Determine change in length, change in diameter and change in volume. Take $E= 210 KN/mm^2$, 1/m = 0.3

Given:

$E = 210\hspace{0.05cm}KN/mm^2 = 210\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}N/mm^2\\ \frac{1}{m} = \mu = 0.3\\ d = 1\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}mm = 1000\hspace{0.05cm}mm\\ l = 3\hspace{0.05cm}m = 3\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}mm\\ t = 10\hspace{0.05cm}mm\\ P = 3\hspace{0.05cm}MPa = 3\hspace{0.05cm}N/mm^2$

To fFind: $\delta L = \delta v = \delta d = ?$

Solution:

Circumferential stress, $\hspace{0.05cm}\sigma_c = \frac{Pd}{2t} = \frac{3\hspace{0.05cm}\times\hspace{0.05cm}1\hspace{0.05cm}\times\hspace{0.05cm}10^3}{2\hspace{0.05cm}\times\hspace{0.05cm}10} = …