810 likes | 973 Views
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: Vereenvoudigingen experimenteel model
E N D
Cursus Betonvereniging25 Oktober 2005Design-by-TestingBeslistheorieTijdsafhankelijk 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 = + R m u S R R
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 1 = + + R m t S 1 - R n 1 R n
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 + R m u 1 R R n
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%)
€0 Rationeel beslissen: ijscoman P{zon} = P{regen} = 0.5 regen ijs zon €1000 regen €2000 patat zon €-500
€0 Rationeel beslissen: ijscoman P{zon} = P{regen} = 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
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
Utility space Results space
Dijkhoogte bepaling • Op bord uitwerken
Tijdsafhankelijke faalkansen • Door veroudering is onderhoud noodzakelijk: • Onderhoudsmodellen
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 t t t t Time start Modellering Lifetime distribution for one component 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
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):
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.
Maintenance in Civil Engineering • Many design and build projects in the past • Nowadays many maintenance projects
State dependent Good Detoriation Model? Consequence of failure Corrective Maint. Contains Effect of Loading? Use dependent Time dependent Load dependent large no small yes no yes
Hydraulic Engineering • corrective maintenance is not advised in view of the risks involved • preventive maintenance • time based • failure based • load based • resistance based