A short block of steel crushes when overloaded; a long slender strut suddenly bows sideways and collapses at a much lower load. That sideways instability is buckling, and Leonhard Euler's 1757 analysis still governs slender-column design today. This article derives the critical load and applies it.
The Physical Idea
Buckling is not a strength failure — it is a stability failure. Below a certain axial load the straight column is stable: nudge it and it returns. At the critical load Pcr, a bent shape becomes possible with no extra load, and the column buckles. The material may still be well below its yield stress.
Derivation (Pin-Ended Column)
Take a pin-ended column of length L under axial load P. In a slightly bent position with lateral deflection y, the bending moment at any section is M = −P·y. The beam equation gives:
EI · d²y/dx² = M = −P y → d²y/dx² + (P/EI) y = 0This is simple harmonic in form, with solution y = A sin(kx) + B cos(kx), where k² = P/EI. Boundary conditions y(0) = 0 gives B = 0; y(L) = 0 gives A sin(kL) = 0. A non-trivial solution (A ≠ 0) requires:
sin(kL) = 0 → kL = nπ → k = nπ/LThe lowest (n = 1) load is the critical one:
P_cr = π² E I / L²Effective Length for Different End Conditions
Other end conditions are handled by replacing L with the effective length Le:
P_cr = π² E I / (Le)²| End Conditions | Effective Length Le |
|---|---|
| Both ends pinned (hinged) | 1.0 L |
| Both ends fixed | 0.5 L |
| One fixed, one free (cantilever) | 2.0 L |
| One fixed, one pinned | 0.7 L |
Fixing the ends shortens the effective length and dramatically raises the buckling load — fixing both ends quadruples Pcr compared with pinned ends.
Slenderness Ratio and Critical Stress
Buckling always occurs about the axis of least moment of inertia. Writing I = A·r² where r is the radius of gyration:
σ_cr = P_cr / A = π² E / (Le/r)²The ratio λ = Le/r is the slenderness ratio. The critical stress depends only on E and λ — not on material strength. That is why a slender aluminium and a slender steel strut of the same λ buckle at stresses in the ratio of their E values.
Worked Example 1 — Buckling Load of a Strut
A solid circular steel strut, 40 mm diameter, 2.0 m long, pinned at both ends. E = 200 GPa. Find the critical buckling load and check validity.
- I = πd⁴/64 = π × 40⁴ / 64 = 1.2566 × 10⁵ mm⁴
- A = πd²/4 = π × 40² / 4 = 1256.6 mm²; r = √(I/A) = √(125660/1256.6) = 10 mm
- Le = L = 2000 mm; λ = Le/r = 2000/10 = 200 (very slender — Euler valid)
- Pcr = π² E I / Le² = 9.8696 × 200 000 × 125 660 / 2000² = 62.0 kN
- σcr = Pcr/A = 62 000/1256.6 = 49.3 MPa < yield (250 MPa) ✔ Euler governs.
Worked Example 2 — Effect of End Fixity
For the same strut, both ends fixed. Find the new critical load.
- Le = 0.5 × 2000 = 1000 mm
- Pcr = π² × 200 000 × 125 660 / 1000² = 248 kN (four times the pinned value)
Limits of Euler's Formula
- Valid only when σcr < the proportional limit — i.e. for slender columns (λ greater than about 80–100 for mild steel).
- Short columns fail by crushing at the yield/crushing stress.
- Intermediate columns fail by combined action; empirical formulas such as Rankine-Gordon or the IS 800 buckling curves are used in practice.
Common Mistakes
- Using the major-axis I instead of the least I — columns buckle about the weak axis.
- Forgetting to apply the effective-length factor for the actual end conditions.
- Applying Euler to stocky columns where yielding, not buckling, governs.