[MUSIC] Welcome to Module 7 of Mechanics and Materials II. Today's an exciting part of the course. We start to actually solve thin-walled pressure vessel problems. We're going to look at some real world thin wall pressure vessels, and let's start with this real world example, which is a pressure vessel I showed before. And it's got a diameter of roughly 2.5 inches. The wall thickness I'm going to say is about one-sixteenth of an inch. We´re going to assume that the maximum internal pressure that´s expected is going to be 200 PSI, or pounds per square inch. And we don´t want this thing to fail, and so we're going to go ahead and find the factor of safety with respect to yielding. And if you don't remember factor of safety, again, go back, review the modules from my first part one course in mechanics of materials. First of all, we need to make sure we can treat this example, this real-world steel cylinder as a thin wall pressure vessel, and so we're going to look at the ratio of D to t. In this case D is 2.5 inches, the diameter, the thickness is one-sixteenth of an inch, and so you find that ratio is equal to 40. If you recall back from one of my earlier modules as long as the ratio is greater than or equal to 20, we can go ahead and treat the problem as a thin walled pressure vessel. So we're okay. Treat as thin walled pressure vessel. And so when we look at the factor of safety, that's going to be in regard to the, whoops that's going to be in regard to the highest stress. What is the highest stress we're going to experience? What you should say, the highest stress we're going to experience is the hoop stress. Let's go ahead and calculate the hoop stress in this vessel. We found that the hoop stress was equal to PD over 2t. The pressure in this case is 200 pounds per square inch. The diameter is 2.5 inches. And the thickness is, so it's two times the thickness, which is one-sixteenth of an inch. If you multiply that out, you find that the hoop stress, that largest expected hoop stress will be 4,000 PSI, or pounds per square inch. My question to you is this is steel. Go ahead and do some research on your own and find out what's a good value for the yield stress in steel in terms of PSI. And what you'll find in most references is that the yield stress for steel, Is about, for most steels around 36,000 psi. So you can see it's definitely stronger than the stresses that we're going to experience. How much of a factor of a safety do we have? Well, factor of safety was defined as the failure stress over the actual stress being experienced. In this case, the failure stress is also the same as the strength of the material. Another way of expressing it. Over the max computed actual stress is the max computed stress that we've found, so this is just a couple of ways of expressing the factor of safety. In this case the failure stress or the strength of the material is 36,000 pounds per square inch, PSI. And the actual stress or the max computed stress that we're going to experience is 4,000 PSI, and so our factor of safety then is equal to 9. And so my question to you then is, is that a good factor of safety for a steel cylinder like this, pressure vessel? And if you recall back to my first course in mechanics and materials, you see that the typical values for factor of safety were for buildings greater than or equal to two. Automobiles greater than or equal to three, spacecraft or aircraft, 1.2 to 2.5 or larger, boilers and pressure vessels greater then 8.5. If they were to fail, a pressure vessel, that's a pretty catastrophic failure, so you want that factor of safety to be quite large, and so since ours was 9, it's above 8.5 and we're in pretty good shape. Again, other factors of safety lifting equipment or hooks, again, if it fails it's a catastrophic failure, so you want it to be 8 or 9 or higher. And bolts, 8.5 or higher. In general you want to use, again, a higher factor of safety for brittle materials to avoid catastrophic failure. You also want to use, you can use a lower factor of safety when the material properties are well known. And as a review you also want a higher factor of safety for uncertain environments where you don't know what kind of stresses you're experiencing. And so in this case we're in pretty good shape. Now that's a pretty good design, and so we'll come back next time and do another real world engineering example. [MUSIC]