Cursus Betonvereniging25 Oktober 2005

Design-by-TestingBeslistheorie

Tijdsafhankelijk falen

Pieter van Gelder

TU Delft

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

Voorbeeld• Nieuw anker voor bevestiging gevelelementen.

• Onder horizontale (wind-)belasting

Mogelijke bezwijkmechanismen:• spreidanker in beton bezwijkt• anker zelf bezwijkt• ankerdoorn breekt uit

Voorbeeld

• Sterkte anker meten in proefopstelling.• Resultaten (in N):• 4897• 2922• 3700• 4856• 3221

Wat is de karakteristieke waarde (5%)?

Statistische zekerheid

• Situatie:• Sterkte R normaal verdeeld

• Veel metingen

• Formule voor sterkte: u : standaard normaal verdeelde variabele

mR: steekproefgemiddelde

SR: standaarddeviatie uit steekproef

RR SumR

Tabel normale verdeling

Statistische onzekerheid• Situatie:• Sterkte R normaal verdeeld• Weinig metingen (n)• Gemiddelde onbekend• Standaarddeviatie onbekend

• Bayesiaanse statistiek:

n

11StmR R1nR

n : aantal metingentn-1 : standaard student verdeelde variabele met n-1 vrijheidsgraden

mR: steekproefgemiddelde

SR: standaarddeviatie uit steekproef

Student t verdeling

Statistische onzekerheid• Situatie:• Sterkte R normaal verdeeld• Weinig metingen (n)• Gemiddelde onbekend• Standaarddeviatie bekend

• Bayesiaanse statistiek:•

n : aantal metingenu : standaard normaal verdeelde variabele

mR: steekproefgemiddelde

R: bekende standaarddeviatie

n

11umR RR

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

• Gegeven:

• 100 metingen

• steekproefgemiddelde 100 kN

• Onbekende standaarddeviatie, uit steekproef: 15 kN

• Vraag:

• Bereken de karakteristieke waarde (5%)

Voorbeeld

0 20 40 60 80 100 120 140 160 180 2000

0.005

0.01

0.015

0.02

0.025

0.03

0.035

sterkte (kN)

kan

sdic

hth

eid

(1

/kN

)m=100 kN, s = 15 kN

veel metingen3 metingen m onbekend, s bekend3 metingen m onbekend, s onbekend

Voorbeeld

0 20 40 60 80 100 120 140 1600

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

sterkte (kN)

kan

sm=100 kN, s = 15 kN

veel metingen3 metingen m onbekend, s bekend3 metingen m onbekend, s onbekend

P=0.05

Beslistheorie

Rationeel beslissen: ijscoman

ijs

regen

zon

patat

regen

zon

€ 0

€ 1000

€ 2000

€ -500

P{zon} = P{regen} = 0.5

Rationeel beslissen: ijscoman

ijs

regen

zon

patat

regen

zon

€ 0

€ 1000

€ 2000

€ -500

P{zon} = P{regen} = 0.5

0 * 0.5 + 100 * 0.5 = 500

2000 * 0.5 - 500 * 0.5 = 750

Verwachte opbrengst:

Irrationeel beslissen

• Risico-avers voorbeeld uitwerken op bord

Definitie van risico

• Risico = kans x gevolg

Matrix of risks

• 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)

Utility space

θ1 θ2 θ3

a1 -2 % 1 % 4 %a2 1 % 1 % 1 %

risk neutral risk averse risk seekingθ1 θ2 θ3 θ1 θ2 θ3 θ1 θ2

a1 0 0.5 1 0 0.75 1 0 0.25a2 0.5 0.5 0.5 0.75 0.75 0.75 0.25 0.25

Results space

Likelihood of the circumstances

The expected value of the return of action a1: “buying shares” amounts to: 0.2 (-2 %) + 0.3 (1 %) + 0.5 (4 %) = 1.9 %.This is larger than the 1 % return of action a2: “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.

•Haringvliet outlet sluices

Lifetime distribution for one component

•

Modellering Modellering

Timestartt t t t

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 / 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:   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

Good Detoriation

Model?

Statedependent

ContainsEffect of Loading?

Consequenceof

failure

CorrectiveMaint.

Usedependent

Timedependent

Loaddependent

yes

yessmall

large no

no

Hydraulic Engineering

• corrective maintenanceis not advised in view of the risks involved

• preventive maintenance

• time based

• failure based

• load based

• resistance based

resistance

loadtimefailure

Ro

resistance

loadtime

Ro

resistance

load

cum. loadtime

Ro

load

time

repairRo

Resistance

based

Load

based

Time

based

Failure

based

Rmin

repair

repair

repair

Δt

Dike Settlement

S.L.S h0 – A ln t = h(t)

U.L.S. h(t) – HW

timetopt

R,Sh0

hmin

S

Condition based maintenance

bad

good

Inspection

Repair

Maintenance

• A case study

• Some concepts

Maintenance strategies Maintenance strategies

of large numbersof large numbers

of similar componentsof similar components

in hydraulic structuresin hydraulic structures

IntroductionIntroduction

• Maintenance replacement

IntroductionIntroduction

– Maintenance replacement

• Large numbers of similar components

IntroductionIntroduction

– Maintenance replacement

• Large numbers of similar components

IntroductionIntroduction

– Maintenance replacement

• Large numbers of similar components

• Same lifetime-distribution

• Same age

• Same function

ModellingModellingCase study

Conclusions

• Variables of a replacement scenario

– Start date of the (start) replacements

– Replacement interval (t)

– Number of preventive ( ) replacements

Modellering Modellering

Timestartt t t t

ModellingModellingCase study

Conclusions

• Finding the optimal strategy

– Balance between risk costs and costs of preventive replacements

– Replacement capacity

– Capacity of the supplier

Modellering Modellering

Modelling

Case studyCase studyConclusions

• Probability of failure for different scenarios

Casestudie Casestudie

0,00

0,02

0,040,06

0,08

0,10

0,12

0,140,16

0,18

0,20

2014

2016

2018

2020

2022

2024

2026

2028

2030

2032

2034

2036

2038

2040

2042

Start date of scenario (years)

Pro

babi

lity

of s

yste

m fa

ilure

(ent

ire s

cena

rio)

4 prev. repl.

8 prev. repl.

ReliabilityReliability MaintainabilityMaintainability

AvailabilityAvailability

The Concept of Availability

Maintainability

Maintainability is the probability that a Maintainability is the probability that a process or a system that has failed will process or a system that has failed will be restored to operation effectiveness be restored to operation effectiveness

within a given time.within a given time.

M(M(tt) = 1 - ) = 1 - ee--tt

where where is repair (restoration) rate is repair (restoration) rate

Availability

Availability is the proportion of the Availability is the proportion of the process or system “Up-Time” to the process or system “Up-Time” to the total time (Up + Down) over a long total time (Up + Down) over a long

period.period.

Availability =Availability =Up-TimeUp-Time

Up-Time + Down-TimeUp-Time + Down-Time

UpUp

DownDownA1A1 A3A3A2A2

tt

B1B1 B2B2 B3B3

Up:Up: System up and runningSystem up and runningDown:Down: System under repairSystem under repair

System Operational States

MTTF is defined as the mean time of the occurrence of the MTTF is defined as the mean time of the occurrence of the first failure after entering service.first failure after entering service.

MTTF = MTTF = B1 + B2 + B3B1 + B2 + B333

UpUp

DownDownA1A1 A3A3A2A2

tt

B1B1 B2B2 B3B3

Mean Time To Fail (MTTF)

MTBF is defined as the mean time between successive MTBF is defined as the mean time between successive failures.failures.

MTBF =MTBF = (A1 + B1) + (A2 + B2) + (A3 + B3)(A1 + B1) + (A2 + B2) + (A3 + B3)33

UpUp

DownDownA1A1 A3A3A2A2

tt

B1B1 B2B2 B3B3

Mean Time Between Failure (MTBF)

Mean Time To Repair (MTTR)

MTTR =MTTR = A1 + A2 + A3A1 + A2 + A333

MTTR is defined as the mean time of restoring a process or MTTR is defined as the mean time of restoring a process or system to operation condition.system to operation condition.

UpUp

DownDownA1A1 A3A3A2A2

tt

B1B1 B2B2 B3B3

Availability

Availability is defined as:Availability is defined as:

A =A = Up-TimeUp-TimeUp-Time + Down-TimeUp-Time + Down-Time

Availability is normally expressed in terms of MTBF and Availability is normally expressed in terms of MTBF and MTTR as:MTTR as:

A =A = MTBFMTBFMTBF + MTTRMTBF + MTTR

Reliability/Maintainability Measures

(Failure Rate)(Failure Rate) = 1 / MTBF = 1 / MTBF

R(t) = R(t) = ee--tt

Reliability R(t)Reliability R(t)

Maintainability M(t)Maintainability M(t)

(Maintenance Rate)(Maintenance Rate) = 1 / MTTR = 1 / MTTR

M(t) = 1 - M(t) = 1 - ee--tt

Types of Redundancy

• Active Redundancy

• Standby Redundancy

Active Redundancy

A

B

DividerDivider

OutputOutputInputInput

Both A and B subsystems are operative at all timesBoth A and B subsystems are operative at all times

Div

Note: the dividing device is a Series ElementNote: the dividing device is a Series Element

Standby Redundancy

A

B

SW

SwitchSwitch

OutputOutputInputInput

StandbyStandby

The standby unit is not operative until a failure-sensing device The standby unit is not operative until a failure-sensing device senses a failure in subsystem A and switches operation to senses a failure in subsystem A and switches operation to subsystem B, either automatically or through manual selection.subsystem B, either automatically or through manual selection.

Series System

A1 A2 An

ppss = p = p11 + p + p22 +……. + p +……. + pn n - (-1)- (-1)nn joint probabilities joint probabilities

ppss : : Probability of system failureProbability of system failure

ppii : : Probability of component failureProbability of component failure

For identical and independent elements:For identical and independent elements:

ppss ~ 1 - (1-p) ~ 1 - (1-p)nn < np (>p) < np (>p)

InputInput OutputOutput

Parallel System

B

A

ppss = p = p11.p.p22 … … ppnn

Multiplicative RuleMultiplicative Rule

OutputOutputInputInput

ppss : : Probability of system failureProbability of system failure

Series / Parallel System

B1

A1

B2

A2

CInputInput OutputOutput

System with RepairsLet MTBF = Let MTBF = and system MTBF = and system MTBF = ss

s = ( 3s = ( 3 + + )/ ( 2)/ ( 22 2 ))

For Active Redundancy (Parallel or duplicated system)For Active Redundancy (Parallel or duplicated system)

s = s = / 2/ 222 = MTBF = MTBF22 / 2 MTTR / 2 MTTR

B

AOutputOutputInputInput

<< <<

For Standby RedundancyFor Standby Redundancy

s = ( 2s = ( 2 + + )/ ()/ (2 2 ))

s = s = / / 22 = MTBF= MTBF22 / MTTR / MTTR

A

B

SW

SwitchSwitch

OutputOutputInputInput

StandbyStandbyNote: The standby Note: The standby system is normally system is normally inactive.inactive.

Note: The switch is a Note: The switch is a series element, neglect series element, neglect for now.for now.

System without Repairs

s = ( 3s = ( 3 + + )/ ( 2)/ ( 22 2 ))

For Active RedundancyFor Active Redundancy

s = (3/2) s = (3/2) where where = 1/ = 1/s s MTBFMTBF

For Standby RedundancyFor Standby Redundancy

s = ( 2s = ( 2 + + )/ ()/ (2 2 ))

s = 2s = 2where where = 1/ = 1/s s MTBFMTBF

For systems without repairs, For systems without repairs, = 0 = 0

s = 3s = 3/ ( 2/ ( 22 2 ) = 3) = 3/ ( 2/ ( 2 ) )

s = 2s = 2/ / 2 2 = 2/ = 2/

MTBFMTBF22 / 2 MTTR / 2 MTTR

MTBFMTBF22 / MTTR / MTTR

MTBFMTBF

MTBFMTBF

ActiveActive

StandbyStandby

With RepairsWith Repairs Without RepairsWithout RepairsTypeType

Summary

Redundancy techniques are used to increase the system MTBF