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Column Buckling Calculator (Euler's Critical Load)

Calculate a slender column's critical (Euler) buckling load and factor of safety against an applied load, for pinned, fixed, or free end conditions.

Critical (Euler) buckling load
493,480.2 N
Factor of safety
4.93
% of critical load applied
20.3%
Effective length (K x L)
2,000 mm (K = 1)

Where this falls

20.3 falls in the "Comfortable margin (factor of safety > 3)" range.

How it works

Euler's critical load is the axial compressive load at which a slender column suddenly buckles sideways rather than simply compressing: Pcr = π² × E × I ÷ (K × L)², where E is the material's elastic (Young's) modulus, I is the cross-section's second moment of area, L is the column's actual length, and K is an effective-length factor set by how its two ends are restrained. K × L (the 'effective length') is the length of an equivalent pinned-pinned column that would buckle at the same load.

K = 1.0 for pinned-pinned ends (free to rotate but not translate) is the baseline case. K = 0.5 for fixed-fixed ends (both rigidly clamped) halves the effective length, which — since Pcr is inversely proportional to length squared — quadruples the critical load versus a pinned-pinned column of the same actual length. K = 0.7 for fixed-pinned (one end clamped, one free to rotate), and K = 2.0 for fixed-free (one end clamped, the other completely free, like a flagpole), which quarters the critical load versus pinned-pinned.

The applied axial load you enter is compared against the critical load as a percentage and a factor of safety (critical load ÷ applied load). A factor of safety below roughly 2 leaves little margin before buckling in practice, and one below 1 means the applied load already exceeds the theoretical buckling point.

FAQ

Why does the same load produce such different critical loads for different end conditions?

Because Euler buckling depends on the effective length squared, not the actual length — a flagpole-style fixed-free column with K = 2.0 behaves, for buckling purposes, like a pinned-pinned column twice as long, and buckling load falls off with the square of length. Restraining both ends (fixed-fixed, K = 0.5) does the opposite, acting like a much shorter column.

Does this account for material yielding, not just buckling?

No — Euler's formula assumes purely elastic behavior and applies to long, slender columns where buckling happens before the material would yield under simple compression. Short, stocky columns typically fail by crushing/yielding at a lower load than Euler's formula predicts, which needs a separate check (e.g. Johnson's parabolic formula) not modeled here. Real structural columns should always be verified against the applicable design code by a qualified engineer.

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