[SOUND]. Welcome to module 10 of Mechanics of Materials II. The learning outcome for today is to develop the expression for Torsional Shearing. And we're going to focus on Circular Bar Torsion, as opposed to having other types of cross sections. And by doing that we're going to make the assumptions that our bar is in pure torsion. It's only going to be subjected to a torque or a moment. And we're going to have a circular cross section which means that the cross sections are going to remain plane and we're going to assume small angles and I've exaggerated the dimensions and the angles in my graph here just so that it's easier to see. But as far as cross sections remaining plane, you can see with this circular pool noodle here that as I torque this okay, if I cut a cross section and that's what these lines are, that they remain plane. They remain in the plane. Other types of cross sections and that's why we're limiting ourselves to circular cross sections, other types of cross sections like this square cross section will actually warp and so as you turn these, the cross section does not remain plain. Parts of it warp in and out of the plain. So here's our bar again, our circular bar undergoing torsion. And we see that we have this angle phi, if we look at it from the side, which is the angle of twist, and then this distance s should be right at this corner here, and that's the arc length of the twist. And we also have an angle if we look at it from the side which we'll call gamma. And again, for small angles and small deflections, this distance s can be assumed to be the same as this distance s. And then we can relate the two. So the first thing I would like you to do is on your own try to relate s and phi. And the relationship is by, S the arc length is equal to the radius times the angle fe. For this other section I would like you to relate S to gamma. And in this case, S is the opposite side, L is the adjacent side, so opposite over adjacent is the tangent of the angle gamma. Since we're using small angles, gamma is approximately equal to tangent gamma and again that's opposite over adjacent or s over L. So torsional shear strain at the outer surface can be found by showing these two or relating these two together, we have gamma equals s over L where s is r phi. And this is the maximum torsion sheer strain because the maximum's going to occur out here at the corner, or out on the outer surface. Inside you can see as you go toward the radius the angle gamma will shrink down to 0. So now let's look at a small element. So we're going to slice out a small element of my bar undergoing torsion. And it's not going to have the full radius r, it's going to be less than that. It's going to have a radius which we're going to call rho. It's going to have an angle d phi and it's going to have a length dx. And we're going to find the rate of twist as the angle of twist per unit length. So the angle of twist is d phi, per unit length dx given the symbol theta. And so now we had gamma max, the shear strain max is equal to r phi over L or I can substitute in and instead of phi over L for the full member, now I'm going to just use the small element d phi over dx. Which again, for the small element, the angle is d phi. And the length is dx. And d phi over dx is the rate of twist, so I can substitute that in and we come up with this relationship. So here's that relationship again. And if I want to find gamma not just at the outer surface but at anywhere, if I go in towards the center, you can see that gamma gets less. And it's going to equal rho times the rate of twist, where rho is the radial distance from the center. So as rho gets larger, we get more and more gamma, or larger and larger gamma, until we get to the outside. And so I can substitute in for the rate of twist, the rate of twist is equal to gamma max over R. And so I end up with rho over R times gamma max which means that the sheer strains now very linearly with rho. As we go from the inside at the center all the way to the outer surface. And now these relationships, are very important, and you'll notice that up until this point as I discuss strain I haven't mentioned anything about material property. So, this theory, this development could be for material that was operating in the elastic or inelastic range. It could be material that is homogeneous or heterogeneous. The only thing we have specified is that it has to be at small angles or small deformations. because we made the assumption that tangent gamma was equal to gamma. And so, we'll pick up with this point here next time. [SOUND]
