Euler Buckling of a Pinned Column — Worked Example

1. Problem

A 7075-T6 aluminium tube strut: OD = 30 mm, wall thickness = 2 mm, effective length L_eff = 600 mm (pinned both ends, K = 1.0). Applied compressive load P = 25 kN. Check Euler buckling margin and compare with FE eigenvalue buckling.

2. Given

3. Required

• Section properties (A, I, r)
• Slenderness ratio and column regime
• Euler buckling load P_cr
• Margin of safety
• FE eigenvalue buckling factor comparison

4. Assumptions

• Perfect geometry (no imperfections)
• Concentric loading (no eccentricity)
• Linear elastic material behaviour
• Pinned-pinned end conditions (K = 1.0)
• Buckling occurs about the weak axis (isotropic section — any axis)
• No local buckling (wall is thick enough relative to diameter)

5. Governing Equations

Euler buckling load:
         π² · E · I
P_cr  =  ───────────
           (K · L)²

Slenderness ratio:
  λ  =  K·L / r       r = √(I/A)

Transition slenderness:
  λ_trans  =  √(2π²E / Fcy)

Buckling stress:
  σ_cr  =  π²E / λ²

Margin of safety:
  MS  =  (P_cr / P) − 1

6. Calculation

Step 1: Section properties.

A  =  π/4 × (OD² − ID²)  =  π/4 × (900 − 676)  =  π/4 × 224  =  175.9 mm²

I  =  π/64 × (OD⁴ − ID⁴)  =  π/64 × (810,000 − 456,976)  =  π/64 × 353,024  =  17,330 mm⁴

r  =  √(I/A)  =  √(17,330 / 175.9)  =  √98.5  =  9.92 mm

6. Calculation (continued)

Step 2: Slenderness ratio and column regime.

λ  =  K·L / r  =  1.0 × 600 / 9.92  =  60.5

λ_trans  =  √(2π² × 71,700 / 503)  =  √(1,417,000 / 503)  =  √2817  =  53.1

Since λ = 60.5 > λ_trans = 53.1  →  Euler (long column) governs.

6. Calculation (continued)

Step 3: Critical buckling load and margin of safety.

P_cr  =  π² × 71,700 × 17,330 / 600²
     =  9.87 × 71,700 × 17,330 / 360,000
     =  1.226 × 10¹⁰ / 360,000
     =  34,030 N  =  34.0 kN

σ_cr  =  P_cr / A  =  34,030 / 175.9  =  193.4 MPa

MS  =  (34.0 / 25.0) − 1  =  +0.36

RF  =  MS + 1  =  1.36

7. Result

P_cr = 34.0 kN. MS = +0.36 (RF = 1.36). The strut meets the Euler buckling requirement. Also check compressive yield: σ = P/A = 25,000/175.9 = 142 MPa < Fcy = 503 MPa (MS = +2.54, not governing).

8. Check

• σ_cr = 193 MPa < Fcy = 503 MPa — confirms elastic buckling (valid for Euler)
• Dimensional check: π² × GPa × mm⁴ / mm² = N — consistent
• FE eigenvalue buckling: a beam model with 20 elements gives a load factor of 1.35 (P_cr = 33.8 kN) — within 0.6% of the hand calculation
• The buckling mode shape from FEA shows the expected half-sine wave for a pinned-pinned column

9. Interpretation

The strut passes the Euler buckling check with a margin of +0.36. The buckling stress (193 MPa) is well below the compressive yield strength (503 MPa), confirming that elastic buckling is the correct failure mode. The margin is modest — if the load uncertainty is high or the column has geometric imperfections, a knockdown factor should be applied. For a real column, an imperfection factor of 0.8 would reduce RF to 1.09.

The FE eigenvalue result (33.8 kN vs 34.0 kN analytical) confirms the hand calculation. The small difference is due to the FE beam formulation including shear flexibility, which slightly reduces the effective stiffness.

10. Limitations

• Perfect geometry assumed — real columns have imperfections that reduce the buckling load
• No initial crookedness or load eccentricity
• Euler buckling is elastic — if the buckling stress exceeds Fcy, the Johnson-Euler formula must be used
• No local wall buckling check (D/t = 15 — generally safe for metals, but check for very thin walls)
• No post-buckling behaviour — the column fails catastrophically at P_cr

11. Related Resources

• Euler Buckling Reference (Reference Library)
• Section Properties for Common Shapes (Reference Library)
• How to Assess Column Buckling
• How to Interpret an Eigenvalue Buckling Analysis
• Buckling Load Calculator