[SOUND] This is Module 17 of Mechanics of Materials Part II. Today's learning outcome is to solve for the maximum shear stress again for elastic torsion, but this time for a straight cylindrical shaft that is non-prismatic. And what I mean by non-prismatic, a prismatic beam is a straight engineering member like we solved in the last two modules with the same cross-section throughout its length. The non-prismatic, it doesn't have to have the same cross-section throughout its length, and so let's look at a real world example again. Here's the simple model of a, we can have a simple model of a torsion bar for a track vehicle. You can see as the suspension goes up and down, there's some torque on this bar up here and maybe we can model it like as shown. And the actual example might be different materials and different torques. But this is the type of process you'd go through for analyzing this situation. And so I've got a Monel alloy, I've given you the G steel out here, I've given some torques that are being applied, and the last five foot of the section from A to B is actually hollow. And so we'll go ahead with that as our model, and this is our worksheet. We want to determine the maximum sheer stress in each section, and then, next module we'll look at the angle of twist at the end of D with respect to A. And so how would I begin? And what you should say is, well since I'm going to need to have the torques for the elastic torsion formula, I better find the torque in each of those sections if I want to find the maximum shear stress in each of those sections. So let's start with a coordinate system here. And so I'll call this x, y, and z coming out. And I'll first start with section, a free body diagram for section C D, and so from D to C, or C to D, we have our cylinder. I'm going to go ahead and cut it. I have my applied torque out here at D which is five foot kips in the negative z direction. I've got my resisting torque, which is going to be in the opposite direction. So it's going to be positive T sub CD, and I can then sum forces, or excuse me sum moments about the z direction and set it equal to zero. By the way I could've assumed Tcd to be in the opposite direction, it would have come out negative. So I would have known the direction anyway, we've seen that in all my previous courses. And so if I sum moments now, I've got minus five foot kips, and the foot kips are on the diagram so I'm not going to write it again, plus Tcd equals zero. Or Tcd, the torque in section CD, is equal to five foot kips, and that's going to be this direction or in the positive K direction. So it's five foot kips k. Okay, let's do the same thing for section, from B to C, and so what I would like you to first do is draw the free body diagram. And now let's do it together. And so I've got. My section. I'm going to cut between B and C here. And so at the end I have my 5 foot kip torque. Our moment applied, and then at point C here, this is point D, this is point C, and we're cutting between B and C, so B would be out here. So at point C, we have 15 foot kips in the positive z direction. And then I'm going to just go ahead and assume Tbc to be in the positive C direction. So some moments about z again, and find the torque from B to C. And if you do that, you should come up with TBC = -10, so it's going to be actually in the negative K direction foot kips, and we'll do finally the same for section AB. Actually, when I cut my member out here in the hallow section between A and B, I actually don't have any other torques being applied to the left than we've seen before, so we should come up with the same torque in section AB that was in section BC. And so 5 foot kips, and then I have 15 foot kips here. And finally, I have my torque from A to B. And recall now we are cutting where we have a little bit of the hollow area in here, but that's not going to change the torque itself, and so some moments and find out what the torque is. If you do that you get, torque from A to B is again -10 foot kips in the k direction. It's in the minus k direction, 10 foot kips in the negative k direction. Okay, so we now have our torque in each of the sections we want to find the maximum shear in each section, I'll do section CD, and then you can do the other one. So for section Tcd, I know it's elastic torsion, so I'm going to use the elastic torsion formula, so I've got tau = T rho / J. And so for, Tau cd max, the maximum torque in this section from c to d. Well we found out what the torque is, that's 5 foot kips, and we're going to change off everything from feet to inches. So I am going to have 12 inches per foot to convert that foot into inches, and then I've got my row. My row at the outer surface for the max shear stress in section CD is going to be half of 5 inches or 2.5 inches. And then we're going to divide by J. J is pi / two * the radius. For this section, the radius is 2.5 inches and it is to the fourth. And so we end up with, if you multiply that out, we get 2.44 and then it's four sixths, I'll just round off to three significant figures. And it's going to be kips per square inch. One thing we should check is to make sure, since we're assuming elastic torsion, we're using the elastic torsion formula, we need to make sure that we stayed in the elastic range. If you look for, this is Monel alloy. If you look up the Monel alloy its maximum, or excuse me, its elastic shear stress maximum is about 30 ksi. So tao elastic max shear for Monel is approximately 30 ksi, which is much, much greater than 2.44 ksi, which we're seeing so certainly the elastic formula applies in this case, in this section. Okay, I'd like you to do the same thing now for sections BC and AB, and then come on back and see how you did. And for section BC we had the torque 10, we're looking at the maximum shear stress, we're just taking the magnitude of the torque, and we convert the foot units to inches. We have our radius, which is four. This is a solid cylinder, so its J is pi over the radius to the fourth divided by 2 pi, times by the radius to the fourth divided by 2. Or this is our maximum shear stress in section BC. Similarly, for section AB, the only change here is that now we have J for a hollow circular cross-section, so I'm going to have to subtract out the inner part that's hollow. And I end up with this maximum shear stress in this case. And these are both for the steel section. If you check the maximum elastic shear stress for steel, you see that they're below that elastic yield limit or proportional limit, and so the shearing proportional limit, so the elastic torsion formula, again, applies for these two sections. And so we have now found the maximum shear stress of each section. We'll come back next module and determine the angle of twist with respect to D with respect to the A end. And I'll see you then. [SOUND]