Problem 8.29

Since we are told that 20 kW of power are transmitted to shaft AB at D, the torque at D can be computed:

Therefore, the force exerted tangentially on gear D is 424.4/0.1 m = 4.24 x 10³ N. The torques at gears F and G are proportional, since 8 kW and 12 kW are taken off at each:

Therefore, the forces are:

at gear F ==>

at gear G ==>

These forces have been moved to the centerline of the shaft, with T, M_x, and M_y representing the torques that show up as a result of that translation. Assuming that the motor M rotates the driving gear clockwise,

Only the reactions at the left end are necessary. Taking moments about an axis coincident with B_y, and using the forces found above,

Taking moments about an axis coincident with B_x,

The bending moments at the ends are zero. The bending moment M_y at F is 150A_x = 658 N-m, whereas M_x = 150A_y, or 127.3 N-m. The plots are below:

The slope of the M_x vs x line does not change until point G is reached (since the only force that can change it is at G). Anticipating the need to have this moment at D as well, I compute both values now:

Since the moment goes to zero at end B, I now have the data to plot M_x completely. M_y is another matter, since there are two forces, F_F and F_D, which cause the slope to change. Near D, the moment is

From this point on, the moment M_y falls linearly to 403.2 at G, and to zero at B.

In this case, the equation for the shearing stress is

Thus, I need to determine the first two terms of the radicand at C and at D to determine which is larger:

But T changes with position also. These calculations, sufficient for problem 25, are insufficient here: T_AC = 0; T_CD = - 169.77; T_DG = 424.4 - 169.77 = 254.7; T_GB = 0. I need to compute

Therefore, the calculation of the radius must be done just to the left of D:

The diameter of the shaft should be 45.2 mm.