Problem: The solid shaft AB rotates at 450 rpm and transmits 25 hp from the motor M to machine tools connected to gears F and G.  The allowable shearing stress is 8 ksi.  Ten hp is taken off at gear F and 15 hp is taken off at gear C.  Determine the smallest permissible diameter of shaft AB.

Solution: Apparently there are no losses to friction, since we have 25 hp of input power and the same (10 + 15 hp) as output. Converting the frequency of 450 rpm to rps, 450/60 = 7.5 rev/s. The applied torque at gear D, then, is

(I will not always show all of the digits but, to avoid round off errors, they have been carried through the calculations to the end.)

The force on gear D is found by:

At C, where the power takeoff is 10 hp, I compute the torque as before:

Computing the force on either of the gears at C,

Repeating this process at gear E,

This ends the preliminary calculations, which were necessary to obtain a free body diagram of shaft AB. (I have assumed clockwise rotation of AB.) Now I move all forces so that they act on shaft AB, including the necessary torques resulting from that translation:

Summing the forces and the moments, I get:

Now I have sufficient information to draw the shear and bending moment (and torque) diagrams:

The shaft must be designed for the largest value of

Considering the above diagrams, and looking at values...

just to the right of D

Just to the left of E:

For an allowable shearing stress of 8 ksi, using equation 8.7 on page 501,

¹ This is problem 7.161 from Beer & Johnston, 2nd Edition, Mechanics of Materials, Mc-Graw-Hill.