Quiz #2 A Mighty Fine Review

Transcription

1 Quiz #2 A Mighty Fine Review February 27: A reliable adventure; a day like all days filled with those events that alter and change the course of history and you will be there!

2 What is a Quiz #2? Three problems to work from chapters 5, 6 and 7 Each question pertains to one of the 3 chapters The Chapter 5 question Series and redundant configurations High and low level redundancy K out n redundancy The Chapter 6 question Standby and load-sharing systems The Chapter 7 Question Static reliability (cases 1,2 and 3) Dynamic reliability (periodic or random loads) What a great quiz!

3 A chapter 5 question An electronic subassembly is composed of two primary components (failure modes); a rectifier (component A) and an inducer (component B), that operate in series. The two components have the following reliability functions: t 3 0.2x10 t 7000 B R () t = e ; R () t = e A Based upon the reliability at 10 3 operating hours, which of the following configurations is preferred? (a) Two subassemblies operating in parallel (high-level active redundancy with 2 A-B s ). R A (1000) =.8187 ; R B (1000) =.9703; Rsys(1000) = (.8187) (.9703) =.7944 R redund (1000) = 1 ( ) 2 =

4 A chapter 5 question R A (1000) =.8187 ; R B (1000) =.9703; (b) One subassembly configured in low-level active redundancy with 2 A s and 2 B s. R redund (1000) = [1 ( ) 2 ] [1 ( ) 2 ] =.9663

5 A chapter 5 question (c) A 2 out of 4 system having four identical subassemblies with component reliabilities as given in part (a). Rsys(1000) = (.8187) (.9703) =.7944 R(1000) =.7944, N = 4, k = 2; R 2/4 (1000) =.9706 Component Reliability = System Reliability N = 4 k<=n k =

6 A chapter 5 question (d) Both component A and B operating as a k out of n redundant system where for component A, N = 4 and k = 2; and for component B, N = 6 and k = 5 With no redundancy, R A (1000) =.8187 ; R B (1000) =.9703 R A 2/4 (1000) =.9794 and R B 5/6 (1000) =.9878 Rsys(1000) = (.9794) (.9878) =.9675 Component Reliability = System Reliability N = 4 k<=n k = Component Reliability = System Reliability N = 6 k<=n k =

7 A chapter 6 question A subassembly is composed of two components, component A and component B, which operates in series and has constant failure rates in failures per year. Find the system reliability at the end of the second year for each configuration described below. Find the MTTF. λ A =.2 λ B =.15 λ s = =.35; R(2) = e -2(.35) =.4966; MTTF = 2.857

8 A chapter 6 question (a) Two identical subassemblies configured as a standby system with standby component A and B failure rates of.05 and.03 respectively. (High level standby redundancy) λ =.35 and λ - =.08; R(2) = and MTTF = 5.18 standby with switching failures section 6.3 λ1 λ2 λ2 switching failure - p enter: intermediate λ1+(λ2-) λ1+(λ2-) -λ2 λ1+(λ2 ) results t P1(t) P2(t) P3(t) R(t) MTTF

9 A chapter 6 question (a) One subassembly with two component A s and two component B s. One component A and one component B are in standby with standby failure rates of.05 and.03 respectively. (Low level standby redundancy) Comp. A standby: λ =.2 and λ - =.05; R(2) = Comp. B standby: λ =.15 and λ - =.03; R(2) = ; R sys = (0.9254) (0.9565) =.8851 λ1 λ2 λ2 switching failure - p enter: t P1(t) P2(t) P3(t) R(t) λ1 λ2 λ2 switching failure - p enter: t P1(t) P2(t) P3(t) R(t)

10 A chapter 6 question (b) Two identical subassemblies operating as a load-sharing system with shared-load component failure rates of.12 and.09 (components A and B respectively). (High level load-sharing redundancy) λ =.21 and λ + =.35; R(2) = and MTTF = λ1 λ2 λ1+ λ2+ enter: intermediate λ1+λ2 λ1+λ2 λ1+ λ1+λ2 λ2+ results t P1(t) P2(t) P3(t) R(t) MTTF

11 A chapter 7 question Complete the following table: Stress (X) Strength (Y) Frequency of Load Static Reliability R a. Constant = 600 psi b. Exponential with mean = 120 volts c. lognormal with median = 300 lbs. s =.45 Weibull with beta = 2.1 and theta = 2500 psi Exponential with mean = 2500 volts constant = 1200 lbs. Random (Poisson) - averaging once every 300 days Periodic every other day Random at a mean rate of 2 per day Dynamic reliability R(100 days)

12 A chapter 7 question part (a) Table 7.2 Random Strength and Constant Stress load frequency is once every Constant Stress s = 600 Strength Distribution Exponential μ Weibull β θ Normal μ σ value 100 value value Static Reliability Dynamic periodic random periodic random periodic reliability loads loads time periods loads loads time periods loads time MTTF = MTTF = periods random loads MTTF = 300 time period s

13 A chapter 7 question part (b) Table 7.2 Random Stress and Strength load frequency is once every 2 Stress Distribution Exponential μ value 120 Strength Distribution Exponential μ value 2500 Static Reliability Dynamic periodic loads random loads reliability time periods time periods MTTF =

14 A chapter 7 question part (c) Table 7.2 Random stress and Constant Strength Stress Constant Strength k =1200 load frequency is once every 0.5 time periods Distribution Lognormal m s value Static Reliability Dynamic reliability periodic loads time periods random time loads periods MTTF =

15 A chapter 7 question Stress (X) Strength (Y) Frequency of Load Static Reliability R a. Constant = 600 psi b. Exponential with mean = 120 volts c. lognormal with median = 300 lbs. s =.45 Weibull with beta = 2.1 and theta = 2500 psi Exponential with mean = 2500 volts constant = 1200 lbs. Random (Poisson) - averaging once every 300 days Periodic every other day Random at a mean rate of 2 per day Dynamic reliability R(100 days)

16 Let s do another problem A Component has three parts in series having the distributions shown. Version c Part Distribution Stand-alone & active Failure rates (λ s) in years standby load-sharing A Exponential B Exponential C Exponential system Exponential

17 High Level a. If there are two of each of the three parts (2-A s, 2-B s, and 2-C s), determine the MTTF and component reliability at t = 4 years if the component is configured in high level redundancy and operates as a standby system and as a load-sharing system. Assume no switching failures.

18 High Level standby with switching failures section 6.3 λ1 λ2 λ2 switching failure - p enter: t P1(t) P2(t) P3(t) R(t) MTTF

19 High Level Load-Sharing - section 6.2 λ1 λ2 λ1+ λ2+ ver a enter: t P1(t) P2(t) P3(t) R(t) MTTF

20 Low-Level b. Compute the component reliability if the parts are configured in low level redundancy (that is each part operates as a standby system and as a load-sharing system).

21 The answers-you work the problem standby-system load-sharing Rsys(4) MTTF Rsys(4) MTTF a. high level redundancy b. low level redundancy Part A Part B Part C

22 Yet another one An aircraft must fly a 9-hour mission. (a) The aircraft has 3 engines. The two front engines, mounted under each wing, each has a failure rate of.0039 failures per flying hour when all three engines are operating and each has a failure rate of.02 failures per flying hour when the rear engine has failed. If either of the two front engines fail, then the other front engine must immediately be shut down otherwise the aircraft cannot be stabilized in flight. The rear engine has a failure rate of.0054 failures per flying hour when all three engines are operating and a failure rate of.012 failures per flying hour when the front engines have shut down. Find R(9) and the MTTF of the engine system for this situation.

23 Yet another one (a) Load sharing system with the two front engines in series. Therefore λ = 2(.0039) =.0078, λ =.0054, λ = 2(.02) =.04, λ = Load-Sharing - section 6.2 λ1 λ2 λ1+ λ2+ enter: intermediate λ1+λ2 λ1+λ2 λ1+ λ1+λ2 λ2+ results t P1(t) P2(t) P3(t) R(t) MTTF

24 Yet another one (b) The aircraft has 3 engines. At least two of the three engines must operate for the aircraft to complete its mission. Each engine has a failure rate of.0054 failures per flying hour. Find R(9) and the MTTF of the engine system for this situation. Answer: k out of n system where k = 2, n = 3, R = R(9) = exp( x 9) = Component Reliability = System Reliability N = 3 k<=n k = MTTF = + =

25 Yet another one (c) The aircraft has four engines two under each wing. The two outer engines (from the fuselage) each has a failure rate of.0054 failures per flying hour while the two inner engines each has a failure rate of.0039 failures per flying hour. The aircraft can complete its mission only if either the two outer engines are operating or the two inner engines are operating. Find R(9) and the MTTF of the engine system for this situation. high level redundant system: ( ) ( ) R = e =.9074; R = e =.9322 R outer eng sys (9) (9) inner eng = 1 (1.9074)(1.9322) =.9937 ( ( ) ) ( ) ( ) ( ) ( ) ( ) R = 1 1 e 1 e = e + e e sys MTTF t t t t t = + = ( ) ( ) ( + )

26 Yet another one (d) The aircraft has four engines two under each wing. The two outer engines (from the fuselage) each has a failure rate of.0054 failures per flying hour while the two inner engines each has a failure rate of.0039 failures per flying hour. The aircraft can complete its mission only if at least one engine on each side is operating. Find R(9) of the engine system for this situation. (d) low level redundancy: (.0039)( 9 ) (.0054)( 9) R = e =.9655; R = e =.9526 inner eng outer eng R sys = 1 ( )( ) =

27 Yet another one (e) The aircraft has four engines two under each wing. The two outer engines (from the fuselage) each has a failure rate of.0054 failures per flying hour while the two inner engines each has a failure rate of.0039 failures per flying hour. The aircraft can complete its mission only if all four engines are operating. Find the MTTF of the engine system for this situation. 4 engines in series: λ s = 2(.0054) + 2(.0039) = MTTF = = (e) R(9) =.8459 MTTF = 53.76

28 The Answers (a) (b) (c) (d) (e) R(9) MTTF I got all the questions correct!

29 An integral problem An important component is manufactured by the Stress n Strength Company. (a) If the component has a fixed design strength of 40 psi and will be subjected to a random stress having the following probability density function (PDF), what is its static reliability, R? 1 fx ( x) =, 0 x Answer: 40 F x (40) = =.80 50

30 An integral problem (b) If the component is subjected to a fixed stress of 25 psi and has a design strength that is a random variable with the following probability density function (PDF), find its static reliability R. 2y f ( y) =, 0 x 50 y ( 50) y 25 3 answer : Fy( y) = ; 1 F (25) = 1 = = y ( 50) ( 50)

31 A double integral problem (c) If both the design strength and the applied stress are random variables having the above distributions, what is the static reliability, R? 50 y y y 2y 2y 2 R = dx dy = dy = = = ( ) Students happy to work a double integral problem

32 safety factor (d) If the safety factor (SF) is defined as the mean strength divided by the mean stress, what is the safety factor for the component in (c)? u x u y 50 = = y 2y 2 = dy = = 2 2 ( 50) = ( 50) 3( 50) 50 0 SF = =

33 Use the formula (e) If both stress and strength have exponential distributions with means as determined in (d), what is the static reliability? u y R = = = u + u y x.5714

34 Finally (f) If the component discussed in (b) is replaced at the mean rate of once every 6 months, Poisson distributed, and subjected to the fixed stress of 25 psi, what is the reliability at 2 years and what is the MTTF? ( ) α ( ) Rt () = e = e 1 R t t α = 2 / yr.; R =.75 ( 1.75)( 2)( 2) R(2) = e =.3678 MTTF 1 = = ( )( ) 2 yr.

35 Reliability students excited about the second quiz