Let's solve the problem step by step: --- ## **Given Data** - Shear Modulus, \( G = 26 \) GPa \( = 26 \times 10^9 \) Pa - Shaft diameter, \( d = 60 \) mm \( = .06 \) m - Allowable shear stress, \( \tau_{allow} = 80 \) MPa \( = 80 \times 10^6 \) Pa - Lengths: \( AB = BC = 1.20 \) m --- ## **Step 1: Maximum Allowable Torque \( T \)** The maximum shear stress in a solid circular shaft is: \[ \tau_{max} = \frac{Tc}{J} \] Where: - \( T \) = applied torque - \( c \) = outer radius (\( = d/2 \)) - \( J \) = polar moment of inertia for a solid shaft (\( = \frac{\pi}{32} d^4 \)) Set \( \tau_{max} = \tau_{allow} \): \[ \tau_{allow} = \frac{T_{max} \cdot c}{J} \] \[ T_{max} = \frac{\tau_{allow} \cdot J}{c} \] Calculate \( J \) and \( c \): \[ c = \frac{d}{2} = \frac{.06}{2} = .03 \text{ m} \] \[ J = \frac{\pi}{32} d^4 = \frac{\pi}{32} (.06)^4 = \frac{\pi}{32} (.00001296) \approx 1.274 \times 10^{-6} \text{ m}^4 \] Plug in the values: \[ T_{max} = \frac{80 \times 10^6 \times 1.274 \times 10^{-6}}{.03} \] \[ T_{max} = \frac{101.92}{.03} \times 10^ \] \[ T_{max} = 3397 \text{ Nm} \] **Maximum allowable torque:** \[ \boxed{T_{max} = 3.40 \text{ kN}\cdot\text{m}} \] --- ## **Step 2: Angle of Twist from Disk A to Disk C (\( \theta_{AC} \))** The angle of twist in a shaft is: \[ \theta = \frac{T L}{J G} \] For this system, sum the twists in each segment: ### **Torque in Each Segment** - Segment AB: torque = \( \frac{2}{3}T \) - Segment BC: torque = \( \frac{1}{3}T \) ### **Total Angle of Twist** \[ \theta_{AC} = \theta_{AB} + \theta_{BC} \] \[ \theta_{AC} = \frac{\left(\frac{2}{3}T\right) L_{AB}}{J G} + \frac{\left(\frac{1}{3}T\right) L_{BC}}{J G} \] \[ = \frac{T}{J G} \left( \frac{2}{3}L_{AB} + \frac{1}{3}L_{BC} \right) \] Plug in \( L_{AB} = L_{BC} = 1.2 \) m: \[ \theta_{AC} = \frac{T}{J G} \left( \frac{2}{3} \times 1.2 + \frac{1}{3} \times 1.2 \right) \] \[ = \frac{T}{J G} \left( .8 + .4 \right) \] \[ = \frac{T}{J G} (1.2) \] Plug in \( T = 3397 \) Nm, \( J = 1.274 \times 10^{-6} \) m\(^4\), \( G = 26 \times 10^9 \) Pa: \[ \theta_{AC} = \frac{3397 \times 1.2}{1.274 \times 10^{-6} \times 26 \times 10^9} \] Calculate denominator: \[ 1.274 \times 10^{-6} \times 26 \times 10^9 = 1.274 \times 26 \times 10^{3} = 33.124 \times 10^3 = 33,124 \] \[ \theta_{AC} = \frac{3397 \times 1.2}{33,124} = \frac{4076.4}{33,124} \approx .123 \text{ radians} \] **Convert to degrees:** \[ .123 \times \frac{180}{\pi} \approx 7.05^\circ \] --- ## **Final Answers** ### (a) **Maximum allowable torque:** \[ \boxed{T_{max} = 3.40 \ \text{kN}\cdot\text{m}} \] ### (b) **Corresponding angle of twist (A relative to C):** \[ \boxed{\theta_{AC} = .123 \ \text{radians} \approx 7.05^\circ} \]