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Cursus Betonvereniging 25 Oktober 2005 Design-by-Testing Beslistheorie Tijdsafhankelijk falen Pieter van Gelder TU Delft.

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Presentatie over: "Cursus Betonvereniging 25 Oktober 2005 Design-by-Testing Beslistheorie Tijdsafhankelijk falen Pieter van Gelder TU Delft."— Transcript van de presentatie:

1 Cursus Betonvereniging 25 Oktober 2005 Design-by-Testing Beslistheorie Tijdsafhankelijk falen Pieter van Gelder TU Delft

2 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

3 Voorbeeld Nieuw anker voor bevestiging gevelelementen. Onder horizontale (wind-)belasting Mogelijke bezwijkmechanismen: spreidanker in beton bezwijkt anker zelf bezwijkt ankerdoorn breekt uit

4 Voorbeeld Sterkte anker meten in proefopstelling. Resultaten (in N): Wat is de karakteristieke waarde (5%)?

5 Statistische zekerheid Situatie: Sterkte R normaal verdeeld Veel metingen Formule voor sterkte: u : standaard normaal verdeelde variabele m R : steekproefgemiddelde S R : standaarddeviatie uit steekproef RR SumR 

6 Tabel normale verdeling

7 Statistische onzekerheid Situatie: Sterkte R normaal verdeeld Weinig metingen (n) Gemiddelde onbekend Standaarddeviatie onbekend Bayesiaanse statistiek: n 1 1StmR R1nR   n : aantal metingen t n-1 : standaard student verdeelde variabele met n-1 vrijheidsgraden m R : steekproefgemiddelde S R : standaarddeviatie uit steekproef

8 Student t verdeling

9 Statistische onzekerheid Situatie: Sterkte R normaal verdeeld Weinig metingen (n) Gemiddelde onbekend Standaarddeviatie bekend Bayesiaanse statistiek: n : aantal metingen u : standaard normaal verdeelde variabele m R : steekproefgemiddelde  R : bekende standaarddeviatie n 1 1umR RR 

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

11 Voorbeeld Gegeven: 3 metingen: 88, 95 en 117 kN Onbekende standaarddeviatie Vraag: Bereken de karakteristieke waarde (5%)

12 Voorbeeld Gegeven: 100 metingen steekproefgemiddelde 100 kN Onbekende standaarddeviatie, uit steekproef: 15 kN Vraag: Bereken de karakteristieke waarde (5%)

13 Voorbeeld

14

15 Beslistheorie

16 Rationeel beslissen: ijscoman ijs regen zon patat regen zon € 0€ 0 € 1000 € 2000 € -500 P{zon} = P{regen} = 0.5

17 Rationeel beslissen: ijscoman ijs regen zon patat regen zon € 0€ 0 € 1000 € 2000 € -500 P{zon} = P{regen} = * * 0.5 = * * 0.5 = 750 Verwachte opbrengst:

18 Irrationeel beslissen Risico-avers voorbeeld uitwerken op bord

19 Definitie van risico Risico = kans x gevolg

20 Matrix of risks Small prob, small event Small prob, large event Large prob, small event Large prob, large event

21

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

23 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.

24 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.

25 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.

26 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 ).

27 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: a 1 = investing in shares a 2 = investing in bonds θ 1 = company profit # 5 % θ 2 = 5 % < company profit # 10 % θ 3 = company profit > 10 % ω 1 = return (0 % ‑ 2 %) = ‑ 2 % per annum ω 2 = return (3 % ‑ 2 %) = 1 % per annum ω 3 = return (6 % ‑ 2 %) = 4 % per annum

28 Decision tree (example 4.1)

29 Utility space Results space

30 Likelihood of the circumstances

31 From discrete to continuous decision models

32 Dijkhoogte bepaling Op bord uitwerken

33 Tijdsafhankelijke faalkansen Door veroudering is onderhoud noodzakelijk: –Onderhoudsmodellen

34 Levensduur T: is een stochastische variabele

35 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 , June 1998, Trondheim, Norway.

36 Haringvliet outlet sluices Lifetime distribution for one component ModelleringModelleringModelleringModellering Time start tt tt tt tt Replacement strategies of large numbers of similar components in hydraulic structures

37 Voorbeeld “leeftijd van mensen”: stochastische variable L mens L mens ~ N(78,6) of EXP(76,8) P(L mens >90)=...? P(L mens >90| L mens >89)= P(L mens >90)/P(L mens >89)=... Uitwerken op bord Vervolgens: Modelvorming voor algemene situatie

38 Verwachte resterende levensduur als functie van reeds bereikte leeftijd

39 Hazard rate population in S-Africa: f(t) / [1 - F(t) ]

40 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)

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

42 Bathtub Curve

43 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 (- λ t) This results in the simple exponential probability law for the Reliability function.

44 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

45 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.

46 β values, 0.5, 1.0, and 1.5.

47

48 Maintenance in Civil Engineering –Many design and build projects in the past –Nowadays many maintenance projects

49 Good Detoriation Model? State dependent Contains Effect of Loading? Consequence of failure Corrective Maint. Use dependent Time dependent Load dependent yes small largeno

50 Hydraulic Engineering corrective maintenance is not advised in view of the risks involved preventive maintenance time based failure based load based resistance based

51 resistance load time failure RoRo resistance load time RoRo resistance load cum. load time RoRo load time repair RoRo Resistance based Load based Time based Failure based R min repair ΔtΔt

52 Dike Settlement S.L.Sh 0 – A ln t = h(t) U.L.S.h(t) – HW time t opt R,S h0h0 h min S

53 Condition based maintenance bad good Inspection Repair

54 Maintenance A case study Some concepts

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

56 Introduction Maintenance  replacement

57 Introduction –Maintenance  replacement Large numbers of similar components

58 Introduction –Maintenance  replacement Large numbers of similar components

59 Introduction –Maintenance  replacement Large numbers of similar components Same lifetime-distribution Same age Same function

60 Modelling Case study Conclusions Variables of a replacement scenario – Start date of the (start) replacements – Replacement interval (  t) – Number of preventive ( ) replacements ModelleringModelleringModelleringModellering Time start tt tt tt tt

61 Modelling Case study Conclusions Finding the optimal strategy – Balance between risk costs and costs of preventive replacements – Replacement capacity – Capacity of the supplier ModelleringModelleringModelleringModellering

62 Modelling Case study Conclusions Probability of failure for different scenarios CasestudieCasestudieCasestudieCasestudie

63 ReliabilityMaintainability Availability The Concept of Availability

64 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 -  t where  is repair (restoration) rate

65 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

66 Up Down A1A3A2 t B1B2B3 Up:System up and running Down:System under repair System Operational States

67 MTTF is defined as the mean time of the occurrence of the first failure after entering service. MTTF = B1 + B2 + B3 3 Up Down A1A3A2 t B1B2B3 Mean Time To Fail (MTTF)

68 MTBF is defined as the mean time between successive failures. MTBF = (A1 + B1) + (A2 + B2) + (A3 + B3) 3 Up Down A1A3A2 t B1B2B3 Mean Time Between Failure (MTBF)

69 Mean Time To Repair (MTTR) MTTR = A1 + A2 + A3 3 MTTR is defined as the mean time of restoring a process or system to operation condition. Up Down A1A3A2 t B1B2B3

70 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

71 Reliability/Maintainability Measures (Failure Rate)  = 1 / MTBF R(t) = e - t Reliability R(t) Maintainability M(t) (Maintenance Rate)  = 1 / MTTR M(t) = 1 - e -  t

72 Types of Redundancy Active Redundancy Standby Redundancy

73 Active Redundancy A B Divider OutputInput Both A and B subsystems are operative at all times Div Note: the dividing device is a Series Element

74 Standby Redundancy A B SW Switch OutputInput 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.

75 Series System A1A1 A2A2 AnAn p s = p 1 + p 2 +……. + p n - (-1) n joint probabilities p s :Probability of system failure p i :Probability of component failure For identical and independent elements: p s ~ 1 - (1-p) n p) InputOutput

76 Parallel System B A p s = p 1.p 2 … p n Multiplicative Rule OutputInput p s :Probability of system failure

77 Series / Parallel System B1 A1 B2 A2 C InputOutput

78 System with Repairs Let MTBF =  and system MTBF =  s  s = ( 3 +  )/ ( 2 2 ) For Active Redundancy (Parallel or duplicated system)  s =  / 2 2 = MTBF 2 / 2 MTTR B A OutputInput <<  << 

79 For Standby Redundancy  s = ( 2 +  )/ ( 2 )  s =  / 2 = MTBF 2 / MTTR A B SW Switch OutputInput Standby Note: The standby system is normally inactive. Note: The switch is a series element, neglect for now.

80 System without Repairs  s = ( 3 +  )/ ( 2 2 ) For Active Redundancy  s = (3/2)  where  = 1/ s  MTBF For Standby Redundancy  s = ( 2 +  )/ ( 2 )  s = 2  where  = 1/ s  MTBF For systems without repairs,  = 0  s = 3  / ( 2 2 ) = 3  / ( 2 )  s = 2 / 2 = 2/  s = 2 / 2 = 2/

81 MTBF 2 / 2 MTTR MTBF 2 / MTTR  MTBF  MTBF Active Standby With Repairs Without Repairs Type Summary Redundancy techniques are used to increase the system MTBF


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