Ali EL HAJJ · 2021. 1. 14. · Inverse Problem for Crack Propagation in Wood Ali EL HAJJ XLIM, UNR...

Inverse Problem for Crack Propagation in Wood

Ali EL HAJJ

XLIM, UNR 7252 Université de Limoges CNRS

Defense of Master ACSYON

Supervisors: Olivier Ruatta - Loic Bourdin

EL HAJJ (XLIM) Defense of Master ACSYON September 2017 1 / 37

Overview

1 Introduction

2 Defining the Problem

3 The Linear Case

4 The Bilinear Case

5 Field Work

6 Conclusions


Introduction


Introduction

A crack is the surface geometric (S) separation on a continuousmediom (Ω), having 2 modes of fracture.

Develop a model to analyze propagation of cracks in wood, todetect the "next" crack point.Inverse problem is taken into consideration.Find parameters of forces attacking a wood platform, that leads toa specific type of deformation.


Defining The Problem


Defining the ProblemThe Approach

Our approach: Determine parameters of forces leading to a particulardisplacement of points on the plank.


Defining the ProblemWilliams’ Series

The elastic displacement fields, uk1 and uk

2 , of point Mk , are expressedin the form of an infinite Williams’ series :

uk1 =

∞∑i=1

[Ai

Ir(i/2)k fi(κ, θk ) + Ai

IIr(i/2)k gi(κ, θk )

]+ T1 − Rxk

2

uk2 =

∞∑i=1

[Ai

Ir(i/2)k li(κ, θk ) + Ai

IIr(i/2)k zi(κ, θk )

]+ T2 − Rxk

1

κ : Kolosov constant.Ai

I : Force coefficient of rupture in mode I.Ai

II : Force coefficient of rupture in mode II.rk and θk : Polar coordinates.


Defining the ProblemWilliams’ Series

T1 and T2 : Horizontal and vertical movements parameters.R : Rotation movement parameter.fi , gi , zi , and li : Polar functions.

fi(κ, θ) = κ cos(

iθ

2

)− i

2cos

(( i2− 2).θ)

+( i

2+ (−1)i

). cos

(iθ

2

)gi(κ, θ) = −κ sin

(iθ

2

)+

i2

sin(( i

2− 2).θ)−( i

2− (−1)i

). sin

(iθ

2

)li(κ, θ) = κ sin

(iθ

2

)+

i2

sin(( i

2− 2).θ)−( i

2+ (−1)i

). sin

(iθ

2

)zi(κ, θ) = κ cos

(iθ

2

)+

i2

cos(( i

2− 2).θ)−( i

2− (−1)i

). cos

(iθ

2

)


Defining the ProblemThe Deformation

463 points having the following deformation:

Initial points before attacking. Final points after attacking.EL HAJJ (XLIM) Defense of Master ACSYON September 2017 9 / 37

The Linear Case


The Linear CasePlan

Restrict the Truncation order N to be ≤ 50; to be discussed later .

All coefficients are known except the forces’ parameters.

Must find A1I ,A

1II , ...,A

NI ,A

NII that leads to a given displacement.

Matrix form to be found, in order to apply SVD.

Small values of M and N is great in practice, and assure a morestable solution.


The Linear CaseWilliams’ Series Review

uk

1 =N∑

i=1

[Ai



]+ T1 − Rxk

2

uk2 =

N∑i=1

[Ai



]+ T2 − Rxk

1

with


iθ

2

)− i

2cos

(( i2− 2).θ)

+( i

2+ (−1)i

). cos

(iθ

2


(iθ

2

)+

i2

sin(( i

2− 2).θ)−( i

2− (−1)i

). sin

(iθ

2


(iθ

2

)+

i2

sin(( i

2− 2).θ)−( i

2+ (−1)i

). sin

(iθ

2


(iθ

2

)+

i2

cos(( i

2− 2).θ)−( i

2− (−1)i

). cos

(iθ

2

)EL HAJJ (XLIM) Defense of Master ACSYON September 2017 12 / 37

The Linear CaseMatrix Form

The Williams’ series can be written in a matrix form as BX = U where :

B is a 2M × 2N matrix.U is a 2M dimensional vector.X is the 2N dimensional vector of unknowns.

Ready to apply SVD.

If we get A1I < 0, set it zero and re-project on the set of positive

first coordinate solutions.


The Linear CaseMatrix Form

B =

r1/21 f1(κ, θ1) r1/2

1 g1(κ, θ1) · · · rN/21 fN(κ, θ1) rN/2

1 gN(κ, θ1)...

......

r1/2M f1(κ, θM) r1/2

M g1(κ, θM) · · · rN/2M fN(κ, θM) rN/2

M gN(κ, θM)

r1/21 l1(κ, θ1) r1/2

1 z1(κ, θ1) · · · rN/21 lN(κ, θ1) rN/2

1 zN(κ, θ1)...

......

r1/2M l1(κ, θM) r1/2

M z1(κ, θM) · · · rN/2M lN(κ, θM) rN/2

M zN(κ, θM)

uT = (u1

1 , ...,uM1 , .....,u

1N , ...,u

MN )

X T = (A1I ,A

1II , ...,A

NI ,A

NII )

.EL HAJJ (XLIM) Defense of Master ACSYON September 2017 14 / 37

The Linear CaseNumerical Results

M=8 and N=10:error = 8.9442e-14 solution = 0.3620

0.3591..

0.00300.0005

M=10 and N=15:error = 1.1423e-12 solution = 0.1282

0.1004..

-0.00040.0000


The Linear CaseNumerical Results

M=100 and N=15:error = 0.7318 solution= 0

0.0859..

-0.00003-0.000006

M=200 and N=50:error = 0.3618 solution= 0.0002

0.0061..

0.00000.0000


The Linear CaseNumerical Results : Given Deformation and Deformation via Inverse-direct Problem

M=8 and N=10

M=100 and N=15

M=10 and N=15

M=200 and N=50EL HAJJ (XLIM) Defense of Master ACSYON September 2017 17 / 37

The Linear CaseThe Truncation Order

Study the error with respect for several values of M.

Our goal is to maintain the most stable solution.

Restrict N ≤ 300.

Fix M, vary N from 2 to 300, and calculate the error correspondingto each N.

Large error with bad behaivour when N exceeds 50, and smallone when N ≤ 50, as we will see in the next figures.


The Linear CaseThe Truncation Order

M=25

M=200

M=50

M=450EL HAJJ (XLIM) Defense of Master ACSYON September 2017 19 / 37

The Bilinear Case


The Bilinear CaseWhy GloptiPoly and Polynomial Optimization

Few amount of data should be taken in practice to avoidcomplexity in systems.

GloptiPoly deals with small-scaled problems.

Polynomial optimization is a powerful method for "globally" solvinga problem of minimizing a polynomial over a semi-algebraic set.

GloptiPoly builds a family of LMI relaxations convergingmonotonically to the solution.


The Bilinear CasePlan

All coefficients are given except κ and the forces’ parameters

Small values of M and N to be taken.

Must find κ and A1I ,A

1II , ...,A

NI ,A

NII that leads to a given

displacement, with 5/3 ≤ κ ≤ 3 and A1I ≥ 0.

Matrix form to be found, to be illustrated on GloptiPoly via leastsquare approach.


The Bilinear CaseWilliams’ Series Review

uk

1 =N∑

i=1

[Ai



]+ T1 − Rxk

2

uk2 =

N∑i=1

[Ai



]+ T2 − Rxk

1

with


iθ

2

)− i

2cos

(( i2− 2).θ)

+( i

2+ (−1)i

). cos

(iθ

2


(iθ

2

)+

i2

sin(( i

2− 2).θ)−( i

2− (−1)i

). sin

(iθ

2


(iθ

2

)+

i2

sin(( i

2− 2).θ)−( i

2+ (−1)i

). sin

(iθ

2


(iθ

2

)+

i2

cos(( i

2− 2).θ)−( i

2− (−1)i

). cos

(iθ

2

)EL HAJJ (XLIM) Defense of Master ACSYON September 2017 23 / 37

The Bilinear caseNew Form of Polar Functions

For clarity, for i = 1, ...,N, the polar functions can be written as:

fi = κ. cos(

iθ

2

)+ ai

gi = −κ. sin(

iθ

2

)+ bi

li = κ. sin(

iθ

2

)+ ci

zi = κ. cos(

iθ

2

)+ di


The Bilinear caseMatrix Form

For simplicity, consider the case where M = 3 and N = 2.

the Williams’ series can be written in a matrix form as AX = Uwhere:

A is a 2M × 2N matrix.U is a 2M dimensional vector.X is the 2N dimensional unknown vector.

all having the forms:


The Bilinear caseMatrix Form

A =

r 1/21 cos

(θ2

)r 1/21 .a1 −r 1/2

1 sin(θ2

)r 1/21 .b1 r1 cos(θ) r1.a2 −r1 sin(θ) r1.b2

r 1/22 cos

(θ2

)r 1/22 .a1 −r 1/2

2 sin(θ2


r 1/23 cos

(θ2

)r 1/23 .a1 −r 1/2

3 sin(θ2


r 1/21 sin

(θ2

)r 1/21 .c1 r 1/2

1 cos(θ2

)r 1/21 .d1 r1 sin(θ) r1.c2 r1 cos(θ) r1.d2

r 1/22 sin

(θ2

)r 1/22 .c1 r 1/2

2 cos(θ2


r 1/23 sin

(θ2

)r 1/23 .c1 r 1/2

3 cos(θ2


X T =

(κA1

I A1I κA1

II A1II κA2

I A2I κA2

II A2II

)UT =

(u1

1 u21 u3

1 u12 u2

2 u32

)


The Bilinear caseIllustration on GloptiPoly

Polynomial optimization problem to be solved by GloptiPoly:

p∗ = infx∈Rn

p(x)

s.t g1(x) ≥ 0, ...,gm(x) ≥ 0

where p,g1, ...,gm ∈ R[x ].

Let x = (κ,A1I ,A

1II , ...,A

NI ,A

NII ) ∈ R2N+1.

Using the least square approach, we get the polynomialoptimization problem:

min p(x) = ||AX − u||

s.t g1(x) = κ− 53≥ 0

g2(x) = 3− κ ≥ 0

g3(x) = A1I ≥ 0


The Bilinear caseNumerical Results

M=2 and N=2:

error = 0.0302 param = 0.0002 kappa= 2.94300.00020.1348-0.0288-0.0227

M=8 and N=3:

error = 0.0637 param = 0.0530 kappa= 1.7251-0.2420

.

.0.0263


Field Work


Field WorkThe Approach

Analyze and explore closely the crack’s growth.

It involves the studying of the direct problem of our task.

Interpret the displacements of 22 points taken on a plank of woodw.r.t consecutive time steps t .

The approach is studied using the opening mode operation.


Field WorkThe Approach

Test plank with initial crack and testingpoints.

Opening mode operation.


Field WorkAbstract Results

t=100: No notable growth. t=900: A slight opening.


Field WorkAbstract Results

t=1100: More and more widening. t=1290: The complete deformation.


Field WorkNumerical Results

Displacements of points at several time references.

Points located initially near the rupture will have a significantmovement at the end of the operation, e.g. points 6 and 12.Way far points from the crack will score no remarkabledisplacement, e.g. points 18 and 22.


Conclusions


Conclusions

Results contribute to predicting their integrity and elasticity limit.

Limited truncation order and less input data to numerically reachmore stable solution in the Linear case.

Power of GloptiPoly appears in practice, with the advantage ofachieving the global solution.

Identifying explicitly the displacement of several points by themode opening operation, according to their proximity orremoteness from the crack.


Thank You!!!Any Questions ?


Ali EL HAJJ · 2021. 1. 14. · Inverse Problem for Crack Propagation in Wood Ali EL HAJJ XLIM, UNR...

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