Taking internal pressure as 1 MPa, E= 210 GPa, μ= 0.3. Determine: I. Thickness of the wall on the hemispherical portion. II. Change in volume of the vessel.

Given:

$L = 1000\hspace{0.05cm}mm,\hspace{0.25cm}t_1 = 6\hspace{0.05cm}mm,\hspace{0.25cm}\frac{1}{m} = \mu=0.3,\hspace{0.25cm}d = 0.5\hspace{0.05cm}m = 0.5\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}mm,\hspace{0.25cm}P = 1\hspace{0.05cm}MPa\hspace{0.05cm}(OR)\hspace{0.05cm}N/mm^2$

To Find $t_2 = \delta v = ?$

Solution

CASE 1: $t_2 =?$

$\frac{t_2}{t_1} = \frac{m - 1}{2m - 1}\\ \frac{t_2}{6} = \frac{3.33 -1}{2\hspace{0.05cm}\times\hspace{0.05cm}3.33 - 1}\\ t_2 = 2.46\hspace{0.05cm}mm$

CASE 2: $\delta v = ?$

change in vol ($\delta v$ ) = change in vol. cylindrical portion + change in vol. in hemispherical portion

$\delta v = \delta v_1 + \delta v_2$

I] Change in volume in cylindrical portion ($\delta v_1$)

$e_v = \frac{\delta v_1}{v_1}\\ e_L + 2e_c = \frac{\delta v_1}{v_1}\\ [\frac{\sigma_L}{E} -\mu\frac{\sigma_c}{E}] + 2[\frac{\sigma_c}{E} - \mu\frac{\sigma_L}{E}] = \frac{\delta v_1}{v_1}\\ v_1 = \textit{area X length}\\ \hspace{0.25cm} = \frac{\pi}{4}d^2L\\ \hspace{0.25cm} = \frac{\pi}{4}\hspace{0.05cm}\times\hspace{0.05cm}500^2\hspace{0.05cm}\times\hspace{0.05cm}1000\\ \hspace{0.25cm} = 196.349\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}mm^3\\ \sigma_c = \frac{Pd}{2t_1} = \frac{1\hspace{0.05cm}\times\hspace{0.05cm}0.5\hspace{0.05cm}\times\hspace{0.05cm}10^3}{2\hspace{0.05cm}\times\hspace{0.05cm}6} = 41.68\hspace{0.05cm}N/mm^2\\ \sigma_L = \frac{Pd}{4t_1} = \frac{1\hspace{0.05cm}\times\hspace{0.05cm}0.5\hspace{0.05cm}\times\hspace{0.05cm}10^3}{4\hspace{0.05cm}\times\hspace{0.05cm}6} = 20.83\hspace{0.05cm}N/mm^2\\ \textit{substituting the respective values}\\ \delta v_1 = 74\hspace{0.05cm}\times\hspace{0.05cm}10^3$

II] Change in vol. in hemispherical portion ($\delta v_2$)

$e_v = \frac{\delta v_2}{v_2}\\ 3e_c = \frac{\delta v_2}{v_2}\\ 3[\frac{\sigma_c}{E} - \mu\frac{\sigma_c}{E}] = \frac{\delta v_2}{v_2}\\ \textit{Here;} t_2 = 2.45\\ \sigma_c = \frac{Pd}{4t_2} = \frac{1\hspace{0.05cm}\times\hspace{0.05cm}500}{4\hspace{0.05cm}\times\hspace{0.05cm}2.45} = 51.02\hspace{0.05cm}MPa\\ v_2 = \frac{\pi}{6}\hspace{0.05cm}\times\hspace{0.05cm}d^3 = \frac{\pi}{6}\hspace{0.05cm}\times\hspace{0.05cm}500^3 = 65.44\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}mm\\ \textit{Substituting the values}\\ \delta v_2 = 33.25\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}mm^3$

$\delta v = \delta v_1 + \delta v-2\\ \hspace{0.25cm} = 74\hspace{0.05cm}\times\hspace{0.05cm}10^3 + 33.25\hspace{0.05cm}\times\hspace{0.05cm}10^3\\ \delta v = 107.25\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}mm$