Euler Buckling: Why Slender Columns Fail by Instability

A steel strut one metre long will carry an enormous load. Make the same strut three metres long, from the same steel, with the same cross section, and it folds up under a small fraction of that. Nothing about the material changed. Here are two columns cut from one bar. The short one fails the way you expect. Push hard enough, the stress reaches the yield strength of the steel, and the material gives way. Now the long one. Push on it, and long before the stress anywhere in it comes near yield, it does this. It bows sideways. The material is still perfectly elastic. Unload it and it springs straight again. But as a structural member it has failed, because it will not take any more load than that. And the gap between the two is not small. Triple the length, and the load the column can take falls by a factor of nine. Nothing in the material noticed. This is buckling, and it is a failure of stability rather than a failure of strength. The question is what load it happens at, and why the length matters so much when the material does not. Here is the case we will solve. A straight column of length L, pinned at both ends, so each end is free to rotate but held against sideways movement. It carries an axial load P through the centroid. The material is linearly elastic with Young's modulus E, and the cross section has a second moment of area I about the axis it bends about. Those are the only four quantities the answer can depend on. The answer we are heading for is this. The critical load is pi squared E I over L squared, and the shape the column takes at that load is a single smooth half sine wave, zero at both pins and largest at midheight. I want both of those to fall out of the mechanics rather than be handed to you. And notice already what that formula does not contain. There is no yield stress in it anywhere. The load at which a slender column buckles has nothing to do with how strong the material is. Only with how stiff it is, and with how the length and the cross section combine. So let the column bend, write equilibrium for the bent shape, and see what the mathematics demands.

Suppose the column is not straight. Not because something pushed it out of line, but suppose it simply is, by some small amount, and ask whether that bent shape can stand in equilibrium. The dashed grey line is where the axis used to be. Take a height x above the bottom pin, and call the sideways movement of the axis at that station v of x. Now cut the column there, and keep the piece below the cut. At the bottom, the axial load P acts along its original line of action, because that pin has not moved sideways at all. And there is no sideways reaction at the pin either, because there is no sideways load anywhere on the column. So the base hands the free body exactly one force, P, at zero offset from the original axis. That is the whole free body. One axial force at the base, and the internal shear and moment at the cut. Nothing else is acting on the piece. Take moments about the cut section. The load P is acting a distance v of x away from that section, so it applies a moment of P times v of x. The internal bending moment has to balance it exactly. So M of x plus P v of x is zero, and the bending moment at any section is minus the axial load, times how far that section has strayed from the line of action. Watch what that means as the cut moves. Near the ends the deflection is small, so the moment is small. At midheight the deflection is largest, and so is the bending moment. So the bending moment along the column has the same shape as the deflection itself, scaled by P. That is unusual. In ordinary beam bending you are handed the loading and you go and find the moment. Here the moment depends on the answer. And there is the loop that makes buckling what it is. Deflection produces bending moment. Bending moment produces curvature. Curvature produces more deflection. The load P is the gain around that loop. Now bring in the bending relation you already have. For small deflections, E I times the second derivative of the deflection equals the bending moment at that section. That is the small deflection form, and it is the honest limit of what we are doing here. As long as the bow stays small compared with the length, the curvature is well approximated by v double prime. Substitute the moment we just found. E I v double prime equals minus P v. Move everything to one side, and there is the governing equation. Divide through by E I, and give the group P over E I a name. Call it k squared. Then the equation reads v double prime plus k squared v equals zero, and k carries units of one over length. That is a second order linear equation with constant coefficients, and you already know every function that satisfies it. Solving it is what turns this into a statement about the load.

v double prime plus k squared v equals zero. Every solution of that is a combination of a sine and a cosine of k x, so write the general one down with two constants in it. Now the end conditions. The bottom is pinned, so it cannot move sideways there. v of zero is zero. Put x equal to zero into the general solution: the sine term vanishes on its own, and what is left is B. So B is zero, and that leaves v equals A sine k x. The top is pinned too, so v of L is zero as well. Put x equal to L in, and you get A times sine of k L equals zero. Look hard at that line, because everything is in it. A product of two things is zero, so at least one of them has to be zero, and there are exactly two ways for that to happen. The first way is that A is zero. That gives v identically zero, the perfectly straight column. And notice it is available at every value of P. The straight configuration is always an equilibrium, and it never stops being one. The second way is that the sine itself is zero. Sine of k L vanishes when k L is a whole number of pi. And that is a condition on the load, because k squared is P over E I. Square both sides, substitute for k squared, and P L squared over E I equals n squared pi squared. So P equals n squared pi squared E I over L squared. These are not loads the column happens to reach. They are the only loads at which anything other than the straight shape is possible at all. Below the smallest of them, A has to be zero. The smallest is n equal to one. Pi squared E I over L squared. That is the Euler critical load, and it came out of demanding that a bent shape can stand in equilibrium, not out of any strength calculation. And the shape that goes with it is A sine of pi x over L. One half of a sine wave. Zero at both pins, largest at midheight, and no point of inflection anywhere in between. Now the part that catches people out. A is still undetermined. The mathematics has fixed the shape and said nothing whatever about how big it is. Here is the same mode at three different amplitudes. Every one of them satisfies equilibrium exactly, and every one of them does it at exactly the same load. That is the signature of the thing we are looking at. The higher values of n are genuine solutions too. n equal to two is a full sine wave, with a stationary point at midheight, and it needs four times the load. n equal to three needs nine times. In a bare column they never happen, because the column has already gone at n equal to one. But brace it sideways at midheight, and you have forbidden the first mode. The second one is what you get instead. Put all of that on one picture. Load runs up the vertical axis. The amplitude of the bow runs across the horizontal one, so the whole vertical axis is the straight column. Load it up from nothing. Below the critical load there is exactly one equilibrium at every value of P, and it sits on that axis at zero amplitude. At the critical load, a second branch appears. Every amplitude on that horizontal line is an equilibrium, at the one load, and the straight solution has stopped being unique. That splitting is called a bifurcation, and it is exactly what we mean when we say the column has buckled. Nothing broke. No stress reached any limit. The column simply ran out of reasons to stay straight. Now, that whole derivation used one particular pair of end conditions. Change them and the numbers change. What is worth seeing is that they change in a way you can read straight off the deflected shape.

Everything in that derivation came from two facts about the ends. They could not move sideways, and they could not carry bending moment, because a pin cannot. Change either one and the answer changes. But look at what the pinned answer really is. The moment is P times the deflection, and the deflection is zero at both pins. So the half wave runs between two points where the bending moment is zero. So here is the statement that replaces the table. The critical load is pi squared E I divided by the square of the distance between adjacent points of zero bending moment. Call that distance the effective length. The reason is simple. Between two such points, the deflected curve is a half sine wave with no moment at either end. That is precisely the pin ended problem we already solved, sitting inside a longer column, and it does not care what happens beyond those two points. And zero bending moment means zero curvature, so those points are the inflection points of the deflected shape. You can find them by looking at where the curve changes the way it bends. Take a column built in at both ends, so neither end can rotate. Load it until it buckles, and it takes this shape: vertical where it meets each support, bulging one way in the middle, and curling back near each end. Find the inflection points. The curve bends one way near the middle and the other way close to each support. It changes over here, and again here, at a quarter of the length in from each end. So the distance between them is half the length of the column. That middle half is a half sine wave with zero moment at both ends, and it has no idea it is not a pinned column of length L over two. Put L over two in place of L. The critical load is four times what the pinned column would carry. The factor K is one half, and it is not a number from a table. It is where the curvature vanishes. Now the other extreme. Built in at the bottom, completely free at the top. This is a flagpole. It buckles into a quarter wave: vertical at the base, and leaning out at the top with no moment there at all. There is only one point of zero moment on the column, and it is the free end itself. So where is the other one? It is not on the column. Reflect the shape through the fixed base. The reflected curve meets the real one with the same slope, so the two together make one smooth curve, and it is exactly the pinned half sine wave, of total length two L. So the effective length is two L, the factor K is two, and the critical load is a quarter of the pinned value. A flagpole is four times weaker than the same member pinned at both ends. One case is left, and it is the only one that is not a round number. Pinned at one end and built in at the other. The pin gives zero moment, so one of the two points is the pin itself. The other one is the single inflection point up here. Where it sits comes out of a transcendental equation rather than a fraction, and it lands at about seven tenths of the length. So K is roughly zero point seven. Here they all are together. In every row, K is nothing more than the fraction of the column that lies between adjacent points of zero moment, and the critical load is pi squared E I over that length squared. One warning worth carrying. These are the idealised values. A real connection is never a perfect pin and never perfectly fixed, so design codes pull these numbers back toward one. But the geometry is the reason they are what they are.

We have a critical load. To compare columns of different sizes we want a critical stress, so divide P critical by the cross sectional area. The cross section now appears twice in that expression, once as I and once as A. Those two are not independent, and the length that ties them together has a name. Define r, the radius of gyration, as the square root of I over A. It is the distance from the bending axis at which you could put the whole area and get the same second moment. For this rectangle, it is here. Substitute I equals A r squared. The area cancels, top and bottom, and what is left is pi squared E over the square of K L divided by r. That group is the slenderness ratio. Effective length over radius of gyration. It is dimensionless, it is the only geometry the answer needs, and it is what engineers actually mean when they call a column slender. So carry those two results across, and plot the critical stress against slenderness. It is a hyperbola: double the slenderness, and the critical stress falls by a factor of four. At a slenderness of two hundred, this steel column buckles at about fifty megapascals. Bring it in to a hundred and forty, and it is around a hundred. At a hundred, about two hundred. Now draw in the yield strength. Two hundred and fifty megapascals for a common structural steel. And look at what the Euler curve does on the left of the picture. It climbs straight through it. That part of the curve is a lie. It predicts a buckling stress well above the stress at which the material gives way. The column would have yielded long before it got there, and the derivation assumed elastic behaviour throughout. The crossing point is where the two mechanisms trade places. Set pi squared E over lambda squared equal to sigma y and solve. Lambda critical is pi times the square root of E over sigma y, and for this steel that is about eighty nine. So above about eighty nine, elastic buckling governs and the strength of the material barely matters. Below it, the column is stocky enough to reach yield first, and now the strength is what matters and the stiffness barely does. But that crossing is a sharp corner, and real columns do not have corners. Two things round it off, and both of them make the column weaker rather than stronger. First, every real column is slightly crooked to start with. So it begins bending from the very first increment of load, instead of waiting for a threshold and then choosing to bend. Second, a rolled section cools unevenly and locks in residual stresses. Part of the cross section is already close to yield before you apply any load at all, so it softens early and the whole member follows. The consequence is this red curve. Real columns sit below both idealisations, and the gap is worst right around the crossing, in the intermediate range, which is where most practical columns actually live. So codes do not use either straight idealisation. They use a fitted curve of this shape, calibrated against tests, with Euler as the upper bound it approaches once the column is slender enough. So, to gather it up. A slender column fails because the straight shape stops being the only equilibrium available, at a load of pi squared E I over the effective length squared. The end conditions enter only through where the points of zero moment sit. And the slenderness ratio tells you which failure you are actually designing against, with Euler an upper bound that gets more honest the more slender the column is.