probabilistic fatigue

7/27/2019 Probabilistic Fatigue

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Introduction

Professor Darrell F. Socie

Department 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 Socie

Mechanical Engineering

1206 West Green

Urbana, Illinois 61801

Office: 3015 Mechanical Engineering Laboratory

[email protected]

Tel: 217 333 7630

Fax: 217 333 5634


2/177


Fatigue Calculations

Who really believesthese numbers ?


SAE Specimen

Suspension

Transmission

Bracket

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


3/177


Analysis Results

1 10 100

48 Data Points

COV 1.27

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

CumulativeProbability

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 %


Analytical Life / Experimental Life

Material Analysis

Strain-Life back calculation of specimen lives


4/177


Probabilistic Models

Probabilistic models are no better than theunderlying deterministic models

They require more work to implement

Why use them?


Quality and Cost

Taguchi

Identify factors that influence performance

Robust design reduce sensitivity to noise

Assess economic impact of variation

Risk / Reliability

What is the increased risk from reduced testing ?


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Risk

10-9

10-8

10-7

10-6

10-5

10-4

10-3

Risk

Time, Flights etc

Acceptable risk


Reliability

99 %

80 %

50 %

10 %

1 %

0.1 %

ExpectedFailures

106

Fatigue Life

105104103


6/177


Risk Contribution Factors

OperatingTemperature

Analysis Uncertainty

Speed

Material

Properties

ManufacturingFlaws


Uncertainty and Variability

customers

materials manufacturing

usage

107

Fatigue Life, 2Nf

1

10 102 103 104 105 106

0.1

10-2

1

10-3

10-4StrainAmplitude

time

50%

100 %

F

ailures

Strength

Stress


7/177


Deterministic versus Random

Deterministic from past measurements the future positionof a satellite can be predicted with reasonable accuracy

Random from past measurements the future position ofa car can only be described in terms of probability andstatistical averages


Deterministic Design

Stress Strength

SafetyFactor

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


8/177


Probabilistic Design

Stress Strength

Reliability = 1 P( Stress > Strength )


3Approach3 contains 99.87% of the data

If we use 3 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

== 5.4103.5)Sss(P)failure(P 6I

For 3 variables, each at 3 :

= 7.5102.1)failure(P 8


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Benefits

Reduces conservatism (cost) compared toassuming the worst case for every designvariable

Quantifies life drivers what are the mostimportant variables and how well are theyknown or controlled ?

Quantifies risk



Introduction

Basic Probability and Statistics

Statistical Techniques

Analysis Methods

Characterizing Variability

Case Studies

FatigueCalculator.com

GlyphWorks


10/177









Introduction



Analysis Methods


Case Studies


GlyphWorks


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Deterministic versus Random

Deterministic from past measurements the future positionof a satellite can be predicted with reasonable accuracy

Random from past measurements the future position ofa car can only be described in terms of probability andstatistical averages


Random Variables

Discrete - fixed number of outcomes

Colors

Continuous - may have any value in thesample space

Strength


12/177


Descriptive Statistics

Mean or Expected Value Variance / Standard Deviation

Coefficient of Variation

Skewness

Kurtosis

Correlation Coefficient


Mean or Expected Value

Central tendency of the data

( )N

x

XExMean

N

1ii

x

=====


13/177


Variance / Standard Deviation

Dispersion of the data

( )N

)xx(

XVar

N

1i

2i

=

=

)X(Varx=

Standard deviation


Coefficient of Variation

x

xCOV

=

Useful to compare different dispersions

= 10 = 1

COV = 0.1

= 100 = 10

COV = 0.1


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Skewness

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

( )3

N

1i

3i

N

)xx(

XSkewness

=

=


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 havekurtosis greater than 3; distributions that are lessoutlier-prone have kurtosis less than 3.

( )4

N

1i

4i

N

)xx(

XKurtosis

=

=


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Covariance

A measure of the linear association betweenrandom variables

[ ])Y()X(E)Y,X(COV YXxy ==


Correlation Coefficient

0 X

Y

0 X

Y

0 X

Y

0 X

Y

0 X

Y

0 X

Y

= 1 = 0 = -1

0 < < 1 ~0 ~ 0

[ ]

YX

YX )Y()X(E

=


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Probability

Basic probabilityConditional probability

Reliability


Basic Probability

( ) 1AP0

( ) certain1AP =

( ) eimposssibl0AP =

The probability of event A occurring:


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Conditional Basic Probability

( ) ( ) ( ) ( )BandABorA

BAPBPAPBAP IU +=

( ) ( )

( )APBAP

ABP 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 0.03and the probability of bar 2 failing is 0.04.

What is the probability of the structure failing?


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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 ==


19/177


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

Standard

Deviations

Probability


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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 )

( ) ( ) 26.099865.0AP1000

==74 % failures


Statistical Distributions

Uniform

Normal

LogNormal

Gumble

Weibull

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

A useful on-line reference:


21/177


Cumulative Distribution Function

( ) ( )xXPXFx =

( ) 0Fx =

( ) 1Fx =

0

1

a b

Fx(X)


Probability Density Function

( ) ( )dxXfbXaPb

a

x= Strength )

Strength

Stress


Joint Probability Density

( )

( )

+

=

2

y

y

y

y

x

x

2

x

x2

2yx

xy

yyx2

x

12

1exp

12

1y,xf

correlation coefficient

Normal distributions


34/177


2D - Joint Probability Density

Contours ofequal probability

y

xx

Y

-3x

3Y

12

34

Stress

Strength


Acceptable Risk

Safety

P(failure) ~ 10-6 10-9

Economic

P(failure) ~ 10-2 10-4


35/177


Extreme Values

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


Key Points

Basic Nomenclature

Random VariablesStatistical Distribution

Cumulative distribution function

Mean

Variance

ProbabilityMarginal

Conditional

Joint


36/177









Introduction



Analysis Methods


Case Studies


GlyphWorks


37/177



Normal Distributions LogNormal Distributions

Monte Carlo

Sampling

Distribution Fitting


Failure Probability

)Sss(P I

Let be the stress and S the fatigue strength

Given the distributions of and S find theprobability of failure

Stress, Strength, S

s


38/177


Normal Variables

Linear Response Function

=

+=n

1iiio XaaZ

Xi ~ N( i , Ci )

=

+=n

1iiioz aa

= =n

1i

2

i

2

iz a


Calculations

Z = S -

z = S -

22

Sz +=

~ N( 100 , 0.2 ) = 20

z

= 200 -100 = 100

S ~ N( 200 , 0.1 ) S = 20

Stress, Strength, S

Safety factor of 2

2.282020 22z =+=

Let Z be a random variable:


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Failure Probability

Failure will occur whenever Z


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Fatigue Data

bf

'f

)N2(2

S=

102 103 104 105 106 107

Fatigue Life101

1000

100

10

1

StressAmplitude

'

f

b

1

'f

f2

SN2

=

b1


LogNormal Variables

=

=n

1i

a

ioiXaZ

as are constant and Xi ~ LN( xi , Ci )

==

n

1i

a

io

i

XaZmedian

( )=

+=n

1i

a2

XZ 1C1CCOV2

i

i


41/177


Calculations

Stress, Strength, S

~ LN( 250 , 0.2 ) = 50

~ LN( 1000 , 0.1 ) = 100

b

1

'f

f2

SN2

=

'f

b = -0.125

2

S8'

f

8

f2

SN2Z

==

8'f

8

f2

SN2Z

==

( ) ( ) 1COV1COV1COV2

f

2 8282

SZ ++=


Results

'fS/2 2Nf Pe rce ntile Life

x 250 1000 355,368 99.9 17,706,069

COVx 0.2 0.1 4.72 99 4,566,613

95 1,363,200

lnx 5.50 6.90 11.21 90 715,589

X 245 995 73,676 50 73,676

x 50 100 1,676,831 10 7,586

lnx 0.198 0.100 1.774 5 3,982

1 1,189

b = -0.125 0.1 307


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Monte Carlo Methods

( ) ( )

+

=

+cbf

'f

'f

b2

f

2'ff N2N2

EE

2

SK

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

'f

'fand

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


43/177


Computer Simulation

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

2. Count the number of 3s

Let Xi = 1 if 30 otherwise

( ) =

=n

1i

iXn

13P


EXCEL

=ROUNDUP( 6 * RAND() , 0 )

=IF( A1 = 3 , 1 , 0 )

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

5 0 0

3 1 0.5

4 0 0.333333

4 0 0.25

5 0 0.2

6 0 0.166667

1 0 0.142857

3 1 0.25

3 1 0.333333

6 0 0.3


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Results

0.1

0.2

0.3

0.4

1 100 200 300 400 500

trials

probability


Evaluate

x

yP( inside circle )

4

rP

2=

= 4 P

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

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

1

-1

1

-1


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0

1

2

3

4

1 100 200 300 400 500

trials


Monte Carlo Simulation

Stress, Strength, S

~ LN( 250 , 0.2 ) = 50

~ LN( 1000 , 0.1 ) = 100'f

b = -0.125

2

S

b

1

'f

f2

SN2

=

Randomly choose values of S andfrom their distributions'f

Repeat many times


46/177


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(),ln,ln)

=NORMINV(RAND(),,)

Normal

Log Normal


47/177


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

'

f

2

S2N

f


Simulation Results

103

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

104 105 106 107 108

Monte Carlo Analytical

Mean 11.25 11.21

Std 1.79 1.77


Life


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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: x

Population variance: x2

Sample mean: X

Sample variance: s2

( ) xXE =( ) 2x2sE =


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50/177


Translation

90% confidence

If we sampled a population many times toestimate the mean, 90% of the time the truepopulation mean would lie between thecomputed upper and lower limit.


Confidence Interval - mean

Lower limit of

xx

1n,n

stX

Upper limit of

xx

1n,n

stX +

For a normal distribution:


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Probabilistic Fatigue 2003-2005 Darrell Socie, University of Illinois at Urbana-Champaign, All Rights Reserved 100of 352

Sample Size

0 5 10 15 20 25 30

Sample size, n

0

0.5

1.0

1.5

2.0

xx

1n,n

stX

n

t 1n,

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 mean

Upper limit below mean

Sample Number

x = 100 x = 10


52/177


Confidence Interval - variance

Upper limit of

2

x1n,1

2

2

xs)1n(

For a normal distribution:


Sample Size

0 5 10 15 20 25 300

2

4

6

8

10

Sample size, n

2

x1n,1

2

2

xs)1n(

1n,12

)1n(

95%

50%

90%


53/177


Maximum Load Data

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

100 1000

Median 431

COV 0.34

200 500

Maximum force from 42 drivers


90% confidence COV 0.41


Maximum Load Data

Maximum Load

0 200 400 600 800 1000 1200

0.001

0.002

0.003

0.004

Pr

obabilityDensity

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


54/177


Choose the Best Distribution

0 200 400 600 800 1000

Maximum Load

Normal

LogNormal

Gumble

Weibull

0.001

0.002

0.003

ProbabilityDensity

15 samples from a Normal Distribution


Goodness of Fit Tests

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

0 200 400 600 800 1000


99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

0 200 400 600 800 1000


99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

100 1000200 500


99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %

100 1000200 500


99.9 %

99 %

80 %

50 %

10 %

1 %

200 400 600 800CumulativeProbability

99.9 %

99 %

80 %

50 %

10 %

1 %

200 400 600 800CumulativeProbability

Normal

LogNormal

Gumble

200

99.9 %99 %

80 %

50 %

10 %

1 %

0.1 %

1000500


200

99.9 %99 %

80 %

50 %

10 %

1 %

0.1 %

1000500


Weibull


55/177


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


Distributions

Normal Strength

Dimensions

LogNormal Fatigue Lives

Large variance in properties or loads

Gumble Maximums in a population

Weibull Fatigue Lives


56/177


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

=


57/177


58/177


Results

0

10

20

30

40

50

60

6 12 18 24 30 36

Z

Frequency

Central limit theorem states that the result should benormal 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 employedfor analysis even though the underlying populationdistribution is unknown.


59/177


Key Points

All variables are random and can be characterizedby a statistical distribution with a mean and variance.

The final result will be normally distributed even ifthe individual variable distributions are not.

Analysis Methods







60/177



Introduction Basic Probability and Statistics


Analysis Methods


Case Studies


GlyphWorks


Reliability Analysis

1 32

0.2

0.4

0.6

0.8

ProbabilityDensity

0.2

0.4

0.6

0.8

ProbabilityDensity

1 32

0.1

0.2

0.3

0.4

-3 -2 -1 0 1 2 3

ProbabilityDensity

Stressing Variables

Strength Variables0.2

0.4

0.6

0.8

Proba

bilityDensity

1 32

P( Failure )

Analysis

?


61/177


Probabilistic Analysis Methods

Monte CarloSimple

Hypercube sampling

Importance sampling

Analytical

First order reliability method FORM

Second order reliability method SORM


Limit States

Limit States

Equilibrium

Strength

Deformation

Wear Functional


62/177


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


63/177


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


64/177


Transform Variables

u1

u2

x

x1

xu

=

u1 = 0

u1 = 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 )

018.0

N

N)f(P f ==

u1 and u2 normally distributed


65/177


66/177


Stratified Sample

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

-3

-2

-1

0

1

2

3

4u1

u2

Each box should haveone sample drawn at randomfrom the underlying pdf

100 trials

25 simulations

= 0.02

= 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

= 0.017 = 0.038

250 calculations


67/177


Importance Sampling (continued)

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

-3

-2

-1

0

1

2

3

4

u1

u2

= 2

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

0183.0PN

N)f(P S

S

f ==

138 simulations


Analytical Methods Strategy

Develop a response function

Transform the set of N variables to a set ofuncorrelated u variables with =0 and =1

Locate the most likely failure point

Approximate the integral of the jointprobability distribution over the failure region


68/177


Analytical Methods Outline

Response Surface

Transform Variables

Limit State Concept

Most Probable Point

Probability Integration

FORM

SORM


Response Surface Methodology

Response surface methodology is a wellestablished collection of mathematical andstatistical techniques for applications where theresponse of interest is influenced by several

variables.

+= )x,x,x,x(f)X(Z n321


69/177


Response Surface

X1 X2

Z

Fit a surface through the points

Solution points


Mathematical Representation

+++

+++

++++=

323312211

2

33

2

22

2

11

332211o

xxCxxCxxC

xBxBxB

xAxAxAA)X(Z Linear

Incomplete Quadradic

Complete Quadradic

Solutions

Linear N + 1

Complete Quadradic N(N + 1)/2

Incomplete Quadradic 2N + 1


70/177


71/177


72/177


2D - Joint Probability Density

Contours ofequal probability

x

y

Stress

Strength


Transform Variables

x

x1

xu

=

u = 0u = 1

Any distribution can be mapped into a normal distribution

u1

u2

d ( )2d5.0expP

The mathematics and behavior of normal distributionsare well understood


73/177


74/177


Failure Probability

g = 0u1

u2

Strength )

How do you really get these distributions ?

Design Load

How do you establish a design load ?

How do you validate the design ?


133/177


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 surveys

k l c % r kl%

1 Unloaded 27

1 Highway 10

2 Good Road 25

3 Mountain 40

4 City 25

2 Half Load 581 Highway 5

2 Good Road 30

3 Mountain 30

4 City 35

3 Fully Loaded 15

1 Highway 15

2 Good Road 25

3 Mountain 40

4 City 20

12 Customer UsageCategories


134/177


Owner Behavior

Instrumented Vehicles

-3

-2

-1

0

1

2

3

0 -1 -2 -33 2 1

To Load, kN0 -1 -2 -33 2 1

To Load, kN

FromL

oad,

kN

Extensive field testing foreach customer usagecategory produces a largenumber of histograms.

Let the usage histogrambe denoted Ukl


Extrapolation

-3

-2

-1

0

1

2

3

0 -1 -2 -33 2 1

To Load, kN0 -1 -2 -33 2 1

To Load, kN

FromL

oad,kN

-3

-2

-1

0

1

2

3

0 -1 -2 -33 2 1

To Load, kN

FromLoad,

kN

-3

-2

-1

0

1

2

3

0 -1 -2 -33 2 1

To Load, kN

FromLoad,

kN

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

Ukl U*kl


135/177


Virtual Customers

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

= *klklki UrcNU

[ ]iU rainflow histogram for an individual carN kilometers

ck car loading, %

rkl car usage, %

*klU

distributions of rainflow histograms fordifferent car usage classifications


Design Loads

Initial design is done on the basis of a singleconstant 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.


136/177


Equivalent Fatigue Loading

101

102

103

104

105

106

Feq

m

1

6

m

eq10

FF

=

1Nn

f

=


Distribution of Feq

FF

Fn

Fn= F+ FDesign Customer:


137/177


138/177


What Strength is Needed?

F

F

Fn S

S


Some Statistics

Z = S - F

Suppose we want a probability of failure of 1 in 50,000

5f 10x2P

=

( ) 2F

2S

FS

Z

Zf1 1.4P +

=

==

Fn = S( 1 COVS )

Fn = F( 1 + COVF )


139/177


Mean Component Strength

( ) 2F

2S

FS

Z

Zf1 1.4P +

=

==

( )2

F

F

2

S

n

S

Fn

S

f1

COV1

COVCOV

F

COV1

1

FP

++

+

=


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

Relia

bility

n

S

F

= 3COVF = 0.2COVS = 0.1


140/177


Mean Strength

FF

Fn SS

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

85.2COV

F1

S

S

n

=

=


Design

Design calculations are made with loads Fn Large database relates Fn to vehicle parameters

so that a new vehicle can be designed fromhistorical measurements

Fatigue calculations are made with materialproperties standard deviations from themean


141/177


Component Testing

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

Component strength not life !


Interpreting the Test Data

Component tests are done with small sample sizes

2

x1N,1

2

2

xs)1N(

Confidence limits

( )2

F

F2

1N,1

2

S

n

S

Fn

S

f1

COV1

COV1NCOV

F

COV11

FP

++

+

=


142/177


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

FromL

oad,

kN

-3

-2

-1

0

1

2

3

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

FromL

oad,

kN

-3

-2

-1

0

1

2

3

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

FromL

oad,

kN

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

Ukl U*kl

How does this process work?


143/177


144/177


Probability Density

-3

-2

-1

0

1

2

3

0 -1 -2 -33 2 1

To Load, kN

FromLoad,kN


Kernel Smoothing

-3

-2

-1

0

1

2

3

From

Load,kN

0 -1 -2 -33 2 1

To Load, kN


145/177


Sparse Data

-3

-2

-1

0

1

2

3

FromLoad,kN

0 -1 -2 -33 2 1

To Load, kN


Exceedance Plot of 1 Lap

0

1

2

3

4

1 10 100 1000

Cumulative Cycles

LoadR

ange,kN

weibull distribution


146/177


147/177


Results

-3

-2

-1

0

1

2

3

0 -1 -2 -33 2 1

To Load, kN

FromL

oad,

kN

-3

-2

-1

0

1

2

3

FromL

oad,

kN

Simulation Test Data

0 -1 -2 -33 2 1

To 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 LapsLoadRa

nge,kN


148/177


149/177


Issues

In the first problem the number of cycles isknown but the variability is unknown andmust be estimated

In the second problem the variability is knownbut the number and location of cycles isunknown and must be estimated


Assumption

0

16

32

48

1 10 102 103 104 105

Cumulative Cycles

LoadR

ange

0

16

32

48

0

16

32

48

1 10 102 103 104 105

On average, more severe users tendto have more higher amplitude cyclesand fewer low amplitude cycles


150/177


Translation

-3

-2

-1

0

1

2

3

FromLoad,kN

0 -1 -2 -33 2 1

To Load, kN


Damage Regions

-3

-2

-1

0

1

2

3

FromLoad,kN

0 -1 -2 -33 2 1

To Load, kN


151/177


ATV Data - Most Damaging in 19

-3

-2

-1

0

1

2

3

FromL

oad,

kN

-3

-2

-1

0

1

2

3

FromL

oad,

kN


0 -1 -2 -33 2 1

To Load, kN0 -1 -2 -33 2 1

To Load, kN


ATV Exceedance

0

1

23

4

5

6

1 10 100 103 104

Cycles

LoadR

ange,kN

Actual Data

Simulation


152/177


Airplane Data - Most Damaging in 334

-30

-20

-10

0

10

20

30

0 -10 -20 -3030 20 10

To Stress, ksi

FromS

tress,

ksi

-30

-20

-10

0

10

20

30

FromS

tress,

ksi


0 -10 -20 -3030 20 10

To Stress, ksi


Airplane Exceedance

0

10

20

30

40

1 10 100 1000Cycles

StressRange,ksi

Simulation

Actual Data


153/177


154/177


155/177


156/177


Modeling Variability

Products: Z = X1 X2 X3 X4 .. Xn

Z LogNormal as n increases

Central Limit Theorem:

COVX

0 0.5 1.0 1.5 2.0

0.5

1.0

1.5

lnX lnX ~ COVX

COVX is a good measure of variability


Standard Deviation, lnx

COVX1 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


157/177


Strain Life Analysis


Material Properties


158/177


159/177


160/177


161/177


29 Individual Data Sets

'

fMedian 820

300 1000

COV 0.25

2000

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %


b

-0.12 -0.08 -0.04

Mean -0.09

COV 0.2599.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %



29 Individual Data Sets (continued)

0.1 1 10

Median 0.57

COV 1.1599.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %


'f

-0.8 -0.6 -0.4 -0.2

c

Mean -0.62

COV 0.23

99.9 %

99 %

90 %

50 %

10 %

1 %

0.1 %



162/177


163/177


Results (continued)


Results (continued)


164/177


165/177


Strain Amplitude 1000 250


Deterministic Analysis


166/177


Deterministic Analysis (continued)


Deterministic Analysis (continued)


167/177


Deterministic Analysis Results


Probabilistic Analysis


168/177


170/177



Introduction Basic Probability and Statistics


Analysis Methods


Case Studies


GlyphWorks


GlyphWorks / Rainflow


171/177


172/177


173/177


StatOutput.xls (continued)

Channel # 6 Total Channels 1

Variable Probabilistic Deterministic

NumCases 11 Load History Variables 0.89 -5.42

Median 45634 ScaleFactor 0.89 -5.42

COV 1.91 Offset 0.00 0.00

Stress Concentrators 0.05 -5.51

Probability (%) Life Kf 0.05 -5.51

99 817179 Material Properties 0.25 -14.54

90 223648 E 0.00 -4.37

50 45634 Sfp 0.23 3.42

10 9311 b 0.00 -4.39

1 2548 efp 0.00 1.42

c 0.00 -9.86

np 0.00 -2.32Kp 0.10 1.56

Analysis Variables 0.37 1.00

Uncertainty 0.37 1.00

Life Distribution Sensitivity


StatOutput.xls (continued)

Variable Distribution Value

Scale

Parameter Value

Scale

Parameter

ScaleFactor Normal 100 0.25 103 0.18

Offset Normal 0 0.10 -0.01 0.08

Kf Uniform 3 0.05 3.0 0.04

E None 208000 0.00 208000 0.00

Sfp Log-Normal 1000 0.10 1054 0.11

b None -0.1 0.00 -0.10 0.00

efp None 1 0.20 1.0 0.00

c None -0.5 0.00 -0.50 0.00

np None 0.2 0.00 0.20 0.00

Kp Log-Normal 1200 0.10 1194 0.10

Uncertainty Log-Normal 1 0.50 0.93 0.48

Inputs Outputs


174/177


StatStress.flo


Rainflow Extrapolation


175/177


Duration Extrapolation


Results Files


176/177


Rainflow Duration


Percentile Extrapolation

probabilistic fatigue

Documents