Columns and Buckling
Compression members, the Euler load, slenderness, end conditions, the Johnson curve for short columns and design margins.
What buckling is
A short, stocky member in compression fails by crushing at the yield stress: its strength is simply Sy × A. A long, slender column fails much earlier, by buckling: at a certain load it suddenly bows sideways, and once bent, the bending makes it bend more and it collapses, even though the average stress is well below yield. It is a failure of stability, not of strength, and it is dangerous because it happens without warning. Any compression member (a strut, a leg of a frame, a piston rod, a thin plate or the web of a beam) must be checked for it.
The Euler load
Pcr = π² E I ÷ (K L)² σcr = π² E ÷ (K L ÷ r)² r = √(I ÷ A)
Leonhard Euler found that the critical load of an ideal straight, pinned-pinned column depends on the stiffness E I (not on the strength of the material), and on the inverse square of the length. Doubling the length divides the buckling load by four. The ratio K L ÷ r is the slenderness ratio: columns with a large slenderness (above about 100 for steel) buckle by the Euler formula; stockier columns need the empirical correction below. The member buckles about the axis with the smallest I (the weakest direction).
End conditions and the effective length
The Euler formula is for pinned ends. Other supports change the length over which the column bends: the effective length K L. A column fixed at the base and free at the top (a flagpole) is like half of a pinned column of twice the length (K = 2), and carries only a quarter of the load; fixed at both ends it behaves like a pinned column of half the length (K = 0.5) and carries four times. Real supports are never perfectly fixed or pinned, so the AISC recommended design values, which allow for supports that are not fully rigid, are K = 0.65 (fixed–fixed), 0.80 (fixed–pinned), 1.0 (pinned–pinned) and 2.1 (fixed–free).
Slenderness and the Johnson curve
Euler’s formula gives stresses above the yield strength for stocky columns, which is impossible. The Johnson (parabolic) formula bridges between the yield strength for very short columns and the Euler curve: σcr = Sy − (Sy² ÷ (4 π² E)) (K L ÷ r)², used below the transition slenderness (K L ÷ r)c = √(2 π² E ÷ Sy). For steel with Sy = 350 MPa the transition is at 106, where both formulas give Sy ÷ 2 = 175 MPa. Real columns, with imperfections and residual stress, fall below both curves: design codes (AISC, Eurocode 3) use column curves with reduction factors and safety factors.
Worked examples
Worked example: a slender steel bar
- A 20 mm round steel bar, 800 mm long, pinned at both ends. I = π d⁴ ÷ 64 = 7 854 mm⁴, A = 314.2 mm², r = √(I ÷ A) = 5.0 mm (= d ÷ 4). Slenderness K L ÷ r = 800 ÷ 5.0 = 160: well above the transition, so Euler applies.
- Pcr = π² × 200 000 × 7 854 ÷ 800² = 24.2 kN, a stress of only 77 MPa, against 110.0 kN if it were a short stub: the bar buckles at about 22 percent of its crushing load.
- With a safety factor of 3 the allowable load is 8.1 kN. If the bar were fixed at the base and free at the top (K = 2), the effective length doubles: Pcr = 6.06 kN, a quarter.
- Doubling the diameter to 40 mm multiplies I by 16 and so the buckling load by 16, while the area (the crushing load) only rises by 4: a tube is again the answer (see Section Properties and Structural Profiles).
Worked example: a square tube column
- A 40 × 40 × 3 mm steel tube (Sy = 350 MPa), 1 500 mm long, pinned. I = 101 972 mm⁴, A = 444 mm², r = 15.2 mm; slenderness = 99, just below the transition value 106: use the Johnson formula.
- Euler would give Pcr = π² E I ÷ L² = 89.5 kN. Johnson: σcr = 350 − (350² ÷ (4π² × 200 000)) × 99² = 198 MPa, so Pcr = 198 × 444 = 87.9 kN. The two are close near the transition; the Johnson value is the one to use below it. With a factor of 2.5 the safe load is about 35 kN.
Eccentric loads: the secant formula
If the load is applied off the center of the column by an eccentricity e (and every real load is, a little), the column bends from the start and the maximum stress is given by the secant formula: σmax = (P ÷ A) [1 + (e c ÷ r²) sec((K L ÷ 2r) √(P ÷ (E A)))], with c the distance to the outer fiber and sec the secant (1 ÷ cos). It shows a sharply rising stress as the load approaches the Euler load.
Worked example: the tube with an off-center load
The same tube carries P = 20 kN applied 10 mm off the center. e c ÷ r² = 10 × 20 ÷ 229.7 = 0.87. The angle is (1 500 ÷ (2 × 15.2)) × √(20 000 ÷ (200 000 × 444)) = 0.743 rad and its secant is 1.358. So σmax = (20 000 ÷ 444) × (1 + 0.87 × 1.358) = 98.3 MPa. The axial stress alone is only 45.0 MPa: the eccentricity more than doubles the stress. Check this against the yield strength divided by the safety factor.
Local buckling of thin walls and plates
The walls of a tube, the flanges and web of an I-beam, and plates in compression can buckle locally as thin sheets, long before the member as a whole. A flat plate of width b and thickness t, simply supported on its long edges, buckles at σcr = k π² E ÷ (12 (1 − ν²)) × (t ÷ b)², with k = 4 for supported edges (about 0.43 for a free edge). For a 3 mm steel plate of width 40 mm: σcr = 4 × π² × 200 000 ÷ (12 × (1 − 0.09)) × (3 ÷ 40)² = 4 067 MPa: no problem here, but a thin 1 mm plate of the same width would buckle at 452 MPa (nine times less). Codes limit the width-to-thickness ratio; stiffeners (ribs, beads, flanges, folded edges) raise the buckling stress.
Designing against buckling
- Shorten the effective length: add a brace or a support at mid-height (it halves K L and quadruples the Euler load); fix the ends rather than pin them if the supports are stiff enough.
- Increase r for the same area: use a tube or a box instead of a bar; the buckling load rises with I, so with the square of the section’s size at equal weight.
- Equalize the stiffness in both directions: a round or square tube buckles equally in every direction; an I-beam or an angle buckles about its weak axis first.
- Allow for imperfections: initial bow, eccentric loads and residual stresses reduce the buckling load; use a safety factor of 2 to 3 for slender columns, higher for uncertain conditions, and follow the code column curves for structures.
- Limit the slenderness: K L ÷ r ≤ 200 for main compression members, 120 to 150 for important ones.
- Check local buckling of thin walls and the web crippling at load points.
- Remember the load path: the connection details decide whether the ends behave as pinned or fixed.
| Member | Slenderness K L ÷ r | Governing mode |
|---|---|---|
| Short, stocky (below 30) | Small | Yielding or crushing: P = Sy A |
| Intermediate (30 to the transition, about 106 for steel) | Moderate | Inelastic buckling: the Johnson or code curves |
| Long and slender (above about 106 for Sy = 350 MPa) | Large | Elastic Euler buckling |
Where buckling shows up in machines
- Lead screws and ball screws in compression: a 12 mm screw 500 mm long with one end fixed and the other supported (K ≈ 0.7) has Pcr = π² E I ÷ (0.7 L)² = 16.40 kN: low for a thin screw (see Mechanical Transmission: Screws, Belts and Gearboxes).
- Piston rods of long-stroke pneumatic and hydraulic cylinders: the maker gives a stroke-and-load curve (see Pneumatic and Hydraulic Actuators).
- Struts, legs, props and tripods, 3D-printer and CNC frames loaded in compression, and the thin walls of enclosures.
- Beams: the compression flange of a deep narrow beam buckles sideways (see Beam Bending and Deflection).
- Thin shells and tubes in axial compression and external pressure: handbooks give the buckling pressure.
Related
- Structures and Strength of Materials: the overview of this guide.
- Section Properties and Structural Profiles: I and r for the profiles.
- Beam Bending and Deflection and Stress, Strain and Safety Factors: bending and the safety factor.
- Worked Design Checks: a column check from the loads to the safety factor.
FAQ
Why does the buckling load not depend on the strength of the steel?
In the Euler range the failure is a loss of stability governed by the stiffness E I and the length. A stronger steel has the same E, so the same buckling load. Only short, stocky columns that fail by yielding benefit from a higher yield strength.
About which axis does a column buckle?
About the axis with the smallest second moment of area (the least radius of gyration), unless it is braced in that direction. Check both axes, with the effective length of each.
What safety factor should I use for a column?
Generally 2 to 3 on the critical load for slender columns, because small imperfections reduce the real buckling load. Structural codes provide column curves with built-in factors.
Can I stop buckling with a brace?
Yes: a lateral brace at mid-height forces the column to buckle in two half-waves, so the effective length halves and the critical load quadruples. The brace must be stiff and strong enough.
References
- Euler, L., Methodus inveniendi lineas curvas, 1744 (the original analysis); Timoshenko, S. P. and Gere, J. M., Theory of Elastic Stability.
- Beer, F. P., Johnston, E. R. and DeWolf, J. T., Mechanics of Materials, McGraw-Hill: columns; AISC 360 and EN 1993-1-1 (column curves).
- Shigley, J. E. and Mischke, C. R., Mechanical Engineering Design, McGraw-Hill: Euler, Johnson and the secant formulas.
Last reviewed: October 2026. The formulas are for ideal columns; real structures need the design code and a generous safety factor.