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Cylindrical vessel 3 m long: change in volume

Short answer

Circumferential stress 90 MPa, longitudinal 45 MPa. Volumetric strain 8.55×10⁻⁴, initial volume 8.4823×10⁸ mm³. Change in volume ≈7.252×10⁵ mm³ or 725.2 cm³ (increase).

A cylindrical vessel 3 m long and 600 mm diameter with 10 mm thick plates is subjected to an internal pressure of 3 MPa. Calculate the change in volume of the vessel. Take E=200 GPaE = 200\ \text{GPa} and Poisson's ratio = 0.3 for the vessel material.

Data: d=600d = 600 mm, L=3000L = 3000 mm, t=10t = 10 mm, p=3 MPap = 3\ \text{MPa}, E=200 GPa=2×105 MPaE = 200\ \text{GPa} = 2\times10^5\ \text{MPa}, ν=0.3\nu = 0.3.

Stresses

σc=pd2t=3(600)2(10)=90.0 MPa,σl=pd4t=45.0 MPa\sigma_c = \frac{pd}{2t} = \frac{3(600)}{2(10)} = 90.0\ \text{MPa}, \qquad \sigma_l = \frac{pd}{4t} = 45.0\ \text{MPa}

Strains

εc=σc−νσlE=90.0−0.3(45.0)2×105=3.825×10−4,εl=σl−νσcE=9.000×10−5\varepsilon_c = \frac{\sigma_c-\nu\sigma_l}{E} = \frac{90.0 - 0.3(45.0)}{2\times10^5} = 3.825\times10^{-4}, \qquad \varepsilon_l = \frac{\sigma_l-\nu\sigma_c}{E} = 9.000\times10^{-5} εv=2εc+εl=8.550×10−4\varepsilon_v = 2\varepsilon_c + \varepsilon_l = 8.550\times10^{-4}

Change in volume

V=π4(600)2(3000)=8.4823×108 mm3,δV=εvV=7.2524×105 mm3V = \frac{\pi}{4}(600)^2(3000) = 8.4823\times10^{8}\ \text{mm}^3, \qquad \delta V = \varepsilon_vV = 7.2524\times10^{5}\ \text{mm}^3

Answer: δV≈725.2 cm3\delta V \approx 725.2\ \text{cm}^3 (725,237 mm³, an increase).

✓ Answer checked by Claude Opus 5.5 (high effort)

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