In this lesson I'd like to turn our attention to the calculations that are involved in looking at a single crystal of a body-centered cubic material. First of all, we know that slip occurs on close packed planes and in closed packed directions. And in the first illustration, what I have up there are two dihedral planes and those planes are the (110) planes. And on those planes are the <111> directions, which are the directions of slip in the body-centered cubic. So we have two sets of planes here. We have two sets of planes that are coming in the other direction, and again, on each of those planes we have <111> directions. And here's our last grouping. So when we look at this, we now have a total of six planes that are of the type (110). And each of the planes contains two <111> directions. So consequently, what we would say is we have a 12 slip systems that is 6 planes and two directions on each plane. If we go back to the FCC, in the FCC we have also 12 slip systems but in that case, what we have are the 4 phases to the tetrahedron. So we have four planes, and then for each of those planes we have the three directions. So that's going to give us a total of 12 slip systems. So the number of slip systems here are similar. Now, the other thing that becomes important, so there are our 6, our 2 directions, and this then represents the lattice points that are located on those (110) planes. Now, I want you to go back and think about the close pack geometry associated with the FCC. Remember they touch. And they're touching everywhere, so we have the highest packing density. In the case of BCC, we have a packing density which is less than the associated packing density in the FCC system. And consequently, that's going to give us some slightly different characteristics with respect to the body-centered cubing material. Now we're going to go ahead and do a calculation in the BCC system. We're given, this material has a stress axis, in which the crystal is lined up in the [010] direction. So that's the direction of our stress with respect to the crystal. And we know that slip occurs on a particular direction, and in this case we're examining the [1-bar 11] direction. And it's occurring on a particular plane, namely the (110) plane. What we would like to do, is to calculate the stress necessary to initiate slip if we recognize that the critical resolved shear stress in this particular material is 30 MPa. Now, what I would like to draw your attention to before we get further along in this calculation, is the fact that the magnitude of the critical resolve sheer stress in the BCC structure is much higher than it was in the case of the FCC. So in BCC, the characteristic is we're going to have a higher critical resolve sheer stress. The question then becomes, why is that? Well, in part we can ascribe some of that to the packing density associated with the BCC structure as compared to the FCC structure. Okay, so now let's continue on with our calculation. So, we'll go back and we'll look at the equation for the critical resolve sheer stress. Now we're going to determine what the relationship between the slip direction and the associated stress axis are. And we find that that cosine of theta is equal to 1 over the square root of 3. Now when we calculate with the value of the cosine of phi is, again, we look at the dot products of the slip plane, and we look at which is converted into a slip direction. Remember the plane was the (110) and the normal to the plane is the (110). So, consequently we can determine what the orientation relationship is between the slip plane normal and the applied stress. And we find that by taking the dot product of those two vectors and dividing by their magnitudes. And that turns out to be a value of 1 over the square root of 2. So we put this in our equation, and now what we see is that we get a value for the applied stress that's necessary in order to initiate slip in this particular material. Namely a value of 73.5 MPa. So, when we do the calculation here, you should go back and review what the results were, associated with the applied stressed necessary to initiate slip in the FCC. And you're going to see that it is much larger value. Now one of the things that particularly interesting about the body-centered cubic crystals is that you can have, because of the fact that the slip plane is not as close packed and it's not the closest pack. It is the closest pack plane in the BCC, but it is not as closely packed as it is in the FCC. It turns out that there are a total of three different types of planes where you can have slip in body-centered cubic. So for example, we'eve been describing slip in the (110) plane on the (110) plane in the <111> direction. And we can also have slip on the (1 2-bar 1) plane in that same direction and on the (2 3-bar 1) plane in the same direction. So, we had the opportunity of having more slip systems operative. Now, it turns out the fact that we have additional slip systems that are possible, leads to a very important behavior with respect to body-centered cubic materials. This is true of all body-centered cubic materials. That there exists in BCC materials, a phenomenon referred to as the ductile-to-brittle transition temperature. That means the behavior above a certain temperature, the material winds up being ductile and the material deforms easily. At lower temperatures, that turns out that that's not the case. Now the question then becomes, why and how can we relate this to these different slip systems that we have in BCC. Well first of all, what we know about the BCC system is that the packing density is much lower in the BCC than in the FCC. Which means that we can get some thermal assistance associated with the deformation process as we go to higher and higher temperatures. So, more slip systems become operative, and consequently, the material can behave in a much more ductile way. So, what we'll find is, at high temperatures, with regard to all body-centered cubic materials, we're going to see a ductile behavior. And at some lower temperature, we're going to see a brittle behavior. And that's ultimately going to lead to some issues when we start using materials such as steels when the temperature can become very low. For example, in the Arctic where the steel is used in construction, the engineer needs to be aware of what the actual temperature may be in the environment. And what the ductile to brittle transition temperature is, in the particular steel that's being used. Thank you.