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University of California, Los Angeles - ENGR 202solution-manual-an-introducing-to-reliability-and-engineering

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CHAPTER 11 11.1 a) A r r r t e t r t 2 2 0 2 2 1 − − + = + + + − → λ λ λ λ b g b g ( ) A e 30 2 1 1)30 1 02 02 1 02 30 = 1 8 + + + − = . − + . . . ... . . . (.02 . b g b g 784 A r r = A + → = + = λ . . . . 1 1 02 8333 b) A P P r r r = + = + + L P MN OPQ + − 1 2 1 1 2 2 1 1 1 λ λ λ λ 1 A = + + LMN OPQ + + + LMN OPQ = + = − − 1 02 1 02 1 02 1 1 02 1 02 1 8065 2 8065 9678 2 2 1 2 2 1 . . . . . . . . . . . . . . c) (note: assumes two repair crews are available. For a single repair crew see problem 11.9) A A A s i s = − − → = − − = 1 1 1 1 8333 9722 ( ) ( . ) . 2 2 11.2 A MTTR A A xMTBF MTBF i i i i = = ≤ i i − 5 . . . 95 98979 and 1 = 0103 prop: hrs avion: hrs struc: hrs elec: hrs env: hrs MTBF MTTR MTBF MTTR MTBF MTTR MTBF MTTR MTBF MTTR = + = → ≤ = → ≤ = + = → ≤ = + = → ≤ = → ≤ 10 000 1 1 17 892 9 2 3333 34 2000 1 1 21 1771 18 2 870 1 1 21 773 7 9 10 000 10 3 , ( / . ) . . . ( / . ) . ( / . ) . , . Γ Γ Γ 11.3 From Eq 11.19: A P P r r r r r = + = + + LMN OPQ + + + LMN OPQ = + + LMN OPQ − − 1 2 1 1 2 2 1 1 1 1 2 2 1 2 1 1 1 1 1 10 5 1 10 1 5 5 λ λ λ λ λ λ λ / . / ( / ) . − + + + LMN OPQ = + = 1 10 − 5 1 1 10 5 1 10 1 5 5 7813 2 7813 9376 2 1 /. /. / ( / ) . . . . . 11.4 From Eq. 11.19: A P P r r r r r = + = + + LMN OPQ + + + LMN OPQ = + + LMN OPQ + LMN OPQ = − − − 1 2 1 1 2 2 1 1 1 1 2 2 1 1 1 1 1 1 2 12 1 1 2 34 λ λ λ λ λ λ λ 11.5 a) A r r = + = + = + = λ 1 2 5 1 2 5 1 4 4 1 / . 80 / . . . . . b) A e 2 4 2 1)2 4 1 1 4 1 2 = 1 80 1264 9264 + + + . − = − + + = . . . . . . . . (.4 . b g b g 11-1Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed. Waveland Press, Inc., Copyright © 2009 11-2 c) A r r r e e ( ) r . . . . . . . 2 ( ) . ) 4 4 1 1 4 1 = 2 5(2 8736 + + + = + + + = − + − λ λ λ λ d) Two components in series: a) A A s = = = 2 2 . . 80 64 , b) A A s = = = 22 . . 9264 8582 2 , c) A A s = = = ( ) . . 2 8736 7632 2 2 e) Two components in parallel: a) A A s = − − = − − = 1 1 1 1 8 96 ( ) ( . ) . 2 2 , b) A A s = − − = − − = 1 1 1 1 9264 9946 ( ) ( . ) . 2 2 2 c) A A s = − − = − − = 1 1 2 1 1 8736 9840 ( ( )) ( . ) . 2 2 f) A P P = + = + + LMN OPQ + + + LMN OPQ = + = − − 1 2 2 2 1 2 2 1 1 1 4 14 14 1 1 4 14 . 7619 25 7619 9524 . .. .. .. .. . . (. ) . 11.6 State Condition 1 primary operating (λ1=2) 2 primary failed and secondary operating (λ2=2) 3 both failed (r=3/2) 2 3 2 43 2 4 1 2 1 1 2 43 1 6 3 8 6 6 17 1 3 3 1 1 2 2 1 1 2 3 1 1 1 P P P P P P P P P P P P P = → = = → = + + = + + FGH IJK = → = L + + MN OPQ = − and A P P = + = + F 1 2 17 6 12HG17 6IKJ = = 17 9 .529 [Show More]

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