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

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Diagram shows a shaft A B C containing two gears at B and C which is connected to a motor A where the torques of 20 Newton meter and T subscript c is applied on B and C and the distance A B is 1400 millimeter and B C is 1000 millimeter. The x axis is taken along the axis of the gears and the y axis is taken along the motor support at A and perpendicular to the x axis.

(T_{C} - 20 N - m) \frac{L_{1}}{J_{1} G_{1}} + \frac{T_{C} L_{2}}{J_{2} G_{2}} = 0.01745 rad

T_{C} [\frac{L_{1}}{J_{1} G_{1}} + \frac{L_{2}}{J_{2} G_{2}}] = 0.01745 rad + (20 N - m) \frac{L_{1}}{J_{1} G_{1}}

T_{C} = \frac{0.01745 rad + (20 N - m) \frac{L_{1}}{J_{1} G_{1}}}{\frac{L_{1}}{J_{1} G_{1}} + \frac{L_{2}}{J_{2} G_{2}}}