# Thermoelastic Analysis for Two Collinear Cracks in an Orthotropic Solid Disturbed by Antisymmetrical Linear Heat Flow

Authors: Bing Wu; Jun-gao Zhu; Daren Peng; Rhys Jones; Shi-hu Gao; Yang-yang Lu
Journal: Mathematical Problems in Engineering
 (2017)

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## Abstract

The problem of two collinear cracks in an orthotropic solid under antisymmetrical linear heat flow is investigated. It is assumed that there exists thermal resistance to heat conduction through the crack region. Applying the Fourier transform, the thermal coupling partial differential equations are transformed to dual integral equations and then to singular integral equations. The crack-tip thermoelastic fields including the jumps of temperature and elastic displacements on the cracks and the mode II stress intensity factors are obtained explicitly. Numerical results show the effects of the geometries of the cracks and the dimensionless thermal resistance on the temperature change and the mode II stress intensity factors. Also, FEM solutions for the stress intensity factorK are used to compare with the solutions obtained using the method. It is revealed that the friction in closed crack surface region should be considered in analyzing the stress intensity factor K.


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## Body

## 1. Introduction

In engineering problems, the thermoelastic analysis for a cracked material has attracted much interest [1–5]. The thermal stress analysis for a cracked isotropic or anisotropic material has attracted much attention. A large number of related papers have been published to address fracture behaviors of isotropic and anisotropic solid. For example, the singularity of the stress fields near the crack-tip with a specified temperature or heat flux loading is studied by Sih [6]. The stress intensity factors of a central crack in an orthotropic material under a uniform heat flow are given by Tsai [7]. The thermoelasticity problem of two collinear cracks embedded in an orthotropic solid has been considered by Chen and Zhang [8]. By using the J-integral obtained from the finite element solutions, the stress intensity factor has been computed by Wilson and Yu [9]. Two alternative approaches for analyzing the nonlinear interaction between two equal-length collinear cracks subjected to remote tensile stress on infinity are developed by Chang and Kotousov [10]. By using a two-dimensional dual boundary element method, the stress intensity factors of a cracked isotropic material under the transient thermoelastic loadings have been calculated by Prasad et al. [11]. The diffraction of plane temperature-step waves by a crack in an orthotropic thermoelastic solid has been investigated by Brock [12]. The steady-state thermoelasticity problem of a cracked fiber-reinforced slab under a state of generalized plane deformation is studied [13]. Using the hyperbolic heat conduction theory and the dual-phase-lag heat conduction model, the transient temperature and thermal stresses around a partially insulated crack in a thermoelastic strip under a temperature impact and the transient temperature field around a partially insulated crack in a half plane are obtained by Hu and Chen [14, 15]. The thermal-medium crack model proposed by Zhong and Lee [16] is applied to investigate the problem of a penny-shaped crack in an infinite isotropic material [17]. Development of a unified model for the steady-state operation of single-phase natural circulation loops is made by Basu et al. [18]. An unsymmetrical end-notched flexure test is described and its suitability for interfacial fracture toughness testing is evaluated [19]. An infinite functionally graded medium with a partially insulated crack subjected to a steady-state heat flux away from the crack region as well as mechanical crack surface stresses is considered [20]. The influence of an axisymmetric partially insulated mixed-mode crack on the coupled response of a functionally graded magnetoelectroelastic material subjected to thermal loading is investigated [21]. The elastic-static problem of a partially insulated axisymmetric crack embedded in a graded coating bonded to a homogeneous substrate subjected to thermal loading is considered [22]. In order to investigate the crack problems in elastic material, many mathematical methods have been widely used such as the integral equation method [23], the linear sampling method [24], and the Trefftz method [25].In the abovementioned works and other works [26–32], the partially insulated boundary condition is mainly used and the closed-form solutions have been obtained for the thermoelastic field around the penny-shaped crack under the loading of uniform heat flow.Under the consideration of the antisymmetrical linear heat flow, the partially insulated boundary condition is applied to address the problem of two collinear cracks in an orthotropic solid and FEM solutions for the stress intensity factorK are used to investigate the accuracy of this methodology in this paper. Applying the Fourier transform technique and integral equation methods, the crack-tip thermoelastic fields involving the jumps of temperature, the elastic displacements across the cracks, and the mode II stress intensity factors are given in explicit forms. Numerical results show the heat flux to the crack surfaces and the mode II stress intensity factors are dependent on the geometries of cracks and the dimensionless thermal resistance. The thermal resistance in the heat conduction through the crack region is of much importance in analyzing the thermoelastic problem of a cracked material with a thermal loading. Furthermore, FEM solutions for the stress intensity factor K are used to compare with the solutions obtained using the method. It is revealed the friction in closed crack surface region has an effect on analyzing the stress intensity factor K and should be considered.


## 2. Statement of the Problem

As shown in Figure1, two collinear cracks for an infinite transversely orthotropic material are embedded in an orthotropic solid. We use Cartesian coordinates system xoy and suppose that the two collinear cracks are situated at the segment of a<x<b. With respect to the state of plane stress, the following constitute equations are(1)σxx=c11∂u∂x+c12∂v∂y-β1θ,σyy=c12∂u∂x+c22∂v∂y-β2θ,σxy=c66∂u∂y+∂v∂x,c11=Exx1-vxyvyx,c22=Eyy1-vxyvyx,c12=Exxvyx1-vxyvyx=Eyyvxy1-vxyvyx,β1=c11αxx+c12αyy,β2=c12αxx+c22αyy, where u and v denote the components of elastic displacement, θ stands for the temperature change, vxy and vyx are the Poisson ratios, Exx and Eyy are Young’s moduli, c66=Gxy is the shear modulus, and αxx and αyy are the coefficients of linear expansion. Applying the following equations:(2)∂σxx∂x+∂σyy∂y=0,∂σxy∂x+∂σyy∂y=0,we have(3)c11∂2u∂x2+c66∂2u∂y2+c12+c66∂2v∂x∂y-β1∂θ∂x=0,c66∂2v∂x2+c22∂2v∂y2+c12+c66∂2u∂x∂y-β2∂θ∂y=0.Furthermore, applying thermal equilibrium equation leads to(4)λ2∂2θ∂x2+∂2θ∂y2=0,λ=λxλy, where λx and λy are the thermal conductivities of a cracked orthotropic material. From the theory of Fourier heat conduction, one has(5)qx=-λx∂θ∂x,qy=-λy∂θ∂y. Due to the antisymmetry of the heat flux acting on two collinear cracks, the thermal field in the region x>0 is only studied. With application of the crack-face boundary conditions, one has(6)σyy+x,0=σyy-x,0=0,σxy+x,0=σxy-x,0=0,a<x<b.For the thermal loading, the antisymmetrical linear heat flow is acting on the crack surfaces; namely,(7)qy+x,0=qy-x,0=-q0x,a<x<b, where q0 is the prescribed constant, where(8)qy+x,0=-λy∂θ+x,0∂y,qy-x,0=-λy∂θ-x,0∂y.Considering the thermal resistance Rc in the heat conduction through the crack region, the thermoelastic boundary conditions on the crack surfaces are written as(9)qy+x,0=qy-x,0=-q0-qcx,a<x<b,where(10)qc=1Rcθ+x,0-θ-x,0.qc denotes the heat flux to the crack surfaces. The limiting value Rc→0 or Rc→∞ represents perfect conduction or perfect insulation on the crack surfaces. In addition, the continuity of elastic displacement and temperature on the crack-free parts of x-axis yields(11)v+x,0=v-x,0,0<x<a,x>b,(12)u+x,0=u-x,0,0<x<a,x>b,(13)θ+x,0=θ-x,0,qy+x,0=qy-x,0,0<x<a,x>b.Figure 1
Two collinear cracks in an orthotropic material under antisymmetrical linear heat flow.


## 3. Solution Procedure

### 3.1. Temperature Field

The temperature field is not dependent on the elastic strain; the temperature field can be solved firstly. By applying the Fourier transform technique, the solution of (4) can be defined as(14)θ±x,y=∫0∞A±ξe-ξδ±λysin⁡ξxdξ.A±ξ is unknown and will be solved. δ+=1 for y>0 and δ-=-1 for y<0. Hereafter the superscripts+ and − denote the physical quantities of the upper (i.e., y>0) and lower (i.e., y<0) region, respectively. From (5), we obtain(15)qx±x,y=-λx∫0∞ξA±ξe-ξδ±λycos⁡ξxdξ,qy±x,y=λλyδ±∫0∞ξA±ξe-ξδ±λysin⁡ξxdξ.Using (8) and the second relation in (13), we obtain(16)A+ξ=-A-ξ.Using the boundary conditions (7), one has(17)∫0∞ξA+ξsin⁡ξxdξ=-q0-qcxλλy,a<x<b.Application of the first relation in (13) leads to(18)∫0∞A+ξsin⁡ξxdξ=0,0<x<aorx>b.In order to solve (17) and (18), we give a definition of the auxiliary function(19)ωx=∂θ+x,0∂x. Applying the inverse Fourier transform, one obtains(20)A+ξξ=2π∫abωscos⁡ξsds.Substituting (20) into (17) leads to(21)2π∫abωsds∫0∞sin⁡ξxcos⁡ξsdξ=-q0-qcxλλy,a<x<b. From the known result [33](22)2∫0∞sin⁡ξxcos⁡ξsdξ=1x+s+1x-s. Equation (21) can be rewritten as(23)1π∫abωs2s2-x2ds=q0-qcλλy,a<x<b. Then, introducing s2=s∗, x2=x∗, 2sds=ds∗, a2=a∗, and b2=b∗, (23) can be written as(24)1π∫a∗b∗ω∗s∗1s∗-x∗s∗ds∗=q0-qcλλy,a<x<b. By virtue of the singular integral equation theory, we can obtain the solution of (21); namely,(25)ω∗x∗x∗=1πx∗-a∗b∗-x∗∫a∗b∗s∗-a∗b∗-s∗x∗-s∗q0-qcλλyds∗+Cx∗-a∗b∗-x∗, where C is the constant determined by(26)12∫a∗b∗ω∗s∗s∗ds∗=0. By some computations, we can get C=0. Furthermore, applying (19) and (25) yields(27)θ+x,0=∫a∗x∗ω∗s∗2s∗ds∗. Substituting (25) into (27), we get the temperature change on the cracks as(28)θ+x,0=-q0-qcλλyb2-a2x2-a2b2-x2,a<x<b. When a=0, the temperature change on the cracks of a single crack of length 2b is given as(29)θ+x,0=-q0-qcxλλyb2b2-x2,0<x<b.


### 3.2. Elastic Field

In what follows, we pay an attention to the solutions of the partial differential equations (3). u±x,y and v±x,y are written as two parts(30)u±x,y=u1±x,y+u2±x,y,v±x,y=v1±x,y+v2±x,y.u1±x,y and v1±x,y correspond to the general solutions under θ±x,y=0. u2±x,y and v2±x,y are the particular solutions under a thermal flux loading. With application of the Fourier transform, u1±x,y and v1±x,y can be expressed as(31)u1±x,y=∑j=12∫0∞Aj±ξe-ξδ±γjycos⁡ξxdξ,v1±x,y=∑j=12∫0∞ηjδ∓Aj±ξe-ξδ±γjysin⁡ξxdξ.Aj±ξj=1,2 are unknown to be solved. γjj=1,2 are the roots of equation as follows: Reγj>0(32)c22c66γ4+c122+2c12c66-c12c22γ2+c11c66=0, where ηjj=1,2 can be calculated as follows:(33)ηj=c11-c66γj2c12+c66γj. In addition, u2±x,y and v2±x,y are chosen as(34)u2±x,y=∫0∞A∗±ξe-ξδ±λycos⁡ξxdξ,v2±x,y=∫0∞δ∓B∗±ξe-ξδ±λysin⁡ξxdξ. Substituting (34) into (3) leads to(35)A∗±ξB∗±ξ=A±ξξM1M2,β1β2λ=c11-c66λ2-c12+c66λc11+c66λc66-c22λ2M1M2.From (31) and (34), the components of stress can be written as(36)σxx±x,y=-∑j=12∫0∞c11-c12γjηjξAj±ξe-ξδ±γjysin⁡ξxdξ-c11M1-c22λM2-β1∫0∞A±ξe-ξδ±λysin⁡ξxdξ,(37)σyy±x,y=-∑j=12∫0∞c12-c22γjηjξAj±ξe-ξδ±γjysin⁡ξxdξ-c12M1-c22λM2-β2∫0∞A±ξe-ξδ±λysin⁡ξxdξ,(38)σxy±x,y=-c66∑j=12∫0∞δ±γj+ηjξAj±ξe-ξδ±γjycos⁡ξxdξ-c66∫0∞δ±M1λ+M2A±ξe-ξδ±λycos⁡ξxdξ. It is seen that the elastic field was only induced by the loading -q0x. We only consider the antisymmetrical linear heat flow -q0x in this paper, and it follows that(39)σyy±x,0=0,x>0.From (37) and (39), one arrives at(40)∑j=12c12-c22γjηjAj+ξ=c22λM2+β2-c12M1A+ξξ. Utilizing the second relation of (6) and (12) yields(41)∫0∞∑j=12Aj+ξ+M1A+ξξcos⁡ξxdξ=0,0<x<a,x>b,(42)∑j=12∫0∞γj+ηjξAj+ξcos⁡ξxdξ+∫0∞M1λ+M2ξA+ξcos⁡ξxdξ=0,a<x<b. To solve the dual integral equations, we introduce a new auxiliary function φx.(43)φx=∂u+x,0-u-x,0∂x. By virtue of theory of the inverse Fourier transform, one gets(44)∑j=12Aj+ξξ+M1A+ξ=-2π∫abφssin⁡ξsds. Making use of (40) and (44) yields(45)A1+ξξ=c22λM2+β2-c22γ2η2M1c22γ2η2-γ1η1A+ξ+c12-c22γ2η2c22γ2η2-γ1η12π∫abφssin⁡ξsds,A2+ξξ=c22γ1η1M1-c22λM2-β2c22γ2η2-γ1η1A+ξ+c22γ1η1-c12c22γ2η2-γ1η12π∫abφssin⁡ξsds. Application of (42) and (45) leads to(46)2π∫abφsds∫0∞cos⁡ξxsin⁡ξsdξ=K∫0∞A+ξcos⁡ξxdξ,a<x<b, with(47)N1=γ1+η1c22γ2η2M1-c22λM2-β2+γ2+η2c22λM2+β2-c22γ1η1M1+c22M1λ+M2γ2η2-γ1η1,N2=γ1+η1c12-c22γ2η2+γ2+η2c22γ1η1-c12,K=N1N2. With application of the inverse Fourier transform and (28), one obtains(48)K∫0∞A+ξcos⁡ξxdξ=2Kπ∫abθ+s,0ds∫0∞sin⁡ξscos⁡ξxdξ. Recalling the known result [34](49)∫abs-ab-sx-sds=π22x-a-b. Inserting (28) into (48), one gets(50)K∫0∞A+ξcos⁡ξxdξ=2Kq0-qcπλλyb2-a2∫abb2-s2s2-a2x2-s22sds=p2x2-a2-b2, where(51)p=Kq0-qcλλyb2-a2. From (46) and (50), we can get(52)φ∗x∗=1πx∗-a∗b∗-x∗∫a∗b∗s∗-a∗b∗-s∗x∗-s∗2s∗-a∗-b∗pds∗+Cx∗-a∗b∗-x∗, Considering the following condition:(53)∫a∗b∗φ∗s∗2s∗ds∗=0. By further calculations, one obtains(54)C=-pa2-b224+2pb2a2+b2Ek3Fk-4pa2b23. With application of (43), one has(55)u+x,0-u-x,0=∫axφsds=∫a∗x∗φ∗s∗2s∗ds∗. Substituting (52) into (55), we have(56)u+x,0-u-x,0=-23pb2a2+b2Eα-EkFkFα+23pxx2-a2b2-x2,a<x<b, where k=b2-a2/b2 and F· and E· are the complete elliptic integrals of the first and second kind, respectively. Fα,k and Eα,k are the incomplete elliptic integrals of the first and the second kinds with(57)α=arcsinb2-x2b2-a2. Applying (38) and (45), the shearing stresses near the crack-tips are expressed as(58)σxy±x,0=c66N2πc22γ2η2-γ1η1∫ab2sφss2-x2ds+c66N1c22γ2η2-γ1η1∫0∞A+ξcos⁡ξxdξ. Substituting (50) and (52) into (58), one has(59)σxy±x,0=c66pN2c22γ2η2-γ1η14a2b23+b2-a222-2b2a2+b2Ek3Fk-2x2-a2-b22×12a2-x2b2-x2+c66p2c22γ2η2-γ1η14N2+N12x2-a2-b20<x<a,σxy±x,0=-c66pN2c22γ2η2-γ1η14a2b23+b2-a222-2b2a2+b2Ek3Fk-2x2-a2-b22×12x2-a2x2-b2+c66p2c22γ2η2-γ1η14N2+N12x2-a2-b2x>b.


### 3.3. Stress Intensity Factors

According to (10), (16), and (28), one has(60)qc=2hcq0x2-a2b2-x22hcx2-a2b2-x2+λλyb2-a2.The quantityhc=λy/Rc is the dimensionless thermal resistance. One can see from (60) that the heat flux to the crack surfaces is dependent on the geometries of the cracks, the material properties, and applied thermal loading. Furthermore, the crack opening displacement has great influences on analyzing the thermoelastic problem. From the obtained results, it is easy to determine(61)Δux,0=u+x,0-u-x,0,KSHDinn=limx→a+⁡π2x-aΔux,0,KSHDout=limx→b-⁡π2b-xΔux,0. From (56), one arrives at(62)KSHDinn=-2pb2a2+b23bEkaFk-abπab2-a2+2pa3πab2-a2,KSHDout=2pb2a2+b23EkFk-1πbb2-a2+2pb3πbb2-a2. From (62), it is seen that the stress intensity factors are expressed in the closed forms. In addition, to consider the stress fields near the crack-tips, the mode II stress intensity factors are defined as(63)Kinn=limx→a-⁡2πa-xσxy±x,0,Kout=limx→b+⁡2πx-bσxy±x,0. One has(64)Kinn=c66pN22c22γ2η2-γ1η14a2b23-a2-b222-2b2a2+b2Ek3Fk×πab2-a2,Kout=-c66pN22c22γ2η2-γ1η14a2b23-b2-a222-2b2a2+b2Ek3Fk×πbb2-a2. It is easy to find that the mode II stress intensity factors are dependent not only on the geometries of the cracks and the heat flux to the crack surfaces qc but also on the material properties.In order to verify (64), we consider a limiting case as a→0, corresponding a single crack with the length of 2b. It is not difficult to find(65)K0=c66p′N2b2bπ2c22γ2η2-γ1η112+2Ek3Fk, where(66)p′=Kq0-qcλλyb2.


## 3.1. Temperature Field

The temperature field is not dependent on the elastic strain; the temperature field can be solved firstly. By applying the Fourier transform technique, the solution of (4) can be defined as(14)θ±x,y=∫0∞A±ξe-ξδ±λysin⁡ξxdξ.A±ξ is unknown and will be solved. δ+=1 for y>0 and δ-=-1 for y<0. Hereafter the superscripts+ and − denote the physical quantities of the upper (i.e., y>0) and lower (i.e., y<0) region, respectively. From (5), we obtain(15)qx±x,y=-λx∫0∞ξA±ξe-ξδ±λycos⁡ξxdξ,qy±x,y=λλyδ±∫0∞ξA±ξe-ξδ±λysin⁡ξxdξ.Using (8) and the second relation in (13), we obtain(16)A+ξ=-A-ξ.Using the boundary conditions (7), one has(17)∫0∞ξA+ξsin⁡ξxdξ=-q0-qcxλλy,a<x<b.Application of the first relation in (13) leads to(18)∫0∞A+ξsin⁡ξxdξ=0,0<x<aorx>b.In order to solve (17) and (18), we give a definition of the auxiliary function(19)ωx=∂θ+x,0∂x. Applying the inverse Fourier transform, one obtains(20)A+ξξ=2π∫abωscos⁡ξsds.Substituting (20) into (17) leads to(21)2π∫abωsds∫0∞sin⁡ξxcos⁡ξsdξ=-q0-qcxλλy,a<x<b. From the known result [33](22)2∫0∞sin⁡ξxcos⁡ξsdξ=1x+s+1x-s. Equation (21) can be rewritten as(23)1π∫abωs2s2-x2ds=q0-qcλλy,a<x<b. Then, introducing s2=s∗, x2=x∗, 2sds=ds∗, a2=a∗, and b2=b∗, (23) can be written as(24)1π∫a∗b∗ω∗s∗1s∗-x∗s∗ds∗=q0-qcλλy,a<x<b. By virtue of the singular integral equation theory, we can obtain the solution of (21); namely,(25)ω∗x∗x∗=1πx∗-a∗b∗-x∗∫a∗b∗s∗-a∗b∗-s∗x∗-s∗q0-qcλλyds∗+Cx∗-a∗b∗-x∗, where C is the constant determined by(26)12∫a∗b∗ω∗s∗s∗ds∗=0. By some computations, we can get C=0. Furthermore, applying (19) and (25) yields(27)θ+x,0=∫a∗x∗ω∗s∗2s∗ds∗. Substituting (25) into (27), we get the temperature change on the cracks as(28)θ+x,0=-q0-qcλλyb2-a2x2-a2b2-x2,a<x<b. When a=0, the temperature change on the cracks of a single crack of length 2b is given as(29)θ+x,0=-q0-qcxλλyb2b2-x2,0<x<b.


## 3.2. Elastic Field

In what follows, we pay an attention to the solutions of the partial differential equations (3). u±x,y and v±x,y are written as two parts(30)u±x,y=u1±x,y+u2±x,y,v±x,y=v1±x,y+v2±x,y.u1±x,y and v1±x,y correspond to the general solutions under θ±x,y=0. u2±x,y and v2±x,y are the particular solutions under a thermal flux loading. With application of the Fourier transform, u1±x,y and v1±x,y can be expressed as(31)u1±x,y=∑j=12∫0∞Aj±ξe-ξδ±γjycos⁡ξxdξ,v1±x,y=∑j=12∫0∞ηjδ∓Aj±ξe-ξδ±γjysin⁡ξxdξ.Aj±ξj=1,2 are unknown to be solved. γjj=1,2 are the roots of equation as follows: Reγj>0(32)c22c66γ4+c122+2c12c66-c12c22γ2+c11c66=0, where ηjj=1,2 can be calculated as follows:(33)ηj=c11-c66γj2c12+c66γj. In addition, u2±x,y and v2±x,y are chosen as(34)u2±x,y=∫0∞A∗±ξe-ξδ±λycos⁡ξxdξ,v2±x,y=∫0∞δ∓B∗±ξe-ξδ±λysin⁡ξxdξ. Substituting (34) into (3) leads to(35)A∗±ξB∗±ξ=A±ξξM1M2,β1β2λ=c11-c66λ2-c12+c66λc11+c66λc66-c22λ2M1M2.From (31) and (34), the components of stress can be written as(36)σxx±x,y=-∑j=12∫0∞c11-c12γjηjξAj±ξe-ξδ±γjysin⁡ξxdξ-c11M1-c22λM2-β1∫0∞A±ξe-ξδ±λysin⁡ξxdξ,(37)σyy±x,y=-∑j=12∫0∞c12-c22γjηjξAj±ξe-ξδ±γjysin⁡ξxdξ-c12M1-c22λM2-β2∫0∞A±ξe-ξδ±λysin⁡ξxdξ,(38)σxy±x,y=-c66∑j=12∫0∞δ±γj+ηjξAj±ξe-ξδ±γjycos⁡ξxdξ-c66∫0∞δ±M1λ+M2A±ξe-ξδ±λycos⁡ξxdξ. It is seen that the elastic field was only induced by the loading -q0x. We only consider the antisymmetrical linear heat flow -q0x in this paper, and it follows that(39)σyy±x,0=0,x>0.From (37) and (39), one arrives at(40)∑j=12c12-c22γjηjAj+ξ=c22λM2+β2-c12M1A+ξξ. Utilizing the second relation of (6) and (12) yields(41)∫0∞∑j=12Aj+ξ+M1A+ξξcos⁡ξxdξ=0,0<x<a,x>b,(42)∑j=12∫0∞γj+ηjξAj+ξcos⁡ξxdξ+∫0∞M1λ+M2ξA+ξcos⁡ξxdξ=0,a<x<b. To solve the dual integral equations, we introduce a new auxiliary function φx.(43)φx=∂u+x,0-u-x,0∂x. By virtue of theory of the inverse Fourier transform, one gets(44)∑j=12Aj+ξξ+M1A+ξ=-2π∫abφssin⁡ξsds. Making use of (40) and (44) yields(45)A1+ξξ=c22λM2+β2-c22γ2η2M1c22γ2η2-γ1η1A+ξ+c12-c22γ2η2c22γ2η2-γ1η12π∫abφssin⁡ξsds,A2+ξξ=c22γ1η1M1-c22λM2-β2c22γ2η2-γ1η1A+ξ+c22γ1η1-c12c22γ2η2-γ1η12π∫abφssin⁡ξsds. Application of (42) and (45) leads to(46)2π∫abφsds∫0∞cos⁡ξxsin⁡ξsdξ=K∫0∞A+ξcos⁡ξxdξ,a<x<b, with(47)N1=γ1+η1c22γ2η2M1-c22λM2-β2+γ2+η2c22λM2+β2-c22γ1η1M1+c22M1λ+M2γ2η2-γ1η1,N2=γ1+η1c12-c22γ2η2+γ2+η2c22γ1η1-c12,K=N1N2. With application of the inverse Fourier transform and (28), one obtains(48)K∫0∞A+ξcos⁡ξxdξ=2Kπ∫abθ+s,0ds∫0∞sin⁡ξscos⁡ξxdξ. Recalling the known result [34](49)∫abs-ab-sx-sds=π22x-a-b. Inserting (28) into (48), one gets(50)K∫0∞A+ξcos⁡ξxdξ=2Kq0-qcπλλyb2-a2∫abb2-s2s2-a2x2-s22sds=p2x2-a2-b2, where(51)p=Kq0-qcλλyb2-a2. From (46) and (50), we can get(52)φ∗x∗=1πx∗-a∗b∗-x∗∫a∗b∗s∗-a∗b∗-s∗x∗-s∗2s∗-a∗-b∗pds∗+Cx∗-a∗b∗-x∗, Considering the following condition:(53)∫a∗b∗φ∗s∗2s∗ds∗=0. By further calculations, one obtains(54)C=-pa2-b224+2pb2a2+b2Ek3Fk-4pa2b23. With application of (43), one has(55)u+x,0-u-x,0=∫axφsds=∫a∗x∗φ∗s∗2s∗ds∗. Substituting (52) into (55), we have(56)u+x,0-u-x,0=-23pb2a2+b2Eα-EkFkFα+23pxx2-a2b2-x2,a<x<b, where k=b2-a2/b2 and F· and E· are the complete elliptic integrals of the first and second kind, respectively. Fα,k and Eα,k are the incomplete elliptic integrals of the first and the second kinds with(57)α=arcsinb2-x2b2-a2. Applying (38) and (45), the shearing stresses near the crack-tips are expressed as(58)σxy±x,0=c66N2πc22γ2η2-γ1η1∫ab2sφss2-x2ds+c66N1c22γ2η2-γ1η1∫0∞A+ξcos⁡ξxdξ. Substituting (50) and (52) into (58), one has(59)σxy±x,0=c66pN2c22γ2η2-γ1η14a2b23+b2-a222-2b2a2+b2Ek3Fk-2x2-a2-b22×12a2-x2b2-x2+c66p2c22γ2η2-γ1η14N2+N12x2-a2-b20<x<a,σxy±x,0=-c66pN2c22γ2η2-γ1η14a2b23+b2-a222-2b2a2+b2Ek3Fk-2x2-a2-b22×12x2-a2x2-b2+c66p2c22γ2η2-γ1η14N2+N12x2-a2-b2x>b.


## 3.3. Stress Intensity Factors

According to (10), (16), and (28), one has(60)qc=2hcq0x2-a2b2-x22hcx2-a2b2-x2+λλyb2-a2.The quantityhc=λy/Rc is the dimensionless thermal resistance. One can see from (60) that the heat flux to the crack surfaces is dependent on the geometries of the cracks, the material properties, and applied thermal loading. Furthermore, the crack opening displacement has great influences on analyzing the thermoelastic problem. From the obtained results, it is easy to determine(61)Δux,0=u+x,0-u-x,0,KSHDinn=limx→a+⁡π2x-aΔux,0,KSHDout=limx→b-⁡π2b-xΔux,0. From (56), one arrives at(62)KSHDinn=-2pb2a2+b23bEkaFk-abπab2-a2+2pa3πab2-a2,KSHDout=2pb2a2+b23EkFk-1πbb2-a2+2pb3πbb2-a2. From (62), it is seen that the stress intensity factors are expressed in the closed forms. In addition, to consider the stress fields near the crack-tips, the mode II stress intensity factors are defined as(63)Kinn=limx→a-⁡2πa-xσxy±x,0,Kout=limx→b+⁡2πx-bσxy±x,0. One has(64)Kinn=c66pN22c22γ2η2-γ1η14a2b23-a2-b222-2b2a2+b2Ek3Fk×πab2-a2,Kout=-c66pN22c22γ2η2-γ1η14a2b23-b2-a222-2b2a2+b2Ek3Fk×πbb2-a2. It is easy to find that the mode II stress intensity factors are dependent not only on the geometries of the cracks and the heat flux to the crack surfaces qc but also on the material properties.In order to verify (64), we consider a limiting case as a→0, corresponding a single crack with the length of 2b. It is not difficult to find(65)K0=c66p′N2b2bπ2c22γ2η2-γ1η112+2Ek3Fk, where(66)p′=Kq0-qcλλyb2.


## 4. Numerical Results and Discussion

In this section, numerical results are carried out to research the effects of the dimensionless thermal resistance of the cracks surfaces and the geometries of the cracks on the temperature change and the mode II stress intensity factors under the antisymmetrical linear heat flow. A ceramic material of Tyrannohex is chosen and the corresponding material properties are given in [35]. That is, one has (67)Exx=135.0GPa,Eyy=87.0GPa,Gxy=50.0GPa,vxy=0.15,vyx=0.09667,αxx=0.32×10-5/°C,αyy=0.32×10-5/°C,λx=3.08W/m°C,λy=2.81W/m°C.Figure2 shows the variations of qc/q0 versus x/b with a/b=0.25 and hc=0,1,2,and4, respectively. Obviously, when hc=0, we have qc=0, corresponding to fully thermally insulated cracks. When hc=1, it means that if the heat conduction through the cracks region is the same as that of the orthotropic material, the cracks approximate to thermally permeable ones. Moreover, one can find it is feasible to reduce the thermal stress concentrations near the crack-tips by the way of filling the matrix material into a crack in engineering problems. When hc=2 or hc=4, it shows the heat conduction through the crack region is twice or 4 times as much as thermal conductivities of an orthotropic material. When hc→∞, we have qc=q0, which means fully thermally permeable cracks. It is revealed that applying the thermally insulated assumption hc=0 will underestimate much the heat flux to the crack surfaces and applying the fully thermally permeable assumption hc→∞ will overestimate much the heat flux to the crack surfaces. The observation reveals that the effects of the heat conduction through the crack region on the crack-tip thermoelastic fields cannot be disregarded.Figure 2
The variations ofqc/q0 versus x/b with a/b=0.25 and hc=0,1,2,and4, respectively.Figure3 shows the variations of K/K0 versus x/b with hc=1 and a/b=0.25,0.5,and0.75, respectively. Figure 4 shows variations of K/K0 versus x/b with a/b=0.25 and hc=1,2,and4, respectively. Hereafter K0 stands for the mode II stress intensity factor for a single and fully thermally insulated crack with the length of 2b under the antisymmetrical linear heat flow. K/K0 tends to zero as hc increases. Obviously, it is unrealistic to assume that the crack is fully thermally permeable. From the above results, it is revealed that the fully thermally insulated or fully thermally permeable assumption on the crack-face is only a limiting case of a real crack. It is concluded that the material properties and the thermal resistance in the heat conduction through the crack region are of much importance and should be considered in analyzing the thermoelastic problem of a cracked material with a thermal loading. EM solutions for the stress intensity factor K are used to compare with the solutions obtained using the method outlined above.Figure 3
The variations ofK/K0 versus x/b with hc=1 and a/b=0.25,0.5,and0.75, respectively.Figure 4
The variations ofK/K0 versus x/b with a/b=0.25 and hc=1,2,and4, respectively.Figures6 and 7 present the temperature profile and thermal stress analysis results, respectively. Another model is employed to check the mesh density convergence, with a local mesh four times finer than the mesh used in this analysis. The maximum difference between these meshes is less than 1%. For crack-tip A, the difference between present solution KII and those obtained using FEM is less than 2%. However, the present solution KII for crack-tip B differs from the FEM solutions by up to 20%. Upon examination, the differences in these analyses are due to the assumption of crack being open during the calculation of stress intensity factors. This assumption is not validated by FEM analysis. To overcome this shortcoming, the friction in closed crack surface region should be considered, in order to obtain more accurate values for KII at tip B. As this is beyond the scope of the present paper, it will be a topic to consider for future studies.


## 5. Conclusions

Considering the effects of the thermal resistance, the problem of two collinear cracks in an orthotropic solid under antisymmetrical linear heat flow is addressed. Applying the Fourier transform technique, the thermoelastic fields involving the jumps of temperature, the elastic displacements across the cracks, and the mode II stress intensity factors are given in closed forms. Numerical results are carried out to show the heat flux to the crack surfacesqc depends on the geometries of the cracks, the material properties, and applied thermal loading, and the mode II stress intensity factors are dependent on the heat flux to the crack surfaces qc, the material properties, and the geometries of the cracks. The fully thermally permeable or fully thermally insulated cracks are the limiting case of the partially insulated cracks. The observations reveal that the thermal resistance in the heat conduction through the crack region is of much importance and should not be neglected in analyzing the thermoelastic problem of a cracked material with a thermal loading. Also, FEM solutions for the stress intensity factor K are used to compare with the solutions obtained using the method; it is revealed that the friction in closed crack surface region should be considered. As this is beyond the scope of the present paper, it will be a topic to consider for future studies.



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*Source: 5093404-2017-11-19.xml*