Closed-Form Solutions for a Mode-III Moving Interface Crack at the Interface of Two Bonded Dissimilar Orthotropic Elastic Layers

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.

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 𝑐_44(𝑗)=𝑐_55(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 𝑐_44(𝑗),𝑐_55(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 thickness ℎ 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 have𝑈_𝑗=𝑉_𝑗=0,𝑊_𝑗=𝑊_𝑗(𝑋,𝑌,𝑡),(2.1) where 𝑈_𝑗,𝑉_𝑗,𝑊_𝑗 are displacement components in the 𝑋,𝑌, and 𝑍 directions, respectively, and 𝑗=1,2. Then 𝜎_𝑋(𝑗)=𝜎_𝑌(𝑗)=𝜎_𝑍(𝑗)=𝜎_{𝑋𝑌(𝑗)}=0,𝜎_{𝑋𝑍(𝑗)}=𝑐_55(𝑗)𝜕𝑊_𝑗𝜕𝑋,𝜎^{𝑌𝑍(𝑗)}=𝑐_44(𝑗)𝜕𝑊_𝑗𝜕𝑌,(2.2) where 𝑐_44(𝑗) and 𝑐_55(𝑗) are the shear moduli in 𝑌𝑍 and 𝑋𝑍 planes, respectively, for both materials. The equation of motion for both layers is 𝜕²𝑊_𝑗𝜕𝑋²+𝜕²𝑊_𝑗𝜕𝑌^2𝑗=1𝐶^2𝑗𝜕²𝑊_𝑗𝜕𝑡²,(2.3)where𝑌_𝑗=𝑌√𝛽_𝑗,𝛽_𝑗=𝑐^44(𝑗)𝑐_55(𝑗),𝐶_𝑗=𝑐_44(𝑗)𝜌_𝑗,(2.4)and 𝐶_𝑗 is the shear wave speed and 𝜌_𝑗 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 transformation𝑥=𝑋−𝑣𝑡,𝑦_𝑗=𝑌_𝑗,𝑧=𝑍.(2.5) Equation (2.3) now becomes 𝑠^2𝑗𝜕²𝑊_𝑗𝜕𝑥²+𝜕²𝑊_𝑗𝜕𝑦^2𝑗=0,(2.6) where 𝑠^2𝑗=1−𝑣²𝐶^2𝑗.(2.7)

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 as𝑊_𝑗(𝑥,𝑦)=𝐹_𝑐𝐴^𝑗exp−𝜉𝑦_𝑗𝑠_𝑗+𝐵^𝑗exp𝜉𝑦_𝑗𝑠_𝑗;𝜉→𝑥,(3.1) where 𝑠_𝑗 is taken as (2.7) and 𝐹_𝑐𝐴^𝑗(𝜉);𝜉→𝑥=2𝜋^∞0𝐹𝐴_𝑗(𝜉)cos(𝜉𝑥)𝑑𝜉,_𝑠𝐵^𝑗(𝜉);𝜉→𝑥=2𝜋^∞0𝐵_𝑗(𝜉)sin(𝜉𝑥)𝑑𝜉.(3.2) The nonzero stresses are given by 𝜎_{𝑦𝑧(𝑗)}(𝑥,𝑦)=𝑠^𝑗𝑐_44(𝑗)√𝛽_𝑗𝜎𝐹_𝑐𝜉𝐵^𝑗exp𝜉𝑦_𝑗𝑠_𝑗−𝜉𝐴^𝑗exp−𝜉𝑦_𝑗𝑠_𝑗;𝜉→𝑥,(3.3)_{𝑥𝑧(𝑗)}(𝑥,𝑦)=−𝑐_55(𝑗)𝐹_𝑠𝜉𝐵^𝑗exp𝜉𝑦_𝑗𝑠_𝑗+𝜉𝐴^𝑗exp−𝜉𝑦_𝑗𝑠_(𝑗);𝜉→𝑥.(3.4)

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 surfaces 𝑌=±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 as𝜎𝜎_𝑦𝑧(1)𝑥,0⁺=−𝑝(𝑥),||𝑥||<𝑎,(3.5)_𝑦𝑧(1)𝑊𝑥,0⁺=𝜎^𝑦𝑧₍₂₎𝑥,0⁻,−∞<𝑥<∞,(3.6)₁𝜎𝑥,0⁺=𝑊²𝑥,0⁻,||𝑥||>𝑎,(3.7)_𝑦𝑧(1)𝑥,+ℎ=𝜎^𝑦𝑧₍₂₎(𝑥,−ℎ)=0,(3.8)𝑝(𝑥)𝑌=±ℎ𝑝(𝑥) where 𝑍 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 layer 𝑋 be rigidly clamped and displaced by equal amount 𝜎_𝑦𝑧(1)(𝑥,0)=−𝑐⁴⁴⁽¹⁾𝜎ℎ𝑝(𝑥),||𝑥||<𝑎,(3.9)_𝑦𝑧(1)(𝑊𝑥,0)=𝜎_𝑦𝑧(2)(𝑥,0),−∞<𝑥<∞,(3.10)₁𝑊(𝑥,0)=𝑊₂(𝑥,0),||𝑥||>𝑎,(3.11)₁(𝑥,ℎ)=𝑊₂(𝑥,−ℎ)=0.(3.12) in opposite directions which produce an antiplane shear motion in the 𝐵₁(𝜉)=𝐴₁(𝜉)exp−2𝜉ℎ₁𝑠₁,𝐵²(𝜉)=𝐴₂(𝜉)exp2𝜉ℎ₂𝑠₂,(4.1)-direction and while the crack moves in the positive ℎ₁=ℎ√𝛽₁,ℎ₂=ℎ√𝛽₂.(4.2)-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 𝑊𝑊₁(𝑥,𝑦)=2𝐹_𝑐𝐴¹(𝜉)𝑒^{−𝑠1ℎ1𝜉}cosh𝜉𝑠₁ℎ¹−𝑦₁;𝜉→𝑥,−∞<𝑥<∞,0<𝑦≤ℎ,(4.3)₂𝜎(𝑥,𝑦)=2𝐹_𝑐𝐴²(𝜉)𝑒^{𝑠2ℎ2𝜉}cosh𝜉𝑠₂ℎ²−𝑦₂;𝜉→𝑥,−∞<𝑥<∞,−ℎ≤𝑦<0,(4.4)_𝑦𝑧(1)(𝑥,𝑦)=−2𝑠¹𝑐₄₄₍₁₎√𝛽₁𝜎𝐹_𝑐𝜉𝐴¹(𝜉)𝑒^{−𝑠1ℎ1𝜉}sinh𝜉𝑠₁ℎ¹−𝑦₁;𝜉→𝑥],−∞<𝑥<∞,0<𝑦≤ℎ,(4.5)_𝑦𝑧(2)(𝑥,𝑦)=2𝑠²𝑐₄₄₍₂₎√𝛽₂𝐹_𝑐𝜉𝐴²(𝜉)𝑒^{𝑠2ℎ2𝜉}sinh𝜉𝑠₂ℎ²+𝑦₂;𝜉→𝑥,−∞<𝑥<∞,−ℎ≤𝑦<0.(4.6)𝐴₂=−𝑠¹√𝛽₂sinh𝜉𝑠₁ℎ₁𝑐⁴⁴₍₁₎𝑠₂√𝛽₁sinh𝜉𝑠₂ℎ₂𝑐⁴⁴₍₂₎𝑒^{−𝜉ℎ1𝑠1−𝜉ℎ2𝑠2}𝐴¹(𝜉).(4.7)𝑊₁=𝑥,20𝐹⁺_𝑐−𝑊²𝑥,0⁻𝑒^{−𝜉ℎ1𝑠1}𝐴₁(𝜉)sinh𝜉𝑠₂ℎ₂cosh𝜉𝑠¹ℎ₁sinh𝜉𝑠²ℎ₂+𝑃sinh𝜉𝑠¹ℎ₁cosh𝜉𝑠²ℎ₂;𝜉→𝑥,(4.8)𝑃=𝑠¹√𝛽₂𝑠₂√𝛽₁𝑐₄₄₍₁₎𝑐₄₄₍₂₎.(4.9)

4. Solution of Problem A

From (3.3) and (3.6), we find that𝐹_𝑐𝜉𝐴¹(𝜉)𝑒^{−𝜉ℎ1𝑠1}sinh𝜉𝑠₁ℎ₁;𝜉→𝑥=√𝛽₁2𝑠₁𝑐₄₄₍₁₎𝐹𝑝(𝑥),0<𝑥<𝑎,_𝑐𝐴¹(𝜉)𝑒^{−𝜉ℎ1𝑠1}sinh𝜉𝑠₂ℎ₂cosh𝜉𝑠¹ℎ₁sinh𝜉𝑠²ℎ₂+𝑃sinh𝜉𝑠¹ℎ₁cosh𝜉𝑠²𝑎ℎ<²𝑥<;𝜉∞→.𝑥=0,(4.10) where 𝑠₁√𝛽₁=𝑠²√𝛽₂,(4.11)

From (3.1), (3.3), and (4.1), we find that 𝑃=𝑐⁴⁴⁽¹⁾𝑐₄₄₍₂₎.(4.12)(1+𝑃)𝑒^{−𝜉ℎ1𝑠1}cosh𝜉𝑠₁ℎ₁𝐴¹(𝜉)=𝐶(𝜉),(4.13)𝐹_𝑐𝜉𝐶(𝜉)tanh𝜉𝑠¹ℎ₁;𝜉→𝑥=√𝛽₁2𝑠₁𝑐⁻¹⁴⁴₍₁₎𝐹+𝑐⁻¹⁴⁴₍₂₎𝑝(𝑥)=𝑝¹(𝑥),0<𝑥<𝑎,_𝑠𝐶(𝜉);𝜉→𝑥=0,𝑎<𝑥<∞.(4.14)𝐶(𝜉)=1𝜉𝜋2^𝑎0Φ(𝜏)sin(𝜉𝜏)𝑑𝜏,(4.15)

From boundary condition (3.7), we find that Φ(𝜏)=−2𝑐sinh(2𝑐𝜏)𝜋²sinh²(𝑐𝑎)−sinh²(𝑐𝜏)^1/2∫^𝑎0𝑝₁(𝑥)sinh²(𝑐𝑎)−sinh²(𝑐𝑥)^1/2𝑐=𝜋sinh²(𝑐𝑥)−sinh²(𝑐𝜏)𝑑𝑥,0<𝜏<𝑎,2ℎ₁𝑠.(4.16)

From (4.3), (4.4), and (4.7), we find that 𝑝(𝑥)=𝑝₀where𝑝₀ Now, the boundary conditions (3.5) and (3.7) lead to the following dual integral equations: Φ(𝜏)=√𝛽₁𝑐⁻¹⁴⁴₍₁₎+𝑐⁻¹⁴⁴₍₂₎𝑝⁰sinh(2𝑐𝜏)𝑠₁𝜋²cosh(𝑎𝑐)sinh²(𝑐𝑎)−sinh²(𝑐𝜏)^1/2𝐿(𝜏)=Π𝜋𝐹𝜋2,tanh(𝑎𝑐)−𝐿(𝜏),2,sinh²(𝑐𝑎)sinh²(𝑐𝑎)−sinh²(𝑐𝜏)tanh(𝑎𝑐),(4.17) 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 𝐹 so that ΠIf we take𝜎_𝑦𝑧(1)(𝑥,𝑦)=𝑝⁰sinh(2𝑐𝑥)𝜋cosh(𝑎𝑐)sinh²(𝑐𝑎)−sinh²(𝑐𝑥)^1/2𝐹𝜋2,tanh(𝑎𝑐)−𝐿(𝜏),𝑎<𝑥<∞,(4.18) then the dual integral equations (4.10) can be written in the form Δ𝑊(𝑥)==𝑊√⁽¹⁾𝑥,0⁺−𝑊⁽²⁾𝑥,0⁻𝛽₁𝑐⁻¹⁴⁴₍₁₎+𝑐⁻¹⁴⁴₍₂₎𝑝⁰sinh(2𝑐𝜏)𝜋𝑠₁cosh(𝑎𝑐)sinh²(𝑐𝑎)−sinh²(𝑐𝑥)^1/2^𝑎𝑥𝐹𝜋2,tanh(𝑎𝑐)−𝐿(𝜏)𝑑𝜏,0<𝑥<𝑎.(4.19) Following [4], the solution of the the dual integral equations (4.14) can be written in the form 𝐾₃=lim_𝑥→𝑎+[√2(𝑥−𝑎)𝜎_𝑦𝑧(1)(𝑥,0)]=2𝑝⁰𝜋tanh𝑐(𝑐𝑎)^1/2𝐹𝜋2,tanh(𝑐𝑎).(4.20)where𝑥=𝑎 For the particular case 𝑎+𝛿𝑎(𝛿𝑎≪𝑎) when 𝐺_III=lim_𝛿𝑎→012𝛿𝑎^0𝛿𝑎𝜎^𝑦𝑧₍₁₎(𝑟,0)Δ𝑊(𝛿𝑎−𝑟,0)𝑑𝑟,(4.21) is a constant, we find that 𝑟 where 𝐺_III=√𝛽₁𝑠𝑐^−144(1)+𝑐⁻¹⁴⁴₍₂₎𝑝²⁰tanh(𝑐𝑎)𝜋𝑐𝐹𝜋2,tanh(𝑐𝑎)².(4.22) and 𝐵₁(𝜉)=−𝐴₁(𝜉)exp−2𝜉ℎ₁𝑠₁,𝐵²(𝜉)=−𝐴₂(𝜉)exp−2𝜉ℎ₂𝑠₂.(5.1) 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 by𝑊𝑊₁(𝑥,𝑦)=2𝐹_𝑐𝑒^{−𝜉ℎ1𝑠1}𝐴₁(𝜉)sinh𝜉𝑠₁ℎ¹−𝑦₁;𝜉→𝑥0≤𝑦≤ℎ,𝑥>0,(5.2)₂𝜎(𝑥,𝑦)=2𝐹_𝑐𝑒^{𝜉ℎ2𝑠2}𝐴₂(𝜉)sinh𝜉𝑠₂ℎ²+𝑦₂;𝜉→𝑥−ℎ≤𝑦≤0,𝑥>0,(5.3)_𝑦𝑧(1)(𝑥,𝑦)=−2𝑠¹𝑐₄₄₍₁₎√𝛽₁𝜎𝐹_𝑐𝜉𝐴¹(𝜉)𝑒^{−𝑠1ℎ1𝜉}cosh𝜉𝑠₁ℎ¹−𝑦₁;𝜉→𝑥,0≤𝑦≤ℎ,𝑥>0,(5.4)_𝑦𝑧(2)(𝑥,𝑦)=2𝑠²𝑐₄₄₍₂₎√𝛽₂𝐹_𝑐𝜉𝐴²(𝜉)𝑒^{𝑠2ℎ2𝜉}cosh𝜉𝑠₂ℎ²+𝑦₂;𝜉→𝑥,−ℎ≤𝑦≤0,𝑥>0.(5.5)and the crack sliding displacement is𝐴₁(𝜉)𝑠¹𝑐₄₄₍₁₎𝑒^{−𝜉ℎ1𝑠1}cosh𝜉𝑠₁ℎ₁√𝛽₁=−𝑠²𝑐₄₄₍₂₎𝑒^{𝜉ℎ2𝑠2}cosh𝜉𝑠₂ℎ₂√𝛽₂𝐴₂(𝜉).(5.6) The stress intensity factor can be written in the form 𝐹_𝑐𝜉𝐶¹(𝜉)coth𝑠₁ℎ₁𝜉;𝜉→𝑥=√𝛽₁2ℎ𝑠₁𝑐⁻¹⁴⁴₍₁₎𝐹+𝑐⁻¹⁴⁴₍₂₎𝑐⁴⁴₍₁₎𝑝(𝑥),0<𝑥<𝑎,_𝑐𝐶¹(𝜉);𝜉→𝑥=0,𝑎<𝑥,(5.7)

Assuming that under applied loading the crack tip advances along the crack plane from (1+𝑃)𝑒^{−𝑠1ℎ1𝜉}𝐴₁(𝜉)sinh𝜉𝑠₁ℎ₁𝐴¹(𝜉)=𝐶₁(𝜉).(5.8) to 𝑝(𝑥)=𝑝₀, then the energy release rate per unit length during this process is given by 𝐶₁(𝜉)=𝜋2𝜉⁻¹^𝑎0Ψ(𝑢)sin(𝑢𝜉)𝑑𝑢,(5.9) where Ψ(𝑢)=√𝛽₁sin(𝑐𝑢)𝑐₄₄₍₁₎𝑐⁻¹⁴⁴₍₁₎+𝑐⁻¹⁴⁴₍₂₎𝑝⁰𝜋ℎ𝑠₁sinh²(𝑐𝑎)−sinh²(𝑐𝑢)^1/2.(5.10) denotes the distance from the crack tip.

From (4.21), we find that 𝜎_𝑦𝑧(1)𝑥,0⁺=−𝑐⁴⁴⁽¹⁾𝑝₀ℎ[1−sinh(𝑐𝑥)sinh²(𝑐𝑥)−sinh²(𝑐𝑎)^1/2],𝑥>𝑎.(5.11)

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 if𝑥=𝑎 Using (5.1), (3.1), and (3.3), we find that 𝐾_III=lim_𝑥→𝑎+√2(𝑥−𝑎)𝜎_𝑦𝑧(1)(𝑥,0)=𝑐⁴⁴⁽¹⁾ℎ𝑝⁰tanh𝑐(𝑐𝑎)^1/2,(5.12)Δ𝑊(𝑥)=𝑊₍₁₎𝑥,0⁺−𝑊⁽²⁾𝑥,0⁻=𝑝⁰√𝛽₁𝑠₁1+𝑐𝑐⁴⁴⁽¹⁾₄₄₍₂₎1−𝜋2sin⁻¹ccoosshh((𝑐𝑐𝑥𝑎)).(5.13)𝐺_III=lim_𝛿→012𝛿^𝛿0𝜎_𝑦𝑧(1)(𝑟)Δ𝑤(𝛿−𝑟)𝑑𝑟=12𝑠₁ℎ𝑐^−144(1)+𝑐⁻¹⁴⁴₍₂₎𝑝²⁰𝑐²⁴⁴₍₁₎tanh(𝑐𝑎).(5.14)𝑥=𝑎+𝑥₁ From (5.4) and (5.5) and the boundary condition (3.10), we find that 𝑎→∞

Using (4.11) and (5.6), the boundary conditions (3.9) and (3.11) lead to the dual integral equations 𝜎_𝑦𝑧(1)𝑥,0⁺=𝜎^𝑦𝑧₍₁₎𝑥,0⁻=−−𝑐⁴⁴⁽¹⁾𝑝₀ℎ1−11−𝑒⁻^{𝜋𝑥1/ℎ},𝑥¹Δ𝑊𝑥₁>0,=𝑝⁰√𝛽₁𝑠₁1+𝑐𝑐⁴⁴⁽¹⁾₄₄₍₂₎1−𝜋2sin⁻¹𝑒^{𝜋𝑥1/2ℎ},𝑥¹𝐾<0,_III=lim_𝑥1→0√2𝑥₁𝜎_𝑦𝑧(1)𝑥¹,0⁺=𝑐⁴⁴₍₁₎𝑝₀2ℎ𝜋.(6.1) where 𝜈²=𝑐⁴⁴⁽¹⁾𝑐₄₄₍₂₎𝑐⁵⁵⁽¹⁾𝑐₄₄₍₂₎−𝑐₅₅₍₂₎𝑐₄₄₍₁₎𝜌¹𝑐₅₅₍₁₎𝑐²⁴⁴⁽²⁾−𝜌₂𝑐₅₅₍₂₎𝑐²⁴⁴⁽¹⁾.(7.1)

Following [4], the solution of the dual integral equations (5.7) for 𝑐₄₄₍₁₎,𝑐₅₅₍₁₎,𝜌₁ can be written in the form𝑐₄₄₍₂₎,𝑐₅₅₍₂₎,𝜌₂where𝜈=0

We can easily find that 𝑐₄₄₍₁₎𝑐₅₅₍₁₎=𝑐⁴⁴⁽²⁾𝑐₅₅₍₂₎.(7.2) The stress intensity factor at is given by and the crack sliding displacement is The energy release rate per unit length during the process is given by

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 taking in (5.11), (5.12), and (5.13) and then letting , so that we have

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

If the velocity of the crack is known for particular values of the constants , we can substitute the values of constants 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 assume . Then, from (7.1), we find that

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.