[SOUND] [MUSIC] Welcome to module 16 of Mechanics of Materials part II. Today's learning outcome is to solve for the angle of twist when we have elastic torsion of a straight cylindrical shaft. And we started this problem last time, we had a steel bar, we modeled it as being maybe a drive shaft for a real world example. And we determined the maximum shear stress in the structure and now we want to determine the angle of twist of the free end out here with respect to the fixed end which is on the right. And so how would you go about doing that? And what you should say is, well we did the theory for that. We found the angle of twist was equal to the torque over the length divided by the modulus of rigidity and the polar moment of inertia, J. And so we found the torques in each of the sections last time, so we'll look at the angle of twist In each section one at a time. We also found j, and we need g which is the modulus of rigidity. And I'd like you to go ahead and research what is a reasonable value for the module of rigidity for steel. And what you should come up with is most references most deals around 80 giga pascals or 80,000 mega pascals is a pretty good number. All right, so here's our angle of twist formula. We have our torques in each section we have our J now and we have an assumed G. Let's first look at the angle at twist of point B with respect to A. So I've got the angle of twist of B with respect A. So from A to B is equal to the torque in A to B is, 5.25* 10 to the 5th, Newton millimeters. The length from A to B is 800 millimeters. We said that G was equal to 80, that's the module's rigidity. The 80,000 megapascals or newtons per millimeter squared. Since megapascals and newton per millimeter squared are the same. And we had J which was 56,150 millimeters to the 4th. And if I run those numbers I get 0.0935 and all the units cancel out. It's dimensionless or in terms of an angle, we'll call it radians and since the twist, the torque is to this direction. We're going to go ahead and say that, the angle twist is in that direction which would be the positive Z direction. I want you to go back and do the same thing now for the section, the twist of c with respect to b, and then c with respect to d, and then add them all together and let's see what we come up with. And so from b to c, we have this value. And from B to C, if I hold out B and look at C, that's going to twist it in the negative Z direction. And then finally from C to D I have 0.0267 if I look at just that section. D with respect to C is going to twist in the positive z direction and if I add those all together plus minus plus I get 1.035 radians in the positive z direction. And that's the angle of twist of the free end with respect to the fixed end over here on the right. And so that's how you do the problem. And that completes this section, this module. [SOUND]