Find the relative torque setup at the ends A and B of the shaft. If the diameter of the shaft is 40 mm. Find the maximum shear stress analysis in the shaft and angle of twist at the point where the torque is applied. Assume shear modulus for the shaft as $0.84 x 10^5$ Mpa.

$\theta = \theta_{AC} = \theta_{CB}\\ T = T_{AC} + T_{CB}\\ J = \frac{\pi}{32} D^4 = \frac{\pi}{32}\hspace{0.05cm}\times\hspace{0.05cm}40^4 = 251.32\hspace{0.05cm}\times\hspace{0.05cm}10^3$

$\frac{T}{J} = \frac{G\theta}{L}\\ \theta = \frac{TL}{JG}\\ \textit{Torque} (T) = T_{AC} + T_{CB}\\ 1000 = T_{AC} + T_{CB}....(1)\\ \textit{Twist} \theta_{AC} = \theta_{CB}\\ [\frac{TL}{GJ}]_{AC} = [\frac{TL}{GJ}]_{CB}\\ T_{AC} = \frac{5}{3}T_{CB}\\ T_{AC} = 1.67 T_{CB}$

From equation (1), $1000 = T_{AC} + T_{BC}\\ 1000 = 1.67 T_{CB} + T_{CB}\\ T_{CB} = 374.5\hspace{0.05cm}Nm\\ T_{AC} = 625.47\hspace{0.05cm}Nm$

CASE II

$\theta_c = \frac{T_{AC} L_{AC}}{GJ}\\ \hspace{0.25cm} = \frac{625.46\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}\times\hspace{0.05cm}3000}{0.84\hspace{0.05cm}\times\hspace{0.05cm}10^5\hspace{0.05cm}\times\hspace{0.05cm}251.32\hspace{0.05cm}\times\hspace{0.05cm}10^3}\\ \hspace{0.25cm} = 8.88\hspace{0.05cm}\times\hspace{0.05cm}10^{-2}\hspace{0.05cm}rad = 5.08^\circ$

$\frac{T_{AC}}{J} = \frac{\tau}{R}\\ \frac{625.46\hspace{0.05cm}\times\hspace{0.05cm}10^3}{251.32\hspace{0.05cm}\times\hspace{0.05cm}10^3} = \frac{\tau}{20}\\ \tau = 49.77\hspace{0.05cm}N/mm^2$

$\frac{T_{CB}}{J} = \frac{\tau}{R}\\ \frac{374.5\hspace{0.05cm}\times\hspace{0.05cm}10^3}{251.32\hspace{0.05cm}\times\hspace{0.05cm}10^3} = \frac{\tau}{20}\\ \tau = 29.80\hspace{0.05cm}N/mm^2$

$\textit{Maximum shear stress} \tau_{max} = 49.77\hspace{0.05cm}N/mm^2$