Effective-Length Worked Example
Calculation of the Euler critical load for a fixed-pinned column, showing the effect of end restraint on the effective length and buckling load.
Problem
An aluminium column (E = 70 GPa, sigma_y = 280 MPa) has a rectangular cross-section 30 mm x 50 mm and length 2.0 m. The ends are fixed-pinned. Calculate the effective length and the Euler critical load.
Given
- E = 70 x 10^9 Pa
- sigma_y = 280 x 10^6 Pa
- Cross-section: b = 0.03 m, h = 0.05 m
- Length L = 2.0 m
- End conditions: fixed-pinned, K = 0.7
Step 1 — Cross-section properties
I_x = b * h^3 / 12 = 0.03 * (0.05)^3 / 12 = 0.03 * 1.25 x 10^-4 / 12 = 3.125 x 10^-7 m^4 (buckling about weak axis) I_y = h * b^3 / 12 = 0.05 * (0.03)^3 / 12 = 0.05 * 2.7 x 10^-5 / 12 = 1.125 x 10^-7 m^4 A = b * h = 0.03 * 0.05 = 1.5 x 10^-3 m^2 The smaller I governs: I_min = 1.125 x 10^-7 m^4
Step 2 — Effective length
L_e = K * L = 0.7 * 2.0 = 1.4 m
Step 3 — Slenderness check
r = sqrt(I_min / A) = sqrt(1.125 x 10^-7 / 1.5 x 10^-3) = sqrt(7.5 x 10^-5) = 0.00866 m Slenderness = L_e / r = 1.4 / 0.00866 = 161.7 lambda_t = pi * sqrt(70 x 10^9 / 280 x 10^6) = pi * sqrt(250) = pi * 15.81 = 49.7 Since 161.7 > 49.7, Euler buckling governs.
Step 4 — Euler critical load
P_cr = pi^2 * E * I_min / L_e^2 = pi^2 * 70 x 10^9 * 1.125 x 10^-7 / (1.4)^2 = 9.8696 * 70 x 10^9 * 1.125 x 10^-7 / 1.96 = 9.8696 * 7875 / 1.96 = 77,723 / 1.96 = 39,654 N ≈ 39.7 kN
Result
The Euler critical load is approximately 39.7 kN. If the column were pinned-pinned (K = 1.0), the critical load would be pi^2 * E * I / L^2 = 19.4 kN. The fixed-pinned end condition increases the buckling load by approximately 100% through the reduced effective length.
Assumptions and limitations
- The K = 0.7 factor assumes idealised fixed-pinned end conditions
- Partial rotational restraint in practice gives K between 0.7 and 1.0
- Imperfections and residual stress reduce the actual buckling load
- Buckling occurs about the weak axis (smaller I)