In the last class we talked about what is called as stress tensor and we had completed
our discussion on tensor. We saw that it is symmetric and so on. The question that can
come to your mind is that in your earlier classes you would have probably studied about
shear stress and normal stress. In fact what probably you would have studied is, in any
plane you can find out what is called as the normal stress and shear stress. How is that related
to what we have studied? There can be a question in your mind or you have studied about principal
stress and so on. We will come to the principal stress in a minute, but before that let us
see what this normal stress and shear stresses are?
Please note that we started our discussion by defining what is called as a stress vector.
The normal stress and shear stress are nothing, but are obtained by resolving my stress vector,
whatever I obtained, from the formula which I derived for it from stress tensor, either
perpendicular in which case I get normal stress or along the plane in which case I get the
shear stress. We are right in the same path as what you would have studied. It is not
that whatever you have studied, it is not right or something like that. We are on the
same path, but we are putting it in a more rigorous fashion, that is all. So that is
called as normal stress and shear stress.
The other question which we just started is if I have a bar and if I do a tensile test,
I get what is called as yield strength of the material sigmay. I have just now shown
that stress tensor has 6 quantities; 9, but because of my relationship that sigmaij is
equal sigmaji, I have 6 quantities. The question that may come to your mind is I have only
one value as a material property, I have 6 values as stresses. Now, how am I going to
tackle this situation or in other words the question is can I somehow combine these 6
chaps into one fellow? Some sort of an equivalent criteria for a uniaxial, please note that
this is uniaxial situation. Can I get an equivalent uniaxial stress and compare that with the
yield strength? Yes, it is possible.
The corresponding theory relies on or is called as Mises stress. Let us go back to our slide
what we were looking at in our previous classes. Look at that slide now. You see that, what
we are seeing is Mises stress. We did a contact analysis of a wheel and axel together and
what we were seeing is what is called as a Mises stress. Let us look at the next slide
and see what we are looking at.
Again see clearly that it is Mises stress distribution, It is possible the theory of
which I will explain. It is possible to get one equivalent stress called the Mises stress
and then compare it with the yield strength. For example whatever stress value I have in
this particular transparence or slide rather, I will straight away go and compare it with
the yield strength of the material. It is about 37 Kgf per mm square or 370 Newton per
mm square. I will straight away compare whatever value I am getting there to the yield strength
of the material, I have obtained from my uniaxial test. Let us look at next transparency or
next slide.
Here again you see that we are looking at Mises or Von-Mises stress, again from Mises
theory. From all these examples, we noticed that we look at what is called as the Mises
theory or we look at what is called as the Mises stress. 612
In your earlier classes, you would have called this theory as failure theories and probably
you would have called this as distortion energy theory. You would have studied a number of
failure theories. You would have studied what is called as maximum shear stress theory called
Tresca’s theory. This is due to a person called Tresca and you would have studied distortion
energy theory and which we will now replace it with or call this as Mises theory. This
term failure theory is a misnomer and unfortunately students do not pick that up very clearly.
Failure does not mean fracture of materials. Lots of people make this mistake. Failure
theories do not mean that when Tresca’s criteria or Mises criteria is satisfied, the
material is going to break into two, no.
For design purposes it has long been assumed, for 100 years now, that if the material starts
yielding, it will not serve the intended purpose. This is what was first thought about and hence
they say that failure means, this failure means not fracture but failure to do the intended
purpose. That is what is meant by failure theories. It is not fractured; it is not failed
means it is not fracture, but it is failed to do the intended purpose for which you have
designed the component and hence people started calling this as failure theories. But the
current terminology today is not failure theory but they are called as yield theories or these
theories since they signify the yielding of the material, yield theories are nothing but
yield criteria, criteria to signify that the material is going to yield. In much more simpler
terms, failure theories signify the criteria for the material to yield. That is what we
usually call this as; I mean that is what we mean when we call things as failure theories.
In order to go into the details of this, we need two things. One is not yet defined or
one is not yet explained, Tresca. One is that we have to define principal stress and the
other we will do is to decompose the stress tensor in order to arrive at the theory due
to Mises. In most of the finite element packages, people look at what are called as Mises stresses
and hence it is very important we understand what is the basis of defining Mises stresses
and what exactly it means or else we will not be able to interpret the results. Before
we go to Mises, let us look at what are the principal stresses? All of you know, I am
sure, what principal stresses are, but we are going to do the same thing in slightly
different setting. What is this setting?
We have been looking at a complicated component with certain boundary conditions and certain
loading conditions and so on and we were always looking at a point inside this. Now, the question
is that if I consider all the planes passing through this point, can I get a plane or more
than a plane where or on which the stresses or stress vector is wholly normal, completely
normal. In other words can I get a plane or more than one plane where shear stresses are
absent or to put it more clearly, can I get planes where my stress vector, please note
this stress vector, is in the same direction as that of n. So, that is the question. Yes,
that is possible and such planes are called as principal planes.
How do I write this mathematically? I write it mathematically by saying that Ti is equal
to sigmani; Ti is equal to sigmani. This sigma is the magnitude of the stress vector which
is totally normal to a plane whose normals are defined by these n’s here. Please note
when I resolve here T1 T2 T3, they are along 1 2 and 3 direction. Suppose, I assume that
this plane is the if if I assume that this plane is the principal plane, then if that
is the normal then what I mean to say is the stress vector will be along that and what
I mean by T’s, please do not get confused, is how I am going to resolve T in three directions?
That is what is meant by T1 T2 and T3.
Let us go one step forward and write that equation with the result which we know for
T. What is that result? Sigmaji nj is equal to sigma ni. Compare this with for example
an equation of the form Ax is equal to lamda x; say, A is a matrix, x is a vector and Ax
is equal to lamda x. What kind of problem is this? Correct; so it is an eigen value
problem. Compare this equation sigmaji nj is equal to sigma ni with this equation. Do
you see that there is a similarity between the two? Yes; so what does sigma actually
signify? You can treat this is as actually a simple matrix, in order to understand it.
Though the concept may be slightly deeper than that but nevertheless, right now it is
easy to just treat this n as a vector, two vectors and you can look at this as a matrix.
So, sigma n is equal to some sigma into, please note there is a difference between these two
sigma’s, into n.
What is this? This is nothing but an eigen value problem of this sigma or principal stresses
are nothing but the eigen values of this stress tensor, sigma. They are nothing but the eigen
values of the stress tensor, sigma. We have shown lot of properties for sigma. It is symmetric,
many more properties we will define later and hence it is possible to say that this
sigma has eigen values and hence say that they are principal stresses or principal stresses
come about because there is a property called eigen value for this matrix sigma or a tensor
sigma. Though we started from here and came here we can always go from here, back to here
and say that yes, if I say I get eigen values what does it mean? It means that they are
nothing but the principal stresses.
Now, let us write down ni in terms of deltaij sigma sorry nj; deltaij nj. ni becomes deltaij
nj, so that, that equation can now be written as sigmaji minus, I am bringing that to the
left hand side, minus sigma deltaij nj is equal to zero.
What else do you get from that equation or how many equations are there by the way? 3
equations. If you want you can write it down as sigma11 minus sigma n1. Let me write down
the first equation; sigma11 minus sigma into n1 plus what is the second part? Sigma; you
know now, sigma12 is equal to sigma21 and so on and so it does not matter because that
makes it easier. You can say sigma12 n2 plus sigma13 n3 is equal to zero and you can write
down two more equations. In order that we have by the way Before we forget, what is
this deltaij by the way? What is this remember?
Kroneker delta and what is that? It is nothing but an identity matrix. Please remember that
and now what does this tell us, from our very fundamental linearization? I can have two
types of solutions. One is trivial solution for n1, n2, n3, which means that all of them
are zero and hence this equation is equal to zero. But if I want to have non- trivial
solution, what is the condition I should have? If I have to have a non-trivial solution then,
exactly; so, the determinant of the coefficients should be equal to zero. So, I can say determinant
of sigma minus sigma I is equal to zero. So, sigma minus, please note the difference between
this sigma and this sigma; always it is like that, but I am sure that the squiggle below
will tell you which is a tensor and which is a scalar. What do I get from this equation?
From this equation, I can get, a cubic equation. I mean from determinant sigma minus sigma
I is equal to zero, I can get a cubic equation and the cubic equation can be written as sigma
cube minus I1 sigma squared minus I2 sigma minus I3 is equal to zero.
What I have done is, I have just left some of the steps; you can easily fill it up. That
determinant is now expanded and written, I get a cubic equation, which is given by this
particular form where I1 is equal to sigmaii. What is sigmaii? Trace of sigma, correct;
trace of this matrix, called sigma. I2 is equal to half of sigmaij sigmaij minus I1
squared and I3 is equal to determinant of this matrix sigma. I hope you all understand
how I got it? I just got it by taking the determinant of that determinant of sigma,
minus sigma I. So, I have got this.
What are I1, I2 and I3? I1, I2 and I3 are called as invariants. In solid mechanics,
we have nice names. The names signify or have lot of meaning to it. So, when I say invariants,
it has a lot of meaning to it. So, it means it does not change; invariant, it does not
change. It does not change with respect to what? Why do you call it as invariant? Is
it with respect to plane? Look at this equation and tell me? Let us just slightly go back
because it is a very important concept which people usually do not understand. Look at
this equation here. I have got this equation and I want to find out sigma and sigma which
is the principal stress I would say, I will not call it a sigma; let me say that I want
to find out the principal stress and what is it that I have been given? I have been
given the stress tensor, sigmaij.
The stress tensor is a function of how I have chosen x1 x2 and x3 or coordinates. So, the
final equation which I should get and I should solve should be such that the principal stresses
should not be a function of coordinates; correct. So, it should be independent of the coordinates
which I have selected or else I can get whatever I want because principal stresses have to
be determined without the interference of this coordinates. What does it mean? It means
that this equation which I am going to solve in order to get my sigmas should be invariant
to the coordinate system. Hence the invariants signify or say that they are independent of
the coordinate system which you have used in order to get the cubic equation. Hence
they are called as invariants.
Once I have this, it is not very difficult. It is very straight forward to find out I1,
I2 and I3 and once I know what are I1, I2 and I3, I can very easily substitute them
in this equation, solve this cubic equation. You know so many methods to solve them. What
is the simplest method you can use? Trial and error will be too difficult. Newton graphs
and Newton’s methods can be used in order to solve them. There are very well programme,
I am not going to the details of it. If you have difficulty, let us see it. But I think
it is a straight forward method to solve the cubic equation. Once I solve the cubic equation,
I can get n’s. So, first of all when I solve the cubic equation what is that I am going
to get? I am going to get three values of sigma.
I am going to get sigma1, sigma2 and sigma3. What are these things sigma1, sigma2 and sigma3?
They are the eigen values of the matrix sigma. Corresponding to this, there should be eigen
vectors; there should be eigenvectors. In this case what are these eigen vectors? They
are the principal planes. They are the n’s. So, for each of these eigen value, I will
get one set of n’s, which I can get by solving the equation which we have written there,
just now we saw that equation, along with the fact that ni ni or n1 squared plus n2
squared plus n3 squared is equal to 1. By solving these two, we are going to have or
we will get an equation or we will get set of equations to get ni’s. So, n’s are
eigen vectors and please go back to your theory on principal planes.
What is another important theory which you would have studied? Principal planes; do they
have a relationship, principal planes? Perpendicular to each other - this is what you would have
used. Principal planes are perpendicular to each other. In more sophisticated way, you
say that they are orthogonal. Why because eigen vectors are orthogonal. So, you say
that whatever you have studied before comes beautifully from our theory that we can define
stress as a tensor and so on. Hence these principal planes are orthogonal. Once I define
principal planes, I can get the maximum shear stress. I am not going into the details, you
know it already, that the maximum shear stress can be defined as how do we define maximum
shear stress?
Let us say that the maximum shear stress is tau; correct. So, that is equal to sigma1
minus sigma3 by 2, after arranging sigma in a descending order; sigma1 being more than
sigma2 and sigma2 being more than sigma3. The shear stress, maximum shear stress is
given by sigma1 minus sigma3 by 2 and we also know that in the planes where maximum shear
stress exists, there are normal stresses and so on. Again this is the fundamental strength
of materials. Tresca’s theory simply states that when maximum shear stress reaches a value,
which is obtained by an experiment where the shear stress reaches a critical value or when
tau reaches a critical value, then we have yielding.
In a uniaxial situation, what is that? sigmay by 2; so, a critical stress in the uniaxial
situation is sigmay by 2. When tau reaches sigmay by 2, we say that yielding has occurred
according to Tresca’s theory. But Tresca’s theory is valid and it is quite good. But
there are certain difficulties, which we will not discuss right now, but it can also be
used. More important theory that is used for metals is what is called as the Mises theory.
Whatever we saw just now, we saw that principal stresses can be arrived at very naturally
from our definition of stress and is quite useful as well if you are going to use Tresca’s
theory or certain brittle materials where you can use the maximum stress theory as well.
You can say that the maximum stress is one which will signify my failure and so on. I
am sure all of you have studied this in your design courses. So, now I am not going to
repeat that.
On the other hand, let us go to what is called as Mises theory. Mises said that the stress
tensor can be decomposed, more importantly, addictively decomposed into two parts which
he called as dilatational and deviatoric. He said that the stress tensor can be addictively
decomposed into a dilatational part and a deviatoric part. A dilatational part is also
called as the hydrostatic part and a deviatoric part. What he physically said is that the
stress tensor can be split into two parts, one responsible for volume changes, the other
for distortion. So, deviatoric is also called as distortion; so, one part responsible for
volume changes and the other part responsible for distortion and that part is called as
deviatoric part.
He defined the dilatational part, because note the word hydrostatic part or the dilatational
part, by a tensor which he wrote like this where, what is p? p is equal to, correct, sigmaii by 3.
He said that the whole of this sigma tensor can be split into a dilatational part and
a deviatoric part, which comes out, as I said, it is additive decomposition. So, what should
be the other fellow here? He should be sigma11 minus p sigma12 sigma13 sigma21 sigma22 minus
p sigma23 sigma31 sigma32 sigma33 minus p. I hope you are Is it clear May be I will write
it further away. When I add these two, I will get my original sigma matrix. Why did he do
this? He did this because he found out, he and many others found out, that of these two
parts, it is the deviatoric part which is responsible for yielding and the dilatational
part does not play a role in yielding. Yielding is not signified by the hydrostatic pressure
which is superimposed on the deviatoric part. He said that that is the by number of experiments,
empirical; that is how he arrived at it.
He looked at this tensor. This is the tensor again and said that this should have three
invariance. He called this invariance as J1, J2 and J3. He called this invariance as J1,
J2 and J3. What is J1? Look at that; zero. J1, J2, J3 are nothing but obtained from my
equation here, this invariance; same thing because the three invariance are always present.
Please note this carefully. These invariance are always present when I look at a second
order tensor. When I say that tensor, second order tensor and I want you to write down
three invariance, these are the invariance.
In the same fashion, the deviatoric part also has invariance defined and these invariance
are called as J1, J2 and J3. What happens to J1, trace of sigma, here? Let me call this
matrix as s, in order to distinguish this from sigma. Yes, any question? Any question?
Yeah! So, we will call this matrix, what I am saying is this matrix is called s, so that
I need not write this down every time. I can say that this matrix sigma is equal to p plus
s. That is all; sigma is equal to p plus s. The first invariant of my s matrix, zero,
that is called J. I do not want to again write as I, because I already have I defined for
sigma and hence I write this as J. So, J1 is equal to zero. What is my J2? Come back
here; come back to this equation here. What happens to this?
That goes off; zero, so, half, instead of sigma, what do I write? Correct; half sij
sij. So, J2 is equal to half sij sij. What is sij sij? Just recapture what we did? Summation
about i and j. That is the doubt? Clear and J3, what is J3? Determinant of s. J3 is equal
to determinant of s. The logic is very simple. If at all there has to be a failure or yielding
in this case, sorry we will not use the term failure; if at all there is going to be yielding,
then the yielding should be independent of my coordinate system. I cannot have a coordinate
system and say that in this coordinate system there is yielding another coordinate system
it will not yield and so on. My yielding theory logically should be a function of J2 and J3.
My yielding theory should be function of J2 and J3. Just to recapitulate, this whole thing
here, what we did was to split this sigma into two parts. By experiments, it was shown
that dilatational part does not have much effect and deviatoric part has an effect,
only has an effect on yielding.
We took this up for further study. Then we found out that there are three invariance
J1, J2 and J3, three invariance and out of which J1 goes off by our definition of the
deviatoric stress and hence we have J2 and J3. The yielding is a function of J2 and J3.
That is what is important, it is a function of J2 and J3. this is where Up till this point
both Mises and Tresca are on the same line.
Mises said that he found, rather from experiments, that look that is not very important and for
metals, note this carefully for metals, he found that if I write yielding to be only
function of J2 that is sufficient for me to define yielding. What he said is, to put it
simply, when J2 reaches a critical value, when J2 reaches a critical value, material
starts flowing; flowing means yielding. His whole emphasis was in the fact that J2 reaches
a critical value. In fact he defined the function of J2 to be just J2 and then said that J2
has to reach a critical value. This theory, ensuing theory from Mises is also called by
the name J2 flow theory.
Someone says I am following J2, in fact many manuals say that what we follow is a J2 flow
theory. What it really means is that they are following what is called as Mises theory.
What is this critical value? That is our next question. If at all I know what is that I
know from experiments. Usually there are two sides to the equation, one side the left hand
side which is J2. We will keep this like that so that if there is a future reference, we
will look at it. On one side I have J2.
J2 has to reach a critical value which means that I have two sides of the equation. Right
hand side, I should have something to do with material. J2 depends upon what? J2 depends
upon the loading, geometry, boundary conditions and all that because J2 ultimately depends
upon sigma and sigma depends upon my loading conditions and so on. The left hand side of
this equation depends upon my loading conditions, boundary conditions and what not, geometry
and all that. The right hand side should depend upon the material property. Let us look at
J2. J2 is square of stress. Let me call this as say k squared; it reaches the critical
value called k squared.
What is this critical value? You know people We talk about all the time critical value.
What is this critical value? This can also be looked at some criteria or people call
this also as a criteria function. What is critical value? What is it in my hands? I
can take a specimen and I can load it, look at yielding. That is in my hands. The critical
value as far as I am concerned should point out the yielding of the material or should
be related to my sigmay, which I can see, monitor, measure or whatever it is. So, the
critical value should depend upon sigmay. What is that I am going to do? I am going
to apply whatever theory I have developed so far for yielding for a uniaxial case and
see whether I can get the value of k which should ultimately depend upon sigmay. That
is what we are going to do now.
Any questions so far? The logic is clear? This is a very, very important part not only
in design but also in finite element because ultimately when you interpret the result,
you are going to interpret only in terms of Mises stress. Usually people interpret finite
element results without understanding what Mises stress is. They will just look at the
Mises stress and then start interpreting. But actually you should understand that we
are applying our well known, within quotes, failure theories or yield theories and that
is how we are doing the whole thing. May be this is not the way you would have studied
it in your earlier classes. You would have just got one expression for distortion energy
or whatever it is, but this is the logical way in which, solid mechanics or continue
mechanics people develop it. It has lot of implications later.
We will stop with just one sentence as to what is a stress tensor for a uniaxial case.
This is the uniaxial case. What is the stress tensor? Let me have say x1 in this direction,
x2 and x3, in which case how do I write the uniaxial stress tensor? Let me call this as,
say, sigma1 or sigma11 if you want. So, it will just be sigma11 00 000 000 period. This
will be sigma11 00 000 000. Remember that this is only the stress tensor, not the deviatoric
stress tensor. What is p in this case? Sigma11 by 3 and hence I have to subtract that from
this in order to get the deviatoric part. We will stop here and proceed to the next
part in the next class.