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Vertical standpipe wall thickness

Short answer

Water pressure at base ρgh=0.2943 N/mm². Required thickness from hoop stress formula t=pd/(2σ_allow)=0.2943×4000/(2×120)=4.91 mm, provide about 5 mm.

A vertical thin walled standpipe is 4 m in diameter and stands 30 m high. If the allowable working stress in tension is 120 N/mm2120\ \text{N/mm}^2, what is the required wall thickness of the pipe?

A standpipe is a vertical pipe filled with water. The pressure is greatest at the base, so the wall must be thickest there; the hoop stress governs because it is twice the longitudinal stress.

Data: d=4d = 4 m =4000= 4000 mm, h=30h = 30 m, σallow=120 N/mm2\sigma_{allow} = 120\ \text{N/mm}^2, ρw=1000 kg/m3\rho_w = 1000\ \text{kg/m}^3, g=9.81 m/s2g = 9.81\ \text{m/s}^2.

Water pressure at the base

p=ρgh=1000(9.81)(30)=294 300 N/m2=0.2943 N/mm2p = \rho g h = 1000(9.81)(30) = 294\,300\ \text{N/m}^2 = 0.2943\ \text{N/mm}^2

Thickness (hoop stress)

σc=pd2t⇒t=pd2σc=0.2943×40002×120=4.91 mm\sigma_c = \frac{pd}{2t} \Rightarrow t = \frac{pd}{2\sigma_c} = \frac{0.2943\times4000}{2\times120} = 4.91\ \text{mm}

Answer: required thickness ≈4.91 mm\approx 4.91\ \text{mm}; provide about 5 mm (the thickness could be reduced higher up where the pressure is less).

✓ Answer checked by Claude Opus 5.5 (high effort)

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