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Chapter 1

A rate-independent ferroelectriccomposite model

1.1 Introduction

The main focus of this section is about the implementation of previously discussedrate-independent ferroelectric model in the analysis of the constitutive behaviour of 1-3 ferroelectric composites. In this section, the simulation results for ferroelectric andferroelastic hysteresis as well as the buttery hysteresis for ferroelectric ceramics underuniaxial loading along the ber direction are discussed. And also the ferroelectric hys-teresis and the buttery hysteresis results are compared with the experimental results.

In this analysis, the matrix material of the composite is assumed to be a linear elasticmaterial without any electro-mechanical coupling. In this analysis, for a given loadingcondition i.e. the given strain and electric eld, all the state variables and its conjugatesare decomposed into matrix part and the ber part using an appropriately assumed

homogenization method for a given ber volume fraction. Then the state variables forthe ber part is considered to check the switching condition. In the case of switching, theinternal variable will be updated and followed to this the state variables for the ber andmatrix will be updated. And this steps will be repeated iteratively until the switchingcriteria gets satised. And then from the decomposed stress and electric displacement,the stress and the electric displacement will be calculated.

1.2 Homogenization methods

In this analysis, three types of homogenization methods are studied. They are Voigtassumption, Reuss assumption and the combination of Voigt and Reuss assumption.

1.2.1 Voigt assumption

In this assumption, both the ber and the matrix material are parallel in all direction.So the composite is assumed to have the same strain and electric eld in both the ber

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and matrix material in all the direction.

E =f

f E = m

m E (1.1)

Along with this relation, the stresses and electric displacements for matrix, ber andcomposites are related as follows

D = vf f

f D + vm m

m D (1.2)

where f , m , f D , m D , f , m , f E , m E , v f , vm are stress in ber, stress in matrix, elec-tric displacement in ber, electric displacement in matrix, strain in ber, strain inmatrix, electric eld in matrix, electric eld in matrix, volume fraction of the ber andvolume fraction of the matrix respectively, Combined with the constitutive relation forboth ber and matrix along with this relation, for a given strain and the electric eld,the stress, the strain, the electric eld and the electric displacement for ber and ma-trix material and also the stress and the electric displacement for the composite can beobtained.

1.2.2 Reuss assumption

In this assumption, the matrix and the ber are arranged series to each other in all thedirection. So the composite is assumed to have the same stress and electric displacement

in both the ber and matrix material in all the direction.

D =f

f D = m

m D (1.3)

Along with this relation, the strains and electric elds for matrix, ber and compositesare related as follows

E = vf f

f E + vm m

m E (1.4)

Similar to the previous assumption, Combined with the constitutive relation for bothber and matrix and with the given equation for this assumption, for a given strain andthe electric eld, the stress, the strain, the electric eld and the electric displacementfor ber and matrix material and also the stress and the electric displacement for thecomposite can be obtained.

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1.2.3 Combined assumption

In 1-3 composites, the material is arranged parallel along the ber direction and arrangedseries in the plane perpendicular to the ber direction. Due to this arrangement, in thecombined assumption, Voigt assumption is used along the direction of ber orientationand the Reuss assumption is used in the remaining direction. So for the relation betweenthe stress and the electric displacement for matrix and ber along the ber direction,Eq.( 1.2) is used and for the remaining direction, Eq.( 1.4) is used. similarly for therelation between the strain and the electric displacement for matrix and ber along theber direction, Eq.( 1.1) is used and for the remaining direction, Eq.( 1.3) is used. Withthese relation along with the constitutive relation, for a given strain and the electriceld, the stress, the strain, the electric eld and the electric displacement for ber andmatrix material and also the stress and the electric displacement for the composite canbe obtained.

1.3 Formulation

The following algorithm explains the frame work behind this analysis.

start values: n +1 , E n +1 , f P in , f Sin , f E in

Predictor step

f P in +1 = f P in , f Sin +1 = f Sin , f E

in +1 = f E

in

compute: n +1 , f n +1 , m n +1 , D n +1 , f D n +1 , m D n +1 ,

f n +1 , m n +1 , f E n +1 , m E n +1 , f n +1 , f D n +1 , f P n +1(using any assumed homogenization method)if f n +1 tol then stop

Corrector step

solve the following equations, iteratively, along with equations for homogenizationassumption

f R a = 0 , f R b = 0 , f R c = 0 , f R d = 0

where f R (k )

a , f R (k )b , f R

(k )c , f R

(k )d are dene from equation()

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1.4 Results

In this section, the results of 1-3 composite model with various volume fraction of ber is analysed in combined homogenization method. The model is analysed with

the triangular cyclic loading of electric eld and the response of ferroelectric behaviorand the buttery curves are calculated. These responses are then compared with theexperimental results.

2000 1500 1000 500 0 500 1000 1500 20000.4

0.3

0.2

0.1

0

0.1

0.2

0.3

0.4

Electric field

e l e c

t r i c d i s p

l a c e m e n

t

vf=1

vf=0.8

vf=0.65

vf=0.35

Fig. 1.1: Longitudinal dielectric displacements vs. longitudinal electric eld.

2000 1500 1000 500 0 500 1000 1500 20002

1.5

1

0.5

0

0.5

1

1.5x 10

3

electric field

s t r a

i n

vf=1

vf=0.8

vf=0.65

0.35

Fig. 1.2: Longitudinal strains vs. longitudinal electric eld.

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Experiment vs Simulation for monolithic pzt

2500 2000 1500 1000 500 0 500 1000 1500 2000 2500

0.4

0.3

0.2

0.1

0

0.1

0.2

0.3

0.4

Fig. 1.3: Longitudinal dielectric displacements vs. longitudinal electric eld.

25 00 2 00 0 1 500 10 00 50 0 0 5 00 10 00 15 00 2 000 2 50 02.5

2

1.5

1

0.5

0

0.5

1

1.5x 10

3

Fig. 1.4: Longitudinal strains vs. longitudinal electric eld.

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