A guiding principle followed by effective designers is to utilize as much available quan- titativeinformation as possible in making design decisions. Therefore, if probabilistic descriptions are available in the form of statistical data for describing strength distribu- tions, loading distributions, or variations in environment, manufacturing, inspection, and/or maintenance practices, these data should be utilized to keep the probability of failure low, or, to state if differently, to keep the relibility above a prescribed level of acceptability.
Reliability may be defined as the probability that a machine or machine part will per- form its intended function without failure for its prescribed design lifetime. If the proba- bility of failure is denoted by , the reliability, or probability of survival, is . Thus reliability is a quantitative measure of survival success, typically based on distribution functions verified by experimental data.
Implementation of the probabilistic design approach requires that the distribution function (probability density function) be known or assumed for both the stress at the crit- ical point (and all factors influencing stress) and the strength at the critical point (and all factors influencing strength).
If the probability density function for stress, , and the probability density function for strength, , are known, they may be plotted as shown in Figure 2.16. By definition, the reliability is the probability that the strength exceeds the stress, or
(2-90) This corresponds to the area in Figure 5.1 that lies outsidethe shaded interferencearea.
The shaded area represents probability of failure, that is, . The probability of failure is sometimes called unreliability.
Three probability density functions of particular interest to a designer are the normal, thelog-normal, and the Weibulldistribution functions. Only one of these, the normal dis- tribution, will be illustrated here, but similar developments are available for the other two.33The probability density function for the normaldistribution is
(2-91) for - q 6 x 6 q
f1x2 = 1
sN22pe-12Ax-sNmNB
2
f1x2
P{s Ú S}
R = P{S 7 s} = P{S - s 7 0}
f1S2 f1s2
R = 1-P{failure}
P{failure}
f(),
f(S) f() f(S)
S
Stress Strength
, S Interference area (failure region) Figure 2.16
Probability density functions for stress and strength, showing interference area (failure region).
33See, for example, refs. 1, 28, 29, or 30.
Reliability: Concepts, Definitions, and Data 77
where is a random variable such as stress or strength, is the estimated populationmean, and is the estimated populationstandard deviation, where
(2-92) and
(2-93) In these expressions, equals the number of items in the population. As a matter of inter- est, the square of the standard deviation, , is defined as the variance. Both variance and standard deviation are measures of dispersion, or scatter, of a distribution. Conventional notation for describing a normal distribution is
(2-94) which is to be read “ is distributed normally with mean and standard deviation .” As shown in Figure 2.17(a), a normal distribution has the well-known bell shape, is symmet- rical about its mean, and tails off to infinity in both directions. The corresponding normal cumulative distribution function, , is plotted in Figure 2.17(b), where
(2-95) and
(2-96) is defined as the standard normal variable, which has a normal distribution with mean of 0 and standard deviation of 1. is any specified value of the random variable . Any nor- mal distribution with mean and standard deviation can be transformed into a standard normal distribution using (2-96). Because all normal distributions can be fully defined by and , they are called two-parameterdistributions. Table 2.9 gives values for the cumu- lative distribution function sN F1X2for the standard normal distribution.
mN
sN
mN X0 X
X= x - mN sN F1X2 = P{X … X0} =
L
X
- q
1 22pe-
t2 2 dt F1X2
sN mN
x
x =d N1mN,sN2 sN2 n
sN = A
1 n - 1a
n
i=1
1xi - m22 mN = 1
na
n
i=1
xi
sN x mN
f(x)
f(x) = F(x)
x 0
(a) (b)
x 0
1
1 2
1
1 2
2 e– x –
2
; –∞ < x < ∞
Figure 2.17
Plots of the probability density function and the cumulative distribution function for a normaldistribu- tion. (a) Probability density function. (b) Cumulative distribution function.
TABLE 2.9 Cumulative Distribution Function F(X)for the Standard Normal Distribution, Where
X .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753
.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389
1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621
1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830
1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015
1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177
1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319
1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441
1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545
1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633
1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706
1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767
2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817
2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857
2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890
2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916
2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936
2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952
2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964
2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974
2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981
2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986
3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990
3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .9993
3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995 .9995
3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .9997
3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998
3.5 .9998 .9998 .9998 .9998 .9998 .9998 .9998 .9998 .9998 .9998
3.6 .9998 .9998 .9999 .9999 .9999 .9999 .9999 .9999 .9999 .9999
3.7 .9999 .9999 .9999 .9999 .9999 .9999 .9999 .9999 .9999 .9999
3.8 .9428 .9431 .9433 .9436 .9438 .9441 .9443 .9446 .9448 .9450 3.9 .9452 .9454 .9456 .9458 .9459 .9461 .9463 .9464 .9466 .9467 4.0 .9468 .9470 .9471 .9472 .9473 .9474 .9475 .9476 .9477 .9478 4.1 .9479 .9480 .9481 .9482 .9483 .9483 .9484 .9485 .9485 .9486 4.2 .9487 .9487 .9488 .9488 .9489 .9489 .9490 .9500 .9502 .9507
F1X2ⴝ L
X ⴚˆ
1
22Pe 1ⴚt2>22dt
An important theorem of statistics states that when independent normally distributed random variables are summed, the resulting sum is itself normally distributed with mean equal to the sum of the individual distribution means and standard deviation equal to the square root of the sum of the individual distribution variances. Thus when and are bothnormal probability density functions, the random variable used in (2-90) is also normally distributed with a mean of
(2-97) and standard deviation
(2-98) The reliability then may be expressed as34
(2-99) where
(2-100) is the standard normal variable [see (2-96) and Table 2.9].
t = y - mNy sNy
e- t2 2dt mNS-mNs 1sN2S+s2s
R= P{y 7 0} = P{S - s 7 0} = 1 2pL
q -
R
sNy = 2sN2S + sN2s mNy = mNS - mNs
y = 1Sf-1s2s2 f1S2
Reliability: Concepts, Definitions, and Data 79
34See, for example, ref. 29.
Example 2.13 Existing Safety Factor and Reliability Level
An axially loaded, straight, cylindrical bar of diameter d12 mm is made of 2024-T4 aluminum. With experimental data for the material tested under conditions that closely correspond to actual operating conditions indicate that the yield strength is normally dis- tributed with a mean value of 330 MPa and standard deviation of 34 MPa. The static load on the bar has a nominal value of 30 kN, but due to various operational procedures, and excitations from adjacent equipment, the load has been found to actually be a normally distributed random variables with standard deviation of 2.2 kN.
a. Find the existing factor of safety for the bar, based on yielding as a failure mode.
b. Find the reliability level of the bar, based on yielding as a failure mode.
Solution
a. Based on yielding, the existing safety factor is where and
whence the existing safety factor is
nex = 330 265 = 1.25 snom = Pnom
A = 4(30000)
p(0.012)2 = 265 MPa Syp-nom = 330 MPa
nex = Syp-nom>snom,
System Reliability, Reliability Goals, and Reliability Allocation
Efforts by a designer toward achieving the failure prevention objective at the design stage often include the tasks of determining the required reliability levels for individual compo- nents or subsystems that will assure the specified reliability goalfor the whole machine as- sembly. The process of assigning reliability requirements to individual components to attain the specified system reliability goal is called reliability allocation. Although the al- location problem is complex, the principles may be demonstrated in a straighforward way by making certain simplifying assumptions.
As a practical matter, designers are well advised to work with a reliability engineer- ing specialist at this stage. Reliability specialists can provide valuable insights in selecting distributions that are well suited to the application, data interpretation, experimental de- sign, reliability allocation, and analytical and simulation techniques for reliability assess- ment and system optimization.
However, the task of setting appropriate reliability goals lies primarily in the province of the designer, in cooperation with engineering and company management. The daunting fact that a reliability of 100 percent can never be achievedforces careful consideration of potential failure consequences. Choosing reliability levels appropriate and acceptable for the particular machine application being proposed and selection of reliability goals are often difficult tasks.35It is alleged, for example, that engineers seeking an appropriate and
b. From given data and . The estimated
mean stress may be calculated as
The estimated standard deviation is
hence
These calculations neglect statistical variability in dimensions used to calculate area.
The lower limit of the reliability integral in (2-99) may be calculated as
From Table 2.9, since , the reliability corresponding to may be read as
Thus one would expect that 97.2 percent of all installations would function properly, but 28 of every 1000 installations would be expected to fail. A decision must be made by the designer about the acceptability of this failure rate for this application.
R = 0.9719
X = -1.91 R = 1 - P
X= - mNS - mNs
2sNS2 +sNs2 = - 330 - 274 2342 + 1.9452
65 = -1.91 s = N(274 MPa, 1.945 MPa)
sNs= 2200
113.1 * 10-6 = 1.945 MPa mNs = P
A = 30,000
113.1 * 10-6 = 265 MPa Syp =d
N(330 MPa, 34 MPa) P =d
N(30 kN, 2.2 kN) Example 2.13
Continues
35See also 1.8.
acceptable reliability goal during early pioneering efforts in spacecraft design decided that the probability that a toreadorcan avoid being gored in a bull fight would also serve as an acceptable reliability goal for spacecraft design. The aerospace industry specifies a relia- bility of “five-nines” (R 0.99999) in many cases, while the “standard” reliability of rolling element bearings is usually taken to be 0.90. Table 2.10 illustrates probability- of-failure criteria used by some industries. Clearly the selection of an appropriate reliabil- ity goal depends upon both the consequences of failure and the cost of achieving the goal.36
For the purpose of making a simplifiedreliability analysis at the design stage, the re- labilities of components and subsystems may be assumedconstant over the design life of the machine. To assess the effects of component or subsystem failure on overall system performance, a functional block diagramis a helpful tool. For a simplified analysis, each block in the diagram represents a “black box” that is assumed to be in one of two states:
failedorfunctional. Depending upon the arrangement of components in the machine as- sembly, the failure of a component will have different effects upon the reliability of the overall machine.
Figure 2.18 illustrates four possible arrangements of components in a machine as- sembly. The seriesconfiguration of Figure 2.18(a), in which all components must function properly for the system to function, is probably the most common arrangement encoun- tered in practice. For a series arrangement the system reliability is found to be
(2-101) RS = q
n
i=1Ri
RS R
Reliability: Concepts, Definitions, and Data 81
36For example, in the 1980s Motorola inaugurated the practice of setting allowable failure rates to a very low value, by design, and named the process Six Sigma. (Six sigma corresponds to a failure rate goal of no more than about 3.4 failures per million units. By contrast, U.S. industry as a whole operates around three sigma, or a failure rate of about 66,000 units per million.) Adopting the Six Sigma process, General Electric Company estimated that spending about $500 million in 1999 on Six Sigma would produce a return of approximately
$2 billion. See ref. 31.
TABLE 2.10 Reliability-Based Design Goals as a Function of Perceived Hazard Level1
Suggested Criteria for Potential Consequences Designated Hazard Design-Acceptable
of Failure Assessment Category Probability of Failure
Some reduction in safety margin or Minor functional capability.
Significant reduction in safety Major margin or functional capability.
Device no longer performs its Hazardous function properly; small numbers of
serious or fatal injuries may occur.
Complete equipment failure, unable Catastrophic to correct the situation; multiple
deaths and/or high collateral damage may occur.
1Excerpted in part from ref. 11.
… 10-9 10-7-10-9 10-5-10-7
… 10-5
where the right-hand side is the product of the individual component reliabilities. If it is assumed that the failure probability is identical for all components, (2-101) may be rewrit- ten as
(2-102) which may be expanded by the binomial theorem to give, neglecting higher-order terms,
(2-103) From this it may be noted that, for a series arrangement, the system reliability decreases rapidly as the number of series components increases. In general, for series systems, the system reliability will always be less than or equal to the reliability of the least reliable component. This is sometimes called the “weakest link” failure model.
Theparallelarrangement of Figure 2.18(b), in which the system continues to function until all components have failed, has a system reliability of
(2-104) provided that all components are active in the system and that failures do not influence the reliabilities of the surviving components. In practice, these assumptions may be invalid if, for example, a standby redundant component is not activated unless the on-line component fails, or if the failure rate of surviving components increases as failures occur. If it is as- sumed that the failure probability is identical for all components, (2-104) may be rewritten as (2-105) From (2-105) it may be noted that, for a parallel arrangement, the system reliability increasesas the number of parallel components increases. As a practical matter, however, designing a parallel system may be difficult and costly, and for a given component relia- bility, the gain in system reliability slows down rapidly as more parallel components are added. Nevertheless, the procedure of adding a second component or subsystem in paral- lel is the basis for the very important fail-safedesign concept discussed in 1.8.
Virtually all arrangements of components in a machine assembly may be modeled as combinations of series and parallel configurations. Figure 2.18(c), for example, is a series arrangement of two parallel subsystems, and Figure 2.18(d) shows a parallel arrangement of two series subsystems. For Figure 2.18(c) there is redundancy at the component level while for Figure 2.18(d) there is redundancy at the subsystem level. System analysis has
RS = 1 - qp RS = 1 - q
p
i=111 - Ri2 n
RS = 1 - nq RS = 11 - q2n q
…
…
1
1 2
2 n p
A B
C D
(b) (a)
(c)
A B
C D (d)
Parallel arrangement of p components.
Series arrangement of n components.
Series-parallel arrangement of components (component level redundancy).
Parallel-series arrangement of components (subsystem-level redundancy).
Figure 2.18
Block diagrams illustrating var- ious design arrangements of components in a machine sys- tem, from a reliability analysis point of view.
shown that for a system with a given number of components, component-level redundancy gives a higher system reliability than subsystem-level redundancy.
The task of simplified reliability allocation may be accomplished by utilizing a tech- nique called equal apportionment. This technique models the system reliability as a series ofmsubsystems, each having equal reliability . Each subsystem may include one or more components in series or parallel arrangements. The specified system reliability then is
(2-106) and the required component reliability to achieve the specified system reliability is
(2-107) 1Ri2req’d = c1RS2specified
system
d1>m 1RS2specified
system
= q
m
i=1
Ri
Ri
Reliability: Concepts, Definitions, and Data 83
Example 2.14 Component Reliability Required to Meet Specified Overall Machine- Reliability Goals
A machine assembly of four components may be modeled as a series-parallel arrangement similar to that shown in Figure 2.18(c). It has been determined that a system reliability of 95 percent is necessary to meet design objectives.
a. Considering subsystems A-BandC-D, what subsystem reliability is required to meet the specified 95 percent reliability goal for the machine?
b. What component reliabilities would be required for A,B,C, and Dto meet the 95 per- cent reliability specification for the machine?
Solution
a. Using (2-107), the required subsystem reliability is
b. For subsystem A-B, the components AandBare in parallel, hence (2-104) may be used to give
or
Hence the system reliability goal of 0.95 can be met only if the subsystemrelia- bilities are at least 0.975. However, because of the redundancy within the subsystems, componentshaving reliabilities as low as 0.84 can meet the reliability requirements of the system.
Reliability Data
A designer striving to achieve the failure prevention objective by implementing reliability methodology at the design stage quickly confronts the need for pertinent,valid,quantita- tivedata relating to both strength and loading. Distributional data defining mean values
Ri = 1 - 21 - RA-B = 1 - 21 - 0.975 = 0.84 RA-B = 1 - 11 - Ri22
Rreq’d = 10.9521>2 = 0.975
and variances in all parameters that affect strength and loading are required. A sufficient body of pertinent distributional design data rarely exists. Valid and useful new data are costly and time consuming to collect. At the design stage it is typically necessary to rely on existing experimental material strength data interpreted statistically, and extrapolated statistical operational field data for loads and environments, gathered from similar service applications.
Historically, experimental material strength data have often been gathered without re- porting information now known to be statistically important; in many cases archival infor- mation, or published histographic information, can be reinterpreted statistically37 to recover the reliability indices of interest. In addition to probability of failure, other indices of use might include average time between failures, average duration of system “down- time,” expected loss of revenue due to failure, or expected loss of manufacturing output due to failure. In the final analysis, to be useful to a designer at the early design stage, prob- abilistic descriptions of such indices must have the potential to be quantitatively linked to strength and/or loading distributions. In critical applications, to support design improve- ment activities, it may become necessary to generate specific statistical data during the de- velopment stage or during field service monitoring. A review of the fundamental design steps in Table 1.1 serves to reiterate this concept.