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

The pitch angles are

\gamma =\tan ^{-1}\left(\frac{3}{9} \right)=18.4°           \Gamma =\tan^{-1} \left(\frac{9}{3} \right)=71.6°

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

V=\frac{2\pi r_pn}{12}=\frac{2\pi (1.293)(600)}{12}=406  ft/min

Therefore the transmitted load is

W_t=\frac{33  000H}{V}=\frac{(33  000)(5)}{406}=406  lbf

and from Eq. (13–38)

W_r = W_t tan Φ cos γ

W_a = W_t tan Φ sin γ                                  (13.38)

, with Γ replacing γ, we have

W_r = W_t tan Φ cos Γ = 406 tan 20° cos 71.6° = 46.6 lbf

W_a = W_t tan Φ sin Γ = 406 tan 20° sin 71.6° = 140 lbf

where [laetx]W_t[/latex] acts in the positive z direction, W_r in the -x direction, and W_a in the -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

R_G = 3.88i – (2.5 + 1.293) j = 3.88i – 3.793j

We shall also require a vector from D to C:

R_C = -(2.5 + 3.625)j = -6.125j

Then, summing moments about D gives

R_G \times W + R_C × F_C + T = 0                              (1)

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

(3.88i – 3.793j) × (-46.6i – 140j + 406k) = (-6.125j) × (F^x_{C}i + F^y_{c}j+F^z_{c}k)+Tj=0                      (2)

After the two cross products are taken, the equation becomes

(-1540i-1575j-720k)+(-6.125F^z_{c}i+6.125F^x_{c}k)+Tj=0

from which

T 5 1575j lbf . in            F^x_{c} = 118 lbf            F^z_{c} = -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

(F^x_{D}i+D^z_{D}k)+(118i+F^y_{C}j-251k) + (-46.6i – 140j + 406k) = 0           (5)

First we see that F^y_{C} = 140 lbf, and so

F_{C} = 118i + 140j – 251k lbf

Then, from Eq. (5),

F_{D} = -71.4i – 155k lbf

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