WEBVTT 00:00:00.035 --> 00:00:03.377 A steel strut one metre long will carry an enormous load. 00:00:03.460 --> 00:00:06.570 Make the same strut three metres long, from the same steel, 00:00:06.653 --> 00:00:11.216 with the same cross section, and it folds up under a small fraction of that. 00:00:11.413 --> 00:00:13.607 Nothing about the material changed. 00:00:14.818 --> 00:00:17.208 Here are two columns cut from one bar. 00:00:17.291 --> 00:00:19.718 The short one fails the way you expect. 00:00:20.031 --> 00:00:20.981 Push hard enough, 00:00:21.064 --> 00:00:25.628 the stress reaches the yield strength of the steel, and the material gives way. 00:00:26.424 --> 00:00:27.876 Now the long one. 00:00:27.980 --> 00:00:28.907 Push on it, 00:00:28.990 --> 00:00:33.484 and long before the stress anywhere in it comes near yield, it does this. 00:00:33.680 --> 00:00:35.272 It bows sideways. 00:00:35.697 --> 00:00:38.145 The material is still perfectly elastic. 00:00:38.228 --> 00:00:43.032 Unload it and it springs straight again. But as a structural member it has failed, 00:00:43.115 --> 00:00:45.624 because it will not take any more load than that. 00:00:46.328 --> 00:00:48.509 And the gap between the two is not small. 00:00:48.592 --> 00:00:53.318 Triple the length, and the load the column can take falls by a factor of nine. 00:00:53.421 --> 00:00:55.489 Nothing in the material noticed. 00:00:56.285 --> 00:00:57.538 This is buckling, 00:00:57.621 --> 00:01:01.174 and it is a failure of stability rather than a failure of strength. 00:01:01.486 --> 00:01:03.644 The question is what load it happens at, 00:01:03.727 --> 00:01:07.292 and why the length matters so much when the material does not. 00:01:08.689 --> 00:01:10.768 Here is the case we will solve. 00:01:10.964 --> 00:01:16.153 A straight column of length L, pinned at both ends, so each end is free to rotate 00:01:16.236 --> 00:01:18.489 but held against sideways movement. 00:01:18.593 --> 00:01:21.937 It carries an axial load P through the centroid. 00:01:22.734 --> 00:01:26.447 The material is linearly elastic with Young's modulus E, 00:01:26.530 --> 00:01:31.535 and the cross section has a second moment of area I about the axis it bends about. 00:01:31.848 --> 00:01:35.215 Those are the only four quantities the answer can depend on. 00:01:35.826 --> 00:01:38.184 The answer we are heading for is this. 00:01:38.380 --> 00:01:42.396 The critical load is pi squared E I over L squared, 00:01:42.479 --> 00:01:47.342 and the shape the column takes at that load is a single smooth half sine wave, 00:01:47.425 --> 00:01:50.398 zero at both pins and largest at midheight. 00:01:51.612 --> 00:01:55.781 I want both of those to fall out of the mechanics rather than be handed to you. 00:01:55.978 --> 00:01:59.345 And notice already what that formula does not contain. 00:01:59.449 --> 00:02:01.969 There is no yield stress in it anywhere. 00:02:02.510 --> 00:02:05.134 The load at which a slender column buckles has nothing 00:02:05.134 --> 00:02:07.596 to do with how strong the material is. 00:02:07.793 --> 00:02:12.287 Only with how stiff it is, and with how the length and the cross section combine. 00:02:12.483 --> 00:02:13.863 So let the column bend, 00:02:13.946 --> 00:02:18.324 write equilibrium for the bent shape, and see what the mathematics demands. 00:02:19.437 --> 00:02:21.656 Suppose the column is not straight. 00:02:21.852 --> 00:02:25.728 Not because something pushed it out of line, but suppose it simply is, 00:02:25.811 --> 00:02:31.060 by some small amount, and ask whether that bent shape can stand in equilibrium. 00:02:31.856 --> 00:02:35.258 The dashed grey line is where the axis used to be. 00:02:35.455 --> 00:02:37.636 Take a height x above the bottom pin, 00:02:37.719 --> 00:02:42.271 and call the sideways movement of the axis at that station v of x. 00:02:43.067 --> 00:02:46.644 Now cut the column there, and keep the piece below the cut. 00:02:46.957 --> 00:02:51.100 At the bottom, the axial load P acts along its original line of action, 00:02:51.183 --> 00:02:54.249 because that pin has not moved sideways at all. 00:02:54.790 --> 00:02:57.691 And there is no sideways reaction at the pin either, 00:02:57.774 --> 00:03:01.069 because there is no sideways load anywhere on the column. 00:03:01.152 --> 00:03:05.202 So the base hands the free body exactly one force, P, 00:03:05.285 --> 00:03:08.281 at zero offset from the original axis. 00:03:09.078 --> 00:03:11.018 That is the whole free body. 00:03:11.214 --> 00:03:16.312 One axial force at the base, and the internal shear and moment at the cut. 00:03:16.764 --> 00:03:18.936 Nothing else is acting on the piece. 00:03:19.732 --> 00:03:21.797 Take moments about the cut section. 00:03:21.880 --> 00:03:25.559 The load P is acting a distance v of x away from that section, 00:03:25.642 --> 00:03:28.719 so it applies a moment of P times v of x. 00:03:28.916 --> 00:03:32.481 The internal bending moment has to balance it exactly. 00:03:33.185 --> 00:03:38.885 So M of x plus P v of x is zero, and the bending moment at any section is minus 00:03:38.885 --> 00:03:44.087 the axial load, times how far that section has strayed from the line of action. 00:03:44.884 --> 00:03:47.601 Watch what that means as the cut moves. 00:03:47.798 --> 00:03:51.816 Near the ends the deflection is small, so the moment is small. 00:03:52.012 --> 00:03:56.483 At midheight the deflection is largest, and so is the bending moment. 00:03:56.966 --> 00:04:01.515 So the bending moment along the column has the same shape as the deflection itself, 00:04:01.598 --> 00:04:03.131 scaled by P. 00:04:03.328 --> 00:04:07.657 That is unusual. In ordinary beam bending you are handed the loading 00:04:07.740 --> 00:04:09.608 and you go and find the moment. 00:04:09.691 --> 00:04:12.048 Here the moment depends on the answer. 00:04:12.659 --> 00:04:15.678 And there is the loop that makes buckling what it is. 00:04:15.875 --> 00:04:18.581 Deflection produces bending moment. 00:04:18.685 --> 00:04:21.158 Bending moment produces curvature. 00:04:21.262 --> 00:04:23.829 Curvature produces more deflection. 00:04:24.025 --> 00:04:26.800 The load P is the gain around that loop. 00:04:27.597 --> 00:04:30.790 Now bring in the bending relation you already have. 00:04:30.987 --> 00:04:33.390 For small deflections, E I times 00:04:33.390 --> 00:04:37.919 the second derivative of the deflection equals the bending moment at that section. 00:04:38.715 --> 00:04:40.838 That is the small deflection form, 00:04:40.921 --> 00:04:43.790 and it is the honest limit of what we are doing here. 00:04:43.986 --> 00:04:46.957 As long as the bow stays small compared with the length, 00:04:47.040 --> 00:04:50.558 the curvature is well approximated by v double prime. 00:04:51.610 --> 00:04:53.945 Substitute the moment we just found. 00:04:54.141 --> 00:04:57.347 E I v double prime equals minus P v. 00:04:57.543 --> 00:05:01.596 Move everything to one side, and there is the governing equation. 00:05:02.392 --> 00:05:07.409 Divide through by E I, and give the group P over E I a name. 00:05:07.594 --> 00:05:09.348 Call it k squared. 00:05:09.544 --> 00:05:14.895 Then the equation reads v double prime plus k squared v equals zero, 00:05:14.978 --> 00:05:17.893 and k carries units of one over length. 00:05:18.689 --> 00:05:22.519 That is a second order linear equation with constant coefficients, 00:05:22.602 --> 00:05:25.772 and you already know every function that satisfies it. 00:05:25.969 --> 00:05:29.499 Solving it is what turns this into a statement about the load. 00:05:30.678 --> 00:05:34.007 v double prime plus k squared v equals zero. 00:05:34.091 --> 00:05:37.818 Every solution of that is a combination of a sine and a cosine 00:05:37.818 --> 00:05:42.172 of k x, so write the general one down with two constants in it. 00:05:42.783 --> 00:05:47.985 Now the end conditions. The bottom is pinned, so it cannot move sideways there. 00:05:48.181 --> 00:05:50.342 v of zero is zero. 00:05:50.434 --> 00:05:53.708 Put x equal to zero into the general solution: 00:05:53.893 --> 00:05:58.120 the sine term vanishes on its own, and what is left is B. 00:05:58.464 --> 00:06:02.745 So B is zero, and that leaves v equals A sine k x. 00:06:02.829 --> 00:06:06.615 The top is pinned too, so v of L is zero as well. 00:06:06.719 --> 00:06:12.339 Put x equal to L in, and you get A times sine of k L equals zero. 00:06:13.252 --> 00:06:16.387 Look hard at that line, because everything is in it. 00:06:16.584 --> 00:06:21.086 A product of two things is zero, so at least one of them has to be zero, 00:06:21.170 --> 00:06:24.317 and there are exactly two ways for that to happen. 00:06:25.114 --> 00:06:27.599 The first way is that A is zero. 00:06:27.703 --> 00:06:32.034 That gives v identically zero, the perfectly straight column. 00:06:32.231 --> 00:06:35.598 And notice it is available at every value of P. 00:06:35.795 --> 00:06:40.939 The straight configuration is always an equilibrium, and it never stops being one. 00:06:41.642 --> 00:06:44.766 The second way is that the sine itself is zero. 00:06:44.963 --> 00:06:48.725 Sine of k L vanishes when k L is a whole number of pi. 00:06:48.829 --> 00:06:53.462 And that is a condition on the load, because k squared is P over E I. 00:06:54.259 --> 00:06:55.580 Square both sides, 00:06:55.664 --> 00:07:02.015 substitute for k squared, and P L squared over E I equals n squared pi squared. 00:07:02.119 --> 00:07:06.531 So P equals n squared pi squared E I over L squared. 00:07:07.444 --> 00:07:10.379 These are not loads the column happens to reach. 00:07:10.463 --> 00:07:11.659 They are the only loads 00:07:11.659 --> 00:07:15.131 at which anything other than the straight shape is possible at all. 00:07:15.745 --> 00:07:19.113 Below the smallest of them, A has to be zero. 00:07:19.910 --> 00:07:22.499 The smallest is n equal to one. 00:07:22.812 --> 00:07:26.273 Pi squared E I over L squared. 00:07:26.469 --> 00:07:31.055 That is the Euler critical load, and it came out of demanding that a bent shape 00:07:31.055 --> 00:07:35.224 can stand in equilibrium, not out of any strength calculation. 00:07:36.020 --> 00:07:42.534 And the shape that goes with it is A sine of pi x over L. One half of a sine wave. 00:07:42.638 --> 00:07:44.191 Zero at both pins, 00:07:44.275 --> 00:07:48.768 largest at midheight, and no point of inflection anywhere in between. 00:07:49.565 --> 00:07:51.865 Now the part that catches people out. 00:07:51.968 --> 00:07:53.986 A is still undetermined. 00:07:54.070 --> 00:07:58.645 The mathematics has fixed the shape and said nothing whatever about how big it is. 00:07:59.349 --> 00:08:02.240 Here is the same mode at three different amplitudes. 00:08:02.437 --> 00:08:06.582 Every one of them satisfies equilibrium exactly, and every one of them does it 00:08:06.582 --> 00:08:08.475 at exactly the same load. 00:08:09.089 --> 00:08:11.993 That is the signature of the thing we are looking at. 00:08:12.905 --> 00:08:16.296 The higher values of n are genuine solutions too. 00:08:16.748 --> 00:08:21.436 n equal to two is a full sine wave, with a stationary point at midheight, 00:08:21.520 --> 00:08:23.645 and it needs four times the load. 00:08:23.749 --> 00:08:26.432 n equal to three needs nine times. 00:08:27.345 --> 00:08:29.490 In a bare column they never happen, 00:08:29.574 --> 00:08:32.849 because the column has already gone at n equal to one. 00:08:32.952 --> 00:08:37.446 But brace it sideways at midheight, and you have forbidden the first mode. 00:08:37.898 --> 00:08:40.209 The second one is what you get instead. 00:08:40.913 --> 00:08:43.120 Put all of that on one picture. 00:08:43.432 --> 00:08:45.860 Load runs up the vertical axis. 00:08:46.172 --> 00:08:49.176 The amplitude of the bow runs across the horizontal one, 00:08:49.260 --> 00:08:52.175 so the whole vertical axis is the straight column. 00:08:53.227 --> 00:08:54.923 Load it up from nothing. 00:08:55.538 --> 00:08:57.569 Below the critical load there is exactly 00:08:57.569 --> 00:09:03.143 one equilibrium at every value of P, and it sits on that axis at zero amplitude. 00:09:03.939 --> 00:09:07.145 At the critical load, a second branch appears. 00:09:07.341 --> 00:09:11.553 Every amplitude on that horizontal line is an equilibrium, at the one load, 00:09:11.637 --> 00:09:14.633 and the straight solution has stopped being unique. 00:09:15.546 --> 00:09:17.900 That splitting is called a bifurcation, 00:09:17.984 --> 00:09:21.711 and it is exactly what we mean when we say the column has buckled. 00:09:21.815 --> 00:09:23.008 Nothing broke. 00:09:23.092 --> 00:09:25.005 No stress reached any limit. 00:09:25.089 --> 00:09:28.190 The column simply ran out of reasons to stay straight. 00:09:28.986 --> 00:09:33.175 Now, that whole derivation used one particular pair of end conditions. 00:09:33.259 --> 00:09:35.767 Change them and the numbers change. 00:09:36.080 --> 00:09:37.729 What is worth seeing is that they change 00:09:37.729 --> 00:09:40.620 in a way you can read straight off the deflected shape. 00:09:41.703 --> 00:09:45.488 Everything in that derivation came from two facts about the ends. 00:09:45.685 --> 00:09:49.351 They could not move sideways, and they could not carry bending moment, 00:09:49.435 --> 00:09:51.259 because a pin cannot. 00:09:51.455 --> 00:09:54.579 Change either one and the answer changes. 00:09:55.492 --> 00:09:58.209 But look at what the pinned answer really is. 00:09:58.406 --> 00:10:03.492 The moment is P times the deflection, and the deflection is zero at both pins. 00:10:03.689 --> 00:10:08.217 So the half wave runs between two points where the bending moment is zero. 00:10:09.014 --> 00:10:12.289 So here is the statement that replaces the table. 00:10:12.485 --> 00:10:14.912 The critical load is pi squared E 00:10:15.109 --> 00:10:16.328 I divided by the square 00:10:16.328 --> 00:10:20.160 of the distance between adjacent points of zero bending moment. 00:10:20.357 --> 00:10:23.167 Call that distance the effective length. 00:10:23.964 --> 00:10:25.578 The reason is simple. 00:10:25.775 --> 00:10:27.351 Between two such points, 00:10:27.435 --> 00:10:32.034 the deflected curve is a half sine wave with no moment at either end. 00:10:32.230 --> 00:10:36.096 That is precisely the pin ended problem we already solved, sitting 00:10:36.096 --> 00:10:41.264 inside a longer column, and it does not care what happens beyond those two points. 00:10:42.060 --> 00:10:45.168 And zero bending moment means zero curvature, 00:10:45.252 --> 00:10:48.852 so those points are the inflection points of the deflected shape. 00:10:49.165 --> 00:10:53.078 You can find them by looking at where the curve changes the way it bends. 00:10:54.475 --> 00:10:58.771 Take a column built in at both ends, so neither end can rotate. 00:10:59.084 --> 00:11:02.335 Load it until it buckles, and it takes this shape: 00:11:02.439 --> 00:11:04.584 vertical where it meets each support, 00:11:04.668 --> 00:11:08.651 bulging one way in the middle, and curling back near each end. 00:11:09.564 --> 00:11:11.469 Find the inflection points. 00:11:11.572 --> 00:11:16.554 The curve bends one way near the middle and the other way close to each support. 00:11:16.751 --> 00:11:21.930 It changes over here, and again here, at a quarter of the length in from each end. 00:11:22.540 --> 00:11:25.850 So the distance between them is half the length of the column. 00:11:25.954 --> 00:11:30.351 That middle half is a half sine wave with zero moment at both ends, 00:11:30.435 --> 00:11:34.650 and it has no idea it is not a pinned column of length L over two. 00:11:35.447 --> 00:11:37.859 Put L over two in place of L. 00:11:37.943 --> 00:11:41.752 The critical load is four times what the pinned column would carry. 00:11:41.948 --> 00:11:46.326 The factor K is one half, and it is not a number from a table. 00:11:46.523 --> 00:11:48.869 It is where the curvature vanishes. 00:11:50.683 --> 00:11:52.190 Now the other extreme. 00:11:52.274 --> 00:11:56.718 Built in at the bottom, completely free at the top. This is a flagpole. 00:11:56.802 --> 00:11:58.889 It buckles into a quarter wave: 00:11:58.973 --> 00:12:03.594 vertical at the base, and leaning out at the top with no moment there at all. 00:12:04.205 --> 00:12:09.453 There is only one point of zero moment on the column, and it is the free end itself. 00:12:09.650 --> 00:12:11.485 So where is the other one? 00:12:11.682 --> 00:12:13.494 It is not on the column. 00:12:13.919 --> 00:12:16.590 Reflect the shape through the fixed base. 00:12:16.694 --> 00:12:19.791 The reflected curve meets the real one with the same slope, 00:12:19.875 --> 00:12:22.321 so the two together make one smooth curve, 00:12:22.405 --> 00:12:27.120 and it is exactly the pinned half sine wave, of total length two L. 00:12:27.917 --> 00:12:30.015 So the effective length is two L, 00:12:30.099 --> 00:12:35.046 the factor K is two, and the critical load is a quarter of the pinned value. 00:12:35.149 --> 00:12:39.806 A flagpole is four times weaker than the same member pinned at both ends. 00:12:41.202 --> 00:12:45.406 One case is left, and it is the only one that is not a round number. 00:12:45.719 --> 00:12:48.599 Pinned at one end and built in at the other. 00:12:48.795 --> 00:12:53.313 The pin gives zero moment, so one of the two points is the pin itself. 00:12:54.109 --> 00:12:57.094 The other one is the single inflection point up here. 00:12:57.290 --> 00:13:00.007 Where it sits comes out of a transcendental equation 00:13:00.007 --> 00:13:04.385 rather than a fraction, and it lands at about seven tenths of the length. 00:13:04.581 --> 00:13:07.194 So K is roughly zero point seven. 00:13:07.898 --> 00:13:09.791 Here they all are together. 00:13:09.988 --> 00:13:11.076 In every row, 00:13:11.160 --> 00:13:15.282 K is nothing more than the fraction of the column that lies between adjacent points 00:13:15.282 --> 00:13:21.169 of zero moment, and the critical load is pi squared E I over that length squared. 00:13:21.965 --> 00:13:24.114 One warning worth carrying. 00:13:24.310 --> 00:13:26.854 These are the idealised values. 00:13:27.468 --> 00:13:29.904 A real connection is never a perfect pin 00:13:29.988 --> 00:13:34.934 and never perfectly fixed, so design codes pull these numbers back toward one. 00:13:35.386 --> 00:13:38.777 But the geometry is the reason they are what they are. 00:13:39.816 --> 00:13:41.533 We have a critical load. 00:13:41.616 --> 00:13:45.318 To compare columns of different sizes we want a critical stress, 00:13:45.401 --> 00:13:49.036 so divide P critical by the cross sectional area. 00:13:49.739 --> 00:13:55.334 The cross section now appears twice in that expression, once as I and once as A. 00:13:55.417 --> 00:14:00.026 Those two are not independent, and the length that ties them together has a name. 00:14:00.823 --> 00:14:06.443 Define r, the radius of gyration, as the square root of I over A. 00:14:06.547 --> 00:14:08.671 It is the distance from the bending axis 00:14:08.671 --> 00:14:12.596 at which you could put the whole area and get the same second moment. 00:14:13.211 --> 00:14:15.371 For this rectangle, it is here. 00:14:16.167 --> 00:14:18.955 Substitute I equals A r squared. 00:14:19.151 --> 00:14:21.206 The area cancels, top and bottom, 00:14:21.206 --> 00:14:26.083 and what is left is pi squared E over the square of K L divided by r. 00:14:26.880 --> 00:14:29.400 That group is the slenderness ratio. 00:14:29.503 --> 00:14:32.337 Effective length over radius of gyration. 00:14:32.534 --> 00:14:36.595 It is dimensionless, it is the only geometry the answer needs, 00:14:36.678 --> 00:14:40.917 and it is what engineers actually mean when they call a column slender. 00:14:41.621 --> 00:14:43.813 So carry those two results across, 00:14:43.896 --> 00:14:46.704 and plot the critical stress against slenderness. 00:14:46.787 --> 00:14:49.502 It is a hyperbola: double the slenderness, 00:14:49.585 --> 00:14:52.720 and the critical stress falls by a factor of four. 00:14:53.517 --> 00:14:55.315 At a slenderness of two hundred, 00:14:55.398 --> 00:14:58.998 this steel column buckles at about fifty megapascals. 00:14:59.194 --> 00:15:02.559 Bring it in to a hundred and forty, and it is around a hundred. 00:15:02.642 --> 00:15:05.012 At a hundred, about two hundred. 00:15:05.924 --> 00:15:08.235 Now draw in the yield strength. 00:15:08.548 --> 00:15:12.601 Two hundred and fifty megapascals for a common structural steel. 00:15:12.914 --> 00:15:16.058 And look at what the Euler curve does on the left of the picture. 00:15:16.141 --> 00:15:18.058 It climbs straight through it. 00:15:18.854 --> 00:15:20.968 That part of the curve is a lie. 00:15:21.072 --> 00:15:23.661 It predicts a buckling stress well above the stress 00:15:23.661 --> 00:15:25.798 at which the material gives way. 00:15:25.994 --> 00:15:27.469 The column would have yielded long 00:15:27.469 --> 00:15:31.986 before it got there, and the derivation assumed elastic behaviour throughout. 00:15:32.782 --> 00:15:36.370 The crossing point is where the two mechanisms trade places. 00:15:36.567 --> 00:15:42.152 Set pi squared E over lambda squared equal to sigma y and solve. 00:15:42.465 --> 00:15:47.852 Lambda critical is pi times the square root of E over sigma y, and for this steel 00:15:47.852 --> 00:15:49.606 that is about eighty nine. 00:15:50.403 --> 00:15:54.513 So above about eighty nine, elastic buckling governs and the strength 00:15:54.513 --> 00:15:56.708 of the material barely matters. 00:15:56.904 --> 00:16:00.420 Below it, the column is stocky enough to reach yield first, 00:16:00.503 --> 00:16:03.952 and now the strength is what matters and the stiffness barely does. 00:16:04.749 --> 00:16:09.393 But that crossing is a sharp corner, and real columns do not have corners. 00:16:09.706 --> 00:16:11.144 Two things round it off, 00:16:11.227 --> 00:16:14.444 and both of them make the column weaker rather than stronger. 00:16:15.147 --> 00:16:18.677 First, every real column is slightly crooked to start with. 00:16:18.781 --> 00:16:22.926 So it begins bending from the very first increment of load, instead of waiting 00:16:22.926 --> 00:16:25.736 for a threshold and then choosing to bend. 00:16:26.533 --> 00:16:31.631 Second, a rolled section cools unevenly and locks in residual stresses. 00:16:31.827 --> 00:16:36.667 Part of the cross section is already close to yield before you apply any load at all, 00:16:36.750 --> 00:16:39.839 so it softens early and the whole member follows. 00:16:40.752 --> 00:16:43.109 The consequence is this red curve. 00:16:43.422 --> 00:16:46.137 Real columns sit below both idealisations, 00:16:46.220 --> 00:16:50.200 and the gap is worst right around the crossing, in the intermediate range, 00:16:50.283 --> 00:16:53.477 which is where most practical columns actually live. 00:16:54.389 --> 00:16:57.954 So codes do not use either straight idealisation. 00:16:58.151 --> 00:17:02.458 They use a fitted curve of this shape, calibrated against tests, with Euler 00:17:02.458 --> 00:17:06.047 as the upper bound it approaches once the column is slender enough. 00:17:06.843 --> 00:17:08.841 So, to gather it up. 00:17:08.945 --> 00:17:10.338 A slender column fails 00:17:10.338 --> 00:17:13.935 because the straight shape stops being the only equilibrium available, 00:17:14.018 --> 00:17:18.036 at a load of pi squared E I over the effective length squared. 00:17:18.832 --> 00:17:23.175 The end conditions enter only through where the points of zero moment sit. 00:17:23.488 --> 00:17:25.322 And the slenderness ratio tells you 00:17:25.322 --> 00:17:27.805 which failure you are actually designing against, 00:17:27.888 --> 00:17:32.080 with Euler an upper bound that gets more honest the more slender the column is.