Euler buckling: critical load, effective length factor K, slenderness and calculator
Published November 11, 2026 — the Euler critical load, the 6 end conditions and their K factor (theoretical and recommended), slenderness, the limit of Euler's formula and the Johnson parabola, two worked examples (in mm and inches) and a calculator.
In short: a column under compression buckles at the Euler critical load Pcr = π²·E·I / (K·L)², I being the smallest area moment of inertia of its section and K a factor set by the end conditions: 1 between two pins, 2 for a cantilever (fixed-free), 0.5 between two fixed ends. Doubling the length divides the critical load by 4. The formula only holds for a slender column: below a slenderness of about 133 for S235 steel, it overestimates the load.
1. Buckling calculator
Pick the end conditions, the section and its dimensions, the length and the material. Yield strength and applied load are optional: the first gives Euler's range of validity and the Johnson load, the second the safety factor.
Default values: 20 mm steel round bar, 1 m long, between two pins (K = 1): I = 7,854 mm⁴, r = 5 mm, slenderness 200, Euler critical load 16.3 kN. The details are in the worked examples.
2. End conditions and K factor
The effective length is Le = K·L: the length of the half-wave of the buckled shape. Theoretical values assume perfect fixity; since no fixing is perfect, the commentary of the AISC 360 specification gives more cautious recommended values. "Guided" means an end that cannot rotate but can move sideways.
The last column compares the theoretical critical load with that of the pinned-pinned column of the same length: it varies as 1/K².
3. Slenderness, Euler's limit and the Johnson parabola
The slenderness ratio λ = K·L / r, with the radius of gyration r = √(I/A), sums up how prone the column is to buckling. The Euler critical stress is σcr = π²·E / λ². It only makes sense well below the yield strength Sy: for a stocky column, yielding, not elastic buckling, limits the load.
The Johnson parabola starts at Sy for a very short column and meets the Euler curve, tangentially, at λt, where the critical stress is Sy/2. With E = 210,000 MPa, λt is about 133 for S235 and 108 for S355 steel. Above it, Euler applies; below it, use Johnson. Structural steel yield strengths are in the S235, S275 and S355 reference.
4. Worked examples
Ø 20 mm steel round bar (metric units)
- I = π × 20⁴ / 64 = 7,854 mm⁴; A = 314.2 mm²; r = √(I/A) = 5 mm.
- Length 1,000 mm between two pins (K = 1): λ = 200; Pcr = π² × 210,000 × 7,854 / 1,000² = 16.3 kN (σcr = 51.8 MPa). λ > 133: Euler applies for S235.
- The same bar fixed at the base and free at the top (K = 2): λ = 400, Pcr = 4.07 kN, four times less.
- Shortened to 300 mm (K = 1): λ = 60, below λt. Euler would give 181 kN (σ = 576 MPa, above Sy); Johnson gives σcr = 235 − 235² × 60² / (4π² × 210,000) = 211 MPa, i.e. 66.3 kN.
Steel tube in inches (US units)
- Tube 1.5 in OD, 0.120 in wall: I = 0.1248 in⁴; A = 0.520 in²; r = 0.490 in.
- Length 72 in, pinned both ends, E = 29,000,000 psi: λ = 147; Pcr = 6,890 lbf (σcr = 13,240 psi).
- For A36 steel (Sy = 36,000 psi), λt = 126: λ = 147 is above it, so Euler applies.
5. In practice: weak axis, safety, codes
- The weak axis: with equal end conditions, the column buckles about the axis of smallest moment of inertia. A 10 × 40 flat bar buckles about its short side (I = 40 × 10³ / 12 = 3,333 mm⁴), sixteen times less stiff than on edge. Moments of inertia of common sections are in the area moment of inertia reference.
- Safety: the critical load is an instability limit, reached without warning. A real column is never perfectly straight nor loaded exactly on its axis, and its failure load is below Pcr: apply a safety factor well above 1, set by the code or the specification.
- Codes: for a steel structure, Eurocode 3 (buckling curves and reduction factor χ) or AISC 360 in the US replace Euler and Johnson, accounting for imperfections and residual stresses. Euler and Johnson remain the right approach for a machine part, a link or a cylinder rod, and for a first sizing.
- Levers: shorten the column or add an intermediate brace (the critical load varies as 1/L²), fix the ends, or move material away from the axis (a tube rather than a solid bar of the same weight).
Printable version: the buckling formulas as a PDF (A4, no sign-up): K factors, Euler and Johnson formulas.
6. FAQ
What is the Euler critical load formula?
Pcr = π²·E·I / (K·L)², with E Young's modulus, I the smallest area moment of inertia of the section, L the column length and K the factor set by the end conditions (1 for two pins, 2 for a fixed-free column).
Which K factor should I use?
In theory: 0.5 fixed-fixed, 0.7 fixed-pinned, 1 pinned-pinned, 2 fixed-free. Since a real fixed end is never perfect, the AISC commentary recommends 0.65, 0.80, 1.0 and 2.1 for the same cases.
When is Euler's formula valid?
For a slender column whose critical stress stays below the yield strength. With the Johnson parabola, the limit is at a slenderness λ = π·√(2·E/Sy), about 133 for S235 steel and 108 for S355. Below it, Euler overestimates the load.
What is the slenderness ratio of a column?
The ratio of the effective length K·L to the radius of gyration r = √(I/A) of the section. The larger it is, the more easily the column buckles: the Euler critical stress is π²·E / λ².
About which axis does a column buckle?
About the axis of smallest moment of inertia, with equal end conditions. A 10 × 40 flat bar buckles about its short side, with I = 40 × 10³ / 12, not about the long one.