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The fundamental solutions for transversely isotropic

piezoelectricity and boundary element method

Haojiang Dinga,*, Jian Liangb

aDepartment of Civil Engineering, Zhejiang University, 310027, Hangzhou, People's Republic of ChinabDepartment of Mechanics, Zhejiang University, 310027, Hangzhou, People's Republic of China

Received 18 February 1997; accepted 12 November 1998

Abstract

In this paper, we rst supplement two groups of simplied general solution based on previous work. Those results

are in terms of harmonic functions and t for the cases of multiple eigenvalues. Then by trial and error, we obtain

the fundamental solutions for three cases for a piezoelectric innite media by giving the expressions of harmonic

functions. Finally, by use of those solutions, we implement a boundary element method program to perform

numerical calculations. The numerical results agree well with the analytical ones. # 1999 Elsevier Science Ltd. All

rights reserved.

Keywords: Piezoelectricity; Boundary element method; Harmonic functions

1. Introduction

For the equilibrium problem of transversely isotro-

pic media, Hu [2] obtained his solution by introducing

two potential functions F and c to express the displa-

cements. For the case of s16s2 (s1 and s2 are the eigen-

values of transversely isotropic materials), Hu [3] later

gave a general solution in terms of three harmonic

functions, and has solved a series of problems. Pan

and Chou [4] got the fundamental solutions for the in-

nite media for the cases of s16s2 and s1=s2 by apply-

ing the expressions of three potential functions. For

the equilibrium problem of piezoelectricity, Chen [5]

and Chen and Lin [6] expressed the innite body

Green's functions and their rst and second derivatives

as the contour integrals over the unit circle by using

the triple Fourier transform. Dunn [7] gave an explicit

solution for the Green's functions for an innite trans-

versely isotropic piezoelectric solid by taking the

Radon transform, coordinator transformation and

evaluation of residues in sequence. Lee and Jiang [8]

obtained a fundamental solution for an innite plane

by using the double Fourier transform. Lee and Jiang

[8] proposed a boundary element formulation based on

a weighted residual statement, giving a fundamental

solution for planar piezoelectric media. Lu and

Mahrenholtz [9] extended the variational boundary el-

ement model to piezoelectricity based on a modied

functional with six kinds of independent variables. But

none of them has performed numerical calculations.

Wang and Zheng [10] obtained a general solution in

terms of four harmonic functions in the case of distinct

eigenvalues. Ding et al. [1] have studied the general

solution systematically for transversely isotropic piezo-

electricity, and in the case of distinct eigenvalues, their

results are simplied and reduce to the Wang's sol-

ution [10], but the simplied general solutions for the

other two cases of multiple eigenvalues have not

appeared yet in the literature. In this paper, based on

the work of Ding et al. [1], we rst supplement these

results which are in terms of four harmonic functions.

Computers and Structures 71 (1999) 447455

0045-7949/99/$ - see front matter # 1999 Elsevier Science Ltd. All rights reserved.

P I I : S 0 0 4 5 - 7 9 4 9 ( 9 8 ) 0 0 2 3 7 - 5

* Corresponding author.

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As an application, we then obtain the fundamental sol-

ution of piezoelectric innite media for three cases by

giving three groups of expressions of the harmonic

functions. Finally we use those fundamental solutions

to complement a boundary element method programas well as perform numerical calculations. The numeri-

cal results agree well with the analytical ones.

2. The simplied general solutions

The constitutive relations of the transversely isotro-

pic piezoelectricity are

sx c11du

dx c12

dv

dy c13

dw

dz e31

df

dz

sy c12du

dx c11

dv

dy c13

dw

dz e31

df

dz

sz c13du

dx c13

dv

dy c33

dw

dz e33

df

dz

tyz c44

dv

dz

dw

dy

e15

df

dy

txz c44dudz

dw

dx

e15df

dx

txy c66

du

dy

dv

dx

Dx e15

du

dz

dw

dx

e11

df

dx

Dy e15dv

dz

dw

dy e11

df

dy

Dz e31du

dx e31

dv

dy e33

dw

dz e33

df

dz1

where si (tij), Di, u(n,w ) and f are the components of

stress, electric displacement, mechanical displacement

and electric potential, respectively; cij, eij and eij are the

elastic, piezoelectric and dielectric constants, respect-

ively. Ding et al. [1] have obtained the general sol-

utions for the basic equations of transversely isotropic

piezoelectricity as follows:

u dc0dy

m1L m2 d2

dz2

d 2Fdxdz

v dc0dx

m1L m2

d2

dz2

d 2F

dydz

w

c11e11L2

m3L

d 2

dz2 c44e33

d4

dz4

F

f

c11e15L

2 m4Ld 2

dz2 c44e33

d 4

dz4

F 2

where

L d2

dx 2

d 2

dy23

and the functions c0 and F satisfy the following

equations:2L

d 2

dz20

3c0 0 4

P3i1

2L

d 2

dz2i

3F 0 5

where zi=siz (i=0,1,2,3), and eigenvalues s20=c66/c44,

s 2i (i = 1,2,3) are the three roots of the following

equation

as6

bs4

cs2

d 0 6

where the coecients a,b,c,d are combinations of ma-

terial constants. Ding et al. [1] pointed out that F

admits the presentation

case a, when s1,s2,s3 are distinct: F F1 F2 F3 7

case b, when s1 T s2 s3:F F1 F2 z2F3 8

case c, when s1 s2 s3:F F1 z1F2 z21F3 9

where Fi (i=1,2,3) satisfy the following equations, re-spectively,2L

d 2

dz2i

3Fi 0, i 1,2,3 10

It will be very easy to nd a particular solution of Fi.

That means Fi may be obtained one by one. So Eq.

(2), the general solutions, must be simplied for each

case, respectively. For case a, Ding et al. [1] gave the

following simplied form:

u 3i1

dcidx

dc0dy

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v 3i1

dcidy

dc0dx

wm 3i1

aim dcidzi

m 1,2 11

where w1 represents the mechanical displacement com-

ponent w, w2 represents the electric potential, and the

coecients aim are given by

ai1 c11e11 m3s

2i c44e33s

4i

m1si m2s3i

ai2 c11e15 m4s

2i c44e33s

4i

m1si m2s3i

i 1,2,3 12

and the functions ci (i=0,1,2,3) satisfy2L

d2

dz2i

3ci 0 i 0,1,2,3 13

For cases b and c, the simplied results can be

obtained by substituting Eqs. (8) and (9), respectively,

into Eq. (2). After a series of work having been done,

the explicit forms ofci are found one by one for some

particular load cases to obtain the solutions. The next

section we describe the details.

3. The fundamental solutions for innite media

3.1. Solutions to the problem of combination of point

force P in z direction and point charge Q

This is an axisymmetric problem; assume c0=0 and

case a

ci Ai signz lnRi sijzj i 1,2,3 14

case b

ci signzBi lnRi sijzj i 1,2

c3 B31

R215

case c

c1 signzC1 lnR1 S1jzj

ci Ci

R1

i 2,3 16

where

Ri

x 2 y2 siz

2

qi 1,2,3

and Ai, Bi, Ci (i=1,2,3) are undetermined coecients.

Substituting Eqs. (14)(16) into the corresponding ex-

pressions of simplied general solutions yields the dis-placements and electric potential. Hence, by utilizing

the constitutive relations, the representations of the

stress components can be obtained. For case a, those

components are expressed as

u signz3i1

Aix

RiRi sijzj

v signz

3

i1

Aiy

RiRi sijzj

wm 3i1

aimAi

Ri

sx c11 c12 signz

3i1

Ai

41

RiRi sijzj

x 2

R3i Ri sijzj

x 2

R2

i Ri sijzj2 5

3i1

xiAizi

R3i

sy c11 c12 signz

3i1

Ai

41

RiRi sijzj

y2

R3i Ri sijzj

y2

R2i Ri sijzj

2 53i1

xiAizi

R3i

txy 2c66xy signz

3i1

Ai

41

R3i Ri sijzj

1

R2i Ri sijzj2

5

txm 3i1

o im AixR3i

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tym 3i1

o imAiy

R3i

sm 3i1

WimAiziR3i

17

where m = 1,2, s1,s2,tx1,tx2,ty1 and ty2 represent, re-

spectively, sz,Dz,txz,Dx,tyz and Dy; the coecients oimand Wim (i=1,2,3) are given by

o i1 c44si ai1 e15ai2, o i2 e15si ai1 e11ai2

Wi1 c33ai1 e33ai2si c13,

Wi2 e33ai1 e33ai2si e3118

Continuity of displacements u, v and stresses sx,sy,txyacross the interface z=0 implies that

3i1

Ai 0 19

Consider the equilibrium conditions of a layer between

two planes of z=2e, yielding

I

I I

I szx,y,e szx,y, e dx dy P 0 20I

I

II

Dzx,y,e Dzx,y, e

dx dy Q

0 21

Substituting Eq. (17) into Eqs. (20) and (21) yields

4p3i1

Wi1Ai P 0 22

4p3i1

Wi2Ai Q 0 23

The constants Ai(i=1,2,3) can be determined by Eqs.

(19), (22) and (23):

A1

PW22 W32 QW21 W31aDa

A2

PW32 W12 QW31 W11

aDa

A3 A1 A2

PW12 W22 QW11 W21aDa

Da 4p

W11 W31W22 W32 W21 W31W12

W32

24

For the other two cases, the undetermined constants

can be derived by the similar method:

B1 QW41 PW42aDb B2 QW41 PW42aDb

B3

QW11 W21 PW12 W22aDb

Db 4pW12W41 W41W22 W11W42 W21W4225

and

C1 0, C2 QW51 PW52aDc

C3 QW41 PW42aDc

Dc 4pW42W51 W41W5226

where

W41

c33a21 a41 e33a22 a42

s2

W42

e33a21 a41 e33a22 a42

s2

W51 c332a41 a51 e332a42 a52s1W52

e332a41 a51 e332a42 a52

s1 27

3.2. Solution to the problem of point force T in the x

direction

For case a, assume

c0 D0y

R0 s0jzj

ci Dix

Ri sijzji 1,2,3 28

for case b, assume

c0 H0y

R0 s0jzj

ci Hix

Ri sijzji 1,2

c3 signz H3x

R2R2 s2jzj29


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for case c, assume

c0 G0y

R0 s0jzj

c1 G1x

R1 s1jzj

ci signzGix

R1R1 s1jzji 2,3 30

where Di, Hi, Gi (i= 0,1,2,3) are undetermined con-stants. The components of displacement and stress can

be obtained by utilizing the simplied general solutions

and constitutive relations. For case a, they are:

u D0

41

R0 s0jzj

y2

R0R0 s0jzj2

5

3i1

Di

41

Ri sijzj

x 2

RiRi sijzj2

5

v D0xy

R0R0 s0jzj2

xy3i1

Di

RiRi sijzj2

wm signzx3i1

aimDi

RiRi sijzj

sx c11 c12D0x

41

R0R0 s0jzj2

2y2

R20R0 s0jzj3

y2

R30R0 s0jzj2

5

x3i1

Di

@xi

R3i c11 c12

43

RiRi sijzj2

2x

2

R2i Ri sijzj3 x

2

R3i Ri sijzj2

5A

sy c11 c12D0x

41

R0R0 s0jzj2

2y2

R20R0 s0jzj3

y2

R30R0 s0jzj25

x3i1

Di

@xi

R3i c11 c12

41

RiRi sijzj2

2y2

R2i Ri sijzj3

y2

R3i Ri sijzj2

5A

txy c66D0y

41

R30

2

R0R0 s0jzj2

4x 2

R20R0 s0jzj3

2x 2

R30R0 s0jzj2

5

2c66y3i1

Di

41

RiRi sijzj2

2x 2

R2i Ri sijzj3

x 2

R3i Ri sijzj2

5

txm o 0m signzD04 1

R0R0 s0jzj

y2

R30R0 s0jzj

y2

R20R0 s0jzj2

5

signz3i1

o imDi

41

RiRi sijzj

x 2

R3i Ri sijzj

x 2

R2i Ri sijzj2

5

tym o 0m signzD0xy

41

R30R0 s0jzj

1

R20R0 s0jzj2

5 signzxy

3i1

o imDi

41

R3i Ri sijzj

1

R2i Ri sijzj2

5

sm 3i1

Wim DixR3i

31


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where

o01 c44s0 o02 e15s0

xi c13ai1 e31ai2si c12 i 1,2,3 32

As before, the continuity of the components wm, txmand tym on z=0 yields

3i1

aimDi 0 m 1,2 33

o0mD0 3i1

o imDi 0 m 1,2 34

and the equilibrium condition givesII

II

txzx,y,h txzx,y, h

dx dy T

0 35

Substituting the 7th equation of Eq. (31) into Eq. (35)

yields

2pc44s0D0 2p3i1

o i1Di T 0 36

Eqs. (34) can be further reduced to a simple one with

the expression ofoim substituted:

3i0

siDi 0 37

Now the undetermined coecients Di (i=0,1,2,3) can

be derived from Eqs. (33), (36) and (37):

D0 Ta4pc44s0

D1 a21a32 a31a22TaDd



Dd 4pc44

s1a21a32 a31a22 s2a31a12

a11a32 s3a11a22 a21a12X 38

By the similar method, we have

H0 Ta4pc44s0

H1 a22a41 a21a42TaDh



Dh 4pc44

s1a22a41 a21a42 s2a11a42

a12a41 a12a21 a11a22

39

and

G0 Ta4pc44s0

G1 a41a52 a42a51TaDg



Dg 4pc44s1a12a51 a42a51 a11a52 a41a52 40

4. Boundary integral formulation

Letting S be the boundary of piezoelectric materialdomain, the boundary conditions of the basic

equations of piezoelectricity are given by

sijnj "t i on St

ui "u i on Su

and

Dini "o on So

f "

f on Sf 41

where ti represents the surface traction, o is the surface

charge, ni is the unit outward normal vector, and the

overbar indicates a prescribed value. The boundary

sets satisfy

St Su So Sf S,St Su So Sf X 42

Based on the extended Somigliana equation, the

boundary integral formulation is obtained:

Cu S

UtdS

S

Tu dS

V

Ub dV 43

where C is the coecient matrix, the general displace-


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ment u, surface traction t and body force b are

u

PTTR

u

v

w

f

QUUS, t

PTTR

txtytz

o

QUUS, b

PTTR

fxfyfzq

QUUS 44

and the two matrices composed by fundamental sol-

utions are

U

PTTR

u11 u12 u

13 f

1

u21 u22 u

23 f

2

u31 u32 u

33 f

3

u41 u42 u

43 f

4

QUUS,

T

P

TTRt11 t

12 t

13 o

1

t21 t22 t

23 o

2

t31 t32 t

33 o

3

t41 t42 t43 o4

Q

UUS

45

where u ij and t

ij (i,j=1,2,3) are, respectively, displace-

ment components and surface traction at a eld point

x in the xj coordinate directions due to a unit load act-

ing in one of the xi directions at a source point x on

the boundary, u 4j and t

4j (j= 1,2,3) are, respectively,

displacement components and surface traction at x in

the xj coordinate directions due to a unit electric

charge at x, f i and o

i (i= 1,2,3) are, respectively,

electric potential and surface charge at x due to a unit

load acting at x in one of the xi directions, and f

4 and

o

4 are, respectively, electric potential and surfacecharge at x due to a unit electric charge at x. In the

absence of body forces, Eq. (43) can be rewritten in

the form

Cxxxuxxx

S

Uxxx,xtx dS

S

Txxx,xux dS 46

If the boundary is in discretization with eight-node iso-

parameteric quadratic elements, Eq. (46) becomes

Cixxxuixxx ePS

8

i1

1

1

1

1

UtiNijJj dz dZ

ePS

8i1

11

11

TuiNijJj dz dZ 47

where Ni is shape function, J is Jacobian matrix, uiand ti represent, respectively, the general displacement

and surface traction at discrete grid, and Ci is a 4 4

coecient matrix, which by applying rigid-body

motion and constant electric potential, can be derived

from the basic equations and Eq. (47)

for nite media

Cixxx ePS

8i1

11

11

TNijJj dz dZ 48

for innite media

Cixxx I4 ePS

8i1

11

11

TNijJj dz dZ 49

where I4 is a 44 unit matrix. By utilizing above for-

mulae and fundamental solutions, we have performed

some numerical calculations. Table 1 gives the values

of material constants in our numerical examples.

4.1. Example 1

A piezoelectric column under uniaxial tension or

uniform electric displacement. Its size is aab. Two

Table 1

Material properties for PZT-4 and PZT-5H ceramics

PZT-4 PZT-5H

Elastic constants (N/m2) c11=12.61010 c11=12.610

10

c12=7.781010 c12=5.51010

c13=7.431010 c13=5.310

10

c33=11.51010 c33=11.710

10

c44=2.561010 c44=3.5310

10

Piezoelectric constants (C/m2) e31=5.2 e31=6.5

e33=15.1 e33=23.3

e15=12.7 e15=17.0

Dielectric constants (C/Vm) e11=6.463109 e11=1.5110

8

e33=5.622109 e33=1.3010

8

Eigenvalues s0=0.970261 s0=1.002828

s1=1.104405 s1=1.154188

s2=1.046767 s2=1.035093

+0.363468i +0.406737i

s3=1.046767 s3=1.0350930.363468i 0.406737i


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load cases are considered. Its boundary conditions are

given by

when z 2ba2:

txz tyz 0

sz p, Dz 0 load case 1

sz 0, Dz D0 load case 2

when z 0:

f 0

when x 2aa2:

sx txz Dx 0

when y 2aa2:

sy tyz Dy 0 50

If we assume

c0 0, ci

1

2ai

x2

y2

2z2i

i 1,2,3 51

the analytical results can be obtained by substituting

Eq. (51) into Eq. (11):

u s13 d31d33ak33 px

w

s33 d233ak33

pz

f d33ak33 pz 52

where [sij] (i,j=1,2,3) is the matrix of exibility factors,

the inverse matrix of elasticity constants [cij] (i,j =1,2,3), and

d31 e31s11 s12 e33s13

d33 2e31s13 e33s33

k33 e33 2e231s11 s12 4e33e31s13 e

233s33 53

In numerical calculation, we use six eight-node

quadratic elements, and set p=100N/m2, D0=1010 C/

m2. Because of the linear relationship, the comparison

between the analytical results and numerical ones isonly needed at an arbitrary node, say at node (a/2,a/

2,b/2). Table 2 lists those results. The ``decoupled''

items mean analytical results of a transversely isotropic

column, whose elastic constants and boundary con-

ditions are the same as PZT-4 or PZT-5H columns.

Table 2 shows that the absolute values of coupledcases are less than those of decoupled cases. In fact, by

further study of the constants listed in Table 1, it can

be veried that,

s11 s12 b 0, s13`0

d31`0, d33 b 0, k33 b 0 54

By this relation, Eq. (52) shows clearly that the piezo-

electric eects always cause the absolute values to be

less in the load case of uniaxial tension.

4.2. Example 2

Piezoelectric innity with spheroidal cavity. Assume

there are load cases applying at the innity: (1) uniax-

ial tension sIz , (2) all-around tension sr = sy = sI

r ;

and (3) uniform electric displacement Dz=DI

z . We use

242 nodes and 80 quadratic elements to perform the

numerical calculation. The stress and induction con-

centration at cavity surface are tabulated in Table 3.

The analytical results of PZT4 material are cited

directly from Kogan et al. [12], while the decoupled

results are calculated by Chen's solutions [11] and for

PZT-4 material.If we apply the load cases 1 and 3 only, the circum-

ferential stress component s ez at the equator and spy at

the poles are expressed by linear combination of our

numerical results

sez 1X870sIz 0X025D

Iz a10

10

sp

y 0X91sIz 0X127D

Iz a10

10 for PZT-4

material55

and

sez 1X9258sIz 0X0167D

Iz a10

10

sp

y 0X652sIz 0X0696D

Iz a10

10 for PZT-5H

material56

Assume only DIz is changeable in Eqs. (55) and (56).

If DIz =0, Eqs. (55) and (56) yield

sez

b

s

p

y 57

The position of the maximum principal stress locatesat the equator, but when DIz increases to a large


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enough amount, Eqs. (55) and (56) yieldsez `spy 58It can be concluded from Eqs. (57) and (58) that chan-

ging the electric loads can make the position of the

maximum principal stress move along the cavity sur-face.

5. Conclusions

The fundamental solutions for transversely isotropic

piezoelectricity are obtained in closed forms. Those

solutions are expressed very concisely by assuming the

expressions of the harmonic functions.

Through those numerical examples, the boundary el-

ement method and the fundamental solutions are veri-

ed. By comparing with the results of coupled

problems and decoupled problems, this paper shows

clearly that the piezo-electric constants have great in-

uence over the mechanical components.

Acknowledgements

This work was supported by the National Natural

Science Foundation of China.

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Table 2

Piezoelectric column

U/a (1010) w/b (1010) f/b

Load case Case 1 Case 2 Case 1 Case 2 Case 1 Case 2

PZT-4 Numerical 1.160 0.589 4.115 1.32 1.321 0.00425

Analytical 1.164 0.559 4.117 1.321 1.321 0.00425

Decoupled 2.997 8.220

PZT-5H Numerical 0.784 0.291 3.557 0.716 0.7163 0.00227

Analytical 0.785 0.291 3.558 0.716 0.7162 0.00227

Decoupled 1.703 5.816

Table 3

Stress and induction concentration

Load case Position PZT-5H PZT-4 Ref.[12] Ref.[11]

1: sz/sI

z r=a,z=0 1.926 1.870 1.88 2.01

2: sr/sI

r r=0,z=a 2.431 2.885 2.75a 2.35

3: Dz/DI

z r=a,z=0 1.509 1.510 1.51

a

We get the value of 2.887 after recalculating the Kogan et al. [12] solution.


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