A single-row deep groove ball bearing is subjected to a 30 second work cycle that consists of the following two parts:
| Part II | Part I | |
| 20 | 10 | duration (s) |
| 15 | 45 | radial load (kN) |
| 6.25 | 12.5 | axial load (kN) |
| 1440 | 720 | speed (rpm) |
The static and dynamic load capacities of the ball bearing are 50 and 68 kN respectively. Calculate the expected life of the bearing in hours.
\text { Given } C_{o}=50 kN \quad C=68 kN .
Step I Equivalent load for complete work cycle
For Part I,
\left(\frac{F_{a}}{F_{r}}\right)=\left(\frac{12.5}{45}\right)=0.278 .
and \left(\frac{F_{a}}{C_{o}}\right)=\left(\frac{12.5}{50}\right)=0.25 .
From Table 15.4, e = 0.37
Table 15.4 X and Y factors for single-row deep groove ball bearings
| e | \left(\frac{F_{a}}{F_{r}}\right)>e | \left(\frac{F_{a}}{F_{r}}\right) \leq e | \left(\frac{F_{a}}{C_{0}}\right) | ||
| Y | X | Y | X | ||
| 0.22 | 2.0 | 0.56 | 0 | 1 |
0.025 |
| 0.24 | 1.8 | 0.56 | 0 | 1 | 0.040 |
| 0.27 | 1.6 | 0.56 | 0 | 1 | 0.070 |
| 0.31 | 1.4 | 0.56 | 0 | 1 | 0.130 |
| 0.37 | 1.2 | 0.56 | 0 | 1 | 0.250 |
| 0.44 | 1.0 | 0.56 | 0 | 1 | 0.500 |
\therefore \quad\left(\frac{F_{a}}{F_{r}}\right)<e .
∴ X = 1 Y = 0
P_{1}=F_{r}=45000 N .
\text { also } \quad N_{1}=\frac{10}{60} \times(720)=120 N .
For Part II,
\left(\frac{F_{a}}{F_{r}}\right)=\left(\frac{6.25}{15}\right)=0.417 .
and \left(\frac{F_{a}}{C_{o}}\right)=\left(\frac{6.25}{50}\right)=0.125 .
From Table 15.4, e = 0.31 (approximately)
\therefore \quad\left(\frac{F_{a}}{F_{r}}\right)>e .
Assuming linear interpolation,
Y=1.6-\frac{(1.6-1.4)}{(0.130-0.07)} \times(0.125-0.07) .
=1.42 \text { and } X=0.56 .
P_{2}=X F_{r}+Y F_{a}=0.56(15000)+1.42(6250) .
= 17 275 N.
N_{2}=\frac{20}{60} \times(1440)=480 rev .
N_{1}+N_{2}=120+480=600 rev .The number of revolutions completed in one work cycle of 30 second duration is 600. Therefore, 1200 revolutions will be completed in one minute.
Or,
n = 1200 rpm
From Eq. (15.13),
P_{e}=\sqrt[3]{\left[\frac{N_{1} P_{1}^{3}+N_{2} P_{2}^{3}}{N_{1}+N_{2}}\right]} (15.13).
P_{e}=\sqrt[3]{\left[\frac{N_{1} P_{1}^{3}+N_{2} P_{2}^{3}}{N_{1}+N_{2}}\right]}=\sqrt[3]{\left[\frac{120(45000)^{3}+480(17275)^{3}}{600}\right]} .
= 28 167.89 N.
\text { Step II Bearing life }\left(L_{10 h}\right) L_{10}=\left(\frac{C}{P_{e}}\right)^{3}=\left(\frac{68000}{28167.89}\right)^{3}=14.069 \text { million rev. }L_{10 h }=\frac{L_{10} \times 10^{6}}{60 n}=\frac{14.069 \times 10^{6}}{60(1200)}=195.4 h .