[MUSIC] This is Module 13 of Mechanics of Materials Part II. And I want you to stay with me now, hang with me. We've been doing a lot of theory for torsion the last several modules, and we're soon to be getting into some real world examples. So hang in there. But today's learning outcome we talked about the polar moment of inertia last time for circular cross-sections, and today we're going to learn how to calculate it so when we get to actual examples, we'll be able to find that value. So here's where we left off, we related the applied torque to the torsion that's experienced, came up with the elastic torsion formula, and it included something we call the polar moment of inertia, or J. So here again is the expression for the polar moment of inertia, and we're going to take a cross section so that we can calculate what that polar moment of inertia is. My dA in this case is going to be this differential area that goes around my cross section at a distance rho from the center, and it's going to have a thickness or width of D rho. And so that dA is going to be, for the angle it's going to be two pi times the radius, or two pi times rho in this case. And the thickness is D rho, and so dA is equal to two pi times rho, D rho. I can now substitute that in, and I have rho squared, two pi rho, D rho and the D rho, I'm going from zero. I'm going to integrate all the way out to the outer surface, r. And so if I rearrange this, I take the two pi out, because it's a constant integral from zero to r of rho Q, now D rho. You can do that integration on your own, come on back. And when you do that integration, you should get rho to the fourth over four, so that's two over four or one-half. The rho is evaluated from r to zero so it becomes r to the fourth, or pi r to the fourth over two. So that's the J for a circular cross section. And so for a solid circular cross section we found that J is equal to pi r to the fourth over two. I can also put it in terms of the diameter of my cross section because the diameter divided by two is the same as the radius, and so in terms of the diameter it's pi D to fourth over 32. Sometimes we don't use a solid cross section. We can also have a circular cross section, but it can be hollow. And for a hollow cross section, we just take the outer radius minus the inner radius. So if this were a hollow bar we could take the J of this total outside radius minus the inside. And that shown here pi over two times r to the 4th outside minus r to the 4th inside or this is also an expression in terms of the diameter D. And so what we see is, that sometimes, often times, hollow circular cross sections can be more efficient for torsion because in those cases, like I said before you've got more mass further from the axis about which we're twisting. And we can save even though we're losing a little bit of resistance from the inside. We can save on materials and cost and it may be more efficient just to use a hollow circular cross section which may have enough J to provide the kind of support we need for the engineering structure. So now we know how to calculate the polar moment of inertia, J. [MUSIC]