Transcript Cursus Betonvereniging 12 Oktober 2004 Tijdsafhankelijk falen

Cursus Betonvereniging
25 Oktober 2005
Design-by-Testing
Beslistheorie
Tijdsafhankelijk falen
Pieter van Gelder
TU Delft
Sterkte - design by testing
• NEN 6700, par. 7.2 Experimentele modellen
• Rekening houden met:
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•
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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
Voorbeeld
• Nieuw anker voor bevestiging gevelelementen.
Onder horizontale (wind-)belasting
•
Mogelijke bezwijkmechanismen:
• spreidanker in beton bezwijkt
• anker zelf bezwijkt
• ankerdoorn breekt uit
Voorbeeld
•
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Sterkte anker meten in proefopstelling.
Resultaten (in N):
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4897
2922
3700
4856
3221
Wat is de karakteristieke waarde (5%)?
Statistische zekerheid
• Situatie:
• Sterkte R normaal verdeeld
• Veel metingen
• Formule voor sterkte:
R = m R + u SR
u : standaard normaal
verdeelde variabele
mR: steekproefgemiddelde
SR: standaarddeviatie uit
steekproef
Tabel normale verdeling
u
 (-u )
u
 (-u )
u
 (-u )
Statistische onzekerheid
• Situatie:
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•
•
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Sterkte R normaal verdeeld
Weinig metingen (n)
Gemiddelde onbekend
Standaarddeviatie onbekend
• Bayesiaanse statistiek:
R = m R + t n -1 S R 1 +
1
n
n : aantal metingen
tn-1 : standaard student
verdeelde variabele
met n-1 vrijheidsgraden
mR: steekproefgemiddelde
SR: standaarddeviatie uit
steekproef
Student t verdeling
Statistische onzekerheid
•
Situatie:
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Sterkte R normaal verdeeld
Weinig metingen (n)
Gemiddelde onbekend
Standaarddeviatie bekend
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Bayesiaanse statistiek:
1
R = mR + us R 1+
n
n : aantal metingen
u : standaard normaal
verdeelde variabele
mR: steekproefgemiddelde
sR: bekende standaarddeviatie
Voorbeeld
• Gegeven:
• 3 metingen: 88, 95 en 117 kN
• Bekende standaarddeviatie 15 kN
• Vraag:
• Bereken de karakteristieke waarde (5%)
Voorbeeld
• Gegeven:
• 3 metingen: 88, 95 en 117 kN
• Onbekende standaarddeviatie
• Vraag:
• Bereken de karakteristieke waarde (5%)
Voorbeeld
•
•
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Gegeven:
100 metingen
steekproefgemiddelde 100 kN
Onbekende standaarddeviatie, uit steekproef: 15 kN
• Vraag:
• Bereken de karakteristieke waarde (5%)
Voorbeeld
m=100 kN, s = 15 kN
0.035
veel metingen
3 metingen m onbekend, s bekend
3 metingen m onbekend, s onbekend
kansdichtheid (1/kN)
0.03
0.025
0.02
0.015
0.01
0.005
0
0
20
40
60
80
100
120
sterkte (kN)
140
160
180
200
Voorbeeld
m=100 kN, s = 15 kN
1
veel metingen
3 metingen m onbekend, s bekend
3 metingen m onbekend, s onbekend
0.9
0.8
0.7
kans
0.6
0.5
0.4
0.3
0.2
0.1
0
P=0.05
0
20
40
60
80
100
sterkte (kN)
120
140
160
Beslistheorie
Rationeel beslissen: ijscoman
P{zon} = P{regen} = 0.5
regen
€0
zon
€ 1000
regen
€ 2000
zon
€ -500
ijs
patat
Rationeel beslissen: ijscoman
P{zon} = P{regen} = 0.5
regen
€0
ijs
Verwachte opbrengst:
0 * 0.5 + 100 * 0.5 = 500
zon
€ 1000
regen
€ 2000
patat
2000 * 0.5 - 500 * 0.5 = 750
zon
€ -500
Irrationeel beslissen
• Risico-avers voorbeeld uitwerken op bord
Definitie van risico
• Risico = kans x gevolg
Matrix of risks
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Small prob, small event
Small prob, large event
Large prob, small event
Large prob, large event
Evaluating the risk
• Testing the risk to predetermined
standards
• Testing if the risk is in balance with the
investment costs
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.
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.
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.
A decision model
 A: the set of all
possible actions
(a), of which one
must be chosen;
 N: the set of all
(natural)
circumstances (θ);
 Ω: the set of all
possible results
(ω), which are
functions of the
actions and
circumstances:
ω = f(a, θ).
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
• θ1 = company profit # 5 %
• θ2 = 5 % < company profit # 10 %
• θ3 = company profit > 10 %
• ω1 = return (0 % - 2 %) = -2 % per annum
• ω2 = return (3 % - 2 %) = 1 % per annum
• ω3 = return (6 % - 2 %) = 4 % per annum
Decision tree (example 4.1)
Results space
a1
a2
θ1
θ2
θ3
-2 %
1 %
1 %
1 %
4 %
1 %
Utility space
ris k n e u tra l
a1
a2
ris k a v e rs e
ris k s e e k in g
θ1
θ2
θ3
θ1
θ2
θ3
θ1
θ2
0
0.5
0.5
0.5
1
0.5
0
0.75
0.75
0.75
1
0.75
0
0.25
0.25
0.25
Likelihood of the circumstances
T he expected value of the return of action a 1 : “buying shares” am ounts to: 0.2 (2 % ) + 0.3 (1 % ) + 0.5 (4 % ) = 1.9 % .
T his is larger than the 1 % return of action a 2 : “buying bonds”.
From discrete to continuous
decision models
Dijkhoogte bepaling
• Op bord uitwerken
• Tijdsafhankelijke faalkansen
• Door veroudering is onderhoud
noodzakelijk:
– Onderhoudsmodellen
Levensduur T:
is een stochastische variabele
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.
M
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•Haringvliet outlet sluices
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Lifetime distribution
for one component
t
t
start
t
t
Time
Replacement strategies of large numbers of similar components in hydraulic structures
Voorbeeld “leeftijd van mensen”:
stochastische variable Lmens
• Lmens ~ N(78,6) of EXP(76,8)
• P(Lmens >90)=...?
• P(Lmens >90| Lmens >89)= P(Lmens
>90)/P(Lmens >89)=...
• Uitwerken op bord
• Vervolgens: Modelvorming voor algemene
situatie
Verwachte resterende levensduur als
functie van reeds bereikte leeftijd
Hazard rate population in S-Africa:
f(t) / [1 - F(t) ]
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)
Integrating this expression
between 0 and T yields an
expression relating the
Reliability function R(t) and the
Hazard Rate h(t):
Bathtub Curve
Constant Hazard Rate
• The most simple Hazard Rate model is to
assume that: h(t) = λ , 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 (- λt)
• This results in the simple exponential
probability law for the Reliability function.
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) ] = β t β -1 / t β
• For the specific case of: β = 1.0 , the
Hazard Rate h(t) reverts back to the
constant failure rate model described
above, with: t = 1/ λ . The specific
value of the β parameter determines
whether the hazard is increasing or
decreasing.
β values, 0.5, 1.0, and 1.5.
β values, 0.5, 1.0, and 1.5.
Maintenance in Civil Engineering
– Many design and build projects in the past
– Nowadays many maintenance projects
Consequence
of
failure
large
small
Corrective
Maint.
Good
Detoriation
Model?
no
State
dependent
yes
no
Time
dependent
Contains
Effect of
Loading?
Use
dependent
yes
Load
dependent
Hydraulic Engineering
• corrective maintenance
is not advised in view
of the risks involved
• preventive maintenance
• time based
• failure based
• load based
• resistance based
repair
Ro
resistance
Failure
based
failure
repair
Ro
Time
based
resistance
Δt
repair
Ro
Load
based
Ro
Rmin
Resistance
based
load
time
repair
load
time
resistance
load
cum. load
time
load
time
Dike Settlement
S.L.S
h0 – A ln t = h(t)
U.L.S.
h(t) – HW
R,S
h0
hmin
S
topt
time
Condition based maintenance
Inspection
good
Repair
bad
Maintenance
• A case study
• Some concepts
Maintenance strategies
of large numbers
of similar components
in hydraulic structures
Introduction
• Maintenance  replacement
Introduction
– Maintenance  replacement
• Large numbers of similar
components
Introduction
– Maintenance  replacement
• Large numbers of similar
components
Introduction
– Maintenance  replacement
• Large numbers of similar
components
• Same lifetime-distribution
• Same age
• Same function
M
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Modelling
Case study
Conclusions
• Variables of a
replacement scenario
– Start date of the
(start)
replacements
– Replacement interval
(t)
– Number of preventive
( )
replacements
t
t
start
t
t
Time
M
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Modelling
Case study
Conclusions
• Finding the optimal
strategy
– Balance between risk costs
and costs of preventive
replacements
– Replacement capacity
– Capacity of the supplier
C
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s
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Modelling
different scenarios
4 prev. repl.
8 prev. repl.
Start date of scenario
(years)
2042
2040
2038
2036
2034
2032
2030
2028
2026
2024
2022
2020
2018
2016
0,20
0,18
0,16
0,14
0,12
0,10
0,08
0,06
0,04
0,02
0,00
2014
Conclusions
• Probability of failure for
Probability of system failure
(entire scenario)
Case study
The Concept of Availability
Reliability
Maintainability
Availability
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
Availability
Availability is the proportion of the
process or system “Up-Time” to the total
time (Up + Down) over a long period.
Availability =
Up-Time
Up-Time + Down-Time
System Operational States
B1
Up
Down
A1
B2
A2
Up: System up and running
Down: System under repair
B3
A3
t
Mean Time To Fail (MTTF)
MTTF is defined as the mean time of the occurrence of the
first failure after entering service.
MTTF =
B1
Up
Down
B1 + B2 + B3
3
A1
B2
A2
B3
A3
t
Mean Time Between Failure
(MTBF)
MTBF is defined as the mean time between successive
failures.
MTBF =
B1
Up
Down
(A1 + B1) + (A2 + B2) + (A3 + B3)
3
A1
B2
A2
B3
A3
t
Mean Time To Repair (MTTR)
MTTR is defined as the mean time of restoring a process or
system to operation condition.
MTTR =
B1
Up
Down
A1 + A2 + A3
3
A1
B2
A2
B3
A3
t
Availability
Availability is defined as:
A=
Up-Time
Up-Time + Down-Time
Availability is normally expressed in terms of MTBF and
MTTR as:
A=
MTBF
MTBF + MTTR
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
Types of Redundancy
• Active Redundancy
• Standby Redundancy
Active Redundancy
A
Input
Output
Div
B
Divider
Both A and B subsystems are operative at all times
Note: the dividing device is a Series Element
Standby Redundancy
A
Input
Output
SW
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.
Series System
Input
A1
A2
An
Output
ps = p1 + p2 +……. + pn - (-1)n joint probabilities
For identical and independent elements:
ps ~ 1 - (1-p)n < np (>p)
ps :
pi :
Probability of system failure
Probability of component failure
Parallel System
A
Input
Output
B
Multiplicative Rule
ps = p1.p2 … pn
ps :
Probability of system failure
Series / Parallel System
A1
Input
A2
Output
C
B1
B2
System with Repairs
Let MTBF = q
and system MTBF = qs
A
Input
Output
B
For Active Redundancy (Parallel or duplicated system)
qs = ( 3l + m )/ ( 2l2 )
qs = m / 2l2 = MTBF2 / 2 MTTR
l << m
A
Input
Output
SW
B
Switch
Note: The switch is a
series element, neglect
for now.
Standby
Note: The standby
system is normally
inactive.
For Standby Redundancy
qs = ( 2l + m )/ (l2 )
qs = m / l2
= MTBF2 / MTTR
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
Summary
Type
With Repairs
Without Repairs
Active
MTBF2 / 2 MTTR
1.5 MTBF
Standby
MTBF2 / MTTR
2 MTBF
Redundancy techniques are used to increase the system MTBF