Hi, this is Module 20 of Mechanics of Materials part IV. Today's learning outcome is to solve a column problem again to determine if it fails in yielding or in buckling. And so I just wanted to give you a couple of real world examples of buckling. This shows I-beams with local flange buckling which is quite common. If you're going to have buckling we don't want to have buckling that's a bad thing. But if you do see buckling the local flange buckling is something that you might see. Here's another example. This is the Windsor Tower fire that what took place in Madrid. And you can see that there was some unprotected steel perimeter columns and that they buckled. If we look here that you can see that they actually buckled as the fire took place. And so those are some real world examples. Let's go ahead now and do a worksheet. Let's look at a steel column that is made of a W30 305 by 74 wide flange I-beam. It's pinned at both ends for rotation about the X-axis where the X-axis is out of the page, and so bending in this direction. And then it's fixed at both ends against rotation about the Y-axis and the Y-axis is shown here. And we're determine the load P that will cause failure. And so here's a top view of a 305 by 74 wide flange I-beam. If you look it up in tables, first of all you can find the yield stress for steel. You can find the modulus of elasticity for steel. And you can find from tables for this W 305 by 74 I-beam that the area is as shown. The I about the area moment of inertia about the X-axis is 164 times 10 to the 6th millimeters to the 4th. And about the Y-axis, its 23.4 times 10 to the 6th millimeters to the 4th. And so the first thing I'm going to do is I'm going to check for yielding or the compressive load that would cause yielding. And so we know, if we go all the way back to my mechanics of materials course one. We know that the normal stress, and it's going to be compression in this case, is P over A. And so we have for P yield, we're looking for what will cause yielding, that's equal to sigma yield times the area. So P yield is equal to 250 mega Pascal's and we'll put everything in terms of Newton's and millimeters. And so that's the yield stress in steel. So we'll say Newton per millimeter squared is the same as a mega Pascal. And then we're going to multiply by the area, which is 9,485 millimeters squared. And if I do that calculation, I'll find that P yield, P that's going to cause yielding due to compression, is equal to 2,371,250 Newtons, or P yield is equal to, to three significant digits, 2370 kilonewtons. Okay, so that's that case. Now let's go ahead and look at buckling about the x-axis. We know that our buckling critical load, so we have P critical, and this is going to be about the x-axis, is equal to pi squared EI over L effective squared. And so P critical about the x-axis is going to be equal to pi times, let's go ahead and put the values in, steel's Young's modulus is 200,000 MPA. And we're going to convert to Newton's per millimeter squared which is the same as MPA. Then we're going to multiply that by I. I about the x-axis is 164 times 10 to the 6th millimeters to the 4th. And then we're going to divide that by the effective length squared. For buckling about the x-axis, we have pin-pin conditions. So L effective is the length itself. So in this case, that would be 6 meters or 6,000 millimeters squared. If I do that mathematics, I'll get the critical buckling load about the x-axis will be 8,992,306 Newtons, or in terms of kilonewtons, that's in three significant figures, 8,990 kilonewtons. Okay, one more case to check. What would that be? Okay, we need to check buckling about the y-axis. So again, we have our Euler buckling load P critical now about the y-axis is equal to pi squared EI over L effective squared. So P critical about the y-axis is equal to pi squared times E 200,000 megapascals, convert megapascals to Newtons per millimeter squared. And then I about the y-axis is given as 23.4 times 10 to the 6th millimeters to the 4th. Then we have to take and divide that by L effective squared. L effective about the y-axis now is fixed end conditions. And so the L effective will be for that case 0.5 times the total length, which is 0.5 times 6,000 millimeters. And we're going to square all that. And if we run the numbers we get P critical about the y-axis for buckling ends up equaling 5,132,194 Newton's or 5,130 kilonewtons. And so now we've checked for yielding and we've checked for buckling about the x and the y-axis. Let's look at those total results. Which one is going to govern now in this case? Which one is going to be the governing case for failure? And well, obviously, it's going to be the lowest P that we're going to experience when we get failure. And in this case that's due to yielding. And so this is a short compressive member. It's going to yield before it fails. But this same approach could be used for any column in this situation. And so you have a good analysis technique for solving these column problems.
