euromem : FROM UNCERTAINTIES TO PARTIAL … · Transformation T From physical space X ......
Transcript of euromem : FROM UNCERTAINTIES TO PARTIAL … · Transformation T From physical space X ......
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euromem : FROM UNCERTAINTIES TO PARTIAL SAFETY FACTORS CALIBRATION: SAFETY FACTORS CALIBRATION:
APPLICATION TO TENSILE MEMBRANE STRUCTURES - Discover the birth of a Eurocode
A TRAINING SCHOOL OF COST ACTION TU130329 Sep-1 Oct 2015 NANTES (France)
Franck SCHOEFS, Professor, Universit de Nantes
Deputy director of Scientific Interest Network MRGENCI
Basis of reliability analysis: Reliability Computation
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- Concept of load combination factors and partial Safety
factors.
- Illustration on simple examples where G writes R S with R
and S known (pdf).
Yesterday
and S known (pdf).
- Safety assessment for more complex cases (no analytical
solution or non-linearity of the solution: R=A2 Y).
- Practical calibration of partial safety factors
Today
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1. Definition of Pf.
Basis of reliability analysis: Reliability Computation
2. How to compute a safety estimate?
3. Another estimates: safety index.
4. Orders of magnitude in the reference case.
5. Examples.
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Aim: Evaluate the probability of failure (tension yield limit)
Resistances R Limit State : G(R,S)=R - S
Sollicitations S
> 0 : Safety
0 : Failure
Problem statement
example: beam in tension
1. Definition of Pf
Difficulties and stakes
Numerical Estmation of a probability of 10-5 (ULS)
Example for R deterministic (1st dimension in probabilistic space)
Prob. S
[ Mpa ]
R
Probability of failureS
4
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If 2 random variables
S
R
P*: design point (highest probability of occurrence
on the limit state)
2. How to compute a safety
estimate?
5
Df,i
Df,i
fRi,Si Pf,t
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fR,S (r,s) = fR(r)*fS(s)
r
s Df
r
r
s Df
r
III.2
If pdf of R and S known and if they are independant
2. How to compute a safety
estimate?
Pf = Df fR,S(r,s)dr ds = 1- -+ FS(r) fR(r)dr = -+ FR(s) fS(s)ds
r r
fS(s)fR(r)
E(R)E(S)
FR(s)
r,s
fS(s)ds
r
s Df
s
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here, S = (Nx , My , Mz)R = (Np , Mp)
Mz
Mp
Np
R
S
III.2
a xNpN
b y2
M + z2MpM
=0
(< 0: failure)
Real study cases
2. How to compute a safety
estimate?
Geometry of beams:
Length: 5 to 20 mDiameter : 0.5 to 1.5 m
Environnement marin
Nx
Np
s
LOADING Np
Displacement
N0
N0Np1
1
s x
y
z
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N simulations (computations) : looking for failure cases Xi s.a. = {XiRn|G(Xi)0}
Computation of G(Xi) : analytical/physical response surface / FEM/ SFEM
Pf = card( )/N: thats an approximation of
For these more complex cases (general case)
Monte carlo simulation (non explicit form of G)
2. How to compute a safety
estimate?
f
8
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Pf = card( )/NError: Shooman
formula
%erreur = fP~
1200
Monte carlo simulation
(non explicit form)
2. How to compute a safety
estimate?
% e
rror
%erreur = f
f
P~
N
P1200
( ) ( )f
f
fPN
PPCOV
.
1~ =
9
Number of simulations
For Pf = 10-3
Huge computational cots
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Safety index (Rjanytsine 1949)
3. Another estimate: safety index
Failure domain
1. Transformation TFrom physical space X
to standard space U
2. Distance btw O and limitstate (IN STANDARD
Most probable point = design pointStandard space U:
Random variables are standard: Normally
distributed, centered, normalized Failure
domain
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Hasofer-Lind sefety index: = OP* BUT Pf cannot be deduced direclty (Eurocodes are thus writen by considering )
Algorthms: rackwith-Fliesser / GRaCE
Hyperplane:
approximation
Physical space:
state (IN STANDARDSPACE)
3. APPROXIMATION of Pf
P*: design point (larger probability of occurrence on the
limit state)
: normal cdf
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Safety index (Rjanytsine 1949)
3. Another estimate: safety index
Standard space U:
Random variables are standard: Normally
distributed, centered, normalized
Failure domain
Failure domain
1. Transformation TFrom physical space X
to standard space U
2. Distance btw O and limitstate (IN STANDARD
Most probable point = design point
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P*: design point (larger probability of occurrence on the
limit state)
: normal cdf
Hasofer-Lind sefety index: = OP* BUT Pf cannot be deduced direclty (Eurocodes are thus writen by considering )
Algorthms: rackwith-Fliesser / GRaCE
Hyperplane:
approximation
Physical space:
state (IN STANDARDSPACE)
3. APPROXIMATION of Pf
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What is the design point?
S
R
P*: design point (highest probability of occurrence
on the limit state)
3. Another estimate: safety index
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Df,i
Df,i
fRi,Si Pf,t
Ri,Si t
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Safety index, methods
Pf(-))))
3. Another estimate: safety index
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: multinormal cumulative dentity function
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Pf
See Eurocode 0
Simplified relationship btw Pf-
4. Orders of magnitude in the
reference case: (linear limit states)
Source : Corus society
report
Pf(-))))
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5. Examples
Academic example: cantilever beam
Limit state on displacement : G(Z)= L/30-(5/48)*(P*L^3/E*I)
First study : sensibility / l
Quantity Mean. m STD
P 4 [kN] 1 [kN]Mfr * 20 [kN.m] 2 [kN.m]L 5 [m] 0
Reference study case: l = 5 m
PL
P
Results obtained with COMREL 7.10Prgm POUTRECONSOLV0.ITI
L 5 [m] 0
I 10-4 [m4] Cov : 20 %
E 2.107 [kPa] Cov : 10 %
FORM-beta= 4.123
Ll. value, -FORM, Param. Sens., Param. Elas.2 4.872 -0.1304 -0.5351E-013 4.705 -0.2053 -0.13094. 4.459 -0.2898 -0.26005 4.123 -0.3821 -0.46346 3.695 -0.4735 -0.76897 3.182 -0.5482 -1.2068 2.609 -0.5918 -1.8159 2.010 -0.6014 -2.69310 1.416 -0.5840 -4.126
P -0.19826E 0.09958I 0.97508PAR1 0.00000Sum of a^2 1.00000
Representative Alphas of Variables FLIM(1) [POUTRECONSOLV0.PTI]
Z
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5. Examples
Academic example: cantilever beam
Pf
l. value, -FORM, Param. Sens., Param. Elas.2 4.872 -0.1304 -0.5351E-013 4.705 -0.2053 -0.13094. 4.459 -0.2898 -0.26005 4.123 -0.3821 -0.46346 3.695 -0.4735 -0.76897 3.182 -0.5482 -1.2068 2.609 -0.5918 -1.8159 2.010 -0.6014 -2.69310 1.416 -0.5840 -4.126
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5. Examples
Academic example: cantilever beam
Limit state on displacement : FLIM(1){poutre console appui}= l/30-(5/48)*(P*l^3/E*I)
Second study : sensibility / std(P)
(P) P -0.37746E 0.13849
Representative Alphas of Variables FLIM(1) [POUTRECONSOLV0.PTI]
1 [kN] 4,12
2 [kN] 3,9
3 [kN] 3,59
l. value, -FORM, Param. Sens., Param. Elas.2 4.853 -0.1682 -0.6930E-013 4.613 -0.3232 -0.21024 4.193 -0.5182 -0.49445 3.591 -0.6689 -0.93146 2.896 -0.7007 -1.4527 2.220 -0.6412 -2.0228 1.625 -0.5459 -2.6879 1.129 -0.4477 -3.56810 0.7264 -0.3609 -4.968
E 0.13849I 0.91561PAR1 0.00000Sum of a^2 1.00000
P -0.52156E 0.17872I 0.83429PAR1 0.00000Sum of a^2 1.00000
Representative Alphas of Variables FLIM(1) [POUTRECONSOLV0.PTI]
STD(P)= 3 kN
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III.2
G= L/30-(5/48)*(P*L^3/E*I)
5. Examples
Academic example: cantilever beam
Quantity Mean. m
P 4 [kN] 1 [kN]
Mfr * 20 [kN.m] 2 [kN.m]
l 5 [m] 0
I 10-4 [m4] Cov : 20 %
E 2.107 [kPa] Cov : 10 %
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Industrial Example: corrosion of a pipe
5. Examples
Source : Corus society
report
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5. Examples
Industrial Example: corrosion of a pipe
Analyze correlations too
Source : rapport socit
Corus
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Probabilistic Design (conservatism of codes or absence of code)
Reliability assessment f structures or structural components designed with a given code ( -> )
Calibration of partial safety coefficients of a code ( -> )
New structures
Formulation and Interest of partial safety factor code format
Summary
Calibration of partial safety coefficients of a code ( -> )
Existing damaged structural assessment
Requalification of existing structures based on decisional appraoches
Inspection Maintenance Repair -> IMR optimisation on a given lifetime
Existing structures
Structural reliability Methods
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Mthodes danalyse de risqueMthodes quantitatives
Tutorial (in French)-20 24 heures en format long-12 16 heures en format court-12 16 heures en format court
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Objectif : analyse de risque / RBI
Structures Tipode utilises comme torchres
Crte de houle
Direction de la houle
P1 P2
Torchre
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M2 TPM - Universit de Nantes
Environnement interaction fluide-structure
Structure fissure dbouchante
Donnes de houle
Efforts rpartis
Fonction de
rponse f1
Fissure de longueur a
matrice de
rigidit
Fonction de rponse
f2
Fonction de rponse f3
Dplacement ou nergie
potentielle totale
Approche physique-mcaniste (modles / instrumentation)
Matriau (pathologie / Inspection) Fiabilit
Risque
Performance/Consquence
IMR
RBI
IllustrationsA : Structures ptrolires
Variables de base Distribution Conditionnement c.o.v
Troncature
H, Hauteur de vague extrme Gumbel sachant (Hs,Tstat, ) 8 % [m-2;m+5]
T, Priode de vague extrme Log-Normale sachant H 10% [m-k ;m+k]k = 3
Coefficients hydrodynamiques CD, CM, CX, CX,
Normale - 35% [m-p;m+p]p = 2Longueur de fissure a Exponentielle - 35% [- ; 2 R]
Probabilits
IMR
Processus, analyse stat.
Fiabilit dpendant
du temps
Lois de dommages
(Paris)
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M2 TPM - Universit de Nantes
Modlisation des donnes dentre
-Jeu de donnes de houle 2 niveaux :
*Non Expert CS : identifier une distribution H et une distribution T : puis analyse
hyp corrlation. Soit excell et tests aux moindres carrs sur CdF soit Matlab et tests
cdf ou pdf : diffrences.
*Expert CS : identifier une distribution jointe (proj sur chaos)
-Jeu de donnes de fissures
identifier une distribution H et une distribution T : puis analyse hyp corrlation. Soit
IllustrationsA : Structures ptrolires
identifier une distribution H et une distribution T : puis analyse hyp corrlation. Soit
excell et tests aux moindres carrs sur CdF soit Matlab et tests cdf ou pdf :
diffrences.
05
1015
2025
0
5
10
15
20
250
0.005
0.01
0.015
0.02
0.025
0.03
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M2 TPM - Universit de Nantes
Modlisation des structures
-Structure tripode:
* Non Expert CS : poutre console. Fissure = ressort + solution RdM.
*Expert CS : modle EF + modle EF fissure
IllustrationsA : Structures ptrolires
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M2 TPM - Universit de Nantes
Fiabilit :
* Non Expert CS : monte carlo + FORM + donnes tripodes (rsultats)
* Expert CS: monte carlo + FORM + EFS
-Analyse de sensibilit ltat limite
IllustrationsA : Structures ptrolires
Limiter le dplacement au sommet :
G = Uc U6
avec Uc = P5P6 / 600 = 0.041 m
= 0
= 90 Structure non-fissure
d
Limiter lnergie de dformation (ou son volution) :
G = c d
Pour comparer la fiabilit : c Uc
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M2 TPM - Universit de Nantes
RBI
* Niveau I et II (courts): Fourniture des PoD PFA (II) ou Analyse des
graphes (I)
*Niveau III (long) : fourniture des ROC
-Sensibilit aux modles de cot (inspection + cout relatif), la
connaissance experte
-Fiabilit dpendant du temps / optimisation des inspections (fatigue ou hyp fi = f(t), ..)
Illustrations
A : Structures ptrolires
=0.1 20 =0.1 40
0 0.2 0.4 0.6 0.8 1
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
=0.1 20
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.05
0.06
0.07
0.08
0.09
0.1
0.11
r
i1
ccc =
f
r2
ccc =
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
=0.1 40
0 0.2 0.4 0.6 0.8 1