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RELIABILITY ENGINEERING SURVIVABILITY ENGINEERING

RELIABILITY ENGINEERING SURVIVABILITY ENGINEERING. Prof. Dr. Ahmed Farouk Abdul Moneim. COURSE CONTENTS. 1) When systems are considered Operationally Feasible?. 2) What is Reliability?. 3) Why do we study Reliability Engineering?. 4)Failure Data and their analyses.

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RELIABILITY ENGINEERING SURVIVABILITY ENGINEERING

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  1. RELIABILITYENGINEERINGSURVIVABILITY ENGINEERING Prof. Dr. Ahmed Farouk Abdul Moneim

  2. COURSE CONTENTS • 1) When systems are considered Operationally Feasible? • 2) What is Reliability? • 3) Why do we study Reliability Engineering? • 4)Failure Data and their analyses • 5) Failure Rate distributions along Life Time of a Product • 6) Main Concepts in the study of Failure phenomena • 7) Fundamental Expressions of Reliability of a single component • 8) Systems Reliability • 9) Complex Systems Reliability by Monte Carlo Simulation • 10)Reliability of Repairable Systems and introduction to Systems Availability • 11) Reliability in Engineering Design

  3. A System is considered SUCCESSFUL and OPERATIONALLY FEASIBLE If and only if it is: RELIABLE 2) MAINTAINABLE 3) MANABLE 4) SUPPORTABLE

  4. What is RELIABILITY RELIABILTY is the PROBABILITY that a System performs Its FUNCTIONSas INTENDED for a SPECIFIED PERIOD OF TIME in a CERTAIN SPECIFIED ENVIRONMENT ENVIRONMENT : Installation Weather, Altitude, Terrain PreventiveMaintenance Users Transportation Storage Time may be measured Chronologically (seconds, minutes ,…, days months, years) or in Distances, Revolutions,…

  5. WHY DO WE STUDY RELIABILITY ENGINEERING 1) To be able to evaluate RELIABILITY and hence we could CONTROL it 2) To Classify the system components as regards to their CRITICALITY 3) To decide, in the design stage, the size of REDUNDANCY 4) To be able to deploy RELIABILITY as a QUALITY characteristic into the different components of the system 5) To decide the Quantity of SPARE PARTS required by the system to SURVIVE a predefined period of time 6) To be able to design proper WARRANTY Policies 7) To decide proper MAINTENANCE Policies

  6. FUNDAMENTAL EXPRESSIONS OF RELIABILITY

  7. Given NComponents at the beginning of a survival test (t = 0). number of failed components up to time t The Ratio of number of FAILED Components NFto the TOTAL Number N is Is the FAILURE function or the PROBABILITY of Failure of a component in the time period from 0 to time t Is Rate of Change of the FAILURE function with respect to time It is called FAILURE PROBABILITY DENSITY FUNCTION number of Survived motors up to time t The Ratio of number of SURVIVED Components NSto the TOTAL Number N is

  8. The Event of FAILURE and Event of SURVIVAL of a component are Mutually Exclusive EVENTS.Therefore,the SUM of their probabilities equals 1 is the SURVIVAL function (RELIABILITY) or the PROBABILITY of SURVIVAL of a component in the time period from 0 to time t Consider again the Failure Probability Density Function: then This is the Unconditional Failure frequency related to the Initial Number of components under test If on the other hand the Failure Frequency is related to the number of components survived (not failed) at time t, we obtain the Conditional frequency as follows: Multiply denominator and numerator by N, we find Therefore, Is the FAILURE or HAZARD RATE

  9. AN IMPORTANT SPECIAL CASE IF THE FAILURE RATE IS CONSTANT h(t)=λ As examples: All transistors, Diodes, Micro Switches, …

  10. SUMMARY Failure Probability Density Function Failure PDF Failure Rate Failure Probability Function Survival Probability Function or RELIABILITY

  11. Failure Data and their analyses A life test is conducted on N = 200 Transistors for a period of 20 weeks . The test results are given:

  12. GENERAL FORMULA FOR MEAN TIME TO FAILURE The Time To Failure TTF of any component or system is a Random Variable with probability density function f (t) Accordingly, the Mean Time To Failure MTTF can be evaluated as follows: f (t) But t Substituting, MTTF Integrating by parts we get, Then finally,

  13. GENERAL FORMULA FOR MEAN RESIDUAL LIFE t -TO f (t) TO f (t) dt t 0 MRL Prof. Dr. Ahmed Farouk Abdul Moneim

  14. COMPONENTS WITH CONSTANT FAILURE RATE Case 1 We have already shown that In case of CONSTANT Failure Rate Therefore, the probability density function f (t) in case of Constant Failure Rate will be Is known asEXPONENTIAL Distribution This distribution f(t) In this case t

  15. Case 2 COMPONENTS WITH INCREASING FAILURE RATE Example

  16. Case 3 COMPONENTS WITH DECREASING FAILURE RATE Example As a general rule

  17. BATH TUB DISTRIBUTIONandPRODUCT LIFECYCLE

  18. Hazard rate h (t) Infant Mortality Useful Life Wear out DFR IFR CFR time Start of Life Start of commissioning Start of Deterioration End of Life

  19. WEIBULL DISTRIBUTION • It is a Universal distribution • It fits all trends of Failure Rate by changing a single parameter β

  20. Derivation of WEIBULL distribution Consider the following Failure Rate as function of time Failures / unit time As m=0, we get the case of Constant Failure Rate (CFR) As m>0, we get case of Increasing Failure Rate (IFR) As m<0, we get case of Decreasing failure rate (IFR) k is Constant Consider a Characteristic Time η as the time interval during which the mean number of failures is ONE Put m+1 = β, then we find h(t) as follows: As β=1, we get the case of Constant Failure Rate (CFR) As β> 1, we get case of Increasing Failure Rate (IFR) As β < 1, we get case of Decreasing failure rate (DFR)

  21. Failure Rate h (t) β is the shape factor η is the Characteristic Life Probability Density Function f f (t) t

  22. Mean Time To Failure

  23. Variance of Time To Failure

  24. SUMMARY OF WEIBULL Distribution

  25. SOLVED EXAMPLES

  26. Example 1 The pdf of time to failure of a product in years is given • Find the hazard rate as a function of time. • Find MTTF • If the product survived up to 3 years, find the mean residual life MRL • Find the design life for a reliability of 0.9 ___________________________________________________ 1) Find the Reliability R(t) MRL f(t) 0 3 10 t 2) Find the Hazard rate h(t) 3) Find MTTF Since h(t) is Increasing function with time Then the product is in deteriorating h(t) 4) Find MRL t 10 5) Find Design Life at R=0.9

  27. Example 2 Ten V belts are put on a life test and time to failure are recorded. After 1600 hours six of them failed at the following times: 476, 529, 550, 921, 1247 and 1522. Calculate MTTF, reliability of a belt to survive to 900 hours. Design life corresponding to Reliability of 0.94 based on: a) CFR b) Weibull distribution MTTF=(476+529+550+921+1247+1522+4*1600)/10 =1164.495 hrs Based on CFR CONTINUED

  28. (1) Based on Weibull Variance=σ2=[(476-1164.945)2+(529-1164.945)2+(550-1164.945) 2+(921-1164.945)2 +(1247-1164.945)2+(1522-1164.945)2+4*(1600-1164.945)2)/(10-1) =158719.1 (2) Square (1) and divide (2)/(1), we find the following equation in ONE Unknownβ. CONTINUED Prof. Dr. Ahmed Farouk Abdul Moneim

  29. The above equation is TRANSCENDENTAL and could be solved by Trial and Error Method or by Excel GOAL SEEK Using Excel GOAL SEEK, we find: β = 3.2 η will be found from (1) as follows: CONTINUED

  30. TRIAL & ERROR 1 2 2 Required Value =1.117 1.273 3 1.1321 1.1176 3.2 Then β = 3.2

  31. Example 2 Continued Then Notice the difference between Exponential and Weibull Distributions

  32. f(t) F = 0.06 545.16 Time 72 F = 0 .06

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