How to use this calculator
- Enter or choose elastic modulus E.
- Enter or choose least second moment of area I.
- Enter or choose unbraced length L.
- Enter or choose effective-length factor K.
- Enter or choose cross-sectional area A.
- The critical buckling load updates as you type. Use Copy result or Share to keep it.
Method
P_cr = π² E I ÷ (K L)²; r = √(I ÷ A); slenderness = K L ÷ r; σ_cr = P_cr ÷ A
- E
- elastic modulus
- I
- least second moment of area
- K
- effective-length factor
- L
- unbraced length
- A
- cross-sectional area
- r
- radius of gyration
Worked example
- EI = 200 × 10⁹ × 500 × 10⁻⁸ = 1 × 10⁶ N·m².
- P_cr = π² × 10⁶ ÷ (1 × 5)² = 9.8696 × 10⁶ ÷ 25 = 394,784 N = 394.78 kN.
- r = √(500 ÷ 20) = 5 cm, so KL/r = 500 ÷ 5 = 100.
- σ_cr = 394,784 N ÷ 2,000 mm² = 197.4 MPa, below fy = 250 MPa, so elastic buckling governs.
A 5 m pinned-end steel column with I = 500 cm⁴ buckles elastically at about 395 kN.
What the result means
The Euler load is an upper bound for a perfect column. If σ_cr approaches or exceeds the yield strength (stocky columns), failure is inelastic and design codes use column curves that give lower capacities. Apply a safety factor before using the load.
Assumptions and limits
- Perfectly straight, centrally loaded, linear-elastic prismatic column.
- The effective-length factor K represents the end restraint you enter.
- Real columns have initial crookedness and residual stresses, so the true capacity is lower than P_cr, especially at intermediate slenderness.
- Not a code design check.
Questions
Which axis should I use?
The one with the smaller I (and its matching length and K), because the column buckles about its weakest axis.
Why does K = 0.5 give four times the load?
Load varies with 1 ÷ (KL)², so halving the effective length quadruples P_cr.
Please note: Quantities are estimates. Confirm requirements with your engineer or supplier before ordering materials.