A single-row deep groove ball bearing has a dynamic load capacity of 40500 N and operates on the following work cycle:

(i) radial load of 5000 N at 500 rpm for 25% of the time;

(ii) radial load of 10000 N at 700 rpm for 50% of the time; and

(iii) radial load of 7000 N at 400 rpm for the remaining 25% of the time.
Calculate the expected life of the bearing in hours.

Question

A single-row deep groove ball bearing has a dynamic load capacity of 40500 N and operates on the following work cycle:

(i) radial load of 5000 N at 500 rpm for 25% of the time;

(ii) radial load of 10000 N at 700 rpm for 50% of the time; and

(iii) radial load of 7000 N at 400 rpm for the remaining 25% of the time.
Calculate the expected life of the bearing in hours.

Accepted Answer

Given C = 40500 N
Step I Equivalent load for complete work cycle
Consider the work cycle of one minute duration.
The values of load P and revolutions N are tabulated as follows:

Revolutions N in element time	Speed (rpm)	Element time (minute)	P (N)	Element No
125	500	0.25	5000	1
350	700	0.5	10000	2
100	400	0.25	7000	3
575		1.00		Total

From Eq. (15.13),

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

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

[latex] \left.=\sqrt[3]{\left[\frac{125(5000)^{3}+350(10000)^{3}+100(7000)^{3}}{575} ight]} ight] [/latex].

= 8860.06 N.

Step II Bearing life [latex] \left(L_{10 h} ight) [/latex]
According to the load life relationship,

[latex] L_{10}=\left(\frac{C}{P_{e}} ight)^{3}=\left(\frac{40500}{8860.06} ight)^{3}=95.51 ext { million rev. } [/latex].

[latex] L_{10 h }=\frac{L_{10} imes 10^{6}}{60 n}=\frac{95.51 imes 10^{6}}{60(575)}=2768.45 h [/latex].

Question 15.10: A single-row deep groove ball bearing has a dynamic load cap...

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