Calculate the critical buckling load Fkrit of a slender compression member from elastic modulus E, smallest second moment of area I and buckling length lk using Fkrit = π²EI/lk². Above this load the member deflects sideways even though the compressive stress may still be far below yield – buckling is a stability failure, not a strength failure.
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Calculate the critical buckling load Fkrit of a slender compression member from elastic modulus E, smallest second moment of area I and buckling length lk using Fkrit = π²EI/lk². Above this load the member deflects sideways even though the compressive stress may still be far below yield – buckling is a stability failure, not a strength failure.
A pinned-pinned steel member (Euler case II, β = 1) with l = 2,000 mm, E = 210 GPa and I = 100 cm⁴ has lk = 2,000 mm and therefore Fkrit = π²·210,000 N/mm²·10⁶ mm⁴/(2,000 mm)² ≈ 518 kN. The same member fixed at one end and free at the other (case I, β = 2, lk = 4,000 mm) carries only a quarter of that, about 130 kN.
Ideal straight member under centric load, homogeneous material, constant cross-section, linear-elastic behaviour up to buckling and small deflections. The result is valid only while the corresponding buckling stress σkrit = Fkrit/A stays below the material's proportional limit, i.e. above the limit slenderness λ₀; for stocky members the Euler formula overestimates the capacity. Initial curvature, load eccentricity, residual stresses, inelastic buckling, torsional buckling, lateral-torsional buckling and the safety factors required by codes are excluded.
A slender compression member can deflect sideways long before its compressive stress reaches yield. This calculator determines the critical Euler buckling load Fkrit and also rearranges the equation for elastic modulus, second moment of area and buckling length.
Buckling is a stability failure: at the critical load Fkrit a neighbouring, buckled equilibrium position exists alongside the straight one. Below Fkrit the member returns to its straight shape after a disturbance; above it, it does not. Fkrit is therefore a bifurcation load (an eigenvalue of the buckling equation), not a stress limit. Crucially, only the flexural rigidity EI and the buckling length lk appear – material strength does not enter the formula at all.
Stand a ruler upright on a table and press down on it: it does not break, it suddenly flicks sideways once a certain force is reached. Hold the same ruler at mid-height and it takes far more force, because you have halved its buckling length – exactly what lk expresses.
Fkrit = π² · E · I / lk²
General: Fkrit = π²·E·I/lk²Buckling length: lk = β·lEuler case I (fixed/free): β = 2, Fkrit = π²EI/(4l²)Euler case II (pinned/pinned): β = 1, Fkrit = π²EI/l²Euler case III (fixed/pinned): β ≈ 0.699, Fkrit ≈ 2.04·π²EI/l²Euler case IV (fixed/fixed): β = 0.5, Fkrit = 4π²EI/l²| Symbol / input | Meaning |
|---|---|
| Buckling load Fkrit | The critical compressive force at which the member buckles out of its straight position (bifurcation load). It is a stability limit of the ideal member, not a permissible working value: codes require a safety factor against buckling, and initial curvature or eccentric load application reduce the actual capacity further. |
| Elastic modulus E | Elastic modulus of the member material, i.e. its stiffness in the linear-elastic range. Source: material tables – structural steel ≈ 210 GPa, aluminium alloys ≈ 70 GPa, grey cast iron ≈ 100 GPa. The buckling load depends only on E, not on yield strength: a high-strength steel buckles at exactly the same load as a mild steel of identical geometry. |
| Smallest second moment of area I | The second moment of area about the axis the member will buckle about. If the member can deflect in any plane the smallest principal value I₂ governs – a member always buckles about its weak axis. Source: a section table, a formula collection, or the principal-moments-of-inertia calculator for arbitrary cross-sections. |
| Buckling length lk | The reduced buckling length lk = β·l, not the member length l. The factor β follows from the end conditions (Euler cases): case I fixed/free β = 2, case II pinned/pinned β = 1, case III fixed/pinned β ≈ 0.699, case IV fixed/fixed β = 0.5. Because lk enters squared, choosing it correctly is the single most effective lever in the whole calculation. |
E is the material's elastic modulus (structural steel ≈ 210 GPa, aluminium ≈ 70 GPa), taken from a material table. I is the second moment of area about the axis the member deflects about; if it can buckle in any plane, use the smallest principal value I₂. lk is the reduced buckling length lk = β·l with the end-condition factor β of the matching Euler case – not the member length.
First identify the end conditions at both ends and read off β: β = 2 (one end fixed, other free), β = 1 (both ends pinned), β ≈ 0.699 (one end fixed, other pinned), β = 0.5 (both ends fixed). Form lk = β·l and enter it with E and the smallest I. Then check whether the Euler formula applies at all: compare σkrit = Fkrit/A with the proportional limit, or the slenderness λ with the limit slenderness λ₀. Finally apply the required safety factor against buckling – Fkrit is not a permissible load.
A pinned-pinned steel member with l = 2,000 mm, E = 210 GPa and I = 100 cm⁴ = 10⁶ mm⁴ is Euler case II, so β = 1 and lk = 2,000 mm. This gives Fkrit = π²·210,000 N/mm²·10⁶ mm⁴/(2,000 mm)² = 518,154 N ≈ 518 kN. With one end fixed and the other free (case I, β = 2, lk = 4,000 mm) the buckling load drops to a quarter, about 130 kN. With both ends fixed (case IV, β = 0.5, lk = 1,000 mm) it rises fourfold to about 2,073 kN. These ratios 0.25 : 1 : 2.04 : 4 are precisely the four Euler buckling loads.
Fkrit grows linearly with E and I but falls quadratically with buckling length: doubling lk leaves a quarter of the load. To strengthen a compression member, the biggest gain therefore comes from a shorter buckling length (an intermediate support, a stiffer end restraint) or from placing material far from the centroid (a tube instead of a solid bar) – not from a stronger material. A high-strength steel buckles at exactly the same load as mild steel of identical geometry.
The calculator converts internally to coherent SI units: E in pascal, I in m⁴, lk in metres, Fkrit in newtons. You may enter E in GPa or MPa, I in cm⁴, mm⁴ or in⁴, and lengths in mm, cm, m or inches. Note for hand calculations: with E in N/mm², I in mm⁴ and lk in mm, Fkrit comes out directly in newtons.
Connecting rods, spindles and push rods, columns and struts, compression members in trusses and scaffolding, threaded rods and tie bars in compression, hydraulic cylinder rods, cross-beams and props in machine and plant design – wherever a slender part carries compression.
Valid for an ideal straight member under centric load with constant cross-section and linear-elastic material at small deflections. The formula loses validity as soon as σkrit = Fkrit/A exceeds the proportional limit, i.e. for stocky members with λ < λ₀, where Euler gives unsafely high values. Initial curvature and load eccentricity, residual stresses, inelastic buckling, torsional buckling, lateral-torsional buckling, tapered sections, elastic intermediate supports and the safety factors prescribed by codes are not covered.
Common mistake: The most common error is entering the member length l instead of the buckling length lk = β·l, which can miss the buckling load by a factor of four. Nearly as common is using the larger second moment of area although the member buckles about its weak axis. Fkrit is also not a permissible working load but the failure limit of the ideal member. And finally: buckling has nothing to do with yield strength – switching to a stronger material with the same E gains nothing against buckling.
Connecting rods, spindles and push rods, columns and struts, compression members in trusses and scaffolding, threaded rods and tie bars in compression, hydraulic cylinder rods, cross-beams and props in machine and plant design – wherever a slender part carries compression.
E is the material's elastic modulus (structural steel ≈ 210 GPa, aluminium ≈ 70 GPa), taken from a material table. I is the second moment of area about the axis the member deflects about; if it can buckle in any plane, use the smallest principal value I₂. lk is the reduced buckling length lk = β·l with the end-condition factor β of the matching Euler case – not the member length.
Valid for an ideal straight member under centric load with constant cross-section and linear-elastic material at small deflections. The formula loses validity as soon as σkrit = Fkrit/A exceeds the proportional limit, i.e. for stocky members with λ < λ₀, where Euler gives unsafely high values. Initial curvature and load eccentricity, residual stresses, inelastic buckling, torsional buckling, lateral-torsional buckling, tapered sections, elastic intermediate supports and the safety factors prescribed by codes are not covered.