Mathematical Problems in Engineering

Mathematical Problems in Engineering / 2008 / Article

Research Article | Open Access

Volume 2008 |Article ID 930820 |

B. M. Singh, J. Rokne, R. S. Dhaliwal, "Closed-Form Solutions for a Mode-III Moving Interface Crack at the Interface of Two Bonded Dissimilar Orthotropic Elastic Layers", Mathematical Problems in Engineering, vol. 2008, Article ID 930820, 10 pages, 2008.

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

Academic Editor: Francesco Pellicano
Received27 Feb 2008
Accepted10 Nov 2008
Published03 Feb 2009


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 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 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 2ğ‘Ž 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 have𝑈𝑗=𝑉𝑗=0,𝑊𝑗=𝑊𝑗(𝑋,𝑌,𝑡),(2.1) where 𝑈𝑗,𝑉𝑗,𝑊𝑗 are displacement components in the 𝑋,𝑌, and 𝑍 directions, respectively, and 𝑗=1,2. Then ğœŽğ‘‹(𝑗)=ğœŽğ‘Œ(𝑗)=ğœŽğ‘(𝑗)=ğœŽğ‘‹ğ‘Œ(𝑗)=0,ğœŽğ‘‹ğ‘(𝑗)=𝑐55(𝑗)𝜕𝑊𝑗𝜕𝑋,ğœŽğ‘Œğ‘(𝑗)=𝑐44(𝑗)𝜕𝑊𝑗𝜕𝑌,(2.2) 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 𝜕2𝑊𝑗𝜕𝑋2+𝜕2𝑊𝑗𝜕𝑌2𝑗=1𝐶2𝑗𝜕2𝑊𝑗𝜕𝑡2,(2.3)where𝑌𝑗=𝑌√𝛽𝑗,𝛽𝑗=𝑐44(𝑗)𝑐55(𝑗),𝐶𝑗=îƒŽğ‘44(𝑗)𝜌𝑗,(2.4)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 transformation𝑥=𝑋−𝑣𝑡,𝑦𝑗=𝑌𝑗,𝑧=𝑍.(2.5) Equation (2.3) now becomes 𝑠2𝑗𝜕2𝑊𝑗𝜕𝑥2+𝜕2𝑊𝑗𝜕𝑦2𝑗=0,(2.6) where 𝑠2𝑗𝑣=1−2𝐶2𝑗.(2.7)

3. Solution of Equilibrium Equation

The solution of the equilibrium (2.6) may be written as𝑊𝑗(𝑥,𝑦)=𝐹𝑐𝐴𝑗exp−𝜉𝑦𝑗𝑠𝑗+𝐵𝑗exp𝜉𝑦𝑗𝑠𝑗;𝜉→𝑥,(3.1) where 𝑠𝑗 is taken as (2.7) and 𝐹𝑐𝐴𝑗=(𝜉);𝜉→𝑥2ğœ‹î€œâˆž0𝐴𝑗𝐹(𝜉)cos(𝜉𝑥)𝑑𝜉,𝑠𝐵𝑗=(𝜉);𝜉→𝑥2ğœ‹î€œâˆž0𝐵𝑗(𝜉)sin(𝜉𝑥)𝑑𝜉.(3.2) The nonzero stresses are given by ğœŽğ‘¦ğ‘§(𝑗)𝑠(𝑥,𝑦)=𝑗𝑐44(𝑗)√𝛽𝑗𝐹𝑐𝜉𝐵𝑗exp𝜉𝑦𝑗𝑠𝑗−𝜉𝐴𝑗exp−𝜉𝑦𝑗𝑠𝑗,ğœŽ;𝜉→𝑥(3.3)𝑥𝑧(𝑗)(𝑥,𝑦)=−𝑐55(𝑗)𝐹𝑠𝜉𝐵𝑗exp𝜉𝑦𝑗𝑠𝑗+𝜉𝐴𝑗exp−𝜉𝑦𝑗𝑠(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 asğœŽğ‘¦ğ‘§(1)𝑥,0+||𝑥||ğœŽ=−𝑝(𝑥),<ğ‘Ž,(3.5)𝑦𝑧(1)𝑥,0+=ğœŽğ‘¦ğ‘§(2)𝑥,0−𝑊,−∞<𝑥<∞,(3.6)1𝑥,0+=𝑊2𝑥,0−,||𝑥||ğœŽ>ğ‘Ž,(3.7)𝑦𝑧(1)𝑥,+ℎ=ğœŽğ‘¦ğ‘§(2)(𝑥,−ℎ)=0,(3.8)𝑝(𝑥)𝑌=Â±â„Žğ‘(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 ğœŽğ‘¦ğ‘§(1)𝑐(𝑥,0)=−44(1)ℎ||𝑥||ğœŽğ‘(𝑥),<ğ‘Ž,(3.9)𝑦𝑧(1)(𝑥,0)=ğœŽğ‘¦ğ‘§(2)𝑊(𝑥,0),−∞<𝑥<∞,(3.10)1(𝑥,0)=𝑊2||𝑥||𝑊(𝑥,0),>ğ‘Ž,(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−2ğœ‰â„Ž1𝑠1,𝐵2(𝜉)=𝐴2(𝜉)exp2ğœ‰â„Ž2𝑠2,(4.1)-direction and while the crack moves in the positive ℎ1=â„Žâˆšğ›½1,ℎ2=â„Žâˆšğ›½2.(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 𝑊1(𝑥,𝑦)=2𝐹𝑐𝐴1(𝜉)𝑒−𝑠1ℎ1𝜉cosh𝜉𝑠1ℎ1−𝑦1𝑊;𝜉→𝑥,−∞<𝑥<∞,0<ğ‘¦â‰¤â„Ž,(4.3)2(𝑥,𝑦)=2𝐹𝑐𝐴2(𝜉)𝑒𝑠2ℎ2𝜉cosh𝜉𝑠2ℎ2−𝑦2î‚î‚„ğœŽ;𝜉→𝑥,−∞<𝑥<∞,âˆ’â„Žâ‰¤ğ‘¦<0,(4.4)𝑦𝑧(1)(𝑥,𝑦)=−2𝑠1𝑐44(1)√𝛽1𝐹𝑐𝜉𝐴1(𝜉)𝑒−𝑠1ℎ1𝜉sinh𝜉𝑠1ℎ1−𝑦1î‚ğœŽ;𝜉→𝑥],−∞<𝑥<∞,0<ğ‘¦â‰¤â„Ž,(4.5)𝑦𝑧(2)(𝑥,𝑦)=2𝑠2𝑐44(2)√𝛽2𝐹𝑐𝜉𝐴2(𝜉)𝑒𝑠2ℎ2𝜉sinh𝜉𝑠2ℎ2+𝑦2;𝜉→𝑥,−∞<𝑥<∞,âˆ’â„Žâ‰¤ğ‘¦<0.(4.6)𝐴2𝑠=−1√𝛽2sinh𝜉𝑠1ℎ1𝑐44(1)𝑠2√𝛽1sinh𝜉𝑠2ℎ2𝑐44(2)ğ‘’âˆ’ğœ‰â„Ž1𝑠1âˆ’ğœ‰â„Ž2𝑠2𝐴1(𝜉).(4.7)𝑊1𝑥,0+−𝑊2𝑥,0−=2ğ¹ğ‘î‚ƒğ‘’âˆ’ğœ‰â„Ž1𝑠1𝐴1(𝜉)sinh𝜉𝑠2ℎ2cosh𝜉𝑠1ℎ1sinh𝜉𝑠2ℎ2+𝑃sinh𝜉𝑠1ℎ1cosh𝜉𝑠2ℎ2,;𝜉→𝑥(4.8)𝑠𝑃=1√𝛽2𝑠2√𝛽1𝑐44(1)𝑐44(2).(4.9)

4. Solution of Problem A

From (3.3) and (3.6), we find that𝐹𝑐𝜉𝐴1(𝜉)ğ‘’âˆ’ğœ‰â„Ž1𝑠1sinh𝜉𝑠1ℎ1=√;𝜉→𝑥𝛽12𝑠1𝑐44(1)𝐹𝑝(𝑥),0<𝑥<ğ‘Ž,𝑐𝐴1(𝜉)ğ‘’âˆ’ğœ‰â„Ž1𝑠1sinh𝜉𝑠2ℎ2cosh𝜉𝑠1ℎ1sinh𝜉𝑠2ℎ2+𝑃sinh𝜉𝑠1ℎ1cosh𝜉𝑠2ℎ2;𝜉→𝑥=0,ğ‘Ž<𝑥<∞.(4.10) where 𝑠1√𝛽1=𝑠2√𝛽2,(4.11)

From (3.1), (3.3), and (4.1), we find that 𝑐𝑃=44(1)𝑐44(2).(4.12)(1+𝑃)ğ‘’âˆ’ğœ‰â„Ž1𝑠1cosh𝜉𝑠1ℎ1𝐴1(𝜉)=𝐶(𝜉),(4.13)𝐹𝑐𝜉𝐶(𝜉)tanh𝜉𝑠1ℎ1=√;𝜉→𝑥𝛽12𝑠1𝑐−144(1)+𝑐−144(2)𝑝(𝑥)=𝑝1𝐹(𝑥),0<𝑥<ğ‘Ž,𝑠𝐶(𝜉);𝜉→𝑥=0,ğ‘Ž<𝑥<∞.(4.14)1𝐶(𝜉)=𝜉𝜋2î€œğ‘Ž0Φ(𝜏)sin(𝜉𝜏)𝑑𝜏,(4.15)

From boundary condition (3.7), we find that Φ(𝜏)=−2𝑐sinh(2𝑐𝜏)𝜋2sinh2(ğ‘ğ‘Ž)−sinh2(𝑐𝜏)1/2âˆ«ğ‘Ž0𝑝1(𝑥)sinh2(ğ‘ğ‘Ž)−sinh2(𝑐𝑥)1/2sinh2(𝑐𝑥)−sinh2𝜋(𝑐𝜏)𝑑𝑥,0<𝜏<ğ‘Ž,𝑐=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)=𝛽1𝑐−144(1)+𝑐−144(2)𝑝0sinh(2𝑐𝜏)𝑠1𝜋2cosh(ğ‘Žğ‘)sinh2(ğ‘ğ‘Ž)−sinh2(𝑐𝜏)1/2𝐹𝜋2,𝜋,tanh(ğ‘Žğ‘)−𝐿(𝜏)𝐿(𝜏)=Π2,sinh2(ğ‘ğ‘Ž)sinh2(ğ‘ğ‘Ž)−sinh2,(𝑐𝜏)tanh(ğ‘Žğ‘)(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 takeğœŽğ‘¦ğ‘§(1)𝑝(𝑥,𝑦)=0sinh(2𝑐𝑥)𝜋cosh(ğ‘Žğ‘)sinh2(ğ‘ğ‘Ž)−sinh2(𝑐𝑥)1/2𝐹𝜋2,tanh(ğ‘Žğ‘)−𝐿(𝜏),ğ‘Ž<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−=√𝛽1𝑐−144(1)+𝑐−144(2)𝑝0sinh(2𝑐𝜏)𝜋𝑠1cosh(ğ‘Žğ‘)sinh2(ğ‘ğ‘Ž)−sinh2(𝑐𝑥)1/2î€œğ‘Žğ‘¥î‚ƒğ¹î‚€ğœ‹2,tanh(ğ‘Žğ‘)−𝐿(𝜏)𝑑𝜏,0<𝑥<ğ‘Ž.(4.19) Following [4], the solution of the the dual integral equations (4.14) can be written in the form 𝐾3=limğ‘¥â†’ğ‘Ž+[√2(ğ‘¥âˆ’ğ‘Ž)ğœŽğ‘¦ğ‘§(1)(𝑥,0)]=2𝑝0𝜋tanh(ğ‘ğ‘Ž)𝑐1/2𝐹𝜋2,tanh(ğ‘ğ‘Ž).(4.20)where𝑥=ğ‘Ž For the particular case ğ‘Ž+ğ›¿ğ‘Ž(ğ›¿ğ‘Žâ‰ªğ‘Ž) when 𝐺III=limğ›¿ğ‘Žâ†’012ğ›¿ğ‘Ž0ğ›¿ğ‘Žî‚ƒğœŽğ‘¦ğ‘§(1)(𝑟,0)Δ𝑊(ğ›¿ğ‘Žâˆ’ğ‘Ÿ,0)𝑑𝑟,(4.21) is a constant, we find that 𝑟 where 𝐺III=√𝛽1𝑠𝑐−144(1)+𝑐−144(2)𝑝20tanh(ğ‘ğ‘Ž)𝐹𝜋𝜋𝑐2,tanh(ğ‘ğ‘Ž)2.(4.22) and 𝐵1(𝜉)=−𝐴1(𝜉)exp−2ğœ‰â„Ž1𝑠1,𝐵2(𝜉)=−𝐴2(𝜉)exp−2ğœ‰â„Ž2𝑠2.(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 by𝑊1(𝑥,𝑦)=2ğ¹ğ‘î‚ƒğ‘’âˆ’ğœ‰â„Ž1𝑠1𝐴1(𝜉)sinh𝜉𝑠1ℎ1−𝑦1𝑊;𝜉→𝑥0â‰¤ğ‘¦â‰¤â„Ž,𝑥>0,(5.2)2(𝑥,𝑦)=2ğ¹ğ‘î‚ƒğ‘’ğœ‰â„Ž2𝑠2𝐴2(𝜉)sinh𝜉𝑠2ℎ2+𝑦2î‚„ğœŽî‚î‚;ğœ‰â†’ğ‘¥âˆ’â„Žâ‰¤ğ‘¦â‰¤0,𝑥>0,(5.3)𝑦𝑧(1)(𝑥,𝑦)=−2𝑠1𝑐44(1)√𝛽1𝐹𝑐𝜉𝐴1(𝜉)𝑒−𝑠1ℎ1𝜉cosh𝜉𝑠1ℎ1−𝑦1î‚„ğœŽî‚î‚„;𝜉→𝑥,0â‰¤ğ‘¦â‰¤â„Ž,𝑥>0,(5.4)𝑦𝑧(2)(𝑥,𝑦)=2𝑠2𝑐44(2)√𝛽2𝐹𝑐𝜉𝐴2(𝜉)𝑒𝑠2ℎ2𝜉cosh𝜉𝑠2ℎ2+𝑦2;𝜉→𝑥,âˆ’â„Žâ‰¤ğ‘¦â‰¤0,𝑥>0.(5.5)and the crack sliding displacement is𝐴1𝑠(𝜉)1𝑐44(1)ğ‘’âˆ’ğœ‰â„Ž1𝑠1cosh𝜉𝑠1ℎ1√𝛽1𝑠=−2𝑐44(2)ğ‘’ğœ‰â„Ž2𝑠2cosh𝜉𝑠2ℎ2√𝛽2𝐴2(𝜉).(5.6) The stress intensity factor can be written in the form 𝐹𝑐𝜉𝐶1𝑠(𝜉)coth1ℎ1𝜉=√;𝜉→𝑥𝛽12â„Žğ‘ 1𝑐−144(1)+𝑐−144(2)𝑐44(1)𝐹𝑝(𝑥),0<𝑥<ğ‘Ž,𝑐𝐶1(𝜉);𝜉→𝑥=0,ğ‘Ž<𝑥,(5.7)

Assuming that under applied loading the crack tip advances along the crack plane from (1+𝑃)𝑒−𝑠1ℎ1𝜉𝐴1(𝜉)sinh𝜉𝑠1ℎ1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−1î€œğ‘Ž0Ψ(𝑢)sin(𝑢𝜉)𝑑𝑢,(5.9) where √Ψ(𝑢)=𝛽1sin(𝑐𝑢)𝑐44(1)𝑐−144(1)+𝑐−144(2)𝑝0ğœ‹î‚€â„Žğ‘ 1sinh2(ğ‘ğ‘Ž)−sinh2(𝑐𝑢)1/2.(5.10) denotes the distance from the crack tip.

From (4.21), we find that ğœŽğ‘¦ğ‘§(1)𝑥,0+=−𝑐44(1)𝑝0ℎ[1−sinh(𝑐𝑥)sinh2(𝑐𝑥)−sinh2(ğ‘ğ‘Ž)1/2],𝑥>ğ‘Ž.(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=ğ‘Ž Using (5.1), (3.1), and (3.3), we find that 𝐾III=limğ‘¥â†’ğ‘Ž+√2(ğ‘¥âˆ’ğ‘Ž)ğœŽğ‘¦ğ‘§(1)=𝑐(𝑥,0)44(1)â„Žğ‘0tanh(ğ‘ğ‘Ž)𝑐1/2,(5.12)Δ𝑊(𝑥)=𝑊(1)𝑥,0+−𝑊(2)𝑥,0−=𝑝0√𝛽1𝑠1𝑐1+44(1)𝑐44(2)21−𝜋sin−1cosh(𝑐𝑥).cosh(ğ‘ğ‘Ž)(5.13)𝐺III=lim𝛿→012𝛿𝛿0ğœŽğ‘¦ğ‘§(1)1(𝑟)Δ𝑤(𝛿−𝑟)𝑑𝑟=2𝑠1â„Žî‚€ğ‘âˆ’144(1)+𝑐−144(2)𝑝20𝑐244(1)tanh(ğ‘ğ‘Ž).(5.14)𝑥=ğ‘Ž+𝑥1 From (5.4) and (5.5) and the boundary condition (3.10), we find that ğ‘Žâ†’âˆž

Using (4.11) and (5.6), the boundary conditions (3.9) and (3.11) lead to the dual integral equations ğœŽğ‘¦ğ‘§(1)𝑥,0+=ğœŽğ‘¦ğ‘§(1)𝑥,0−=−−𝑐44(1)𝑝0ℎ11−1−𝑒−𝜋𝑥1/ℎ,𝑥1𝑥>0,Δ𝑊1=𝑝0√𝛽1𝑠1𝑐1+44(1)𝑐44(2)21−𝜋sin−1𝑒𝜋𝑥1/2ℎ,𝑥1𝐾<0,III=lim𝑥1→0√2𝑥1ğœŽğ‘¦ğ‘§(1)𝑥1,0+=𝑐44(1)𝑝02.â„Žğœ‹(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.


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Copyright © 2008 B. M. Singh et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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