[SOUND] Hi and welcome back. In today's module we're going to continue talking about the Distortion Energy Theory. The learning outcomes for today's module are to be able to understand and apply the maximum Distortion Energy Theory to relatively simple problems. So let's get started. Last time we talked about the Von Mises Equation, which is part of the Maximum Distortion Energy Theory. And here you can see that the sigma prime stands for sigma effective, and that's equal to the equation that you can see to the side. That equation is for a triaxial stress state and the stresses have been calculated in the x, y, and z directions. Sometimes students find the second equation as an easier form to remember or to understand. And keep in mind that the failure criteria for Von Mises Theory or the Distortion Energy Theory is that when your effective stress is greater than your yield strength, you have hit yield. So your component is no longer in a safe design at a lope. So last time we left off, one quick thing. It's a good idea, for quiz two, to start making an equation sheet for easy reference. So we've had a lot of equations so far in this course. And quiz two will be a lot easier if you have a sheet with all of your equations just there, easy to reference for you guys. Definitely make sure to put the Von Mises Equations on that sheet. So last time we left off with example one, where we had this aluminum cube in a triaxial stress state. If we look we can see that the aluminum cube has a strain at failure or a strain at fracture greater than 0.05 inches per inch. And that means that it's behaving in a ductile manner and therefore we can use the Von Mises Theory, or the Distortion Energy Theory. Another thing to note is that your yield strength and tension is equal to your yield strength and compression. Which again means that the Distortion Energy Theory is valid. And what they wanted us to find was the effective stress in this cube and the factor of safety. So I'm going to go ahead and give you three possible answers. You guys should have worked this out at home. So why don't you take a look at your answer, and then see if it matches up with any of these. And what you should have found is that the equivalent stress was 58.3. So the effective stress in this cube was 58.3 MPa. So let's go ahead and calculate that. So the first thing to note here is always a good idea to write down your stresses and their signs. So we know we have a sigma x, and it's positive, of 20 MPa. We know we have a sigma y, it's acting in the negative direction so it's -40 MPa. We have a sigma z in the positive z direction of ten MPa. And how xy or yx is equal ten MPa counter clockwise. Which is positive because this is when a counter clockwise shear stress looks like. This is a positive shear stress. Positive and then a negative shear stress would look like this. So you can see we're positive here. Okay so the equation for the equivalent stress is equal to one over the square root of two times sigma x minus sigma y squared. Plus sigma y minus sigma z squared plus sigma z minus sigma x squared. Plus six times the tao xy squared plus tao yz squared plus tao zx squared. And this is all to the one half. And if we look through here we'd see we only have tao and xy direction. So these other sheer stresses are going to fall out and go to zero. And that means that our equation boils down to, equals one over the square root of two times 20 minus. Now here's the important part, you need to keep your signs, keep track of your signs. So we have a -40 here plus -40 minus ten, I'm sorry, ten minus 20 plus six times the square root of. I'm sorry the square of ten. And all of this is to the one half. And when you solve this what you'll find is that the sigma effective is 58.3. Okay, so the other question here was, what is the factor of safety? And so, here we know that our yield strength and tension is equal to our yield strength and compression. Our factor of safety is going to be the strength, 75 MPa. Divided by your effective stress which is 58.3 MPa. And so your factor of safety is 1.28, which means that you have not had any yield in this design yet. And so you're operating in a relatively safe environment with this load level and design level. Okay, so that's a relatively straightforward Von Mises example. Let's look at one that's a little but more complicated. So here we have a rod and it's subjected to a torque, T = 100 pounds of force per inch. We have a load P and it's in the negative x direction of 250 pounds. And a load F in a positive y direction of 25 pounds of force. So the rod has been machined out of 4130 steel bar. It has a diameter d = 0.5 inches. And the strain at fraction is again greater than 0.05 inches per inch. So they want you to find the effective stress. At point A which is right here, point A. And they want you to also find the factor of safety. So a couple things to keep in mind, this is a complex learning situation so let's look and make sure we understand all the types of loads that are happening. And stresses that are happening at point A. So here you can take a look. This is part of MIL handbook 5J. And what it shows you are the different strengths for different types of steel. You'll remember on the problem it said it had a 4130 steel bar, and you can see that the leftmost column is 4130 steel. Keep in mind that for MIL handbook 5J they use the nomenclature F subscript ty for yield strength. And so you should be able to figure out the yield strength of this component. And then, So let's go ahead and think about the stresses present at point A. So take a moment and look at all the different loads and figure out what stresses would be present at point A? Okay, so what you should come up with is that point A has a normal axial load. It's a compressive load due to the load P. So it will have a normal axial compressive stress at point A. There's also a normal bending stress due to the force F. The force F is going to go ahead and push this beam up. Which is going to cause this beam to deform upwards and you can see that point A will be in compression. Here is your neutral axis of the beam. And so you can see that you are at the edge of the neutral, I'm sorry the edge of the object far away from neutral access. So, your bending stress is maximum and your transverse shear is zero. So, you have this normal bending stress happening at point A, and you also have a shear torsion stress. Because the torque is coming around this way if you guys use the right hand rule you'll see that the torque is negative which will give you a negative torsional stress. But it's still causing the shear at point A. So at this point you should go and try to work through this problem as far as you can on your own. And then in the next module we'll solve the entire problem. Remember to use the Von Mises Theory. And that's it for today, I'll see you next time. [MUSIC]