A slender column fails not by crushing but by suddenly bowing sideways – buckling. Euler’s formula predicts this for long columns; for stockier ones yielding interacts with buckling and Johnson’s formula is used instead.
How to use the Buckling Calculator
- End conditions (effective length factor K)
- Column length L
- Area moment of inertia I
- Cross-section area A
- Material
- Young’s modulus E
- Yield strength σ_y
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Choose the end conditions, enter the length, the area moment of inertia (about the weak axis) and the cross-section area, and choose a material.
Inputs explained
| Field | What to enter |
|---|---|
| End conditions (effective length factor K) | Choose one: Fixed–fixed (theory 0.5), Fixed–fixed (design 0.65), Fixed–pinned (theory 0.7), Fixed–pinned (design 0.8) … |
| Column length L | Units: m, cm, mm, km, in, ft … |
| Area moment of inertia I | Units: mm⁴, cm⁴, m⁴, in⁴ |
| Cross-section area A | Units: mm², cm², m², in² |
| Material | Choose one: Aluminium 6061-T6, Aluminium 2024-T3, Aluminium 1100-H12, Steel AISI 4340 … |
| Young’s modulus E | Units: GPa, MPa, Pa, psi, Mpsi |
| Yield strength σ_y | Units: MPa, kPa, Pa, GPa, psi, ksi |
What each result means
- Radius of gyration r = sqrt([Area moment of inertia I]÷[Cross-section area A])×1000
- Slenderness ratio K·L/r = [End conditions (effective length factor K)]×[Column length L]÷sqrt([Area moment of inertia I]÷[Cross-section area A])
- Transition slenderness = sqrt(2×pi^2×[E]÷[sy])
- Euler critical load = pi^2×[E]×[Area moment of inertia I]÷([End conditions (effective length factor K)]×[Column length L])^2
- Johnson critical load = [Cross-section area A]×([sy] – ([sy]×[Slenderness ratio K·L÷r]÷(2×pi))^2÷[E])
Example values
How the main result changes when Column length L changes and everything else stays at the example values above:
| Column length L (m) | Critical buckling load (N) |
|---|---|
| 1.5 | 471,503.4171 |
| 2.25 | 435,882.6885 |
| 3 | 386,013.6684 |
| 3.75 | 321,896.3569 |
| 4.5 | 243,693.9358 |
| 6 | 137,077.8389 |
Formula
Euler: P_cr = π²·E·I / (K·L)² · Slenderness λ = K·L/r, r = √(I/A)
Johnson (λ < λ_c): P_cr = A·[σ_y − (σ_y·λ/(2π))²/E], λ_c = √(2π²E/σ_y)
Worked example
Inputs — End conditions (effective length factor K): Pinned–pinned (theory 1.0); Column length L: 3 m; Area moment of inertia I: 2,500,000 mm⁴; Cross-section area A: 2,000 mm²; Material: Structural steel A36.
Results — Radius of gyration r = 35.3553 mm; Slenderness ratio K·L/r = 84.8528; Transition slenderness = 125.6637; Euler critical load = 548,311.35562 N; Johnson critical load = 386,013.6684 N; Critical buckling load = 386,013.6684 N; Governing formula = Johnson (intermediate column).
Key terms
- Slenderness ratio
- Effective length divided by radius of gyration.
Frequently asked questions
What is the effective length factor?
A multiplier on the length that accounts for how the ends are supported.
Why use design K values?
Real connections are never perfectly fixed, so codes recommend slightly larger K.
Which axis should I use for I?
The weakest axis – the column buckles about it.
Why switch to Johnson's formula?
For short columns Euler overestimates the load; yielding governs instead.