Question 11.35: The worm shaft shown in part a of the figure transmits 1.2 h...

Shafts subjected to thrust can be constrained by bearings, one of which supports the thrust. The shaft floats within the endplay of the second (roller) bearing. Since the thrust force here is larger than any radial load, the bearing absorbing the thrust (bearing A) is heavily loaded compared to bearing B. Bearing B is thus likely to be oversized and may not contribute measurably to the chance of failure. If this is the case, we may be able to obtain the desired combined reliability with bearing A having a reliability near 0.99 and bearing B having a reliability near 1. This would allow for bearing A to have a lower capacity than if it needed to achieve a reliability of  \sqrt{0.99}. To determine if this is the case, we will start with bearing B.

Bearing B (straight roller bearing)

Try a reliability of 1 to see if it is readily obtainable with the available bearings.

Eq. (11-6):      C_{10}=1.2(0.339)\left\{\frac{900}{0.02+4.439[\ln (1 / 1.0)]^{1 / 1.483}}\right\}^{3 / 10}=10.1  kN

The smallest capacity bearing from Table 11-3 has a rated capacity of 16.8 kN. Therefore, we select the 02-25 mm straight cylindrical roller bearing.

Bearing at A (angular-contact ball)
With a reliability of 1 for bearing B, we can achieve the combined reliability goal of 0.99 if bearing A has a reliability of 0.99.

Trial #1:
Tentatively select an 02-85 mm angular-contact with C_{10}=90.4  kN \text { and } C_{0}=63.0  kN.

Table 11-1:    Interpolating, X_{2}=0.56, Y_{2}=1.88

Eq. (11-9):       F_{e}=0.56(0.957)+1.88(2.47)=5.18  kN

Eq. (11-6):

Trial #2:
Tentatively select a 02-90 mm angular-contact ball with C_{10}=106  kN \text { and } C_{0}=73.5  kN.

Table 11-1:     Interpolating, X_{2}=0.56, \quad Y_{2}=1.93

Select an 02-90 mm angular-contact ball bearing

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Eq. (11-6): C_{10}=a_{f} F_{D}\left[\frac{x_{D}}{x_{0}+\left(\theta-x_{0}\right)\left(\ln 1 / R_{D}\right)^{1 / b}}\right]^{1 / a}

Eq. (11-9): F_{e}=X_{i} V F_{r}+Y_{i} F_{a}