[MUSIC] Welcome to Module 8 of Mechanics and Materials II. Again, with today's learning outcome, we're going to solve another thin-walled pressure vessel problem. So to put that theory to work that we developed earlier in the course. And so in this case, our second worksheet is a steel cylindrical pressure vessel. It's got the dimensions as shown here below. It has a wall thickness of 15 millimeters, and we're told, on this plane AA that we see, that the normal stress on that plane, which is perpendicular to the surface of the vessel, is going to be 100 megapascals in tension. And the angle theta that's shown is going to be 30 degrees, and we're to determine, given that information, what is the air pressure in the pressure vessel. So let's go ahead and draw a stress block. And I will have longitudinal stresses, and I will have hoop stresses. But in this case, we're told, the information that we're given, is only that we have. And you should try this problem on your own, go ahead and continue on without me doing it. And then join me in progress as we go along. You may start me and stop me as I go through the problem, trying to keep ahead of me and test yourself out. But the information that we're given, and which we want to look and use, is that on a plane cut AA, as shown in our figure, we have a normal stress that's equal to a 100 megapascals in tension. And so we're going to have a cut here along a plane. This plane we'll call AA, and this is 30 degrees. And we know that the tension there on that plane is 100 megapascals in tension. So let's go ahead and draw my reoriented stress block. And so here's my reoriented stress block on a 30 degree angle. So this is AA, this is 30 degrees. And now we're told, the information that we're given, is on that plane, we have 100 megapascals in tension. And so we can now use that information to go ahead and draw a good Mohr circle, and so try to draw a Mohr circle on your own. See if you can continue with the problem and solve it. If you can't, come back and let's work it together. And so here's my axis for my Mohr circle. Say tau Is on the y axis. And sigma is on my x axis. I've got, like I've drawn before, I've got my longitudinal stress here, which is my vertical face. So this is sigma long, which is equal to, we calculated before to be pD / 4t. We found that the hoop stress is twice as much. So it's two times longer out here. Sigma hoop is equal to pD / 2t. And so I can now, using those two points, draw my Mohr circle. I've got my maximum sheer stress, which is also the radius of my circle, is equal to the hoop stress, minus the longitudinous stress, divided by 2. And that's equal to, if you calculate that out, pD / 8t. We don't know what p is, that's what we want to find. We do know what D and t is, and we also know that on a face AA, we've got 100 megapascals in tension. Remember, a 30 degree turn on my stress block, in this case in the counterclockwise direction, is going to be twice that angle on my Mohr's circle. And so, if I go twice that angle on my Mohr's circle, or 60 degrees. And I draw a line out here. I know that my normal stress is going to be 100, and I'm also going to have some shear stress that I didn't draw on here, but that is unknown on the A face. And so now I've got some additional information that's going to be very useful to me. Remember I know what this radius is as well. That's equal to tau max which is pD / 8t. And so if I look at this, I can write an equation now. If I drop a vertical down here, this is going to be 100 shear stress, and so we've got 100 megapascals is equal to. Okay, we go out a distance pD / 4t, and then we go out another distance which is equal to the radius, which is pD / 8t. And then we go out another distance, which is how far? And what you should say, okay, that's r times cosine 60 degrees, but r is equal to pD / 8 t. And that's times cosine of 60 degrees and now we have an expression. We know what D is, we know what t is. The only unknown that we have in this equation is p and that's the air pressure we want to solve for. So I've 100 MPa = pD / t, and if you add one-fourth plus one-eighth plus one-eighth times cosine 60 degrees, you get 0.4375. I can put in my diameter now and my thickness. So I'm going to have p times the diameter. The diameter's given as 750 millimeters. The thickness is given as, in the problem statement, 15 millimeters. That's going to be multiplied by 0.4375. And so I have an expression I can solve for p. P ends up being, the air pressure in my pressure vessel, ends up being 4.557, to three significant figures, MPa. And so we've solved our problem. So as a result of these first several modules in this course, you should have a pretty good handle on the analysis and potentially the design of thin-walled pressure vessels. And we'll pick up next time and start the torsion part of the course. [MUSIC]