For the shaft application defined in Prob. 3–74, p. 138, perform a preliminary specification for tapered roller bearings at C and D. A bearing life of 10^8 revolutions is desired with a 90 percent combined reliability for the bearing set. Should the bearings be oriented with direct mounting or

Question

For the shaft application defined in Prob. 3–74, p. 138, perform a preliminary specification for tapered roller bearings at C and D. A bearing life of [latex]10^{8}[/latex] 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 C? 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 CD is from the axial component of the force transmitted through the bevel gear, and is directed toward bearing C. By observation of Fig. 11-14, direct mounted bearings would allow bearing C to carry the thrust load.

From the solution to Prob. 3-74, the axial thrust load is [latex]F_{a e}=362.8 lbf[/latex], and the bearing radial forces are [latex]F_{C x}=287.2 lbf , F_{C z}=500.9 lbf , F_{D x}=194.4 lbf ext {, and } F_{D z}=307.1 lbf ext {. }[/latex]

Thus, the radial forces are

[latex]\begin{aligned}&F_{r C}=\sqrt{287.2^{2}+500.9^{2}}=577 lbf \&F_{r D}=\sqrt{194.4^{2}+307.1^{2}}=363 lbf\end{aligned}[/latex]

The induced loads are

Eq. (11-15): [latex]F_{i C}=\frac{0.47 F_{r C}}{K_{C}}=\frac{0.47(577)}{1.5}=181 lbf[/latex]

Eq. (11-15): [latex]F_{i D}=\frac{0.47 F_{r D}}{K_{D}}=\frac{0.47(363)}{1.5}=114 lbf[/latex]

Check the condition on whether to apply Eq. (11-16) or Eq. (11-17), where bearings C and D are substituted, respectively, for labels A and B in the equations.

[latex]F_{i C} \leq ? \geq F_{i D}+F_{a e}[/latex]

[latex]181 lbf <114+362.8=476.8 lbf[/latex] , so Eq.(11-16) applies

Eq. (11-16a):

[latex]\begin{aligned}F_{e C} &=0.4 F_{r C}+K_{C}\left(F_{i D}+F_{a e} ight) \&=0.4(577)+1.5(114+362.8)=946 lbf >F_{r C,} ext { so use } F_{e C}\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{10^{8}}{90\left(10^{6} ight)}=1.11 \&R=\sqrt{0.90}=0.949\end{aligned}[/latex]

Eq. (11-7): [latex]F_{R C}=1(946)\left(\frac{1.11}{4.48(1-0.949)^{1 / 1.5}} ight)^{3 / 10}=1130 lbf[/latex]

Eq. (11-16b): [latex]F_{e D}=F_{r D}=363 lbf[/latex]

Eq. (11-7): [latex]F_{R D}=1(363)\left(\frac{1.11}{4.48(1-0.949)^{1 / 1.5}} ight)^{3 / 10}=433 lbf[/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 11.41: For the shaft application defined in Prob. 3–74, p. 138, per...

Related Answered Questions