42154443 Reliability in Fatigue

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Reliability in fatigueOn the choice of distributions in the load-strength model

Tomas Torstensson

9th January 2004


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Reliability in fatigue

On the choice of distributions in the load-strength model

Abstract

In this thesis the influence of the choice of distributions in the load-strength model is considered. Accurate predictions of the failure probabilityis very useful when aiming at the most cost effective design of a compo-nent. Two distributions for load and strength are evaluated, the lognormaldistribution and the Weibull distribution. From the load-strength modelthe failure probability can be determined which is the probability that thecomponent in question fails within a specific time. The main conclusion isthat the lognormal distribution should be used rather than the Weibull dis-

tribution, especially when the data available is limited. A possible way ofupdating the model with observed failure rates using Bayesian methods isalso suggested.


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Tillforlitlighet inom utmattning

Val av fordelningar i last-styrka-modellen

Sammanfattning

I detta exjobb undersoks vilken paverkan olika fordelningsval har palast-styrka-modellen. Noggranna forutsagelser av felsannolikheten ar myck-et anvandbara for kostnadseffektiv dimensionering av komponenter. Tvafordelningar for lasten och styrkan studeras, lognormalfordelningen ochWeibullfordelningen. I last-styrka-modellen kan felsannolikheten beraknas,d.v.s. sannolikheten att den aktuella komponenten gar sonder inom en visstid. Huvudslutsatsen ar att lognormalfordelningen bor anvandas snarare anWeibullfordelningen, i synnerhet vid begransad tillgang pa data. Ett mojligt

satt att uppdatera modellen med felutfall med hjalp av Bayesianska metoderforeslas ocksa.


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Acknowledgments

I would like to thank my supervisors Par Johannesson, Jacques de Mareand Thomas Svensson for their extensive counseling and encouragement. Thefrequent meetings have made it possible to get feedback and advice in orderto make progress with the project. I wish to thank my examiner GunnarEnglund. I am also grateful to Bengt Johannesson at Volvo Trucks andBertil Jonsson at Volvo Articulated Haulers. I had access to their reportsand measurement data and my Masters thesis could not have been carriedout without that material. I wish to thank Fraunhofer-Chalmers ResearchCentre for giving me this opportunity to do my Masters thesis and it hasbeen a pleasure to work here. A special thanks to Marlo who reviewed themanuscript. Finally I would like to thank Ebba for her love and support.


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Contents

1 Introduction 6

2 Background 8

2.1 Normal distribution . . . . . . . . . . . . . . . . . . . . . . . . 82.2 Lognormal distribution . . . . . . . . . . . . . . . . . . . . . . 9

2.3 Three parameter Weibull distribution . . . . . . . . . . . . . . 92.4 A target customer . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.4.1 Duty based on normal distribution . . . . . . . . . . . 102.4.2 Duty based on lognormal distribution . . . . . . . . . . 102.4.3 Duty based on Weibull distribution . . . . . . . . . . . 10

2.5 Failure probability . . . . . . . . . . . . . . . . . . . . . . . . 112.5.1 Entire population . . . . . . . . . . . . . . . . . . . . . 112.5.2 The target customer . . . . . . . . . . . . . . . . . . . 11

2.6 Distributions for duty and capacity . . . . . . . . . . . . . . . 112.7 Models for duty and capacity . . . . . . . . . . . . . . . . . . 12

2.7.1 Estimation of capacity . . . . . . . . . . . . . . . . . . 132.7.2 Estimation of duty . . . . . . . . . . . . . . . . . . . . 13

2.8 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.9 Duty intensity . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Applications in industry 15

3.1 Volvo Trucks . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.1.1 Load-strength model . . . . . . . . . . . . . . . . . . . 153.1.2 Load-strength simulations . . . . . . . . . . . . . . . . 16

3.2 Volvo Articulated Haulers . . . . . . . . . . . . . . . . . . . . 163.2.1 Estimation of capacity . . . . . . . . . . . . . . . . . . 16

3.2.2 Estimation of duty . . . . . . . . . . . . . . . . . . . . 18

4 Inference based on data from Volvo Trucks 19

4.1 The lognormal distribution . . . . . . . . . . . . . . . . . . . . 194.2 The Weibull distribution . . . . . . . . . . . . . . . . . . . . . 20

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4.2.1 Estimators based on percentiles . . . . . . . . . . . . . 214.2.2 Estimators based on Maximum Likelihood . . . . . . . 224.2.3 Estimators based on an iterative procedure . . . . . . . 234.2.4 Estimators based on elimination . . . . . . . . . . . . . 24

4.2.5 Estimators based on Maximum Likelihood with twoparameters . . . . . . . . . . . . . . . . . . . . . . . . 254.2.6 Density and distribution fits . . . . . . . . . . . . . . . 25

4.3 Comparing the estimation methods . . . . . . . . . . . . . . . 274.3.1 Simulations . . . . . . . . . . . . . . . . . . . . . . . . 274.3.2 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . 28

4.4 Distribution of the duty . . . . . . . . . . . . . . . . . . . . . 284.5 Failure probability . . . . . . . . . . . . . . . . . . . . . . . . 29

4.5.1 Simulations . . . . . . . . . . . . . . . . . . . . . . . . 304.5.2 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . 314.5.3 Sensitivity analysis . . . . . . . . . . . . . . . . . . . . 31

4.5.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . 324.6 Qualitative analysis of duty distribution . . . . . . . . . . . . 32

4.6.1 Duty based on three parameter Weibull distribution . . 324.6.2 Duty based on lognormal distribution . . . . . . . . . . 334.6.3 Duty based on two parameter Weibull distribution . . . 334.6.4 Failure probability for different driven distances . . . . 33

5 Analysis of Volvo Articulated Haulers model 36

5.1 Model properties . . . . . . . . . . . . . . . . . . . . . . . . . 365.2 General truncation of the normal distribution . . . . . . . . . 37

5.3 Simulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.5 Failure probability . . . . . . . . . . . . . . . . . . . . . . . . 405.6 Density functions for capacity and duty . . . . . . . . . . . . . 41

6 Feedback using Bayesian estimates 42

6.1 Distributions of capacity and duty . . . . . . . . . . . . . . . . 426.2 Bayesian estimates . . . . . . . . . . . . . . . . . . . . . . . . 436.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

7 Conclusions and discussion 47

7.1 Future research . . . . . . . . . . . . . . . . . . . . . . . . . . 49

Appendix 49

A Estimated parameters with different methods 50

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B Results from Weibull inference 52

C Proportion p0 for different n 56

D Sensitiveness in failure probability 58

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List of Figures

4.1 Probability density fX (x), City. . . . . . . . . . . . . . . . . . 254.2 Cumulative distribution FX (x), City . . . . . . . . . . . . . . 264.3 Probability density fX (x), Highway . . . . . . . . . . . . . . . 264.4 Cumulative distribution FX (x), Highway . . . . . . . . . . . . 274.5 Density functions, fD, when = 0.65, = 10

5, = 6 106. . 294.6 Density functions, fD, when = 1.30, = 10

5, = 6

106. . 30

4.7 Failure probability Pf for different s, City. . . . . . . . . . . . 344.8 Failure probability Pf for different s, Highway. . . . . . . . . . 34

5.1 Distribution fits for C when n = 10000. . . . . . . . . . . . . . 395.2 Density fits for C when n = 10000. . . . . . . . . . . . . . . . 395.3 Density functions for C and D. . . . . . . . . . . . . . . . . . 41

6.1 Density functions for C and D before and after update. . . . . 45

C.1 Proportion p0 for different n when = 0.65. . . . . . . . . . . 57C.2 Proportion p0 for different n when = 1.30. . . . . . . . . . . 57

D.1 Failure probability Pf for different , varying around = 0.65. 59D.2 Failure probability Pf for different , varying around = 1.30. 59D.3 Failure probability Pf for different when = 0.65. . . . . . . 60D.4 Failure probability Pf for different when = 1.30. . . . . . . 60D.5 Failure probability Pf for different when = 0.65. . . . . . . 61D.6 Failure probability Pf for different when = 1.30. . . . . . . 61

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List of Tables

4.1 The entities mE and sE when = 0.65. . . . . . . . . . . . . . 314.2 The entities mE and sE when = 1.30. . . . . . . . . . . . . . 31

5.1 Weibull parameter values for different distribution fits. . . . . 405.2 Failure probability for different capacity distributions. . . . . . 40

B.1 Estimation of when = 0.65 and n = 5. . . . . . . . . . . . 52B.2 Estimation of when = 0.65 and n = 5. . . . . . . . . . . . 53B.3 Estimation of when = 0.65 and n = 5. . . . . . . . . . . . 53B.4 Estimation of when = 0.65 and n = 10. . . . . . . . . . . 53B.5 Estimation of when = 0.65 and n = 10. . . . . . . . . . . . 53B.6 Estimation of when = 0.65 and n = 10. . . . . . . . . . . . 54B.7 Estimation of when = 1.30 and n = 5. . . . . . . . . . . . 54B.8 Estimation of when = 1.30 and n = 5. . . . . . . . . . . . 54B.9 Estimation of when = 1.30 and n = 5. . . . . . . . . . . . 54B.10 Estimation of when = 1.30 and n = 10. . . . . . . . . . . 55B.11 Estimation of when = 1.30 and n = 10. . . . . . . . . . . . 55

B.12 Estimation of when = 1.30 and n = 10. . . . . . . . . . . . 55

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Chapter 1

Introduction

The load-strength model is a tool for reliability analysis in fatigue.The dam-age that an external cyclic load causes to a material is called the fatigue of

the material. The load itself can be characterized by its local maxima andminima. A good way to examine the load is the rainflow count method,see Johannesson [3], which gives the load amplitudes. Some definitions areneeded in order to describe the model

C = Capacity (strength) ,

D = Duty (load) .

The capacity can, e.g., represent the strength of a vehicle component andthe duty is then the load that the component is exposed to. The basis of themodel is that failure occurs (the component breaks) if the duty exceeds thecapacity, i.e. D > C. In lecture notes from a course for Swedish industry, see

Johannesson and de Mare [2], the load-strength model and several applica-tions are described. The duty naturally depends on the time or distance thecomponent has been used. The idea is that both capacity, C, and duty, D,are modelled as random variables. The scatter in the strength of the com-ponents is modelled by C and the scatter in the load is modelled by D. Thescatter in the strength is easy to understand (material and manufacturingproperties), but the scatter in the load is more complex. It consists of severalparts. Duty varies based on how the vehicle is driven, the road conditions,and so on. Therefore it is much harder to find and motivate a suitable modelfor the duty than the capacity. The failure probability is the probability that

failure occursPf = P(D > C) .

The model can be used in several ways. One way of using the model is inthe case where there is a restriction on the failure probability for a critical

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component. The objective could also be to find the minimum life cycle cost.The total cost is a sum of the manufacturing and operation cost. When thefailure probability decreases (stronger component) the manufacturing costincreases and for the operation cost the relationship is reversed. This meansthat an optimal design failure probability can be found which minimizes thelife cycle cost for the component. Then the component can be adjusted bychanging the manufacturing procedure in order to satisfy that condition.

The load-strength model have been used by PSA Peugeot Citroen, seeThomas et al. [7], and Volvo Construction Equipment, see Olsson [5] andSamuelsson [6]. They use different assumptions for the distributions of thecapacity and load. Volvo Construction Equipment uses a model in whichcapacity and duty is based on the three parameter Weibull distribution. PSAPeugeot Citroen models both capacity and duty with the normal distribution.

In Chapter 2 the background of the load-strength model is describedand the model and definitions are introduced. The use of the load-strength

model in industry is described in Chapter 3. In Chapter 4 the estimationof parameters in the lognormal and Weibull distribution are examined. Acouple of methods for estimation of the parameters in a three parameterWeibull distribution are examined. The basis of the estimation is data fromVolvo Trucks. Also the quantitative differences that depends on the choice ofdistribution are studied. The model of Volvo Articulated Haulers is analyzedin Chapter 5. In Chapter 6 feedback using Bayesian methods is examined.Finally in Chapter 7 conclusions are drawn and proposals for how to use theload-strength model in the future are suggested.

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Chapter 2

Background

The capacity, C, and the duty, D, are assumed to be continuous randomvariables. A continuous random variable X is defined by its density function,

fX (x), and its distribution function, FX (x).

P(a X b) =b

a

fX (x) dx = FX (b) FX (a)

fX (x) =d

dxFX (x)

In this thesis principally two distributions will be considered, the lognormaldistribution and the three parameter Weibull distribution (used by Volvo).The reason why these distributions are used is discussed in Section 2.6. Insome examples in this chapter the normal distribution (used by PSA Peugeot

Citroen) will also be considered.

2.1 Normal distribution

If X N(, 2) the density function is

fX (x) =1

2

2exp

(x )

2

22

, < x < .

The distribution function can not be determined explicitly

FX (x) =

x

fX (y) dy .

If X N(0, 1) (standard normal) then the distribution function is denotedby (x).

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2.2 Lognormal distribution

If X LN(, 2) it means that log X N(, 2), where log is the naturallogarithm. The density function is

fX (x) = 1x

2

2exp

(log x )222

, x > 0 .

Also for this distribution the distribution function can not be expressed ex-plicitly

FX (x) =

x0

fX (y) dy .

2.3 Three parameter Weibull distribution

If X W( , , ) then the density function is

fX (x) =

x

1exp

x

, , > 0, x > .

For this distribution there is a explicit expression for the distribution function

FX (x) =

x

fX (y) dy = 1 exp

x

.

In case = 0 the distribution is called a two parameter Weibull.

2.4 A target customer

The 100p% customer zp is defined as a quantile in the duty distribution, seeJohannesson and de Mare [2]

P(D zp) = pP(D > zp) = 1

p

where 1 p is the probability of finding a customer more extreme than zp.How to calculate an extreme customer depends of course on the distributionof the duty.

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2.4.1 Duty based on normal distribution

Assume that the duty is normally distributed with mean value mD =300 MPa and standard deviation D = 60 MPa. The 90% customer is sought,and p = 0.90 yields

z0.90 = mD + 0.10 D = mD + 1.28 D = 377 MPa.

2.4.2 Duty based on lognormal distribution

If the duty is lognormal the 100p% customer is found in the following way

P(D zp) = P(log D log zp) = P

log D

log zp

=

log zp

= p

which gives

log zp

= p

zp = e+p .

Assuming that the mean and variance is the same as in the example withthe normal distribution yields z0.90 = 379 MPa.

2.4.3 Duty based on Weibull distribution

Assume that the duty is inversely proportional to a Weibull distributed ran-dom variable, i.e. D = 1/Y where Y

W(,, 0).

FY(y) = 1 exp

y

, y 0, , > 0 .

The 100p% customer can be determined directly from the definition

P(D zp) = P

1

Y zp

= P

Y 1

zp

= 1 FY

1

zp

= exp

1

zp

= p .

Extracting zp gives the explicit expression

zp =1

( logp)1/ .If the mean value and the standard deviation is the same as in the examplewith the normal distribution, then z0.90 = 372 MPa.

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2.5 Failure probability

There are different kinds of failure probabilities. One of them is the proba-bility that a failure occurs considering the entire population. Another is theprobability that failure occurs for the 100p% customer.

2.5.1 Entire population

The probability that a failure occurs for the entire population is denoted byPf and is calculated as

Pf = P(D > C).

2.5.2 The target customer

Given the 100p% customer the duty D = zp which implies that

Pf, p = P(zp > C).

This means that once the 100p% customer is known only the distribution ofthe capacity is needed.

In case that the capacity is normally distributed it is rather simple tocalculate the failure probability. Let z0.90 = 400 MPa and assume thatC N(500, 502). The failure probability for the target customer is then

Pf = P(C < zp) = P

C mC

C C.

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2.7.1 Estimation of capacity

By performing experiments with varying load amplitudes Si, a sequenceof a total of MC load cycles before the component breaks is received. Sincefailure corresponds to d = 1, C can be extracted

1 =

MCi=1

1

CSki

C =

MCi=1

Ski .

Both the mean and the standard deviation of C can be estimated byrepeating the experiment a number of times.

2.7.2 Estimation of duty

The duty, D, is estimated from a load process and an observation is deter-mined by the formula

D =

MDj=1

Skj

where MD is the total number of load cycles. The mean and the standarddeviation of D can be estimated from load measurements on different cus-tomers.

2.8 Applications

In a typical application only the relation between C and D is of interest. Fur-thermore, a value for the Wohler exponent is chosen, e.g. Volvo ConstructionEquipment (VCE) often uses k = 3 since the components are welded. In thatcase C and D are determined from the expressions

CV C E =

MCi=1

S3i , DV CE =

MDj=1

S3j .

2.9 Duty intensity

Since the duty is accumulated linearly it is possible to determine the dutyintensity. If e.g. observations of the duty are given but they correspond to

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different distances they must obviously be normalized in some way. Theobservations are transformed to the duty intensity, D [duty/km], to make itpossible to use them for further inference. The duty intensity is calculated asthe duty divided by the driven distance. The duty at a certain distance, s,is just D = s

D. If the duty is observed at different times the duty intensity

have the unit [duty/h] . Then the duty at a certain time, t, is calculated asD = t D. The duty intensity will be explained more carefully in the contextin which it is used in the thesis.

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Chapter 3

Applications in industry

3.1 Volvo Trucks

Volvo Trucks has made extensive trials in order to predict the life distributionby means of the load strength model.

3.1.1 Load-strength model

1. Analysis of fatigue data regarding strength.The analysis includes a description of fatigue data variation and fatiguemodelling by the life distribution. The capacity, C, is calculated fromrig tests and is determined by the formula

C = i

ni (Si)k

where the ni is the number of cycles until a failure occurs at the loadlevel Si and

i

ni is the total number of cycles until failure occurs.

2. Analysis of operational loading.The accumulated duty, D, is calculated from the load spectrum and isdetermined by the formula

D =

i

ni (Si)k.

Duty values are divided by the distance in km to get a comparable unit[duty/km], i.e. the duty intensity D. The Wohler exponent k = 4 areused in the calculations which differs from the value k = 3 that VolvoConstruction Equipment uses. That value is justified from examinationof the Wohler curve.

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3. Life predictions and modelling.The values for duty/km and capacity are examined and fitted inferenceis based on a three parameter Weibull distribution. Then the life canbe predicted by means of simulation techniques. There are two moreassumptions except those already stated.

The duty and the capacity are assumed to be independent. Duty isactively accumulated until it reaches capacity (failure occurrence).

The duty of the component is based on the lateral loading only,even though loading may exist in vertical and longitudinal direc-tion as well.

3.1.2 Load-strength simulations

In order to estimate the duty, a number of field measurements have been

performed at different markets, e.g Sweden, Norway, Germany and Brazil.The distribution of the capacity has been examined from rig tests and athree parameter Weibull inference has been made. The results are shownin diagrams with Level crossing spectrum, Range spectrum and Powerspectrum. Some other types of diagrams are plotted too, e.g. accumulatedfailure rate against driven distance and accumulated driven distance againsttime [year].

3.2 Volvo Articulated Haulers

Bertil Jonsson at Volvo Articulated Haulers has written a report in which heuses the load strength model in a slightly different way. The load is estimatedby driving for an hour, and the strength is estimated from the Paris equation.

3.2.1 Estimation of capacity

In order to estimate the capacity the Paris equation is used

da

dN= B (K)k

where a is the crack length. The number of cycles is denoted by N. Thecrack growth rate is dadN. The random variable B corresponds to materialproperties and crack geometry of the component. The range of the stressintensity factor is denoted by K and k = 3 is a constant. Later it will turnout that it is equal to the Wohler exponent.

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If the load amplitude is denoted by S, and using the fact that

K = S f(a)

where f(

) is a function that is determined by measuring the stress intensity

factor for different crack lengths, the expression can be written in the form

da

dN= B (S f(a))k.

Rewriting and integration the expression above yields

aca0

da

B (S)k f(a)k=

Nc0

dN

1

B

aca0

da

f(a)k = Nc (S)k

where a0 is the initial crack length and ac (20 mm here) is the crack lengthdefined as failure. The number of cycles until failure is denoted by NC.Extracting Nc gives

Nc =1

B

acA0

da

f(a)k (S)k

where A0 is with capital letter since it is considered to be a random variable.

Comparing this expression to the other one for N

N = C(S)k

implies that C can be identified and written as

C = g(B, A0) =1

B

acA0

da

f(a)k

i.e. a function g() of the random variables B and A0.The entities B and A0 are considered to be normally distributed andexperiments have been made in order to estimate mean and standard de-

viation. The distribution of C can then be estimated by making computersimulations. Volvo Articulated Haulers then fits a three parameter Weibulldistribution to these capacity values.

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3.2.2 Estimation of duty

The duty is determined by driving in different ways (normal and forced driv-ing) for one hour and after that making an assumption of how common thedifferent types of driving styles are. Once that is made an estimation of the

distribution of the duty intensity, D, can be determined. In these calcula-tions the Wohler exponent k = 3 which is a value that is often chosen inthis context. The duty is assumed to be based on a three parameter Weibulldistribution which is fitted to the observations.

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Chapter 4

Inference based on data from

Volvo Trucks

In this section estimation of parameters in the lognormal and the Weibulldistribution will be examined. The basis of the examination will be the datafrom Volvo Trucks. The different estimation methods have been implementedin Matlab. All numerical calculations in this thesis have been carried outthrough the use ofMatlab.

4.1 The lognormal distribution

Assume that the random variables X1, X2, . . ., Xn i.i.d. (independent andidentically distributed) LN(, 2). Let = (1, 2) = (,

2). The densityfunction for Xi will then be

fX (x,) =1

x

2

2exp

(log x 1)

2

22

, x > 0 .

Given that x1, x2, . . ., xn is a random sample from X1, X2, . . ., Xn, theparameter vector can be estimated. Since log Xi is normally distributedthe MLE (maximum likelihood estimate) of will be the same as for thenormal distribution, but with the logarithm of the observations, i.e.

1 = 1n

ni=1

log xi ,

2 =1

n

ni=1

(log xi 1)2 .

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The estimate 2 is slightly biased, but the estimate

2 =1

n 1n

i=1

(log xi 1)2

is unbiased. The unbiased estimate, 2, is used further on in this thesis.

4.2 The Weibull distribution

Assume that the random variables X1, X2, . . ., Xn i.i.d. W( , , ) aregiven. = (1, 2, 3) = ( , , ). The density function for Xi is

fX (x,) =

x

1exp

x

, , > 0, x >

where is the shape parameter, is the scale parameter, and is the locationparameter. Note that is a threshold. Given a sample x1, x2, . . . , xn fromX1, X2, . . . , X n the task is to find an estimator of. The likelihood functionis

L() =n

i=1

fX (xi) =

n ni=1

xi

1 ni=1

exp

xi

Normally L or log L is maximized in order to find the MLE of, but the prob-lem with the three parameter Weibull distribution is that it has a thresholdwhich means that L() can be singular due to the factor

(xk )1 where xk = min1in

xi

Since > xi i it is only the factor with the minimum xi that has to beexamined. The expression can be examined by letting tend to xk

< 1 : limx

k

(xk )1 =

= 1 : limx

k

(xk )1 = 1

> 1 : limx

k

(xk )1 = 0

This irregularity causes difficulties in the determination of the parametersand thus it would be an advantage to use other methods to estimate at leastone of them. It turns out that it is often most convenient to estimate thethreshold first.

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4.2.1 Estimators based on percentiles

The distribution function of a three parameter Weibull distributed variableis

FX

(x) = 1

expx

, x > .The 100p% population percentile, xp is determined from the relationship

FX (xp) = p

which gives

1 exp

xp

= p

x

= log(1 p)xp = + ( log(1 p))1/ .

Let y1, y2, . . . , yn be the ordering of a sample x1, x2, . . . , xn, i.e. yi = x(i).Then the sample distribution function that corresponds to the ith orderedobservation is

pi =i

n + 1

and then the corresponding 100pi percent sample percentile ti is given by

ti = ynpi

where x is the smallest integer that is larger than or equal to x.The idea is that three sample percentiles yield a system of three equationswith the three unknown parameters. Let the estimation of the parametersbe , and . Then the system is given by

ts = + [ log(1 ps)]1/ = ynps s = i,j,kwhere 0 < pi < pj < pk < 1.If pj is chosen such that

log(1

pj ) =

{[

log(1

pi)][

log(1

pk)]}1/2

the estimation of will be, see Zanakis [8]

= log

log(1 pk) log(1 pi)

log

tk ti

.

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where the estimate also is found with help from the sample percentiles

=titk t2j

ti + tk 2tj .

If this estimated value exceeds the smallest observed value, y1, then it is notpermissible and = y1 should be used instead, but the probability that sucha situation occurs is very small.The values for pi and pk can be chosen such that the asymptotic variance ofthe estimator is minimized, see Dubey [1], which yields

pi = 0.16731,

pk = 0.97366.

A efficient estimator of can be found by using the 1st, 2nd and nth ordered

observation of the sample. When the estimation of is determined it is aneasy task to find the estimation of . Finally the estimators are

=y1yn y22

y1 + yn 2y2 , = + y0.63n , = log

log(1 pk) log(1 pi)

log

tk ti

.

The advantage of these estimators are both their simplicity and accuracy,especially when n is small which suits the load-strength application since it

often involves small samples. These estimates will be denoted by , and .

4.2.2 Estimators based on Maximum Likelihood

Consider the two parameter Weibull distribution, which corresponds to =0. The distribution function is

FX (x) = 1 exp

x

, x > 0 .

Let X W(,, 0) and Y W( , , ). Assume that is known or es-timated. Then it is possible to transform the three parameter Weibull dis-tributed random variable into a two parameter such one. The advantage ofthis approach is that there will be no singularity problem since the thresholdhas already been estimated.

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Let Z = Y and examine the distribution function of ZFZ(z) = P(Z z) = P(Y z) = P(Y z + ) = FY(z + ) = FX (z)

which implies that Z and X are identically distributed.

Since the MLE can be determined for the two parameter Weibull distribu-tion, a way to estimate the parameters could be to first estimate accordingto the rule

=y1yn y22

y1 + yn 2y2and then determine the MLE of and by maximizing the likelihood func-tion for the sample values {xi } , i = 1, . . . , n. The estimates of theparameters according to this method will be denoted by , and .

4.2.3 Estimators based on an iterative procedure

A natural estimation of the location parameter is = x(1). The disadvan-tage with this estimation is that it will always exceed the true value. It istherefore of interest to examine the expectation of X(1)

E(X(1)) =

x fX(1)(x)dx .

Consequently the distribution of X(1) must be determined

FX(1)(x) = P(X(1) x) = 1 P(X(1) > x) = 1 P(X1 > x, X2 > x, . . . , X n > x)= {independent} = 1 P(X1 > x) P(X2 > x) . . . P(Xn > x)

= 1 (P(X > x))n

= 1 (1 FX (x))n

= 1

1

1 exp

x

n

= 1 exp

n

x

= 1 exp

x

where

n1

=

1

=

n1/.

The calculations above show that X(1)

W(, , ). The expectation of

X(1) can be determined since the expectation of a two parameter Weibulldistributed variable is known. Assume that Y W(,, 0). Then it followsthat

E(Y) =

+ 1

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where () is the gamma function that is defined by

(x) =

0

tx1etdt .

The random variable X is simply a translation of Y, units to the right

which implies thatE(X) = E(Y) + .

Finally the expectation of X(1) can be determined

E(X(1)) = +

+ 1

= +

n1/

+ 1

.

This means that the bias of the estimate = X(1) is

n1/

+ 1

.

Therefore = x(1)

n1/

+ 1

could be an appropriate bias corrected estimate of .

Therefore it would be possible to estimate the parameters iteratively ac-cording to the scheme

1. Starting guesses for and are determined by a direct method.

2. An estimation of is determined by the rule above.

3. The parameters and are determined by a two parameter Maximum

Likelihood distribution fit to the values {Xi }, i = 1, . . . , n.By looping over point 2 and 3 the parameters will eventually converge andthe loop is terminated when they change less than a certain tolerance in oneiteration.

4.2.4 Estimators based on elimination

Since an estimate of can be determined

= x(1) n1/

+ 1

a reasonable method to calculate the parameters would be to substitute inthe likelihood function. Now there is only two free parameters left which canbe estimated with ML. When that is done is determined by the expressionabove using the MLE of and .

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0 1 2 3 4

x 105

0

0.5

1

1.5

2

x 105

Mix

Zan

Iter

VTC

ML2

Elim

Figure 4.1: Probability density fX (x), City.

4.2.5 Estimators based on Maximum Likelihood withtwo parameters

If the threshold = 0 the distribution is the usual Weibull distribution. Thelikelihood function can be maximized directly, for estimating and in thiscase, and no singularity problem occurs.

4.2.6 Density and distribution fits

Since real observations were available from Volvo Trucks it is interestingto apply the proposed methods to this data. Note that Volvo Trucks usesthe assumption that 1/D is three parameter Weibull distributed. One way

to visualize the result from the inference is to plot the fitted density ordistribution function for each of the methods (see Figures 4.1-4.4). (Theresulting Weibull density and distribution functions have been plotted.) Theobservations are represented by diamonds () in the figures. The estimatedparameter values for the different methods are tabulated in Appendix A.

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0 2 4 6

x 105

0

0.2

0.4

0.6

0.8

1

MixZanIterVTCML2ElimEmpir

Figure 4.2: Cumulative distribution FX (x), City

0 5 10 15

x 105

0

1

2

3

4

5

6

7

x 104

Mix

Zan

Iter

VTC

ML2Elim

Figure 4.3: Probability density fX (x), Highway

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0 0.5 1 1.5 2

x 104

0

0.2

0.4

0.6

0.8

1

MixZanIterVTCML2ElimEmpir

Figure 4.4: Cumulative distribution FX (x), Highway

4.3 Comparing the estimation methods

Assume is an estimate of. The usual way to estimate the accuracy of anestimate is to calculate the mean squared error (m.s.e.)

m.s.e.() = E(( )2) = V() + (E( ))2 = V() + b()2.

4.3.1 Simulations

The different estimation methods can be compared by simulations. Thecapacity C W(2, 2.06 1011, 1.56 1011) where the parameters are takenfrom the Volvo Trucks report. In these simulations only the duty intensity,D, is varied. The design distance (a reasonable distance for the life of avehicle) is set to 1 000 000 km which means that s = 1 000 000 and the dutyD = s D as usual. This means that D = s D = sY where Y W( , , ).In order to compare the methods four different sets of parameters for Y havebeen chosen.

1. = 0.65, = 105, = 6 106, n = 52. = 0.65, = 105, = 6 106, n = 103. = 1.30, = 105, = 6 106, n = 5

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4. = 1.30, = 105, = 6 106, n = 10where n is the number of observations. Both the parameters and the numberof observations have values that are similar to the ones that were determinedfrom the real observations. This is very important since it means that the

conclusions from the simulations are in some way also true for real cases. Foreach setting of parameters 10 000 iterations have been carried out to get anaccurate estimate of the m.s.e. The results are found in tables in AppendixB. The column V() (%) means the proportion in percent of the m.s.e,

i.e. 100 V()m.s.e.() . The remaining part of the m.s.e. is due to the bias of theestimates and it is shown in the column b()2 (%).

4.3.2 Conclusions

The direct method (Zan) and the method that first uses the direct methodin order to estimate and then computes MLE of the other two parameters(Mix) are the best methods (smallest m.s.e.). For almost every setting ofparameters these methods are first and second best. It is not possible toconclude which of the two methods is best when looking only at the simula-tions (see Appendix B). The direct method is simple to use which makes itmore suitable for use in industrial applications.

4.4 Distribution of the duty

Since the duty depends on the driven distance a design distance must bechosen. A reasonable design distance is 1 000 000 km which means that s =1 000 000. The duty D = sD. The distribution ofD can be determined sincethe distribution of 1/D is known. Let D = 1/Y where Y W(Y, Y, Y).

D = s DfD(d) =

1

sfD

d

s

FD(d) = P(D d) = P

1

Y d

= P

Y 1

d

= 1 FY

1

d

fD(d) =

fY

1d

1d2

=1d2

fY

1d

which implies that

fD(d) =YY

s

d2

sd YY

Y1exp

s

d YY

Y, 0 < d C) = P s

Y> C

= P(CY < s) =

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0 0.5 1 1.5 2 2.5

x 1011

0

1

2

3x 10

11

distance = 5e+05

distance = 1e+06

distance = 1.5e+06

Figure 4.6: Density functions, fD, when = 1.30, = 105, = 6 106.

If the duty is lognormal, i.e. D LN(, 2) then the calculation of the failureprobability is slightly different.

Pf = P(D > C) =

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where lg is the logarithm to the base 10. One problem is how to deal withthe cases when p = 0 since then Eexp = . If those cases are dealt withseparately it is possible to determine the accuracy of the two methods when

p = 0. The probability that the estimated Pf is less or equal to zero isdenoted by p0. The mean (mE) and standard deviation (sE) of the errorEexp can be determined by simulating a number of times, in this case 10 000.When = 0.65, Pf = 1.4490 104 and when = 1.30, Pf = 1.0634 105.

It is also of interest to examine how the proportion p0 changes when the

n=5 n=10mE sE p

0 mE sE p

0

Zan 0.105 1.032 0.4373 -0.5007 0.9605 0.2626Mix -0.0457 1.063 0.4386 -0.3364 0.9155 0.2627

Table 4.1: The entities mE and sE when = 0.65.

n=5 n=10mE sE p

0 mE sE p

0

Zan 0.9168 1.6 0.7507 -0.7226 1.508 0.7808Mix 0.5553 1.657 0.7619 -0.4021 1.463 0.7772

Table 4.2: The entities mE and sE when = 1.30.

sample sizes increases. This relation is plotted in Figures C.1 and C.2 which

are found in Appendix C. The direct method was used because it is veryfast, especially for bigger samples.

4.5.2 Conclusions

It is hard to draw any conclusions from the simulations that are summarizedin Tables 4.1 and 4.2 since the probability that Pf = 0 is so high. Thismeasure of the error, Eexp would probably be better in a situation wherethe failure probability always is positive, e.g. when either the capacity or theduty is lognormal.

4.5.3 Sensitivity analysis

Since the accuracy in determining the failure probability Pf depends on theaccuracy in the estimates of the parameters, it is of interest to examine how

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big that influence is. One way to do that is to vary one of the parametersand keeping the other two fixed and then calculate the failure probability foreach combination. The figures are found in Appendix D. The m.s.e. for thedirect method is included in the captions.

4.5.4 Conclusions

Since the estimations of the parameters are dependent the figures in Ap-pendix D do not show the true relationship, but nevertheless they give somequalitative information. Anyhow it is clear that the failure probability de-pends mostly on the value of . Therefore the method of determining , ifthere should be a threshold at all, will have a big influence on the final result.Probably it would be better to use a more robust model.

4.6 Qualitative analysis of duty distributionThere are two properties that have to be examined in order to decide howto model the duty. The first of them, which will be discussed in this sectionis the qualitative property of the model. The other of the two properties arecomposed of the computational properties, such as stability and accuracy.

4.6.1 Duty based on three parameter Weibull distri-bution

Since the duty D = sY

where Y

W(Y, Y, Y) as a consequence there

will be an upper limit for D which will be equal to sY . The question is ifit is reasonable to have an upper limit for the duty? This means that it isimpossible to exceed a certain duty limit no matter how the driver drivesor what the road conditions are. This seems strange, but maybe it could be

justified if the upper limit is so high that the probability to cause a duty closeto this upper limit would be very low. One way to examine this phenomenonmore strictly mathematically is to determine certain quantiles for the dutydistribution.

P(D zp) = p, which gives

zp =

s

Y + Y ( logp)1/YIn order to be able to compare the results for different duty distributionsthe observations from the Volvo Construction Equipment report have beenused (City and Highway). The parameters that were estimated by Volvo

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Construction Equipment (Weibull++) are used here, see Appendix A. Twoimportant quantiles are the median z0.50 and the 95% quantile z0.95. Thedesign distance s is 1 000 000 here. The quantiles and the upper limits aregiven below.

1. City: z0.50 = 8.53 1010, z0.95 = 15.5 1010, sY = 15.6 10102. Highway: z0.50 = 6.74 1010, z0.95 = 7.76 1010, sY = 7.81 1010

It is clear that if it were be an upper limit it should not be as close to z0.95as it is in this case. It is unreasonable that so many drivers have duty valuesthat are close to the upper limit. Therefore this model seems inappropriate.

4.6.2 Duty based on lognormal distribution

In this case there is no upper limit. The quantiles in the lognormal distribu-

tion can be found numerically for the two cases.1. City: z0.50 = 6.98 1010, z0.95 = 25.0 1010

2. Highway: z0.50 = 2.83 1010, z0.95 = 8.77 1010

4.6.3 Duty based on two parameter Weibull distribu-tion

Here there is no upper limit. The quantiles are given by the same expressionas for the three parameter Weibull distribution, but with Y = 0.

1. City: z0.50 = 6.22 1010

, z0.95 = 34.7 1010

2. Highway: z0.50 = 2.60 1010, z0.95 = 15.3 1010

4.6.4 Failure probability for different driven distances

A model that describes duty in a good way should give reasonable results fordifferent driven distances (s). One way to visualize this is to determine Pffor different s and the resulting graphs are found in Figures 4.7 and 4.8. Thecurves that are denoted by drivers corresponds to the failure probability forone certain driver, i.e. one duty intensity value (di). From this the empirical

distribution function for the duty is generated and it is determined by therelations

P(D < s di) = 0P(D s di) = 1

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0 1 2 3 4 5 6

x 106

106

105

104

103

102

101

100

Pf

VTClognML2drivers

driven distance

Figure 4.7: Failure probability Pf for different s, City.

0 1 2 3 4 5 6 7 8

x 106

106

105

104

103

102

101

100

driven distance

Pf

VTClognML2drivers

Figure 4.8: Failure probability Pf for different s, Highway.

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In Figures 4.7 and 4.8 it is clear that the upper limit results in strangeproperties for the failure probability. It is zero until a certain distance andthen it suddenly increases very fast. This is not reasonable because the failureprobability should increase in a smoother way as it does for the other twodistributions. It is also interesting to note how much this upper limit differsin the two cases.

There is also a big difference between the lognormal fit and the Weibullfit without upper limit (corresponds to two parameter Weibull distribution).This is due to the fact that the density function for the lognormal distribu-tion decreases faster for larger values compared to the Weibull distribution,especially for smaller s. That is why the failure probability differs with or-ders of magnitude for small s. Consequently the load-strength model is verysensitive with respect to the choice of distribution. Therefore the model mustbe compared with real outcomes before it can be used.

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Chapter 5

Analysis of Volvo Articulated

Haulers model

The model by Volvo Articulated Haulers is described more precisely in Sec-tion 3.2. One interesting aspect of this approach is the model for calculatingcapacity. It is stated that

C =1

B

acA0

da

f(a)k

where the random variables and parameters are described in Section 3.2. Thefunction f() has been determined by tests

f(a) = 0.1388 a2

+ 0.35 a + 5.4 .

Since the assumption is that A0, the initial crack length, and B, a mate-rial parameter, are normally distributed tests have been done in order todetermine the mean and variance. The results of these tests are that

A0 N(mA0 , 2A0) = N(10, 1.78) ,B N(mB , 2B) = N(1.832 1013, 2.098 1027) .

5.1 Model properties

In the model by Volvo Articulated Haulers it is assumed that the initial crack,A0, should be greater than zero and less than the crack length by failure, i.e.

0 A0 20 .

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Therefore it is of interest to examine this property for the model that ischosen. Since the initial crack is normally distributed it can happen thatA0 takes values outside the interval [0, 20]. Therefore the correspondingprobabilities are of importance

P(A0 < 0) = P

A0 mA0A0

< mA0A0

= 1

mA0A0

= 1

101.78

= 3.31 1014

and by symmetry

P(A0 > 20) = P(A0 < 0).

Since the probability that A / [0, 20] is negligible this model error is probablynot a big problem. The material parameter, B, should be greater than zero.

Since B is normally distributed it can take values less than zero and thereforethat probability must be determined

P(B < 0) = P

B mB

B C) = P(log D

log C > 0) = D + log t C

2D + 2C

and with the numerical values

Pf =

11.97 + log 5 000 20.97

1.139 + 0.3386

= (0.3972) = 0.346.

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This value can be compared to the failure probability when the duty wasbased on a three parameter Weibull distribution which gave Pf = 0.335 andthis means that the transformation to a lognormal duty seems reasonable.

6.2 Bayesian estimates

For each machine included in the set of observed machines a random variableis associated

i = 1{Di>Ci}, i = 1, 2, . . . , n .

This means that machine number i is broken if i = 1, and it works if i = 0.The total number of machines that are broken after a certain time can thenbe expressed in the following way

Sn =n

i=1

i .

Since the probability that i = 1 simply is equal to the failure probability itwill hold that

Sn Bin(n, p), p = Pf.

In the Volvo Articulated Haulers report the total number of machines is n =917 and the number of broken machines after 5 000 h is 18 which correspondsto 1.96%

2% of the population. In this case this would yield the observation

s917 = 18.Now the Bayesian method can be applied. First, assume that the distributionof C is much more accurate than the distribution of D. This is a reasonableassumption since in general, more information about the capacity than theduty is available. Therefore the distribution of C is kept fixed, but thedistribution of D is modified. Since it is easier if only one parameter is freethe new model will be

C LN(20.97, 0.3386)D

LN(, 1.139)

where the parameter is assumed to have normal prior distribution. Supposethat

N(11.97, 1)

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The prior mean value of is just the estimation of D that was determinedfrom the duty observations. The prior variance of has here been set to1, which seems reasonable. It is not evident what variance to use. Oneidea could be to take into account the spread in the estimate of D, butthis is hard to carry out since the lognormal distribution of the duty wastransformed from a Weibull distribution and not fit from data. The issue ofchoosing the variance is actually a question about how much influence theprior distribution will have on the posterior distribution, i.e. the final model.A small prior variance means that the duty observations will have a greatinfluence on the model, and a large variance means that the failure rate willhave a greater influence. Therefore this issue must be carefully examined indeveloping this type of Bayesian methods for the load-strength model.

Due to the Bayesian method a reasonable estimation of D would beE(|s197 = 18). In general

E(|Sn = sn) = f|Sn(|sn) d .The density function f|Sn(|sn) can be found via Bayes theorem, see Lind-gren [4]

f|Sn(|sn) =fSn|(sn|) f()fSn|(sn|) f()d

.

Since Sn is binomial it will hold that

fSn|(sn|) = P(Sn = sn| = ) =

n

sn

psn(1 p)nsn, sn = 0, 1, . . . , n

where

p =

+ log 5 000 20.97

1.139 + 0.3386

.

The normalization integral can then be solved numerically

fSn|(sn|)f()d =

917

18

12.451.216

181

12.45

1.216

899

12 1 exp

( 11.97)

2

2 1

d = 0.001427 = C1 .

Now the posterior distribution of can be determined

f|Sn(|sn) = C

917

18

12.451.216

181

12.45

1.216

899

12 1 exp

( 11.97)

2

2 1

.

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0 0.5 1 1.5 2 2.5

x 109

0

1

2

3

4

5

6

7x 10

9

CD, before updateD, after update

Figure 6.1: Density functions for C and D before and after update.

Since the posterior distribution is known both the mean and variance can bedetermined numerically

E(|S917 = 18) = 9.963V(|S917 = 18) = E(2|S917 = 18) (E(|S917 = 18))2 = 99.267 9.96262

= 0.0132.

Let be the estimation of D. The prior estimate = E() = 11.97 andthe posterior estimate = E(

|S917 = 18) = 9.96 with variance 0.0132.

This means that the parameter D in the load-strength model has decreasedfrom 11.97 to 9.963 due to the failure rate. It is of interest to compare thedistribution before and after the update. In Figure 6.1 the density functionsare plotted and it can be observed that the distribution of the duty has movedto the left which means that the duty values in general are much lower afterthe update. The model after the update gives the new failure probability

Pf =

9.963 12.45

1.216

= (2.0455) = 0.0204.

which means that the failure probability has decreased from 35% to 2%.

The reason that the new distribution fits well to the observation of the actualfailure probability is a combination of the fact that the number of machinesthat is observed (n = 917) is large which gives a high accuracy in the pro-portion 2% and that the variance where set to 1. This means that the load-strength model adapts almost completely to the failure rate. Therefore it

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would be interesting to see how much the result would differ if it instead wasone out of 50 machines that were broken. This is still 2% but the accuracyis much lower.

Assume that the prior distribution of is the same as before, i.e. N(11.97, 1) and that the observation s50 = 1 is given. Then the normalizationintegral is

fSn|(sn|)f()d =

50

1

12.451.216

11

12.45

1.216

49

12 1 exp

( 11.97)

2

2 1

= 0.02690.

In this case the posterior mean and variance of will be

E(|S50 = 1) = 10.228

V(|S50 = 1) = E(2|S50 = 1) (E(|S50 = 1))2 = 104.76 10.2282= 0.156.

This means that the estimate has decreased from 11.97 to 10.23. Thisupdated parameter value corresponds to the failure probability

Pf =

10.23 12.45

1.216

= (1.826) = 0.0339.

This value is somewhat larger than 2% which was the result in the other

case. Therefore the observation s50 = 1 had less influence on the model thanthe observation s917 = 18, just as predicted.

6.3 Conclusions

Bayesian estimation is a powerful tool for this kind of application since itmakes it possible to improve the model as new data becomes available. Fur-thermore it can be adjusted so that the observations that are most accuratehave a major influence on the resulting model. It would be interesting tohave failure rates at different times as then the time development of the

load-strength model could be studied. Probably this sort of feedback is oneof the things that can improve the model enough to make it useful in indus-trial applications.

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Chapter 7

Conclusions and discussion

The main focus of this report is examining the properties of the load-strengthmodel with respect to the choice of distributions for capacity and duty. Two

distributions for duty and capacity have been examined, the Weibull andthe lognormal distribution. The estimation methods were evaluated basedon the data measurements from Volvo Trucks. Since it was not obvioushow to estimate the parameters if an entity is modelled as three parameterWeibull distributed, different methods were considered and the effectivenesswas determined by extensive simulations. A very simple direct method wasone of the two most effective ones and therefore it is recommendable to usethat one. Since the estimation method is general this method could also beuseful in other applications where the three parameter Weibull distributionis used.

It is also important to study the properties of the distributions in the

context of the load-strength model. The assumption that one over the dutyintensity 1/D is three parameter Weibull distributed leads to some strangeproperties for the distribution of the duty, D, if the shape parameter < 1.By plotting the density function when < 1 one can see that a ratherhigh proportion of the probability mass is close to the upper limit which isobviously unreasonable. Results suggest that a reasonable condition whenusing the three parameter Weibull distribution, for modelling D, is that theshape parameter > 1. Furthermore, the Weibull assumption implies anupper limit for the duty and a lower limit for the capacity which means thata safe distance is established. If one were to drive less than this distance

the failure probability is zero. Then, the failure probability increases ratherrapidly as you drive on past that safe distance. In contrast if both capacityand duty are assumed to be lognormal there will always be an overlap ofthe distributions, also for very short distances, which means that the failureprobability will increase more smoothly as the distance increases.

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The failure probability depends on the upper tail of the duty distributionand the lower tail of the capacity distribution. Many observations are neededin order to determine the tail of a distribution with high accuracy. Therefore,the number of observations must be increased considerably in order to obtainreasonable accuracy in the calculation of the failure probability. The fact thatthe failure probability depends mostly on the tails means that the choice ofdistribution has a great impact on the final result. This is true even if there isno threshold involved. For example, the upper tail of the duty distribution isin general significantly thinner if the lognormal distribution is used comparedto the tail if it is based on a two parameter Weibull distribution.

In the Volvo Articulated Haulers report a model for determining the dis-tribution of the capacity is used in which no rig test is needed. It is ouropinion that it would be very interesting to examine this method more rig-orously. From this model one can easily obtain a large number of capacityvalues and then a distribution fit can be carried out. The lognormal dis-

tribution seems to describe the capacity distribution better than the threeparameter Weibull distribution in this case. We think that the distributionsof the initial crack length a0 and the parameter B could be determined moreprecisely which in turn would improve the model. Further on we think themodel must be compared to rig tests on the same type of components in orderto find out if it works properly. If the model were to give a good descriptionof the capacity it would be very useful since it is cheaper than a rig test andit is easier to apply.

Feedback has been carried out in order to use the failure rate measuredin the field. A natural approach is the Bayesian method. Even though onlya simple example has been examined in this report the result of this exam-

ple shows how powerful this method is. In this example the model adaptsvery closely to the failure rate. This is good since this is a direct observa-tion of what we want to predict. If capacity and duty values are observedthey are only indirect observations of the failure probability. Observationsof the failure rate could be useful in many ways. First, the model couldbe updated with the Bayesian model. Secondly, this information could beused for the improvement of the original load-strength model. Then it can,e.g., be examined if the load-strength model is unbiased with respect to thefailure probability. Examinations of these kind of properties can be the basisfor how to weigh the capacity and duty observations in comparison to the

failure rates. In addition, if the times or distances when components failwere observed it would be very useful in determining the distributions forcapacity and duty. Then time or distance properties of the model could alsobe further examined and the ability to predict the future failure rate couldbe checked.

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It is concluded based on the research presented here that the load-strengthmodel must be used with great care and the user must be aware of the factthat subjectivity in the choice of distribution has a huge impact on the result,especially if few observations are available. Even if many observations areavailable the result can differ significantly due to the fact that the relevanttails have different properties depending on the choice of distribution. Ac-cording to the study in this report the lognormal distribution should be usedrather than the three parameter Weibull distribution for modelling both thecapacity and duty.

7.1 Future research

When using the model it is our opinion that the accuracy in the estimation ofthe distribution parameters should be included which would give a confidenceinterval for the calculated failure probability. Doing this would gain insightinto what extent you can trust the result.

In general the model can be compared to the use of safety factors. Safetyfactors are tools for handling component design. They are not very accu-rate, but they will continue to be used as long as there is no other methodthat works better. Hopefully it will turn out that the load-strength model,if used properly, gives more precise and accurate results. Since statistics areinvolved in the load-strength model it is very important that accuracy in thepredictions are rigorously examined. We think that it is a great challengeto further develop this model. The fact that recent progress in computertechnology makes it possible to collect huge amounts of data from field mea-

surements creates a situation where statistical methods can be increasinglyuseful.

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Appendix A

Estimated parameters with

different methods

The Weibull distributions fits are carried out on the duty observations fromCity and Highway driving and the resulting parameter values are tabulatedfor the different estimation methods.

Estimates based on Maximum Likelihood and directmethod (Mix)

City 0.634 8.81

106 6.57

106

Highway 0.669 2.47 105 1.40 105

Estimates based on percentiles, direct method (Zan)

City 0.51 2.00 105 6.57 106Highway 0.670 2.68 105 1.40 105

Estimates based on iteration (Iter)

City 1.393 1.89 105 1.25 106Highway 1.160 3.77 105 8.62 106

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Estimates determined by VTC, Weibull++

City 0.53 1.06 105 6.41 106Highway 0.85 3.12

106 1.28

105

Estimates based on ML with 2 parameters (ML2)

City 1.514 2.05 105Highway 1.469 4.94 105

Estimates based on elimination (Elim)

City 0.508 4.50

106 6.30

106

Highway 0.793 2.46 105 1.23 105

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Appendix B

Results from Weibull inference

The different estimation methods are compared with respect to the meansquared error (m.s.e.). Two different parameter sets are used

1. = 0.65, = 105, = 6 106

2. = 1.30, = 105, = 6 106

and then the m.s.e. is determined for the estimation of , and , respec-tively. For each combination 10 000 samples have been used in order to givea high accuracy in the calculation of the m.s.e.

m.s.e.() V() (%) b()2 (%) RankZan 0.1315 95.9 4.1 1Mix 0.1773 97.5 2.5 2

Iter 0.3742 64.6 35.4 4ML2 1.343 22.1 77.9 5Elim 0.1852 99.9 0.1 3

Table B.1: Estimation of when = 0.65 and n = 5.

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Appendix C

Proportion p0 for different n

The entity p0 is the proportion of cases when the numerical calculation ofthe failure probability, Pf, gives the result zero even though the actual value

is not zero. This reflects how sensitive the calculation of Pf is when the dis-tribution of the duty has an upper limit and the distribution of the capacityhas a lower limit (threshold).

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0 10 20 30 40 50 60 70 80 90 1000

0.1

0.2

0.3

0.4

0.5

n

p0

Figure C.1: Proportion p0 for different n when = 0.65.

0 10 20 30 40 50 60 70 80 90 1000.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

n

p0

Figure C.2: Proportion p0 for different n when = 1.30.

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Appendix D

Sensitiveness in failure

probability

The failure probability is determined by varying one of the parameters atthe time and keeping the other two fixed. Since the parameter estimates arenot independent this does not give the true variation but nevertheless somequalitative information. The parameters that are varied corresponds to thedistribution of the duty.

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0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.210

5

104

103

Pf

Figure D.1: Failure probability Pf for different , varying around = 0.65Zan: m.s.e.()=0.1315 (n = 5), m.s.e.()=0.0571 (n = 10).

0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.710

6

105

104

Pf

Figure D.2: Failure probability Pf for different , varying around = 1.30Zan: m.s.e.()=0.448 (n = 5), m.s.e.()=0.2424 (n = 10).

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0 0.5 1 1.5 2

x 105

105

104

103

102

Pf

Figure D.3: Failure probability Pf for different when = 0.65Zan: m.s.e.()=2.092 1010 (n = 5), m.s.e.()=6.219 1011 (n = 10).

0 0.5 1 1.5 2

x 105

106

105

104

103

102

Pf


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2 2.5 3 3.5 4 4.5 5 5.5 6 6.5

x 106

109

108

107

106

105

104

103

102

101

100

Pf


2 2.5 3 3.5 4 4.5 5 5.5 6 6.5

x 106

1011

1010

109

108

107

106

105

104

103

102

101

Pf


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Bibliography

[1] Dubey, S.D. (1967). Some percentile estimators for Weibull parameters,Technometrics 9.

[2] Johannesson, P. and de Mare, J. (2002), (2003-12-19). Prob-lemdriven statistikkurs, Utmattning, belastning och tillf orlitlighet,

[3] Johannesson, P. (1999). Rainflow analysis of switching markov loads,Doctoral Thesis in Mathematical Sciences, Lund Institute of Technology,ISBN: 91-628-3784-2.

[4] Lindgren, B.W. (1998). Statistical theory, Fourth edition, Florida, Chap-man & Hall, ISBN: 0-412-04181-2.

[5] Olsson, K.E. (1989). Fatigue reliability prediction, Scandinavian Journalof Metallurgy 18.

[6] Samuelsson, J. (1997). Fatigue design of construction equipment, Volvo

Technology Report.

[7] Thomas, J.J., Perroud, G., Bignonnet, A. and Monnet, D. (1999). Fa-tigue design and reliability in the automotive industry, In Fatigue Designand Reliability, ESIS publication 23, edited by Marquis, G. and Solin,J.

[8] Zanakis, S.H. (1979). A Simulation Study of Some Simple Estimators forthe Three-Parameter Weibull Distribution, J. Statist. Comput. Simul.,Vol. 9.

42154443 Reliability in Fatigue

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