[MUSIC] Hi and welcome back. In today's module, we're going to continue talking about bending stresses and we'll work through an example problem. This is still part of Unit 2: Static Failure. The learning objective for today's module is to understand how to calculate bending stresses in an object. And again, this is intended as review. So if you've never seen bending stress before, I ask that you please refer to Dr. Whiteman's modules on bending stresses. So last time we were working through this worksheet problem, an example problem where we had a rod, OA attached to a rod, AB. We assumed that rod AB is strong enough, and no analysis is necessary on that rod. And we had a load F coming in along the X axis and a load P coming in along the -y axis. And what, we were trying to do is determine the bending stress at Point O. So, the first step is we need to think of what is the bending moment that causes the bending stress at point O? Because bending stress is sigma is equal to My over I and it's usually this is going to be the hardest thing to figure out. So what causes the bending stress? Well let's take a look at our model again. So if we have the model fixed on this left side and we push down at point B, what you can see is that the beam is going to deflect. And it's feeling tensile stresses here at the top, compressive stress is here at the bottom. This load F is not causing any of the bending stress. The bending stress is coming from the load P at point B. So let's take a look and think about that in terms of a free body diagram, so we see P as causing the bending stress. So, P is pushing down at point B, if we think of this in terms of a free body diagram. Then, rod AB is going to push down on rod OA, so we have this reactionary force where rod AB is pushing down on rod OA. And rod OA is going to say, you know what, I'm going to push right back on you and so we get a reactionary force here from rod OA, pushing back on AB. But the one that we're really worried about is this reactionary force right here, which is causing a moment through the rod OA. So this is causing this moment right here. So essentially, our moment is coming from point P. Okay, so now that we have considered that, let's go ahead and continue on in our example problem. So what I've done is I've basically put load P acting at point A because we know essentially that's what is causing the moment. Just to simplify and keep track of forces instead of using the very long thing RAB ROA. So, let's start out by figuring out what the moment is at point O. So our stress equals Mc over I. We're trying to find M and we know it's caused by point P. So I'm going to draw a free body diagram. And what I see here is that I have point P pushing down, which is 500 newtons. My rod is fixed along this side, and there's going to be a reactionary force here, let's call it R1 pushing up on the bar. And then to make sure that the bar doesn't start to spin in this direction, we actually have a moment acting on the bar here, M. So if I go ahead and I draw the bending moment diagram, what I'll see Is my reactionary force is going to come up, and it's going to come up by 500 newtons. It's going to be the equal and opposite force to this and then it's going to come over and drop down by 500 newtons. So that's my shear diagram and then when I look at my bending moment diagram, I'm going to have a negative bending moment, right? So, using the right hand rule, I can see this is a negative bending moment so I'm going to start down here. My moment is going to be 500 newtons times the distance at the point of interest. And here I'm at point O, so I'm going to be at negative 250 newton meters at point O and then it's going to come up and that's going to be my bending moment diagram. So now what we've determined, and here's a shortcut. Because this beam isn't constrained anywhere between the point that you want the stress and the point where the load is being applied, you can do a little bit of a simpler bending moment where you just say M is equal to force times distance. Which is equal to, in this case, 500 Newtons times 0.5 meters or 250 Newton meters and it's a negative bending moment. Again, that only works if you're not constrained anywhere else on the beam between where you apply the load and where you're calculating the moment. So then, if we look at our stress equation -My over I, our next thing is going to be y, which is the distance from the neutral axis, which we can see this neutral plane. Right here is our neutral axis, right here is our neutral plane. And the distance from the neutral axis to the point of interest is just going to be d over 2. It's in the positive y direction, so this is going to be a positive y. It's really critical that you keep track of the signs. So your I is going to be the area moment of inertia. It will be pi d to the 4th over 64. You should have learned to calculate area moments of inertia and your solid mechanics or deformable bodies class and there are a number of charts online. If you just google area moments of inertia of a cylinder or of a rod, that you'll find these charts for easy reference. So then we go ahead and we plug all of this in being sure to keep track of the sign. So negative times -250 newton dot meters times d over 2. And this times the reciprocal of I which is 64 over pi d to the 4th. So what we see is that these ds cancel, we can do some canceling here and we get -[(-250N.m) times 32 divided pi d cubed. Which is = 32(-250 N.m divided by pi times d, which is 0.04 cubed and we get our bending stress is equal to positive 39.8 mega pascals. Okay, so a couple things to keep in mind here. So when we look at point this load at point A pushing down, we can see it's pulling the filaments of this bar on top of the bar and it's squishing the filaments at the bottom of the bar. So we would guess that our stresses are in tension at the top. And that's what we calculated here with this positive value, our stresses are positive. Another thing to notice is that we said our bending stress, which is occurring again in the x direction. That's where you're stretching the beam filaments in the x direction, was right around 40 megapascals. And when we looked at our axial stress, also acting in the x direction, it was 0.8 mPa. So even though the axial load was higher, the bending stress dominates, which is a really common thing. So, it's always important to keep an eye on your bending stress because the values can get quite large. And it's critical to understand that this bending stress is acting in the x direction. A lot of students think it's acting in the y direction because that's the direction the load is moving in. But the filaments are getting stretched in the x direction, and that's the direction the bending stress is acting in. Okay, so we've completed that example for this module. Next time, we're going to get into transverse shear stresses, which also occur in bending. So that's it for today, I'll see you next time. [MUSIC]