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Euler Buckling Load of a Pinned Column

MediumEuler bucklingcolumns
A pinned–pinned steel column is 4 m long with E = 200 GPa and I = 5 × 10⁶ mm⁴. Its Euler buckling load is most nearly:
  1. A154 kN
  2. B196 kN
  3. C2467 kN
  4. D617 kN
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Answer: D. 617 kN

Equation: Euler buckling load Pcr = π²EI/(KL)². Values: pinned–pinned so K = 1; L = 4 m; E = 200 × 10⁹ Pa; I = 5 × 10⁶ mm⁴ = 5 × 10⁻⁶ m⁴. Step 1: π²EI = 9.870 × 200×10⁹ × 5×10⁻⁶ = 9.870 × 10⁶ N·m². Step 2: (KL)² = 16 m². Step 3: Pcr = 9.870×10⁶/16 = 617 kN. 154 kN uses K = 2 (fixed–free); 2,467 kN uses K = 0.5 (fixed–fixed); 196 kN uses π instead of π².
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