1. **Problem Statement:** (a) A steel tube has an outer diameter $d_o = 75$ mm and inner diameter $d_i = 60$ mm. The maximum allowable shearing stress is $\tau_{max} = 88$ MPa. (i) Calculate the maximum torque $T$ that can be applied to the tube. (ii) Calculate the shear stress at the inner surface of the steel tube. 2. **Formulas and Important Rules:** - The polar moment of inertia $J$ for a hollow circular shaft is given by: $$J = \frac{\pi}{32} (d_o^4 - d_i^4)$$ - The maximum shear stress in a circular shaft under torque $T$ is: $$\tau = \frac{T c}{J}$$ where $c$ is the outer radius. - Shear stress varies linearly from zero at the center to maximum at the outer surface. 3. **Calculations:** (i) Calculate $J$: $$d_o = 75 \text{ mm}, \quad d_i = 60 \text{ mm}$$ $$J = \frac{\pi}{32} (75^4 - 60^4)$$ Calculate powers: $$75^4 = 31640625, \quad 60^4 = 12960000$$ So, $$J = \frac{\pi}{32} (31640625 - 12960000) = \frac{\pi}{32} \times 18680625$$ $$J = \frac{3.1416}{32} \times 18680625 \approx 1832630.5 \text{ mm}^4$$ (ii) Maximum shear stress $\tau_{max} = 88$ MPa occurs at outer radius $c = \frac{d_o}{2} = 37.5$ mm. Using formula for torque: $$T = \frac{\tau_{max} J}{c} = \frac{88 \times 1832630.5}{37.5}$$ Calculate: $$T = \frac{161101084}{37.5} = 4296028.9 \text{ Nmm} = 4296.03 \text{ Nm}$$ (iii) Shear stress at inner surface radius $r_i = \frac{d_i}{2} = 30$ mm: Shear stress varies linearly with radius: $$\tau_i = \tau_{max} \times \frac{r_i}{c} = 88 \times \frac{30}{37.5} = 88 \times 0.8 = 70.4 \text{ MPa}$$ **Final answers:** - Maximum torque $T = 4296$ Nm - Shear stress at inner surface $\tau_i = 70.4$ MPa