AGMA 07FTM19-2007 How to Determine the MTBF of Gearboxes《如何确定齿轮箱的MTBF》.pdf

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1、07FTM19How to Determine the MTBF of Gearboxesby: Dr. G.G. Antony, Neugart USA LPTECHNICAL PAPERAmerican Gear Manufacturers AssociationHow to Determine the MTBF of GearboxesDr. Gerhard G. Antony, Neugart USA LPThe statements and opinions contained herein are those of the author and should not be cons

2、trued as anofficial action or opinion of the American Gear Manufacturers Association.AbstractMean Time Between Failures (MTBF) became a very frequently used value describing reliability ofcomponents,assemblies,andsystems. MTBFwasoriginallyintroducedandusedmainlyinconjunctionwithelectroniccomponents,

3、andsystems.ThedefinitionandapplicationoftheMTBFformechanicalcomponentsis not broadly available, used, or recognized. There is no clear standard which would describe how todetermineforinstancetheMTBFofagearbox.Ontheotherhand,moreandmorecustomersandconsultantsare requesting an MTBF number from the man

4、ufacturer for mechanical devices, such as a gearbox.Electronic components of modern drive-trains generally consist of an electronic controller and electronicdrive,motorandgearbox.Thedrive/controllermanufacturernormallyhasanMTBFvalue,sodomeanwhilesomemotormanufacturerssupplyMTBFvaluesfortheirproduct.

5、InthefieldofgearsitisdifficulttoobtainanMTBF from the manufacturer due to the lack of applicable, generally recognized definitions and standards.The paper shall evaluate, compare and suggest ways in determining a gearbox MTBF based on the alreadyestablished, proven, design, calculation standards and

6、 test methods used in the gear design.Copyright 2007American Gear Manufacturers Association500 Montgomery Street, Suite 350Alexandria, Virginia, 22314October, 2007ISBN: 978-1-55589-923-31How to Determine the MTBF of GearboxesDr. Gerhard G. Antony Neugart USA LPIntroductionInthepastdecades,“MeanTimeB

7、etweenFailures”(MTBF) has become a very frequently and broadlyused characteristic measure of reliability forcomponents, systems and devices used mainly inconjunction with electrical and electronicequipments.From the engineering point of view, assessing thelifeandreliabilityofproductsisavitalpartofpr

8、oductdesign, development, and selection. Life and reli-ability of a product are also important characteris-tics for the user (customer) in comparing the wid-gets (gearboxes) to assess their useful value or lifefor a certain application. The reliability of aproductbecomes a frequently used marketing

9、and salesfeature.The life characteristics of different products andcomponentsdependof awiderangeof factorsfromtypeandconditionofmaterialtotypeof exposuretoloads,magnitudeoftheloadsandothereffectssuchas environment. Products are designed for a cer-tain purpose, function, duty, load etc. The life andr

10、eliability is a characteristic statistical value and itcanbeonlyapproachedandassessedbystatisticalmethods.Todayanincreasingnumberofmanufacturerscom-bine a mechanical device, such as a gearbox, withan electromechanical, such as an electric motor,andelectronic drives, logic controllers sensor intoacom

11、pact integrated “mechatronics” product. The(Mean Time Between Failures) MTBF value ofmany electronic components and systems is ob-tainable from the manufacturer. The design life ofmechanical components and systems is mainlybasedontheendurancecharacteristicsof thecom-ponents on the statistical life e

12、xpectancy under acertainloadsuchastheL10designlife. Canthereli-ability of amechanicaldevicesuchas agearbox beexpressed in terms of MTBF? How many hoursMTBF equals a known L10 life to? The here pre-sentedpaperisanattempttofindsomeanswersforthese and similar MTBF related questions.Lifeandreliabilityre

13、latedissuesarebasedontests,observation of populations of “widgets” applyingmathematical (statistical) evaluations, approxima-tions,usingappropriatefunctionsandformulas. Itiscommon practice to define characteristic values,and choose “scientific” statistical methods to ana-lyze and evaluate them. It i

14、s fairly easy to definecertain characteristics of selected number of testspecimens apply statistical methods on anobserved population/group and come up with a“scientific statistical conclusion”. However to haverealistic,meaningfulconclusionsthedefinitionsandthe evaluation methods must be clear and t

15、ranspa-rent. Thispaperisanattempttoexpressthelifeandreliability of gearboxes in terms of MTBF withoutgoing into an in-depth discussion of statisticalmethods. As mentioned MTBF is a widely usedcharacteristicvaluetoquantifyreliabilityofelectron-ic components and systems, but it is not commonlyused for

16、 reliability assessment of mechanicalcomponents and systems.ForthecorrectinterpretationoftheMTBFvalueitisnecessary to understand some basic concepts ofprobability of failures and the methods of theirevaluation.Failure rateThe fundamental first step in determining thereliability and life of a widget

17、is to observe a repre-sentative set of samples, a “population” of widgetsand record the failures over a certain time frame.Thecollecteddatawillshowacertainnumberoffail-ures over an observed time period. An absolutenumber of failures has no real practical meaning; italways has toberelatedtotheobserve

18、dpopulationsize. This is expressed by the failure rate.Failurerateistherelativefrequencyatwhichacom-ponent or system fails in a given time i.e., failures/minutes, hours, years or within a certain time re-lated measure such as distance i.e., failures/miles(inautomotivefield),orperoperatingcyclessucha

19、sfailuresin1millionrevolutions(bearings)etc. Aswecan see, it is not an absolute number of units failedbut rather a relative number, related to the size ofthe observed tested number of units, population ofproducts, see figure 1.2Figure 1. Failure rate, FRIt would be more exact to call it the “relativ

20、e failurerate”; because the value is related to the overallobservedpopulation, it assumes avaluebetween0(0% failures) and 1 (100% failures per hour). Thisrelative failure rate can be recorded at regular timeintervals (important inwherewewant findout if andhowthefailurerateis changingover thelifetime

21、)orjustrecorditforapredefinedperiodsuchasthe“de-signlife” thewidget, - this wouldmean, weassumethat the failure rate is constant during this period.Example iPod:(- source: “AppleInside”July272006reporting,un-der headline “iPods built to last 4 years” )“Apple spokeswoman Natalie Kerris recentlytold t

22、he Chicago Tribune that iPods have a fail-urerateof less then5%, whichshesaidis”fairlylow”comparedwithotherconsumerelectronics.However, a survey conducted by “Macintouch”last year found that out of nearly 9,000 iPodsowned by more than 4,000 respondents, morethan1,400oftheplayershadfailed. Thesurveyc

23、oncluded that the failure rate was 13.7 %,stemming from an equal mix of hard drive andbattery related issues”Remark:Basedonthenumbers,theactualfailurerateshouldyieldtoFR=1400/9000=0.155or15.5%fortheob-served time interval, which can be interpreted that15.55% 13.7% = 1.8% failed for some other rea-so

24、n than the listed. Other explanations are alsopossible however the survey does not list any rea-sons.Is Mrs. Kerris from Apple correct with the 5% valueoverthedesignlifeoraretheconclusionsofthesur-vey with 13.7% correct?Here are some important questions before we takesides:Are both talking about the

25、 same type of failures?Does Apple consider the necessity to replace thebatteryafailure? HowmanyunitsweresurveyedbyApple? Is the population of 9000 samples in thesurveyrepresentative? (Appleshipsinexcessof10million units a quarter.) What was the real usagetime or reference time of both observations?

26、Doesthe statement “built to last 4 years” means 4 years24/7usage,orforinstanceonly4hoursadayusagesay 6 days a week (duty cycle)?None of the above two failure rates gives anyspecific information about these important basicfactors, neither about the conditions under whichthe data were collected. Let u

27、s assume that thesurveywas basedona12hourdaily usageover a2year period, the failure rate, should be calculatedas:3= 0.000017757hourFR =1400 failures9000 units (2 years 365 days(12 hrs)The example above highlights the main difficulty ofthereliability/lifecalculationandthemainsourceformisrepresentatio

28、nsnamelytheselectionofthepop-ulation,theobservedtimeintervalandfailuremode,etc.Example K-gearbox:The right angle bevel-helical K-boxes have 2years warranty200000 K boxes are shipped/yearWarranty repair, return, failure 1%(200000 2 0.01) = 4000Estimatedaverageoperatinghours 8hours/day= 0.000001712hou

29、rFR =4000 warranty returns400000 units (2 yr warranty) (365 days) (8hrs)Electronic components and system such as a sim-pleLEDorcomplexprocessorchipsareusedinmil-lions of computers or other devices under exactlydefined and controlled conditions (certain voltageand clock frequency rate, temperature et

30、c.) On theother hand, gearboxes are subjected to a widerange,andfarlesscontrolledconditions, loads,andenvironments. Also the population size is substan-tiallyhigherforelectroniccomponentsthanforgear-boxes. Electronic components are routinely labtested in high volumes. It is economically impracti-cal

31、tolifetestalargepopulationofgearboxesoroth-er mechanical components, securing exact sameload conditions etc., to find the mortality rate. Fail-ure rate values result mainly from “field tests” i.e.,observations in real world applications.On the other hand, gearboxes are designed basedon well-establis

32、hed, statically proven methodsfrequently regulated by standards such as AGMA,ISO, DIN, etc. Many components such as the vitalbearings have well known statistically provenreliability, and lifecharacteristics. The shafts gearsfasteners etc are all based on the endurance limithence there is theoretical

33、ly no life limitation underthe nominal load. This issue will be discussedfurther in relation to the two proposed methods ofgauging gearbox MTBF.Bathtub curveMost components follow the characteristic plot offailure rate over time, (shown in figure 2).Figure 2. Bathtub curve4Theplotofthefailurerateove

34、rtimeformostengine-ered components and systems resembles theformofabathtub,hencethename“bathtubcurve”. Ithas3 characteristic areas; the “infant mortality” periodwith decreasing failure rates, followed by a almostflat, nearly constant failure rate period frequentlycalledthe“usefullifeperiod”andfinall

35、ythethirdpartwith increasing failure rate the “wear out”. The fail-urerateoflivingcreatures,suchashumansalsore-sembles the bathtub curve.Electronic components have a very distinctive in-fantmortality;tominimizethisimpactonthereliabil-ity in practical applications, electronic componentsare frequently

36、 subjected to a “burn in” which sepa-rates the early failures from the population. On theother hand, the wear out of solid state electroniccomponents is far less significant.Mechanical components, such as gears and gear-box-components behave differently. There is nosignificant infant mortality; howe

37、ver, the wear outcan be significant. For obvious reasons, in mostpractical applications, the useful life period is of in-terest, not the reliability during the infant mortalityperiodorduringtheperiodexceedingthedesignlifenamely the wear out period.Probability density function, PDF, andcumulative dis

38、tribution function, CDFTheprobabilityofanoccurrence,heretheprobabili-ty of a certain failure rate, is mathematically de-scribed, approximated and analyzed by defining asuitable Probability Density Function, PDF. Themost common and well known PDF is the normalprobability distribution (Gauss distribut

39、ion) applica-ble to many natural phenomena (Figure 3). Thearea under the PDF i.e., the integral of the PDF iscalled the Cumulative Distribution Function, CDF.However the Gauss normal distribution function isnot applicable to “the bathtub curve type” distribu-tions (figure 4).Whereasthenormalprobabil

40、itydistributionfunctionhas the same basic shape for all parameters, theWeibull3or2parameterdistributionfunctionallowsto ”model”(describe) widely different shapes ofPDFs depending upon the shape parameter .Weibull is well known to us “gear-people” familiarwith the bearing design and associated “B-lif

41、e” ra-tings. Mr. Waloddi Weibull suggested bearingsshould be compared at a life corresponding to 10%failure probability, the B10 (L10) life.F(t) = 1 etor R(t) = etFigure 3. Gauss normal probability distribution function5Figure 4. Weibull probably distribution functionF(t) the Weibull cumulative dist

42、ribution functionCDF(herethewidelyused2parameterdistribution)providestheprobabilityoffailure. R(t)is“reliability”,the complement of F(t) where:t failure time, characteristic life,or scale factor shape parameter or slopee EulersnumberorNapiersconstant(thebase for natural logarithms)For the 3 characte

43、ristic areas of a bath tub curve- theinfantmortality,decreasingfailurerateofthebathtub curve, corresponds to beta values 1.In the Weibull probability plot which is using an ad-justed logarithmic scale, the distribution functions,have the shape of a simple line where the slope isequal to the paramete

44、r .Furthermore at a time t = , 63.21% of the popula-tion will fail independent from the value, sinceF(t) t= !1-1/e = 0.6321.In the Weibull plot the horizontal line at 0.6231 fail-ureratehasaspecialmeaning(figure5). Forfailureprobability distributions with =1, the t value corre-spondingtotheintersect

45、ionpointoftheF(t)lineandthehorizontal“0.6321line”canbeinterpretedasthemean time between failures. Note this is only cor-rect when =1i.e., constant failurerateusefulliferegionwhichisthescopeofmostpracticalconsider-ations.AlsonoteF(t) = 63.21 %failureprobability meansR(t) = 36.78% survival probability

46、6Figure 5. Weibull plotMean Time Between Failures, MTBFItshouldbeemphasizedthat inallpracticalapplica-tions of widgets the reliability duringthe “usefullife”(alsocalledthedesignlife)iswhatmatters. Thispe-riod is characterized by =1 in the Weibulldistribution.ThebasicdefinitionofMeanTimeBetweenFailur

47、esissimpleandlogical,justcompareittothedefinitionoffailurerate,FR. TheMTBFistheactuallyrecipro-cal value of the FR MTBF = 1/FR (figure 6.)Figure 6. MTBF7Lets calculate the MTBF for the 2 examples pre-sented above.Example iPod:TheAppleiPodexample= 56 314hrsMTBF =9000 2 years 365 days 12 hrs1400 failu

48、resObviously we cannot expect that an iPod will last56314 hrs equivalent to over 6 years uninterruptedoperation.Example K-Gearbox:Right angle helical bevel K-boxes have 2 yearswarranty200000 K boxes are shipped/year Warrantyrepair, return, failure 1% i.e., 200000 unit 2 years 1% = 4000 unitsEstimate

49、d average operating hours: 8 hours/day= 584 000 hrsMTBF =(200000 units)(2 yrs)(2 yrs warranty)(365 days)(8 hrs)4000 warranty MTBFHere again the expectation that a K-box will lastabout 67 years under continuous operation wouldbe a false interpretation of the MTBF value.Comparing the 2 values we can certainly say theK-box is about 10timesmorereliablethenaniPod.The above examples calculated the MTBF basedonfieldsurveydatausinganumberofassumptions.As pointedout intheparagraphabout FailureRate,the populating size, the