1. **State the problem:** We have a composite cylindrical shaft with an aluminum core radius $r_a = 30$ mm and a steel shell outer radius $r_s = 40$ mm. A torque $T$ is applied at end A causing twisting. The maximum shearing stress in the steel shell is given as $\tau_{max,steel} = 150$ MPa. We need to find the maximum shearing stress in the aluminum core $\tau_{max,al}$. 2. **Given data:** - Aluminum core radius $r_a = 30$ mm = 0.03 m - Steel shell outer radius $r_s = 40$ mm = 0.04 m - Length $L = 2$ m (not directly needed for stress calculation) - Maximum shear stress in steel $\tau_{max,steel} = 150$ MPa - Shear modulus of steel $G_s = 77.2$ GPa = $77.2 \times 10^3$ MPa - Shear modulus of aluminum $G_a = 27$ GPa = $27 \times 10^3$ MPa 3. **Concepts:** - The shaft is composite and subjected to the same angle of twist $\theta$ because it is bonded. - The torque is shared between steel and aluminum: $T = T_s + T_a$. - The shear stress $\tau$ in a circular shaft under torque is $\tau = \frac{T r}{J}$ where $J$ is the polar moment of inertia. - For a solid circular shaft, $J = \frac{\pi r^4}{2}$. - For the steel shell (hollow shaft), $J_s = \frac{\pi}{2} (r_s^4 - r_a^4)$. - For the aluminum core (solid shaft), $J_a = \frac{\pi}{2} r_a^4$. 4. **Calculate polar moments of inertia:** $$ J_a = \frac{\pi}{2} (0.03)^4 = \frac{\pi}{2} \times 8.1 \times 10^{-7} = 1.27 \times 10^{-6} \text{ m}^4 $$ $$ J_s = \frac{\pi}{2} ((0.04)^4 - (0.03)^4) = \frac{\pi}{2} (2.56 \times 10^{-6} - 8.1 \times 10^{-7}) = \frac{\pi}{2} \times 1.75 \times 10^{-6} = 2.75 \times 10^{-6} \text{ m}^4 $$ 5. **Relate angle of twist $\theta$ for each material:** $$ \theta = \frac{T_s L}{G_s J_s} = \frac{T_a L}{G_a J_a} $$ Rearranged: $$ \frac{T_s}{G_s J_s} = \frac{T_a}{G_a J_a} $$ 6. **Express $T_a$ in terms of $T_s$:** $$ T_a = T_s \frac{G_a J_a}{G_s J_s} $$ 7. **Total torque:** $$ T = T_s + T_a = T_s + T_s \frac{G_a J_a}{G_s J_s} = T_s \left(1 + \frac{G_a J_a}{G_s J_s}\right) $$ 8. **Calculate the ratio:** $$ \frac{G_a J_a}{G_s J_s} = \frac{27 \times 10^3 \times 1.27 \times 10^{-6}}{77.2 \times 10^3 \times 2.75 \times 10^{-6}} = \frac{34.29 \times 10^{-3}}{212.3 \times 10^{-3}} = 0.1615 $$ 9. **Calculate $T_s$ in terms of $T$:** $$ T_s = \frac{T}{1 + 0.1615} = \frac{T}{1.1615} $$ 10. **Maximum shear stress in steel:** $$ \tau_{max,steel} = \frac{T_s r_s}{J_s} \Rightarrow T_s = \frac{\tau_{max,steel} J_s}{r_s} $$ Calculate $T_s$: $$ T_s = \frac{150 \times 2.75 \times 10^{-6}}{0.04} = 1.031 \times 10^{-2} \text{ MNm} = 10,312.5 \text{ Nm} $$ 11. **Calculate total torque $T$:** $$ T = 1.1615 \times T_s = 1.1615 \times 10,312.5 = 11,978 \text{ Nm} $$ 12. **Calculate torque in aluminum:** $$ T_a = T - T_s = 11,978 - 10,312.5 = 1,665.5 \text{ Nm} $$ 13. **Calculate maximum shear stress in aluminum:** $$ \tau_{max,al} = \frac{T_a r_a}{J_a} = \frac{1,665.5 \times 0.03}{1.27 \times 10^{-6}} = 39,327,559 \text{ Pa} = 39.3 \text{ MPa} $$ **Final answer:** The maximum shearing stress in the aluminum core is approximately **39.3 MPa**.