Euler's and Rankine's theory of columns

Why it matters

Compression members in machines often fail by buckling long before the material is crushed: a long ball screw pushing a load, the rod of an extended linear actuator or hydraulic cylinder, a slender strut in a robot frame or a 3D-printer gantry post. Buckling is sudden and depends on stiffness, length and end fixing rather than strength, so a part that passes a stress check can still collapse. Screw and actuator catalogues give buckling charts based on exactly the theory below.

Key ideas

Buckling is instability, not overstress. A perfectly straight column under an axial load stays straight until the load reaches a critical value. At that load the straight shape is no longer the only equilibrium position: a slightly bent shape is also in equilibrium, and any disturbance grows. The critical load depends on E·I and on length and end conditions, not on yield strength (for long columns).

Euler's assumptions. The column is initially perfectly straight, the load is exactly axial, the material is homogeneous, isotropic and linear-elastic up to buckling, the cross-section is uniform, self-weight is neglected and deflections are small. Failure is by buckling alone.

Euler's formula and effective length. For a pin-ended column the critical load is π²·E·I/L². Other end conditions are handled by an effective length L_e, the distance between points of zero moment (inflection points) in the buckled shape:

• both ends pinned: L_e = L;
• one end fixed, other free (flagpole): L_e = 2L;
• both ends fixed: L_e = L/2;
• one end fixed, other pinned: L_e = L/√2 ≈ 0.7L. Practical design codes recommend slightly larger values because real fixity is never perfect; take them from your code or data book.

Which I? A column buckles about the axis with the least second moment of area, so use I_min (and the least radius of gyration k_min). A circular or tubular section has no weak axis, which is why tubes are efficient columns.

Slenderness ratio. λ = L_e/k, with k = √(I/A). The Euler critical stress is σ_cr = π²·E/λ². Euler's theory is valid only when σ_cr is below the proportional limit, that is when λ ≥ π·√(E/σ_p). For mild steel with σ_p ≈ 250 MPa and E = 200 GPa this limit is about 89. Below it, Euler's formula overestimates the load (as λ → 0 it predicts an infinite load).

Short, intermediate and long columns. Short columns fail by crushing at P_c = σ_c·A. Long columns fail by elastic buckling (Euler). Intermediate columns fail by a mix of yielding and buckling, and real columns have initial crookedness and eccentric loads that reduce their strength further.

Rankine-Gordon formula. An empirical formula that blends the two limits: 1/P_R = 1/P_c + 1/P_e. It gives P_R ≈ P_c for very short columns and P_R ≈ P_e for very long ones. In the usual working form P_R = σ_c·A/[1 + a·(L_e/k)²], where a is Rankine's constant (theoretically σ_c/(π²·E); in practice an empirical value for the material from a data book, about 1/7500 for mild steel and 1/1600 for cast iron for pinned ends).

Eccentric loading. If the load acts at eccentricity e, bending starts immediately and the maximum stress is given by the secant formula; the column fails at a load below the Euler value. Keep actuators and screws well aligned.

Formulas

P_e = π²·E·I / L_e²

• P_e: Euler critical load (N); E: Young's modulus (Pa); I: least second moment of area (m⁴); L_e: effective length (m).

L_e = L (pinned-pinned), 2L (fixed-free), L/2 (fixed-fixed), L/√2 (fixed-pinned)

• L: actual length (m).

k = √(I/A), λ = L_e / k

• k: least radius of gyration (m); A: area (m²); λ: slenderness ratio (dimensionless).

σ_cr = π²·E / λ², Euler valid for λ ≥ π·√(E/σ_p)

• σ_cr: Euler critical stress (Pa); σ_p: proportional limit (Pa).

1/P_R = 1/P_c + 1/P_e, P_c = σ_c·A

• P_R: Rankine load (N); σ_c: crushing (compressive yield) stress (Pa).

P_R = σ_c·A / [1 + a·(L_e/k)²]

• a: Rankine's constant (dimensionless; material-dependent, from a data book).

Worked examples

Example 1 (standard): ball-screw buckling check. Given: a steel ball screw of root diameter 16 mm and unsupported length 800 mm pushes a load. One end is held in a fixed (angular-contact pair) bearing, the other in a simple support bearing, so take fixed-pinned. E = 200 GPa, factor of safety 3. Find the buckling load and the allowable thrust, and check Euler validity.

1. L_e = 0.7 × 800 = 560 mm.
2. I = π·d⁴/64 = π × 16⁴/64 = 3217 mm⁴; A = π × 16²/4 = 201.1 mm²; k = d/4 = 4 mm.
3. λ = 560/4 = 140, above about 89, so Euler applies.
4. P_e = π²·E·I/L_e² = π² × 200 000 × 3217/560² = 20 250 N.
5. Critical stress 20 250/201.1 = 100.7 MPa, well below the proportional limit. Allowable thrust = 20 250/3 = 6750 N. Answer: P_e ≈ 20.2 kN; allowable thrust ≈ 6.75 kN. If the far end were left free (fixed-free), L_e would be 1600 mm and P_e would fall by a factor of (1600/560)² ≈ 8.2.

Example 2 (GATE level): Euler versus Rankine for a tubular strut. Given: a steel tube, outer diameter 50 mm, inner 40 mm, length 2.5 m, both ends pinned. E = 200 GPa, σ_c = 320 MPa, Rankine's constant a = 1/7500. Find the Euler and Rankine loads.

1. A = π/4 × (50² − 40²) = 706.9 mm².
2. I = π/64 × (50⁴ − 40⁴) = 181 100 mm⁴; k = √(I/A) = 16.01 mm (also √(D² + d²)/4).
3. λ = 2500/16.01 = 156.2.
4. Euler: P_e = π² × 200 000 × 181 100/2500² = 57 210 N.
5. Rankine: P_c = 320 × 706.9 = 226 200 N; P_R = σ_c·A/[1 + a·λ²] = 226 200/(1 + 156.2²/7500) = 226 200/4.252 = 53 200 N.
6. For comparison, 1/P_R = 1/P_c + 1/P_e with the computed Euler load gives 45 660 N; the two forms agree only if a = σ_c/(π²·E) = 1/6170, so the empirical constant matters. Answer: Euler ≈ 57.2 kN, Rankine ≈ 53.2 kN (with a = 1/7500). For this slender strut the two are close; for a short strut Rankine would be far lower than Euler.

Common mistakes

• Using the actual length instead of the effective length for the given end conditions.
• Using the larger I (or k) of a rectangular section; buckling occurs about the weak axis.
• Applying Euler's formula to short or intermediate columns, where it overestimates the load.
• Confusing a fixed-free column (L_e = 2L) with fixed-fixed (L_e = L/2): a factor of 16 in load.
• Forgetting that Rankine's constant depends on the material (and on how end conditions are included).
• Mixing mm and m in π²·E·I/L_e².

For GATE ME

Expect Euler loads for the four standard end conditions, ratios of critical loads when end conditions or dimensions change, slenderness ratio and the limiting slenderness for Euler validity, Rankine loads, and which axis a rectangular column buckles about. Practise ratio reasoning: P_e is proportional to d⁴ for a solid round and to 1/L_e².

Quick check

1. A pinned column has P_e = 100 kN. What is P_e if both ends are fixed instead?
2. What is the effective length of a fixed-free column of length 2 m?
3. Effective length 4 m, least radius of gyration 0.05 m. Find λ.
4. P_e = 150 kN and P_c = 300 kN. Find the Rankine load.
5. About which axis does a 20 mm × 40 mm rectangular strut buckle?

Answers: 1. 400 kN. 2. 4 m. 3. 80. 4. 1/(1/150 + 1/300) = 100 kN. 5. The axis parallel to the 40 mm side (the axis with the smaller I).