Delivered power (to the machine tool at
F) is computed by multiplying the torque in shaft EF times the angular
speed: 
.
(Note that the torque in EF is the same as the torque on gear E, so I am using E as the subscript rather than EF.)
Where two gears mesh, the forces are equal but opposite; the tangential speeds are the same:
Thus, 
Of course, this would have to be true
if there are no frictional losses. The power transmitted is 
,
where 
.
To determine the torque at B, I have to consider the allowable shearing stress:
But these torques are related through the equations developed above. For TB = 60.3, TE would have to be 377 N-m - too large! So far, then, we would have to choose a TE of 323.3 N-m.
However, checking the relation between TC and TE, if TE is 323.3 N-m, TC = 150.8 N-m. This is larger than the 117.8 that is allowed in shaft CD, so TC = 117.8 N-m is the limiting torque.
Using the relation between TB and TC, for TC = 117.8 N-m, TB = 47.1 N-m.
Now I can find the power:
