In the last class we stopped at deriving the equilibrium equation. We said that the equilibrium
equations are similar to what you have studied in your earlier classes on say engineering
mechanics and sigma F is equal to zero was one of the equations. We said that there are
two equations, sigma F is equal to zero and sigma M is equal to zero and we said we will
concentrate first on sigma F is equal to zero. Before that there were one or two questions
which I think we should get to know the answer, because quite an interesting question was
their initially, before we just ended the class. What was the question?
I have two tensors, say sigma and delta; the two tensors. I actually operated on these
two tensors and said that I get a scalar. If you remember I said that sigmaij deltaij
is equal to sigmaii. I operated on these two tensors. The question was what is this operation
and what are the types of operations that are possible and what is the result? Is it
that when I operate on these two tensors, on two tensors I will get only a scalar? The
question was very much valid because I said that these can be also looked at as a matrix.
If I just say multiplication of matrix, I should in fact get another matrix or another
second order tensor. If this is the understanding, yes, you are correct, but the only difference
is that there are number of operations that are possible with two tensors. This operation
for example is called as a dyadic operation.
What is this operation? Let us understand this, let us not get confused about it. Suppose
I have two vectors; let us go from what we know. Suppose I have two vectors say u and
v. All of you know that I can do two operations with u and v. A dot product can be obtained
which is say u dot v and what is this? What is the result of this? Scalar; what is that
you do here? Actually it is u1 v1 plus u2 v2 plus u3 v3 and how do I express this in
indical notation? ui vi. So, one operation with two vectors gives me a scalar.
Let us look at another operation. Let us look at this operation? What is this operation?
Cross product; what is the result of this operation? What is the result of this operation?
Another vector; so, by operating with the same vectors on one hand I can get a scalar,
on another hand, I can get a vector. We saw how to express this also in terms of indical
notation yesterday. Similarly you can look at a number of operations with tensor also.
What did we do here? Actually you have ui vj. I made j is equal to i in order to get
this scalar product. In fact it is not that there are only two operations with vectors.
I can have another operation called u dyadic v. The result is a second order tensor; the
result is second order tensor. The second order tensor, say for example, let me call
this as some tensor C, where this operation indicates that cij is equal to uivj; cij is
equal to uivj. You can see that Before we proceed to this first equation, you can see
that with the same two vectors, I can get a scalar by operating it; by operating it
with another symbol and another rule we get a vector and yet another operation gives me
a second order tensor. Now the question is what are these operations? what are we trying
to do here?
These operations, in later classes will mimic certain physical phenomena. In order to understand
what this means, this particular operation means, you can look at for example force and
displacement. If you want to get say work, done what will you do? Yeah! So, you will
look at dot product. You would see that when I have these vectors, especially in this course
with the physical meaning, particular operations will give rise to certain physical quantity
or phenomena. This is independent of the tensor analysis and so on, but for engineers it is
easier to say that this dot product between these two vectors gives me this quantity;
cross product gives me this product and so on.
Now coming back to this equation in the same sense that I can have different operations;
I can have different operations defined on a second order tensor as well. This has sigmaij
deltakl. So I can get a fourth order tensor by an operation which is called as a dyadic
operation; by a dyadic operation, such that ijkl, let me say this results in say D. Let
me say that this results in D such that Dijkl is equal to sigmaij deltakl. Whenever I put
this operation due to some physical phenomena that is taking place, the result will be a
fourth order tensor whose components will be given by this operation. After all vectors,
tensors and other things that we are going to use has some physical meaning. Stress as
We understand stress; even though I defined it very rigorously, we know what is stress
and what it means and so on. This kind of intuitive feeling you would have developed
already in the previous classes.
The question is why are you defining so very rigorously all this kind of things? Why are
you now taking us to a tensor? Why are you defining stress as a second order tensor?
This is another question, why do you want to define? Why not we stick to what we know
as a matrix. Now this takes you to a slightly higher mathematical plane. What is that and
that is going to form, that kind of plane is going to form the bases of your further
understanding of subjects like this. What is that plane and what is that understanding?
When I prove or show that stress, I have not yet done it; that requires some more rigour
but nevertheless, when I show that this stress is a second order tensor, I am bestowing on
that certain qualities of a second order tensor.
What does it mean? If I keep defining, sigma is a second order tensor, epsilon is a second
order tensor, epsilon dot is a second order tensor and so on, a number of quantities,
that means that these guys will take the qualities of a mathematical, say, understanding called
tensor or mathematical definition called tensor. A second order tensor has certain properties,
independent whether it is stress is a second order tensor or whatever, it has certain properties.
It has certain Eigen values, Eigen vectors and so on. So, all these things will come
to the second order tensor. We are going to see in this class or maybe I hope it is in
this class that we are going to see principle stresses; how are they defined?
You would have studied it in one fashion. But once I define second order tensor is the
one which is going to be characterized by stress, then I can say that these are nothing
but Eigen values of this tensor. I need not go and study principle stress, principle strain,
principle so and so, as long as I say these guys are second order tensors. So, they get,
they inherit all the properties of a second order tensor, once I define a physical quantity
to be a second order tensor. In the same sense, when I say vector immediately you would go
and tell me or you would tell me immediately that a vector has a magnitude and direction.
What is that you have done? What is that we are looking at if, when I say displacement?
It is a vector or velocity is a vector or force is a vector. What is that I am doing?
I am saying that these quantities will follow the rules of a vector. These quantities will
follow dot product; these quantities will follow cross product and so on.
Once I say these things, I need not worry about saying or looking at rules for these
operations. One of the operations will give me what is called as the fourth order tensor.
Another operation called contraction operation, where I will make, say two quantities to be
the same say i and k to be the same or j and l to be the same and so on, will give me a
second order tensor. Sigmaij deltajl which means i imply summing up about j. Let us not
go into details because you can very systematically define what is called as transpose and so
on. Let us not worry about that right now, so that we do not deviate too much away from
what we are doing. But nevertheless since this question was asked we can see that I
can now define a contraction called contraction operation on a fourth order tensor. What did
I do? I simply said let me give a rule to you by making two of these indices to be the
same. As soon as I define this rule, I also follow all the summation notation which means
that the resulting operation or the resulting quantity is a, fantastic; it is a second order
tensor.
I can give you one more rule. Next rule is that take the trace of this second order tensor.
That means that second order tensor is a matrix; sum of the diagonal terms. What that does
mean here? That means that make the other two quantities, other two indices also to
be the same.
So, I can sigmaij deltaji, that is another operation; another operation. Now you will
see that why am I doing all these? Why is that this becomes important? You will see
that like we had F, what is this? Force and a displacement define what we called as a
physical quantity, work, I am going to say that I will operate with sigma and epsilon
or epsilon dot rate of strain.
I am going to define strain may be in the next class. with if I going to If I can operate
with these quantities, I will get some other physical quantities, some other physical quantities
like for example stress power. These operations we will carry out later, we will make use
of such operations with the tensors. There are number of operation with tensors. In the
same fashion, they are nothing but an extension of what you already know as operations with
vectors. We will come back to this topic after some time. But we will get back to the problem
which we were dealing with in the last class.
Let me recapitulate this problem. We were trying to get at the equilibrium equation.
We had said that there are two types of forces. The surface force has resulted in a stress,
what is called as surface traction and body forces. Body force is defined as F per unit
volume of the body. If it is going to be defined per unit mass, then you have to correspondingly
bring into account the density of the body and so on; rho of F has to be brought.
This is the equation and that is going to be equal to zero. We had defined this in terms
of an indical notation as well and said that, that is how we had written it in terms of
indical notation. We also said that we can replace F, sorry T in terms of sigma so that
this term can be written as sigmaji nj. The next thing That is where we stopped in the
last class, just summarized it and next thing we said that we want to replace the surface
integral by means of a volume integral and we want to use divergence theorem. That is
where we stopped and what is divergence theorem? We had already defined what is divergence
theorem?
Say, take a quantity phi; phi i comma i dv is equal to surface integral phii ni dS. This
is what we had talked about in the last class. I had already said that this coma means it
is differentiation with respect to xi and so. Now, applying this equation to my last
step here in this equation, the last step of the previous equation for this place at
this place, what do I get? Sigmaji nj will become now, what is that it will become? Sigmaji comma, that is important,
j now volume integral dV plus integral V Fi dV is equal to zero.
Bringing them under one case or one integral symbol, I can write this as sigmaji coma j
plus Fi dV is equal to zero. What is that you learn from this? What is that you learn
from this or what can you conclude from this? Look at this equation carefully. It means
that the whole body is under equilibrium. When I say whole body is under equilibrium,
if I take a small chunk from this body, a small chunk from this body and apply this
equation, it has to be under equilibrium. This equation has to be valid for arbitrary
volumes of the body. I take small volume, I substitute it in terms of traction forces,
this has to be valid. So, this has to be valid for arbitrary volumes of the body or in other
words this has to be valid at every point of the body.
I can consider infinitesimal volume around a point and say that I take this out, apply
the same procedure and I will get the same equation. For every infinitesimal volume,
at every point that you can define, this equation has to be valid. What does that mean? That
means that when this total guy is going to be zero, it is not zero on an average sense.
It is not zero in an average sense but it is zero at every point in other words zero
because this chap here is going to be zero at every point.
I get sigmaji coma j plus Fi is equal to zero. So, that is the equation which actually I
get by considering equilibrium equation or by considering the equilibrium is called as
an equilibrium equation. Please note that I started with sigma F is equal to zero. How
many equations are there by the way? There are three equations. Correct; there are three
equations. Please note this carefully, students tend to make a mistake that this i is not
a repeated index. This i is not a repeated index. This i sits in one part of the term,
this i is another part of the term. The repeated indices are valid only if it is grouped like
this.
So, you should not say that this is one equation because both j is repeated as well as i is
repeated. Please note ultimately the summation operation signifies a tensor operation; summation
operation signifies a tensor operation. Hence this means that I have to apply i is equal
to 1, i is equal to 2, i is equal to 3 and the summing up is valid for j. So, you can
say that then, i is equal to 1, I get, what is the type of equation I will get? dow sigma11
by dow x1 plus dow sigma21 by dow x2 plus dow sigma31 by dow x3 plus F1, which is the
body force in one direction is equal to zero. Like this you can write down three equations.
This is the equilibrium equations; very, very fundamental equations. What all you do is
to solve this or variant of this; you either solve this directly or variant of this. Through
that procedure only we are going to get stiffness matrices and so on.
Let us now look at the next part of the equilibrium equation. Just to recapitulate so that you
are not lost in these whole things, what all we have been doing, please note that we started,
where did we start? We started with a small anomaly in the definition of stress which
we know. We had In all our earlier classes or in the beginning classes of your engineering
you would have learnt stress to be force per unit area. We thought or we saw that there
is an anomaly between that definition and the result what you saw in the first class
and hence we started defining stress. We defined it much more rigorously; we did not compromise
on your understanding. We said, yes, your understanding is right, but I have to define
that in a slightly more rigorous fashion.
Hence I defined stress taking a point and taking a plane. From that we define state
of stress, from that we came to equilibrium equation. One small factor is left, but I
am going to do this derivation, not because it is very difficult, though the result is
extremely important. I can do that in a much less rigorous fashion, but I am going to do
this in a very rigorous fashion basically for you to understand the use of indical notation.
It is going to be a small jugglery here and there and hence that will enhance your understanding
of the indical notation and hence I am going to do that only from that point of view. We
will come back to this figure now and put a small say add some things which we have
not done right now I hope. Any question on this?
Now, let me add to this figure what is called a position vector. Let me define a position
vector say to that point to be say r. What is r, a position vector, which means that
it is defined by means of x1 x2 and x3. They are nothing but the coordinates of that point.
So, that is the position vector. What is that I am going to do now? I am going to say that
sigma M is equal to sorry sigma M is going to be zero and what is the effect of sigma
M? What is the effect of sigma M? That is what I am going to do now. How do I do that?
What is that I am going to do? I am going to take a moment about, say, the O, origin.
What is the result or how do I take the what is the moment or how do I get it? By defining
the cross product.
You see that there is a vector operation called cross product which you are using because
you have understood what moment is and when I do a cross product, I will get this moment.
See, lots of difficult things are simple or made simple by taking a cross product. Once
you understand what the cross product is, you can take moment without much problem.
So this cross product is associated with this kind of activity, determining moment or physical
phenomena.
So, let me write that for the body forces and then for the surface forces; that is equal
to zero. Yeah! Any question? Now let me express this equation in terms of indical notation.
What is the indical notation we had used? Please note that carefully for cross product
or please look at it what is the indical notation that we have used? Correct, fantastic! So
epsilonijk, epsilonijk; here rj Fk; rj Fk dV plus same thing epsilonijk rj Tk n dS is
equal to zero. Tk is the stress vector. Now I want to replace this by our stress tensor.
You know the operation I am going to do, same thing as what we did before. What is that
I do? This quantity, I am going to replace this by sigma. How do I write that? How am
I going to write this? That is the crux of the whole thing.
Let me make a, I would say mistake. Can I write this as sigmajk nj? Can I write like
this? Because, Tk Immediately you have a tendency because, I have been using j there, so, immediately
or Remember we wrote Ta is equal to sigmaji nj; that is how we wrote. So, when I say k
here, immediately the tendency for students is to write jk nj; that will be the tendency
to write. Now you replace it with that. So, let me replace that with that. Let us see
what happens?
Sigmajk nj dS; this is n. Look at that equation. Does it appear nice to you? Does it appear
nice? What is the problem? Fantastic; so, j is repeated. I am not very happy with that
or it is not the right way to do things. I know that these two indices have to be repeated
and I know that they are dummies sitting there and I also know that if I use any other index,
I will mean the same thing. So, what is that I do? I am saying that this is wrong and so
let me replace it with another index say l.
Hope it is clear; nl dS is equal to zero; nl dS is equal to zero. Any question? Yeah!
Yeah! How is j repeated? See, when I put j here it does not matter, whether it is in
a same variable or in an independent variable, as long as they are in the same conglomeration.
For example we said sigmaij deltaij or ji, when they are in the same conglomeration or
same assembly, to understand that, then I said that you have to add it because j gets
repeated. That is sigmaji delta sigmaij deltaji or deltaij I said you have to add it. Sigmaij
epsilonij, you have to add it or sum it up rather; summation notation is valid. It is
not that the summation is valid, only if it gets repeated for the same variable or the
same tensor; it is not that the sigmaij should become sigmaii.
Say, for example ui vi; i gets repeated in u and v. The result is a scalar. Same fashion,
here it does not matter where it gets repeated. So, it becomes epsilonijk rj sigmalk nl; looks
quite complex but once you understand, it becomes very easy to understand. Take a minute
and look at this and see what the result of this kind of operation is or what results
from this operation? What are the fellows who get repeated there? k gets repeated, j
gets repeated, l gets repeated and what is that ultimately you have? i, which means that
what is it? It is a vector. That is the result of a cross product.
Now, we go ahead and apply divergence theorem to this particular part of the equation because
I want to convert this surface integral into volume integral. Let us write the first step
here. Integral epsilonijk; I am going to again, you know be aware; I am going to ask you question.
I am not going to write that so that you will become familiar with it; Fk dV plus plus let
us see what you have to offer to me. Where? How am I going to write this? How am I going
to write that? Because of the fact that this differentiation, I mean differentiation with
xl does not give me, it is zero so I can write this as epsilonijkl sorry ijk rj. No; both
of them are there know, no both of them are there. So, rj, correct sigmalk coma l dV is
equal to zero; sorry. Let me bring them together and see what are the things that I am going
to get with it? So, let me write this as V epsilonijk into rj into I have to differentiate
this rj coma l sigmalk plus rj sigmalk coma l. So, the first part I am taking; rj into
sigmalk coma l plus Fk, dV we will put it at the end plus sigmalk, correct, rj coma
l the whole thing dV is equal to zero. Is every one with me? What does this give me?
Actually we have done quite complex operations. I think now it is so very logical and straight
forward. I do not think there is any difference, I mean difficulty in this. Now, look at that
and tell me how I can simplify this. How I can simplify this? What is the first term?
That is the first term is zero. Why? Because, it is the result of my equilibrium equation;
very good. I want you to participate because, then you will understand whatever we have
done; fantastic. So, this term goes to zero. I have only this term. Now, let me look at
that term rj coma l. What is rj coma l? No; dow r1 by dow x1 or dow r1 by dow x2 and so
on. So, what would be the result of this? What will be the result of this? No, no; please
note this carefully. What is r? r is nothing but the position vector; position vectors
consists of r1 r2 and r3. Note this; r1, r2 and r3. r1 corresponds to the position in
the x1 direction, independent of x2 and x3. There are in other words In this euclidean
space we need three quantities to define, r1 r2 and r3. So, r1 They are all independent.
So, r1 is the quantity which is independent of 2 and 3; 2 independent of 1 and 3,3 independent
of 1 and 2. What is that I will get out of this? What are the quantities that are involved
here? They are say dow x1 by dow1 which means j and l is 1; dow x1 by sorry dow r1 by dow
x2 dow r1 by dow x3 and so on. Since the r’s, it depends upon only the corresponding x,
so what will be dow r1 by dow x1? That will be 1. But dow r1 by dow x2 is zero; correct,
because it is independent. So, many people what they do is, which may be also easier
to understand is, instead of r they will write x also; or in other words the position vector
x will be written as xi ei.
I did not do that in the beginning because, it may be slightly confusing. But I think
now it will become clear so that when I now differentiate dow x1 by dow x1, it will be
1 and x1 by x2 will be zero and so on. It is also a nice way of looking at it. The whole
thing comes from the fact that this coordinates are independent. So, if you want you can replace
it, this r by x as well so that you do not get confused.
I hope you understand that and you can replace it by x as well without losing the understanding
of what is position vector? So, this term now is equal to how do I write this now? So,
xjl is equal to 1 when j is equal to l and is equal to fantastic, one minute and equal
to zero when j is not equal to l.
Fine; so this is equal to deltajl. Is every one with me? So, this xj coma l is equal to
deltajl is one thing where you will repeatedly be using in mechanics; lot of places you will
use this. It comes from the fact that x1 x2 and x3 are independent quantities and x1 depends
upon only the corresponding x and so on. Extending a same argument like what I did in the last
derivation that this chap, this fellow who is inside has to be independent of the volume
which I am considering, this term inside the bracket has to be equal to zero. Whatever
term that is inside the bracket has to be equal to zero; in this bracket. What is that
I will get out of this? I will say that epsilonijk sigma; before we proceed what is sigmalk?
Let us simplify this. Sigmalk xj coma l, what will that become? This is deltajl; this is
what we saw here. What will this become? What will this or what will sigmalk deltajl should
become? What is this? What does this indicate? This indicates that this is equal to 1 only
when j is equal to l. When j is equal to l it has, one minute, j is equal to l, this
is equal to 1. When j is not equal to l this becomes zero. So, sigmalk deltajl can be written
as sigma, no, no; when j is equal to l only this will have a value or else it will not
have a value. So, when j is equal to l only, it has a value or else it will not have a
value. So, l will be replaced by j. That is it. So, this whole term sigmalk deltajl will
become sigmajk.
If you have a doubt, you just write it down in your note book; expand this and you will
understand because deltajl, see, this is again one of the operations which you will quite
often do. When you multiply some quantity with deltajl or delta quantity, since delta
has a value only when these two are the same, immediately what you can do is go and replace;
remove delta, go and replace this index with this index, so, sigmajk. Extending my previous
argument which I am not going to repeat epsilonijk sigmajk is equal to zero. So, when I do this,
I mean you expand it. I am not going to do that now; thing is quite simple now for you.
When I do that I get to the fact that sigma is a symmetric tensor or sigmaij is equal
to sigmaji.
So, sigmaij is equal to sigmaji. What does it mean? It means that in the tensor sigma11
sigma12 sigma13 sigma21 sigma22 sigma23 sigma31 sigma32 and sigma33, this is what you have
to define and this is equal to this, this is equal to this and this is equal to this.
As I told you, it had lot of simplifications using indical notation and once such a derivation
you go through it again; may be you can go through it two or three times. Once you go
through this derivation you will understand completely the operations you can do with
indical notations. Patiently go through this and you will understand, then you will not
have any doubts about indical notations. This is the reason why I did this.
So, with these two derivations we move over to another important property, two more properties.
The ultimate aim, before we go there I mean, we will close this class by looking at the
ultimate aim of what I am trying to do. The ultimate aim is to interpret results whether
it is design or manufacture or whatever it is. What is my ultimate aim?
Now, all of us know that some tensile tests are done with different specimens; standard
tensile tests are done and you have what is called as yield strength of the material,
sigmay or in other words what we find from tests is given by one quantity or yield strength
of the material. The question is that look, you have defined 9 quantities and called this
by various names, tensors and all that. Now, as a designer my interest is to tell or to
determine when my material will fail; that is my interest. You have given me 9 quantities
here. But, from the material result or test result I have only one quantity. How am I
going to link up this with this? Which are the 9 quantities that I have to take or can
you now combine and tell me what that quantity is which I have to take? That is the question
and that is what I am going to answer in the next class.