## Step 1: **Determine the Internal Torque in the Aluminum Segment** ### **Explanation:** - The system consists of three segments: Br (Bronze, left), Al (Aluminum, middle), Br (Bronze, right). - **Torques applied:** - At point B: **10 kN·m (clockwise)** - At point C: **10 kN·m (counterclockwise)** - **Fixed ends at A and D**. Let's perform a **torque balance** for the aluminum segment (between B and C): - **At B:** +10 kN·m (enters Al from left) - **At C:** -10 kN·m (leaves Al to right) - Both torques act in opposite directions, so the **net internal torque in the Aluminum section (BC) is 10 kN·m**. ### **Result after Step 1:** > **The internal torque in the aluminum segment (BC) is \( T_{Al} = 10\, \text{kN}\cdot\text{m} = 10,000\, \text{N}\cdot\text{m} \).** --- ## Step 2: **Calculate the Maximum Shear Stress in the Aluminum Segment** ### **Explanation:** The formula for **maximum shear stress** in a solid circular shaft is: \[ \tau_{max} = \frac{Tc}{J} \] Where: - \( T = 10,000\, \text{N}\cdot\text{m} \) (from Step 1) - \( c = \) outer radius \( = \frac{d}{2} = \frac{60}{2} = 30\, \text{mm} = .03\, \text{m} \) - \( J = \) polar moment of inertia for a solid shaft: \[ J = \frac{\pi}{32} d^4 = \frac{\pi}{32} (.06)^4 \] Let's calculate \( J \): \[ J = \frac{\pi}{32} (.06)^4 = \frac{3.1416}{32} \times .00001296 \approx .000001273\, \text{m}^4 \] Now, plug into the shear stress formula: \[ \tau_{max} = \frac{10,000 \times .03}{.000001273} \] \[ \tau_{max} = \frac{300}{.000001273} \] \[ \tau_{max} \approx 235,729,000\, \text{N/m}^2 = 235.7\, \text{MPa} \] ### **Result after Step 2:** > **The maximum shear stress in the aluminum segment is \( \boxed{235.7\, \text{MPa}} \).** --- ## **Final Answer** ### **Maximum Shear Stress in Aluminum Segment:** \[ \boxed{235.7\, \text{MPa}} \]