Bayesian inference to determine elastoplastic parameter distributions

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Bayesian inference to determine elastoplastic parameter

distributions

Hussein Rappel, Lars Beex, Ludovic Noels, St´ephane Bordas

hussein.rappel@uni.lu

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Introduction

1 F F 2 F F 20 F F · · · ·

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Introduction

Objective

Find the distribution from which the parameters of the specimens are

coming with limitednumber of specimens

π

(P

ar

ame

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A simple approach

Least squares method σ = E  J = 1₂ nm P i =1 (σi− E i)2 E = argmin E J (E ) σ E 1

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A simple approach

Now that we have{E1, · · · , E20} Emean=P20i =1 E20i s(standard deviation)= r P20 i =1(Ei−Emean) 2 20−1

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Bayesian inference

Original belief

Observations

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Bayesian inference

posterior z }| { π(x |y ) = prior z }| { π(x )× likelihood z }| { π(y |x ) π(y ) | {z } evidence =⇒ π(x |y ) ∝ π(x ) × π(y |x ) y := observation x := parameter π(x ) := original belief

π(y |x ) := given by the mathematical model that relates y to x π(y ) := is a constant number

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Bayesian inference

π(x |y ) ∝ π(x ) × π(y |x ) Parameters Material model Measurements (stress) Measurement noise

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Full Bayesian hierarchical model

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Full Bayesian hierarchical model

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Bayesian updating-Least squares

(1) −→ Least squares (2) −→ Bayesian updating π(xd|xm) ∝ π(xd|xm)π(xd) Parameters Material model Measurements (stress) Measurement noise (1) (2) PDF's parameters PDF

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Linear elasticity with Beta distribution as parameter

distribution

E = {E1, · · · , E20} and Ei ∼ Beta(α, β, a, c) = (Ei−a) α−1_(c−E i)β−1 (c−a)α+β+1_B(α,β)

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Linear elasticity with Beta distribution as parameter

distribution

Bayesian inference

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Linear elasticity with Beta distribution as parameter

distribution

Bayesian inference

π(α, β, a, c|Ei) ∝ π(Ei|α, β, a, c)π(α)π(β)π(a)π(c)

and for 20 specimens:

π(α, β, a, c|E) ∝Q20

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Linear elasticity with Beta distribution as parameter

distribution

π(α, β, a, c|E) =Q20 i =1Beta(Ei; α, β, a, c)N(α, sα2)N(β, sβ2)N(a, sa2)N(c, sc2) where N(x , s2) = √1 2πsexp _{(x −x )}2 2s2

If the priors are not informative much more specimens will be needed to make the problem numerically tractable

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Linear elasticity with Beta distribution as parameter

distribution

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Linear elasticity-perfect plasticity

E = {E1, · · · , E20} Ei ∼ Beta(α1, β1, a1, c1)

σy 0 = {σ_{y 01}, · · · , σ_{y 020}}

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Copula

Copulas are tools enable us to model dependence of several random variables in terms of their marginal distribution.

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Copula

for two parameters

h(x1, x2) = ch(F1(x1), F2(x2), ρ)f1(x1)f2(x2)

h := the joint PDF of the random variables

f1(x1) and f2(x2) := the marginal PDFs of the random variables

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Linear elasticity-perfect plasticity-Modular Bayesian

Identify copula parameter (ρ) using

both E and σy0 Identify (α1,β1,a1,c1)MAP

using E

Identify (α2,β2,a2,c2)MAP

using σy0 Find E & σy0

using Least squares method

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Linear elasticity-perfect plasticity-Modular Bayesian

Find E & σy0 using Least squares method

π(α1,β1,a1,c1,α2,β2,a2,c2,ρ|E,σy0)∝

∏

Beta(α1,β1,a1,c1)Beta(α2,β2,a2,c2)ch(FE(E),F_σy0(σy0))

×

priors

i=1 20

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Linear elasticity-perfect plasticity-Modular Bayesian

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Summary

We used Bayesian inference for identification of the material

parameters distribution with limited number of specimens

As the number of measurements for each specimen is large enough the effect of measurement uncertainty is smaller than the parameter variability

This means that we can break the full Bayesian problem to a least squares and Bayesian inference problem

To be able to overcome sampling problems we had to use modular

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Future study

Studying the effect of parameter variability and parameters dependence in presence of geometrical randomness

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A special acknowledgement

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Introduction

A0 L0 F F stress: σ = _AF 0 strain: = ∆L_L 0 σ

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Copula

Sklar’s theorem

H(x1, x2) = CH(F1(x1), F2(x2))

H := the joint cumulative distribution function (CDF) of the random

variables

F1(x1) and F2(x2) := the marginal CDFs of the random variables

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Copula

Probability density function

h(x1, x2) = ch(F1(x1), F2(x2))f1(x1)f2(x2)

h := the joint probability density function (PDF) of the random variables

f1(x1) and f2(x2) := the marginal PDFs of the random variables

c = _∂F∂2C 1∂F2

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Linear elasticity-perfect plasticity

π(E , σy 0) = Beta(α1, β1, a1, c1)Beta(α2, β2, a2, c2)ch(FE(E ), Fσy 0(σy 0))

FE(E ) =RaE1Beta(t; α1, β1, a1, c1)dt

Fσy 0(σy 0) = Rσy 0

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Linear elasticity-perfect plasticity

c_hGauss(FE(E ), Fσy 0(σy 0)) = 1 √ (|R|)exp " Φ−1(FE) Φ−1(Fσy 0) #T (R−1− I)( " Φ−1(FE) Φ−1(Fσy 0) # ! , R = " 1 ρ ρ 1 #

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Linear elasticity-perfect plasticity

Parameters to be identified

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Linear elasticity-perfect plasticity-Modular Bayesian

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