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).

Hint:

Write the hoop strain in terms of both the hoop and axial stress using Hooke's law for biaxial stress.

Answer 1

Correct Answer: 840

Step 1: Recall the two stresses at the outer wall.
Internal pressure $p$ creates hoop stress $\sigma_h = pD/(2t)$ and axial stress $\sigma_l = pD/(4t) = \sigma_h/2$, with radial stress ignored since the wall is thin compared to the diameter.

Step 2: Check what pressure the hoop stress alone would allow.
If Poisson's effect is ignored, $\varepsilon_h \approx \sigma_h/E$, so $\sigma_h = E\varepsilon_h = 210\times10^9 \times 0.00034 = 7.14\times10^7$ Pa, giving $p = 2t\sigma_h/D = 2(0.015)(7.14\times10^7)/3 = 7.14\times10^5$ Pa. This rough number ignores that the axial stress also shrinks the hoop strain a little through Poisson's ratio.

Step 3: Bring in the axial stress correction.
The true hoop strain is $\varepsilon_h = (\sigma_h - \nu\sigma_l)/E = (\sigma_h - \nu\sigma_h/2)/E = \sigma_h(1-\nu/2)/E$. So the actual allowable $\sigma_h$ is higher by a factor $1/(1-\nu/2) = 1/0.85$ compared to Step 2's rough number: $\sigma_h = 7.14\times10^7/0.85 = 8.4\times10^7$ Pa.

Step 4: Convert hoop stress back to pressure.
$p = \dfrac{2t\sigma_h}{D} = \dfrac{2(0.015)(8.4\times10^7)}{3} = \dfrac{2.52\times10^6}{3} = 8.4\times10^5$ Pa $= 840$ kPa.

Final Answer:
Correcting for the Poisson effect from the axial stress raises the allowable pressure from the naive uniaxial estimate up to the true biaxial value. \[ \boxed{p = 840.0 \ \text{kPa}} \]