[SOUND] Hi and welcome back. In today's module we're going to continue and actually wrap up Distortion Energy Theory. And the learning outcome for today's module is to be able to apply the Distortion Energy Theory to a complex problem. So, this is the problem that we talked through last time. A rod, it has a torque of 100 pounds of force per inch being applied to the end. A load P applying to the end of 250 pounds in the negative x-direction, and a force F being applied up at the end and a positive y-direction of 25 pounds of force. Because this is a ductile material, we can apply the von Mises theory. And we can see that because our strain at fracture is greater than 0.05. In addition, in some cases you might run across a bar like this and it might have changes in diameter or a hole, some sort of stress concentration points. And you don't need to use stress concentrations here because it's a ductile material. So we're not going to use any stress concentrations. So let's go ahead and get started on solving this problem. So, we were also given this chart which is out of the Mil Handbook 5J. And it gives you the strength of various materials. Just to note that in Mil Handbook 5J their designation for yield strength is F, capital F, subscript ty which is referring to tensile yield. So that's equal to the capital Sy that you typically are used to seeing. So we already talked about how there's three types of stresses. At point A we have an axial compressive stress that's being caused by P. Normal stress, that's a bending compressive being caused by the load F, and a shear stress which is a negative and due to torsion caused by T. So let's work through solving for each of these stresses individually. And then we'll go ahead and combine them in the von Mises theory. So we're going to start with the easiest stress. And that's our normal axial compressive stress. So that equation, the stress is going to be in the x-direction sigma x = P over A where P is equal to 250 lbF And A is equal to pi times the diameter .5 inches squared divided by 4, and if you solve for that what you find is that this is 1.3 KSI And note that is occurring in the negative X direction so we need a negative sign. So -1.3 KSI, that's the normal stress. So we can go ahead and also solve for the bending stress here. So the bending stresses are going to occur at points A like this, where we're going to have stresses compressing point A and then a neutral axis, and then stresses and tension at the bottom. So at A we can see we're in compression and that our bending stress is coming from this load F. So, I can see the stresses are acting in the x-direction, the equation for bending stress is negative MC over I. We're going to simplify this equation to be relevant for a cylinder. So we've done this a couple of times, where we set c equal to d over 2. And I is Pi d to the fourth divided by 64. And when we simplify that out we get 32M over pi d cubed. And that's our bending stress equation. Now because our moment is actually going to be positive and our diameter is going to be positive, the negative sign is going to hold which make sense because we know we're in compression here. So we know this is going to be a negative stress. And so really the only question we have left is, what is the bending moment at this point A? And because we have no other loads or constraints between point A and the place where we're applying the force F we can just say our moment is force times distance. Which in this case is going to be 25 pounds of force times the distance of 20 inches. And that is divided by Pi times 0.05 inches cubed and we get sigma x is equal to, let's see 40.7 KSI and that's negative, so negative 40.7. Okay, so we figured out our stresses, our normal stresses due to axial loading and due to bending loading. Let's go ahead and figure out our stress now due to shear or due to torsion, which we know our point A is in the XZ plane at the top of the rods so tau XZ is going to be equal to Tr over J. And you guys have done this a few times now. So you know that our J is going to be Pi d to the fourth divided by 32. And our R is going to be d over 2, so that's going to simplify to 16T over Pi d cubed. And so when we work through that we have been directly given a torque here. So 100 pounds of force per inch times 16 divided by Pi times 0.5 inches cubed is equal to 4.1 KSI. So now, if I looked at my stresses, I can see all of the stresses start to add up and so what I have is sigma X due to axial is equal to negative 1.3 KSI. I have a sigma X due to bending which is equal to negative 40.7 KSI, and I have a tau Xz, which is equal to 4.1 KSI. And actually, that is negative as well, because the torque is in the negative direction. So when I plug this into my effective stress, you can see immediately I don't have any stress in the y, or in the z, so a lot of these components are going to fall out. And then I only have a shear stress in the Xz plane, so these components fall out. So what I end up with is a sigma prime is equal to 1 divided by the square root of 2 times sigma x squared plus sigma x squared plus 6 tau Xz squared, and this is all to the one half. I have parentheses there, that's not really necessary but that's okay. So this is going to simplify down to. I can pull out, this is going to be a 2 here, right? So I'm going to add these two together, so I can pull out a square root of 2, and that will simplify this equation down to sigma x squared plus 3 tau xz squared to the one-half is equal to my effective stress. So what I'm going to do is I'm going to plug this in. And so what we get is sigma effective is equal to negative 40.7 plus negative 1.3 right, they're in the same direction so you just add them together. Squared plus 6 times negative 4.1 squared all to the one half. And I get an effective stress that is equal to 42.6 KSI. So it's really the bending stress that is driving this equation, or this failure mode. The bending stress is by far the largest component. And note that it was the smallest load here by far but it's still driving it. Okay so, we figured out the effective stress at point A. Now what we can do is try to figure out the factor of safety and when we look at this table from the military standard, what we see is for a bar that is greater than a half inch, the yield strength of 4130 steel is 70 KSI. So we can plug that in to our equation of yield strength divided by effective strength, and our n is going to be 70 KSI divided by 42.6 KSI. So, our n is equal to 1.6. So, that's our factor of safety. All right, so, that's it for today's example. This raps up the von Mises theory and the Distortion Energy Theory. Next time we'll get into how to deal with brittle materials and static loading. See you next time. [SOUND]