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.

Question

A single-row deep groove ball bearing is subjected to a 30 second work cycle that consists of the following two parts:

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.

Accepted Answer

[latex] ext { Given } C_{o}=50 kN \quad C=68 kN [/latex].

Step I Equivalent load for complete work cycle
For Part I,

[latex] \left(\frac{F_{a}}{F_{r}} ight)=\left(\frac{12.5}{45} ight)=0.278 [/latex].

and [latex] \left(\frac{F_{a}}{C_{o}} ight)=\left(\frac{12.5}{50} ight)=0.25 [/latex].

From Table 15.4, e = 0.37

Table 15.4 X and Y factors for single-row deep groove ball bearings

e	[latex]\left(\frac{F_{a}}{F_{r}} ight)>e[/latex]		[latex]\left(\frac{F_{a}}{F_{r}} ight) \leq e[/latex]		[latex]\left(\frac{F_{a}}{C_{0}} ight)[/latex]
e	Y	X	Y	X	[latex]\left(\frac{F_{a}}{C_{0}} ight)[/latex]
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

[latex] herefore \quad\left(\frac{F_{a}}{F_{r}} ight)

∴ X = 1 Y = 0

[latex] P_{1}=F_{r}=45000 N [/latex].

[latex] ext { also } \quad N_{1}=\frac{10}{60} imes(720)=120 N [/latex].

For Part II,

[latex] \left(\frac{F_{a}}{F_{r}} ight)=\left(\frac{6.25}{15} ight)=0.417 [/latex].

and [latex] \left(\frac{F_{a}}{C_{o}} ight)=\left(\frac{6.25}{50} ight)=0.125 [/latex].

From Table 15.4, e = 0.31 (approximately)

[latex] herefore \quad\left(\frac{F_{a}}{F_{r}} ight)>e [/latex].

Assuming linear interpolation,

[latex] Y=1.6-\frac{(1.6-1.4)}{(0.130-0.07)} imes(0.125-0.07) [/latex].

[latex] =1.42 ext { and } X=0.56 [/latex].

[latex] P_{2}=X F_{r}+Y F_{a}=0.56(15000)+1.42(6250) [/latex].

= 17 275 N.

[latex] N_{2}=\frac{20}{60} imes(1440)=480 rev [/latex].

[latex] N_{1}+N_{2}=120+480=600 rev . [/latex]

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),

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

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

[latex] =\sqrt[3]{\left[\frac{120(45000)^{3}+480(17275)^{3}}{600} ight]} [/latex].

= 28 167.89 N.

[latex] ext { Step II Bearing life }\left(L_{10 h} ight) [/latex]

[latex] L_{10}=\left(\frac{C}{P_{e}} ight)^{3}=\left(\frac{68000}{28167.89} ight)^{3}=14.069 ext { million rev. } [/latex]

[latex] L_{10 h }=\frac{L_{10} imes 10^{6}}{60 n}=\frac{14.069 imes 10^{6}}{60(1200)}=195.4 h [/latex].

Question 15.12: A single-row deep groove ball bearing is subjected to a 30 s...