[MUSIC] Hi, and welcome back. In today's module, we're going to continue going over the Brittle Coulomb Mohr theory. This is the last mini subject in Unit 2, Static Failure. The learning outcomes for today's module, are to understand how to apply the Brittle Coulomb Mohr theory to an actual problem. And so this is the problem we left off with last time. You can see we have this cylinder of bone. Our x direction is going up, our y direction is going to the left, and you can see this cylinder of bone has had a torque and had a bending moment applied to it. We also note that the bone is behaving in a brittle manner. And that the yield strength in compression, is not equivalent to the yield strength, or I'm sorry. The ultimate strength in compression is not equivalent to the ultimate strength and tension. And this is a perfect time, or a perfect example for the Brittle Coulomb Mohr theory because it's brittle, and because it has unequivalent strengths. So let's go ahead and take a look at this problem. So, what I want you to notice is we're looking for the factor of safety at point A. And point A is on the very left of the cylinder, and it's on the xz plane. So we have this torque, it's positive being applied to the cylinder. And then we have a negative bending moment, because z is coming out of the page, the z axis is coming out of the page. So our bending moment using the right hand rule is negative. And it's probably going to pull this point A into tension. So if we look through all of that, I think that the easiest place to start here is to start with the torsional shear stress. Since point a is on the xz plane, we know we have, our shear stress is the xz direction. And the equation for shear stress is T, or torsional shear stress, is Tr/J. In this case, because we have a cylinder, our r is just the diameter of the cylinder divided by 2, and our J Is the polar moment of inertia, which is 32 divided by pi d to the fourth. Which is 16 times, so it's going to simplify down to 16 times our torque, which is given to us as 80 Newton meters divided by pi times the diameter, which is 0.024 meters cubed. And what we find is that our torsional shear stress is 29.5 megapascals. And it's positive. Now our bending stress, so first off, note that our bending stress is going to act along this x direction. In fact, it's probably going to look like something like this, and then we'll have our neutral axis coming right through here. And then our stresses would get in compression as we work to the other side of this cylinder. So the equation for bending stress in the x direction is -mc over I. Our c is simply the distance between the neutral axis to our point of interest, which is going to be the radius, which is the diameter divided by 2. And then our I is going to be 64 over pi d to the 4th. And so this is going to simplify down to -32m/pi d cubed. And what we find here is, remember that our bending moment is negative. So 100 Newton meters, -100 Newton meters And our diameter is 0.024 cubed. So I want you to note one thing. Our diameter, our c value, so our radius, is acting in the positive y direction. So it would have a positive sign here. If we were looking at the other side, it would be acting in the negative direction, which would add another negative sign there. And when we would end up with a final, negative answer. Here, we're going to have a positive answer, because we have a negative bending moment in a negative equation. So we end up with a positive stress of 73.7 megapascals, which is what we expected, that it would be in tension. Okay, so now we have both of our stresses acting in the x direction, and then a shear stress acting in the xz plane. And we're going to go ahead and we're going to put these into the principle stress equation. So, we figured out our bending stress in the x direction was 73.7, and it was positive. And we figured out that our tau, our shear stress due to torsion on the xd plane, is 29.5, also positive. Okay, so if go ahead and I plug into the principal stress equation, I have my sigma 1 and my sigma 2 is equal to, in this case, sigma x plus sigma z, divided by 2, plus or minus the square root of sigma x minus sigma z divided by 2 squared, plus tau x z squared. So my sigma 1 sigma 2 is going to be equal to, note that I don't have any stress in the z direction, so I set these to 0. So my sigma x is 73.7 divided by 2, plus or minus the square root of 73.7 divided by 2 squared plus 29.5 squared. And what I end up with is my principal stresses point A, is 83.9 megapascals and -10.2 megapascals. Now note, I'm not determining the direction that they're occurring in, I don't actually care what direction they're occurring in. I just want to know what the maximum stresses are at point A. So the principal stresses here are giving me maximum normal stresses, which are 83.9 and -10.2. So I can immediately see that when I line up these stresses, I'm going to get 83.9 is greater than or equal to 0, which is greater than or equal to -10.2. And that means I'm in case 2. So here is my factor of safety equation. Okay, so I'm going to go ahead and use this equation, and what I'm going to find is that 1 over n is equal to 83.9 divided by the ultimate strength and tension, right? Because I want to compare a tensile stress to a tensile strength. So 120 megapascals minus -10.3. And I want to compare my compressive stress to a compressive strength over 170 megapascals. And what I end up with, is n is equal, I'm sorry. You all can do that math. So n is equal to 1.3, which is great because it means no failure. Okay, now we've worked through all of unit 2. Including, so we looked at, we did a review of the different states of stresses. We looked at the stress concentration factors. And we also learned distortion energy theory, which is also called von Mises Theory. And we learned the Brittle Coulomb Mohr theory. So you should go ahead and work through all of the example problems, and then go ahead and take Quiz 2. It's a great idea for Quiz 2 if you have an equation sheet with, for example, the von Mises equations, the Brittle Coulomb Mohr equations, your bending stress equations, some moments of inertia or second moments of area equations. That'll be really helpful for you to reference throughout Quiz 2. That's it for today, I'll see you next time. [MUSIC]