[MUSIC] Welcome back to Mechanics of Materials II. So far in the course, we've gone through thin-walled pressure vessels and the theory for torsional shearing stress. We looked at elastic torsion in the last several modules, elastic torsion of straight cylindrical shafts. And now we're going to start to look at inelastic torsion of straight cylindrical shafts. And so today's learning outcome for Module 20 is to develop the relationships used in the analysis and design of inelastic torsion in straight cylindrical shafts. So this is a review from early in the course. Here I have a circular bar and pure torsion. We found that the torsional shear strain at the outer surface was the maximum, gamma max. It was equal to the radius times the angle of twist over the length, or for a small cut, r times d phi dx, where d phi dx was equal to theta, the rate of twist. And so we found that shear strains vary linearly with rho. So here is the shear strain versus rho, and so at any particular point, the shear strain was equal to rho times theta, where rho is less than r. And I can substitute in then for theta of gamma maxover r, and I get this result. Rho, again, is the radial distance from the center out in my shaft to whatever point I want to find the shear strain. And the angle of twist again is, as a reminder, is d phi dx or over the entire length of the shaft, phi over L. And so now I have gamma is equal to rho times theta. I'll substitute in theta, phi over L, and I come up with this result. And that should be on all of our review, if you need to, go back to the earlier modules to see the more explicit development. And so up until now when we talk about strains, we haven't specified anything about material property. So the material could be elastic or inelastic, it could be homogeneous or heterogeneous. The only thing we've specified is that we have to have small angles of twist because we assume that tangent gamma was approximately equal to gamma. Okay, when we do a stress-strain diagram, we've looked at stress-strain diagrams for normal stress and normal strain. This is a typical shear stress-shear strain diagram, and you can see that, for most materials, there's not such a well-defined yield point of where we transfer from the elastic region to the plastic region. But we're going to go ahead and make the assumption again, like we did for normal stress-strain, that we can work with it as an idealized elastoplastic material. So that when I hit some yield stress and strain, the material becomes perfectly plastic. That assumption simplifies the calculations, and it's quite accurate for most problems. And so here again is my bar inelastic or in pure torsion. Here's my idealized model of the material. And so, as I twist this, if I go beyond the elastic range, a portion of my outer bar can become plastic. And so I've got a demo here of a drive shaft that has been twisted until it experienced, in this case, inelastic failure. You can see how one, this line where it's kind of twisted itself and actually failed. So there's been inelastic yielding. So here again is my diagram of shear-strain versus rho. At the outer surface radius, we get the maximum strain. Anywhere in between, the strain is equal to the distance from the center out or rho times theta, and so I can set up a relationship here. Gamma max divided by r outer is equal to theta, which is the same as gamma over rho. So this gamma over rho relationship holds all the way through, and so it's going to hold when we hit the yield strain. The yield strain is going to occur at a distance out, which we're going to call the radius, where the elastic region ends. And so here we go, here's a cross section. We go out, it's in the elastic range, and then once we hit our elastic, it goes beyond where the yield stress for the material, the yield shear stress, and the rest of the region is plastic. So the inner region is elastic, and the outer region is what we'll call fully plastic. In the elastic region, tao was equal to G times gamma. That was Hooke's Law for shear. And so in the elastic region, I have the shear stress, and the elastic region is equal to the modulus of rigidity times the strain in the elastic region. In the plastic region, however, the shear stress in the plastic region is going to be the yield stress. It's not going to go up any higher. It's going to be flattened out and be plastic the whole way at a constant, we're going to assume it to be a constant value. Okay, so now let's relate the applied torque to the stresses and geometry of the problem. So the total torque will be the torque that's going to be picked up by the elastic portion of the cross-section plus the torque for the plastic portion of the cross-section. And for the elastic portion, we've done that, we've used the elastic torsion formula. Here it is. And so T elastic, the torque elastic, will equal to the torque yield. Excuse me, not the torque yield. The shear stress yield times J. The polar moment of inertia for the elastic region divided by r of the elastic region. And that's just a rearrangement of my elastic torsion formula. Now, we're going to go ahead and find the T for the plastic region as well. Here's my cross-section. I'm going to integrate a small differential element over the plastic region. And so for this element, it's going to be at a distance rho from the center, and the force acting on it is going to be the shear stress times dA. And so, again, we're going to actually integrate and find the total torque from the r elastic to the r outer portion of the cross-section. And so tau plastic is the integral of the force, dF, which is tau yield for the plastic region times dA times its moment arm which is rho, and we're going to integrate that over the entire plastic area. And so here's a blowup of my differential element for a small slice. One side is going to be d rho, the other side by arc length is going to be rho d phi. And so I have T plastic, the torque plastic. I'm going to integrate all the way around my cross-section, so it's going to be integrated from 0 to 2 pi. I'm going to integrate from r elastic to r outer where we're in the plastic range. We've got rho times tau yield times dA, dA is d rho times rho d phi. And so rho d phi times, d rho times rho d phi or rho d phi times d rho. And so I can now pull out tau yield because it's a constant in the plastic range. There's no other functions of phi in here, so I can integrate from 0 to 2 pi, and that comes out to be 2 pi. So I have 2 pi tau yield pulled out, and then I integrate from r elastic to r outer of rho squared d rho. You can do that integral, and we find that the torque for the plastic region is equal to what is shown. And we'll use this development for finding or solving problems for inelastic torsion of straight cylindrical shafts. And we'll pick up with that next time. [MUSIC]