Budynas-Nisbett:Shigley's Ill.Design of Mechanical 11.Rolling-Contact T©The McGraw--Hill Mechanical Engineering Elements Bearings Companies,2008 Design,Eighth Edition 554 Mechanical Engineering Design The rating life is a term sanctioned by the ABMA and used by most manufactur- ers.The rating life of a group of nominally identical ball or roller bearings is defined as the number of revolutions (or hours at a constant speed)that 90 percent of a group of bearings will achieve or exceed before the failure criterion develops.The terms minimum life,Lio life,and Bio life are also used as synonyms for rating life.The rating life is the 10th percentile location of the bearing group's revolutions-to-failure distribution. Median life is the 50th percentile life of a group of bearings.The term average life has been used as a synonym for median life,contributing to confusion.When many groups of bearings are tested,the median life is between 4 and 5 times the Lio life. 11-3 Bearing Load Life at Rated Reliability When nominally identical groups are tested to the life-failure criterion at different loads,the data are plotted on a graph as depicted in Fig.11-4 using a log-log trans- formation.To establish a single point,load Fi and the rating life of group one(Lio) are the coordinates that are logarithmically transformed.The reliability associated with this point,and all other points,is 0.90.Thus we gain a glimpse of the load-life func- tion at 0.90 reliability.Using a regression equation of the form FLI/a =constant (11-1) the result of many tests for various kinds of bearings result in ·a=3 for ball bearings .a=10/3 for roller bearings(cylindrical and tapered roller) A bearing manufacturer may choose a rated cycle value of 10 revolutions (or in the case of the Timken Company,90(10)revolutions)or otherwise,as declared in the manufacturer's catalog to correspond to a basic load rating in the catalog for each bearing manufactured,as their rating life.We shall call this the catalog load rating and display it algebraically as Cio.to denote it as the 10th percentile rating life for a particular bearing in the catalog.From Eq.(11-1)we can write F LVe FaLila (11-2) and associate load F with Cio,life measure LI with Lio,and write C1oL=FLe where the units of L are revolutions. Figure 11-4 log F Typical bearing load-ife log-log curve. 0 logL
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 11. Rolling−Contact Bearings © The McGraw−Hill 555 Companies, 2008 554 Mechanical Engineering Design log L log F 0 Figure 11–4 Typical bearing load-life log-log curve. The rating life is a term sanctioned by the ABMA and used by most manufacturers. The rating life of a group of nominally identical ball or roller bearings is defined as the number of revolutions (or hours at a constant speed) that 90 percent of a group of bearings will achieve or exceed before the failure criterion develops. The terms minimum life, L10 life, and B10 life are also used as synonyms for rating life. The rating life is the 10th percentile location of the bearing group’s revolutions-to-failure distribution. Median life is the 50th percentile life of a group of bearings. The term average life has been used as a synonym for median life, contributing to confusion. When many groups of bearings are tested, the median life is between 4 and 5 times the L10 life. 11–3 Bearing Load Life at Rated Reliability When nominally identical groups are tested to the life-failure criterion at different loads, the data are plotted on a graph as depicted in Fig. 11–4 using a log-log transformation. To establish a single point, load F1 and the rating life of group one (L10)1 are the coordinates that are logarithmically transformed. The reliability associated with this point, and all other points, is 0.90. Thus we gain a glimpse of the load-life function at 0.90 reliability. Using a regression equation of the form F L1/a = constant (11–1) the result of many tests for various kinds of bearings result in • a = 3 for ball bearings • a = 10/3 for roller bearings (cylindrical and tapered roller) A bearing manufacturer may choose a rated cycle value of 106 revolutions (or in the case of the Timken Company, 90(106) revolutions) or otherwise, as declared in the manufacturer’s catalog to correspond to a basic load rating in the catalog for each bearing manufactured, as their rating life. We shall call this the catalog load rating and display it algebraically as C10, to denote it as the 10th percentile rating life for a particular bearing in the catalog. From Eq. (11–1) we can write F1L1/a 1 = F2L1/a 2 (11–2) and associate load F1 with C10, life measure L1 with L10, and write C10L1/a 10 = F L1/a where the units of L are revolutions
56 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 555 Further,we can write C10(LRnR60)1/a Fp(LDnp60)1/a catalog rating,Ibf or kN- desired speed,rev/min rating life in hours desired life,hours rating speed,rev/min desired radial load,Ibf or kN Solving for Cio gives LDnD60 l/a C10=FD LRnR60 (11-3) EXAMPLE 11-1 Consider SKF,which rates its bearings for 1 million revolutions,so that Lio life is 60LgnR=106 revolutions.The LRnR60 product produces a familiar number.Timken, for example,uses 90(10)revolutions.If you desire a life of 5000 h at 1725 rev/min with a load of 400 Ibf with a reliability of 90 percent,for which catalog rating would you search in an SKF catalog? Solution From Eq.(11-3), C10=FD LDnD60 I/a =400 5000(1725)607/3 106 =32111bf=14.3kN LRIR60 If a bearing manufacturer rates bearings at 500 h at 33 rev/min with a reliability of 0.90,then LRng60=500(33)60=106 revolutions.The tendency is to substitute 10 for LgnR60 in Eq.(11-3).Although it is true that the 60 terms in Eq.(11-3)as displayed cancel algebraically,they are worth keeping,because at some point in your keystroke sequence on your hand-held calculator the manufacturer's magic number (10 or some other number)will appear to remind you of what the rating basis is and those manufacturers'catalogs to which you are limited.Of course,if you evaluate the bracketed quantity in Eq.(11-3)by alternating between numerator and denominator entries,the magic number will not appear and you will have lost an opportunity to check. 11-4 Bearing Survival:Reliability versus Life At constant load,the life measure distribution is right skewed as depicted in Fig.11-5. Candidates for a distributional curve fit include lognormal and Weibull.The Weibull is by far the most popular,largely because of its ability to adjust to varying amounts of skewness.If the life measure is expressed in dimensionless form asx=L/L1o,then the reliability can be expressed as [see Eq.(20-24),p.970] b R=exp (= (11-4) where R=reliability x life measure dimensionless variate,L/L1o xo=guaranteed,or"minimum,"value of the variate
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 11. Rolling−Contact Bearings 556 © The McGraw−Hill Companies, 2008 Rolling-Contact Bearings 555 Further, we can write C10(L RnR60) 1/a = FD(L DnD60) 1/a catalog rating, lbf or kN desired speed, rev/min rating life in hours desired life, hours rating speed, rev/min desired radial load, lbf or kN Solving for C10 gives C10 = FD L DnD60 L RnR60 1/a (11–3) EXAMPLE 11–1 Consider SKF, which rates its bearings for 1 million revolutions, so that L10 life is 60L RnR = 106 revolutions. The L RnR60 product produces a familiar number. Timken, for example, uses 90(106) revolutions. If you desire a life of 5000 h at 1725 rev/min with a load of 400 lbf with a reliability of 90 percent, for which catalog rating would you search in an SKF catalog? Solution From Eq. (11–3), C10 = FD L DnD60 L RnR60 1/a = 400 5000(1725)60 106 1/3 = 3211 lbf = 14.3 kN If a bearing manufacturer rates bearings at 500 h at 331 3 rev/min with a reliability of 0.90, then L RnR60 = 500(331 3 )60 = 106 revolutions. The tendency is to substitute 106 for L RnR60 in Eq. (11–3). Although it is true that the 60 terms in Eq. (11–3) as displayed cancel algebraically, they are worth keeping, because at some point in your keystroke sequence on your hand-held calculator the manufacturer’s magic number (106 or some other number) will appear to remind you of what the rating basis is and those manufacturers’ catalogs to which you are limited. Of course, if you evaluate the bracketed quantity in Eq. (11–3) by alternating between numerator and denominator entries, the magic number will not appear and you will have lost an opportunity to check. 11–4 Bearing Survival: Reliability versus Life At constant load, the life measure distribution is right skewed as depicted in Fig. 11–5. Candidates for a distributional curve fit include lognormal and Weibull. The Weibull is by far the most popular, largely because of its ability to adjust to varying amounts of skewness. If the life measure is expressed in dimensionless form as x = L/L10, then the reliability can be expressed as [see Eq. (20–24), p. 970] R = exp − x − x0 θ − x0 b (11–4) where R = reliability x = life measure dimensionless variate, L/L10 x0 = guaranteed, or “minimum,’’ value of the variate
Budynas-Nisbett:Shigley's Ill.Design of Mechanical 11.Rolling-Contact T©The McGraw-Hil Mechanical Engineering Elements Bearings Companies,2008 Design,Eighth Edition 556 Mechanical Engineering Design Figure 11-5 log F Constant reliability contours. Rated line Point A represents the catalog rating C1o at x =L/L10 =1. Point B is on the target reliability design line RD,with a load of Co.Point Dis a R安0.90 point on the desired reliability Fo contour exhibiting the design R=Rp life xp=Lo/Lio at the design Design line load Fp. logx X10 XD Dimensionless life measurex =characteristic parameter corresponding to the 63.2121 percentile value of the variate b=shape parameter that controls the skewness Because there are three distributional parameters,xo.6.and b,the Weibull has a robust ability to conform to a data string.Also,in Eq.(11-4)an explicit expression for the cumulative distribution function is possible: F=1-R=1 (=] (11-5) EXAMPLE 11-2 Construct the distributional properties of a 02-30 mm deep-groove ball bearing if the Weibull parameters are xo=0.02.(0-xo)=4.439,and b=1.483.Find the mean, median,10th percentile life,standard deviation,and coefficient of variation. Solution From Eq.(20-28),p.971,the mean dimensionless life ux is Answer x=0+(0- r(+)=02+439r(+s） =4.033 The median dimensionless life is,from Eg.(20-26)where R=0.5, Answer 1)b 00=0+(0-o(n 。1/1.483 =0.02+4.439(h05 =3.487 The 10th percentile value of the dimensionless life x is 1 、1/1483 Answer x0.10=0.02+4.439n ÷1 (as it should be) 0.90
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 11. Rolling−Contact Bearings © The McGraw−Hill 557 Companies, 2008 556 Mechanical Engineering Design log x log F C10 FD xD x10 Dimensionless life measure x B A Rated line Design line D R = 0.90 R = RD Figure 11–5 Constant reliability contours. Point A represents the catalog rating C10 at x = L/L10 = 1. Point B is on the target reliability design line RD , with a load of C10. Point D is a point on the desired reliability contour exhibiting the design life xD = LD /L10 at the design load FD. θ = characteristic parameter corresponding to the 63.2121 percentile value of the variate b = shape parameter that controls the skewness Because there are three distributional parameters, x0, θ, and b, the Weibull has a robust ability to conform to a data string. Also, in Eq. (11–4) an explicit expression for the cumulative distribution function is possible: F = 1 − R = 1 − exp − x − x0 θ − x0 b (11–5) EXAMPLE 11–2 Construct the distributional properties of a 02-30 mm deep-groove ball bearing if the Weibull parameters are x0 = 0.02, (θ − x0) = 4.439, and b = 1.483. Find the mean, median, 10th percentile life, standard deviation, and coefficient of variation. Solution From Eq. (20–28), p. 971, the mean dimensionless life μx is Answer μx = x0 + (θ − x0) 1 + 1 b = 0.02 + 4.439 1 + 1 1.483 = 4.033 The median dimensionless life is, from Eq. (20–26) where R = 0.5, Answer x0.50 = x0 + (θ − x0) ln 1 R 1/b = 0.02 + 4.439 ln 1 0.5 1/1.483 = 3.487 The 10th percentile value of the dimensionless life x is Answer x0.10 = 0.02 + 4.439 ln 1 0.901/1.483 . = 1 (as it should be)
Budynas-Nisbett:Shigley's Ill.Design of Mechanical 11.Rolling-Contact ©The McGraw-Hil Mechanical Engineering Elements Bearings Companies,2008 Design,Eighth Edition Rolling-Contact Bearings 557 The standard deviation of the dimensionless life is given by Eq.(20-29): Answer =(0-x r(+)-r+)] =449[(+)-r+s) =2.753 The coefficient of variation of the dimensionless life is Answer Cx= - =0.683 11-5 Relating Load,Life,and Reliability This is the designer's problem.The desired load is not the manufacturer's test load or catalog entry.The desired speed is different from the vendor's test speed,and the reliability expectation is typically much higher than the 0.90 accompanying the catalog entry.Figure 11-5 shows the situation.The catalog information is plotted as point A, whose coordinates are (the logs of)Cio and x1o =Lio/Lio=1,a point on the 0.90 reliability contour.The design point is at D,with the coordinates(the logs of)Fp and xp,a point that is on the R=Rp reliability contour.The designer must move from point D to point A via point B as follows.Along a constant reliability contour (BD),Eq.(11-2)applies: Faxua=FDx .1/a from which =n() (a) Along a constant load line (AB),Eq.(11-4)applies: Rp exp [(-] Solving for xB gives xg=和+0-o)(血尼66 Now substitute this in Eq.(a)to obtain 1/a 1/a FB=FD XD XD =FD xo+(0-x0)(In 1/RD)1/6 However,Fg C10.so C10=FD XD 1/a xo+(0-xo)(In 1/Rp)1/5 (11-61
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 11. Rolling−Contact Bearings 558 © The McGraw−Hill Companies, 2008 Rolling-Contact Bearings 557 The standard deviation of the dimensionless life is given by Eq. (20–29): Answer σˆx = (θ − x0) 1 + 2 b − 2 1 + 1 b 1/2 = 4.439 1 + 2 1.483 − 2 1 + 1 1.4831/2 = 2.753 The coefficient of variation of the dimensionless life is Answer Cx = σˆx μx = 2.753 4.033 = 0.683 11–5 Relating Load, Life, and Reliability This is the designer’s problem. The desired load is not the manufacturer’s test load or catalog entry. The desired speed is different from the vendor’s test speed, and the reliability expectation is typically much higher than the 0.90 accompanying the catalog entry. Figure 11–5 shows the situation. The catalog information is plotted as point A, whose coordinates are (the logs of) C10 and x10 = L10/L10 = 1, a point on the 0.90 reliability contour. The design point is at D, with the coordinates (the logs of) FD and xD, a point that is on the R = RD reliability contour. The designer must move from point D to point A via point B as follows. Along a constant reliability contour (B D), Eq. (11–2) applies: FB x1/a B = FD x1/a D from which FB = FD xD xB 1/a (a) Along a constant load line (AB), Eq. (11–4) applies: RD = exp − xB − x0 θ − x0 b Solving for xB gives xB = x0 + (θ − x0) ln 1 RD 1/b Now substitute this in Eq. (a) to obtain FB = FD xD xB 1/a = FD xD x0 + (θ − x0)(ln 1/RD)1/b 1/a However, FB = C10, so C10 = FD xD x0 + (θ − x0)(ln 1/RD)1/b 1/a (11–6)
Budynas-Nisbett:Shigley's Ill.Design of Mechanical 11.Rolling-Contact T©The McGraw-Hill 559 Mechanical Engineering Elements Bearings Companies,2008 Design,Eighth Edition 558 Mechanical Engineering Design As useful as Eq.(11-6)is,one's attention to keystrokes and their sequence on a hand- held calculator strays,and,as a result,the most common error is keying in the inappropriate logarithm.We have the opportunity here to make Eq.(11-6)more error- proof.Note that 1 In =In- RD 1=n(l+py+=p时=1-Rn 1-Pf where pf is the probability for failure.Equation(11-6)can be written as 71/a C10÷fFD XD Lx0+(0-x0)1-RD)b R≥0.90 (11-7刀 Loads are often nonsteady.so that the desired load is multiplied by an applica- tion factor af.The steady load af Fp does the same damage as the variable load Fp does to the rolling surfaces.This point will be elaborated later. EXAMPLE 11-3 The design load on a ball bearing is 413 Ibf and an application factor of 1.2 is appro- priate.The speed of the shaft is to be 300 rev/min,the life to be 30 kh with a reliability of 0.99.What is the Cio catalog entry to be sought (or exceeded)when searching for a deep-groove bearing in a manufacturer's catalog on the basis of 106 revolutions for rat- ing life?The Weibull parameters are xo =0.02.(-xo)=4.439,and b=1.483. Solution XD= L=60Lnnp=60(30000)300 L10 60LRBR =540 106 Thus,the design life is 540 times the Lio life.For a ball bearing,a=3.Then,from Eq.(11-7). 540 1/3 Answer C10=(1.2)(413) =66961bf 0.02+4.439(1-0.99)1/1.483 We have learned to identify the catalog basic load rating corresponding to a steady radial load Fp.a desired life Lp,and a speed np. Shafts generally have two bearings.Often these bearings are different.If the bear- ing reliability of the shaft with its pair of bearings is to be R,then R is related to the individual bearing reliabilities RA and Rg by R=RARB First,we observe that if the product RARg equals R,then,in general,RA and Rg are both greater than R.Since the failure of either or both of the bearings results in the shutdown of the shaft,then A or B or both can create a failure.Second,in sizing bear- ings one can begin by making RA and Rg equal to the square root of the reliability goal,R.In Ex.11-3,if the bearing was one of a pair,the reliability goal would be 0.99,or 0.995.The bearings selected are discrete in their reliability property in your problem,so the selection procedure "rounds up,"and the overall reliability exceeds the goal R.Third,it may be possible,if RA>R,to round down on B yet have the product RARg still exceed the goal R
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition III. Design of Mechanical Elements 11. Rolling−Contact Bearings © The McGraw−Hill 559 Companies, 2008 558 Mechanical Engineering Design As useful as Eq. (11–6) is, one’s attention to keystrokes and their sequence on a handheld calculator strays, and, as a result, the most common error is keying in the inappropriate logarithm. We have the opportunity here to make Eq. (11–6) more errorproof. Note that ln 1 RD = ln 1 1 − pf = ln(1 + pf +···) . = pf = 1 − RD where pf is the probability for failure. Equation (11–6) can be written as C10 . = FD xD x0 + (θ − x0)(1 − RD)1/b 1/a R ≥ 0.90 (11–7) Loads are often nonsteady, so that the desired load is multiplied by an application factor af . The steady load af FD does the same damage as the variable load FD does to the rolling surfaces. This point will be elaborated later. EXAMPLE 11–3 The design load on a ball bearing is 413 lbf and an application factor of 1.2 is appropriate. The speed of the shaft is to be 300 rev/min, the life to be 30 kh with a reliability of 0.99. What is the C10 catalog entry to be sought (or exceeded) when searching for a deep-groove bearing in a manufacturer’s catalog on the basis of 106 revolutions for rating life? The Weibull parameters are x0 = 0.02, (θ − x0) = 4.439, and b = 1.483. Solution xD = L L10 = 60L DnD 60L RnR = 60(30 000)300 106 = 540 Thus, the design life is 540 times the L10 life. For a ball bearing, a = 3. Then, from Eq. (11–7), Answer C10 = (1.2)(413) 540 0.02 + 4.439(1 − 0.99)1/1.483 1/3 = 6696 lbf We have learned to identify the catalog basic load rating corresponding to a steady radial load FD, a desired life L D, and a speed nD. Shafts generally have two bearings. Often these bearings are different. If the bearing reliability of the shaft with its pair of bearings is to be R, then R is related to the individual bearing reliabilities RA and RB by R = RA RB First, we observe that if the product RA RB equals R, then, in general, RA and RB are both greater than R. Since the failure of either or both of the bearings results in the shutdown of the shaft, then A or B or both can create a failure. Second, in sizing bearings one can begin by making RA and RB equal to the square root of the reliability goal, √ √ R. In Ex. 11–3, if the bearing was one of a pair, the reliability goal would be 0.99, or 0.995. The bearings selected are discrete in their reliability property in your problem, so the selection procedure “rounds up,” and the overall reliability exceeds the goal R. Third, it may be possible, if RA > √R, to round down on B yet have the product RA RB still exceed the goal R