MecMovies 4.0 : M6.8: Maximum Torque Based on Twist Angle

From equilibrium, the internal torques $T_{1}$ and $T_{2}$ were expressed in terms of the unknown external torque $T_{C}$ as:

$T_{1} = T_{C} - 20 N - m T_{2} = T_{C}$

Substitute these expressions into the equation for the rotation angle at $C$ :

$ϕ_{C} = (T_{C} - 20 N - m) \frac{L_{1}}{J_{1} G_{1}} + \frac{T_{C} L_{2}}{J_{2} G_{2}}$

$= 1^{\circ} = 0.01745 rad$

The rotation angle at $C$ is specified to be $1^{\circ}$ ...

... and the rotation angle must be converted to radians for this calculation.