Question 3.11.8: The shaft depicted in Fig. 11–18a carries a helical gear wit...

The reactions in the xz plane from Fig. 11–18b are

R_{zA} = \frac{3980\left(50\right)}{150}= 1327 N

R_{zB} = \frac{3980\left(100\right)}{150}= 2653 N

The reactions in the xyplane from Fig. 11–18c are

R_{yA} = \frac{1770\left(50\right)}{150}+\frac{169 000)}{150}= 1716.7 = 1717 N

R_{yB} = \frac{1770\left(100\right)}{150}-\frac{169 000)}{150}= 53.3 N

The radial loads F_{rA} \ and \ F_{rB} are the vector additions of R_{yA} \ and \ R_{zA} \ , \ and \ R_{yB} \ and \ R_{zB} , respectively:

F_{rA} = \left(R^2_{zA}+ R^2_{yA} \right)^{{1}/{2}} = \left(1327^2 + 1717^2 \right)^{{1}/{2}} = 2170 N

F_{rB} = \left(R^2_{zB}+ R^2_{yB} \right)^{{1}/{2}} = \left(2653^2 + 53.3^2 \right)^{{1}/{2}} = 2654 N

Trial \ 1 :With direct mounting of the bearings and application of the external thrust to the shaft, the squeezed bearing is bearing  A as labeled in Fig. 11–18a. Using K of 1.5 as the initial guess for each bearing, the induced loads from the bearings are

F_{iA} = \frac{0.47 F_{rA}}{K_A}= \frac{0.47 \left(2170\right)}{1.5}= 680 N

F_{iB} = \frac{0.47 F_{rB}}{K_B}= \frac{0.47 \left(2654\right)}{1.5}= 832 N

Since F_{iA} is clearly less than F_{iB} + F_{ae} , bearing A carries the net thrust load, and Eq. (11–16)

is applicable. Therefore, the dynamic equivalent loads are

F_{eA} = 0.4 F_{rA} + K_A \left(F_{iB} + F_{ae}\right) = 0.4\left(2170\right) + 1.5\left(832 + 1690\right) = 4651 N

F_{eB} = F_{rB} = 2654 N

The multiple of rating life is

Estimate R_D as \sqrt{0.99} = 0.995 for each bearing. For A, from Eq. (11–7) the catalog entry

C_{10} should equal or exceed

C_{10} = \left(1\right)\left(4651\right)\left[ \frac{2.67}{\left(4.48\right)\left(1-0.995\right)^{{2}/{3}} } \right]^{{3}/{10}}= 11 486 N

From Fig. 11–15, tentatively select type TS 15100 cone and 15245 cup, which will work: K_A = 1.67 \ , \ C_{10} = 12 100 N.

For bearing B , from Eq. (11–7),

the catalog entry C_{10} should equal or exceed

C_{10} = \left(1\right)2654 \left[\frac{2.67}{\left(4.48\right)\left(1 − 0.995\right)^{{2}/{3}}}\right]^{{3}/{10}}= 6554 N

Tentatively select the bearing identical to bearing A, which will work:

K_B = 1.67 \ , \ C_{10} = 12 100 N.

Trial \ 2: Repeat the process with K_A = K_B = 1.67 from tentative bearing selection.

F_{iA}= \frac{0.47 F_{rA}}{K_A}=\frac{0.47 \left(2170\right)}{1.67}= 611 N

F_{iB}= \frac{0.47 F_{rB}}{K_B}=\frac{0.47 \left(2654\right)}{1.67}= 747 N

Since F_{iA} is still less than F_{iB} + F_{ae} , Eq. (11–16)

is still applicable.

F_{eA} = 0.4F_{rA} + K_A \left(F{iB} + F_{ae}\right) = 0.4 \left(2170\right) + 1.67\left(747 + 1690\right) = 4938 N

F_{eB} =F_{rB} = 2654 N

For A, from Eq. (11–7) the corrected catalog entry C_{10} should equal or exceed

C_{10}= \left(1\right) \left(4938\right)\left[\frac{2.67}{\left(4.48\right)\left(1-0.995\right)^{{2}/{3}} } \right]^{{3}/{10}} =12195 N

Although this catalog entry exceeds slightly the tentative selection for A, we will keep it since the reliability of bearing B exceeds 0.995. In the next section we will quantitatively show that the combined reliability of A and B will exceed the reliability goal of 0.99.  For bearing B \ , \ F_{eB} = F_{rB} = 2654 N. From Eq. (11–7),

C_{10}= \left(1\right) \left(2654\right)\left[\frac{2.67}{\left(4.48\right)\left(1-0.995\right)^{{2}/{3}} } \right]^{{3}/{10}} = 6554 N

Select cone and cup 15100 and 15245, respectively, for both bearing

A and B. Note from Fig. 11–14 the effective load center is located at a = −5.8 mm, that is, 5.8 mm into the cup from the back. Thus the shoulder-to-shoulder dimension should be 150 − 2\left(5.8\right) = 138.4 mm. Note that in each iteration of Eq. (11–7) to find the catalog load rating, the bracketed portion of the equation is identical and need not be re-entered on a calculator each time.