[MUSIC] Hi, and welcome back to mechanics of materials part two. So far in the course we've completed the entire section on thin walled pressure vessels and internal pressure, and we've now looked at the theory for torsional shearing stress and strain, and we've come up with the elastic torsion formula. And now we get into the exciting part where we start to solve some real world examples, for torsional problems. And so, today is learning module 15, and the learning outcome is to solve for the maximum shear stress for an elastic torsion of a straight cylindrical shaft. And so here's an example it's a steel bar. And it's fixed on the right hand side over here. And we want to determine the maximum shear stress. And it has some torque applied to it. And so as a real world example, this is just a small drive shaft. And it's a cylindrical shaft and it could be subjected to different kinds of torques and so the example that we're going to go through is very applicable to these kinds of situations. And so the first thing I have as a question to you is, how would you go about starting to determine the maximum sheer stress in the structure. And what you should say is, well we've developed a theory for that. We're going to use the elastic torsion formula. We're looking at elastic torsion here of straight cylindrical shafts. And so if you recall, that is Tau. The shear stress is equal to the torque. Times row the radius out to the point that we're looking at the sheer stress divided by the polar moment of inertia. And so we know that the maximum sheer stress is going to occur where row is the maximum, and that's going to be at the outer surface. Of the cylindrical shaft, and so max tau occurs at outer surface. And so therefore in this case the diameter is 27.5, so rho would be half of that. So it's 27.5 divided by 2 or 13.75 mm. And that stays constant throughout the entire length of the shaft because it stays at the same diameter. And so J therefore, the polar moment inertia is also constant throughout the shaft, so let's write that down. So, J is constant throughout the length. And so my question to you is we know rho now. We know J. Tau max is going to occur where T is max. And so, let's write that down. Tau max occurs in our shaft, somewhere along it wherever the max T occurs. Where T is maxed. The torque and so how do I find out where T is max? So what we want to do now is we want to find which of these sections from c to d or from b to c or from a to b does the maximum torque occur? And so to do that I'm going to go ahead and draw a free body diagram, let's start with section C D. So here's my shaft. I'm going to cut it between section C and D. I've got an implied torque of 4 times ten to the fifth Newton millimeters and then I have my resisting torque T sub CD. To find out what T sub CD is I'll sum moments in the z direction, set it equal to zero. By the right hand rule 4 times 10 to the 5th is in the positive Z direction, so I've got 4 times 10 to the 5th. And TCD is in the opposite direction. I'm leaving units off because the units are shown on the diagrams. So I've got minus TCD equals 0, Or TCD equals 4 times 10 to the fifth Newton-millimeters. Okay, now we want to do the same process for finding the torque in section from B to C, and I'd like you to do that on your own, and then let's come back and do it together, and so we have the FBD from B to C. Here's my shaft. I'll cut it between B and C. I've got the applied torque at the end, 4 x 10 to the 5th, and again, I'm going to leave off units. And I have in the opposite direction 2.5 times 10 to the 5th. And then I have resisting torque inside that section when I cut it which is TBC. Once again, I'm going to sum moments in the z direction, so it equal to zero. So I have 4 times 10 to the fifth. And then the 2.5 by the right hand rules in the opposite directions, so it's going to be -2.5 times 10 to the fifth. And then minus TBC equals 0. Where TBC equals 1.5 time 10 to the fifth newton millimeters. And, so now I've found the torque in section CD and in section BC, the last section we need to find is the torque in section from a to b and I'd like you to go ahead and do that on your own and then come on. And back. And so, to do the torque, or find the torque in section AB, we must first draw the free body diagram, and here I've drawn that free body diagram, I went ahead and saw moments in the z direction. And I came up with T AB is equal to 5.25 times 10 to the fifth Newton millimeters. Remember we found T CD to be 4 times 10 to the fifth Newton millimeters and T from B to C to be 1.5 times 10 to the the fifth Newton millimeters. And so the largest torque where the largest sheer stress is going to occur is in section AB because it has the largest magnitude. Okay, to wrap it up, we have our elastic torsion formula. We know what row is. We're going to find J, which is constant throughout the cylinder. We have now found where the Max T occurs, the torque occurs in section AB. So row E we found to be 13.75 millimeters here's the largest torque in section AB. This is a solid circular cross section. So I want you to go ahead and calculate J on your own. Come on back. And when you calculate J, this is how we found the polar moment of inertia. And so we've got J=pi over 2. And we're looking at the outer radius, so r is equal to 13.75. And we're going to raise that to the fourth power. And that ends up equalling 56,150 mm to the fourth. So, I've got all my terms to find tau max now. So tau max equals max t. Which we found was 5.25 times 10 to the fifth Newton millimeters. We have our row at the outer surface which is 13.75 millimeters, and we're going to divide that by J, which we found to be 56,150 millimeters to the fourth, and that gives me Newton's, or 129 Newtons per millimeter squared. Newtons per millimeter squared is the same as megapascal, so this is 129 megapascals, and that's our solution. So we have been able to find the maximum shear stress in this Straight cylindrical shaft for elastic torsion, and we will come back next time, and we will determine the angle of twist. And, we'll see you then. [MUSIC]