# Mechanics of Materials: Lesson 67 - Beam Column Buckling Example

https://youtu.be/H_B4f-MjNKM?si=O4QqFdxWzhkS_HIS

## Untitled chapter

Hey gang, welcome back. We got a little Mechanics of Material Column Buckling Example today. We're going to be continuing our talk about Column Buckling. About Oilers Buckling Equation. 

Remember some things we talked about last time. These were end, fixity. I think it's eye. There's an eye right there. Fixity. Conditions. That's how our beam is going to buckle is connected on the ends. 

The n-fixity conditions. 

Pin pins 1. Which is also called ideal. Fix fixed, pin fixed, and fixed free. Let's see if we can use these to solve this. 

This looks like a static problem, doesn't it? 

Use a factor of safety for buckling of 1.5, find the max value for W. This distributed load. 

How big can that distribute load be? 

They give us yield stress, and they give us the modularity for this material. 

We have a cross section here. It's 80 in this direction, which means it's 45 into the board, the thin direction. Let's talk about that buckling for a second. As that load increases, I've got a yardstick here. As that load increases, and it's kind of like this, wide side, thin side. As that load increases, I don't know if you can see me bending that or not. But it wants to bend. I'm just doing pin pin with my fingers here. It's allowed to swivel to rotate here at my fingers. It wants to bend into the board. Let's just call that for fun. Let's just call this the y-axis, let's call it that way, arrow. 

Then the x-axis is coming out at us here. The x-axis is in this direction here. If we have bending like this, it's bending into the board, if you will. That's what we call bending around the y-axis. That is the weak axis of bending for this beam, because it wants to always fail on that axis. It doesn't want to fail that way, because it's thicker this way. It doesn't want to fail that way. It will, if somehow I can hold it and prevent it and make it buckle that way. I can't do it with my fingers on this. 

There are two things that change here. Number one is the moment of inertia definitely changes. You have i, x and you have i, y. Again, i, y is going to be what we call the weak axis, i, x is going to be the strong axis. Let's find one of the most confused things that students get on this, get this wrong, if they get these two things backwards, so what they should be using for the base, they use for the height, and what they should use for the height, they use as the base. Be really careful. I'm telling you, y'all, it helps you so much. If you can get one of those little metal 12 inch rulers, the little flexible ones, and on the test, you can kind of flex it, you can go, okay, let's see, this is the base, that's the height. You can kind of see that it'll help you, okay. So let's talk about, so this is, these two things here, I've got to get that sorted out. That's, and I think that's pretty easy to do. The other, the thing that, the other thing that this most commonly messed up on these is what happens here and what happens there, okay. Now, in buckling, well, let me draw a little exaggerated buckle here, okay. Buckling in this direction, whoa, okay. There's two, there's that one, and the buckling end of the board will just be a straight line, wouldn't it? Because you can't see it, because it's going into the board, but the other one is the one that's bending that way, right. In that direction, in that bow direction there, when it bows, right. That pin gets to act like a pin, it allows the thing to pivot around a point, right. So in this direction here, which is bending around the X axis, remember, okay. It's going to be pin pin, but what about bending, it's into the board, or out at you now, okay. What about bending in that direction? Well, that pin swivels this way, but it doesn't swivel this way, right. In that direction, actually, it's fixed, isn't it? It's fixed and fixed, okay. And now, the n-fixity conditions change. So for bowing or buckling around the X axis, it's going to be pin pin, and for buckling around the Y axis, into the board, if you will, is going to be fixed fixed, okay. So those values right there are going to change for Euler's buckling equation. Now, do we remember Euler's buckling equation, okay. It's this, p critical, the critical force that's going to make that thing fail, is equal to pi squared EI divided by KL quantity squared. Now, another big mistake I see students making, don't do this, don't be Johnny Weeksauce here, right. Is it KL quantity squared? Sometimes I see you just doing L squared here, or putting the K on the outside. But the K goes inside the parentheses here, okay. That's a big one. Now, most of the time, beams are pin pin, and we call that the ideal connection, right. And K is equal to 1, and so that just kind of goes away. And so it's L squared. So sometimes you'll see this equation written just like this, which is an L squared on the bottom. But that's assuming you have ideal connections, okay. And do we always, well no, okay. So we're going to, what we're going to have to do, is we're going to have to evaluate both conditions here. What does it take to make it buckle this way, and what does it take to make it buckle that way? We're going to get two P criticals, one for each one of those scenarios, and then we're going to have to choose which one is going to happen. Well, which one happens first? The biggest one or the smallest one? Well, the smallest one is going to happen first, right. Because once I get to that load, it's going to buckle in that direction. And so we're going to get two P critical values, we're going to select the smallest one to then go and calculate what is W, okay. Is that a lot? We got this, man. We got it. We got it. Let's do I first. Let's calculate I, okay. So here we go. Let's do this guy right here. We'll do around the X first, okay. So when it buckles around the X, okay. When it buckles around the X, right. The base, the base is always parallel to the axis, okay. So we're doing skinny side into the board. So the base is going to be what? Well, it's going to be 45 is the base. Let's do 112, shall we? Remember this for rectangular shapes? 112 b h, which is 80 cubed, okay. Now it doesn't want to go this way, right. This way. It doesn't want to go that way. It's why I call this strong. The strong is always going to have of these two numbers, the biggest one is going to be cubed. That's going to make the biggest. That makes him strong. Well, if we're at the end, okay. And so the weak axis of bending is going to be 112. There's the 80 and look who's cubed now. Well, the 45, right. That's going to be a smaller I. That kind of makes sense, doesn't it. Strong axis, weak axis. Big number, smaller number, okay. Let's see what that number is, okay. On, on, 80 cubed times 45 equals divided by 12 equals. And remember, these are in millimeters, so it's going to be millimeters to the fourth. Oh, it's big number. Are you afraid of big numbers? I'm not afraid. Don't be scared. So 45 cubed times 80 equals divided by 12 equals 600 and 7,500, okay. So let's calculate P critical for bending around the X axis. And remember, bending around the X is this way. So for bending around the X, remember, this one was pin, pin. And this guy down here was fixed, fixed. 

Now you might say, I already know it's going to bend around the weak axis because that's where it wants to bend. Maybe, but you know what, the end conditions change everything. I mean, it makes it, it changes everything. Because it's going to change this number down here, big town, and it's square. So we better check it out. So let's do X first, bending around the X. So the P critical force to do that is going to be equal to pi squared times E, which is down here, okay. Now we're going to do this. We're going to do 200,000 Newton's over millimetres squared, right. I turned it into mega pass gals from gigas, okay. Times I, which is this number, 1,920,000 millimetres to the fourth, okay. Divided by K, what is K for pin, pin, it's 1 times the length, the unsupported length of the beam, which is 3 meters, which we're going to call 3,000 millimetres, okay. And then we're going to square that whole thing, okay. So let's see what that is equal to, okay. So pi squared, pi squared equals times 200,000 equals times 1,920, 1,2,3 equals divided by 3000, and squared, right, equals bam. 421.1, now that's Newton's right, everything's in Newton's. I got 421,103, or I'm going to call it 421.1 kilo Newton's, okay. Stone, coal, kilo Newton's, yo. Okay, so here we go, next one, pi critical. Now we're going to do bending around the Y, why? Why? Because we have to, okay. 200,000 times this value, 607, 500 divided by, oh, here we go. Now this one's fixed fixed, so it's 0.5, so 0.5 times 3000 times all of that squared. Why? Let's see, so here we go. 

pi squared equals times 2,000, 1,2,3 equals times 607, 500 equals divided by 3,000 times 0.5, that's 1500, right? 1500 squared equals bam. 532.95, so 533. Keller Newton's, okay, uh-oh, so what does that mean? This thing does not fail around the weak axis. It fails around the strong axis, right? Wow, those in conditions changed everything, didn't they? Okay, so this is the number we're going to work with. Now that number is the absolute number that that thing is going to fail at. You don't want to put 422 kilo Newton's on there, it'll break. It'll fail, it'll buckle, right? Well, but what about this? Yes, it says use a factor of safety, a buckling of 1.5, which means we don't want it to fail at 421. We want to fail at something less than that, don't we? Because we want to make it, we want to, we want to lie to them like the Ford F150, right? It's a half-ton truck. You should be able to put a thousand pounds in the bed, it's a half a ton, right? And if that's what Ford says, half a ton. But I guarantee you they got a factor of safety, if you've ever been able to home depot on the weekend, and people are putting like 3,000 pounds of concrete in the back of one of those, right? They make a factor of safety on that because people are, they will break stuff, right? So how do I make that number, a smaller number with 1.5? Multiply by 1.5 or divide it by 1.5? Divide, right? So 421.1 divided by 1.5 equals 280.73. So our real buckling number we want to use is 280.73, okay? Kill the noons. And that's got a factor of safety, a 1.5 on it. It really fails at 421. But we're going to tell the public, yeah, 280 so it's going to hold you. But we got a 1.5 factor of safety built into that, right? So what we want to do now is that is the number that when pushing down on here will make this thing buckle. So how big does double you need to be for that? 

Well, what you have is this. We've got this distributed load. We've got a pin connection over here. This is static, isn't it, y'all? 

Okay, and I've got this distributed load up here on the top. And how big is that distributed load? It's 5W, isn't it? Where did that come from? 5 times W pounds per feet or whatever it is. I mean, not pounds per feet. It'd be kilo nothing's per meter, wouldn't it, okay? And then over here, it's going to take 280.73 to make me buckle. So how big is W, right? 

Ooh, that looks like this, doesn't it? Take the moment at A. And what do we get? Remember, that's a rectangular distributed load. So we're going to put that right in the middle. Right, the 5W rotates me positive. So 5W times how far away? 2.5, right? Minus 280.73 times how far away? 5, right? The 5's divided away, right? And what do I get? So W is equal to 280.73 kilo noons divided by 2.5, which is equal to divided by 2.5 equals 112.3. So that's what W would need to be is 112.3 kilo noons per meter. Remember, it's a distributed load. That's the maximum that it could be so that that thing doesn't buckle. 

Now there's one more thing that you probably want to go check, and that is this, right? They gave me, why did they give me sigma yield right here, gang? Why? Well, remember, a beam can fail two ways. It can fail from buckling when you're in compression or it can fail from just straight deforming, right? It just crushes it, right? So we could check that pretty quickly, right? We got 250 megapascals, 250, Newton's over millimeter squared is equal to the crushing force that would take it to fail that way, divided by the area which is what, 80 by 45. And to be honest with you, I don't know that I've ever worked one of these problems where that actually failed because of this before buckling. But we check it anyway, right? So the force has got to be healthy. So this is going to be 250 times 80 times 45 is 900,000. That's Newton's right. So the force that it's going to take to crush it is 900,000 Newton's or 900 kilonewtons. But it buckles it 421, right? So is it ever going to buckle? Is it ever going to have a compressive failure? No, okay? So check mark, boom, boom. We are good there, okay? All right gang, we're getting awfully close to the end of the semester. I hope this helps, and I'll see you on the next video.