Chapter - For the deep-groove 02-series ball bearing with R = 0.90, the design life x D, in multiples of rating life, is The design radial load F D is From Eq. (-6), x D = 30 000(300)(60) 0 6 = 540 F D =.2(.898) = 2.278 kn 540 C 0 = 2.278 0.02 + 4.439[ln(/0.9)] /.483 = 8.59 kn Table -2: Choose a 02-30 mm with C 0 = 9.5 kn. Eq. (-8): /3 [ 540(2.278/9.5) 3 ].483 0.02 R = exp 4.439 = 0.99-2 For the Angular-contact 02-series ball bearing as described, the rating life multiple is x D = The design load is radial and equal to 50 000(480)(60) 0 6 = 440 F D =.4(60) = 854 lbf = 3.80 kn Eq. (-6): 440 C 0 = 854 0.02 + 4.439[ln(/0.9)] /.483 = 9665 lbf = 43.0kN /3 Table -2: Select a 02-55 mm with C 0 = 46.2 kn. Using Eq. (-8), [ 440(3.8/46.2) 3 ].483 0.02 R = exp 4.439 = 0.927 R.G. Budynas, J.K. Nisbett, Shigley - Progetto e costruzione di macchine, 3e, 204 McGraw-Hill Education (Italy) srl
-3 For the straight-roller 03-series bearing selection, x D = 440 rating lives from Prob. -2 solution. F D =.4(650) = 230 lbf = 0.279 kn ( ) 440 3/0 C 0 = 0.279 = 9.kN Table -3: Select a 03-55 mm with C 0 = 02 kn. Using Eq. (-8), [ 440(0.28/02) 0/3 ].483 0.02 R = exp = 0.942 4.439-4 We can choose a reliability goal of 0.90 = 0.95 for each bearing. We make the selections, find the existing reliabilities, multiply them together, and observe that the reliability goal is exceeded due to the roundup of capacity upon table entry. Another possibility is to use the reliability of one bearing, say R. Then set the reliability goal of the second as R 2 = 0.90 R or vice versa. This gives three pairs of selections to compare in terms of cost, geometry implications, etc. -5 Establish a reliability goal of 0.90 = 0.95 for each bearing. For a 02-series angular contact ball bearing, 440 /3 C 0 = 854 0.02 + 4.439[ln(/0.95)] /.483 = 35 lbf = 50.4kN Select a 02-60 mm angular-contact bearing with C 0 = 55.9 kn. [ 440(3.8/55.9) 3 ].483 0.02 R A = exp = 0.969 4.439 For a 03-series straight-roller bearing, 440 3/0 C 0 = 0.279 = 05.2kN 0.02 + 4.439[ln(/0.95)] /.483 Select a 03-60 mm straight-roller bearing with C 0 = 23 kn. [ 440(0.28/23) 0/3 ].483 0.02 R B = exp = 0.977 4.439 R.G. Budynas, J.K. Nisbett, Shigley - Progetto e costruzione di macchine, 3e, 204 McGraw-Hill Education (Italy) srl
Form a table of existing reliabilities R goal R A R B 0.92 0.90 0.927 0.94 0.872 0.95 0.969 0.977 0.947 0.906 The possible products in the body of the table are displayed to the right of the table. One, 0.872, is predictably less than the overall reliability goal. The remaining three are the choices for a combined reliability goal of 0.90. Choices can be compared for the cost of bearings, outside diameter considerations, bore implications for shaft modifications and housing modifications. The point is that the designer has choices. Discover them before making the selection decision. Did the answer to Prob. -4 uncover the possibilities? To reduce the work to fill in the body of the table above, a computer program can be helpful. -6 Choose a 02-series ball bearing from manufacturer #2, having a service factor of. For F r = 8kNand F a = 4kN Eq. (-5): x D = 5000(900)(60) 0 6 = 270 270 /3 C 0 = 8 = 5.8kN 0.02 + 4.439[ln(/0.90)] /.483 Trial #: From Table (-2) make a tentative selection of a deep-groove 02-70 mm with C 0 = 37.5kN. Table -: F a C 0 = 4 37.5 = 0.07 F a /(VF r ) = 0.5 > e X 2 = 0.56, Y 2 =.46 Eq. (-9): F e = 0.56()(8) +.46(4) = 0.32 kn Eq. (-6): ( ) 270 /3 C 0 = 0.32 = 66.7kN> 6.8kN Trial #2: From Table -2 choose a 02-80 mm having C 0 = 70.2 and C 0 = 45.0. Check: F a C 0 = 4 45 = 0.089 R.G. Budynas, J.K. Nisbett, Shigley - Progetto e costruzione di macchine, 3e, 204 McGraw-Hill Education (Italy) srl
Table -: X 2 = 0.56, Y 2 =.53 Eq. (-6): Selection stands. F e = 0.56(8) +.53(4) = 0.60 kn ( ) 270 /3 C 0 = 0.60 = 68.5 kn < 70.2kN Decision: Specify a 02-80 mm deep-groove ball bearing. -7 From Prob. -6, x D = 270 and the final value of F e is 0.60 kn. 270 /3 C 0 = 0.6 = 84.47 kn 0.02 + 4.439[ln(/0.96)] /.483 Table -2: Choose a deep-groove ball bearing, based upon C 0 load ratings. Trial #: Tentatively select a 02-90 mm. From Table -, interpolate for Y 2. C 0 = 95.6, F a C 0 = 4 62 = 0.0645 C 0 = 62 kn F a /C 0 Y 2 0.056.7 0.0645 Y 2 0.070.63 Y 2.7 0.0645 0.056 =.63.7 0.070 0.056 = 0.607 Y 2 =.7 + 0.607(.63.7) =.66 F e = 0.56(8) +.66(4) =.2 kn 270 C 0 =.2 0.02 + 4.439[ln(/0.96)] /.483 = 88.6 kn < 95.6kN Bearing is OK. Decision: Specify a deep-groove 02-90 mm ball bearing. /3 R.G. Budynas, J.K. Nisbett, Shigley - Progetto e costruzione di macchine, 3e, 204 McGraw-Hill Education (Italy) srl
-8 For the straight cylindrical roller bearing specified with a service factor of, R = 0.90 and F r = 2 kn -9 x D = 4000(750)(60) = 80 0 6 ( ) 80 3/0 C 0 = 2 = 57.0kN y R y O O z R z O T P z R y A " 2 P y A F 20 R z A 2 3 " 4 B x T Assume concentrated forces as shown. P z = 8(24) = 92 lbf P y = 8(30) = 240 lbf T = 92(2) = 384 lbf in T x = 384 +.5F cos 20 = 0 384 F = = 272 lbf.5(0.940) M z O = 5.75P y +.5R y A 4.25F sin 20 = 0; thus 5.75(240) +.5R y A 4.25(272)(0.342) = 0 R y A = 4.73 lbf y M O = 5.75P z.5r z A 4.25F cos 20 = 0; thus 5.75(92).5R z A 4.25(272)(0.940) = 0 R z A = 43 lbf; R A = [( 43) 2 + ( 4.73) 2 ] /2 = 43 lbf F z = R z O + P z + R z A + F cos 20 = 0 R z O R z O + 92 43 + 272(0.940) = 0 = 34.7 lbf R.G. Budynas, J.K. Nisbett, Shigley - Progetto e costruzione di macchine, 3e, 204 McGraw-Hill Education (Italy) srl
F y = R y O + P y + R y A F sin 20 = 0 R y O R y O So the reaction at A governs. + 240 4.73 272(0.342) = 0 = 42 lbf Reliability Goal: 0.92 = 0.96 A 02-35 bearing will do. R O = [( 34.6) 2 + ( 42) 2 ] /2 = 46 lbf F D =.2(43) = 496 lbf x D = 30 000(300)(60/0 6 ) = 540 540 C 0 = 496 0.02 + 4.439[ln(/0.96)] /.483 = 4980 lbf = 22.6 kn Decision: Specify an angular-contact 02-35 mm ball bearing for the locations at A and O. Check combined reliability. /3-0 For a combined reliability goal of 0.90, use 0.90 = 0.95 for the individual bearings. y O z 20 R O 6 A F A 0 B R B C F C 20 x x 0 = 50 000(480)(60) 0 6 = 440 The resultant of the given forces are R O = 607 lbf and R B = 646 lbf. At O: F e =.4(607) = 850 lbf 440 Ball: C 0 = 850 0.02 + 4.439[ln(/0.95)] /.483 = 262 lbf or 50. kn Select a 02-60 mm angular-contact ball bearing with a basic load rating of 55.9 kn. At B: F e =.4(646) = 2304 lbf 440 3/0 Roller: C 0 = 2304 0.02 + 4.439[ln(/0.95)] /.483 = 23 576 lbf or 04.9 kn Select a 02-80 mm cylindrical roller or a 03-60 mm cylindrical roller. The 03-series roller has the same bore as the 02-series ball. /3 R.G. Budynas, J.K. Nisbett, Shigley - Progetto e costruzione di macchine, 3e, 204 McGraw-Hill Education (Italy) srl
- The reliability of the individual bearings is R = 0.999 = 0.9995 y R y O R z O O z 300 A F z A F C F y A 400 C R y E 50 R z E E x From statics, -2 Given: R y O = 63.4N, Rz O = 07 N, R O = 95 N R y E = 89.N, Rz E = 74.4N, R E = 96 N 60 000(200)(60) x D = = 4320 0 6 4340 C 0 = 0.96 0.02 + 4.439[ln(/0.9995)] /.483 = 8.9kN A 02-25 mm deep-groove ball bearing has a basic load rating of 4.0 kn which is ample. An extra-light bearing could also be investigated. F ra = 560 lbf or 2.492 kn F rb = 095 lbf or 4.873 kn /3 Trial #: Use K A = K B =.5 and from Table -6 choose an indirect mounting. 0.47F ra K A 0.47(2.492).5 <? > 0.47F rb K B ( )(0) <? > 0.47(4.873).5 0.78 <.527 Therefore use the upper line of Table -6. F aa = F ab = 0.47F rb =.527 kn K B P A = 0.4F ra + K A F aa = 0.4(2.492) +.5(.527) = 3.29 kn P B = F rb = 4.873 kn R.G. Budynas, J.K. Nisbett, Shigley - Progetto e costruzione di macchine, 3e, 204 McGraw-Hill Education (Italy) srl
Fig. -6: f T = 0.8 Fig. -7: f V =.07 Thus, a 3l = f T f V = 0.8(.07) = 0.856 Individual reliability: R i = 0.9 = 0.95 Eq. (-7): [ 40 000(400)(60) (C 0 ) A =.4(3.29) 4.48(0.856)( 0.95) 2/3 (90)(0 6 ) =.40 kn [ 40 000(400)(60) (C 0 ) B =.4(4.873) 4.48(0.856)( 0.95) 2/3 (90)(0 6 ) = 6.88 kn From Fig. -4, choose cone 32 305 and cup 32 305 which provide F r = 7.4kN and K =.95. With K =.95 for both bearings, a second trial validates the choice of cone 32 305 and cup 32 305. ] 0.3 ] 0.3-3 O z y R z O 82. 6" R y O A 226 20 T 4" T B 45 2" R z C R = 0.95 = 0.975 T = 240(2)(cos 20 ) = 2706 lbf in F = 2706 = 498 lbf 6 cos 25 C x R y C In xy-plane: MO = 82.(6) 20(30) + 42R y C = 0 In xz-plane: R y C R y O = 8 lbf = 82 + 20 8 = lbf MO = 226(6) 452(30) 42R z c = 0 R z C R z O = 237 lbf = 226 45 + 237 = 2 lbf R O = ( 2 + 2 2 ) /2 = 2 lbf R C = (8 2 + 237 2 ) /2 = 298 lbf F eo =.2(2) = 34.4 lbf F ec =.2(298) = 357.6 lbf x D = 40 000(200)(60) 0 6 = 480 R.G. Budynas, J.K. Nisbett, Shigley - Progetto e costruzione di macchine, 3e, 204 McGraw-Hill Education (Italy) srl
480 (C 0 ) O = 34.4 0.02 + 4.439[ln(/0.975)] /.483 = 438 lbf or 6.398 kn 480 (C 0 ) C = 357.6 0.02 + 4.439[ln(/0.975)] /.483 = 3825 lbf or 7.02 kn Bearing at O: Choose a deep-groove 02-2 mm. Bearing at C: Choose a deep-groove 02-30 mm. There may be an advantage to the identical 02-30 mm bearings in a gear-reduction unit. -4 Shafts subjected to thrust can be constrained by bearings, one of which supports the thrust. The shaft floats within the endplay of the second (Roller) bearing. Since the thrust force here is larger than any radial load, the bearing absorbing the thrust is heavily loaded compared to the other bearing. The second bearing is thus oversized and does not contribute measurably to the chance of failure. This is predictable. The reliability goal is not 0.99, but 0.99 for the ball bearing. The reliability of the roller is. Beginning here saves effort. Bearing at A (Ball) F r = (36 2 + 22 2 ) /2 = 25 lbf = 0.957 kn F a = 555 lbf = 2.47 kn Trial #: Tentatively select a 02-85 mm angular-contact with C 0 = 90.4kNand C 0 = 63.0kN. F a C 0 = 2.47 63.0 = 0.0392 x D = Table -: X 2 = 0.56, Y 2 =.88 25 000(600)(60) 0 6 = 900 F e = 0.56(0.957) +.88(2.47) = 5.8 kn F D = f A F e =.3(5.8) = 6.73 kn 900 C 0 = 6.73 0.02 + 4.439[ln(/0.99)] /.483 = 07.7kN> 90.4kN Trial #2: Tentatively select a 02-95 mm angular-contact ball with C 0 = 2 kn and C 0 = 85 kn. F a C 0 = 2.47 85 = 0.029 /3 /3 /3 R.G. Budynas, J.K. Nisbett, Shigley - Progetto e costruzione di macchine, 3e, 204 McGraw-Hill Education (Italy) srl
Table -: Y 2 =.98 F e = 0.56(0.957) +.98(2.47) = 5.43 kn F D =.3(5.43) = 7.05 kn 900 C 0 = 7.05 0.02 + 4.439[ln(/0.99)] /.483 = 3 kn < 2 kn O.K. Select a 02-95 mm angular-contact ball bearing. Bearing at B (Roller): Any bearing will do since R =. Let s prove it. From Eq. (-8) when ( ) af F 3 D x D < x 0 R = /3 C 0 The smallest 02-series roller has a C 0 = 6.8kNfor a basic load rating. ( ) 0.427 3 (900) <? > 0.02 6.8 0.048 < 0.02 R = Spotting this early avoided rework from 0.99 = 0.995. Any 02-series roller bearing will do. Same bore or outside diameter is a common choice. (Why?) -5 Hoover Ball-bearing Division uses the same 2-parameter Weibull model as Timken: b =.5, θ = 4.48. We have some data. Let s estimate parameters b and θ from it. In Fig. -5, we will use line AB. In this case, B is to the right of A. For F = 8 kn, (x) = 5(2000)(6) 0 6 = 3.8 This establishes point on the R = 0.90 line. log F 2 F 00 39.6 8 A R 0.90 B R 0.20 2 0 0 3.8 72 0 0 00 x 2 log x R.G. Budynas, J.K. Nisbett, Shigley - Progetto e costruzione di macchine, 3e, 204 McGraw-Hill Education (Italy) srl
The R = 0.20 locus is above and parallel to the R = 0.90 locus. For the two-parameter Weibull distribution, x 0 = 0 and points A and B are related by: x A = θ[ln(/0.90)] /b () x B = θ[ln(/0.20)] /b and x B /x A is in the same ratio as 600/5. Eliminating θ Solving for θ in Eq. () θ = b = Therefore, for the data at hand, ln[ln(/0.20)/ ln(/0.90)] ln(600/5) =.65 x A [ln(/r A )] = = 3.9 /.65 [ln(/0.90)] /.65 [ ( ) x.65 ] R = exp 3.9 Check R at point B: x B = (600/5) = 5.27 [ R = exp Note also, for point 2 on the R = 0.20 line. ( ) 5.27.65 ] = 0.20 3.9 log(5.27) log() = log(x m ) 2 log(3.8) (x m ) 2 = 72-6 This problem is rich in useful variations. Here is one. Decision: Make straight roller bearings identical on a given shaft. Use a reliability goal of (0.99) /6 = 0.9983. Shaft a FA r = (2392 + 2 ) /2 = 264 lbf or.75 kn F r B = (5022 + 075 2 ) /2 = 86 lbf or 5.28 kn Thus the bearing at B controls 0 000(200)(60) x D = = 720 0 6 0.02 + 4.439[ln(/0.9983)] /.483 = 0.080 26 ( ) 720 0.3 C 0 =.2(5.2) = 97.2kN 0.080 26 Select either a 02-80 mm with C 0 = 06 kn or a 03-55 mm with C 0 = 02 kn R.G. Budynas, J.K. Nisbett, Shigley - Progetto e costruzione di macchine, 3e, 204 McGraw-Hill Education (Italy) srl
Shaft b F r C = (8742 + 2274 2 ) /2 = 2436 lbf or 0.84 kn F r D = (3932 + 657 2 ) /2 = 766 lbf or 3.4 kn The bearing at C controls 0 000(240)(60) x D = = 44 0 6 C 0 =.2(0.84) ( ) 44 0.3 = 22 kn 0.0826 Select either a 02-90 mm with C 0 = 42 kn or a 03-60 mm with C 0 = 23 kn Shaft c The bearing at E controls F r E = (32 + 2385 2 ) /2 = 2632 lbf or.7 kn F r F = (472 + 895 2 ) /2 = 987 lbf or 4.39 kn x D = 0 000(80)(60/0 6 ) = 48 ( ) 48 0.3 C 0 =.2(.7) = 94.8kN 0.0826 Select a 02-80 mm with C 0 = 06 kn or a 03-60 mm with C 0 = 23 kn -7 The horizontal separation of the R = 0.90 loci in a log F-log x plot such as Fig. -5 will be demonstrated. We refer to the solution of Prob. -5 to plot point G (F = 8 kn, x G = 3.8). We know that (C 0 ) = 39.6kN,x =. This establishes the unimproved steel R = 0.90 locus, line AG. For the improved steel (x m ) = 360(2000)(60) 0 6 = 43.2 We plot point G (F = 8 kn, x G = 43.2), and draw the R = 0.90 locus A m G parallel to AG log F 2 F 00 55.8 39.6 A m A Improved steel Unimproved steel R 0.90 8 0 R 0.90 G 3 G 3 3.8 43.2 0 0 0 00 2 x log x R.G. Budynas, J.K. Nisbett, Shigley - Progetto e costruzione di macchine, 3e, 204 McGraw-Hill Education (Italy) srl
We can calculate (C 0 ) m by similar triangles. log(c 0 ) m log 8 log 39.6 log 8 = log 43.2 log log 3.8 log ( ) log 43.2 39.6 log(c 0 ) m = log 3.8 log + log 8 8 (C 0 ) m = 55.8kN The usefulness of this plot is evident. The improvement is 43.2/3.8 = 3.3 fold in life. This result is also available by (L 0 ) m /(L 0 ) as 360/5 or 3.3 fold, but the plot shows the improvement is for all loading. Thus, the manufacturer s assertion that there is at least a 3-fold increase in life has been demonstrated by the sample data given. -8 Express Eq. (-) as F a L = C a 0 L 0 = K For a ball bearing, a = 3 and for a 02-30 mm angular contact bearing, C 0 = 20.3kN. At a load of 8 kn, life L is given by: L = K F a For a load of 30 kn, life L 2 is: K = (20.3) 3 (0 6 ) = 8.365(0 9 ) = 8.365(09 ) 8 3 =.434(0 6 )rev L 2 = 8.365(09 ) 30 3 = 0.30(0 6 )rev In this case, Eq. (7-57) the Palmgren-Miner cycle ratio summation rule can be expressed as l + l 2 = L L 2 Substituting, 200 000.434(0 6 ) + l 2 0.30(0 6 ) = Check: -9 Total life in revolutions Let: l = total turns f = fraction of turns at F f 2 = fraction of turns at F 2 l 2 = 0.267(0 6 )rev 200 000.434(0 6 ) + 0.267(06 ) 0.30(0 6 ) = O.K. R.G. Budynas, J.K. Nisbett, Shigley - Progetto e costruzione di macchine, 3e, 204 McGraw-Hill Education (Italy) srl
From the solution of Prob. -8, L =.434(0 6 )revand L 2 = 0.30(0 6 )rev. Palmgren-Miner rule: from which Total life in loading cycles l L + l 2 L 2 = f l L + f 2l L 2 = l = f /L + f 2 /L 2 l = 0.40/[.434(0 6 )]+0.60/[0.30(0 6 )] = 45 585 rev 6 min 0 min/cycle 4 min at 2000 rev/min = 8000 rev at 2000 rev/min = 2 000 rev 20 000 rev/cycle Total life in hours 45 585 rev = 22.58 cycles 20 000 rev/cycle ( 0 min )( ) 22.58 cycles = 3.76 h cycle 60 min/h -20 While we made some use of the log F-log x plot in Probs. -5 and -7, the principal use of Fig. -5 is to understand equations (-6) and (-7) in the discovery of the catalog basic load rating for a case at hand. Point D F D = 495.6 lbf log F D = log 495.6 = 2.70 x D = 30 000(300)(60) 0 6 = 540 log x D = log 540 = 2.73 K D = FD 3 x D = (495.6) 3 (540) = 65.7(0 9 ) lbf 3 turns log K D = log[65.7(0 9 )] = 0.82 F D has the following uses: F design, F desired, F e when a thrust load is present. It can include application factor a f, or not. It depends on context. R.G. Budynas, J.K. Nisbett, Shigley - Progetto e costruzione di macchine, 3e, 204 McGraw-Hill Education (Italy) srl
Point B x B = 0.02 + 4.439[ln(/0.99)] /.483 = 0.220 turns log x B = log 0.220 = 0.658 ( ) /3 ( ) xd 540 /3 F B = F D = 495.6 = 6685 lbf x B 0.220 Note: Example -3 used Eq. (-7). Whereas, here we basically used Eq. (-6). Point A log F B = log(6685) = 3.825 K D = 6685 3 (0.220) = 65.7(0 9 ) lbf 3 turns F A = F B = C 0 = 6685 lbf log C 0 = log(6685) = 3.825 x A = log x A = log() = 0 K 0 = F 3 A x A = C 3 0 () = 66853 = 299(0 9 ) lbf 3 turns (as it should) Note that K D /K 0 = 65.7(0 9 )/[299(0 9 )] = 0.220, which is x B. This is worth knowing since K 0 = K D x B log K 0 = log[299(0 9 )] =.48 log F F 4 0 4 6685 B A 3 0 3 495.6 D 0.658 540 2 0 2 0. 0 0 0 2 2 0 3 3 x log x Now C 0 = 6685 lbf = 29.748 kn, which is required for a reliability goal of 0.99. If we select an angular contact 02-40 mm ball bearing, then C 0 = 3.9kN= 769 lbf. R.G. Budynas, J.K. Nisbett, Shigley - Progetto e costruzione di macchine, 3e, 204 McGraw-Hill Education (Italy) srl