Mumbai university > mechanical engineering > sem 3 > strength of materials

$\eta_L = 80\% = 0.8\ \eta_c = 50\% = 0.5\ E = 200GPa = 200\hspace{0.05cm}\times\hspace{0.05cm}10^3\ \mu = 0.286\ d = 300\hspace{0.05cm}mm\ l = 2500\hspace{0.05cm}mm\ t = 6\hspace{0.05cm}mm\ P = 2\hspace{0.05cm}MPa = 2\hspace{0.05cm}N/mm^2$ Circumferential stress $\sigma_c = \frac{Pd}{2t.\eta_L} = \frac{300\hspace{0.05cm}\times\hspace{0.05cm}2}{2\hspace{0.05cm}\times\hspace{0.05cm}6\hspace{0.05cm}\times\hspace{0.05cm}0.8} = 62.5\hspace{0.05cm}N/mm^2$ Longitudinal stress $\sigma_L = \frac{Pd}{4t.\eta_c} = \frac{2\hspace{0.05cm}\times\hspace{0.05cm}300}{4\hspace{0.05cm}\times\hspace{0.05cm}6\hspace{0.05cm}\times\hspace{0.05cm}0.5} = 50\hspace{0.05cm}N/mm^2$ For $\delta d\ e_c = \frac{\sigma_c}{E} - \mu\frac{\sigma_L}{E} = \frac{62.5}{200\hspace{0.05cm}\times\hspace{0.05cm}10^3} - 0.286\frac{50}{200\hspace{0.05cm}\times\hspace{0.05cm}10^3}\ e_c = 2.41\hspace{0.05cm}\times\hspace{0.05cm}10^{-4}\ e_c = \frac{\delta d}{d}\ \delta d = 0.0723\hspace{0.05cm}mm$ For $\delta L\ e_l = \frac{\sigma_L}{E} - \mu\frac{\sigma_c}{E}\ e_L = 1.606\hspace{0.05cm}\times\hspace{0.05cm}10^{-4}\ e_L = \frac{\delta L}{L}\ \delta L = 0.4015\hspace{0.05cm}mm$ For $\delta v\ e_v = e_L + 2e_c\ \hspace{0.25cm} = 6.426\hspace{0.05cm}\times\hspace{0.05cm}10^{-4}\ e_v = \frac{\delta v}{V}\ \delta v = 113556.79\hspace{0.05cm}mm^3$