“Using Weibull Analysis to Solve REAL Engineering Problems”

Jim Brenemanweibullman@gmail.com

ASQ-RRD Newsletter Editor & Nominating ChairSAE Fellow- Reliability

1. What Happens if a Weibull distribution doesn’t fit the data?(Comparing the Weibull to other possible distributions)

2. Weibull Confidence Bounds and their use in Reliability.3. Regression with Life data

(Modeling S/N data, …)

4. Sudden Death Testing

a follow-on to:“We just had a failure, Will Weibull Analysis Help?”

and

Special Topics in Weibull Analysis

Weibull Definitions

• η (Eta)= characteristic Life• ≈ mean time to failure• β (Beta) = “slope” of line*• ≈ failure mode type• Weibull equation:•• If t=η:

• Bxx life=age at which xx% of the fleet fail• B1 life= age at which 1% of the fleet fail• B10 life= age at which 10% of the fleet fail

% 1t

Cumulative failed e

β

η

− = −

1% ( ) (1 ) 100% 63.2%Cumulative failed at t xe

η= = − =

100001000100

9580

50

20

5

2

1

0.01

Time(hrs)

Perc

ent

63.2%B10

η

Δy

Δx

* On special 1-1 paper only

But first, a quick review

Steady StateFailure rate

Failu

re R

ate

Operating Time (hours, cycles, months, seconds)

β < 1 β = 1 β > 1

The Weibull Distribution can describe each portion of the Bathtub curve

Typical failure modes: Typical failure modes Typical failure modes:

* Inadequate burn-in* Misassembly* Some quality problems

* Independent of time* Maintenance errors* Electronics* Mixtures of problems

* LCF* TMF* HCF* Stress rupture* Corrosion

But first, a quick review

Your Weibull plot looks like one of the following:

The data has a gradual convex (or concave) bendon Weibull paper.

Solution:Needs a t0 correction (positive or negative)or a different distribution.

The data has a “dogleg” bend or cusp when plotted on Weibull paper.Solution:Break the data apart… two (or more) failure modes (use an inspection of the parts to tell you where)

Most of unfailed data

Failures are mostly low-time parts or Serial numbers of failed parts are close togetherSolution:Batch problem… look for the cause-supplier quality, maintenance, bases/customers.

What Happens if a Weibull distribution doesn’t fit the data?

• Batch Problems• Look for design changes (Class 1 or 2)2.

Inspection differences• Multiple failure modes

• Identify different modes and separate into distinct Weibull analyses

• Positive Start Time, Parts can’t fail until used later (happens occasionally)

• Use a t0 adjustment

Batch problemsor Positive t0

or Log Normal

Time-to-Failure

ProbabilityCDF(%)

Batch problemsor Positive t0

or Log Normal

Time-to-Failure

ProbabilityCDF(%)

Negative t0or

classic Bi-Weibull

Time-to-Failure

ProbabilityCDF(%)

Negative t0or

classic Bi-Weibull

Time-to-Failure

ProbabilityCDF(%)

• Multiple failure modes (Most Likely Cause)• Identify different “stressors”1 and

separate into distinct Weibull analyses• Batch Problems

• Look for design changes (Class 1 or 22), Inspection differences

• Negative Start Time, parts degrade before being used –shelf aging (this is rare)

• Use a t0 adjustment

Curved Weibulls

2 Class I changes affect an item's fit, form or function. These are changes that affect an item's specifications, weight, interchangeability, interfacing, reliability, safety, schedule, cost, etc. Class II changes are changes to correct documentation or changes to hardware not otherwise defined as a Class I change.

1 “Stressor” could be different failure modes, or same failure mode at two different locations, two different suppliers, etc

Curved Weibulls & the positive t0 correctionThe data appears curved on Weibull paper, AND you have ruled out batch Problems and multiple failure modes, NOW What?

( ) ( )( )01 t tF t eβη− −= −

• Time does not start at zero.• Early failures are

impossible.• Last minute checks:

• Early failure data not reported?• Run-in time not accounted for?

• Then it’s time for the 3 parameter Weibull:

1 00001 0001 001 010.1

9580

50

20

5

21

0.01

TTF

Perc

ent

Step 1: A “back-of-the-envelope” method to find the t0 correction.

• To estimate t0, read the time from the curved line when it “lines out”.

Use this procedure to check the reasonableness of your MINITAB® Weibull

Think K&E #48“Ships curve”

~18

Step 2: Use software (e.g. MINITAB) to fit the data to a three-parameter Weibull

t0 = 17.5

β = 0.96

η = 544.1

Example: B1 =

• Estimate β and η in the usual way, from the ‘straightened’ line.

• Shift all points by subtracting t0.

• Add back t0 when calculating B1 life.

4.5 +17.5 =22 hours

3 parameter Weibull caveats:

• The Weibull plot should show curvature.• There should be a physical explanation of why failures:

• Cannot occur before positive t0(e.g. bearing spalling)

• Or, age on the shelf before entering service negative t0(while this does not occur that often,

improperly desiccated bearings are an example)

• A significant increase in the correlation coefficient (r) should be observed in going from 2-parameter to 3-parameter Weibull(Ranked Regression).

(After ruling out batch problems and multiple Failure modes)

Curved Weibulls & the lognormal distribution…

• Sometimes the Weibull distribution may not be the best choice.• Sometimes time-to-failure data follow the lognormal distribution.

Consider the following data froma lab coverplate study program:(Cycles to failure)

Coverplate failures(cycles)19892160256927582813297930163283329435033853391642944462517857165984637865567000

Comparing Weibull & lognormal distribution fits

Lognormal wins!

OR DOES IT!!

SOMETIMES, you could be fooled!When you look into the “Physics ofFailure” for this failure mode it turns out that the lognormal can be derived for this failure mode3,

so yes, Lognormal does WINS!! 3 See Ireson, Reliability Handbook, 2-8, 1966.

Confidence Intervals give:

* A plausible range of values for a population parameter(e.g. β, η, Bx life, R(t), FailureRate(t)).

MINITAB can generate all of these confidence bounds and more.

* The precision of an estimate.(When the number of failures in a large sample is low, the confidence interval will be wide to reflect the uncertainty of the observation.)

However, confidence bounds are more useful in comparing two data sets (e.g. two suppliers, two locations, two treatments, etc.).

Weibull Confidence Bounds and their use in Reliability.

Comparing two Weibull distributions …The two datasets might be from:

1. New design vs old design2. Different geographic sources3. Different fleets or customers4. Different product usage 5. Different production lots6. Different vendors7. Different alloys (e.g. material life comparison)

Comparing two Weibull distributions …Two methods:

1. 80% confidence bounds do not overlap at B10 life. This will be give you 90%confidence the two sets of data are different if the bounds do not overlap.(MINITAB™ can be used here)

2. Maximum likelihood (β,η) contours do not overlap.(Need Supersmith™, Reliasoft’s Weibull++ ™, SAS™ … MINITAB® doesn’t

have that capability yet)

Comparing two Weibull distributions …• Suppose we have asked two possible suppliers for a particular component

to test 8 each of their production of the same component to failure.• The Weibull analysis for each sample results in the following:

• Weibull distribution for supplier A: β=2.64, η=872)• Weibull distribution for supplier B: β=2.73, η=1931)

• Should we conclude that there is a real (statistically significant) difference in the reliability performance of the two suppliers?

• Method 1: Bounds do not overlapUsing 2-sided 80% CI’s on B10 for each sample, declare a difference if their CI’s do not overlap.

For A: B10 = 372, 2S-80%CI = (248, 559)For B: B10 = 847, 2S-80%CI = (570, 1254)

1 00001 0001 00

99

90807060504030

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5

32

1

2.640 872.2 8 0.228 >0.2502.726 1 931 8 0.1 87 >0.250

Shape Scale N AD P

Data

Perc

ent

Supplier ASupplier B

Variable

Demo Part lives from Supplier A and BWeibull - 80% 2-sided CI for each supplier

Comparing two Weibull distributions …

2-sided 80% CIs

Declare a difference, since the CI bounds do not overlap.

Comparing two Weibull distributions

• Using Supersmith™: Calculate 90%CI η, βcontours for each sample.

• Declare the sample to be different since there is no overlap in the contours.

Supplier BSupplier A

90% Conf. Contour for MLE Comparison of Two Samples

Method 2: Using Maximum Likelihood contours

Regression with Life Data• Regression with Life Data performs a regression with one or more

predictors. The model can include factors, covariates, interactions, and nested terms. This model will help you understand how different factors and covariates affect the lifetime of your part or product.

• In Minitab Regression with Life Data differs from regression commands in that it accepts censored data and has the capability to use different distributions (e.g. Weibull, log-normal) for the response since most life data is NOT normally distributed.

jeb18

Regression with Life DataIn its simplest form, let’s look at some superalloy data

that is a function of stress alone:

jeb19

Stress(ksi) Cyclestofail Code145.9 5733 185.2 13949 1

116.4 15616 187.2 56723 1

100.1 12076 185.8 152680 199.8 43331 1113 18067 1

120.4 9750 186.4 156725 185.6 112968 286.7 138114 289.7 122372 2

114.8 21300 1144.5 6705 191.3 112002 1

142.5 11865 1100.5 13181 1118.4 8489 1118.6 12434 1

118 13030 180.8 57923 287.3 121075 180.6 200027 180.3 211629 184.3 155000 1

Now, let’s fit:Cyclestofail=f(stress)

Assuming Cycles to fail is lognormallyDistributed.

LCF data from strain-controlled testingFailure code (1=failed, 2=runout)

From Nelson , Accelerated Testing, 1990, p 272.

Superalloy Cyclestofail=f(stress)

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Superalloy Cyclestofail=f(stress, stress2)

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60

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90

100

110

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150

160

0 50000 100000 150000 200000 250000

Stress(ksi)

Calcw/Stress^2

Stre

ss(k

si)

Cycles to Failure

Actual cycles to failure

Life =f(Temp, Stress) example• Life data regression can be done

for life as a function of more than one factor. For example, suppose we have:

• Life=f(Stress,Temp) (see tab “Life=f(temp,stress)”)

• Using MINITAB’s Regression with Life data, fit this data, trying both lognormal and Weibull life assumptions.

jeb24

Temp(deg C) Stess(ksi) Hrs to failure Censor180 200 959 1180 200 1065 1180 200 1065 1180 200 1087 1180 200 1087 2180 200 1087 2180 200 1087 2180 200 1087 2180 250 216 1180 250 315 1180 250 455 1180 250 473 1180 250 473 2180 250 473 2180 250 473 2180 250 473 2180 300 241 1180 300 315 1180 300 332 1180 300 380 1180 300 380 2180 300 380 2180 300 380 2180 300 380 2180 350 241 1180 350 241 1180 350 435 1180 350 455 1180 350 455 2180 350 455 2180 350 455 2180 350 455 2

Temp(deg C) Stess(ksi) Hrs to failure Censor170 200 439 1170 200 904 1170 200 1092 1170 200 1105 1170 200 1105 2170 200 1105 2170 200 1105 2170 200 1105 2170 250 572 1170 250 690 1170 250 904 1170 250 1090 1170 250 1090 2170 250 1090 2170 250 1090 2170 250 1090 2170 300 315 1170 300 315 1170 300 439 1170 300 628 1170 300 628 2170 300 628 2170 300 628 2170 300 628 2170 350 258 1170 350 258 1170 350 347 1170 350 588 1170 350 588 2170 350 588 2170 350 588 2170 350 588 2

From Zelen, "Factorial Experiments in Life Testing",Technometrics, pp269-288, 1959.

Regression with Life Data: Hrs to failure versus Temp(deg C), Stess(ksi)

Response Variable: Hrs to failure

Censoring Information CountUncensored value 32Right censored value 32Censoring value: Censor = 2Estimation Method: Maximum LikelihoodDistribution: Weibull

Regression Table

Standard 95.0% Normal CIPredictor Coef Error Z P Lower UpperIntercept 18.6781 3.53440 5.28 0.000 11.7508 25.6054Temp(deg C) -0.0381557 0.0130746 -2.92 0.004 -0.0637814 -0.0125300Stess(ksi) -0.0336215 0.0145797 -2.31 0.021 -0.0621972 -0.0050458Stess(ksi)*Stess(ksi) 0.0000500 0.0000262 1.91 0.056 -0.0000014 0.0001014Shape 2.93802 0.449028 2.17752 3.96412

Log-Likelihood = -242.546

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Life =f(Temp, Stress) example- Weibull

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210-1-2-3-4-5

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90807060504030

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Loc 0.0000000Scale 1Mean -0.57721 6StDev 1 .28255Median -0.36651 3IQR 1 .57253Failure 32Censor 32AD* 84.578

Table of Statistics

Standardized Residuals

Perc

ent

Probability Plot for SResids of Hrs to failure

Censoring Column in Censor - ML EstimatesSmallest Extreme Value

Life =f(Temp, Stress) example- Weibull

Sudden Death TestingIn Sudden Death Testing the total sample to be tested (from a recommended as few as nine units to more than 15 units, depending on cost)• Is divided into groups of three or more. • All groups being of equal size. • All units in each group are (preferably) tested simultaneously. • When the first unit in a group fails, the group has failed (weakest link) and testing

is stopped on the remaining units in the group. Hence the name “Sudden Death” testing. Sudden Death testing can reduce test times by 60% or more depending on the failure mode/part.

Saving experimental test time and/or calendar time (i.e. $$).

Example:Suppose you have 40 bearings and divide them into 5 groups of 8 bearings each.The median rank of the first failure in each group of 8 in a Weibull distribution is:

1

i-0.3Median Rank=N+0.4

1-0.3so, for the 1st failure in 8, MR = 0.0838+0.4

Therefore, the failure time each of the first failures in the 5 groups will be on the Sudden death 8.3% line. From this we can

=

extrapolate to the Weibull for the entire population of 40 bearings (the remaining bearings in each group are censored at the time of the first failure in the group).

Example 1:

Suppose the first failure of 8 in each of the five groups failed at: 65, 120, 155, 200, 300 hours.A Weibull plot can be produced with the 5 failures and 35 censored times.The MINITAB data input would look like:

Combined 5 failures with 35 censored times. Here Censor=1 indicates failed time, Censor=0 indicates a censored time. Freq indicates how many points at the time.

data from Kececioglu(Reliability & Life Testing, Vol 2, 1993)

Bearingfailure(Hours) Freq Censor

65 1 1120 1 1155 1 1200 1 1300 1 1

65 7 0120 7 0155 7 0200 7 0300 7 0

Sudden Death vs “Total Life” TestingExample 2:A company tests 16 bearings to failure out of every batch of 100,000 bearings. To compare the efficiency of Sudden Death testing relative to testing all 16 to failure the 16 bearings were divided into 4 sets of 4 each. Each set of 4 was tested until the first failure occurred, but then the remaining three were all tested to failure as well. This same procedure was followed for the other three sets of 4 bearings. At the end they had results of a Sudden Death test AND results of having tested all bearings, in order to compare the results.

Sudden Death vs “Total life test”

Set 1 Set 2 Set 3 Set 418993 107470 97783 11486252196 105354 93105 6991073178 88331 180174 9760427520 182614 25562 72751

Analyzing the data as Sudden Death testing gives B10 life of 27546 cycles.

Sudden Death vs “Total Life” Testing

Testing all 16 bearings to failure gives B10 life=29041cyclesConclusion: In this case Sudden Death testing produced a slightly smaller B10 life. However, the difference in potential testtime savings is significant: Sudden death: 608,389 total test cycles Testing all to failure test time= 1,407,407 cycles A test time saving of ~800,000 cycles.

Sudden death testing isn’t ALWAYS this good, in this case it was. The small number of bearings tested overall and, in each set, gives a wider confidence bound. So, the more items overall, and hence in each group, the closer Sudden Death testing will be (and of course save considerable test time). However, for expensive test articles, and hence smaller numbers of articles available (~15 or fewer) Sudden Death testing is not an Optimal choice.

A final thought:“The purpose of models is

often not to fit the data but to sharpen the

questions.”