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An Introduction to Reliability and Maintainability Engineering CHAPTER 8 8 For Component 1: MTTF = 10,000 Γ(1+1/2) = 8862 Cost1 = 840 [ 1 + (P/A,.03,20) 2,000 / 8862 ] = $ 3660 Cost2 = 870 [1 + (P/A,03,20) 2,000 / 10,000 ] = $ 3458 Note: (P/A,.03,20) = 8 R( t ) = e − (1+.03) 20 − 1 = 14 .03(1+.03) 2...

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An Introduction to Reliability and Maintainability Engineering

CHAPTER 8

8.1 For Component 1: MTTF = 10,000 Γ(1+1/2) = 8862.3 Cost1 = 840 [ 1 + (P/A,.03,20) 2,000 / 8862.3 ] = $ 3660.19 Cost2 = 870 [1 + (P/A,03,20) 2,000 / 10,000 ] = $ 3458.60 Note: (P/A,.03,20) = −

8.2 R(t ) = e

(1+ .03)20 − 1 = 14. 877 . 03( 1+. 03) 20

FG t IJ β HθK −

RC (1) = e

FG 1 IJ.91 H 3.5 K =.7263

FG IJ H K

1.8 − 1 5

RS (1) = e

FG IJ

8. − 1

RA (1) = e H 4 K =.7190 RLS (1) = e Rsys = RC RA RS RLS =.7236×.7190×.9463× .8465 =.4183

FG IJ HK

1 − 1 6

= .9463 = .8465

Reliability Goal= y = 4 .995 =.998748

.998748− .7263 x100 = 37.51 .7263 .998748− .7190 %A = x100 = 38.91 .7190 ln(1 − R ) n 8.3 R = 1 − (1− Ri ) → n = , ln(1 − Ri ) nA = nLS

ln(1−.998748) = 5.2647 → 6 units ln(1−.7190) ln(1 −.998748) = = 3566 → 4 units . ln(1−.8465)

8.4 λ ( t ) =

FG IJ HK

β t θ θ

λ ≈ AFR =

β −1

1 1−0

and

z

1

0

AFR =

% LS nC =

ln(1− .998748) = 51577 → 6 units , . ln(1 −.7263 )

, nstr =

ln(1−.998748) = 2 .27 → 3 units ln(1−.9463)

1 t2 − t1

z

t2

λ ( t ' )dt '

t1

L β FG 1 IJ t' O M θ H θ K PP βF 1 I t ' dt ' = M G J θ Hθ K MM β PP Q N

1

β

β −1

.998748 −.9463 x100 = 5.54 .9463 . 998748−. 8465 = x100 = 17.99 .8465

%S =

%C =

β −1

=

β −1

1

FG 1 IJ = 1 H K θ

θ θ β −1

0

1 = .3198 . 35 . 91 1 . λ A ≈ AFR A = .8 = 3299 4

λ C ≈ AFR C =

λ S ≈ AFRS = λ LS ≈ AFRLS

λinew = wi λ* = wi (− ln(.995 )) = wi (.005102 ); wi =

λi

41

=

4

∑λ i=1

β

i

1 =.0552 1.8 5 1 = 1 =.1667 6

λi .3198+ .3299+ .0552+ 1667 .

=

λi .8716

An Introduction to Reliability and Maintainability Engineering

λC

new

λ Anew

IJ.005012 =.00184 GHF..3198 8716K F .3299I =G H .8716JK.005012 =.00190

λS

=

F 1I =G J Hλ K F 1 JI θ =G H .00184K F 1 I θ =G H .00190JK

new

IJ.005012 =.00032 GHF ..0552 8716 K F .1667 IJ .005012 =.00096 =G H .8716 K

=

λ LSnew

1/ β i

θi

i

1/.91

C

A

1 I GHF.00032 JK

. yrs = 10132

θS =

= 2520.9 yrs

θ LS =

1/.8

1 /1 .8

= 87. 4 yrs

FG 1 IJ H . 00096K

1/1

= 10417 . yrs

8.5 Total parts count=153+28+34=215 RA: Reliability allocated to ith component= R *ni / N =.99 ni / 215

F GH

ni / N

1 1 − R* λi = − ln 1 − wi ti Component system board hard drive DC power pack

8.7

Resistor 1 2 3

MTTFi = 1 / λi , Rw: Rel. accounting for importance= e − λi ti

RA .99287 .99869 .99841

λι 3.767x10-6 1.457x10-6 7.96x10-7

MTTF 267,666 686,342 1,256,861

Rw .99249 .99854 .99841

Operating Wattage/Rated Wattage 180/200=.9 180/225=.8 180/300=.6 SYSTEM FAILURE RATE IN 10-6 OPERATING HOURS R1 R2 R3 .037x73=2.71 .033x73=2.409 .026x73=1.898 .033x73=2.409 .029x73=2.117 .023x73=1.679 .026x73=1.898 .023x73=1.679 .019x73=1.387

Fan Size small medium large

Fan Size smalll medium large

Fan Size

I, JK

R1 50+73(1)=123 90+73(1)=163 160+73(1)=233

SYSTEM COST R2 50+73(1.2)=137.6 90+73(1.2)=177.6 160+73(1.2)=247.6

R3 50+73(2)=196 90+73(2)=236 160+73(2)=306

MTTF/COST=(1/ SYSTEM FAILURE RATE) / SYSTEM COST R1 R2 R3

42

An Introduction to Reliability and Maintainability Engineering

small medium large

3010 2547 2261

3017 2660 2405

2688 2524 2356

The largest MTTF per dollar cost occurs for R2 and the small fan. −6 The system reliability is: e −( 2.409×10 )( 10, 000) = .9762 .

8.15 A: µ = MTTF = 12Γ(1+1/1.7) = 10.71 A: 225 + ( P / A,.05 ,10 )40 (1 /10 .71) − ( P / F ,.05,10 )60 100 + 4300 = 26 ,000 B: 245 + ( P / A,.05,10) 35(.11) − ( P / F ,.05,10) 40 100 = 25,017 note: (P/A,.05,10) = 7.72 and (P/F,.05,10) = .6139

8.16

m = SF ◊ K

q=

Æ

b

m

G 1+ 1/ b

g

=

m G (2.25)

=

SF ◊ K 1133 .

1.133 I FG K IJ -FG J H H qK P (Y < K ) = 1 - e = 1 - e SF K b

.8

-

SF = 12 . Æ P( Y < K) =.615, SF = 2.0 Æ P(Y < K) =.470, SF = 4 Æ P(Y < K ) =.305 8.17 N = (1.23 x 1028 ) (35)-14.85 = (1.23) (1.1765) x 105 = 144709.5 or 144,709,500 cycles to failure. Yrs to failure = 144,709,500 cycles to failure / [ 20 cycles/sec x 3600 sec/hr x 350 hrs / yr ] = 5.74 yrs Material selected does not support design life.

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