The left bearing is compressed by the axial load, so it is properly designated as bearing A.
\begin{aligned}&F_{r A}=875 lbf \\&F_{r B}=625 lbf \\&F_{a e}=250 lbf\end{aligned}
Assume K = 1.5 for each bearing for the first iteration. Obtain the induced loads.
Eq. (11-15): F_{i A}=\frac{0.47 F_{r A}}{K_{A}}=\frac{0.47(875)}{1.5}=274 lbf
Eq. (11-15): F_{i B}=\frac{0.47 F_{r B}}{K_{B}}=\frac{0.47(625)}{1.5}=196 lbf
Check the condition on whether to apply Eq. (11-16) or Eq. (11-17).
F_{i A} \leq ? \geq F_{i B}+F_{a e}
274 lbf <196+250 lbf , so Eq.(11-16) applies
We will size bearing B first since its induced load will affect bearing A, but it is not affected by the induced load from bearing A [see Eq. (11-16)].
From Eq. (11-16b), F_{e B}=F_{r B}=625 \text { lbf. }
Eq. (11-3):
\begin{aligned}&F_{R B}=a_{f} F_{D}\left[\frac{L_{D}}{L_{R}}\right]^{3 / 10}=1(625)\left[\frac{90000(150)(60)}{90\left(10^{6}\right)}\right]^{3 / 10} \\&F_{R B}=1208 lbf\end{aligned}
Select cone 07100, cup 07196, with 1 in bore, and rated at 1570 lbf with K = 1.45.
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(625)}{1.45}=203 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) \\&=0.4(875)+1.5(203+250)=1030 lbf \\F_{e A} &>F_{r A}\end{aligned}
Eq. (11-3):
\begin{aligned}&F_{R A}=a_{f} F_{D}\left[\frac{L_{D}}{L_{R}}\right]^{3 / 10}=1(1030)\left[\frac{90000(150)(60)}{90\left(10^{6}\right)}\right]^{3 / 10} \\&F_{R A}=1990 lbf\end{aligned}
Any of the bearings with 1-1/8 in bore are more than adequate. Select cone 15590, cup 15520, rated at 2480 lbf with K = 1.69. Iterating with the new value for K, we get F_{e A}=1120 lbf \text { and } F_{r A}=2160 lbf \text {. } The selected bearing is still adequate.
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Eq. (11-3): C_{10}=F_{R}=F_{D}\left(\frac{L_{D}}{L_{R}}\right)^{1 / a}=F_{D}\left(\frac{ L _{D} n_{D} 60}{ L _{R} n_{R} 60}\right)^{1 / a}
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\left\{\begin{array}{l}F_{e A}=0.4 F_{r A}+K_{A}\left(F_{i B}+F_{a e}\right) \\F_{e B}=F_{r B}\end{array}\right.
Eq. (11-17): \text { If } \quad F_{i A}>\left(F_{i B}+F_{a e}\right) \quad\left\{\begin{array}{l}F_{e B}=0.4 F_{r B}+K_{B}\left(F_{i A}-F_{a e}\right) \\F_{e A}=F_{r A}\end{array}\right.
