So let's go through our review of some mechanical properties here of steels quickly. It is a lot of material on the E-Class site.

And this is just so we won't go through all this material. So I just wanted to highlight a few essential points.

And maybe some of you will not take the class I teach in the fall on mechanical properties of steels. And in that case, it may be interesting to have this data as background data.

But I do want the number of slides that I do want to discuss because they're important. One of them is related to the fact that, as I stressed during the last lecture,

that whatever you know about or learn about ferrite or ferritic steels does not necessarily apply to austenitics. And one of them is the fact that you can assume that austenitics will, in general,

be softer than ferritic steels when it comes to yielding, but not when it comes to strain hardening. So they strain harden much faster than ferritic steels. So as a consequence, you get more elongation, higher

strengths at rupture. So their formability is usually perceived as being better because of this work hardening. And again, as I said, the key reason for this behavior

is related to the fact that the dislocations in austenite are characterized by planar glide. Now, planar glide, in general, doesn't necessarily have to be due to low stacking fault energies.

But in this particular case, for austenitic stainless steels, which tend to have low stacking fault energies, that's the main reason. The low stacking fault energies forces the dislocations to stay on slip planes, particular slip planes.

They cannot cross slip easily. And that gives you the need to have higher stresses, apply higher stresses, for the dislocations to cut through or pass these obstacles.

And so you get strain hardening. In the case of ferritic steels, which all of them have very high, extremely high, stacking fault energies, you don't have that problem in dislocations. And counter obstacles, they'll just happily move to another slip plane by cross slip.

And so you won't get this strain hardening effect. OK? So difference here between BCC and FCC. OK? And so the question is, is it possible to turn a BCC metal into behavior of FCC metal?

No. But you can engineer the microstructure to improve things. And so in general, that's the reason why people develop multiphase microstructures.

So in this table here, this is just a table that can help you as a reference to compare FCC and BCC slip. Now there's one thing here that I do want to mention. And forget the crystallography here,

but it's the fact that of twinning, yes? It's remember that when we do test, again, ferritic steels at temperatures that are lower

and lower, sub-zero temperature, what we see is a very strong pronounced increase in the flow strength or the yield strength, tensile strength, yes?

And this effect is a function of the strain rate that we apply. So if you have a low strain rate, you're going to have something like this. If you have a very high strain rate, thousands per second, that's a very high strain rate, you

can't do this with a conventional tensile testing machine. So it increases, but the effect remains that there is a strong temperature dependence. And I told you that the cleavage, this flow stress

could increase and go beyond the cleavage stress, and then the material will become brittle, basically. Now very closely associated with this is twinning. So at low temperatures, the ferritic steels

will display twinning. And that is a different deformation mode. And it's believed that the cleavage is very often

preceded by twinning. When two twins interact with each other, they'll create a microscopic crack, which will lead to cleavage of your steel. But the twinning, again, is characteristic of low temperatures deformation or very high strain

rates deformation in ferritic steel. So usually, if you do a normal room temperature or stress strain measurement, you're

never going to see twins. However, if you take a Sharpie tested sample at low temperatures, you will see that in these low temperature samples, you will see twins in BCC. So the observation of twins usually

tells you that the material has been deformed at low temperature or at very high strain rates. Even if you don't know anything about this material and you have to do a failure analysis

and you observe twins in this microstructure, you know that it's either been tested or subjected to high strain rates or deformations at very low temperatures. Twinning is not rare.

So it's relatively rare to see twins in BCC or ferritic steel. But it's not rare to see twinning in FCC. And it's closely related to the fact that the partial dislocations that

form a dissociated dislocation are actually partials that are responsible for the twinning. But again, that's more detailed stuff. I don't want to go into detail. This is just an example. So these are twins in gamma iron.

Or this is actually not in gamma iron. This is stainless steel. And you can see these twins, they form these very straight bands here. These are not deformations. These are recrystallization twins. And so if you get a sample, you don't know what it is.

But you see these grains with all these parallel lines. You know it cannot be a ferritic steel. Ferritic steels also have twins. The twins are very special. They're always extremely thin. They're never recrystallization twins.

They're never recrystallization. They're always deformation twins. So this is an example here of twins, many twins, in a ferritic steel that's been tested at low temperature

during impact loading. And some people will call them Neumann bands, after the person who first observed them. All right. Good. So important to know that dislocations

that we introduce in the microstructure are surrounded by deformation, elastic deformation of the lattice, and also by stresses. And the profile of the stresses

are relatively simple. If I have a pure edge dislocation here, this is edge dislocation in ferrite. You have an area where we have such called compression area, and a tension area, or a dilatation area. In the case of a screw dislocation,

it can be very much more complex, where the compression and the tension areas are changed. So this is in compression, this in dilatation, compression, dilatation, et cetera. So that means that because of this elastic energy,

the strain energy in the lattice, dislocations carry energy. Increase the energy of the crystal. And that also means that the dislocations will interact with each other. They can attract each other.

So when dislocations get close to each other, these dilatation and compression areas can interact with each other. And this location can push each other apart, or attract each other. For instance, even two simple edge dislocations

here, in this geometry. OK. If you plot the ratio of x over y, there will be regions here where you'll have attraction.

So this region here, where this is 45 degrees. So if the dislocation is on this side of this line, you will tend to have attractive interaction. If they're on the other side, for instance, a dislocation is here, you will have repulsion.

And so this interaction is important, for instance, because it gives rise to recovery structures. When you recrystallize a material, before you recrystallize it, the dislocations

will rearrange themselves in low energy configurations, which is basically dislocations under the influence of their own stress and strain field will rearrange themselves to low energy configurations.

So that's when you leave them to their own, let them do what they want to do, and you don't apply any external stress.

However, when you apply stresses, dislocations will move around, and they will interact. And this interaction can be very different depending on the situation. You can't say there is one interaction. But in general, what we say is

that dislocations will act as obstacles to other dislocations, yes? And it doesn't necessarily mean that the dislocations will hang on to each other.

Sometimes it's something different that happens. So for instance, this is an important interaction in BCC iron. That's what we call the formation of sessile jogs.

So the idea is the following. It's that when dislocations cut each other, they can pass each other. But where they pass, they change the structure of the dislocation.

And they can make a little piece of the dislocation can now be prevented from moving. That's what is called a sessile jog. It's a little piece of dislocation that cannot move anymore. So it's basically becoming a pinning point.

It's become a pinning point. And that's the actual obstacle. It's not the other dislocation. It's the fact that this other dislocation made a jog on the dislocation. So this is the idea. Say you have here a dislocation here.

It's an edge dislocation. And here you have another dislocation on another glide plane. Yes? And this dislocation moves, for instance, this way. And it will cut this dislocation. So in this particular case, when it cuts it,

I form a jog here. So you see? Because when a dislocation passes, there is a little shear. So this part of the crystal is sheared. So you form a jog. And it's very tiny. This jog, the magnitude of it is OK.

This is another situation here. This is dislocation. Now it is dislocation on this plane that moves. So you see it's in the other direction. It's in this direction.

And it cuts the same dislocation. Yes? In this case, both dislocations end up with a kink. Yes? A kink. And these are screw jogs. We call these screw jogs. And these are edge jogs. And the edge jogs have as, if I magnify this,

they look like this. All right. And I've got this Burgers vector here, this Burgers vector here. So it's an edge jog type. So now you can, this interaction,

this jog formation. So when you have two screw dislocations that interact, you will form two edge jogs, like these, as here. And the Burgers factor, yes?

So the Burgers factor, in this case, you can see, is a Burgers factor of this plane, but not of this plane. Now the problem is that this little jog lies in this plane. So it cannot glide anymore.

So this jog here, this edge jog, acts as a very strong pinning point, called sessile jog. Now one of the things you'll notice is in order to have this type of jog formation, I have to have dislocations on different glide planes.

Yes? That's why this interaction is called forest dislocation interaction. It's as if a dislocation moves through a forest of dislocation on other glide planes. And forest dislocation interaction is the reason, one of the main reasons,

for strain hardening. It's one of the main strain hardening contributions. When lots of dislocations move, yes, I can have strain, yes?

But because this strain, yes, is in very specific directions, on very specific planes, if I look at a single crystal, yes, and I have a lot of dislocations passing

on glide planes, yes? There will be a displacement, yes, a sideways displacement. Now in general, for instance, if you test a single crystal, the force that you apply will always be determined by the machine, yes?

The machine doesn't go, the machine stays like this. And if this crystal is embedded in a matrix, same thing. If even though your crystal moves this way, the surroundings prevent this from happening, yes?

So this displacement does not occur. So the way the crystal will make up for this is by rotations, yes? And it is this rotation, of course, and this rotation is made possible by other slip

systems, yes? But this rotation results in texture formation, OK, texture formation. So I'm not going to go into this, yes? So this part of the slides, just have a look at them,

yes? It gives you background information on texture, yes? What is important for you to know is that texture can be due to deformation, but also due to transformation.

So because when you transform austenite to ferrite, yes, if your austenite is textured in a certain way, that will carry through to the ferrite, OK?

So there's lots of information about recrystallization and how texture is measured. So just have a look at it. Certainly, if you're not familiar with textures and pole

figures, there are a few very simple pictograms that allow you to understand what, in principle, ODFs are, et cetera. But again, that's not the point of the course here.

So the reason why there is texture when you do transformation, that's because there are orientation relationships between the ferrite and the austenite. And the one that's most important

and that you can use in practice most cases is the so-called KS orientation relationships between ferrite and austenite. And that says that the close-packed planes will be parallel and that the close-packed directions will

be parallel, yes? So that means that if you have a single ferrite, sorry, single austenite grain, single austenite orientation, then you have 24 possible, yes, equivalent ferrite

orientations. So in general, if this turns into ferrite,

and so this would be variant one, variant two, variant three, variant four, variant two, et cetera. So in general, transformation will decrease texture,

because when you go through the transformation, the crystal has 24 choices. So that's one of the reasons why when you do a recrystallization annealing,

you never go into transformation region, because you don't want this effect of randomization. So for instance, you have a ferritic steel. You cold roll it heavily to get the right one-one-one texture, yes?

You do recrystallization annealing, but you don't transform the steel. You want recrystallization and grain growth, so you get the one-one-one oriented grain. But you don't transform the microstructure to gamma, even though it might recrystallize faster,

or give you a softer material. You cannot do this, because you will lose your texture. So in general, transformations lead to randomization of the orientation. And so again, there are some interesting data here

that I will not go into. And there were, yes.

So on Thursday, the texture bit, you don't have to, there won't be any questions on the texture bit. OK? OK? So don't go and learn that segment of this material,

right? OK? So it's optional. The texture, crystallographic texture bit, there's quite a lot of slides. I don't want you to go and study that, yes? It's just for your information, and in particular, for the people who may not take 669.

They have something, if ever they need it, they can look up. And also, with this part of the material, we're just going to focus on the essential bits. And there's a lot of additional information

that you may want to look up. And if you're interested in the theories, et cetera, that will be discussed in the fall in the course I teach. But again, you just need to know the essential bits. And so it's about strengthening.

Strengthening and the common ways we strengthen steel grades, yes? Now strengthening is basically dislocation accumulation,

yes? That's basically what you do, yes? And so the strength mechanisms that work are the ones that you make a lot of dislocations. You make sure they get stuck at obstacles, yes?

And you prevent them from being annihilated, yes? Now that may sound like a very strange concept. Well, how can you annihilate a dislocation? Well, one of the ways you annihilate dislocations is through their interaction with other dislocations.

I'll give you a simple example. For instance, if I have, say this is a dislocation loop, yes? And this is the same dislocation loop, yes? OK? And they have the same Burgers factor, yes? The same Burgers factor here, here.

So basically the same dislocation. They're on the same glide plane, so they're next to each other. And I'm applying stress, yes? And so they become larger, yes? And at one time they will meet each other, yes? They will meet each other here. OK? So what happens here, yes?

What I would like to have in terms of strengthening is, well, that I get as more dislocations as possible. However, in this case, if I were to look at the dislocation core here, I would see that one of the dislocation

is a dislocation that looks like this in the core, and the other one is a dislocation that looks like this. OK? I'm not going to prove this. I'm just going to say this. So you have an extra half plane here, extra half plane here, and they move towards each other.

When they meet, the dislocation is gone. It's just a perfect crystal, yes? And the dislocation is gone. So this piece of dislocation, suddenly, it's gone. So it's dislocation annihilation.

So here, I had four times L, and here four times L in terms of, yes? And when they meet, I lose this part.

OK? So we'll just go into how this is done, this strengthening.

And let's just see, in terms of steels, what are typical mechanisms that we use to strengthen steels, and then give some examples of steels, examples.

So this is an important table, because it shows you how these mechanisms are actually used. So one of the things you can do is solute atoms. That's basically alloying, yes?

And these solute atoms can be immobile. They can work as immobile obstacles, and those are usually for steels or substitutional elements. But they can also be mobile solutes, and those are interstitial elements, yes?

We can strengthen crystals by dislocations. One of the important mechanisms was this jog formation that I explained. So the mechanism is just other dislocations interacting with a dislocation. Grain boundaries act as very effective obstacles to dislocation,

because it interrupts the glide plane at the boundary. And here, the important thing we will work with is the grain size. With precipitation hardening,

I basically introduce particles in the matrix, yes? Very important, particles in the matrix. Particles at grain boundaries don't do anything more than what the grain boundary already did, yes? So very important that if you're doing precipitation hardening,

they should be in the lattice. Now, these particles can be soft particles, yes? That means they can be cut by the dislocations, or they can be hard particles, yes? That work as very strong obstacles to dislocation glide,

and the dislocation cannot pass them. We can harden by the use of multiphase microstructures, so we have to add additional phases, and we can have structure strengthening, yes? And that's where the...

Basically, we use transformations or twinning, mechanical twinning, mechanical transformation, which mechanically induce transformation to get strengthening. You can get martensitic transfer, it can occur in a single phase steel, it can occur in multiphase steels,

or we can do deformation twinning, also in single phase or in multiphase steels. So some examples. The first for solute atoms, we can add phosphorus, silicon, or manganese to steel. Mobile obstacles are used when we make bake hardening steels,

bake hardening steels for automotive applications. Dislocations are used as strengthening mechanism, strain hardening for all the steels, yes? Grain boundaries, of course, occur in all steels, because they're polycrystalline aggregates,

and also in the microalloyed steels, because in the microalloyed steel, we can engineer the grain size so that it's much smaller than usual. Precipitation hardening, soft particles for steels, well, copper.

Copper steels will contain soft copper precipitates that can give you precipitation strengthening, or you can have very hard microalloyed steels, or mar-aging steels, yes, which contain carbide, nitrites,

or some inter-metallics, yes, to harden the steel. Multiphase microstructure are very common nowadays in a certain application. You have the standard ferrite pearlite steels, which are strengthened basically by the presence of cementite

in the pearlite. Dual-phase steels are strengthened by the presence of martensite phase in the microstructure, and you have duplex structure, and one of the most famous ones is duplex stainless steels, where you have a mixed ferrite plus austenite microstructure.

In the unstable, what we call unstable austenitic stainless steel, that means that when you deform them, part of the microstructure turns into martensite. We have strain-induced martensite formation. In trip steel, we also get strain-induced transformation,

but only in the retained austenite phase, and in twip steels, also structure strengthening, I have strain-induced twinning will give you the strengthening mechanism.

So what's the picture? What's the picture of the interaction of dislocation with point-like obstacles in the lattice? Well, it depends on the interaction, how strong the interaction is.

Depending on the situation, you can have a non-local, or what we call diffuse interaction between the obstacles, these dots here are the obstacles, and the dislocation. In this case, the dislocation kind of wiggles its way

between the obstacles. Some obstacles attract, some obstacles repel the dislocation, and the wavelength here of the dislocation

can be very much larger than the distance between the obstacles. And the obstacle can be weak, like in this case, and I have a long wavelength, or the obstacles can be very strong, the interaction can be strong,

so then the dislocation is much more curvy, because it's got a much more stronger influence from each obstacle. So this is for a diffuse situation. Dislocation kind of wiggles its way between the obstacles. In the case of the point-like interaction,

I have a very strong attractive interaction between the point defect, the solute, the obstacle, what have you, and the dislocation. And this is a very localized interaction.

You can see the dislocation is really at the obstacle. And this can be a weak obstacle. That means that the dislocation bends as I increase the stress. The radius here increases, and then when I reach a maximum force

on the dislocation, it just jumps to the next obstacle, basically. And this is a case of a weak obstacle, where this radius stays very large. Or this radius can be very small, much smaller,

of the order of half the spacing here, and then before it will be released. And then you also have situations where the obstacle moves to the dislocation.

So there we have mobile obstacles, and dislocations are immobilized. When does this happen? Well, for instance, during aging, you can have carbon atoms moving to dislocations,

and they pin. But if you take a ferritic steel that contains some carbon in solution, and you do a test at 100 degrees, then the carbon is mobile enough to catch up with the dislocations.

It's mobile enough because when dislocations move, they move, they stop at obstacles, which are always in the lattice, and then they break free of these obstacles. But during the time where they are stuck, we call that waiting time.

They wait. The force has to increase before they get released. So during that waiting time, carbon atoms can hop to the dislocation and pin them. That phenomenon is also aging, but we call it dynamic aging,

and dynamic strain aging, because it's related to dislocations. So obstacles do not necessarily have to be immobile. You can have mobile obstacles and stationary dislocation.

Again, so these strengthening mechanisms are very essential because the yield strength of pure iron, 30 to 40 megapascal. So that's not enough.

So let's look at some fundamentals here. I'm not going to go into too much detail. But basically, if you have a localized obstacle, there's basically a potential around this obstacle.

And so the force on the dislocation is the derivative of this potential. So it goes up and then down. So that means that as the dislocation approaches the defect, it's attracted to it. So here it's attracted to it.

And then once it's at the obstacle, if I want to move it further, then I have to go in the other way. I have to pull it away. So it's stuck at the obstacle. So in other words, you have this force displacement curve

that characterizes these obstacles. So if I have a very strong interaction like this, the red curve means I will have a much stronger obstacle

to overcome because the pinning force is much higher. So the obstacle attracts. And then when you want to move it beyond the obstacle, you have to increase the applied force.

Now, there is a very simple relation, which I will not derive here today, is between the force and the line tension of the dislocation. So the force is related to the line tension

and the sign of this angle here. OK? And so let me skip this. So this is what I want to show you. The way you have to think about it

is this obstacle here exerts a retaining force, F here, on the dislocation. It's like a point force here, yes? And as I apply a stress on my dislocation,

yes, and the force is stress times Burgers vector. You don't need to, yes? I get a force from the dislocation,

excuse me, yes, from the dislocation that counteracts this force of the obstacles, which is due to the effect of these two line tension vectors. So if I make the sum of these two in this direction, yes?

At every time, at every time, I have equilibrium between this vector and this vector here. That's basically what this equation said. But let's now have a look at typical interactions.

And typical interactions that are point-like are with solid solutions, yes? Solid solutions. And so I can have substitutional solid solution, yes, that gives me strengthening.

For instance, silicon and phosphorus, manganese, titanium, niobium, et cetera, yes? So that means that when you add these elements, they will replace iron in the microstructure, in the lattice.

And then I have interstitial alloying elements, such as carbon and nitrogen. And they will sit at octahedral positions. So what happens is that when you add this element, it does two of these elements. Locally, I have a modulus change, yes?

The lattice becomes softer, or the lattice becomes stronger, yes? That's one thing. The other thing that happens is the lattice parameter, yes, doesn't stay equal to the lattice parameter

of pure iron, yes? In the case of silicon and phosphorus, it decreases. In the case of the other elements, it increases, but with amounts that are different from atom to atom.

So these two effects, what we call a modulus and the lattice distortion, yes, are the two main reasons why solutes act as obstacles

for dislocation motion. Now, it's important for you to realize that the lattice distortion, yes, is also specific. For instance, with these elements, silicon, phosphorus, manganese, titanium, et cetera,

I get a dilatational distortion of the lattice. So the lattice is, you can basically think of these atoms as spherical impurities. And they will either expand the lattice locally, yes, or shrink it locally, yes?

Interstitial atoms, yes, are very small atoms, yes? The way they distort the lattice is by sitting in octahedral interstitial locations.

When they do this, they give an asymmetric distortion of the lattice, yes? So the lattice is, for instance, for a case of carbon, is extended in one direction and compressed in the other direction, yes? And so instead of having a spherical dilatation,

I get this elliptical distortion of the lattice. That's why we talk about an elastic dipole, yes? And this, actually, this particular type of distortion

makes it so that these interstitial interact very strongly with dislocations, yes? In fact, much more strongly than most substitutional elements. And that is the reason why carbon and nitrogen

have such a huge solid solution strengthening effect on the lattice, not because they like to go to dislocations. It's because of the particular lattice distortion

that they give, okay? Now, if the atom... Say we have molybdenum in BCC iron, yes? That atom is oversized, yes?

It creates a dilatational, pure dilatational field, yes? So the atom is in compression, yes? And it compresses the lattice around it, yes? So I have a stress, a hydrostatic stress, yes?

So the molybdenum atom is fully in compression, yes? Inside the atom, the stress is constant. And outside, yes? I also have a hydrostatic stress, but it doesn't carry on forever.

It decreases radially, yes? As one over R to the third, yes? So an atom will therefore have a limited range of influence, okay?

Because you have a radially decreasing compressive stress around an atom. So you have to imagine, for instance, this is a large molybdenum atom, which is oversized and gives you this, okay?

Okay, so as I said, when an edge dislocation is put in the lattice, you have stresses around it, yes? That are very different, yes? And the solute atoms will interact with this dislocation

via the lattice distortion and via the change in the modulus that you have here locally.

Now, you can show that from a theoretical point of view, yes, again, when I have atoms in solution, yes,

the strengthening effect from these atoms is usually proportional to the square root of their concentration, yes?

Square root of their concentration, yes. Now, it is when you go into actual data, yes? Well, first of all,

the square root relation is just one of the theories. There are other theories. I'll show you one or two, I think, in a moment. But what you find is that the atoms, when you alloy them, they don't have a very simple behavior as alloying elements.

So for instance, one of the things that you have is the size of these atoms, yes? The interstitials are very small, yes? But as I said, they have a very pronounced hardening effect

because of the way they distort the lattice. Most of the alloying elements that we use for solid solution hardening are within a band of plus or minus 15% away from the lattice parameter of iron, which is here, yeah?

So from 0.21 to about 0.29 nanometers, yes? And the elements niobium and molybdenum are generally the largest atoms, yes, in this. And the elements phosphorus and silicon

are on the other scale, yes? So most of these elements here give me lattice expansion, yes? And the silicon and the phosphorus give it a local lattice contraction, okay? And in general, you know,

the misfit that you get is larger, the larger the atom is, as you expect. And you can see here that the case of phosphorus, it's negative. Silicon, it's negative. And for manganese, it's positive.

Okay? Right. So you can determine a misfit parameter, yes, on the basis of the size of the atoms or more accurately on the basis of the measurements of lattice parameters. So you can add enough silicon

or enough molybdenum to steel and measure the lattice parameter. You'll see, in one case, the lattice parameter will be reduced. In the other one, it'll increase. And that allows you to determine this misfit parameter, yes? And you see, it's a very small parameter, right?

So you have to think of this delta as a strain, yes? So we're talking about very small amounts of strain, okay? In the lattice parameter, okay?

Now, we can also show that if you know this parameter, yes, and a number of simplifications hold, and surprisingly, it's very often the case

that you can apply them. If you know this misfit parameter, the strength of the interaction, yes, will be proportional to this misfit, yeah? And it's multiplied by g.

Now, delta is not very large, 10 to the minus 4, but g is very high, yes? So together, this product comes out to be quite impressive, yes? So what we have here is, and again, I really glossed over the theory here,

is that the solid solution hardening is proportional to this misfit, yes, and is proportional to the square root of the density or concentration of my solute.

That's basically what theories give, yeah? So this delta just is basically, if you want to correct theoretical,

the correct way of defining it is 1 over a dA dC i, where A is the lattice parameter of ferrite, for instance, and dC i is the concentration of an alloying element, yes? And so we have a size effect, and as I said, solutes also have a modulus effect,

so you can measure the modulus and the change of the modulus with the concentration of your alloying element, yes? And these two parameters are the most important ones. But in general, the size effect will be the most dominant, yes?

So that this formula here is actually quite a handy way to do a back-of-the-envelope calculation of what the hardening will be, solid solution hardening will be for a certain element in steel.

Oh, and here, by the way, is an alternative formula for the strengthening effect from solute. And you can see I don't have one half, and here I have also not one, but four thirds,

and that's a little bit of a problem with the theory of solute solution strengthening for steel, is that the theories are not advanced enough at this stage for us to really be able to use them

in practice very easily, and they're different theories, yes? And we don't know which one applies to which element, etc. So what we can say is, well, there's a concentration effect,

and there is an effect due to the lattice distortion. The more lattice distortion, the higher the strengthening. The higher my concentration, the higher the strengthening. Now, so if you're an engineer,

you do need to know, you know, what is the impact of the strengthening. So what people do in general, yes, is you just say, well, you know, if I don't have a good theory, I'll just assume I have a linear relation,

and I'll just use experimental data to make the relation between the concentration and the solute solution, solute solution strengthening for a specific element. So you just say strengthening from solid solution in a tensile test is equal to the concentration

of a certain element times the solid solution effect per weight percent, and you just sum this for all the elements in your steel composition. So for instance, if that first element is phosphorous, you just say...

then your solution effect is 680 megapascals per mass percent. And for silicon, 83 megapascal per mass percent. And for manganese, 32 megapascal per mass percent. So that's at first view seems to be

a very simple way, and convenient way, and quick way to do things. There is a problem, however, is that many people do the experiments, and they don't come out the same data.

So everybody agrees that this is a very high strengthening, and that this is number two, and this is number three. But the variations can be very important. And again, the problem is that a lot of these solid solution

strengthening measurements are not done in a scientific way. So people don't make nice single crystals, and don't make pure, very pure binary alloys. They just use steels. And of course, there will be some carbon, there will be some oxygen, all kinds of elements

that may influence the outcome of this experiment. And so it's a challenge, also, of how you can do this in practice. Just for your information, these are very good values

of solid solution strengthening for substitution. And this is a huge value, a very, very large value. So obviously, that's one of the reasons why you will see use of phosphorus.

That's because it's got a very large strengthening effect at low concentrations. That's also why often silicon and manganese are added to, for instance, constructional steels, because they give you a lot of strengthening for a low amount

of alloying addition. However, can you add 1% of phosphorus to steel? No, that would be a very bad idea, because it may become very strong, but other properties such as toughness, et cetera, will just go down the drain.

So you cannot do this. Yes. At best, maybe 700, 500 PPM, 800 PPM, that's about the most you can add to be on the safe side for terms of toughness.

Same with silicon. Silicon, it's much less than phosphorus, but it's still close to 100 megapascal per weight percent that you add. So here again, if you add one mass percent of silicon,

you will see a very bad impact on toughness. And so the most you can add for strengthening purposes is about half a percent. Half a percent is safe, yes, and does not have any negative impact on toughness.

Manganese is ideal, because if you add it, you get strengthening. And it doesn't have any negative effect on toughness.

And it doesn't have any negative effect on toughness. Actually, it improves the toughness. Important thing, again, that I want to say, OK, this is for ferrite, yes. For austenitic alloys and steel,

the situation is even worse, with very little data available. And these strengthening effects do not hold for austenites. For instance, for austenitic steels, this is close to zero, or even negative.

Yes? So this is not a strengthener in austenite. In austenitic steel. Silicon, however, is a very strong strengthener in austenitic steel also. In general, the solute solution

strengthening of alloying elements in austenite is much lower than in ferrite. You don't have these large effects. This is an important figure here,

because you can see what the effect is in practice for the difference. So you have alloy concentration.

And here you have the yield strength. And then you have yield strength and tensile strength. And you can see that silicon, manganese, nickel, titanium are in the same range, yes, are strong strengtheners.

Titanium is also a nice strengthener. We don't use it, because it's kind of an expensive alloying element. And cost considerations are always very important for ferritic steels.

Of course, if you have data like this, you could say, well, it's very simple to check theories, because if you have a theory that tells you the strengthening from solid solution

is proportional to the square root of C, then if you do a natural log plot of this, then this data versus this data

should have a slope of a half. And there you see directly, so this is actually done here, this is slope one, this is slope 4 third,

so half is somewhere in between. You can see that basically you have all kinds of slopes. And it's actually very difficult to decide what theories apply, because within the ranges where

we can do the alloying, there's very little difference between a slope of 2 thirds, 3 halves, 1 half. And that's a big challenge, is that the data that we need to check theories for steel when it comes to steels

is not there yet. So that's one of the challenges, because this is some data here. This is an example here. We'll go through, for instance, yield strength of a martensitic structure, you

see as a function of the nickel content. If you use the data that's directly quenched data, there is a lot of substructure effect from the martensite.

If you recrystallize the martensite, you see a nice linear line as a nickel to the 1 half. This is solid solution strengthening of BCC iron by carbon. In this particular case, you get a nice square root of carbon dependency.

Yes? Same here for in FCC, this is for FCC alloy. This is a data for nitrogen, and this is for a BCC alloy

for carbon. You can see that you more or less have a linear relation. Yes? For nitrogen and carbon, yes?

And this is an example here of solid solution strengthening effect. For instance, phosphorus, I have values that give you 12,050 megapascal per percent, and the lowest is around 500. And you see the same thing here with silicon,

very high values, extremely high values, and then much lower values, et cetera. So that's very common to see this very wide thing. So again, I'll discuss this in more detail in the course in the fall, but the best thing to do

is to leave out the values for which the research was not solid enough in terms of information on the steel, knowledge about steel composition,

and then take the median value, yes? Median value of not the mean value, but the median value in that. That gives a pretty good approach. There are people today who are trying to work things out

from first principles, using molecular dynamics, things like this. There are lots of physics is involved, very fundamental physics that's not solved yet,

so we cannot, at this stage, rely on this kind of research to build, to design alloys. But I'm pretty confident one day, it'll be possible.

So this is a tensile strength increase. Solid solution hardening, just to finish here, to wrap up, for the interstitials is very large. So you have values that go into the 5,000.

Nitrogen here, 5,500. Boron, very, very high. And so the obvious thing to do is to alloy with these elements.

The trouble is, what's the problem with carbon in ferrite? Well, you cannot dissolve it in ferrite. It has no, at room temperature, it basically has no solubility. So the first thing it tries to do is to go to dislocations, or it goes to grain boundaries, or it forms carbides.

So yes, it's a fabulous effect, but you can't really use it very often. Somewhat lesser problems with the use of nitrogen, yes, but it's very difficult to alloy nitrogen, yes.

So that's a challenge. And there are a number of issues with boron that I'm not going to go into, the main one being related to the hardenability. But again, I want to stress the fact

that the solid solution hardening from interstitials is very, very pronounced. OK, we'll continue this on Thursday. Thank you for your attention.