 Yeah, this one works optional, so as far as it even came. But this is a good one. We started our initial look at these subjects, what now, 14 weeks ago or something. We looked first at axial loading and our, I don't even remember if we specifically stated it, but the objects being loaded, we're relatively short in that their length dimension was not terribly bigger than their width dimension. I don't remember that we actually said it. But that, if we did look at it more specifically, for ductile materials under this kind of loading, they would tend to deform something like this. And we did look at that a bit when we looked at the different strains and the different directions with Poisson's ratio. So a ductile material would tend to deform like that. A brittle material like a ceramic like brick or concrete would tend to, if pushed far enough, would tend to not really deform before it would finally just fracture and have a catastrophic failure. At least catastrophic if nothing more in terms of the part itself is no longer abuse. What we're going to look at now is axial loadings where the length is much greater than the width and that the mode of failure is not necessarily just one of some kind of catastrophic destruction, but more, quite possibly, something where the piece buckles rather than just deforms it. In terms of whatever's on the end of it loading in that way, it could be the same type of failure. If this was a column support for a building or a deck, it could be that this slope there is enough to cause what would be termed a failure of the structure. Maybe it doesn't come apart and maybe once the load's reduced, everything returns to normal as it happens to happen with elastic solvents. But it could be just this angle imposed on the structure is enough to cause what would be considered failure of the structure. If it's your Aunt Martha sitting on the edge of that deck at your wedding reception and she tumbles off the deck, you could consider that a failure because, well, first thing her lawyers are going to do is write you out of the will. So even if it's just deformation there, we want to prevent it. So we're going to look now at loading and possible failures of columns. Our target of concern is going to be what will term this critical load where that's the smallest possible load, sorry, the greatest possible load we can put on and not go into failure, however that might be defined. We'll define it based upon the normal stresses in the material because that's one way we can quantify this as we look at just single members of a structure, not looking at the tire structure. But it's certainly possible that even if the member itself doesn't fail, we can consider the structure itself as in failure. So we're going to model this column to start with as a simply pinned two-member linkage. So it's two pieces of equal length pinned together. We need to allow the top to move up and down so we can imagine that something like that will suffice. That way the top is free to move up and down as part of the deformation, but not free to move side to side because certainly you could imagine if you're loading a long skinny member and the top can go anywhere it wants, then the thing's just going to fall over. We don't even have any real loading on the piece. We just pushed it aside. So we've got to keep it so that it's stable side to side, but free to move up and down. And then we'll apply some load to that and try to see what we can estimate for it's possible failure. So as it's loaded, then it'll deform something like this, of course, as being an equilateral triangle because the two pieces are of equal length. Within the limits of our ability to sketch these things, it'll deform something like that. Just like the last couple weeks of concern to us is this amount of deflection from the neutral position. And just like we did with the beams earlier, we'll call that a little v. Now because these are elastic solids, there's a tendency of the solid to try to return itself to the neutral position, which we don't have modeled here yet. So what we'll do is imagine that there's a spring that becomes depressed as our model for a column deforms. And it's that spring trying to restore things that allows us to make a model of this business. So if that was length L, and this is L over 2, we'll use this part of the deflection there as theta. Remember, as usual, these pictures are grossly exaggerated just so we were able to see them. They're actually much, much smaller than this. The deflections and the length are much, much smaller. So let's see. Let's label this point A. If we look at a free body diagram of point A, we see something like, well, let me load it back in there, although I don't have anything going on here. So there's some load P there. This is a two-force member, but we don't, well, we will use that. So that puts that as the load there. And that's what P, well, let's see, we'll start it. Let's try it now in a second. We've also got this spring restoring force there. And then, of course, there's load from the bottom leg as well. So we have a free body diagram there on the pieces. This, that component is equivalent to P. That angle is theta. That's right, you know? Yeah. OK, yeah, we're OK. And so the horizontal balance here, this is then P tan theta. And we know then that the balance of the restoring force, the elasticity of the column itself that we're modeling with this spring must be 2 P tan theta. Now, we could also have modeled it as not this linear spring there. We could also have modeled it as having some kind of torsional spring there that would try to return it to its normal. It's just a matter of how an author of a book wants to look at it, which way he thinks his students will see it the best. Ours happens to use the linear spring and not the torsional spring as the model. But the results in the end are going to be the same. So now we've got this little bit of business there. We're going to have to use some of our very fair approximations such that we're talking about very small angles here. So if measured in radians, then the angle is equal to the tangent of the angle itself. So we're going to be able to be done with that. And the deflection, remember that's this piece there. The actual distance that the point was deflected will be approximately L over 2 theta, theta again in radians. That's just it's such a small angle that the arc length is essentially equal to the total displacement as it is. So we can put those together in the general displacement. Remember our spring formula, the spring forces, a spring constant times whatever the amount of deflection is. So we can put those two together. We get then a formula that allows us to put all the pieces together to that point. Now we want that to be less than or equal to the actual force that we calculated would be the restoring force as it is, which is 2 p theta. That's using both approximations there for small angles. Make sure everything's OK. No. Let's see, we want this spring to be greater than that. If it's not greater than that, then we're going to lose our piece. So this is the restoring force of the spring. This is the force due to the load. We want the restoring force of the spring to be greater than the load, otherwise it will fail. So this is restoring side. This is load side. So that makes sense now that we got it as greater than. If the load is greater than the restoring force, then it's going to fail. So we've then got a critical, a lower, actually an upper limit on our load that we call p critical equal to something like k L. Let's see, 2 comes over. Theta is canceled. We get k L over 4 as our design limit. Yeah, that's where we everything. That looks OK. That's just the turning point. Anything on either side of that, if our restoring force is greater than our load, then we have stability. And if our restoring force is less than the load being applied, we have unstability. In other words, the column cannot withstand those loads. And it's going to continue to buckle farther. As it goes farther, it deflects more. And sooner or later, it either fails or the restoring force becomes great enough. So we're going to use that p critical to be our limit. And we want to find just what it can do if it goes to any more than that, then we're going to lose, by whatever means, the integrity of the column. OK, so we've gotten down to pretty simple. You can imagine it depends upon the length of the solid. You've done this before, too long skinny things are much more wobbly than our short skinny things. We can even see that with my audio visual aid. A short piece is much more stable than is a long piece. It takes much less to make this deflect. The veins aren't even sticking out in my neck when I do this one. But when I do this one, it's ready to get deflect as much. They're supposed to be looking at the veins in my neck now. You can see it's much more stable. Even though they're same material, same cross-section, it's much easier to make the long one deflect. So the greater the L, the lower the more we have to worry about it, as we'll see in a little bit. OK, so let's continue our model. We've now got a way we can model the entire piece, not have to use that two bar linkage model again, where as we load this, now failure will come something like that. We want to model that now by putting these two things together. So we'll make an imaginary cut somewhere right in the middle. And we know we've got loads like that. But those are offset and make a couple. Going that way, so we need moment, internal moment, going the other way to counterbalance that moment that we have now with, again, this being the deflection at center column of V. Now we can put those two together just like we did in the last couple weeks. This is really no more than a beam in simple bending. We know then that that moment is related to that amount of bending by one of our bending formulas. And that's just what the spring needs to oppose. So that's got to equal minus p, pV. That's the size of this couple. And it has to oppose that to the minus sign. All right, we can rearrange things a little bit. And you'll like where this is going, because it means that not all of the classes you take are useless to us. So we can rearrange this a little bit, bring it over the pV, dividing through by EI. We can get down to that. Which looks like what? Looks like a differential equation. Specifically a linear, meaning the coefficient terms are constant, second-order differential equation. So most likely, you all have it already solved in your head, because you're taking d for q or d for q right now, aren't you? So everybody's got it solved, I assume. I know that as an undergraduate, I would solve things that quickly. Just by pretty much glancing at it. Yeah, just in fact, maybe I just don't even put up the solution. So quickly done. Anyway, I don't know. I have to look up this stuff and follow it along. So we've got that then as our linear first-order, sorry, second-order differential equation with solutions, something like C1 sine lambda x plus C2 cosine lambda x as a solution. Look fairly familiar. How do we find out what C1 and C2 are? Yeah, boundary conditions, some kind of known situation, such as when v equals 0, then we know that, I guess the other way around, at x equals 0, v equals 0. Because then I wrote down the same thing. At x equals 0, we know v equals 0. That's the business of us having that upper end in a track. We also know that x equals l, v equals 0 again, because we've got the bottom pinned. Unfortunately, that leads to C1 equals C2 equals 0, which is a trivial solution. It just says that there's no equation left. So that's not a sufficient solution for us. So what we'll do is just keep C2 equal to 0, because that is certainly a solution. And then just solve on the remainder part, which is simply another solution to these. It's not as if we're ignoring parts of the solution. So this will give us then the solution, the square root of p over e i x, no, l, yeah, l, actually using the lower limit. That equals 0 is now our solution. In terms of l, that's the only way that we'll get a solution in here where the sine of that equals 0, which will also satisfy our second boundary condition. Doesn't matter then what C1 is. If the second term is 0, it will be a sufficient solution. So that then translates to in the type of formalism we use for differential equations. Now that n is significant in that. It gives us possible modes of failure. Our beam could fail simply like that. That's n equals 1. It could also fail like this, which would be n equals 2, and so on. This second mode of failure, we could guarantee if we fix the beam at a midpoint, drawing's a little off a midpoint. But if we fix the beam at a midpoint so that it can't deflect side to side, then this is the more likely, in fact, the first mode of failure isn't even possible if we fix it at the end. What also happens, though, because of that is that the critical load for that failure is much greater in the second case, in fact, four times greater in the second case than it is in the first. Because solving for the critical load back here at this point then gives us that. And we see that if n is greater, if n is greater then our load goes up. Actually, it's n squared, doesn't it? Which will give us the fourth. We need. Yeah, because we put the L over. Then square, which will square the n. So if we have the second failure mode, then n equals 2, which gets squared. We go to a four times greater possible critical load. And you can see this in certain structures where, as there are silver columns, they might actually pin them together with a stringer of some kind to help prevent the middle from deflecting, which then allows the end to go up, allows then a greater critical load as a possibility there. So we'll use n equals 1 for most of what we're looking at here so that we find the minimum load that could cause failure that way. Anything worse than that is of protection. All right, so let's relate this down to the stresses. We're going to see in a deformed column like that. So the normal stresses, I think our book actually will call them the critical stress because then based upon the critical loads that we're determining here, remember that's the maximum load we can withstand and avoid failure. So now we'll relate it to the stresses. Was there a question? Yeah, would you use this for something like a foundation wall or something? Well, not necessarily for a wall. You might want to do it if these are the, you know, the, or the heart-free foundation. No, not really. I'm thinking more of just a timber, a wall made out of wood that you, the typical height of a wall is 8 feet. That's a long run without any support. So there are typically stringers put in there. But if you pour concrete around these, then they're not really columns anymore. Is a column a self made out of concrete? Yeah, yeah, you could except concrete won't fail by buckling like this. Concrete will fail. Remember that the stress strain diagram for concrete was something like that and then it just fails. It doesn't go through very much bendy. All right, so we're going to set n equals to 1 since that's the load of failure that supports the least critical load. Then we want to find the maximum critical load within that constriction. So we're going to use the limit here on the stresses as something like this. And then there might also be a factor of safety that we'll put in here, too, which we actually will do. OK. What's the pain in the arm? Oh, I've got the A. Good, because I need that right now. Since I over A, it's purely a matter of the geometry of the cross section. And we've seen it before, not in this class, but we've seen this before in dynamics and physics one. Our book gives it the symbol R, but we know it as the radius of gyration. Dynamics, we were using the symbol small k. Purely geometry, which means if you're buying beams like those I beams in the back of the book, that will all be specified and probably be right in the table. And so we've got then that I over A is, no, no, not the A, the L squared. Because what we'll do now is take that little piece there and define what's known as the slenderness ratio. We don't have any symbol for it, just define it, define as L over R, which as you get older goes down. Because not only do you shrink as a person, you tend to fill out as a person. So on human beings, L over R is constantly increasing. Not in my case, though, of course. So we've got then I squared over E times E times the inverse square of that slenderness ratio, which has no particular symbol to it. Pi squared E, all right. So I think we're OK. This almost seems to be missing because I have A, right? Pi squared R, L times E. Oh, we're missing. I think that's OK. I don't see exactly where my answer pert came in. Maybe we'll find it in a second with the notes. Is A's an area? Pardon me? Is that A just an area? Yeah, the cross-sectional area. In dynamics, we can find it as the restitution is I over a moment. This is area, first moment of area that we're using, not necessarily the first moment of mass like we're using. So the unit's different. Dynamics. For I, they're still in units. Well, yeah, they would be because there's mass moment of inertia, this is area moment of inertia. In dynamics, we use mass moment of inertia and actually a mass radius of gyration because of that. So they're similar, not exactly the same. If you remember, we first used it like that. But if the density is constant and the thickness is constant, then you get to what we had in statics, which is a first moment of area. But that's what's constant density and constant thickness. So they're related but not interchangeable directly. And so our stress as a function of this slenderness ratio is known as Euler's buckling equation. And it's going to look like something like that since it's 1 over the inverse square. So let's look at a little problem. All right. For maybe building a deck or something, it's very common that the columns are square, usually 4 by 4 because that's what's available. But let's determine you can't custom cut columns. They don't have to be 4 by 4. This is just a lot cheaper when you go to Lowes and get them. So we're going to look at the possibility of using square columns for a couple different Lowes. So for a wood, Young's modulus is about 13 gigapascals. And we want to find A, the dimension A, which is just the length of one side of a wood column, such that we stay under a critical stress, which we get from over here, of 12 megapascals, a factor of safety of 2.5, a column length of 2 meters, about 60, and for two different types of two different Lowes, 100 kilonewtons or 200 kilonewtons. So we have everything we're given, except we'll have to back out of the area. And since we're assuming it's square, then it's going to get a bit simpler as we step through it. OK, we'll apply the factor of safety. We have to apply it somewhere. It's usually specified. We'll apply it to the load. So we're going to design this for a critical Low. And we'll start with the first 100 kilonewtons times the 2.5. So we're going to design it for a critical load of 250 kilonewtons. That will over design it. We'll put in a lot more material in the column to withstand the 250, expecting only 100 will be on it. So we could solve for a critical moment of area from our equation. And all the pieces that are in there, pi squared e, that will give us a critical area and allows us to solve them for a. And then once we've got that, then we check that against the allowable stress that was given and determine if it was below that limit. So once the work's done, all we have to do is watch our units as usual, because we've got all those pieces now. So you check that with a proper unit. That should be in units of length to the fourth. So let's use meters, OK? So we've got all those pieces. Don't forget for a square shape like this, the moment of area is 1 to the power of a to the fourth. And down here in the area, of course, be a squared. So we can solve for a critical moment of area. From that, get the length, proposed length of a side. Double check it and make sure it's below the limit of the allowed normal stress. Anybody want to have this with the right units and meters? Travis, you look like you have it. What is that? 1.95 times 10 to the power of 6. And that's not what I have. Yeah, I have 7.8 times 10 to the power of 3 for I in here. Yeah, I've got 7.014 times 10 to the power of 2. What? That's what I have to do. Calculator's different than mine. Hey, you're out. Fellas, this is on tape, which will probably make the 5 to 10 times better. Give you that window. You got this now, Travis? OK. What did you forget to square? I used 100,000 instead of 200,000. Yeah, we're applying factor safety to the load. OK, from that, you should get a limit, then, a squared that you get in the controls. Actually, since we're we want to know the real normal stresses and check those, not the limit here. So this should actually be 100. Because if we put the 250 in there, we'd be putting the factor safety in twice. 2.1, you know what I got. What did you get for the dimension A? I got. Where did it go? 09834. 098. Using this, then backing A out of it, you got 098. Is that meters? So you might specify 100 millimeters instead of 98 millimeters. No, that's not millimeters. That's meters. We'd be using spaghetti if that was millimeters. It's terrible when it rains. I'll see you for no reason. OK, so maybe 100 millimeters makes this calculation real easy. Comes out something like about 10 megapascals, which is below the 12. So you use the factor of safety. Yeah, it's already been applied. If we apply it again here, then we've applied it twice. All right. You guys, well, I don't guess we need to, will you? I guess you could. Double check it for the 200 kilonewton load just to see what happens when it doesn't quite work great. Just redo it for the 200 kilonewton load. Do it there as well, remembering the factor of safety. We only want to apply it once, and then we'll see what we get there. So force goes up a little bit. We're going to need greater moment of aerated withstand that. So A goes up, and then see what happens to the stresses, whether they go up too much or not enough. The load's increased. We're also increasing the area, so it might still be OK. All right, this one's easy. That just doubles. 6 times 10 to the minus 3 meters to the fourth. That should give dimension on the side of, anybody have to have it? Yeah, 117. Yeah, 117 millimeter, something like that. Of course, you can't buy custom made wood with SI dimensions in America. You're going to have to buy this wood from Europe. Like a 770 millimeter. Yeah. We'll have to get out of here. Earl can handle it. It's slapped for a pretty hard word. Earl and Ace hardware can handle anything we send them. So you should get 14.6. I think that's what you said, John, anyway. And that's greater than the 12, so this, if we really do need the 200 kilonewton limit, we're going to have to design it based on the critical stresses rather than the critical buckling. It's close enough, though. You'll be gone by the time. No, because you build these in the summer. That's when your mother-in-law's in town. Sorry, Nana. Just kidding. She's watching this. OK, for a picture of what we have going on here, we have that Euler's buckling equation. That's just the equation we came up for the stresses a little bit earlier. But now we're designing for a particular limit and we can't go above that. So that's essentially what we're doing now. We don't want to stay below that critical limit. Stay somewhere down on the buckling curve, somewhere down there. OK, with that one, David, you're happy with that? That's for square timbers. And part of the reason square timbers are used for posts most commonly is because there is a bias in the orientation of the solid itself. If you use a column like this, this kind of column is much more likely to buckle with respect to this axis. Let me double check just what our book calls that, the y-axis. Yeah. Then it is with respect to the x dimension. And you know that, too, from the very same demonstration piece, as I push down on this, it's much more likely to bend perpendicular to the smallest dimension than it is. Can't even make it bend. Can't a little bit. Not easy. This way, it buckles very easily. And that's because of the aspect ratio of the cross section is much different. So square columns are probably most commonly used because they can be put in in any one of the two orientations possible without concern of which way they might buckle. I noticed they were doing that on the new wing of Lens Falls National on South Street. Doing what? They had the timbers. No, they had support there on the steel, but the support columns were square. The columns were square? Yeah, they were using the support building burn up. Yeah, well, whatever one gives a sufficient moment of area, two is economical enough to use. And three, are those going to be exposed? No, but they already tear your beams. Yeah, OK. Sometimes, because a square beam might look better than does a niving exposed, it depends. If you're doing a group hug, you want all that stuff out because it just looks cooler. So we'll investigate the possibility now with a beam like this. Oh, there's the dimensions I'm looking for. This will be two inches down here, four inches up there. We'll use Douglas Furr as a Young's modulus of about 1.9 times 10 to the sixth psi. Good thing it's in psi. We'd hate to have to use wood that has values. And the wood we grow in America doesn't come in SI units. Laub will stress of 3.78 times 10 to the third psi. 5 foot long, I guess 5 foot high is the better term for a column like this. OK. And oh, a factor of safety on allowable load of 1.5. So once we've figured out the load put in a factor of safety, that way you can have 1 and 1 half mothers-in-law on the deck. All right. So find slenderness ratio, this L over R, and then find the allowable load once we've determined that. Now, the concern is that it's more likely to buckle in one direction than the other. Well, experience tells us which it is, but it doesn't mean it's not something that we can double check. Ixx, that's the one we've been using mostly, which the 112th BH cubed, where B is the 2 inches, the 4 inches. Now, remember, we're not necessarily using that number. We have the radius of gyration as the piece down here. So we've got to determine what the radius of gyration is. Then we need to do the same thing in the y direction. Just to double check this in both directions. We know from experience which one to worry about in particular, but double check it anyway. Make sure it gives us what we expect it to give us. Get our numbers. We have the radius of gyration apart. A is the same for both of them. So whichever one's the smaller, it's going to be the one that's the greater concern. And it reduces to H over the square root of 12, B over the square root of 12. So it tells you right there which one's the lesser. And then once we've got that, we can do zone A for the critical load. It's obviously, hopefully it's obvious that we're going to need to use yy there as the critical radius of gyration. So that will give us a pre-critical load. And the allowable load is that times our factor of c. So we would expect this to fail at one place. We'll put even greater limits on that. We need it right. I want to reduce that. What, Chris? Why are we using a smaller radius of gyration? Because it's more likely to fail in that direction. That lowers our critical load. If we design it with the greater one, we'll have a higher critical load. If we use that, then it could be over what the lower critical load is. It does get a bit confusing when we have two things going on at once. Sometimes we want the greater in one place. Sometimes we want the lesser. Remember, this is our critical load. It's the upper limit that we would expect to be finding. So we may have a critical load yet. Check with somebody. That's me. Come on, let's talk to somebody. David's got it. For r, y, y, I have 5, 7, 7 inches. So the slenderness ratio is 5 feet over that, symbol for that. And this ratio, of course, is unitless. So that's about 104. Is that OK so far? That's easy for it. Oops, oops, oops. One of those days, this is May. Units wouldn't have worked out on the area, which is what, 8 square inches? So far, we have units up there. Oh, yeah, and this is unitless. So that'll give us units of pound. These are the slenderness ratios, 104 unitless and that's squared, a lot of good. For that, I don't use the pi. What's somebody else got for that? I have 13.9 kips and then we apply the factor of safety to that, reduce that by 1.5 just to make sure, just to make sure we'll only allow a lesser little bit than that one. That comes out to be 9.2. You have 11.1 for which one? For this one, you have to square pi. You have to square the slenderness ratio. Sure, if you're doing something else wrong. This cap. Everybody's talking about 13.9. Phil, you got 13.9? I don't, safety don't. All right. Notice that based on the ultimate limit, which is the business we looked at in the very first couple weeks, if you're looking at the stress, we use the ultimate limit stress for this part of it, then we find that the allowable load is much, much greater, meaning this is much more likely to fail by buckling than fail due to just simple normal stress. The type of thing we looked at in the first week was much greater danger than failing by buckling. Do you get it finally? You need it. You want a slide roll? Yeah, I'll do better off. Yeah. The batteries don't go down on those. Where did that question come from? The P to the critical. This one? Yeah. Where'd that come from? Right there on my nose. I don't think I had actually that equation. Remember, we had this critical stress equation, and that's where we brought in the slenderness ratio. This equation was pi squared d pi squared e over l r squared. If you solve this for the critical stress, then you get the equation we have over there. I just didn't take that extra step. You got it fixed finally? That was square root of 8. Square root of 8. All right. Slightly different, but related problem when we get started. Imagine using an I-beam for a column. We'll use our regular model. We'll modify this a little bit for more possibilities. On Monday. But imagine this. What we've got is the beam is stabilized at midpoint, which forces it into the second mode of failure. However, since it's an I-beam, it depends upon which way the I-beam is turned. So we're going to look at it as if we're looking down straight down at there. We can see the I-beam looks something like that. We'll be looking straight down at it, and we would see these supports if we look down there. But they're all the way down at the midpoint. So it's oriented in that way as opposed to it being an H-beam with two midsupports. So it's supported that way. Which means in the plane pictured, possible mode of failure is the second mode. In the plane, the opposite plane, where those supports are in that direction, it's now more likely to fail in the first mode. And there's a different moment of area for the two directions. So we have to make sure we've got that. So for this picture, we'll put y in that direction. Is that right? No, that's x. Because then this will be the xx direction that agrees with this one. And for it turned on its side, a 90 degree review, that'll be yy. So we could actually put the hidden line of the flange. Sorry, the web as we see it there, if that helps a little bit. So we're looking right down the y-axis, right at the beam like that. So we're seeing the two new lines of the flanges there. So we have to investigate both of those modes of failure. This is all other things being much less likely, because it'll fail in this way first. But it's not a symmetric beam in the two directions. So we have to look at both of the possible modes of failure. So here's some of the other information you need for this. Steel beam. So we'll use 29 times 10 to the third KSI as Young's modulus. An allowable stress of 42 KSI. Factor of safety, 2.5 on the buckling equation. Not too big a deal. We apply that then to the, where's the other piece of, oh, the type of beam it is. So once we find out the normal stress, then we'll apply it to there to apply it to where it was buckling equation. The other piece you need to know is this beam is a W8 by 24. Yeah, W8 by 24. Oh, also that it's 25 feet taller. I think that's all the pieces then you need. So determine a critical load based on the two possible failure modes, using that particular beam. Because it's more likely to bend in one direction than the other. These are two very different moments of inertia. And they're right in the table. You don't have to actually calculate those. But they're significantly different. And they're right there in the tables in the book. And they differ by a factor of about 4, which is surprising. But remember, these deformations are very small. So we're not used to seeing them in our practical experience, even imagining them. You'll find in the book that one has a moment of inertia about that axis of 82.8 and about the other axis of 18.3. That's the concern for this mode. This mode accepts the bends in this direction, then we're looking at moments in that direction. Also, the effective length is half on that one. But that's either putting in the N equals 2 or the half either way. All right, so that takes us to the end. We'll look at that one before Monday.