Answer Happy

Table 2.2/2 Dynamic load factors for combined loading Single-row radial bearings • : - 1 Y = 0 Table 2.2/2 Bearing A Bearing B Bunge х Y X Y 1.6 0:42 11.15 13 161. 160 0. RUS, AMS 041 00 0.44 14 0.014 2.00 0.000 0.004 XT 0.50 1.55 00 0.04 561 2.28 1.15 0.10 0.2 0.20 0334 0.00 00 6.44 0.2 114 0.5 0.30 1.05 1.6 14 2.1 270 1.0 . 02 04 0.6 200020 65 0607 20 303 05.19 2000 1012 2.5 888888888822233& 0,4 0.6 ELMBO 72 73 OJOJ 3) 320X300/22 X XOX 3x00X OX 11 12 X 40 X 1.07 155 0.30 0,31 . 0.63 0.37 0.60 0,4 0.40 04 DO 0.0 140 0.30 31 TX 140 0.4 F. 04 10 0.50 0.44 84 0,33 050 0.41 1.55 8.30 1.5 1.75 10 15 14 1.4 14 2.0 19 1.18 F FB 88 30 0.4 24 Fun A shaft application is supported by two tapered roller bearings, which are mounted as above. Bearing Ais selected as 30214 whereas Bearing Bis chosen as 32009x. Using the necessary loading data given below, calculate the equivalent axial load [kN] that acts on (and supported by) the relevant bearing. F. = 4.00 KN FA5.00 KN F = 5.60 KN a 5.203
13 1.15 0.30 0.43 0.4 88808 313 01.13 0.5 0.41 0.6 0.4 02 0.42 . EL M, DO 72 73 O2, O3 320 M X 320/22 X XX 00.00 X *OX MEX 12X40X 302... 04 0.22 0. 0.3 BUM 0.63 0.31 0.60 0.40 0. 0/0 10 2,5 05 1.07 1. 1,35 1.55 13 1.00 1.35 1.75 888 88888888 10...12 0.2 1.14 0.98 0.39 0.44 0,30 0.44 9.30 0.44 0.35 0.50 0.41 48 484858039 sec 0.4 04 15.10 0.0 0.30 031 323 08 FB FA 0.4 04 32 1.5 "FF Sub A shatt application is supported by two tapered roller bearings, which are mounted as above. Bearing Ais selected as 30214 whereas Bearing B is chosen as 32009X. Using the necessary loading data given below, calculate the equivalent axial load (kN] that acts on (and supported by the relevant bearing. F. = 4,00 KN Fra = 5.00 KN Fis = 5.60 KN a. 5.203 b. 5.806 05.017 d. 5.612 e. 5.429
Main dimensions Designa- tion Load rating Double-row radial bearings F Bearing type so to > d D 00 Cined Comod Ciso Corso kN mod KN mm 10 4,41 30 30 35 9 14 11 1200 2200 1300 SURAS 5,73 1,38 7,14 1,61 7.71 1,89 1,38 5,49 1,61 5,93 1,89 给 12 32 32 37 10 14 12 1201 2201 1301 5,75 1.50 7.40 1.81 7,86 2,06 4,42 1,50 5,70 1,81 6,04 2,08 15 35 35 11 14 13 17 1202 2202 1302 2302 7.77 2,06 7,47 1.99 10,5 2,96 12,7 3.36 5,98 2,06 5,75 1.91 8,10 2,96 9,73 3,36 巴马总总会与马与老太高容易乌兹兹 e х Y х Y 42 see single-row deep groove ball bearings, but (F//co) 0,1 MAR 32, 33 0,86 1 0,73 0,62 1,17 72 BG, 73 BG")1,14 1 0,55 0,57 0,93 12 00 0,33 1,95 01 0,35 1,8 02 0,33 1,95 3,0 0,30 2,1 0,31 20 05 0,29 2,15 0,65 3,35 06 0,25 1 2,5 3,85 07 0,24 2,65 4,1 08 0,22 2,9 10 0,21 2,95 11 0,20 3,2 12 13 0,19 3,25 14 22 0,18 3,5 的95%8叫听听$%%n&= CON OON อาญต่อ 0. เกศ เก้ 12 16 14 19 09 1203 2203 1303 2303 9,942,70 7,65 2.70 10,2 2,88 7,852.88 13,0 3,74 10,0 3,74 15,2 4,15 11.7 4.15 4,6 ကို قافي في 20 47 10,8 12,4 12,9 52 52 47 14 18 15 21 14(40) 15(44) 1204 2204 1304 2304 11204 11304 3.42 3,63 4,01 5,49 3,42 4,01 8,33 3.42 9,53 3,63 9,93 401 14,7 5,49 8,33 3.42 9,93 4,01 19,1 10,8 12,9 A self-aligning ball bearing is selected for a particular machine construction. Using the data below and the table above, [hours) of the bearing. Type: 1200 F = 2.10 KN
0336 08 0.43 13 1.15 108 0.4 8 0.40 0.5 0.15 0.6 0.4 3130513 33 0.24 042 . LM, DO 72 73 OJ, O3) 320 X 300/22 X XIX 00X00 x 10 x X 12X40X 302 04 0.22 0.30 03 M 0.63 0.21 0.00 0.4 0. 040 8888 8 888 29888 10.12 0.2 1.14 0.95 0.39 0,44 0,38 0.44 0.30 0.44 0.35 0.41 2,5 09 1.07 1. 1,35 1.55 10 1. 1.35 1.75 0.4 98584858 sees 0.4 15.10 0.0 323 08 0. 0.30 21 FA FB 0.4 04 FSO A shaft application is supported by two tapered roller bearings, which are mounted as above. Bearing Als selected as 30214 whereas Bearing B is chosen as 32009. Using the necessary loading data given below, calculate the equivalent axial load [kN] that acts on (and supported by the relevant bearing. F. = 4,00 KN Fra = 5.00 KN Fra = 5.60 KN a. 5.203 b. 5.806 05.017 d. 5.612 e 5.429