Chapter 2. 1 From Equation 2.10: P(A 1 F) ˆ P(A 1)P(F A 1 ) S i P(F A i )P(A i ) The denominator is

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1 Chapter 2 1 From Equation 2.10: P(A 1 F) ˆ P(A 1)P(F A 1 ) S i P(F A i )P(A i ) The denominator is 0:3 0:0001 0:01 0:005 0:001 0:002 0:0002 0:04 ˆ 0:00009 P(A 1 F) ˆ 0:0001 0:3 ˆ 0:133: 0:00009 Similarly 0:005 0:01 P(A 2 F) ˆ 0:00009 ˆ 0:555: 0:005 0:01 P(A 3 F) ˆ 0:00009 ˆ 0:111: 0:04 0:0002 P(A 4 F) ˆ 0:00009 ˆ 0:088: 2 n ˆ 25 p ˆ 0:02. From Equation F(0) ˆ 25! 0!25! 0:020 0:98 25 ˆ 0:98 25 ˆ 0:6035 F(1) ˆ 25! 1!24! 0:021 0:98 24 ˆ 25 0:02 0:98 24 ˆ 0:3079 F(>1) ˆ 1 (F(0) F(1))ˆ 0:0886

2 PRACTICAL RELIABILITY ENGINEERING±SOLUTIONS MANUAL 3 3 n ˆ 25 p ˆ 0:2. From Eqn 2.37 F(0) ˆ 25! 0!25! 0:2 0 0:8 25 ˆ 0:8 25 ˆ 0: !24! F(1) ˆ 0:21 0:8 24 ˆ 25 0:2 0:8 24 ˆ 0:0236 F( > 1) ˆ 1 (F(0) F(1))ˆ 0: P ˆ 0:02 n ˆ 25 np ˆ 0:5 f(0) ˆ 0:50 0! exp ( 0:5) ˆ exp ( 0:5) ˆ 0:6065 f(1) ˆ 0:51 1! exp ( 0:5) ˆ 0:3032 f( > 1) ˆ 1 (0:6065 0:3032) ˆ 0:0903 p ˆ 0:2 n ˆ 25 np ˆ 5 f(0) ˆ 50 exp ( 5) ˆ exp ( 5) ˆ 0:0067 0! (use your calculator ± this is off the end of Appendix 2) f(1) ˆ 51 exp ( 5) ˆ 0:0337 1! f( > 1) ˆ 1 (0:0067 0:0337) ˆ 0:9596 Note that answers are close to exact (binomial) results when p is small (ˆ 0:02), but very inaccurate when it is large (0.2). A common `rule of thumb' is not to use the Poisson approximation for p > 0:1.

3 4 CHAPTER 2 5 (a) A ˆ ``Batch is Good'' B ˆ ``> 1 failure in 25'' P(A) ˆ 0:9(given) P(B A) ˆ 0:0886(from Question2(c)) SP(B E i )P(E i ) ˆ P(B A)P(A) P(B A)P(A) P(A) ˆ 0:1 P(B A) ˆ 0:9726 (from Question 3(c)) From Equation :9 0:0886 P(A B) ˆ (0:0886 0:9) (0:9726 0:1) ˆ 0:45 (i.e. there is an almost 50% chance of wrongly rejecting a good batch!) (b) A ˆ ``Batch is Bad'' B ˆ ``0 or 1 failure in 25'' P(A) ˆ 0:1 (given) P(B A) ˆ 0:0038 0:0236 ˆ 0:0274 (from Question 4(a) and (b)) SP(B E i )P(E i ) ˆ P(B A)P(A) P(B A)P(A) P(A) ˆ 0:9 P(B A) ˆ 0:6035 0:3079 ˆ 0:9114 (from Question 5(a) and (b)) From Equation :1 0:0274 P(A B) ˆ (0:0274 0:1) (0:9114 0:9) ˆ 0: (a) Pages 38±39. (b) For ``Poisson Process'', failures occur at random at rate l. In some time interval range t we expect lt failures. For a single item at risk, probability that it doesn't fail ˆ Prob of zero failures in Poisson Dist ˆ P(x) ˆ e m m x x!

4 PRACTICAL RELIABILITY ENGINEERING±SOLUTIONS MANUAL 5 St m ˆ lt and x ˆ o P(o) ˆ e lt ˆ Reliability R(t) Hence distribution function F(t) ˆ 1 R(t) ˆ 1 e lt and density function f(t) ˆ d dt (F(t)) ˆ le lt which are the equations of the exponential distribution. (i.e. if number of failures in given t follows the Poisson distribution, the distribution of times to failure is exponential) (c) R(t) ˆ exp 200 ˆ 0: Over 15 hours the reliability of a unit is exp ( 15=200) ˆ 0:928. Substitute in the binomial formula (2.23) with p ˆ 0:072, q ˆ 0:928. (a) f(0) ˆ 0:928 3 ˆ 0:7992 (b) f(1) ˆ 3! 1!2! 0:072:0:928 2 ˆ 0:1860 probability of 0 or 1 failures (ˆ prob. of ``not more than one'') ˆ 0:7992 0:1860 ˆ 0:9852 (c) f(2) ˆ 3! 0:072 2 : 0:928 ˆ 0:0144 2! 1! probability of 0 or 1 or 2 failures (ˆ prob of ``not more than two'') ˆ 0:7992 0:1860 0:0144 ˆ 0:9996 (Alternatively, probability of ``not more than 2'' ˆ 1 (probability of 3) ˆ 1 0:072 3 ˆ 0:9996) These answers can also be obtained from Figure 2.16 putting n ˆ 3 and p ˆ 0:072.

5 6 CHAPTER 2 8 (a) l ˆ 2=1053 ˆ 0:0019 failures per hour (b) u ˆ 1053=2 ˆ 526:5 hours (c) From Appendix 3, for n ˆ 2(2 1) ˆ 6, and a ˆ 0:9, x 2 ˆ 10:6 i.e. u L ˆ =10:6 ˆ 198:7 Using Equation 2.43 for u L ˆ 500, T ˆ :6=2 ˆ 2650 hours. so further testing required (without further failures) ˆ ˆ 1597 hours. 9 (a) Pages 31±32, 46±47. (b) F(t) ˆ 1 exp ( T b ) R Page Using either Equations 2.13 and 2.16, or calculator x ˆ 110:48 s ˆ 23:52 At x ˆ 150, z ˆ ( :48)=23:52 ˆ 0:446 From Appendix 1, F( 0:446) ˆ 0:6758(by interpolation) so R(100) ˆ 0:6758: From p 46, f(100) ˆ 1=s(2p) 1=2 exp ( ((t m) 2 =2s 2 )) ˆ 1=23:52 (2p) 1=2 exp ( (10:48 2 =2 23:52 2 )) ˆ 0:0783 so h(100) ˆ f (100)=R(100) ˆ 0:116 failures per hour: Similarly, at x ˆ 150, z ˆ ( :48)=23:52 ˆ 1:68 From Appendix 1, F(1:68) ˆ 0:9535, so R (150) ˆ 0:0465 f (150) ˆ 1=23:52 (2p) exp ( ((39:52 2 =2 23:52 2 )) ˆ 0:0170 so H(150) ˆ F(150)=R(150) ˆ 0:366 failures per hour. (increasing hazard suggests normal is reasonable assumption) KS test requires F(x) at every failure time using fitted mean and SD. These are given in tabular form

6 PRACTICAL RELIABILITY ENGINEERING±SOLUTIONS MANUAL 7 t z F(z) i/n Diff The largest difference is (at x ˆ 116.7). From Appendix 5, we need a value of to reject the `fit' null hypothesis even at 10% significance. As our statistic is much lower than this, we conclude that there is insufficient evidence to reject the hypothesis that the data are normally distributed. 11 (a) z ˆ (45 47:2)=1:38 ˆ 1:59 from Appendix 1, proportion < 45 ˆ 1 0:944 ˆ 0:056 (b) P(0) ˆ 0:944 5 ˆ 0:75 ; P(at least 1) ˆ 1 P(0) ˆ 0:25 P(1) ˆ 5 0:56 0:944 4 ˆ 0:222 ; P(fewer than 2) ˆ P(0) P(1) ˆ 0:972 P(5) ˆ 0:056 5 ˆ 5: (c) (i) Observed proportion is 4% compared with normal dist estimate of 5.6%. This is probably well within sampling error. Also, in practice real data is usually more truncated in the tails than the normal (it has to be, as no real measure can go from 1 to 1). (ii) As we want a 90% interval, this will be bounded by individual upper and lower 95% limits. Upper 90% limit is value such that the probability of the test result (4 failures in 100) or lower is i.e. we require p such that P(4 or fewer failures) ˆ From Figure 2.16 this is at approximately Similarly, for lower limit we require p such that P(4 or more failures) ˆ 0.05, i.e. P(3 or fewer failures) ˆ From Figure 2.16 this is approximately i.e. the 90% confidence interval for p is from to (1.3% to 8.7%).

7 8 CHAPTER 2 (d) We estimate from the sample that the proportion failed is 4%, and if I state that the true value is somewhere between 1.3% and 8.7%, there is only a 10% chance that I am wrong. (e) We use the distribution of minimum values (i.e. of lowest extreme values). Standard deviation ˆ 0:88 Nm ˆ 1:283s i.e. s ˆ 0:686 Mean ˆ 45:5 ˆ m 0:577s m ˆ 45:5 0:577 0:686 ˆ 45:986 From Equation 2.36 At 45 Nm, y ˆ (45 45:896)=0:686 ˆ 1:306 F(45) ˆ 1 exp [ exp ( 1:306)] ˆ 0:237 At 44 Nm, y ˆ (44 45:896)0:686 ˆ 2:764 F(44) ˆ 1 exp [ exp ( 2:764)] ˆ 0: The data given are X i (inter-arrival times). We require the cumulative elapsed arrival times, x i. These are: 96, 177, 282, 316, 408, 489, 578, 716, 791, 947, 1152, 1263, Note that the test ends in a failure, so we must use (n 1) in place of n (i.e. 12 instead of 13) x 0 ˆ 1440 SX i ˆ :: ˆ 7215 so (7215=12) (1440=2) U ˆ 1440 (1=(12 12)) ˆ 118:75=120 ˆ 20:99

8 PRACTICAL RELIABILITY ENGINEERING±SOLUTIONS MANUAL 9 The negative sign indicates that any trend is a reduction in failure rate. However, reference to Appendix 3 shows that a value of 0:99 has a probability of (1±0.8389) ˆ 0.16 of arising by chance even if there is no underlying trend, so even at a significance level of 15% we cannot reject the hypothesis of `no trend'. Note that it would be wise to plot a graph of cumulative failures against X i first, so as to obtain a subjective impression of the trend before refining matters with a significance test. 13 Pages 22±23, 45, 48± Page From Appendix 6, the median rank of the first failure in the sample of 10 is 6.7% cumulative failure. The 95% rank is 25.9%. 35,000 cycles for the first failure represents only a relatively small margin, so extrapolating backwards from the 95% point indicates that the population is quite likely to include more than 10% with lives that that are less than 30,000 cycles. In other words, the test does not give reasonable s-confidence that the specification has been achieved. This is, of course, only a statistical interpretation. Practical considerations are included in the answers to Question (a) There is likely to be a finite failure-free life for a bolt in fatigue loading. Therefore the fit to a 2±parameter distribution should be examined further. Actual fatigue load and fatigue properties data should be obtained, fatigue life should be estimated, and further tests considered. Alternatively, a stronger bolt could be used. See the discussion of fatigue in Chapter 8. (b) (c) Probably the best approach would be to strengthen the component by redesigning it to reduce stress values, particularly stress concentrations. What does the failure evidence show in terms of crack initiation points (Chapter 8)? Also, the in-use load pattern is likely to be very variable for such an item, so the test results may not accurately represent actual conditions. The mechanism(s) of failure should be ascertained (e.g. filament degradation, vacuum loss, etc.), analysed, and compared with the interpreted failure data. It would not be practicable to determine how to improve the durability without knowledge of the underlying failure mechanism(s). (d) Again, the main mechanism(s) (wear? fatigue?) should be identified (Chapter 8). As in a., a failure-free life should be expected. Life might be increased by reducing contact or internal stresses, improvements in lubrication, reducing machining tolerances, surface treatment, etc.

9 10 CHAPTER 2 (e) The definition of failure is important, as well as the mechanism(s) (Chapter 9). Is failure defined as a light output level, or complete failure? For such a low-cost item, it should be possible to test a much larger sample to increase confidence in the results. 17 Page (1) In any landing situation, if the landing loads are high (high aircraft weight, hard landing, crosswind, etc.) they are likely to affect more than one tyre. (2) Worn tyres are more likely to fail than new ones. The pattern of wear might not be random, resulting in multiple failures. (3) One failing tyre could damage others, causing them to fail also. 1. and 3. do not apply to car tyres. Also, when one car tyre fails, the wheel/tyre combination is always changed before the others are put at risk by further driving.