How this instrument works
Buckling is not a strength failure but a stability failure: a slender column under axial load reaches a point where the perfectly straight, compressed shape stops being the lowest-energy configuration, and the slightest sideways nudge sends it bowing outward with no further increase in load. Euler's formula, Pcr = π²EI ⁄ (KL)², gives the exact load at which that switch happens for an ideal column — perfectly straight, loaded dead-centre, and still within its elastic range.
Every term earns its place. E, the elastic modulus, is stiffness, not strength — buckling resistance comes from how hard the material resists bending, never from how much stress it can absorb before yielding. I, the second moment of area, measures how that stiffness is spread around the cross-section's centroid; a flat ruler bends easily edge-on and barely at all flat-on because I differs by a factor of the width-to-thickness ratio squared. K rescales the real length L into an effective one, folding the end restraints into a single number: 1 for a column free to rotate at both ends, 0.5 when both ends are fixed against rotation, 2 for a flagpole fixed at the base and free at the top.
The formula assumes an idealised column, and every real one falls short of it. A slight initial crookedness, an off-centre load, or a weld's residual stress all reduce the load a real member can carry below Pcr, which is why codes divide it by a factor of safety before calling it an allowable load. The formula also has a lower bound: below a certain slenderness ratio, KL ⁄ r, the material yields before it buckles, and the smooth Euler curve has to hand off to an empirical one — Johnson's parabola or a code buckling curve — for short, stocky columns.
- Enter Elastic modulus — about 200 GPa for structural steel, 70 GPa for aluminium; the unit menu accepts Pa, GPa, ksi or psi directly.
- Enter Moment of inertia, the section's second moment of area about the axis it will buckle around — usually the weaker one — in m⁴ or mm⁴.
- Set Effective length factor from the end conditions: 1 for pinned-pinned, 0.5 fixed-fixed, 2 fixed-free, about 0.7 fixed-pinned.
- Enter Column length, the unsupported length between the points that restrain sideways movement.
- Read Critical buckling load in newtons, kilonewtons or pounds-force, then apply your code's factor of safety before treating it as a working load.
Worked example — a 100 mm square steel strut, pinned at both ends
Take a solid steel strut of square cross-section, 100 mm on a side, 3 m long, pinned at both ends so it can rotate freely but not translate — a common bracing member in a roof truss. Its second moment of area comes from the rectangle formula I = bd³ ⁄ 12 = (0.1)(0.1)³ ⁄ 12 = 8.33 × 10⁻⁶ m⁴, the same value about either axis, since the section is square. Steel's elastic modulus is E = 200 GPa = 200,000,000,000 Pa, and because the ends are pinned, K = 1.
Feed those four numbers into the formula: Pcr = π² × 200,000,000,000 × 0.00000833 ⁄ (1 × 3)² = 16,442,760.93 ⁄ 9 = 1,826,973.44 N — about 1,827 kN, or 1.83 meganewtons. That is the theoretical load at which the strut snaps sideways, independent of whatever grade of steel it is cut from, since ordinary structural steels all share very nearly the same 200 GPa modulus. A designer would then divide by a factor of safety — commonly around 1.92 to 2.0 for this kind of check — before calling something in the 900–950 kN range a safe working load.
Questions
How do I pick the effective length factor K?
K encodes how the column's ends are restrained. Use 1 for both ends pinned (free to rotate, not to translate), 0.5 for both ends fixed against rotation, 2 for a cantilever fixed at the base and free at the top, and about 0.7 for one end fixed and one pinned. Design codes such as AISC recommend slightly higher, conservative values than these theoretical ones, since real end connections are never perfectly rigid or perfectly free.
Why doesn't a stronger grade of steel resist buckling better?
Because buckling is governed by stiffness, not strength. Euler's formula uses only the elastic modulus E, and ordinary structural steels — mild, high-strength, or quenched-and-tempered — all share an E very close to 200 GPa regardless of yield strength. A higher-grade steel raises the load the column can carry before it yields, but does nothing for Pcr; only a stiffer material, a larger or better-shaped section, or a shorter effective length raise buckling resistance.
Does Euler's formula work for short, stocky columns?
Not accurately. It assumes the column stays elastic right up to the moment it buckles, which holds only above a critical slenderness ratio, KL ⁄ r, where r is the radius of gyration. Below that ratio the material yields before the Euler load is reached, so the formula overpredicts capacity. Short columns are instead checked against an inelastic buckling curve — Johnson's parabola or the AISC and Eurocode design curves — which blend into the Euler curve at the transition slenderness.
Why does the formula ignore the column's weight and strength?
Because neither one enters an elastic stability problem. The instant of buckling is decided purely by stiffness (E and I) and by how far the ends are apart in an effective sense (KL); mass and yield stress play no role in this particular calculation. Weight matters for the gravity loads applied elsewhere in a design, and strength matters once a section is also checked for crushing or for buckling below the elastic range — but not inside this formula.
Where does the π² in Euler's formula come from?
From solving the differential equation for a bent column, EI y'' = −Py, whose solution is a sine wave. For a pin-ended column of length L, the deflection must return to zero at both ends, which happens only at discrete loads where L times the square root of P ⁄ EI equals a whole multiple of π. The lowest such load — the one that actually governs — comes from setting that product equal to exactly π, giving P = π²EI ⁄ L².