Cursus Betonvereniging 25 Oktober 2005 DesignbyTesting Beslistheorie Tijdsafhankelijk falen

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Cursus Betonvereniging25 Oktober
2005Design-by-TestingBeslistheorieTijdsafhanke
lijk falen

• Pieter van Gelder
• TU Delft

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Sterkte - design by testing

• NEN 6700, par. 7.2 Experimentele modellen
• Rekening houden met
• Vereenvoudigingen experimenteel model
• Onzekerheden m.b.t. lange-duur effecten
• Representatieve steekproeven
• Statistische onzekerheden
• Wijze van bezwijken (bros/taai)
• Eisen m.b.t. detaillering
• Bezwijkmechanismen

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Voorbeeld

• Nieuw anker voor bevestiging gevelelementen.

• Onder horizontale (wind-)belasting
• Mogelijke bezwijkmechanismen
• spreidanker in beton bezwijkt
• anker zelf bezwijkt
• ankerdoorn breekt uit

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Voorbeeld

• Sterkte anker meten in proefopstelling.
• Resultaten (in N)
• 4897
• 2922
• 3700
• 4856
• 3221
• Wat is de karakteristieke waarde (5)?

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Statistische zekerheid

• Situatie
• Sterkte R normaal verdeeld
• Veel metingen
• Formule voor sterkte

u standaard normaal verdeelde
variabele mR steekproefgemiddelde SR
standaarddeviatie uit steekproef


S
u
m
R
R
R
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Tabel normale verdeling
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Statistische onzekerheid

• Situatie
• Sterkte R normaal verdeeld
• Weinig metingen (n)
• Gemiddelde onbekend
• Standaarddeviatie onbekend
• Bayesiaanse statistiek

n aantal metingen tn-1 standaard student
verdeelde variabele met n-1
vrijheidsgraden mR steekproefgemiddelde SR
standaarddeviatie uit steekproef
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1
S
t
m
R
-
R
1
n
R
n
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Student t verdeling
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Statistische onzekerheid

• Situatie
• Sterkte R normaal verdeeld
• Weinig metingen (n)
• Gemiddelde onbekend
• Standaarddeviatie bekend
• Bayesiaanse statistiek

n aantal metingen u standaard normaal
verdeelde variabele mR steekproefgemiddelde sR
bekende standaarddeviatie
1



s
1
u
m
R
R
R
n
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Voorbeeld

• Gegeven
• 3 metingen 88, 95 en 117 kN
• Bekende standaarddeviatie 15 kN
• Vraag
• Bereken de karakteristieke waarde (5)

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Voorbeeld

• Gegeven
• 3 metingen 88, 95 en 117 kN
• Onbekende standaarddeviatie
• Vraag
• Bereken de karakteristieke waarde (5)

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Voorbeeld

• Gegeven
• 100 metingen
• steekproefgemiddelde 100 kN
• Onbekende standaarddeviatie, uit steekproef 15
  kN
• Vraag
• Bereken de karakteristieke waarde (5)

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Voorbeeld
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Voorbeeld
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Beslistheorie
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Rationeel beslissen ijscoman
Pzon Pregen 0.5
regen
ijs
zon
1000
regen
2000
patat
zon
-500
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Rationeel beslissen ijscoman
Pzon Pregen 0.5
Verwachte opbrengst
regen
ijs
0 0.5 100 0.5 500
zon
1000
regen
2000
patat
2000 0.5 - 500 0.5 750
zon
-500
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Irrationeel beslissen

• Risico-avers voorbeeld uitwerken op bord

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Definitie van risico

• Risico kans x gevolg

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Matrix of risks

• Small prob, small event
• Small prob, large event
• Large prob, small event
• Large prob, large event

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(No Transcript)
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Evaluating the risk

• Testing the risk to predetermined standards
• Testing if the risk is in balance with the
  investment costs

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Decision-making based on risk analysis

• Recording different variants, with associated
  risks, costs and benefits, in a matrix or
  decision tree, serves as an aid for making
  decisions. With this, the optimal selection can
  be made from a number of alternatives.

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Deciding under uncertainties

• Modern decision theory is based on the classic
  Homo Economicus model
• has complete information about the decision
  situation
• knows all the alternatives
• knows the existing situation
• knows which advantages and disadvantages each
  alternative provides, be it in the form of random
  variables
• strives to maximise that advantage.

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But in reality

• The decision maker
• does not know all the alternatives
• does not know all the effects of the
  alternatives
• does not know which effect each alternative has.

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A decision model

• A the set of all possible actions (a), of which
  one must be chosen
• N the set of all (natural) circumstances (?)
• O the set of all possible results (?), which are
  functions of the actions and circumstances
  ?  f(a, ?).

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Example 4.1

• Suppose a person has EUR 1000 at his disposal and
  is given the choice to invest this money in bonds
  or in shares of a given company.
• The decision model consists of
• a1 investing in shares
• a2 investing in bonds
• ?1 company profit  5 
• ?2  5  lt  company profit  10 
• ?3  company profit gt 10 
• ?1  return (0  - 2 )  -2  per annum
• ?2  return (3  - 2 )   1  per annum
• ?3  return (6  - 2 )   4  per annum

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Decision tree (example 4.1)
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Utility space
Results space
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Likelihood of the circumstances
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From discrete to continuous decision models
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Dijkhoogte bepaling

• Op bord uitwerken

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• Tijdsafhankelijke faalkansen
• Door veroudering is onderhoud noodzakelijk
• Onderhoudsmodellen

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Levensduur Tis een stochastische variabele
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J.K. Vrijling and P.H.A.J.M. van Gelder, The
effect of inherent uncertainty in time and space
on the reliability of flood protection, ESREL'98
European Safety and Reliability Conference 1998,
pp.451-456, 16 - 19 June 1998, Trondheim, Norway.
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• Haringvliet outlet sluices

Modellering

Lifetime distribution for one
component
Replacement strategies of large numbers of
similar components in hydraulic structures
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Voorbeeld leeftijd van mensen stochastische
variable Lmens

• Lmens N(78,6) of EXP(76,8)
• P(Lmens gt90)...?
• P(Lmens gt90 Lmens gt89) P(Lmens gt90)/P(Lmens
  gt89)...
• Uitwerken op bord
• Vervolgens Modelvorming voor algemene situatie

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Verwachte resterende levensduur als functie van
reeds bereikte leeftijd
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Hazard rate population in S-Africa f(t) / 1 -
F(t)
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T time to failure

• The Hazard Rate, or instantaneous failure rate is
  defined as
• h(t) f(t) / 1 - F(t) f(t) / R(t)
• f(t) probability density function of time to
  failure,
• F(t) is the Cumulative Distribution Function
  (CDF) of time to failure,
• R(t) is the Reliability function (CCDF of time to
  failure).
• From     f(t) d F(t)/dt , it follows that
• h(t) dt d F(t) / 1 - F(t) - d R(t) / R(t)
  - d ln R(t)

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Integrating this expression between 0 and T
yields an expression relating the Reliability
function R(t) and the Hazard Rate h(t)
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Bathtub Curve
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Constant Hazard Rate

• The most simple Hazard Rate model is to assume
  that h(t) ? , a constant. This implies that
  the Hazard or failure rate is not significantly
  increasing with component age. Such a model is
  perfectly suitable for modeling component hazard
  during its useful lifetime.
• Substituting the assumption of constant failure
  rate into the expression for the Reliability
  yields
• R(t) 1 - F(t) exp (- ?t)
• This results in the simple exponential
  probability law for the Reliability function.

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Non-Constant Hazard Rate

• One of the more common non-constant Hazard Rate
  models used for evaluation of component aging
  phenomenon, is to assume a Weibull distribution
  for the time to failure
• Using the definition of the Hazard function and
  substituting in appropriate Weibull distribution
  terms yields
• h(t) f(t) / 1 - F(t) ß t ß -1 / t ß

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• For the specific case of  ß 1.0 , the Hazard
  Rate h(t) reverts back to the constant failure
  rate model described above, with  t 1/  ? .
  The specific value of the ß parameter determines
  whether the hazard is increasing or decreasing.

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ß values, 0.5, 1.0, and 1.5.
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ß values, 0.5, 1.0, and 1.5.
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Maintenance in Civil Engineering

• Many design and build projects in the past
• Nowadays many maintenance projects

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large
no
small
yes
yes
no
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Hydraulic Engineering

• corrective maintenance
• is not advised in view
• of the risks involved
• preventive maintenance

• time based
• failure based
• load based
• resistance based

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repair
Ro
resistance load
Failure based
failure
time
repair
Ro
resistance load
Time based
?t
time
repair
Ro
resistance load cum. load
Load based
time
repair
Ro
load
Rmin
Resistance based
time
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Dike Settlement
S.L.S h0 A ln t h(t) U.L.S. h(t) HW
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Condition based maintenance
Inspection
good
Repair
bad
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Maintenance

• A case study
• Some concepts

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Maintenance strategies
of large numbers
of similar components
in hydraulic structures
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Introduction

• Maintenance ? replacement

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Introduction

• Maintenance ? replacement
• Large numbers of similar components

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Introduction

• Maintenance ? replacement
• Large numbers of similar components

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Introduction

• Maintenance ? replacement
• Large numbers of similar components
• Same lifetime-distribution
• Same age
• Same function

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Modellering

• Modelling
• Case study
• Conclusions

• Variables of a replacement scenario
• Start date of the (start) replacements
• Replacement interval (?t)
• Number of preventive ( ) replacements

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Modellering

• Modelling
• Case study
• Conclusions

• Finding the optimal strategy
• Balance between risk costs and costs of
  preventive replacements
• Replacement capacity
• Capacity of the supplier

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Casestudie

• Modelling
• Case study
• Conclusions

• Probability of failure for different scenarios

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The Concept of Availability
Reliability
Maintainability
Availability
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Maintainability
Maintainability is the probability that a process
or a system that has failed will be restored to
operation effectiveness within a given time.
M(t) 1 - e-mt
where m is repair (restoration) rate
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Availability
Availability is the proportion of the process or
system Up-Time to the total time (Up Down)
over a long period.
Up-Time Up-Time Down-Time
Availability
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System Operational States
B1
B2
B3
Up
t
Down
A1
A3
A2
Up System up and running Down System under
repair
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Mean Time To Fail (MTTF)
MTTF is defined as the mean time of the
occurrence of the first failure after entering
service.
B1 B2 B3 3
MTTF
B1
B2
B3
Up
t
Down
A1
A3
A2
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Mean Time Between Failure (MTBF)
MTBF is defined as the mean time between
successive failures.
(A1 B1) (A2 B2) (A3 B3) 3
MTBF
B1
B2
B3
Up
t
Down
A1
A3
A2
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Mean Time To Repair (MTTR)
MTTR is defined as the mean time of restoring a
process or system to operation condition.
A1 A2 A3 3
MTTR
B1
B2
B3
Up
t
Down
A1
A3
A2
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Availability
Availability is defined as
Up-Time Up-Time Down-Time
A
Availability is normally expressed in terms of
MTBF and MTTR as
MTBF MTBF MTTR
A
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Reliability/Maintainability Measures
Reliability R(t)
(Failure Rate) l 1 / MTBF R(t) e-lt
Maintainability M(t)
(Maintenance Rate) m 1 / MTTR M(t) 1 - e-mt
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Types of Redundancy

• Active Redundancy
• Standby Redundancy

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Active Redundancy
A
Output
Input
Div
B
Divider
Both A and B subsystems are operative at all times
Note the dividing device is a Series Element
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Standby Redundancy
A
SW
Output
Input
B
Switch
Standby
The standby unit is not operative until a
failure-sensing device senses a failure in
subsystem A and switches operation to subsystem
B, either automatically or through manual
selection.
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Series System
A1
A2
An
Input
Output
ps p1 p2 . pn - (-1)n joint probabilities
For identical and independent elements
ps 1 - (1-p)n lt np (gtp)
ps Probability of system failure pi
Probability of component failure
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Parallel System
A
Output
Input
B
Multiplicative Rule
ps p1.p2 pn
ps Probability of system failure
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Series / Parallel System
A1
A2
Input
Output
C
B1
B2
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System with Repairs
Let MTBF q and system MTBF qs
A
Output
Input
B
For Active Redundancy (Parallel or duplicated
system)
qs ( 3l m )/ ( 2l2 )
l ltlt m
qs m / 2l2 MTBF2 / 2 MTTR
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A
SW
Output
Input
B
Switch
Standby
Note The switch is a series element, neglect for
now.
Note The standby system is normally inactive.
For Standby Redundancy
qs ( 2l m )/ (l2 )
qs m / l2 MTBF2 / MTTR
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System without Repairs
For systems without repairs, m 0
For Active Redundancy
qs ( 3l m )/ ( 2l2 )
qs 3l / ( 2l2 ) 3 / ( 2l )
qs (3/2) q where q 1/l qs 1.5 MTBF
For Standby Redundancy
qs ( 2l m )/ (l2 )
qs 2l/ l2 2/ l
qs 2q where q 1/l qs 2 MTBF
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Summary
With Repairs
Without Repairs
Type
MTBF2 / 2 MTTR
1.5 MTBF
Active
MTBF2 / MTTR
2 MTBF
Standby
Redundancy techniques are used to increase the
system MTBF