Shear & Bending Moment Diagrams: 1 Shear Simple Spans

So this series of videos will be about shear and bending moment diagrams, and we need those to size a beam. It's very important because the maximum shear, wherever it occurs, and that's why we're doing this shear diagram, wherever the maximum shear occurs, that dictates to me how big the cross-section needs to be in square inches. the width, the depth, or else maybe a steel beam profile, whatever it is, I'm going to need that many square inches based on the maximum shear. The maximum moment, on the other hand, dictates also the shape. It dictates something called the section modulus, which I've gotten into in another video. But the section modulus is essentially the depth. So the shear dictates the area width times depth, and then the bending moment maximum is going to favor depth over width. Excellent. So in this video, I'm going to look at five simply supported beams. To keep it constant and to be able to compare these to each other, I'm keeping the span at 24 feet and the total load at 12 kips. So in the first example, it's 12 kips, in the second one, it's 12 non-symmetrical, then 2 6s, then 3 4s, then a uniform 12, and I want to see what happens with the shear diagram. Definition of the shear diagram, it's basically a plot of the sum of forces going from left to right on a beam. Please note the word sum. I'm not going to plot 4 kips, the reaction 4 kips, no, I'm going to plot their summation from left to right. Very good, let's get started. First, by calculating the reactions. In the first case, I have 12 kips symmetrical. That gives me two reactions. In the third case, it's also symmetrical, and the total load is 12, and the reactions are 6 and 6. The same in the fourth case and in the fifth case. Now, I skipped case number two because it was not symmetrical. But to get the reactions, we use the shortcut I discussed in a previous video. 8 over the total and 16 over the total of the load. So a third of 12, one reaction is 4 kips, the other reaction is 8 kips. The load is closer to the left, you get the 8 kips, you get the 4 kips. Very good. So starting with the first case, let's plot something called a shear diagram, And to plot that, I'm going to assume loads are positive. Sorry, reactions are positive and upward. Loads are downward and negative. So going from left to right in the first case, the first load I encounter is plus 6 kips. I go upward, 6 kips. Then in the next 12 feet, there's no new load. So I'm going to stay at 12 kips. Here we go. And then I lose suddenly at this one location, I encounter a minus 12 kips. I am at plus 6 and I'm about to lose 12 kips. I go down to, oops, I go down to minus 6 kips. Suddenly at this one location, to the left of it you have plus 6, to the right of it you have minus 6. There was a sudden drop of 12 kips. Then there is no change in load for the last 12 feet. And then I encounter a reaction of plus upward, 6 kips. It takes me back to 0. So the maximum shear in this problem is 6 kips. Be it plus or minus, it doesn't matter. In the shear diagram, I'm looking for the absolute value of shear, the largest number. Excellent. If the loading is symmetrical, the shear diagram is symmetrical. Excellent. Let's go to the next one. The next one says your reaction is 8 kips. Its arrowhead is upward, so we go up 8 kips. And we travel over. There's no new loads, so we stay at 8 kips. Let me put the 8 kips here. And we suddenly lose 12 kips. We are at plus 8, and we lose 12 kips. I'm down at minus 4. minus 4 kips, and then there's no new loads. So I continue, and then I encounter a plus 4 kips, and it takes me back up to zero. And so this is the shear diagram for this loading. The maximum shear is the largest of the two reactions. In this case, the maximum shear is 8 kips. In the previous case, the maximum shear was 6 kips. Excellent! So, The loading was not symmetrical, the shear diagram is clearly not symmetrical as well. Excellent. Let's look at the next one. Summing forces, I go up 6 skips. Then there's not much going on, so I continue at a constant. I was at plus 6 skips. And then I lost suddenly at this one location 6 skips from plus 6, minus 6. I ended up at 0. And then there was no change. And then there was another minus 6. Now I'm at minus 6 skips. And then it's constant. And then the reaction takes me back up to 0. Please note that I'm plotting loads plus reactions. I'm not plotting individual forces. I'm plotting their summation. Very good. Maximum shear is at the support. Maximum shear is equal to 6 skips. Plus or minus, doesn't matter. Load is symmetrical. Shear diagram is symmetrical. There was no change in load between concentrated loads. That's when we see a horizontal, a constant. The shear is not changing. Looking at problem number four, the numbers here are just for illustration. I'm not trying to complicate your life. Just explain the example, explain the concept using easy numbers. So the reaction takes me up 6 kips. Then there is no change in the next 6 feet. So I stay at plus 6 kips. And then I encounter 4 kips down. I am at plus 6. I lose 4. And now I am at plus 2. Plus 2 kips. Constant. No change between the first 4 and the second 4. And then I lose 4 kips again. Now I'm at minus 2. Plus 2 kips, minus 4. for the load, I'm at minus 2. Then there is no change, and then there's a loss of 4 again, and now I'm at minus 6 skips. And for the last 6 feet, there's no change, so it remains constant. And then the reaction brings me back up to 0. Every shear diagram must start at 0, must end at 0, regardless if it's simply supported, overhanging, whatever it is, the value at the very end has to be 0. Excellent. Symmetrical loading, symmetrical shear diagram. Now let's come to a uniform load. It does something a little bit different. With the previous example, there was no change between loads and reactions. It was constant. The value of shear was constant. Okay. Now, with a uniform load, I need to find the coefficient, or the increment of the uniform load. It's basically how much is it dropping per foot, because that is going to give me a sloping straight line, and I need its slope, which is little w. So little w is 12 kips, and it's dispersed over 24 feet. So it's equal to 0.5 kip per every linear foot. So what's going on here is if you travel one foot, you're going to lose 0.5 kips. You travel another foot, you're going to lose 2 times 0.5, 3 times 0.5, 4 times 0.5. So every foot you travel, you lose 0.5 kips. So let's see. Plotting the shear diagram, I go up 6 kips. That's the reaction. At that one location, there is a burst upward of 6 kips. Then I start losing 0.5 kips per every foot. So 6, 5.5, 5, 4.5, 4, 3.5, etc. Once I travel 12 feet, at 12 feet, I'm going to lose from 6 skips. I'm starting with 6 skips. I'm going to lose 12 feet times 0.5 kip per foot. I'm going to lose 6 skips. So I'm going to end up at 0 at mid-span. So this guy is doing something like that. And then once you pass the mid-point or the center span, Then you start losing minus 0.5, minus 1, minus 1.5, etc. So this line continues until it reaches the other end. And at the other end, I would have lost 24 feet at 0.5 kip per foot. So I would have dropped 12 kips. I started at plus 6 and I lost 12 kips. Now I find myself at minus 6 kips. and then the reaction brings me back up to zero. Excellent. So these are the first five cases and the shear diagrams for simply supported beams. So in summary, the shear diagram is a plot of the sum of forces, not the individual forces, along the span of a beam. Loads take you down, reactions take you up. The shear at any location along the span is going to dictate how much square inches of cross-section you need. So, beam number 2 is going to need 8 kips worth of area versus beam number 1 is going to need only 6 kips worth of area. Beam number 2 is going to have a larger cross-section than beam number 1. The maximum shear is always at the support. In a simply supported beam, the maximum shear is equal to the reaction 6 kips, 6 kips, 6 kips, 6 kips is the maximum shear. When the reactions are not equal, it's the larger of the two. 8 kips is going to require more area than 4 kips. Excellent. So also we said it's the absolute value. If this were minus 16, I would have taken minus 16. I would have taken 16 rather as an absolute value and done the shear for that. but in this case the larger value was positive. If the load diagram is symmetrical, then clearly the shear and bending moment diagram will be symmetrical. What I did not say leads to the next video, but I'm going to say it now and then elaborate in the next video. The sum of areas in the shear diagram should equal to 0. This positive area should equal to this negative area. This positive area should equal to this negative area. This area should equal to this one. This one plus this one should equal to this one plus this one. This area should equal to this area. Very good.