Well, OK, so whatever. So first announcement is about a makeup class.

So it's not about this makeup class. It's a makeups class about this course. So we missed one class last week.

And next week, on Tuesday, I'll be absent. I have some personal issues to attend. So I have to travel to Europe for a few days. So what we'll do is we'll group these two makeup classes.

And I suggest to do it this week on Friday afternoon. You wanted to go to Seoul? Well, forget it. You have to leave later.

3 to 5. I don't think. I think there's no English class. And that's about the only time I'd like to do it. And I don't want to postpone it too much.

3 to 5 on Friday. Yeah, and say to your friends who may or may not show up. Then the other thing, when we were meeting last time,

we were discussing. We had seen an example of how to use the Beach Curler

formula for two parallel edge dislocations. We've done it for two parallel screw dislocations. And we just finished going through the math for the two parallel edge dislocations.

And it was towards the end of the class, so maybe things were not so clear at that time. So basically, what we do is we assume we have one dislocations here in the origin of your coordinate axis.

And you consider another dislocation parallel to it, an edge dislocation. And this edge dislocation can have the same burgers vector, or basically they both look the same,

or they can have a different burgers vector. In this case, if we, in both cases, consider, in the three cases, consider our line direction in the board. So you know line direction in the board, yes,

extra half plane up here. That means a finish start right hand convention. Burgers vector is here. This one's the same dislocation. In this case, the extra half plane is pointing downward, so the burgers vector points to the right.

So this one stays here, and then we put these dislocations anywhere in this space. So the core of this dislocation has an x and a y-coordinate.

That's basically what defines its position. So basically, there are some places where the dislocation, interesting places where the dislocation can be. And so this is the diagonal here at 45 degrees.

Then if the dislocation of this core is here, x is equal to y. Obviously, so x over y is 1 anywhere along this line. If this location is on this side of the diagonal, x is necessarily larger than y. So x over y is larger than 1.

And above the diagonal, x over y is smaller than 1. And if it's exactly on this line, so you have a dislocation that's exactly here, well, here x over y is 0. So in this diagram, what you basically do

is the position of the dislocation, you actually use the x over y ratio. So this is the x over y ratio, and this is the force on the dislocation. And if this force that you calculate is positive, you have repulsion. If it's negative, you have attraction.

And so the results tells you that if you have the same sign, and you're in the region from 1 to infinity

for x over y. So if you're in this region here, you have a positive force. It's kind of repulsive. The dislocation will. However, if I get close enough to this diagonal,

to x equal to y, so x over y 1, I see that the repulsive force gradually decreases and becomes 0. So if the edge dislocations do this, I have 0 force.

And then I come in a range where the dislocations attract each other. They're attractive. In this region here, this region here,

there is attraction. So the dislocation will want to move closer to each other. Now you have to realize that there may be an attractive force between these two,

but this dislocation cannot move up and down. So it will glide as close as possible to this guy. And if they have opposite sign, it's just the reverse. Opposite signs, yes?

The red region, they will always attract. In this region, they will pull apart. So again, it's interesting because it explains to us why it is that if you deform a material,

and you heat it up slightly, so that means you give the dislocations enough thermal energy, they will move with respect to each other and form low energy dislocation configurations.

In other words, they will tend to move. I may go back, back in the other direction. Oops, there we go. They will tend to move to situations where the force between them is zero or is very low.

And that is the principle of recovery, basically. So what does this graph tell you? Well, for instance, it tells us that if you

have the same sign, the edge dislocations in this region here will be attracted towards this one. And the edge dislocation in this part of the diagram,

located in this part of space, will be repulsed. So as a consequence, like sign dislocations, we'll have a tendency to align like this. Go on top of one another, where the force is zero.

So that means going to this point, x over y equals zero. And when you have that, edge dislocations like this, you have a low angle tilt boundary. And this would be, for instance, the result of a deformation

that's followed by recovery. And you can see the amount of dislocation hasn't diminished. But the energy has diminished. The strain energy has diminished. In the reverse, say we have different Burgers factors.

Then in this part of space, the interaction will be repulsive. And in this part of space, the interaction will be attractive. So you will tend to see dislocations

with different Burgers factors form an equilibrium configuration that looks like this. So the line along the 45 degree diagonal, again, to minimize the.

And so what does this correspond to? This corresponds to this point here, the other zero force point. And you can have this side. Or this is also a low energy configuration, an alternative one. OK? When this happens, again, a stress effect

is the amount of dislocation doesn't decrease. It's the same amount of dislocation. But it's not recrystallization, what we're seeing here. OK, so last week I realized it was a little bit too

fast with my explanations here. And I needed to go back a little bit.

So remember last week we calculated the energy of a dislocation line.

And we noticed that, so you had basically these three terms, a term gb square divided by 4 pi. Then a term where you see this cosine alpha,

that basically tells me that the dislocation energy is a function of whether it's an edge dislocation, or a screw dislocation, or a mixed dislocation. And then a term that is related to how you integrate, how you determine the energy, basically.

And although it's an interesting physical problem, this calculation, it turns out that the impact of this factor is not so large. So I told you this natural logarithm there at the end is about 5, and it's not very sensitive to the things

we will do with this term. And then you have l, which is the length of the dislocation. The longer dislocation is, the more energy it has. So that allowed us to determine, to tell you that, or introduce a factor which we call the line tension, is basically the energy per unit length.

And that's if you take the average, the orientations, if you take this natural logarithm equals to 5, et cetera, you calculate 4 pi factor. This is the energy per unit length of dislocations.

And I told you that this b squared criterion was very important, because it tells you whether a specific dislocation will change

into another dislocation, or whether two dislocations will come together and form a new dislocation. And we like to talk about this as dislocation reactions. Obviously they're not reactions like chemical reactions, right? They're like mechanical effects, all right?

So the best way to introduce this dissociation is we have to talk about FCC crystals, because you'll see in a moment in BCC crystals like ferrite, like everyday steels we have,

we don't get dissociation of dislocations. So necessarily have to talk about FCC iron, or austenitic steels. So in austenitic steels, if we look at the crystallography of an edge dislocation,

so we have here a glide plane, which are 111 planes in FCC. And then we have drawn here an extra half plane in that particular crystal structure.

And so this extra half plane is just a 110 type plane inserted here. And so if I, again, I take my line direction

into the screen, my extra half plane is coming from up. So my Burgers factor is to the left. It's a convention. Now if you do this, you have to look at the crystal

structure. The 110 planes, they're actually stacked. It's not always the same plane. It actually consists of a plane I type, B type, A type, B type, AB, AB, AB. So the stacking of the, so when you introduce an extra half plane,

and you don't want to disturb the stacking sequence, you actually introduce two planes, two 110 planes. Now what can happen, and what does

happen in many austenitic steels, is that these two 110 planes, you can see, they can move away from each other. They can move away from each other

as separate edge, smaller edge dislocations, smaller edge dislocations. And when doing this, you will keep the right stacking sequence of 110 planes above and below the slip plane here.

We are not talking about the stacking in this direction, because when we do this, we'll see in a moment

that we create a stacking fault in the stacking of the 111 planes. So we have, when this happens, we have now an edge dislocation on this side, an edge dislocation on this side, a dislocation on this side.

And for this particular Burgers vector, a upon 2 onto 1 bar 10. I'll show you in a moment how this works. These are the two partials that we get. 1 over 6, 1 bar 21. 1 over 6, 2 bar 1, bar 1.

And the defect that we create here is a stacking fault, and we'll see in a moment why, intrinsic stacking fault. So first of all, this is an application of the b square criteria, because this will only

happen if the dislocation, the b square of the starting dislocation, and the b square of the reaction dislocation, if you want, this b square is larger than b1 square plus b2 square, where b1 is this one and b2 is this one, Burgers vector.

OK, so let's first do this little calculation, all right? So the b square criteria, right, so again, we're not talking this situation here,

crystallographic situation, does not hold for ferritic steel, so BCC. It's FCC, and so necessarily for austenitic steels, which have an FCC structure. So this is what we call the Burgers vector

of undissociated dislocation, and this is the Burgers vector of these two new dislocations, these two new dislocations. And this is the reaction. We write it like a chemical reaction, but it's not a chemical reaction, so don't think about it as a chemical reaction.

a upon 2, 10 bar 1 is a upon 6, 2 bar 2 plus a upon 6, 11 bar 2. So I calculate the length of this dislocation, basically squared. I find a square over 2, and I make the sum of this, the length square of this factor,

and the length square of this factor. It's very simple, and you have the way. I show you the way how to do it here, and you find a square divided by 3. So this is 0.5 times a square, and this is 0.3 times a square. So this is obviously less than this,

and so yes, the dissociation will occur on the basis of this b squared criteria. And so what the line energy tells you is that yes, on the base of b squared, this should happen.

However, when you do this, and these two dislocations are created, in between these two dislocations, you create a stacking fault. You create a stacking fault. So what do we mean is that the stacking fault in a normal FCC is ABC, ABC, ABC, like this.

I'll show you in a moment how this works out in the crystal. And then there are actually two ways in which,

so if you remember, this is what we had, the two extra half planes here. I'm going to call this number one and this number two. The dissociation can go in principle in two ways. I can have, or I can have, looks the same,

except that in this case, the dissociation that was here is on this side, and two is here. And in the other case, we get this.

They're reversed. They're reversed. So depending on the way this dissociation occurs, we will have a stacking that looks as if we have removed plane B. Or we

will have a stacking which looks like we have inserted a plane C. And so it will look like ABCA, C-A-B-A. And C-A-C-A, that is the stacking you get in hexagonal in epsilon iron.

So we call this a nucleus, if you want, of HCP iron. It can also work out differently if the dissociation has happened. So that the stacking fault looks

like you have inserted a C plane, then you have ABCA. Normally you have B, ABC, et cetera, A, C, B, C, A, B. So in this case, if you look here, along this line,

you have A, then mirrored with respect to this A plane, C, C, B, B. So you have made a twin here, a twin. Now, of course, you don't insert planes. And so what happens is you shift the crystals.

You make a shift. So if you do this, and you can try to do it rather easily by just looking at the stacking. So what you're basically looking at here is the stacking of 111 planes viewed along a 110 direction.

And so you can see you have stacking A, B, C, A, B, C. So you see here, three planes above this atom is repeated.

This one is repeated here. This one is repeated there. So these positions have names. Now, you can do a shift, for instance.

If we shift now this B atom to this position here. So it comes right over the C atom. I'm just doing one shift. I'm pushing this. Then, of course, this B atom is not called B anymore.

It's not called B because it's above a C atom. So I call it a C atom now. A, B. So A, B, C, A, C, A, B, C, yes. So I didn't insert anything. I didn't put it. I shifted my crystal as if a small partial dislocation

had passed. That's basically what happens when you have a dislocation passing.

And if I do another shift, now I have this one, and I shift A to this position, then A comes on top of B. So the name of A is not called A anymore.

It's called B because we're talking about stacking here. So it's a B type stacking. This B becomes C. And we have formed A, B, C, A, C, B, A. And you can see the twin. So these different stackings are not

because we actually add a layer of atoms. It's just because of the shifts. You get different shifts whether the partials are this way or that way, the associated atom. OK? So whatever you do in between these two dislocations,

you create a stacking fault. And now, what is interesting is that if you start with, say, an edge dislocation,

your two partials will not be edge dislocations, not but they will have an edge component. You see, a Burgers vector, and say I have a dislocation line, yes, and a Burgers vector

like this, yes? This angle is whatever. Yes, it's not 0. It's not pi over 2. But I can always decompose it in an edge component perpendicular to the line and a screw component, yes?

So when you do this disassociation, the two partials have edge components, yes? And the extra half planes, yes, and the extra half planes are coming from the same direction, right?

So if I go back to where we started with this morning, I have an edge dislocation at the origin, and I have another edge dislocation here on the same plane, yes? So in what regions are we here?

We're in this region, right? In the red region, yes? And that region is repulsion. So when the dislocation dissociate, they instantly push each other away because their edge components are repulsive, OK?

And we know from this graph that this repulsion is always repulsive, yes? But there's something that limits the distance they travel away from each other, and that's the stacking fault because as they push each other away,

we generate a stacking fault, yes? And the energy of the stacking fault is positive. So that means that there will be a sweet spot where they stop moving away from each other when their repulsion

energy is equal to the energy created by the stacking fault, generated at the stacking fault, OK? And so we can calculate this, right?

So let's, so here, see? So the force between the two dislocations, so what do we have is basically we're going to calculate the force, yes? So this is one of the partial dislocations, yes?

And I multiply with this. So this is, if you want, the shear component of the dislocation at the origin.

So this is tau times b, yes? This is basically the peach curler equation for this particular case, OK? So I've just applied this, and we know the distance. They're at d, d away from each other.

And I basically plug in the only thing I need to plug in, which are the two vectors, Burgers vectors. And I have to make a dot product, dot product here. And I find this, g times a square divided by 24 pi times d.

So again, shear modulus plays a role. And the distance, one over the distance. So the closer they are to each other, the larger the repulsive force. And the repulsive force dies as one over d.

And this force is balanced by an attractive force due to the stacking fault energy. That means that when these dislocations, this dislocation moves to the left, and this dislocation moves, OK, this is not the Burgers vector, it moves to the right, I create more.

And more stacking fault, yes? OK, and I call the energy of the stacking force gamma. So if this force is equal to stacking fault energy, I will get this, I get equilibrium.

So the distance at which these two partial dislocations stop is given by this. g a squared divided by 24 pi times gamma. So the distance between the two partials

is proportional to one over gamma. So if the energy of the stacking fault is very high, the dissociation distance will be small. If the stacking fault energy is very low, the dislocation can be very widely separated.

Let's just, because the distance is proportional to one over the stacking fault energy. So if the stacking fault energy is large, I will have very small dissociation.

If the stacking fault energy is low, I have a very wide dislocation, yeah? And that has very, very important implications, yes? And if the dislocation, the stacking fault energy, is extremely high, then there's no dissociation.

Where does this, like in steels, in austenitic steel, ferritic steel, when do dislocations start to decide we don't dissociate, yes? Well, we'll see.

But a value for the stacking fault energy where there is no more, where you can be sure there won't be any dissociation, about 100 millijoules per square meters, yeah? OK?

So millijoules is a joules energy per meter squared, surface energy, gamma is. All right. So well, this is an example here, right? Just for instance, the top here is austenitic steel, yes?

You see these four white lines, yes? So the two top ones are two partials, yes, belonging to a dislocation. So they have two Burgers factors, they're indicated, BP1, partial 1, and BP2, partial 2, yes?

And there is another dislocation that's dissociated, yes? So now, the equation we've just seen, there is a more sophisticated version of it, which takes into account the orientation

of the undissociated dislocation, whether a dislocation is edge, or screw, or mixed, yes? But it's basically the same idea. So the distance between the two parts is proportional to 1 over the stacking fault energy. So if you use this equation, yes, use this equation,

for instance, yes? You measure this by experiments. So you get d from experiments. How do I measure this? Well, very simple. I measure the distance between these two dislocations,

like this, yeah? And then I plot this as a function of the angle. So this is TM work. I can determine what the Burgers factor is. And I can determine if I know what the Burgers factor is. I can see what the line direction is.

I see here the line direction is here, and here it's in this direction. So I can determine theta, yes? And so if I know theta from here, I can plot all the measurements I made, yes? So these are measurements.

Measurements of d values, yeah? And I see here, and I plot this here. And then I plot this equation here for different values of the stacking fault energy, yeah? And then by fitting the data to these calculations here,

these lines here, that are obtained for different stacking fault energies, I can basically determine what the stacking fault energy is approximately. You can see the scatter can be quite considerable,

but the stacking fault energy for this particular material, 18% manganese steel, is around 30 millijoules per square meters. And so you can assume that all the dislocations

will be dissociated, OK? All right, the dislocation, the stacking fault energy, yes, the stacking fault energy, is determined by two parameters, yes? And the first parameter is composition,

and the second parameter is temperature. So this is an example here for stainless steels, yeah? You know that austenitic stainless steels are very widely used in many applications, and you

know that the main alloying elements in this type of steel are chrome and nickel, yeah, in austenitic. And the chrome is added to give the steel a corrosion

resistance, and the nickel is added to give the steel the austenitic structure, not for corrosion, right? And the chrome is not added to get austenite. These two elements have different reasons to be added. Let's put it this way. Now, they influence the stacking fault energy.

So for instance, if you look at the number of steels, yes, where you don't vary the nickel content too much, right, 13% to 16%, so a relatively narrow range, and you plot the stacking fault energy

as a function of the chrome content, which varies 13% to 25%, you see that there is a decrease in the stacking fault energy. So adding chrome will result in wider stacking faults, yeah? You look at the effect of nickel. Again, you use alloys where you select them

so that you have a relatively narrow range of chrome contents, so 17% to 19% of chrome. And then you vary, you plot the stacking fault energy as a function of nickel content, yeah? And so you see that the nickel content increasing

will increase the stack. So that has an effect of reducing the stacking fault width. So stacking fault energy is strongly influenced by the composition.

Why is that? And I also said the stacking fault energy is temperature dependent. Why is that? Well, first of all, I'll make our lives simple

by first saying that people have looked at, in metals, how do stacking faults, what do stacking faults look like? By looking at what dislocation is on what side of the stacking fault.

And they find that the stacking fault is usually, what they say, we call intrinsic. And when we have an intrinsic stacking fault, it basically means it looks like we have removed a B plane or it's the HCP type stacking fault

that prevails in practice. OK. So basically, our stacking fault looks like this, like a little sliver of austenite inside the FCC structure.

So if that's the case, you can basically use thermodynamics to calculate the stacking fault energy. Because you basically have to say, well, when I create a stacking fault, I basically

transform austenite into HCP. And that's one thing I do. And then second, I create interfaces. I create austenite epsilon iron interfaces. So this little, this is the stacking fault region,

used to be austenite. So I transform it to epsilon iron. And so that's one thing. And then the other thing is here, it used to be a gamma interface is transformed

to a gamma epsilon interface, here and here. So with this, we can calculate this reaction just using delta G, the change in free energy

from thermodynamics. And I don't derive this equation here, but the relation is stacking fault energy is 2 times the molar surface density times the free energy

change for the reaction, in this case, transformation gamma to epsilon, plus a term related to the surface energies. So I have a actually very simple formula which, if I know this interfacial energy value,

and if I know this, I can basically calculate what the stacking fault energy is. This term here is basically the molar surface density of 1, 1, 1. This should be gamma planes, of course.

And this is the formula. You just apply this, and you get this parameter. So easy to calculate if you have the lattice parameter of your alloy. So let's calculate an example here.

So what we need is a lattice parameter. We need Avogadro's number. We need the driving force for the gamma to epsilon transformation. That's 200 joules per mole. Where does this come from? This basically comes from either thermodynamic measurements

or from programs that allow you, like Thermocalc, that allow you to get this data, this thermodynamic data. So basically from the literature, if you want. Then the interfacial energy. That's probably one of the weakest point in this theory.

10 millijoules per square meters. That's a low interface energy. And that's what we usually use for this kind of interfaces.

The shear modulus appears in the formula. And of course, the Burgers factor of the partial dislocations. That's A upon 2, 1, 1, 2. So if you apply this, you find 0.15 nanometers.

And then Poisson's ratio. So the stacking fault energy. So this equation here. I just plug in all these parameters. Of course, making sure I've got the right dimensions here. And I find 32 millijoules per square meter

for this particular set of data. So if I know the stacking fault energy, I can determine how much is the separation between two screw dislocations. That's basically this equation.

Separation for screw dislocation. Theta is 0. For edge dislocation, theta is pi over 2, or 90 degrees.

And so I find 3 nanometers for a screw dislocation and 7 nanometers for a edge dislocation, about double. So the distance between screw dislocation and edge dislocations is different. And the edge dislocations are more widely separated

than screw dislocations for the same original Burgers factor and for the same stacking fault energy. So obviously, composition has an effect on free energy.

Because you can have epsilon phase, and you can have gamma phase. But depending on what alloy you have, you have different free energies. And of course, the free energy of a phase is temperature dependent.

So what we see is that the effect of the temperature is that the stacking fault energy increases when the temperature increases.

So you get less, the dislocation separation is smaller at higher temperatures. So now let's talk about alpha iron, ferritic steels.

What's the problem, or what's the situation there? Why did I choose to focus on austenitic steels when I talked about stacking fault? Well, the simple reason is because the stacking fault energy of alpha iron and of ferritic steel is huge.

So I don't know how high it is. So the only way you can actually guess how high it is is by calculating it.

Because with austenitic steels, you can just experimentally look at the separation and then determine the stacking fault energy. In BCC and ferritic steels, you can never see a separation, so you have to assume it's very high. So what you can do is, and what people do nowadays

because of the computational tools we have, they will determine what we say the generalized stacking fault energy. So what they basically do, they take, for instance, the BCC lattice, and they shear the lattice.

They shear the lattice. And they can determine what is the increase of the energy they get as you shear the lattice. And you can computationally spoken, if you do DFT calculations, you can do any shift you want

and calculate things that will, of course, not be seen in nature or experimentally. So you can do this shift, and you can determine for every position, every lattice mismatch, basically, what the energy is. And if you do that, for instance,

this is for a shear on a cube plane. This is for a shear on a 1-0 plane. And this is for a shear, also a 1-1-0 plane, but this shear is along a 1-1-1 direction, and this one is along a cube edge direction.

And you see, I had told you 100 millijoules per square meter, that's kind of high already. This goes up to 1,000, 500, et cetera. So the stacking fault energies in BCC iron,

and in ferrits, it's huge. So we never see dissociation in BCC iron and ferritic steels. And that is a very essential property of BCC iron, as we will see in a moment.

You can do the same for FCC. And I just want to introduce this. Again, you plot the energy that you create when you shift a crystal, top part of a crystal,

in a 1-1-2 direction. And you compute the energy as a function of the position. And so when you do this, so for instance,

I shift this B atom here to the right. So first, I see that when I create a stacking fault,

this B atom here will go to a position where it's on top of a, where it's shifted. And it comes in between.

In this symmetric position here, it's difficult to see, but it's kind of equally separated from these two A atoms. This has a slightly higher energy. We call this the energy of the unstable stacking fault. And then when we continue to shift B,

it becomes a C stacking plane. And that has a slightly higher energy. And that's the intrinsic stacking fault or the stable stacking fault. And you can see that the structure here is an HCP crystal, an HCP sliver of HCP material in the FCC material.

And then if I would continue to push this C atom to the right, it would now face this A atom. And it would have an AA condition. That's a very high energy situation, very high energy.

And of course, I get an increase in the stacking fault. And then as I continue, the energy will decrease. So that's a new concept that people are using instead of just the stacking fault energy, which is this. They use the generalized stacking fault energy,

which allows you to, for instance, consider this. Because the way the dislocations will dissociate will require to go through this unstable stacking fault situation.

More details later, perhaps, if we get that far. All right, OK. So we will need this a little bit in the future. But I just want to make sure I introduce you

some concept, which is very widely used, yes, to describe dislocations in FCC austenitic steels

and in BCC ferritic steels, yes? And so you know that both these crystal structures have cubic unit cells. But it's very, very inconvenient to use cubic unit cells

to describe dislocations, yes? Instead, we use these geometrical shapes here. This is a tetrahedron, yes, tetrahedron. And this is a rhombic dodecahedron.

Dodeca means 12 faces. OK, so first, let's do, perhaps, this tetrahedron, yes? And when we use this tetrahedron

to work with dislocations in FCC crystals, we call it the Thompson tetrahedron, because that's the guy who just proposed the use of this system. And it's very useful. So basically, where does the tetrahedron fit

in the FCC unit cell? Well, here. So do you see the unit cell? And then I put one of the corners of the tetrahedron in the origin, and this is what I have, OK? So you can see, going from here to here, what is the vector,

yes? If I go, for instance, from this atom, which I call A, to this atom, which I call B, yes? I, so this is x, this is y, and the atoms of the unit

cells are here. This is the unit cell. And so I have, at the level higher, I have this atom here, and I have this atom here.

And I forgot this atom here. So the tetrahedron is seen from the z direction down, yes, is like this, and then this line here. So this is atom A, this is atom B, this atom I call D,

and this atom I call C. So this factor here, what is it? Well, we know that this is the length, this is A, 0, 1, 0.

And this length here, to this here, is A, 1, 0, 0. So from here to here is A upon 2, 1, 0, 0.

And from here to here is A upon 2, 0, 1, 0. OK? So this factor here is the sum of this factor and this factor. So it's the sum of A on 2, 0, 1, 0, plus this factor.

And this factor is the negative of this factor, so it's A over 2, minus 1, 0, 0. So A over 2, minus 1, 0, 0. So AB, the vector AB, is equal to the sum of these two

vectors, A over 2, minus 1, 1, 0. So I basically have a tetrahedron where all the edges are Burgers factors,

undissociated Burgers factors. So this would be point A, this is B. So if I go, as I said.

from A to B, this vector here is A over 2, 1, bar 1, 1, 0. Is this right? Yeah. OK. All right.

Good. So now we do another thing. OK, and this point here is called point C. Yes. This plane here is crystallographically 1, 1, 1, but we also call it the ABC plane, ABC plane.

This factor here, A upon 2, bar 1, 1, 0, we can call it that way, but we also call it the AB vector. So with the use of the advantage if you wonder why people make things confusing and using AB

instead of using the crystal. Well, you can use whatever you want, basically, but this is shorter than this. That's basically why people use it. It's much more, it also gives you a shorthand notation of the dislocations.

Right, and then there are other important points. There's another important point on the Thomson tetrahedron.

So if I take point B here, yes, and I connect it with this point here. So I basically take the diagonal, yes. It will intersect my triangle here in a special point, which

I call point delta. And I can do this for all the other planes. For instance, if I B, I do the same. I connect with the diagonal, the atom on the diagonal, in the diagonal direction, I find point beta.

And why is that important? So I have a point here in the middle. That's delta. Yeah, so basically, this is point D.

If I let down the normal on this plane from point D, I find delta. Now, the interesting thing is that these vectors here

are actually the vectors for the partial dislocations. So I can see that, for instance, this vector can be created by the sum of A delta, delta B.

So AB A delta plus delta B. And of course, these vectors here have crystallographic notations

equivalent. And so this is shown here. So this ABC, this delta here, so B delta, A delta, these are the corresponding normal crystallographic

notations for these vectors. A upon 6, minus 1, 2, minus 1. A upon 6, 2, minus 1, minus 1. And then of course, there is one in this direction also.

That's the Thomson tetrahedron. The Thomson tetrahedron is something for FCC metals and alloys. So if you're working on ferritic steels, you don't need all of this because the slip planes

and the Burgers factors are different. And that's why in, this is for gamma iron. For alpha iron and for ferritic steels,

the configurations of the glide planes and the Burgers factors are different. It's one thing. And the second thing is, as I already said, there are no partial dislocations. So there are no equivalent A delta, delta B type

of vectors. So how does it work there?

In a BCC crystal, the glide planes are 1, 1, 0. Now, I don't know how much you know about the formation in BCC

crystals, but maybe you've had some classes about this in the past. And then people say, well, yes, it's 1, 1, 0, but it's also 1, 1, 2. And it can be 1, 2, 3, et cetera. So at this stage, don't worry about it. We'll talk about this in more detail. But for all practical purposes, in steels, again, in steels,

1, 1, 0. You should consider 1, 1, 0 as the glide planes. But we'll go back into all the details as we continue. So what do we do in this case?

Well, we do the same thing as we did for the Thomson tetrahedron. Because what the Thomson tetrahedron did is basically just cut out a shape which is bound by the slip planes, all the possible slip planes in the crystal structure. So what we do is we do the same thing now for BCC,

and we cut out the slip planes. And we find something similar, where the surfaces are slip planes. The faces are slip planes. And the edges are Burgers factors.

So here, this is the shape. It's a little bit more complex. And so we have the Burgers factors. And here, 1, 1, 0 type planes. So you can see.

OK. So do you see the square? You see the square? So you're looking down one of the cube directions, like an x direction, right?

And so if I myself, yeah. So there's another cube direction, et cetera. So you just have the symmetry of the crystal also.

So you see the hexagon? You see the hexagon? Yeah? OK. So in what direction is the cube hexagonal? Does a cube have a hexagonal symmetry? That's in 1, 1, 1 directions, right? So you can see here, when you see the hexagonal shape,

these edges also are parallel in the same direction. And these are indeed 1, 1, 1 directions. And so what it basically means is

that our slip planes are 1, 0 planes, and our slip directions, our Burgers factors, are a upon 2, and also in this case, a upon 2, 1, 1, 1 directions.

Right. So if I have, so now, I made a drawing of this, right? So one of the, so for instance, myself,

it's actually not right. So here, this is, yes, it's one of the, could be a 1, 1, 0 plane, yes?

And let's say, yes, we have a dislocation with this Burgers factor. So if I have a loop here, a dislocation loop, for instance, like this, this part here

will be a screw dislocation. Here is also a screw dislocation. And this part here will be edge type. This is the Burgers factor. All right. OK. So that's what we have in the case of the BCC iron.

No dissociation, OK? And we'll see that, I'm going to stop here, that that will lead to very different behaviors, OK?

And from a mechanical point of view, I think. We'll stop here. Right. Let's stop here. I'll talk about this on Thursday. So you don't have to worry about this for the quiz.

Good. For the guys who came in late, Friday we have makeup. Friday? 3 to 5. Yes.

And 3 to 5. Yeah, and Thursday morning quiz, as usual.