All right. So we had started discussing the strengthening mechanism

due to solid solution in, again, as we are doing in this course, focusing on alpha iron, gamma iron, and steels, the alloys of these things. And so we were going through the effects that solutes have when they strengthen the steel.

So the two main effects are size effects, and what we call the modulus effect. So the size effect is very simple to understand.

You introduce an atom that may be, for instance, much larger than the iron atom. For instance, you introduce tungsten in a ferrite atom. It's going to generate a distortion in the lattice,

and that will have, as we will see, strengthening effect. Second, or you can add an element that's just a contrary. For instance, silicon or aluminum, they're smaller,

and they will give a lattice contraction. And we'll use a parameter called delta to describe this strain, this volume strain that you get around a solute. The other thing is when you add an element to steel,

locally you change the elastic properties. What you actually do is change the electronic properties and the binding, but you can understand this in terms of making the material locally

softer or harder. And this will also have an impact on the strength. For instance, you can have a very simple example is a vacancy. It's not really an alloying element,

but the vacancy has basically no modulus. Modulus equals zero. So that will have very strong influence on the strength of the material, because of this big difference in modulus. And then there are a number of other less,

so why is this, doesn't want to move, there we go. So the size effect, we use a parameter delta,

which is 1 over a dA dC, A being the lattice parameter. Yes, and so dA dC is the dependency or the change of the lattice parameter with the concentration of the solute. And if you divide this by A, it's basically a strain.

The modulus, we do the same thing. It's 1 over G dG dC. Relative change of the modulus with the concentration of the solute. And then, as I said, you have other effects which may or may not have an influence. We have what we call the chemical interaction.

And when we refer to chemical interaction, we typically refer to the influence or the effect that solutes have on extended dislocations. And so this would be an effect that only occurs when we have stacking faults. So in low stacking fault material.

So it doesn't really happen in ferrite or ferritic steels. And the reason is that solutes may be migrating to the stacking faults

and having an influence on the motion of the dislocation in this way. Sometimes when you add enough solutes, you can have a phenomenon of ordering.

For instance, if you add silicon to ferrite, first the silicon will be randomly arranged. And then as you add more and more, past 10 or 12 atomic percents of silicon,

similar things happen with aluminum, you get an ordered phase. Silicon and iron will form a new crystalline structure because the silicon is ordered. And you form what is called a DO3 structure, for instance.

OK. Now, which has a chemical composition of Fe3Si or Fe3Al. But what you also have in alloys is that when you form Fe3Si or Fe3Al,

you have long range order. Well, that kind of order can occur at lower concentration. And then we talk about short range order. So you basically have patches of ordering.

And of course, ordering can only have effects when you have enough alloying present. So more alloyed, yes. And basically, what we're talking about is a preferred local arrangement of iron atoms and solutes.

Now, when you have a normal crystal that doesn't have ordering, when a dislocation passes through the crystal, before the dislocation, you have perfect crystal. And after the dislocation has passed,

the crystal is perfect too. There's no change. However, when you have an ordered structure, most of the time when you have a dislocation passes through, this ordered structure is not kept after the dislocation has passed. So there is an energy difference

before and after the passage of this dislocation. And this also leads to strengthening effects. Because short range order is present, the passage of the dislocation partially removes order across the glide plane. And you basically randomize the crystal.

And that has an effect on the stress you need to move the next dislocations through that same region. In steels, yes, we have a special kind

of strengthening effects, which are due to the very strong relation between interstitials,

interaction issues, strong interaction between interstitials and dislocations. And we're talking about carbon and nitrogen interstitials mainly. And we have a phenomenon that's called snook ordering. Snook ordering is a very simple effect.

It is due to the fact that carbon and nitrogen interstitials have a high diffusivity. So at room temperature, carbon and nitrogen typically make one jump from one octahedral place

to another octahedral place. So you can have effects that are due or interactions that are due to this very short range diffusion. And that is what snook ordering is about. You have dislocations passing through the lattice.

And at room temperature, the carbons that are in the vicinity of the core of the dislocation can move to an energetically more favorable place.

And thereby, they slow down. They form obstacles to the dislocations. We'll talk about this in more detail because it is important. What does it mean, for instance? You take a ferritic steel, a carbon steel,

and you measure the stress strain curve. But you don't go all the way. You stop after a few percents of strain. Yes? So originally, there is no yield point or anything.

You have a nice, continuous yielding behavior. So you deform a little bit, a few percent, 3%, 4%. You stop. Yes, so you remove the load. And after a few seconds, you measure again. And lo and behold, you have a yield point there.

Yes? That yield point is the proof of this snook ordering. All right? And we also have electronic interaction.

So when we consider a dislocation in a metal, you wouldn't think much about electronic effects. But we have stresses, yes, in the vicinity of the dislocations.

And that disturbs the normal density distribution of electrons. There are now regions with different electronic densities. I'm not going to describe that in detail. But so you can imagine different densities,

electronic densities. And then when you add an atom, which has a different valence, valency than iron, that atom will interact electrically,

as it were, with this new electronic distribution around the core of the dislocation. So electronic effects are changes in electronic density.

You may wonder, why do we talk about electronic density? There are some materials where the dislocations are actually charged. Like if you think about ionic crystals, like sodium chloride,

you also have dislocations there. There, necessarily, they have a charge. At the end of the dislocations, you have an electric charge. In the case of the metals, you don't really have a charge, but you have the distribution of electrons

is altered. So electrons move away from regions that are in compression. And they go and reach, as it were, to regions that are in tension. And so you get a dipole, an electrical dipole.

And a dipole can interact with a charged particle. And that would be the charged particle is an element, a solute that you have added, which has a valency different than one of that of iron. Is that an important effect? No.

Basically, of these six effects, the main one, of which we know a lot, is the size effect. Even if you wanted to take electronic effects into account, you'd have lots of difficulties finding any data about it.

So let's start with the simple approach, which is an engineering approach with solute strengthening. Well, it's an empirical approach. We just assume that the strengthening is proportional to the solute concentration in mass percent.

There's no strong theoretical support for this approach, yes? OK. There are theories which give a linear relation between the amount of strengthening

and the concentration of the solute. But there are other theories also. So there's no strong support for this approach. But in practice, it works. And the reason why it works is, and I'll make this point later on, is that we

don't have much experimental evidence to support any of the available theories when it comes to solid solution strengthening in steels. So and the results are good enough very often

because the levels of alloyings are so low. And so we assume a linear relation. And when we do this, we see that interstitial solutes like carbon, nitrogen, boron, they

have very high strengthening effects in ferritic steels. Phosphorus, silicon, manganese are the next ones that have an intermediate strengthening effect. And then you have elements like nickel and moly have weak effects. And in certain cases, chromium can even

have a softening effect. Now anything you know about solid solution strengthening in ferrite does not necessarily hold for austenite, yes? So manganese is a solid solution strengthener

in ferrite, yes? But it actually, it softens austenite. So when you have an austenitic steel, you add manganese. It doesn't get harder. In fact, there are many cases where manganese makes it a softer material. So be very careful, OK?

But as a rule, interstitials increase the strength. So carbon and nitrogen have high strengthening effects in austenitic steels. And then silicon, titanium, and moly have a strong intermediate strengthening effect.

Manganese and nickel do not have a pronounced solid solution strengthening effect in austenite. So we have lots of data to rely on. For instance, this is iron-phosphorus graph. You can see nice linear relation.

With silicon, there's a lot more scatter in the data. But we can draw a straight line through it. Iron-aluminum, again, straight line, which supports this empirical approach of saying, well, strengthening is proportional to the weight

or mass content of the alloying element, yeah? And the same thing for austenitic steels. For instance, an example here, a recent example for a twip steel, add some aluminum, you see a linear increase in the strength.

And so you can derive from this 20 megapascals per mass percent of aluminum that you add, yes? And so you can use this in practice. However, when you do this, and this

is a list of some of the solid solution strengthening factors that I have found in the literature, yes? So you see two things. First of all, there's lots of data out there.

So many people have looked at this. Second, if you look at the data, it's very puzzling. Take, for instance, interstitial hardening effects, yes? Carbon, 5,544 megapascal per weight percent

is one of the measurements, one of the report mentioned. 2,263 is another value, yes? So a huge difference, obviously, yes? The same for nitrogen. I mean, I go 5,015, yes?

Some elements, you have very little information, yes? For boron, I only find one value. Phosphorus, same thing. It goes 1,200, 500. Silicon, 140, 60, 32, 80, huge differences.

Lots of scatter. Well, this, of course, reflects the fact that when people do these measurements, very often they do this on steels, yes? And so the concentration of the elements

is never measured perfectly in the sense that, for instance, the case of this carbon here, are we sure that the person who did the measurements checked that there was no carbon present as cementite

or a transition carbide? How sure are we that the carbon was not segregated? For instance, to grain boundaries. We all, we don't know this information, right? But it's very important, of course, yeah? So ideally, yes, the best measurements

are the ones that are made on single crystals and perfect binary solutions, yes? So there shouldn't be any precipitation, yeah? OK? And obviously, in the case of carbon, it's really difficult to control because carbon is basically insoluble in ferrite.

So how can you get reliable data? So it's a challenge, all right? So what do I do? It's an empirical approach. And it's not based on theory.

So what do you do in practice? Because it's not that these measurements are bad. Oh, this is a bad measurement, and this is a good measurement. They're all good measurements. They're based on the author measuring different manganese contents and put a line through it.

And that's what that person obtains. So there's nothing wrong with that. So what I do is I collect these, and I choose the median value, yes? So not the mean value, the median. That's the value that's in the middle of the row, right?

So if you have five data points, like here, I have seven. The middle one, 830 megapascal, that's the one I use, yes? Here, I have an even number of values.

So I choose the value that's midway between these two, OK? The median value. That way, I kind of use a statistical approach that minimizes the effect of outliers. Because there may be data out there that's really not good,

yes, but I have no way to judge. In order to minimize the impact of outliers, I use median values, yeah? And if you use these values, you find data that's actually very often used, or close to what's very often used, yeah? OK, and here you have data for austenite.

And of course, the strengthening elements here, with the exception of carbon and nitrogen, are different. You have silicon, niobium, titanium are the strengtheners, and not phosphorus, silicon, and manganese, as in the case of ferrite.

Another thing that's important is here. It's basically the same data you've just seen, except I compare them to each other. So it's the same y-axis, solid solution strengthening, ferrite, and austenite.

So what do you see? You see that the general level is lower. You see? The interstitial atoms strengthening, the substitutional atoms, the level of strengthening in austenitic steels is much lower. So you cannot expect in austenite, austenitic steels

and austenite in general, to get the same amount of strengthening that you get in ferrite or ferritic steels. For instance, just an example, for phosphorus,

we have about 800 megapascal per weight percent. There is not a single substitutional element in austenite that has this type of strengthening effect. So much lower solid solution strengthening in austenite than in ferrite.

So how do you work? How do I use this approach? It's very simple. I have an example here. Say you have an IF steel, a titanium IF steel. And this decomposition has 30 ppm of carbon, 30 ppm of nitrogen, 0.15 manganese, 0.05 silicon,

and 120 ppm of phosphorus. So first of all, think. Is there any carbon or nitrogen in this steel?

You remember, nitrogen and carbon have huge impact in terms of strengthening. In these steels, the effect of carbon and nitrogen on the strength is a big zero. Why? Because we've added titanium.

The titanium stabilizes the nitrogen. And it also stabilizes the carbon. So there is no carbon and no nitrogen in solution. So there is also no strengthening from nitrogen and carbon.

So always be careful with this. OK? So there's no solute strengthening from carbon and nitrogen in these steels. However, there may be an excess titanium. Usually, in order to make sure all the carbon is precipitated as carbon, titanium carbide,

or as all the nitrogen is precipitated as titanium nitride, we add enough titanium. So do we have some excess titanium? So this excess titanium is in solution. And you need to take care of that amount in solution,

not the total amount. The amount in solution, the amount that doesn't form the nitrides or the carbides. So typically, this excess is about 200 to 300 ppm. And in this case, we'll just say it's 200 ppm. So how do you calculate the yield strength?

For instance, the yield strength is the friction stress. What's the friction stress? That's our piral stress. Piral stress is temperature dependent. It's the thermal part. It's basically the critical resolved shear stress times 2.

I like to use 39, but you could use 40. I mean, there's this experimental. But OK, that's about the value. And then you multiply the strengthening effect.

So for manganese, I had in my table here. Let's go back here, if I'm right. So you have here 43.9.

That's this median value. So here, 43. I multiply with the weight percent of manganese. I do the same for silicon, phosphorus, titanium, and aluminum. In my composition, I sum this all up, 63. Now, 63 is very low. 63 is a very low value for the yield strength.

But that's the value. That's the value. Takes into account the lattice friction and the solid solution strengthening. It's low. Usually, when you do a measurement on an IF steel,

and there, it's very important. When it's not temper rolled, it's very low. It's around 100 to 150. Yeah, it's very soft. That's still far away from the 63. Why is that higher in practice?

Well, first of all, we didn't take into account any of the other strengthening effects, which we'll discuss. We form precipitates when we make titanium carbide, titanium nitride. That has a strengthening effect. And of course, we have grain boundaries, which act as a strengthening also.

So the grain size is not taken into account. Precipitates are not taken into account. And these will be, we'll see. And of course, there are a few, there are some dislocations. It's not a very sizable density.

So dislocations, grain boundaries, and precipitates will contribute to this yield strength also. OK, but that we haven't discussed. But this is the contribution of the lattice friction and the solid solution strengthening in the material.

63 megapascal. So of course, this is a very, very lightly alloyed material. OK, if I had half a percent of silicon and 1% of manganese, you'd have a much higher value.

OK, all right, but now let's go back to the theory now. OK, so now we're going to try to understand where does the strengthening come from and look at theoretical models. Well, first of all, you have to know

that we'll focus on only two contributions, the size effect and the modulus effect. And as I've told you, the size effect is the most important one. First of all, what's with the size effect?

So it really depends on whether we're talking about an interstitial atom or a substitutional atom. And it also depends on whether we're talking about alpha iron or gamma iron, whether we're talking about ferritic steel or austenitic steel.

So let's start with interstitial atoms, the carbon in iron. So let's look at an elementary volume, yes, of the size of the lattice, the unit cell in the lattice,

little elementary volume, yes. And we assume now that we insert a carbon atom in this position, yes, in the lattice.

Then we find out that this spherical volume is distorted into an ellipsoid, yes. So it becomes stretched in this direction, and it's compressed into two perpendicular directions, yes.

This kind of lattice straining is called a tetragonal distortion. And it gives rise to strain ellipsoid. That's what happens when you introduce carbon in the ferrite lattice.

And you've heard this before. This is what happens in reality. This is the cubic unit cell of alpha iron. When I put a carbon in the octahedral position, the lattice is stretched in this direction.

And when these two atoms are pushed away, these four atoms move in. So what used to be a spherical space, yes, in this lattice becomes an ellipsoid. If I have a substitutional, yes, so this

is for interstitial impurity. This is for a substitutional, let me backtrack. Let's keep, I'm still discussing interstitials here. But I'm discussing interstitials in gamma iron.

When you put carbon in the lattice of gamma iron, yes, and you consider a little space, spherical space around the position where you do this, what happens is the lattice expands isotropically, yes?

So I don't get a ellipsoid. I get isotropic distortion. So what happens? Well, that's the case of carbon in gamma iron. When I put carbon now in an octahedral position in gamma iron, I push up these two atoms.

And I also push away these four atoms, yes, in the plane perpendicular to this axis. So in the case of substitutional atoms,

always, whether it's alpha iron or gamma iron, I always get isotropic distortion. So either this elementary volume expands or contracts, depending on the relative size of the solute. So for instance here, I've changed the central iron

atom in my BCC unit cell by another atom that's larger. The lattice will expand in three directions, yes? I put in gamma iron a solute atom here, a larger solute than iron atom in the lattice,

substitutional position, and the lattice expands isotropically, yeah? So that would mean that in alpha iron, when I have interstitial, I get tetragonal distortion.

And when I get substitutional, I get isotropic distortion, yes? In the case of gamma iron, when I put in carbon, I get isotropic distortion. And I put in substitutional alloying,

I get also isotropic distortion. Is that true? Not really. You can get also tetragonal distortions in gamma iron and in austenitic steels. And the reason is because carbon will often

form a complex. It will associate itself with another point defect, another solute, or a vacancy, for instance, yes? And that will give you a tetragonal distortion

around the carbon atom. So in alpha iron, the carbon atom will always give rise to tetragonal distortion. In the case of gamma iron, I will get a distortion

if the carbon atom is associated with a substitutional atom or with other lattice defects, yes? So of course, you're going to ask me the question, well, how do you know this really? We talk about the atomic level

and the amount of distortions are tiny, yes? So how do you measure this? Well, it's very simple. You just measure lattice parameters, yeah? So for instance, if I take gamma iron, austenite, and I add carbon, yes?

I add carbon to gamma iron. What do I find? I find that the lattice, the unit cell just increases, becomes larger. So it's isotropic. If I do the same with carbon, oops, this should be,

this is alpha iron here. Please, change this. This is an arrow here. This should be alpha iron, please. What you get is, and you can see this effect in martensite.

Why do we need to make martensite? Because the solubility of carbon is zero in ferrite. So you have to go make martensite. What you see in martensite is that as you add carbon, you get a tetragonal distortion, of course.

And that's because carbon gives it tetragonal distortion of the octahedral space, yes? And you can actually use these two parameters to calculate very precisely what the strains are. For substitutional atoms, it's very simple.

If you add aluminum to iron, yes? If you add silicon to iron, chrome to iron, BCC iron, the lattice expands or contracts homogeneously. And so you can use the slope of this line to compute the change of the lattice parameter

with the concentration. So you can, and it's basically the slope of this here. It gives you, for instance, 0.03 nanometers per atom. And for silicon, you can do the same, yes?

And as you see here in this graph, you can have homogeneous isotropic expansion, but you can also have isotropic compression, yes? That depends on the relative size of the atoms, yes? And so here you have a list.

There's a lot is known about this. But even here, if you collect all the data that you can find in the literature, there is a range of values available.

And you kind of have to choose amongst these values, yes? For this delta. So what this delta is is basically the linear strain, yes, of the lattice parameter with addition

of the alloying element. So that is an important parameter, yes? And so now we need, in order to understand solid solution strengthening, we need to understand

what's the interaction and how do we describe this interaction, and what are the important parameters in this interaction? So first of all, so for the sake of simplicity,

to make the geometry simple, what I have here are a row of atoms, yes? And here, these cusps here are at this location that has run into these atoms.

We just think of them now as obstacles, yes? And we'll see how the interaction actually works. Just think about them as obstacles, yes? So this location will bow out between the two solutes, yes?

And this bowing out will continue until we reach a critical point and the dislocation is released by the obstacle, yes? OK, so let's say, so what can we measure here, yes?

So we have this, so we have my dislocation

is running through the crystal, yes? It arrives here, it bows out, yes? And it continues to bow out, yes? Like this, and then at a certain value of my external, you know, the external stress

is tau times b on the dislocation, yes? If I have a certain value of distress, I will have a certain value of the radius. As I increase the stress, yes, it will bow out more

and at a certain critical value, it will be released. And so that's at a certain critical value, yes? So what do we have, what are the forces that are in place

when we have this interaction? OK, so you have to think about the obstacle, yes, OK?

And you have to think about the dislocation, OK? So what I'm going to do is I'm going to use a pair of scissors, yes? Pair of scissors, I'm going to cut this off here, yes? And I'm going to cut this also here, same pair of scissors,

yeah? Cut this off, yes? So in order to keep the balance here, yes? I need to restore a force, yes, along the piece

that I've cut off, yes? And I need to consider the force of the obstacle that balances this, yeah? Now what is this force? Well, this force is nothing else than the tension that

works on the line. That's the dislocation line tension. That's a constant value, yes? And what's working in this direction is the force of the obstacle on the dislocation, which we call f.

Now what happens when you increase or rather decrease the radius of curvature here, when you are making the cusps larger, yes?

The cusp is larger, yes, like this. The force actually becomes larger. So let me draw here a vector diagram. It's better.

So this is for, so I have t here, t here. The force of the obstacle has to balance this, yes? When the stress on the dislocation increases,

and I have a more pronounced cusp here, the t is now, the dislocation line tension is oriented this way. This is now the force that the dislocation has to,

the obstacle has to exert on the dislocation. And when this f value reaches, or when the sum reaches a maximum, yes, then the obstacle

releases the dislocation, yes? So what is important here is this angle, phi, yeah? And if this is f max here, say, so f is now f max,

this angle here, now the dislocation will break away from the obstacle, yes? This angle is the critical angle, yes? Phi c, yes? And of course, if the critical angle is this one,

if this would be the critical angle, yes, max, so in other words, f max is this, yes? So this would be a weak obstacle, whereas this is a strong obstacle, yes?

So weak obstacles have phi c's are small, the critical angles are small. Strong obstacles have phi c very large, yes? So if I would look into a material with dislocation, and I could make a snapshot of dislocations

moving through a material with strong solute dislocation interactions, the dislocations would look like this, yeah? This would be strong.

If I would look into a material where I have a solute which gives me weak interactions, weak obstacles, the dislocations would look like this, yes? Because the breakaway angles, phi, would be very large, yes?

Here, the break angles, this angle here, would be close to pi over 2, 90 degrees, right? Very small angles. In this case, yes, phi critical is close to 0, yes?

So the shape of the dislocation tells me something about the strength of the interaction, OK? So let's put this in a little bit of formulas here. Well, first of all, we have, let me see if I can get the pen

here, OK? So first of all, we have the force on the dislocation. So that is tau times b times the length of the dislocation. So that's the force on the dislocation, OK? And so if I concentrate on the situation here,

yes, I have basically, yes, the line tension going here, the line tension going there. The sum of these two is what?

If we know that this angle is phi, it's 2t sinus phi. So 2t sinus phi, yes, balances the external force, yes? And so I can rewrite this equation very simply

by using what we know t to b, g b squared divided by 2, right? And so this equilibrium of these three vectors

can be expressed like this, OK? But I also know what the force is on the dislocation, because I apply it externally, right? So in terms of the externally applied force, I have tau times b times l.

And OK, now if we look at the breakaway moment, so we have the critical shear stress, b times l, is f max. And f max is simply this equation here, where phi is the critical break off angle.

So now I write this as tau c is equals to g b divided by l times cosine of the critical angle. OK, so what is this? This is the shear stress applied to this.

OK, so that's nothing else. That's coming from the externally applied stress, right? So this tells me something about the strength of the material, yes? g and b are materials parameters. There's the modulus and the Burgers factor. So that's a given.

So what are the two important parameters? l and this angle, the critical breakaway angle. So phi c tells me something about the strength of the interaction. Like I said there, phi is large.

It's a weak interaction. Phi is small. It's a strong interaction. Yes? It's large, weak interaction, small, strong interaction. And then the other thing we need to know is the obstacle spacing. But basically, if I have information on these two things, I can calculate strengthening.

At no moment in this discussion did I ever say, was there anything specific about the obstacle? Actually, the obstacle in this particular chapter

are solutes, but they could be particles. They could be dislocations. They could be anything, really. So this is actually, this equation here, it's actually a very fundamental equation when it comes to strengthening mechanisms in steel

and actually in any crystalline solid. So let's, just so you have a feeling for numbers here.

So let's calculate one of these angles. Say we know some parameter values. What is the typical angle? So this angle is determined by the maximum resisting

force of the solute F and by the line tension of the dislocation. That's basically. So OK, well, we're going to try to say something about solid solution strengthening. So let's say we have a silicon or manganese and ferrite.

And we're going to look at the misfit effect on the strengthening. So we need to use the parameter delta. And for silicon and manganese, if you go through the list,

0.02, 0.03, that's the typical value for delta. And let's take 0.03 as a value. So what is the force now? For a single solute, for a single manganese atom

will exert on a dislocation. Now this you have to accept from me, 1 5th GB squared times delta. I haven't shown, proven it now, we'll show it later. But we need to have some formula for F max if we want to calculate a critical angle. So let's assume, and actually this is a very useful formula.

If you are ever doing this kind of calculation, this is a good formula to get an idea, a very reasonable idea of the force that a solute will have on the dislocation.

Anyway, so you have this formula. And then you plug in all these data values. So this is the value of G. This is the value of B. And this is delta. And you calculate this.

And you find a very, very small force of about 3 times 10 to the minus 11 newtons. Now, second thing we need to calculate is the line tension. Because the phi c is the arc cosine of F max divided by 2t.

So the line tension is GB squared over 2. So that's easy. So this is 1 over 2. This is G. And this is again B squared. So tau is 2.46 times 10 to the minus 11 newtons.

So I can calculate phi c. So this calculate here is 1.56 radians, or 89.66. So that's only one degree off 90 degrees, or half a degree

off 90 degrees. So that means that your dislocations will be interacting with the manganese or silicon atom. They will bend out a little bit. And then they'll quickly be released by the obstacle.

So it's what we call a weak interaction. And you will usually find this very close to 90 degrees. So this location will have a relatively straight shape in the crystal. And so we have weak obstacles.

Weak obstacles in the sense that you don't have these very large cusps being formed. Don't forget, this is a one single atom. Of course, the strengthening comes

from adding more atoms to get more solute solutions. Strengthening. So now, how do we calculate this f? So obviously, you can feel when we calculate f, we need to have phi c, the critical angle.

We need to have f and t. t is the line tension. So that's no problem. I only need G and B to calculate this. To calculate f max, that's a little bit more of a challenge here. So I need to say something about the interaction between the dislocation and the solute.

OK, so say for instance, here is the lattice. And here is the dislocation. And in this lattice, I've put in an atom, a misfitting atom. So there will be, around this atom, there will be a displacement.

This is a misfitting atom. It's larger than the iron atom. So around it, the lattice will be squeezed. So I can calculate what the distribution is of the displacements of the atoms.

Yes? I can calculate the shear strains at these interfaces. If you can think of shells surrounding these atoms and the shear stress that applies. I'm not going to go too much in the math here because.

The math is rather lengthy and more interested in giving you the results so you can work with those. But anyway, so that's the idea. So the interaction here will depend very much on actually where you put the dislocation also.

So because if the dislocation passes on this plane, yes, I will get a very different interaction than if the dislocation passes a little bit away from the atom. And if it's very far away, it will not feel the solute.

What we'll first try to do is we'll try to determine the interaction energy. What's the interaction energy? Between a dislocation and a misfitting obstacle. So this is just a schematic here.

So this is a solute here. And say the interaction energy looks like this. It's a well, a potential well. So basically, this means that the dislocation is attractive. It doesn't have to be this way, it can be repulsive also.

So the dislocation comes here, is attracted to this potential well. And then when you want to move it out of the potential, you have to increase the stress on the dislocation to pull it out of this attraction.

But that's not the force, of course, because that's just the interaction energy. So the force is the derivative of this energy. Of course, the dislocation, so the obstacle

has an influence on the dislocation. And of course, there is a vice versa. The dislocation has an effect on the obstacle. So if I want the force on the dislocation, I have to do minus the derivative of the energy.

So that looks like this. So this and this. So if it's attractive interaction, the dislocation will get, when it comes close to the defect, it's going to be pulled towards the defect, to the point. And then when the dislocation needs

to go beyond the solute, it will have to be pulled loose of the point defect or the solute. So that's the shape of the force.

Force distance relation here. So let's do a little bit of math here, because we do need to do this. We have the distortion around a substitutional alloying element. And it is this distortion that interacts with the stress

field of dislocation. And so this interaction leads to an increase or a decrease of the elastic strain energy of the lattice. So in this discussion, or in these theories, we assume that the atoms are elastic spheres

with a certain radius. It's inserted into a spherical lattice hole created by removing an iron atom with a radius RA. And it's the relative size difference between substitutional atom and iron

that is of importance. So we use this parameter delta. And delta is nothing else than this ratio here is the difference between the radius of the solute

and iron atoms divided by R. So it's the radial strain if you want. So you can rewrite this. RS is 1 plus delta R iron. And this is the important parameter.

This basically gives us an idea of the amount of strain that you have, volume strain. But it's a linear parameter. So you can calculate the volume difference between the solute and the solvent atom because it's the volume strain that's important here.

And you get what's called this misfit volume. So that's basically calculating the change in volume that needs to be accommodated in the lattice.

So that is given here. We assume, or actually we know, that delta is a very small number. So we can simply do this simplification here. And that gives me this value for the misfit volume.

OK. So if I have an isotropic point defect, I get this misfit volume. Say it's a large, it's a large atom.

So you will have a hydrostatic pressure there. Hydrostatic pressure at the interface between the substitutional atom and the iron matrix. And so I get elastic strain energy, elastic strain energy,

which is equal to the work done by pressure against the volume change. So this would be this energy. So it's P times this delta V. I

didn't say at this stage what this pressure is. So it's basically P times this. So that is elastic strain energy. So now let's backtrack a little bit here and say, do we have a hydrostatic stress associated with dislocations?

Yes, yes we do. The hydrostatic stress is minus. In this particular case, we use minus

because it's easier for us to work with. Sigma xx plus sigma yy plus sigma zz divided by 3. Where these are the principal stresses. You remember this value, right? That's the, in the theory of plasticity,

we've seen this, right? That's the stress along the yield surface, yes? That's your mean stress, yes? OK. So we can compute this. Let's have a look at how much this is. How would you compute it?

Well, you look for sigma xx, sigma yy, sigma zz. In the previous lectures, yes, for an edge dislocations, these are the formulas, yes? And if you make the sum of these three, yes, and you divide by three, this is what you find.

This is the hydrostatic stress associated with an edge dislocation. And it's a function of the position, yes, in space around the dislocation. All right. So, well, why don't we put some numbers in, right?

OK. So this is the formula that was on the previous page. For reasons that will be clear in a moment, I'm going to simplify this formula by using, instead of x and y, I'm going to use x

is equal to r cosine theta, and y is equal to r sine theta, yes? I'm just going to use this transformation because it

simplifies the formula to this here, yeah? So this y divided by x squared plus y squared now becomes sine theta over r, yes? And that allows me to compute this p value simply. So let's say we're looking at alpha iron. We have an edge dislocation here, yes?

And we want to know, well, what would be the maximum hydrostatic pressure I can get? Around the core. Well, I'll just choose theta equals to 1. That's, sorry, sine theta equals to 1, yes?

That's definitely going to give me a large value. I'm going to choose r, a certain distance away from dislocation, about 10b. That's the distance, yeah? And g is 81.5 gigapascal. So that I can calculate p. And I find 1.6 gigapascal.

So it's a pretty large hydrostatic stress. So what happens now? I have a piece of material that

wants to expand on one side, yes? And I have a hydrostatic stress somewhere else that can provide this deformation, yes? So if they come together, the energy

can be locally diminished by having the hydrostatic pressure provide the work to expand the lattice.

So the interaction energy is the hydrostatic pressure provided by the dislocation times the volume change.

And so it's basically what we derived just now. So times, so that is the hydrostatic pressure. And that is the delta v. And I can express this in terms of x and y,

or in terms of r and the angle theta. So right, and I can also use here, instead of using this r square r to the third,

the radius of the iron atom to the third, I can use a delta v parameter. Delta v parameter is nothing else then. The lattice volume changed due to one solute atom.

So it's this here. And you can see 4 delta this. So it's an alternative way of using this, the same equation. So let's, OK, so let's, I just want to take one more minute.

So for instance, what does this mean? I have manganese interacts with an edge dislocation in alpha iron. So this is my interaction energy. I have all these parameters. I know what g is. I know what b is. I know what the Poisson ratio is.

I know what the radius is of r, my iron atom. And I know what delta is for manganese. So I plug this all in into this energy equation. So the energy is now here. This is the result. This is the interaction energy.

It's a potential well. And I now compute, for instance, in this particular case, I will compute the force on the solute. That's the derivative of the interaction energy.

So the derivative of this line here is negative, of course. It reaches a maximum, a minimum, a maximum, and like this. So this is the force-distance relation. And in this particular case, we've chosen y is equal to b. So y is the distance between the position of the atom is here.

And the dislocation moves this way.

So you can see, obviously, that depending on what this distance is, I'll have another profile. OK? But I can calculate from these different profiles what the maximum value will be. Yes? And that will be f max. Yes?

Anyway, we're over time. We'll continue with this on Thursday.