analysis of reliability of engineering design

8/14/2019 Analysis of Reliability of Engineering Design

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Analysis Of Reliability Of EngineeringDesign

Reliability is a new concept, barely three decades old and is primarily due to

the complexity, sophistication and automation inherent in moderntechnology. It has been established that the reliability, which is a measure ofquality, is an essential element at each stage of the equipmentmanufacturing procedure through design and production to final delivery tothe user.Reliability can be defined in the following ways:

1. The reliability of a component (or a system) is the probability that thecomponent (or a system) will not fail for a time t.

2. The reliability of a system is called its capacity for failure freeoperation for a definite period of time under given operatingconditions.

3. Reliability is the probability of an item performing its intended

function over a given period of time under the operating conditionsencountered.

Some important terms:

i. Failure rate ()It is the ratio of total number of failures during the test interval to thetotal test time.

= f/t

ii. Mean time between failure (m)It is the reciprocal of the constant failure rate.

m = 1/ = t/f

iii. Mean time to failure (MTTF)If we have a life-test information on n items with failure times t1, t2..

tn Then nMTTF = 1/n ti

i=1

iv. Reliability functionLet a fixed number (N) of components be repeatedly tested. There will

be, after a time t, n components, which survive the test and m components,which fail.The Reliability (or probability of survival) expressed as a fraction at any timet, during the test is

R (t) = n/NThe probability of failure or unreliability at any time t, can be expressed as

F (t) = m/N

It is clear that at any time, t,

R (t) + F(t) = 1

because R (t) & F (t) are mutually exclusive events.

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v. Density function, f(t)The probability that a random trial yields a value of t within the interval

from t1 to t2 , ist2 f (t) dtt1

then f(t) is the density function for the continuous random variable.

vi. Distribution function, F (t)The distribution function, F (t) is the probability that in a random trial,

the random variable is not greater than t, thus,t

F (t) = f (t) dt -

F (t) is recognized as unreliability function, when speaking of failures.

vii. Reliability, R (t)The reliability is given by

R (t) =1-F (t)In integral form, R (t) can be expressed as

R (t) = f (t) dtt

viii. Hazard Rate, Z (t):It is also called as instantaneous failure rate, defined as the ratio of

density function, f (t) to the reliability function, R (t).Z (t) = f (t)/R (t).

1. A shaft on two supports is shown. The distance between the supportsis L. Assume that the axis for the left bore in the housing is the

reference axis (which means that the left bore surface is taken as thedatum). Then the position of the shaft axis with respect to the boreaxis is determined by the eccentricities of the bearing outer ring, balls,inner ring and shaft plus the magnitude of a clearance in the bearing(clearance between the balls and the rings after the bearing ismounted on the shaft). The latter, however is neglected here for thesake of simplicity. Assume that the system of balls in a bearing can betreated as a ring, the tolerances of which are equal to those for anindividual ball.

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Figure 1.1

Then the eccentricity vector for the left support l is determined by theequation

4l = i 1

i=1

where 1 , 2 , 3 and 4 are the eccentricities for the outer ring, balls, innerring and shaft respectively. Similarly the eccentricity for the right support isdetermined by

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r = i . 2 i=1

where 1 , 2 , 3 , 4 and 5 are the eccentricities for the bore, outer ring,balls, inner ring and shaft respectively. (It is assumed here that the housingsfor the two bearings cannot be bored with one setup, which means that theright bore may have a displaced geometric axis with respect to the left). Thestatistical means for the eccentricity magnitude in equations 1 and 2 arerelated to the tolerance by equations 3 and the Rayleigh distributionparameters are found from equation 4.

e = / 4 , .. 3

where e - mean of eccentricity magnitude - tolerance

i = [(2/ )^ ] x [i /4] .. 4

where i - distribution parameter

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- tolerance for each outside diameter of a ring

The resultant vector of eccentricities with respect to the datum is

= l + r .. 5

The distribution of [ ] is given by

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2 = i2 . 6i=1

The probability that a misalignment angle = e/L exceeds a specificmagnitude * is

*P( > *) = f(e) de . 7

e*

where e* = *L

As an illustration, consider a numeric solution. The following nominaldimensions are given :the shaft diameter at both supports is ds = 40 mm,housing diameter at both supports db = 80 mm. Assume that the same typeof bearing is used at both supports and neglect the effect of ball tolerances.Consider for simplicity a transition type fit for both housing and shaftinterfaces with the bearing rings. The following dimensions are assigned :

Shaft at two supports

ds = 40 +0.002-0.014

Bore of the bearing

db = 40 +0.016-0

Outer diameter of the bearing

Db = 80 +0.003-0.016

Diameter of the housing

Dh = 80 +0.019-0

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Then the corresponding tolerances are

Shaft s = 0.016 mm

Bore of the bearing b1 = 0.016 mm

Outer diameter of the bearing b2 = 0.019 mm

Diameter of the housing h1 = 0.019mm

The calculation of corresponding i , according to eqn. 4 and eqn. 6 gives

2 = 0.006415 = 0.08

For L = 300 mm the probability of exceeding various angles of misalignmentis calculated using equation 7

The probability P ( > 0) = 1, while the probability P ( > 0.001) = 0.00088.In this particular case the probability of a significant misalignment is small.

This illustrates how the tolerances of various dimensions of differentcomponents in a system affect the probability of exceeding a specific angleof misalignment. However, the reliability of a bearing operating at a givenspeed and load under the condition of misalignment is a property of thebearing itself. Thus the bearing reliability is conditional, the condition being inthis case the probability of a specific position of the shaft. Note also that inthis example two bearings are operating in similar conditions as far asmisalignment is concerned. This may be a common cause for failure, otherparameters being independent.

2. A single lip seal separating a rotating shaft from the stationary case isshown. This type of a seal is usually used for retaining lubricants inmachines having rotating and/or oscillating shafts. Various polymermaterials are used for seal elements. The function of the seal isdetermined by its ability to maintain a constant radial pressurebetween the sealing element and the shaft. Anything causing adecrease in this pressure affects leakage and thus seal failure.Consider factors affecting the radial lip seal performance.

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OIL

Figure 2.1 Lip Seal On Rotating Shaft

1. Temperature : It affects oil viscosity and properties of thesealing material. If temperature increases, the viscosity of the oilgoes down, and the oil film between the lip and the shaftbecomes thinner. This leads to increased friction losses and thusto increase local temperature. The variation of temperatureaffects the sealing material as well: Thermal expansion duringtemperature rise and stress relaxation during temperature fall,speed up the process of material ageing. Thus temperaturecontrol is important from the point of view of seal reliability.Temperature at the lip of the seal is affected by

Ambient temperature Temperature of the oil Friction coefficient between seal and shaft Initial preload by a garter spring

Conductivity of materials

2. Wear: Wear at the contact point of the seal directly affects itsfunctioning. The rate of the wear depends on

Coefficient of friction at the seal- shaft interface Pressure exerted by the garter spring Pressure due to the seal-shaft fit Pressure caused by the oil Shaft surface finish Shaft surface hardness Speed of rotation Shaft oscillations Abrasion resistance of the seal material

3. Fatigue: Fatigue manifests itself in a gradual deteriorationof the seal material, which loses its elastic properties. As a

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result, a seal cannot follow the shaft displacements, an eventthat constitutes seal failure. Factors affecting fatigue are:

Fluctuating temperature Chemical environment Oscillating shaft

Pressure at the seal-shaft interface

Not all the above factors are equally important in seal reliability. Toassign some weight to these factors would require more specific data aboutthe design and the seal. Note that seal reliability is affected by the process ofassembly as well, since it can simply be damaged while assembled. Inpractical situations, experience accumulated in industry is used as a guide.

This shows how complicated it is to access the reliability of acomponent in an operating system. Again, there are factors associated withthe component itself and with the system in which this component operates,and factors that are external to the system. Given the complexity of physical

processes and uncertainties associated with these processes and the system,true reliability is impossible to predict for a component in many situationspreviously unencountered. True reliability will reveal itself when the productis in service. In the mean time, at the detail design stage, use of designexperience and analytical and experimental methods will increase therelative level of reliability.

3. A brake hoist is shown. Two springs are exerting forces on thecorresponding shoes, thus keeping the brake normally locked. To unlock thebrake, a hydraulic actuator is used.

a. Draw a reliability diagram of the brake considering only components

indicated.b. Draw a reliability diagram if the brake is normally unlocked butbecomes locked when hydraulic pressure is decreased. Explain thedifference between the two cases.

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1

246

3

6 5 34 2 1

5

Figure 3.1 Brake Hoist

In the first case, the brake is normally locked, thus both the springs areexerting forces on the corresponding shoes. To unlock the brake, thehydraulic actuator that is used must retract both the springs and therebyboth the shoes, so the reliability diagram consists of all the components inseries system. Therefore if one of the components fails, the brake does notunlock.

In the second case, the brake is normally unlocked but becomes lockedwhen hydraulic pressure is decreased. This is possible even if one brake shoefails to lock the other shoe locks the brake. But for the above situation to be

true all the components that lock the individual brake shoes must be inseries.

Figure 3.2 Reliability Diagram

Figure 3.3 Reliability Diagram

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4. Three possible configurations of gear reduction boxes are shown.Assuming that the input output characteristics of all boxes areidentical,

a. Discuss the reliability advantage and disadvantage of everyconfiguration and

b. Draw the reliability diagram for each configuration considering onlysuch components as gears

Figure 4.1 Gear Configurations

In the first case, the failure of any one of the elements of thesystem will lead to failure of the system. So all the components in thesystem (the bearing 1, input shaft, gears, output shaft and bearing 9are in series. The above system has a reliability advantage over theothers by having the least number of components.

Figure 4.2 (a)

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1 2 3 4 5 6 7

1

2 3

4

6

5

7

9

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Figure 4.3 (b)

Figure 4.4 (c)

In the second case the reliability is increased by having a redundantset of gears from the input shaft to the midddle shaft. So reliability in speedreduction is enabled to a greater extent than the previous case. But sincethere is only one set of gears from the middle shaft to the final output shaft,failure of this set of gears may also lead to a system failure. So reliability isreduced due to the above mentioned reason.

In the third case , reliability is reduced at the intial stage itself wherethere is only one set of gears from the input shaft to the middle shaft. Afterthis the reliability is increased with the redundant set of gears from middleshaft to the output shaft.

5. Two possible configurations of spur gear - support system are shown:a. Discuss the reliability advantages and disadvantages of both

configurationDraw the reliability diagram of the gear-bearings-shaft systemDiscuss the influence of the system on the bearings reliabilty

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1 2 3 45 6

7 8

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S

B

B

G

Figure 5.1 Spur Gear Support System

Figure 5.2 Reliability Diagram (A)

Figure 5.3 Reliability Diagram (B)

In both the cases, the presence of a redundant bearing enables thesystem to function even in the event of failure of one of the bearings. Thereliability of the spur gear support system depends on the position of thebearing and the shape of the gear.

In the first case, the cross-section of the gear is irregular therebyallowing a difference in contact pressure. i.e., both the bearings are notworking under similar load conditions.

In the second case, the load on the bearings is same, so reliability ismore. Further, the end support provided by the side of the bearing is placedin such a way that its failure will not affect the working of the system to agreat extent unlike the first case.

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S

B

B

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6. Two concentric springs make up a support. Both the springs are knownto be in a state of random failure with the rates h1 = 0.005failures/kilocycle and h2 = 0.008 failures/kilocycle for the inner andouter springs respectively. Each of the two springs is capable of

withstanding the fluctuating load. However, if the inner spring fails, thefailure rate of the outer one becomes h2* = 0.012 failures/kilocycle; ifthe outer spring fails the failure rate of the inner one becomes h1* =0.0075 failures/kilocycle. What is the reliability function of the supportfrom 0 to 400 kilocycle?

Figure 6.1 Concentric Springs

Failure rates

Inner spring h1 = 0.005 failures/kilocycleOuter spring h2 = 0.008 failures/kilocycle

t = 400 kilocycles

Probability of both the springs operating successfully

P1 (t) = exp [- (h1+ h2) t]= exp [- (0.005 + 0.008)400 ]= 0.00551

Probability of outer spring failing

P2 (t) = { exp ( - h1* t) - exp [ - (h1+ h2 ) t ]}[ h1 / ( h1 + h2 - h1*)]= 0.04427 * 1.4545= 0.06439

Probability of the inner spring failing

P3 (t) = { exp ( - h2* t) - exp [ - (h1+ h2 ) t ]}[ h1 / ( h1 + h2 - h2*)]

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= 0.002719 * 5= 0.01359

Reliability function

R(t) = P1 (t) + P2 (t) + P3 (t)

= 0.00551 + 0.06439 + 0.01359= 0.083

7. The failures of 15 shock absorbers were recorded. The results (inKilocycles) were as follows :2.5,3.5,4,6.5,7,9.5,10.5,14,17,19,21,24,27,32 and 36. If it known thatthe shock absorbers were in a state of random failure, what would the95 and 99 percent two sided confidence limits on the mean life be?

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The estimation of the mean life is * = 1/15 ti i=1

= 1/15 (2.5+3.5++32+36)

= 15.6 Kilocycles

The chi-square values from tables are

95 percent confidence ( = 0.05), 2n = 30

2 [1-(/2), 2n] = 2 (0.975, 30) = 16.791

2 [(/2), 2n] = 2 (0.025, 30) = 46.979


2 [1-(/2), 2n] = 2 (0.995, 30) = 13.787

2 [(/2), 2n] = 2 (0.005, 30) = 53.632

Then the 95 percent confidence limits for are

(U) = 2n / { 2 [1-(/2), 2n]}= (30 *15.6) / 16.791= 27.8

(L) = 2n / { 2 [(/2), 2n]}= (30 *15.6) / 16.791= 9.999 10

Then the 95 percent confidence limits for are 10.0 27.8

Then the 99 percent confidence limits for are

(U) = 2n / { 2 [1-(/2), 2n]}= (30 *15.6) / 13.787= 33.94

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(L) = 2n / { 2 [(/2), 2n]}= (30 *15.6) / 53.632= 8.72

Then the 99 percent confidence limits for are 8.72 33.94

It is seen how an increase in the confidence level (1-) widens the confidenceinterval. A probability that the true mean is within the specified limits ishigher if the range is wider.

8. For the data given in the previous problem determinea. 95 percent low confidence limit for the 10 Kilocycle reliability andcompare it with the estimated reliability for this number of cycles.b. 95 percent two sided confidence limit for the number of cycles atwhich 15% of the shock absorbers will have failed.

a. The estimated reliability is given by

R*(t) = exp (- t / *)

R*(10) = exp (- 10 / 15.6)= 0.526

where as the lower confidence limit is

R (10) = exp (- 10 / (L))= exp (-10/10)= 0.368

Thus there is a 95% chance that the true reliability after10 Kilocycleswould be as low as 0.368, which is much smaller than the estimatedvalue 0.526.

b. The estimated number of cycles for the reliability level of 0.85 is

tp* = * ln [(1-p)-1]t0.15 = 15.6 ln[(1-0.15) -1]

= 2.53 Kilocycles

The confidence limits for tp are

(L) ln [(1-p)-1] tp (U) ln [(1-p)-1]

10 ln [(1-0.15)-1] tp 27.8 ln [(1-0.15)-1]

1.6 Kilocycles tp 4.5 Kilocycles

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9. Consider the data given in problem 7. Do they fit the exponentialdistribution and if so, what is the value of the parameter ? Let us take

ti = 5 Kilocycles. Note that in this case n = 15, carry out thecalculations and plot the results on the weibull paper.

Table 9.1 Parameter values:

tiKilocycles

Ns(ti) F(ti) R(ti) f(ti) h(ti)

< 5 12 0.176 0.824 - -5-10 9 0.370 0.630 0.0388 0.06210-15 7 0.5 0.5 0.026 0.05215-20 5 0.63 0.37 0.026 0.07

20-25 3 0.76 0.240 0.026 0.10825-30 2 0.825 0.175 0.013 0.07430-35 1 0.89 0.11 0.013 0.11835-40 0 0.955 0.045 0.013 0.288Where F(ti) = 1 [(Ns(ti)+ 0.7)/(N +0.4)] , here N = 15

R(ti) = 1- F(ti)f(ti) = d[R(ti)- R (ti+1)]/dth(ti) = f(ti) / R (ti)

It is seen that a straight line approximates the failure data fairly accurately,except for the last point. For the comparison h(t) is calculated in the abovetable, and the results are shown in the plot. It is seen that h(t) has a small

upward trend if it was not the last point (the out-of-sequence point can bedue to the small size of the recorded data).The simplest way to find is touse the equation

= - t/ (ln R (t))= - t/ (ln 1-F (ti)),

for any point t =ti that lies on the straight line. Let us take t i =20 Kilocyclesand the corresponding F (ti) = 0.63

Then, = - t/ (ln 1-F (ti)),

= 20/0.994= 20.1 Kilocycles

This estimation gives larger than the mean * = 15.6 Kilocycles in problem7. At the same time for * = 15.6, the constant hazard value h* = 0.064 andapproximates only the initial part of the failure times well.

This discrepancy between the two values of estimated by twomethods, graphical and numerical may be due to the errors in plotting and

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interpretation of data on graph, to the assumption that the data can bedescribed by exponential distribution (in fact, it cannot be), or both.

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Figure 9.1 Weibull Plot

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10. Failures of 20 shafts operating at a constant stress level were reported.The decimal logarithm, log K, of the no. of cycles to failure was distributed asfollows: 4.2, 4.4, 4.58, 4.6, 4.7, 4.76, 4.81, 4.86, 4.88, 4.96, 4.98, 4.99, 5.02,5.12, 5.17, 5.25, 5.36, 5.4, 5.49 & 5.78. 1. Take the interval t = (log K) =0.2 and plot the reliability, failure density and hazard functions. Does it looklike log K is normally distributed? 2. Assume that log K is a normally

distributed variable and that the mean and standard deviation can beapproximated by the average and an unbiased variance, respectively. Whatno. of failures should be expected in a population of 100 products with similarshafts in similar conditions during the first 100 kilocycles? 3. What are the 95percent confidence limits for the no. of failures during the first 100 kilocycles?

The test and reliability data are given in the table as follows. Plots of R (t i),f (ti), h (ti) are as shown in the figure below. Although h(t) function isincreasing, which means that qualitatively a normal distribution for log Kwould not be contradictory, the f(t) function for the assumed t does notgive an indication that the normal distribution is a good fit.

The average value of log K is 20

*log K = 1/20 (log K)i i=1

= 4.96

and the variance of log K is20

S2log K = 1/19 (log K - *log K)2i=1

= 0.146

and according to the assumption = S log K= 0.382.

The number of failures during 100 Kilocycles is

Nf (number of cycles < 105) = N0F(100) = N0(z100)

Table 10.1 Parameter Values:

Ti = log K Ns F(ti) R(ti) f(ti) h(ti)4.2-4.4 18 0.083 0.917 - -4.4-4.6 16 0.181 0.819 0.49 0.6

4.6-4.8 14 0.279 0.721 0.49 0.684.8-5.0 8 0.574 0.426 1.48 3.475.0-5.2 5 0.721 0.279 0.74 2.655.2-5.4 2 0.868 0.132 0.74 5.605.4-5.6 1 0.917 0.083 0.25 3.015.6-5.8 0 0.966 0.034 0.25 7.35

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Where F(ti) = 1 [(Ns(ti)+ 0.7)/(N +0.4)] , here N = 20R(ti) = 1- F(ti)f(ti) = d[R(ti)- R (ti+1)]/dth(ti) = f(ti) / R (ti)

Where

z100 = (log 105 log K*) / = 0.105

From the normal table (B-1) (0.105) = 0.5596, so Nf = 56 shafts. At thesame time, if F(ti = 5.0) = 0.574 is used from the test data directly,without implying a normal distribution, the no. of failed shafts is 57, whichis in a good agreement with the result based on mathematical model

The confidence interval on the population mean is given by

*log K t /2 , v [ Slog K / 20 ] log K *log K+ t /2 , v [ Slog K / 20 ]

For = 0.05 and v = n-1 = 19 the t value from the tables( B -3) is t 0.025,19= 2.093, then

4.781 log K 5.138

The confidence interval on the variance is

[(n-1)S2 log K] / 2 /2 , v 2 [(n-1)S2 log K] / 2 1 - /2 , v

For = 0.05 and v= n-1, the 2 values are (table B-2)

2 0.025, 19 = 32.852 and 2 0.975, 19 = 8.907

then,0.084 2 0.311

The corresponding limits for z100 are

-0.247 z100 0.756

From the normal table( B-1 ), ( -0.247) = 0.401 and ( 0.756) =0.7764

Thus there is a 95 percent confidence that the no. of failed shafts may bewithin the limits

40 N f 78

This indicates that the margin of uncertainty if N f = 56 would be taken as atarget in planning the maintenance schedules.

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Figure 10.1 Reliability, failure, and hazard functions for shafts

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11. Consider the data in problem 9 for 20 failed shafts, and (1) Make aweibull probability plot, (2) estimate the parameters of the distribution, (3)find out the nature of the failure rate, and (4) find out whether thedistribution is close to normal, an assumption made in problem 9.

(1) The plot is constructed on the basis of data in table 10.1 assuming =

0. It looks like a straight line fits the experimental points, and thus theWeibull distribution can be assumed to be appropriate in this case.

(2) The estimation of can best be accomplished by taking any twopoints that lie on a straight line (F1, t1) and (F2, t2).

Then

= [ln ln (1- F2)-1 - ln ln (1- F1)-1 ] / [ln t2- ln t1]

Taking from table 10.1(F1, t1) =(0.083,4.3) and (F2, t2) = (0.966,5.7) gives

= [1.218- (-2.466)] / [1.74-1.46]= 13.08

The is found from the plot for F(t) = 0.632; t = = 5.0. Here = 0, sincethe line is sufficiently straight.

(3) Since > 1, the failure rate is increasing.

(4) The distribution would be close to normal if 3.44. In this case it ismuch greater. Thus the assumption made in problem 9 does not seem tobe satisfactory

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Figure 11.1 Weibull Plot

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12. Failures of dc motors in specified intervals of days are given in table

12.1. Assuming that the dc motors were in a state of random failure,what is the 95 percent, two-sided confidence interval for the mean life?Determine the number of days at which 10 percent motors will havefailed.

Table 12.1 Failure Of Dc Motors:

Interval0-200 200-400 400-600 600-800 800-1000

Number offailures

5 3 2 2 1

Table 12.2 Probability of Failure:

Days Mid point Number of failures

Probability offailure (Pi)

0-200 100 5 0.384200-400 300 3 0.230400-600 500 2 0.153600-800 700 2 0.153800-1000 900 1 0.077

Expected mean time to failure isn

m = ti Pi i=1

= {100(0.384) + 300(0.230) + 500(0.153) + 700(0.153) +900(0.077)}

m = * = 360.3


2 [1-(/2), 2n] = 2 (0.975, 26) = 13.844

2 [(/2), 2n] = 2 (0.025, 26) = 41.923

= 2n / {2 [1-(/2), 2n]}

= (26 *360) / 13.844

= 676.1 days

= 2n / {2 [(/2), 2n]}

= (26 *360) / 41.923

= 223.2 days

Then the 95 percent confidence limits for are 223.2 * 676.1

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2. 10% failure P =0.1

Time to failure tp = * ln [(1-p)-1]

= 360 ln (1-01) 1

= 37.92 days

analysis of reliability of engineering design

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