Physical Metallurgy of Steels - Part 10
Bhadeshia, Harry
A series of 12 lectures on the physical metallurgy of steels by Professor H. K. D. H. Bhadeshia. Part 10 deals with time-temperature-transformation (TTT) diagrams, overall transformation kinetics, and then moves on to introduce TRIP (transformation induced plasticity) steels.
 So, just in case any of you missed that is a homework assignment which you should hand in tomorrow. It should take you no more than half an hour to do. Just look at how we derived the growth rate of ferrite from austenite and follow the procedure. Draw a composition profile at the interface and then balance two terms which is the rate at which carbon is partitioned or absorbed against the rate at which it is taken away by diffusion. Okay. Okay. So, today I'm going to cover two things, time to transmission diagrams and I'm going to make a start on the transformation induced plasticity skills. But first, time to temperature transformation diagrams where you saw the TTT diagram or typical TTT diagram will have two C curves, one for the reconstructive transformations and the other one for the displacive transformations. And diagrams like these are used widely in industry. Okay. But what actually determines where those curves fall on the temperature versus time plot? That's what we are after. How do we actually calculate such a curve? We've done with growth. I'm going to briefly introduce you to nucleation because it's the combination of nucleation and growth which gives us a time temperature transformation diagram. Now most of you will have done nucleation theory but I'm going to go through it again briefly with you. Nucleation is as follows. Supposing I'm observing my austenite, okay, and I've got as much resolution as I need and I'm observing it at temperature. And you will suddenly see that in some regions, a new structure and chemical composition will be created by random fluctuation, all right? So the system is homogeneous but suddenly you will see a fluctuation of structure and chemical composition which is consistent with ferrite. Okay. So it's composition of that region and the crystal structure is consistent with ferrite. So it's a random fluctuation atoms are moving about and you will just find a region which has the structure and composition of ferrite. But of course if you've created ferrite in austenite you will also have an interface between the two phases, right? And that interface is badly fitting so there's a certain energy, it's a defect. And that fluctuation will not be able to grow until it gets to a size where the cost of creating interface is less than the cost of creating, than the gain from the creation of ferrite. So that's basically nucleation theory that by random fluctuation we get regions which have the right structure and right composition for the product phase and if that fluctuation is large enough then it will grow into a phase. So we can express that very simply, let's imagine that we've got a spherical nucleus forming in our austenite. So this is the free energy curve for austenite and free energy curve for ferrite. This is the equilibrium temperature and below that you can see that the free energy of ferrite is smaller than that of austenite and that's where we might be able to see this fluctuation. So in the creation of this sphere I will get a reduction in free energy so this term is negative, delta G chemical is negative because you can see that ferrite has a lower free energy than austenite below the equilibrium temperature and this is simply the 4 upon 3 pi cubed is simply the volume of the nucleus. But when we create a spherical nucleus we also create a surface and the area of the surface is 4 pi r squared and the cost of creating that interface is sigma which is in joules per meter squared. It's the interfacial energy per unit area. So that is positive, this is negative and we might also have some strain because the volume change or shape change or whatever. The strain energy scales with the volume of your material. So this is chemical free energy change once the transformation to happen, strain opposes it and the creation of interface opposes it. So that's the balance, that's the free energy change delta G when we create a nucleus of size r, radius r. If I plot that expression, so if I plot on the vertical axis the free energy so delta G and it's 0 here, negative and positive and along here is the radius r then the chemical free energy change will tend to promote the transformation and it will be something like this. So this is 4 upon 3 pi r cubed into delta G chemical that tends to favor and it varies with r cubed and r cubed will be smaller than r squared for very small values of r. So the surface energy term will be denoted at small radii. So this is the 4 pi r squared into sigma interfacial energy term and the strain energy term will follow a curve which varies with r cubed. I'm not going to plot that on here just for the sake of simplicity. Now the net effect of these two terms is to produce a curve which looks like this. So it isn't until the radius becomes greater than r star that the free energy change delta G actually decreases and this height here is called the activation energy G star. So G star is the activation energy of nucleation. So simply because you are below the equilibrium transformation temperature doesn't mean that the transformation happens instantaneously. You've actually got to overcome the cost of the interface. So you need fluctuations which are greater than r star for them to develop into successful nuclei. Now in order to find G star and r star, sorry that r star is incorrect. Rub that out. Why doesn't it rub out? Let me just, r star is actually here at this point. Now in order to find the value of G star and r star, this is the first equation is simply the free energy change when you form a particle of radius r. If I differentiate that with respect to r and set this to 0, then I get r star varies with the interfacial energy. So if the interfacial energy is 0, then any fluctuation will grow into our product phase but it never is 0 in practice. And of course if I have a larger driving force, then the critical size of fluctuation which is successful will be smaller. And if I now substitute r star into the first equation, then I can get the value of G star which varies with the cube of the interfacial energy. So it's very sensitive to the interfacial energy. Larger interfacial energy will make it much more difficult for nucleation to happen and it varies with the square of the chemical driving force. So if I increase my driving force, then the activation barrier becomes smaller. Now this is just a summary of the variation of interfacial energy and the chemical free energy change plus strain energy as a function of the particle radius and this is the critical size r star and here we have the activation energy. We can now say the probability of a successful event is given by exponential minus G star upon KT, the usual thermally activated Arrhenius type equation. And nu over there is the jump frequency, is the attempt frequency. So it's attempting to get over the barrier. Not all of those attempts are successful. The probability of a successful attempt is that exponential minus G star upon KT and is simply the number density of nucleation sites per unit volume. And we have a second exponential term here which is exponential minus Q upon KT. And that is the activation energy for the transfer of atoms across the interface because you know you've got to go from this crystal structure of austenite to that of ferrite when you transfer an atom. That has a certain barrier associated with it and that's a constant value Q. G star will vary with your transformation conditions. If you transform at a larger under cooling, G star will be smaller because the driving force is greater. So this is the equation for nucleation rate per unit volume. So now we have the growth rate. Yeah, we've done many growth rate calculations. And we have the nucleation rate. So we should be able to calculate the volume fraction as a function of time, temperature, composition, etc. Right? We can calculate the volume fraction that gives us the TTT diagram. So let's just see the problems that arise when we want to calculate volume fraction. So imagine that I have two particles here. So here we have two particles which have grown inside the austenite. And that is at a certain time T. And I look at those particles a small time interval later, then they will have become bigger and also I might have nucleated new particles. So these two particles here are new particles. So is there anything wrong with that diagram on your right hand side? What do you think is wrong with that diagram? Yeah, I've got this particle and this particle are new particles. These two have grown. What is wrong with that diagram? Go ahead, be brave. I think you know the answer. Yeah, it doesn't make sense to have a bi-nucleating inside ferrite which has already grown, right? And similarly we've got this problem of overlap. So once we start to look at volume fraction, we've got to think about particles touching each other. And how can we take account of particles touching each other? And this is a good illustration. This is a glass crystallizing, like window glass. If you heat it long enough, then it will crystallize. And these are particles of crystalline glass growing. And you can see there's a problem here, they touch. They don't start growing through each other, but you're stopping them from growing. And eventually when all of this is filled, it will look like an equiaxed grain structure, you know, the sort of thing that you see when you look in your microscopes. So the problem is that we need to take account of what's known as hard impingement. That means growing particles physically touching each other. And we cannot allow nucleation to happen in regions which have already transformed. Okay, so we'll just write that down. So hard impingement. Impingement means hitting each other, particles nucleated from different positions, positions touch. So let me draw that diagram again. I've got a region here of the pair phase. Let's say this is the time t equals tau, and this is at a time t equals tau plus delta tau. And I've got two particles here. A short time later, they have grown, and I might have nucleated two new particles. Now let's ignore at first the fact that some particles are growing in other particles or that particles have touched. If I calculate the change in the fraction of ferrite in going from tau to tau plus d tau, it will be wrong. But let's do that anyway. Let's call that the change in extended volume. That means we allow particles to grow through each other. So the change in extended volume, change in extended volume, dp, and we'll call our transformation product beta and e for extended. So this is the beta phase growing. So this will be the wrong volume we have calculated because obviously you can't transform regions which have already transformed. And the real change in the volume of beta we'll write as dv beta. So this is the true change. And obviously they're not going to be equal. But if I multiply the right hand side here by the probability of finding untransformed material, then that corrects for the extended volume because what we want is the change in volume which falls into untransformed material. And what is the probability of finding untransformed material? So this will not equal this. If I multiply the extended volume by the probability of finding untransformed material, the probability of finding untransformed material is equal to what? What's the probability of finding untransformed material when there is no transformation? What's the probability of finding untransformed material when half of it is transformed? Half, right? So it's simply given by 1 minus the volume fraction of beta. So that's 1 minus volume of beta divided by the total volume. Because v beta over v is the volume fraction of beta. Everyone happy with that? So we can write that the real change in volume is equal to the change in extended volume multiplied by 1 minus v beta upon v. So this is a really important equation. It allows you to look at many particles growing together and calculating a volume fraction. And this is called the Avrami equation. I'm going to write that out again on the next page, but rearranging to get the extended volume on the left hand side. So we have dv beta extended is equal to dv beta divided by 1 minus v beta upon v. Just rearrange the previous equation so that I have dv beta on the left hand side. And now I can integrate that so that I get the extended volume is equal to, now, what's the integral of dx over x? Yep, louder, louder. Yeah, logarithm. So we will have logarithm, leave some space after equals, yeah? Logarithm of 1 minus v beta upon v. And we have to put a minus v there because we've got 1 upon minus v, right? So this is the relationship between the true volume of beta and the extended volume of beta. It's a very, very simple equation, but extremely powerful. This is possibly the most useful kinetic equation that I know, far better than phase field theory or any fancy methods for calculating volume fractions. So let's assume that we now have completed the theory for impingement of particles. Now let's do a calculation. If we are observing a system transforming, and if I plot the size of a particle, size of particle versus the time t, you will see a particle nucleated. And let's assume that the growth rate is constant, then this particle which nucleated at this point, which we'll call tau 1. And then there might be another particle which nucleates at a different time, tau 2, and then another one at tau 3, and so on. This is what you would actually observe. Particles have started at different times. Before the time period tau 1, there was no particle 1. So the size is 0 at that point, and then it starts to grow, and we are assuming a constant growth rate. This is a constant growth rate, which we will call G. Then the volume of a particle nucleated at a particular point will be, let's say, a small v with a subscript tau, and we'll assume that the growth is isotropic. That means the same in all directions, and that will be 4 upon 3 pi G cubed, the growth rate cubed, times t minus tau. Because before the point tau, there is no particle. It only increases in size beyond the point tau here. Before that, it doesn't exist, and cubed. That's an individual particle, but of course we will have many particles nucleating at different times. So in the small time interval, the number of particles that will form is the nucleation rate times the volume, because it's a nucleation rate per unit volume, times the volume, times the time interval. So the number of particles nucleated in interval d tau is equal to the nucleation rate times the volume, because it's the nucleation rate per unit volume times the volume, multiplied by the number density of nucleation sites into d tau. So n is the number density of nucleation sites. You happy with that? I'll move on to the next page. So the change in extended volume then is equal to the nucleation rate per unit volume times the volume times the number density of nucleation sites multiplied by the volume of each particle, which is given by 4 upon 3 pi g cubed t minus tau cubed times the time interval d tau. There's a number of particles that have been created in that time interval times the volume of each of those particles. What all I have to do now is integrate this from time equals 0 to any particular value of time. So if I integrate this from t equals 0 to t equals a particular time tau, then I get the extended volume of beta is equal to i Bn into one third pi g cubed into t to the power of 4. Show that for yourself later on, but the 4 disappears because we've integrated to get a 4 in the time exponent. No, it should be d tau because we are integrating tau from t equals 0 to a value of tau. So tau will disappear in this. t is the real time. Tau is the point at which a particular particle comes into existence. So that's the extended volume. And we already have a relationship between the extended volume and the real volume. If I go back, here is the extended volume as a function of the real volume. So we can simply substitute for the extended volume. So instead of extended volume, I write minus V log of 1 minus V beta upon V is equal to 2. And if I unlock this, then I will get 1 minus the volume fraction of beta is equal to exponential minus i Bn upon 3. Sorry, there is no V there into oops. And that defines the volume fraction as a function of time. The driving force comes into the growth equation, into the nucleation equation, and the number density of nucleation sites. So if nucleation happens at grain boundaries, then the grain size determines the number density of nucleation sites. So everything is in that equation. And we've taken account of hard impingement. If I plot that equation out, then this will be the typical shape. So I've got time here and V beta upon V is the volume fraction. And it's what's known as a sigmoidal curve going from 0 to 1. So that's how the volume fraction evolves with time. And the incubation period tore for a particle to come into existence is not fixed. It depends on at what point that particle came into existence. So when people talk about time-temperature transformation diagrams, they make the mistake of saying that the incubation period is between the C curve and the vertical axis. That is not an incubation period. That is the time required to achieve a detectable amount of transformation. So it depends on your technique. If you have a dilatometer which has a 1% volume fraction measurement accuracy, and I take 1% here, then you start to plot your curve at the time corresponding to 1%. I can plot another curve which corresponds to 5%, 10%, and so on. And that will generate the whole of the TTT diagram by calculation. Now here in this example, I took the growth rate to be constant. So can you tell me which phase transmission in steel gives you a constant growth rate? So one which doesn't involve diffusion, alternatively, where the composition of the product phase is the same as that of the parent. So can you tell me of reconstructive transformation where the composition of the parent is the same as the product? Perlite. It's a mixture of cementite and ferrite with the same average composition as the austenite. So the perlite colony will grow at a constant rate. So this would be a model for perlite. And if I obtain an experimental curve of the volume fraction versus time, and I get an exponent to time of 4, then that indicates that the growth rate is constant. It doesn't prove it because there might be another mechanism which also leads to an exponent of 4. And the exponent of 4, 3 comes from the growth, because we have growth rate cubed, isotropic growth, and 1 from the constant nucleation rate that we have assumed. So supposing now that we have parabolic growth, that means that the size varies with T to the power of half, then what do I expect the exponent to be? So parabolic growth rate and a constant nucleation rate. What would the time exponent be? Sorry? Almost there. So we've got size varying with T to the half. So what will volume vary with? Volume will vary with the cube of the size. So 3 upon 2, right? T to the half cubed, 3 upon 2. And then 1 will come from the fact that we have nucleation times time. So the exponent that you will get is 5 upon 2. Yeah? Is everyone happy about that? So parabolic growth will lead to a different time exponent. So again, if you measure experimentally and you see that the time exponent is 5 upon 2, that indicates that the growth rate is parabolic with time and the nucleation rate is constant. So it's an indication, but it may be ambiguous because another mechanism might also lead to T to the 5 upon 2. If you look in Christian's theory for transformations in metals and alloys, you'll find a table of different combinations of nucleation and growth functions which give you different values of the time exponent. So if you analyze your deletometer curves more carefully, you'll be able to study also the mechanism of your transformation product using this analysis. Okay? I'm just going to repeat all that using my slides. Yeah? So this is the, what happens after a small time interval. These particles have grown. So the dark blue regions there represents the increment of transformation. And we've also nucleated another two particles. But the calculation of the change in volume fraction is going to be wrong because we've got overlap of phases. This particle here should not form an area which has already transformed. So we correct for that by multiplying the incorrect change in volume fraction by the probability of finding untransformed material. And that really is the essence of Avrami theory that tells you how to convert extended volume into real volume. So you don't need to worry about impingement between particles. The extended volume is very much easier to calculate because you just calculate the growth rate and the nucleation rate, add it all up, and then make a correction here. Okay? And of course, this is just an integration of that equation. This is showing you, observing the system, you would see a variety of particle sizes growing from the matrix because they start at different values of the incubation period tour. And the volume of a single particle, assuming a constant growth rate, will be given by G cubed into T minus tour cubed times 4 upon 3 pi because we are assuming isotropic growth. You don't need to assume isotropic growth. You can have G1, G2, G3 as growth rates being different in different directions. It's extremely flexible theory. And then we arrive at our final Avrami equation. And it contains everything. It contains the growth rate, nucleation rate, time, temperature, everything. Now one bad thing that you will find in the literature is that they express the Avrami equation as the second equation, which is just saying, look, there's a constant care and there's a time exponent n. So they use it completely empirically. And they don't realize that they are using it empirically. They think that is the Avrami theory. So I don't know if you remember, but there was a talk given, a seminar given, where the person was comparing a phase field model with overall transformation kinetics and then said, look, oh, this equation is completely empirical. But what that person didn't realize is that he is using it empirically. The equation itself is perfectly OK if you put in the right terms. It's a very, very powerful equation, much more powerful than phase field theory. Right. So these are just typical curves. And to find the time exponent, you take a double logarithm of the volume fraction. It's fairly straightforward mathematics. And then you plot your time, temperature, transmission diagram. What I'd like you to tell me now is this was one phase growing from the austenite. Right. Just one phase. What happens if I have more than one phase? So supposing that I have ferrite growing from austenite and in another region, I also have cementite growing because our carbon concentration is very high. So independently growing. How can I modify the Arami equation to take account of more than one phase growing? It's very, very important. Because when we do tempering or create power plant steels, we get many different kinds of carbides forming, not just one at a time. And for more than 60 years after the Arami equation, there was no theory for this. So I want you to create that theory right now. How can I modify this equation to deal with more than one phase at a time? So just think in that diagram on the right, in addition to the blue particles, I also have some red particles. What would I have to do? Well, think about how this probability term changes. What does that represent? That just represents the amount of untransformed material, right? So if I have another phase forming, how do I modify that probability term? Yeah? Yeah? Yeah. Perfect. Yeah. So if I have a phase theta forming, then I have V alpha plus V theta at the top in the probability term. But of course, we only have alpha and alpha on both sides. So I need a second equation, which says dV theta equals 1 minus V alpha plus V theta over V into dV theta. And then I solve those equations simultaneously. If I have six phases, I have six equations. It makes a remarkable difference to the accuracy of your predictions if you do it properly. So this was solved just a few years ago when we was to model precipitation kinetics in power plant steels, where you get many different kinds of particles forming like cementite, molybdenum carbide, M23C6, and so on. So the way you would modify that equation for multiple phases is very simple. So we have dV alpha is equal to 1 minus V alpha plus V theta over the total volume into dV alpha extended. And we need a second equation, which is dV theta is equal to 1 minus V alpha plus V theta upon the total volume times dV theta extended. And you need to solve these equations simultaneously, which is not a problem. What is more, you know, you can do this numerically. You do it step by step. Or you can do it analytically if you know the dependence of theta and alpha. But numerically is the best procedure because supposing that the composition of the matrix changes while the phases are growing, then you can alter the composition of the matrix in each step, and that will influence the growth and nucleation rates. So that is called soft impingement, where particles alter the composition of the matrix and therefore they will grow or nucleate at a different rate. So it's the overlap of diffusion fields. They haven't touched, but the overlap of diffusion fields, which is called soft impingement, as opposed to actually physically touching. So this is the most powerful theory to deal with any number of reactions happening at the same time. And last year in GIFT, we applied this, Unju Song, applied this to oxidation. So we have a steel containing silicon and manganese and so on. And even though silicon is an extremely strong oxidizing agent, the combination of silicon with oxygen produces a much bigger free energy change than iron with oxygen, the iron oxide forms first. And the reason is that there is a competition. There is much more iron than silicon, and also the silicon has to diffuse longer distances. And equations like these automatically predict that. So you can download this from our website and you can see exactly this theory being applied for oxidation. It doesn't necessarily have to be precipitation and so forth. So time-temperature transformation diagrams are extremely useful. And if you use Avrami theory properly, that means not empirically, then it is the perfect theory for modeling phase transformation kinetics. And of course, once you have a time-temperature transformation diagram, you can easily convert it into a continuous cooling transformation diagram because a continuous cooling curve is simply a set of isothermal steps. So you can add them all up. Now, I want to start, make a start today on trip steels. And I'm not going to teach you things which you probably already know, that it's a wonderful steel which you use in cars to save the world, to reduce the weight and so on. But actually, you're wrong. The weight of cars has actually increased over the last 10 years. It hasn't decreased. In spite of all your research. So why is that? Why is the weight of a car increased? The average weight of a car in Europe has increased, not decreased. Why is that? Safety. Safety. That's the thing. So you see this bar here. This is a picture which I have taken. This bar is there so that if you get a side impact, then the deflection is limited so that you are safe. Your car will be a total write-off because this spreads the deformation across the body of the car. Yeah? So you will be safe, but your car will no longer be usable or repairable. So the additional safety features have meant that the weight of a car has actually increased, not decreased. So all the thesis that you see in GIFT which start off by saying, oh, we've got to save fuel and save the world and so on, this isn't going to save the world. It's simply improving your safety at this stage. But of course you can also reduce the weight. So this is another piece of steel in a car where you are joining up two pieces of steel using a laser, laser weld. And then you make the component from a composite steel. So this is called a tailored blank. This is the blank. That means it hasn't been formed. But you don't require the same properties in every region of your component. So you choose the steel which has the right strength, right ductility, cost, et cetera, and form the component out of a composite. So this is now actually very common in the manufacture of cars. And it does lead to a reduction in weight because you don't need the same thickness even in all regions of your car. So here, for example, this part here was made from that tailored blank which I showed you in the previous slide. This is a mini, but actually it's a BMW. It's made by BMW. So we are not succeeding actually in reducing the weight of a car in terms of what the customer buys and in terms of safety regulations. If we didn't have trip steels, the problem would be much worse. That's the main point to write in your thesis introductions. We justify your work. OK. Now, I'm going to start by distinguishing between a trip steel and a trip-assisted steel. Do you know the difference? What's the difference between a trip steel and a trip-assisted steel? Exactly right. So the original work on trip steels was by Zaké and his group. And it involved 100% austenite, a steel which is 100% austenitic. So obviously to get a steel which is 100% austenitic, you have a large concentration of alloying elements. So for example, this is an example of a very, very expensive material in the context of steels because we have about 30%, 8% of nickel. I don't think Posco will be very happy if I suggest that we make such a steel, right? But if you think about carbon nanotubes and so on, they are far, far more expensive and they are not worried about the cost. So we have to actually charge much more for our steels. They are much better than carbon nanotubes and yet we charge a lot less. OK. So let's take this steel. First, I'm only going to talk about trip steels. Trip-assisted means you have a small volume fraction of austenite and other phases present. So you're getting an assistance from the trip, but it's not a trip steel. But this has a martensite start temperature of minus 80 degrees centigrade and you can see that the transformation produces a shape change. So this is the surface relief produced by the formation of martensite in this material and that is the deformation which is the transformation plasticity. Transformation plasticity means plastic strain caused by a phase change. So there's a crystal structure change, but it's still a physical deformation. Now what I want to do is just to introduce you to how to calculate the strain along any particular direction. OK. So we've got a transformation happening. We know the nature of the shape change. An invariant plane strain with a shear parallel to the habit plane and the volume change normal to the habit plane. And the value of the shear strain is of the order of 0.26 whereas the volume change delta is about 0.03, 3%. So imagine that this blue region here is a square of austenite of dimension 1. And I define my coordinates here as z1 along the horizontal axis. Z3 is normal to the horizontal plane and z2 is poking out of the plane of the board. And this set of axes is called orthonormal because I've said that the austenite has dimensions 1 and these axes are all at 90 degrees to each other. So we call this an orthonormal set of axes. So what I want to do is I want to define a deformation matrix so that if I multiply it matrix by any direction I will get the change in that vector due to this deformation. Okay so we've got an orthonormal set of coordinates. Set 1, set 2 and set 3. So set 1 is along here and set 3 along there. So these are unit vectors and this is my austenite. And when it transforms, change is shape where this is the shear strain and here is the deletion strain, the volume change. So z1 has a length of 1 and z2 and z3 also have lengths of 1 and they are all mutually perpendicular. So I will call z1 the 1, 0, 0 direction. Similarly z2 is the 0, 1, 0 direction and z3 is the 0, 0, 1 direction. So the vector z1 we can write as 1, 0, 0, z2 is equal to 0, 1, 0 and z3 is 0, 0, 1. Straightforward. Okay so when this deformation happens what happens to the vector z1? What are its coordinates when the deformation happens? Yeah go on. Be brave. So does it change? So the vector z1 lies here and I am shearing. Does it change? Yeah. It doesn't change, right? So it remains as 1, 0, 0 by this deformation. So I will define my deformation matrix zpz. The first vector z1 remains as 1, 0, 0. How about z2 which is coming out of the plane of the board? Again it lies in the invariant plane so it is not going to change. So it is 0, 1, 0 and how about z3? What happens to the component of z3 in this direction? So initially z3 has no component along z1, right? But by the shear what happens? So this has changed into this. So what is the component along z1? S, right? So I write down S here and along z2 there is no deformation along z2 and along z3, yeah, 1 plus delta correct. So that is the matrix which defines the deformation. If I multiply that matrix now by any other vector then I will be able to predict what happens to that vector as a function of the deformation. So supposing I want to calculate what happens when I multiply zpz by a vector 1, 0, 1, okay? That just arbitrarily take any direction, 1, 0, 1. Then 1, 0, 1 is a column vector, okay? It is a column. So if I multiply the matrix by the column what do I get as the first row by column? So 1 times 1 plus 0 times 0 plus S times 1 is 1 plus S, right? So become 1 plus S and then 0, 1, 0 times that is just 0 and this times this will be 1 plus delta. So as a consequence of the deformation 1, 0, 1 becomes a new vector 1 plus S, 0 and 1 plus delta. It will have changed direction, it will have changed magnitude. To work out its magnitude I just take the sum of squares and square root because we've got an orthonormal coordinate system, right? So I want you to prove to yourself for the next lecture that given that S is equal to 0.026, so S is equal to 0.26 and delta is equal to 0.03. What is the change in length? Let me just remind myself. Yeah, that you get 14% elongation, okay? So prove to yourself that the elongation is 14% when 1, 0, 1 is deformed by this deformation, okay? So I'm going to use this information in the next lecture. So I want to discover, you know, supposing that all of my orthonite transforms into martensite under stress. What is the maximum elongation I can get from trip? Okay? In order to do that I need to work out the elongation along the stress axis which could have arbitrary coordinates, yeah? So I take my deformation matrix multiplied by a vector, it gives me the new vector which will have a different length and if I take its length divided by the original length then I get the elongation, right? Okay, that's all for today.
 So, just in case any of you missed that is a homework assignment which you should hand  in tomorrow.  It should take you no more than half an hour to do.  Just look at how we derived the growth rate of ferrite from austenite and follow the procedure.  Draw a composition profile at the interface and then balance two terms which is the rate  at which carbon is partitioned or absorbed against the rate at which it is taken away  by diffusion.  Okay.  Okay.  So, today I'm going to cover two things, time to transmission diagrams and I'm going  to make a start on the transformation induced plasticity skills.  But first, time to temperature transformation diagrams where you saw the TTT diagram or  typical TTT diagram will have two C curves, one for the reconstructive transformations  and the other one for the displacive transformations.  And diagrams like these are used widely in industry.  Okay.  But what actually determines where those curves fall on the temperature versus time plot?  That's what we are after.  How do we actually calculate such a curve?  We've done with growth.  I'm going to briefly introduce you to nucleation because it's the combination of nucleation  and growth which gives us a time temperature transformation diagram.  Now most of you will have done nucleation theory but I'm going to go through it again  briefly with you.  Nucleation is as follows.  Supposing I'm observing my austenite, okay, and I've got as much resolution as I need  and I'm observing it at temperature.  And you will suddenly see that in some regions, a new structure and chemical composition will  be created by random fluctuation, all right?  So the system is homogeneous but suddenly you will see a fluctuation of structure and  chemical composition which is consistent with ferrite.  Okay.  So it's composition of that region and the crystal structure is consistent with ferrite.  So it's a random fluctuation atoms are moving about and you will just find a region which  has the structure and composition of ferrite.  But of course if you've created ferrite in austenite you will also have an interface  between the two phases, right?  And that interface is badly fitting so there's a certain energy, it's a defect.  And that fluctuation will not be able to grow until it gets to a size where the cost  of creating interface is less than the cost of creating, than the gain from the creation  of ferrite.  So that's basically nucleation theory that by random fluctuation we get regions which  have the right structure and right composition for the product phase and if that fluctuation  is large enough then it will grow into a phase.  So we can express that very simply, let's imagine that we've got a spherical nucleus  forming in our austenite.  So this is the free energy curve for austenite and free energy curve for ferrite.  This is the equilibrium temperature and below that you can see that the free energy of ferrite  is smaller than that of austenite and that's where we might be able to see this fluctuation.  So in the creation of this sphere I will get a reduction in free energy so this term is  negative, delta G chemical is negative because you can see that ferrite has a lower free  energy than austenite below the equilibrium temperature and this is simply the 4 upon  3 pi cubed is simply the volume of the nucleus.  But when we create a spherical nucleus we also create a surface and the area of the  surface is 4 pi r squared and the cost of creating that interface is sigma which is  in joules per meter squared.  It's the interfacial energy per unit area.  So that is positive, this is negative and we might also have some strain because the  volume change or shape change or whatever.  The strain energy scales with the volume of your material.  So this is chemical free energy change once the transformation to happen, strain opposes  it and the creation of interface opposes it.  So that's the balance, that's the free energy change delta G when we create a nucleus of  size r, radius r.  If I plot that expression, so if I plot on the vertical axis the free energy so delta  G and it's 0 here, negative and positive and along here is the radius r then the chemical  free energy change will tend to promote the transformation and it will be something like  this.  So this is 4 upon 3 pi r cubed into delta G chemical that tends to favor and it varies  with r cubed and r cubed will be smaller than r squared for very small values of r.  So the surface energy term will be denoted at small radii.  So this is the 4 pi r squared into sigma interfacial energy term and the strain energy term will  follow a curve which varies with r cubed.  I'm not going to plot that on here just for the sake of simplicity.  Now the net effect of these two terms is to produce a curve which looks like this.  So it isn't until the radius becomes greater than r star that the free energy change delta  G actually decreases and this height here is called the activation energy G star.  So G star is the activation energy of nucleation.  So simply because you are below the equilibrium transformation temperature doesn't mean that  the transformation happens instantaneously.  You've actually got to overcome the cost of the interface.  So you need fluctuations which are greater than r star for them to develop into successful  nuclei.  Now in order to find G star and r star, sorry that r star is incorrect.  Rub that out.  Why doesn't it rub out?  Let me just, r star is actually here at this point.  Now in order to find the value of G star and r star, this is the first equation is simply  the free energy change when you form a particle of radius r.  If I differentiate that with respect to r and set this to 0, then I get r star varies  with the interfacial energy.  So if the interfacial energy is 0, then any fluctuation will grow into our product phase  but it never is 0 in practice.  And of course if I have a larger driving force, then the critical size of fluctuation which  is successful will be smaller.  And if I now substitute r star into the first equation, then I can get the value of G star  which varies with the cube of the interfacial energy.  So it's very sensitive to the interfacial energy.  Larger interfacial energy will make it much more difficult for nucleation to happen and  it varies with the square of the chemical driving force.  So if I increase my driving force, then the activation barrier becomes smaller.  Now this is just a summary of the variation of interfacial energy and the chemical free  energy change plus strain energy as a function of the particle radius and this is the critical  size r star and here we have the activation energy.  We can now say the probability of a successful event is given by exponential minus G star  upon KT, the usual thermally activated Arrhenius type equation.  And nu over there is the jump frequency, is the attempt frequency.  So it's attempting to get over the barrier.  Not all of those attempts are successful.  The probability of a successful attempt is that exponential minus G star upon KT and  is simply the number density of nucleation sites per unit volume.  And we have a second exponential term here which is exponential minus Q upon KT.  And that is the activation energy for the transfer of atoms across the interface because  you know you've got to go from this crystal structure of austenite to that of ferrite  when you transfer an atom.  That has a certain barrier associated with it and that's a constant value Q.  G star will vary with your transformation conditions.  If you transform at a larger under cooling, G star will be smaller because the driving  force is greater.  So this is the equation for nucleation rate per unit volume.  So now we have the growth rate.  Yeah, we've done many growth rate calculations.  And we have the nucleation rate.  So we should be able to calculate the volume fraction as a function of time, temperature,  composition, etc.  Right?  We can calculate the volume fraction that gives us the TTT diagram.  So let's just see the problems that arise when we want to calculate volume fraction.  So imagine that I have two particles here.  So here we have two particles which have grown inside the austenite.  And that is at a certain time T.  And I look at those particles a small time interval later, then they will have become  bigger and also I might have nucleated new particles.  So these two particles here are new particles.  So is there anything wrong with that diagram on your right hand side?  What do you think is wrong with that diagram?  Yeah, I've got this particle and this particle are new particles.  These two have grown.  What is wrong with that diagram?  Go ahead, be brave.  I think you know the answer.  Yeah, it doesn't make sense to have a bi-nucleating inside ferrite which has already grown, right?  And similarly we've got this problem of overlap.  So once we start to look at volume fraction, we've got to think about particles touching  each other.  And how can we take account of particles touching each other?  And this is a good illustration.  This is a glass crystallizing, like window glass.  If you heat it long enough, then it will crystallize.  And these are particles of crystalline glass growing.  And you can see there's a problem here, they touch.  They don't start growing through each other, but you're stopping them from growing.  And eventually when all of this is filled, it will look like an equiaxed grain structure,  you know, the sort of thing that you see when you look in your microscopes.  So the problem is that we need to take account of what's known as hard impingement.  That means growing particles physically touching each other.  And we cannot allow nucleation to happen in regions which have already transformed.  Okay, so we'll just write that down.  So hard impingement.  Impingement means hitting each other, particles nucleated from different positions, positions  touch.  So let me draw that diagram again.  I've got a region here of the pair phase.  Let's say this is the time t equals tau, and this is at a time t equals tau plus delta  tau.  And I've got two particles here.  A short time later, they have grown, and I might have nucleated two new particles.  Now let's ignore at first the fact that some particles are growing in other particles or  that particles have touched.  If I calculate the change in the fraction of ferrite in going from tau to tau plus d  tau, it will be wrong.  But let's do that anyway.  Let's call that the change in extended volume.  That means we allow particles to grow through each other.  So the change in extended volume, change in extended volume, dp, and we'll call our  transformation product beta and e for extended.  So this is the beta phase growing.  So this will be the wrong volume we have calculated because obviously you can't transform regions  which have already transformed.  And the real change in the volume of beta we'll write as dv beta.  So this is the true change.  And obviously they're not going to be equal.  But if I multiply the right hand side here by the probability of finding untransformed  material, then that corrects for the extended volume because what we want is the change  in volume which falls into untransformed material.  And what is the probability of finding untransformed material?  So this will not equal this.  If I multiply the extended volume by the probability of finding untransformed material, the probability  of finding untransformed material is equal to what?  What's the probability of finding untransformed material when there is no transformation?  What's the probability of finding untransformed material when half of it is transformed?  Half, right?  So it's simply given by 1 minus the volume fraction of beta.  So that's 1 minus volume of beta divided by the total volume.  Because v beta over v is the volume fraction of beta.  Everyone happy with that?  So we can write that the real change in volume is equal to the change in extended volume  multiplied by 1 minus v beta upon v.  So this is a really important equation.  It allows you to look at many particles growing together and calculating a volume fraction.  And this is called the Avrami equation.  I'm going to write that out again on the next page, but rearranging to get the extended  volume on the left hand side.  So we have dv beta extended is equal to dv beta divided by 1 minus v beta upon v.  Just rearrange the previous equation so that I have dv beta on the left hand side.  And now I can integrate that so that I get the extended volume is equal to, now, what's  the integral of dx over x?  Yep, louder, louder.  Yeah, logarithm.  So we will have logarithm, leave some space after equals, yeah?  Logarithm of 1 minus v beta upon v.  And we have to put a minus v there because we've got 1 upon minus v, right?  So this is the relationship between the true volume of beta and the extended volume of  beta.  It's a very, very simple equation, but extremely powerful.  This is possibly the most useful kinetic equation that I know, far better than phase  field theory or any fancy methods for calculating volume fractions.  So let's assume that we now have completed the theory for impingement of particles.  Now let's do a calculation.  If we are observing a system transforming, and if I plot the size of a particle, size  of particle versus the time t, you will see a particle nucleated.  And let's assume that the growth rate is constant, then this particle which nucleated at this  point, which we'll call tau 1.  And then there might be another particle which nucleates at a different time, tau 2, and  then another one at tau 3, and so on.  This is what you would actually observe.  Particles have started at different times.  Before the time period tau 1, there was no particle 1.  So the size is 0 at that point, and then it starts to grow, and we are assuming a constant  growth rate.  This is a constant growth rate, which we will call G.  Then the volume of a particle nucleated at a particular point will be, let's say, a  small v with a subscript tau, and we'll assume that the growth is isotropic.  That means the same in all directions, and that will be 4 upon 3 pi G cubed, the growth  rate cubed, times t minus tau.  Because before the point tau, there is no particle.  It only increases in size beyond the point tau here.  Before that, it doesn't exist, and cubed.  That's an individual particle, but of course we will have many particles nucleating at  different times.  So in the small time interval, the number of particles that will form is the nucleation  rate times the volume, because it's a nucleation rate per unit volume, times the volume, times  the time interval.  So the number of particles nucleated in interval d tau is equal to the nucleation rate times  the volume, because it's the nucleation rate per unit volume times the volume, multiplied  by the number density of nucleation sites into d tau.  So n is the number density of nucleation sites.  You happy with that?  I'll move on to the next page.  So the change in extended volume then is equal to the nucleation rate per unit volume times  the volume times the number density of nucleation sites multiplied by the volume of each particle,  which is given by 4 upon 3 pi g cubed t minus tau cubed times the time interval d tau.  There's a number of particles that have been created in that time interval times the volume  of each of those particles.  What all I have to do now is integrate this from time equals 0 to any particular value  of time.  So if I integrate this from t equals 0 to t equals a particular time tau, then I get  the extended volume of beta is equal to i Bn into one third pi g cubed into t to the  power of 4.  Show that for yourself later on, but the 4 disappears because we've integrated to get  a 4 in the time exponent.  No, it should be d tau because we are integrating tau from t equals 0 to a value of tau.  So tau will disappear in this.  t is the real time.  Tau is the point at which a particular particle comes into existence.  So that's the extended volume.  And we already have a relationship between the extended volume and the real volume.  If I go back, here is the extended volume as a function of the real volume.  So we can simply substitute for the extended volume.  So instead of extended volume, I write minus V log of 1 minus V beta upon V is equal to  2.  And if I unlock this, then I will get 1 minus the volume fraction of beta is equal to exponential  minus i Bn upon 3.  Sorry, there is no V there into oops.  And that defines the volume fraction as a function of time.  The driving force comes into the growth equation, into the nucleation equation, and the number  density of nucleation sites.  So if nucleation happens at grain boundaries, then the grain size determines the number  density of nucleation sites.  So everything is in that equation.  And we've taken account of hard impingement.  If I plot that equation out, then this will be the typical shape.  So I've got time here and V beta upon V is the volume fraction.  And it's what's known as a sigmoidal curve going from 0 to 1.  So that's how the volume fraction evolves with time.  And the incubation period tore for a particle to come into existence is not fixed.  It depends on at what point that particle came into existence.  So when people talk about time-temperature transformation diagrams, they make the mistake  of saying that the incubation period is between the C curve and the vertical axis.  That is not an incubation period.  That is the time required to achieve a detectable amount of transformation.  So it depends on your technique.  If you have a dilatometer which has a 1% volume fraction measurement accuracy, and I take  1% here, then you start to plot your curve at the time corresponding to 1%.  I can plot another curve which corresponds to 5%, 10%, and so on.  And that will generate the whole of the TTT diagram by calculation.  Now here in this example, I took the growth rate to be constant.  So can you tell me which phase transmission in steel gives you a constant growth rate?  So one which doesn't involve diffusion, alternatively, where the composition of the product phase  is the same as that of the parent.  So can you tell me of reconstructive transformation where the composition of the parent is the  same as the product?  Perlite.  It's a mixture of cementite and ferrite with the same average composition as the austenite.  So the perlite colony will grow at a constant rate.  So this would be a model for perlite.  And if I obtain an experimental curve of the volume fraction versus time, and I get an  exponent to time of 4, then that indicates that the growth rate is constant.  It doesn't prove it because there might be another mechanism which also leads to an exponent  of 4.  And the exponent of 4, 3 comes from the growth, because we have growth rate cubed, isotropic  growth, and 1 from the constant nucleation rate that we have assumed.  So supposing now that we have parabolic growth, that means that the size varies with T to  the power of half, then what do I expect the exponent to be?  So parabolic growth rate and a constant nucleation rate.  What would the time exponent be?  Sorry?  Almost there.  So we've got size varying with T to the half.  So what will volume vary with?  Volume will vary with the cube of the size.  So 3 upon 2, right?  T to the half cubed, 3 upon 2.  And then 1 will come from the fact that we have nucleation times time.  So the exponent that you will get is 5 upon 2.  Yeah?  Is everyone happy about that?  So parabolic growth will lead to a different time exponent.  So again, if you measure experimentally and you see that the time exponent is 5 upon 2,  that indicates that the growth rate is parabolic with time and the nucleation rate is constant.  So it's an indication, but it may be ambiguous because another mechanism might also lead  to T to the 5 upon 2.  If you look in Christian's theory for transformations in metals and alloys, you'll find a table  of different combinations of nucleation and growth functions which give you different values  of the time exponent.  So if you analyze your deletometer curves more carefully, you'll be able to study also  the mechanism of your transformation product using this analysis.  Okay?  I'm just going to repeat all that using my slides.  Yeah?  So this is the, what happens after a small time interval.  These particles have grown.  So the dark blue regions there represents the increment of transformation.  And we've also nucleated another two particles.  But the calculation of the change in volume fraction is going to be wrong because we've  got overlap of phases.  This particle here should not form an area which has already transformed.  So we correct for that by multiplying the incorrect change in volume fraction by the  probability of finding untransformed material.  And that really is the essence of Avrami theory that tells you how to convert extended volume  into real volume.  So you don't need to worry about impingement between particles.  The extended volume is very much easier to calculate because you just calculate the growth  rate and the nucleation rate, add it all up, and then make a correction here.  Okay?  And of course, this is just an integration of that equation.  This is showing you, observing the system, you would see a variety of particle sizes  growing from the matrix because they start at different values of the incubation period  tour.  And the volume of a single particle, assuming a constant growth rate, will be given by G  cubed into T minus tour cubed times 4 upon 3 pi because we are assuming isotropic growth.  You don't need to assume isotropic growth.  You can have G1, G2, G3 as growth rates being different in different directions.  It's extremely flexible theory.  And then we arrive at our final Avrami equation.  And it contains everything.  It contains the growth rate, nucleation rate, time, temperature, everything.  Now one bad thing that you will find in the literature is that they express the Avrami  equation as the second equation, which is just saying, look, there's a constant care  and there's a time exponent n.  So they use it completely empirically.  And they don't realize that they are using it empirically.  They think that is the Avrami theory.  So I don't know if you remember, but there was a talk given, a seminar given, where the  person was comparing a phase field model with overall transformation kinetics and then said,  look, oh, this equation is completely empirical.  But what that person didn't realize is that he is using it empirically.  The equation itself is perfectly OK if you put in the right terms.  It's a very, very powerful equation, much more powerful than phase field theory.  Right.  So these are just typical curves.  And to find the time exponent, you take a double logarithm of the volume fraction.  It's fairly straightforward mathematics.  And then you plot your time, temperature, transmission diagram.  What I'd like you to tell me now is this was one phase growing from the austenite.  Right.  Just one phase.  What happens if I have more than one phase?  So supposing that I have ferrite growing from austenite and in another region, I also have  cementite growing because our carbon concentration is very high.  So independently growing.  How can I modify the Arami equation to take account of more than one phase growing?  It's very, very important.  Because when we do tempering or create power plant steels, we get many different kinds of  carbides forming, not just one at a time.  And for more than 60 years after the Arami equation, there was no theory for this.  So I want you to create that theory right now.  How can I modify this equation to deal with more than one phase at a time?  So just think in that diagram on the right, in addition to the blue particles, I also  have some red particles.  What would I have to do?  Well, think about how this probability term changes.  What does that represent?  That just represents the amount of untransformed material, right?  So if I have another phase forming, how do I modify that probability term?  Yeah?  Yeah?  Yeah.  Perfect.  Yeah.  So if I have a phase theta forming, then I have V alpha plus V theta at the top in the  probability term.  But of course, we only have alpha and alpha on both sides.  So I need a second equation, which says dV theta equals 1 minus V alpha plus V theta  over V into dV theta.  And then I solve those equations simultaneously.  If I have six phases, I have six equations.  It makes a remarkable difference to the accuracy of your predictions if you do it properly.  So this was solved just a few years ago when we was to model precipitation kinetics in power  plant steels, where you get many different kinds of particles forming like cementite,  molybdenum carbide, M23C6, and so on.  So the way you would modify that equation for multiple phases is very simple.  So we have dV alpha is equal to 1 minus V alpha plus V theta over the total volume into  dV alpha extended.  And we need a second equation, which is dV theta is equal to 1 minus V alpha plus V theta  upon the total volume times dV theta extended.  And you need to solve these equations simultaneously, which is not a problem.  What is more, you know, you can do this numerically.  You do it step by step.  Or you can do it analytically if you know the dependence of theta and alpha.  But numerically is the best procedure because supposing that the composition of the matrix  changes while the phases are growing, then you can alter the composition of the matrix  in each step, and that will influence the growth and nucleation rates.  So that is called soft impingement, where particles alter the composition of the matrix  and therefore they will grow or nucleate at a different rate.  So it's the overlap of diffusion fields.  They haven't touched, but the overlap of diffusion fields, which is called soft impingement,  as opposed to actually physically touching.  So this is the most powerful theory to deal with any number of reactions happening at  the same time.  And last year in GIFT, we applied this, Unju Song, applied this to oxidation.  So we have a steel containing silicon and manganese and so on.  And even though silicon is an extremely strong oxidizing agent, the combination of silicon  with oxygen produces a much bigger free energy change than iron with oxygen, the iron oxide  forms first.  And the reason is that there is a competition.  There is much more iron than silicon, and also the silicon has to diffuse longer distances.  And equations like these automatically predict that.  So you can download this from our website and you can see exactly this theory being applied  for oxidation.  It doesn't necessarily have to be precipitation and so forth.  So time-temperature transformation diagrams are extremely useful.  And if you use Avrami theory properly, that means not empirically, then it is the perfect  theory for modeling phase transformation kinetics.  And of course, once you have a time-temperature transformation diagram, you can easily convert  it into a continuous cooling transformation diagram because a continuous cooling curve  is simply a set of isothermal steps.  So you can add them all up.  Now, I want to start, make a start today on trip steels.  And I'm not going to teach you things which you probably already know, that it's a wonderful  steel which you use in cars to save the world, to reduce the weight and so on.  But actually, you're wrong.  The weight of cars has actually increased over the last 10 years.  It hasn't decreased.  In spite of all your research.  So why is that?  Why is the weight of a car increased?  The average weight of a car in Europe has increased, not decreased.  Why is that?  Safety.  Safety.  That's the thing.  So you see this bar here.  This is a picture which I have taken.  This bar is there so that if you get a side impact, then the deflection is limited so  that you are safe.  Your car will be a total write-off because this spreads the deformation across the body  of the car.  Yeah?  So you will be safe, but your car will no longer be usable or repairable.  So the additional safety features have meant that the weight of a car has actually increased,  not decreased.  So all the thesis that you see in GIFT which start off by saying, oh, we've got to save  fuel and save the world and so on, this isn't going to save the world.  It's simply improving your safety at this stage.  But of course you can also reduce the weight.  So this is another piece of steel in a car where you are joining up two pieces of steel  using a laser, laser weld.  And then you make the component from a composite steel.  So this is called a tailored blank.  This is the blank.  That means it hasn't been formed.  But you don't require the same properties in every region of your component.  So you choose the steel which has the right strength, right ductility, cost, et cetera,  and form the component out of a composite.  So this is now actually very common in the manufacture of cars.  And it does lead to a reduction in weight because you don't need the same thickness  even in all regions of your car.  So here, for example, this part here was made from that tailored blank which I showed you  in the previous slide.  This is a mini, but actually it's a BMW.  It's made by BMW.  So we are not succeeding actually in reducing the weight of a car in terms of what the customer  buys and in terms of safety regulations.  If we didn't have trip steels, the problem would be much worse.  That's the main point to write in your thesis introductions.  We justify your work.  OK.  Now, I'm going to start by distinguishing between a trip steel and a trip-assisted steel.  Do you know the difference?  What's the difference between a trip steel and a trip-assisted steel?  Exactly right.  So the original work on trip steels was by Zaké and his group.  And it involved 100% austenite, a steel which is 100% austenitic.  So obviously to get a steel which is 100% austenitic, you have a large concentration  of alloying elements.  So for example, this is an example of a very, very expensive material in the context of  steels because we have about 30%, 8% of nickel.  I don't think Posco will be very happy if I suggest that we make such a steel, right?  But if you think about carbon nanotubes and so on, they are far, far more expensive and  they are not worried about the cost.  So we have to actually charge much more for our steels.  They are much better than carbon nanotubes and yet we charge a lot less.  OK.  So let's take this steel.  First, I'm only going to talk about trip steels.  Trip-assisted means you have a small volume fraction of austenite and other phases present.  So you're getting an assistance from the trip, but it's not a trip steel.  But this has a martensite start temperature of minus 80 degrees centigrade and you can  see that the transformation produces a shape change.  So this is the surface relief produced by the formation of martensite in this material  and that is the deformation which is the transformation plasticity.  Transformation plasticity means plastic strain caused by a phase change.  So there's a crystal structure change, but it's still a physical deformation.  Now what I want to do is just to introduce you to how to calculate the strain along  any particular direction.  OK.  So we've got a transformation happening.  We know the nature of the shape change.  An invariant plane strain with a shear parallel to the habit plane and the volume change normal  to the habit plane.  And the value of the shear strain is of the order of 0.26 whereas the volume change delta  is about 0.03, 3%.  So imagine that this blue region here is a square of austenite of dimension 1.  And I define my coordinates here as z1 along the horizontal axis.  Z3 is normal to the horizontal plane and z2 is poking out of the plane of the board.  And this set of axes is called orthonormal because I've said that the austenite has dimensions  1 and these axes are all at 90 degrees to each other.  So we call this an orthonormal set of axes.  So what I want to do is I want to define a deformation matrix so that if I multiply  it matrix by any direction I will get the change in that vector due to this deformation.  Okay so we've got an orthonormal set of coordinates.  Set 1, set 2 and set 3.  So set 1 is along here and set 3 along there.  So these are unit vectors and this is my austenite.  And when it transforms, change is shape where this is the shear strain and here is the deletion  strain, the volume change.  So z1 has a length of 1 and z2 and z3 also have lengths of 1 and they are all mutually  perpendicular.  So I will call z1 the 1, 0, 0 direction.  Similarly z2 is the 0, 1, 0 direction and z3 is the 0, 0, 1 direction.  So the vector z1 we can write as 1, 0, 0, z2 is equal to 0, 1, 0 and z3 is 0, 0, 1.  Straightforward.  Okay so when this deformation happens what happens to the vector z1?  What are its coordinates when the deformation happens?  Yeah go on.  Be brave.  So does it change?  So the vector z1 lies here and I am shearing.  Does it change?  Yeah.  It doesn't change, right?  So it remains as 1, 0, 0 by this deformation.  So I will define my deformation matrix zpz.  The first vector z1 remains as 1, 0, 0.  How about z2 which is coming out of the plane of the board?  Again it lies in the invariant plane so it is not going to change.  So it is 0, 1, 0 and how about z3?  What happens to the component of z3 in this direction?  So initially z3 has no component along z1, right?  But by the shear what happens?  So this has changed into this.  So what is the component along z1?  S, right?  So I write down S here and along z2 there is no deformation along z2 and along z3, yeah,  1 plus delta correct.  So that is the matrix which defines the deformation.  If I multiply that matrix now by any other vector then I will be able to predict what  happens to that vector as a function of the deformation.  So supposing I want to calculate what happens when I multiply zpz by a vector 1, 0, 1, okay?  That just arbitrarily take any direction, 1, 0, 1.  Then 1, 0, 1 is a column vector, okay?  It is a column.  So if I multiply the matrix by the column what do I get as the first row by column?  So 1 times 1 plus 0 times 0 plus S times 1 is 1 plus S, right?  So become 1 plus S and then 0, 1, 0 times that is just 0 and this times this will be  1 plus delta.  So as a consequence of the deformation 1, 0, 1 becomes a new vector 1 plus S, 0 and 1  plus delta.  It will have changed direction, it will have changed magnitude.  To work out its magnitude I just take the sum of squares and square root because we've  got an orthonormal coordinate system, right?  So I want you to prove to yourself for the next lecture that given that S is equal to  0.026, so S is equal to 0.26 and delta is equal to 0.03.  What is the change in length?  Let me just remind myself.  Yeah, that you get 14% elongation, okay?  So prove to yourself that the elongation is 14% when 1, 0, 1 is deformed by this deformation,  okay?  So I'm going to use this information in the next lecture.  So I want to discover, you know, supposing that all of my orthonite transforms into martensite  under stress.  What is the maximum elongation I can get from trip?  Okay?  In order to do that I need to work out the elongation along the stress axis which could  have arbitrary coordinates, yeah?  So I take my deformation matrix multiplied by a vector, it gives me the new vector which  will have a different length and if I take its length divided by the original length  then I get the elongation, right?  Okay, that's all for today.
Physical_Metallurgy_of_Steels_-_Part_10_0_4823.png Physical_Metallurgy_of_Steels_-_Part_10_4824_7175.png Physical_Metallurgy_of_Steels_-_Part_10_7176_7295.png Physical_Metallurgy_of_Steels_-_Part_10_7296_7367.png Physical_Metallurgy_of_Steels_-_Part_10_7368_9191.png Physical_Metallurgy_of_Steels_-_Part_10_9192_11243.png Physical_Metallurgy_of_Steels_-_Part_10_11244_11531.png Physical_Metallurgy_of_Steels_-_Part_10_11532_11603.png Physical_Metallurgy_of_Steels_-_Part_10_11604_12659.png Physical_Metallurgy_of_Steels_-_Part_10_12660_13355.png Physical_Metallurgy_of_Steels_-_Part_10_13356_16271.png Physical_Metallurgy_of_Steels_-_Part_10_16272_16439.png Physical_Metallurgy_of_Steels_-_Part_10_16440_17423.png Physical_Metallurgy_of_Steels_-_Part_10_17424_18815.png Physical_Metallurgy_of_Steels_-_Part_10_18816_19991.png Physical_Metallurgy_of_Steels_-_Part_10_19992_20051.png Physical_Metallurgy_of_Steels_-_Part_10_20052_20519.png Physical_Metallurgy_of_Steels_-_Part_10_20520_21539.png Physical_Metallurgy_of_Steels_-_Part_10_21540_22475.png Physical_Metallurgy_of_Steels_-_Part_10_22476_23939.png Physical_Metallurgy_of_Steels_-_Part_10_23940_30143.png Physical_Metallurgy_of_Steels_-_Part_10_30144_33155.png Physical_Metallurgy_of_Steels_-_Part_10_33156_33755.png Physical_Metallurgy_of_Steels_-_Part_10_33756_33827.png Physical_Metallurgy_of_Steels_-_Part_10_33828_37319.png Physical_Metallurgy_of_Steels_-_Part_10_37320_37391.png Physical_Metallurgy_of_Steels_-_Part_10_37392_39371.png Physical_Metallurgy_of_Steels_-_Part_10_39372_41015.png Physical_Metallurgy_of_Steels_-_Part_10_41016_43895.png Physical_Metallurgy_of_Steels_-_Part_10_43896_44231.png Physical_Metallurgy_of_Steels_-_Part_10_44232_54299.png Physical_Metallurgy_of_Steels_-_Part_10_54300_54359.png Physical_Metallurgy_of_Steels_-_Part_10_54360_55103.png Physical_Metallurgy_of_Steels_-_Part_10_55104_56039.png Physical_Metallurgy_of_Steels_-_Part_10_56040_56123.png Physical_Metallurgy_of_Steels_-_Part_10_56124_56423.png Physical_Metallurgy_of_Steels_-_Part_10_56424_57047.png Physical_Metallurgy_of_Steels_-_Part_10_57048_57095.png Physical_Metallurgy_of_Steels_-_Part_10_57096_58847.png Physical_Metallurgy_of_Steels_-_Part_10_58848_58931.png Physical_Metallurgy_of_Steels_-_Part_10_58932_59123.png Physical_Metallurgy_of_Steels_-_Part_10_59124_59183.png Physical_Metallurgy_of_Steels_-_Part_10_59184_59231.png Physical_Metallurgy_of_Steels_-_Part_10_59232_59303.png Physical_Metallurgy_of_Steels_-_Part_10_59304_59351.png Physical_Metallurgy_of_Steels_-_Part_10_59352_59399.png Physical_Metallurgy_of_Steels_-_Part_10_59400_59711.png Physical_Metallurgy_of_Steels_-_Part_10_59712_60227.png Physical_Metallurgy_of_Steels_-_Part_10_60228_60275.png Physical_Metallurgy_of_Steels_-_Part_10_60276_63623.png Physical_Metallurgy_of_Steels_-_Part_10_63624_63683.png Physical_Metallurgy_of_Steels_-_Part_10_63684_63731.png Physical_Metallurgy_of_Steels_-_Part_10_63732_63779.png Physical_Metallurgy_of_Steels_-_Part_10_63780_68627.png Physical_Metallurgy_of_Steels_-_Part_10_68628_68711.png Physical_Metallurgy_of_Steels_-_Part_10_68712_70943.png Physical_Metallurgy_of_Steels_-_Part_10_70944_72143.png Physical_Metallurgy_of_Steels_-_Part_10_72144_73043.png Physical_Metallurgy_of_Steels_-_Part_10_73044_73115.png Physical_Metallurgy_of_Steels_-_Part_10_73116_73343.png Physical_Metallurgy_of_Steels_-_Part_10_73344_73523.png Physical_Metallurgy_of_Steels_-_Part_10_73524_73571.png Physical_Metallurgy_of_Steels_-_Part_10_73572_73619.png Physical_Metallurgy_of_Steels_-_Part_10_73620_74195.png Physical_Metallurgy_of_Steels_-_Part_10_74196_74267.png Physical_Metallurgy_of_Steels_-_Part_10_74268_74639.png Physical_Metallurgy_of_Steels_-_Part_10_74640_74999.png Physical_Metallurgy_of_Steels_-_Part_10_75000_75263.png Physical_Metallurgy_of_Steels_-_Part_10_75264_75323.png Physical_Metallurgy_of_Steels_-_Part_10_75324_75371.png Physical_Metallurgy_of_Steels_-_Part_10_75372_75599.png Physical_Metallurgy_of_Steels_-_Part_10_75600_75923.png Physical_Metallurgy_of_Steels_-_Part_10_75924_76019.png Physical_Metallurgy_of_Steels_-_Part_10_76020_76295.png Physical_Metallurgy_of_Steels_-_Part_10_76296_78527.png Physical_Metallurgy_of_Steels_-_Part_10_78528_78599.png Physical_Metallurgy_of_Steels_-_Part_10_78600_78755.png Physical_Metallurgy_of_Steels_-_Part_10_78756_80123.png Physical_Metallurgy_of_Steels_-_Part_10_80124_89771.png
So, just in case any of you missed that is a homework assignment which you should hand in tomorrow. It should take you no more than half an hour to do. Just look at how we derived the growth rate of ferrite from austenite and follow the procedure. Draw a composition profile at the interface and then balance two terms which is the rate at which carbon is partitioned or absorbed against the rate at which it is taken away by diffusion. Okay. Okay. So, today I'm going to cover two things, time to transmission diagrams and I'm going to make a start on the transformation induced plasticity skills. But first, time to temperature transformation diagrams where you saw the TTT diagram or typical TTT diagram will have two C curves, one for the reconstructive transformations and the other one for the displacive transformations. And diagrams like these are used widely in industry. Okay. But what actually determines where those curves fall on the temperature versus time plot? That's what we are after. How do we actually calculate such a curve? We've done with growth. I'm going to briefly introduce you to nucleation because it's the combination of nucleation and growth which gives us a time temperature transformation diagram. Now most of you will have done nucleation theory but I'm going to go through it again briefly with you. Nucleation is as follows. Supposing I'm observing my austenite, okay, and I've got as much resolution as I need and I'm observing it at temperature. And you will suddenly see that in some regions, a new structure and chemical composition will be created by random fluctuation, all right? So the system is homogeneous but suddenly you will see a fluctuation of structure and chemical composition which is consistent with ferrite. Okay. So it's composition of that region and the crystal structure is consistent with ferrite. So it's a random fluctuation atoms are moving about and you will just find a region which has the structure and composition of ferrite. But of course if you've created ferrite in austenite you will also have an interface between the two phases, right? And that interface is badly fitting so there's a certain energy, it's a defect. And that fluctuation will not be able to grow until it gets to a size where the cost of creating interface is less than the cost of creating, than the gain from the creation of ferrite. So that's basically nucleation theory that by random fluctuation we get regions which have the right structure and right composition for the product phase and if that fluctuation is large enough then it will grow into a phase. So we can express that very simply, let's imagine that we've got a spherical nucleus forming in our austenite. So this is the free energy curve for austenite and free energy curve for ferrite. This is the equilibrium temperature and below that you can see that the free energy of ferrite is smaller than that of austenite and that's where we might be able to see this fluctuation. So in the creation of this sphere I will get a reduction in free energy so this term is negative, delta G chemical is negative because you can see that ferrite has a lower free energy than austenite below the equilibrium temperature and this is simply the 4 upon 3 pi cubed is simply the volume of the nucleus. But when we create a spherical nucleus we also create a surface and the area of the surface is 4 pi r squared and the cost of creating that interface is sigma which is in joules per meter squared. It's the interfacial energy per unit area. So that is positive, this is negative and we might also have some strain because the volume change or shape change or whatever. The strain energy scales with the volume of your material. So this is chemical free energy change once the transformation to happen, strain opposes it and the creation of interface opposes it. So that's the balance, that's the free energy change delta G when we create a nucleus of size r, radius r. If I plot that expression, so if I plot on the vertical axis the free energy so delta G and it's 0 here, negative and positive and along here is the radius r then the chemical free energy change will tend to promote the transformation and it will be something like this. So this is 4 upon 3 pi r cubed into delta G chemical that tends to favor and it varies with r cubed and r cubed will be smaller than r squared for very small values of r. So the surface energy term will be denoted at small radii. So this is the 4 pi r squared into sigma interfacial energy term and the strain energy term will follow a curve which varies with r cubed. I'm not going to plot that on here just for the sake of simplicity. Now the net effect of these two terms is to produce a curve which looks like this. So it isn't until the radius becomes greater than r star that the free energy change delta G actually decreases and this height here is called the activation energy G star. So G star is the activation energy of nucleation. So simply because you are below the equilibrium transformation temperature doesn't mean that the transformation happens instantaneously. You've actually got to overcome the cost of the interface. So you need fluctuations which are greater than r star for them to develop into successful nuclei. Now in order to find G star and r star, sorry that r star is incorrect. Rub that out. Why doesn't it rub out? Let me just, r star is actually here at this point. Now in order to find the value of G star and r star, this is the first equation is simply the free energy change when you form a particle of radius r. If I differentiate that with respect to r and set this to 0, then I get r star varies with the interfacial energy. So if the interfacial energy is 0, then any fluctuation will grow into our product phase but it never is 0 in practice. And of course if I have a larger driving force, then the critical size of fluctuation which is successful will be smaller. And if I now substitute r star into the first equation, then I can get the value of G star which varies with the cube of the interfacial energy. So it's very sensitive to the interfacial energy. Larger interfacial energy will make it much more difficult for nucleation to happen and it varies with the square of the chemical driving force. So if I increase my driving force, then the activation barrier becomes smaller. Now this is just a summary of the variation of interfacial energy and the chemical free energy change plus strain energy as a function of the particle radius and this is the critical size r star and here we have the activation energy. We can now say the probability of a successful event is given by exponential minus G star upon KT, the usual thermally activated Arrhenius type equation. And nu over there is the jump frequency, is the attempt frequency. So it's attempting to get over the barrier. Not all of those attempts are successful. The probability of a successful attempt is that exponential minus G star upon KT and is simply the number density of nucleation sites per unit volume. And we have a second exponential term here which is exponential minus Q upon KT. And that is the activation energy for the transfer of atoms across the interface because you know you've got to go from this crystal structure of austenite to that of ferrite when you transfer an atom. That has a certain barrier associated with it and that's a constant value Q. G star will vary with your transformation conditions. If you transform at a larger under cooling, G star will be smaller because the driving force is greater. So this is the equation for nucleation rate per unit volume. So now we have the growth rate. Yeah, we've done many growth rate calculations. And we have the nucleation rate. So we should be able to calculate the volume fraction as a function of time, temperature, composition, etc. Right? We can calculate the volume fraction that gives us the TTT diagram. So let's just see the problems that arise when we want to calculate volume fraction. So imagine that I have two particles here. So here we have two particles which have grown inside the austenite. And that is at a certain time T. And I look at those particles a small time interval later, then they will have become bigger and also I might have nucleated new particles. So these two particles here are new particles. So is there anything wrong with that diagram on your right hand side? What do you think is wrong with that diagram? Yeah, I've got this particle and this particle are new particles. These two have grown. What is wrong with that diagram? Go ahead, be brave. I think you know the answer. Yeah, it doesn't make sense to have a bi-nucleating inside ferrite which has already grown, right? And similarly we've got this problem of overlap. So once we start to look at volume fraction, we've got to think about particles touching each other. And how can we take account of particles touching each other? And this is a good illustration. This is a glass crystallizing, like window glass. If you heat it long enough, then it will crystallize. And these are particles of crystalline glass growing. And you can see there's a problem here, they touch. They don't start growing through each other, but you're stopping them from growing. And eventually when all of this is filled, it will look like an equiaxed grain structure, you know, the sort of thing that you see when you look in your microscopes. So the problem is that we need to take account of what's known as hard impingement. That means growing particles physically touching each other. And we cannot allow nucleation to happen in regions which have already transformed. Okay, so we'll just write that down. So hard impingement. Impingement means hitting each other, particles nucleated from different positions, positions touch. So let me draw that diagram again. I've got a region here of the pair phase. Let's say this is the time t equals tau, and this is at a time t equals tau plus delta tau. And I've got two particles here. A short time later, they have grown, and I might have nucleated two new particles. Now let's ignore at first the fact that some particles are growing in other particles or that particles have touched. If I calculate the change in the fraction of ferrite in going from tau to tau plus d tau, it will be wrong. But let's do that anyway. Let's call that the change in extended volume. That means we allow particles to grow through each other. So the change in extended volume, change in extended volume, dp, and we'll call our transformation product beta and e for extended. So this is the beta phase growing. So this will be the wrong volume we have calculated because obviously you can't transform regions which have already transformed. And the real change in the volume of beta we'll write as dv beta. So this is the true change. And obviously they're not going to be equal. But if I multiply the right hand side here by the probability of finding untransformed material, then that corrects for the extended volume because what we want is the change in volume which falls into untransformed material. And what is the probability of finding untransformed material? So this will not equal this. If I multiply the extended volume by the probability of finding untransformed material, the probability of finding untransformed material is equal to what? What's the probability of finding untransformed material when there is no transformation? What's the probability of finding untransformed material when half of it is transformed? Half, right? So it's simply given by 1 minus the volume fraction of beta. So that's 1 minus volume of beta divided by the total volume. Because v beta over v is the volume fraction of beta. Everyone happy with that? So we can write that the real change in volume is equal to the change in extended volume multiplied by 1 minus v beta upon v. So this is a really important equation. It allows you to look at many particles growing together and calculating a volume fraction. And this is called the Avrami equation. I'm going to write that out again on the next page, but rearranging to get the extended volume on the left hand side. So we have dv beta extended is equal to dv beta divided by 1 minus v beta upon v. Just rearrange the previous equation so that I have dv beta on the left hand side. And now I can integrate that so that I get the extended volume is equal to, now, what's the integral of dx over x? Yep, louder, louder. Yeah, logarithm. So we will have logarithm, leave some space after equals, yeah? Logarithm of 1 minus v beta upon v. And we have to put a minus v there because we've got 1 upon minus v, right? So this is the relationship between the true volume of beta and the extended volume of beta. It's a very, very simple equation, but extremely powerful. This is possibly the most useful kinetic equation that I know, far better than phase field theory or any fancy methods for calculating volume fractions. So let's assume that we now have completed the theory for impingement of particles. Now let's do a calculation. If we are observing a system transforming, and if I plot the size of a particle, size of particle versus the time t, you will see a particle nucleated. And let's assume that the growth rate is constant, then this particle which nucleated at this point, which we'll call tau 1. And then there might be another particle which nucleates at a different time, tau 2, and then another one at tau 3, and so on. This is what you would actually observe. Particles have started at different times. Before the time period tau 1, there was no particle 1. So the size is 0 at that point, and then it starts to grow, and we are assuming a constant growth rate. This is a constant growth rate, which we will call G. Then the volume of a particle nucleated at a particular point will be, let's say, a small v with a subscript tau, and we'll assume that the growth is isotropic. That means the same in all directions, and that will be 4 upon 3 pi G cubed, the growth rate cubed, times t minus tau. Because before the point tau, there is no particle. It only increases in size beyond the point tau here. Before that, it doesn't exist, and cubed. That's an individual particle, but of course we will have many particles nucleating at different times. So in the small time interval, the number of particles that will form is the nucleation rate times the volume, because it's a nucleation rate per unit volume, times the volume, times the time interval. So the number of particles nucleated in interval d tau is equal to the nucleation rate times the volume, because it's the nucleation rate per unit volume times the volume, multiplied by the number density of nucleation sites into d tau. So n is the number density of nucleation sites. You happy with that? I'll move on to the next page. So the change in extended volume then is equal to the nucleation rate per unit volume times the volume times the number density of nucleation sites multiplied by the volume of each particle, which is given by 4 upon 3 pi g cubed t minus tau cubed times the time interval d tau. There's a number of particles that have been created in that time interval times the volume of each of those particles. What all I have to do now is integrate this from time equals 0 to any particular value of time. So if I integrate this from t equals 0 to t equals a particular time tau, then I get the extended volume of beta is equal to i Bn into one third pi g cubed into t to the power of 4. Show that for yourself later on, but the 4 disappears because we've integrated to get a 4 in the time exponent. No, it should be d tau because we are integrating tau from t equals 0 to a value of tau. So tau will disappear in this. t is the real time. Tau is the point at which a particular particle comes into existence. So that's the extended volume. And we already have a relationship between the extended volume and the real volume. If I go back, here is the extended volume as a function of the real volume. So we can simply substitute for the extended volume. So instead of extended volume, I write minus V log of 1 minus V beta upon V is equal to 2. And if I unlock this, then I will get 1 minus the volume fraction of beta is equal to exponential minus i Bn upon 3. Sorry, there is no V there into oops. And that defines the volume fraction as a function of time. The driving force comes into the growth equation, into the nucleation equation, and the number density of nucleation sites. So if nucleation happens at grain boundaries, then the grain size determines the number density of nucleation sites. So everything is in that equation. And we've taken account of hard impingement. If I plot that equation out, then this will be the typical shape. So I've got time here and V beta upon V is the volume fraction. And it's what's known as a sigmoidal curve going from 0 to 1. So that's how the volume fraction evolves with time. And the incubation period tore for a particle to come into existence is not fixed. It depends on at what point that particle came into existence. So when people talk about time-temperature transformation diagrams, they make the mistake of saying that the incubation period is between the C curve and the vertical axis. That is not an incubation period. That is the time required to achieve a detectable amount of transformation. So it depends on your technique. If you have a dilatometer which has a 1% volume fraction measurement accuracy, and I take 1% here, then you start to plot your curve at the time corresponding to 1%. I can plot another curve which corresponds to 5%, 10%, and so on. And that will generate the whole of the TTT diagram by calculation. Now here in this example, I took the growth rate to be constant. So can you tell me which phase transmission in steel gives you a constant growth rate? So one which doesn't involve diffusion, alternatively, where the composition of the product phase is the same as that of the parent. So can you tell me of reconstructive transformation where the composition of the parent is the same as the product? Perlite. It's a mixture of cementite and ferrite with the same average composition as the austenite. So the perlite colony will grow at a constant rate. So this would be a model for perlite. And if I obtain an experimental curve of the volume fraction versus time, and I get an exponent to time of 4, then that indicates that the growth rate is constant. It doesn't prove it because there might be another mechanism which also leads to an exponent of 4. And the exponent of 4, 3 comes from the growth, because we have growth rate cubed, isotropic growth, and 1 from the constant nucleation rate that we have assumed. So supposing now that we have parabolic growth, that means that the size varies with T to the power of half, then what do I expect the exponent to be? So parabolic growth rate and a constant nucleation rate. What would the time exponent be? Sorry? Almost there. So we've got size varying with T to the half. So what will volume vary with? Volume will vary with the cube of the size. So 3 upon 2, right? T to the half cubed, 3 upon 2. And then 1 will come from the fact that we have nucleation times time. So the exponent that you will get is 5 upon 2. Yeah? Is everyone happy about that? So parabolic growth will lead to a different time exponent. So again, if you measure experimentally and you see that the time exponent is 5 upon 2, that indicates that the growth rate is parabolic with time and the nucleation rate is constant. So it's an indication, but it may be ambiguous because another mechanism might also lead to T to the 5 upon 2. If you look in Christian's theory for transformations in metals and alloys, you'll find a table of different combinations of nucleation and growth functions which give you different values of the time exponent. So if you analyze your deletometer curves more carefully, you'll be able to study also the mechanism of your transformation product using this analysis. Okay? I'm just going to repeat all that using my slides. Yeah? So this is the, what happens after a small time interval. These particles have grown. So the dark blue regions there represents the increment of transformation. And we've also nucleated another two particles. But the calculation of the change in volume fraction is going to be wrong because we've got overlap of phases. This particle here should not form an area which has already transformed. So we correct for that by multiplying the incorrect change in volume fraction by the probability of finding untransformed material. And that really is the essence of Avrami theory that tells you how to convert extended volume into real volume. So you don't need to worry about impingement between particles. The extended volume is very much easier to calculate because you just calculate the growth rate and the nucleation rate, add it all up, and then make a correction here. Okay? And of course, this is just an integration of that equation. This is showing you, observing the system, you would see a variety of particle sizes growing from the matrix because they start at different values of the incubation period tour. And the volume of a single particle, assuming a constant growth rate, will be given by G cubed into T minus tour cubed times 4 upon 3 pi because we are assuming isotropic growth. You don't need to assume isotropic growth. You can have G1, G2, G3 as growth rates being different in different directions. It's extremely flexible theory. And then we arrive at our final Avrami equation. And it contains everything. It contains the growth rate, nucleation rate, time, temperature, everything. Now one bad thing that you will find in the literature is that they express the Avrami equation as the second equation, which is just saying, look, there's a constant care and there's a time exponent n. So they use it completely empirically. And they don't realize that they are using it empirically. They think that is the Avrami theory. So I don't know if you remember, but there was a talk given, a seminar given, where the person was comparing a phase field model with overall transformation kinetics and then said, look, oh, this equation is completely empirical. But what that person didn't realize is that he is using it empirically. The equation itself is perfectly OK if you put in the right terms. It's a very, very powerful equation, much more powerful than phase field theory. Right. So these are just typical curves. And to find the time exponent, you take a double logarithm of the volume fraction. It's fairly straightforward mathematics. And then you plot your time, temperature, transmission diagram. What I'd like you to tell me now is this was one phase growing from the austenite. Right. Just one phase. What happens if I have more than one phase? So supposing that I have ferrite growing from austenite and in another region, I also have cementite growing because our carbon concentration is very high. So independently growing. How can I modify the Arami equation to take account of more than one phase growing? It's very, very important. Because when we do tempering or create power plant steels, we get many different kinds of carbides forming, not just one at a time. And for more than 60 years after the Arami equation, there was no theory for this. So I want you to create that theory right now. How can I modify this equation to deal with more than one phase at a time? So just think in that diagram on the right, in addition to the blue particles, I also have some red particles. What would I have to do? Well, think about how this probability term changes. What does that represent? That just represents the amount of untransformed material, right? So if I have another phase forming, how do I modify that probability term? Yeah? Yeah? Yeah. Perfect. Yeah. So if I have a phase theta forming, then I have V alpha plus V theta at the top in the probability term. But of course, we only have alpha and alpha on both sides. So I need a second equation, which says dV theta equals 1 minus V alpha plus V theta over V into dV theta. And then I solve those equations simultaneously. If I have six phases, I have six equations. It makes a remarkable difference to the accuracy of your predictions if you do it properly. So this was solved just a few years ago when we was to model precipitation kinetics in power plant steels, where you get many different kinds of particles forming like cementite, molybdenum carbide, M23C6, and so on. So the way you would modify that equation for multiple phases is very simple. So we have dV alpha is equal to 1 minus V alpha plus V theta over the total volume into dV alpha extended. And we need a second equation, which is dV theta is equal to 1 minus V alpha plus V theta upon the total volume times dV theta extended. And you need to solve these equations simultaneously, which is not a problem. What is more, you know, you can do this numerically. You do it step by step. Or you can do it analytically if you know the dependence of theta and alpha. But numerically is the best procedure because supposing that the composition of the matrix changes while the phases are growing, then you can alter the composition of the matrix in each step, and that will influence the growth and nucleation rates. So that is called soft impingement, where particles alter the composition of the matrix and therefore they will grow or nucleate at a different rate. So it's the overlap of diffusion fields. They haven't touched, but the overlap of diffusion fields, which is called soft impingement, as opposed to actually physically touching. So this is the most powerful theory to deal with any number of reactions happening at the same time. And last year in GIFT, we applied this, Unju Song, applied this to oxidation. So we have a steel containing silicon and manganese and so on. And even though silicon is an extremely strong oxidizing agent, the combination of silicon with oxygen produces a much bigger free energy change than iron with oxygen, the iron oxide forms first. And the reason is that there is a competition. There is much more iron than silicon, and also the silicon has to diffuse longer distances. And equations like these automatically predict that. So you can download this from our website and you can see exactly this theory being applied for oxidation. It doesn't necessarily have to be precipitation and so forth. So time-temperature transformation diagrams are extremely useful. And if you use Avrami theory properly, that means not empirically, then it is the perfect theory for modeling phase transformation kinetics. And of course, once you have a time-temperature transformation diagram, you can easily convert it into a continuous cooling transformation diagram because a continuous cooling curve is simply a set of isothermal steps. So you can add them all up. Now, I want to start, make a start today on trip steels. And I'm not going to teach you things which you probably already know, that it's a wonderful steel which you use in cars to save the world, to reduce the weight and so on. But actually, you're wrong. The weight of cars has actually increased over the last 10 years. It hasn't decreased. In spite of all your research. So why is that? Why is the weight of a car increased? The average weight of a car in Europe has increased, not decreased. Why is that? Safety. Safety. That's the thing. So you see this bar here. This is a picture which I have taken. This bar is there so that if you get a side impact, then the deflection is limited so that you are safe. Your car will be a total write-off because this spreads the deformation across the body of the car. Yeah? So you will be safe, but your car will no longer be usable or repairable. So the additional safety features have meant that the weight of a car has actually increased, not decreased. So all the thesis that you see in GIFT which start off by saying, oh, we've got to save fuel and save the world and so on, this isn't going to save the world. It's simply improving your safety at this stage. But of course you can also reduce the weight. So this is another piece of steel in a car where you are joining up two pieces of steel using a laser, laser weld. And then you make the component from a composite steel. So this is called a tailored blank. This is the blank. That means it hasn't been formed. But you don't require the same properties in every region of your component. So you choose the steel which has the right strength, right ductility, cost, et cetera, and form the component out of a composite. So this is now actually very common in the manufacture of cars. And it does lead to a reduction in weight because you don't need the same thickness even in all regions of your car. So here, for example, this part here was made from that tailored blank which I showed you in the previous slide. This is a mini, but actually it's a BMW. It's made by BMW. So we are not succeeding actually in reducing the weight of a car in terms of what the customer buys and in terms of safety regulations. If we didn't have trip steels, the problem would be much worse. That's the main point to write in your thesis introductions. We justify your work. OK. Now, I'm going to start by distinguishing between a trip steel and a trip-assisted steel. Do you know the difference? What's the difference between a trip steel and a trip-assisted steel? Exactly right. So the original work on trip steels was by Zaké and his group. And it involved 100% austenite, a steel which is 100% austenitic. So obviously to get a steel which is 100% austenitic, you have a large concentration of alloying elements. So for example, this is an example of a very, very expensive material in the context of steels because we have about 30%, 8% of nickel. I don't think Posco will be very happy if I suggest that we make such a steel, right? But if you think about carbon nanotubes and so on, they are far, far more expensive and they are not worried about the cost. So we have to actually charge much more for our steels. They are much better than carbon nanotubes and yet we charge a lot less. OK. So let's take this steel. First, I'm only going to talk about trip steels. Trip-assisted means you have a small volume fraction of austenite and other phases present. So you're getting an assistance from the trip, but it's not a trip steel. But this has a martensite start temperature of minus 80 degrees centigrade and you can see that the transformation produces a shape change. So this is the surface relief produced by the formation of martensite in this material and that is the deformation which is the transformation plasticity. Transformation plasticity means plastic strain caused by a phase change. So there's a crystal structure change, but it's still a physical deformation. Now what I want to do is just to introduce you to how to calculate the strain along any particular direction. OK. So we've got a transformation happening. We know the nature of the shape change. An invariant plane strain with a shear parallel to the habit plane and the volume change normal to the habit plane. And the value of the shear strain is of the order of 0.26 whereas the volume change delta is about 0.03, 3%. So imagine that this blue region here is a square of austenite of dimension 1. And I define my coordinates here as z1 along the horizontal axis. Z3 is normal to the horizontal plane and z2 is poking out of the plane of the board. And this set of axes is called orthonormal because I've said that the austenite has dimensions 1 and these axes are all at 90 degrees to each other. So we call this an orthonormal set of axes. So what I want to do is I want to define a deformation matrix so that if I multiply it matrix by any direction I will get the change in that vector due to this deformation. Okay so we've got an orthonormal set of coordinates. Set 1, set 2 and set 3. So set 1 is along here and set 3 along there. So these are unit vectors and this is my austenite. And when it transforms, change is shape where this is the shear strain and here is the deletion strain, the volume change. So z1 has a length of 1 and z2 and z3 also have lengths of 1 and they are all mutually perpendicular. So I will call z1 the 1, 0, 0 direction. Similarly z2 is the 0, 1, 0 direction and z3 is the 0, 0, 1 direction. So the vector z1 we can write as 1, 0, 0, z2 is equal to 0, 1, 0 and z3 is 0, 0, 1. Straightforward. Okay so when this deformation happens what happens to the vector z1? What are its coordinates when the deformation happens? Yeah go on. Be brave. So does it change? So the vector z1 lies here and I am shearing. Does it change? Yeah. It doesn't change, right? So it remains as 1, 0, 0 by this deformation. So I will define my deformation matrix zpz. The first vector z1 remains as 1, 0, 0. How about z2 which is coming out of the plane of the board? Again it lies in the invariant plane so it is not going to change. So it is 0, 1, 0 and how about z3? What happens to the component of z3 in this direction? So initially z3 has no component along z1, right? But by the shear what happens? So this has changed into this. So what is the component along z1? S, right? So I write down S here and along z2 there is no deformation along z2 and along z3, yeah, 1 plus delta correct. So that is the matrix which defines the deformation. If I multiply that matrix now by any other vector then I will be able to predict what happens to that vector as a function of the deformation. So supposing I want to calculate what happens when I multiply zpz by a vector 1, 0, 1, okay? That just arbitrarily take any direction, 1, 0, 1. Then 1, 0, 1 is a column vector, okay? It is a column. So if I multiply the matrix by the column what do I get as the first row by column? So 1 times 1 plus 0 times 0 plus S times 1 is 1 plus S, right? So become 1 plus S and then 0, 1, 0 times that is just 0 and this times this will be 1 plus delta. So as a consequence of the deformation 1, 0, 1 becomes a new vector 1 plus S, 0 and 1 plus delta. It will have changed direction, it will have changed magnitude. To work out its magnitude I just take the sum of squares and square root because we've got an orthonormal coordinate system, right? So I want you to prove to yourself for the next lecture that given that S is equal to 0.026, so S is equal to 0.26 and delta is equal to 0.03. What is the change in length? Let me just remind myself. Yeah, that you get 14% elongation, okay? So prove to yourself that the elongation is 14% when 1, 0, 1 is deformed by this deformation, okay? So I'm going to use this information in the next lecture. So I want to discover, you know, supposing that all of my orthonite transforms into martensite under stress. What is the maximum elongation I can get from trip? Okay? In order to do that I need to work out the elongation along the stress axis which could have arbitrary coordinates, yeah? So I take my deformation matrix multiplied by a vector, it gives me the new vector which will have a different length and if I take its length divided by the original length then I get the elongation, right? Okay, that's all for today.
transformation induced plasticity skills. But about time-temperature transformation diagrams, TTT diagram by calculation. Now here in this model with overall transformation kinetics and that diagram on the right, in addition to the context of steels because we have about 30%, 8%