1. **Problem statement:** A thin cylinder with internal diameter $d_i=800$ mm, length $L=5$ m, wall thickness $t=5$ mm is subjected to internal pressure $p=50$ MN/m². Given Poisson's ratio $\nu=0.3$ and Young's modulus $E=210$ GPa, find: (a) Hoop and longitudinal stresses. (b) Change in internal diameter and length. (c) Principal stresses when a torque of 80 Nm is applied. 2. **Convert units:** $d_i=800$ mm = 0.8 m, $t=5$ mm = 0.005 m, $p=50$ MN/m² = $50\times10^6$ Pa, $E=210$ GPa = $210\times10^9$ Pa. 3. **(a) Hoop and longitudinal stresses:** For thin cylinders, hoop stress $\sigma_h = \frac{p d_i}{2 t}$ and longitudinal stress $\sigma_l = \frac{p d_i}{4 t}$. Calculate: $$\sigma_h = \frac{50\times10^6 \times 0.8}{2 \times 0.005} = \frac{40\times10^6}{0.01} = 4\times10^9 \text{ Pa} = 4000 \text{ MPa}$$ $$\sigma_l = \frac{50\times10^6 \times 0.8}{4 \times 0.005} = \frac{40\times10^6}{0.02} = 2\times10^9 \text{ Pa} = 2000 \text{ MPa}$$ 4. **(b) Change in internal diameter and length:** Radial strain $\epsilon_r$ is negligible for thin walls. Hoop strain $\epsilon_h = \frac{\sigma_h - \nu \sigma_l}{E}$ Longitudinal strain $\epsilon_l = \frac{\sigma_l - \nu \sigma_h}{E}$ Calculate: $$\epsilon_h = \frac{4\times10^9 - 0.3 \times 2\times10^9}{210\times10^9} = \frac{4\times10^9 - 0.6\times10^9}{210\times10^9} = \frac{3.4\times10^9}{210\times10^9} = 0.01619$$ $$\epsilon_l = \frac{2\times10^9 - 0.3 \times 4\times10^9}{210\times10^9} = \frac{2\times10^9 - 1.2\times10^9}{210\times10^9} = \frac{0.8\times10^9}{210\times10^9} = 0.00381$$ Change in diameter $\Delta d = d_i \times \epsilon_h = 0.8 \times 0.01619 = 0.01295$ m = 12.95 mm Change in length $\Delta L = L \times \epsilon_l = 5 \times 0.00381 = 0.01905$ m = 19.05 mm 5. **(c) Principal stresses with torque:** Torque $T=80$ Nm causes shear stress $\tau = \frac{T c}{J}$ where $c$ is outer radius and $J$ is polar moment of inertia. Outer diameter $d_o = d_i + 2t = 0.8 + 2 \times 0.005 = 0.81$ m Outer radius $c = 0.81/2 = 0.405$ m Inner radius $r_i = 0.8/2 = 0.4$ m Polar moment of inertia for thin cylinder $J = 2 \pi r_i^3 t = 2 \pi \times (0.4)^3 \times 0.005 = 2 \pi \times 0.064 \times 0.005 = 0.00201$ m$^4$ Shear stress: $$\tau = \frac{80 \times 0.405}{0.00201} = \frac{32.4}{0.00201} = 16119 \text{ Pa} = 0.0161 \text{ MPa}$$ 6. **Calculate principal stresses:** Principal stresses are: $$\sigma_1 = \frac{\sigma_h + \sigma_l}{2} + \sqrt{\left(\frac{\sigma_h - \sigma_l}{2}\right)^2 + \tau^2}$$ $$\sigma_2 = \frac{\sigma_h + \sigma_l}{2} - \sqrt{\left(\frac{\sigma_h - \sigma_l}{2}\right)^2 + \tau^2}$$ Calculate: $$\frac{\sigma_h + \sigma_l}{2} = \frac{4000 + 2000}{2} = 3000 \text{ MPa}$$ $$\frac{\sigma_h - \sigma_l}{2} = \frac{4000 - 2000}{2} = 1000 \text{ MPa}$$ $$\sqrt{1000^2 + 0.0161^2} \approx 1000 \text{ MPa}$$ So, $$\sigma_1 = 3000 + 1000 = 4000 \text{ MPa}$$ $$\sigma_2 = 3000 - 1000 = 2000 \text{ MPa}$$ The torque shear stress is negligible compared to normal stresses. **Final answers:** (a) Hoop stress = 4000 MPa, Longitudinal stress = 2000 MPa (b) Change in diameter = 12.95 mm, Change in length = 19.05 mm (c) Principal stresses approximately 4000 MPa and 2000 MPa