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.
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),
P_{e}=\sqrt[3]{\left[\frac{N_{1} P_{1}^{3}+N_{2} P_{2}^{3}+\cdots}{N_{1}+N_{2}+\cdots}\right]} (15.13).
P_{e}=\sqrt[3]{\left[\frac{N_{1} P_{1}^{3}+N_{2} P_{2}^{3}+\cdots}{N_{1}+N_{2}+\cdots}\right]}\left.=\sqrt[3]{\left[\frac{125(5000)^{3}+350(10000)^{3}+100(7000)^{3}}{575}\right]}\right] .
= 8860.06 N.
Step II Bearing life \left(L_{10 h}\right)
According to the load life relationship,
L_{10}=\left(\frac{C}{P_{e}}\right)^{3}=\left(\frac{40500}{8860.06}\right)^{3}=95.51 \text { million rev. } .
L_{10 h }=\frac{L_{10} \times 10^{6}}{60 n}=\frac{95.51 \times 10^{6}}{60(575)}=2768.45 h .