1. **Problem Statement:** Calculate the maximum shear stress in the hollow shaft segment CD subjected to a torque $T_C = 28$ kNm, with inner radius $R_i = 22$ mm and outer radius $R_o = 52$ mm. 2. **Relevant Formula:** The maximum shear stress $\tau_{max}$ in a hollow circular shaft under torque $T$ is given by: $$\tau_{max} = \frac{T c}{J}$$ where: - $T$ is the applied torque (in N·mm), - $c$ is the outer radius (in mm), - $J$ is the polar moment of inertia for a hollow circular shaft: $$J = \frac{\pi}{2} (R_o^4 - R_i^4)$$ 3. **Convert Units:** - Torque $T_C = 28$ kNm = $28 \times 10^3$ Nm = $28 \times 10^6$ Nmm - Outer radius $c = R_o = 52$ mm - Inner radius $R_i = 22$ mm 4. **Calculate Polar Moment of Inertia $J$:** $$J = \frac{\pi}{2} (52^4 - 22^4) = \frac{\pi}{2} (7311616 - 234256) = \frac{\pi}{2} \times 7077360$$ $$J \approx 1.5708 \times 7077360 = 11109500.5 \text{ mm}^4$$ 5. **Calculate Maximum Shear Stress $\tau_{max}$:** $$\tau_{max} = \frac{T c}{J} = \frac{28 \times 10^6 \times 52}{11109500.5}$$ $$\tau_{max} \approx \frac{1.456 \times 10^9}{11109500.5} \approx 131.1 \text{ MPa}$$ 6. **Final Answer:** The maximum shear stress in the shaft CD is approximately **131.1 MPa**.