In Ex. 11–3, the minimum required load rating for 99 percent reliability, at xD =LD/L10 = 540, is C10 = 6696 lbf = 29.8 kN. From Table 11–2 a 02-40 mm deepgrooveball bearing would satisfy the requirement. If the bore in the application had o be 70 mm or larger (selecting a 02-70 mm deep-groove ball

Question

In Ex. 11–3, the minimum required load rating for 99 percent reliability, at [latex]x_D ={L_D}/{L_{10}}= 540 \ , \ is \ C_{10} = 6696 \ lbf = 29.8 [/latex] kN. From Table 11–2 a 02-40 mm deepgroove ball bearing would satisfy the requirement. If the bore in the application had to be 70 mm or larger (selecting a 02-70 mm deep-groove ball bearing), what is the resulting reliability?

Accepted Answer

From Table 11–2,

Table 11–2 Dimensions and Load Ratings for Single-Row 02-Series Deep-Groove and Angular-Contact Ball Bearings
Bore, mm	OD, mm	Width, mm	Fillet Radius, mm	Shoulder Diameter, mm		Load Ratings, kN
				Shoulder Diameter, mm		Deep Groove		Angular Contact
				[latex]d_S[/latex]	[latex]d_H[/latex]	[latex]C_{10}[/latex]	[latex]C_0[/latex]	[latex]C_{10}[/latex]	[latex]C_0[/latex]
10	30	9	0.6	12.5	27	5.07	2.24	4.94	2.12
12	32	10	0.6	14.5	28	6.89	3.10	7.02	3.05
15	35	11	0.6	17.5	31	7.80	3.55	8.06	3.65
17	470	12	0.6	19.5	34	9.56	4.50	9.95	4.75
20	47	14	1.0	25	41	12.7	6.20	13.3	6.55
25	52	15	1.0	30	47	14.0	6.95	14.8	7.65
30	62	16	1.0	35	55	19.5	10.0	20.3	11.0
35	72	17	1.0	41	65	25.5	13.7	27.0	15.0
40	80	18	1.0	46	72	30.7	16.6	31.9	18.6
45	85	19	1.0	52	77	33.2	18.6	35.8	21.2
50	90	20	1.0	56	82	35.1	19.6	37.7	22.8
55	100	21	1.5	63	90	43.6	25.0	46.2	28.5
60	110	22	1.5	70	99	47.5	28.0	55.9	35.5
65	120	26	1.5	74	109	55.9	34.0	63.7	41.5
70	125	24	1.5	79	114	61.8	37.5	68.9	45.5
75	130	25	1.5	86	119	66.3	40.5	71.5	49.0
80	140	26	1.5	93	127	70.2	45.0	80.6	55.0
85	150	28	2.0	99	136	83.2	53.0	90.4	63.0
90	160	30	2.0	104	146	95.6	62.0	106	73.5
95	170	32	2.0	110	156	108	69.5	121	85.0

for a 02-70 mm deep-groove ball bearing, [latex]C_{10} = 61.8 \ \ kN =13 888[/latex] lbf. Using Eq. (11–19)

[latex]R\doteq 1-\left\{\frac{x_D\left(\frac{a_fF_D}{C_{10}} ight)^a-x_0 }{ heta -x_0} ight\}^b \ \ R \ge 0.90 [/latex],

recalling from Ex. 11–3 that [latex]a_f = 1.2 \ , \ F_D = 413 \ lbf \ , \ x_0 = 0.02 \ , \left( heta − x_0 ight) = 4.439, and \ b = 1.483 [/latex] , we can write

[latex]R\doteq 1-\left\{\frac{\left[540\left[\frac{1.2\left(413 ight) }{13 888} ight]^3-0.02 ight] }{4.439} ight\}^{1.483} =0.0999 963[/latex]
which, as expected, is much higher than 0.99 from Ex. 11–3.

Question 3.11.9: In Ex. 11–3, the minimum required load rating for 99 percent...

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