Question:
A cylindrical pressure vessel made of steel has diameter of 3 m and wall thickness of 15 mm. For steel, Young's modulus and Poisson's ratio are 210 GPa and 0.3, respectively. The cylinder is designed such that the allowable normal strain at the outer cylindrical surface is equal to 0.00034. The permissible pressure in the tank is ________ kPa (rounded off to 1 decimal place).
A cylindrical pressure vessel made of steel has diameter of 3 m and wall thickness of 15 mm. For steel, Young's modulus and Poisson's ratio are 210 GPa and 0.3, respectively. The cylinder is designed such that the allowable normal strain at the outer cylindrical surface is equal to 0.00034. The permissible pressure in the tank is ________ kPa (rounded off to 1 decimal place).
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Write the hoop strain in terms of both the hoop and axial stress using Hooke's law for biaxial stress.
Updated On: Aug 14, 2026
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Verified By Collegedunia
Correct Answer: 840
Solution and Explanation
Step 1: Write the thin-cylinder stress formulas.
For a thin cylindrical vessel of diameter \(D\) and wall thickness \(t\) under internal pressure \(p\), the hoop stress is \(\sigma_h = \dfrac{pD}{2t}\) and the longitudinal stress is \(\sigma_l = \dfrac{pD}{4t}\), so \(\sigma_l = \sigma_h/2\).
Step 2: Write the hoop strain using Hooke's law for biaxial stress.
At the outer surface the radial stress is negligible for a thin shell, so this is a biaxial stress state. The circumferential strain is \(\varepsilon_h = \dfrac{1}{E}(\sigma_h - \nu\sigma_l) = \dfrac{1}{E}\left(\dfrac{pD}{2t} - \nu\dfrac{pD}{4t}\right) = \dfrac{pD}{2tE}\left(1 - \dfrac{\nu}{2}\right)\).
Step 3: Solve for the pressure.
Rearranging, \(p = \dfrac{2tE\varepsilon_h}{D(1 - \nu/2)}\). With \(t = 0.015\) m, \(E = 210 \times 10^9\) Pa, \(D = 3\) m, \(\nu = 0.3\) (so \(1-\nu/2 = 0.85\)) and \(\varepsilon_h = 0.00034\):
\(p = \dfrac{2 \times 0.015 \times 210 \times 10^9 \times 0.00034}{3 \times 0.85} = \dfrac{2.142 \times 10^6}{2.55} = 8.4 \times 10^5\) Pa.
Final Answer:
Converting to kPa gives the permissible pressure directly from the allowable strain. \[ \boxed{p = 840.0 \ \text{kPa}} \]
For a thin cylindrical vessel of diameter \(D\) and wall thickness \(t\) under internal pressure \(p\), the hoop stress is \(\sigma_h = \dfrac{pD}{2t}\) and the longitudinal stress is \(\sigma_l = \dfrac{pD}{4t}\), so \(\sigma_l = \sigma_h/2\).
Step 2: Write the hoop strain using Hooke's law for biaxial stress.
At the outer surface the radial stress is negligible for a thin shell, so this is a biaxial stress state. The circumferential strain is \(\varepsilon_h = \dfrac{1}{E}(\sigma_h - \nu\sigma_l) = \dfrac{1}{E}\left(\dfrac{pD}{2t} - \nu\dfrac{pD}{4t}\right) = \dfrac{pD}{2tE}\left(1 - \dfrac{\nu}{2}\right)\).
Step 3: Solve for the pressure.
Rearranging, \(p = \dfrac{2tE\varepsilon_h}{D(1 - \nu/2)}\). With \(t = 0.015\) m, \(E = 210 \times 10^9\) Pa, \(D = 3\) m, \(\nu = 0.3\) (so \(1-\nu/2 = 0.85\)) and \(\varepsilon_h = 0.00034\):
\(p = \dfrac{2 \times 0.015 \times 210 \times 10^9 \times 0.00034}{3 \times 0.85} = \dfrac{2.142 \times 10^6}{2.55} = 8.4 \times 10^5\) Pa.
Final Answer:
Converting to kPa gives the permissible pressure directly from the allowable strain. \[ \boxed{p = 840.0 \ \text{kPa}} \]
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