Free engineering calculators

First version

Column buckling

Elastic critical load, first buckling mode and effective length.

m
MPa
m⁴
K=1
Critical load4,386.5 kN
Effective length6 mK = 1

Engine vs formula: π²EI/(KL)² · Relative difference 1.3 × 10−5 %

Open in Stabileo ↗

Sparse Cholesky solver (96 free DOFs)

Residual 1.48e-14 (96 free DOFs)

Assumptions

Linear elastic, small deflection, prismatic Euler–Bernoulli member. SI units: m, MPa, m⁴, kN, kN/m; E is converted to kN/m². Self-weight excluded. Elastic buckling only; not a code design check. Maxima are absolute values; x is measured from the left.

Buckling uses 32 frame elements. For fixed–pinned ends, K = 0.7 is the conventional rounded value; the exact value is about 0.6992. The engine resolves the end conditions directly, so the roughly 0.24% difference is mainly due to this rounding, not the finite-element discretization.

The mode amplitude is arbitrary, magnified for visibility; it is not a post-buckling displacement. The member axis is horizontal.

Closed form

Pcr = π²EI/(KL)²

Worked examples

L = 6 m; E = 200000 MPa; I = 0.00008 m⁴; EI = 16000 kN·m²

K = 0.5 → Pcr = 17545.9633797 kN

K = 0.7 → Pcr = 8952.02213251 kN

K = 1 → Pcr = 4386.49084493 kN

K = 2 → Pcr = 1096.62271123 kN

References

Timoshenko & Gere, Theory of Elastic Stability, 2 ed. (1961); Young & Budynas, Roark’s Formulas for Stress and Strain, 7 ed. (2002), Chapter 8.

Technical details

Production Stabileo solver · planar Euler–Bernoulli frame elements.

Total nodal DOFs: 99

Force equilibrium residual: 1.2 × 10−14 kN