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Unformatted text preview: SOL (14. )
Known: A No. 203 radial ball bearing carries a radial load of 201‘.) lb and a thrust load of 151] lb at 18m rpm.
Find: Determine the bearing Bll} life. Schematic and Given Data: No. 203 Radial Bearing
Fr = zooN, Ft=150 N
90% reliability mmwmﬂm
Blil life = ‘3' Assumptions:
1. Table 14.2 accurately gives the bearing capacity. 2. Ball bearing life varies inversely with the l0f3 power ofthe load (i.e.. Eq. (14.521)
is aoourate}. 3. The life given is for a 90% reliability.
4. The load E; can be found from Eq. (14.3]. Analysis:
I. From Table 14.1, for a 208 bearing the bore is 40 mm. From Table 14.2. for Ln = 90 x 105 rev and a zoo series bearing. C = 9.40 W =
2112.3 lb. From Fig. 14.13, for 90 percent reliability, Kr = 1.0. From Table 14.3, K; = 1.0 for a steady load. The ratio FtlFr= 150 Ibﬂﬂﬂ lb = {US The equivalent load from Eq. (14.3} is F: = Fr [1 + 1.115({Fu‘Fr} (135)] = 2001M! + 1.115({15ﬁf200}  0.35)] = 289.2 lb
From Eq. (14.524. L = KrLR{ClFeKal333 Substituting Tvalues into Eq. (14.53,); N WH owew 3.33
L=90><106rev 2112:3111 =ﬁ76xlﬂwrov 239.2 lb =l 57'5 “10mm“ J: 9387'53 hr (60 IniufhrHIEUO revlmitt] I Comment: The life of 938363 hours corresponds to about ill? years of continuous
operation where the bearing runs 24 hoursfday and 7 daysfweekua long life! SOLUTION (14.9)
Known; A radial contact ball hearing has a given radial load. Find: Determine the radial load change required to (a) double the life and (h) triple the
life. Schematic and Given Data: Assumptions: 1. Ball bearing life varies inverselyr with the 105'? power of the lead.
2. The life given it for a 90% reliability. Maigret: 1. Let L1 and F: be the original life and load for the bearing. Let L: and F2 he the
new life and load. 2. Since Lir'Lz = [inFli‘m, F251 = (Lil2W”
3. To deuhle the life, L2 = 2L], and F2113] = (liZF'im = 0.312 I
4. Te triple the life, L2 = 3L1, and szF] = (UHm” = {1119 I Comment: To double the hearing life the radial load must be reduced te {IHSIZ of its original value; to triple the bearing life the radial lead must be reduced In [1.719 of its
original value SOLUTION (14.14)
Known: A hearing has a life of 5000 hr for 90% reliability. Find: Estimate the lives for 50% reliability and 99% reliability.
Schematic and Given Data: EN 90% reliabilityr 50% reliability 99% reliabilitg.r LI=SUUﬂhI L2? L=?
90% 50% 99% Assumption: Bearing life varies inversely with the 101’3 power of the load (Le, Eq.
(14.2a) is suitable). Analysis:
1. From Eq. (14.23), L = KrLR{CfFr)333
2. For identical bearings with the same LR, C, and Fr, __...L— = 4 = _.L__ KrLﬂC’Frls'ﬂ 90% KlLﬁlﬂanoa 50% KrLﬂClFf‘ﬂ 9%
01' L_ = L = L I Enema Kr sole Krliiim 3. From Fig. 14.13, for 90% reliability, r = 1.0; for 99% reliability, K; = 0.21; for
50% reliability, Kr = 5.0. 4 Lone = Lgott = Leora
' 1.0 5.0 0.21 5. PW Lyme = 5000 hours, L501}, ={5)(50{10} = 25,000 hours I
and Lena; = (UJIKSUOU) 1: 1,050 hours I Comment: A higher reliability requirement (fewer bearing failures) means a shorter
life. SOLUTION {14.210}
Known: Radial and angular bearings are to operate for 501.11] hr with 98% reliability with known combinations of radial and thrust loads. Find: Select suitable (a) radial and {biangular beatings for each load combination. Schematic and Given Data:
Case (21} Ft: 1.5 1:11, Case {b} Ft: 3.0m Radial Bearing Angular Hearing 98% reliability L = 51ml} hr life
lightmoderate impact.
Pr = 3 ltN Aﬁumpﬁons:
l. The inner ring of the bearing fits with enough interference to prevent relative motion during operation.
1. The internal fits between the halls and their races are correct. 3. Bearing misalignment is 15' or less. Analysis: From Table 14.3. for Iight~moderate impact. Kn = 1.5. L = 501.11] hours = {5001] hoursHlDUﬂ reviminﬂﬁrﬂ minlhr) = 3 X 103 rev.
From Table 14.3, with 98% reliabilityr we have Kr = {1.33. The life corresponding to rated capacity. LR : 90 x 105.
From Eq. (14.5b).
or... = FeKa (urchin3 = Fdl.5}(300!{0.33)(90}}°3 = ano Fe. PPS“FL— s. For F; = 1.5 kN. FrlFr = 0.5.
T. For the radial bearing, Eq (143) gives F3 2 (3.0}(1 + 1.1 15(1115D = 3.51} RN.
8. The required value of rated capacity for the application. Croq = 3(3 .513} = 10.5 1:14. 9. Select bearing No. 211. I
11]. For the angular bearing, Eq. [14.11)ng Fe : F; = 3.!) kN. ll. Creq = {3){311} = 9 kN. 12. Select 40 mm bore bearing No. 203. I {MI
13. For Fl = 3.0 W. FdFr = 1.1]. 14. For the radial ligating, Eq. (14.3) gives F: = {3.D)(l + l.115{0.ﬁ§}) = 5.]? W.
15. Cmq = [3)(5.1T}= 15.52 Hi. 16. Select bearing No. 213. I
17. For the angular mfg—ﬁg, Eq. [14.4) gives E: = {3.U}(1.ﬂ + 0.870(032l) = 3.84 kN.
13. CM : [3)(334) = 11.51 Hi. 19. Select bearing No. 211. l sownou (14.2w) Known: A printing roll is driven by a gear. The bottom surface of the roll is in conta
With a similar roll that applies a uniform {upward} loading. Find: Select identical 200 series ball bearings for A and B.
Schematic and Given Data: Gear,
120 mm dia. Decisionsﬂssumptions:
1. Use a design life of 30,000 hours as suggested by Table 14.4.
2. A reliability of 90% is desired. 3. The application factor1 Ka = 1.1 {T able 14.3, favorable gearing}. Analysis: _
1. Since the printing roll is in static equilibrium. the summation of torques equals zero.
[2' Torque = 0]: R1175)  1200{cos 20)(60} = 0
Hence. Rn = 901.6 N
2. Also, the summation of moments, the summation of horizontal forces, and the
summation of vertical forces should each be zero. Horizontal forces:
[2 Ma = 0]: 901.6{210} + 1200(cos 2031620) ~ Bh(420) = 0
Hence, 311 = 1346.9 N [E H: =9]: 90!.6+1200(00520]1346.9 Ar. =0
Hence. Ah = 132.3 N m:
[2 M. = D]: 4(300}(210} — 120mm 2031520) ~ 13.4420} = 0
Hence, By = 91.9 N
[2 F" = G]: A. — 91.9 + 411300)  120mm 20} = 0
Hence, Au = 697.? N
3. The. bearing radial loads are: Bearing A: F=1./182. 32+ 697. t: = 721 N BearingB: F= 1I’18446.9 +9192 =1849N 4. Since the. radial load on hearing B is greater than on hearing A, the bearing
selection will be based On bearing B. 5. From Eq. {14.5b), cm, = Felia {LilierD3
where Fc = F1 : 1849 N
Also, K3 = 1.1 (Table 14.3. favorable gearing} and Kr = 1.[} for 90% reliability. L 350W? ﬁﬂminlm [109 I“ =63ﬂx105mv min 1 hr 1 .5 .3
6. Therefore. Cm = 13490.1 M... = 3545 N
{luau x 105} 7. From Table 14.21 56131125 mm bore bearing 295. I Comment: Th: shaft size. requirement r1121}r nmsimte 11313 of a larger bore bearing. ...
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 Fall '08
 ZHOU

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