A ball bearing is operating on a work cycle consisting of three parts—a radial load of 3000 N at 1440 rpm for one quarter cycle, a radial load of 5000 N at 720 rpm for one half cycle, and radial load of 2500 N at 1440 rpm for the remaining cycle. The expected life of the bearing is 10 000 h.
Calculate the dynamic load carrying capacity of the bearing.

Question

Accepted Answer

[latex] ext { Given } L_{10 h }=10000 h [/latex].

Step I Equivalent load for complete work cycle
Considering the work cycle of one minute duration,

[latex] N_{1}=\frac{1}{4}(1440)=360 rev [/latex].

[latex] N_{2}=\frac{1}{2}(720)=360 rev [/latex].

[latex] N_{3}=\frac{1}{4}(1440)=360 rev [/latex].

The average speed of rotation is given by,

[latex] n=N_{1}+N_{2}+N_{3}=1080 rpm [/latex].

From Eq. (15.13),

[latex] P_{e}=\sqrt[3]{\left[\frac{N_{1} P_{1}^{3}+N_{2} P_{2}^{3}+N_{3} P_{3}^{3}}{N_{1}+N_{2}+N_{3}} ight]} [/latex] (15.13).

[latex] P_{e}=\sqrt[3]{\left[\frac{N_{1} P_{1}^{3}+N_{2} P_{2}^{3}+N_{3} P_{3}^{3}}{N_{1}+N_{2}+N_{3}} ight]} [/latex]

[latex] =\sqrt[3]{\left[\frac{360(3000)^{3}+360(5000)^{3}+360(2500)^{3}}{1080} ight]} [/latex].

= 3823 N
Step II Dynamic load carrying capacity of bearing

[latex] L_{10}=\frac{60 n L_{10 h }}{10^{6}}=\frac{60(1080)(10000)}{10^{6}} [/latex].

[latex] =648 ext { million rev. } [/latex]

From Eq. (15.7),

[latex] C=P\left(L_{10} ight)^{1 / 3} [/latex] (15.7).

[latex] C=P\left(L_{10} ight)^{1 / 3}=3823(648)^{1 / 3}=33082 N [/latex].

Question 15.11: A ball bearing is operating on a work cycle consisting of th...

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