[MUSIC] Hi, this is Module 11 of Mechanics of Materials Part II. Last time we looked at shearing strain for torsional loaded members. Now we're going to develop the expression for torsional shearing stress. And so you remember last time we looked at the strain. We said that the torsional strain at the outer surface was shown as here. And that it varied linearly. And so that, as we go out a distance Rho from the center the shearing strain becomes larger and larger until it reaches a max at the outside. And that Rho, again, was the radial distance from the center. So we saw that shearing strains very linearly with row. And so it starts out with zero at the very center where Rho is equal to zero and it goes to gamma max at where Rho is equal to r at the outer surface. Now for shear strain this is recalled back from my first, part one of my Mechanics Material course where we talked about pure shear as being the change in an angle between the perpendicular reference axis. And so we call the shear strain an angular distortion, or what we call the shear distortion. And this was the diagram that we came up with and shear strain was defined as gamma one plus gamma two. Or, since this is a right angle, after we have this distortion, this angular distortion, it´s equal to pi over two, or 90 degrees minus this angle beta. And I did a demo in that first part one course and here was my little block, and if I put shearing stresses on it, we saw that we get this angular distortion, or this sheer strength. And we see now that we get the same thing for my circular bar torsion. Here's my little block here, and as I torque this thing, you see you have the exact same angular torsion. And so, going back to my slide here. All I've done is take this diagram and I've turned it just slightly so that instead of having gamma one and gamma two, I now have put all the gamma up here, but I still have that same angular or sheer distortion. Let's also refer back to my first course in Mechanics Materials or part one when we talked about Hooke's Law in Shear. It was valid again in the linear elastic region and it stated that we had tau which is the shearing stress is equal to G, which was the modules of rigidity times the shearing strain. So again, G is the modulus of rigidity or the shear modulus. Here's our block undergoing our little element with shear stresses on it and we get shear strain. And we found out for our circular bar torsion that the maximum shear strain occurs at the outer surface and it's equal to this. We saw that strain was literally proportional to the distance we were from the center of the circular bar. And the distance we measured from the center is Rho. And if I now take Hooke's Law in Shear and I look at it for Tau max, the maximum shear stress. That's equal to G times the maximum shear strength. And I can put in for gamma max, R times theta, which I did here. If I look, however, at just tau, not tau max, but I look at the shear stress anywhere within my circular bar, it's equal to G times gamma. And we know that gamma varies with the distance from the center Rho and so that's equal to Rho times Theta or the rate of twist. Now I can substitute in Theta here is equal to tau max over Gr. And then when I do that the G's cancel and I end up with the sheer stress at any where in my circular bar is equal to the distance I am out from the center over the radius of the bar times tau max. So just like sheer strains, sheer stress is also very linearly with Rho, the distance from the center out from the center towards the outer surface. And so that's where we'll pick up next time. [MUSIC]