# Closed-Form Solutions for a Mode-III Moving Interface Crack at the

Authors: B. M. Singh; J. Rokne; R. S. Dhaliwal
Journal: Mathematical Problems in Engineering
 (2008)

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## Abstract

An integral transform technique is used to solve the elastodynamic problem of a crack of fixed length propagating at a constant speed at the interface of two bonded dissimilar orthotropic layers of equal thickness. Two cases of practical importance are investigated. Firstly, the lateral boundaries of the layers are clamped and displaced in
equal and opposite directions to produce antiplane shear resulting in a tearing motion along the leading edge of
the crack, and secondly, the lateral boundaries of the layers are subjected to shear stresses. The analytic solution
for a semi-infinite crack at the interface of two bonded dissimilar orthotropic layers has been derived. Closed-form
expressions are obtained for stressing the intensity factor and other physical quantities in all cases.


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## Body

## 1. Introduction

First of all,
Sih and Chen [1]
solved the problem of a Griffith crack in an orthotropic layer under antiplane
shear. They reduced the solution to a Fredholm integral equation of the second
kind, and values for stress intensity factors were obtained by solving the
Fredholm integral equation numerically. Singh et al. [2], and Tait and Moodie
[3] obtained
closed-form solutions for a finite length crack moving with constant velocity
in a strip. In [2], an
integral transform method was used, while in [3] complex variable techniques
were used. Closed-form solutions for a finite length crack moving at constant
velocity in an orthotropic layer of finite thickness
were obtained by Danyluk and Singh [4], and that work was an
extension of the work discussed in [1–3]. Making use of complex variable methods, Georgiadis
[5] solved the problem
of a cracked orthotropic strip, and the problem of steadily moving crack in an
orthotropic material under antiplane shear stress has been studied by Piva
[6, 7]. Recently, Li [8] obtained a closed-form
solution for a mode-III interface crack between two bonded dissimilar elastic
layers. This paper is concerned with a mode-III moving crack interface between
two bonded orthotropic dissimilar elastic layers whose closed-form solution has
been obtained. Furthermore, the exact results for a semi-infinite interface
crack in two bonded elastic orthotropic elastic layers have been obtained
directly from those of a finite length crack results through a limiting
process.The results of this paper are more general than those
of the paper of Li [8]. If we assume that the velocity of the moving crack is
zero and assuming the constants c44(j)=c55(j)=μ(j)(j=1,2), we get the results of the paper
of Li [8], where μ(j) are the shear moduli of the upper
and lower layers and c44(j),c55(j) are defined in the paper.The standard method for solving mixed boundary value
problems is to reduce the solution into Fredholm integral equation of the
second kind, where approximate solutions can be found. The aim of this paper is
to obtain closed forms or exact solutions of the problems.


## 2. Basic Equations and Formulation of the Crack Problem

Consider two elastic layers of equal thicknessh occupying the region −∞<X<∞, −h<Y<h, −∞<Z<∞,
where 0XYZ is a fixed rectangular coordinate system. We
assume that a crack of length 2a is moving at a constant velocity v in the X-direction at the interface of the two layers
as shown in Figure 1. The purpose of this investigation is to determine the
effect of orthotropy of the materials on the initial direction of propagation
of the crack which is moving with constant speed. Assuming that there is a
single nonvanishing displacement component in the Z-direction, we have(2.1)Uj=Vj=0,Wj=Wj(X,Y,t),
where Uj,Vj,Wj are displacement components in the X,Y,
and Z directions, respectively, and j=1,2.
Then
(2.2)σX(j)=σY(j)=σZ(j)=σXY(j)=0,σXZ(j)=c55(j)∂Wj∂X,σYZ(j)=c44(j)∂Wj∂Y,
where c44(j) and c55(j) are the shear moduli in YZ and XZ planes, respectively, for both materials. The
equation of motion for both layers is
(2.3)∂2Wj∂X2+∂2Wj∂Yj2=1Cj2∂2Wj∂t2,where(2.4)Yj=Yβj,βj=c44(j)c55(j),Cj=c44(j)ρj,and Cj is the shear wave speed and ρj is the constant density of the material. For a
crack moving with constant speed in the X-direction, it is convenient to introduce the
Galilean transformation(2.5)x=X−vt,yj=Yj,z=Z.
Equation (2.3) now becomes
(2.6)sj2∂2Wj∂x2+∂2Wj∂yj2=0,
where
(2.7)sj2=1−v2Cj2.Figure 1
Two bonded layers of piezoelectric ceramic materials with crack at interface.


## 3. Solution of Equilibrium Equation

The solution of
the equilibrium (2.6) may be written as(3.1)Wj(x,y)=Fc[Ajexp(−ξyjsj)+Bjexp(ξyjsj);ξ→x],
where sj is taken as (2.7) and
(3.2)Fc[Aj(ξ);ξ→x]=2π∫0∞Aj(ξ)cos(ξx)dξ,Fs[Bj(ξ);ξ→x]=2π∫0∞Bj(ξ)sin(ξx)dξ.
The nonzero stresses are given by
(3.3)σyz(j)(x,y)=sjc44(j)βjFc[ξBjexp(ξyjsj)−ξAjexp(−ξyjsj);ξ→x],(3.4)σxz(j)(x,y)=−c55(j)Fs[ξBjexp(ξyjsj)+ξAjexp(−ξyjs(j));ξ→x].Now, we consider the two basic problems involving
shear stress loading and displacement conditions on the surface of the layer.Problem A.
Let the antiplane shear stress be applied to the
surfacesY=±h, then the equivalent problem in this instance involves the application of a
shear stress −p(x) to the crack surfaces at Y=0. The boundary conditions can then be written as(3.5)σyz(1)(x,0+)=−p(x),|x|<a,(3.6)σyz(1)(x,0+)=σyz(2)(x,0−),−∞<x<∞,(3.7)W1(x,0+)=W2(x,0−),|x|>a,(3.8)σyz(1)(x,+h)=σyz(2)(x,−h)=0,  where p(x) is an even function. Problem A consists of
solving (2.6) together with (3.5)–(3.8).Problem B.
Let the lateral boundaries of the layerY=±h be rigidly clamped and displaced by equal
amount p(x) in opposite directions which
produce an antiplane shear motion in the Z-direction and while the crack moves in the
positive X-direction at a constant speed. In order to
use the integral transform technique, it is necessary to solve an alternative
but equivalent problem.
The equivalent conditions are(3.9)σyz(1)(x,0)=−c44(1)hp(x),|x|<a,(3.10)σyz(1)(x,0)=σyz(2)(x,0),−∞<x<∞,(3.11)W1(x,0)=W2(x,0),|x|>a,(3.12)W1(x,h)=W2(x,−h)=0.


## 4. Solution of Problem A

From (3.3) and (3.6), we find that(4.1)B1(ξ)=A1(ξ)exp(−2ξh1s1),B2(ξ)=A2(ξ)exp(2ξh2s2),
where
(4.2)h1=hβ1,h2=hβ2.From (3.1), (3.3), and (4.1), we find that
(4.3)W1(x,y)=2Fc[A1(ξ)e−s1h1ξcoshξs1(h1−y1);ξ→x],−∞<x<∞,0<y≤h,(4.4)W2(x,y)=2Fc[A2(ξ)es2h2ξcoshξs2(h2−y2);ξ→x],−∞<x<∞,−h≤y<0,(4.5)σyz(1)(x,y)=−2s1c44(1)β1Fc[ξA1(ξ)e−s1h1ξsinhξs1(h1−y1);ξ→x],−∞<x<∞,0<y≤h,(4.6)σyz(2)(x,y)=2s2c44(2)β2Fc[ξA2(ξ)es2h2ξsinhξs2(h2+y2);ξ→x],−∞<x<∞,−h≤y<0.From boundary condition (3.7), we find that
(4.7)A2=−[s1β2sinh(ξs1h1)c44(1)s2β1sinh(ξs2h2)c44(2)e−ξh1s1−ξh2s2]A1(ξ).From (4.3), (4.4), and (4.7), we find that
(4.8)W1(x,0+)−W2(x,0−)=2Fc[e−ξh1s1A1(ξ)sinh(ξs2h2){cosh(ξs1h1)sinh(ξs2h2)+Psinh(ξs1h1)cosh(ξs2h2)};ξ→x],where(4.9)P=s1β2s2β1c44(1)c44(2).
Now, the boundary conditions
(3.5) and (3.7) lead to the following dual integral equations:
(4.10)Fc[ξA1(ξ)e−ξh1s1sinh(ξs1h1);ξ→x]=β12s1c44(1)p(x),0<x<a,Fc[A1(ξ)e−ξh1s1sinh(ξs2h2){cosh(ξs1h1)sinh(ξs2h2)+Psinh(ξs1h1)cosh(ξs2h2)};ξ→x]=0,a<x<∞.
Closed-form solution of the dual integral equations (4.10) is difficult to
obtain and only approximate solution of these dual integral equations can be
obtained by changing them into a Fredholm integral equation of the second kind.
For obtaining closed form, we assume that
(4.11)s1β1=s2β2,
so that
(4.12)P=c44(1)c44(2).If we take(4.13)(1+P)e−ξh1s1cosh(ξs1h1)A1(ξ)=C(ξ),
then the dual integral equations (4.10) can be written in the form
(4.14)Fc[ξC(ξ)tanh(ξs1h1);ξ→x]=β12s1(c44(1)−1+c44(2)−1)p(x)=p1(x),0<x<a,Fs[C(ξ);ξ→x]=0,a<x<∞.
Following [4], the solution of the the
dual integral equations (4.14) can be written in the form
(4.15)C(ξ)=1ξπ2∫0aΦ(τ)sin(ξτ)dτ,where(4.16)Φ(τ)=−2csinh(2cτ)π2(sinh2(ca)−sinh2(cτ))1/2∫0ap1(x)(sinh2(ca)−sinh2(cx))1/2sinh2(cx)−sinh2(cτ)dx,0<τ<a,c=π2h1s. For the particular case p(x)=p0 when p0 is a constant, we find that
(4.17)Φ(τ)=β1((c44(1)−1+c44(2)−1)p0sinh(2cτ))s1π2cosh(ac)(sinh2(ca)−sinh2(cτ))1/2[F(π2,tanh(ac))−L(τ)],L(τ)=Π(π2,sinh2(ca)sinh2(ca)−sinh2(cτ)tanh(ac)),
where F and Π are elliptic integrals of the first and third
kinds, respectively, as defined in the table of
Gradshteyn and Ryzhik [9].The stress distribution along the crack is given
by(4.18)σyz(1)(x,y)=p0sinh(2cx)πcosh(ac)(sinh2(ca)−sinh2(cx))1/2[F(π2,tanh(ac))−L(τ)],a<x<∞,and the crack sliding
displacement is(4.19)ΔW(x)=W(1)(x,0+)−W(2)(x,0−)=β1((c44(1)−1+c44(2)−1)p0sinh(2cτ))πs1cosh(ac)(sinh2(ca)−sinh2(cx))1/2∫xa[F(π2,tanh(ac))−L(τ)]dτ,0<x<a.
The stress intensity factor can be written in the form
(4.20)K3=limx→a+[2(x−a)σyz(1)(x,0)]=2p0π[tanh(ca)c]1/2F[π2,tanh(ca)].Assuming that under applied loading the crack tip
advances along the crack plane fromx=a to a+δa(δa≪a), then the energy release rate per unit length during this process is given by
(4.21)GIII=limδa→012δa∫0δa[σyz(1)(r,0)ΔW(δa−r,0)]dr,
where r denotes the distance from the crack tip.From (4.21), we find that
(4.22)GIII=β1s(c44(1)−1+c44(2)−1)p02tanh(ca)πcF[π2,tanh(ca)]2.We find that the shear stress displacement and
intensity factors obtained above are in agreement with the corresponding results
in Danyluk and Singh [4].The energy release rate (4.22) is new, which has not
been obtained in [4].


## 5. Solution of Problem B

For this problem, we find that the boundary condition
(3.12) will be satisfied if(5.1)B1(ξ)=−A1(ξ)exp(−2ξh1s1),B2(ξ)=−A2(ξ)exp(−2ξh2s2).
Using (5.1), (3.1), and (3.3), we find that
(5.2)W1(x,y)=2Fc[e−ξh1s1A1(ξ)sinh(ξs1(h1−y1));ξ→x]0≤y≤h,x>0,(5.3)W2(x,y)=2Fc[eξh2s2A2(ξ)sinh(ξs2(h2+y2));ξ→x]−h≤y≤0,x>0,(5.4)σyz(1)(x,y)=−2s1c44(1)β1Fc[ξA1(ξ)e−s1h1ξcosh[ξs1(h1−y1)];ξ→x],0≤y≤h,x>0,(5.5)σyz(2)(x,y)=2s2c44(2)β2Fc[ξA2(ξ)es2h2ξcosh[ξs2(h2+y2)];ξ→x],−h≤y≤0,x>0. From (5.4) and (5.5) and the
boundary condition (3.10), we find that
(5.6)A1(ξ)s1c44(1)e−ξh1s1cosh(ξs1h1)β1=−s2c44(2)eξh2s2cosh(ξs2h2)β2A2(ξ).Using (4.11) and (5.6), the boundary conditions
(3.9) and (3.11) lead to
the dual integral equations
(5.7)Fc[ξC1(ξ)coth(s1h1ξ);ξ→x]=β12hs1(c44(1)−1+c44(2)−1)c44(1)p(x),0<x<a,Fc[C1(ξ);ξ→x]=0,a<x,
where
(5.8)(1+P)e−s1h1ξA1(ξ)sinh(ξs1h1)A1(ξ)=C1(ξ).Following [4], the solution of the dual integral equations (5.7) for p(x)=p0 can be written in the form(5.9)C1(ξ)=π2ξ−1∫0aΨ(u)sin(uξ)du,where(5.10)Ψ(u)=β1sin(cu)c44(1)(c44(1)−1+c44(2)−1)p0π(hs1)[sinh2(ca)−sinh2(cu)]1/2.We can easily find that(5.11)σyz(1)(x,0+)=−c44(1)p0h[1−sinh(cx)[sinh2(cx)−sinh2(ca)]1/2],x>a.
The stress intensity factor at x=a is given by
(5.12)KIII=limx→a+[2(x−a)σyz(1)(x,0)]=c44(1)hp0[tanh(ca)c]1/2,
and the crack sliding displacement is
(5.13)ΔW(x)=W(1)(x,0+)−W(2)(x,0−)=p0β1s1(1+c44(1)c44(2))[1−2πsin−1(cosh(cx)cosh(ca))].
The energy release rate per unit
length during the process is given by
(5.14)GIII=limδ→012δ∫0δσyz(1)(r)Δw(δ−r)dr=12s1h(c44(1)−1+c44(2)−1)p02c44(1)2tanh(ca).


## 6. Solution for a Semi-Infinite Interface Crack

For the case of clamped boundaries of the layers, the closed-form solution for a semi-infinite
interface crack is obtained by takingx=a+x1 in (5.11), (5.12), and (5.13) and then letting a→∞,
so that we have
(6.1)σyz(1)(x,0+)=σyz(1)(x,0−)=−−c44(1)p0h[1−11−e−(πx1/h)],x1>0,ΔW(x1)=p0β1s1(1+c44(1)c44(2))[1−2πsin−1(eπx1/2h)],x1<0,KIII=limx1→0[2x1σyz(1)(x1,0+)]=c44(1)p02hπ.The result for stress intensity factor for the special
case of a stationary crack in an infinite strip coincides with the
corresponding result obtained by Georgiadis [5] and Rice [10].


## 7. Conclusions

The closed-form solution provided in this paper
is of less importance due to condition (4.11).
Condition (4.11) can be written in the form
(7.1)ν2=c44(1)c44(2)[c55(1)c44(2)−c55(2)c44(1)][ρ1c55(1)c44(2)2−ρ2c55(2)c44(1)2].If the velocity of the crack is known for particular
values of the constantsc44(1),c55(1),ρ1, we can substitute the values of constants c44(2),c55(2),ρ2 so that (7.1) is satisfied. Due to this, the
solutions of (7.1) exist for some layered specified orthotropic materials.For the stationary crack, we assumeν=0. Then, from (7.1), we find that(7.2)c44(1)c55(1)=c44(2)c55(2).Equation (7.2) is very simple and hence we can easily
find the solutions for stationary crack problems for upper and lower layers of
orthotropic materials from (7.2), which has practical value. The solutions are
already obtained by Li [8] for stationary crack in isotropic elastic layers,
while due to condition (7.2), we can find solutions for stationary crack in
orthotropic layers.



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*Source: 930820-2009-02-03.xml*