We've had the opportunity to talk a little bit about internal moments that are created when a beam is placed in a condition of bending, where the beam is going to want to bend. And we've also had the opportunity to talk a little bit about something known as the moment of inertia, which takes the geometry of the beam into account for how much it resists that bending. Well in this particular segment we're going to talk about how to apply bending stress on a beam to help design the beams that you might be using in a structural system. Let's consider the first the design of a simply supported but somewhat complexly loaded beam. Here I have a beam. It's supported here with a pin support at point A. We'll give it a roller support. That's a point B. We'll have some distances along here. 5 feet, 8 feet, 2 feet, and 5 feet. And those will distinguish between some loading, again a relatively complex loading. Let's go ahead and scale that a little bit better of this beam. And the loads are going to be 500 pounds, 1,000 pounds, and 200 pounds on the cantilevered end. Hopefully by now we know how to begin a problem like this, or the types of information we can gather from the loading. The first thing we're going to want to do is go ahead and determine reactions at the supports. The first thing we're going to want to do is determine the reactions at the supports. So we expect to have reactions rA, y, and rA, x, and a reaction here rB, y. Notice the simplest. The sum of forces in the x direction tell us that rA, x is going to have to be equal to 0, since it's the only force that's being applied in the x direction. For a sum of forces in the y direction, we recognize that rA, y plus rB, y must be equal to the sum. Those are the only two that are pointing up, must be equal to the sum of all those pointing down, which is a total of 1,700 pounds. Now we'll go ahead and take moments about a particular point. Let's take the moments around point A. Here's point A, so we can eliminate rA, y, and rA, x from the equation. We'll take the moments about that point, and we'll start by doing the, we'll set the clockwise moments equal to the counterclockwise moments. The clockwise moments are going to be 500 pounds applied at 5 feet. We'll add 1,000 pounds applied at 13 feet, which notices the total distance out to 1,000 pounds, and we'll add 200 pounds applied at a total of 20 feet. That sum is going to be equal to the moment that's supplied by rB, y, the reaction in the y direction at B that is applied at a distance of 15 feet from our rotation point. If we solve for rB, y, we end up with 2,500 plus 13,000 plus 4,000 pound feet divided by 15 feet. The units and feet cancel, and we get a value for rB, y of 1,300 pounds, which means that the value for rA, y is equal to 400, so that the total adds up to the 1,700 pounds expressed in the sum of forces in the y direction. Now that we've found those reactions, let's create a shear diagram, and from that create the bending moment diagram. First let's go ahead and modify our reactions. Here is our 400 pound reaction, and here is our 1,300 pound reaction.  So I'll begin by moving up a value of 400, so the shear on this first segment is equal to 400, then we move down, and the shear value there is going to be 400 minus 500, so this is at a level of 400. This is 400 minus the 500, so a level of negative 100. We continue with that negative 100 till we reach the place where we add the 1,000 pound load, so that's going to drop this down to negative 1,100 until we reach the application, the 1,300 pound reaction force in an upward direction, which puts us at a net shear of 200, and finally we see that that bounces back out to zero at the end of the beam as we push down the 200. So we have two areas of positive shear and a large area of negative shear that's sort of connected there, and this is our shear diagram representing our internal shear forces. Well, we can use that shear diagram to then create a moment diagram for internal bending moment, make sure the various points of interest line up, and as we've demonstrated, the slope of the moment diagram is going to be equal to the shear, and that's going to be applied over that entire length. Remember, we sort of generated this before by considering a formula where we take a little piece and cut it off and calculate what the moment would have to be at any particular place along that piece. Well, we recognize that we have a pin at the far left side at point A, so we have a moment of zero there, and we similarly have a moment of zero on the other end where we have a free end. So let's go ahead and see if we can determine what's happening over each of these sections. Starting with the moment of zero, we slant up with a shear of 400 pounds over a distance of 5 feet. So 400 pounds applied over a distance of 5 feet, that 400 times 5 is going to give us a level here of 2,000 pound feet for the internal moment at that point. Then we're going to drop down at a much lower rate at negative 100, actually that's drawn at a little bit too steep of a rate. We're going to drop down at 100 pounds over a distance of 8 feet. Well, if you start at 2,000 and you subtract 100 over the 8 feet, you're going to subtract a total of 800. So we drop down to a level of 1200 pound feet of internal moment at this location. Now we're similarly dropping, but this time much faster at a slope of negative 1100 pounds applied over 2 feet. So that 1200 minus 2200 is going to give us a value here of negative 1000 pound feet. And then our shear starts pushing us up again at a slope of 200 applied over 5 feet. Well, 200 times 5 is going to give us the 1000 feet and bring us back to zero. So this is the shape of our bending moment diagram. And all the areas here above the line are representing positive bending moment with tension on the bottom and compression on top. And on this right hand side we get the opposite where we have tension on the top and compression on the bottom. Now that we've created the moment diagram, the main reason for the moment diagram is to identify the places where the internal bending moment is greatest. Notice we have two locations. We have the location with the greatest positive internal bending moment there at 2,000 pound feet and the location with the greatest negative at negative 1000 pound feet. If we have a beam that is not symmetrical, for example, a triangle beam or some other strange shape that isn't symmetrical, then we would want to consider each of these separately. If however, our beam is something that is symmetrical, for example, the shape of something like an I beam, then we can go ahead and consider just one of the two forms. We'll go ahead and consider just one. We'll assume a symmetrical shape. We're going to use a box beam and we're going to then identify the maximum moment, the maximum internal bending moment. M max is this value of 2,000 pound feet. So now that we've identified that internal bending moment and the maximum value that it has anywhere along the beam, now we must consider the geometry of the beam in order to convert internal moment to axial stress. So we're going to consider the design of a small box beam. Without a lot of information, let's go ahead and choose a beam that is 2 inches square and has a thickness of a quarter inch. And because our bending is applied to the top of the beam, our bending moment can be considered as bending around the edges of the beam. And we want to consider the moment, the second moment of area or the area moment of inertia around this horizontal axis that we'll label xx. If we were somehow bending the beam side to side, we would consider it around the vertical axis yy. But we're pushing down from the top, so we'll use this horizontal axis. We go and do some research and look up a formula for the geometry of a small box beam. And there we see the formula for a hollow rectangular cross section. And notice in this case, rb and rd are the same because it's a square. And I see the formula there, ixx, that includes b and d. And then in interior dimensions, b and d in both cases are divided by 12. So because b and d are equal, I'm basically going to have my d to the fourth minus d to the fourth for the interior divided by 12. So let's copy this formula over here. We have this second moment of area. We're going to basically have b outside d outside cubed minus b inside d inside cubed divided by 12. Well, again, both of these have a measure of two inches for the outside, two inches to the fourth divided by 12 minus one and a half inches on the inside. Notice again, we're taking the thickness twice, once on the top and once on the bottom, one and a half inches on the inside to the fourth power divided by 12. If I do those calculations, I get 16 minus 5.0625 over 12 or .911 inches to the fourth. So how do we use this value? What does this value mean? Well, that value helps give us a sense for how much this particular beam is going to resist that bending moment. And if you recall, in our discussion of internal stresses, we found a relationship, sigma equals my over i, a relationship for bending. Well, the i we're going to use here is that i along the x-axis. The moment we're going to use is that maximum moment we calculated from the beam, m max. And then what we want to do for our value y is we're going to take the maximum value, the maximum distance from the x-axis, from the neutral axis in the middle. In this case, that's going to be half of the height of the beam. So we take that maximum distance. And if we take both of these maximum values, that will give us the maximum location of the maximum value of axial stress. And that value will occur at the top or bottom of the beam in that location that we identified as having the maximum moment. That will be the place that the beam is most likely to fail under the particular loading that we've analyzed. So let's put in the numbers that we have here, shall we? We know that the maximum moment is 2,000 pound feet. Our distance to the top or the bottom of the beam from the neutral axis is one inch. And then we use that value that we just calculated for the second moment of area or for the area moment of inertia of 0.911 inches to the fourth. Oops, we're missing a piece here. We have a value that says a foot. So we're going to have to use a multiplier of 12 inches per foot so that we can reduce that particular, let's do this without the parentheses, shall we? Multiplier of 12 inches per foot. In that case, the value of foot cancels out. And two of our inch values, we'll cancel with two of the inch values here, leaving us with units of pounds per square inch. So our maximum internal stress is equal to 26,300 pounds per square inch or 26.3 kilo pounds per square inch, abbreviated KSI. Well, what do we do with that value? That tells us under this loading that the most stress is located at that particular point underneath the peak moment at the top and bottom of the beam. In one case, it's under compression. In one case, it's in tension. And it has a value of 26.3 thousand pounds per square inch. Well, if I'm considering design, what I then ask myself is what material am I using? If I select an example material, let's say, for example, the material I've chosen is ASTM structural steel, A36 steel that has a yield stress of 36 KSI. That's the point at which you would actually yield and start to deform permanently. Well, if we know that yield stress, it looks like our maximum stress is less than that yield stress. Is it enough less than that yield stress? That's a hard question. You can say, certainly, it's less, so we're safe. But this is where we take into account a factor of safety into our design. I'm going to take the yield stress that I'm designing for, this 36 KSI, and divide it by the stress that we actually predict to experience given the load, 26.3 KSI. When I do so, the ratio between those two is 1.37, and this value is what we call a factor of safety. It means for even for our design, failure would be expected to be at 1.37 times whatever we design for. Usually that's not a significantly large factor of safety. Usually we'd like a factor of safety of at least two, and often the factor of safety will be on the order of three or four. So if I wanted to change this design, I might need to do so to get a factor of safety of a greater value. Let's say, for example, I just so as that I wanted to get a factor of safety of above two, then I would have to reconsider the shape of the design. Let's go ahead and do so now. I'm going to keep the same shape, but knowing that I want a greater area moment of inertia, I'm going to make the beam a little bit taller. I don't want to add a lot of extra expense and a lot of extra weight of the material. So I don't want to make a significantly large difference, but let's see if we make the beam just a little bit bigger instead of being two inches by two inches. We'll instead make it two inches by three inches, maintaining the same thickness. Notice this is going to increase this distance Y as well as increasing the moment of inertia. If I do the calculations, I get two inches times three inches cubed minus 1.5 inches times 2.5 inches cubed, each of these divided by our factor of 12. And when I complete those calculations, I get a new moment of inertia, area moment of inertia of 2.55 inches to the fourth. I then put that into my maximum bending stress formula, MY over I, noting I still have the same maximum moment 2,000 pound feet, but now I have a new value in the denominator 2.55 inches to the fourth and a new value for Y, what was one inch is now an inch and a half from the center, 1.5 inches and then our conversion of 12 inches per foot. When I calculate that value, I get 14.1 KSI, which is significantly lower. And when I compare that to the 36 KSI to find my factor of safety, I get a value of 2.55. Not a very large factor of safety, but it is above our design choice of 2.


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