 Let's give some examples of stress. And by stress, I'm meaning material stress, the type that's inside a material, not necessarily the stress you feel when you've forgotten to do your homework. So let's take a look at an example here, a real life example of a trailer hitch. So here I have a picture of a lovely trailer that all of you are going to be considering purchasing when you reach your late 60s, early 70s, and are thinking of retiring. And it's being towed here by a van of some sort. And in order to tow that around, we're going to need to be able to attach the two of them together. So we're going to consider this trailer hitch down in the lower left here, this little piece here that's going to be used to haul it around. Notice it's consists of a metal box sort of shape. And here's the picture of the metal box over here on the left. And we're going to say that it's 2 inches on a side. And it has a thickness of about 0.25 inches. So let's go back and sort of sketch what this looks like. Think about it this in terms of a free body diagram. So let's sketch my trailer here. My trailer has some wheels on it. It has the little hitch that's sort of extending from it. OK. And if we think about the free body diagram just roughly, we know we have some sort of weight of the trailer. The trailer is being held up by the wheels. When we're pulling the trailer to get the trailer started to accelerate the trailer, we're going to need to sort of pull on it. And it's going to be resisting that pull with its own inertia and or with the drag of the wheels on the ground. So at the point where that trailer hitch is connected to the car, there are going to be some reaction forces. There's going to be an axial force. We'll call that Rx. There's going to be a tangential or a shear force. We'll call that Ry. And there's going to be some sort of moment applied there. And I'm going to go ahead and give values to these. We're going to arbitrarily sort of say that the amount necessary to sort of balance things, the reaction here to hold up the trailer is 500 pounds. Notice most of the weight is probably being held up here by the wheels. But some of it's going to need to be balanced there. And we'll say that the pulling, that it takes 1,000 pounds of force to sort of get this thing moving. And then finally we'll say, again, somewhat arbitrarily that the moment that's being applied at this point right here is going to have a value of 10 pound feet. And again, that's probably going to be calculated by some balance of moments around this point. What's remaining after those moments are balanced. So assuming we have those values, let's go ahead and talk about the stresses that are associated with each of these pieces. So let's start with axial stress. If you remember, that was sigma is equal to p over a. So now we need to identify what p is. Now p is going to be perpendicular to a surface. So if I look closely at the hitch, let me think about the hitch here. Let's sort of look at the hitch in a three dimensional sketch here. So there's the side of the hitch. rp is going to be the, if we're slicing across that hitch there, we're basically pulling along that direction. So that's going to be, correspond with our rx, or our 1,000 pounds. So in this case, we have 1,000 pounds that we're pulling perpendicular to that surface. Now we also got to think a little bit about the dimensions of that surface. And if I remember correctly, we had dimensions that were 2 inches by 2 inches. And we had a thickness here, a thickness of roughly 0.25 inches. So we're going to go ahead and approximate this here. It's not a perfect approximation, but we're going to say that the area here that we're applying this to is going to be equal to, well, let's look at these pieces. Each of these pieces is about 2 times 0.25. Actually, we don't even need to approximate. So we have 2 inches by 0.25, and there are two of those areas. Actually, let's keep it simple. We'll just assume that there are four areas that are 2 by 0.25. Notice that's an approximation because we're going to end up counting each of these corners twice, but we're going to keep them relatively small. Just keep this simple times now. So there's 2 inches times a quarter an inch, but there's roughly four of them, and that's going to be our approximate area of the hitch. And that value there, let's see here, is going to be 2 square inches, or 2 inches squared. Well, that's simple enough. The axial stress that we're looking at then is going to be equal to 500 pounds per square inch, which we abbreviate with PSI. That's an example of calculating the axial stress. Again, we notice that the area that we're applying the force to is perpendicular to the direction of the force. Now let's consider shear stress. Tau is equal to V over A. In this case, we're going to use the same A, the same area, but now we're considering an application of forces that is parallel to that surface. So our application is going to be parallel. Notice that's vertically in this particular case, which corresponds with our reaction force, Ry of 500 pounds. So we take our 500 pound shear at that particular point. Notice the shear might vary if we move further down the line. We're assuming that there's a point at which the two are connected. We divide that by the 2 square inches, and we get 250 PSI of shear stress. Finally, let's go ahead and consider the bending moment. And if we remember, that was a significantly more complex equation where the bending moment, the axial force and bending, sigma equals My over I. Well, we don't have to solve for a moment. It's given to us in this problem. There's the moment there. But we do have to figure out what this Y and this I are. So let me draw this here over on the left-hand side. I have this box. And because it's symmetric vertically here, notice we're going to be bending along this vertical axis. I'm going to go ahead and put the distance here. The neutral axis is going to be in the middle, 1 inch from the top. And usually, our largest Y is the one we use in this case, because it's going to vary depending on where we are. But we really want to know what the maximum value is. So we're going to actually look for the maximum axial stress, which is going to be whatever the furthest Y is. In this case, either way, either direction we go in, it's going to be 1 inch. So we're going to use a Y of 1 inch. We know what our moment is. Our moment is 10 pound feet. Notice everything else is in inches, so we're going to have to do a conversion of 12 inches per foot so we can cancel out feet and get an answer in inches here. And then we have a Y of 1 inch. But now we also need that value I, that moment of inertia. Well, one of the easiest things to do with moment of inertia is go look up a reference. Let's go check out the internet, shall we? If you do some searches for area moments of inertia, you will often find charts that help you with common geometries for calculating the moment of inertia. Here's an example of a chart, and you'll see that there is a hollow rectangular section, which we have. Notice this is a more general description for a hollow rectangular section, which basically says that our width and our height are different. But in this case, we have the same value of D and B, where that B is our width and D is our depth here. So we're going to use them at the same sort of value there. Let's go ahead and call those W for width. If we do that, notice we've got some equations here that include the outside B and D and the inside B1 and D1. Let's go ahead and see if we can record those here. And in this case, the formula it gave us was IXX. IXX effectively means that we're doing a moment of inertia around an x-axis. Notice you could similarly do a moment of inertia around the y-axis if, for some reason, you were bending it instead of lifting it up and down, you were twisting it side to side. For example, if you were carrying what happens when you turn the car around a corner and you're sort of bending around a vertical axis instead. But let's look at this formula here. It's basically going to tell us that the outside distance we'll call that D outside to the fourth divided by 12 minus D inside to the fourth divided by 12. That's the formula for the area moment of inertia for this box. Well, we know that our D outside is 2 inches, and our D inside is going to be 2 inches minus the thickness twice, so that's going to be 1.5 inches, because the thickness is a quarter inch on each side. So if I do that math, I get 2 inches. To the fourth power minus 1.5 inches to the fourth power divided by 12. If I do that math, I get 0.497 inches to the fourth or roughly half an inch to the fourth. If I plug that in down here, 0.5 inches to the fourth, you'll notice we have an inch on the top, another inch on the top, and we can cancel out those two with two of these. And our final answer is going to be in pounds per square inch. Well, 10 times 12 divided by 0.5 is going to give us a value of 240 psi. Now, the one thing about that 240 psi is that indicates the amount of bending stress at that point y that we calculated. In other words, it indicates the bending stress at the top and or bottom edges of the hitch. That on one part of the hitch, we have a bending stress of 240 psi compressive, and on the other part of the hitch, we have a bending stress of 240 psi in tension. So if we're trying to figure out how much we have total, these are going to be the maximum locations, we would consider adding 500 psi in the place where it's in tension and subtracting 500 psi where it's in tension. So if it was in compression on the top, we would say 500 minus 240 psi. And on the bottom, it would be 500 plus 240 psi. So what's the point? Why do we need these internal stresses? Well, the key reason why we care about internal stresses, the main reason is that there is something known as yield stress. Yield stress is a material property for design that indicates when the design is going to fail. And when you're designing, you hope that whatever axial stress you have is less than the yield stress of the material property you are using. If it's greater than that yield stress, then that material is going to deform and change what it looks like. Perhaps break, perhaps stretch in a fashion that is permanent. And then you will no longer be able to use the original design as it was intended. Let's look at one more example. Consider, for example, a steel bolt hanging from a roof. That steel bolt will have a certain cross-sectional area. Term cross-sectional area means we're cutting across the bolt. When we look at the area across the bolt, we'll say it as a cross-sectional area A. And let's say that that bolt has a diameter equal to half an inch. Well, if we know what material the bolt is made out of, perhaps it's made out of steel. And we might be able to look up in a chart somewhere what the yield strength of steel is. Here's an example yield strength for a certain type of steel might be 50,000 psi pounds per square inch, which means that the steel can withstand internal stresses of 50,000 pounds per square inch before it deforms or before it breaks. And we'll discuss with that in more detail later. So well, if that's the case, we recognize that sigma is equal to force over area. Well, if I apply a weight to this bolt, the axial force is going to be equal to that weight. So the internal stress is going to be equal to the weight divided by the area. Well, if I want to know how much weight can I hold, what's the maximum weight I can hold? It's going to be equal to the yield stress times the area. That means if I hold exactly that weight, that's the point at which it's going to start failing. Well, that's equal to 50,000 psi times the area. And our area in this case would be pi r squared. So pi times a radius, which is a quarter of an inch squared. Or if I do the math, 9,817 pounds, noticing that the square inch and the inches squared are going to cancel out. This means for this particular bolt, if it's made out of steel, if I put 9,817 pounds on that bolt, I'm in danger of failing. Usually, in structural design, we include something called a factor of safety. In other words, we want to design with some number that's significantly less. In other words, that our design stress times some factor will use it as ks here, must be less than the yield stress. So for example, if I decided to use a factor of safety of 4, then I would end up dividing my maximum weight by this value of 4. And instead say that the maximum load we could handle is going to be 2,454 pounds. And by using that factor of safety, we know that we should be well below the point of failure of the bolt. So this is why we're interested in internal stresses. There are basically a way that we can compare our loads and their effect on our structural design to the materials that we have created the structural design out of.