Buckling

Slender members in compression can suddenly bow sideways at loads far below their crushing strength. Learn Euler's critical load, effective length, slenderness ratio, Johnson's formula and design against instability.

Press down on a plastic ruler standing on end: at some load it suddenly bows out sideways. It hasn't been crushed — it has become unstable. That's buckling, and it's one of the most dangerous failure modes because it gives little warning.

Columns, struts, piston rods, jack screws, thin panels and even drink cans are designed against buckling.

An analogy

A pencil pushed end-to-end between your fingers flips sideways; a short stub of pencil doesn't. Slenderness, not strength, is the problem.

Stability vs strength

A short, stocky column fails by crushing (σ = S_y). A long, slender one fails by buckling at a load that depends only on E, I and length — not on strength. High-strength steel doesn't help a slender column.

End conditions and effective length

Fixing the ends stops rotation and raises capacity. A fixed–fixed column carries 4× the load of a pinned–pinned one; a flagpole (fixed–free) carries only ¼. Real 'fixed' ends are never perfect, so codes use slightly larger K values.

Slenderness ratio

λ = KL/k. Euler's formula is valid only when σ_cr is below the proportional limit — for mild steel roughly λ > 100–120. Below that, use Johnson's parabola or code column curves. Buckling happens about the weakest axis (smallest k).

Imperfections and design codes

Real columns are slightly crooked and loaded off-centre, so they bend from the start and fail below P_cr. The secant formula and code buckling curves (Perry–Robertson, AISC, EC3 curves a–d) include imperfections and residual stresses. Plates and shells (local buckling) are even more imperfection-sensitive.

Formulas

• Euler load: P_cr = π² E I / (K L)²
• Critical stress: σ_cr = π² E / λ²
• Effective length factors: pinned–pinned K = 1, fixed–free K = 2, fixed–pinned K = 0.7, fixed–fixed K = 0.5
• Johnson (intermediate columns): σ_cr = S_y − (S_y λ / 2π)² / E

Worked example: Euler load of a steel strut

Given: Solid Ø50 mm, L = 3 m, pinned both ends; E = 200 GPa

Find: P_cr and slenderness.

• I = π × 50⁴/64 = 306 796 mm⁴
• P_cr = π² × 200 000 × 306 796/3000² = 67.3 kN
• k = d/4 = 12.5 mm → λ = 3000/12.5 = 240 (Euler valid)

Answer: P_cr ≈ 67 kN

Worked example: Effect of fixing the ends

Given: Same strut, both ends fixed (K = 0.5)

Find: New P_cr.

• L_e = 1.5 m
• P_cr = 67.3 × 4 = 269 kN

Answer: ≈ 269 kN (4× higher)

Common mistakes

• Wrong: Using the strong-axis I. — Right: Buckling occurs about the axis with the smallest I (largest slenderness).
• Wrong: Applying Euler to stocky columns. — Right: Check λ; for short columns use Johnson or yield.

Applications

• Hydraulic cylinders: Long-stroke piston rods are checked for buckling when fully extended.
• Building columns & bracing: Braces and floors reduce effective length of steel columns.
• Thin-walled structures: Aircraft stringers, cans and silos are limited by local and shell buckling.

Key takeaways

• P_cr = π²EI/(KL)².
• End fixity changes K from 0.5 to 2.
• Slenderness λ = KL/k decides Euler vs Johnson.
• Check the weakest axis and include imperfections.

Going further

• Eigenvalue buckling in FEM: Linear buckling analysis finds load multipliers and mode shapes; nonlinear analysis with imperfections gives realistic collapse loads.
• Lateral-torsional buckling: Beams buckle by twisting sideways — governed by unbraced length and warping stiffness.
• Post-buckling: Plates can carry load after buckling (effective width); cylindrical shells lose capacity abruptly and are imperfection-sensitive (knock-down factors).