# The Analysis of a Cracked Material under Combined Unsymmetric Thermal Flux and Symmetric Linear Mechanical Loading

Authors: B. Wu; Q. A. Huang; D. Peng; X. Y. Yang; Z. Feng
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

An improved partially permeable crack model is put forward to deal with the problem of a single crack embedded in an orthotropic or isotropic material under combined unsymmetric thermal flux and symmetric linear mechanical loading. With the application of the Fourier transform technique (FTT), the thermoelastic field is given in a closed form. Numerical results show combined unsymmetric linear thermal flux, symmetric linear mechanical loading, and dimensionless thermal conductivity, and the coefficient has influences on fracture parameters. For the improved partially permeable crack, the mode II stress intensity factor and the energy release rate might be zero or positive under combined unsymmetric thermal flux and symmetric linear mechanical loading. Therefore, closure of the crack tip region need not be considered under combined unsymmetric thermal flux and symmetric linear mechanical loading when making use of fracture parameters as a criterion.


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## Body

## 1. Introduction

An elastic solid’s expansion and contraction from changes in temperature is inhibited for the external constraints and the mutual constraints between the internal parts of the solid, resulting in thermal stress [1, 2]. So, it is essential to address the thermoelastic field of cracked solid in practical engineering [3–5]. A great many of studies have been published to consider the fracture behaviors of various cracks which are embedded in an orthotropic solid under thermal loading [6–14]. By making use of the J-integral, Wilson and Yu obtained the stress intensity factor [11]. Chen and Zhang addressed the thermoelasticity problem of cracked orthotropic solid [12]. Rizk investigated the problem of the cracked orthotropic semifinite material [15]. Wu et al. applied the Fourier transform technique to compute the II stress intensity factors of two collinear cracks under antisymmetrical linear thermal ﬂux [16]. Afterwards, Wu et al. gave the solution of a single crack in the closed form under quadratic thermomechanical loading by using the Fourier transform technique [17].In the abovementioned studies, the partially permeable crack model becomes more widely available [18–21]. According to the mathematical intuition, an improved partially permeable crack model in Figure 1(a) can be raised as(1)Q1c=−hcΔT+εQ1,where Q1c and Q1 denote the heat flux per thickness to the crack surface and initial heat flux; hc stands for the thermal conductivity on the crack region. The coefficient ε is regarded as a constant and in line with the complex and real situation, i.e., the nonoptimal cases of complicated crack surface. In general, ε proves to be negative or positive relying on the thermoelastic field.An improved partially permeable crack model is presented to consider the problem of a single crack embedded in an orthotropic or isotropic material under combined unsymmetric linear thermal flux and symmetric linear mechanical loading. With application of the Fourier transform technique, the thermoelastic partial differential equations (PDE) are converted to singular integral equations. The thermoelastic field is given in the closed form by solving singular integral equations. The obtained results reveal physical quantities that have influences on fracture parameters for cracked solid under combined unsymmetric linear thermal flux and symmetric linear mechanical loading.


## 2. Problem Statement

A single crack embedded in an orthotropic or isotropic material under combined unsymmetric linear thermal flux and symmetric linear mechanical loading is shown in Figures1(a) and 1(b).Figure 1
(a) A single crack embedded in an orthotropic or isotropic material under unsymmetric linear thermal flux. (b) A single crack embedded in an orthotropic or isotropic material under symmetric linear mechanical loading.
(a)(b)The constitute equations are given based on the state of plane stress [22].(2)σxx=c11∂u∂x+c12∂v∂y−β1T,(3)σyy=c12∂u∂x+c22∂v∂y−β2T,(4)σxy=c66∂u∂y+∂v∂x,where(5)c11=Exx1−vxyvyx,c22=Eyy1−vxyvyx,c12=Exxvyx1−vxyvyx=Eyyvxy1−vxyvyx,β1β2=c11c12c12c22αxxαyy,where σyy and σxy are the components of stress, Exx and Eyy represent Young’s moduli, T is the temperature, vxx and vyy denote Poisson’s ratios, u and v indicate the components of elastic displacement, c66=Gxy represents the shear modulus, and αxx and αyy are the coefficients of linear expansion. Using the following equation,(6)∂σxx∂x+∂σxy∂y=0,∂σxy∂x+∂σyy∂y=0,one obtains(7)c11∂2u∂x2+c66∂2u∂y2+c12+c66∂2v∂x∂y=β1∂T∂x,(8)c66∂2v∂x2+c22∂2v∂y2+c12+c66∂2u∂x∂y=β2∂T∂y.Based on the Fourier heat conduction, one has(9)Qx=−λx∂T∂x,Qy=−λy∂T∂y,where Qx and Qy are the components of the heat flux; λx and λy are the coefficient of heat conduction. Furthermore, based on equilibrium equation, one has(10)∂Qx∂x+∂Qy∂y=0.Using the thermal equilibrium equation, one has(11)λ2∂2T∂x2+∂2T∂y2=0,where(12)λ=λxλy.In this study, the crack-face boundary conditions are used as follows.(13a)QyIx,0−QyIIx,0=−Q1x2a2,−a<x<a,(13b)σyyIIx,0=σyyIx,0=−σ1ax,−a<x<a.The superscriptsI and II denote the physical quantities of y > 0 and y < 0 region, respectively. Q1 and σ1 are the prescribed constants. By using the improved partially permeable crack model, the crack-face boundary conditions can be given as(14a)σxyIIx,0=σxyIx,0=0,0<x<a,(14b)QyIx,0−QyIIx,0=−Q1−Q1cx2a2,0<x<a,σyyIIx,0=σyyIx,0=−σ1ax,0<x<a.Based on the antisymmetry of thermal flux and symmetry of mechanical loading, only half of the field (i.e.,x > 0) is considered under combined unsymmetric thermal flux and symmetric linear mechanical loading.(15)σxyIIx,0=σxyIx,0=0,0<x<a,(16)QyIx,0−QyIIx,0=−Q1−Q1cx2a2,0<x<a,(17)σyyIIx,0=σyyIx,0=−σ1ax,0<x<a,where(18)Q1c=−hcTIx,0−TIIx,0+εQ1.Moreover, factors such as stresses, elastic displacements, temperature, and thermal flux across the crack-free region of the horizontalx-axis comply with the following conditions.(19)σxyIx,0=σxyIIx,0,σyyIx,0=σyyIIx,0,x>aorx<−a,(20)uIx,0=−uIIx,0,vIx,0=−vIIx,0,x>aorx<−a,(21)TIx,0=TIIx,0,QyIx,0=QyIIx,0,x>aorx<−a.


## 3. Solution Procedure

### 3.1. Temperature Field

Due to equation (11) is irrelevant to the elastic strain, the temperature field can be solved out first. Considering the antisymmetry of thermal flux, the solution of equation (11) is written with application of FTT.(22)TI,IIx,y=∫0+∞g±ξe−ξδ±λysinξxdξ,where g±ξ is unknown and will be solved. δ+=1 and δ−=−1 denote y>0 and y<0, respectively. From (9), one gets(23)QxI,IIx,y=−λx∫0∞ξg±ξe−ξδ±λycosξxdξ,QyI,IIx,y=λyλ∫0∞δ±ξg±ξe−ξδ±λysinξxdξ.Making use of the second relation in equation (21), one gets(24)g+ξ=−g−ξ.By the aid of the first relation in equations (21) and (17), one attains(25)∫0+∞g+ξsinξxdξ=0,x>a,(26)∫0+∞g+ξsinξxdξ=−Q1−Q1cx4a2λλy,0<x<a.For the solvation of equations (25) and (26), the auxiliary function γx is defined as(27)γx=∂TIx,0−TIIx,0∂x.Using inverse Fourier transform, one has(28)g+ξξ=1π∫0aγscosξsds.Substituting equations (28) into (26), one gets(29)2π∫0aγsds∫0+∞sinξxcosξsdξ=−Q1−Q1cx2a2λλy.From the known result [23],(30)∫0+∞cosξssinξxdξ=121x−s+1x+s.Equation (29) can be expressed as follows:(31)1π∫−aaγs1s−xds=Q1−Q1cx2a2λλy.Equation (31) is a singular integral equation including the Cauchy kernel [24], and the solution can be depicted as(32)γx=1πa2−x2∫−aaa2−s2x−sQ1−Q1cs2a2λλyds+Ca2−x2.Considering the following condition,(33)∫−aaγsds=0.After some computations, one hasC = 0. Equation (32) can be given as(34)γs=Q1−Q1c4a2λλya2−x22x2−a2.In addition, with application of equations (27) and (34), the temperature change on the crack is given as(35)TIx,0−TIIx,0=−Q1−Q1c4a2λλyxa2−x2,which is consistent with that in [16] for a single crack of length 2a.


### 3.2. Elastic Field

In order to solve equations (7) and (8), u1I,IIx,y and vI,IIx,y are written as [25](36)uI,IIx,y=u1I,IIx,y+u2I,IIx,y,vI,IIx,y=v1I,IIx,y+v2I,IIx,y,where u1I,IIx,y and v1I,IIx,y correspond to the general solutions under TI,IIx,y=0. u2I,IIx,y and v2I,IIx,y are the peculiar solutions under thermal flux. With application of Fourier transform technique, u1I,IIx,y and v1I,IIx,y can be written as(37)u1I,IIx,y=∑j=12∫0+∞gj±ξe−ξδ±ρjycosξxdξ,(38)v1I,IIx,y=∑j=12∫0+∞ηjδ±gj±ξe−ξδ±ρjysinξxdξ,and gj±ξ will be solved. ρj,j=1,2Reρj>0 are chosen as the roots of the following character equation.(39)c22c66ρ4+c122+2c12c66−c12c22ρ2+c11c66=0,where(40)ηj=c11−c66ρj2c12+c66ρj.Furthermore,u2I,IIx,y and v2I,IIx,y are chosen as(41)u2I,IIx,y=∑j=12∫0+∞g∗±ξe−δ±ξλycosξxdξ,(42)v2I,IIx,y=∑j=12∫0+∞δ±L∗±ξe−δ±ξλysinξxdξ.Substituting equations (41) and (42) into (7) and (8), one has(43)g∗±ξL∗±ξ=K1K2g±ξξ,where(44)K1K2=c11−c66λ2−c12+c66λc12+c66λc66−c22λ2−1β1β2λ.Based on equations (2)–(4), (37), (38), (41), and (42), the stresses are rewritten as(45a)σxxI,IIx,0=−∑j=12∫0+∞c11−c12γjηjξgj±ξsinξxdξ−c11M1−c12λM2−β1∫0+∞g±ξsinξxdξ,(45b)σyyI,IIx,0=−∑j=12∫0+∞c12−c22γjηjξgj±ξsinξxdξ−c12M1−c22λM2−β2∫0+∞g±ξsinξxdξ,(45c)σxyI,IIx,0=−c66∑j=12∫0+∞δ±γj+ηjξgj±ξcosξxdξ+∫0+∞δ±M1λ+M2g±ξcosξxdξ.To obtain the explicit solution of the problem, two parts are divided. One part is what is induced by mechanical loading−σ1x/a. The other part is what is resulted from thermal flux −Q1x/2a2. The solution procedure is neglected for simplicity under mechanical loading −σ1x/a. The elastic displacement on the crack is obtained with application of the Fourier transform technique (FTT) [17].(46)vIx,0=σ1η1ρ2−η2ρ1xa2−x22aω1ρ2+η2+O1.Furthermore, one can obtain the stresses field as follows:(47)σyyI,IIx,0=2xπax2−a2σ1+O1.Then, the elastic field under linear thermal flux will be given in the closed form. With application of thermal flux, it fits the condition as follows.(48)σyyIx,0=σyyIIx,0=0,0<x<a.With application of equations (48) and (45b), one arrives at(49)∑j=12c12−c22ρjηjgj+ξ=c22λM2+β2−c12M1g+ξξ.With application of equation (15) and the first relation in equation (20), the dual integral equations are given:(50)∫0+∞∑j=12gj+ξcosξx+M1g+ξξcosξxdξ=0,x>a,(51)∫0+∞∑j=12ρj+ηjξgj+ξcosξxdξ+∫0+∞M1λ+M2g+ξcosξxdξ=0,0<x<a.To give the solvation of equations (50) and (51), the auxiliary function χx can be defined as(52)χx=∂uIx,0∂x.Applying equation (52) and inverse Fourier transform, one has(53)∑j=12gj+ξξ+M1g+ξ=−2π∫0aχssinξsds.Utilizing equations (49) and (53), one obtains(54)g1+ξξ=c22λM2+β2−c22ρ2η2M1c22ρ2η2−γ1η1g+ξ−c22ρ2η2−c12c22ρ2η2−ρ1η12π∫0aχssinξsds,(55)g2+ξξ=c22ρ1η1M1−c22λM2−β2c22ρ2η2−ρ1η1g+ξ−c12−c22ρ1η1c22ρ2η2−ρ1η12π∫0aχssinξsds.With application of equations (53)–(55), one obtains(56)2∫0aχsds∫0+∞sinξscosξxdξ=πω2∫0+∞g+ξcosξxdξ,0<x<a,where(57)ω2=H1H2,with(58)H1=ρ1+η1c22ρ2η2M1−c22λM2−β2+ρ2+η2c22λM2+β2−c22ρ1η1M1+c22M1λ+M2ρ1η1−ρ2η2,H2=ρ1+η1c22ρ2η2−c12+ρ2+η2c12−c22ρ1η1.With the knowledge of equations (30), and (56) can be rewritten as(59)1π∫−aaχsx−sds=ω2∫0+∞g+ξsinξxdξ.With application of equation (22) and the inverse Fourier transform, one has(60)∫0+∞g+ξcosξxdξ=1π∫−aaT+s−T−sx−sds=−Q1−Q1c16a2λλy2x2−a2.Based on the singular integral theory with the Cauchy kernel, equation (59) is obtained as(61)χx=ω2Q1−Q1c16a2πλλya2−x2∫−aaa2−s2x−s2s2−a2ds+Ea2−x2.After some calculations,E = 0. Equation (61) can be given as(62)χx=ω2Q1−Q1c8a2λλya2−x2x3−a2x.With application of equation (52), one obtains(63)uIx,0=∫−axχsds=ω2Q1−Q1c24a2λλya2−x2a2−x2a2−x2.From equations (45c), (54), (55), and (63), one can obtain the shearing stresses as follows.(64)σxyI,IIx,0=c66H1Q1−Q1c16a2λλyc22ρ2η2−ρ1η1a2x−x3x2−a2.


### 3.3. Crack Tip Field

Using equations (18) and (35), one has(65)Q1c=EcQ1xa2−x2+4εQ1a2λEcxa2−x2+4a2λ.In particular, whenε = 0, one has(66)Q1c=EcQ1xa2−x2Ecxa2−x2+4a2λ.The quantityEc=hc/λy stands for the dimensionless thermal conductivity. In equation (65), when Ec⟶∞ or Ec⟶0, Q1c⟶Q1 or Q1c⟶εQ1. The observations reveal εQ1 can be considered to be created by the thermal flux or other large loading such as mechanical loading.Furthermore, the stress intensity factor is of much importance for analysis of the behavior of cracked solid. The mode I stress intensity factor and the mode II stress intensity factor are given as(67)KI=limx⟶a+2πx−aσyyIx,0,(68)KII=limx⟶a+2πx−aσxyIx,0.Next, the corresponding intensity factors to characterize opening crack displacement are defined as(69)KvCOD=limx⟶a+π2a−xvIx,0,(70)KuCOD=limx⟶a+π2a−xuIx,0.With application of equations (47), (64), (46), (63), (67), (68), (69), and (70), one has(71)KI=2σ1πaπ,KII=0,KvCOD=σ1η1ρ2−η2ρ1πa2πω1ρ2+η2,KuCOD=0.The energy release rate (G) has an important effect on the analysis of crack growth, and the energy release rate is defined as [25](72)G=limδ⟶01δ∫0δσxyIx+a,0uIa−x,0+σyyIx+a,0vIa−x,0dx.


## 3.1. Temperature Field

Due to equation (11) is irrelevant to the elastic strain, the temperature field can be solved out first. Considering the antisymmetry of thermal flux, the solution of equation (11) is written with application of FTT.(22)TI,IIx,y=∫0+∞g±ξe−ξδ±λysinξxdξ,where g±ξ is unknown and will be solved. δ+=1 and δ−=−1 denote y>0 and y<0, respectively. From (9), one gets(23)QxI,IIx,y=−λx∫0∞ξg±ξe−ξδ±λycosξxdξ,QyI,IIx,y=λyλ∫0∞δ±ξg±ξe−ξδ±λysinξxdξ.Making use of the second relation in equation (21), one gets(24)g+ξ=−g−ξ.By the aid of the first relation in equations (21) and (17), one attains(25)∫0+∞g+ξsinξxdξ=0,x>a,(26)∫0+∞g+ξsinξxdξ=−Q1−Q1cx4a2λλy,0<x<a.For the solvation of equations (25) and (26), the auxiliary function γx is defined as(27)γx=∂TIx,0−TIIx,0∂x.Using inverse Fourier transform, one has(28)g+ξξ=1π∫0aγscosξsds.Substituting equations (28) into (26), one gets(29)2π∫0aγsds∫0+∞sinξxcosξsdξ=−Q1−Q1cx2a2λλy.From the known result [23],(30)∫0+∞cosξssinξxdξ=121x−s+1x+s.Equation (29) can be expressed as follows:(31)1π∫−aaγs1s−xds=Q1−Q1cx2a2λλy.Equation (31) is a singular integral equation including the Cauchy kernel [24], and the solution can be depicted as(32)γx=1πa2−x2∫−aaa2−s2x−sQ1−Q1cs2a2λλyds+Ca2−x2.Considering the following condition,(33)∫−aaγsds=0.After some computations, one hasC = 0. Equation (32) can be given as(34)γs=Q1−Q1c4a2λλya2−x22x2−a2.In addition, with application of equations (27) and (34), the temperature change on the crack is given as(35)TIx,0−TIIx,0=−Q1−Q1c4a2λλyxa2−x2,which is consistent with that in [16] for a single crack of length 2a.


## 3.2. Elastic Field

In order to solve equations (7) and (8), u1I,IIx,y and vI,IIx,y are written as [25](36)uI,IIx,y=u1I,IIx,y+u2I,IIx,y,vI,IIx,y=v1I,IIx,y+v2I,IIx,y,where u1I,IIx,y and v1I,IIx,y correspond to the general solutions under TI,IIx,y=0. u2I,IIx,y and v2I,IIx,y are the peculiar solutions under thermal flux. With application of Fourier transform technique, u1I,IIx,y and v1I,IIx,y can be written as(37)u1I,IIx,y=∑j=12∫0+∞gj±ξe−ξδ±ρjycosξxdξ,(38)v1I,IIx,y=∑j=12∫0+∞ηjδ±gj±ξe−ξδ±ρjysinξxdξ,and gj±ξ will be solved. ρj,j=1,2Reρj>0 are chosen as the roots of the following character equation.(39)c22c66ρ4+c122+2c12c66−c12c22ρ2+c11c66=0,where(40)ηj=c11−c66ρj2c12+c66ρj.Furthermore,u2I,IIx,y and v2I,IIx,y are chosen as(41)u2I,IIx,y=∑j=12∫0+∞g∗±ξe−δ±ξλycosξxdξ,(42)v2I,IIx,y=∑j=12∫0+∞δ±L∗±ξe−δ±ξλysinξxdξ.Substituting equations (41) and (42) into (7) and (8), one has(43)g∗±ξL∗±ξ=K1K2g±ξξ,where(44)K1K2=c11−c66λ2−c12+c66λc12+c66λc66−c22λ2−1β1β2λ.Based on equations (2)–(4), (37), (38), (41), and (42), the stresses are rewritten as(45a)σxxI,IIx,0=−∑j=12∫0+∞c11−c12γjηjξgj±ξsinξxdξ−c11M1−c12λM2−β1∫0+∞g±ξsinξxdξ,(45b)σyyI,IIx,0=−∑j=12∫0+∞c12−c22γjηjξgj±ξsinξxdξ−c12M1−c22λM2−β2∫0+∞g±ξsinξxdξ,(45c)σxyI,IIx,0=−c66∑j=12∫0+∞δ±γj+ηjξgj±ξcosξxdξ+∫0+∞δ±M1λ+M2g±ξcosξxdξ.To obtain the explicit solution of the problem, two parts are divided. One part is what is induced by mechanical loading−σ1x/a. The other part is what is resulted from thermal flux −Q1x/2a2. The solution procedure is neglected for simplicity under mechanical loading −σ1x/a. The elastic displacement on the crack is obtained with application of the Fourier transform technique (FTT) [17].(46)vIx,0=σ1η1ρ2−η2ρ1xa2−x22aω1ρ2+η2+O1.Furthermore, one can obtain the stresses field as follows:(47)σyyI,IIx,0=2xπax2−a2σ1+O1.Then, the elastic field under linear thermal flux will be given in the closed form. With application of thermal flux, it fits the condition as follows.(48)σyyIx,0=σyyIIx,0=0,0<x<a.With application of equations (48) and (45b), one arrives at(49)∑j=12c12−c22ρjηjgj+ξ=c22λM2+β2−c12M1g+ξξ.With application of equation (15) and the first relation in equation (20), the dual integral equations are given:(50)∫0+∞∑j=12gj+ξcosξx+M1g+ξξcosξxdξ=0,x>a,(51)∫0+∞∑j=12ρj+ηjξgj+ξcosξxdξ+∫0+∞M1λ+M2g+ξcosξxdξ=0,0<x<a.To give the solvation of equations (50) and (51), the auxiliary function χx can be defined as(52)χx=∂uIx,0∂x.Applying equation (52) and inverse Fourier transform, one has(53)∑j=12gj+ξξ+M1g+ξ=−2π∫0aχssinξsds.Utilizing equations (49) and (53), one obtains(54)g1+ξξ=c22λM2+β2−c22ρ2η2M1c22ρ2η2−γ1η1g+ξ−c22ρ2η2−c12c22ρ2η2−ρ1η12π∫0aχssinξsds,(55)g2+ξξ=c22ρ1η1M1−c22λM2−β2c22ρ2η2−ρ1η1g+ξ−c12−c22ρ1η1c22ρ2η2−ρ1η12π∫0aχssinξsds.With application of equations (53)–(55), one obtains(56)2∫0aχsds∫0+∞sinξscosξxdξ=πω2∫0+∞g+ξcosξxdξ,0<x<a,where(57)ω2=H1H2,with(58)H1=ρ1+η1c22ρ2η2M1−c22λM2−β2+ρ2+η2c22λM2+β2−c22ρ1η1M1+c22M1λ+M2ρ1η1−ρ2η2,H2=ρ1+η1c22ρ2η2−c12+ρ2+η2c12−c22ρ1η1.With the knowledge of equations (30), and (56) can be rewritten as(59)1π∫−aaχsx−sds=ω2∫0+∞g+ξsinξxdξ.With application of equation (22) and the inverse Fourier transform, one has(60)∫0+∞g+ξcosξxdξ=1π∫−aaT+s−T−sx−sds=−Q1−Q1c16a2λλy2x2−a2.Based on the singular integral theory with the Cauchy kernel, equation (59) is obtained as(61)χx=ω2Q1−Q1c16a2πλλya2−x2∫−aaa2−s2x−s2s2−a2ds+Ea2−x2.After some calculations,E = 0. Equation (61) can be given as(62)χx=ω2Q1−Q1c8a2λλya2−x2x3−a2x.With application of equation (52), one obtains(63)uIx,0=∫−axχsds=ω2Q1−Q1c24a2λλya2−x2a2−x2a2−x2.From equations (45c), (54), (55), and (63), one can obtain the shearing stresses as follows.(64)σxyI,IIx,0=c66H1Q1−Q1c16a2λλyc22ρ2η2−ρ1η1a2x−x3x2−a2.


## 3.3. Crack Tip Field

Using equations (18) and (35), one has(65)Q1c=EcQ1xa2−x2+4εQ1a2λEcxa2−x2+4a2λ.In particular, whenε = 0, one has(66)Q1c=EcQ1xa2−x2Ecxa2−x2+4a2λ.The quantityEc=hc/λy stands for the dimensionless thermal conductivity. In equation (65), when Ec⟶∞ or Ec⟶0, Q1c⟶Q1 or Q1c⟶εQ1. The observations reveal εQ1 can be considered to be created by the thermal flux or other large loading such as mechanical loading.Furthermore, the stress intensity factor is of much importance for analysis of the behavior of cracked solid. The mode I stress intensity factor and the mode II stress intensity factor are given as(67)KI=limx⟶a+2πx−aσyyIx,0,(68)KII=limx⟶a+2πx−aσxyIx,0.Next, the corresponding intensity factors to characterize opening crack displacement are defined as(69)KvCOD=limx⟶a+π2a−xvIx,0,(70)KuCOD=limx⟶a+π2a−xuIx,0.With application of equations (47), (64), (46), (63), (67), (68), (69), and (70), one has(71)KI=2σ1πaπ,KII=0,KvCOD=σ1η1ρ2−η2ρ1πa2πω1ρ2+η2,KuCOD=0.The energy release rate (G) has an important effect on the analysis of crack growth, and the energy release rate is defined as [25](72)G=limδ⟶01δ∫0δσxyIx+a,0uIa−x,0+σyyIx+a,0vIa−x,0dx.


## 4. Numerical Results

This section presents a research on the influence of combined unsymmetric linear thermal flux, symmetric linear mechanical loading, the dimensionless thermal resistance and the coefficient on the heat flux to the crack surface, the mode II stress intensity factor (KII), and the energy release rate (G). For convenience, the isotropic material constants are chosen as shown in [26] (Table 1).Table 1
Constants of steel.
ExxGPaEyyGPaGxyGPavxyvyxαxx10−5/C°αyy10−5/C°λxw/m°Cλyw/m°C20520579.20.30.3031.141.1448.648.6Figure2 shows Q1/Q1c versus Ec with x/a=0.25, 0.5, 0.75, 0.99 for ε = 0.01. The cases of Ec = 20, Ec = 40, Ec = 50, and Ec = 60 indicate the heat conduction is 19 times, 39 times, 49 times, and 59 times, respectively, stronger than thermal conductivity of the orthotropic material. Ec = 0 or Ec⟶∞ corresponds to the thermally fully impermeable or permeable case.Figure 2
Q1/Q1c versus Ec with x/a=0.25, 0.5, 075, 0.99 for ε = 0.01.Figure3 presents Q1/Q1c versus ε with x/a=0.25, 0.5, 075, 0.99 for Ec = 1. It is seen from Figure 3 when ε increases, Q1/Q1c increases. The case of Ec = 1 correlates with the situation that the crack region is identical to that of the isotropic material.Figure 3
Q1/Q1c versus ε with x/a=0.25, 0.25, 075, 0.99 for Ec = 1.Figure4 presents KII versus Q1. In the case of purely unsymmetric thermal flux, KII remains zero.Figure 4
KII versus Q1.Figure5 shows G, as computed using equation (72), for the improved partially permeable crack with a variety of crack lengths. In the case of purely unsymmetric thermal flux or symmetric tension stress, the value of G may turn zero or positive. It means that symmetric linear mechanical loading rather than unsymmetric thermal flux has an effect on G.Figure 5
G versus tension stress with 2a = 0.25, 0.5, 1, 2 mm for Q1=10,20,30,40J/m2×s.It is revealed that dimensionless thermal conductivity, the coefficient, and the size of crack have great effects on the heat flux per thickness to the crack surface, the mode II stress intensity factor, and the energy release rate. In the case of purely unsymmetric thermal flux, the mode II stress intensity factor and the energy release rate is zero. The mode II stress intensity factor and the energy release rate may be zero or positive under combined unsymmetric thermal flux and symmetric linear mechanical loading. The mode II stress intensity factor and the energy release rate may be zero or positive under combined unsymmetric thermal flux and symmetric linear mechanical loading. Contrary to the situations under symmetric thermal flux and symmetric mechanical loading [17], such consideration need not be taken when using the energy release rate as a criterion.


## 5. Conclusions

An improved partially permeable crack model is put forward to deal with the problem of a crack embedded in an orthotropic or isotropic solid under combined unsymmetric thermal flux and symmetric linear mechanical loading. Using the Fourier transform technique, the jumps of temperature, elastic displacements on the crack, and so on are obtained in the closed form. Numerical results show combined unsymmetric linear thermal flux, symmetric linear mechanical loading, dimensionless thermal conductivity, and the coefficient have great influences on fracture parameters. For the extended partially permeable crack, the mode II stress intensity factor and the energy release rate may be zero or positive under combined unsymmetric thermal flux and symmetric linear mechanical loading. Therefore, closure of the crack tip region need not be considered under combined unsymmetric thermal flux and symmetric linear mechanical loading when making use of fracture parameters as a criterion.



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