[SOUND] Hi and welcome back. In today's module we'll be doing a quick review of axial stress and torsional stress. And this will be still in Unit 2: Static Failure. So the learning outcomes for today's module is one to understand how to calculate axial stress in an object and two to understand how to calculate torsional stress in an object. This module is intended to be a quick review. We're assuming that you've already seen this in a mechanics of materials or deformable bodies class. If you've never taken those classes before, this is not in-depth enough for you to completely learn the material so you should probably go back and review for axial stress Dr. Whiteman's lecture, you can find here. And then for torsional stress, you can find a number of lectures by Dr. Whiteman here at this link. So again, this is really intended as a review in case you haven't seen this material for a year or two. Okay, so let's get started. So if we think of stress, what we can think about it as a pressure. It's a state at a point, again, it's due to mechanical thermal loading. Typically we can calculate it, the most common equation is sigma = F / A. And the units are in psi or pascals. So on our top bar here, we have a load being applied on the exterior. And what we can imagine is these internal pressures from this load or these internal stresses. So when you think of the stress the way I typically think of it is as a pressure inside the object. Hence the PSI units, right, force over area. So, a couple of things to keep in mind. So, there are normal stresses, which we refer to as sigma. So sigma is for normal stresses. And those are acting perpendicular to the surface of the object and there are also shear stresses which we refer to as tau in this class and those are acting parallel to the surface of interest. So normal stresses act and can deform like classically normally stresses are found in a axial load and you can think of them as stretching the object. And shear stresses kind of distort the object. So you can see how the shear stress has kind of caused the object to shift. So if we go through these types of stresses, today what we're going to talk about are axial stresses, which are normal stress, and torsional stresses, which are shear stress. Again, we're going to stick to the assumptions that we're dealing with isotropic materials that are homogeneous in nature, unless specified specifically in a problem that you're working through on a worksheet or in an exam, you can assume that the object or component is at room temperature. We're always going to be dealing with materials that are conforming to Hooke's law, where sigma is equal to E times epsilon or the Young's modulus times the strain. And we're primarily going to be within the elastic range. So for axial stresses in this course, we're going to always assume you have axial centric loading. And in this case then your stress will be your force divided by your area. In this course, we'll also have tension as positive. So tension the stress is moving in the positive X direction, and therefore has a positive value, where compressive stresses moving in the negative X direction will have a negative value. And just to remember if we look at the distribution, or a reminder if we look at the distribution in stresses, so if I were to cut this cylinder here, axial stresses are uniform in nature, so they would be the same value across the surface, like that. Okay, so then if we go ahead and start looking at torsional stresses, a reminder that torsional stresses are really twisting. So axial stresses are this pulling or this pushing, and torsional stresses, you're really twisting the object. This is a shear stress and a common equation is that you have tau is equal to T. T is your torque times rho. Rho is your radial location. So it's the location of interest from the central axis. So if I wanted to know the stress at this point, my rho would be this distance here. If I wanted to know the stress at the surface, my rho would be this whole radius. And then J, J is your polar moment of inertia which you should have learned how to calculate back in mechanics and dynamics. Now a couple things to think about for torsional stresses, these twisting stresses is that they are not uniform across the bar. So if we look at the bar, I'm sorry it's a rod, so let's draw a big cylinder here. And we start to look at, so here's going to be the center of the cylinder so here's my central axis going through here. So if I look at the stress distribution, it's going to be practically zero at the center. Very low stresses because your rho value is very, very small, right, and that drops your stress down. And the further you get from the center, the larger your rho value becomes, the larger your stresses become. And so at the surface, your torsional stresses are maximum. So the distribution looks something like this. And so often what engineers will do, is they'll just worry about torsion along the top of the component because that's where the stress is maximum. Which makes sense right? We're typically worried about the max stress and therefore a lot of times you'll see this equation simplified down to Tr/j where r is the radius of your part, right. So I'm going to give you guys another example to kind of check your knowledge here and make sure you're still remembering your mechanics of materials, material well. So here we have this rod. The rod, we're going to call it to be OA. So here's point O, and here's point A, and this is the rod that we're worried about. And it's attached to another rod right here, this rod here I'm going to call AB. And so, what I'm going to assume is that this rod is strong enough. It's not going to deflect. It's not going to deform. And so we're not going to worry about calculating anything in this rod in part of the problem. In fact, what we're going to calculate are the stresses here at point O. This is point O. This is point A. Okay, so this rod OA has this constant diameter across it of 4 centimeters. It has a force, F, right here, which is 1,000 newtons, and it's applied in the negative x direction, so it's causing a compression in this rod. And then it has a force p applied to the end of rod ab here and that's also causing some stresses to occur in this rod OA. And you can assume that rod OA is made of a ductile material. So what I want you guys to do is find the torsional shear stress at point O and to find the axial stress at point O. And this is going to be in worksheet three That you'll go back or if you pull up that worksheet, you can go and work through this problem. And then get as far as you can, or get all the way through, and we'll go through the answer in the next module. Alright, that's it for today. I'll see you in the next module. [SOUND]