# Theoretical Analysis of Two Collinear Cracks in an Orthotropic Solid under Linear Thermal Flux and Linear Mechanical Load

Authors: Bing Wu; Bao-Yin Zhu; Hao-Meng Song; Qiong-Ao Huang
Journal: Advances in Mathematical Physics
 (2022)

---

## Abstract

This paper studies the problem of a cracked orthotropic solid subject to linear thermal flux and linear mechanical load. The proposed extended partially insulated crack model is employed to simulate two collinear cracks. Taking advantage of Fourier transform technique and superposition theory, the closed form of some physical quantities and fracture parameters is obtained. Some simple examples are employed to demonstrate dimensionless thermal conductivity (Rc) between the upper and below crack regions, and the proposed coefficient (ε) has great effects on some physical quantities and fracture parameters.


---

## Body

## 1. Introduction

Multicomponent composite materials are widely used in the material industry. However, considering the complex factors involving working environment, internal and external loads, and production process, it is inevitable to contain a series of various cracks in these solids. The appearances of different kinds of cracks will reduce the capacity of cracked structures and even bring about severe accidents. Therefore, it is vital to do some research on fracture analysis of a cracked solid by utilizing the theory of thermal elasticity for the purpose of safety [1–3]. With the rapid growth of thermoelasticity theory, a great deal of treatises and papers was published to investigate fracture characters of cracked solids [4–6]. The fracture parameters of an orthotropic material containing a central crack under heat flow were obtained by Tsai [7]. The closed form of fracture parameters of cracked orthotropic solids was calculated by Ju and Rowlands [8]. The closed form of some physical quantities of two collinear cracks was studied by Chen and Zhang [9]. The transient thermal problem of a cracked orthotropic plate was taken into account by Noda [10]. Some physical quantities of a cracked orthotropic semi-infinite medium were given by Rizk [11]. On the other hand, the thermoelastic problems of orthotropic functionally graded solids brought about the widespread attention. For example, the fracture parameters of orthotropic functionally graded solids under mechanical load were given explicitly by Kim and Paulino [12]. The problem of a cracked solid subject to plane temperature-step waves was investigated by Brock [13]. The equivalent domain integral was formulated to study the fracture problems subject to thermal stresses by Dag [14]. The problems of cracked orthotropic solids subject to symmetrical thermomechanical loads with application of Fourier transform technique (FTT) and superposition principle were studied by Wu et al. [15].Subject to thermal load, the analysis of fracture behavior for cracked solids which were often regarded as orthotropic or isotropic has generated enormous publicity [16–20]. To simulate two collinear cracks, a partially insulated crack model prevailed [21–23].
(1)Q1c=−hcΔT,where the definitions of Q1c, hc, and ΔT have been given in detail [3]. The case of hc⟶0 or hc⟶∞ denotes a fully thermally impermeable or permeable state.The following extended partially insulated crack model is also put forward by virtue of mathematical intuition.(2)Q1c=−hcΔT+εQ1,where Q1 presents initial heat flux. The coefficient ε is considered a constant. Whether it is negative or positive is mainly relies on the portion of thermal flux and mechanical load. Clearly, the crack model proposed in (2) returns to (1) when ε=0.The reasons of introducing constantεQ1 in (2) are as follows. First, the value of hc does not precisely address the cracks with thermal resistance. Second, the constant εQ1, which is introduced as an adjustment factor, conforms to the complex situation and meets the abnormal state of crack surface.This paper employs an extended partially insulated crack model to discuss two collinear cracks under linear thermal flux and linear mechanical load. The thermoelastic field is given in explicit form based on the proposed extended partially insulated crack model, Fourier transform, and superposition theory. The results show the effects of dimensionless thermal conductivity (Rc) between the upper and below crack regions and the proposed coefficient (ε) on Q1c and KII and S. It is revealed the boundary conditions of crack surface, thermal properties of crack, and the raised coefficient should be paid attention to the analysis of crack growth under thermal load in numerical results.


## 2. Problem Statement

Two collinear cracks in an orthotropic solid are taken into account as shown in Figure1. They are located at a<x<b.Figure 1
Two collinear cracks subject to linear thermal flux and linear mechanical load.Making use of the state of plane stress [3], we obtain
(3)σx=c11∂u∂x+c12∂v∂y−β1T,(4)σy=c12∂u∂x+c22∂v∂y−β2T,(5)τxy=c66∂u∂y+∂v∂x,where
(6)c11=Exx1−vxyvyx,c22=Eyy1−vxyvyx,c12=Exxvyx1−vxyvyx=Eyyvxy1−vxyvyx,β1β2=c11c12c12c22αxxαyy,where the definitions of u, v, σx, σy, τxyT, vxx, vyyExx, Eyy, c66=Gxy, αxx, and αyy have been given in [3].
(7)∂σx∂x+∂τxy∂y=0,∂τxy∂x+∂σy∂y=0.One obtains(8)c11∂2u∂x2+c66∂2u∂y2+c12+c66∂2v∂x∂y=β1∂T∂x,(9)c66∂2v∂x2+c22∂2v∂y2+c12+c66∂2u∂x∂y=β2∂T∂y.Making use of the Fourier heat conduction leads to(10)Qx=−λx∂T∂x,Qy=−λy∂T∂y,where the definitions of Qx, Qy, λx, and λy have been given in [3]. Furthermore, based on the equilibrium equation, one has
(11)∂Qx∂x+∂Qy∂y=0.Taking advantage of the thermal equilibrium equations brings out(12)λ2∂2T∂x2+∂2T∂y2=0,where
(13)λ=λxλy.The crack face boundary conditions are depicted as(14)QyIIx,0−QyIx,0=−Q1x2b−a2,a<x<b,(15)σyIIx,0=σyIx,0=σ1xb−a+σ0,a<x<b,Hereafter, the subscript ‘I’ or ‘II’ denotes the physical quantity of the upper (y>0) or below (y<0) part. Q0, Q1, σ0, and σ1 stand for the prescribed constants. Based on (14) and (15), thermal flux is composed of only one part (−Q1x/2b‐a2) and mechanical loading is divided into two parts (σ1x/b−a and σ0). As linear thermal flux and linear mechanical load are antisymmetrical and symmetrical, respectively, the thermoelastic field of the region (x>0) is only dealt with. The crack-surface boundary conditions are expressed with the application of the improved partially insulated crack model.
(16)τxyIIx,0=τxyIx,0=0,a<x<b,(17)QyIIx,0−QyIx,0=−Q1−Q1c2b−a2xa<x<b,(18)σyIIx,0=σyIx,0=σ1xb−a+σ0,a<x<b.According to Equations (17) and (18), the solutions under thermal flux (−Q1x/2b‐a2) and mechanical loading (σ0) have been given explicitly in [24, 25]. Next, we depict the boundary conditions of crack-surface subject to linear mechanical load (σ1x/b−a).
(19)τxyIx,0=τxyIIx,0=0,a<x<b,(20)σyIIx,0=σyIx,0=σ1xb−a,a<x<b,where
(21)Q1c=−hcTIx,0−TIIx,0+εQ1.Besides, some physical quantities conform to the following conditions:(22)τxyIx,0=τxyIIx,0,σyIx,0=σyIIx,0,x>bor0<x<a,uIx,0=−uIIx,0,vIx,0=−vIIx,0,x>bor0<x<a,TIx,0=TIIx,0,QyIx,0=QyIIx,0,x>bor0<x<a.


## 3. Solution Procedure

### 3.1. Temperature Field

According to [25], one obtains the explicit form of temperature difference on crack faces as
(23)TIx,0−TIIx,0=−Q1−Q1c2b−a2λλyx2−a2b2−x2,a<x<b.


### 3.2. Elastic Field

To achieve the goal of explicit form in Equations (8) and (9), uI,IIx,y and vI,IIx,y are expressed according to [26].
(24)uI,IIx,y=∑i=12uiI,IIx,y,vI,IIx,y=∑i=12viI,IIx,y,where the definitions of ujI,IIx,y and vjI,IIx,yj=1,2 have been given in [26].
(25)u1I,IIx,y=∑j=12∫0+∞gjI,IIξe−ξδ±γjysinξxdξ,(26)v1I,IIx,y=∑j=12∫0+∞ηjδ±gjI,IIξe−ξδ±γjycosξxdξ.Hereafter,δ+=1 or δ−=−1 denotes y>0 or y<0. gjI,IIξ need to solve. The definitions of γjj=1,2 have been given in [26].
(27)c22c66γ4+c122+2c12c66−c12c22γ2+c11c66=0,where
(28)ηj=c11−c66γj2c12+c66γj.Furthermore,u2I,IIx,y and v2I,IIx,y are chosen as
(29)u2I,IIx,y=∑j=12∫0+∞g∗I,IIξe−δ±ξλysinξxdξ,(30)v2I,IIx,y=∑j=12∫0+∞δ±L∗I,IIξe−δ±ξλycosξxdξ.Taking advantage of Equations (29), (30), (8), and (9), we have
(31)g∗I,IIξL∗I,IIξ=M1M2gI,IIξξ,where
(32)M1M2=c11−c66λ2−c12+c66λc12+c66λc66−c22λ2−1β1β2λ.By the aid of Equations (3)–(5), (25), (26), (29), and (30), the components of stress are in the form of the following expressions:
(33)σxI,IIx,0=∑j=12∫0+∞c11−c12γjηjξgjI,IIξcosξxdξ+c11M1−c12λM2−β1∫0+∞gI,IIξcosξxdξ,(34)σyI,IIx,0=∑j=12∫0+∞c12−c22γjηjξgjI,IIξcosξxdξ+c12M1−c22λM2−β2∫0+∞gI,IIξcosξxdξ,(35)τxyI,IIx,0=−c66∑j=12∫0+∞δ±γj+ηjξgjI,IIξsinξxdξ+∫0+∞δ±M1λ+M2gI,IIξsinξxdξ.In order to get the explicit solution of this considered problem, we depict the dual integral equations as(36)τxyIx,0=τxyIIx,0=0,x>0,(37)vIx,0=−vIIx,0=0,0<x<aorx>b.Using Equations (36) and (37), one gets
(38)gjIξ=gjIIξ,g2Iξ=−γ1+η1γ2+η2g1Iξ.Applying Equations (20) and (37), one obtains
(39)∫0+∞ξg1Iξcosξxdξ=0,0<x<aorx>b,(40)∫0+∞ξg1Iξcosξxdξ=σ1xℓ1b−a,a<x<b,where
(41)ℓ1=c12−c22γ1η1−γ1+η1γ2+η2c12−c22γ2η2.In order to solve Equations (39) and (40), the auxiliary function ϕx is introduced as
(42)ϕx=∂vIx,0∂x.Applying inverse Fourier transform leads to(43)g1Iξξ=−2γ2+η2η1γ2−η2γ1π∫abϕssinξsds.Inserting Equation (43) into (40), one has
(44)2π∫abϕsds∫0∞sinξscosξxdξ=−σ1η1γ2−η2γ1ℓ1γ2+η2b−ax.Recalling the known result [27],
(45)∫0∞cosξxsinξsdξ=ss2−x2.Based on Equation (45), Equation (44) can be expressed as
(46)1π∫ab2sϕss−xds=−σ1η1γ2−η2γ1ℓ1γ2+η2b−ax.It is convenient to introduces2=s¯,x2=x¯,2sds=ds¯, a2=a¯,b2=b¯, and ϕ¯s¯=ϕs. Equation (46) is rewritten as
(47)1π∫a¯b¯ϕ¯s¯s¯−x¯ds¯=σ1η1γ2−η2γ1ℓ1γ2+η2b−ax¯.According to the singular integral containing the Cauchy kernel4 [28], the solution of Equation (47) is obtained
(48)ϕ¯x¯=1πx¯−a¯b¯−x¯∫a¯b¯s¯s¯−a¯b¯−s¯x¯−s¯σ1η1γ2−η2γ1ℓ1γ2+η2b−ads¯.In the application of Equation (42), the closed form of elastic displacement is obtained
(49)vIx,0=∫a¯x¯ϕ¯t¯2t¯dt¯.Inserting Equation (48) into (34), the stress field is obtained as
(50)σyI,IIx,0=−σ1πb−aa2−x2b2−x2×2x2x2−a2−b2bFλ+2bx2Eλ+O1,0<x<a,σyI,IIx,0=−σ1πb−aa2−x2b2−x2×2x2x2−a2−b2bFλ+2bx2Eλ+O1,x>b.Fλ and Eλ denote the first and second kinds of complete elliptical integrals, respectively, where
(51)λ=b2−a2b.For simplicity, the detailed procedure of reduction under thermal flux is omitted. The shear stresses are obtained according to [25].
(52)τxyI,IIx,0=c66H2Pc22γ2η2−γ1η1x2−a2b2−x2×4a2b23+a2−b222+2b2a2+b2Eλ3Fλ−2x2−a2−b22+O1,0<x<a,τxyI,IIx,0=c66H2Pc22γ2η2−γ1η1x2−a2b2−x2×4a2b23+a2−b222+2b2a2+b2Eλ3Fλ−2x2−a2−b22+O1,x>b,where
(53)P=Q1−Q1cℓ2b−a2b+aλλy,ℓ2=H1H2,H1=γ1+η1c22γ2η2M1−c22λM2−β2+γ2+η2c22λM2+β2−c22γ1η1M1+c22M1λ+M2γ1η1−γ2η2H2=γ1+η1c22γ2η2−c12+γ2+η2c12−c22γ1η1.By superposition theory, the exact solutions of the physical quantities are obtained subject to linear thermal flux (−Q1x/2b−a2) and linear mechanical load (σ1x/b−a+σ0).


### 3.3. Crack-Tip Field

Using Equations (2) and (23), one obtains the closed form of heat flux to the crack surface.
(54)Q1c=2εQ1b−a2λ+Q1Rcx2−a2b2−x22b−a2λ+Rcx2−a2b2−x2.We define the value ofRc=λy/hc to stand for the dimensionless thermal resistance between crack faces. It is easily found Equation (54) is different from that in [25]. When Rc=0 and Rc⟶∞, one obtains Q1c=εQ1 or Q1c⟶Q1, meaning partially thermally insulated or fully conductive cracks. When Rc=0 and ε=0, one has Q1c=0, meaning fully thermally insulated cracks.


## 3.1. Temperature Field

According to [25], one obtains the explicit form of temperature difference on crack faces as
(23)TIx,0−TIIx,0=−Q1−Q1c2b−a2λλyx2−a2b2−x2,a<x<b.


## 3.2. Elastic Field

To achieve the goal of explicit form in Equations (8) and (9), uI,IIx,y and vI,IIx,y are expressed according to [26].
(24)uI,IIx,y=∑i=12uiI,IIx,y,vI,IIx,y=∑i=12viI,IIx,y,where the definitions of ujI,IIx,y and vjI,IIx,yj=1,2 have been given in [26].
(25)u1I,IIx,y=∑j=12∫0+∞gjI,IIξe−ξδ±γjysinξxdξ,(26)v1I,IIx,y=∑j=12∫0+∞ηjδ±gjI,IIξe−ξδ±γjycosξxdξ.Hereafter,δ+=1 or δ−=−1 denotes y>0 or y<0. gjI,IIξ need to solve. The definitions of γjj=1,2 have been given in [26].
(27)c22c66γ4+c122+2c12c66−c12c22γ2+c11c66=0,where
(28)ηj=c11−c66γj2c12+c66γj.Furthermore,u2I,IIx,y and v2I,IIx,y are chosen as
(29)u2I,IIx,y=∑j=12∫0+∞g∗I,IIξe−δ±ξλysinξxdξ,(30)v2I,IIx,y=∑j=12∫0+∞δ±L∗I,IIξe−δ±ξλycosξxdξ.Taking advantage of Equations (29), (30), (8), and (9), we have
(31)g∗I,IIξL∗I,IIξ=M1M2gI,IIξξ,where
(32)M1M2=c11−c66λ2−c12+c66λc12+c66λc66−c22λ2−1β1β2λ.By the aid of Equations (3)–(5), (25), (26), (29), and (30), the components of stress are in the form of the following expressions:
(33)σxI,IIx,0=∑j=12∫0+∞c11−c12γjηjξgjI,IIξcosξxdξ+c11M1−c12λM2−β1∫0+∞gI,IIξcosξxdξ,(34)σyI,IIx,0=∑j=12∫0+∞c12−c22γjηjξgjI,IIξcosξxdξ+c12M1−c22λM2−β2∫0+∞gI,IIξcosξxdξ,(35)τxyI,IIx,0=−c66∑j=12∫0+∞δ±γj+ηjξgjI,IIξsinξxdξ+∫0+∞δ±M1λ+M2gI,IIξsinξxdξ.In order to get the explicit solution of this considered problem, we depict the dual integral equations as(36)τxyIx,0=τxyIIx,0=0,x>0,(37)vIx,0=−vIIx,0=0,0<x<aorx>b.Using Equations (36) and (37), one gets
(38)gjIξ=gjIIξ,g2Iξ=−γ1+η1γ2+η2g1Iξ.Applying Equations (20) and (37), one obtains
(39)∫0+∞ξg1Iξcosξxdξ=0,0<x<aorx>b,(40)∫0+∞ξg1Iξcosξxdξ=σ1xℓ1b−a,a<x<b,where
(41)ℓ1=c12−c22γ1η1−γ1+η1γ2+η2c12−c22γ2η2.In order to solve Equations (39) and (40), the auxiliary function ϕx is introduced as
(42)ϕx=∂vIx,0∂x.Applying inverse Fourier transform leads to(43)g1Iξξ=−2γ2+η2η1γ2−η2γ1π∫abϕssinξsds.Inserting Equation (43) into (40), one has
(44)2π∫abϕsds∫0∞sinξscosξxdξ=−σ1η1γ2−η2γ1ℓ1γ2+η2b−ax.Recalling the known result [27],
(45)∫0∞cosξxsinξsdξ=ss2−x2.Based on Equation (45), Equation (44) can be expressed as
(46)1π∫ab2sϕss−xds=−σ1η1γ2−η2γ1ℓ1γ2+η2b−ax.It is convenient to introduces2=s¯,x2=x¯,2sds=ds¯, a2=a¯,b2=b¯, and ϕ¯s¯=ϕs. Equation (46) is rewritten as
(47)1π∫a¯b¯ϕ¯s¯s¯−x¯ds¯=σ1η1γ2−η2γ1ℓ1γ2+η2b−ax¯.According to the singular integral containing the Cauchy kernel4 [28], the solution of Equation (47) is obtained
(48)ϕ¯x¯=1πx¯−a¯b¯−x¯∫a¯b¯s¯s¯−a¯b¯−s¯x¯−s¯σ1η1γ2−η2γ1ℓ1γ2+η2b−ads¯.In the application of Equation (42), the closed form of elastic displacement is obtained
(49)vIx,0=∫a¯x¯ϕ¯t¯2t¯dt¯.Inserting Equation (48) into (34), the stress field is obtained as
(50)σyI,IIx,0=−σ1πb−aa2−x2b2−x2×2x2x2−a2−b2bFλ+2bx2Eλ+O1,0<x<a,σyI,IIx,0=−σ1πb−aa2−x2b2−x2×2x2x2−a2−b2bFλ+2bx2Eλ+O1,x>b.Fλ and Eλ denote the first and second kinds of complete elliptical integrals, respectively, where
(51)λ=b2−a2b.For simplicity, the detailed procedure of reduction under thermal flux is omitted. The shear stresses are obtained according to [25].
(52)τxyI,IIx,0=c66H2Pc22γ2η2−γ1η1x2−a2b2−x2×4a2b23+a2−b222+2b2a2+b2Eλ3Fλ−2x2−a2−b22+O1,0<x<a,τxyI,IIx,0=c66H2Pc22γ2η2−γ1η1x2−a2b2−x2×4a2b23+a2−b222+2b2a2+b2Eλ3Fλ−2x2−a2−b22+O1,x>b,where
(53)P=Q1−Q1cℓ2b−a2b+aλλy,ℓ2=H1H2,H1=γ1+η1c22γ2η2M1−c22λM2−β2+γ2+η2c22λM2+β2−c22γ1η1M1+c22M1λ+M2γ1η1−γ2η2H2=γ1+η1c22γ2η2−c12+γ2+η2c12−c22γ1η1.By superposition theory, the exact solutions of the physical quantities are obtained subject to linear thermal flux (−Q1x/2b−a2) and linear mechanical load (σ1x/b−a+σ0).


## 3.3. Crack-Tip Field

Using Equations (2) and (23), one obtains the closed form of heat flux to the crack surface.
(54)Q1c=2εQ1b−a2λ+Q1Rcx2−a2b2−x22b−a2λ+Rcx2−a2b2−x2.We define the value ofRc=λy/hc to stand for the dimensionless thermal resistance between crack faces. It is easily found Equation (54) is different from that in [25]. When Rc=0 and Rc⟶∞, one obtains Q1c=εQ1 or Q1c⟶Q1, meaning partially thermally insulated or fully conductive cracks. When Rc=0 and ε=0, one has Q1c=0, meaning fully thermally insulated cracks.


## 4. Fracture Parameters

It is important that the stress intensity factors including the mode-I and mode-II should be defined as the analysis of cracked growth.(55)KIInn=limx⟶a−2πa−xσyI,IIx,0,KIOut=limx⟶b+2πx−bσyI,IIx,0,(56)KIIInn=limx⟶a−2πa−xτxyI,IIx,0,KIIOut=limx⟶b+2πx−bτxyI,IIx,0.Based on Equation (55), one can obtain
(57)KIInn=πab2−a22a2bFλ−Eλσ1πb−a,KIOut=πbb2−a22bb2Eλ−a2Fλσ1πb−a.Whena=0, it means the mode-I stress intensity factor of a single crack with 2b. It is easily found that
(58)KIOut=2πbσ1π.According to Equation (56) and Reference [25], one has
(59)KIIInn=c66H2P2c22γ2η2−γ1η14a2b23−a2−b222−2b2a2+b2Eλ3Fλπab2−a2,KIIOut=−c66H2P2c22γ2η2−γ1η14a2b23−a2−b222−2b2a2+b2Eλ3Fλπbb2−a2.The importance of strain energy in a unit volume of the solid is illustrated for nonisothermal [29, 30].
(60)dWdV=Sr.where the definitions of Sand r have been given in [24]. For the orthotropic solid, Equation (60) can also be given based on the above concepts of energy density function
(61)Sr=c22σxI2+c11σyI2−2c12σxIσyI2c11c22−c122+τxyI22c66.The following strain energy density factor on the crack line is defined to study crack growth in fracture mechanics [24].
(62)SInn,Out=14πc22ℓ2+c11−2c12ℓc11c22−c122KIInn,Out2+1c66KIIInn,Out2,where
(63)ℓ=1ℓ1∑j=12−1jγ1+η1γj+ηjc11−c12γjηj.


## 5. Numerical Results

For the sake of simplicity, some numerical examples are employed to demonstrateRc and ε have great effects on Q1c,KII,andS subject to linear thermal flux(−Q1x/2b−a2) and linear mechanical load (σ0x/b−a). The orthotropic material like Tyrannohex is selected as in [31] (Table 1).Table 1
Tyrannohex.
ExxMPaEyyMPaGxyMPavxyvyxαxx10−7/°Cαyy10−7/°Cλxw/m°Cλyw/m°C13500087000500000.150.0966732323.082.81Figure2 shows Q1c/Q1 versus x/b with Rc=1,2,3,4 for ε=0.01 and a/b=3/4. Q1c/Q1 increases with an increase of Rc for a fixed x/b as shown in Figure 2. The case of Rc=1 implies the heat conduction between the upper and blow crack faces is the same as that of external material of crack. The cases of Rc=2,3,4 imply to the heat conduction between the upper and blow crack faces are twice, triple, and quadruple as much as thermal conductivities of the external material of crack. Figure 3 displays Q1c/Q1 versus x/b with ε=−0.02,0,0.02,0.04 for Rc=2 and a/b=3/4. As the dimensionless thermal resistance Rc increases, Q1c/Q1 increases. The constant ε is considered to be an adjustment quantity. The bigger the value of constant ε, the greater the heat flux per thickness through crack. It means making use of the extended partially insulated crack model involving the greater Rc or ε will overestimate the heat flux per thickness to the crack surface. Furthermore, it is suitable to decease or increase stress field near outer and inn cracks tip by the way of filling certain materials into the region between the upper and below crack faces according to Figures 2 and 3.Figure 2
Q1c/Q1 versus x/b with Rc=1,2,3,4 for ε =0.01 and a/b=3/4.Figure 3
Q1c/Q1 versus x/b with ε=−0.02,0,0.02,0.04 for Rc=2 and a/b=3/4.Figure4 displays KIIInn/KII0Inn or KIIOut/KII0Out versus x/b with a/b=0.25,0.5,0.75 where KII0Inn and KII0Out denote KIIInn and KIIOut for Q1c=0, respectively. KIIInn/KII0Inn or KIIOut/KII0Out decreases when Rc increases for a fixed x/b. Figure 5 displays KIIInn/KII0Inn or KIIOut/KII0Out versus x/b with Rc=1,2,3,4 for ε=0.01 and a/b=3/4. As the dimensionless thermal resistance Rc increases, the mode-II stress intensity factors decrease. The bigger the value of constant ε, the smaller the mode-II stress intensity factors. It means making use of the extended partially insulated crack model involving the greater Rc and ε will underestimate the mode-II stress intensity factors. The obtained results reveal that the crack face boundary conditions, the thermal properties of crack, and the raised coefficients have great influences on the heat flux per thickness to the crack surface and the mode-II stress intensity factors.Figure 4
KIIInn/KII0Inn or KIIOut/KII0Out versus x/b with Rc=1,2,3,4 for ε =0.01 and a/b=3/4.Figure 5
KIIInn/KII0Inn or KIIOut/KII0Out versus x/b with ε=−0.02,0,0.02,0.04 for Rc=2 anda/b=3/4.In order to present the influence of the thermal properties of crack onSInn,Out, the value of S0 is easily defined.
(64)S0=c22ℓ2+c11−2c12ℓ4c11c22−c112σ02,which denotes the strain energy density factor of a crack with 2b under mechanical load σ0. Figures 6 and 7 show SInn/S0 and SOut/S0 versus Q1 with ε=−0.03,0,0.03 for Rc=2, σ0=1 MPa, and a/b=1/4. Figures 6 and 7 respond to the two cases of strain energy density factor near outer and inn cracks for partially thermally insulated cracks. It is easily seen that SInn/S0 and SOut/S0 are made up of the mode-II stress intensity factor under thermal flux (−Q1x/2b−a2) and the mode-I stress intensity factor induced by mechanical load (σ0x/b−a). The corresponding SInn/S0 and SOut/S0 increase with an increase of thermal flux and mechanical load. The strain energy density factor on the crack line is greatly influenced by the adjustment quantity ε. The bigger the value of constant ε, the smaller SInn/S0 and SOut/S0. So, applying the bigger value of constant ε will underestimate SInn/S0 and SOut/S0.Figure 6
SInn/S0 versus Q1 with ε=−0.03,0,0.03 for Rc=0, σ0=1 MPa, anda/b=1/4.Figure 7
SOut/S0 versus Q1 with ε=−0.03,0,0.03 for Rc=0, σ0=1 MPa, anda/b=1/4.From the above figures, it is revealedRc and ε have significant impacts on the analysis of a cracked solid. In other words, some physical quantities (i.e., Rc and ε) should be given enough attention to the analysis of the thermoelastic field.


## 6. Conclusions

This paper addresses two collinear cracks in an orthotropic solid under linear thermal flux and linear mechanical load in this paper. Some physical quantities and fracture parameters are obtained in explicit forms with application of the proposed extended partially insulated crack model, Fourier transform, and superposition theory. The results show thatRc and ε have vital effects on Q1c and some fracture parameters. The obtained results reveal the boundary conditions of crack face, thermal properties of crack, and the raised coefficients should be concerned about the analysis of a cracked solid under the thermal load.



---

*Source: 8371954-2022-02-09.xml*