Let's break down the problem step by step: ### Given Data: - **Material:** Aluminum alloy 6061-T6 - **Power, \( P \):** 60 hp - **Speed, \( N \):** 1000 rev/min - **Factor of Safety (FS):** 2 - **Yield Strength of 6061-T6 (\( S_y \)):** Typically about 40,000 psi (you may want to confirm with your textbook or data sheet) - **To Find:** Minimum shaft diameter (\( d \)) in inches --- ### Step 1: Convert Power to Torque \[ P = 2\pi N T / 60 \] Convert horsepower to ft-lbf/s: \[ 1 \text{ hp} = 550 \text{ ft-lbf/s} \] So, \[ P = 60 \text{ hp} \times 550 = 33,000 \text{ ft-lbf/min} \] But let's keep all units consistent: \[ P = 60 \text{ hp} \times 550 \text{ ft-lbf/s} = 33,000 \text{ ft-lbf/s} \] At 1000 rev/min: \[ P = \frac{2\pi N T}{60} \] \[ T = \frac{P \cdot 60}{2\pi N} \] Convert \( P \) to in-lbf/s for easier calculation (1 ft = 12 in): \[ P = 60 \text{ hp} \times 550 \text{ ft-lbf/s} \times 12 = 396,000 \text{ in-lbf/s} \] \[ T = \frac{P}{2\pi N/60} \] \[ T = \frac{60 \times 550}{2\pi \times 1000/60} \] Or, use: \[ T (\text{in-lb}) = \frac{63025 \times P(\text{hp})}{N(\text{rpm})} \] \[ T = \frac{63025 \times 60}{1000} = 3781.5 \text{ in-lbf} \] --- ### Step 2: Maximum Shear Stress For a solid shaft: \[ \tau_{max} = \frac{T r}{J} \] where \( J = \frac{\pi d^4}{32} \) and \( r = \frac{d}{2} \): \[ \tau_{max} = \frac{T \cdot (d/2)}{\frac{\pi}{32} d^4} = \frac{16 T}{\pi d^3} \] --- ### Step 3: Factor of Safety \[ \tau_{allowable} = \frac{S_y}{2 \cdot FS} \] (For shear, using maximum shear stress theory, divide by 2) \[ \tau_{allowable} = \frac{40,000}{2 \times 2} = 10,000 \text{ psi} \] --- ### Step 4: Solve for \( d \) \[ \frac{16 T}{\pi d^3} = \tau_{allowable} \] \[ d^3 = \frac{16 T}{\pi \tau_{allowable}} \] \[ d = \left( \frac{16 \times 3781.5}{\pi \times 10,000} \right)^{1/3} \] Calculate: \[ d^3 = \frac{60,504}{31,415.93} = 1.926 \] \[ d = (1.926)^{1/3} = 1.23 \text{ in} \] --- ### **Final Answer:** \[ \boxed{1.23 \text{ in}} \] Express your answer as **1.23 in** (to three significant figures).