 when we plotted the stress strain diagram which is the result of an uniaxial tension test, we had noted that the curve is linear portion of the curve which is from o to y does not continue straight onwards, but actually bends. And we said this is because of plasticity due to slip and I am talking about say for instance room temperature test on a material like aluminum. And we wanted to understand how does this slip take place and what is that which is weakening the crystal just in fact trying to stop this curve from going upward that means the materials at for instance strain as this much is not this strong, but is actually weaker by this amount. So, at the heart of this is the weakening of crystal in the presence of dislocations. Now, suppose I want to understand plastic deformation or permanent deformation, I have two crystals, two parts of a crystal for instance this is part one and this is part two. And I can share one part with respect to the other by applying a shear force. Now, when I apply such a shear force then the crystal may be sheared, but if you look at the kind of stresses required to cause this kind of a plastic deformation it turns out to be of the order of G P A which you will site again in the next slide. And it is important to note that all plastic deformation by slip require shear stresses at the microscopic scale. Even though in uniaxial tension test at the macroscopic level we are actually applying a tensile force, but at the microscopic level shear stresses are required for plastic deformation to take place by slip. But if you notice we already seen that the typically crystals yields of the order of mega pascals could be tens or hundreds of mega pascals that means that something is weakening the crystal drastically. And in 1930s Taylor and Orovan and Olyani postulated that it is dislocations a theoretical concept at that stage which is weakening the crystal. And later on TM observations in the 1950s proved that it is actually indeed dislocations which are weakening the crystal and which is at the heart of plasticity at the micro scale. And it is important to reiterate the fact that even though I am doing uniaxial tension test as we did before the except for the vertical and horizontal axis all the incline planes or vertical horizontal planes all the other incline planes actually feel a shear stress in them. And therefore, plasticity by slip is possible. Now, if you look at just to reiterate some of the important things we pointed out in the previous slide the shear modulus of crystals is in the range of about 20 to 150 GPA. And if you use a complete shearing of one part of the crystal with respect to the other the maximum shear required for that comes out the order of about g by 2 pi which means it is about 3 to 30 MPa. And actual yield stress in shear is of the order of 10 MPa or the order of those that number this implies the crystal is actually much weaker compared to the theoretical shearing of the crystal theory shearing mechanism which was proposed originally for shearing of the crystal. This implies that something at the heart of it is actually weakening the crystals. And this is dislocations which are weakening the crystals this also implies suppose I make a crystal which is completely free of dislocations. And this has been done and this has been done in small cylindrical specimens which are known as whiskers. And the radius of these whiskers in the micron scale then it has been observed that actually you can approach near to the theoretical shear stress. In these cases when you load a whisker what really happens is that dislocations have to be nucleated. And therefore, when the dislocation nucleates then this strength of the crystal drastically falls down. In fact whiskers of tin can have yield strength in shear of about 10 power minus 2 g which is about 1000 times stronger than a bulk tin crystal. Now, later on in the context of nano crystal we will see how some of these can be realized. And how this some of these dislocation free concepts comes spontaneously when you go to the nano scale. Therefore, dislocations are weakening my crystal and if there are no dislocations to cause plasticity and which I mean plasticity plasticity by slip then the crystal can actually be very very strong. Let us mention of you take up a few more points regarding motion of dislocations and plasticity by slip. Plastic deformation by slip occurs by motion of dislocations and they are leaving the crystal. It is very very important that the dislocation actually leaves the crystal. Dislocation may move under an externally applied stress that implies that I can actually move a dislocation by applying an external force or it may so happen there are some internal stresses or spontaneous reasons why dislocation may also move. At the local level shear stresses on the slip plane can only drive dislocation that means normal stresses are not responsible for moving dislocations. The minimum stress required to move a dislocation is called the pearls nebaro stress or sometimes some other context it is also called the pearls stress or the lattice friction stress. This is the bare minimum shear stress I have to apply at the atomic level on the planes on the slip planes. So, that plastic deformation can be initiated. In other words if I apply on the macroscopically of course, I may apply a tensile force, but then using the Schmidt factor I can actually calculate the shear stress on the slip plane which is nothing but a resolved shear stress on the plane. And when this resolved shear stress on the slip plane exceeds this pearl stress or the lattice friction stress or the pearls nebaro stress then the dislocation can move and actually initiate plasticity. Dislocations may also move under the influence of other internal stress fields like for instance those arising from other dislocations, those arising from coherent precipitates, those arising due to phase transformation, those arising due to thermal mismatch at interfaces etcetera. So, internal stresses may also be responsible for moving dislocations, but in any case the pearl stress must be exceeded if slip has to be initiated at the microscopic scale. The value of pearl stress is different for pure edge and pure screw components of a dislocation. And this is very very important because you could have a crystal which is as we shall see later free of edge dislocations, but not free of screw dislocations. The first step in plastic deformation can be considered as a step created when the dislocation moves and leaves the crystal. We will have a graphic of this in the next slide, but essentially the dislocation has to move and leave the crystal. And I am illustrating this concept for the case of an edge dislocation. An edge dislocation can be thought of as an extra half plane in a crystal or equivalently a missing half plane in a crystal as below, but nevertheless it is the edge dislocation itself is neither the extra half plane or the missing half plane. It is the region between, it is the line between the two which I have drawing schematically here in this diagram. And this is the dislocation going into the plane of the board. And when I apply a shear force as shown here and of course here if you consider region of a crystal then this externally applied or the resulting shear stress on the dislocation can cause it to move. And as you can see here the dislocation is actually tending to move in this direction. And with that means at this point of time you can if you label this plane one and plane two of these green atoms below. Then this atom this red plane which is just color differently for easy identification is bonded to these, but at a later step you can see that it is actually bonded between two and three planes the green planes. Therefore, a dislocation is moving under the shear stress and as it moves you can see that it is getting closer to this surface or you might call this surface s. Then finally you can see that it leaves the surface creating a small step. And the value of this step is the burgers vector b which is a measure of the strength of the dislocation. Therefore, if I have to cause plastic deformation then dislocations have to plastic deformation by slip. Then dislocations have to move under the shear stress and finally of course we have to leave the crystal. This is of course the free surface of a crystal if I am talking about a single crystal, but then in a polycrystalline material it could actually go and reside in a grain boundary or may initiate a slip in the neighboring grain. It is important to note that this is the first stage where all the energy of the dislocation has been dissipated and the step has been created because this dislocation is associated with stress fields and these stress fields cost energy to the crystal. But when this step is formed all that energy dissipated and only of course a little extra energy comes because of the formation of the step here which is in the form of a surface energy. Now it is important to note that this is one small step for a dislocation, but a giant leap for plasticity. This is lies at the heart of all plastic deformation this small step which we are considering here. When as I pointed out even this dislocation leaves the crystal a step of height b is created with this all the energy and stress stored in the crystal due to the dislocation is relieved. Therefore, now the dislocation expresses itself as a vector what you might call a vector of strength b on this as a surface of the crystal. If you look at dislocations in infinite crystals then they have a certain stress field and if I am only talking about a single isolated dislocation in an infinite crystal such a dislocation fields no forces. Of course you might want to when you apply external forces that force may translate into shear stresses at the slip plane level and it may feel the shear stresses. But in the absence of external loading such a dislocation would feel no forces. But suppose I consider a dislocation in a semi infinite crystal when I mean a semi infinite crystal I imply that the dislocation is. So, there is material on the left hand side there is no material on the right hand side and the dislocation is positioned at a distance d from the free surface. This implies that this dislocation is not located in an symmetric way with respect to the material there is a certain asymmetry with respect to the right and left. Of course I will assume that the top and bottom are infinite for now and therefore, such a dislocation is asymmetrically position with respect to the material. And such a dislocation actually feels and force towards a free surface and this force is called the image force. And therefore, dislocations in finite crystals feel this can only feel such kind of forces. And this force is called the image force because to calculate this force what is done is that of course you have this material here which is in blue color. And this is my free surface which is drawn here which is marked in b the original free surface. Then to calculate the force that this dislocation feels towards the free surface what I do is that I put an additional material on the right hand side. I put a fictitious dislocation exactly at a distance d in this fictitious material of course of the same material properties as the blue material. And now because this is now a positive edge dislocation this is a negative edge dislocation these positive and negative edge dislocations feel a force. And that is why this is called an image force in other words it is the attractive force of a dislocation towards a free surface. And since I use a concept of an image dislocation to calculate the force I call it the image force. Of course the reason that the dislocation feels a force towards the free surface is that now this is attraction free surface. That means there are no tractions and it is a perfectly free surface. And this is my material and this is my vacuum or air and here is my dislocation. Now the energy of the dislocation would decrease if it is position little closer to the free surface. In other words the dislocation is at position p1 and it moves to position p2. Then the energy of the dislocation is lower because if I look at the stress fields of this dislocation then it is intense closer to the dislocation and it is weaker further away. Here the dislocation is closer to the free surface implies that more of the intense region the dislocations have been relieved that means the energy of the dislocation is going to be lower. And therefore continuously the energy of the dislocation will be lower as I go towards the free surface. And since gradient of an energy is actually the force that means the dislocation will feel a force towards this free surface which is called the image force. Now it is important to note that this image force varies as 1 by d. The image force here can be written as a function of 1 by d where g, b, etcetera and nu are the material properties. G is the shear modulus, nu is the Poisson's ratio and b is the strength of the burgers vector or the modulus of the burgers vector. So image force goes as 1 by d. This is of course the classical formulae. I have illustrated it for the case of an edge dislocation but it is essentially true also for a case of a screw dislocation that it will feel a force towards the free surface. However in the case of an edge dislocation though the force does not change but there are issues related to the cancellation of the tractions when you use an image construction. Nevertheless we now what you call concentrate on the force and notice that when a dislocation is positioned in a semi-infinite crystal it feels a force towards the free surface which is called an image force. Then the value of the image force gets intense and intense because now this goes asymptotically with d as the dislocation is positioned closer to the free surface. Now of course if I want to consider a nano crystal and ask myself this position what will happen if I put this dislocation in a nano crystal. In a nano crystal or a small size crystal it is obvious that there are multiple surfaces which are in proximity with the dislocations. Now suppose this is my dislocation in the body then originally of course there was only one free surface here which I called S but now there is an additional free surface on the left hand side you can see here and this is my glide plane of the dislocation where the dislocation can slip the middle plane then I now have two surfaces in this case S 1 and S 2. This implies I have to now construct two images with respect to these free surface to actually calculate my image force. So, therefore there will be one so this is my positive edge dislocation there will be one negative edge dislocation on this side of the surface at a distance of d and there will be one more image on the left hand side at a distance of of course the same distance here from this distance to the surface if I call this now as x then this will also be x and therefore I have two images and the net force can be thought of as a super position of these two forces arising from these two surfaces which are parallel to this or perpendicular to the slip plane. This force of course can be calculated by super position using the formula which we saw in the previous slide and therefore I can write my image force as g minus g b square by 4.1 minus nu divided by 1 by d minus 1 by l minus d l minus d being this distance the other dislocation is located. So the minus sign comes in between because one dislocation is pulling it towards this surface the other dislocation pulling it to the left and therefore the net will be a what you call super position with the negative sign in between. Now, this force I am talking about here is what you might call the glide component of the image force because now this lies perpendicular to the slip plane, but there are two other surfaces that means this glide force this force which is coming from these images would have a tendency to make the dislocation slip on the slip plane. But now suppose I am talking about reference to these top and bottom surface I got a top surface and I have got a bottom surface. Now obviously these would also constitute give rise to images like I can draw these schematically these images like this one like this and one of course at the same distance here below at a distance like this has to be above. So I can draw images but this would be called what you might call the climb component of the image force let me draw them clearly here. So now I have can think of a dislocation sitting here and this is my distance d and of course this is my h minus d. So I can construct an image here at a distance of. So this is positive this is negative and one more at a distance d. So there are two images coming again, but this image force will make this dislocation actually climb that means it will move perpendicular to its slip plane and you know climb requires actually vacancy diffusion. And therefore it is not possible normally at low temperatures. Therefore to summarize this slide in bulk crystals or in infinite crystal a dislocations field no forces arising automatically that means no force in the absence is a single dislocation in infinite body there are no forces. If it is a single dislocation in a semi infinite body then you have the image force coming from a single image. If it is a finite crystal then you have multiple images I have just showed you and therefore if a dislocation is positioned in a nano crystal then you will have multiple images contributing to the net force on the dislocation. And it is important to note that in the case of a bulk crystal if the dislocation is far away from the free surface the force will be too small and I need not worry about it. While in a case of a nano crystal I cannot ignore the images because they are located because the dislocation is going to be in proximity of the free surfaces. Continuing with the topic of dislocations in nano crystals if I notice the energy of a dislocation it goes as g v square. And that means it is proportional to g v square there are other constants as in this equation like of course the Poisson's ratio this two is coming from something known as the core energy core is the region very close to the dislocation line which is where in the theory of elasticity breaks down wherein there is extra free volume in the crystal. And this gamma 0 is the control volume or the volume of the crystal we are talking about and therefore it goes the energy goes logarithmically with the size of the crystal. Typically in calculation this gamma 0 or r is about taken as about 70 burgers vectors. But in the case of nano crystals it is important to note that this energy formula is not valid anymore because the presence of the dislocation can actually lead to the deformation of the crystal itself because we notice before now when I have a edge dislocation in the crystal I can think of it as an extra half plane. This introduction of extra half plane actually introduces if I am plotting my sigma x x contours if I plot my sigma x x contours then I will notice that there is a compressive region these are iso stress contours above and there is a tensile stress below. And therefore, so this is compressive this is tensile and these stresses would lead to the distortion of the crystal. In of course, if it is a infinite crystal I do not have to worry about these distort the distortion to the crystal because of that. But in the case of a nano crystal it is clear suppose I am not talking about the cylindrical nano crystal as shown below here. And this is now my sigma x x plot which I showed you you can notice that these are the sigma x x butterflies I can see here. So, this is where my extra half plane would be. So, this is where my extra half plane should schematically lie and therefore, these are compressive regions in the crystal and these are tensile regions in the crystal. But the mere presence of the dislocation as you can see is actually distorting the crystal. So, this region you can see here is this deformation you are seeing this is the undeformed crystal the dotted line and the deformed crystal is by the solid line. You can see the crystal is getting deformed the entire shape of the crystal is getting altered by the mere presence of this edge dislocation. Now of course, to draw this whole body this is actually cylindrical in three dimensions. So, let me draw the full geometry for you here on the board. So, you got a cylinder going in the third dimension like this and my dislocation is located here to be consistent with the geometry there I will actually draw the extra half plane below. So, this is my dislocation and now my body is getting distorted in the presence of this dislocation in a cylindrical body. Therefore, if I now compare my infinite body energy case which is given by the formula here or the large body case the energy with the size of the domain actually goes like this. But you can see that there is a relaxation in the and therefore, there I am taking two sizes of crystal. Suppose, I am taking a small crystal then you can see that this is now my energy plot for the free crystal I mean the small size crystal and therefore, for a given diameter say about I am talking about 140 burgers vector there is a reduction in the energy and this reduction in energy is coming of course, from two factors one is the fact that there is lesser amount of material which is being strained. But additionally there is actually also an additionally a domain deformation which implies that there is going to be a relaxation in energy. It is important to note in a infinite crystal the entire half space above the dislocation suppose now I am talking about a infinite crystal. So, this is my dislocation and infinite crystal this is going to be entire half space above this dotted line is going to be compressive and this is going to be tensile. Now, if you compare this with the plot which I have shown you right here you can see that this is not true anymore for a finite crystal and smaller the crystal these effects will start to dominate that actually here even though this region above is tensile here. But, if you go further away you see that these are regions of compression that means this region above is now compressive. That means though the top down symmetry is approximately maintained as in the case for an infinite crystal you see that the sign switches as you go towards a top free surface. And therefore, it the behavior is different from that of a bulk crystal or any infinite crystal where entire half space is purely of one sign of stress. So, to summarize my this slide dislocations in nano crystals have a lower energy as compared to dislocations in infinite crystals because of two reasons. One of course that the smaller amount of material is available to be strain which means there is going to be reduction in energy. But also additionally there is going to be domain deformations or which is going to lead to a relaxation in the strain or the energy of the crystal. Additionally we also notice that infinite crystals there is actually switch in the sign of the stress field as you go away from the dislocation core. We will soon take up an extreme example of these kind of different deformations and interesting effects observed in specific kind of nano crystals as in the coming slides. So, to summarize few of the points we just now seen in and to consider new points in a nano crystal one or more free surface may be in the proximity of the dislocation. As the dislocation is position closer and closer to a free surface then the image force strength increases. At a certain stage the image force may actually exceed the pearl stress. Pearl stress as we defined before is the inherent lattice friction for the motion of dislocation. And this pearl stress force can be calculated from the pearl stress by multiplying it with the burgers vector. And when this happens it implies that the dislocation can spontaneously leave the crystal without application of any external stress. So, this is a very important point that I am making my crystals smaller and smaller. And therefore, there are proximity of free surfaces which are possible. And the crystal now the dislocation will feel image force which will lead the crystal to become spontaneously completely dislocation free without application of any external stresses when my pearls stress is exceeded by the or the pearls force is exceeded by the image force. So, this of course, will happen as I pointed out when the image force actually exceeds the pearls nabara stress or the pearl stress for all dislocations. To make this calculation of course, I have to make sure that the least image force is experienced. Suppose, I have a I just for the now take a symmetrical body like a this is of course, the z direction the crystal goes. And I take a symmetrical body right at the center of the body the dislocation will experience no image force. Because, the two all the phases are equally distant, but suppose I consider a dislocation at a distance b from the center. Then if the force experience because of the image forces now you see the image force strength is going to be very very weak. Because, this surface is not very different in distance from this surface even if this little weak force can be exceeded by the is larger than the pearls stress. Then all dislocation suppose you consider dislocation here obviously, it is closer it will leave if this is the worst case scenario or near worst case because center of dislocation of course, feels no force. If this dislocation can leave the crystal then I assume that all the dislocation would have left the crystal and this dislocation crystal will become spontaneously dislocation free. And for aluminum single crystals and nickels this size is has been found to be of the order of about tens of nanometers where in the crystal is found to become spontaneously dislocation free. So, we will have a few more things to say about this in the coming slides. And but the beautiful thing is that now if it becomes spontaneously dislocation free the nano crystal may approach the theoretical strength of the crystal that means now you can produce extremely strong nano crystal merely because the dislocations are leaving the crystal. And of course, I am assuming now these are a glissile dislocations which are which have the ability to move then I can have a dislocation free crystal which is extremely strong. A few points have to be told about the pearls nabarro stress or the pearls stress this of course, is the original formula I am citing here going back to pearls nabarro. Here you see that it goes exponentially with the width of something known as the width of the dislocation and burgers vector. There are modified formulae and there are atomistic calculations which have superseded this formula, but nevertheless this gives you a nice intuitive physical field. And it is important note that this burgers vector of course, is a result of the crystallography that means even if I did not have a dislocation this is the shortest lattice translation vector of the crystal. The width of the dislocation is a measure of the bonding in the crystal that means you have a covalently bonded crystal then the width will be small and the pearl stress will be large. We have metallic crystal then the it will be more relaxed kind of a crystal ionic crystal again would have a small is expected to have a small width. So, this is a reflection of the bonding this is a reflection of the crystallography and put together they give you the pearl stress. And if the pearl stress is larger then image forces would find it difficult to overcome the pearl stress. So, if you take a mid crystal like aluminum the critical size expected to be of the order of when they when of course, here I am citing not a completely dislocation free crystal, but an edge dislocation free crystal. Because, screw dislocations have a higher pearl stress and additionally they have experience a smaller image force that means you have to go to smaller sizes to actually get rid of screw dislocations. And you can get rid of edge dislocations at a larger size. And for a material like aluminum it is about the edge dislocation it will become spontaneously edge dislocation free at a size over 36 nanometer that means these numbers lying in the nano scale. So, nickel it is even smaller about 20 nanometer for niobium which is a b c c material and therefore, which has an higher pearl stress the sizes you have to go to even smaller sizes to actually make it completely edge dislocation free. And even when niobium at 8 nanometer is expected to become edge dislocation free screw dislocations will persist in the crystal. And silicon of course, covalently bonded diamond cubic structure it is got a very large pearl stress. And there are no experimental evidence so far to actually determine when it becomes completely spontaneously dislocation free. Work done on titanium and palladium thin film show that the grain size in the range of about 6 to nanometer the grains are dislocation free. So, experimentally they are seen in thin films that if I am taking my titanium or palladium thin film the region of about 6 to 9 nanometer then it becomes spontaneously dislocation free. And it also showed that many grains had twins and stacking faults. We will have a little more to say about the stacking fault part. The twins part is it is clear that may be the deformation mechanism is switching from plasticity by slip to plasticity by twinning which also we will talk about a little more in some of the coming slides. And additionally we should note that regarding the stacking fault part that in nano crystals the partials may leave the crystal leaving a stacking fault. That means the entire dislocation may not leave the crystal, but actually the partials may leave the crystal. And I am talking about a partial here I am referring actually to something known as a shockly partial. Because we know energetically the fundamental lattice fraction lattice translation vector in say an aluminum material which is a slip plane 1 1 1 and the burglars vector 1 bar 1 0 lying on this 1 1 1 plane. Then such a dislocation can actually lower its energy by splitting into 2 partials both of which lie on the 1 1 1 plane. And these burglars vectors of these partial are 1 6 1 2 1 type. So, if you have 1 6 1 bar 2 1 bar added to 1 6 2 bar 1 bar 2 bar 1 1. Then I get my fundamental dislocation, but these partials have lower energy because I pointed out the energy of a dislocation goes as the square of the burglars vector. So, this if a in other words suppose I am talking about a dislocation with 2 b strength. Then the square of that will be 4 b square while if I have that splits into 2 b then I get b square plus b square which is 2 b square that means the crystal has lower its energy. Now, regarding splitting of partial now suppose I have a crystal and now I am drawing the slip plane in this crystal. So, this is my slip plane in the crystal and I put a dislocation. So, what might happen now I am talking about a. So, let me now consider this to be the 1 1 1 plane in a FCC material like aluminum. Now, this dislocation instead of remaining like a full dislocation with a burglars vector like would tend to split into partials and the region between the 2 partials would be my stacking fault. Therefore, I have my 1 1 1 plane of the crystal this is my full dislocation given by this burglars vector b equal to half 1 bar 1 0 and this is my partial of the type in which I call the shockly partial which is of the type 1 2 1 to make sure it lies on the slip plane I can call it 1 bar 2 1 bar. Now, these partials will also be attracted towards the free surfaces this to this surface and this to this surface. So, in then what might happen is that you will actually land up in the end configuration as a crystal on the right hand side. So, of course these when these dislocations leave the crystal these partials you will have a step created on the surface which has a strength of a partial and not of the full dislocation, but I will be left with a stacking fault. So, I will be left with a stacking fault in the crystal and this tells me and stacking fault as you know is a fault in the stacking sequence suppose you talk about an FCC aluminum then you know that the packing sequence goes as a b c a b c a b c and suppose the packing sequence is change such that the you it becomes a b c a c a b c then such a thing is called a stacking fault. Therefore, you can see that in the case of nano crystals not only the full dislocation will leave the crystal and make it completely dislocation free, but additionally there is a possibility that the partial may leave the crystal and therefore, you are left with a stacking fault in the material and this is been observed actually experimentally in titanium and palladium crystals by Professor Anganathan and his co-workers and published in script material 2001. So, in addition of course it is obvious based on the argument just have made that suppose even if you have a poly crystalline material or a single crystal material the surface regions of the poly crystal may become dislocation free. So, close to the surface region now suppose I have a crystal and I have my surface region suppose this is now my crystal there will be a small region close to the surface where in the pearl stress is exceeded. So, this crystal itself could be of the order of microns or 100s of microns, but this small region which is in the nano scale can actually become spontaneously dislocation free because of the presence of image forces. So, there are importance. So, if you are working in a bulk crystal there is some importance of image forces, but this region is very small compared to the whole crystal we tend to ignore this region, but definitely in the case of nano crystals it is very clear that I cannot ignore image forces and beautifully so that I now can recover my theoretical shear strength because dislocations may spontaneously leave the crystal making it completely dislocation free and giving me the theoretical shear strength. We had mentioned before that when you are talking about nano crystals then I cannot ignore domain deformations. Now, we will take up an extreme example of domain deformation and this work theoretically goes back to the work of Eshelby in 1950s where we talked about dislocation and thin plate and here domain deformations sort of cannot be ignored and they become a significant part of the entire structure. And here if you look at the thickness of the plate it is of the order of about 20 burgers vectors. So, this thickness is of the order of 20 b thinner plates will bend more thicker plates will bend less in the presence of an edge dislocation. It is important to note that originally of course Eshelby also study dislocations in thin cylinders and if in and what the kind of dislocation he talked about was actually screw dislocation in thin cylinders and these cylinders will actually tend to twist in the presence of a screw dislocation. So, now there are two type of domain deformations possible I can talk about a dislocation in a plate like this the one has drawn or I can talk about a dislocation in a cylinder and now I am talking about a screw dislocation in the middle of a cylinder. Now, in the presence of this screw dislocation actually the domain will actually twist the cylinder will tend to twist. Therefore, in nano crystals there are important effects directly coming because of the presence of the dislocations. Now, the important thing of course note is that the energy of the plate is going to be constantly relaxed or there will be a reduction in the energy because of the presence of the dislocation. Additionally you can see the stress values are also relaxed. We will talk about certain interesting effects which also come about because of the presence of these dislocations. Now, if you talk about a dislocation in a bulk crystal we saw that the energy actually decreases as a dislocation goes towards the free surface. This implies that the dislocation is going to be unstable and now when I am talking about unstable I am talking about mechanical instability because dislocations are anyhow thermodynamically unstable defects. The reason dislocations are thermodynamically unstable is because we know that the criteria for stability of in condensed matter system at constant temperature and pressure is the Gibbs free energy. Now, putting a dislocation actually costs it energy and therefore, the enthalpy of the system is going to increase. But, nevertheless putting a dislocation gives the crystal a configurational benefit. Even though this term is highly positive this term being negative and there is going to be a entropic benefit by putting a dislocation. We would assume that there is going to be a temperature T such that this term can be offset by this term and the overall Gibbs free energy is negative in the present delta G is negative in the presence of an dislocation. But, unfortunately this temperature happens to be above the melting point of all known crystals and therefore, dislocations are thermodynamically unstable defects. But, the other point we are talking about here is the mechanical stability which means that the energy of the dislocation actually decreases when we go towards the free surface. That means suppose I take a crystal and my dislocation is here the energy constant decreases as we go towards the free surface and therefore, they are mechanically unstable. Of course, here in this stability I am ignoring the fact that there is a pulse oscillation sitting on top of it. Because, I know that the dislocation sits in a pulse energy value and if I ignore that then I am talking about the including the pulse value of course, it is metastable otherwise it is mechanically unstable. But it is interesting to note that dislocations can actually become thin in these thin cylinders and thin plates they can actually be stable. So, I shall be at shown in 1950s that a screw dislocation can be stable in a cylinder. It is a very interesting point mechanically stable in a thin cylinder and now of recent time we have also shown that dislocations can be mechanically stable in plates like this. That means that irrespective of a force have a dislocation in the center of this plate. I can put it here or here or here and there is a region of stability within the crystal. Of course, if it is very close to the free surface it will again be mechanically unstable but there could be a region of stability of the dislocation. In other words a dislocation position here would feel a force towards the center and not towards the free surface. So, that is an interesting point. So, I have a dislocation in a normal crystal suppose I draw normal crystal like this your dislocation here will feel a force towards the free surface. But here if I put a dislocation somewhere here then it will actually feel a force towards the center of the domain rather than or the center mid plane of the domain rather than away towards the free surface. And of course, this you have to demarcate a region where this is stable it is not that entire crystal will be stable. But nevertheless the important point to note is that these special nano crystals, these special nano crystals that means having a specific geometry dislocations can be mechanically stable. So, this is very interesting that in bulk crystals dislocations are always mechanically unstable or force you want to very technically correct their metastable. But here in these crystals you can actually see that they can be mechanically stable in thin plates and thin cylinders. Now, an additional interesting thing is that normally as we pointed out in the previous case that when you have a dislocation then it feels an image force towards the free surface. That means that suppose I take a crystal and plot the energy of the crystal as the dislocation is positioned towards the free surface the energy would decrease as shown by curve like this. So, it is decreasing always and finally of course when it leaves the crystal it creates a step. Now, if you talk about the dislocation in a thin cylinder like this and I am positioning my dislocation for instance here. Then I move it to a next position here then here and or in the left or in the center of the domain and I track my energy of the crystal now. Then it is seen that the energy practically does not change that is this material now is some kind of an analog of a nuclear is some kind it exists in a state of a neutral equilibrium we all know that if we take a ball a ball and put it on a plane then this ball is in a state of neutral equilibrium. Now, exactly equivalent to that this dislocation now is residing in a state of neutral equilibrium with respect to its position in a thin plate like this. So, this is a nano crystalline plate which has certain dimensions as we have seen here before that the length is of the order of 100 burgers vector and the thickness of the 20 burgers vector in such a thin plate not only that the dislocation bends the domain not only there is a slight relaxation with respect to the energy in a bulk crystal because of this bending additional in addition to the fact that the amount of material strain is small. But more interestingly so that the crystal the dislocation now resides in a state of neutral equilibrium with respect to its configurations and these has been termed as neutral equilibrium material structures and these are very special structures which were discovered very recently. Therefore, you can see that there are important effects coming because of the presence of dislocations in nano crystals that means in bulk crystals dislocations of certain characteristics when you go down to nano crystals there are additional effects which come in and which have to be taken into account a few of this we have already pointed out like the case of the image force the case of the domain bending and very interesting case of this for a given specific geometry the case of what you might call neutral equilibrium. And even more interestingly suppose I now have to move a crystal move the dislocation along the slip plane which is now my central plane this central plane I move the dislocation then I can introduce a concept known as reversible plastic deformation due to elasticity because now this resides in a state of neutral equilibrium. Therefore, if I move my dislocation from point to point so I am in a thin crystal here and the thickness is of the order of 20 Burgers vectors then my dislocation if it is position in different places then the energy does not change but you know the bend crystal bends differently that means because of the presence of the dislocation you have got now plastic deformation because what is plastic deformation plastic deformation is deformation in the absence of externally applied forces. Here even in the absence of any externally applied forces you see that I am exaggerating here the picture the crystal actually bends because of the dislocation which I just showed you. And this imply this bending is permanent I do not have to apply any external stresses this is merely because of the presence of the dislocation it is permanent therefore it can be called plastic deformation. And this plastic deformation changes differently as I move the dislocation suppose I put it right in the middle of the plate then it will bend a little differently compared to the case where it is position towards a free surface. Therefore, it is exploring various configuration and this is permanent deformation but since it is existing without change in energy therefore you can call it reversible that means the two states are equivalent in terms of energy. So, it is strictly reversible and it is plastic and this essential deformation is coming from the elastic stress field dislocation. So, you can come up with strange effects like as we know that reversible and plastic do not go hand in hand there in fact oxymorons plastic and elastic or oxymorons but you get very interesting effects like reversible plastic deformation due to elasticity coming because of I am talking about dislocations in thin plates. So, dislocations may weaken the crystal but then when I am worried when I am talking about dislocations in nano crystals there are interesting effects which come because of the presence of dislocations in nano crystals. So, we saw that if dislocations can weaken a crystal then we can ask the obvious question how can I strengthen the crystal and one mechanism we had already seen that if you remove the dislocations then the crystal becomes strong. But this is a not practical approach as we saw that we can only make crystals as thin as about microns in diameter which are actually can be made dislocation free and additionally dislocations can always nucleate during loading and therefore this is not a practical viable solution to strengthen a crystal. There are other mechanisms which can actually use which can be used actually to increase the strength of dislocation and this arises from the fact that any impediment to the motion of dislocation increases the strength of the material. So, if I can stop these dislocations from moving freely they I make glissile dislocations less glissile then I would actually increase the strength of the crystal. So, what are the mechanisms I have got for increasing the strength of the crystal few of them interesting here solid solution strengthening that means I input solute atoms and this can be substitutional or interstitial and this will actually increase in some sense the inherent resistance to the motion of dislocations leading to the increase in strength of the material. So, more I can add the volume fraction till of course the solubility limit is exceeded then I would increase my strength of the material. Further we have already seen that precipitation hardening and dispersoid strengthening can actually increases strength of the crystal and we had noted that dispersoid strengthening can also be used at high temperatures to increases strength of the material. For a dislocation mechanism is another mechanism in which we can increase the strength of the material this essentially implies if I increase my dislocation density and which automatically happens when you do plastic deformation that implies the during plastic deformation the dislocation is actually multiplying and therefore the density is increasing. But surprisingly that when there is it is like if there is one car it can easily drive through the road, but the many cars you get a traffic jam. Similarly if you have many dislocations then one dislocation is actually moving by cutting across the other dislocations present in the slip plane and this would give rise to an hardening effect which is coming from dislocation density. And finally the grain boundary hardening which is a very important effect and in this context we will take up a little more detail when we talk about the hall patch relation and more importantly the break down of the hall patch relation especially in nano crystals. So, the important strengthening mechanisms are solid solution strengthening wherein I can use an interstitial solute like carbon or a substitutional solid like nickel in aluminum or nickel in copper. I can do precipitate or dispersoid strengthening wherein largest precipitates especially finally distributed can give rise to sort of impede the motion of a dislocation giving extra strengthening. I can use forest dislocation strengthening that means increase the dislocation density or I can make a fine grain material because grain boundary is a region when finally dislocation ceases to exist. Of course, it when a dislocation reaches a grain boundary either it may be absorbed by the grain boundary or may become a dislocation associated with the grain boundary or it may initiate slip in the next grain adjoining grain in which case it continues to slip through the crystal or through the poly crystal. An alternate mechanism which also can give plastic deformation which is not that common as normal slip. Suppose, I take a rod of aluminum and bend it normally I am expecting that the plasticity I am observing at room temperature the permanent deformation I am getting is because of slip. But then suppose I deform iron B C C iron at low temperatures then their dislocations by slip it becomes difficult and actually this twinning which predominates. Twinning typically occurs in low stacking fault materials like copper and brass. Suppose, I look at a microstructure of copper or brass I will actually observe lot of twins and many of this can easily even be observed in an optical microscope as straight lines because twins are nearly or atomically sharp. In high stacking fault energy middle aluminum twinning is actually difficult the reason being that in a if you look at a twin and I am now talking about a mirror twin then the second nearest neighbors are affected. And if the second nearest the first nearest neighbors are the same in either side of the twin boundary but the second nearest neighbors are affected. And so in the case of a stacking fault and therefore, the stacking fault energy is low then you expect easy twinning. Twinning can occur either during annealing in which case they are called annealing twins or they can occur during deformation and in fact they are can be reason for actually giving rise to plastic deformation in which case they are called deformation twins. Twinning can become an important mechanism of plastic deformation and can give additional ductility to materials where in slip may be limited. And one such example I showed you was the case of I told you about is the B C C metals like iron at low temperatures. And in fact there have been examples like the twip steel which is called trans twinning induced plasticity where twinning steels have been developed keeping the precisely the twinning aspect in the focus. That means not only slip that means dislocation activity can give you plastic deformation, but twinning is also an important mechanism by which you can get actually some ductility and some plasticity. Though this is usually much smaller compared to dislocation activity for suppose I am I can get 20 percent ductility because of slip then typically the twinning value would be 1 to 2 percent which is much smaller than that by what they might call by slip. Process parameters like temperature and strain rate and material parameters like stacking for energy grain size etcetera determine the mechanism operative for plastic deformation. And I am here now only contrasting two mechanisms I am taking I am not taking a gamete of all the mechanisms which I stated before I am only talking between twinning and slip. Then if I have to look at my process parameters like temperature strain rate and I have to look at material parameters like stacking for energy grain size to determine which mechanism is going to be predominant in a given material. For a given strain rate at higher temperature stood slip would be favorable as compared to twinning and the reason is that slip is thermally activated. In other words suppose I do plastic deformation and slip is the mechanism I am talking about at higher temperature my dislocation actually can jump from one pulse value to another by thermal activation that means I need to apply lesser stress to cause this plastic deformation. And therefore slip would be preferred at higher temperatures on increasing the strain rate which means that there is less time for stress and temperature and thermal activation. Then the stress required for both slipping slip and twinning increase, but the transition temperature from slip to twinning is shifted to higher temperatures. So, we will see this effect as a graph below. And here I am talking about strain rate variations of about 10 orders of magnitude that means the strain rate is not merely from 1 per second to 2 per second it is orders of magnitude variations strain rate if I have to change. So, that I can actually see a transition from slip to twinning and the grain sizes here I am talking about variation in grain size also in addition to the temperature and strain rate. And here of course there is a case of a single crystal there is a large grain size material which is of the order of about 10 microns and there is a small grain size material which is of the order of about also both this is 100s of microns and this is 10s of microns. That means I am now in a single diagram I am trying to study the deformation mechanism the competing deformation mechanism slip and twinning with as a function of the temperature of the strain rate. Now it is seen clearly from this graph that for a given strain rate as we reduce the grain size slip becomes preferred. But before that let us understand for a single grain size let me take a grain size like g s 1 and I am tracking this curve like this here. Now what is seen that below this curve the region below this curve is a region where I would observe twinning and the region above this curve all this region of space above this curve I would actually observe slip. In other words suppose I am operating as a certain strain rate here which could be for instance 10 power 2 or 10 power minus 2 and I actually heat the crystal. Then I would observe that when I cross this line here that the mechanism changes from twinning to slip that means at higher temperatures you observe slip at lower temperatures you observe twinning. Now for a given strain rate suppose I am at a given temperature and I increase my strain rate which means that I am actually giving lesser time for thermal activation to take place then you observe that the mechanism changes from slip to twinning. So for other grain sizes so if you increase the strain rate then you cross this line and that means you actually go from slip to twinning as you increase the strain rate. Now so we understood two effects the effect of strain rate the effect of temperature for a given grain size that means I increase my strain rate I switch from slip to twinning I increase my temperature I switch from twinning to slip. So this is the basic mechanism but what happens when I reduce my grain size reduction in grain size which is of course in the opposite direction of this arrow this is increasing grain size this arrow shows then this is increasing grain size. So if I decrease my grain size this is in some sense like decreasing increasing temperature. So decreasing in grain size is some in some viewing point we can actually think of it as increasing temperature. Because now suppose I take a grain size 1 2 material which could be of the order of hundreds of microns. So I am sitting here now the obviously the mechanism at point p is going to be slip because it is lying above this curve. So I am specifically tracking one curve here which is now my this curve which of course I will draw little wiggly curve around this. So I am talking about grain size 2 and with respect to grain size 2 I am talking about a point p. So all points lying above this g s 2 line are actually points where in slip would take place. So at point p you expect slip to take place for this given strain rate and temperature. Now similarly if I am talking about grain size 1 on the other hand this point say I call it m point m for grain size 1 would show slip because it lies above grain size 1 curve. In other words if for grain size 2 point m is going to exhibit twinning because it lies below the curve. But for grain size 1 point m is actually going to show that means that processing conditions like temperature and strain rate are actually going to if for grain size 1 give rise to actually slip. In other words when I am decreasing my grain size it is actually as if increasing my temperature that means that means the mechanism is switching from twinning to slip. So point m would have been twinning for grain size 2 but is actually slip for grain size 1 which means that when I decrease my grain size the mechanism switches from twinning to slip. That means that it is a function of as I just pointed out which mechanism the system chooses is a function of the process parameters which we just now concerned the curve. But additionally also of the material parameters like grain size and stacking fault energy obviously the stacking fault energy increases then the region of twinning which is shown here the zone of twinning would increase that means I have to go to twinning would be less preferred and this region would decrease and therefore I would usually observe slip in materials with high stacking fault energy this region would go down this curves would go down corresponding. So to summarize it is seen for a given strain rate as we reduce the grain size slip becomes preferred even at lower temperature slip that means slip occupies more region of the temperature strain rate space. So to summarize these slides that we have 2 important mechanisms of plastic deformation twinning and slip wherever because of low temperature or high strain rates slip is suppressed then we have twinning typically gives lower plasticity then slip but then there are special steels like the flip steels which have been developed to exploit the twinning aspect of plasticity. We have to note that it is the combination of process parameters like temperature and strain rate and material parameters like stacking fault energy and grain size which tell you for a given strain rate and temperature what kind of a mechanism would you observe. The next we will talk about some of these aspects of twinning later on in when you take up some examples but we will talk of some more basics about some phenomena like super plasticity and also about creep before we proceed further. We had seen the importance of twinning in the context of plastic deformation especially at low temperatures and high strain rates. Before we leave this topic of twinning let us take up a few additional points. Twinning detwining is the main mechanism of deformation in shape memory alloys and as you know shape memory alloys nowadays have been investigated for very important applications in diverse fields including biomedical applications. We also have noted already that twinning typically does not contribute to large deformations but twinning can lead to large orientation changes and this is of particular help in the formation of deformation structures which is observed when we deform HCP materials like titanium, magnesium, etcetera. So to summarize this slide though twinning typically does not give large deformations it may help by giving large rotations which plays an important role in the formation of texture of HCP materials and additionally in important materials of not only good academic interest but also of in terms of applications they play an important role in the deformation of shape memory alloys.