Wall thickness of a cylindrical shell of 800 mm internal diameter and having the internal volume of 1 m³ is 10 mm. If the shell is subjected to an internal pressure of 1.5 MPa, what will be the increase in the capacity of the cylinder?

[Assuming water is incompressible; Poisson ratio = 0.3; modulus of elasticity = 200 GPa]

Concept:

We use strain relations in a thin cylindrical shell to determine the increase in capacity under internal pressure.

Given:

• Internal diameter, \( D = 800 \, \text{mm} \)
• Internal radius, \( r = \frac{D}{2} = 400 \, \text{mm} \)
• Wall thickness, \( t = 10 \, \text{mm} \)
• Internal pressure, \( p = 1.5 \, \text{MPa} \)
• Modulus of elasticity, \( E = 200 \times 10^3 \, \text{MPa} \)
• Poisson's ratio, \( \nu = 0.3 \)
• Initial volume, \( V = 1 \, \text{m}^3 = 1 \times 10^9 \, \text{mm}^3 \)

Step 1: Calculate circumferential (hoop) stress and strain

\( \sigma_c = \frac{pr}{t} = \frac{1.5 \times 400}{10} = 60 \, \text{MPa} \)

Longitudinal stress, \( \sigma_l = \frac{pr}{2t} = \frac{1.5 \times 400}{2 \times 10} = 30 \, \text{MPa} \)

Hoop strain, \( \varepsilon_c = \frac{\sigma_c}{E} - \nu \times \frac{\sigma_l}{E} = \frac{60}{200 \times 10^3} - 0.3 \times \frac{30}{200 \times 10^3} \)

\( \varepsilon_c = 3 \times 10^{-4} - 0.045 \times 10^{-3} = 2.55 \times 10^{-4} \)

Step 2: Calculate longitudinal strain

\( \varepsilon_l = \frac{\sigma_l}{E} - \nu \times \frac{\sigma_c}{E} = \frac{30}{200 \times 10^3} - 0.3 \times \frac{60}{200 \times 10^3} \)

\( \varepsilon_l = 1.5 \times 10^{-4} - 0.9 \times 10^{-4} = 0.6 \times 10^{-4} \)

Step 3: Calculate volumetric strain

\( \varepsilon_v \approx 2 \varepsilon_c + \varepsilon_l = 2 \times 2.55 \times 10^{-4} + 0.6 \times 10^{-4} = 5.7 \times 10^{-4} \)

Step 4: Calculate increase in volume

\( \Delta V = \varepsilon_v \times V = 5.7 \times 10^{-4} \times 1 \times 10^9 = 570000 \, \text{mm}^3 \)