The physics
A straight member under compression stays straight as long as the load is small. At the critical load Pcr a second equilibrium appears: a bent shape in which the bending stiffness exactly balances the moment of the compressive force about the deflected axis. Any disturbance then grows, and the member buckles sideways, often far below the load at which the material would yield.
With the deflection v(x) (up positive) and a compressive force P, equilibrium of a slightly bent element gives the buckling equation. For constant P in a part of the member:
Its general solution has four constants, fixed by two conditions at each end:
The conditions form a homogeneous system A(k)·C = 0. A bent shape exists only when det A(k) = 0, the characteristic equation: sin kL = 0 for a pin-ended member, cos kL = 0 for a cantilever, tan kL = kL for a fixed–pinned one. Its smallest root gives the critical load:
The effective length Le = KL is the length of a pin-ended member with the same critical load. It is the distance between the points of zero moment (inflection points) of the mode shape.
| Symbol | Meaning | SI unit |
|---|---|---|
| P | Compressive axial force (P = −N) | kN |
| P_cr | Critical (buckling) load | kN |
| α_cr | Critical load factor P_cr / P: all loads × α_cr buckle the member | — |
| E, I | Young's modulus; second moment of area about the buckling axis | GPa, cm⁴ |
| k | Wave number √(P/EI) | 1/m |
| K, L_e | Effective length factor and effective length K·L | —, m |
| r | Radius of gyration √(I/A) | mm |
| λ | Slenderness L_e / r | — |
| f_y | Yield strength | MPa |
Euler's cases and supports
- Pinned at both ends: sin kL = 0, kL = π, K = 1.
- Fixed at one end, free at the other: cos kL = 0, kL = π/2, K = 2.
- Fixed and pinned: tan kL = kL, kL = 4.4934, K = 0.699.
- Fixed at both ends (one end free to slide along the axis): kL = 2π, K = 0.5.
- Fixed and guided (the guided end can sway but not rotate): K = 1. Pinned and guided: K = 2.
Interior supports, hinges, and lateral springs split the member into parts, each with its own four constants. A spring brace of stiffness k at mid-length of a pin-ended member raises the critical load until k = 16π²EI/L³; a stiffer brace forces the second mode, with a node at the brace, and adds nothing more. Axial forces applied part-way along make the compression piecewise constant: only the compressed parts drive buckling, and tension stiffens.
Stability criteria
- Equilibrium: the straight member is stable for P < Pcr, neutral at P = Pcr, and unstable above it. In energy terms, the second variation of the potential energy, ½∫(EI·v″² − P·v′²)dx, is positive for every possible shape only below Pcr.
- Validity of Euler's formula: the Euler stress σE = π²E/λ² must stay below the yield strength, so λ must exceed λp = π√(E/fy). Residual stresses and crookedness make real columns fail earlier.
- Inelastic buckling: the Johnson parabola σcr = fy − fy²λ²/(4π²E) applies below λc = π√(2E/fy), where it touches Euler's curve. Alternatively, EN 1993-1-1 gives χ = 1/(Φ + √(Φ² − λ̄²)) with Φ = ½[1 + α(λ̄ − 0.2) + λ̄²], λ̄ = √(Afy/Ncr), and the curve's imperfection factor α (a₀ 0.13, a 0.21, b 0.34, c 0.49, d 0.76).
- Safety: the safety factor n = Pb/P is compared with the required one (Johnson), or the utilisation P/Nb,Rd with 1 (Eurocode 3, γM1 = 1.0, the applied load taken as the design load).
- Second order: with lateral loads or an initial bow, deflections and moments grow roughly by 1/(1 − P/Pcr). The largest stress P/A + MII·c/I shows why real members fail below Pcr.
Method and validation
The scene uses the beam tool's finite elements (cubic Hermite elements, at most L/160 long) with the consistent geometric stiffness of each element, which accounts for the axial force. The critical load factors are the eigenvalues of (K + λKG)·φ = 0. They are bracketed by a Sturm count (the number of negative pivots of the factorised matrix equals the number of buckling loads below λ), so no mode is skipped, and refined by bisection. The mode shapes come from inverse iteration. The second-order analysis solves (K + KG)·d = F with the initial bow included.
Automated tests compare the results with Euler's cases (agreement to about 10⁻⁹), the higher modes n²π²EI/L², a continuous column, the bracing threshold 16π²EI/L³, and Timoshenko's beam-column formulas for a uniform and a central point load and for an initially bowed column. The step-by-step solution is checked against the finite elements for every preset and hundreds of random members.
Assumptions and limitations
- Straight, prismatic, linear-elastic member; Euler–Bernoulli bending (no shear deformation).
- Small deflections: the critical load is a bifurcation point. What happens after buckling (the elastica) is not modelled.
- Buckling in the plane of the drawing only, about the chosen axis, with the same supports for both axes.
- Loads keep their direction as the member bends (dead loads, not follower loads).
- Lateral-torsional, torsional, and local (plate) buckling are not checked.
- The member is drawn horizontally; its self-weight, if included, acts across it.