1. **Problem Statement:** Calculate the maximum shear stress developed in the steel shaft in regions CF and BC. 2. **Given Data:** - Diameter of shaft, $D = 25\ \text{mm} = 0.025\ \text{m}$ - Radius, $C = 12.5\ \text{mm} = 0.0125\ \text{m}$ - Power delivered by motor, $P = 12\times10^3\ \text{W}$ - Rotational speed, $n = 50\ \text{rev/min}$ - Power removed by gears: $P_A = 3\times10^3\ \text{W}$, $P_B = 4\times10^3\ \text{W}$, $P_C = 5\times10^3\ \text{W}$ 3. **Formulas and Important Rules:** - Torque $T$ is related to power $P$ and angular velocity $\omega$ by: $$T = \frac{P}{\omega}$$ - Angular velocity $\omega$ in rad/s is: $$\omega = \frac{2\pi n}{60}$$ - Maximum shear stress $\tau_{max}$ in a circular shaft is: $$\tau_{max} = \frac{T c}{J}$$ where $c$ is the outer radius and $J$ is the polar moment of inertia: $$J = \frac{\pi}{2} c^4$$ 4. **Calculate Angular Velocity $\omega$:** $$\omega = \frac{2\pi \times 50}{60} = \frac{100\pi}{60} = \frac{5\pi}{3} \approx 5.236\ \text{rad/s}$$ 5. **Calculate Torque in region CF:** Power at F is $12\times10^3$ W, so $$T_{CF} = \frac{P}{\omega} = \frac{12\times10^3}{5.236} \approx 2291.83\ \text{N.m}$$ 6. **Calculate Maximum Shear Stress in CF:** Calculate $J$: $$J = \frac{\pi}{2} (0.0125)^4 = \frac{\pi}{2} \times 2.4414\times10^{-7} = 3.832\times10^{-7}\ \text{m}^4$$ Calculate $\tau_{max}$: $$\tau_{max,CF} = \frac{T_{CF} \times c}{J} = \frac{2291.83 \times 0.0125}{3.832\times10^{-7}} \approx 74.8\times10^{6} \ \text{Pa} = 74.8\ \text{MPa}$$ 7. **Calculate Torque in region BC:** Power removed by gears A, B, and C is $3+4+5=12$ kW, so power transmitted in BC is: $$P_{BC} = 12 - (3+4+5) = 0\ \text{W}$$ This suggests no torque transmitted beyond C, but assuming the problem means torque at B and C separately: Torque at C (after removing 3 kW at A and 4 kW at B): $$P_{BC} = 12 - 3 - 4 = 5\times10^3\ \text{W}$$ Calculate torque at BC: $$T_{BC} = \frac{P_{BC}}{\omega} = \frac{5\times10^3}{5.236} \approx 954.93\ \text{N.m}$$ 8. **Calculate Maximum Shear Stress in BC:** Calculate $J$ for radius $c=0.0125$ m (same as above): $$J = 3.832\times10^{-7}\ \text{m}^4$$ Calculate $\tau_{max}$: $$\tau_{max,BC} = \frac{T_{BC} \times c}{J} = \frac{954.93 \times 0.0125}{3.832\times10^{-7}} \approx 31.2\times10^{6} \ \text{Pa} = 31.2\ \text{MPa}$$ **Final Answers:** - Maximum shear stress in region CF: $\boxed{74.8\ \text{MPa}}$ - Maximum shear stress in region BC: $\boxed{31.2\ \text{MPa}}$