Question 13.8: The bevel pinion in Fig. 13–36a rotates at 600 rev/min in th...

The pitch angles are

The pitch-line velocity corresponding to the average pitch radius is

Therefore the transmitted load is

which acts in the positive z direction, as shown in Fig. 13–36b. We next have

where W_{r} is in the −x direction and W_{a} −y direction, as illustrated in the isometric sketch of Fig. 13–36b.

In preparing to take a sum of the moments about bearing D, define the position vector from D to G as

We shall also require a vector from D to C:

Then, summing moments about D gives

R _{G} \times W + R _{C} \times F _{C}+ T = 0 (1)

When we place the details in Eq. (1), we get

\begin{array}{l}(3.88 i -3.793 j ) \times(-46.6 i -140 j +406 k ) \\+(-6.125 j ) \times\left(F_{C}^{x} i +F_{C}^{y} j +F_{C}^{z} k \right)+T j = 0\end{array} (2)

After the two cross products are taken, the equation becomes

from which

T =1575 j lbf \cdot \text { in } \quad F_{C}^{x}=118 lbf \quad F_{C}^{z}=-251 lbf (3)

Now sum the forces to zero. Thus

F _{D}+ F _{C}+ W = 0 (4)

When the details are inserted, Eq. (4) becomes

\left(F_{D}^{x} i +F_{D}^{z} k \right)+\left(118 i +F_{C}^{y} j -251 k \right)+(-46.6 i -140 j +406 k )= 0 (5)

First we see that F_{C}^{y}=140 lbf, and so

Then, from Eq. (5),

These are all shown in Fig. 13–36b in the proper directions. The analysis for the pinion shaft is quite similar.