314 Maximum Shear and Relative Gear Rotation of a Compound Steel Shaft

$T = \dfrac{P}{2\pi f}$

$T_A = \dfrac{-35(1000)}{2\pi (4)} = -1392.6 \, \text{N}\cdot\text{m}$

$T_B = \dfrac{-20(1000)}{2\pi (4)} = -795.8 \, \text{N}\cdot\text{m}$

$T_C = \dfrac{55(1000)}{2\pi (4)} = 2188.4 \, \text{N}\cdot\text{m}$
 

Relative to C:
 

 

$\tau_{max} = \dfrac{16T}{\pi d^3}$

$\tau_{AB} = \dfrac{16(1392.6)(1000)}{\pi (55^3)} = 42.63 \, \text{MPa}$

$\tau_{BC} = \dfrac{16(2188.4)(1000)}{\pi (65^3)} = 40.58 \, \text{MPa}$

∴ $\tau_{max} = \tau_{AB}= 42.63 \, \text{ MPa}$             answer
 

$\theta = \dfrac{TL}{JG}$

$\theta_{A/C} = \dfrac{1}{G}{\Large \Sigma} \dfrac{TL}{J}$

$\theta_{A/C} = \dfrac{1}{83\,000} \left[ \dfrac{1392.6(4)}{\frac{1}{32}\pi (55^4)} + \dfrac{2188.4(2)}{\frac{1}{32}\pi (65^4)} \right]\,(1000^2) $

$\theta_{A/C} = 0.104\,796\,585 \, \text{rad}$

$\theta_{A/C} = 6.004^\circ$           answer