 That's why you get paid a lot. All right, we're doing our last little bit of subject matter here. We're going to look at what happens as things are structural support members as calling. It's not unlike what we started with. In fact, I'm not real sure. Well, part of why, you'll see in a second why we couldn't have done this in the first week because we did start with this type of loading in the first week. That's where we first introduced, of course, our notion of the normal stress. And we'll use that again this morning, but we didn't look at anything more than just that normal stress. We assumed that the only deformation would be if the material was ductile, then it would deform something like that. And if it was brittle, it would deform something like, well, I don't even need that outline. If it was brittle, it would deform something like this. It would just fracture along the line and slip, that's what concrete does. Our best example of a brittle material that's used in compressive support like that. And we even looked at why it would be that angle. Remember that the maximum shear stress in axial loading was at 45 degrees. Remember back when we did that, so that's very much what that's here. It's not as nice and neat as that. It might be a couple 45 degree angles in a couple of directions, making it look like a pretty jagged thing. But you go look at a piece of concrete or brick that fractures in axial loading like this. And it's just amazing when you look closely, however, it almost looks like it's, you'd expect it to be a crystalline structure when everything aligns, all the atoms line up nicely in a crystalline structure and then you get fractures along those planes. You almost think that that's the kind of thing going on. But we're gonna change a little bit now in our look at this. Those are not the only possible modes of failure. This is a failure in that things might not fit right even if the load is taken off and the thing goes back to its original shape. It might be such that that deformation is too great to be managed. Certainly that's a catastrophic time of failure. But the type of failure we're gonna look at now is very apparent with much longer members in axial loading because what they tend to do then is bubble. And so we're gonna look at that kind of possibility because we now know what can happen to material when it's in a bending load like this, which is something we didn't have in the first week when we were doing axial loading that led to our normal stress concern. So we'll take a look now at what the bending loads are and what kind of moments are being caused and what kind of normal stresses that leads to using our bending, the bending formulas we came up with in the last month or so that we've been working on that kind of problem. We just had transverse loads to cause the bending. Now we're gonna look at axial loads causing this bending. So to get things going, our model is going to be something like this. We'll imagine the top is supported such that it can roll up and down in this channel sort of thing. So there's no possibility of sideways deflection of this load. This load starts in this direction in this place and stays there. The only possible thing it can do of course is this all gets a little bit shorter but the load can't deflect sideways which is gonna be a completely different problem. That's not buckling. That's some kind of structural failure of a cross member that would have held it there. So our initial model is that we can allow the top of it to roll up and down as needed but just can't deflect sideways. Then the piece itself can deflect but we're gonna model this deflection as a two piece member pinned at the bottom, pinned in the middle with a spring here to represent the elastic nature of the material. So that if we release the load then the material that the object will return to its neutral position on its own. So this will be our first model for this type of loading. Fairly simple but leads to perfectly usable results which most of our simple models do. Thank God. I mean thank whoever. We'll take a free body diagram of this joint. So there's a little piece there and of course it's pushing on the spring. So the spring's gonna push back on it. We'll just call that F. It will be done with that for now because of course there isn't actually a spring there. And then in response to both the load and the spring force, there are of course then two forces there symmetrically directed. These are just two force members so we know that the forces lie directly along those. Not quite symmetric there in the drawing in the cartoon. Oh, I'll never call cartoons again. Maybe someday I'll tell you why. So we've got that equilibrium condition free body diagram of the forces on that joint. And it's this deflection there that we're concerned with. So we'll even use the same symbol we had for deflection. The last week or two we've been talking about that. So that's gonna represent our deflection there. So let's put some little things together. Let's see a column of length L. So that's L over two itself. So then we can put the rest of a little bit of model together if we want. So let's see. This piece here, that will be P itself, the vertical component. And if this angles theta, then this length here is P tan theta. And V will be for very small angles. We've done this kind of approximation before. V is very much like the arc length. And so we'll be very close for very small angles to something like L over two theta. We can finish with our little bit of a picture here. Let's see. F is K new, or KV, where K is the representative spring constant of that spring. We don't know what it is. We don't actually have a spring there, but we're gonna be done with that in a second, as you can imagine, on these type of things we've done before. And so putting the deflection in, oh, Bobby, putting the deflection in, we've got K theta, so something like that there. All right, so that's the spring force. The spring force we're using to model the elastic bending of this column under axial load in some kind of buckling response, but we're gonna try to avoid buckling failure, of course. And that leftward horizontal component must equal two of these rightward horizontal components, two because there's one top and bottom in each of the members. So this must equal then itself two P tan theta. That's just saying the right side's got to equal the left side for equilibrium. All right, we can tidy things up a little bit. Again, use our small angle approximation that tan theta is very close to theta itself. Again, remember, in radians. And then we can then say, so it equals, we get the two over, we get K, the theta cancels tan theta, we get K L over four. That about right? Which is what we're gonna call the critical limit. If P is less than that, we're okay. And if P is greater than that, then we have failure. So we want to, being of a gambling nature, we want to hit right at P as the critical limit and then take our chances that it won't fail and make sure that we're out of the country if and when it does. So we're gonna look at that as our critical limit for failure. Trouble is, of course, that we've got that spring constant in there when there isn't really a spring in the problem. So we have to deal with that here very, very quickly. All right, fair enough for a simple model to get us started for the buckling response of a column. Yeah. That's not supposed to be over four? Oh, over four, yeah, sorry. Yes, it is supposed to be over four. What's your problem there? All right, there we go. See, it's over four. You can check the tape. See, it's over four. All right, so that's where we got away. We're gonna be on the verge of failure, throwing in a good factor of safety and everything will be happy after that. So a little bit of different look at our simple model here. That looks like a guitar, I think. Now we'll go back to the more realistic response of this thing. It's simply pinned at the bottom under some kind of axial load. And this is all supposed to be symmetric there. So let's take a look at this more like what we've done in the past where we now take a look at the interior forces, the moments that are being caused there and the bending response to those. So this, even as a sub-piece, must itself be in equilibrium. So we know that there's a normal force there, P, because of the deflection, though, that sets up a couple that must be wrapped into with some more of a couple that way. We must have a moment interior response moment, something like that. Okay, we understand that kind of bending moment because we've been looking at that for a couple weeks now. We know that it's a factor of the flexural rigidity. Remember that has to do with the material chosen for the beams and whatever the area moment of inertia of the beam itself is times the second derivative of the deflection curve itself where X will just take from one end, let's see. So, oh, and that has got to equal minus P times whatever the deflection is minus because that's our convention for that direction. Okay, so far, everybody comfortable with that? So far, nice to, that's why we couldn't do this buckling kind of thing in the first week when we first look at axial loads. We didn't have this, it took us a while to get there, but now we can solve that, so let's see. So, V squared V DX squared plus over E is one of the solutions for that. We wanna see what would give us a zero bending moment for that type of thing. And then it looks right, all the minus signs in place. Yeah, we're okay. This should be exciting to about seeing this kind of thing. What is this thing? Second order homogeneous linear differential equation which you have probably already solved in your mind. Have you not? Do we, did you, you were there already with that? Come on, you're all on VVQ, aren't you? You see in that form, now you're okay with it? You're not excited to see that? Everybody's always, all students are always excited to see differential equations up here in their work. Because now they say, now I get it, now I know why you're making me do that. Okay, which solutions for the deflection C1 sine lambda x or lambda squared is the P over DI. That's one of the solutions you use for this type of thing. You've been here, even if the recall of it isn't immediate. So we gotta get rid of the constants by using boundary conditions. We really only have two boundary conditions involved in deflection. At x equals zero, the deflection is zero. Remember I had it purposely put in that, in that sort of a roller channel that made it look like the top of a guitar. And then also at the bottom it was pinned so there'd also be deflection there. I mean no deflection there. Gives us probably one of the most conveniently simple solutions ever that's zero equals zero and we're all done and we go home early. Because what more can you do with such a perfect, clean, simple solution is zero equals zero. And so what we'll do is we'll let one of them go to zero and then use the other one instead. So our book has chosen to let this one go to zero and use the solution to that one which then gives us the possibility. What's the, what x equals zero, the sign of zero is zero itself. But we get around that with different modes to it. P over E i. I guess to better say it, we'll take that boundary condition on that one and then just use this one. So we get that, we get then different modes of the solution itself where n can be integers from one to however many. The solution for n equals one is a buckling response of that. And this, in fact, this is very much the same type of thing you saw in physics too when you did vibrations, you didn't do vibrations in physics too or you didn't, you didn't remember. Anyway, this is one of the harmonic modes. If you shake a rope at the slowest frequency, you can get it to do that. If you change the frequencies, then you can get the next modal responses where the second one, this response, these are all buckling type responses. And it's very easy to see the first response with a simple column that buckles easily without much force at all. It'll go into the first mode of buckling, well, not failure. I'm not gonna push it that hard then I have to pay for this. I could greater muscle, get it into the second mode. You can see that the natural response is the first mode, the second mode is gonna require a lot more force. Also, it would require some kind of support at the center here to induce that. If this was a column in a building and there was a second floor here and then the top floor here and that was a load, you might get that type of response but it's a very difficult one to even model in a simple way there. So that gives us our first solution then where n equals one. And so now we have our critical limit. We can bring all this a little bit back together. Where am I? Gotta get it right. Pi squared EI over L squared. So the critical force, of course, has something to do with whatever the column itself is. And again, the area moment of inertia plays a critical part in this. But it's inversely proportional to the length which means a shorter column has a greater critical limit. You can put more load on a short column than on a long one. And that's also easy to demonstrate with a really long column. It takes almost nothing on an old decrepit professor like me has no trouble getting some bending going here. But a column of half the length, it's significantly harder. Can you see the strain in my muscle fibers there? Yeah! Do you see the doughnuts? Yeah. It's much harder. Easily stand up. Give it a try. You'll be amazed at how little it takes to buckle this and how even young, I don't know, if I trust you. Give it a try. I won't tell Mary Leslie if you break it. Don't. Can you see? Can you see? No. That one. There's nothing to it. Oh, yeah. Yeah. Bobby, you want to give her a try? Oh, good. Ready? I'm breaking it. No? How often do I bring out a visual aid? Give me the column, the column's dying for it. Forget it, you did. I'm breaking it. Bam! You got it. There's an activity here right there. It was good, wasn't it? Weren't you glad when you finally did it? Yeah. I don't know. It's a lot closer. What? It's not there, it's a lot closer. Ha ha ha. You guys, I'm gonna put these over here so you can't get to them. You've lost your one chance. You had a chance to do it. Pat stepped up, took the chance, enjoyed it. But you guys couldn't do that. All right, so we have now finally our critical load limit. We go above this for a column of this description. We greatly increase the chance of failure. We stay below that. In fact, if we stay well below this critical limit, we're in much better shape to avoid buckling of failure by that axial load. What's the lambda value that you have up there? Is that just lower? What do you guys call that in differential equations? Yeah, maybe you use a different symbol than lambda. You use omega, so you could have omega squared in the differential equation of solutions. Does it mean that we're gonna see it? It means exactly what it meant in any of those differential equations where it's a collection in a way of the pertinent variables I'd forget the term. You thought that we would see that that would focus on oscillations like the shoulder there? Yeah, while this isn't oscillating, it looks exactly like a flexible member in an oscillation. So let me just copy that back down there. Our critical load limit for buckling, but don't forget, this is all still just axial loading. So we also have a critical limit on the normal stress that we got back in the very first week. We can't violate either one of these. It's not one of those cases where if one's okay, it's all right if you see the other. You can't exceed either one because if we protect against buckling, but then it fails by normal stress failure, what good is protecting it from buckling when the thing fails anyway? Let's see, we've got one last little piece to put in here. It's gonna be a little bit of switch from dynamics using the radius of gyration like we do in dynamics. However, this, we're gonna give the symbol not K, not, which we use in dynamics because we ought to use K for something in here so we don't mix that up. So our book happens to use R, and so the radius of gyration. However, remember our moment of inertia in dynamics, we have a mass moment of inertia. This is an area moment of inertia. So in dynamics we did I over M and here we do I over A since it's an area moment. For constant thickness, constant density, solids, this comes out to be the same thing anyway because the density and thickness would cancel out of it and most of the things we're looking at are constant density and constant thickness. So then we put that in and we get our last little piece to it that I guess college designers like to use. So we have I coming out, one of the Ls coming out because we define this as the slenderness ratio which if you're watching the Royal Wedding this morning, the new princess, the Duchess of Cambridge has a very high slender dance ratio. And her sister too, two very, very slender young ladies. All right, so our symbol for it in this class is nothing. We just separate it out, high squared D over L over I, I'm getting this right. No, sorry, L over, where'd the R go? I over, oh, I got A in here, I need A in there. So we're a little short on the, yeah, we're a little short on geometry. That's okay, no, what I want to do is, yeah, that's okay, we'll do that but then we'll make another shift for it. This is all just algebra which is the general place where I get screwed up anyway. So let's get that radius of gyration in there. That's what we're trying to introduce here. And then we need an A in here. So there, I think that's okay. Is that all right with the algebra? Pah, L over R squared, yeah, pi squared E, okay, that's right. That's the piece we needed there. I, R squared over L squared times, we had pi squared so we need to take the squared out so we're good. All right, there we go. So it all just comes down to geometry of the column. Everything okay with that? Do I get all the algebra straight finally? What? I think that top right one's doing it all. This one? Yeah, the R squared's doing it top. That's not okay? Because the R squared does come up top. Once we take that invert and multiply, I think it's okay. If not, there's a little tiny algebra thing there. Because I know that last one's the right bit we needed. But the whole thing comes down to geometry, radius of dry areas, another cross section, length, which we've already seen is a big deal. We knew that the area has always been a concern with this. So let's give it a test, see what it means to us. So look at a Douglas fir column, saying just exactly the type of thing you put on a deck and dimensions on the piece, a two by four. Very common structural element in a deck. Put some axial, I mean some coordinate directions on it. X will be axial, of course, with Y up, Z across, just like we've done before. Douglas fir has a Young's modulus. Of course, we're gonna need that of 1.9 times 10 to the 6 PSI. And a normal stress limit. Remember, we've got to look at both the buckling and the normal stress limits. Here's 10 to the third. Let's figure out what we can do with that kind of column. All right, so let's find the slenderness ratio and the allowable load with a factor of safety of one and a half. Let's see, what's the slenderness ratio? Throw it down, didn't ya? You were thinking of the future queen. She was so pretty. And it was amazing as she came up in her limousine and the reporters pointed this out. She was holding her bouquet with one hand and waving with the other. That's how relaxed she was. Because if you're not relaxed, you need two hands on the bouquet. You didn't watch that, this one, you went up at 4 a.m.? Well, another historic, you missed Love Canal and now you missed the Royal Wedding. Love Canal, dang. You guys, you're taking part in it. So the slenderness ratio, remember, is the ratio of the length to the radius of gyration to the cross-sectional area. And then remember, R is the square root of I over A. So, what's the matter? So the slenderness ratio is I over L. I over L, is that what I wrote down? Who do you've been listening to? No, it says right here, L over R. You're taking notes in pencil, aren't you? Don't you, after all these weeks with me, you wouldn't take notes in pen. That'd be the dumbest thing you could possibly do. Sorry, it's wrong, it's L over R. I didn't mean L over, well, whatever I said, I didn't mean it. It doesn't even have a symbol, so we'll give it SR for slenderness ratio. All right, sorry about that. I'm just, she was so pretty. Finally, we saw her, we got to see her dress. She run along the limousine, do you have the veil over her head? You can barely see her shoulders, and they're saying, oh, we finally see the dress. Look how pretty it is. Just barely see her shoulders through the veil, through the window as she zips by. Those are reporters. They need to get a life. Apparently, I do too. All right, so it seems like a fairly simple thing to do, except there's a little bit of a concern in that the slenderness ratio that we now understand and has to do with the moment of inertia area, the moment of inertia with respect to the y-axis is different than that for the z-axis. So which of the two slenderness ratios is the smaller? Because remember, the slenderness ratio itself is on the top, so if we have the smaller one of those, then we know that's the smaller of the critical load limits. Y does not equal IZ, then we need to find out which one of those is smaller so we know which radius of gyration is smaller so that we have the best protection, the greater the slenderness ratio, the better the protection. Is it, yeah, remember the moment of inertia for a rectangular solid is one-twelfth BHQ. So whichever one has the smaller dimension in there is the worst direction, and that's, of course, the smaller of the two to four. So the IZ is much smaller than I, sorry, IY is much smaller than IZ. I don't even have to calculate it out. Well, you put the numbers in and answer that for yourself, but IY is going to be the smaller the smaller as Jake said, that's what he said, Jake, right? Remember perpendicular to the axis is the base that has the smaller, I mean, is the height, so that has the smaller moment of inertia because the gyration with respect to the Y axis. Watch your units, of course, the length is in feet, the radius of gyration using what we have here would be inches, then we need that in the critical calculations themselves. Remember the critical load need the slender and the ratio in there. And the greater that is, the greater our load, critical load is. What do we, what do I do now? Do something else up? It was well, I followed. But what? So you take it over some SR? Yeah, SR is on the bottom, I don't know, pi over SR if you want, the book doesn't use a symbol for the slenderness ratio. Who wouldn't have caught that inverse with the units? Because the units would still work out slender ratio and unit list. We have the slenderness ratio for us to use. Okay, we want the smaller one because the inverse of it is the bigger we get a better protection limit. Confirm 104, later on, that's what I got too. Now, you can see that indeed that's the right response because if the weaker direction is the y direction a moment in that direction is going to cause it to bend towards the bigger side and that's exactly what a column does. This will deflect side to side but will not deflect towards us because the greater slender or the lesser slenderness ratio is about the y axis, which in this case would be across the face like that. So the bending moment then is that way and that's just how these things tend to fail. What, Jake? Colin, you got 104? 52, I had 104. Who's the right? Oliver and Ursa? You got 104, Doobie? So it's a tie, Pat? 104. What do you have for the radius of gyration? Radius gyration, I over, square root of I over A, 112th, IZ. This is I, right? This is I, right? Then over two inches times four inches. Remember, you're in the book, I think it's like that. Actually, you have to be very careful that you're getting the right orientation and then the right B over H cubed or B times H cubed, the critical limit. It was 1.5 times the critical load you tied? Not times it because that increases it. That makes you think that you can put in a greater critical load than you can. So you want to divide the critical load into this. Remember, this is the critical limit. Anything over that will cause buckling and then under it is safe so you want, you know you want to go to the lower number. You can calculate this straight out and then just divide by the critical limit, critical load, or you can produce it by the factor of safety there if you wish. We're looking for the maximum force that we can support without buckling. So the lowest possible calculation that comes out for that is the safer. Up to including zero, I guess. But so we have all those numbers now. Let's take a quick check. Let's look at the allowable, which is just simply that, that load now over the area of the column. It uses 9.2's already in there. 0.2, a little bit over that. Is it 4.2? 1.2. Oh, 1.2. Sorry, I said 4.2. Which is way under this. So if we'd only looked at that, we would have put in much more load than the beam could have with buckling. So if we'd done this back on the first week, without looking at buckling, just looking at this, we would have put in a much greater load, but almost a little bit over three times the load it actually could have taken that would have caused it to buckle if we'd only looked at the allowable stress. All right, we have all the pieces there. Close enough, no sense starting another problem.