 Hi everyone. I want to do another example of looking at the loading scenario for a simple part, analyzing the stress, and then putting it on more circle in order to find the max stresses. So I'll go ahead and bring up my screen, and I've kind of got the basic layout drawn here. I've actually used this in another video where we have a bolt and it's screwed into something partway, and then we've got a wrench or something on it where we're pushing with 100 pounds of force on the end of that wrench. And we can see the dimensions on here. The bolt is sticking out four inches. It's got a half inch diameter shaft, and my wrench is 12 inches long. Pretend the bolt is straight out of the table. So we want to understand what's the state of stress in this part. Now, I've mentioned before that we need to pick a location of interest, and I've already indicated that on the drawing with that little square, basically saying, well, this is our stress element that we're interested in. And the reason we might pick that location is probably obvious to you at this point, but it's because we're at the base of the bolt, right? And we know that bending stresses contribute significantly to any stress situation. So we need to find a location that's going to have the highest bending while the bending arm is right here. So whatever gives us the longest bending arm is going to be the likely point of problems. So I'll go ahead and start with this. Our first step is to identify what state of loading we have. So we're looking again at this little square spot. And we need to remember our primary loads, P, V, M, and T. And we need to decide which of these exist. So is there an axial load, axial being vertical in the direction of the shaft of the bolt? In this case, we're saying that this 100 pound load pushes into or out of the screen. So we would say, no, we don't have any axial load in this case. Is there a shear component to the load at that location? And in this case, we would say yes, there is. Because for this free body diagram to have equal and opposite forces, there needs to be something pushing back opposing that 100 pound force in the into out of the board direction. So we would say yes, we have a shear. Is there a bending moment? Well, we already kind of described that yes, because this force is at a distance perpendicular to the force and around the the the axis that's perpendicular to the shaft of the bolt. Then yes, we have a bending moment. And finally, is there a torsion? Yes, this one's probably the most most obvious because we're we're trying to turn that bolt, right? So we have to apply a torque to it. And we have a force applied at a distance perpendicular to the the the axis that passes through the the vertical axis of the shaft. So that 12 inch distance is our our torque arm. So yes, we do have a torque. And shear, this one's pretty obvious that it's 100 pounds, because it's got to be equal and opposite to our applied load of 100 pounds. Moment, we just talked about. It's 100 pounds with a moment arm of four inches. So multiply those and we get 400 inch pounds of moment. Torque is that 100 pounds. And the distance that we're torquing around that long axis, it's at a distance of 12 inches. So we get 1200 inch pounds of torque that we're applying. Great. So we can go ahead and and turn these then into stress calculations. First, I'm just going to take a quick snapshot of our free body diagram. Our stress element sits here like this. And we've got this surface that we've cut right at the location of that stress element. And I do this just to make some of the visualization easier. So I'm going to pull out my stress element. Now, for this example problem, I'm going to actually ignore that shear, direct shear stress of tau V. And, you know, as you start doing more problems like this, you come to, you know, realize that tau V is usually relatively small compared to tau T. So I'm going to ignore it in this case, we may not always ignore it. But just for this example, I will. So the first thing we have to decide is what direction our stresses are on the stress element. So for our bending moment, we're pushing into the board as it's as it's drawn here into the screen. And what does that cause on our stress element? A tension or a compression? Because we're pushing into and bending it away from us as we look at the screen, that stress element is going to be in tension. So we can go ahead and draw our stress tension. And I know that this is sigma sub B corresponding to bending. And again, because we don't have any axial, I have no sigma sub A. So I don't even need to worry about that. Now for shear, I again need to look at my stress element. And what I can see is that I'm applying a shear in a, well, I guess if I'm looking from the top, I'm applying it in a counterclockwise direction. So counterclockwise direction as I look at the top. And this arrow kind of gives us a sense of how this stress is applied to our stress element. So as it's shearing across, it's shearing across the top of my stress element as I've drawn it here. And if I know this one arrow, then I know the rest of the arrows that go on my stress element for the shear. All right. So now I need to actually calculate what these values are. We have a basic equation, My by I for bending stress. And I can start plugging things in. I already calculated M up here. So I have my 400 inch pounds. My Y is my maximum distance from the neutral axis. Now in this case, my neutral axis is going to pass right through the middle of my shaft. My shaft has a diameter of one half inch. So that means I have a quarter inch maximum distance from that neutral axis to the outer surface. The other thing we would want to be careful of is making sure that remember how we talked about how our normal stress is applied is zero at the neutral axis and maximum at the inner and outer tension and compression sides. We just need to make sure we're not at the neutral axis. Now in this case, our neutral axis would pass through a plane that is right in the middle of the bolt as you look at it in the screen. So the middle this way. And that means that if my stress elements were located on the left or the right, then I would have zero bending stress. But because it's located on this side, we don't have to worry about it. Now, I'm realizing that I misspoke a little bit or should have added in something about the shear load before. And that is that, you know, yes, we often neglect direct shear tau sub V. But also it doesn't really apply in this case, because recall that shear stress is distributed across a surface in a parabolic. And therefore, it's max at the neutral axis and zero at the outer edges. So our shear stress would actually be zero where we've located the stress element. If we had located the stress element on the sides, we would have no bending stress, but we would have that max four thirds V over a shear stress. So just kind of pointing that out. So it's kind of for completeness. All right. So back to my bending moment calculation. I've got M, I've got Y, and now I need I. I again is area moment of inertia, which we would typically look up the equation. I happen to know that for a circular shaft, it's one fourth pi r to the fourth, rather than right r, but r to the fourth. But I'm going to erase that and just plug in r, which is 0.25 inches to the fourth. And if I go ahead and calculate all this out. Well, one thing I can see is that, you know, I have inches to the fourth on the bottom, which is going to cancel with these two inches units, leaving inches squared on the bottom. So it's going to be pounds per inch of squared or PSI, which is what we would expect for a stress. And I can calculate all this out and I get 32595 PSI. Great. Now let's go ahead and do our tau sub t, for which we have the equation t r by j. I already calculated torque above. So that's 1200 inch pounds are exactly the same thing in this scenario of a quarter inch. And j actually, interestingly enough, j is is two times I. So in this case, it's one half pi r squared. And if I calculate that, I get 48892.4 PSI. Great. So I know my stresses as they're applied to my stress element. Scroll down a little bit. Now the next step would be to construct my Mohr circle. So I'll go ahead and put in a set of axes, sigma x prime, tau x, y prime, prime. And I can again compare my stresses with where they would be plotted on this chart. So if I look at my top side, I have a positive bending stress. And I have a clockwise rotation, positive shear stress. So those values, I'm going to come out here positive, positive, something like this. And it's going to have that 32595 comma 48892. My x face has no normal stress, and only a shear stress. And that shear stress is causing counterclockwise rotation. So that indicates a negative. So negative on the shear, zero on the normal. So it's going to be out here at 048892, where that's negative. And I get my two points, right? So I can go ahead and rough in my circle. I can't draw a circle to save my life. And I see something like, like this. Now, if I use my calculations from before, I could find my center is going to be 16297.5. And in this case, I know that relatively easily, because it's just going to be half of this 32595, because my other normal stress is zero. So this plus zero over two. And I can calculate then my max. Now, obviously, because I can't draw, my max looks like it's less than this 32, but it's actually, you know, slightly out here somewhere. So my max, if I take this and I add in whatever I calculate my radius to be, let's see, which I would get if I plug in my calculation, it's 51537.1 psi. So I'm just using my radius equation that we presented before. Then I get sigma max is equal to my center plus my radius. So I end up with 67 830 834.6 psi. And tau max is that radius. So 51537.1 psi. And those are typically the main values that I'm going to be interested in when I'm trying to understand my max state of stress. Now, you know, if we think about again that that three dimensional Mohr circle, in this case, I can see because my my Mohr circle crosses the vertical axis, my two inner circles are over here and over here. So my this already is my maximum circle. So I don't need to worry about that. I could figure out my orientation. And if I dash in this line, I could figure out my orientation to get to my max stress circle, sorry, two theta p. And if I did that, I end up getting that two theta p equals negative 71.6 degrees. If I plug that into my equation, may or may not be of interest, you know, I divide that by two to get 3035 and a half or so degrees of rotation for my max stress element. So I can calculate that if I if I need to know that orientation. Okay, so that's an example of taking taking a rigid body under load, picking a location that we're interested in, figuring out which primary loads apply, figuring out which stresses apply, and then turning that stress into a Mohr circle, which gives us a maximum state of stress at some, you know, orientation that we didn't know in advance. So that's a pretty useful thing to have in our toolkit for understanding what the max state of stress is. And we need that so that we can actually do a failure analysis, determine if this part's going to be expected to fail based on the failure criteria that we'll talk about in our in our next video. All right, thanks.