[MUSIC] So from today's lecture, we are going to talk about the one dimensional defect of dislocations, okay. So, from the study of this section we will be able to answer these few questions. First, we will be able to understand why metals and metal alloys can go through a lot of plastic deformation while keeping their crystal structure. And secondly, we will be able to understand why plastic deformation occurs at stresses or forces. Which are much lower than the theoretical strength of a perfect crystal calculated by the chemical bonding strength of the crystal. And thirdly we will be able to know why and how the strengths of metals. Can be changed to a very large degree by without changing their chemical composition. So here is your deform metal, where if you look at the metal under optical microscope, you will see a lot of traces of deformation. Essentially these deformation traces are corresponding to the one dimensional defects of dislocations. Okay, so to understand the dislocations, we'll need to study some basics of mechanical properties. And we will go in to define stress and strengths, supposed we have a cylindrical sample of this. And which have a cross-sectional area of A0 and we load the sample by force F on the upper and lower surface of the sample. And then we just defined this nominal tensile stress as the force divided by the cross sectional area of A0. And in response to the force, we kind of elongate your sample and correspondingly. We define the nominal tensile strain as the displacement or shape change in this tensile direction by its initial length l0. So essentially, your length change U divided by your initial total length of your sample, l 0 would be your nominal tensile strength. And relevant with the nominal tensile strain there will be a nominal lateral strain. So that you are elongating your sample in the pulling direction, but it contracts in the lateral direction. And another important parameter for the deformation is so called Poisson's ratio, which is the minus lateral strain divided by tensile strength. There is a minus sign here, because you will have positive tensile strength but negative lateral strength. The negative sign here or the minus sign here is to make your Poisson's ratio a positive number. A second loading state of your material is this shear loading, suppose you have a piece of material. And you apply some force in parallel to the upper surface and lower surface of your material. And the shear stress is calculated by the force, again, divided by the cross sectional area of the force that acting on this area. So it's a shear stress and in response to this shear stress, you will have a shear strain. So you're kind of doing this deformation and the shear strain is quantified by the displacement. In the direction of the force divided by the distance or the initial dimension of your sample vertical to your loaded plane. And the engineering shear strain is defined as a set by W divided by the initial length here. Which is essentially the tan theta of this angle and you will know that for small strains tan theta can be approximated by theta. So these are the definitions of stress and strengths. In the most general three dimensional case you will have six components of normal and shear stresses. So these are your three components of normal stresses, your three components of shear stresses. In response to the six stress values, you will have a correspondingly six values of your strengths. So these are your normal strains, these are your shear strengths. And you will have a 6 by 6 complex matrix correlating the stress matrix and your strain matrix. Unfortunately, although you will have in principle 6 by 6, 36 independent parameters on this matrix. With the increase in crystal symmetry, the number of independent parameters in this matrix will reduce. For example, if you have an isotropic material, you will have only two independent parameters in all of these 36 parameters. And if you have a cubic crystal lattice, you will only have three independent parameters out of these 36 parameters. Okay, now we will be able to study how the plastic deformation takes place in metals and here is your deformed metal. So apparently the plastic deformation or shear deformation will take place. Across certain crystallographic planes and certain crystallographic orientations. And this from the atomic view it's like when you apply this shear stress to two rows of items. These items will just shear relative to each other, making these shear traces. So this is a side view of your deformation traces, and here is your front view of your deformation traces. And now, with the knowledge on chemical bonding and crystal structure. We will be able to calculate the theoretical strength of this shear deformation. Okay, so here is a quick review on what you have learned on the interatomic potential. So, here shows interatomic potential U as a function of interatomic spacing. And from the chapters on crystal structure and interatomic chemical bonding. You know that there exists an existing interatomic spacing so that your interatomic potential is a minimum. And this will corresponds to a zero forced position so that when you add this equilibrium interatomic spacing, the interatomic force is zero. And above this interatomic spacing, the items tend to attract each other. And below this spacing, the interatomic interaction will kind of repulse each other. So, if you work in the small region near the equilibrium position, this force as a function of interatomic spacing curve. Will be like some linear behavior right, and this linear region comes to your elastic deformation. Where your normal stress is linearly proportional to a normal strain, by this Young's modulus. And similarly, your normal or your shear stress is linearly proportional to your shear strain, by this shear modulus. And the slope of this curve actually corresponds to your Young's modulus and shear modulus. Okay, with this knowledge we will be able to calculate the theoretical strength of a crystal. For example, if you have two rows of lattice items, and you have a shear stress like this. So, theoretically, the way to make the plastic deformation or shear deformation happen is that. You break the chemical bonds in between the sheared part on the top and the lower part on the bottom, right? So essentially you just need to sum up all the chemical bonds here, where, which you need to break during this deformation. So essentially, whereas there's force as a function of interatomic spacing curve. For simplicity, you can most easily zoom that the force high as a sinusoidal shape, so that the force can be expressed by this equation. So there is a maximum force here, and here is your interatomic spacing A. And you have a parameter x here which describes how much you deform right? And by definition their shear strain is your x divided by a right, and shear modulus. Is somewhere you can predict from your chemical bonding strengths, right. So from this equation you will be able to calculate the theoretical strength of your crystal, right? So, in this framework, you will calculate the theoretical strengths or the maximum shear stress. On the order of your shear modulus divided by 2 pi or on the order of your shear modulus divided by 10, so that is your theoretical strength. Shear modulus is something you can measure in the laboratory, right and then you can ask me the theoretical strengths of your crystal. And if more realistic force is used, here we just use the sinusoidal form. If more realistic force curve is used, the maximum shear stress or the theoretical strength of your crystal is still on the order of G over 30, okay. So, this comes to a contradictory, okay, so experimental measured strength of metallic crystals. Is only on the order of 10 to the 4 minus 4 or 10 to the minus 8 shear modulus. Recall that theoretically we should have a shear strength of a crystal on the order of G over 10, but we only have this tiny, tiny strength of your crystal. So some has to explain this, so some has to explain why the theoretical prediction. Is just a several orders of magnitude higher than the real material strength. Then to explain this discrepancy in 1934, three talented scientists proposed this idea of one dimensional defects of dislocations. So, with the presence of dislocations you don't have to break all the chemical bonds at the same time. And you only need to break a limited number of chemical bonds at a certain time and here shows what they proposed. They proposed that some one dimensional defects like this have atomic plane inserted into this crystal lattice. So when you do this shear deformation, only this half extra plane of items will move across your crystal. Instead of you're breaking all the chemical bonds here and making the plastic deformation to take place. So here is the schematic of the geometry of dislocations, and here shows an example illustration how the dislocation moves actually. So it's like a worm moving on the ground, so you have one extra row of items here. And with shear stress, and this extra row of items will move from your left hand side gradually to your right hand side. So at the same time, you only need to break the chemical bonds of this. One row of or one column of items instead of you breaking all the chemical bonds here. So here shows a kind of animation of how this location moves, so when you apply a shear stress enough for the motion of dislocation. Which is called the critical resolve the shear stress or CRSS, the stress is large enough to move dislocations. Then the dislocations will move across the crystal right, like this and eventually. Your dislocation will move to the surface of your crystal causing a surface step here. Okay, here shows the animation where you apply a shear stress and essentially instead of breaking all the chemical bonds here. This half extra play of items just to move from your right hand side to your left hand side. Okay, so that's the reason why you have a material actual crystal material strength. Much lower than the theoretical strength calculated from your chemical bonding. And essentially the critical result of shear stress or CRSS is much smaller than theoretical strengths of your crystal. So, today we have talked about the concept of dislocations. Thank you very much. [MUSIC]
