Given:

$r = 800\hspace{0.05cm}mm,\hspace{0.5cm}d = 1600\hspace{0.05cm}mm\\ t = 8\hspace{0.05cm}mm\\ P = 0.6\hspace{0.05cm}MPa = 0.6\hspace{0.05cm}N/mm^2\\ \frac{1}{m} = \mu = 0.25$

To Find $\sigma_c\hspace{0.25cm}\&\hspace{0.25cm}\delta d$

Solution

$\sigma_c = \frac{Pd}{4t}\\ \hspace{0.25cm} = \frac{0.6\hspace{0.05cm}\times\hspace{0.05cm}1600}{4\hspace{0.05cm}\times\hspace{0.05cm}8}\\ \sigma_c = 30\hspace{0.05cm}N/mm^2$

$e_c = \frac{\sigma_c}{E} - \mu\frac{\sigma_c}{E} = \frac{30}{200\hspace{0.05cm}\times\hspace{0.05cm}10^3} - 0.25\frac{30}{200\hspace{0.05cm}\times\hspace{0.05cm}10^3}\\ \hspace{0.25cm} = 1.125\hspace{0.05cm}\times\hspace{0.05cm}10^{-4}\\ e_c = \frac{\delta d}{d}\\ \delta …