Title: Cursus Betonvereniging 25 Oktober 2005 DesignbyTesting Beslistheorie Tijdsafhankelijk falen
1Cursus Betonvereniging25 Oktober
2005Design-by-TestingBeslistheorieTijdsafhanke
lijk falen
- Pieter van Gelder
- TU Delft
2Sterkte - 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
3Voorbeeld
- Nieuw anker voor bevestiging gevelelementen.
-
- Onder horizontale (wind-)belasting
- Mogelijke bezwijkmechanismen
- spreidanker in beton bezwijkt
- anker zelf bezwijkt
- ankerdoorn breekt uit
4Voorbeeld
- Sterkte anker meten in proefopstelling.
- Resultaten (in N)
- 4897
- 2922
- 3700
- 4856
- 3221
- Wat is de karakteristieke waarde (5)?
5Statistische 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
6Tabel normale verdeling
7Statistische 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
1
1
S
t
m
R
-
R
1
n
R
n
8Student t verdeling
9Statistische 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
10Voorbeeld
- Gegeven
- 3 metingen 88, 95 en 117 kN
- Bekende standaarddeviatie 15 kN
- Vraag
- Bereken de karakteristieke waarde (5)
11Voorbeeld
- Gegeven
- 3 metingen 88, 95 en 117 kN
- Onbekende standaarddeviatie
- Vraag
- Bereken de karakteristieke waarde (5)
12Voorbeeld
- Gegeven
- 100 metingen
- steekproefgemiddelde 100 kN
- Onbekende standaarddeviatie, uit steekproef 15
kN - Vraag
- Bereken de karakteristieke waarde (5)
13Voorbeeld
14Voorbeeld
15Beslistheorie
16Rationeel beslissen ijscoman
Pzon Pregen 0.5
regen
ijs
zon
1000
regen
2000
patat
zon
-500
17Rationeel 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
18Irrationeel beslissen
- Risico-avers voorbeeld uitwerken op bord
19Definitie van risico
20Matrix of risks
- Small prob, small event
- Small prob, large event
- Large prob, small event
- Large prob, large event
21(No Transcript)
22Evaluating the risk
- Testing the risk to predetermined standards
- Testing if the risk is in balance with the
investment costs
23Decision-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.
24Deciding 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.
25But 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.
26A 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, ?).
27Example 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
28Decision tree (example 4.1)
29Utility space
Results space
30Likelihood of the circumstances
31From discrete to continuous decision models
32Dijkhoogte bepaling
33- Tijdsafhankelijke faalkansen
- Door veroudering is onderhoud noodzakelijk
- Onderhoudsmodellen
34Levensduur Tis een stochastische variabele
35J.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.
36- Haringvliet outlet sluices
Modellering
Lifetime distribution for one
component
Replacement strategies of large numbers of
similar components in hydraulic structures
37Voorbeeld 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
38Verwachte resterende levensduur als functie van
reeds bereikte leeftijd
39Hazard rate population in S-Africa f(t) / 1 -
F(t)
40T 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)
41Integrating this expression between 0 and T
yields an expression relating the Reliability
function R(t) and the Hazard Rate h(t)
42Bathtub Curve
43Constant 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.
44Non-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
45- 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.
46ß values, 0.5, 1.0, and 1.5.
47ß values, 0.5, 1.0, and 1.5.
48Maintenance in Civil Engineering
- Many design and build projects in the past
- Nowadays many maintenance projects
49large
no
small
yes
yes
no
50Hydraulic Engineering
- corrective maintenance
- is not advised in view
- of the risks involved
- preventive maintenance
- time based
- failure based
- load based
- resistance based
51repair
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
52Dike Settlement
S.L.S h0 A ln t h(t) U.L.S. h(t) HW
53Condition based maintenance
Inspection
good
Repair
bad
54Maintenance
- A case study
- Some concepts
55Maintenance strategies
of large numbers
of similar components
in hydraulic structures
56Introduction
- Maintenance ? replacement
57Introduction
- Maintenance ? replacement
- Large numbers of similar components
58Introduction
- Maintenance ? replacement
- Large numbers of similar components
59Introduction
- Maintenance ? replacement
- Large numbers of similar components
- Same lifetime-distribution
- Same age
- Same function
60Modellering
- Modelling
- Case study
- Conclusions
- Variables of a replacement scenario
- Start date of the (start) replacements
- Replacement interval (?t)
- Number of preventive ( ) replacements
61Modellering
- Modelling
- Case study
- Conclusions
- Finding the optimal strategy
- Balance between risk costs and costs of
preventive replacements - Replacement capacity
- Capacity of the supplier
62Casestudie
- Modelling
- Case study
- Conclusions
- Probability of failure for different scenarios
63The Concept of Availability
Reliability
Maintainability
Availability
64Maintainability
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
65Availability
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
66System Operational States
B1
B2
B3
Up
t
Down
A1
A3
A2
Up System up and running Down System under
repair
67Mean 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
68Mean 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
69Mean 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
70Availability
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
71Reliability/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
72Types of Redundancy
- Active Redundancy
- Standby Redundancy
73Active 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
74Standby 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.
75Series 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
76Parallel System
A
Output
Input
B
Multiplicative Rule
ps p1.p2 pn
ps Probability of system failure
77Series / Parallel System
A1
A2
Input
Output
C
B1
B2
78System 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
79A
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
80System 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
81Summary
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