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Hollow steel tube strut: critical, safe load

Short answer

With both ends hinged, L_e = 5 m. I = 3.571×10⁵ mm⁴, critical load P_cr = π²EI/L² = 29.60 kN. Safe load with factor of safety 3 is 9.87 kN.

A hollow mild steel tube is 5 m long and 4 cm internal diameter. Thickness of tube is 8 mm and it is used as a strut with both ends hinged. Determine critical load and safe load on the strut. Take E=2.1×105 N/mm2E = 2.1\times10^5\ \text{N/mm}^2 F.O.S = 3

Given: L=5L = 5 m = 5000 mm, di=40d_i = 40 mm, t=8t = 8 mm, so do=40+2×8=56d_o = 40 + 2\times8 = 56 mm. Both ends hinged (Le=LL_e = L), E=2.1×105 N/mm2E = 2.1\times10^5\ \text{N/mm}^2, FOS = 3.

Section properties

I=π64(564−404)=π64(9834496−2560000)=3.571×105 mm4I = \frac{\pi}{64}(56^4 - 40^4) = \frac{\pi}{64}(9834496 - 2560000) = 3.571\times10^5\ \text{mm}^4 A=π4(562−402)=1206.4 mm2A = \frac{\pi}{4}(56^2 - 40^2) = 1206.4\ \text{mm}^2

Radius of gyration r=I/A=17.2r = \sqrt{I/A} = 17.2 mm, so λ=5000/17.2=290.6\lambda = 5000/17.2 = 290.6, which is very slender, so Euler's formula applies.

Critical load

Pcr=π2EIL2=π2×2.1×105×3.571×10550002=29,604 N=29.60 kNP_{cr} = \frac{\pi^2EI}{L^2} = \frac{\pi^2\times2.1\times10^5\times3.571\times10^5}{5000^2} = 29{,}604\ \text{N} = 29.60\ \text{kN}

Safe load

Psafe=PcrFOS=29.603=9.87 kNP_{safe} = \frac{P_{cr}}{FOS} = \frac{29.60}{3} = 9.87\ \text{kN}

Answer: critical load = 29.60 kN; safe load = 9.87 kN.

✓ Answer checked by Claude Opus 5.5 (high effort)

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