So today, we will be starting a new chapter on precipitation hardening in steels.

So those of you who've ever taken a introductory mechanical metallurgy or material science class on mechanical properties of materials will know that that is a very common way used

to harden alloys, and steel is also, we also use precipitation hardening in steel. In fact, in all types of steels,

we apply the method to increase strength. In steels in general, with the carbon steels, the most famous application is high strength low alloy steels.

Although there, we have to say that the amount of strengthening you get from the precipitation is not that large, mainly because we have very low levels, low volume fractions of precipitates.

The reason why the HSLA steels are strong is a combination of both grain refinement and precipitation hardening.

The other steels that are precipitation hardened are typically special steels, stainless steels in particular. So we will talk a little bit about different classes

of steel, different types of steel, like maracnesitic stainless steels, and austenitic stainless steel, ferritic stainless steels that are precipitation hardened. So a little bit of a departure from the types of steels we've been discussing up to now. So in general, if you want to picture yourself

what is happening during the precipitation strengthening, so imagine a single crystal or a grain inside steel, and we apply externally a certain force, F.

And we have here a screw dislocation propagating from right to left, right to left. And we have, by a certain trick of microstructure control, we've managed to get particles embedded in the lattice.

And the dislocation meets these particles on its way through the lattice and interacts with these particles. And in general, you can say that there are two mechanisms.

If the particles are very large or if they are very hard, and they can, in that case, be very small, then we have a process which is called bypassing.

Dislocations do not go through the particle, but bypass the particle. And in the process of doing so, as they bypass the particle, they leave behind a dislocation loop around the precipitate.

And every time a new dislocation passes through, it leaves another dislocation loop around the precipitate. So that's one thing that can happen. Another thing that can happen is that we have small particles that are soft or, yes,

screw dislocation passes now, and it cuts through the particle. And in order to do that, there

needs to be some kind of coherency between that particle and the matrix so that the slip can be transferred through the particle. So high level of coherency will be required to manage, to cut the particle.

So and then if we go back now a step to this bypassing mechanism, yes,

lack of coherency between the particle and the matrix can also be a way to force dislocations to bypass the particle. So we have basically two extremes to these pictures,

large or hard or incoherent particles, or particles that are all these things at the same time. Or we can have small, soft, coherent particles that can be sheared as the dislocation pass.

And you know, steels are very complex, so precipitates occur often. The precipitates that will impact the strength via cutting or bypassing have necessarily to be in the lattice.

So if you precipitate something in a grain boundary, grain boundary is by itself an obstacle. Having the precipitate in the grain boundary doesn't help.

So it's important if you want to achieve precipitation strengthening, the particle needs to be in the lattice. So some examples, first of all, up there in the corner, vanadium carbide is an example of a precipitate in a high strength low alloy steel.

So you can see a ferrite grain clearly, and then the particles, the vanadium carbide particles embedded in the grain. Very common precipitates in steel is carbide,

it's cementite. And so that particular image is of a martensite lathe which contains small precipitates of cementite.

And you can see indirectly that the cementite is coherent with the ferrite lattice because it's got this needle shape.

And you can see the needles are either at this specific, somewhere, 70 or 80 degree angle or horizontal. So that tells you there is a high degree of coherency. There's going to be a lattice correspondence orientation

relationship between the cementite and the matrix. Example here of a other precipitate

that we see in high strength low alloy steels. These are carbides and nitrides of niobium and titanium. And another way to precipitation harden ferrite is by copper additions.

We'll talk about this a little bit more in depth today. And you can see here, this is a micrograph. You can see all these dots here are basically copper precipitates. Again, inside, you can see the grain here.

Grain boundaries here, yes. You can see it's inside the grain. That's a necessary condition for the precipitation hardening. So when you look at this, what can you see?

Well, obviously, there's going to be very important parameters to precipitation strengthening. Obviously, the properties of the particles, whether they're hard or soft. For instance, copper is definitely a much softer phase

than ferrite. These nitrites are extremely hard in comparison to ferrite. So very different behavior. So that's one thing, the properties of the precipitate. Second, the volume fraction. You can see here volume fraction of my copper

here is much larger than the volume fraction I have there of my vanadium carbide, volume fraction. And this is even higher density, the carbide in the martensite. And then, obviously, the size of these particles.

The cementite particles are, let me see which one. I think in this case, actually, these carbides here are the smallest of these examples.

So the size of the particle will have influence. And of course, as always, in dislocation strengthening mechanism, if you have an obstacle, you remember, the condition for breakaway from an obstacle

is that the externally applied stress is proportional to a critical breakaway angle. And 1 over L, and L being the distance between the obstacles.

Yes? So the density, the radius, the size of the particles, and the inter-particle spacing are important parameters. Now, one of the things you have to realize

is that the radius, the density, the inter-particle spacing, and even the structure of these precipitates,

these are not single numbers. This very much depends on how you engineer the microstructure. So take, for instance, a very nice example of precipitation hardening in steel

is when you add a few percent of copper to ferrite. So this work is recently done at TFT here. And it shows what happens when you precipitate copper, when you form copper, in the ferrite matrix.

So originally, the copper, which is in supersaturation, so you basically form it from a supersaturated solution, won't go too much into the microstructure, the way you achieve the microstructure, but I'll say a few words about this probably

later on today. So you start making very tiny particles, which are more clusters of copper atoms. And these clusters, copper is FCC metal. Well, in this case, it's actually a BCC precipitate,

basically. And it's an alloy. It does contain iron, yes, but it's definitely BCC, yes? And obviously, that's not the natural state of copper, yes? And it will evolve to FCC, pure FCC copper particles.

But it will do this in steps, yes, in steps where you make a special crystal structure, which is called 9R, which is orthorhombic, which is still iron-copper alloy.

And you can see the microstructure up here, the lattice, high-resolution TEM lattice image up there, which is a magnification of this small particle, yes?

Small particles, about five nanometers at this stage. So this has grown from a tiny cluster to a larger particle. And then this thing continues to grow, yes? And there is, again, a crystal structure change,

yes, to FCC copper. All the while, the iron atoms are expelled from the lattice, from the lattice of this growing particle. And you get nicely twins. You can see the twins in these particles. And eventually, the particle becomes larger.

It coarsens. If you wait long enough, you will have ripening processes. The particles become very larger, very much larger, at the expense of smaller particles. And you end up with FCC copper. And it's basically pure copper particles.

So depending on how far I've carried out the aging, at what temperature I've done the aging, I will get different particles, different types of particles. And that will impact the precipitation strengthening.

So that's an important, if you were ever involved in research in this area, never forget this. It's not because you do one heat treatment that you have to get a single radius, composition,

distribution, and inter-particle spacing for your precipitates. So it involves a little bit more work. And we'll discuss this. OK. Yeah. Let us now have a look at the precipitates.

And to connect with the previous slide, let's look at the crystallography of these precipitates. We have, when we try to engineer precipitation hardening

in a grain, inside a grain, we always have to make sure that there is some crystallographic relation possible between the precipitate and the matrix.

And so just to highlight the similarity between the matrix, which can be BCC iron or FCC gamma iron, I show here on this slide the matrix crystallography

and then the crystallography of the main types of precipitates that we see for steels. So take, for instance, a BCC iron, so alpha iron.

Crystal structure also goes by the name A2, right? Well, one of the precipitates that you can use for precipitation is the B2 structure, or cesium chloride, the same as cesium chloride. I don't know if you're familiar with that.

And that's the structure of FEL, or nickel aluminum. And you can see it's basically very similar to BCC iron, except that the central atom is now replaced by another atom, aluminum.

So you can have iron aluminum. You can also have nickel aluminum by adding nickel and aluminum to your iron in a certain way. We'll talk about details in a moment. And then we have the important DO3 structure.

And if you look carefully, and I've tried to visualize this as best as I could in this crystal structure, you can see that it's nothing else than a combination of the A2 structure and the B2

structure. You can see, you see the four A2 blocks and four B2 blocks. You can see these cesium chloride unit cells, and the normal BCC unit cell.

And that's the big unit cell. DO3. And a very important representative of the DO3 structure is iron 3 aluminum. So don't mix iron 3 aluminum with nickel 3 aluminum.

They have nothing in common. Iron 3 aluminum. Then we can also go one step further and complicate the structure of DO3 one more step

by having the four unit cells in the DO3, which still looked like A2. We now, in the center, we put another atom.

For instance, on the dark atoms, titanium. The lighter atom, silicon. And I get a structure which is called the L2-1 structure. And in steels, you can have the L2-1 structure with precipitates Fe titanium silicon.

Just do some faces. I'm not sure if everybody gets it. So you see this unit cell here, right? So you've got it here once, and here once,

and here in the back, and here in the front. So these three are B2 units. And then the other units are 1, 2, 3, and 4 here are 8 A2 units. So that makes the DO3. And now if I have basically two different unit cells,

and there are two B2. So this is a combination of A2 plus B2. And this is a combination of two types of B2s together.

And so one atom would be silicon, and the other atom would be titanium. So you can see that simply by making the lattice a little bit more complex, having larger unit cells,

I can make structures that will pretty much be able to match crystallographically their matrix without too much lattice strain. So that's for BCC.

For FCC, so our starting structure is, again, the basic unit cell for FCC gamma iron. That's A1 here. I can do the same thing as what I did here. I can have a crystal structure, for instance,

titanium aluminum, TiAl. When I replace the atoms on the mid-horizontal plane here by aluminum, and then the other 10 atoms are titanium,

then I get TiAl. One step further, so that would be the equivalent. Once the first nickel-3 aluminum, I get this by having all the atoms on the six side planes

to be aluminum. Sorry, nickel in this case, because you want three. So you can see the formula. You have six atoms divided by two, so you have three nickels.

And you have eight atoms divided by eight. You have one aluminum atom. And then we can also have carbides. And carbides have also a basic structure, very similar,

if you look at this, very similar to gamma iron, except that in between the main atoms, yes, I put in, yes, interstitially, as it were. It's not an interstitial. But if I put carbon atoms here interstitially,

I would make the structure of the carbides. And all the carbides look like this. And we can form carbides not only in FCC, but also in ferrite, in BCC iron.

But you can see the connection between these common precipitates in BCC and FCC steels. And that's really the reason why you can precipitate them in the lattice. Because there's going to be some lattice strain,

but the lattice strain's not that large. So you can precipitate them in the lattice rather than in the grain boundaries. When the mismatch is very high, you will get precipitation in grain boundaries. And that's not what you want, basically.

So let's just start asking ourselves, is there a simple way to relate volume fraction of precipitates? The radius and the distance between the particles.

The mean distance between the particles. That's a very useful thing to have, this kind of relation. And so a nice derivation.

And then I'll also add a correction to the formula. But it's a very nice derivation. Because certainly when you're involved in experiments, you may know what the volume fraction will be of precipitates just by calculation.

You know how much, for instance, niobium you've added, how much carbon you've added. So you can calculate what the volume fraction of niobium carbide will be. You can also, for instance, TM analysis, get to know the radius of the particles. So if you know these two things, then you

can calculate the third thing, the mean inter-particle distance. And if you know the mean inter-particle distance, you have a way to determine the strengthening effect that you are trying to achieve

or that you have achieved. OK, so a relation between the volume fraction of precipitates, the mean precipitation size, and the average spacing between the precipitates. So what we do, you consider a thin slab of the grain,

and that slab has a thickness of the mean grain diameter, 2 RP. So it's a slab that's as thick as the mean particle diameter, precipitate diameter, and it's got a surface area of A, which we don't necessarily

need to specify this. So the whole volume is 2R times A. And if I multiply this with the volume fraction of precipitates, it gives me the volume

of precipitate in that slab. The volume of precipitate in the slab is 2RPA times F. So once we have this, we're in business. So this is the volume of precipitates.

So then I can calculate the number of precipitates in this slab. And it's very simple. I have the dimensions of the precipitates, which

I assume to be circular. 4 thirds pi RP to the third divided

by the volume of precipitates. This should be 1 over, actually. So I'll correct this when I put it on the E class.

OK, so anyway, this is the answer. And I can determine the areal density of particles. So that is the number. Yes, this is correct here. This is the number of precipitates divided by A.

That gives me how many precipitates I have per unit area. Yes, OK. Now, if I assume, yes, so the areal density tells me

how much, basically, how much of the area

I assign to one precipitate. So this is one precipitate on this for this area. So that's the areal density. Of course, if I know the areal density, 1 over the areal density and the square root from that

gives me the inter-particle spacing. And this is what I do here. The average distance between the particles is 1 over the square root areal density, and it's this.

This is an important formula that, as I said, connects the average distance between particles, which is an essential feature here, with the radius of the particles and the volume fraction of the particles.

So now we can go on, first of all, by making a correction to this formula. If you ever use this formula in practice for HSLA steel,

just go ahead, it's fine formula, because we have very low precipitate fractions. But if you're studying precipitation hardened

steels, which are not of the HSLA type, the volume fractions are much larger in this case, much larger than in the case of HSLA steels. Then you better use a better approximation for the inter-particle distance.

And that is because this formula up here overestimates the particle spacing. So I'm not going to prove this. This is the equation.

It's actually, you can see that the original equation is in there, and the correction is very simple. It's just numerical values. So you can as well use this formula.

So the way it looks like is that, I think I should have volume fraction.

So if the volume fraction is larger than 0.1, then the top formula overestimates the distance. So you need to have to use the corrected formula.

For HSLA steels, our volume fractions are of the order of 10 to the minus 3. So well within the area where the formula on top applies. So you can use that formula.

In other cases, particularly precipitation hardened stainless steels, et cetera, you should perhaps use this formula. And here I give an example. When f is large, that's the case, for instance, for precipitation hardened austenitic steels.

This is a very common grade, A286. This is a better approximation. So let's have a look at this formula first.

Just simply look at the formula. Yes? And let's not consider many practical considerations. So let's plot L as a function of the radius of the precipitates.

Yes? OK. At constant volume fraction. Right? So when you have a constant volume fraction and you change the radius, the particles become larger.

But you also have fewer particles. You cannot have the particles all becoming larger. So you're basically discussing coarsening. So you have small particles to start with. Yes?

And these particles become larger at constant volume fraction, so I get less particles. All right? So what happens to the distance between these particles? It gets larger. OK?

OK, now let's look at the top one is for certain volume fraction. So here we have L values are large. And as I increase the radius, the particles become larger.

I get less particles, obviously. And you can see that the distance inter-particles becomes larger. And if we increase the volume fraction, so here I have a larger volume fraction. Obviously, I have twice as many particles.

Automatically, the distance becomes smaller. And again, if I coarsen the particles at same volume fraction, at this higher volume fraction, I get an increase of the inter-particle spacing.

The problem is that when you actually do precipitation-hardening treatment, both these parameters change.

At the beginning, you have no precipitates. And then you precipitate particles. And at the beginning, you have nucleation phenomena.

And after that, you have growth phenomena, usually diffusion-controlled growth. And then eventually, you get coarsening phenomena. So nucleation, you have little particles, tiny particles far apart.

And the particles grow. Eventually, of course, you do your precipitation treatment at a single temperature. So eventually, you reach the equilibrium volume fraction.

So the volume fraction of particles increase. But as your aging time runs, your volume fraction will saturate.

Particles will keep on coarsening. The radius continues to increase. So obviously, the distance between these particles, well, let's see what happens.

Is it a simple function of time or not? Well, obviously, because of the way f changes and r changes. Yes. So for instance, let's avoid 0 because 0 is not such a nice. But here at the beginning, I have tiny particles.

Our p is very, very small. And f is increasing. f is increasing. So I've got small particles. f is increasing. And l is r, small, divided by something that's increasing.

So l is decreasing. OK, so that's one thing. Let's look at long times now where the particles are coarsening. In this case, f has reached its saturation.

f doesn't change anymore. So f is constant. And the only thing that happens to l is it increases because the radius increases. So here, l goes down with time. And here, l goes up with time.

So it's clear that there's going to be a soft point where l is minimum, where we reach a dip. And then the coarsening causes l to increase again.

So why is that so important for the mechanics? Well, because l is in the denominator. So when l is small, this will reach its highest value. So this spot here, where l is minimum, is very important.

Yes? And it will correspond to the situation at which we call peak aging, peak aging. That's where you get the maximum strengthening. And again, it's a function of time. And why is it a function of time?

Because the volume fraction and rp are a function of time. So now let's have a look at some real situations. So precipitation strengthening models.

It's kind of interesting. When you look at, so again, I remind you, you have two situations. You have the precipitate cutting can be cut, or the precipitate can be bypassed. When you cut a precipitate, many things

can happen, depending on the dislocations, the matrix dislocations, the dislocations in the precipitate, the crystallography of the precipitate, you name it. It's complex.

One of the things that's really important is that the strengthening when you cut particles is proportional to the square root of volume fraction times diameter of particle. And there are a number of strengthening mechanisms.

And we'll discuss some of them. And some of them, we'll calculate examples. You can have strengthening due to coherency. You can have strengthening due to what's called chemical hardening, order hardening, stacking fault hardening, and modulus hardening.

Yes? These are different ways in which precipitates cause hardening. In the case of a particle, the properties of the particle themselves are unimportant. The dislocations just don't go through them. So it's very simple, actually.

There's only one mechanism. So in this case, we have hard, large particles. We cannot be sheared. So the properties of dislocations or faults in the precipitates are irrelevant.

And in this case, the strengthening is proportional to square root of f and r divided by rp. So we have the cutting mechanism. And the other one is the bypassing mechanism. And here we have square root f times rp.

And here we have square root f divided by rp. So you can see the fundamentally different mechanism. In this case, it's good to have larger particles. In this case, it's bad to have large particles.

In both cases, it's good to have a high volume fraction of particles. So here, this is always good to have a high volume

fraction of precipitate particles. OK, so let's now discuss an example of precipitate cutting.

And we'll talk about modulus hardening. And we'll derive the formula which describes the hardening.

Well, the word already says it. What is this strengthening due to? It's due to a difference in elastic modulus and shear modulus and Young's modulus, et cetera, between the matrix and the precipitate. And the formula gives us delta tau.

That's the strengthening effect from a precipitate, which has a different modulus from the matrix. It's 0.8g times p divided by l. And then the square root of 1 minus the ratio

of the precipitate Young's modulus divided by the matrix Young's modulus square. So if you look at this formula, you can already see something very interesting. First of all, if Ep, the precipitate modulus,

is equal to Em, there's no hardening. OK, that's obvious. The other thing is if Ep is 0, which means void, there's nothing there, the hardening will have a maximum value.

Yes? So it gives you something, the hardening, that's a little bit counterintuitive. It basically means that when a particle is soft, I get hardening. When it's softer from an elastic point of view, I will get hardening.

So what I basically have, the way you have to think about it, is this location passes through the precipitate. I have different shear modulus in both phases.

And so the dislocation properties will be different in the particle. That's basically what modulus hardening is all about. So first of all, what we're trying to do

is what you're always trying to do when you are considering strengthening mechanisms is you basically look at this graph. So this is just a review of something we discussed earlier.

You look at the dislocation, which is held up at these obstacles. And in this case, these obstacles are not solute atoms, but they are precipitates. And again, the precipitate exerts a, they would hold the dislocations from moving.

And this is balanced, this force is balanced by the line tension of the dislocation. And we know that if this angle here is phi over 2,

and this angle is phi, we basically have f is 2 times t, 2 times the line tension times cosine phi over 2. That's a fundamental equation. This can be rewritten in terms of the geometry.

If you can use phi over 2 or theta over 2 here, it's the same. And you get then 2t sinus theta over 2. So now the other thing we know is

that the situation is caused by an externally applied stress. Otherwise the dislocation would just be standing straight, minding their own business, and not running into these obstacles.

So the force on the dislocation which causes the situation to happen is tau times b times the length of the dislocation segment. So it's tau times b. And if the dislocation segment has a radius r,

and the distance between the particles is l, then we know we can write tau b times l, tau b times l as tau b times 2r sinus theta over 2.

So now if I combine this and this, I get this. So tau times b is f over l. F is this here, the force of the particle.

And l is the distance between these particles. So we will now apply this to copper and iron.

Copper and iron, we know from analysis of the strengthening mechanism that this is a modulus hardening mechanism.

So we look at cutting of copper precipitate, which is softer than the matrix, softer in terms of modulus. And we look at the precipitate strengthening that result.

So basically we have the same as what we just looked at on the slide. A line, dislocation line, a counter, randomly dispersed copper particles. And they have a mean distance l. We apply an external stress, which

gives us a shear stress on the glide plane. And the dislocations bow between the neighboring precipitates. And they have a curved shape with a certain radius r. And then what we get is we get a balance, whereby

the precipitate exerts a dragging force on the dislocation. And that's balanced by the line tension. And so we can write, that's what we just saw. F is 2t times sine phi. Right, this should be phi over, no, it

is a little bit of confusion here, I think. This is, this phi over 2 is phi, I should, OK. I'll make sure this, so in terms

of what you're going to see in the next slide, this angle here is not called phi over 2. This angle is called phi. So I'll correct this. So 2t sine phi, where t is the line tension.

And we know we can write it by gb square over 2. We also know that this is not perfectly correct, because it will depend on whether we

are looking at the screw dislocation or edge dislocation. But we'll just use this for the time being. And when I have breakaway, that's when this angle reaches a critical value

equal to the phi c, 2t phi c. And if I substitute the formula for t in here, I get this equation. So f max is gb square cosine phi c.

So what I want to say is that I have a situation that goes from this to a situation that goes from this. So as I increase the tau, tau is increased,

you go from here to here. The t vectors always stay the same. Remember, the only thing that changes is their sum becomes larger as the bending

between the advance of the dislocation between the two pinning point increases. So in other words, this angle here, t to c.

Sorry, theta decreases. So you can see here it's this value at every time f is equal to t, 2t cosine phi. But you can see that as the angle decreases, the cosine goes up.

And I reach a steadily higher f. When I reach f max, which defines the obstacle strength, I also reach a critical angle.

And the dislocation passes the obstacle. So that's well known from earlier. So of course, this force, the situation

can also relate it to the externally applied force. So this f max is then tau, the strengthening, times b times l is equal to this f max. So I can determine that strengthening is equal to gb over l times cosine of the critical angle.

So how do I look at this in terms of the critical angle is basically what defines the obstacle strength.

Because t is always the same for all my dislocations. So what defines the obstacle strength is this critical angle. So now I am going to look at precipitates.

The shear modulus of the precipitate is shear modulus gp. The shear modulus of the matrix is gm. And so we have different line tensions in the matrix, tm, and in the precipitate.

The precipitate should be p. And so very simply put, the line tension in the precipitate is alpha gp times b square. And in the matrix, it's alpha gm times b square,

where alpha is 1 half. And now you can express the force equilibrium at the matrix precipitate interface. And I'll show you how this is done. But it basically, this force equilibrium states that the tp times sinus theta p

is equal to tm times sinus theta. We'll see this in a moment what it means. This is how it works. Let me go one slide further. So the dislocation is first outside the particle.

And it moves into it. As it moves into it, this location assumes this shape outside the particle. And inside the particles, it makes a straight line.

And then the same situation on the outside. So and if you look inside and outside the particle, I have different.

different, yes, different line tensions, okay? Okay. You can see here. And I also have different, the different angles here. This angle theta P here is the angle that this location

makes with the normal to the interface here, yes? And theta M is the angle that the external line tension makes with the normal to the particle here, okay? Now, what's important

here is that as the line, this location line moves through this, the particle, eventually, say very close to breakaway, yes, this angle here, yes, you see when this point moves

to this, to this here, yes, so close to breakaway, this angle here goes to 90 degrees, yes? Yes?

So as I approach breakaway, breakaway being the dislocation comes out of the particle, this angle theta P prime is close to 90 degrees. And theta M here, yes, again, as this line

basically moves up to here, yes, theta M and theta C move towards each other, yes?

It's because this, the normal to the surface moves upward here, right? So when they meet, yes, theta C and theta M become, move closer to each other. At breakaway, at breakaway

that's when the dislocation, this dislocation goes out of the, you see it, when it goes out there's a small dislocation line segment, so at breakaway it comes out and so that's

where this holds. Okay, but another thing that's important here, the force equilibrium at the matrix interface yields that is expressing the fact that the components of the line

tension in this direction here should be equal. So theta M times sinus of, Tm times sinus of theta M should be equal to Tp times sinus theta P. So if the angle is large,

Tp should be small and if the angle is small, Tm should be large. So that's, you

see there is a difference in Tm and Tp, yes, and that causes this difference in Tm and, theta M and theta P and these two components of T, the small arrows here, should be equal and opposite and that is expressing, because I need to express force equilibrium at the

interface. Okay, so now if I use this, I express equilibrium at this equilibrium and I express this equilibrium at breakaway conditions. So that's where theta P goes

to 90 degrees, yes, and theta M reaches the critical angle for breakaway. I get this, theta P is equal to theta M times sinus critical angle. Okay, this is the same

in three steps. Okay, and now it's very simple. I use the equation, the fact that sine square plus cosine square is equal to one and the fact that there is a simple relation

between the shear modulus and the Young's modulus and the shear modulus, yes, namely that the shear modulus is the Young's modulus divided by two minus one, two times one minus

the Poisson ratio. So if I combine these two things with the equation I just had here, right, so I need to use sinus, I need to have sinus critical angle, okay, right, so this is

the critical angle, that's square root one minus sinus square of the critical angle and it ends up being this simple equation. GB divided by L times the square root of one minus the

square of the ratio of the particle modulus over the matrix modulus. And so this is the

basic theory, there is a more refined theory, assuming random array of precipitates, and it basically gives me a slightly lower value for delta theta. And I can plot this, yes, if you

plot the, in delta theta, the only important factor is this here, so if you plot the square root parameter as a function of the ratio of EP over EM, going from zero to one, yes,

you find that it goes from one for a void to zero when matrix and precipitate parameters are the same. And I can determine what the modulus ratio is for copper and iron, yes,

it's about 0.6, so that means that this parameter will be close to one, about 0.8, close to a very high value. And I can also, using the formulas we just discussed, determine

what the breakaway, this breakaway angle looks like. So say if we have a particle

that has a modulus that's very similar to that of iron, yes, then the angle here, the breakaway angle before the dislocation breaks away will be, this angle here, will be very

close to 180 degrees, yes? So that's what you get. If EP is equal to EM, the breakaway angle is close to, yes, on the contrary, if this is a void, yes, so basically EP is

zero, then the angle before you have breakaway becomes, this angle here, becomes very small. It goes close to zero, and that's what you see here. And so what's the breakaway angle that you'll have, or you should see for, so this angle here, you

should see for a ratio of particle modulus over matrix modulus, that's 0.62, yes, equal to 0.62. You can see it should be between 60 and 80 degrees, so a little bit less than

80 degrees, about 70 degrees. That should be the breakaway angle. So let's see what we get in the last few minutes in practice for iron-copper. So you make measurements

and you extract from your measurement the strengthening effect due to precipitation. So first thing you have to do, yes, is remove the effect of copper solid solution, because

it so happens that when you add copper for precipitation hardening, you also have copper in solution. So you have copper precipitates and some of the copper remains

in solution. So the copper that remains in solution has a solid solution hardening effect. So for instance, if you add 2% mass percent of copper, you can see you have an increase in about 75 megapascal. So always very careful when you add alloying

elements and you're trying to evaluate how much strengthening you get from precipitation strengthening, make sure that you also take into account the fact that that element

may, in solution, also cause solid solution hardening. So in this case, you must remove the solid solution hardening due to copper. So let's have a look at the data. So first

of all, how do you do this in practice? Well, it's very simple. This is a, the principle is simple. This is the iron-copper phase diagram, yes. It's the iron-rich part of

it. So here we have a few percents, goes to up to three and a half mass percent of copper. So let's say we have 3% of copper. We fully, so the first step is we solution treat the alloy. So in this case, it means it's going to gamma phase region. So solubility

at the solution temperature is high. So all the copper is in solution. Then the thing you do is you quench it. You quench it to room temperature. And so the solubility

at the, at this temperature is very low. So when we age the material, we reheat it to 500 degrees C. You can see the solubility is very low, yes. Copper will precipitate.

First nucleate and form clusters, then the clusters will grow and you'll get this structural change also. So you can look at the precipitation sequence as a function of time, yes. So you

can look at the precipitate radius as a function of aging time. So you can see it here. Particles are coarsening, yes. You can also see the precipitate spacing, how the precipitate spacing changes with time. So here I had particles that go from 5 nanometers, 15 nanometers.

You can see how much time it takes to get 5 nanometers. About 28 minutes is 3,000 seconds, yes. The precipitation spacing increases, goes from about 50 nanometers to

250 as the aging proceeds, yes. The, and it depends on the composition also. You can see here that at one mass percent, I stick at around 300 nanometers, yes. And if I

look at the hardness, I see that for one mass percent, the peak aging is relatively weak, yes, relatively weak and occurs at around 10,000 seconds, yes. And the peak

aging for two mass percent occurs earlier at around, I would say here, that's 200, between 200 and 300 seconds, yes. So a few minutes. That's where I have the peak aging.

So why don't I see this, the precipitate spacing changing? Well, that's because where these measurements starts, yes, I'm already past the peak aging, yes. I'm already past

the peak. I'm already, this is data for two mass percent, I'm already past the peak aging and we're looking at what's called over-aging and the particles are basically coarsening, yes. Okay, I can do something similar, yes, but instead of having in the

x-axis the precipitate, the aging time, I can have the precipitate radius. I'm seeing them over time here. We'll just continue on Thursday because it would take me too

long. Anyway, thank you for your attention and we'll see each other on Thursday to continue.