[MUSIC] Hi, this is module 22 of Mechanics of Materials part two. Today's learning outcome is to solve another problem for inelastic torsion of straight cylindrical shafts. We solved the problem last time, we'll do another one this time to solidify your knowledge of this theory and process. And so we're doing inelastic torsion of a straight cylindrical shaft. Now again, let's look at a real world problem. Here's a suspension system for a race car, and you know as it goes up and down, this part here is put into torsion, or torque. And so it's a torsion bar suspension. You want to decide to do some tests regarding a new torsion bar design. And you start this initiative by doing an analysis. And so what we're going to do is, we're going to take a little model here, we're going to say it's fixed on the right-hand side. It's 600 millimeters long. And it's going to be composed of an inner core of aluminum and an outer core of steel perfectly bonded together. And so, first of all I've got my idealized shear stress-strain diagrams for steel and aluminum. Okay, because I have two different materials this time. And what I want to do is I have the composite torsion bar. It's aluminum core, it's surrounded by steel, but they're perfectly bonded together. So, there's no slippage at the interface, they're perfectly bonded together. A torque is applied. Such that the aluminium just begin to yield at its outer surface. And we want to determine the torque, T, required for that aluminium to just begin yielding, so we can start with the relationship where we know that the strain is linearly proportional to the distance from the center. And so we have the maximum strain out here in the steel at a radius of 30 millimeters is proportional to the yield strain of aluminum, which is 0.005 from my graph over it's radius, which is 25 millimeters. So, let's go ahead and put those values in. What I don't know is what gamma max steel, is that strain in the steel is going to be. Over r outer, which is 30. Millimeters and that's equal to the strain yield of aluminum, which is .005 for my graph over my r for my aluminum the radius out to where the aluminum is fully elastic, which is 25 Millimeters. And so that gives me a gamma max steel of 0.006 radiance. And so you can see that for my steel, It will have all ready yielded at 0.002. So, when I get to 0.005 it's out here in the plastic range, and it goes out to a full strain of 0.006. So, it is completely plastic in this situation. So, at the interface Gamma steel Is equal to 0.005 because it is perfectly bonded together, so it is the same as the aluminum. And that's greater than the gamma. That's greater than the gamma yield for steel. So, therefore the steel is in the plastic regions. Steel Has Yielded. Okay, so that's now going to allow us to figure out what the torque is going to be for all elastic aluminum, all plastic steel, for the outer composite. So, I want to determine the total torque required for the aluminum to just begin yielding. That total torque will be the fully elastic aluminum core plus the fully plastic steel outer part and so we've done the elastic torsion problem several times now. You should be able to do it on your own, so figure out how much torque will be in the aluminum, which has gone fully elastic. And so come on back once you've done that. We use the elastic torsion formula to do that, it's shown here. The torque is equal to the yield sheer stress of aluminum, which is 140 megapascals. Times j, which is pi over two times the radius to the fourth. The radius for the aluminum is 25 mm divided by the r elastic, since it's going to be elastic all the way out to the interface, r is 25 mm. So, this is the torque that's in the carried by the elastic aluminum core. There it is again. Now we must find the amount of torque sustained by the plastic outer portion of steel and so we can do that together. We've got T Plastic Steel. And we developed this relationship in the theory section for inelastic torsion. Is equal to two thirds pie, tao yield for steel is 165 megapascals And I've got r outer is 30 mm cubed minus r inner for the steel is 25 mm cubed. And if I multiply that out, I get 3,931,000 newton millimeters. And so, now I have the torque elastic aluminum and the torque plastic steel. The total torque is those added together. So, I've got equals 3,436,000 + 3,931,000 = 7,370,000 N-mm, or I can convert that into another unit of kilonewton meters, which may be more common, and that ends up being 7.37 KN-m. And so, that is my answer. That is the torque required for the aluminum to just begin yielding and the steel outer core to be fully plastic. And so that's another good problem. You should have a pretty good handle now on inelastic torsion. And we'll see you next time. [MUSIC]