All right. So, first of all, some organizational points. Everybody handed in? Yes? Okay. It's next week. I think it's the week of December 2nd. This will be the last teaching week

for this course because the week after that I have to do, I'm teaching in India for a week. So, I'll be teaching next week and I'll try to finish the course next week. So, I'll know by Monday whether we need an extra class, but I'll probably be able to

tell by Monday. And so, next week will also be the last quiz as usually on Thursday. Okay. Right. And as we had said, there will be no final exam. I'll just use your quizzes

to get the final results. Right. Okay. So, a few things that I haven't put up the final

slides, version of the slides for the precipitation strengthening on E-Class. But anyway, some slides I amended. One of the slides that I amended is the relation between the,

remember this very interesting relation here that we found which connects the spacing between

the particles and their radius and their volume fraction, right? So, that's a very useful relation to use in practice. Right. So, this is an example here if you use it, if you have volume fraction of 10 to the minus 2, yes? This is, by the way, this is volume fraction,

right? So, maximum value is 1, not 100, okay? So, this is 1%. F0.01 is 1%. So, that means you have L will be 14 times the particle radius, for instance. Now, okay,

this is the basic equation. You can use it, as I said, if you have low volume fractions, that's fine. And of course, there are lots of simplifying assumptions in, when we computed it.

So, and there are more correct equations. So, one of them is shown here, yes? Which is a better approximation. So again, if you apply this formula here, you can see that

for, you get 16 times the radius of the precipitate for the, if you have a volume fraction of 1%. And I also give you an equation that's similar. It gives very similar results

than the previous one. But this is actually the correct one, you know, if you want to be absolutely correct and not make shortcuts. This is the one you should use, yes? And

so, I've taken the notation where the, where we use the diameter instead of the radius, this D is diameter here. And so, say you, by TM measurements, for instance, you are able to

determine the particle sizes, yes? You can calculate the mean value of the particle size, yes? Then you make a column, particle diameter, and then you make a column with the squares

of these, yes? Of the measurements. And then the cube of these measurements. And then you measure the mean value of these three columns, yes? And that gives you the mean value of the diameter, the mean value of the square, and the mean value of the cube of the particles,

yes? And then you substitute this into this and that will give you a, you know, the absolute correct, stereologically correct value, okay? So if you don't want to, you know, like in your PhD, for instance, that's the formula I would use. Of course, the

the problem is, in order to get this, be able to use this value, you need to have, you know, to have measurements of D, you know? And if you don't have that, right, then you use the simple formula which will give you a value that are close enough,

all right? So everything depends on your particular situation, which formula you want to use. Are you going to find big differences? Not, yes? Is it going to matter a lot with the type of data you'll be able to get? Probably not, but it's a nice, you know, if you have, you know, measurements of actual particle diameter that you get, for instance, from

extraction replicas with TM, well, this is a nice way to measure it. Of course, how many should you measure? Nowadays, you can do automated measurements, so you can measure hundreds of particles, so you get good statistics, if you can make good samples, of

course. And, right, this is basically what I wanted to say about this, this formula. Oops, sorry. So we discussed this. I don't have anything different to say. Right. So,

we had said that, you know, if we introduce particles in the matrix, so it's important here, we're talking about particles in the matrix. This is a grain, yes? The particles should be in the matrix, yeah? And two things can happen. Either the particles can pass,

excuse me, cannot pass the particles, yes? And then we have bypassing, or the dislocations can go through the particle, okay? And we had, we were talking about, we were discussing, so the particle gets sheared by the dislocation.

Of course, the force that the obstacle, the particle will have exert on the dislocation will depend on the shearing process. And so we had discussed modulus hardening, and in some depth, I'll review this, but there are other

possible hardening mechanism, stacking fault hardening, order hardening, chemical hardening, and coherency hardening that are possible. So we'll go, we'll first review what we found for modulus hardening. See, look at the system where it's being used, and we've already

introduced that. That's copper and iron, yes? And then we'll discuss a bit more quickly the other hardening mechanism. So just repeating here, modulus hardening is when you have,

the precipitate has a different modulus than the, at this location, yes? And we, we had to review the theory again, okay. I've added this here in the back, it's just, it's kind of theoretical background info is kind of interesting to know

because of the equations you'll get. So you, you know, you're not supposed to learn this in, by heart in any way, but it's a derivation that basically shows you,

right, and it's hidden here, but that in the particular situation that we encounter often, yes, is that the hardening effect that you will get from the cutting here, yes, will be proportional to the, the maximum force that the

obstacle exerts on the dislocation to the power we have, yeah? And so that's shown how you get that specific power in the slide here. So remember that because

just in case you're wondering why when we go through the different formulas, why there's always this power through three halves there, you can find it back here. It's basically simple math, yes, and that is related to this, so the simple model of, which is called Friedel

model, Friedel statistics model, that's when, when a dislocation pass through an array of obstacles, you get some kind of steady state that each time when the dislocation

advance by passing one obstacle and getting caught on the next obstacle, yes? So the dislocations move like that, and so that allows you to do some simple geometrical calculations,

which we call Friedel statistics, right? And then if you apply this, you find that, you find the critical shear stress, which is the maximum retaining force that the obstacle

to the power three halves, okay? And yes, okay? Right, so let's go on. So we've been talking about the copper precipitation now, right, as an example of modular strengthening, and I had

reminded you of the fact that the strengthening effect is related to the reciprocal of the distance, so that's a really important parameter. The smaller I make it, the larger I get strengthening, and the cosine of the breakaway angle, yes? So the breakaway angle tells me something

about f max, basically, yes? And its maximum value is for the breakaway angle zero, yes, is one, okay? And so we had gone through the theory, yes, and the result, yes, and this is

how you have to imagine the dislocation going through this particle when there is a difference in modulus. What it means is that I have a difference in dislocation tension, yes,

this T value here, yes, and finally we get, okay, we get this equation here, right?

This equation here is, all right, and it tells me that the strengthening is proportional to one over L, the distance between the copper precipitates, and then this factor here,

square root of one minus the ratio of the modulus of the particle and the matrix, and all right, okay. Again, so some of you, I don't know, maybe for your research or something, you know, you'll go into the literature and you'll see, well, you know, I can't find this

equation or is this, obviously this equation is a simplification, yes? And one of the things that is simplified here is, and it's very often simplified in the elementary theories, is we don't take into account the fact that the line tension is in practice not equal to GB

square divided by two, right? But it's a more complicated function, it's a function of what type of dislocation I have, et cetera. So this is the actual equation, we've seen it before,

just if you don't remember. So for instance, if you have a screw dislocation, this equation becomes this, right? And there is this factor here, this mysterious factor which is always a problem, is the natural logarithm of R over R zero, right? R being the

the distance basically between dislocations and how far does the influence of a dislocation extend? So it's kind of a really difficult parameter to define. And divided by R zero, that's the size of the core of the dislocation, yes? So in this particular case,

for R, you just take the distance between the particles, that's very reasonable. And for R zero, you find two times the Burgers factor is a good value. Surprisingly enough, you know,

it looks like a factor that will have a big impact, actually it's a factor that doesn't have such a big impact. So it's, that's why working with GB square over two very often works surprisingly well in practice. But anyway, this would be, this would definitely be the equation to use, again, if you would be involved in research. And of course,

there are added complications, the dislocations are curved, yes? So they're not really purely edge or purely screw dislocations. Our lattice, certainly iron, is not isotropic. So

here we assume by using one value for G that the lattice is isotropic, yes? So that's not the case, yes? And then,

when you have a strengthening situation in the lattice, it may not be the only one. And that's very often the case. You have, for instance, as we've seen for copper, copper will have an impact as a solid solution strengthening and as a precipitate, yes? So how do you account for both of these strengthening effects which occur at the same time? And that's in addition

to the way in the models you approach the geometry of the problem. So that's, having said this, you can look at this, the equation we have for copper, for general

modulus strengthening. So this is the delta tau. And you see that for a constant L, G, and B, so if this is constant, yes? The constant value, it depends only on this, on the ratio of the modulus of precipitate and matrix. And you see that if I have a very low

precipitate modulus, so very soft modulus, a very soft precipitate, I'll actually get a very strong hardening. So that's why copper and iron is a strong hardener. It's actually because it's soft, yes? And I can also, from the same model,

get the critical angle, yes? And so if I have very soft precipitate, that means that at breakaway, the dislocations will look like this, right? This angle here,

two times phi c, is close to zero. Whereas if the particle has a modulus very close to that of the matrix, there's not much strengthening, and the angle is very large at breakaway. Now, if we look at copper and iron, yes? You make the ratio of the copper and the iron modulus,

it's about 0.62, yes? So that means that I have a considerable, I expect a considerable amount of strengthening, yes? And a breakaway angle that should be somewhere around 80 degrees

for that particular, if I use that particular theory, okay? Okay, so first of all, you could check, is the theory basically close to what you get in practice?

How good is the theory? So first of all, you see the theory tells me that the strengthening effect is proportional to the square root of the volume fraction, yes? So, and the volume fraction is something I can easily calculate, yes? I don't need to make an experiment,

I just assume all the copper is precipitated, yes? And so, yes, you can see pretty regular line linear relation. The formula, excuse me, the theory also tells us that

the strengthening should be proportional to 1 over L, the inter-particle spacing, yes? And so, this is the situation here, so here I have a reciprocal of the inter-particle,

inter-precipitates spacing for different amounts of volume fractions, and you see I get a nice linear equation. So, you know, the theories that we have for modulus hardening are pretty good. So, this is the reciprocal of the length, so that means on this side

I have coarse particles, yes? So, the precipitation is, oops, this is right, I need to, you probably don't have this slide, but I see here this is not volume fraction, but this is strength effect here. So, on this side we have large particles,

low strength, on this side we have small particles, small spacing, small particle spacing, and we have much higher strength. And these are typical values, right, 50 nanometers to 200

nanometers. So, it's going to, precipitation hardening really needs really fine microstructures, you know, we're not talking about microns here, yes? We're talking about nanometers, about 100 nanometers in distance between precipitates, that's the kind of distances, and the precipitates themselves. Well, let's have a look at some more data here. Okay,

well, right, so, again, we already discussed this, that we have to take, if you want to take data, you need to take off the solid solution hardening by the copper. We also discussed

how, last Monday, how you make this precipitation by supersaturated, and this is kind of a number of results here, that's where we stopped on Monday. So, you have the aging time, and you can measure, of course you can measure yield strength and tensile strength, etc.

An easier way to get lots of data without having to make too many samples is measuring the hardness, right? And you see at 1% of copper, I have a peak hardening here at about,

at about this time here, 10,000 seconds, yes? If I have more copper, yes, I reach the peak aging at about 100, 200, about, this is 200, 400, yeah, 400 seconds, so a few minutes, I get peak aging

around here. Before this, the material is said to be over, under age, so I have an increase in the strength. After that, the material is over aged, yes? Usually, this transition,

yes, in introductory material science mechanics test, is defined as a transition between

particle cutting and particle bypassing. It doesn't have to be that way. So, classically, when this theory is presented, yes, people say, well, you know, when you reach the maximum of hardening, yes, that is because at this stage,

the dislocations, the particle has become so big, yes, that the dislocation will bypass the particle, and then it goes down, yes? It doesn't have to be this way. In particular, in this case, in the case of a copper, you reach the maximum, and after reaching the maximum,

you're still cutting through the particles, right? And it's only at much higher over aging times that you get what is called the bypassing of the particles by the dislocations.

Right, and you can see here, there's some nice data here, of the kinetics of the coarsening, and so you see here, this is five nanometers, yes? So, this is ten nanometers, yes? So, your, the peak aging is obtained when you have, the particles are fine enough,

right? That's important. Okay, and then, of course, you can always translate your strengthening effects that you measure by hardness to a strengthening effect that you

would measure in a tensile test, and this is the, remember, the formula we saw earlier, so you use the hardness in Vickers in mega Pascal, yes? And divide by three. Don't use the formula that you multiply with three, yes? Because that's another

way of using your hardness data. So, if you do this, you see here our strengthening effect. So, this one is not corrected, the data here is not corrected for solid solution hardening, so I need, I have to take my starting value here, and then look at this distance here,

and that is about one giga Pascal, so a thousand mega Pascal divided by three, is about 333 mega Pascal increase in strength by adding two mass percents of copper. So, it's a considerable amount of strengthening, yes? It's considerable amount of strengthening,

right? So, no wonder that people will use copper to strengthen, it doesn't have negative, it doesn't have negative impact on other mechanical properties. For instance, in terms of toughness, it's, the toughness is not influenced negatively by the addition of copper.

The only problem there is, is when you make the steels. At high temperature, the copper tends to form liquid films along grain boundaries, yes? And so, materials can literally

break when you're trying to process, process them. So, actually steel makers hate copper, and we try to make sure that we don't put copper in any of our steels, yes? Unless,

but the method is used, so you, you know, you do have copper added steels, but usually for sheet products, and other, many other, you know, high volume products, you, you stay, you generally stay away from copper additions.

For precipitation strengthening, because of the steel making. Okay, so now, let's have, let's have a look at some other data here. So, in this case, we do have the strengthening effect in megapascal from a tensile test, yes? So, and I've already

converted the data as a function of precipitation, precipitate radius, but in this direction, you could also have aging time. And what you see, I reach a maximum, yes? And then a decrease, overage, the materials overage, yes? So, what, how can I use this data, yes? So,

you can use this data actually cleverly. You can say, okay, in this case, it's only the strengthening, right? So, we took away all the other strengthening contributions, solid solution, the pyro strength, the effect of the grain size, etc., yeah? So, we have pure precipitate strengthening, and nicely, we find the peaks hardening is indeed around 300

megapascal, like we, we calculated for the hardness, yes? Just about. So, what can we do here? We measure this delta sigma, so it's about 300 megapascal, and you know, I can always

convert tensile data into shear data by using my Taylor factor. So, by measuring this, I can calculate the increase in the shear strength from, due to precipitation hardening.

Okay, now, I know how much copper I added, so I can calculate the volume fraction of copper in my, because copper, remind you, is not really soluble, so most of it will form a precipitate, so I can calculate the volume fraction, and I can measure the radius of the precipitates.

Simply, you know, as I said, you make your samples, you put them in the TM, and you just measure, yes? So, you can calculate the critical angle for, for breakaway, so you can do this, and indeed, this is what you find. If the particles are very small, the angle is 180 degrees,

this angle, two times this angle is 180 degrees, yes. And then, as the particle increases

in size, it goes, it decreases, right? Because the angle, am I saying this right? Yes, so here is a soft particle, and then it increases, the angle decreases as the particle becomes larger. And I can calculate the maximum force also, so as the angle decreases,

the force has to increase, okay? This was, remember, copper iron, so the angle, the critical

should be about 80 degrees, yes? So, let's see, if we reach 80 degrees, no, we never reach 80 degrees, yes? We reach 80 degrees, this is 90 degrees, right? So, we're still above, so that means the dislocations can continue to cut the particles, yes, even though we're beyond the peak aging, yeah? What else can I say? Oh yeah, so what's also

important is where do we get the peak aging, and why do we get the peak aging? Well, you see, apparently, it's related to the structural change in the particle.

You remember, I said when the particles form very small clusters, they're BCC, and then they change into other types of crystallography before becoming FCC? Well, you can see that we reach the peak aging when we have tiny clusters of copper iron particles,

yes, which are body-centered cubic. So, that's actually appears to be the reason why we have this peak aging. All right, so just to say a few more things, so BCC copper

is a modulus effect, yes? So, it's a soft particle, so we get the attractive obstacle. In the TM analysis, people have reported that they don't see loops around the particles,

so when the breakaway angles are, so the phi angle here is very, very small, close to zero,

the bypassing should give me particles with loops around them, yes? And for every dislocation, another loop, yeah? So, that's how you can tell experimentally if you have cutting or bypassing of the precipitate. So, the TM showed that there is no bypassing

if the precipitate is less than 35 microns. And the peak is at five microns, so it's way beyond the maximum, the peak aging that you get, that the bypassing process starts.

So, dislocation bypassing, very low angles here, takes place when the particle radius is larger than 35 millimeters, 35 microns. And there is also no work hardening in the overage condition, it basically means the same thing, you don't accumulate dislocations. Right, and then,

so, yeah, so I need to go back here, the way this diagram is done is a little bit confusing, it's because the, so it should go from, let's see, no, it's correct, it should go from

high to low. Right, so what you basically have is, as the particle becomes coarser, yes, the critical angle is, this is not 2 phi, but phi itself, yes, decreases, yes, and I go from

particle cutting eventually to particle bypassing, but only after 35 nanometers, so long after the

peak strength, okay? Right, so let's quickly overview now the strengthening, other strengthening mechanisms, so we have another mechanism is coherency strengthening. Differences in volume between particle and matrix,

matrix volume that it replaces, will give us elastic stresses acting on the matrix, and so we, one of the reasons why that happened is because if there is a difference in lattice parameter, there's actually a difference in lattice and atomic radii, and we can change this

into, or we can connect this with lattice parameters, yes, and then, and define a parameter delta, which controls this coherency hardening, and that's basically lattice strain, so we,

this delta is the lattice parameter of the particle minus lattice parameter matrix divided by the lattice parameter of the matrix, and so we usually, and then we, for the theory, we need to, of coherency strengthening, we have to use a misfit parameter epsilon, yes,

and if you look up in the literature, you find pretty complex equations for this parameter, so it's equal to delta times, for instance, this parameter here, where g is the shear modulus, and this is the Poisson modulus, and the thing is, if you go through this, you see that

the Poisson modulus is typically about 0.3, so this is, so 1 minus 0.3 is 0.7 times 2 is 1.4, yes, and this is 1 plus the modulus is 1.3, so this is, yes,

usually you can just forget this parameter here, okay, so it's very quickly simple to relate,

so this goes away, and this goes away, so you can relate this epsilon to, I made, I forgot the parameter here, to delta, yes, so usually you can simplify this

1 plus 1, okay, all right, so the maximum interaction force by theory, I don't derive it here, is related to the strength of the mismatch, large mismatch is, increases the

interaction force, and then, if you go through the theory, most of the theories end up with a strengthening effect, which is proportional to this coherency effect to the power three-thirds,

the three-halves, excuse me, three-halves, yes, and the square root of f times the mean particle diameter, yes, okay, and so this is what we know to be the case for

precipitates hardening, and this is what I told you, that in the case of precipitation hardening, we get the maximum interaction force to the power two-thirds, so in practice, for instance, yes, this would be a formula which is derived from this one that you would use

in practice, where the reason why it looks so different, it's not very complicated, it's because we have changed, replaced t, the line tension, with the appropriate formula, but at the end of

the day, in practice, so we don't have epsilon, but we have delta to the power three-halves, and the important factor, square root of volume fraction times particle dimension, and you can, you know, using this theory, also determine what will be the maximum, the peak stress, yes,

and at which radius you'll have this peak stress. If you ever go through analysis of this coherency hardening mechanism, yes, you will see that there are different

theories. None of these theories is more correct than the other one, they just use different approaches to solve the problem, in particular, the geometry, how they average the, because this

is a macroscopic value, yes, it's what you measure in a tensile test, for instance, so you need to average out the effects of a distribution of particles on a distribution of dislocation, so you do get parameters that may differ, for instance, here, in, although

most theories agree with this part of the formula, this factor here, this numerical factor may, you know, may vary, slightly less than two to three, yes, so that will impact the results these formulas give, yeah, but again, the idea is, you can use the formulas to guide

you, yes, to what's important to do, yes, but in practice, you use, you work in the reverse, right, you do measurements, and then you see what is the hardening mechanism, yes, what is

the hardening, what's the operating hardening mechanism, yes, and then you apply these formulas to your data to see which one fits the situation. The chemical hardening effect is basically due to the following, you have dislocation passes the particle, yes, it will

shear the particle, yes, and when it's finished shearing the particle, you can see here and here, I've created new interfaces, new interfaces, and these interfaces have energy, right, so the creation of this interface is what causes the

effect of the, the restraining effect, the obstacle effect of the particle on the dislocations, yes, so well, if this particle has a special ordered structure, this shearing may cause

the formation of an anti-phase boundary, and then we have to take care of that, but in this case, we assume that the particle is not ordered, yes, you basically shear the particle,

and you form these two surfaces, yes, okay, so if you do the theory, the maximum force, interaction force, is proportional to the energy of these interfaces, makes sense, right,

and the chemical hardening effect can be calculated, and the theory gives chemical hardening effect is again the maximum interaction force to the power three halves, yes, so as the maximum interaction force is proportional to this surface energy, you,

that's what you find here, it's proportional to the square root of f, and oddly enough, it's one, it's proportional to one over the particle radius, okay, so, so the theory

predicts that, so if you have fixed f, the strengthening will decrease with precipitate size, and this is not, this not observed in age hardening, so that part of, that the impact of

this process on the hardening is probably minimal, because we never observe this, this effect, that the particle gets larger, and we, the stuff gets softer, yes, so that mechanism is probably

unlikely to be of any importance, for instance, you could have thought that when you shear these copper particles, the creation of these interfaces might cause the strengthening, well,

observe a decrease in the strengthening with the radius of the particle, okay, and another strengthening effect is, it takes into account the fact that in certain alloy systems, we have low stacking fault energies, so we have dissociated dislocations, yes, and

the stacking fault strengthening is in fact due to the difference in the stacking fault energy between precipitate and matrix, okay, maximum interaction force is proportional to this difference, yes, so as a consequence, the strengthening effect is proportional

to the stacking fault energy difference to the power that we have, and as expected for age hardening and, or precipitation hardening, it's also proportional to the volume fraction and the radius of the particle, the square root of that, okay, okay, so,

right, so that's a possible situation that can occur, an important, however, in practice, right,

for precipitation hardened alloys, yes, order hardening often occurs, so that's when dislocations glide and they cut through particles which are ordered particles, you will create a lattice disorder and the formation of anti-phase

boundaries, yes, and there's very important group of FCC metals and alloys which are iron-based or nickel-based or cobalt-based which are based on this precipitation hardening

model, so this is, for instance, the A2 structure, so BCC iron, for instance, right, so if I look along the 1-1-0 direction, yes, and I shear the lattice on a 1-1-2 plane, yes,

I shear it over a Burgers vector, A upon 2, 1-1-1, this is before shear, and this is after shear, and what's to say? Well, nothing, right, after this location has passed, I still

have perfect lattice, yes, now if now I look at the cesium chloride or B2 structure of nickel, of iron-aluminum, for instance, and I shear this lattice, then I do not get the same structure,

of course, I have the same structure here and here, but where the shear happened, it does not look the same, for instance, there are no dark atoms here, large dark atoms that are so close to each other normally in the lattice, yes, so I created what's called an anti-phase boundary,

it's not a stacking fault, yes, it's anti-phase boundary, so, right, so the same situation as what I showed for chemical hardening, the dislocation passes

to the particle, and at the interface, we create an interface boundary, and that has a certain energy, so that would be, for instance, I have glide here on FCC, on this plane, yes, and it means that after the passage of the dislocation through, for instance, a nickel-3-aluminum

particle, yes, on this plane, I create a higher energy anti-phase boundary, it's similar to a stacking fault, but it's not a stacking fault, okay, because in a stacking fault, across the stacking fault, you still have normal stacking,

and the atoms across the stacking faults are, you know, where they're supposed to be, if you look at any 2-1-1-1 planes in the lattice, right, it's just the stacking is straight, it's odd, yes, in this case, you have, at the boundary, there's just the wrong atoms

across the boundary, right, so high energies, much higher energies, anti-phase boundary, much higher energies than stacking faults, high energies, so obviously, the maximum force that we get is proportional to this anti-phase boundary energy, yes, and of course, if I compute

the strengthening effect, I find again, the anti-phase boundary energy to the power 3 halves times our well-known square root f times rp relation, yes,

so, and this is an example here where this theoretical equation is used in practice, but you get the same, basically, the same behavior, yes, so very important here,

the very well-known system is the so-called gamma-gamma prime system in iron-based superalloys, yes, and it's very similar to the nickel-based superalloys, they're very similar, yeah, all right, so let's talk a little bit about these, how we make these,

this type of precipitates in superalloys, first of all, really important here, in gamma FCC structure, we do precipitation hardening with nickel-3 aluminum,

yes, nickel-3 aluminum, that precipitate is formed in the matrix, yes, okay, and you can see it requires nickel, all right,

nickel-3 titanium, yes, which you could form, for instance, by adding to an austenitic steel, yes, titanium is a similar precipitate, but it doesn't form

in the matrix, it forms in grain boundaries, yes, so this is not, you have to have the right precipitate, yeah, it has to form in the matrix, otherwise I don't get precipitation, it's the same thing in BCC alloys, in BCC alloys we don't really use much precipitation hardening in the BCC alloys themselves, but in martensitic steels we do use it, yes,

and here we use FeAl, yes, because it forms in the matrix, yes, there are other, you can form other precipitates, for instance, if you add some nickel and

molybdenum to these alloys, you can form nickel-3 molybdenum, these are in the grain boundaries, now does this mean that we always avoid these particles that are in the grain boundaries? No, because sometimes, for instance, if we have applications at high temperatures,

we want to have particles in the matrix, which give me precipitation hardening, and we want to have particles in the grain boundaries, which will prevent grain growth, remember grain size is also a strengthening mechanism, right, so particles in the grain are not necessarily bad, right, as I said, certainly in high temperature applications.

Okay, so let's look at what, how we work with, because precipitation strengthening is not very common in ferritic steels, yes, we don't use age hardening in ferritic steels,

age hardening is very common in aluminum, yes, because, because that's a very efficient way to harden the microstructure, but in steels, and particularly ferritic steels, we don't use age hardening so much, but in stainless steel is a different story,

so we have, for instance, martensitic steels, and so there's a well-known 17-4 pH is precipitation hardened steel, this is a very common precipitation hardened steel, it has an MS temperature at 132 degrees C, so I can austenitize this steel, which contains

17%, by the way, 17% of chrome, 4% of nickel, yes, and if I cool it down quickly enough, I form a martensitic steel, which I can then, so I get quenched martensite,

which may be soft, generally, because it doesn't have much carbon, and then I reheat it, yes, to 500 to 600 degrees C, and there I can do precipitation hardening of my martensite, yeah? With austenitic steels, it's a bit different, yes, so let's first, before I talk about

semi-austenitics, let's look at the, it's the austenitic precipitation hardened steel, those are very stable austenitic steels, so they're always austenitic, yes, so you have to do

the precipitation in the austenite, so you reheat this material, yes, and then you cool it down to room temperature, nothing happens, right, it's austenite before and after, just to kneel to material, okay, and why does nothing happen? Because the MS temperature is far below room temperature, so I do then aging, precipitation hardening for a long time, because everything is

very slow in austenite, yes, and I precipitate things like nickel 3 aluminum, yes, in that microstructure, but the structure remains fully austenitic, and it gets slow precipitation of,

for instance, nickel 3 aluminum in the austenite, okay, so phase diagram here, which tells you typically, in terms of the processing of these materials, people have done very clever

engineering of the structure of these intermetallic compounds, yes, and what the, for instance, the 25 nickel, 15% chromium super alloys, we use,

gamma, pure alloys, we use, we actually use titanium and aluminum to have a mixture of nickel 3 titanium and nickel 3 aluminum titanium, yes, to get the right properties,

okay, and this is an example here, what you get with these alloys, they're ordered alloys, right, so they have a this L12 structure, yes, and you see the strengthening effect here, precipitation strengthening, so at first, as the particles grow,

I reach the peak strengthening, and after that, yes, when the particles get larger, I get bypassing, and you can see the bypassing here, this is a TM picture, for instance,

look at this particle here, you can see the dislocation loops around the particles, yes, and you can see that these are large particles, right, they're far larger than the five nanometers that give me the peak aging value, okay, this is the same here,

an austenitic super alloy, this is a very common one, a 286, yes, so high nickel, high chrome, yes, and you see here again, peak aging, you need long times for aging,

and then the size of the particles at peak aging is again, as you can see, precipitates radius about five nanometers, so these are very, very tiny particles, yes, let me go back now to this here, we also have semi-austenitic

precipitation hardened stainless steels, these are a little bit more complex, yes, in terms of the way you do the aging, so normally, the MS temperature is below

room temperature, so you can anneal this material and shape it, yes, it's important, you can shape it, and then you can change the location of the MS temperature, yes, by the choice of the annealing temperature, so if you do a high annealing temperature,

yes, you don't precipitate much carbon, when you do that, yes, so annealing at high temperature, the, you know, the MS temperature is a function of the carbon content of austenite, so if I have

a lot of carbon, I have very low MS, yes, okay, so if I precipitate a little bit of carbon, the MS temperature in the case of one will increase a little bit, in the case of two, yes, where I form precipitates, the MS temperature can go up two and beyond room

temperature, so that after heat treating two, yes, I will form a fully martensitic microstructure, in the case of one, if I want to form a fully martensitic microstructure, I have to do cryogenic cooling, yes, cooling below with the cooler, like liquid nitrogen, or

yeah, and then I can do these steels and usually get an actual aging treatment after that to form the precipitates, yes, okay, so it's,

precipitation hardening is complex heat treatments, okay, so these are examples that I showed, right, you can have ferritic H-hardenable steels, this is an example

here for Fe2 titanium silicon, precipitation-hardened steel, some data. Again, the maximum peak aging is always for particles that are very, very tiny in dimensions. And we have some more here for BCC steels and some more.

You can just have a look at it. And I just wanted to get to this slide. That is, what do you do when you have a combination of hardening mechanisms in the same precipitation hardening?

For instance, you have a gamma prime precipitate in austenite. So we know when it cuts this, I form anti-phase boundaries. So anti-phase boundaries effects will impact. But that particle itself also has coherency strengths. So there is also a coherency strengthening effect.

So if you want to have an idea of what's the contribution of one hardening effect and the other one, if they both occur, how do I add them together? Do I, for instance, say I have one mechanism one, which gives me a contribution tau 1, and another mechanism gives me

a contribution tau 2, do I sum the two or what? Well, there are theoretical reasons why the best way of adding the effects is using Pythagoras average.

So you take the square of the first one, hardening contribution, square of the second one, take the square root of those. That gives you a value that's close to what you can expect. And then an example, and that holds for this gamma prime

precipitates mechanism one would be coherency strains, and mechanism two would be order strains, for instance, due to order hardening due to anti-phase boundaries. A little bit over time. So I'll stop here.

I will continue Monday, Tuesday morning, excuse me, about the precipitation of nitrites and carbides. And then we'll start probably already on Monday the last chapter, which is about microstructural hardening,

how you strengthen steels by using multi-phase microstructures. OK? See you on Tuesday morning.