probabilistic aspects of fatigue · probabilistic aspects of fatigue ... safety factors. larger...

1

Introduction

Professor Darrell F. SocieDepartment of Mechanical and

Industrial Engineering

© 2003-2005 Darrell Socie, All Rights Reserved

Probabilistic Aspects of Fatigue

•Probabilistic Fatigue © 2003-2005 Darrell Socie, University of Illinois at Urbana-Champaign, All Rights Reserved 1 of 352

Contact Information

Darrell SocieMechanical Engineering1206 West GreenUrbana, Illinois 61801

Office: 3015 Mechanical Engineering Laboratory

[email protected]: 217 333 7630Fax: 217 333 5634

2


Fatigue Calculations

Who really believes these numbers ?


SAE Specimen

Suspension

Transmission

Bracket

Fatigue Under Complex Loading: Analysis and Experiments, SAE AE6, 1977

3


Analysis Results

1 10 100

48 Data PointsCOV 1.27

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

Analytical Life / Experimental Life

Strain-Life analysis of all test data

Non conservative

Conservative


Material Variability

1 10 100

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity


Material Analysis

Strain-Life back calculation of specimen lives

4


Probabilistic Models

Probabilistic models are no better than the underlying deterministic modelsThey require more work to implementWhy use them?


Quality and Cost

TaguchiIdentify factors that influence performanceRobust design – reduce sensitivity to noiseAssess economic impact of variation

Risk / ReliabilityWhat is the increased risk from reduced testing ?

5


Risk

10-9

10-8

10-7

10-6

10-5

10-4

10-3

Ris

k

Time, Flights etc

Acceptable risk


Reliability

99 %80 %

50 %

10 %

1 %

0.1 %

Exp

ecte

d Fa

ilure

s

106

Fatigue Life

105104103

6


Risk Contribution Factors

OperatingTemperature

Analysis Uncertainty

Speed

MaterialProperties

ManufacturingFlaws


Uncertainty and Variabilitycustomers

materials manufacturing

usage

107

Fatigue Life, 2Nf

1

10 102 103 104 105 106

0.1

10-2

1

10-3

10-4

Stra

in A

mpl

itude

time

50%

100 %

Failu

res

Strength

Stress

7


Deterministic versus Random

Deterministic – from past measurements the future position of a satellite can be predicted with reasonable accuracy

Random – from past measurements the future position of a car can only be described in terms of probability and statistical averages


Deterministic Design

Stress Strength

SafetyFactor

Variability and uncertainty is accommodated by introducing safety factors. Larger safety factors are better, but how much better and at what cost?

8


Probabilistic Design

Stress Strength

Reliability = 1 – P( Stress > Strength )


3σ Approach3σ contains 99.87% of the data

If we use 3σ on both stress and strength

The probability of the part with the lowest strengthhaving the highest stress is very small

P( s < S ) = 2.3 10-3

σ≈=≤≥Σ= − 5.4103.5)Sss(P)failure(P 6I

For 3 variables, each at 3 σ:

σ≈= − 7.5102.1)failure(P 8

9


Benefits

Reduces conservatism (cost) compared to assuming the “worst case” for every design variableQuantifies life drivers – what are the most important variables and how well are they known or controlled ?Quantifies risk



IntroductionBasic Probability and StatisticsStatistical TechniquesAnalysis Methods Characterizing VariabilityCase StudiesFatigueCalculator.comGlyphWorks

10

Basic Probability and Statistics








11


Deterministic versus Random

Deterministic – from past measurements the future position of a satellite can be predicted with reasonable accuracy

Random – from past measurements the future position of a car can only be described in terms of probability and statistical averages


Random Variables

Discrete - fixed number of outcomesColors

Continuous - may have any value in the sample space

Strength

12


Descriptive Statistics

Mean or Expected ValueVariance / Standard DeviationCoefficient of VariationSkewnessKurtosisCorrelation Coefficient


Mean or Expected Value

Central tendency of the data

( )N

xXExMean

N

1ii

x

∑====µ=

13


Variance / Standard DeviationDispersion of the data

( )N

)xx(XVar

N

1i

2i∑

=

−=

)X(Varx =σ

Standard deviation


Coefficient of Variation

x

xCOVµσ

=

Useful to compare different dispersions

µ = 10σ = 1

COV = 0.1

µ = 100σ = 10

COV = 0.1

14


Skewness

Skewness is a measure of the asymmetry of the data around the sample mean. If skewness is negative, the data are spread out more to the left of the mean than to the right. If skewness is positive, the data are spread out more to the right. The skewness of the normal distribution (or any perfectly symmetric distribution) is zero.

( ) 3

N

1i

3i

N

)xx(XSkewness

σ

−=

∑=


Kurtosis

Kurtosis is a measure of how outlier-prone a distribution is. The kurtosis of the normal distribution is 3. Distributions thatare more outlier-prone than the normal distribution have kurtosis greater than 3; distributions that are less outlier-prone have kurtosis less than 3.

( ) 4

N

1i

4i

N

)xx(XKurtosis

σ

−=

∑=

15


Covariance

A measure of the linear association between random variables

[ ])Y()X(E)Y,X(COV YXxy µ−µ−==σ


Correlation Coefficient

0 X

Y

0 X

Y

0 X

Y

0 X

Y

0 X

Y

0 X

Y

ρ = 1 ρ = 0ρ = -1

0 < ρ < 1 ρ ~0ρ ~ 0

[ ]YX

YX )Y()X(Eσσ

µ−µ−=ρ

16


Probability

Basic probabilityConditional probabilityReliability


Basic Probability

( ) 1AP0 ≤≤

( ) certain1AP =

( ) eimposssibl0AP =

The probability of event A occurring:

17


Conditional Basic Probability

( ) ( ) ( ) ( )BandABorA

BAPBPAPBAP IU −+=

( ) ( )( )AP

BAPABP I=

( ) ( )( )BP

BAPBAP I=

P(B) given A has occurred

union intersection


Independent Events

1 2

If A and B are unrelated

( ) ( )BPABP =

( ) ( )APBAP =

( ) ( ) ( )BPAPBAP ⋅=I

Suppose the probability of bar 1 failing is 0.03 and the probability of bar 2 failing is 0.04.

What is the probability of the structure failing?

18


Failure Probability

Bar 1 or bar 2 fails

( ) 03.0AP =

( ) 04.0BP =

( ) ( ) ( ) 0012.0BPAPBAP =⋅=I

P( failure) = 0.03 + 0.04 - 0.0012 = 0.0688

( ) ( ) ( ) ( )BAPBPAPBAP IU −+=


Reliability

P( no failure ) = Reliability

( ) AnotofyprobabilitAPDefine

( ) ( )A1PAP −=

( )BAPliabilityRe I=

( ) ( ) ( )BPAPBAP ⋅=I

( ) 9312.096.097.0BAP =⋅=I

For the 2 bar structure

( ) 0688.0liabilityRe1failureP =−=

19


Structural Reliability

Collapse occurs if any member fails

( ) ( )n4321 AAAAAPAP ⋅⋅⋅⋅= IIII

( ) ( ) ( ) ( ) ( ) ( )n4321 APAPAPAPAPAP ⋅⋅⋅⋅⋅⋅⋅=

( ) linkeachfor01.0APSuppose =

( ) ( ) 932.099.0AP 7 ==


Reliability

6

5

4

3

2

1

010-9 10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 1

Sta

ndar

d D

evia

tions

Probability

20


6 Sigma

6 sigma is 1 in a billion ( 0.999999999 Reliability )

Suppose a structure has 1000 bolted joints:

( ) ( ) 999999.0999999999.0AP 1000 ==1 in a million

3 sigma is ( 0.99865 Reliability )

( ) ( ) 26.099865.0AP 1000 ==

74 % failures


Statistical Distributions

UniformNormalLogNormalGumbleWeibull

http://www.itl.nist.gov/div898/handbook/

A useful on-line reference:

21


Cumulative Distribution Function

( ) ( )xXPXFx ≤=

( ) 0Fx =∞−

( ) 1Fx =∞

0

1

a b

Fx(X)


Probability Density Function

( ) ( )dxXfbXaPb

ax∫=≤<

( ) ( ) ξξ= ∫∞−

dfXFx

xx

0

1

a b

fx(X)ab

1−

22


Probability

( ) ( )dxXfxXxPx

xx∫

ε+

ε−

=ε+≤<ε−

0

1

a b

fx(X)ab

1−

x

2ε


Survival Function

0

1

a b

Fx(X)

( )XF1)X(S xx −=

23


Hazard Function

a b

hx(X)

( )XF1)X(f)X(h

x

xx −

=

Instantaneous failure rate


Mean or Expected Value

0

1

a b

fx(X)

( ) ∫∞

∞−

==µ dx)X(fxXE xx

24


Variance

( ) dx)X(f)X(XVar x2

x∫∞

∞−

µ−=

( ) ( ) 2x

2XEXVar µ−=

Probability Density

x

)X(Varx =σ


Normal Distribution

0.1

0.2

0.3

0.4

-3 -2 -1 0 1 2 3Quantile x

Prob

abilit

y D

ensi

ty

-3 -2 -1 0 1 2 3Quantile x

0.5

1.0

Cum

ulat

ive

Prob

abilit

y

σ

µ−−πσ

= 2

2

x 2)x(exp

21)x(f tables

25


Normal Probability Plot

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

0

1

2

3

-1

-2

-3

Sta

ndar

d D

evia

tions

Linear Scale

1NmP+

=


Normal Plot

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

0 200 400 600 800 1000

Mean 432.67COV 0.34

Maximum force from 42 drivers

Cum

ulat

ive

Pro

babi

lity

26


LogNormal Distribution

0.5

1.0

Cum

ulat

ive

Prob

abilit

y

1 32Quantile x

0.2

0.4

0.6

0.8

Prob

abilit

y D

ensi

ty

1 32Quantile x

σ = 0.6

σ = 1.0

σ = 0.6

σ = 1.0

σ

−

πσ= 2

2

x 2

)mxln(

exp2x

1)x(f tables

( )µ= expm


Useful Relationships

( )2xlnxlnx 5.0exp σ+µ=µ

+

µ=µ

2x

xxln

COV1ln

( )2x

2xln COV1ln +=σ

1)exp(COV 2xlnx −σ=

2x

xxlnx

COV1)exp(X

+

µ=µ=

27


Median / Mean ratio

COVX0 1 2 3 4 50

0.2

0.4

0.6

0.8

1.0

X

Xµ


σlnX - COVX

COVX

0 0.5 1.0 1.5 2.0

0.5

1.0

1.5

σ lnX

28


LogNormal Probability Plot

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

0

1

2

3

-1

-2

-3

Sta

ndar

d D

evia

tions

Log Scale


LogNormal Plot

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

100 1000

Median 431COV 0.34

200 500


Cum

ulat

ive

Pro

babi

lity

29


Gumbel Extreme Value

a-3b a+4ba a-3b a+4ba

1.0

0.5

Cum

ulat

ive

Prob

abilit

y

Quantile x Quantile x

0.5

Prob

abilit

y D

ensi

ty

−−

−=b

)ax(expexp)x(Fx

−−

−

−−

=b

)ax(expexpb

)ax(expb1)x(fx


Gumble Probability Plot

Cum

ulat

ive

Pro

babi

lity

99.9 %

99 %

80 %

50 %

10 %

1 %

99.9 %

99 %

80 %

50 %

10 %

1 %

Cum

ulat

ive

Pro

babi

lity

Linear Scale-2

-1

0

1

2

3

4

5

6

7

-ln( -

ln( F

x(X) )

)

30


Gumble Plot

200 400 600 800

Location 362.94Scale 124.61

Maximum force from 42 drivers99.9 %

99 %

80 %

50 %

10 %

1 %

Cum

ulat

ive

Pro

babi

lity


Weibull

1 32Quantile x

Prob

abilit

y D

ensi

ty 1.0

0.8

0.2

0.6

0.4

1 32

Cum

ulat

ive

Prob

abilit

y

Quantile x

c=0.5

c=1.0

c=3.0 1.00.8

0.2

0.6

0.4

c=0.5

c=1.0c=3.0

−−=

c

x bxexp1)x(F

−

=

− c

c

1c

x bxexp

bxc)x(f

31


Weibull Probability Plot

Cum

ulat

ive

Pro

babi

lity

Cum

ulat

ive

Pro

babi

lity

-7

-6

-5

-4

-3

-2

-1

0

1

2ln

( -ln

( Fx(X

) ) )

99.9 % 99 %

80 %

50 %

10 %

1 %

0.1 %

Log Scale


Weibull Plot

200 1000

99.9 % 99 %

80 %

50 %

10 %

1 %

0.1 %

Scale Factor: 473.56Slope: 3.8

500

Cum

ulat

ive

Pro

babi

lity

32


99.9 % Probability

785Weibull1223Gumbel1125LogNormal893Normal

Estimated maximum force from the distributions


Comparison

0 200 400 600 800 1000 1200Maximum Load

0.001

0.002

0.003

0.004

Pro

babi

lity

Den

sity

Normal

LogNormal

Gumble

Weibull c = 3.5

All distributions are approximately normal around the median

33


Joint Probability Density

x

y

fxy(x,y)

µX

µY


Strength

Stress



( )

( )

σµ−

+

σµ−

σ

µ−ρ−

σ

µ−ρ−

−

ρ−σσπ=

2

y

y

y

y

x

x

2

x

x2

2yx

xy

yyx2x12

1exp

121y,xf

ρ correlation coefficient

Normal distributions

34


2D - Joint Probability Density

Contours of equal probability

y

xµx

µY

-3σx

3σY

12

34

Stress

Strength


Acceptable Risk

SafetyP(failure) ~ 10-6 → 10-9

EconomicP(failure) ~ 10-2 → 10-4

35


Extreme Values

For most durability problems, we are not interested inthe “large extremes” of stress or strength. Failure is much more likely to come from moderately high stressescombined with moderately low strengths.


Key Points

Basic NomenclatureRandom Variables

Statistical DistributionCumulative distribution functionMeanVariance

ProbabilityMarginalConditionalJoint

36

Statistical Techniques








37


Statistical Techniques

Normal DistributionsLogNormal DistributionsMonte CarloSamplingDistribution Fitting


Failure Probability

)Sss(P ≤≥Σ I

Let Σ be the stress and S the fatigue strength

Given the distributions of Σ and S find the probability of failure

Stress, Σ Strength, S

s

38


Normal Variables

Linear Response Function

∑=

+=n

1iiio XaaZ

Xi ~ N( µi , Ci )

∑=

µ+=µn

1iiioz aa

∑=

σ=σn

1i

2i

2iz a


Calculations

Z = S - Σ

µz = µS - µΣ

22Sz Σσ+σ=σ

Σ ~ N( 100 , 0.2 ) σΣ = 20µz = 200 -100 = 100

S ~ N( 200 , 0.1 ) σS = 20


Safety factor of 2

2.282020 22z =+=σ

Let Z be a random variable:

39


Failure Probability

Failure will occur whenever Z <= 0

Z = S - Σ

Z = µz – z σz = 0

z = 3.54 standard deviations

P(failure) = 2 x 10-4

2.28100z

Z

Z =σµ

=

For this case only, a safety factor of 2 means a probability of failure of 2 x 10-4. Other situations will require differentsafety factors to achieve the same reliability.


Failure Distribution


What is the expected distribution in fatigue lives?

40


Fatigue Data

bf

'f )N2(2

S∆=σ

102 103 104 105 106 107

Fatigue Life101

1000

100

10

1

Stre

ss A

mpl

itude

'fσ

b1

'f

f 2SN2

σ

∆=

b1


LogNormal Variables

∏=

=n

1i

aio

iXaZ

a’s are constant and Xi ~ LN( xi , Ci )

∏=

=n

1i

aio

iXaZmedian

( )∏=

−+=n

1i

a2XZ 1C1CCOV

2i

i

41


Calculations


~ LN( 250 , 0.2 ) σ = 50

~ LN( 1000 , 0.1 ) σ = 100

b1

'f

f 2SN2

σ

∆=

'fσ

b = -0.1252S∆

8'f

8

f 2SN2Z σ

∆

==−

8'f

8

f 2SN2Z σ

∆==

−

( ) ( ) 1COV1COV1COV2

f

2 8282SZ −++= σ

−

∆


Results

'fσ∆S/2 2Nf Percentile Life

µx 250 1000 355,368 99.9 17,706,069COVx 0.2 0.1 4.72 99 4,566,613

95 1,363,200µlnx 5.50 6.90 11.21 90 715,589

X 245 995 73,676 50 73,676σx 50 100 1,676,831 10 7,586

σlnx 0.198 0.100 1.774 5 3,9821 1,189

b = -0.125 0.1 307

42


Monte Carlo Methods

( ) ( )

εσ+

σ=

∆ +cbf

'f

'f

b2f

2'ff N2N2

EE

2SK

Given random variables for Kf, ∆S, Find the distribution of 2Nf

'f

'f and εσ

Z = 2Nf = ?


Simple Example

Probability of rolling a 3 on a die

1 2 3 4 5 6

fx(x) 1/6

Uniform discrete distribution

43


Computer Simulation

1. Generate random numbers between 1 and 6,all integers

2. Count the number of 3’s

Let Xi = 1 if 30 otherwise

( ) ∑=

=n

1iiX

n13P


EXCEL

=ROUNDUP( 6 * RAND() , 0 )

=IF( A1 = 3 , 1 , 0 )

=SUM($B$1:B1)/ROW(B1)

5 0 03 1 0.54 0 0.3333334 0 0.255 0 0.26 0 0.1666671 0 0.1428573 1 0.253 1 0.3333336 0 0.3

44


Results

0.1

0.2

0.3

0.4

1 100 200 300 400 500

trials

prob

abilit

y


Evaluate π

x

y P( inside circle )

4rP

2π=

π = 4 P

x = 2 * RAND() - 1y = 2 * RAND() - 1

IF( x2 + y2 < 1 , 1 , 0 )

1

-1

1

-1

45


π

0

1

2

3

4

1 100 200 300 400 500

π

trials


Monte Carlo Simulation


~ LN( 250 , 0.2 ) σ = 50

~ LN( 1000 , 0.1 ) σ = 100'fσ

b = -0.1252S∆

b1

'f

f 2SN2

σ

∆=

Randomly choose values of S andfrom their distributions'

fσ

Repeat many times

46


Generating Distributions

0

1

x

Fx(X)

Randomly choose a value between 0 and 1

x = Fx-1( RAND )

RAND


Generating Distributions in EXCEL

=LOGINV(RAND(),lnµ,lnσ)

=NORMINV(RAND(),µ,σ)

Normal

Log Normal

47


EXCEL

893 204 134,677 1102 301 32,180 852 285 6,355 963 173 929,249 1050 283 35,565 1080 265 77,057 965 313 8,227 1073 213 420,456 1052 226 224,000 954 322 5,878 965 240 68,671 993 207 277,192 1191 368 11,967 831 210 59,473

'fσ

2S∆

2Nf


Simulation Results

103

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

104 105 106 107 108

Monte Carlo AnalyticalMean 11.25 11.21Std 1.79 1.77

Cum

ulat

ive

Pro

babi

lity

Life

48


Summary

Simulation is relatively straightforward and simple

Obtaining the necessary input data is difficult


Nomenclature

Population: the totality of the observations

Sample: a subset of the population

Population mean: µx

Population variance: σx2

Sample mean: X

Sample variance: s2

( ) xXE µ=

( ) 2x

2sE σ=

49


Sample Mean and Variance

in EXCEL =AVERAGE(A1:An)

in EXCEL =STDEV(A1:An)

n

xX

n

1∑

=

1n

)Xx(s

n

12

−

−=

∑

Mean

Variance


Confidence Intervals

( ) xXE µ=

What is the probability that a sample X is greater than µx? 50%

P( L <= µx <= U ) = 1 - α

There is a 1 – α chance of selecting a sample in the intervalbetween L and U that contains the true mean of the population

50


Translation

90% confidence

If we sampled a population many times to estimate the mean, 90% of the time the true population mean would lie between the computed upper and lower limit.


Confidence Interval - mean

Lower limit of µ

xx

1n, nstX µ≤− −α

Upper limit of µ

xx

1n, nstX µ≥+ −α

For a normal distribution:

51


Sample Size

0 5 10 15 20 25 30Sample size, n

0

0.5

1.0

1.5

2.0

xx

1n, nstX µ≤− −α

nt 1n, −α

95%

90%


5 Samples from Normal Distribution

0

20

40

60

80

100

120

140

0 10 20 30 40 50 60 70 80 90 100

95 % confidence – 5 samples will be outside confidence limit

Lower limit above meanUpper limit below mean

Sample Number

µx = 100 σx = 10

52


Confidence Interval - variance

Upper limit of σ

2x

1n,12

2xs)1n(

σ≤χ

−−α−

For a normal distribution:


Sample Size

0 5 10 15 20 25 300

2

4

6

8

10

Sample size, n

2x

1n,12

2xs)1n(

σ≤χ

−−α−

1n,12

)1n(−α−χ

−95%

50%

90%

53


Maximum Load Data

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

100 1000

Median 431COV 0.34

200 500


Cum

ulat

ive

Pro

babi

lity

90% confidence COV 0.41


Maximum Load Data

Maximum Load0 200 400 600 800 1000 1200

0.001

0.002

0.003

0.004

Pro

babi

lity

Den

sity

Normal COV = 0.34

LogNormal

Normal COV = 0.41

Uncertainty in Variance is just as important, perhaps more important than the choice of the distribution

54


Choose the “Best” Distribution

0 200 400 600 800 1000Maximum Load

Normal

LogNormal

Gumble

Weibull

0.001

0.002

0.003

Pro

babi

lity

Den

sity

15 samples from a Normal Distribution


Goodness of Fit Tests99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

0 200 400 600 800 1000

Cum

ulat

ive

Prob

abili

ty

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

0 200 400 600 800 1000

Cum

ulat

ive

Prob

abili

ty

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

100 1000200 500

Cum

ulat

ive

Prob

abili

ty

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

100 1000200 500

Cum

ulat

ive

Prob

abili

ty

99.9 %

99 %

80 %

50 %

10 %

1 %

200 400 600 800

Cum

ulat

ive

Prob

abili

ty

99.9 %

99 %

80 %

50 %

10 %

1 %

200 400 600 800

Cum

ulat

ive

Prob

abili

ty

Normal

LogNormal

Gumble

200

99.9 % 99 %

80 %

50 %

10 %

1 %

0.1 %

1000500

Cum

ulat

ive

Prob

abili

ty

200

99.9 % 99 %

80 %

50 %

10 %

1 %

0.1 %

1000500

Cum

ulat

ive

Prob

abili

ty

Weibull

55


Analytical Tests

Stephens, M. A. (1974). EDF Statistics for Goodness of Fit and Some Comparisons, Journal of the American Statistical Association, Vol. 69, pp. 730-737.

Kolmogorov-Smirnov

Chakravarti, Laha, and Roy, (1967). Handbook of Methods of Applied Statistics, Volume I, John Wiley and Sons, pp. 392-394.

Anderson-Darling

Snedecor, George W. and Cochran, William G. (1989), Statistical Methods, Eighth Edition, Iowa State University Press.

Chi-squareany univariate distribution

tends to be more sensitive near the center of the distribution

gives more weight to the tails


DistributionsNormal

StrengthDimensions

LogNormalFatigue LivesLarge variance in properties or loads

GumbleMaximums in a population

WeibullFatigue Lives

56


Statistics Software

www.palisade.com

BestFit Minitab

www.minitab.com


Central Limit Theorem

as n → ∞ is the standard normal distribution

If X1, X2, X3 ….. Xn is a random sample from the population,with sample mean X, then the limiting form of

n/XZ X

σµ−

=

57


Translation

When there are many variables affecting the outcome,The final result will be normally distributed even if the individual variable distributions are not.


ExampleProbability of rolling a die

1 2 3 4 5 6

fx(x) 1/6

Uniform discrete distribution

Let Z be the summation of six dice

Z = X1 + X2 + X3 + X4 + X5 + X6

58


Results

0

10

20

30

40

50

60

6 12 18 24 30 36Z

Freq

uenc

y

Central limit theorem states that the result should be normal for large n

500 trials

CZ = 0.20

XZ = 21.12


Central Limit Theorem

Sums: Z = X1 ± X2 ± X3 ± X4 ± ….. Xn

Z → Normal as n increases

Products: Z = X1 • X2 • X3 • X4 • ….. Xn

Z → LogNormal as n increases

Normal and LogNormal distributions are often employed for analysis even though the underlying population distribution is unknown.

59


Key Points

All variables are random and can be characterized by a statistical distribution with a mean and variance.The final result will be normally distributed even if the individual variable distributions are not.

Analysis Methods





60



IntroductionBasic Probability and StatisticsStatistical TechniquesAnalysis MethodsCharacterizing VariabilityCase StudiesFatigueCalculator.comGlyphWorks


Reliability Analysis

1 32

0.2

0.4

0.6

0.8

Prob

abilit

y D

ensi

ty

0.2

0.4

0.6

0.8

Prob

abilit

y D

ensi

ty

1 32

0.1

0.2

0.3

0.4

-3 -2 -1 0 1 2 3

Prob

abilit

y D

ensi

ty

Stressing Variables

Strength Variables0.2

0.4

0.6

0.8

Prob

abilit

y D

ensi

ty

1 32

P( Failure )

Analysis

?

61


Probabilistic Analysis Methods

Monte CarloSimpleHypercube samplingImportance sampling

AnalyticalFirst order reliability method FORMSecond order reliability method SORM


Limit States

Limit StatesEquilibriumStrengthDeformationWearFunctional

62


Limit State Problems

Z(X) = Z( Q, L, b, h )

Response function

Limit State function

g = Z(X) - Zo = 0

Same as the failure function


Limit States

Stress

Strength

g < 0

g > 0

failures

g = Strength - Stress

63


Monte Carlo

Stress

Strength

g < 0

g > 0

failures

Monte Carlo is simple but what if each of these calculationsrequired a separate FEM model?


Low Probabilities of Failure

Stress

Strength

failures

rare event

Need about 105 simulations for P(Failure) = 10-4

64


Transform Variables u1

u2

x

x1

xuσ

µ−=

µu1 = 0σu1 = 1


Standard Monte Carlo

-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4u1

u2

1000 trials

g = 3 – ( u1 + u2 )

018.0NN)f(P f ==

u1 and u2 normally distributed

65


Multiple Simulations1 252 153 194 215 166 227 168 129 1510 1111 1912 1513 2014 1715 916 2317 1518 2019 1420 2121 1622 1823 2224 1325 20

1000 trials

25 simulations

µ = 0.017

σ = 0.0040

25,000 calculations


Stratified Sampling Methods

-4 -3 -2 -1 0 1 2 3 4

Divide into regions of constant probability and then sample by region

66


Stratified Sample

-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4u1

u2

Each box should have one sample drawn at randomfrom the underlying pdf

100 trials

25 simulations

µ = 0.02

σ = 0.0093

2,500 calculations


Latin Hypercube Sample

-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4u1

u2

10 trials

25 simulations

µ = 0.017

σ = 0.038

250 calculations

67


Importance Sampling (continued)

-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4u1

u2

β = 2

The probability of a sample outside the circle is PS = 0.14( 2 standard deviations )

0183.0PNN)f(P S

S

f ==

138 simulations


Analytical Methods Strategy

Develop a response functionTransform the set of N variables to a set of uncorrelated u variables with µ=0 and σ=1Locate the most likely failure pointApproximate the integral of the joint probability distribution over the failure region

68


Analytical Methods Outline

Response SurfaceTransform VariablesLimit State ConceptMost Probable PointProbability Integration

FORMSORM


Response Surface Methodology

Response surface methodology is a well established collection of mathematical and statistical techniques for applications where the response of interest is influenced by several variables.

ε+⋅⋅⋅⋅= )x,x,x,x(f)X(Z n321

69


Response Surface

X1X2

Z

Fit a surface through the points

Solution points


Mathematical Representation

⋅⋅+++

⋅⋅⋅⋅⋅⋅+++

⋅⋅⋅⋅⋅⋅++++=

323312211

233

222

211

332211o

xxCxxCxxCxBxBxB

xAxAxAA)X(Z Linear

Incomplete QuadradicComplete Quadradic

SolutionsLinear N + 1

Complete Quadradic N(N + 1)/2Incomplete Quadradic 2N + 1

70


Evaluation of Response Surface

Factorial Design

Suppose stress is affected by speed and temperature

high low

high

low

X

XX

X

temperature

spee

d 4 deterministic solutions needed

2n solutions


Evaluation of Response SurfaceFactorial Design

high low

high

low

X

XX

X

temperature

spee

d 9 deterministic solutions needed

3n solutionsX X X

X

X

71


Graphical Representation

X2X1

Z



x

y

fxy(x,y)Stress

Strength

72


2D - Joint Probability DensityContours of equal probability

µx

µy

Stress

Strength


Transform Variables

x

x1

xuσ

µ−=

µu = 0σu = 1

Any distribution can be mapped into a normal distribution

u1

u2

d ( )2d5.0expP −∝

The mathematics and behavior of normal distributions are well understood

73


X to u mapping

f(X)

F(X)F(X)

f(X)

uniform normal


Limit States

Limit states define probability problems

g = Strength(x1) – Stress(x2)

For example:

Prob( g<= 0 ) = Probability of failure = Pf

Analysis focuses on g = 0

74


Failure Probability

g = 0

u1

u2

∫<

=0g

2121uu dudu)u,u(f)Failure(P21


Thought Experiment

Suppose stress and strength had the same distribution

What is the probability of failure?

-3 -2 -1 0 1 2 3x

0.1

0.2

0.3

0.4

Prob

abilit

y D

ensi

ty

75


g function

u1

u2

g = Strength(x1) – Stress(x2)

Pf = 50%


Probable Combinations of u1 and u2

g = 0

u1

u2

Most likely failure point

g functions are not straight lines in N dimensional space

76


Most Probable Point

u1

g = 0u2

β

Probability is related to the minimum distance point β


Safety Index β

Sometimes called reliability index

Unlike safety factors, failure probability is directly related to the safety index

Pf = Φ(-β)

β is a number measured in standard deviations

77


Determining the MPP

Requires an efficient numerical search to find the tangent point of a hypersphere (β-sphere) and the limit state function in u space


Probability Integration

FORMSORM

78


First Order Reliability Model

g = 0

u2

β

exact

Linear approximation

u1

( ) ( )∑=

−+=n

1iiiio *uuaaug

FORM

MPP


FORM

u1

Pf ≈ Φ(-β)u2

β

MPP

u2*

u1*

The joint probability density is rotationally symmetric so it is possible to rotate the coordinate system

79


Second Order Reliability Model

g = 0u2

βexact

Quadradic approximation

u1

( ) ( ) ( ) ( )( )∑ ∑ ∑∑= = = =

−−+−+−+=n

1i

n

1i

n

1ijjii

n

1jij

2iiiiiio *uu*uuc*uub*uuaaug

SORM


SORM

u2

β

MPP

u2*

u1*

( ) ( )∏=

βκ−β−Φ=n

1iif 1P

κi surface curvature

80


Sensitivity Factors

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

Designpoint Change in response

with a change in mean

Change in probabilitywith a change in mean


Sensitivity Factors

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

Designpoint Change in response

with a change in variance

Change in probabilitywith a change in variance

81


Deterministic Sensitivity Factors

iX)X(Z

∂∂

Change in response mean with respect to a change in the input variable mean

Frequently normalized by the means to compare relative importance of variables

X/XZ/)X(Z

i∂∂


Probabilistic Sensitivity Coefficient

iii /

P/PSσσ∂

∂=σ

iii /

P/PSµµ∂

∂=µ

Standard deviation sensitivity

Mean deviation sensitivity

82


Probabilistic Sensitivity Factors

1X

)X(Z 2ii

ii =ασ

∂

∂∝α ∑

αi - probabilistic sensitivityZ(X) - responseXi - input variableσi - standard deviation of Xi


Software

General PurposeDurability

83

Variability








84


Sources of Variabilitycustomers

materials manufacturing

usage

107

Fatigue Life, 2Nf

1

10 102 103 104 105 106

0.1

10-2

1

10-3

10-4

Stra

in A

mpl

itude

Strength

Stress



Variability and Uncertainty

Variability: Every apple on a tree has a different mass.

Uncertainty: The variety of the apple is unknown.

Variability: Fracture toughness of a material

Uncertainty: The correct stress intensity factor solution

85


Sources of Variability

Stress VariablesLoadingCustomer UsageEnvironment

Strength VariablesMaterialProcessingManufacturing ToleranceEnvironment


Sources of Uncertainty

Statistical UncertaintyIncomplete data (small sample sizes)

Modeling ErrorAnalysis assumptions

Human ErrorCalculation errorsJudgment errors

86


Modeling Variability



Central Limit Theorem:

COVX

0 0.5 1.0 1.5 2.0

0.5

1.0

1.5

σ lnX σlnX ~ COVX

COVX is a good measure of variability


Standard Deviation, lnx

COVX

1 2 3

68.3% 95.4% 99.7%

0.05

0.1

0.25

0.5

1

1.05

1.10

1.28

1.60

2.30

1.11

1.23

1.66

2.64

5.53

1.16

1.33

2.04

3.92

11.1

99.7% of the data is within a factor of ± 1.33 of the mean for a COV = 0.1

COV and LogNormal Distributions

87


Variability in Service Loading

Quantifying Loading VariabilityMaximum LoadLoad RangeEquivalent Stress


Maximum Force

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

100 1000

Median 431COV 0.34

200 500

Maximum force from 42 automobile drivers99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

100 1000200 500

Cum

ulat

ive

Pro

babi

lity

Force

88


Maximum Load Correlation

0 200 400 600 800 1000

108

107

106

105

104

103

Maximum Load

Fatig

ue L

ives


Loading Variability

1 10 102 103 104 105

Cumulative Cycles

0

16

32

48

Load

Ran

ge

54 Tractors / Drivers

89


Variability in Loading

1 10

54 Tractors/Drivers

COV 0.53

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

n neq SS ∑∆=∆

Equivalent Load

Cum

ulat

ive

Pro

babi

lity


Mechanisms and Slopes

100

103

104

Stre

ss A

mpl

itude

, MPa

100

Cycles101 102 103 104 105 106 107

110

10-1210-1110-1010-910-810-710-6

Crack Growth Rate, m/cycle

1

10

100

mM

Pa,

K∆

1

3

Crack Nucleation

Crack Growth10

100

103 104 105 106 107 108

Total Fatigue Life, Cycles

1

51

Equ

ival

ent L

oad,

kN

Structures

A combination of nucleation and growth

n = 3

n = 5

n = 10

90


Effect of Slope on Variability99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

n = 10 5 3

Equivalent Load0.1 1 100.1 1 10

n COV3 0.535 0.4310 0.38


Loading History Variability

Test TrackCustomer Service

91


Test Track Variability

0.1 1 10

COV 0.12

40 test track laps of a motorhome99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

Equivalent Load


Customer Usage Variability

0.1 1 10

COV 0.32

42 drivers / cars99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

Equivalent Load

92


Variability in Environment

InclusionsPit depth


Inclusions That Initiated Cracks

Barter, S. A., Sharp, P. K., Holden, G. & Clark, G. “Initiation and early growth of fatigue cracks in an aerospacealuminium alloy”, Fatigue & Fracture of Engineering Materials & Structures 25 (2), 111-125.

COV = 0.27

93


Pits That Initiated Cracks

7010-T7651

Pre-corroded specimens

300 specimens

246 failed from pits

Crawford et.al.”The EIFS Distribution for Anodized and Pre-corroded 7010-T7651 under Constant Amplitude Loading” Fatigue and Fracture of Engineering Materials and Structures, Vol. 28, No. 9 2005, 795-808


Pit Size Distribution

40

30

20

10

100 200 400 500300

Freq

uenc

y

area µm

Mean = 230 COV = 0.32

94


Pit Depth Variability

Pits12 Data PointsMedian 24.37

100

COV 0.33

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

10Pit Depth, µm

Dolly, Lee, Wei, “The Effect of Pitting Corrosion on Fatigue Life”Fatigue and Fracture of Engineering Materials and Structures, Vol. 23, 2000, 555-560

Cum

ulat

ive

Pro

babi

lity


Variability in Materials

Tensile StrengthFracture ToughnessFatigue

Fatigue StrengthFatigue Life

Strain-LifeCrack Growth

95


Tensile Strength - 1035 Steel

25

50

75

100

500 550 600 650 700

Num

ber o

f hea

ts

Tensile Strength, MPa

Metals Handbook, 8th Edition, Vol. 1, p64

Mean = 602COV = 0.045


Fracture Toughness

100

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Median 76.7COV 0.06

60 70 9080

26 Data Points

Cum

ulat

ive

Pro

babi

lity

KIc, Ksi√in

Kies, J.A., Smith, H.L., Romine, H.E. and Bernstein, H, “Fracture Testing of Weldments”, ASTM STP 381, 1965, 328-356

Mar-M 250 Steel

96


Fatigue Variability

102 103 104 105 106 107

Fatigue Life101

1000

100

10

1

Stre

ss A

mpl

itude Fatigue life

Fatig

ue s

treng

th


Fatigue Life Variability

50 100 150 200 250

10

20

30

40

10

20

Num

ber o

f Tes

ts

Life x10 350 100 150 200 250

10

20

30

40

10

20

Num

ber o

f Tes

ts

Life x10 3

Production torsion bars5160H steelOne lot, 71 parts

25 lots, 300 parts

Metals Handbook, 8th Edition, Vol. 1, p219

µx = 123,000 cyclesCOV = 0.25

µx = 134,000 cyclesCOV = 0.27

97


Statistical Variability of Fatigue Life

50

10

2

90

98

104 105 106 107 108

Cycles to Failure

Per

cent

Sur

viva

l

∆s=210∆s=245∆s=315∆s=280∆s=440

Sinclair and Dolan, “Effect of Stress Amplitude on the Variability in Fatigue Life of 7075-T6 Aluminum Alloy”Transactions ASME, 1953

7075-T6 Specimens


COV vs Fatigue Life

440 14,000 0.12315 25,000 0.38280 220,000 0.70245 1,200,000 0.67210 12,000,000 1.39

∆S COVX

98


Variability in Fatigue Strength

( )∏=

−+=n

1i

a2X 1C1CCOV

2i

i

085.0b)N(S2S b

f'f −≈=

∆

( ) 088.0139.11C2

'f

)085.(2S =−+=

−


Strain Life Data for 950X Steel

102 103 104 105 106 107

Fatigue Life101

1

0.1

10-2

10-3

10-4

Stra

in A

mpl

itude

378 Fatigue Tests

29 Data Sets

99


29 Individual Data Sets'fσ

Median 820

300 1000

COV 0.25

2000

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

b

-0.12 -0.08 -0.04

Mean -0.09COV 0.25

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %C

umul

ativ

e P

roba

bilit

y


29 Individual Data Sets (continued)

0.1 1 10

Median 0.57COV 1.15

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

'fε

-0.8 -0.6 -0.4 -0.2

c

Mean -0.62COV 0.23

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

100


Input Data Simulation

( )( ) ( ) ( )( ) ( )bb

ff

bbff ,,Ncf

'f

,,Nbf

'f N2,,LN2

E,,L

2σµ

εεσµσσ σµε+

σµσ=

ε∆


Simulation Results

0.0001

0.001

0.01

0.1

1

10

100

1 10 100 103 104 105 106 107

Stra

in A

mpl

itude

Fatigue Life

101


Correlation

0

0.25

0.5

0.75

1.0

0.01 0.1 1 10'fε

c

0

0.05

0.1

0.15

0.2

102 103 104

'fσ

b

ρ = -0.828 ρ = -0.976


Generating Correlated Data

z1 = ( rand() )Φz2 = ( rand() )Φ

2213 1zzz ρ−+ρ=

)zexp( 1lnln'f '

f'f σσ σ+µ=σ

3bb zb σ+µ=

z1= N(0,1)

102


Correlated Properties

0.0001

0.001

0.01

0.1

1

10

100

1 10 100 103 104 105 106 107

Fatigue Life

Stra

in A

mpl

itude


Curve Fitting

102 103 104 105 106 107

Fatigue Life101

1

0.1

10-2

10-3

10-4

Stra

in A

mpl

itude

cf

'f

bf

'f )N2()N2(

E2ε+

σ=

ε∆

Assume a constant slope to get a distribution of properties

'fσ

103


Property Distribution

0.1 1

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

365 Data Points

Median 0.26

COV 0.42

'fε

378 Data Points

Median 802 MPa

COV 0.12

'fσ

100 1000

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

b = -0.086 c = -0.51


Correlation

0

500

1000

1500

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7'fε

'fσ

104


Simulation

0.001

0.01

0.1

1

10

0.00011 10 100 103 104 105 106 107

Stra

in A

mpl

itude

Fatigue Life


Strength Coefficient

5000

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

500

0

500

1000

1500

0 500 1000 1500 2000 2500

365 Data Points

Median 1002 MPa

COV 0.14

'fσ

'K

'K

ρ = 0.863

105


Crack Growth Data

0

10

20

30

40

50C

rack

Len

gth,

mm

0 50 100 150 200 250 300 350Cycles x103

Virkler, Hillberry and Goel, “The Statistical Nature of Fatigue Crack Propagation”, Journal of Engineering Materials and Technology, Vol. 101, 1979, 148-153


Crack Growth Rate Data

300,000

68 Data PointsMedian 257,000COV 0.07

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

Fatigue Lives Crack Growth Rate

5 10 5010-5

10-4

10-3

10-2

Cra

ck g

row

th ra

te, m

m/c

ycle

Stress intensity, MPa√m

106


Crack Growth Properties

Median 5 x 10-8

COV 0.44

10-7

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

10-6

Median 3.13COV 0.06

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

2 54

mKCdNda

∆=

C m


Beware of Correlated Variables

( )2/m1SC

aaN2m

m

2/m1i

2/m1f

f

−π∆

−=

−−

Nf and C are linearly related and should have the same variability, but

07.0COVfN =

44.0COVC =

because C and m are correlated.

107


Correlation

2.0

2.5

3.0

3.5

4.0

10-8 10-610-7

C

m


Correlated Variables

2YYX

2XZ 2 σ+σρσ+σ=σ

-1 -0.5 0 0.5 1

0.5

1

1.5

2

ρ

σZ

σX = σY = 1

108


Calculated Lives

105 106

Computed fromC and m pairs experimental

105 106

0.1 %0.1 %

99.9 %

99 %

90 %

50 %

10 %

1 %

Cum

ulat

ive

Pro

babi

lity

99.9 %

99 %

90 %

50 %

10 %

1 %

Cum

ulat

ive

Pro

babi

lity experimental

( )∫ ∆=

f

o

a

amf KC

daN


Manufacturing/Processing Variability

Bolt ForcesSurface FinishDrilled Holes

109


Variability in Bolt Force

100 1000

Force200 Data PointsMedian 130COV 0.14

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Bolt Force, kN

Preload force in bolts tightened to 350 Nm

Cum

ulat

ive

Pro

babi

lity


Surface Roughness Variability

100

125 Data PointsMedian 43COV 0.10

Machined aluminum casting

10Surface Finish, µin

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

Surface Finish

110


Drilled HolesFighter Spectrum154 Data PointsMedian 126,750COV 0.22 in life

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

105 Cycles

From: J.P. Butler and D.A. Rees, "Development of Statistical Fatigue Failure Characteristics of 0.125-inch 2024-T3 Aluminum Under Simulated Flight-by-Flight Loading," ADA-002310 (NTIS no.), July 1974.

180 drilled holes in a single plate

COV 0.07 in strength


Analysis Uncertainty

Miners Linear Damage ruleStrain Life Analysis

111


50

90

99.9

99

10

1

0.01

0.01 0.1 1 10 100

Computed Life / Experimental

Cum

ulat

ive

Pro

babi

lity

of F

ailu

re 964 TestsCOV = 1.02

Miners Rule

From Erwin Haibach “Betriebsfestigkeit”, Springer-Verlag, 2002

A safety factor of 10 in life would result in a 10% chance of failure


SAE Specimen

Suspension

Transmission

Bracket

Fatigue Under Complex Loading: Analysis and Experiments, SAE AE6, 1977

112


Analysis Results

1 10 100


99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity


Strain-Life analysis of all test data


Material Variability

1 10 100

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity


Material Analysis

Strain-Life back calculation of specimen lives

113


Modeling Uncertainty

( )∏=

−+=n

1i

a2X 1C1CCOV

2i

i

CU = 1.09

Analysis Uncertainty CU = ?

The variability in reproducing the original strain life data from the material constants is CM ~ 0.44

2M

2N2

U C1C1

C1 f

+

+=+

90% of the time the analysis is within a factor of 3 !99% of the time the analysis is within a factor of 10 !


Variability from Multiple Sources

( )∏=

−+=n

1i

a2X 1C1CCOV

2i

i

Suppose we have 4 variables each with a COV = 0.1

The combined variability is COV = 0.29

Suppose we reduce the variability of one of the variables to 0.05

The combined variability is now COV = 0.27

If all of the COV’s are the same, it doesn’t do any good to reduce only one of them, you must reduce all of them !

114


Variability from Multiple Sources

( )∏=

−+=n

1i

a2X 1C1CCOV

2i

i

Suppose we have 3 variables each with a COV = 0.1and one with COV = 0.4

The combined variability is COV = 0.65

Suppose we reduce the variability of these variables to 0.05

The combined variability is now COV = 0.60

If one of the COV’s is large, it doesn’t do any good to reduce the others, you must reduce the largest one !


Variability SummarySource COV

Service Loading 0.5

Materials 0.1Manufacturing 0.1Surface Finish 0.1

Environment 0.3

Strength

Stress

Fatigue LivesAnalysis Uncertainty

1.01.0

5

StressStrengthlifeFatigue

∝

115



Variability: Every apple on a tree has a different mass.

Uncertainty: The variety of the apple is unknown.

Variability: Multiple samples of the same material

Uncertainty: What is the material


Strain Life Data for 93 Steels

1 10 102 103 106 10710510410-4

10-2

10-3

0.1

10

1

Stra

in A

mpl

itude

1 10 102 103 106 10710510410-4

10-2

10-3

0.1

10

1

Stra

in A

mpl

itude

116


Uncertainty for all Steels

100 1000 10000


99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

'fσ


Uncertainty for Structural Steels

100 1000 10000

COV 0.12

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

'fσ Su 600 – 900 MPa

19 Data Points

117



Fatigue Strength Coefficient

Variability Uncertainty CombinedAll Steels 0.12 0.48 0.75

Structural Steel 0.12 0.12 0.24


Quiz

What is the expected variability ?

At my last seminar everyone hit a golf ball and we recorded the maximum acceleration.

118


Results

12.97.75.8811.115.510.318.111.644.260.37

µσ

COV

Case Studies





119





Case Studies

DARWINSouthwest Research

BicycleLoading Histories

120


A Software Framework for Probabilistic Fatigue Life Assessment

ASTM Symposium on Probabilistic Aspects of Life Prediction

Miami Beach, FloridaNovember 6-7, 2002

R. C. McClung, M. P. Enright, H. R. Millwater*, G. R. Leverant, and S. J. Hudak, Jr.

Southwest Research

Slides 6 – 27 used with permission of of Craig McClung


Motivation

121


UAL Flight 232July 19,1989


Turbine Disk FailureAnomalies in titanium engine disks

Hard AlphaVery rareCan cause failureNot addressed by safe life methods

Enhanced life management processRequested by FAADeveloped by engine industryProbabilistic damage tolerance methodsSupplement to safe life approach

SwRI and engine industry developed DARWIN with FAA funding

122


Probabilistic Damage Tolerance

Probabilistic Fracture Mechanics

Probability of DetectionAnomaly Distribution

Finite Element Stress Analysis

Material Crack Growth Data

NDE Inspection Schedule

Pf vs. Cycles

Risk Contribution Factors


Zone-Based Risk Assessment

Define zones based on similar stress, inspection, anomaly distribution, lifetime

Total probability of fracture for zone:(probability of having a defect) x (POF given

a defect)Defect probability determined by anomaly

distribution, zone volumePOF assuming a defect computed with

Monte Carlo sampling or advanced methods

POF for disk obtained by summing zone probabilities

As individual zones become smaller (number of zones increases), risk converges down to “exact” answer

1

2 3 4

m

5 6 7

123


Fracture Mechanics Model of Zone

m

7

Retrieve stresses along line

Fracture Mechanics Model

hx

hy

x

Y

grad

ient

dire

ctio

n

1

2

3

4

5

Defect

Finite Element Model

(Not to Scale)


Stress Processing

FE Stresses and plate definition

stress gradient

Stress gradient extraction

FE Analysis

0.0 0.2 0.4 0.6 0.8 1.0Normalized distance from the notch tip, x/r

-0.8

-0.4

0.0

0.4

0.8

1.2

1.6

2.0

σ/σo

(σz)relax

(σz)residual

(σz)elastic

Shakedown module

Computed relaxed stressσelastic - σresidual

σ0/σ

Residual stress analysis

3 4 5 6 7 0 1 2 3Load Step

01020304050607080

Hoo

p St

ress

(ksi

)

Rainflow stress pairing

124


Anomaly Distribution# of anomalies per volume of material as function of defect size

Library of default anomaly distributions for HA (developed by RISC)


Probability of Detection CurvesDefine probability of NDE flaw detection as function of flaw sizeCan specify different PODs for different zones, schedulesBuilt-in POD library or user-defined POD

125


Random Inspection Time“Opportunity Inspections” during on-condition maintenance

Inspection time modeled with Normal distribution or CDF table


Output: Risk vs. Flight Cycles

126


Output: Risk Contribution FactorsIdentify regions of component with highest risk


Implementation in IndustryFAA Advisory Circular 33.14 requests risk

assessment be performed for all new titanium rotor designs

Designs must pass design target risk for rotors

Components

Risk

MaximumAllowable

Risk

10-9

RiskReductionRequired

CA B

127


Probabilistic Fracture Mechanics

Material Crack Growth Data

Finite Element Stress Analysis

Anomaly Distribution NDE Inspection

Pf vs. Flights

RPM-Stress Transform Crack Initiation

0

4000

8000

12000

0 1000 2000 3000 4000

Load Spectrum Editing

Code Enhancements

Sensor (RPM) Input

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0 5 10 15 20 25 30 35 40

THRESHOLD STRESS RANGE (KSI)

PRO

BA

BIL

ITY

OF

FAIL

UR

E A

T 20

,000

CYC

LES

0

20

40

60

80

100

120

NU

MB

ER O

F C

YCLE

S PE

R M

ISSI

ON

PfCycles Per Mission

DARWIN for Prognosis Studies


Three Sources of Variability

Anomaly size (initial crack size)FCG properties (life scatter)Mission histories (stress scatter)

128


Hard Alpha Defects in Titanium

Initial DARWIN focus on Hard AlphaSmall brittle zone in

microstructureAlpha phase stabilized by N

accidentally introduced during melting

Cracks initiate quickly

Extensive industry effort to develop HA distribution


Resulting Anomaly DistributionsPost 1995 Triple Melt/Cold Hearth + Vacuum Arc Remelt

1.00E-02

1.00E-01

1.00E+00

1.00E+01

1.00E+02

1.00E+02 1.00E+03 1.00E+04

DEFECT INSPECTION AREA (sq mils)

EXC

EED

ENC

E(p

er m

illio

n po

unds

)

MZ(5-10in Billet)/#1 FBH

MZ(5-10in Billet)/#3 FBH

MZ (12-13in Billet)/#1 FBH

#2/#1 FBH

#2/#2 FBH

#3/#1 FBH

#3/#2 FBH

#3/#3 FBH

129


FCG Simulations for AGARD Data

Use individual fits to generate set of a vs. N curves for identical conditions

Characterize resulting scatter in total propagation life

Lognormal distribution appropriate in most cases

N, cycle0 2000 4000 6000 8000

a, c

m

0.0

0.1

0.2

0.3

0.4

0.5

Corner Crack Specimen∆Ki=18.7 MPa√m, ∆Kf=56.9 MPa√m

AGARD


Engine Usage Variability

Stress/Speed: ∆σ ∝ (RPM)2

Total Cyclic Life (LCF): Nf = Ni + Np

Ni ∝ ∆σ 3-5

Np ∝ ∆σ 3-4

Life/Speed: Nf ∝ (RPM)6

Component life is very sensitive to actual usage

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

10000

0 2500 5000 7500 10000 12500TIM E (SEC)

RPM

(Low

Spe

ed S

pool

)

Air Combat Tactics

Air to Ground & Gunnery

Basic Fighter Maneuvers

Intercept

Peace Keeping

Surface to Air Tactics ( hi alt)

Surface to Air Tactics ( lo alt)

Suppression of Enemy Defenses

Cross Country

130


Usage Variability

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

10000

0 2500 5000 7500 10000 12500TIME (SEC)

RPM

(Low

Spe

ed S

pool

)

Air Com bat Tactics

Air to Ground & Gunnery

Basic Fighter Maneuvers

Intercept

Peace Keeping

Surface to Air Tactics ( hi alt)


Suppression of Enem y Defenses

Cross Country


0

1000

2000

3000

4000

5000

6000

7000

8000

9000

0 2500 5000 7500 10000 12500TIM E (SEC)

RPM

(Low

Spe

ed S

pool

)

Peace Keeping

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

0 2500 5000 7500 10000 12500TIM E (S EC)

RPM

(Low

Spe

ed S

pool

)

Components of Usage Variability:

• Mission type

• Mission-to-mission variability

• Mission mixing variability


0.001

0.01

0.1

1

0 1000 2000 3000 4000 5000 6000 7000 8000

Number of Flight Cycles

Nor

mal

ized

PO

F

Air Combat Tactics

Combat Air Patrol

Air-Ground Weapons Delivery

Air-Air Weapons Delivery

Instrument/Ferry

Initiation and PropagationNo Inspection

Variability in Mission Type

Air-Air Weapons Delivery

Time

RPM

Air-Ground Weapons Delivery

Time

RP

M

131


Web Site: www.darwin.swri.org


Fatigue Design and Reliability in the Automotive Industry

Fatigue Design and ReliabilityESIS Publication 23

J-J. Thomas, G. Perroud, A. Bignonnet and D. Monnet

PSA Peugeot Citroën

132


Fatigue Design at PSA

Fatigue design at PSA is done with a probabilistic approach that includes analysis of customer usage, production scatter, definition of the appropriate design loads and an acceptance testing criterion.


Stress-Strength Interference

Stress Strength


How do you really get these distributions ?

Design Load

How do you establish a design load ?How do you validate the design ?

133


Stress Distribution

Variability in loading has two components, how it is used and how it is driven.

Car UsageHighway, city, fully loaded, empty etc.

Driving Stylepassive, aggressive etc.

The usage of a car is independent of the owners driving styleso that the distributions of car usage and driving style canbe obtained separately.


Car Usage

Customer surveysk l ck % rkl %1 Unloaded 27

1 Highway 102 Good Road 253 Mountain 404 City 25

2 Half Load 581 Highway 52 Good Road 303 Mountain 304 City 35

3 Fully Loaded 151 Highway 152 Good Road 253 Mountain 404 City 20

12 Customer Usage Categories

134


Owner BehaviorInstrumented Vehicles

-3

-2

-1

0

1

2

3

0 -1 -2 -33 2 1To Load, kN

0 -1 -2 -33 2 1To Load, kN

From

Loa

d,kN

Extensive field testing for each customer usage category produces a large number of histograms.

Let the usage histogram be denoted Ukl


Extrapolation

-3

-2

-1

0

1

2

3

0 -1 -2 -33 2 1To Load, kN

0 -1 -2 -33 2 1To Load, kN

From

Loa

d,kN

-3

-2

-1

0

1

2

3

0 -1 -2 -33 2 1To Load, kN

From

Loa

d,kN

-3

-2

-1

0

1

2

3

0 -1 -2 -33 2 1To Load, kN

From

Loa

d,kN

Measured data, e.g. 1,000 km Extrapolated data, 300,000 km

Ukl U*kl

135


Virtual Customers

Thousands of virtual customers can now be generated by combining customer usage with driving style.

[ ] [ ]∑= *klklk

i UrcNU

[ ]iU rainflow histogram for an individual car

N kilometersck car loading, % rkl car usage, %

[ ]*klU distributions of rainflow histograms for

different car usage classifications


Design Loads

Initial design is done on the basis of a single constant amplitude load, Feq

Find a constant amplitude load and number of cyclesthat will produce the same fatigue damage as the customeroperating a car for the design life.

136


Equivalent Fatigue Loading

101 102 103 104 105 106

Feq

m1

6

m

eq 10FF

∆= ∑

1Nn

f

=∑


Distribution of Feq

µF

ασF

Fn

Fn = µF + ασFDesign Customer:

137


Choosing αContours of equal probability

y

xµx

µY

-3σx

3σY

12

34

Stress

Strength

If we use 3σ on both stress and strength

3σ P( s < S ) = 2.3 10-3

σ≈=≤≥Σ= − 5.4103.5)Sss(P)failure(P 6I


Distribution of Strength

106

Sexp

µS

σS

138


What Strength is Needed?

µF

ασF

Fn µS

βσS


Some Statistics

Z = S - F

Suppose we want a probability of failure of 1 in 50,0005

f 10x2P −=

( )2F

2S

FS

Z

Zf

1 1.4Pσ+σ

µ−µ=

σµ

==Φ−

Fn = µS( 1 – β COVS )Fn = µF( 1 + α COVF )

139


Mean Component Strength

( )2F

2S

FS

Z

Zf

1 1.4Pσ+σ

µ−µ=

σµ

==Φ−

( )2

F

F

2

Sn

S

Fn

S

f1

COV1COVCOV

F

COV11

FP

α+

+

µ

α+−

µ

=Φ−


Reliability

1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2

10-1

10-2

10-3

10-4

10-5

10-6

10-7

10-8

10-9

Rel

iabi

lity

n

S

Fµ

α = 3COVF = 0.2COVS = 0.1

140


Mean Strength

µF

ασFFn µS

βσS

µS = 1.4 Fn for a reliability of 2 x 10-5

85.2COV

F1

S

S

n

=µ

−=β


Design

Design calculations are made with loads FnLarge database relates Fn to vehicle parameters so that a new vehicle can be designed from historical measurements

Fatigue calculations are made with material properties β standard deviations from the mean

141


Component Testing

Test load is frequently higher than Fn to get failures near the design life

Component strength not life !


Interpreting the Test Data

Component tests are done with small sample sizes

2x

1N,12

2xs)1N(

σ≤χ

−−α−

Confidence limits

( )2

F

F2

1N,1

2

Sn

S

Fn

S

f1

COV1COV1NCOV

F

COV11

FP

α+

+χ

−

µ

α+−

µ

=Φ

−α−

−

142


Validation

Full scale vehicle simulation done at the end for design final validation


Extrapolation

-3

-2

-1

0

1

2

3

0 -1 -2 -33 2 1To Load, kN

0 -1 -2 -33 2 1To Load, kN

From

Loa

d,kN

-3

-2

-1

0

1

2

3

0 -1 -2 -33 2 1To Load, kN

From

Loa

d,kN

-3

-2

-1

0

1

2

3

0 -1 -2 -33 2 1To Load, kN

From

Loa

d,kN

Measured data, e.g. 1,000 km Extrapolated data, 300,000 km

Ukl U*kl

How does this process work?

143


Service Loading Spectra

-5

5

0

Late

ral F

orce

,kN

Time history

0 1 2 3-3 -2 -1-3

-2

-1

0

1

2

3

To Load, kN

From

Loa

d,kN

0

1

2

3

4

1 10 100 1000Cumulative Cycles

Load

Ran

ge,k

N

RainflowHistogram

ExceedanceDiagram


Problem Statement

Given a rainflow histogram for a single user, extrapolate to longer times

Given rainflow histograms for multiple users, extrapolate to more users

144


Probability Density

-3

-2

-1

0

1

2

3

0 -1 -2 -33 2 1To Load, kN

From

Loa

d,kN


Kernel Smoothing

-3

-2

-1

0

1

2

3

From

Loa

d,kN

0 -1 -2 -33 2 1To Load, kN

145


Sparse Data

-3

-2

-1

0

1

2

3

From

Loa

d,kN

0 -1 -2 -33 2 1To Load, kN


Exceedance Plot of 1 Lap

0

1

2

3

4


Load

Ran

ge,k

N

weibull distribution

146


10X Extrapolation

0

1

2

3

4


Load

Ran

ge,k

N

6

5


-3

-2

-1

0

1

2

3

From

Loa

d,kN

Probability Density

0 -1 -2 -33 2 1To Load, kN

147


Results

-3

-2

-1

0

1

2

3

0 -1 -2 -33 2 1To Load, kN

From

Loa

d,kN

-3

-2

-1

0

1

2

3

From

Loa

d,kN

Simulation Test Data

0 -1 -2 -33 2 1To Load, kN


Exceedance Diagram

0

1

2

3

4

5

6

1 10 100 103 104

Cycles

1 Lap (Input Data)

Model Prediction

Actual 10 LapsLoad

Ran

ge,k

N

148


Problem Statement

Given a rainflow histogram for a single user, extrapolate to longer timesGiven rainflow histograms for multiple users, extropolate to more users


Extrapolated Data Sets

0.1 1 10 102 103 104

Normalized Life

Airplane

TractorATV

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

149


Issues

In the first problem the number of cycles is known but the variability is unknown and must be estimated

In the second problem the variability is known but the number and location of cycles is unknown and must be estimated


Assumption

0

16

32

48

1 10 102 103 104 105

Cumulative Cycles

Load

Ran

ge

0

16

32

48

0

16

32

48

1 10 102 103 104 105

On average, more severe users tend to have more higher amplitude cycles and fewer low amplitude cycles

150


Translation

-3

-2

-1

0

1

2

3

From

Loa

d,kN

0 -1 -2 -33 2 1To Load, kN


Damage Regions

-3

-2

-1

0

1

2

3

From

Loa

d,kN

0 -1 -2 -33 2 1To Load, kN

151


ATV Data - Most Damaging in 19

-3

-2

-1

0

1

2

3

From

Loa

d,kN

-3

-2

-1

0

1

2

3

From

Loa

d,kN


0 -1 -2 -33 2 1To Load, kN

0 -1 -2 -33 2 1To Load, kN


ATV Exceedance

0

1

2

3

4

5

6

1 10 100 103 104

Cycles

Load

Ran

ge,k

N

Actual Data

Simulation

152


Airplane Data - Most Damaging in 334

-30

-20

-10

0

10

20

30

0 -10 -20 -3030 20 10To Stress, ksi

From

Stre

ss,k

si

-30

-20

-10

0

10

20

30

From

Stre

ss,k

si


0 -10 -20 -3030 20 10To Stress, ksi


Airplane Exceedance

0

10

20

30

40

1 10 100 1000Cycles

Stre

ss R

ange

,ksi

Simulation

Actual Data

153


Tractor Data - Most Damaging in 54

-1.5

-1.0

-0.5

0

0.5

1.0

1.5

0 -0.5 -1.0 -2.01.5 1.0 0.5To Strain x 10-3

From

Sta

rinx1

0-3

-1.5

-1.0

-0.5

0

0.5

1.0

1.5

From

Stra

in x

10-

3


0 -0.5 -1.0 -2.01.5 1.0 0.5To Strain x 10-3


Tractor Exceedance

0

1.0

2.0

3.0

1 10 100 103 104 105

Cycles

Stra

in R

ange

x 1

0-3 Simulation

Actual Data

154

www.FatigueCalculator.com








155


www.FatigueCalculator.com


Probabilistic Fatigue Analysis

156


Modeling Variability



Central Limit Theorem:

COVX

0 0.5 1.0 1.5 2.0

0.5

1.0

1.5

σ lnX σlnX ~ COVX

COVX is a good measure of variability


Standard Deviation, lnx

COVX

1 2 3

68.3% 95.4% 99.7%

0.05

0.1

0.25

0.5

1

1.05

1.10

1.28

1.60

2.30

1.11

1.23

1.66

2.64

5.53

1.16

1.33

2.04

3.92

11.1

99.7% of the data is within a factor of ± 1.33 of the mean for a COV = 0.1

COV and LogNormal Distributions

157


Strain Life Analysis


Material Properties

158


Stress Concentration


Results

159


Results (continued)


Results (continued)

160


Results (continued)


Ten SimulationsLife COV

6470 0.9596930 0.8986710 0.6886640 0.9086580 0.8696470 0.9597010 0.7236690 0.9086170 0.7916560 0.971

Mean 6623 0.8674COV 0.038 0.114

161


29 Individual Data Sets'fσ

Median 820

300 1000

COV 0.25

2000

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

b

-0.12 -0.08 -0.04

Mean -0.09COV 0.25

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %C

umul

ativ

e P

roba

bilit

y


29 Individual Data Sets (continued)

0.1 1 10

Median 0.57COV 1.15

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

'fε

-0.8 -0.6 -0.4 -0.2

c

Mean -0.62COV 0.23

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

Cum

ulat

ive

Pro

babi

lity

162


Correlation

0

0.25

0.5

0.75

1.0

0.01 0.1 1 10'fε

c

0

0.05

0.1

0.15

0.2

102 103 104

'fσ

b

ρ = -0.828 ρ = -0.976


Correlated Variables

163


Results (continued)


Results (continued)

164


Uncorrelated Variables


Results (continued)

165


Strain Amplitude± 1000 ± 250


Deterministic Analysis

166


Deterministic Analysis (continued)


Deterministic Analysis (continued)

167


Deterministic Analysis Results


Probabilistic Analysis

168


Probabilistic Analysis (continued)


Probabilistic Analysis (continued)

170





GlyphWorks / Rainflow

171


StatStrain.flo


StatStrain Glyph

172


StatOutput.xls

10000 100000 1000000

99.9 %

99 %

90 %

50 %

10 %

1 %

0 1 %

Life Distribution, Channel 6


StatOutput.xls (continued)

Probabilistic Sensitivity, Channel 6

Load History Variables57%

Stress Concentrators3%

Material Properties16%

Analysis Variables24%

173



Channel # 6 Total Channels 1

Variable Probabilistic DeterministicNumCases 11 Load History Variables 0.89 -5.42

Median 45634 ScaleFactor 0.89 -5.42COV 1.91 Offset 0.00 0.00

Stress Concentrators 0.05 -5.51Probability (%) Life Kf 0.05 -5.51

99 817179 Material Properties 0.25 -14.5490 223648 E 0.00 -4.3750 45634 Sfp 0.23 3.4210 9311 b 0.00 -4.391 2548 efp 0.00 1.42

c 0.00 -9.86np 0.00 -2.32Kp 0.10 1.56

Analysis Variables 0.37 1.00Uncertainty 0.37 1.00

Life Distribution Sensitivity



Variable Distribution ValueScale

Parameter ValueScale

ParameterScaleFactor Normal 100 0.25 103 0.18

Offset Normal 0 0.10 -0.01 0.08Kf Uniform 3 0.05 3.0 0.04E None 208000 0.00 208000 0.00

Sfp Log-Normal 1000 0.10 1054 0.11b None -0.1 0.00 -0.10 0.00

efp None 1 0.20 1.0 0.00c None -0.5 0.00 -0.50 0.00

np None 0.2 0.00 0.20 0.00Kp Log-Normal 1200 0.10 1194 0.10

Uncertainty Log-Normal 1 0.50 0.93 0.48

Inputs Outputs

174


StatStress.flo


Rainflow Extrapolation

175


Duration Extrapolation


Results Files

176


Rainflow Duration


Percentile Extrapolation

177


Results


probabilistic aspects of fatigue · probabilistic aspects of fatigue ... safety factors. larger...

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