How to use it
- Choose Metric (GPa, cm⁴, m, cm²) or US (ksi, in⁴, ft, in²).
- Choose the end conditions, or Custom K to enter your own effective length factor.
- Enter E (steel about 200 GPa or 29,000 ksi, aluminium about 70 GPa), the smallest I of the section, the length and the area.
- Read the critical load, the critical stress, the effective length, the radius of gyration and the slenderness ratio.
Formulas
- Critical load P_cr = π² E I / (K L)²
- Critical stress σ_cr = P_cr / A
- Radius of gyration r = √(I / A)
- Slenderness ratio λ = K L / r
- Equivalently σ_cr = π² E / λ²
Worked example
A steel column, E = 29,000 ksi, I = 50 in⁴, L = 10 ft (120 in), A = 10 in², pinned at both ends (K = 1): P_cr = π² × 29,000 × 50 / 120² = 14,310,926 / 14,400 = 993.81 kip. The critical stress is 993.81 / 10 = 99.38 ksi, r = √(50 / 10) = 2.236 in and the slenderness ratio is 120 / 2.236 = 53.7. That stress is well above the yield strength of ordinary structural steel, so this column would yield before it buckled elastically: Euler’s load is not the limit here.
In metric: E = 200 GPa, I = 100 cm⁴, L = 3 m, pinned: P_cr = π² × 200 × 10⁹ × 10⁻⁶ / 9 = 219.32 kN. Fixed–free (K = 2) the same column carries only a quarter, 54.83 kN.
Tips
- Length matters most: the load falls with the square of the effective length. Doubling L or going from pinned to fixed–free cuts P_cr to a quarter.
- Apply a safety factor. Real columns have initial crookedness and eccentric loads; Euler’s value is an upper bound for an ideal column.
- Get I from the section tables or the moment of inertia calculator, and use the weak axis.