Shear & Bending Moment Diagrams: 6 Shear And Bending Moments For Standard Cases

So to conclude the chapter on shear and bending moment diagrams, I would like to take a look at some symmetrical and some standard loading cases that I have abstracted from the steel manual. Now you don't have this chart during the ARE, but you do have the charts that come from the steel manual, which are basically the same, only these are in color and a little bit nicer. So, looking at this, this is the same, but it shows you the steel manual version versus the cleaned up version. So what I would like to do is derive some of these, and then we'll look at the cases in the steel manual and make sure that they're the same as this chart. Now, case number 1 and case number 7 are the most popular on the ARE and in real life and what have you. Case number 1 is a single concentrated load at mid-span. How much is the maximum shear? How much is the maximum moment? You get those because it's a standard case. You don't draw shear and moment diagrams. It's a standard case. There's formulas to get the maximum shear and maximum moment. So you get those and you size a beam. Case number seven is a uniformly loaded beam, simply supported. And it has a maximum shear, a maximum moment. We get those. We size the beam. Cases five, six, and eight are cantilevered. Cases nine and ten are continuous beams. Somebody's done the math, ran a computer, and gave us some numbers that we just plug in to get the maximum shear, maximum moment, then we size the beam. Very good. Looking at condition 1, a single concentrated load at mid-span, we know that the maximum shear is equal to the reaction. So if I have a load P, then half of it goes here, and half of it goes here. And the maximum shear is at the support. The maximum shear is equal to the reaction. Well, the reaction is P over 2, so the maximum shear is P over 2. So let me quickly draw a shear diagram. I'm just drawing it once to prove the formula. Otherwise, I'm just going to use the formula from now on. So looking at this, we said at zero shear, when the areas go from positive to negative, you get the maximum bending moment. And the maximum bending moment is equal to the area in the shear diagram up to zero shear. So it's the area of this rectangle whose height is P over 2 and whose base is L over 2. So the maximum moment is going to be the area of this rectangle, P over 2 times L over 2, which is PL over 4. So the maximum moment when you have a single concentrated load is PL over 4. The maximum shear is equal to the reaction. Very good. Looking at condition 7, a uniform load, we see that the total load is capital W, the span is L, and so the reaction here is half of W on each side, and that gives me the maximum shear. It's half the load. Now, to get the maximum moment, I'm going to find the area of this triangle up to 0 shear, and this triangle has a height of w over 2. It has a base of l over 2. And the area of a triangle is equal to base times height over 2. So W, capital W, not lowercase w, times l over 2 divided by 2. So that gives me W, l over 8. Where W is capital W, the total uniform load, not the coefficient. Very good. So the maximum moment when you have a uniform load is WL over 8. Let's look at the two cantilevers very quickly. I have a concentrated load at the tip of a cantilever, and it's worth P, and the span of the cantilever is L. So the maximum shear, the reaction is equal to P, because there's only one side, there's no two reactions. so the P goes over here, and the reaction is equal to P, and the maximum shear is equal to P. So I go down a certain amount P, I travel over, no new loads, I go up P. So the maximum bending moment is equal to this area, which is P times L. So the maximum moment is P times L. From now on, we'll just use the equations. For a uniform load, it's like W is over here over L over 2. And this reaction is equal to W, capital W, not lowercase. So the maximum shear is equal to W. And the area of this triangle that gives me the maximum bending moment, which is negative always with a cantilever, it's negative. So it's equal to W times L over 2. WL over 2 is the maximum moment in this case. Very good. I think we're going to stop here. We're going to take a look and make sure that it's the same in the steel manual. So let me do that. Let me say I gave you for a single concentrated load that the maximum moment is PL over 4. And the maximum shear is P over 2. So looking at case number 7 from the steel manual, these tables will be in the reference tab on your exam. They have to give you these formulas. And you have to recognize them and use them successfully. So the reaction is equal to the shear, and it's equal to half the load. Good. The maximum moment is at the point load. Wherever this is, that's where your maximum moment is. And it's equal to PL over 4, which is the same as what we have. Very good. Let's look at a uniform load. For a uniform load, I said that the maximum shear is equal to capital W over 2. They're saying the reaction is equal to the shear, and it's equal to, let me zoom in a little bit, little w, L over 2. Well, little w times L is equal to big W. So we can get rid of this and say W over 2. That's one and the same. It's 1 kip per foot, let's say, over 24 feet. That's 24 kips. So I'm using 24 kips instead of 1 kip per foot times 24 feet. Very good. Now the maximum moment we said earlier is WL over 8, where W is capital W. Here they're saying the maximum moment is at the center, and it's equal to little wl square over 8. Well, wl times l over 8, that's the same as this one. So it's essentially capital W, which is little wl over 8, one and the same. Maximum moment, either little wl square over 8, or else capital W, l over 8. Excellent. By the way, you don't need all this junk. That's for the engineers. We don't need this. We don't need all this information. So let's use the cleaned up versions. Here are the cantilevers. Here's all these other cases. I'm just showing you how to read the tables from the steel manual. And I would much rather for this seminar or these videos to use the symmetrical cases that are cleaned up, that actually came from the steel manual. It's one and the same. Very good. Let's have an example. I'm going to do the following. Let's say we have a beam, and this beam has a span, and it has a uniform load, and on top of that uniform load, it has a concentrated load and two reactions. So there's a reaction, there's another reaction, and there's a span of 24 feet, 12 and 12. And let's say the concentrated load is 12 kips. Let's say that the uniform load here is 18 kips, capital W. How much is little w just in case you need it? We don't need it, but just in case, it's 18 kips spread over 24 feet, or 0.75 kip per foot. So, calculate the maximum shear and maximum moment for this loading. That's the extent of what they would ask you on the ARE. So, I want you to use the formulas. I don't want to draw diagrams, find the area on the shear diagram, get the moment. That's when the problem is not symmetrical or it's not on these tables. Excellent. So, the maximum shear is very simply equal to the reaction. Let's add these two up, divide by 2, we have our maximum shear. We know that the maximum shear is at the support. We know that the maximum shear is equal to the reaction. That's it. 18 plus 12 is 30, divided by 2. 15 kips is my maximum shear. And I will use this 15 kips to size the area. This is going to give me the area in square inches of the cross section. Then I will decide wood, steel, concrete, whatever it is, I'm going to need a certain area based on 15 kips. Excuse me. Now, calculate the maximum moment. I'm going to go to the tables and charts, and I'm going to see that the maximum moment for this case is PL over 4, where this is P. In our case, it's 12 kips, and L is the total span of 24 feet. And I'm going to see that the second case is a uniform load. I can add two cases as long as they are both symmetrical. So I have a capital W of 18 kips and a span of 24 feet. and the maximum moment for this case is capital WL over 8. So the maximum moment for the concentrated load is P, 12 kips, times L, 24 feet, divided by 4, which gives me 72 kip foot. And the maximum moment for the uniform load is W, 18 kips, times 24 feet, divided by 8, which is 54 kip foot, for a grand total of 126 kip foot. That's it. Problem done. Now, if you happen to use little w, then you would have said that the moment is little wl square over 8. So let me go back here, and let's say I'm using the steel manual and you don't like these colored charts. Then it would have been 0.75 kip per foot. times the span square, the 24 now has to be squared, divided by 8. Well, guess what? This must give 54 kip foot. It's one and the same. This is the same as that one. I prefer to use capital W just because deflection, it gets ugly with units if you're using little w. That's why it's safer to use big W. It gives the same answer. Excellent. So calculate the maximum shear, calculate the maximum moment, no more drawing diagrams, let's just use these formulas.