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Hollow steel strut: crippling and safe load

Short answer

Using effective length 0.7 × 5 = 3.5 m and slenderness ratio 213, Euler’s formula applies. Crippling load = 36.83 kN, crippling stress = 42.48 N/mm², safe load with factor of safety 3 = 12.28 kN.

A hollow 5m long mild steel tube has the internal diameter of 4cm and thickness of 6mm. It is used as a strut with one end fixed and the other hinged. Find crippling load, crippling stress and the safe compressive load for the member if E=2×105 N/mm2E = 2\times10^5\ \text{N/mm}^2 and factor of safety is 3.

Given: L=5L = 5 m = 5000 mm, di=40d_i = 40 mm, t=6t = 6 mm so do=52d_o = 52 mm. One end fixed, other hinged: Pcr=2π2EIL2P_{cr} = \dfrac{2\pi^2EI}{L^2} (Le=0.7LL_e = 0.7L). E=2×105 N/mm2E = 2\times10^5\ \text{N/mm}^2, FOS = 3.

Section properties

I=π64(524−404)=2.332×105 mm4I = \frac{\pi}{64}(52^4 - 40^4) = 2.332\times10^5\ \text{mm}^4 A=π4(522−402)=867.1 mm2A = \frac{\pi}{4}(52^2 - 40^2) = 867.1\ \text{mm}^2

r=I/A=16.4r = \sqrt{I/A} = 16.4 mm, and Le/r=3500/16.4=213L_e/r = 3500/16.4 = 213, so the strut is slender and Euler's formula is valid.

Crippling load

Pcr=2π2×2×105×2.332×10550002=36,832 N=36.83 kNP_{cr} = \frac{2\pi^2\times2\times10^5\times2.332\times10^5}{5000^2} = 36{,}832\ \text{N} = 36.83\ \text{kN}

Crippling stress

σcr=PcrA=36,832867.1=42.48 N/mm2\sigma_{cr} = \frac{P_{cr}}{A} = \frac{36{,}832}{867.1} = 42.48\ \text{N/mm}^2

Safe load

Psafe=36.833=12.28 kNP_{safe} = \frac{36.83}{3} = 12.28\ \text{kN}

Answer: crippling load = 36.83 kN; crippling stress = 42.48 N/mm²; safe load = 12.28 kN.

✓ Answer checked by Claude Opus 5.5 (high effort)

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