to be positive
or is it by design or it is by chance? Is it that it is positive only for this problem
or for any other problem? Is the question clear? What do you think will be the answer? One minute; let me give you a clue. In order to answer that question, let us follow
the same logic by applying say unit displacement at say for example 1 and then zero displacements
and so on. Now you can see or you can say whether this
is positive by design and not by chance. Let us look at my first equation, say K1 say
u1 ……. not a unit displacement at least u1 is equal to F1. So, u1 is equal to F1 by K1. u1 is equal to
F1 by K1. What does this tell? If I apply a force F1, then u1 would follow
the force in the same direction as long as K1 is positive. So, if I apply a force I know that the displacement
has to be in the same direction; it cannot be in the opposite direction. So that particular condition is satisfied
as long as K1 is positive. On the other hand if K1 happens to be negative,
what happens? Say for example I take this bar and then I
start applying a force here. It will move in this direction if K1 is negative. That is not possible. The bar has to move in the same direction. What does it indicate? It indicates that K1 has to be positive and
the next thing is that, is it clear? That is very simple thing. On the other hand can K1 be zero; then u1
becomes infinity. So, necessarily these diagonal terms have
to be positive. We will come to a much larger condition of
positive definiteness of stiffness matrices later. Let us not worry about it because we have
lot to learn from the physics of the problem before we go into the mathematics of the problem. I will not define all those things right now. Ultimately we will come to that. What is my next step? What is my next step, what do you think is
the next step? Boundary condition, very good; so, what is
the boundary condition of this problem? This problem states that one is fixed and
that a force is applied at this point and that is the secondary thing; we will come
to that later but first thing is that u1 is fixed. Please note one thing carefully that when
I say boundary condition and say u1 is fixed, I just state that u1 is known. It does not mean many students make a mistake,
that boundary condition means u1 is equal to zero; a particular degree of freedom, u1
or whatever be the degree of freedom that happens to be zero; not necessary. It just means that the displacement for that
degree of freedom is stated. In this problem, it is stated to be zero. Suppose in another problem I say that this
displacement is 0.5 mm; that is also a boundary condition. Boundary condition simply means in this particular
problem displacement happens to be zero. What are the types of boundary conditions
and what are the ways in which to express them, all these things we will see as we go
along. But, for this particular problem it is important
to understand that the displacement happens to be a boundary condition and that, that
displacement happens to be zero. Why I am stating this point is because a same
configuration, as far as geometry is concerned, the same configuration can be looked at as
a beam as well. The only thing is if I now apply a force in
this direction, it is no more a bar, a beam; it becomes a beam which means that it can
bend. A bar can take only axial loads; you know
from strength of materials. On the other hand a beam can take a bending
load. Is that clear? Again, I have different types of displacements,
a displacement, which looks something like this and I have different types of boundary
condition. What are the boundary conditions? Look at the question, what are the boundary
conditions? The previous case for bar we defined what
is the boundary condition? There was only one; now we just modify the
problem and say that what are the boundary conditions? Displacement and rotation that is right. So, displacement as well as rotation, both
is specified at that place. In other words the same node or the same position,
same point is specified in a slightly different fashion; the geometry happens to be the same. But, why is this difference? That is because of the governing differential
equation for these two problems; they happen to be different. Before we go further, before we go to that
point let us now worry only about this problem and get a complete physical picture of finite
element analysis. So, we will continue with this. As you correctly said u1 is specified and
u1 happens to be zero. So, how many unknowns are there? Now two unknowns; u2 and u3 and u1 is already
known and happens to be zero. Since u1 is equal to zero, I eliminate these
two lines or in other words I eliminate that row and the column and restrict my attention
to this particular piece. Now I know something else also in this equation. What is it that I know? What is it that I know in this equation? F3; correct. So, I know that F3 happens to be the applied
force P and what happens to F2? Do I know it or I do not know it? Is it that we do not know? Note that carefully; this is the mistake many
students make. No, no; please note very carefully. Look at the problem again, look at this problem
again. Look at this problem, look at this. Now, where is that node? Node is sitting here. It is nice that you have made that answer
because this is usually the confusion. Please note that what we are doing here is
to put down the external forces that are applied. We are not worried about the corresponding
internal forces that are developed; we are looking at the external forces. These are the external forces. Please do not get confused between that and
the internal forces which are developed due to stresses. This is the usual confusion for students and
hence this is an external force, P is an external force. Correspondingly is there an external force
at node 2? No; so, this happens to be zero. Now I have two equations and two unknowns. Just to summarize, what did we do? We discretized, we divide or in other words
we divided this into a number of elements. Then we wrote the stiffness matrix for each
of these elements. We assembled them, that is what we did here
and then we wrote down boundary conditions carefully and also put down the forces that
happen. One other thing before we go ahead and solve
this problem; it is very important to note from this equation. What is it? What is that you see; another thing that you
see? You please note that at the place where we
give a boundary condition we are not specifying the force. It is not possible to specify both a boundary
condition as well as a force. What does it mean? What is it mean? It means that when I fix a particular node,
the force that acts on the node is decided by the system. So, if I say P, it is very clear from this
problem before even solving it that the reaction for P has to be taken at the place where I
have fixed. So, the force that is going to happen or going
to be present at this place is going to be decided by the force that you give here. I cannot specify both a force as well as a
boundary condition at the same point. When I specify a force at this point, the
displacement at this point is decided by this system, by the stiffness matrices and when
I specify a displacement at this point the force that is required to sustain that effect
is specified again by the system, by equilibrium and so on. We will have more interpretation of this equation
and how to solve them or how to solve this equation in the next class.
