All articles

Column buckling: why slender columns fail before they crush

Push on a ruler and it bows sideways long before the plastic is anywhere near breaking. Euler's buckling load, slenderness and the radius of gyration, how end conditions set the effective length, why stocky columns follow Johnson's parabola instead — and two structures that fell because a compression member buckled.

Press on a ruler

Stand a plastic ruler on a table and press down on its top. It does not squash or shorten in any way you can see. At some load it suddenly bows out sideways, and a little more force folds it right over. The plastic was nowhere near its crushing strength; the ruler failed because staying straight had become unstable.

That is buckling. It is a stability failure rather than a strength failure, it arrives with little warning, and for slender parts in compression it happens at a load far below the one that would crush the material. Every column, strut, brace and compressed chord in a structure has to be checked for it.

Euler's buckling load

In 1744 Leonhard Euler worked out the load at which a perfectly straight, slender column with pinned ends buckles: P = π²·E·I/L². E is the material's modulus, I the second moment of area of the cross-section, and L the length.

Look at what is missing: strength. Euler's load depends on stiffness and geometry alone. A slender rod of high-strength steel buckles at the same load as one of mild steel, because both have a modulus of about 200 GPa; an aluminium rod of the same size buckles at about a third of that load, because its modulus is about 70 GPa. And length is squared, so doubling a column's length cuts its buckling load to a quarter.

Slenderness and the radius of gyration

Divide Euler's load by the area and it becomes a stress: σ = π²·E/(L/r)². Here r = √(I/A) is the radius of gyration — a measure of how far, on average, the section's material sits from its centre. The ratio L/r, the slenderness, is the one number that says how prone a column is to buckle.

This is why columns are so often tubes. Spread the same area of material into a thin-walled tube and r grows, so the slenderness falls and the buckling stress rises — the same weight of steel carries far more as a tube than as a solid rod. A rectangular section is weakest about its thinner direction, which is the way it will always buckle.

End conditions set the effective length

How the ends are held changes the shape the column buckles into, and so the length that acts like a pinned column. That effective length is K·L. A column pinned at both ends has K = 1. Fixed at the base and free at the top, like a flagpole, it bows like half of a pinned column twice as long, so K = 2 and it carries a quarter of the pinned load. Fixed at the base and pinned at the top, K is about 0.7; fixed at both ends, K = 0.5 and the load is four times the pinned value.

Real connections are never perfectly rigid, so design practice uses larger factors for the fixed cases — the AISC steel specification's commentary recommends 2.1, 0.80 and 0.65 in place of 2, 0.7 and 0.5. Assuming a joint is fixed when it can rotate a little is one of the easiest ways to overestimate a column.

Stocky columns yield first: Johnson's parabola

Euler's stress grows without limit as a column gets shorter, but a real column cannot carry more than its yield strength. Worse, real columns are never perfectly straight and contain residual stresses from manufacture, so columns of middling slenderness start to yield before they reach the Euler load and fail at less than it predicts.

In 1893 J. B. Johnson proposed a parabola for this range: σ = σy − (σy/2π)²·(L/r)²/E. It starts at the yield strength for a very short column and meets Euler's curve smoothly at half the yield strength, at a slenderness of √(2π²E/σy) — about 126 for a steel with E = 200 GPa and σy = 250 MPa. Below that, Johnson governs; above it, Euler does. Modern design codes use more refined column curves, but the picture is the same: a plateau near yield, then Euler's hyperbola.

When columns buckle, structures fall

In 1907 the Quebec Bridge, then set to be the longest cantilever span in the world, collapsed during construction and killed 75 workers. A lower chord of one cantilever arm — a huge compression member built up from plates joined by light lattice bars — buckled because its lacing was too weak to make the parts act together, and the arm fell into the St Lawrence River.

In 1978 the roof of the Hartford Civic Center in Connecticut, a steel space frame, fell in under snow a few hours after a crowd had left. Compression members in the frame had been designed with too little resistance to buckling, and once some buckled the rest followed. Both failures came from members that were strong enough to carry their load, but not stable enough.

Le = LK = 1load ×1vs pinned ends

Buckling load is π²EI/(K·L)², so it scales with 1/K². The bracket marks the stretch of column that bends like a pinned column: all of it for pinned ends, the middle half for fixed ends, and for a flagpole just half of a pinned half-wave — which is why it carries a quarter of the load.

1. The length of a slender pinned column is doubled. What happens to its Euler buckling load?

2. Two slender columns are identical except that one is mild steel and the other high-strength steel. Which buckles at the higher load?

3. Why does Johnson's parabola replace Euler's formula for stocky columns?

Column buckling calculator →