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Fundamentals of reliability engineering and applications part3of3

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This is a three parts lecture series. The parts will cover the basics and fundamentals of reliability engineering. Part 1 begins with introduction of reliability definition and other reliability …

This is a three parts lecture series. The parts will cover the basics and fundamentals of reliability engineering. Part 1 begins with introduction of reliability definition and other reliability characteristics and measurements. It will be followed by reliability calculation, estimation of failure rates and understanding of the implications of failure rates on system maintenance and replacements in Part 2. Then Part 3 will cover the most important and practical failure time distributions and how to obtain the parameters of the distributions and interpretations of these parameters. Hands-on computations of the failure rates and the estimation of the failure time distribution parameters will be conducted using standard Microsoft Excel.
Part 3. Failure Time Distributions
1.Constant failure rate distributions
2.Increasing failure rate distributions
3.Decreasing failure rate distributions
4.Weibull Analysis – Why use Weibull?

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  • 1. Fundamentals of Reliability  Fundamentals of Reliability Engineering and Applications Part 3 of 3 E. A. Elsayed ©2011 ASQ & Presentation Elsayed Presented live on Dec 14th, 2010http://reliabilitycalendar.org/The_Reliability_Calendar/Short_Courses/Shliability Calendar/Short Courses/Short_Courses.html
  • 2. ASQ Reliability Division  ASQ Reliability Division Short Course Series Short Course Series The ASQ Reliability Division is pleased to  present a regular series of short courses  featuring leading international practitioners,  academics, and consultants. academics and consultants The goal is to provide a forum for the basic and  The goal is to provide a forum for the basic and continuing education of reliability  professionals.http://reliabilitycalendar.org/The_Reliability_Calendar/Short_Courses/Shliability Calendar/Short Courses/Short_Courses.html
  • 3. Fundamentals of Reliability Engineering and Applications E. A. Elsayed elsayed@rci.rutgers.edu Rutgers University December 14, 2010 1
  • 4. Outline Part 1. Reliability Definitions Reliability Definition…Time dependent characteristics Failure Rate Availability MTTF and MTBF Time to First Failure Mean Residual Life Conclusions 2
  • 5. Outline Part 2. Reliability Calculations1. Use of failure data2. Density functions3. Reliability function4. Hazard and failure rates 3
  • 6. Outline Part 2. Reliability Calculations1. Use of failure data2. Density functions3. Reliability function4. Hazard and failure rates 4
  • 7. OutlinePart 3. Failure Time Distributions1. Constant failure rate distributions2. Increasing failure rate distributions3. Decreasing failure rate distributions4. Weibull Analysis – Why use Weibull? 5
  • 8. Empirical Estimate of F(t) and R(t)When the exact failure times of units is known, weuse an empirical approach to estimate the reliabilitymetrics. The most common approach is the RankEstimator. Order the failure time observations (failuretimes) in an ascending order: t 1 ≤ t 2 ≤ ... ≤ t i −1 ≤ t i ≤ t i +1 ≤ ... ≤ t n −1 ≤ t n 6
  • 9. Empirical Estimate of F(t) and R(t)F (ti ) is obtained by several methods i1. Uniform “naive” estimator n i2. Mean rank estimator n +1 i − 0.33. Median rank estimator (Bernard) n + 0. 4 i −3/84. Median rank estimator (Blom) n +1/ 4 7
  • 10. Exponential Distribution F (t ) = R(t ) = exp [ −λt ] 1− 1− 1- F (t ) exp [ −λt ] = 1 = exp [ λt ]1 − F (t ) 1ln = λt 1 − F (t ) 1⇒ ln ln ln t + ln λ = 1 − F (t )= ln λ + ln tyy= a + bx 8
  • 11. Median Rank Calculationsi t (i) t(i+1) F=(i-0.3)/(n+0.4) R=1-F f(t) h(t)0 0 70 0 1 0.0014 0.00141 70 150 0.067307692 0.9327 0.0013 0.00142 150 250 0.163461538 0.8365 0.001 0.00133 250 360 0.259615385 0.7404 0.0009 0.00134 360 485 0.355769231 0.6442 0.0008 0.00135 485 650 0.451923077 0.5481 0.0006 0.00126 650 855 0.548076923 0.4519 0.0005 0.00127 855 1130 0.644230769 0.3558 0.0004 0.00128 1130 1540 0.740384615 0.2596 0.0002 0.00129 1540 - 0.836538462 0.1635 9
  • 12. Failure Rate Failure Rate 0.002 0.0016Failure Rate 0.0012 0.0008 0.0004 0 0 1 2 3 4 5 6 7 8 9 Time 10
  • 13. Probability Density Function Probability Density Function 0.0016 0.0014 0.0012Probability Density Function 0.001 0.0008 0.0006 0.0004 0.0002 0 0 1 2 3 4 5 6 7 8 9 Time 11
  • 14. Reliability Function Reliability Function 1.2 1 0.8Reliability 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 Time 12
  • 15. Exponential Distribution: Another ExampleGiven failure data:Plot the hazard rate, if constant then use theexponential distribution with f(t), R(t) and h(t) asdefined before.We use a software to demonstrate these steps. 13
  • 16. Input Data 14
  • 17. Plot of the Data 15
  • 18. Exponential Fit 16
  • 19. Exponential Analysis
  • 20. Weibull Model • Definition β −1 βt   t β = exp  −    β > 0, η > 0, t ≥ 0 η η  f (t )    η     β −1  t β  βt  =(t ) λ (t ) 1− F = f= η η  R(t ) =−   exp  (t ) / R(t )  η       18
  • 21. Weibull Model Cont.• Statistical properties ∞ 1 MTTF = η ∫0 t1/ β e−t dt = ηΓ(1 + β )   2 2 2 1   Var η Γ(1 + ) −  Γ(1 + )  =   β  β        Median life = η ((ln 2)1/ β ) 19
  • 22. Versatility of Weibull Model β −1 βtHazard rate: =λ (t ) f= η η  (t ) / R(t )  Hazard Rate Constant Failure Rate Region β >1 0 < β <1 Early Life Wear-Out Region Region β =1 0 Time t 20
  • 23. Weibull Model 21
  • 24. Weibull Analysis: Shape Parameter 22
  • 25. Weibull Analysis: Shape Parameter 23
  • 26. Weibull Analysis: Shape Parameter 24
  • 27. Normal Distribution 25
  • 28. Failure Data Table 1: Failure Data Design A Design BSample # Cycles Sample # Cycles 1 726,044 11 529,082 2 615,432 12 729,957 3 508,077 13 650,570 4 807,863 14 445,834 5 755,223 15 343,280 6 848,953 16 959,903 7 384,558 17 730,049 8 666,686 18 730,640 9 515,201 19 973,224 10 483,331 20 258,006 26
  • 29. Linearization of the Weibull Model   t β  F (t ) =R (t ) =exp  −    1− 1−  η       t β  1- F = exp  −    (t )  η     Taking the log   t β  ln (1- F (t )) =  −     η     1 ⇒ ln ln = ln η β ln t − β 1 − F (t ) 27
  • 30. Linearization of the Weibull Model 1⇒ ln ln = ln η β ln t − β 1 − F (t )This is an equation of straight liney = a + bxUse linear regression , obtain a and b by solving∑= n a + b ∑ x y∑ xy a ∑ x + b ∑ x= 2or by using Excel 28
  • 31. Calculations using Excel• Weibull Plot   t β F (t ) = R(t ) =exp  −    1− 1−  η     1⇒ ln ln = ln η β ln t − β is linear function of ln(time). 1 − F (t ) ˆ• Estimate F (ti ) at ti using Bernard’s Formula For n observed failure time data (t1 , t2 ,..., ti ,...tn ) ˆ (t ) = i − 0.3 F i n + 0.4 29
  • 32. Linearization of the Weibull ModelDesign A Median Cycles Rank Ranks 1/(1-Median Rank) ln(ln(1/(1-Median Rank))) ln(Design A Cycles)384,558 1 0.067307692 1.072164948 -2.663843085 12.8598499483,331 2 0.163461538 1.195402299 -1.72326315 13.088457508,077 3 0.259615385 1.350649351 -1.202023115 13.13838829515,201 4 0.355769231 1.552238806 -0.821666515 13.15231239615,432 5 0.451923077 1.824561404 -0.508595394 13.33007974666,686 6 0.548076923 2.212765957 -0.230365445 13.41007445726,044 7 0.644230769 2.810810811 0.032924962 13.4953659755,223 8 0.740384615 3.851851852 0.299032932 13.53476835807,863 9 0.836538462 6.117647059 0.593977217 13.60214777848,953 10 0.932692308 14.85714286 0.992688929 13.6517591 30
  • 33. Linearization of the Weibull Model Line Fit Plot 1.5 1ln(ln(1/(1-Median Rank))) 0.5 0 -0.5 -1 -1.5 -2 -2.5 -3 12.8 13 13.2 13.4 13.6 13.8 ln(Design A Cycles) 31
  • 34. Linearization of the Weibull ModelSUMMARY OUTPUT Regression StatisticsMultiple R 0.98538223R Square 0.97097815Adjusted R Square 0.96735041Standard Error 0.20147761Observations 10ANOVA df SSRegression 1 10.86495309Residual 8 0.324745817Total 9 11.18969891 β ln η Coefficients Standard ErrorIntercept -57.1930531 3.464488033ln(Design A Cycles) 4.2524822 0.259929377 Beta (or Shape Parameter) = 4.25 Alpha (or Characteristic Life) = 693,380 32
  • 35. Reliability Plot 1.0000 .9000 .8000Survival Probability .7000 .6000 .5000 .4000 .3000 .2000 .1000 .0000 0 200,000 400,000 600,000 800,000 1,000,000 1,200,000 1,400,000 Cycles
  • 36. Input Data
  • 37. Plots of the Data
  • 38. Weibull Fit
  • 39. Test for Weibull Fit
  • 40. Parameters for Weibull
  • 41. Weibull Analysis
  • 42. Example 2: Input Data
  • 43. Example 2: Plots of the Data
  • 44. Example 2: Weibull Fit
  • 45. Example 2:Test for Weibull Fit
  • 46. Example 2: Parameters for Weibull
  • 47. Weibull Analysis
  • 48. Versatility of Weibull Model β −1 βtHazard rate: =λ (t ) f= η η  (t ) / R(t )  Hazard Rate Constant Failure Rate Region β >1 0 < β <1 Early Life Wear-Out Region Region β =1 0 Time t 46
  • 49. Graphical Model Validation• Weibull Plot   t β F (t ) = R(t ) =exp  −    1− 1−  η     1⇒ ln ln = ln η β ln t − β is linear function of ln(time). 1 − F (t ) ˆ• Estimate F (ti ) at ti using Bernard’s Formula For n observed failure time data (t1 , t2 ,..., ti ,...tn ) ˆ (t ) = i − 0.3 F i n + 0.4 47
  • 50. Example - Weibull Plot • T~Weibull(1, 4000) Generate 50 data Weibull Probability Plot 0.99 0.96 0.90 0.75 0.50 0.632Probability 0.25 If the straight line fits 0.10 β the data, Weibull distribution is a good 0.05 model for the data 0.02 0.01 10 -5 10 0 η 10 5 48 Data