Euler Critical Buckling Load

Also known as column buckling load · critical load

Pcr=π2EI(KL)2P_{cr} = \frac{\pi^{2} E I}{(K L)^{2}}

Worked example: pinned 4 m steel column, I = 1e-5 m^4 → 1234 kN — press Try an example to run it live, then adjust anything.

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Euler Critical Buckling Load explained

PcrK LEI

A slender column does not fail by crushing — it goes sideways, suddenly and without warning, at a load that has nothing to do with the material's strength. Leonhard Euler solved the problem in 1744 as an appendix on elastic curves, and the result is startling: only E, I and length appear. A high-strength steel column and a mild steel column of identical shape buckle at exactly the same load. A 4 m pinned steel column with I = 10⁻⁵ m⁴ carries PcrP_{\mathrm{cr}} = π² × 200 × 10⁹ × 10⁻⁵ ÷ 4² = 1.234 × 10⁶ N, about 1234 kN.

K captures the end restraint by converting the real length into the effective length between inflection points: 1.0 pinned-pinned, 0.5 fixed-fixed, 0.7 fixed-pinned, and 2.0 for a flagpole fixed at the base and free at the top — which means a cantilevered column buckles at a sixteenth of the pinned-pinned load. Two traps. Use the least I of the section, since the column buckles about its weak axis, and check slenderness: below KL/r of roughly 100–120 for steel the column yields before Euler's elastic curve is reached, and the formula badly overpredicts. The 1907 Quebec Bridge collapse, which killed 75 workers, came down to compression chords whose buckling capacity had been overestimated on exactly that kind of error.

Euler Critical Buckling Load formula

Pcr=π2EI(KL)2P_{cr} = \frac{\pi^{2} E I}{(K L)^{2}}
Where
  • PcrP_{cr}= Critical buckling load (N)
  • EE= Young's modulus (kPa)
  • II= Least area moment of inertia (mm⁴)
  • KK= Effective length factor
  • LL= Unbraced length (m)