Let's carefully analyze the problem and solve it step-by-step. --- # **Given Data:** - Shaft dimensions: \( 3.2\, \text{in} \times 6.4\, \text{in} \) - Material: Steel with \( G = 11,500\, \text{ksi} \) - Allowable shear stress: \( \tau_{allow} = 12\, \text{ksi} \) - Length of the shaft: \( L = 6.\, \text{ft} \) - Power transmitted: \( P \) (unknown, but related to torque and angular velocity) - The shaft transmits power from a tractor to farm implements. --- # **Part (a): Find the largest torque \( T \) that can be transmitted** ### Step 1: Convert all units to consistent units Since shear stress is given in ksi and dimensions are in inches, keep everything in inches: - Length: \( 6.\, \text{ft} = 6. \times 12 = 72\, \text{in} \) ### Step 2: Cross-sectional area and polar moment of inertia The shaft is rectangular with dimensions: - Width \( b = 3.2\, \text{in} \) - Height \( h = 6.4\, \text{in} \) For torsion, the **polar moment of inertia** \( J \) of a rectangular shaft about its centroid is: \[ J = \frac{b h^3}{3} \quad \text{(approximate for rectangular shaft)} \] But more accurately, the polar moment of inertia for a rectangular shaft is: \[ J = \frac{b h^3}{3} \quad \text{(approximate)} \] Alternatively, for torsion, the maximum shear stress at the outer fibers is: \[ \tau_{max} = \frac{T c}{J} \] where: - \( c \) is the distance from the center to the outer fiber (for rectangular cross-section, the maximum shear stress occurs at the furthest point from the neutral axis). In rectangular sections, the maximum shear stress occurs at the neutral axis at the mid-height or width, but for simplicity, the maximum shear stress is: \[ \tau_{max} = \frac{T \cdot c}{J} \] The maximum shear stress occurs at the outer fiber, which is at the maximum distance from the neutral axis: \[ c = \frac{h}{2} = \frac{6.4}{2} = 3.2\, \text{in} \] --- ### Step 3: Calculate \( J \) \[ J = \frac{b h^3}{3} = \frac{3.2 \times (6.4)^3}{3} \] Calculate \( h^3 \): \[ 6.4^3 = 6.4 \times 6.4 \times 6.4 = 262.144 \] Now, \[ J = \frac{3.2 \times 262.144}{3} \approx \frac{839.1168}{3} \approx 279.7056\, \text{in}^4 \] --- ### Step 4: Find maximum torque \( T_{max} \) Using shear stress formula: \[ \tau_{allow} = \frac{T_{max} \times c}{J} \] Rearranged: \[ T_{max} = \frac{\tau_{allow} \times J}{c} \] Plugging in the numbers: \[ T_{max} = \frac{12\, \text{ksi} \times 279.7056\, \text{in}^4}{3.2\, \text{in}} \] Note: \( 1\, \text{ksi} = 100\, \text{psi} \) \[ T_{max} = \frac{12,000\, \text{psi} \times 279.7056}{3.2} \] Calculate numerator: \[ 12,000 \times 279.7056 \approx 3,356,467.2 \] Now divide: \[ T_{max} \approx \frac{3,356,467.2}{3.2} \approx 1,048,271\, \text{lb·in} \] **Final answer for part (a):** \[ \boxed{ T \approx \mathbf{1,048,271\, \text{lb·in}} } \] --- # **Part (b): Find the angle of twist \( \phi \) in radians at the torque found in part (a)** ### Step 1: Recall the formula for angle of twist: \[ \phi = \frac{T L}{J G} \] where: - \( T \) = torque (from part a) - \( L \) = length of the shaft - \( J \) = polar moment of inertia (already calculated) - \( G \) = shear modulus (\( 11,500\, \text{ksi} \)) - \( \phi \) = angle of twist in radians ### Step 2: Convert \( G \) to consistent units: \[ G = 11,500\, \text{ksi} = 11,500,000\, \text{psi} \] ### Step 3: Convert length \( L \): \[ L = 72\, \text{in} \] ### Step 4: Calculate \( \phi \): \[ \phi = \frac{T \times L}{J \times G} \] Plug in numbers: \[ \phi = \frac{1,048,271\, \text{lb·in} \times 72\, \text{in}}{279.7056\, \text{in}^4 \times 11,500,000\, \text{psi}} \] Calculate numerator: \[ 1,048,271 \times 72 \approx 75,429,912 \] Calculate denominator: \[ 279.7056 \times 11,500,000 \approx 3,218,935,400 \] Now, compute \( \phi \): \[ \phi = \frac{75,429,912}{3,218,935,400} \approx .0234\, \text{radians} \] --- # **Final Answers:** | Part | Result | |---------|------------------------------| | (a) Largest torque \( T \) | **1,048,271 lb·in** | | (b) Twist angle \( \phi \) | **.0234 radians** | --- Let me know if you'd like further clarification!