\begin{aligned}&F_{r A}=560 lbf \\&F_{r B}=1095 lbf \\&F_{a e}=200 lbf \\&x_{D}=\frac{L_{D}}{L_{R}}=\frac{40000(400)(60)}{90\left(10^{6}\right)}=10.67 \\&R=\sqrt{0.90}=0.949\end{aligned}
Eq. (11-15): F_{i A}=\frac{0.47 F_{r A}}{K_{A}}=\frac{0.47(560)}{1.5}=175.5 lbf
Eq. (11-15): F_{i B}=\frac{0.47 F_{r B}}{K_{B}}=\frac{0.47(1095)}{1.5}=343.1 lbf
F_{i A} \leq ? \geq\left(F_{i B}+F_{a e}\right)
175.5 lbf \leq(343.1+200)=543.1 lbf \text {, } so Eq. (11-16) applies.
We will size bearing B first since its induced load will affect bearing A, but is not itself affected by the induced load from bearing A [see Eq. (11-16)].
From Eq. (11-16b), F_{e B}=F_{r B}=1095 lbf.
Eq. (11-7): F_{R B}=1.4(1095)\left(\frac{10.67}{4.48(1-0.949)^{1 / 1.5}}\right)^{3 / 10}=3607 lbf
Select cone 32305, cup 32305, with 0.9843 in bore, and rated at 3910 lbf with K = 1.95.
With bearing B selected, we re-evaluate the induced load from bearing B using the actual value for K.
Eq. (11-15): F_{i B}=\frac{0.47 F_{r B}}{K_{B}}=\frac{0.47(1095)}{1.95}=263.9 lbf
Find the equivalent radial load for bearing A from Eq. (11-16), which still applies.
Eq. (11-16a):
\begin{aligned}&F_{e A}=0.4 F_{r A}+K_{A}\left(F_{i B}+F_{a e}\right) \\&F_{e A}=0.4(560)+1.5(263.9+200)=920 lbf\end{aligned}
F_{e A}>F_{r A}
Eq. (11-7): F_{R A}=1.4(920)\left(\frac{10.67}{4.48(1-0.949)^{1 / 1.5}}\right)^{3 / 10}=3030 lbf
Tentatively select cone M86643, cup M86610, with 1 in bore, and rated at 3250 lbf with K = 1.07. Iterating with the new value for K, we get F_{e A}=702 lbf \text { and } F_{r A}=2312 lbf .
By using a bearing with a lower K, the rated load decreased significantly, providing a higher than requested reliability. Further examination with different combinations of bearing choices could yield additional acceptable solutions.
______________________________________________________________________________________________________________________________________
Eq. (11-7): C_{10} \doteq a_{f} F_{D}\left[\frac{x_{D}}{x_{0}+\left(\theta-x_{0}\right)\left(1-R_{D}\right)^{1 / b}}\right]^{1 / a} \quad R \geq 0.90
Eq. (11-15): F_{i}=\frac{0.47 F_{r}}{K}
Eq. (11-16): \text { If } \quad F_{i A} \leq\left(F_{i B}+F_{a e}\right) \quad \begin{cases}F_{e A}=0.4 F_{r A}+K_{A}\left(F_{i B}+F_{a e}\right) & (11-16 a) \\ F_{e B}=F_{r B} & (11-16 b)\end{cases}
