For the shaft application defined in Prob. 3–76, p. 139, perform a preliminary specification for tapered roller bearings at A and B. A bearing life of 500 million revolutions is desired with a 90 percent combined reliability for the bearing set. Should the bearings be oriented with direct mounting

Question

For the shaft application defined in Prob. 3–76, p. 139, perform a preliminary specification for tapered roller bearings at A and B. A bearing life of 500 million revolutions is desired with a 90 percent combined reliability for the bearing set. Should the bearings be oriented with direct mounting or indirect mounting for the axial thrust to be carried by the bearing at A? Assuming bearings are available with K = 1.5, find the required radial rating for each bearing. For this preliminary design, assume an application factor of one.

Accepted Answer

The thrust load on shaft AB is from the axial component of the force transmitted through the bevel gear, and is directed to the right. By observation of Fig. 11-14, indirect mounted bearings would allow bearing A to carry the thrust load.

From the solution to Prob. 3-76, the axial thrust load is [latex]F_{a e}=92.8 lbf[/latex] , and the bearing radial forces are [latex]F_{A y}=639.4 lbf , F_{A z}=1513.7 lbf , F_{B y}=276.6 lbf ext {, and } F_{B z}=705.7 lbf ext {. }[/latex]

Thus, the radial forces are

[latex]\begin{aligned}&F_{r A}=\sqrt{639.4^{2}+1513.7^{2}}=1643 lbf \&F_{r B}=\sqrt{276.6^{2}+705.7^{2}}=758 lbf\end{aligned}[/latex]

The induced loads are

Eq. (11-15): [latex]F_{i A}=\frac{0.47 F_{r A}}{K_{A}}=\frac{0.47(1643)}{1.5}=515 lbf[/latex]

Eq. (11-15): [latex]F_{i B}=\frac{0.47 F_{r B}}{K_{B}}=\frac{0.47(758)}{1.5}=238 lbf[/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]515 lbf >238+92.8=330.8 lbf[/latex] , so Eq.(11-17) applies

Notice that the induced load from bearing A is sufficiently large to cause a net axial force to the left, which must be supported by bearing B.

Eq. (11-17a):

[latex]\begin{aligned}F_{e B} &=0.4 F_{r B}+K_{B}\left(F_{i A}-F_{a e} ight) \&=0.4(758)+1.5(515-92.8)=937 lbf >F_{r B}, ext { so use } F_{e B}\end{aligned}[/latex]

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

[latex]\begin{aligned}&x_{D}=\frac{L_{D}}{L_{R}}=\frac{500\left(10^{6} ight)}{90\left(10^{6} ight)}=5.56 \&R=\sqrt{0.90}=0.949\end{aligned}[/latex]

Eq. (11-7): [latex]F_{R B}=1(937)\left(\frac{5.56}{4.48(1-0.949)^{1 / 1.5}} ight)^{3 / 10}=1810 lbf[/latex]

Eq. (11-16b): [latex]F_{e A}=F_{r A}=1643 lbf[/latex]

Eq. (11-7): [latex]F_{R A}=1(1643)\left(\frac{5.56}{4.48(1-0.949)^{1 / 1.5}} ight)^{3 / 10}=3180 lbf[/latex]

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Eq. (11-7): [latex]C_{10} \doteq a_{f} F_{D}\left[\frac{x_{D}}{x_{0}+\left( heta-x_{0} ight)\left(1-R_{D} ight)^{1 / b}} ight]^{1 / a} \quad R \geq 0.90[/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 11.42: For the shaft application defined in Prob. 3–76, p. 139, per...

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