Elastic flexural buckling of columns

Euler buckling of slender columns: derivation, effective lengths, axis of buckling, slenderness ratio, limit of validity and effect of imperfections.

Drafted with Aria, reviewed by the AiCanCode.org team. Spotted an error? Use Give Feedback at the bottom of the page.

Why it matters

Aircraft structures are made of slender, thin members: stringers and spar caps on the compression side of a wing, landing-gear side braces, engine-mount struts and control rods. These usually fail not by crushing but by buckling, a sudden sideways instability at a load that can be far below the yield load. Euler's theory gives that load and shows which design choices (length, end fixity, section shape) control it.

Key ideas

Stability, not strength. A perfectly straight, centrally loaded column stays straight as the load increases until the critical load P_cr. At P_cr a bent shape becomes an equilibrium position too (neutral equilibrium); any slightly larger load makes the straight position unstable and the column deflects sideways. Elastic buckling happens with the stress still below yield, and it depends on stiffness (E·I), not on strength.

Euler's derivation (pinned–pinned). For a deflected shape y(x), the moment is M = −P·y, so E·I·y'' + P·y = 0. With y = 0 at both ends, non-trivial solutions exist only when √(P/EI)·L = nπ. The lowest, n = 1, gives P_cr = π²·E·I/L² with a half-sine-wave shape. Higher modes need restraint at intermediate points to appear.

Effective length. Other end conditions are handled with an effective length Le = K·L, the distance between points of zero moment (inflection points) of the buckled shape:

  • Pinned–pinned: Le = L.
  • Fixed–free (flagpole): Le = 2L.
  • Fixed–pinned: Le ≈ 0.7L (0.699L exactly).
  • Fixed–fixed: Le = 0.5L. Design codes use slightly larger values because real fixity is never perfect; take them from your code book.

Which axis? A column buckles about the axis with the smallest E·I/Le², usually the minor principal axis of the section. A rectangle 30 × 60 mm buckles about the axis parallel to its 60 mm side. Tubes and boxes, with equal I in all directions, are efficient struts, which is why aircraft struts are tubes.

Slenderness ratio and critical stress. With I = A·r² (r = least radius of gyration), σcr = P_cr/A = π²·E/(Le/r)². The slenderness ratio λ = Le/r is the single number that controls elastic buckling. Euler's formula is valid only while σcr is below the proportional limit, i.e. for λ above λc = π·√(E/σy) (about 89 for mild steel, about 48 for a high-strength aluminium alloy with σy = 300 MPa).

Short and intermediate columns. Below λc, columns fail by inelastic buckling or crushing. Empirical formulas such as Johnson's parabola or the Rankine–Gordon formula, 1/P = 1/Pcrush + 1/P_Euler, cover that range; their constants come from your data book or code.

Imperfections. Real columns are slightly crooked or eccentrically loaded. They bend from the start, with deflection growing rapidly as P approaches P_cr, and the extra bending stress makes them fail below the Euler load (secant formula). This is why design codes apply reduction factors.

Formulas

P_cr = π²·E·I / Le² P_cr = critical (Euler) load (N), E = Young's modulus (Pa), I = least second moment of area (m⁴), Le = effective length (m).

Le = K·L K = 1 (pinned–pinned), 2 (fixed–free), 0.7 (fixed–pinned), 0.5 (fixed–fixed).

r = √(I / A), λ = Le / r Least radius of gyration (m) and slenderness ratio (–).

σcr = π²·E / λ² Critical stress (Pa); valid only when σcr is below the proportional limit.

λc = π·√(E / σy) Slenderness above which Euler's formula applies; σy = yield or proportional-limit stress (Pa).

1/P = 1/(σc·A) + 1/P_cr Rankine–Gordon interpolation (σc = crushing stress); constants from the data book.

Worked examples

Example 1 (standard). A pin-ended aluminium-alloy strut is a tube of outer diameter 40 mm and inner diameter 36 mm, 1.2 m long. E = 70 GPa, σy = 300 MPa. Find the Euler load and check that Euler's formula applies.

  1. I = π·(D⁴ − d⁴)/64 = π × (40⁴ − 36⁴)/64 = 43 216 mm⁴.
  2. A = π·(D² − d²)/4 = π × (1600 − 1296)/4 = 238.8 mm².
  3. r = √(I/A) = √(43 216/238.8) = 13.45 mm; Le = L = 1200 mm; λ = 1200/13.45 = 89.2.
  4. P_cr = π²·E·I/Le² = π² × 70 000 × 43 216/1200² = 20 734 N.
  5. σcr = 20 734/238.8 = 86.8 MPa.
  6. λc = π·√(70 000/300) = 48.0. Since λ = 89.2 > 48.0 (and σcr ≪ σy), elastic buckling governs.

Answer: P_cr ≈ 20.7 kN; σcr ≈ 86.8 MPa, well below yield, so Euler is valid. Yield alone would need 300 × 238.8 = 71.6 kN, so buckling governs by a factor of 3.5.

Example 2 (GATE level). A steel bar of rectangular section 60 mm × 30 mm and length 2.5 m is fixed at one end and pinned at the other. E = 200 GPa, σy = 250 MPa. Find the critical load, the critical stress, and the shortest length for which Euler's formula remains valid.

  1. Least I is about the axis parallel to the 60 mm side: I = b·h³/12 = 60 × 30³/12 = 1.35 × 10⁵ mm⁴.
  2. A = 1800 mm²; r = √(1.35 × 10⁵/1800) = 8.66 mm.
  3. Fixed–pinned: Le = 0.7 × 2500 = 1750 mm; λ = 1750/8.66 = 202.
  4. P_cr = π²·E·I/Le² = π² × 200 000 × 1.35 × 10⁵/1750² = 87 014 N.
  5. σcr = 87 014/1800 = 48.3 MPa.
  6. λc = π·√(200 000/250) = 88.9. Euler applies down to Le = 88.9 × 8.66 = 770 mm, i.e. L = 770/0.7 = 1.10 m.

Answer: P_cr ≈ 87.0 kN, σcr ≈ 48.3 MPa; Euler's formula is valid for lengths above about 1.10 m. Using the major-axis I (5.4 × 10⁵ mm⁴) would overestimate the load fourfold.

Common mistakes

  • Using the larger I of the section instead of the least I.
  • Using the actual length instead of the effective length, or K = 0.5 for a fixed–free column (it is 2).
  • Applying Euler's formula to short columns where σcr would exceed yield.
  • Forgetting that buckling load depends on E, not on strength; a stronger alloy with the same E buckles at the same load.
  • Mixing mm⁴ with m² in P = π²EI/L².

For GATE AE

Expect numericals on the Euler load for different end conditions, ratios of critical loads when length, diameter or end fixity change, slenderness ratio and the limit of validity of Euler's formula, and the axis of buckling for rectangular, I- and tubular sections. Practise remembering the K values and that P_cr scales with d⁴ for a solid circular strut and 1/Le².

Quick check

  1. How does P_cr change if a pinned column is made fixed at both ends?
  2. How does P_cr change if the length of a column doubles?
  3. What is Le for a fixed–free column of length L?
  4. A solid circular strut's diameter doubles. By what factor does P_cr increase?

Answers: 1. It becomes 4 times larger. 2. It falls to one quarter. 3. 2L. 4. 16.

Interview questions

All Mechanics of Solids interview questions

Try answering each one aloud before you open it.

  1. 1.What is elastic flexural buckling in columns?Concept

    Elastic flexural buckling refers to the sudden lateral deflection or bending of a column under axial compressive load, occurring when the load reaches a critical level. This phenomenon happens without any material yielding, meaning the column remains within its elastic limit. The critical load at which buckling occurs is determined by the column's material properties, length, cross-sectional shape, and end conditions.

  2. 2.Explain the Euler's formula for the critical buckling load of a column.Concept

    Euler's formula for the critical buckling load (P_cr) of a column is given by P_cr = (π²·E·I) / (K·L)², where E is the modulus of elasticity, I is the moment of inertia of the column's cross-section, L is the effective length of the column, and K is the column effective length factor, which depends on the end conditions. This formula applies to long, slender columns that buckle elastically.

  3. 3.What are the typical end conditions for columns, and how do they affect buckling?Concept

    Typical end conditions for columns include pinned-pinned, fixed-fixed, fixed-free, and fixed-pinned. These conditions affect the effective length factor (K) in Euler's formula. For example, a pinned-pinned column has K = 1, a fixed-fixed column has K = 0.5, a fixed-free column has K = 2, and a fixed-pinned column has K = 0.7. The effective length factor influences the critical buckling load, with shorter effective lengths leading to higher critical loads.

  4. 4.Why is the concept of slenderness ratio important in column buckling?Concept

    The slenderness ratio, defined as the effective length (L_e) divided by the radius of gyration (r) of the column, is crucial in determining the buckling behavior. A high slenderness ratio indicates a slender column, which is more prone to buckling under lower loads. It helps in classifying columns as short, intermediate, or long, influencing the design approach and safety considerations.

  5. 5.How does material anisotropy affect the buckling of columns?Application

    Material anisotropy, where material properties vary with direction, can significantly affect column buckling. Anisotropic materials may have different moduli of elasticity in different directions, altering the critical buckling load. For instance, in composite materials, the orientation of fibers can influence the stiffness and strength, requiring careful consideration in design to prevent unexpected buckling.

  6. 6.What happens if a column is eccentrically loaded?Application

    An eccentric load e applies a moment P·e from the start, so the column bends immediately instead of staying straight until a bifurcation point. The lateral deflection grows rapidly as P approaches the Euler load, and the extra bending stress (given by the secant formula, σmax = (P/A)[1 + (e·c/r²)·sec((Le/2r)√(P/EA))]) causes yielding at a load below P_cr. Design codes allow for this, and for initial crookedness, with reduced column strength curves.

  7. 7.Which cross-sections are efficient for columns, and why?Application

    A column buckles about its weakest axis, so efficient sections have a large least radius of gyration for their area, with material far from the centroid in every direction. Thin-walled tubes and box sections are best, which is why aircraft struts and torque tubes are tubes. Wide-flange H-sections are used in buildings because their minor-axis I is reasonable; a narrow I-beam is poor as a column because it buckles about its weak axis, and very thin walls bring the risk of local buckling.

  8. 8.Calculate the critical buckling load for a steel column with a length of 3 meters, pinned at both ends, with a moment of inertia of 8.1 x 10⁶ mm⁴ and a modulus of elasticity of 200 GPa.Numerical

    I = 8.1 × 10⁶ mm⁴ = 8.1 × 10⁻⁶ m⁴ and Le = L = 3 m for pinned ends. P_cr = π²·E·I/Le² = π² × 200 × 10⁹ × 8.1 × 10⁻⁶/3² = 1.78 × 10⁶ N ≈ 1777 kN. You should then check that P_cr/A is below the proportional limit, otherwise the column fails by yielding first and Euler does not apply.

  9. 9.A column with a fixed-free end condition has a length of 4 meters and a radius of gyration of 50 mm. Calculate its slenderness ratio.Numerical

    First, convert the radius of gyration to meters: 50 mm = 0.05 m. The effective length factor K for a fixed-free column is 2. The effective length L_e = K·L = 2·4 m = 8 m. The slenderness ratio is L_e / r = 8 m / 0.05 m = 160.

  10. 10.Explain how lateral-torsional buckling differs from flexural buckling in columns.Concept

    Lateral-torsional buckling involves both lateral deflection and twisting of a member, typically occurring in beams under bending. In contrast, flexural buckling involves only lateral deflection without twisting, primarily affecting columns under axial compression. Lateral-torsional buckling is influenced by the member's cross-sectional shape, length, and loading conditions, while flexural buckling is mainly determined by the column's slenderness and end conditions.

Finished this topic? Mark it so your progress, study plan and readiness keep up.

Stuck on something here?

PreviousTheories of failure