Question 14.5: The shaft shown in Figure 14–15 carries a transverse load of...

The radial loads on the bearings are
F_{rA} = (6800 lb)(4 in/10 in) = 2720 lb
F_{rB} = (6800 lb)(6 in/10 in) = 4080 lb
T_A = 2500 lb

To use Equation (14–7), we must assume values of Y_A and Y_B. Let’s use Y_A = Y_B = 1.75. Then

P_{A}=0.4 F_{r A}+0.5 \frac{Y_{A}}{Y_{B}} F_{r B}+Y_{A} T_{A}               (14-7)

P_{A}=0.4 (2720 lb)+0.5 \frac{1.75}{1.75} (4080 lb)+1.75(2500 lb) = 7503 lb

P_B = F_{rB} = 4080 lb

Using Table 14–4 as a guide, let’s select 4000 h as a design life. Then the number of revolutions would be

L_d = (4000 h)(350 rpm)(60 min/h) = 8.4 × 10^7 rev

The required basic dynamic load rating can now be calculated from Equation (14–3) (C=P_d(L_d/10^6)^{1/k}), using k = 3.33. Note that the exponent 1/k is 1/3.33, resulting in a value of 0.30.

C_A = (7503 lb)(8.4 × 10^7/10^6)^{0.30} = 28400 lb

Similarly,
C_B = (4080 lb)(8.4 × 10^7/10^6)^{0.30} = 15 400 lb

From Table 14–7, we can choose the following bearings.
Bearing A
d = 2.5000 in           D = 5.0000 in
C = 29 300 lb            Y_A = 1.65

Bearing B
d = 1.7500 in            D = 4.0000 in
C = 21 400 lb             Y_B = 1.50
We can now recompute the equivalent loads:

P_{A}=0.40(2720 lb) + 0.5 \frac{1.65}{1.50}(4080 lb) + 1.65(2500 lb) = 7457 lb

P_B = F_{rB} = 4080 lb

From these, the new values of C_A = 28 200 lb and C_B = 15 400 lb are still satisfactory for the selected bearings.