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, $\hspace{0.05cm}\sigma_c = \frac{Pd}{2t} = \frac{3\hspace{0.05cm}\times\hspace{0.05cm}1\hspace{0.05cm}\times\hspace{0.05cm}10^3}{2\hspace{0.05cm}\times\hspace{0.05cm}10} = 150\hspace{0.05cm}N/mm^2$

Longitudinal stress,$\hspace{0.05cm}\sigma_t = \frac {Pd}{4t} = \frac{3\hspace{0.05cm}\times\hspace{0.05cm}1\hspace{0.05cm}\times\hspace{0.05cm}10^3}{4\hspace{0.05cm}\times\hspace{0.05cm}10} = 75\hspace{0.05cm}N/mm^2$

Shear stress,$\hspace{0.05cm}\tau = \frac{Pd}{8t} = \frac{3\hspace{0.05cm}\times\hspace{0.05cm}1\hspace{0.05cm}\times\hspace{0.05cm}10^3}{8\hspace{0.05cm}\times\hspace{0.05cm}10} = 37.5\hspace{0.05cm}N/mm^2$

i)For $\delta d\\ e_c = \frac{\sigma_c}{E} - \mu\frac{\sigma_L}{E} = \frac{150}{210\hspace{0.05cm}\times\hspace{0.05cm}10^3} - \frac{0.3\hspace{0.05cm}\times\hspace{0.05cm}75}{210\hspace{0.05cm}\times\hspace{0.05cm}10^3} = 6.07\hspace{0.05cm}\times\hspace{0.05cm}10^{-4}\\ e_c = \frac{\delta d}{d}\\ 6.07\hspace{0.05cm}\times\hspace{0.05cm}10^{-4} = \frac{\delta d}{1\hspace{0.05cm}\times\hspace{0.05cm}10^3}\\ \delta d = 0.6\hspace{0.05cm}mm$

ii) For $\delta L\\ e_L = \frac{\sigma_L}{E} - \mu\frac{\sigma_c}{E} = \frac{75}{210\hspace{0.05cm}\times\hspace{0.05cm}10^3} - 0.3\frac{150}{210\hspace{0.05cm}\times\hspace{0.05cm}10^3} = 1.42\hspace{0.05cm}\times\hspace{0.05cm}10^{-4}\\ e_L = \frac{\delta_L}{L}\\ 1.42\hspace{0.05cm}\times\hspace{0.05cm}10^{-4} = \frac{\delta_L}{3\hspace{0.05cm}\times\hspace{0.05cm}10^3}\\ \delta_L = 0.4\hspace{0.05cm}mm$

iii) For $\delta_v\\ e_v = e_L + 2e_c\\ \hspace{0.25cm} = 1.3\hspace{0.05cm}\times\hspace{0.05cm}10^{-3}\\ e_v = \frac{\delta_v}{V}\\ 1.3\hspace{0.05cm}\times\hspace{0.05cm}10^{-3} = \frac{\delta_v}{\frac{\pi}{4}d^3}\\ \delta_v = 3.197\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}mm^3$