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Round bar fixed-free: buckling load from lateral deflection

Short answer

From the cantilever test, EI = 15×10³ × 3000³ / (3 × 35) = 3.857×10¹² N·mm². For fixed-free, effective length is 2L = 6000 mm, giving buckling load P_cr = π²EI / (6000)² = 1.057×10⁶ N ≈ 1057 kN.

A round bar fixed at bottom and free at top has a length of 3 m. Determine the buckling load for the bar if the load is applied axially on top. If a horizontal force of 15 kN at top can produces a horizontal deflection of 35 mm.

Given: round bar fixed at the bottom, free at the top, L=3L = 3 m = 3000 mm. A horizontal force H=15H = 15 kN at the top gives a horizontal deflection δ=35\delta = 35 mm.

Step 1: find EIEI from the lateral load test

The bar acts as a cantilever under HH:

δ=HL33EI  ⇒  EI=HL33δ=15×103×300033×35=3.857×1012 N mm2\delta = \frac{HL^3}{3EI} \;\Rightarrow\; EI = \frac{HL^3}{3\delta} = \frac{15\times10^3\times3000^3}{3\times35} = 3.857\times10^{12}\ \text{N mm}^2

Step 2: buckling load (fixed-free)

Effective length Le=2L=6000L_e = 2L = 6000 mm.

Pcr=π2EILe2=π2×3.857×1012(6000)2=1.057×106 NP_{cr} = \frac{\pi^2EI}{L_e^2} = \frac{\pi^2\times3.857\times10^{12}}{(6000)^2} = 1.057\times10^6\ \text{N}

Answer: buckling load Pcr=1057P_{cr} = 1057 kN (about 1.06 MN).

✓ Answer checked by Claude Opus 5.5 (high effort)

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