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lecture3 - ROLLING CONTACT BEARINGS Bearing ~—(6min...

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Unformatted text preview: ROLLING CONTACT BEARINGS Bearing ~— (6min! Scar/am; Mva 1404(9% a $<vi 1i @MJWW. Major bearing types _ R%£Zm3 debcl’ balnfi35 01’ JAQOQQ_ZEaflé3'9T rfiaflgé a A Plain Bearings : I Bearing or Sleeve 7i $1211 a mid/Mm my. ¥%¢ rank. Eaxnfi; aQu_ hr iflg Hang“; %/ £&_ Yflut + 1 Aim/ml 3— t. A jab rt'cam/f «I Wed fir mafia/J2. /mfl&mj KL £00.41 é— d4.L.i/@wé Maé Shaft or Journal 11m??? / gain/091% Rolling contact bearings _ W W Z604 ‘3 WWI/“’M' MY‘Wfi’A ego/mun}: M My? gym/2d mg!» 754% 5 “3'12“; (MM! _ (gojfi Iota/rm? : RM»? {fglrmwfj are SPAM: - MM beam 2 M» , ‘ a 'hé ‘jéfihfiué cifl. ,beghffiy A$wai fiyékafin 130,61. 5604’ng 3M, amt MAI/11' [lg/W Speai / 157M” ./ .lrui ijb. flsééw baa/“My: 92/: Wed @404 7%” speed, A431,, {firmi gym: Slhaft loads - Radial (190d U 4444#—\_ Ball bearing types ROdLa/l / #h’ast OY MoguAl/r Wot (K9014! + Must). RWQX' dLSIy/ntd fiYinél %W «mbi M1 Thrust 1 duly/mi %m [Du/rt kusé £94411. Mug/Mm mind: duty/nut 429* a. mama/L9” 5/ mafia/a» /'/;MJ[’ and: Roller bearing types - may: Mm ( gamma"; {palm 13mm; — sflmwfl rad/42,, — 1‘me may” (mm + mm 3m; _ ”‘ML ( med My. M31 5,093. (I a fry/14,05 Masl’ {(7,061 0f MI boa/v: ' 3 am C001 7477 79mm [0174 Rolling Element Bearing Failure: m any cm % beam-g Mm u 741% M PM A #LZaa/J [J 7% Lia/Lu} {aflm a} PM! I?) 1’41, /Z00Ji 1’: M1. éwaé. flaymj 7% boX/J are Jaizdacéd I5} 2171me Md“ 5 WWW {W3 twig“. Tlu. Cam/f/esmm 31 a, 50/] Way». #u. mm” + wfir ‘pjfi4 Wm 11 Lew/2M 5/7”: (Hr/film sfregseJ)' t ' ! mes-firwgnlwx 1 ($44 Mal 5851(st EXUZL ”75 JA— AJ/AAwA‘k/j 'PWC‘” a Say/42a, Mm. fl‘iflx/m, Im a gar/2a, /0fjm‘, fléwe , Mail: are Cow/(1&1 Just 13% M M1. 2331/ 5W/éu. MAM PmSpaga/iu /3 %z bra/M Sax/41m. ”Maw/u} A V v {A 5,004”? P710122”? Spa/jib??? OY [Stigma dig am. gm 0*/ . 0] W L if ”midi/fed //ij”¢ , mg [I Q W5” 7/ '21 77/273 ffiL‘Lbév _/~ IEEAR/Né— LOAD -, Eypr W15 Mew #41} Ear log (F) a mafJ 0% LilM/éfla‘é Ael‘a/Vujfi 5 74: J éi E [xi/010v m%/Mj A 312115 F: a, F7. slope=l/ a i . 4/ . f L (MJIMLJ ‘ W £7116. 14/»ng L, 9. LL” YUPf ,«f 3 C [ log(L) L1 __ '1 a I»: 7?: Rating Life(L10 ): T/xL W13 % Of Q (arm?!) of @pW/y (Mm/(fig! 50/] 02’ Mix Among: (.5 Would (11 Mt. W54,» 91 Y‘WMJ %f 40/, of (9er of 50,011,142); xM-[Z Lew/054 m. €XC€Ld bp/M #17 M C/VIZA’LQ WWI. (fol 5/ ML , 3 , z” 'a/ EA Basic Load Rating ( C ): TA; Dam. [and «0154?, J dgm a; W132”! MM /gond 49th a. (7% 4/ “174/90,,% 1%, ( 11.5%]; a. ll galfi/M/WS 6044 Emily: %Y a. ”VD/{0V7 1% (9/ W W £14941 faanM/elm' n7 #4 W Val? . (W MAM 7’14}? {1r 5% 50 “M27 Z'Ii/O‘m’ r F:C 2/4 L/oWL./%o./ abaa/mu'gaja/wi 59% jawi' U —--3 L: (.970. M ”rd/2&6“; f/ YCMMLKMJ W '0 F “yr! L9? C: F(L»0>CL T143 )lfjmjyg %firl 5m? $55614” . I/ 7%. ragga] /g92d F 4014M (f 61457 a} a FDA/rm 2150M“? Arab/x / W a yea/(L # {f (/ISLM / M“? C :r 3’5”” m Ca/wé/M. {,2 ijw Wflwy I %V Nwé alas 300,0 5n a1 0. (Ms/Ltd 99M ”6/ 500 Var/mud)». ., L,= (3000 6r)x(500%)( 6c: TM) 901/04 m m c —. F(_;’t;bi—)l/a 90110‘ Basic Static Load Rating ( C0): [m WW 29y gray/1:21;, Mfg!” 42/1,. Lam sin/I; ”170,4: W83 Co 5/771 2» r4441 ,W A WW; 1'; W flab/Ty £5an 404% #4 5W 1; fl/DJ/ ”/9153; 771w (5a m Amt may 5 an) a} aft/mix a a; 44¢. win—£7 “5120214 w/MM awrzypmm’r /29L a para/rm! 5%me1 {9/ ”Mu/z"; p/W % 1000! g #4 5017/ or MHZ/r dad/mail”. _._..- ..._ , - ‘ _.#r...‘._.‘..-_,_ _ _ 7/15-».1 Effect of Thrust Loads: F; :‘ Ofp’iét‘ad mda! ”1390/1, E = fifpégd MYLLSL‘ £304 \/ z a W %wé)r ._v. ' l V ' gum; www’mfigrh7/fl > , . .G 3*» ~ ' * “1/7515“? 0491319» 0 VJ, g 1-2 '\— ['Ml’fllt? gulf”,- KIM/2,1! « a 6’ r~\ Effect of Shock or Impact Forces: V T/u. W M MW aim/t do 2: W afW %" 0M7 thug m’ Wad 4L 5604qu W expen'wo. . W/I"! / 7% ngml @701 WJIL A: maid-Moi 5y 0. 404 afloflmfigm m SAM AAA £7444. //'6. ' Bearing Sdection: (1) Take the free body diagram of the shaft and perform a force analysis. Assume the shaft to be simply supported due to the bearings. (2) From the force analysis, find the appropriate shaft size when it is not given. (3) Calculate the equivalent radial load Fe at each bearing site. Else eqs 11-11 and 11—12. Use at [00.4-070paaoxm $.29” mm W V (4) Calculate the required basic load rating C at each bearing site. C=Fe(L10)1/a where L10 is in millions of shaft revolutions. From fuMul/d 0V H-l-r/ (5),\Select the smallest bearing which has a basic load rating equal to or greater than the one calculated in step 4. (6) Compare the bearing bore to the shaft size and make sure that the bearing will fit. If the bearing does not fit, select a bearing of bigger dimension that will fit. Then check the basic load rating. ‘ Required Basic Load Rating With Different Reliabilities: i Im ML firm/15m: Madras} A aging? W [and I? mm M 150ch an a fcéaérélj 37/ 90X : WM a clue/rid 4% Lb (3” (9W of a glam/.1: Y‘QAfiAIL/L/éff r bwlm m Adm. A,» /m‘d U1)” 24.0113 eywimw. Examples: GLM F; 3 8 KN J F; : L} KN LB : 5000,1’ ”Lb : 900 W/m OZ- Sent: had 550ng Fm; c ? L”, (3 M 07/ baa/rm"? at 5:8)00 +15”qu zoom}? loérw', ,‘ 6o mw'n :5 L” = (50005) " 9°° L}; : 270m [0‘ m M mm? C (i : ~\ 6: E<Llo)ya' MLVL'Q:Z fiéflw ‘\/-' VFW F: MK 0/5 XVFV 4)’F0L MAM X+)/ an”; seflded /yom filéé ”— Z 54ch ML Maj/15% X;}/, 77(14%1Yc/ M W 19 aria/ML V E: 37.5 51: 3—: 5>€ => X1: 54, y,-.II+S F; t? (23w?) ma 1/-2 g l<~ / .:c(?)+ /-95'(4)= /0.2! M F.“ d- C: FL(L,., 14": /0.25?0 27 V3 > ) < O) : éé‘QKN L>"(/,yky\i(/¢omfl1L&//,3) Assam 730m. : 80 ”mm =) C: 70.2 KM / 60'g 4’5kN Fa 4 \ ’6 .——-—-: \._/ Co (I; Of? 7962 289 .03»: i:'r>e F} f "=7? Xz='{&/ y; :I.§3 . 7‘ 73% HA WP“ .rafl) ; /\)’3(41) = /0.'6 KN __——— ‘/ C: a (£46) a : /O:5<Z70)y3: 63’5K", < 70:2 KN =9 9 [yea/mg of 235m a’omnm A)“ Mad + C:7o.?.Ku h———-—-————.______. ...
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