Strong enough and still falls over

Press down on a metre rule stood on its end and it will not crush. It will bow sideways, suddenly, at a load far below anything the material minds — and if the load is removed it springs back, undamaged.

That is buckling, and it is a different kind of failure from everything else on this site. Nothing broke. No stress reached a limit. The column simply stopped being able to stay straight, and the load at which it stops is set by geometry rather than by strength.

Two failure modes, and the lower one wins #

A column has two ways to fail and takes whichever arrives first.

Squashing is the material running out. The load divided by the area reaches the yield stress, and it happens at $P = \sigma_y A$ regardless of length.

Buckling is the column running out of stability. Euler’s formula gives the load at which a perfectly straight, pin-ended column ceases to prefer being straight:

$$P_{\mathrm{cr}} = \frac{\pi^2 EI}{L^2}.$$

The two are plotted against each other in the figure. The horizontal line is squashing, which does not care about length; the falling curve is Euler, which cares about nothing else. The governing capacity is the lower of the two, so a column is stocky and crushes or slender and buckles, and there is a crossover between.

The formula’s contents are worth reading. $E$ and $I$ — stiffness, and how the material is arranged — appear on top; the length appears squared underneath. The yield stress does not appear at all. A column of high-strength steel and one of ordinary steel, identical in shape, buckle at exactly the same load. Buying stronger steel for a slender column buys nothing.

Why the length is squared #

The mechanism is a feedback loop that either converges or does not.

Give a straight column a small sideways nudge. The load, now applied to a bent member, has a lever arm equal to the deflection, so it produces a moment. That moment bends the column further, which increases the lever arm, which increases the moment.

Against it works the column’s own bending stiffness, which resists curvature. The competition has two possible outcomes: the stiffness wins and the column returns to straight, or the load wins and the deflection runs away.

The critical load is where they exactly balance. The restoring moment goes as $EI$ divided by the square of the length — a longer member has more curvature for the same tip deflection — while the disturbing moment goes as the load times the deflection. Setting them equal gives the inverse square, and $\pi^2$ falls out of the shape the buckled column takes, which is a half sine wave.

That inverse square is harsher than anything in strength calculations, where the relationships are linear or at worst quadratic in the right direction. It is why the height of a storey matters so much more to a column than its load does, and why a column spliced mid-height without lateral restraint at the splice is a very different member from the one on the drawing.

Slenderness, which combines all of it #

Comparing columns needs one number rather than three, and slenderness is it.

Writing $I = Ar^2$, with $I$ the second moment of area, defines the radius of gyration $r$, which is the distance at which the whole area could be concentrated to give the same second moment. Dividing the Euler load by the area gives a critical stress:

$$\sigma_{\mathrm{cr}} = \frac{\pi^2 E}{(L/r)^2},$$

and $\lambda = L/r$ is the slenderness ratio. One dimensionless number that contains the length, the size and the shape.

The crossover with yielding happens at $\lambda = \pi\sqrt{E/\sigma_y}$, which for ordinary structural steel is about 75 and for high-strength steel is lower — the stronger material crosses over sooner, because its squash line sits higher and meets the same Euler curve earlier. Another way of saying that strength does not help a slender column.

A column is judged by its largest slenderness, which means its weakest axis. An I-section has two very different radii of gyration, so as a column it can only use the smaller — which is the argument for hollow sections, where every axis is equally good.

Why real columns are worse than both curves #

The Euler load belongs to a perfectly straight column of perfectly uniform material loaded exactly on its axis. Nothing is.

A real column has an initial bow of perhaps a thousandth of its length, residual stresses from rolling or welding, and a load that arrives slightly off centre. Each of those means the column is bending from the first newton rather than from the critical load, and the bending grows as the load rises.

The consequence is visible in the third curve on the first figure: real columns fall below both the squash line and the Euler curve, and the shortfall is worst near the crossover. That is where the two mechanisms interact — the member is bending enough to matter and stressed enough to yield — and it is where most practical columns sit.

Of those three defects one can be computed rather than fitted. Residual stress is not an accident of a particular column: a hot-rolled I-section cools unevenly, the flange tips cool first and finish in compression, and the value is around thirty per cent of yield before the member has been loaded at all. That compression is already present when the load arrives, so the tips reach yield early — and a yielded fibre goes on carrying its share of the load while contributing nothing whatever to stiffness.

The two axes separating like that is the part worth pausing on, because one column curve cannot express it. What is happening is in the flange itself, drawn as the applied stress rises.

Every steel design code contains a set of column curves derived from testing thousands of real members, precisely because neither theoretical line is safe on its own in that region. The curves differ by section type and by axis, because residual stresses differ by how the section was made.

The amplification curve explains the shape of the failure. A column with an initial bow does not reach the Euler load and then buckle — it deflects progressively more as the load rises, and it fails when the combined bending and axial stress reaches yield somewhere. The load makes itself worse, and the buckling load is the asymptote of a process rather than an event.

The critical load without solving anything #

There is a second route to $P_{\mathrm{cr}}$ that needs no differential equation, and it is worth having because it generalises to problems where the equation cannot be solved and because it fails in an instructive direction.

The argument is energy. When a column bows, two things happen at once. The column stores strain energy in bending, an amount $\frac{1}{2}\int EI (y'')^2\,dx$. And the load descends, because a bowed column is shorter end to end than a straight one, doing work $\frac{P}{2}\int (y')^2\,dx$. Below the critical load the strain energy exceeds the work available and the column springs back; above it, the work exceeds the energy required and the deflection grows. The critical load is where they are equal:

$$P_{\mathrm{cr}} = \frac{\int EI\,(y'')^2\,dx}{\int (y')^2\,dx}.$$

Assume the true shape — a half sine wave — and this returns $\pi^2EI/L^2$ exactly. That is a check rather than a result.

The interesting case is assuming the wrong shape. Take a parabola, $y = x(L-x)$, which has the right end conditions and looks entirely plausible. The numerator gives $4EI L$, the denominator $L^3/3$, and the answer is

$$P = \frac{12EI}{L^2},$$

against the true $9.87EI/L^2$. Twenty-two per cent too high.

And it is always too high. Assuming a shape is imposing a constraint on how the column may deform, and a constrained column is a stiffer column, so every approximate mode shape overestimates the critical load. There is no version of this method that errs the other way, which makes the energy method a source of upper bounds — precisely the wrong direction for a safety calculation, and precisely the right direction for a check that can fail. If an assumed shape returns a value below the known answer, the arithmetic is wrong.

That asymmetry recurs across structural mechanics and it is worth naming here, because it is the same principle as the lower-bound argument in masonry seen from the opposite side: guessing a force system that satisfies equilibrium gives a safe underestimate of collapse, and guessing a displacement shape gives an unsafe overestimate. Which of the two a method belongs to decides whether an error in it is conservative or dangerous, and the question is worth asking of any structural approximation before trusting it.

The material property that decides is not strength #

The Euler formula contains $E$ and not $\sigma_y$, and the consequences of that are larger than they look, because the two properties vary very differently between materials.

Every structural steel has essentially the same modulus. Mild steel at 275 MPa yield and high-strength steel at 460 both have $E \approx 205$ GPa, so two columns of identical geometry buckle at identical loads while one is a two-thirds stronger material. All that the extra strength does is raise the squash line, which moves the crossover to a lower slenderness — the stronger steel is buckling-governed over a wider range of members than the weaker one.

Between materials the ratio moves. Aluminium has about a third of steel’s modulus at comparable strengths, so an aluminium column of the same shape buckles at a third of the load, and aluminium structures are stability-dominated to a degree that surprises anyone arriving from steel. Timber is similar in kind. Concrete has a modulus around a sixth of steel’s, which is one reason concrete columns are stocky.

The design instinct that follows is worth stating explicitly because it inverts the usual one. For a strength problem, a stronger material is a smaller member. For a stability problem, a stronger material is the same member, and the only things that help are a larger second moment of area, a shorter unrestrained length, or a stiffer material — the first two being geometry and the third being nearly unavailable, since the modulus of a given class of material is not a variable a designer can buy.

How much worse imperfections can get #

A real column falls perhaps ten or twenty per cent below the theoretical curves, which is bad but manageable. That figure is not a general property of buckling problems; it is one of the mildest cases there is.

The extreme is the thin cylindrical shell in axial compression — a silo, a tank, a rocket casing. Classical theory gives a critical stress, and tests in the 1930s produced failures at twenty to forty per cent of it, with scatter so wide that the results looked like measurement error. They were not. The explanation, given by Koiter in 1945, is that the cylinder has an enormous number of buckling modes at very nearly the same critical load, and a structure whose modes are clustered is savagely sensitive to imperfections, because a dent of almost any shape resembles some available mode closely enough to trigger it.

The practical response was an admission of defeat dressed as a coefficient: knockdown factors, empirical multipliers derived from the lower envelope of test data, used for decades without a theory that predicted them. A dent of the order of the wall thickness — which is within any reasonable manufacturing tolerance — can halve the capacity of a shell.

The general point matters beyond shells. Imperfection sensitivity is governed by how closely spaced the buckling modes are, not by how slender the member is. A column has well-separated modes and is forgiving. A shell has clustered modes and is not. And a structure whose members are all at the same slenderness — a space frame with identical compression chords, say — has borrowed some of the shell’s problem, because whatever triggers one member is present in all of them at once.

What it does to a beam #

Buckling is usually described as a column problem, and the same instability governs beams.

Lateral-torsional buckling is the compression flange behaving as a column while the tension flange holds it back. The remedy is restraint along the compression flange, and the requirement grows exactly as the section is made more efficient — a deeper, narrower beam is a better beam and a worse one.

What the ends and the restraints are doing therefore decides a beam’s capacity as much as a column’s, and a beam adequate on a drawing that shows restraints is not adequate on a site that has not installed them.

It is an eigenvalue, which is why the amplitude is missing #

The one genuinely strange feature of the result — that it says when a column buckles and not by how much — is not a defect of the derivation. It is what kind of problem this is.

The governing equation for a pin-ended column is

$$EI\,v'' + P\,v = 0, \qquad v(0) = v(L) = 0$$

and it is homogeneous: every term contains the unknown $v$. So $v = 0$ satisfies it for every value of $P$ — a perfectly straight column stays straight at any load — and for almost all values of $P$ that is the only solution.

At particular values a second solution appears, and it appears as a family: if $v(x)$ is a solution, so is $2v(x)$, and so is any multiple. The load is determined and the amplitude is not. That is an eigenvalue problem, the critical loads are its eigenvalues and the buckled shapes its eigenvectors, and the missing amplitude is a property of the mathematics rather than an omission.

Two things follow that are worth having.

There is a whole sequence of them. $P_n = n^2\pi^2EI/L^2$ — the second mode is four times the first, the third nine times — each with its own shape carrying $n-1$ internal nodes. A column reaches the first and never sees the others, because it has failed.

Unless the first is prevented. Hold the column at mid-height and the first mode cannot form; the lowest available shape is the second one, and the capacity is four times what it was. That is what a brace buys, and the factor of four is not a coefficient somebody fitted — it is $n^2$ with $n = 2$.

Where the model stops #

Perfectly straight, perfectly axial, perfectly elastic. Covered above, and the reason design uses empirical curves.

Pin ends. The formula as written is for a pin-ended column, which is an idealisation of a support like any other. Other end conditions change the effective length, and the change is large — a factor of sixteen in capacity between the best and worst of the standard four.

Elastic material throughout. Above about half the squash load, parts of a real steel section have yielded because of residual stress, and the effective stiffness is lower than $E$ suggests. The tangent-modulus theory addresses this and is why the real curve dips.

Overall buckling only. A thin-walled section can buckle locally — a flange rippling or a web folding — at a load unrelated to the member’s overall slenderness. Section classification exists to keep local buckling out of the picture, and a class 4 section cannot reach the capacity the Euler formula implies.

One buckling mode. A beam in bending can buckle sideways and twist at once, which is lateral-torsional buckling and has its own theory. An open section in compression can twist rather than bow.

The figures carry a distortion worth naming. The buckled shapes are drawn with a lateral deflection of tens of pixels against a length of a few hundred, an amplitude of perhaps a tenth of the length. A real column at its critical load has deflected by essentially nothing — the whole point of the critical load is that it is where deflection begins to run away, and the amplitude is indeterminate in the linear theory. Every drawing of a buckled column shows a post-critical state that the calculation cannot predict.

The ladder from here #

Later rungs: the Euler derivation from the governing differential equation. Effective length and the four standard cases. The radius of gyration and slenderness. The tangent-modulus and reduced-modulus theories. Imperfections and the Perry–Robertson approach. Column curves and how they were obtained. Local buckling and section classification. Lateral-torsional buckling. Plate and shell buckling, where imperfection sensitivity becomes extreme. And the energy method, which gets the critical load without solving the differential equation at all.

Euler published the result in 1744, and it was ignored by engineers for a century because the columns of the time were stocky enough to crush. It became indispensable the moment iron allowed slender ones.