Calculate the value of G for brass, if the angle of twist over a length of 1 meter is $7.2^\circ$, when the compound bar is subjected to a twisting couple 520 N-m. Also calculate the maximum shear stress in both the material if $G_s= 80 Mpa$.

For composite bar

$\theta = \theta_B = \theta_S\\ T = T_B + T_S$

$\theta = 7.2^\circ = 0.1256\hspace{0.05cm}rad\\ \theta_B = \theta_S = 0.1256\\ T = 520\hspace{0.05cm}Nm = 520\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}Nmm\\ J = \frac{\pi}{32}[D^4 - d^4]\\ \hspace{0.25cm} = \frac{\pi}{32}[32^4 - 19^4]\\ J = 90.14\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}mm^4\\ R = \frac{D}{2} = 16\hspace{0.05cm}mm$

$\frac{T}{J} = \frac{G\theta}{L}\\ T = \frac{GJ\theta}{L}\\ \textit{But}\hspace{0.5cm}T = T_B + T_S\\ 520\hspace{0.05cm}\times\hspace{0.05cm}10^3 = (\frac{GJ\theta}{L})_B + (\frac{GJ\theta}{L})_S\\ G_B = 45.849 \hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}N/mm^2$

For Brass, $\frac{\tau_B}{R_B} = \frac{G\theta_B}{L_B}\\ \tau_B = 92.13\hspace{0.05cm}N/mm^2$

For Steel, $\frac{\tau_S}{R_S} = \frac{G\theta_S}{L_S}\\ \tau_S = 0.09\hspace{0.05cm}N/mm^2$