An outer hub rotates around a stationary shaft, supported by two tapered roller bearings as shown in Fig. 11–23. The device is to operate at 250 rev/min, 8 hours per day, 5 days per week, for 5 years, before bearing replacement is necessary. A reliability of 90 percent on each bearing is acceptable

Question

An outer hub rotates around a stationary shaft, supported by two tapered roller bearings as shown in Fig. 11–23. The device is to operate at 250 rev/min, 8 hours per day, 5 days per week, for 5 years, before bearing replacement is necessary. A reliability of 90 percent on each bearing is acceptable. A free body analysis determines the radial force carried by the upper bearing to be 12 kN and the radial force at the lower bearing to be 25 kN. In addition, the outer hub applies a downward force of 5 kN. Assuming bearings are available with K = 1.5, find the required radial rating for each bearing. Assume an application factor of 1.2.

Accepted Answer

The lower bearing is compressed by the axial load, so it is designated as bearing A.

[latex]\begin{aligned}&F_{r A}=25 kN \&F_{r B}=12 kN \&F_{a e}=5 kN\end{aligned}[/latex]

Eq. (11-15): [latex]F_{i A}=\frac{0.47 F_{r A}}{K_{A}}=\frac{0.47(25)}{1.5}=7.83 kN[/latex]

Eq. (11-15): [latex]F_{i B}=\frac{0.47 F_{r B}}{K_{B}}=\frac{0.47(12)}{1.5}=3.76 kN[/latex]

Check the condition on whether to apply Eq. (11-16) or Eq. (11-17)

[latex]F_{i A} \leq ? \geq F_{i B}+F_{a e}[/latex]

[latex]7.83 kN <3.76+5=8.76 kN ext {, }[/latex] so Eq.(11-16) applies

Eq. (11-16a):

[latex]\begin{aligned}F_{e A} &=0.4 F_{r A}+K_{A}\left(F_{i B}+F_{a e} ight) \&=0.4(25)+1.5(3.76+5)=23.1 kN

[latex]\begin{aligned}L_{D} &=(250 rev / min )\left(\frac{60 min }{ hr } ight)\left(\frac{8 hr }{ ext { day }} ight)\left(\frac{5 ext { day }}{ ext { week }} ight)\left(\frac{52 ext { weeks }}{ yr } ight)(5 yrs ) \&=156\left(10^{6} ight) rev\end{aligned}[/latex]

Assume for tapered roller bearings that the specifications for Manufacturer 1 on p. 608 are applicable.

Eq. (11-3): [latex]F_{R A}=a_{f} F_{D}\left[\frac{L_{D}}{L_{R}} ight]^{3 / 10}=1.2(25)\left[\frac{156\left(10^{6} ight)}{90\left(10^{6} ight)} ight]^{3 / 10}=35.4 kN[/latex]

Eq. (11-16b): [latex]F_{e B}=F_{r B}=12 kN[/latex]

Eq. (11-3): [latex]F_{R B}=1.2(12)\left[\frac{156}{90} ight]^{3 / 10}=17.0 kN[/latex]

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Eq. (11-3): [latex]C_{10}=F_{R}=F_{D}\left(\frac{L_{D}}{L_{R}} ight)^{1 / a}=F_{D}\left(\frac{ L _{D} n_{D} 60}{ L _{R} n_{R} 60} ight)^{1 / a}[/latex]

Eq. (11-15): [latex]F_{i}=\frac{0.47 F_{r}}{K}[/latex]

Eq. (11-16): [latex] ext { If } \quad F_{i A} \leq\left(F_{i B}+F_{a e} ight) \quad\left\{\begin{array}{l}F_{e A}=0.4 F_{r A}+K_{A}\left(F_{i B}+F_{a e} ight) \F_{e B}=F_{r B}\end{array} ight.[/latex]

Eq. (11-17): [latex] ext { If } \quad F_{i A}>\left(F_{i B}+F_{a e} ight) \quad\left\{\begin{array}{l}F_{e B}=0.4 F_{r B}+K_{B}\left(F_{i A}-F_{a e} ight) \F_{e A}=F_{r A}\end{array} ight.[/latex]

Question 11.43: An outer hub rotates around a stationary shaft, supported by...

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