Column Buckling: Euler vs Johnson — Which Formula to Use
Euler / Johnson
Columns and struts fail not by yielding but by suddenly bowing sideways — buckling. A column that is long and slender buckles elastically at a load far below its yield limit; a shorter, stockier column crosses into yielding before it buckles. Getting the critical load right is safety-critical: an under-conservative choice means the column appears stronger on paper than it is in reality.
Two classical formulas divide the slenderness range: the Euler elastic formula governs long columns (slenderness λ above the tangent slenderness λ₁), and the Johnson intermediate parabola covers the transition zone where partial yielding lowers the critical stress before full elastic buckling can develop. The MechanixCalc column buckling calculator selects the governing regime automatically from slenderness, then returns the critical load and safety factor — this guide explains the logic behind that selection.
Slenderness ratio — the key parameter
The slenderness ratio λ = Le/r captures everything about a column's susceptibility to buckling in one number. The effective length Le = K·L accounts for the end conditions: pinned-pinned K = 1.0, fixed-fixed K = 0.5, fixed-free K = 2.0, fixed-pinned K = 0.7. The radius of gyration r = √(I/A) comes from the cross-section geometry — a smaller r relative to the length means the column is more slender and more prone to buckling.
Three regimes follow directly from λ. Very stocky columns (λ ≤ 20) fail by simple compression yielding; the load capacity is just the yield strength times the area. Intermediate columns (20 < λ ≤ λ₁) follow the Johnson parabola. Long columns (λ > λ₁) fail elastically and the Euler formula governs. The tangent slenderness λ₁ = π·√(2E/Sy) marks the boundary where the Euler critical stress exactly equals half the yield strength — this is the cross-over point and is different for every material.
Euler elastic buckling — long columns
For λ > λ₁, the column buckles before the material yields. The Euler formula expresses the critical stress as a function of the elastic modulus and the slenderness squared. Because the stress varies as 1/λ², a doubling of column length reduces the critical load to one quarter. The critical load P_cr = σ_cr · A.
The Euler formula is purely a geometry and stiffness result — material strength does not appear. A long, slender steel column and a long, slender aluminium column of the same slenderness ratio have critical stresses in the ratio of their elastic moduli (~3:1), regardless of their yield strengths.
σ_cr = π² · E / λ²where σ_cr = critical buckling stress (MPa); E = elastic modulus (MPa); λ = Le/r = slenderness ratio; Le = K·L = effective length (mm); K = end-condition factor; r = radius of gyration = √(I/A) (mm)
Johnson intermediate parabola — short-to-medium columns
Between the crushing regime and the Euler zone, partial yielding and geometric imperfections lower the critical stress below the Euler prediction. The Johnson parabola fits this transition by starting at the yield strength (at λ = 0) and blending smoothly down to Sy/2 at the tangent slenderness λ₁, where it meets the Euler curve exactly. Using the Euler formula in this regime would overestimate the critical load and is unconservative.
In practice, most machine structural members — brackets, press frames, hydraulic cylinder rods and machine tool columns — fall in the Johnson regime. The boundary λ₁ for S355 steel (E = 210 000 MPa, Sy = 355 MPa) is about 108; for a softer S250 grade (Sy = 250 MPa) it rises to about 129. Always compute λ₁ for the actual material before deciding which formula applies.
σ_cr = Sy · [1 − Sy · λ² / (4 · π² · E)]where Sy = material yield strength (MPa); λ₁ = π · √(2E / Sy) = tangent slenderness (the Euler–Johnson boundary); at λ = λ₁ the formula gives σ_cr = Sy/2, matching the Euler curve
Choosing the right regime and applying a safety factor
The selection logic is: compute λ and λ₁, compare, then pick the formula. Always use the governing (lower) critical load — never apply Euler when Johnson gives a smaller result for the same slenderness. The MechanixCalc calculator runs this comparison in a fail-closed sequence so a column is never assigned the Euler load when the Johnson parabola would be more conservative.
For a column in service, a minimum safety factor SF = P_cr / P_applied ≥ 3 is commonly required (some codes and applications require higher — machinery frames often use SF ≥ 4). The safety factor must be applied to P_cr, not to the critical stress alone, because the cross-section area also carries uncertainty. Additional checks are needed for I-section columns (lateral-torsional buckling per EN 1993-1-1 §6.3.2), thin-wall sections (local plate buckling per EN 1993-1-5), and any column with an eccentric or combined axial-plus-bending load.
Worked example
Find the critical buckling load of a pinned-pinned mild-steel column with a solid circular cross-section: diameter d = 40 mm, length L = 1000 mm, E = 210 000 MPa, Sy = 250 MPa. Determine the governing regime and the safety factor under an applied load P = 60 kN.
Given
- Diameter d40 mm
- Column length L1000 mm
- End conditionPinned-pinned (K = 1.0)
- Elastic modulus E210 000 MPa
- Yield strength Sy250 MPa
- Applied load P60 kN
Result
- Governing regimeJohnson (λ = 100 < λ₁ = 128.8)
- Critical buckling stress σ_cr174.6 MPa
- Critical buckling load P_cr≈ 219 kN
- Safety factor SF3.65 (P = 60 kN applied)
- Cross-section area: A = π·d²/4 = π·40²/4 = 1256.6 mm².
- Second moment of area: I = π·d⁴/64 = π·40⁴/64 = 125 664 mm⁴.
- Radius of gyration: r = √(I/A) = √(125 664 / 1256.6) = √100 = 10.0 mm. (For a solid circle, r = d/4 exactly.)
- Effective length: Le = K·L = 1.0 × 1000 = 1000 mm. Slenderness ratio: λ = Le/r = 1000/10 = 100.
- Tangent slenderness: λ₁ = π·√(2E/Sy) = π·√(2 × 210 000 / 250) = π·√1680 = π·40.99 = 128.8.
- Regime check: λ = 100 < λ₁ = 128.8, so the Johnson intermediate parabola governs (not Euler).
- Johnson critical stress: σ_cr = Sy·[1 − Sy·λ²/(4π²E)] = 250·[1 − 250·10 000/(4·9.8696·210 000)] = 250·[1 − 2 500 000/8 290 464] = 250·[1 − 0.3016] = 250·0.6984 = 174.6 MPa.
- Critical buckling load: P_cr = σ_cr·A = 174.6 × 1256.6 = 219 400 N ≈ 219 kN.
- Safety factor: SF = P_cr / P = 219 / 60 = 3.65.
Illustrative example — apply your own geometry, material and end conditions. The result assumes a perfectly straight column with concentric loading; real columns require an appropriate safety factor (≥ 3 for most structural applications) and additional checks for lateral-torsional buckling and combined loading. The MechanixCalc calculator computes all regimes and checks in a single pass.
Do it on your own numbers
Compute the critical load, governing regime (Euler / Johnson / short-column), slenderness, and safety factor for your column — plus lateral-torsional buckling and thin-wall local buckling checks. Free 30-minute preview, no sign-up.
Open the Column BucklingFrequently asked questions
Which formula governs — Euler or Johnson?
Compute the slenderness ratio λ = Le/r and the tangent slenderness λ₁ = π·√(2E/Sy). If λ > λ₁, the Euler elastic formula governs; if 20 < λ ≤ λ₁, the Johnson parabola governs; if λ ≤ 20, the column simply crushes at the yield load. The MechanixCalc calculator performs this comparison automatically and reports the governing regime.
Which standard governs this calculation?
The Euler and Johnson formulas are classical mechanics results found in most structural and machine-design references (e.g. Shigley's Mechanical Engineering Design). For structural steel columns the additional lateral-torsional buckling check follows EN 1993-1-1:2005 §6.3.2, and thin-wall local buckling follows EN 1993-1-5:2006 Tables 4.1–4.2. The MechanixCalc calculator covers all three checks.
What effective-length factor K should I use?
Theoretical values: K = 1.0 for pinned-pinned (most common default), K = 0.5 for fully fixed both ends, K = 2.0 for one end fixed and the other free (flagpole), K = 0.7 for one end fixed and the other pinned. In practice the end fixity is rarely perfect, so conservatively use K values at or above the theoretical values — many codes recommend K = 0.65 for built-up fixed-fixed columns rather than the theoretical 0.5.
What safety factor should I apply to the critical buckling load?
A minimum safety factor of SF ≥ 3 on the critical buckling load is common for structural columns; machinery frames and cylinder rods often require SF ≥ 4. The factor accounts for load eccentricity, material imperfections, and residual stresses that are not modelled in the basic Euler/Johnson formula. Always apply the safety factor to P_cr (not to the critical stress) so that both the formula uncertainty and the area uncertainty are covered.
Is the column buckling calculator free?
You can run the full calculation during a free 30-minute preview with no sign-up required. A free 14-day account trial unlocks every calculator with no credit card. The branded PDF engineering report with the governing formula, regime and safety-factor narrative, and the ability to save and reload calculations, are part of a paid plan.
Related
- Beam DeflectionSize the beam spanning between column supports or check combined bending and axial demand.
- Fatigue AnalysisComplement the static buckling check with a fatigue life assessment when the axial load cycles.
- Weld DesignDesign the base-plate or splice welds that carry the buckling-critical axial load into the column.
- Beam deflection formulas explained
- Fillet weld design to EN 1993-1-8