Free-Body diagram of the shaft A B C which cuts through A B shows the torques 600 Newton meter, 200 Newton meter acting on B and C, and a positive internal torque T subscript 1 acts on A B towards A. The length of A B is 100 millimeters, B C is 300 millimeters. The internal torque at A B is 400 Newton meter. A positive internal torque T subscript 2 acts on B C towards B. The internal torque at B C is negative 200 Newton meter. Tau at A B is 130.4 mega Pascal and Tau at B C is negative 65.2 mega Pascal. The rotation angles is 0.0158 radians at B, minus 0.0079 radians at C.

To find the rotation angle at B, add the twist angle in shaft segment (1) to the rotation angle at A:

ϕB=ϕA+ϕ1

   =0+0.0158 rad   =0.0158 rad

To find the rotation angle at C, add the twist angle in shaft segment (2) to the rotation angle at B:

ϕC=ϕB+ϕ2

   =0.0158+(-0.0237)   =-0.0079 rad