Budynas-Nisbett:Shigley's Ill.Design of Mechanical 11.Rolling-Contact T©The McGraw-Hill Mechanical Engineering Elements Bearings Companies,2008 Design,Eighth Edition 564 Mechanical Engineering Design EXAMPLE 11-4 An SKF 6210 angular-contact ball bearing has an axial load F,of 400 lbf and a radial load F of 500 Ibf applied with the outer ring stationary.The basic static load rating Co is 4450 Ibf and the basic load rating Cio is 7900 Ibf.Estimate the Lio life at a speed of 720 rev/min. Solution V=1 and Fa/Co=400/4450 =0.090.Interpolate for e in Table 11-1: Fa/Co e 0.084 0.28 0.090 e from which e=0.285 0.110 0.30 Fa/(VF,)=400/[(1)500]=0.8>0.285.Thus,interpolate for Y2: Fa/Co Y2 0.084 1.55 0.090 Y2 from which Y2=1.527 0.110 1.45 From Eq.(11-9) Fe=X2VF,+Y2Fa=0.56(1)500+1.527(400)=890.81bf With Lp =Lio and Fp Fe,solving Eq.(11-3)for Lio gives Answer 60LRIR 106 7900 L10= =16150h 60nD 60(720) 890.8 We now know how to combine a steady radial load and a steady thrust load into an equivalent steady radial load F that inflicts the same damage per revolution as the radial-thrust combination. 11-7 Variable Loading Bearing loads are frequently variable and occur in some identifiable patterns: Piecewise constant loading in a cyclic pattern Continuously variable loading in a repeatable cyclic pattern ·Random variation Equation (11-1)can be written as FL constant =K (a) Note that F may already be an equivalent steady radial load for a radial thrust load com- bination.Figure 11-9 is a plot of Fa as ordinate and L as abscissa for Eq.(a).If a load level of F is selected and run to the failure criterion,then the area under the F-L trace is numerically equal to K.The same is true for a load level F2:that is,the area under the F2-L2 trace is numerically equal to K.The linear damage theory says that in the case of load level F1.the area from L=0 to L =LA does damage measured by FLA=D
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 11. Rolling−Contact Bearings © The McGraw−Hill 565 Companies, 2008 564 Mechanical Engineering Design EXAMPLE 11–4 An SKF 6210 angular-contact ball bearing has an axial load Fa of 400 lbf and a radial load Fr of 500 lbf applied with the outer ring stationary. The basic static load rating C0 is 4450 lbf and the basic load rating C10 is 7900 lbf. Estimate the L10 life at a speed of 720 rev/min. Solution V = 1 and Fa/C0 = 400/4450 = 0.090. Interpolate for e in Table 11–1: Fa/(V Fr) = 400/[(1)500] = 0.8 > 0.285. Thus, interpolate for Y2: From Eq. (11–9), Fe = X2V Fr + Y2Fa = 0.56(1)500 + 1.527(400) = 890.8 lbf With L D = L10 and FD = Fe, solving Eq. (11–3) for L10 gives Answer L10 = 60L RnR 60nD C10 Fe a = 106 60(720) 7900 890.8 3 = 16 150 h Fa/C0 Y2 0.084 1.55 0.090 Y2 from which Y2 = 1.527 0.110 1.45 Fa/C0 e 0.084 0.28 0.090 e from which e = 0.285 0.110 0.30 We now know how to combine a steady radial load and a steady thrust load into an equivalent steady radial load Fe that inflicts the same damage per revolution as the radial–thrust combination. 11–7 Variable Loading Bearing loads are frequently variable and occur in some identifiable patterns: • Piecewise constant loading in a cyclic pattern • Continuously variable loading in a repeatable cyclic pattern • Random variation Equation (11–1) can be written as FaL = constant = K (a) Note that F may already be an equivalent steady radial load for a radial–thrust load combination. Figure 11–9 is a plot of Fa as ordinate and L as abscissa for Eq. (a). If a load level of F1 is selected and run to the failure criterion, then the area under the F1-L1 trace is numerically equal to K. The same is true for a load level F2; that is, the area under the F2-L2 trace is numerically equal to K. The linear damage theory says that in the case of load level F1, the area from L = 0 to L = L A does damage measured by Fa 1 L A = D.
Budynas-Nisbett:Shigley's Ill.Design of Mechanical 11.Rolling-Contact T©The McGraw-Hill Mechanical Engineering Elements Bearings Companies,2008 Design,Eighth Edition Rolling-Contact Bearings 565 Figure 11-9 Plot of pa as ordinate and L as abscissa for FaL constant. The linear damage hypothesis says that in the case of load F F1,the area under the curve from L =0 to L LA is a measure of the damage D=FaLA.The complete damage to failure is measured by Ciol B. Figure 11-10 A three-part piecewise continuous periodic loading Fa cycle involving loads Fel.Fe2, and Fe3.Fea is the equivalent steady load inflicting the same F damage when run for+ 12+13 revolutions,doing the same damage D per period. Consider the piecewise continuous cycle depicted in Fig.11-10.The loads Fi are equivalent steady radial loads for combined radial-thrust loads.The damage done by loads Fel,F2,and Fe3 is D=Fah Foh+Fl3 ( where li is the number of revolutions at life Li.The equivalent steady load Fa when run for+2+l3 revolutions does the same damage D.Thus D=Fg1+2+3) (c) Equating Eqs.(b)and (c),and solving for Feg,we get 「F凸+F+Fgh] (11-101 l1+l2+3 =∑ig] where fi is the fraction of revolution run up under load Fei.Since li can be expressed as niti,where n is the rotational speed at load Fi and ti is the duration of that speed, then it follows that Feg T∑nF盘a L∑ntJ (11-110 The character of the individual loads can change,so an application factor(af)can be prefixed to each Fei as (afi Fi)a;then Eq.(11-10)can be written r=[∑iaar]= K (11-12)
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 11. Rolling−Contact Bearings 566 © The McGraw−Hill Companies, 2008 Rolling-Contact Bearings 565 L1 L2 0 L F2 a F1 a Fa A B Figure 11–9 Plot of Fa as ordinate and L as abscissa for Fa L = constant. The linear damage hypothesis says that in the case of load F1, the area under the curve from L = 0 to L = L A is a measure of the damage D = Fa 1L A. The complete damage to failure is measured by Ca 10L B . Consider the piecewise continuous cycle depicted in Fig. 11–10. The loads Fei are equivalent steady radial loads for combined radial–thrust loads. The damage done by loads Fe1, Fe2, and Fe3 is D = Fa e1l1 + Fa e2l2 + Fa e3l3 (b) where li is the number of revolutions at life Li . The equivalent steady load Feq when run for l1 + l2 + l3 revolutions does the same damage D. Thus D = Fa eq (l1 + l2 + l3) (c) Equating Eqs. (b) and (c), and solving for Feq, we get Feq = Fa e1l1 + Fa e2l2 + Fa e3l3 l1 + l2 + l3 1/a = fi Fa ei 1/a (11–10) where fi is the fraction of revolution run up under load Fei . Since li can be expressed as ni ti , where ni is the rotational speed at load Fei and ti is the duration of that speed, then it follows that Feq = ni ti Fa ei ni ti 1/a (11–11) The character of the individual loads can change, so an application factor (af ) can be prefixed to each Fei as (af i Fei)a ; then Eq. (11–10) can be written Feq = fi(af i Fei) a 1/a Leq = K Fa eq (11–12) Fa Fe2 a Fe1 a l 1 l 2 l 3 l F a eq Fe3 a Figure 11–10 A three-part piecewisecontinuous periodic loading cycle involving loads Fe1, Fe2, and Fe3. Feq is the equivalent steady load inflicting the same damage when run for l1+ l2 + l3 revolutions, doing the same damage D per period