Solution to Problem 322 Torsion

Based on maximum allowable shear:
$\tau_{max} = \dfrac{16 T}{\pi D^3}$
 

 

For the 1st segment:
$60 = \dfrac{16(450)(1000)}{\pi D^3}$

$D = 33.68 \, \text{ mm}$
 

For the 2nd segment:
$60 = \dfrac{16(1200)(1000)}{\pi D^3}$

$D = 46.70 \, \text{ mm}$
 

Based on maximum angle of twist:
$\theta = \dfrac{TL}{JG}$

$\theta = \dfrac{1}{JG} \Sigma TL$

$4^\circ \left( \dfrac{\pi}{180^\circ} \right) = \dfrac{1}{\frac{1}{32}\pi D^4 (83\,000)} \, [ \, 450(2.5) + 1200(2.5) \, ] \, (1000^2)$

$D = 51.89 \, \text{ mm }$
 

Use   $D = 51.89 \, \text{ mm}$           answer