10
Application of Hankel Transform for
Solving a Fracture Problem of a Cracked
Piezoelectric Strip Under Thermal Loading
Sei Ueda
Osaka Institute of Technology
Japan
1. Introduction
In this chapter, an example of the application of Hankel transform for solving a fracture
problem will be explained. In discussing axisymmetric problems, it is advantageous to use
polar coordinates, and the Hankel transform method is powerful to solve the general
equations in polar coordinates. A brief account of the Hankel transform will be given. Here
f is a function of r , its transform is indicated by a capital F , Jν is the ν th order Bessel
function of the first kind, and the nature of the transformation either by a suffix or by a
characteristic new variable s . It will be assumed without comment that the integrals in
question exist, and that, if necessary, the functions and their derivatives tend to zero as the
variable tends to infinity. The Hankel transform of order ν > −1 / 2 , Hν [ f (r )] or Fν ( s ) , of a
function f (r ) is defined as
Hν [ f (r )] ≡ Fν (s ) = ∫ rJν (sr ) f (r )dr
∞
0
and its inversion formula is
f ( r ) = ∫ sJν (sr )Fν (s )ds
∞
0
Also, integrating by parts twice gives
⎡ d 2 f 1 df ν 2
−
Hν ⎢ 2 +
r dr r 2
⎢⎣ dr
⎤
f ⎥ = −s 2 Fν (s )
⎥⎦
provided that rf ( r ) and rdf (r ) / dr tend to zero as r → 0 and as r → ∞ .
The piezoelectric materials have attracted considerable attention recently. Owing to the
coupling effect between the thermo-elastic and electric fields in piezoelectric materials,
thermo-mechanical disturbances can be determined form measurement of the induced
electric potential, and the ensuing response can be controlled through application of an
appropriate electric field (Rao & Sunar, 1994). For successful and efficient utilization of
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208
Fourier Transform – Materials Analysis
piezoelectric as sensors and actuators in intelligent systems, several researches on piezothermo-elastic behavior have been reported (Tauchert, 1992).
Moreover a better understanding of the mechanics of fracture in piezoelectric materials
under thermal load conditions is needed for the requirements of reliability and lifetime of
these systems. Using the Fourier transform, the present author studied the thermally
induced fracture of a piezoelectric strip with a two-dimensional crack (Ueda, 2006a, 2006b).
Here the mixed-mode thermo-electro-mechanical fracture problem for a piezoelectric
material strip with a penny-shaped crack is considered. It is assumed that the strip is under
the thermal loading. The crack faces are supposed to be insulated thermally and electrically.
By using the Hankel transform (Sneddon & Lowengrub, 1969), the thermal and electromechanical problems are reduced to a singular integral equation and a system of singular
integral equations (Erdogan & Wu, 1996), respectively, which are solved numerically (Sih,
1972). Numerical calculations are carried out, and detailed results are presented to illustrate
the influence of the crack size and the crack location on the stress and electric displacement
intensity factors. The temperature, stress and electric displacement distributions are also
presented.
2. Formulation of the problem
Fig. 1. Penny-shaped crack in a piezoelectric strip
A penny-shaped crack of radius c is embedded in an infinite long piezoelectric strip of
thickness h = h1 + h2 as shown in Figure 1. The crack is located parallel to the boundaries
and at an arbitrary position in the strip, and the crack faces are supposed to be insulated
thermally and electrically. The cylindrical coordinate system is denoted by (r ,θ , z) with its
origin at the center of the crack face and the plane r − θ along the crack plane, where z is
the poling axis. It is assumed that uniform temperatures T10 and T20 are maintained over
the stress-free boundaries. In the following, the subscripts r,θ , z will be used to refer to the
direction of coordinates. The material properties, such as the elastic stiffness constants, the
piezoelectric constants, the dielectric constants, the stress-temperature coefficients, the
coefficients of heat conduction, and the pyroelectric constant, are denoted by c kl , ekl , ε kk ,
λkk ( k, l = 1, 2, 3) , κ r , κ z , and pz , respectively.
The constitutive equations for the elastic field are
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Application of Hankel Transform for Solving a Fracture
Problem of a Cracked Piezoelectric Strip Under Thermal Loading
209
∂uri
∂u
∂φ
u
⎫
+ c12 ri + c13 zi + e31 i − λ11Ti , ⎪
∂r
∂z
∂z
r
⎪
∂u
∂u
∂φ
u
σ θθ i = c 12 ri + c11 ri + c13 zi + e31 i − λ11Ti , ⎪⎪
∂r
∂z
∂z
r
⎪
⎬ ( i = 1, 2)
∂uri
∂uzi
∂φi
uri
σ zzi = c13
+ c13
+ c 33
+ e33
− λ33Ti , ⎪
∂r
∂z
∂z
r
⎪
⎪
∂φi
⎛ ∂uzi ∂uri ⎞
⎪
σ zri = c 44 ⎜
+
⎟ + e15
∂z ⎠
∂r
⎪⎭
⎝ ∂r
σ rri = c11
(1)
where Ti ( r , z) is the temperature, φi ( r , z) is the electric potential, uri ( r , z ) , uzi (r , z) are the
displacement components, σ rri (r , z) , σ θθ i (r , z) , σ zzi (r , z) , σ zri (r , z) (i = 1, 2) are the stress
components. The subscript i = 1, 2 denotes the thermo-electro-elastic fields in 0 ≤ z ≤ h1 and
− h2 ≤ z ≤ 0 , respectively. For the electric field, the constitutive relations are
∂u ⎞
∂φ
⎛ ∂u
Dri = e15 ⎜ zi + ri ⎟ − ε 11 i ,
∂z ⎠
∂r
⎝ ∂r
⎫
⎪
⎪⎪
⎬ (i = 1, 2)
⎪
u
∂u
∂u
∂φ
Dzi = e31 ri + e31 ri + e33 zi − ε 33 i + pzTi ⎪
r
∂r
∂z
∂z
⎭⎪
(2)
where Dri (r , z) , Dzi ( r , z) (i = 1, 2) are the electric displacement components.
The governing equations for the thermo-electro-elastic fields of the medium may be
expressed as follows:
⎛ ∂ 2Ti
κ 2 ⎜⎜
2
⎝ ∂r
+
1 ∂Ti ⎞ ∂ 2Ti
= 0 (i = 1, 2)
⎟+
r ∂r ⎠⎟ ∂z2
⎛ ∂2u 1 ∂uri ∂uri ⎞
∂2u
∂2u
∂2φ
∂T
c11 ⎜⎜ 2ri +
− 2 ⎟⎟ + c44 2ri + ( c13 + c44 ) zi + ( e31 + e15 ) i = λ11 i ,
r ∂r
∂r∂z
∂r∂z
∂r
∂z
r ⎠
⎝ ∂r
⎛ ∂2u 1 ∂uzi ⎞
⎛ ∂2u 1 ∂uri ⎞
⎛ ∂2φ 1 ∂φi ⎞
∂2u
∂2φ
c44 ⎜⎜ 2zi +
⎟⎟ + c33 2zi + ( c13 + c44 ) ⎜⎜ ri +
⎟⎟ + e15 ⎜⎜ 2i +
⎟⎟ + e33 2i
r ∂r ⎠
r ∂r ⎠
∂z
∂z
⎝ ∂r
⎝ ∂r∂z r ∂z ⎠
⎝ ∂r
⎛ ∂2u 1 ∂uzi ⎞
⎛ ∂2u 1 ∂uri ⎞
⎛ ∂2φ 1 ∂φi ⎞
∂2u
∂2φ
e15 ⎜⎜ 2zi +
⎟⎟ + e33 2zi + ( e15 + e31 ) ⎜⎜ ri +
⎟⎟ − ε11 ⎜⎜ 2i +
⎟⎟ − ε33 2i
r ∂r ⎠
r ∂r ⎠
∂z
∂z
⎝ ∂r
⎝ ∂r∂z r ∂z ⎠
⎝ ∂r
where κ 2 = κ r / κ z .
(3)
⎫
⎪
⎪
⎪
⎪
∂Ti ⎪
= λ33
,⎪⎪
(4)
∂z ⎬(i = 1, 2)
⎪
⎪
∂Ti ⎪⎪
= −pz
∂z ⎪
⎪
⎪⎭
The boundary conditions can be written as
∂
⎫
T1 (r , 0) = 0
(0 ≤ r < c ) ⎪
∂z
⎬
T1 (r , 0) = T2 ( r, 0) (c ≤ r < ∞ )⎪⎭
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210
Fourier Transform – Materials Analysis
T1 (r , h1 ) = T10 ,
⎫
⎪
∂
∂
⎪
T1 (r, 0) = T2 ( r, 0),⎬ (0 ≤ r < ∞ )
∂z
∂z
⎪
T2 (r, −h2 ) = T20
⎪⎭
(6)
for thermal loading conditions and
(0 ≤ r < c ) ⎫
σ zz1 ( r , 0) = 0
⎬
uz 1 (r , 0) = uz 2 (r , 0) (c ≤ r < ∞ )⎭
σ zr 1 (r , 0) = 0
(0 ≤ r < c ) ⎫
⎬
ur 1 (r , 0) = ur 2 ( r , 0) (c ≤ r < ∞ )⎭
Dz 1 (r , 0) = 0
(0 ≤ r < c ) ⎫
⎬
φ1 (r, 0) = φ2 (r , 0) (c ≤ r < ∞ )⎭
σ zz1 (r , 0) = σ zz 2 (r , 0), σ zz1 (r , h1 ) = 0, σ zz 2 (r , −h2 ) = 0,⎫
⎪
σ zr 1 (r , 0) = σ zr 2 (r , 0), σ zr 1 (r , h1 ) = 0, σ zr 2 (r , −h2 ) = 0,⎬ (0 ≤ r < ∞ )
Dz1 (r , 0) = Dz 2 (r , 0),
Dz1 (r , h1 ) = 0,
Dz 2 (r , −h2 ) = 0 ⎪⎭
(7)
(8)
(9)
(10)
for the electromechanical conditions.
3. Temperature field
For the problem considered here, it is convenient to represent the temperature as the sum of
two functions.
Ti ( r , z) = T (1) ( z) + Ti(2) ( r , z) ( i = 1, 2)
(11)
where T (1) ( z) satisfies the following equation and boundary conditions:
d 2T (1)
=0
dz2
(12)
T (1) ( h1 ) = T10 , ⎪⎫
⎬
T (1) ( −h2 ) = T20 ⎪⎭
(13)
and Ti(2) ( r , z) (i = 1, 2) is subjected to the relations:
⎛ ∂ 2Ti(2) 1 ∂Ti(2) ⎞ ∂ 2Ti(2)
+
= 0 (i = 1, 2)
⎟+
2
⎜
r ∂r ⎠⎟
∂z2
⎝ ∂r
κ2 ⎜
∂ (2)
d
⎫
T1 (r , 0) = − T (1) (0) (0 ≤ r < c ) ⎪
dz
∂z
⎬
(c ≤ r < ∞ )⎭⎪
T1(2) ( r, 0) = T2(2) (r , 0)
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(14)
(15)
Application of Hankel Transform for Solving a Fracture
Problem of a Cracked Piezoelectric Strip Under Thermal Loading
211
⎫
T1(2) (r , h1 ) = 0,
⎪
∂ (2)
∂ (2)
⎪
T1 (r , 0) = T2 (r , 0), ⎬ (0 ≤ r < ∞ )
∂z
∂z
⎪
⎪
T2(2) (r , − h2 ) = 0
⎭
(16)
It is easy to find from Eqs.(12) and (13) that
T (1) ( z) =
1
{(T10 − T20 ) z + T10 h2 + T20 h1}
h1 + h2
(17)
By applying the Hankel transform to Eq.(14) (Sneddon & Lowengrub, 1969), we have
Ti(2) ( r, z) = ∑ ∫ Dij (s) J 0 ( sr )exp(sτ ij z)ds (i = 1, 2)
∞
2
j =1
0
(18)
where Dij ( s) (i, j = 1, 2) are unknown functions to be solved and τ ij (i, j = 1, 2) are given by
τ 11 = τ 22 = −κ ,⎫
⎬
τ 12 = τ 21 = κ ⎭
(19)
Taking the second boundary condition (15) into consideration, the problem may be reduced
to a singular integral equation by defining the following new unknown function G0 (r )
(Erdogan & Wu, 1996):
{
}
⎧ ∂ (2)
T (r , 0) − T2(2) (r , 0)
⎪
G0 (r ) = ⎨ ∂r 1
⎪⎩
0
⎫
(0 ≤ r < c ) ⎪
⎬
(c ≤ r < ∞ )⎭⎪
(20)
Making use of the first boundary condition (15) with Eqs.(16), we have the following
singular integral equation for the determination of the unknown function G0 (t ) :
∫0t {M0
c
(1)
}
(t, r ) + M0(2) (t, r ) G0 (t )dt = −
2 T10 − T20
κ h1 + h2
(0 ≤ r < c )
(21)
In Eq.(21), the kernel functions M0(1) (t, r ) and M0(2) (t, r ) are given by
⎧
⎫
2 1
⎛r⎞
E
(r < t ),⎪
⎪
2
2 ⎜ ⎟
π t −r ⎝t ⎠
⎪
⎪
M0(1) (t, r ) = ⎨
⎬
⎧
⎫
r
t
t
2
1
⎪
⎪
⎛ ⎞
⎛ ⎞
⎪
⎪
E
K
r
t
(
)
+
>
⎨
⎬
⎜
⎟
⎜
⎟
⎪ π ⎪ t(t 2 − r 2 ) ⎝ r ⎠ rt ⎝ r ⎠ ⎪
⎪
⎭
⎩ ⎩
⎭
∞ ⎧
⎫⎪
⎪ 2 ρ ( s )ρ 2 ( s )
M0(2) (t, r ) = − ∫ s ⎨ 1
+ 1⎬ J 0 ( sr ) J 1 (st )ds
0 ⎪
⎩ κρ0 (s )
⎭⎪
(22)
(23)
where K and E are complete elliptic integrals of the first and second kind, and
ρ k (s) ( k = 0, 1, 2) are given by
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Fourier Transform – Materials Analysis
ρ0 (s ) = ρ 2 ( s){1 − exp( −2 sκ h1 )} − ρ1 (s ){1 − exp( −2sκ h2 )} , ⎫
ρi (s ) = τ i 1 − τ i 2 exp( −2 sκ hi ) (i = 1, 2)
⎪
⎬
⎪
⎭
(24)
Once G0 (t ) is obtained from Eq.(21), the temperature field can be easily calculated as
follows:
Ti(2) ( r, z) = ∑ Tij(2) (r , z) (i = 1, 2)
2
j =1
where
Tij(2) ( r, z) = ∫ Rij (s)R0 (s ) J 0 (sr )exp( sτ ij z)ds (i, j = 1, 2)
∞
0
(25)
(26)
with
R0 (s ) = ∫ tG0 (t ) J 1 (st )dt ,
c
ρ 2 (s)
,
ρ0 (s )
ρ (s)
R21 (s ) = − 1 ,
ρ0 (s)
0
R11 (s ) = −
⎫
⎪
⎪
⎪⎪
ρ (s )
R12 (s ) = 2 exp( −2sκ h1 ), ⎬
ρ0 ( s )
⎪
⎪
ρ1 (s )
exp( −2 sκ h2 ) ⎪
R22 (s ) =
ρ0 (s)
⎪⎭
(27)
On the plane z = 0 , the temperatures Ti(2) (r , 0) (i = 1, 2) are reduced to
Ti(2) (r , 0) =
( −1)i
2
∫r
c
⎧2
∞⎪
( −1)i ⎪⎫
G0 (t )dt + ∫ ⎨ ∑ Rij (s) −
⎬ R0 (s ) J 0 ( sr )ds (i = 1, 2)
0
2 ⎪⎭
⎪⎩ j = 1
(28)
4. Thermally induced elastic and electric fields
The non-disturbed temperature filed T (1) ( z) given by Eq.(17) does not induce the stress and
electric displacement components, which affect the singular field. Thus, we consider the
elastic and electric fields due to the disturbed temperature distribution Ti(2) ( r , z ) (i = 1, 2)
only. It is convenient to represent the solutions uzi (r , z ) , uri (r , z ) and φi (r , z ) (i = 1, 2) as the
sum of two functions, respectively.
(2)
⎫
uzi (r , z) = u(1)
zi ( r , z ) + uzi ( r , z ),
⎪
⎪
⎪
uri (r , z) = uri(1) (r , z) + uri(2) (r , z), ⎬ (i = 1, 2)
⎪
⎪
φi (r , z) = φi(1) ( r, z) + φi(2) (r , z) ⎪⎭
(29)
(1)
(1)
where u(1)
zi ( r , z ) , uri ( r , z ) , φi ( r , z ) ( i = 1, 2) are the particular solutions of Eqs.(4) replaced
(2)
(2)
Ti by Ti , and uzi (r , z) , uri(2) (r , z) , φi(2) (r , z) (i = 1, 2) are the general solutions of
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Application of Hankel Transform for Solving a Fracture
Problem of a Cracked Piezoelectric Strip Under Thermal Loading
213
homogeneous equations obtained by setting Ti = 0 (i = 1, 2) in Eqs.(4). In the following, the
superscripts (1) and (2) indicate the particular and general solutions of Eqs.(4).
Substituting Eqs.(29) into Eqs.(1) and (2), one obtains stress σ rri (r , z ) , σ θθ i (r , z ) , σ zzi (r , z ) ,
σ zri (r , z ) and electric displacement Dri (r , z) , Dzi (r , z) (i = 1, 2) expressions.
Using the displacement potential function method (Ueda, 2006a), the particular solutions
can be obtained as follows:
(1)
σ zzi
(r , z) = ∑ ∫ p1(1)
ij Rij ( s )R0 ( s ) J 0 ( sr )exp( sτ ij z )ds , ⎪
0
⎫
∞
2
⎪
⎪
(1)
σ zri
Rij (s )R0 (s ) J 1 (sr )exp(sτ ij z)ds , ⎪
(r, z) = ∑ ∫ p(1)
2
ij
0
⎪
j =1
⎪
2
⎪
∞ (1)
(1)
Dzi (r, z) = ∑ ∫ p3ij Rij ( s)R0 (s) J 0 ( sr )exp(sτ ij z)ds , ⎪
0
j =1
⎪
⎬ (i = 1, 2)
2
∞ 1 ( 1)
(1)
uzi ( r , z) = ∑ ∫
p4 ij Rij (s )R0 (s ) J 0 (sr )exp(sτ ij z)ds , ⎪⎪
0 s
j =1
⎪
2
⎪
∞ 1 (1)
(1)
uri (r , z) = ∑ ∫
p R (s )R0 (s ) J 1 (sr )exp(sτ ij z)ds , ⎪
0 s 5 ij ij
⎪
j =1
⎪
2
∞ 1
φi(1) (r , z) = − ∑ ∫ p6(1)ij Rij (s)R0 (s) J 0 (sr )exp(sτ ij z)ds ⎪⎪
0 s
j =1
⎭
j =1
∞
2
(30)
where the constants p(1)
kij ( i, j = 1, 2, k = 1, 2,..., 6) are given in Appendix A. The general
solutions are obtained by using the Hankel transform technique (Sneddon & Lowengrub,
1969):
(2)
σ zzi
(r , z) = ∑ ∫ sp1(2)
ij Aij ( s ) J 0 ( sr )exp( sγ ij z )ds , ⎪
6
∞
⎫
⎪
⎪
∞
(2)
σ zri
Aij (s ) J 1 (sr )exp(sγ ij z )ds , ⎪
( r , z) = ∑ ∫ sp2(2)
ij
0
⎪
j =1
⎪
6
⎪
∞
(2)
(2)
Dzi (r , z) = ∑ ∫ sp3 ij Aij (s ) J 0 (sr )exp(sγ ij z )ds , ⎪
0
j =1
⎪
⎬ (i = 1, 2)
6
∞
(2)
⎪
u(2)
zi ( r , z ) = ∑ ∫0 p4 ij Aij ( s ) J 0 ( sr ) exp( sγ ij z )ds , ⎪
j =1
⎪
6
⎪
∞ (2)
(2)
uri (r , z) = ∑ ∫ p5ij Aij (s ) J 1 (sr )exp(sγ ij z )ds , ⎪
0
⎪
j =1
⎪
6
∞
Aij (s ) J 0 ( sr )exp(sγ ij z )ds ⎪
φi(2) (r, z) = − ∑ ∫ p6(2)
ij
⎪
0
j =1
⎭
j =1
0
6
(31)
where Aij (s ) (i = 1, 2, j = 1, 2,..., 6) are the unknown functions to be solved, and the constants
γ ij and p(2)
kij ( i = 1, 2, j, k = 1, 2,..., 6) are given in Appendix B.
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Fourier Transform – Materials Analysis
Similar to the temperature analysis, the problem may be reduced to a system of singular
integral equations by taking the second boundary conditions (7)-(9) into consideration and
by defining the following new unknown functions Gl (r ) ( l = 1, 2, 3) :
{
}
⎧ ∂ (2)
u (r , 0) − u(2)
⎪
z 2 ( r , 0)
G1 (r ) = ⎨ ∂r z1
0
⎩⎪
{
}
{
}
⎫
(0 ≤ r < c ) ⎪
⎬
(c ≤ r < ∞ )⎭⎪
⎧ ∂ ⎡ 1 (2)
⎫
⎤
u (r , 0) − ur(2)
⎪r
2 ( r , 0) ⎥ (0 ≤ r < c ) ⎪
G2 (r ) = ⎨ ∂r ⎣⎢ r r 1
⎦
⎬
⎪
⎪
0
(
)
≤
<
∞
c
r
⎩
⎭
⎧ ∂ (2)
φ (r , 0) − φ2(2) ( r, 0)
⎪−
G3 (r ) = ⎨ ∂r 1
0
⎩⎪
⎫
(0 ≤ r < c ) ⎪
⎬
(c ≤ r < ∞ )⎭⎪
(32)
(33)
(34)
Making use of the first boundary conditions (7)-(9) with Eqs.(10), we have the following
system of singular integral equations for the determination of the unknown functions
Gl (t ) (l = 1, 2, 3) :
∫0 t[{Z11 M0
c
+
∞
{
(1)
}
(t, r )} G (t )]dt = σ
(t, r ) + M11 (t, r ) G1 (t ) + M12 (t, r )G2 (t ) +
∞
Z13
M0(1) (t, r ) +
M13
zz 0 ( r )
(1)
(t, r ) + M 22 (t, r ) G2 (t ) +
∫0 t[ M21 (t, r )G1 (t ) + {Z22 M1
∞
c
+ M23 (t, r )G3 (t )]dt = σ zr 0 ( r ) (0 ≤ r < c )
∫0 t[{Z31 M0
c
+
{
∞
(1)
}
(t, r )} G (t )]dt = D
}
(t, r ) + M 31 (t, r ) G1 (t ) + M 32 (t, r )G2 (t ) +
∞
Z33
M0(1) (t, r ) +
where the kernel functions
(0 ≤ r < c )
3
M1(1) (t, r ) ,
M 33
z 0 ( r ) (0 ≤ r < c )
3
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{Z
{Z
{Z
{Z
(36)
(37)
M kl (t, r ) and the constants Zkl∞ ( k, l = 1, 2, 3) are given by
2
⎧ 4 ⎧ ⎛r⎞
⎫
⎛ r ⎞ ⎫ 2 ⎪⎧ t
⎛r⎞
⎛ r ⎞ ⎪⎫
− K ⎜ ⎟ ⎬ (r < t ), ⎪
E
⎪
⎨K ⎜ ⎟ − E ⎜ ⎟ ⎬ +
⎨ 2
2 ⎜ ⎟
⎝ t ⎠ ⎭ π rt ⎩⎪ t − r ⎝ t ⎠
⎝ t ⎠ ⎭⎪
⎪ π rt ⎩ ⎝ t ⎠
⎪
M1(1) (t, r ) = ⎨
⎬
4 ⎧ ⎛t⎞
⎪
⎛ t ⎞⎫ 2 1
⎛t⎞
⎪
−
+
>
(
)
K
E
E
r
t
⎜ ⎟⎬
⎪
⎪
2 ⎨ ⎜ ⎟
2
2 ⎜ ⎟
π
r
r
r
π
−
t
t
r
⎝
⎠
⎝
⎠
⎝
⎠
⎩
⎭
⎩
⎭
⎧ ∞s
⎪ ∫0
⎪ ∞
⎪⎪ ∫ s
M kl (t, r ) = ⎨ 0
∞
⎪∫ s
⎪ 0
⎪ ∞
⎪⎩ ∫0 s
(35)
∞
kl ( s ) − Zkl
∞
kl ( s ) − Zkl
∞
kl ( s ) − Zkl
∞
kl ( s ) − Zkl
} J (sr )J (st )ds
} J (sr )J (st )ds
} J (sr )J (st )ds
} J (sr )J (st )ds
0
1
0
2
1
1
1
2
( k = 1, 3, l = 1, 3),⎫
⎪
⎪
( k = 1, 3, l = 2), ⎪⎪
⎬
( k = 2, l = 1, 3), ⎪
⎪
⎪
( k = 2, l = 2) ⎪
⎭
(38)
(39)
Application of Hankel Transform for Solving a Fracture
Problem of a Cracked Piezoelectric Strip Under Thermal Loading
215
∞
Zkl (s ) = ∑ p(2)
k 1 j d1 jl ( s ), Zkl = lim Zkl ( s ) ( k, l = 1, 2, 3)
6
s →∞
j =1
(40)
In Eq.(40), the functions d1 jl (s ) ( j = 1, 2,..., 6, l = 1, 2, 3) are given in Appendix C.
The functions σ zz 0 (r ) , σ zr 0 (r ) and Dz0 (r ) , which correspond to the stress and electric
displacement components induced by the disturbed temperature field Ti(2) ( r , z) (i = 1, 2) on
the r -axis in the plate without crack, are obtained as follows:
⎧⎪
⎫⎪
⎫
(2) T
(1)
σ zz0 (r ) = ∫ R0 (s ) ⎨ ∑ p11
j d1 j ( s ) + ∑ p11 j R1 j ( s )⎬ J 0 ( sr )ds , ⎪
∞
6
2
⎪
j =1
⎩⎪ j = 1
⎭⎪
⎪
2
⎧⎪ 6 (2) T
⎫⎪
⎪⎪
∞
(1)
σ zr 0 (r ) = ∫ R0 ( s) ⎨ ∑ p21 j d1 j (s) + ∑ p21 j R1 j (s)⎬ J 1 (sr )ds , ⎬
0
j =1
⎪
⎩⎪ j = 1
⎭⎪
⎪
6
2
⎧
⎫
∞
⎪
⎪
⎪
(2) T
(1)
Dz 0 (r ) = ∫ R0 (s ) ⎨ ∑ p31 j d1 j ( s ) − ∑ p31 j R1 j (s )⎬ J 0 (sr )ds ⎪
0
j =1
⎪⎭
⎩⎪ j = 1
⎭⎪
0
(41)
where the functions d1Tj (s ) ( j = 1, 2,..., 6) are also given in Appendix C. These components
are superficial quantities and have no physical meaning in this analysis. However, they are
equivalent to the crack face tractions in solving the crack problem by a proper
superposition.
To solve the singular integral equations (21) and (35)- (37) by using the Gauss-Jacobi
integration formula (Sih, 1972), we introduce the following functions Φ l (t ) (l = 0, 1, 2, 3) :
⎛c+t⎞
Gl (t ) = ⎜
⎟ Φ l (t ) ( l = 0, 1, 2, 3)
⎝c −t ⎠
1/ 2
(42)
Then the stress intensity factors K I , K II and the electric displacement intensity factor K D
may be defined and evaluated as:
{
}
{
}
⎫
∞
∞
Φ 1 (c ) + Z13
Φ 3 (c ) , ⎪
K I = lim+ {2π (c − r )} 1 / 2 σ zz1 (r , 0) = (π c )1 / 2 Z11
r →c
⎪⎪
∞
Φ 2 (c ),
K II = lim+ {2π (c − r )} 1 / 2 σ zr 1 (r , 0) = (π c )1 / 2 Z22
⎬
r →c
⎪
∞
∞
Φ 1 (c ) + Z33
Φ 3 (c ) ⎪
K D = lim+ {2π (c − r )} 1 / 2 Dz1 (r , 0) = (π c )1 / 2 Z31
⎪⎭
r →c
(43)
5. Numerical results and discussion
For the numerical calculations, the thermo-electro-elastic properties of the plate are assumed
to be ones of cadmium selenide with the following properties (Ashida & Tauchert, 1998).
The values of the coefficients of heat conduction for cadmium selenide could not be found in
the literature. Since the values of them for orthotropic Alumina (Al2O3) are
κ r = 21.25[W/mK] and κ z = 29.82[W/mK] (Dag, 2006), the value κ 2 = κ r / κ z = 1 / 1.5 is
assumed. To examine the effects of the normalized crack size c / h and the normalized crack
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216
Fourier Transform – Materials Analysis
location h1 / h on the stress and electric displacement intensity factors, the solutions of the
system of the singular integral equations have been computed numerically.
c12 = 45.2 × 109 [N / m 2 ], ⎫
⎪
c13 = 39.3 × 10 [N / m 2 ],
c 33 = 83.6 × 10 9 [N / m 2 ], ⎪
⎪
c 44 = 13.2 × 109 [N / m 2 ],
⎪
⎪
e31 = −0.16[C / m 2 ],
e33 = 0.347[ C / m 2 ],
⎪
⎬
2
e15 = −0.138[C / m ],
⎪
ε 11 = 82.6 × 10 −12 [C / Vm], ε 33 = 90.3 × 10 −12 [C / Vm], ⎪
⎪
λ11 = 0.621 × 10 6 [N / Km 2 ], λ33 = 0.551 × 106 [ N / Km 2 ], ⎪
⎪
pz = −2.94 × 10 −6 [CK −1m −2 ].
⎭⎪
c11 = 74.1 × 109 [N / m 2 ],
9
(44)
In the first set of calculations, we consider the temperature field and the electro-elastic fields
without crack. Figure 2 shows the normalized temperature (Ti ( x ) − T20 ) / T0 (i = 1, 2) on the
crack faces (0 ≤ r < c, z → 0 ± ) and the crack extended line (c ≤ r ≤ 2c, z = 0) for h1 / h = 0.25
and c / h = 0.5 , where T0 = T10 − T20 . The maximum local temperature difference across the
crack occurs at the center of the crack.
Figure 3 exhibits the normalized stress components (σ zz0 (r ),σ zr 0 (r )) / λ33T0 and the electric
displacement component Dz 0 (r ) / pzT0 on the r -axis in the strip without crack due to the
temperature shown in Figure 2. The maximum absolute values of σ zz 0 (r ) and Dz 0 (r ) occur
at the center of the crack (r / c = 0.0) , whereas the maximum value of σ zr 0 (r ) occurs at the
crack tip (r / c = 1.0) .
1
(Ti-T20)/T0
0.8
0.6
h1/h=0.25
c/h=0.5
0.4
: i=1 (z→0+)
-
: i=2 (z→0 )
0.2
0
1
r/c
2
Fig. 2. The temperature on the crack faces and the crack extended line for c / h = 0.5 and
h1 / h = 0.25
In the second set of calculations, we study the influence of the crack size on the stress and
electric displacement intensity factors. Figures 4(a)-(c) show the plots of the normalized
stress and electric displacement intensity factors (K I , K II ) / λ33T0 (π c )1 / 2 , K D / pzT0 (π c )1 / 2
versus c / h for h1 / h = 0.25 , 0.5 and 0.75. Because of symmetry, the values of K I and K D
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Application of Hankel Transform for Solving a Fracture
Problem of a Cracked Piezoelectric Strip Under Thermal Loading
217
0.01
h1/h=0.25
Dz0
0
(σzz0, σzr0)/λ33T0
c/h=0.5
0.04
-0.01
-0.02
σzr0
0.02
-0.03
σzz0
0
Dz0/pzT0
0.06
-0.04
0
1
r/c
2
Fig. 3. The stress components σ zz0 , σ zr 0 and the electric displacement component Dz0 on the
r -axis without crack due to the temperature shown in Fig. 2
0.02
0.04
(a)
(b)
h1 /h=0.50
0.03
1/2
0.01
0
KII/λ 33T0 (πc)
K I/λ 33T0 (πc)
1/2
h1 /h=0.25
h1 /h=0.5
-0.01
h1 /h=0.25, 0.75
0.02
0.01
h1 /h=0.75
-0.02
0
0
2
1
c/h
0.02
1
c/h
0
(c)
h1/h=0.75
KD/pZ T0 (πc)
1/2
0.01
h1/h=0.5
0
-0.01
h1/h=0.25
-0.02
0
1
c/h
2
Fig. 4. (a) The effect of the crack size on the stress intensity factor K I . (b) The effect of the
crack size on the stress intensity factor K II . (c) The effect of the crack size on the electric
displacement intensity factor K D
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2
218
for
h1 / h = 0.5
[KD ]h
1 /h = 0.25
are zero, and
= − [ K D ]h
[KI ]h /h =0.25
= − [ K I ]h
Fourier Transform – Materials Analysis
1 /h = 0.75
[KII ]h /h =0.25 = [KII ]h /h =0.75
,
. The absolute value of K I / λ33T0 (π c )1 / 2 for h1 / h = 0.25 and
/h = 0.75
1
1
1
0.75 monotonically increases with increasing c / h , but the value of K II / λ33T0 (π c )1 / 2 and the
1
absolute value of K D / pzT0 (π c )1 / 2 increase at first, reach maximum values and then decrease
with increasing c / h . The value of K I for h1 / h = 0.75 becomes negative so that the contact
of the crack faces would occur. The results presented here without considering this effect
may not be exact but would be more conservative. Since the contact of the crack faces will
increase the friction between the faces and make thermo-electrical transfer across the crack
faces easier, the stress and electric displacement intensity factors would be lowered by these
two factors.
In the final set of calculations, we investigate the influence of the crack location on the
intensity factors. Figure 5 indicates the effect of the crack location on K I , K II and K D for
c / h = 0.5 . As h1 / h increases, the values of K I and K D tend to decrease or increase
monotonically. The value of K II / λ33T0 (π c )1 / 2 decreases if the crack approaches the free
boundaries ( h1 / h → 0.0 or 1.0) , and the peak value of K II / λ33T0 (π c )1 / 2 =0.0277 occurs at
h1 / h = 0.5 .
0.03
0.02
0.01
1/2
0.02
KD
0.01
0
0
KI
-0.01
KD/pzT0(πc)
(KI, KII)/λ33T0(πc)
1/2
KII
-0.01
c/h=0.5
0
0.2
0.4
0.6
0.8
1
-0.02
h1/h
Fig. 5 The effect of the crack location on the stress intensity factors K I , K II and the electric
displacement intensity factor K D
6. Conclusion
An example of the application of Hankel transform for solving a mixed-mode thermoelectro-elastic fracture problem of a piezoelectric material strip with a parallel penny-shaped
crack is explained. The effects of the crack size (c / h ) and the crack location ( h1 / h ) on the
fracture behavior are analyzed. The following facts can be found from the numerical results.
1.
2.
The large shear stress occurs in the strip without crack due to the disturbed temperature
field.
The normalized intensity factors are under the great influence of the geometric
parameters h1 / h and c / h .
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Application of Hankel Transform for Solving a Fracture
Problem of a Cracked Piezoelectric Strip Under Thermal Loading
3.
4.
219
For the case of h1 / h > 0.5 , mode I stress intensity factor becomes negative so that the
contact of the crack faces would occur.
The intensity factors of crack near the free surfaces due to the thermal load are not so
large.
Appendix A
The constants p(1)
kij ( i, j = 1, 2, k = 1, 2,..., 6) are
(
)
)
)
2
⎫
p1(1)
ij = c 13 − c 33 kijτ ij C ij − e33τ ij N ij − λ33 ,
⎪
⎪
1
,
p(1)
c
k
C
e
N
τ
=
+
+
44
15 ij
ij ij ij
2 ij
⎪
⎬ (i, j = 1, 2)
(1)
2
p3ij = e31 − e33 kijτ ij C ij + ε 33τ ij N ij + pz , ⎪
⎪
(1)
(1)
⎪
p(1)
4 ij = − kijτ ijC ij , p5 ij = C ij , p6 ij = N ij
⎭
(
(
(A.1)
where
⎫
⎪
,
mij(1) + kij mij(2) ⎪
⎪
nij(11) + kij nij(12) ⎪⎪
N ij = (1)
, ⎬ (i, j = 1, 2)
mij + kij mij(2) ⎪
⎪
τ ij2 nij(11) + nij(21) ⎪
kij = − 2 (12)
⎪
τ ij nij + nij(22) ⎪⎭
C ij =
b′21 − b′22τ ij2
(A.2)
with
⎫
⎪
⎪
⎪
⎪
( H 1r + H 3r ) b′21 − b′22τ ij2 + H 4 r mij(1) τ ij , ⎪
⎪
⎬ (i , j = 1, 2)
( H 1 z + H 3 z ) b′21 − b′22τ ij2 + H 4 zmij(1) τ ij , ⎪
⎪
H 1r − H 2 rτ ij2 b′21 − b′22τ ij2 + H 4 r mij(2) τ ij , ⎪
⎪
⎪
(2)
2
2
H 1 z − H 2 zτ ij b′21 − b′22τ ij + H 4 zmij τ ij ⎪
⎭
mij(1) = a′41 − a′42τ ij2 ,
mij(2)
nij(11)
nij(21)
nij(12)
nij(22)
=
(
2
a′44τ ij
{
={
= {(
= {(
=
)
(
(
− a′43 τ ij2 ,
)(
)(
)
)
)
)
}
}
}
}
(A.3)
c44ε 33 + e15 e33
c ε + e2
c ε + e2
c ε +e e ⎫
, H1z = 44 11 15 , H2 r = 33 33 33 , H2 z = 33 11 15 33 , ⎪
e33ε 11 − e15ε 33
e15ε 33 − e33ε 11
e33ε 11 − e15ε 33
e15ε 33 − e33ε 11 ⎪
⎬ (A.4)
p e −λ ε
p e −λ ε
c ε +e e
c ε +e e
H3r = 13 33 31 33 , H3z = 13 11 15 31 , H4r = z 33 33 33 , H4 z = z 15 33 11 ⎪
e33ε 11 − e15ε 33
e15ε 33 − e33ε 11 ⎪⎭
e33ε 11 − e15ε 33
e15ε 33 − e33ε 11
H1r =
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220
Fourier Transform – Materials Analysis
a′41 = c11 ,
a′42 = c 44 + ( e15 + e31 )( H 1r + H 3r ), ⎫
⎪
a′43 = c13 + c 44 + ( e15 + e31 )H 1r , a′44 = ( e15 + e31 )H 2 r ,
⎬
⎪
b′21 = λ11 ,
b′22 = −( e15 + e31 )H 4 r
⎭
(A.5)
Appendix B
The constants γ ij (i = 1, 2, j = 1, 2,..., 6) are the roots of the following characteristic equations:
( f 4 g′2 + g4 f 2′ ) γ ij6 + ( f 4 g′0 + f 2 g′2 + g4 f 0′ + g2 f 2′ ) γ ij4 +
+ ( f 2 g′0 + f 0 g′2 + g2 f 0′ + g0 f 2′ ) γ j2 + ( f 0 g′0 + g0 f 0′ ) = 0
(i = 1, 2, j = 1, 2,..., 6)
(B.1)
where ℜ[γ 1 j ] < ℜ[γ 1 j + 1 ] , ℜ[γ 2 j ] > ℜ[γ 2 j + 1 ]( j = 1, 2,..., 5) and
f4
f2
f0
f 2′
f 0′
= c 44 e33 ,
⎫
= (c13 + c 44 )( e15 + e31 ) − c11 e33 − c 44 e15 , ⎪⎪
⎪
= c11 e15 ,
⎬
⎪
= c 33 ( e15 + e31 ) − e33 (c13 + c 44 ),
⎪
= −c 44 ( e15 + e31 ) + e15 (c13 + c 44 )
⎪⎭
(B.2)
g 4 = c 44ε 33 ,
g2
g0
g′2
g′0
⎫
⎪
= −( e15 + e31 )2 − c11ε 33 − c 44ε 11 , ⎪
⎪
= c11ε 11 ,
⎬
= e33 ( e15 + e31 ) + ε 33 (c13 + c 44 ), ⎪⎪
= − e15 ( e15 + e31 ) − ε 11 (c13 + c 44 ) ⎪⎭
The functions p(2)
kij ( s ) ( i = 1, 2, j, k = 1, 2,..., 6) are
(
)
(B.3)
⎫
p1(2)
ij = c 13 aij + γ ij c 33 − e33bij ,
⎪
⎪
γ
=
−
+
1
,
p(2)
c
a
e
b
44
15 ij
ij ij
2 ij
⎪
⎪
p(2)
3 ij = e31 aij + γ ij e310 + ε 330 bij , ⎪
( i = 1, 2, j = 1, 2,..., 6)
⎬
⎪
p(2)
4 ij = 1,
⎪
⎪
p(2)
5 ij = aij ,
⎪
(2)
⎪
p6 ij = bij
⎭
(B.4)
⎫
⎪
+
+ g0
⎪
⎬ (i = 1, 2, j = 1, 2,..., 6)
2
(c 44γ ij − c11 )aij − c13 − c 44 ⎪
bij = −
⎪
e15 + e31
⎭
(B.5)
(
(
)
)
where aij and bij ( i = 1, 2, j = 1, 2,..., 6) are given by
aij =
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g′2γ ij2 + g′0
g 4γ ij4
g2γ ij2
,
Application of Hankel Transform for Solving a Fracture
Problem of a Cracked Piezoelectric Strip Under Thermal Loading
221
Appendix C
The functions dijk (s ) (i = 1, 2, j = 1, 2,..., 6, k = 1, 2, 3) are given by
d1 jk (s ) = q j, k + 9 ( s ), ⎫
⎪
⎬ ( j = 1, 2,..., 6, k = 1, 2, 3)
d2 jk (s ) = q j + 6, k + 9 (s )⎪⎭
(C.1)
( j = 1, 2, 3), ⎫
⎪
( j = 1, 2, 3), ⎪⎪
δ j + 3, k + 6 (s ) = p(2)
j 2 k exp( −sγ 2 k h2 )
⎬ ( k = 1, 2,..., 6)
( j = 1, 2,..., 6), ⎪
δ j + 6, k (s ) = p(2)
j 1k
⎪
( j = 1, 2,..., 6) ⎪⎭
δ j + 6, k + 6 (s ) = − p(2)
j2k
(C.2)
where the functions q j, k (s ) ( j, k = 1, 2,..., 12) are the elements of a square matrix Q = Δ −1 of
order 12. The elements δ j , k (s ) ( j, k = 1, 2,..., 12) of the square matrix Δ are given by
δ j , k ( s) = p(2)
j 1 k exp( sγ 1 k h1 )
The functions dijT (s ) (i = 1, 2, j = 1, 2,..., 6) are
12
⎫
d1Tj ( s ) = ∑ q j, k (s )uk (s ), ⎪
⎪
k =1
⎬ ( j = 1, 2,..., 6)
12
d2T j ( s ) = ∑ q j + 6, k (s )uk (s )⎪⎪
k =1
⎭
(C.3)
where
uk ( s ) = −
uk + 3 ( s ) = −
uk + 6 ( s ) = −
R0 (s ) 2 (1)
∑ p R1 j (s)exp(sτ 1 j h1 )
s j =1 k 1 j
R0 (s ) 2 (1)
∑ p R2 j (s)exp(−sτ 2 j h2 )
s j =1 k 2 j
{
}
R0 (s ) 2
∑ p(1) R1 j (s) − p(1)
k 2 j R2 j ( s )
s j =1 k 1 j
⎫
( k = 1, 2, 3), ⎪
⎪
⎪
⎪
( k = 1, 2, 3), ⎬
⎪
⎪
( k = 1, 2,..., 6)⎪
⎪
⎭
(C.4)
7. References
Ashida, F. & Tauchert, T.R. (1998). Transient Response of a Piezothermoelastic Circular Disk
under Axisymmetric Heating. Acta Mechanica, Vol. 128, pp. 1-14, 0001-5970
Dag, S., Ilhan, K.A. & Erdogan, F. (2006), Mixed-Mode Stress Intensity Factors for an
Embedded Crack in an Orthotropic FGM Coating, Proceedings of the International
Conference FGM IX, 978-0-7354-0492-2, Oahu Island, Hawaii, October 2006
Erdogan, F. & Wu, B.H. (1996). Crack Problems in FGM Layers under Thermal Stresses.
Journal of Thermal Stresses, Vol. 19, pp. 237-265, 0149-5739
www.intechopen.com
222
Fourier Transform – Materials Analysis
Rao, S.S. & Sunar, M. (1994). Piezoelectricity and Its Use in Disturbance Sensing and Control
of Flexible Structures: a Survey. Applied Mechanics Review, Vol. 47, pp. 113-123,
0003-6900
Sih, G.C. (Ed.). (1972). Methods of Analysis and Solution of Crack Problems, Noordhoff,
International Publishing, 978-9048182466, Leyden
Sneddon, I.N. & Lowengrub, M. (1969). Crack Problems in the Classical Theory of Elasticity,
John Wiley & Sons, Inc., 978-0471808459, New York
Tauchert, T.R. (1992). Piezothermoelastic Behavior of a Laminated Plate. Journal of Thermal
Stresses, Vol. 15, pp. 25-37, 0149-5739
Ueda, S. (2006a). The Crack Problem in Piezoelectric Strip under Thermoelectric Loading.
Journal of Thermal Stresses, Vol. 29, pp. 295-316, 0149-5739
Ueda, S. (2006b). Thermal Stress Intensity Factors for a Normal Crack in a Piezoelectric Strip.
Journal of Thermal Stresses, Vol. 29, pp. 1107-1126, 0149-5739
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Fourier Transform - Materials Analysis
Edited by Dr Salih Salih
ISBN 978-953-51-0594-7
Hard cover, 260 pages
Publisher InTech
Published online 23, May, 2012
Published in print edition May, 2012
The field of material analysis has seen explosive growth during the past decades. Almost all the textbooks on
materials analysis have a section devoted to the Fourier transform theory. For this reason, the book focuses
on the material analysis based on Fourier transform theory. The book chapters are related to FTIR and the
other methods used for analyzing different types of materials. It is hoped that this book will provide the
background, reference and incentive to encourage further research and results in this area as well as provide
tools for practical applications. It provides an applications-oriented approach to materials analysis written
primarily for physicist, Chemists, Agriculturalists, Electrical Engineers, Mechanical Engineers, Signal
Processing Engineers, and the Academic Researchers and for the Graduate Students who will also find it
useful as a reference for their research activities.
How to reference
In order to correctly reference this scholarly work, feel free to copy and paste the following:
Sei Ueda (2012). Application of Hankel Transform for Solving a Fracture Problem of a Cracked Piezoelectric
Strip Under Thermal Loading, Fourier Transform - Materials Analysis, Dr Salih Salih (Ed.), ISBN: 978-953-510594-7, InTech, Available from: http://www.intechopen.com/books/fourier-transform-materialsanalysis/application-of-hankel-transform-for-solving-a-fracture-problem-of-a-cracked-piezoelectric-strip-unde
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InTech China
Unit 405, Office Block, Hotel Equatorial Shanghai
No.65, Yan An Road (West), Shanghai, 200040, China
Phone: +86-21-62489820
Fax: +86-21-62489821