Euler's and Rankine's theory of columns

Elastic buckling of columns by Euler's theory with effective lengths, slenderness ratio and its validity limit, and Rankine's empirical formula, with a rod and a tube worked through.

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Why it matters

A slender member in compression can fail by buckling sideways at a stress far below the material's yield strength. Connecting rods, piston rods of hydraulic cylinders, steering tie-rods, jack screws, suspension struts and chassis members in compression must all be checked for it. Euler's theory gives the elastic buckling load of long columns; Rankine's empirical formula bridges the gap to short columns that fail by crushing.

Key ideas

Short versus long columns. A short, stocky compression member fails when the direct stress P/A reaches the crushing (yield) strength. A long, slender one becomes unstable and bows out sideways at a critical load that depends on stiffness (E·I) and length, not strength. Intermediate columns fail by a combination.

Euler's assumptions.

  • The column is initially perfectly straight and the load is exactly axial.
  • The material is homogeneous, isotropic and linear elastic, and stresses stay below the proportional limit.
  • The column is long compared with its cross-section, so failure is by buckling alone; self-weight and shear deformation are ignored.
  • Deflections are small, so E·I·y″ = −P·y can be used.

Euler's critical load. Solving E·I·y″ + P·y = 0 with pinned ends gives a non-zero bent shape only when P = n²·π²·E·I/L². The smallest, n = 1, is the critical load P_e = π²·E·I/L². At this load the straight form is no longer stable.

Effective length. Other end conditions are handled by an effective length L_e, the length of an equivalent pinned–pinned column (the distance between points of inflection of the buckled shape):

  • Both ends pinned: L_e = L.
  • One end fixed, other free: L_e = 2L.
  • Both ends fixed: L_e = L/2.
  • One end fixed, other pinned: L_e = L/√2 ≈ 0.7L. Real supports are rarely perfectly fixed, so design codes give practical L_e values; use those for design.

Axis of buckling. The column buckles about the axis with the smallest I (smallest radius of gyration k), unless end conditions differ between axes. Always use I_min unless told otherwise.

Slenderness ratio and validity. λ = L_e/k with k = √(I/A). The Euler stress is σ_e = π²·E/λ². Euler is valid only when σ_e is below the proportional limit, which sets a minimum slenderness: λ_min = π·√(E/σ_p). For mild steel this is roughly 80–90, depending on the stress value used. Below that, Euler badly overestimates the strength.

Rankine's formula. An empirical combination: 1/P_R = 1/P_c + 1/P_e, where P_c = σ_c·A is the crushing load. It tends to P_c for short columns and to P_e for very long ones. Rearranged: P_R = σ_c·A / (1 + a·λ²), with Rankine's constant a = σ_c/(π²·E) in theory, but in practice a and σ_c are taken from tables for each material. Commonly quoted textbook values are σ_c = 320 MPa and a = 1/7500 for mild steel and σ_c = 550 MPa, a = 1/1600 for cast iron; use your data book's values.

Imperfections and eccentricity. Real columns are slightly crooked and loads slightly eccentric, so they bend from the start and fail below P_e. The secant formula handles eccentric loads; design codes build imperfection effects into column curves. A factor of safety is always applied to the critical load.

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)

k = √(I/A), λ = L_e / k, σ_e = π²·E / λ²

  • k: least radius of gyration (m); λ: slenderness ratio; σ_e: Euler stress (Pa).

λ_min = π·√(E / σ_p)

  • σ_p: proportional limit (or the crushing stress used). Euler applies only for λ above this.

1/P_R = 1/P_c + 1/P_e, P_R = σ_c·A / (1 + a·(L_e/k)²)

  • P_R: Rankine load (N); P_c = σ_c·A: crushing load; σ_c: crushing stress (Pa); a: Rankine constant (from tables).

Solid circle: k = d/4; hollow circle: k = √(D² + d²)/4; rectangle (about the minor axis): k = b/√12.

Worked examples

Example 1 (standard): Euler load and end conditions. Given: a solid steel rod, d = 40 mm, L = 2 m, E = 200 GPa. Find the Euler load for pinned–pinned ends and compare with fixed–free and fixed–fixed.

  1. I = π × 40⁴/64 = 125 664 mm⁴; A = 1256.6 mm²; k = d/4 = 10 mm; λ = 2000/10 = 200 (long column).
  2. Pinned–pinned: P_e = π²·E·I/L_e² = π² × 200 000 × 125 664 / 2000² = 62.0 kN. σ_e = 62 012/1256.6 = 49.3 MPa, well below yield.
  3. Fixed–free, L_e = 4 m: P_e = 62.0/4 = 15.5 kN.
  4. Fixed–fixed, L_e = 1 m: P_e = 62.0 × 4 = 248 kN. Answer: 62.0 kN (pinned), 15.5 kN (fixed–free), 248 kN (fixed–fixed). The 16 : 1 spread shows how much end conditions matter.

Example 2 (GATE level): Euler versus Rankine for a tube. Given: a mild-steel tube, D = 60 mm, d = 50 mm, length 2.5 m, both ends pinned. E = 200 GPa; take σ_c = 320 MPa and a = 1/7500. Find the Euler and Rankine loads and a safe load with factor of safety 3.

  1. A = π/4 × (60² − 50²) = 863.9 mm²; I = π/64 × (60⁴ − 50⁴) = 329 376 mm⁴.
  2. k = √(I/A) = 19.53 mm (check: √(60² + 50²)/4 = 19.53 mm). λ = 2500/19.53 = 128.0.
  3. Euler: P_e = π² × 200 000 × 329 376 / 2500² = 104.0 kN. Validity: λ_min = π·√(200 000/320) = 78.5, and 128 > 78.5, so Euler applies.
  4. Rankine: P_R = σ_c·A / (1 + a·λ²) = 320 × 863.9 / (1 + 16 393/7500) = 276 460 / 3.186 = 86.8 kN.
  5. Safe load = 86.8 / 3 = 28.9 kN. Answer: P_e ≈ 104 kN, P_R ≈ 86.8 kN, safe load ≈ 28.9 kN. Rankine is lower because it also accounts for material crushing.

Common mistakes

  • Using the larger I (or the wrong axis) instead of I_min.
  • Using the actual length instead of the effective length, or the wrong L_e for the end conditions.
  • Applying Euler to a short column with λ below the limit.
  • Mixing mm and m in π²·E·I/L² (keep N, mm and MPa together).
  • Forgetting that Rankine's constants are empirical and material-specific.
  • Assuming a column fails at P_e in service; imperfections reduce the real capacity, so use a factor of safety.

For GATE ME

Expect Euler loads for given end conditions, ratios of buckling loads when end conditions, length or diameter change (P ∝ d⁴/L_e² for solid circles), slenderness ratio and limiting slenderness, Rankine load with given constants, and which axis a rectangular column buckles about. Practise the effective-length table and ratio reasoning.

Quick check

  1. A pinned column's length doubles. What happens to P_e?
  2. A fixed–free column is changed to fixed–fixed. By what factor does P_e change?
  3. Find the slenderness ratio for L_e = 4 m and k = 50 mm.
  4. A rectangular column is 40 mm × 80 mm. About which axis does it buckle?

Answers: 1. It falls to one-quarter. 2. L_e goes from 2L to L/2, so P_e rises 16 times. 3. 80. 4. About the axis parallel to the 80 mm side (least I = 80 × 40³/12).

Interview questions

All Engineering Mechanics and Strength of Materials interview questions

Try answering each one aloud before you open it.

  1. 1.What is Euler's theory of columns?Concept

    Euler's theory of columns is a mathematical approach to determine the critical load at which a slender column will buckle. It assumes that the column is perfectly straight, homogeneous, and has no initial imperfections. The theory is applicable to long columns where buckling occurs before the material yields.

  2. 2.Explain Rankine's theory of columns.Concept

    Rankine's formula is an empirical combination of crushing and buckling: 1/P_R = 1/P_c + 1/P_e, where P_c = σ_c·A is the crushing load and P_e the Euler load. Rearranged it reads P_R = σ_c·A/(1 + a·(L_e/k)²), with Rankine's constant a and σ_c taken from tables for each material. For short columns it approaches the crushing load and for very slender columns it approaches the Euler load, so one formula covers the whole range, including the intermediate columns where Euler overestimates strength.

  3. 3.How does Euler's formula for critical load differ from Rankine's formula?Concept

    Euler's formula is primarily applicable to long, slender columns and is based on the assumption that buckling occurs before any material yielding. It is given by P_cr = (π²EI) / (L²), where E is the modulus of elasticity, I is the moment of inertia, and L is the effective length. Rankine's formula, on the other hand, is a combination of Euler's formula and a material strength-based formula, making it suitable for columns of all lengths. It accounts for both buckling and material failure.

  4. 4.Why is Euler's theory not suitable for short columns?Application

    Euler's theory is not suitable for short columns because it assumes that buckling occurs before the material yields. In short columns, the material may yield before buckling occurs, making Euler's assumptions invalid. Short columns are more likely to fail due to material strength rather than buckling, which is why Rankine's formula is preferred for such cases.

  5. 5.What happens if a column is not perfectly straight or has initial imperfections?Application

    An initially crooked column or an eccentric load produces bending from the moment the load is applied, so lateral deflection grows steadily as the load rises instead of appearing suddenly at the critical load. Bending stresses add to the direct stress, and the column yields or collapses at a load below the Euler value. That is why design uses empirical or code-based column curves, the secant formula for eccentric loads, and a factor of safety on the theoretical buckling load.

  6. 6.Why is the effective length of a column important in Euler's theory?Application

    Euler's load P = π²EI/L_e² depends on the square of the effective length, the length between points of inflection of the buckled shape. End restraint changes it: L for both ends pinned, 2L for one end fixed and the other free, L/2 for both fixed, and L/√2 for fixed–pinned. Going from fixed–free to fixed–fixed therefore raises the buckling load sixteen-fold. Because real joints are never perfectly fixed, design codes specify practical effective lengths that are somewhat longer than the ideal values.

  7. 7.How does the moment of inertia affect the buckling load of a column?Application

    The Euler load is directly proportional to I, so a larger second moment of area raises the buckling load in proportion. A column buckles about the axis with the least I (least radius of gyration), so a rectangular bar bends about its thin direction. That is why tubes and other sections with similar I about every axis are efficient for compression members: they put material far from the centroid in all directions.

  8. 8.Calculate the critical load for a steel column with a length of 3 meters, a modulus of elasticity of 200 GPa, and a moment of inertia of 0.0001 m⁴. Assume both ends are pinned.Numerical

    P_cr = π²·E·I/L² = π² × 200 × 10⁹ × 1 × 10⁻⁴ / 3² = 1.974 × 10⁸ / 9 = 2.19 × 10⁶ N. So the Euler critical load is about 2193 kN. With such a large I for a 3 m length, the column is quite stocky, so you should also check the slenderness ratio, because the crushing load may govern instead.

  9. 9.A column has a critical load of 150 kN according to Euler's theory. If the column's length is doubled, what will be the new critical load?Numerical

    According to Euler's formula, P_cr = (π²EI) / (L²). If the length L is doubled, the new length is 2L. The new critical load P_cr_new = (π²EI) / ((2L)²) = (π²EI) / (4L²) = P_cr / 4. Therefore, the new critical load is 150 kN / 4 = 37.5 kN.

  10. 10.Explain how safety factors are used in the design of columns.Application

    Safety factors are used in the design of columns to account for uncertainties in material properties, loading conditions, and potential imperfections in the column. By applying a safety factor, engineers ensure that the column can support loads greater than the expected maximum load, reducing the risk of failure. The safety factor is typically applied to the calculated critical load, providing a margin of safety in the design.

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