Euler's and Rankine's theory of columns

Why it matters

Slender members in compression (building columns, machine-tool columns, piston and connecting rods, jack screws, truss struts) can fail by suddenly bowing sideways at a load far below the one that would crush the material. Buckling gives little warning, so compression members are checked for stability as well as strength. Euler's formula sets the theoretical limit for long columns, and Rankine's formula covers the practical range of lengths.

Key ideas

Short and long columns. A short, stocky column fails by crushing or yielding when P/A reaches the compressive strength σc. A long, slender column buckles: at the critical load Pcr the straight shape is no longer the only equilibrium, and a small disturbance causes large lateral deflection. Intermediate columns fail by a mix of both.

Euler's assumptions.

1. The column is initially perfectly straight and the load is exactly axial.
2. The material is homogeneous, isotropic and linearly elastic (stress stays below the proportional limit).
3. The cross-section is uniform; self-weight is neglected.
4. Deflections are small; failure is by buckling alone. Real columns have crookedness and eccentricity, so they bend from the start and fail somewhat below Pcr.

Euler's critical load. For a pin-ended column, solving EI y'' = −P y with y = 0 at both ends gives the smallest non-zero load Pcr = π² E I / L². Buckling occurs about the axis with the smallest I (the weakest direction).

Effective length. Other end conditions are handled by an effective length Le, the distance between points of zero moment (inflection points) in the buckled shape:

End conditions	Le	Pcr relative to pinned–pinned
Both ends pinned	L	1
One fixed, one free	2L	1/4
Both ends fixed	L/2	4
One fixed, one pinned	L/√2 ≈ 0.7L	2
Design codes give slightly larger practical values for "fixed" ends because real fixity is imperfect; use the code's table for design.

Slenderness ratio. With radius of gyration r = √(I/A), the critical stress is σcr = π² E/(Le/r)². Euler's formula is valid only while σcr is below the proportional limit, i.e. for Le/r ≥ π √(E/σp). For mild steel (E = 200 GPa, σp ≈ 250 MPa) the limit is about 89. Below it Euler's formula overestimates the strength, approaching infinity as Le/r → 0.

Rankine's (Rankine–Gordon) formula. An empirical combination: 1/PR = 1/Pc + 1/PE, where Pc = σc A is the crushing load and PE the Euler load. Rearranged with an empirical Rankine constant a ≈ σc/(π² E), it becomes the formula below. For short columns PR → Pc; for very long columns PR → PE. Values of σc and a for cast iron, mild steel, wrought iron and timber are tabulated in data books.

Eccentric loading (secant formula) and code column curves (IS 800 for steel) are used in practice; they are introduced in design courses.

Formulas

Pcr = π² E I / Le²

• Pcr = critical (buckling) load (N), E = Young's modulus (N/mm²), I = least second moment of area (mm⁴), Le = effective length (mm).

r = √(I/A), λ = Le / r (slenderness ratio, dimensionless)

σcr = π² E / λ²

λ_limit = π √(E / σp) (lowest slenderness for which Euler applies; σp = proportional limit)

1/PR = 1/Pc + 1/PE

PR = σc A / [1 + a (Le/r)²]

• σc = crushing stress (N/mm²), a = Rankine constant (from data book), A = area (mm²).

Safe load = critical load / factor of safety.

Radii of gyration: solid circle r = d/4; hollow circle r = √(D² + d²)/4; rectangle about the weak axis r = b/√12 (b = smaller side).

Worked examples

Example 1 (standard, Euler). A solid steel strut of 50 mm diameter and 2.5 m length has E = 200 GPa. Find the Euler load for the four standard end conditions and check that Euler applies (σp = 250 MPa).

1. I = π × 50⁴ / 64 = 3.068 × 10⁵ mm⁴; A = 1963.5 mm²; r = d/4 = 12.5 mm.
2. π² E I = 9.8696 × 200 000 × 3.068 × 10⁵ = 6.056 × 10¹¹ N·mm².
3. Pinned–pinned (Le = 2500 mm): Pcr = 6.056 × 10¹¹ / 2500² = 96.9 kN.
4. Fixed–free (Le = 5000 mm): 24.2 kN. Fixed–fixed (Le = 1250 mm): 387.6 kN. Fixed–pinned (Le = 1768 mm): 193.8 kN.
5. Validity: λ for pinned–pinned = 2500/12.5 = 200 > 89, and σcr = 96 900/1963.5 = 49.4 MPa < 250 MPa. ✓ (For fixed–fixed, λ = 100 and σcr = 197 MPa, still below 250 MPa.)

Answer: 96.9 kN (pinned–pinned), 24.2 kN (fixed–free), 387.6 kN (fixed–fixed), 193.8 kN (fixed–pinned).

Example 2 (GATE level, Rankine). A hollow cast-iron column, 200 mm outside and 160 mm inside diameter, 4 m long, is pinned at both ends. Take σc = 550 MPa and a = 1/1600 (data-book values for cast iron). Find the Rankine crippling load and the safe load with a factor of safety of 3. Compare with Euler (E = 100 GPa).

1. A = π/4 (200² − 160²) = 11 310 mm².
2. I = π/64 (200⁴ − 160⁴) = 4.637 × 10⁷ mm⁴; r = √(I/A) = √(200² + 160²)/4 = 64.03 mm.
3. λ = 4000 / 64.03 = 62.47; a λ² = 62.47² / 1600 = 2.439.
4. PR = 550 × 11 310 / (1 + 2.439) = 6.220 × 10⁶ / 3.439 = 1.809 × 10⁶ N.
5. Safe load = 1809 / 3 = 603 kN.
6. Euler: PE = π² × 100 000 × 4.637 × 10⁷ / 4000² = 2.86 × 10⁶ N, much higher than Rankine because at λ ≈ 62 the column is intermediate and crushing also matters.

Answer: Rankine load ≈ 1809 kN; safe load ≈ 603 kN; Euler would give 2860 kN (unconservative here).

Common mistakes

• Using the larger I; buckling happens about the axis of least I.
• Forgetting the effective-length factor, or inverting it (fixed–free is 2L, not L/2).
• Applying Euler to short or intermediate columns where it overestimates strength.
• Mixing units: E in GPa with I in mm⁴. Use N/mm² with mm⁴ and mm.
• Treating Rankine's constant as universal; it depends on material and end conditions.
• Forgetting that Pcr ∝ 1/Le², so halving the effective length multiplies the load by 4.

For GATE PI

Expect Euler load with different end conditions, ratios of critical loads, slenderness ratio and limiting slenderness, Rankine load for given σc and a, and the effect of changing diameter (Pcr ∝ d⁴ for solid circles) or length. Practise the effective-length table until it is automatic.

Quick check

1. A column fixed at both ends has what Euler load relative to the same column pinned at both ends?
2. Slenderness ratio for Le = 4 m and r = 50 mm?
3. If the diameter of a solid circular strut is doubled, how does Pcr change?
4. Euler load 100 kN, crushing load 300 kN: Rankine load?

Answers: 1. Four times. 2. 80. 3. 16 times (I ∝ d⁴). 4. 75 kN.