[MUSIC] This is Module 19 of Mechanics of Materials Part II. Today's learning outcome is to investigate stresses on inclined planes for the case of pure shear due to torsion. So here's our case of circular bar pure shear torsion. If I look at a little element block here and I cut it out, I can see that because of the torque on the right hand side, I'm going to get a shear stress down here, going to get a shear stress up on this side, and that'll complete my stress block. Now I can turn that 180 degrees, and I can see that that's pure shear, so I just flipped it over. And so, now taking that pure shear stress block, which is induced on my circular bar, let's go ahead and draw Mohr's circle. And, for Mohr's circle, if I look at my vertical face here, by my Mohr's circle sign convention from my previous courses, counter clockwise is going to be negative. And so we're going to have down here, the vertical face is going to be 0, and because it has no normal stress and -tau. And for the horizontal face up here, that's clockwise, so that's going to be positive by my Mohr's circle sign convention. So my horizontal face is going to 0 and positive tau, and then I can now draw a line between them for the diameter. I see that the center is at 0, I can draw my Mohr's circle. I can see that it has a diameter of tau all the way around, and so my maximum normal stress is going to occur here, and it's going to be equal to tau. And it's at a 90 degree angle, yeah 90 degree angle on the Mohr's circle, so it's going to be a 45 degree angle on my stress block. And then my sigma 2 principal stress is going to to be -tau. And so recall now, theta sub A, we'll call this A. This is 2 theta sub A down here, 90 degrees. So theta sub A on the shear block is going to be half of 90, or 45 degrees counterclockwise on stress block. And so if I draw that, this is what it looks like. I've got my vertical face, I turn it 45 degrees, 90 degrees on my Mohr's circle, or 45 degrees on my stress block, and I have a positive normal stress of tau and 0 sheer stress. And then if I go another 180 degrees on my Mohr's circle, or 90 degrees on my stress block, I get a normal stress that's in compression and negative. And so no shear stress, at these are the principal stresses. Okay, so here is the result, this is my Mohr's circle. Here is the stress block, and so what I see is that on a 45 degree angle, I'm going to get a max tension and a max compression. And so, chalk is strong in compression, but weak in tension. So, if I take a piece of chalk and I apply it to pure sheer or pure torsion, I can see that it'll actually break on a 45-degree angle. And you can see that, and you can try that on your own. So you can see the 45-degree helical surface, because it's being pulled apart by the tension, and the compression is strong and it holds firm. And so the chalk breaks on a 45-degree helical surface, and this is typical for any brittle material subject to torsion. And in fact, here's another example, this is a thin aluminum tube, a hollow tube. And if I put pure torque on that, I can see that I get in one 45 degree direction's max compression, and so you can see that it buckles, and in the other 45 degree direction it gets max tension so that it tears. And so a perfect example of what happens when I tprque this pure circular bar. And so let's look at one more example, here is a spiral fracture. I talked about this at the beginning of the course, a spiral fracture of a bone. If you get your, let's say like your arm caught somewhere and it gets twisted and broken, that's a torsional break, we call it a spiral fracture. And you can see here, here's the bone length, and here's where the fracture's occurring. Again, it's about an angle of 45 degrees, and so as predicted by the theory. So it's quite interesting, and so that's where I'll leave off for today, and we'll pick up again next module. [MUSIC]