A single-row deep groove ball bearing No. 6002 is subjected to an axial thrust of 1000 N and a radial load of 2200 N. Find the expected life that 50% of the bearings will complete under this condition.

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Accepted Answer

[latex] ext { Given } F_{a}=1000 N \quad F_{r}=2200 N [/latex].

Bearing = No. 6002
Step I X and Y factors
Referring to Table 15.5, the capacities of bearing No. 6002 are,

Table 15.5 Dimensions and static and dynamic load capacities of single-row deep groove ball bearings

Designation	Basic load ratings (N)		Principal dimensions (mm)
Designation	[latex]C_{0}[/latex]	C	B	D	d
61800	630	1480	5	19	10
6000	1960	4620	8	26
6200	2240	5070	9	30
6300	3750	8060	11	35
61801	695	1430	5	21
6001	2240	5070	8	28
6201	3100	6890	10	32
6301	4650	9750	12	37
61802	815	1560	5	24	15
6002	2500	5590	9	32
6202	3550	7800	11	35
6302	5400	11400	13	42
61803	930	1680	5	26	17
6003	2800	6050	10	35
6202	4500	9560	12	40
6303	6550	13500	14	47
6403	11800	22900	17	62
61804	1500	2700	7	32	20
16404	3400	7020	8	42
6004	4500	9360	12	42
6204	6200	12700	14	47
6304	7800	15900	15	52
6404	16600	30700	19	72
61805	1960	3120	7	37	25
16005	4000	7610	8	47
6005	5600	11200	12	47
6205	6950	14000	15	52
6305	11400	22500	17	62
6405	19600	35800	21	80
61806	2080	3120	7	42	30
16006	5850	11200	9	55
6006	6800	13300	13	55
6206	10000	19500	16	62
6306	14600	28100	19	72
6406	24000	43600	23	90
61807	3000	4030	7	47	35
16007	6950	12400	9	62
6007	8500	15900	14	62
6207	13700	25500	17	72
6307	18000	33200	21	80
6407	31000	55300	25	100
61808	3350	4160	7	52	40
16008	7800	13300	9	68
6008	9300	16800	15	68
6208	16600	30700	18	80
6308	22400	41000	23	90
6408	36500	63700	27	110
61809	3800	6050	7	58	45
16009	9300	15600	10	75
6009	12200	21200	16	75
6209	19600	33200	19	85
6309	30000	52700	25	100
6409	45500	76100	29	120
61810	4250	6240	7	65	50
16010	10000	16300	10	80
6010	13200	21600	16	80
6210	19600	35100	20	90
6310	36000	61800	27	110
6410	5200	87100	31	130
61811	5600	8320	9	72	55
16011	12200	19500	11	90
6011	17000	28100	18	90
6211	25000	43600	21	100
6311	41500	71500	29	120
6411	63000	99500	33	140
61812	6100	8710	10	78	60
16012	13200	19900	11	95
6012	18300	29600	18	95
6212	28000	47500	22	110
6312	48000	81900	31	130
6412	69500	108000	35	150
61813	8300	11700	10	85	65
16013	14600	21200	11	100
6013	19600	30700	18	100
6213	34000	55900	23	120
6313	56000	92300	33	140
6413	78000	119000	37	160
61814	9150	12100	10	90	70
16014	19000	28100	13	110
6014	24500	37700	20	110
6214	37500	61800	24	125
6314	63000	104000	35	150
6414	104000	143000	42	180
61815	9800	12500	10	95	75
10615	20000	28600	13	115
6015	26000	39700	20	115
6215	40500	66300	25	130
6315	72000	112000	37	160
6415	114000	1530000	45	190

[latex] C_{o}=2500 N ext { and } C=5590 N [/latex]

[latex] ext { Also, } \quad F_{a}=1000 N \quad F_{r}=2200 N [/latex].

[latex] \left(\frac{F_{a}}{F_{r}} ight)=\left(\frac{1000}{2200} ight)=0.455 [/latex].

[latex] ext { and } \quad\left(\frac{F_{a}}{C_{o}} ight)=\left(\frac{1000}{2500} ight)=0.4 [/latex].

Referring to Table 15.4,

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

The value of Y is obtained by linear interpolation.

[latex] Y=1.2-\frac{(1.2-1.0)}{(0.5-0.25)} imes(0.4-0.25)=1.08 ext { and } [/latex].

X = 0.56

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

[latex] P=X F_{r}+Y F_{a}=0.56(2200)+1.08(1000) [/latex].

= 2312 N
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} \quad 5590=2312\left(L_{10} ight)^{1 / 3} [/latex].

[latex] herefore \quad L_{10}=14.13 ext { million rev. } [/latex]

[latex] ext { Step III Bearing life }\left(L_{50} ight)[/latex].

It can be proved that the life [latex] \left(L_{50} ight) [/latex], which 50% of the bearings will complete or exceed, is approximately five times the life [latex] \left(L_{10} ight) [/latex] which 90% of the bearings will complete or exceed.
Therefore

[latex] L_{50}=5 L_{10}=5(14.13)=70.65 ext { million rev. } [/latex]

Question 15.7: A single-row deep groove ball bearing No. 6002 is subjected ...

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