 So now we're going to define an important structural term known as stress. And we're going to think about that again as being something that's internal to a body, internal to a material that's inside a structure. So so far we've talked about internal forces and we've thought about internal forces as being small pairs of forces that are acting equal and opposite to each other at some location within a body. And so far we've considered three kinds of those forces. We've considered axial forces where we have a combined surface that I've split apart and there's an equal and opposite normal reaction. We've considered shear forces again where we have equal and opposite reactions but this time they are parallel to the surface. Again we're recognizing that these surfaces are mentally separated as the result of slicing through some connected piece of material. And finally we've considered bending moment as the reaction of two moments applied in opposite directions. Now it turns out each of these force pairs is actually a simplification. We consider these as being concentrated or point loads but they really are a simplification of a type of distributed force at this slice surface. That there's not just a single vector but instead there are a series of small distributed forces applied over the entire area that's being sliced through. In the case of our axial force this roughly represents a uniform distributed force. In the case of our shear force this similarly represents a little harder to draw here but similarly represents a uniform distributed force. But in the first case in axial it was normal and in this case it's tangential. Finally if we consider bending moment, bending moment actually can be considered instead of thinking about it as being a twist we can actually think about it as being a series of pulls and pushes. Where we're pulling on the top of the beam and pushing near the bottom of the beam and we have equal and opposite reactions on the other side. In other words we have a non-uniform or actually a linearly varying. It's non-uniform it's not the same but it varies, a linearly varying distributed normal force. So if instead of us thinking about each of these internal forces as being individual forces and individual paired forces we now start thinking them as being a whole sequence of smaller distributed forces. If you remember distributed forces are an application of a force applied to an area, a small area. That area might be the entire surface or it might be some smaller portion of that surface. And that area is matching up on opposite sides. These two areas are effectively the same place we've just separated them so that we can see what's going on. And we're going to go ahead and define, let's use P to represent the axial force pair. We'll use V to represent the shear force pair and we'll use M here to represent the bending moment. Well we're going to go ahead and define something known as axial stress. Axial stress is the relationship between this overall force, internal force and the area over which it's applied. In other words we're going to have units of force like pounds per unit area for example inches squared. And this term we're going to call axial stress. Notice it's sort of like a distributed force except that it's a distributed force pair that you have equal and opposite distributed forces acting on the same area. We have a Greek letter we use for that. This is the Greek letter sigma, a small version of the Greek letter sigma. The larger version looks something like this and is used for other things like the sum of, that's the capital version. We're not using that, we're going to use a small Greek letter sigma. And so our notation for axial stress is the Greek letter sigma and its relationship is related to the axial force as it's applied to an area. For shear force we use a different letter, actually I should say for shear stress we use a different letter. We use the Greek letter tau, but the relationship is very similar. Tau is related to the shear force as it's applied over an area. And notice that area as in the case of the shear force that area is parallel to the shear stress. And again this is going to be in pounds per inch squared or newtons per meter squared units of force per unit area. The bending moment is a little bit more complex. Notice first of all we recognize that the stresses or the forces we're talking about, these distributed forces are still normal. And because they're in a normal direction we're also going to think about them as being axial. So we use the same letter, the same sigma to represent an amount of axial force. But in this case maybe it's the axial force due to bending. We'll go ahead and put a B in there for this particular term. It's a slightly different type of axial stress although it is axial stress. Notice it's varying over the height of the beam or the material that we're looking at. And so I'm going to go ahead and give the formula for axial stress and bending. And it does need to be related to the moment, the bending moment M. In this case the formula looks something like this. Sigma equals MY over I. Well now we've added two new terms here, this term Y and this term I. What exactly are these terms? Well let's take a look a little bit more closely and describe exactly what this relationship is. Before we do so I will make a note that this is still in the units of pounds per square inch or newtons per meter squared. Same type of units as the axial stress. Let's see how we got this sigma equals MY over I. Now this is sort of a simplification. Generally if we want to work on a derivation of this we might want to think about it in terms of calculus. There are more detailed descriptions that use calculus as part of the derivation. But let's see if we can think about it a little more simply for our own purposes here. So I'm going to think about this linearly distributed set of forces as a series of smaller axial forces. And I'm going to label each of these small axial forces. So we'll call this one P1 and this one is P2 and P3 and so on and so forth up to however many we have Pn. So we have a bunch of axial forces Pi. And we're going to apply each of those axial forces to a small area that corresponds with it. Now I'm drawing them along the line here but officially they are all being applied to the same part of the surface. So perhaps I can erase these and think about each of these forces as being applied to an area. And what we're going to do is we're going to define a location here where the force is zero. This is a special location called a neutral axis. This is a place where the internal force and then is zero. It makes sense that this exists because if you're stretching along the top and compressing along the bottom there must be someplace in the middle that's not being either stretched or compressed where there's a balance and that's called the neutral axis. Usually the neutral axis is going to be in the center of whatever your cross section is going to be if that cross section is symmetric. It'll usually be halfway between the top and the bottom if that cross section is symmetric. And we're going to measure up from that neutral axis that the location of each of these little areas is going to be up some distance y. That's the y that we're looking at there but we're going to go ahead and consider this to be yi which are distances from the neutral axis. So notice all of this is meant to represent the effect of one big moment. So if that moment m, we're going to think about that moment m as being equal to a sum of a whole bunch of the little moments I'm going to use the capital sigma here but we're going to add up a force p1 and a moment arm y1 and we'll add it to p2 and the moment arm y2 and we'll add it to p3 and the moment arm y3, et cetera, et cetera. We're going to add up all these pieces so that's what the sigma here means. It means I add up i pieces pi times yi. We're going to add up all the forces times their moment arms and that's going to give us the total moment that's expressed by all these little teeny pieces of force. Well, I'm going to recognize from my definition of axial stress over here that sigma equals p over a. If I think about axial stress by themselves, sigma is equal to p over a as long as we're thinking about this as being force pairs. And if that's the case, let me solve that and I'm going to say that p is equal to a times sigma or sigma times a. Well, let me substitute that in here and recognize that this sum of moments, if I replace each p with sigma that corresponds to it, sigma i and the area that corresponds to it, ai, and then the yi is still there. I now have another way of representing the total moment. If I knew the axial stress at each of these little pieces, here's the axial stress at this piece is going to be sigma one and then sigma two is applied at this piece, etc. Okay, that's the relationship between the force, the little, the part of the force, and the small area that it's being applied to. Well, I'm going to arbitrarily add a value yi here and square this here. Okay, I'm going to multiply by yi and divide by yi, which works just fine as long as I'm not thinking about the neutral axis where it's zero. Okay, but if I multiply and divide by the same value, then I end up multiplying it by one and I don't change my overall value. Here's why this is useful and important. I'm going to recognize that the relationship between sigma and y, notice that it's the p is going to grow in a linear fashion. That there is some slope that relates sigma to y, that sigma gets bigger as y does and it's basically proportional at that point. Particularly if we have a point where when y equals zero, sigma equals zero. So because I have a proportional relationship, I'm going to recognize that for every sigma and every y, they are going to give me some sort of constant value k. So I'm going to temporarily replace this with a k. Sigma i, k, ay, ai, yi squared. Why am I doing this? Well, if I'm summing up a bunch of things and they're all multiplied by the same value, the sum of things that are doubled is the same as double the sum. For example, if I added the numbers one plus three plus five, that's going to be equal to nine. Well, if I doubled each of them, two plus six plus ten, that sum is going to be equal to eighteen, which is the same as doubling the sum at the end. So because of that, I can take this k out and have the new sum. Well, this piece right here, this sum of ai, yi squareds, this is the thing that we call a moment of inertia. Actually, formally, this is the area moment of inertia because we are using the area inside it. We represent that with the letter i, and we'll talk in more detail about how that geometry comes into play later. But if I represent that with i, I recognize that my k is the relationship between the stress and the distance away. We know that's related to our moment. If I resolve that, this is where I get the relationship sigma equals my over i. Where again, i is our area moment of inertia, which is a specific geometric relationship, kind of like area. m is our internal bending moment, and y is the distance we are away from the neutral axis. So no matter where you are on the neutral axis, we can find out what the overall... If we know how far we are from the neutral axis, we can figure out what our overall bending stress is at that particular point.