So, good morning. Last lecture we talked about plasticity and how for plasticity to describe theory, to have a theory of plasticity, we need four elements. An element, first of all, that describes the elastic

stress-strain relation, so we know we have Hooke's law for that. We need a criterion for yielding, we need to know at what stress level do we get plastic

permanent deformation. We need to have a hardening rule because we need to know how the initial yield conditions will change as we deform material. And we also need a flow rule. A flow rule that describes the evolution of the plastic

strain during the deformation. And we'll see that this flow rule, in the case of plasticity, will connect plastic strain increments to the stress.

All right, so, well, Hooke's law, we know Hooke's law, first condition, we already know we have the theoretical elements to do elasticity. For instance, in the case of uniaxial stress, if I express my Hooke's law in terms of principle stresses, we find, if I express

this incremental way, the small amounts of increase in the stress result in small increases in the strain, the epsilon, and I can easily compute this for uniaxial conditions,

straining conditions, and you see the equations here, rather simple. I'm probably using it for a long time already. So, when we discuss plastic deformation, we have two things to consider. First, let's

look at what happens when we deform this little square here, and it becomes this dashed shape here, geometrical shape. So, this deformation, I can cut it up in a number of events. First of all, I can have a volume change separately, and then a shape change.

So, in this case, there is a change in size, you can see, of this square, and here there is no change in the size of this square, because the shearing, I have the material that used to be on the left side here, this wedge of material is now

on the right. So, there's been no change in the volume of this thing. And in addition, this final shape can be rotated and translated to mimic an actual plastic deformation. Let's look at a specific situation where we're asking ourselves

what's the work that we need to do to achieve this volume change and the shape change. Well, let's look at the volume change here. So, if I strain the material in this direction, it will get longer in the x direction, and I will carry out work

which is given by this well-known equation, half of the stress that I apply on this side of the cube times the strain in the x direction. And if I now do a shape change of this cube, I just shear it a little bit

in the x direction using this shear stress T xy, this is the work that I have to provide. In this case, I have clearly a volume change, in this case I don't have a volume change, but I have a shape change.

All right. So, we're going to need to say something about the stresses in a material. Yes. And we are going to do this by looking at a well-known approach

that you probably know from your undergraduate lectures, and that is the Mohr circle approach. And that Mohr circle, most of us have seen it, we've learned how to use it, but very quickly we've really lost the ability

to use it because it's rather complicated in terms of orientation. And you know there are things where if you want to know the stresses, you have to rotate in the direction different from the direction that's on your

graph, and then you have to sometimes divide the angle by two, and it's very and so we tend to forget how it works and we don't use Mohr circle. But there is a simple way to use Mohr circle where you don't have to worry about these things. And so let me explain this, how it works.

So from now on you can use Mohr circle whenever you need it. Well, first of all, one of the reasons why we have problems using Mohr circle is because we need to make conventions about

positive and negative stresses. And the convention we use is for the stress state. So if you look at this little cube here seen on its side, we first of all define two planes. We define this right plane here.

Always we call that the H plane. We call it the H plane. H for horizontal because the normal forces are horizontal and this plane here on top we call the V plane. Vertical because the normal stresses are vertical on this

plane. So you have to imagine that is a little element in a tensile specimen or another piece of steel that's being deformed. Elastically in conditions of plane stress.

The situation is such that we can ignore the stresses that are perpendicular to this plane of projection here, this screen. So that's number one. So we define an H plane and a V plane and then we say that when the normal stresses are pointing

outward, when we're talking about tension, then we have normal stresses that are positive. If we're talking about compression, the stresses, normal stresses compress this little volume then the signs are negative. Makes sense. The convention is more important

when it comes to the shear stresses. When the shear stress on the H plane points up, we say it's a positive shear stress. If it points down, we say it's a negative shear stress. If it

points up in this direction, then in the V plane it points to the right and these are positive, two positive shear stresses by convention. So if it points downward on the H plane,

then we call it a negative shear stress. So let's do one more thing with this convention, the elements of convention, is when we are ready to draw more circle, we have an axis

where we will plot the normal stresses and a Y axis where we will plot the shear stresses

and what is important is that the positive direction points downward. So that's a little bit something you have to remember, positive directions point downward. So what we basically have in this plane here, in this plane stress plane, is any

stress condition can be, is one point. So if I have my stress axis and my shear stress axis, any

point in this graph defines a stress condition. And in order to do this, I need my more circle. And to build up my more circle, that's one

thing. And then the second thing is that once I have my more circle for a specific stress condition, I can of course not only have my stress situation on the faces of a

small cube in the material with this orientation, but for any other orientation. So I can find out in what direction I have only tensile stresses or in what direction I have the highest

shear stresses. So the best way to do this is to give you an example. So say first that you have a stress state, you know the stress state,

so in two dimension plane stress, that means you have an x, a sigma xx, a sigma yy, and a shear stress xy. And shear stress yx is the same in size.

So these three points will help you form or draw a more circle. And we can do this by computing the dimensions, the size of the principal stresses and the maximum shear stress.

And these are the formulas. The principal shear, you have two principal normal stresses, sigma 11, sigma 22, which are given by these formulas and the maximum shear stress is given by the mean value of these two

principal stresses. So if I have a condition of stress state, this point here, it has a sigma xx normal stress, tau

xy shear stress. And then I can build the stress state, obtain the stress state on the other plane here, that is here.

So that is sigma yy and tau xy. So that is positioned here. And the more circle is the circle that goes through these two points. So if I know what sigma 11 and sigma 22 are and tau max are, I can easily find the center point of this circle

and the radius of this circle, because the radius of this circle is equal to this maximum shear stress. OK, so let's let's see how this works out in practice. Say I have

a stress state sigma xx is 80 megapascal, sigma yy is 40 megapascal, tau xy is 60 megapascal. So the stress state that I've measured for this little element here is 80,

it has its coordinates 80 and 60 for this plane. So 60 is on this side, so it's positive. So the arrow points up. Sigma xx is 80, it's positive, so it's pointing normal and to the right on the H plane. On the Y plane

I have minus 40 for the normal stress, so it's a compressive stress, and then a negative, excuse me, the shear stress is here. The negative, that's important, the negative stress. Alright, so now, and you can calculate sigma xx, sigma

11, sigma 22 and tau max. OK, the tau max is here, and this is the situation with the values indicated. So step one, we have our circle. Step two, yes,

we find the pole, the pole point. The pole point is very simply, so the stress state here corresponds to this plane, and the stress state here corresponds to this plane. So then, so this is what we

call the V point, that's the stress state in this situation, and this is the H point that is the stress state on this plane. So we draw a horizontal through the H point and a vertical through the V point.

And that gives us the P point, it's always the P point. And now, if I want to know the stress rate on any elementary volume of surface at other orientations, so I want to know what was the stress state on this plane, or what's

the stress state on this plane, or what is the orientation of the planes where I will have the largest shear stresses, or where is the shear stress. Well, very simply,

if I connect the pole point to this point, yes, to the sigma 11, yes, I find an orientation, yes, where there are no shear stresses, because this condition

of stress, yes, I find a normal stress in the X direction of 104 and in the Y direction

of minus 46. So these, given by these two red points. And there are no shear stresses. And if you were to draw the more circle for these

conditions, you would find that you have these stresses are according along the principle directions. I have principle directions. In what conditions do I have the maximum

shear stresses in the material? Yes, well, here is the maximum shear stress, yes, if I continue in this direction. And you can see that the shear stress here is

minus 48. So as we go from this surface, yes, and we gradually change its orientation, yes, I go from a situation where

there is biaxial stress, I have shear stresses, then I have a situation where, so I have a shear stress like this, then I have the shear stress becomes very small, then there is no shear stress, and then the shear stress becomes negative.

You can see that on this H-plane that the stress is now negative, and indeed here we are in a negative situation. And you see that in this case I do not need to guess this angle or divide it by two. I can just read off

the stresses, the shear stresses and the tensile stresses in this particular case. Please note that when you have maximum shear stresses, you still have, you can still have normal stresses. However, when you are along,

when your stresses and strains are along, oriented along the axis of principal stress, then you have no shear stresses. You only have normal stresses. Well, that was a general case, but now let's look

at an even simpler case, the case where we have a tensile test. So in the case of a tensile test, the stress axis is already along a principal direction. So

in this case, my little element is here and I apply stress in these horizontal directions. It's positive stress. It's not compression. It's tensile stress. And there is no stress in the Y direction.

Yes. And because my stresses are oriented parallel to the principal directions, I have no shear stresses. So sigma XY, sigma 12 is zero. So this is the value of the stress

in the X direction. In the Y direction, it's zero. So I'm here and there's no shear stresses to take into account. Now I'm going to rotate this element and come

to a situation where and try to find at what angle I reach the maximum shear stress. Well, so that means just tilting the cube along the first before I continue. The P point in this case

is here. It's the same as the V point. Yes. So if I now tilt this element, the H plane is here. Yes. And I see that the value of the shear stress increases. You can see it increasing and then reaches a maximum at this point.

Yes. Okay. What is interesting is that this angle is, because these two lengths are the same, this angle is 45 degrees. And indeed, that's what we know is if we have a tensile test, the plane on which the

shear stresses are maximum are inclined at 45 degrees. And you can see here that it works nicely. Again, note that when that element is

oriented to give you the maximum shear stresses, there are still normal stresses acting on the faces. Okay. Right. So how are things in three dimensions?

So if I have a piece of metal that's subjected to a number of stresses, external forces, it can be shown

that the stress state can be represented as an ellipse.

An ellipse. Like this. And the three axis of this ellipse form an orthogonal set which are oriented which give me the directions of the principal stresses.

So I can represent and in the case of a Mohr circle, which you're basically looking at, is a plane

stress condition. And as you know, that you have a Mohr circle. And the Mohr circle is in fact nothing else than a section through this stress ellipse of revolution.

So working with this stress ellipsoid is not so convenient. We can use another view of it. We can look at it in principal stress space. And we can also use stress diagram. So if we look

at principal stress state, we basically have in that space, we have axes which are aligned along the principal directions. And a stress state, that is the stress on a particular

plane here, is defined by this vector from the origin to A. And it has components along the X, the Y and the Z direction. So alternatively, you can think

of little planes. And for a specific stress state, you will have a component normal to that plane and a component, a shear component in that plane for any stress situation.

And we can, in addition to this, make a Mohr circle construction for three dimensions in two dimensions by having a Mohr circle for each set

of two principal stresses. So you have a Mohr circle for instance in this case A, that is sigma one and

sigma three, sigma one and sigma two, and sigma three, sigma two and sigma three. And as long as my stress state, defined by tau n and

sigma n, is within this larger of the three circles, my material will really not deform if or not reach specific maximum

shear stress. So let's perhaps have a look at specific conditions so we know what is meant by these three diagrams, by the stress ellipsoid, the principal stress

space and Mohr stress diagram. The Mohr stress diagram is in two dimensions. So let's look at the stress ellipsoid if we have homogeneous tension or compression. That means the material is compressed or under tension in three directions and the stress or the

pressure is the same. So in this case our stress ellipsoid so we know that the principal stresses are equal, so our ellipse becomes a sphere.

In the stress space, my stress condition, I have negative value equal to p minus p for x1, for along sigma one, along sigma two and along sigma three. So that's point A, basically

is the stress condition in principal stress space. And in the Mohr stress diagram, I should have three circles which are defined by the differences in the principal

stresses. And while they're all equal, so on the x-axis, we just have a single point instead of circles.

Now let's look at the system that's of interest to us. This is the uniaxial tension. There, this uniaxial tension, this is in reality for the stress ellipsoid.

What you can think of it is basically as a line and half the length of this is equal to the applied tensile stress. Sigma two and sigma three are zero. You have a very, very narrow cigar, if you want to, with zero

radius. In principal stress space, sigma one is the applied tensile stress. And then I have zero sigma two and zero sigma three. So it's a point in principal stress space, of course.

And the stress diagram, Mohr's stress diagram, I have a large principal stress equal to sigma t. And the two other principal stresses are zero. So

they're here at the origin and they're zero. So they should fall on each other. Okay, let's make it slightly more complicated. Let's make a biaxial tension situation.

So that means I have now a little volume element where I apply the stresses in two perpendicular directions. So in this case, sigma three is zero. So instead of having an ellipse, I kind of have a

very flat ellipse. And depending on, so if sigma one is equal to sigma two, it would be a circle, a very flat ellipse because sigma three is zero. In the principal stress space, I have now my stress conditions are

sigma one along this principal direction and sigma two along the perpendicular direction. So this c gives me the stress state. And then Mohr's stress diagram, yes, here we have to be careful when we use

Mohr's circles. Never forget that although sigma three is zero, it should be taken into account. So we have three, we have not two principal stresses, but three principal stresses.

So these are our three Mohr's circles. we can, I'm not going to go into it, but obviously the plane

on which we will have the maximum shear stress in this condition is determined by the circle, Mohr's circle, defined by sigma three,

the stress in sigma three which is zero along sigma three that's zero and sigma one. And the angle beta here is equal to,

gives me the orientation of the plane where I will see the largest shear stress. And so interestingly enough, not influenced by the value of sigma two. All right, good. So now we're kind of

more comfortable, let's say, with stresses and that we can define a stress ellipsoid and there is a stress space where we have

as axis, we use the principal directions of the stress and we go back to the problem of shape change and volume change during deformation of a solid piece of steel.

So we have, we're going to look at the energy to deform this piece of steel, we're going to look at the hydrostatic part of this energy which is related to the volume change. And we're going to look at a deviatoric part of the

energy which is related to this change of shape. So if I apply the equation we saw earlier, one half epsilon times sigma

being the energy needed to change the volume of a small cube if I pull it along the x direction. And I generalize that in three dimensions. Again, expressing, using

only tensile, using a coordinate axis where we can ignore, which allows us to ignore shear stresses, yes? I find for the hydrostatic, for the for this energy of the formation of

volume change, one half sigma x times epsilon x sigma y times epsilon y, sigma z times epsilon z, yes? And I can get an equation for this energy which is

only dependent on stresses by substituting for the epsilons here the relation given by Hooke's law. So sigma x is, epsilon x is replaced by this first equation

in Hooke's law and I do the same for epsilon y and epsilon z. And doing this, I find this nice symmetrical equation one over two e sigma x square sigma y square sigma z square minus one over e Poisson

ratio sigma x times sigma y, sigma y times sigma z and sigma y times sigma z. And this should be sigma z times sigma x. So please correct this if you're using these notes, should sigma x

z times sigma x. And I will correct this in the notes that are online. So we're going to work further on this formula but before we do this we're going to introduce what we call deviatoric stresses, yes? And to do this

we first start by defining a mean stress. The mean stress or the average stress is basically the sum of sigma x, sigma y and sigma z divided by three. Very reasonable way to define the mean. And then what we're going to do is define

deviatoric stresses which are basically the stress in a particular principle direction minus the deviatoric. So sigma x is the mean value times the deviation from this mean value. And we call this the deviatoric

stress in the x direction, y direction. Z direction is nothing else than sigma x minus sigma a, sigma y minus sigma a and sigma z minus sigma a. Now that

has interesting property, this approach. Well first of all let's just assume that we only have hydrostatic

stress. That this sigma a is equal to sigma x, sigma y and sigma z. So in this case the

work done by this stress, we have this kind of stress conditions where there is only a hydrostatic stress then the energy for the deformation is given by this equation.

This equation. And we'll come back to that. Now the situation we've explained up to now we just had was very general. Let's see what the deviatoric stresses are in tension. We will have again our simple

tensile test. So the deviatoric stress is sigma x minus one third of sigma x plus sigma y plus sigma z. Where in tension, sigma y and sigma z are zero. So sigma x

prime deviatoric stress is two thirds of sigma x, sigma y prime is one third of sigma x and sigma z prime is one third of sigma x with a minus sign both. So the sum of the deviatoric stresses is zero. Yes. And that's a rather

general. It's here we derived it for a tensile situation but it's actually a general rule. The sum of these deviatoric stresses is zero. And we will use this property in a moment. So that's

a lot of math here and equations but let's have a look in principle stress space what these deviatoric stresses mean. So in stress space, a stress

condition is a point. For instance, this point B is a stress condition. It means it's got an x, a z, a y and a z coordinate in this space. And what we do

when we use the deviatoric approach, deviatoric stress approach is we have the deviatoric stress component in the three direction, in the z direction, in the x direction and the y direction.

Together they form the hydrostatic component to the stress state. And then I have a component, a deviatoric component sigma x prime

which is sigma x minus sigma average along the x direction, a sigma y prime along the y direction and a sigma z prime along the

z direction. And so this vector here that connects A to B, that's the deviatoric stress situation. And you see that B is also equal to sigma x, sigma y, sigma z. And so the sum

of the hydrostatic and the deviatoric stresses are the same as the original stress state. And again you can clearly see that the deviatoric is sigma x, that's this distance here, that's the sum of the deviatoric

stress plus sigma a. So again, in uniaxial tension we only have a stress along the x axis, along this sigma 1. So this would be the applied stress. So I can decompose

this in a stress that has one third the component, one third of sigma x in the z direction, one third of sigma x in the x direction and one third sigma x in the y direction. So that gives me

my hydrostatic stress component. And then the deviatoric is this one here, has components, so the sigma x prime, which is this, a sigma y prime in this direction and a sigma

z prime in this direction. Now what is interesting is that this vector, the deviatoric, is perpendicular to the hydrostatic component. So again, what this illustrates with the uniaxial tension situation is that

any given state of stress can be divided in a hydrostatic component and a deviatoric stress component. And what we'll see is that the hydrostatic component will be responsible

for volume changes and the deviatoric component is the stress component which is responsible for shape changes. But first let's do a little bit of the math to show this

and also see how this is connected to yielding. So if we calculate the volume change, from this little block here, and say this block has a volume one, a unit volume that is the starting volume.

And we strain it in the x, y and z directions and we get a new volume, volume v. So what is the volume change? The volume change is the starting volume times these strains that I applied.

So this is not the volume change but the new volume is one plus, this was one plus the strain of this one plus the strain in this direction, one plus the strain in this direction.

And because epsilon x, epsilon y and epsilon z are very small I can approximate this product by one plus the sum of epsilon x, epsilon y and epsilon z. And having this, the dilatation which I defined

as the volume change divided by the original volume is this sum, epsilon x plus epsilon y plus epsilon z. And that is non-zero in the case of an elastic deformation, I can have a volume change.

But it is zero when we have a plastic deformation. So we are going to look at this sum, this dilatation and express it in terms of the stresses.

And again, we do this using Hooke's law. We know that epsilon x, epsilon y and epsilon z are given by these three equations for x, y and z, again parallel to principle directions. So epsilon x plus y

plus epsilon z is equal to one over x sigma x plus sigma y plus sigma z. And then this other term, two times the Poisson ratio divided by e and the same sum.

We now do one more step. We are going to use, make use of deviatoric stresses. And so when I do this, when I do this, I replace sigma

x by sigma a plus sigma x prime. I find three times sigma a, three times sigma a plus sigma x prime, sigma y prime plus sigma z prime. And this sum, sum of deviatoric

stresses is zero. So this equation is very much simplified to this one. So the volume change, yes, when you do an elastic deformation is equal to, yes, an

elastic deformation of any type, yes, is entirely due to the mean stress only, yes. So, and of course the value of the elastic constant. So it's three times one minus two times the Poisson ratio divided by e times sigma a, the

hydrostatic component of the stress state. And of course, you can see that this here is nothing but the bulk modulus or one over the compressibility. Okay, so we basically

know what the energy is that we need to change the volume. I basically need to have this equation, this here, yes, this

here. Multiply it with sigma a, the stress, and divide by two. And so this is what I get. That is basically the work needed to do the volume change, yes.

Okay, and again, I can change the make use of the relation between the

sigma a, the average stress, and sigma x, sigma y, and sigma z to get this equation for the volume

change energy. And so this allows us now, because we know what is the general energy required for deformation. Yes, for

a deformation. And we know the part of the volume change of that work. So the actual shape change, ud, the energy needed for the shape

change can now be computed as the sum, excuse me, as the difference of the total energy required minus the volume change energy required. Right, so, and if we do this, so

the last part here comes from this equation, and this part here comes from the equation we've seen, we have derived earlier, which at the very beginning, which is this equation here.

One over two sigma, oops, excuse me, one earlier. Yeah, this equation here. That's the total energy for shape

change and volume change. Right, so now we're back here. Yes, we need to do a bit of manipulations, but at the end you find the energy needed for the shape change is given by

this very nice and simple equation. If we, again, do not have to take shear stresses into account. And now we look at one of the many

criteria for the initiation of yield of plastic deformation, and a very famous one is that of formesis, which we can actually use in many cases when we're thinking about

deformation of steel. It says it's a maximum distortion energy criteria, so a maximum shape change energy criteria, and it says that yielding will take place when the amount of distortion I've given to the material reaches a maximum

value, and that value is the same value it would take to distort, to start yielding in a uniaxial test. So having material, the same volume of material, and I

apply stress to it, then if I have applied the work that I do has reached the same level as the work it takes me to let a tensile bar yield,

per unit volume, and in two cases, if I have the same volume, it's of course per unit volume, so if these two values are the same, then I reach yielding. So what do we do? Well, this is the shape change energy, so now we're going to say, well

let's just put in the conditions for a simple uniaxial tensile test, so in this case sigma y is zero and sigma z is zero, so this equation here and sigma x is equal to the yield stress. So

the energy needed to have the material yield in uniaxial deformation is given by one plus nu divided by three epsilon times the yield strength square.

So if I look at these two things, this is for a general deformation, excuse me, general stress state and this is for the uniaxial stress state, I can see that these constants will disappear except for a factor one half, and

I will get this nice yield stress, yield criteria. This factor here, sigma x minus sigma y squared, sigma y minus z squared and sigma z minus sigma x should be here,

not yield stress square, is equal to sigma yield stress. So please correct this in your note, this should be sigma x. If you, this may not be the best moment to introduce

this, oh, I see that we are missing some material here, but that's not a big problem, so let me go back here.

So this equation here, this equation here is nothing but the equation of a cylinder, a cylinder in the

stress space with a axis, axis of the cylinder along the diagonal

of the three axis here. And it basically tells me that if the stress state is inside this cylinder, the material will not yield. If

my stress rate is outside this point, the material will yield, will plastically deform. One of the interesting points to look at in the

stress space is the intersection of this cylinder, yes, with the, this is sigma one, sigma three,

two, yes, the intersection for situations which we call plane stress situations, the yield

surface becomes a yield function in the sigma one, sigma three plane, which

is an ellipse. So because sigma two is zero, we call this condition plane stress, plane stress conditions, yes? Okay, there is no, no stresses

in the third dimension, and our yield surface looks like this. That is very important yield surface, yes,

yield line if you want, because it corresponds to situations that we often encounter in practice with steels. Very often we are dealing with rather thin shells, yes, where there is no stress on, in the

third direction, in the third dimension, mainly stresses in the plane of the shell, or the, there is not much variation of stress in the perpendicular

direction to sigma one and sigma three. Now, let me show you that in the case of steels for instance, for a number of different steels, the

Yielding behavior has been studied in detail. And we find, indeed, that the yielding condition is very close to this equation here, this equation.

And it tells us, basically, that this concept of poniesis, to tell us that the material will yield when the amount of deformation energy

has reached the per-unit volume at the same level as the amount of deformation energy at yielding in a uniaxial tensile test, that concept basically holds for steel most of the time.

Good. Now, if we would look in great detail at these yield surfaces, we would see some interesting parameters.

First of all, stresses along this direction or along this direction are uniaxial stresses. So these points here are nothing else but the yield stresses in simple tensile situations, yield stress.

And then the sections, the different segments here, of this diagram, represent different stress states. In this quadrant, sigma x and sigma y

are positive and in tension. In this quadrant, sigma x is in tension and sigma y is in compression. OK. So this corner of the stress space is easy to study.

That's why you have data in this corner, because we basically have a biaxial tension situation.

And usually what we see is that, again, the points align very well along an ellipse that goes through the yield point. However, if you look into the details,

very often there are some deviations. And we'll come back to this in a later lecture, but this is one type of deviations. If I have, for instance, a sheet metal, cold rolled, highly formable steel is being tested,

and I measure the mechanical properties in samples that were taken at different angles to the rolling direction, as shown in this graph. And in these samples, I measure a property called the R value.

And the R value is nothing else but the ratio of the width strain over the thickness strain during plastic deformation. So if I have a tensile specimen here, for instance,

and I pull it in this direction so it deforms, the thickness strain and the width strain are not the same very often.

And we call this width strain over thickness strain, we call this the R value. And the fact that they're not the same, that property is called normal anisotropy.

So the section isn't reduced at the same rate in the thickness and in the width. So this property can, however, also be sensitive to the direction in which you take the specimen, in the plane of the sheet.

So you also have planar anisotropy. And you can measure this because if you measure it at different, this R value at different angles, you find this quite complex behavior.

The reason why we have this complex behavior, that's due to texturing. And the texturing itself is a result of the processing. So if you cold roll the steel and you do a recrystallization annealing afterwards,

you notice that the R value will increase with amount of reduction that you give and will also, the way in which the material is anisotropic in the plane will also change with the amount of deformation.

This has a big impact on our yield functions because it means that when materials are strongly anisotropic, as a consequence of, for instance, in this case, crystallographic texture, that has an impact on their yielding behavior.

And we have to use adapted yield criteria. But we'll talk about this next time we meet. Thank you very much for your attention.