Solution Manual to Reliability Engineering 2nd Edition - Bradley (2020)

All Chapters Covered




SOLUTIONS

, RELIABILITYENGINEERING –ALIFECYCLEAPPROACH t, t, t, t, t, t,




INSTRUCTOR’S t , MANUAL
CHAPTER 1 t ,




The Monty Hall Problem
t, t, t,




The truth is that one increases one’s probability of winning by changing one’s choice. The
t, t, t, t, t, t, t, t, t, t, t, t, t, t,



easiest way to look at this from a probability point of view is to say that originally there is a
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




probability of ⅓ over every door. So there is a probability of ⅓ over the door originally chosen, and a
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,



combined probability of ⅔ over the remaining two doors. Once one of those two doors is
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




opened, there remains a probability of ⅓ over the door originally chosen, and the other
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,



unopened door now has the probability ⅔. Hence it increases one’s probability of winning the
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,



car by changing one’s choice of door.
t, t, t, t, t, t, t,




This does not mean that the car is not behind the door originally chosen, only that if one were to
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




repeat the exercise say 100 times, then the car would be behind the first door chosen about 33
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




times and behind the alternative choice about 66 times. Prove for yourself using Excel!
t, t, t, t, t, t, t, t, t, t, t, t, t, t,




Another way to prove this result is to use Bayes Theorem, which the reader can source for
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




himself on the internet.
t, t, t, t,




Assignment 1.2: Failure Free Operating Period t , t , t , t , t ,




The FFOP (Failure Free Operating Period) is the time for which the device will run without
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




failure and therefore without the need for maintenance. It is the Gamma value for the
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,



distribution. From the list of failure times 150, 190, 220, 275, 300, 350, 425, 475, the Offset is
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




calculated as 97.42 hours – say 100 hours. This is the time for which there should be no
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,



probability of failure. It will be seen from the graph in the software with Beta = 2 that the
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,



distribution is of almost perfect normal shape and that the distribution does not begin at the
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




origin. The gap is the 100 hours that the software calculates when asked.
t, t, t, t, t, t, t, t, t, t, t, t, t,




When the graph is studied for Beta = 2 it will be seen that there is a downward trajectory in the three
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




left hand points. If this trajectory is taken down to the horizontal axis it is seen to intersect it at about
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,



120 hours. This is the estimation of Gamma. In the days before software this was always the
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




most unreliable estimate of a Weibull parameter and the most difficult to obtain graphically.
t, t, t, t, t, t, t, t, t, t, t, t, t, t,




Assignment 1.3 t ,




When the offset is calculated it is seen to be negative at – 185.59 (say 180). This indicates that the
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




distribution starts before zero on the horizontal axis. This is the phenomenon of shelf life. Some
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




items have failed before being put into service. This can apply in practice to rubber
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




components and paints, for example.
t, t, t, t, t,




1
@@SSee
isis
mmiciicsis
oolala
titoionn

,Assignment 1.4: The Choice between Two Designs of Spring t, t, t, t, t, t, t, t,




DESIGN A t , DESIGN B t,




Number Cycles to Failure t, t, Number Cycles to Failure t, t,




1 726044 1 529082

2 615432 2 729000

3 807863 3 650000

4 755000 4 445834

5 508000 5 343280

6 848953 6 959900

7 384558 7 730049

8 666600 8 973224

9 555201 9 258006

10 483337 10 730008



Using the WEIBULL-DR software for DESIGN A above we get
t, t, t, t, t, t, t, t, t,



β=4 t, t,



Correlation = 0.9943 t, t,




F400k = 8% (measured from the graph in the Weibull printout below Fig M1.4 Set A) Hence
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,



R400k = 92%
t, t, t,




For DESIGN B we get from the WEIBULL-DR software (not shown here)
t, t, t, t, t, t, t, t, t, t, t,



Β=2 t, t,



Correlation = 0.9867 t, t,




F400k = 20% t, t,



Hence R400k = 80% t, t, t,




Hence DESIGN A is better t, t, t, t,




From Fig 1.4.1 Set A we can read in the table that for F = 1% at 90% confidence, the R value is
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,



126922 cycles. For an average use of 8000 cycles per year we get 126922/8000 = 15.86 years A
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t ,



conservative guarantee would therefore by 15 years.
t, t, t, t, t, t, t,




NOTE: The above calculations ignore the γ value. If this is calculated, the following figures
t, t, t, t, t, t, t, t, t, t, t, t, t, t,



emerge as shown in Fig 1.4.2 (the obscuration of some of the figures is the way the current
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,



version of the software prints out)
t, t, t, t, t, t,




DESIGN A t,




β=3 t, t,



γ = 101 828.6 say 100 000
t, t, t, t, t, t,



For F = 1% at 90% confidence, F = 176149
t, t, t, t, t, t, t, t, t,



Dividing by 8000 we get 176149/8000 = 22 years
t, t, t, t, t, t, t, t, t,




2
@@SSee
isis
mmiciicsis
oolala
titoionn

, Fig 1.4.1 Set A t, t, t,




A figure of 22 years or even 15 years for any guarantee is very long indeed. Company policy
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,



would have to be invoked – there are matters to consider in the determination of guarantees
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




other than the test data provided. These matters could include corrosion, user abuse etc. Such
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,



factors are more likely to occur, the longer the operating period. Questions need to be asked
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,



such as is there an industry standard for such guarantees, what are competitors offering as
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




guarantees, etc.
t, t,




A further point to note is that DESIGN B exhibits very peculiar characteristics if the γ value is
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,



taken into account. The β value remains at 2 but the γ value is negative at over 50 000 cycles!
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,




This implies that there is a probability of failure before entering service. This data looks suspect
t, t, t, t, t, t, t, t, t, t, t, t, t, t, t, t,



and further tests should be done to confirm the reliability characteristics of DESIGN B.
t, t, t, t, t, t, t, t, t, t, t, t, t, t,




3
@@SSee
isis
mmiciicsis
oolala
titoionn