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

The pitch angles are

γ = tan^{−1} \left( \frac {3}{9}\right) = 18.4°    Γ = tan^{−1} \left( \frac{9}{3}\right)= 71.6°

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

V =\frac{2πr_{P}n}{12} =\frac{2π(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

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

W_{r} = W_{t}  tan  \phi   cos  Γ = 406  tan  20° cos  71.6° = 46.6  lbf

W_{a} = W_{t}  tan  \phi  sin  Γ = 406  tan  20°  sin  71.6° = 140  lbf

where W_{r} is in the −x direction and W_{a} is 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} ×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)+ T j = 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)+ T j = 0

from which

T = 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 + F^{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.