Fracture Analysis for a Crack in Orthotropic Material Subjected to Combined 2I-Order Symmetrical Thermal Flux and 2J-Order Symmetrical Mechanical Loading

Article Fracture Analysis for a Crack in Orthotropic Material Subjected to Combined 2i-Order Symmetrical Thermal Flux and 2j-Order Symmetrical Mechanical Loading

Bing Wu 1, Daren Peng 2,* and Rhys Jones 2

1 College of Civil Engineering and Architecture, Hebei University, Baoding 071002, China; [email protected] 2 Centre of Expertise Structural Mechanics, Department of Mechanical and Aerospace Engineering, Monash University, Clayton, VIC 3800, Australia; [email protected] * Correspondence: [email protected]

Abstract: The problems of crack formation in orthotropic materials under 2i order polynomial function heat flow and 2j order polynomial function mechanical loading are considered. An extended local insulation crack model is proposed, and fracture analysis is carried out for the above problems. Utilizing Fourier transform technique (FTT) and principle of superposition, the jumps of temperature, elastic displacements on the crack, and so on are obtained. The advantage of this analysis is that the explicit closed form solutions of main parameters in classical fracture mechanics, i.e., the stress intensity factor, the energy release rate, and the energy density have been presented. A simple example is used to demonstrate the method proposed in this paper. The analysis results show that the non-dimensional thermal conductivity and the combined ratio of the heat flux per thickness perpendicular to the crack surface to the mechanical load have a great influence on the calculation of fracture parameters. Only when they meet certain conditions can the correct fracture parameter Citation: Wu, B.; Peng, D.; Jones, R. calculation results be obtained. Fracture Analysis for a Crack in Orthotropic Material Subjected to Keywords: extended partially insulated crack; FTT; fracture parameters; closed form solutions Combined 2i-Order Symmetrical Thermal Flux and 2j-Order Symmetrical Mechanical Loading. Appl. Mech. 2021, 2, 127–146. https://doi.org/10.3390/ 1. Introduction applmech2010008 With the thermal-stress effects in consideration, fracture behavior of engineering structures resulting from applied thermal loading has been induced [1,2]. So, it is of much Received: 30 January 2021 importance to conduct research on thermo-elastic field of various cracks embedded in Accepted: 26 February 2021 a solid in genuine engineering situations [3–5]. A range of papers published have been Published: 4 March 2021 investigating the fracture behaviors of an orthotropic solid with a variety of cracks under thermal loading [6–14]. Tsai considered the stress intensity factors for a cracked orthotropic Publisher’s Note: MDPI stays neutral material under uniform heat flow [10]. Chen addressed thermo-elasticity problem of two with regard to jurisdictional claims in collinear cracks embedded in an orthotropic solid [11]. Wilson and Yu gave the way to the published maps and institutional affil- J-integral approach to compute the thermal stress intensity factor for cracked problems [12]. iations. Prasad et al. calculated the stress intensity factors of a crack in isotropic plate subjected to the transient thermoelastic loading with application of two-dimensional dual boundary element method [13]. Chang et al. gave the closed-form solution of a penny-shaped crack embedded in an infinite linear transversely isotropic triple medium under uniform anti- Copyright: © 2021 by the authors. symmetric heat flux [14]. Hu and Chen obtained the temperature field and thermal stress Licensee MDPI, Basel, Switzerland. of a partially insulated cracked material in a half-plane subjected to temperature impact This article is an open access article by using heat conduction theory [15,16]. Brock investigated the problem of a crack which distributed under the terms and was embedded in a thermo-elastic solid under plane temperature-step waves [17]. Wu conditions of the Creative Commons et al. used Fourier transform technique (FTT) to address the thermo-elastic field of two Attribution (CC BY) license (https:// collinear cracks under anti-symmetrical linear thermal flux [18]. Afterwards, Wu et al. creativecommons.org/licenses/by/ obtained the explicit solution of a crack under combined quadratic thermo-mechanical 4.0/).

Appl. Mech. 2021, 2, 127–146. https://doi.org/10.3390/applmech2010008 https://www.mdpi.com/journal/applmech Appl. Mech. 2021, 2 128

loads by making use of FTT and principle of superposition [19]. Rekik considered the response of an elastic material under thermal loading that had an effect on mixed-mode crack which is axisymmetric and partially insulated [20]. The method of combining differential quadrature method (DQM) and dual-phase-lag (DPL) theory to solve the thermoelastic or viscoelastic fracture analysis of orthotropic materials can be found in References [21,22]. The applications of fracture analysis under thermal ﬁeld in other ﬁelds, such as functionally graded material (FGM) coating and Mageto- Electroelastic solid, are presented in References [23–27]. In the above-mentioned works, the following partially insulated crack model is fre- quently used: Qc = −hc∆T (1a)

where Qc denotes the heat ﬂux per thickness to the crack surface, hc stands for the thermal conductivity on the crack region, and ∆T is the temperature difference on the crack, respectively. The values of hc → 0 and hc → ∞ represent completely thermally insulated and permeable cases on the crack surface, respectively. Furthermore, based on the mathematical intuition, an extended partially insulated crack model in Equation (1a) can be proposed as [28] Qc = −hc∆T + εQ0, (1b)

where Q0 denotes initial heat flux, and the coefficient ε can be an arbitrary constant. Obviously, when ε = 0, the crack model presented in Equation (1b) reduces to that with the crack-face boundary thermal condition of (1a). Moreover, there follow two reasons of introducing the constant εQ0 in Equation (1b). Firstly, the thermal conductivity through the crack region hc is invariably regarded as a constant. Nevertheless, thermal loading has an effect on hc in practical engineering. In other words, the value of hc cannot be used to precisely investigate crack filled with a certain thermal resistance through the crack region. So, the constant εQ0 is regarded as an adjustment factor to be consistent with the complicated and genuine situation, and to model on the non-optimal cases of complicated crack surface. The coefficient ε can turn positive or negative relying on temperature field. An extended partially insulated crack model is proposed to consider the problem of a single crack in an orthotropic solid subjected to combined 2i order polynomial function thermal flux and 2j order polynomial function mechanical loading. With application of FTT, the thermo-elastic partial differential equations (PDE) are transformed to singular integral equations, on which the thermo-elastic fields with principle of superposition are based. The obtained results reveal that dimensionless thermal conductivity and the coefficient have influences on the heat flux per thickness normal to the crack surface and fracture parameters for cracked solid subjected to combined constant thermo-mechanical loading.

2. Problem Statement A single crack embedded in an orthotropic solid is shown in Figure1. And 2 i order polynomial function thermal ﬂux and 2j order polynomial function mechanical loading are considered. Appl. Mech. 2021, 2, FOR PEER REVIEW 3

Appl. Mech. 2021, 2 A single crack embedded in an orthotropic solid is shown in Figure 1. And 2i order129 polynomial function thermal flux and 2j order polynomial function mechanical loading are considered.

FigureFigure 1. 1.A A single single crack crack which which is is embedded embedded in in anan orthotropicorthotropicsolid solidsubjected subjected toto combinedcombined 22ii orderorder polynomialpolynomial function function thermal thermal ﬂux flux and and 2 j2orderj order polynomial polynomial function function mechanical mechanical loading. loading.

TheThe constitute constitute equations equations are are given given based based on on the the state state of of plane plane stress: stress: ∂∂uv σ =+−∂u ∂v β θ , σ =xxc cc11+ c 12− β θ 1 , (2)(2) xx 11 ∂x ∂∂xy12 ∂y 1

∂u ∂∂uv∂v σ =σ c =+−+ c − ββθθ , yy yy 12 cc1222 222 2 , (3)(3) ∂x ∂∂xy∂y ∂u ∂v σxy = c66[ ∂∂+uv], (4) σ =+∂y ∂x , xy c66[] (4) ∂∂yx where Exx Eyy where c11 = , c22 = , (5) 1 − vxyvyx 1 − vxyvyx E v E v xx yx yy Exyyy c12 = Exx = = , (6) c = − , c22 − , (5) 11 1 −vxyvyx 1 v−xyvyx 1 vvxy yx 1 vvxyyx β c c α 1 = 11 12 xx , (7) β2 cE12vEvc22 αyy c ==xx yx yy xy , 12 −− (6) where σyy and σxy represent the components11vvxy yxof stress, vv xyE yx xx and Eyy denote the Young’s moduli, θ is the temperature, vxx and vyy indicate the Poisson’s ratios, u and v represent the components of elastic displacement,β c66 = Gxy representα the shear modulus, and αxx 11112cc xx and αyy are the coefﬁcients of linear expansion.= Utilizing the, following equations: β α (7) 21222cc yy ∂σ ∂σxy ∂σxy ∂σyy xx + = 0, + = 0 (8) σ σ ∂x ∂y ∂x ∂y where yy and xy represent the components of stress, Exx and Eyy denote the one obtains θ 2 2 2 Young’s moduli, is the∂ temperature,u ∂ u vxx and vyy∂ indicatev the∂T Poisson’s ratios, u and c + c + (c + c ) = β (9) 11 ∂x2 66 ∂y2 12 66 ∂x∂y = 1 ∂x v represent the components of elastic displacement, cG66 xy represent the shear mod- ∂2v ∂2v ∂2u ∂T α αc + c + (c + c ) = β (10) ulus, and xx and y66y ∂ arex2 the22 coefficients∂y2 12 of linear66 ∂x ∂expansion.y 2 ∂y Utilizing the following equations: Based on the Fourier heat conduction, it can be obtained as follows:

∂T ∂T Q = −λ , Q = −λ (11) x x ∂x y y ∂y

Appl. Mech. 2021, 2 130

where Qx, Qy denote the components of the heat ﬂux, and λx and λy denote the coefﬁcient of heat conduction. Furthermore, in accordance with equilibrium equation, one leads to

∂Q ∂Qy x + = 0 (12) ∂x ∂y

and assuming thermal equilibrium equation, one obtains

∂2T ∂2T λ2 + = 0 (13) ∂x2 ∂y2

where s λ λ = x (14) λy In this paper, we take advantage of the crack-face boundary conditions as follows:

n x2i QI (x ) − QII (x ) = −(Q − Q ) −a < x < a y , 0 y , 0 0i ci ∑ 2i+1 (15a) i=0 2a

n σ2j σII (x ) = σI (x ) = x2j −a < x < a yy , 0 yy , 0 ∑ 2j (15b) j=0 a The superscripts I and II represent the physical quantities of y > 0 and y < 0 region, respectively. Qi and σ2j denote the prescribed constants. With the symmetry of the thermomechanical loadings in account, we simply consider half of the thermo-elastic ﬁeld n 2i 2i+1 (i.e., x > 0 and −∞ < y < +∞) under the thermomechanical loadings − ∑ Q0ix /2a i=0 n 2j 2j and ∑ σ2jx /a . The following boundary conditions of the crack-surface are depicted as j=0

I II σxy(x, 0) = σxy(x, 0) = 0 0 < x < a (16)

n II I 2i 2i+1 Qy (x, 0) − Qy(x, 0) = −(Q0i − Qci)∑ x /2a 0 < x < a (17a) i=0 n σ2j II (x ) = I (x ) = x2j < x < a σyy , 0 σyy , 0 ∑ 2j 0 (17b) j=0 a From Equations (17a) to (17b), the thermal flux and the mechanical loading consist 2i 2i+1 of several parts. Firstly, the solutions under the thermal flux −Q0ix /2a and mechan- 2j 2j ical loading σ2jx /a will be given. Secondly, utilizing principle of superposition, such expressions as temperature change, elastic displacement on the crack and stresses under n n 2i 2i+1 2j 2j thermal flux − ∑ Q0ix /2a , and mechanical loading ∑ σ2jx /a , will be obtained. i=0 j=0 The thermo-elastic boundary conditions across the crack-surface subjected to the thermal 2i 2i+1 2j flux −Q0ix /2a and mechanical loading σ2j/a are as follows.

I II σxy(x, 0) = σxy(x, 0) = 0 0 < x < a (18)

2i (Q − Q )x σ2j QI (x, 0) − QII (x, 0) = − 0i ci , σII (x, 0) = σI (x, 0) = x2j 0 < x < a (19) y y 2a2i+1 yy yy a2j where I II Qci = −hc(T (x, 0) − T (x, 0)) + εQi0 (20) Appl. Mech. 2021, 2 131

Moreover, such factors as stresses, elastic displacements, and temperature and heat ﬂux across the crack-free region of the horizontal x-axis comply with the following conditions. I ( ) = II ( ) I ( ) = II ( ) > < − σxy x, 0 σxy x, 0 , σyy x, 0 σyy x, 0 x a or x a (21) uI (x, 0) = −uII (x, 0), vI (x, 0) = −vII (x, 0) x > a or x < −a (22) I ( ) = II ( ) I ( ) = II ( ) > < − T x, 0 T x, 0 , Qy x, 0 Qy x, 0 x a or x a (23)

3. Solution Procedure 3.1. Temperature Field The temperature ﬁeld can be solved out ﬁrst now that Equation (13) is irrelevant to the elastic strain. With application of FTT, one can depict the solution of Equation (13) as follows: Z +∞ ± TI,II (x, y) = g±(ξ)e−ξδ λy cos(ξx)dξ (24) 0 g±(ξ) are unknown and will be solved. δ+ = 1 and δ− = −1 denote y > 0 and y < 0, respectively. From (11), one gets

Z ∞ I,II ± −ξδ±λy Qx (x, y) = λx ξg (ξ)e sin(ξx)dξ (25) 0 Z ∞ I,II ± ± −ξδ±λy Qy (x, y) = λyλ δ ξg (ξ)e cos(ξx)dξ (26) 0 Applying the second relation in Equation (23), one gets

g+(ξ) = −g−(ξ) (27)

Utilizing the ﬁrst relation in Equations (23) and (19), one attains

Z +∞ g+(ξ) cos(ξx)dξ = 0 x > a (28) 0

Z +∞ (Q − Q )x2i g+( ) ( x)d = − ci 0i < x < a ξ cos ξ ξ 2i+1 0 (29) 0 4a λλy

For the solvation of Equations (28) and (29), the auxiliary function γ2i(x) is deﬁned as

∂[TI (x, 0) − TII (x, 0)] γ(x) = (30) ∂x Applying inverse Fourier transform leads to

1 Z a g+(ξ)ξ = − γ(s) sin(ξs)ds (31) π 0 Substituting Equation (31) into (29), one gets

2 Z a Z +∞ (Q − Q )x2i (s)ds ( s) ( x)d = 0i ci γ sin ξ cos ξ ξ 2i+1 (32) π 0 0 2a λλy

from the known result [29].

Z +∞ 1 1 1 cos(ξx) sin(ξs)dξ = ( + ) (33) 0 2 s − x s + x Appl. Mech. 2021, 2 132

Equation (32) can be expressed as follows:

1 Z a 1 (Q − Q )x2i (s) ds = 0i ci γ 2i+1 (34) π −a s − x 2a λλy

Equation (34) features a singular integral equation including the Cauchy kernel [28], and the solution can be unraveled as √ 1 Z a a2 − s2 (Q − Q )s2i C (x) = √ 0i ci ds + √ 2i γ 2i+1 (35) π a2 − x2 −a x − s 2a λλy a2 − x2

Considering the following condition:

Z a γ(s)ds = 0 (36) −a

After some computations, one has C2i = 0, and Equation (35) can be given as

(Q − Q )x γ(x) = 0i √ ci i= 0 (37a) 2 2 2aλλyπ a − x

2i−1 j (Q0i − Qci)x (Q0i − Qci) 2i+1 γ(x) = − √ Ij + √ x i ≥ 1 (37b) ∑ 2i+1 2 2 2i+1 2 2 j=0 2a λλyπ a − x 2a λλy a − x where Z a q 2 2 2i−l−1 Il = − (a − s )s ds(1 ≤ l ≤ 2i−1). (38) −a In addition, with application of Equations (30), (37a) and (37b), the temperature change on the crack is given as

(Q − Q ) p TI (x, 0) − TII (x, 0) = − 00 c0 a2 − x2 i = 0 (39a) aλλy

2i−1 (Q − Q )I Z x sl (Q − Q ) Z x s2i+1 TI (x, 0) − TII (x, 0) = − 0i √ci l √ ds + 0i ci √ ds i ≥ 1 (39b) ∑ 2i+1 2 2 2 2 2i+1 2 2 l=0 2a λλyπ a − x −a a − s 2a λλy −a a − s

where

l E(x, l) = R x √ s ds l = 2k < 2i − 1 −a a2−s2 l (− )l/2 l/2−1 l [ − )( ( ))] = 1 arcsin(x) + 1 (−1)m sin l 2m arcsin x/a 2l 2l−1 ∑ l−2m (40a) l/2 m=0 m l (− )l/2 l/2−1 l [−( − ) ] − − π + 1 (−1)m sin l 2m π/2 2l+1 2l ∑ l−2m l/2 m=0 m

l E(x, l) = R x √ s ds l = 2k+1 ≤ 2i − 1 −a a2−s2 (l−1)/2 (− )(l+1)/2 l [( − )( ( ))] = 1 (−1)m cos l 2m arcsin x/a 2l−1 ∑ l−2m (40b) m=0 m (l−1)/2 (− )(l+1)/2 l [( − ) ] − 1 (−1)m cos l 2m π/2 2l−1 ∑ l−2m m=0 m

2i+1 F(x, i) = R x √s ds −a 2− 2 a s (− )i+1 i 2i + 1 [( + − )( ( ))] = 1 (−1)m cos 2i 1 2m arcsin x/a 22i ∑ 2i+1−2m (40c) m=0 m (− )i+1 i 2i + 1 [ ( + − ) ] − 1 (−1)m cos π 2i 1 2m /2 22i ∑ 2i+1−2m m=0 m Appl. Mech. 2021, 2 133

3.2. Elastic Field In order to solve Equations (9) and (10), uI,II (x, y) and vI,II (x, y) are written as [19]

uI,II (x y) = uI,II (x y) + uI,II (x y) vI,II (x y) = vI,II (x y) + vI,II (x y) , 1 , 2 , , , 1 , 1 , (41)

uI,II (x y) vI,II (x y) TI,II (x y) = uI,II (x y) 1 , and 1 , respond to the general solutions under , 0. 2 , I,II and v2 (x, y) are the peculiar solutions subjected to thermal ﬂux loading. With application uI,II (x y) vI,II (x y) of FTT, 1 , and 1 , can be written as

2 Z +∞ ± uI,II (x, y) = g±(ξ)e−ξδ γjy sin(ξx)dξ (42) 1 ∑ j j=1 0

2 Z +∞ ± I,II ( ) = ± ±( ) −ξδ γjy ( ) v1 x, y ∑ ηjδ gj ξ e cos ξx dξ (43) j=1 0 g±( ) (j = ) Re > j ξ have not been solved. γj 1, 2 ( γj 0) are the roots of character equation as follows: 4 2 2 c22c66γ + (c12 + 2c12c66 − c12c22)γ + c11c66 = 0 (44) where 2 c11 − c66γj ηj = (45) (c12 + c66)γj I,II I,II Furthermore, u2 (x, y) and v2 (x, y) are chosen as

2 Z +∞ I,II ∗± −δ±ξλy u2 (x, y) = ∑ g (ξ)e sin(ξx)dξ (46) j=1 0

2 Z +∞ I,II ± ∗± −δ±ξλy v2 (x, y) = ∑ δ L (ξ)e cos(ξx)dξ (47) j=1 0 Substituting Equations (46) and (47) into Equations (9) and (10), one has

∗± ± g (ξ) K1 g (ξ) ∗± = (48) L (ξ) K2 ξ

where 2 −1 K1 c11 − c66λ −(c12 + c66)λ β1 = 2 (49) K2 (c12 + c66)λ c66 − c22λ β2λ Based on Equations (2)–(4), (42), (43), (46) and (47), the components of stress are rewritten as

2 I,II R +∞ ± σxx (x, 0) = ∑ 0 (c11 − c12γjηj)ξgj (ξ) cos(ξx)dξ j=1 (50) R +∞ ± + (c11 M1 − c12λM2 − β1) 0 g (ξ) cos(ξx)dξ

2 I,II R +∞ ± σyy (x, 0) = ∑ 0 (c12 − c22γjηj)ξgj (ξ) cos(ξx)dξ j=1 (51) R +∞ ± + (c12 M1 − c22λM2 − β2) 0 g (ξ) cos(ξx)dξ " 2 I,II R +∞ ± ± σxy (x, 0) = −c66 ∑ 0 δ (γj + ηj)ξgj (ξ) sin(ξx)dξ j=1 (52) R +∞ ± ± i + 0 δ (M1λ + M2)g (ξ) sin(ξx)dξ Appl. Mech. 2021, 2 134

To solve the problem, it is divided into two parts. One part is what is resulted 2j 2j from mechanical loading σ2jx /a . The other part is what is induced by thermal ﬂux 2i 2i+1 −Q0ix /2a . Firstly, the mechanical loading is considered and the dual integral equations are written. I II σxy(x, 0) = σxy(x, 0) = 0 x > 0 (53) vI (x, 0) = −vII (x, 0)= 0 x > a (54) Utilizing Equations (53) and (54), one obtains

+ − + γ1 + η1 − gj (ξ) = gj (ξ), g2 (ξ) = − g1 (ξ) (55) γ2 + η2

Applying the second relation in Equations (19) and (54), one arrives at

Z +∞ + ξg1 (ξ) cos(ξx)dξ = 0 x > a (56) 0

+ 2i Z ∞ σ2jx g+( ) ( x)d = < x < a ξ 1 ξ cos ξ ξ 2i 0 (57) 0 ω1a where γ1 + η1 ω1 = (c12 − c22γ1η1) − (c12 − c22γ2η2) (58) γ2 + η2 The auxiliary function φ(x) is chosen to solve Equations (56) and (57),

∂vI (x, 0) φ(x) = (59) ∂x With knowledge of inverse Fourier transform (IFT), one has

Z a + 2(γ2 + η2) g1 (ξ)ξ = − φ(s) sin(ξs)ds (60) (η1γ2 − η2γ1)π 0

According to the known result of Equation (33), inserting Equation (60) into (57), we have 1 Z a φ(s) σ2j(η1γ2 − η2γ1) ds = x2j 2j (61) π −a s − x ω1(γ2 + η2)a Based on the singular integral theory including the Cauchy kernel, the following solution is given: √ 1 Z a a2 − s2 σ2j(η1γ2 − η2γ1) D2j (x) = √ s2jds + √ φ 2j (62) π a2 − x2 −a x − s ω1(γ2 + η2)a a2 − x2

After some computations, D2j = 0. With application of Equations (59) and (62), one obtains the elastic displacement as √ Z x σ (η γ − η γ ) a2 − x2 vI (x, 0) = φ(s)ds = 0 1 2 2 1 j = 0 (63a) −a ω1(γ2 + η2)

2j−1 R x σ2j(η1γ2−η2γ1)Il R x l σ2j(η1γ2−η2γ1) R x 2j+1 vI (x, 0) = φ(s)ds = − ∑ √ √ s ds + √s ds j ≥ 1 (63b) −a ( + ) a2j a2−x2 −a a2−s2 ω (γ +η )a2j −a a2−s2 l=0 ω1 γ2 η2 π 1 2 2 Furthermore, one can obtain the stress ﬁelds as follows: I,II x σyy (x, 0) = − 1 − √ σ0 j= 0 (64) x2 − a2 Appl. Mech. 2021, 2 135

2j−1 σ ∑ sl I Z a 2j l Z a σ s2j+1 I,II ( ) = − √ l=0 + √ 2j ≥ σyy x, 0 ds ds j 1 (65) −a a2j a2 − s2(s − x) −a a2j a2 − s2(s − x) Then, the explicit solution of elastic field under 2i order polynomial function thermal flux will be obtained. With the application of heat flux, it fits in the following condition.

I II σyy(x, 0) = σyy(x, 0) = 0 0 < x < a (66)

With application of Equations (51) and (66), one arrives at

+ − gj (ξ) = −gj (ξ) (67)

2 + + g (ξ) ∑ (c12 − c22γjηj)gj (ξ) = (c22λM2 + β2 − c12 M1) (68) j=1 ξ With application of Equation (16), one can depict the ﬁrst relation in Equation (22), and the dual integral equations are: " # Z +∞ 2 + + M1g (ξ) ∑ gj (ξ) sin(ξx) + sin(ξx) dξ = 0 x > a (69) 0 j=1 ξ

Z +∞ 2 Z +∞ + + ∑ (γj + ηj)ξgj (ξ) sin(ξx)dξ + (M1λ + M2)g (ξ) sin(ξx)dξ = 0 0 < x < a (70) 0 j=1 0 To achieve the solvation of Equations (69) and (70), the auxiliary function χ(x) can be deﬁned as ∂uI (x, 0) χ(x) = (71) ∂x Applying Equation (69) and inverse Fourier transform leads to

2 Z a + + 2 ∑ gj (ξ)ξ + M1g (ξ) = χ(s) cos(ξs)ds (72) j=1 π 0

Utilizing Equations (68) and (72), one obtains

Z a + c22λM2 + β2 − c22γ2η2 M1 + c22γ2η2 − c12 2 g1 (ξ)ξ = g (ξ) + χ(s) cos(ξs)ds (73) c22(γ2η2 − γ1η1) c22(γ2η2 − γ1η1) π 0

Z a + c22γ1η1 M1 − c22λM2 − β2 + c12 − c22γ1η1 2 g2 (ξ)ξ = g (ξ) + χ(s) cos(ξs)ds (74) c22(γ2η2 − γ1η1) c22(γ2η2 − γ1η1) π 0 With application of Equations (70), (73) and (74), one obtains

Z a Z +∞ Z +∞ + 2 χ(s)ds sin(ξx) cos(ξs)dξ = πω2 g (ξ) sin(ξx)dξ 0 < x < a (75) 0 0 0 where H1 ω2 = (76) H2 With

H = (γ + η )(c γ η M − c λM − β ) + (γ + η ) 1 1 1 22 2 2 1 22 2 2 2 2 (77) (c22λM2 + β2 − c22γ1η1 M1) + c22(M1λ + M2)(γ1η1 − γ2η2)

H2 = (γ1 + η1)(c22γ2η2 − c12) + (γ2 + η2)(c12 − c22γ1η1) (78) Appl. Mech. 2021, 2 136

With the knowledge of Equation (35), Equation (75) can be rewritten as

Z a Z +∞ 1 χ(s) + ds = ω2 g (ξ) sin(ξx)dξ (79) π −a x − s 0

The use of Equation (24) and the inverse Fourier transform leads to

+ + + ( ) ( ) R ∞ g+(ξ) sin(ξx)dξ = − 2 R ∞ χ(s)dsR ∞ sin ξx sin ξs dξ 0 π 0 0 ξ (80) 1 R +∞ x+s = − π 0 χ(s) ln x−s ds from the known result [29].

Z +∞ 1 1 x + s sin(ξx) sin(ξs)dξ = ln (81) 0 ξ 2 x − s

According to the singular integral theory with the Cauchy kernel, Equation (79) is obtained: √ Z a 2 2 Z +∞ ω2 a − t t + s χ(x) = √ dt γ(s) ln ds (82) π a2 − x2 −a x − t 0 t − s With application of Equation (71), one obtains

Z x uI (x, 0) = χ(s)ds (83) −a

From Equations (52), (73), (74) and (80), one can obtain the shearing stresses as follows:

Z a I,II c66 H2χ(s) x + s σxy (x, 0) = − H1γ(s) ln ds (84) πc22(γ2η2 − γ1η1) −a s − x x − s

The shearing stresses are given, which is distinct from the achieved results for cracked solid under combined constant thermal ﬂux and constant mechanical loading [19]. Taking the advantage of principle of superposition, such expressions as temperature change, elastic displacement on the crack, and stresses under combined 2i order polynomial function thermal ﬂux and 2j order polynomial function mechanical loading can be given.

3.3. Crack-Tip Field The behavior of the stress intensity factors (SIF) is of much importance, and the corresponding SIF are given as q I KI = lim 2π(x − a)σyy(x, 0) (85) x→ a+ q I KII = lim 2π(x − a)σxy(x, 0) (86) x→ a+ Next, we deﬁne the corresponding intensity factors to characterize opening crack displacement. r COD π I Kv = lim v (x, 0) (87) x→ a+ 2(a − x) r COD π I Ku = lim u (x, 0) (88) x→ a+ 2(a − x) The energy release rate has an important effect on the analysis of crack growth, and the energy release rate is deﬁned as

Z δ 1 I I I I G = lim σxy(x + a, 0)u (a − x, 0) + σyy(x + a, 0)v (a − x, 0)dx (89) δ→ 0 δ 0 Appl. Mech. 2021, 2 137

The energy density dW/dV featuring positive deﬁniteness is of fundamental signiﬁ- cance to the formulation of mechanics theories and physics of solids. As for non-iso-thermal problem, the strain energy which is reserved in a unit volume of the orthotropic solid is written [2,30]. dW S 1 1 = = σ ε − α θσ (90) dV r 2 ij ij 2 kk kk where S indicates the strain energy density factor, along with r representing the reach to the crack tip. The application of concepts above entails an understanding of the energy density function. For the orthotropic solid, Equation (90) can also be illustrated as

2 + 2 − 2 ! dW S 1 c22σxx c11σyy 2c12σxxσyy σxy = = + 2 (91) dV r 2 2c11c22 − c12 c66

Furthermore, the strain energy density factor on the crack line have an effect on the analysis of crack growth. That is, we have ! 1 c w2 + c − 2c w 1 S = 22 11 12 K2 + K2 (92) 0 4π c c − c2 I c II 11 22 12 66

where 1 2 γ + η w = (−1)j 1 1 (c − c γ η ) (93) ∑ + 11 12 j j w1 j=1 γj ηj

3.4. Explicit Expressions for the Particular Case of Combined the Constant Thermo-Mechanical Loading According to the above results, the thermoelastic ﬁeld in the full plane under combined the constant thermal ﬂux −(Q00 − Qc0)/2a and the constant mechanical loading σ0 can be given inexplicit form. I,II x σ (x, 0) = − 1 − √ σ0 (94) yy x2 − a2 c ω (Q − Q )a σI,II (x, 0) = − 66 2 00 c0√ (95) xy 2 2 2c22λλy(γ2η2 − γ1η1) x − a √ (Q − Q )ω x a2 − x2 uI (x, 0) = − 00 c0 2 (96) 4λλy σ (η γ − η γ ) p vI (x, 0) = 0 1 2 2 1 a2 − x2 (97) ω1(γ2 + η2) With application of Equations (85)–(88) and (94)–(97), one has √ KI = σ0 πa (98) √ c66ω2(Qc0 − Q00) πa KII = (99) 2λλyc22(ρ2η2 − ρ1η1) √ COD (Qc0 − Q00)ω2 πa Ku = (100) 4λλy √ COD σ0(η1γ2 − η2γ1) πa Kv = (101) ω1(γ2 + η2) Equations (94)–(101) are the same as Reference [19]. Making use of Equations (20) and (39a) yields √ 2 2 EcQ00 a − x + εQ00aλ Qc0(x) = √ (102) 2 2 Ec a − x + aλ Appl. Mech. 2021, 2 138

which is the same as Reference [28] for cracked solid under combined constant thermomechanical loading. In particular, when ε = 0, it gives √ 2 2 EcQ00 a − x Qc0(x) = √ (103) 2 2 Ec a − x + aλ which is the same as Reference [19] for cracked solid under combined constant thermomechanical loading. The quantity Ec = hc/λy stands for dimensionless thermal conductivity. In Equation (102), when Ec → ∞ or Ec → 0, Qc0 → Q00 , or Qc0 → εQ00 . The observation reveals εQ00 can be considered to be created by the thermal ﬂux or other large loading, such as mechanical loading.

4. Numerical Results In this section, we take the example of combined constant thermal flux and constant mechanical loading and research the influence of the constant thermo-mechanical loading, the dimensionless thermal resistance and the coefficient on the heat flux to the crack surface and the Mode-II SIF. The Mode-II SIF, the energy release rate (G) and the strain energy density factor on the crack line (S) under the constant thermal flux −(Q00 − Qc0)/2a and the constant mechanical loading σ0 are rewritten as KII0, G0, and S0. The material constants of orthotropic and isotropic are listed in Tables1 and2[31].

Table 1. Material constants of Tyrannohex.

E E G α α λ λ xx yy xy vxy vyx xx yy x y (GPa) (GPa) (GPa) (10−5/0C) (10−5/0C) (w/m0C) (w/m0C) 135 87 50 0.15 0.09667 0.32 0.32 3.08 2.81

Table 2. Constants of Steel.

E E G α α λ λ xx yy xx vxy vyx xx yy x y (MPa) (MPa) (MPa) (10−5/0C) (10−5/0C) (w/m0C) (w/m0C) 205900 205900 79200 0.3 0.303 1.14 1.14 48.6 48.6

Figure2a,b show Qc0/Q00 versus Ec with x/a = 0, 0.25, 0.5, 0.75 for ε = 0.01. It is easily seen that Qc0/Q00 increases with an increase of Ec for ε = 0.01. When Ec increases, it means that the heat ﬂux per thickness to the crack surface increases. So, Qc0/Q00 increases. The constant ε is regarded as an adjustment physical quantity. The higher the value of constant ε, the greater the heat ﬂux per thickness to the crack surface. When Ec = 0 and ε = 0, we have Qc0 = 0, which means a fully thermally impermeable crack. When Ec→∞, we have Qc0 = Q00, which means a fully thermally permeable crack. Figure3a,b show Qc0/Q00 versus ε with x/a = 0, 0.25, 0.5, 0.75 for Ec = 1. It is easily seen that the variations of Qc0/Q00 increase with an increase of ε for Ec = 1. From Figure2 to Figure3, the case Ec = 1 correlates with the situation that the crack region is identical to that of the orthotropic material. The cases Ec = 2, Ec = 3, and Ec = 4 indicate that the heat conduction is one, two and three times stronger than thermal conductivity of the orthotropic material respectively. The case with Ec = 0 and ε = 0 corresponds to the thermally impermeable and the case with Ec→∞ corresponds to the thermally impermeable. Appl. Mech. 2021, 2, FOR PEER REVIEW 16

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(a) (b)

Figure 2. (a) Qc0/Q00 versus Ec with x/a = 0, 0.25, 0.5, 0.75 for ε = 0.01. The orthotropic material listed in Table 1; (b) Qc0/Q00 versus Ec with x/a = 0, 0.25, 0.5, 0.75 forε = 0.01. The isotropic material listed in Table 2.

Figure 3a,b show Qc0/Q00 versus ε with x/a = 0, 0.25, 0.5, 0.75 for Ec = 1. It is easily seen that the variations of Qc0/Q00 increase with an increase of ε for Ec= 1. From Figure 2

to Figure 3, the case Ec = 1 correlates with the situation that the crack region is identical to (a)that of the orthotropic material. The cases Ec = 2, Ec = 3,(b and) Ec = 4 indicate that the heat conduction is one, two and three times stronger than thermal conductivity of the ortho- Figure 2.2. ((aa))Q Qc0//QQ00 versusversus Ec withwith x/ax/a= = 0, 0, 0.25, 0.25, 0.5, 0.5, 0.75 0.75 for forε = 0.01.ε = The0.01. orthotropic The orthotropic material mate listedrial in listed Table 1in;( Tableb) Q 1;/Q (b) c0 00 tropic material respectively. The case with Ec = 0 and ε = 0 corresponds to thec0 thermally00 versusQc0/Q00 Ecversus with Ecx/a with= 0, 0.25,x/a = 0.5,0, 0.25, 0.75 0.5, for 0.75ε = 0.01. forε The= 0.01. isotropic The isotropic material material listed in listed Table 2in. Table 2. impermeable and the case with Ec→∞ corresponds to the thermally impermeable.

Figure 3a,b show Qc0/Q00 versus ε with x/a = 0, 0.25, 0.5, 0.75 for Ec = 1. It is easily seen that the variations of Qc0/Q00 increase with an increase of ε for Ec= 1. From Figure 2 to Figure 3, the case Ec = 1 correlates with the situation that the crack region is identical to that of the orthotropic material. The cases Ec = 2, Ec = 3, and Ec = 4 indicate that the heat conduction is one, two and three times stronger than thermal conductivity of the orthotropic material respectively. The case with Ec = 0 and ε = 0 corresponds to the thermally impermeable and the case with Ec→∞ corresponds to the thermally impermeable.

(a) (b)

FigureFigure 3. 3.( a(a) )Q Qc0//Q00 versusversus εwith withx/a x/a= 0.25,= 0.25, 0.5, 0.5, 0.75, 0.75, 0.9 0.9 for for Ec Ec = 1. = The1. The orthotropic orthotropic material material listed listed in Tablein Table1;( b1;) (Qb) Q/cQ0/Q00 ε c0 00 c0 00 versusversusε with withx/a =x/a 0, =0, 0.25, 0.25, 0.5, 0.5, 0.75 0.75 for for Ec =Ec 1. =The 1. The isotropic isotropic material material listed listed in Tablein Table2. 2. ε FigureFigure4 a,b4a,b display display K KII0II0/K/KII00II00 versusversus Ec, Ec, where where K KII0II0 representsrepresents K KI00I00 whenwhen QQc0c0 = 0= and 0 and ε=0.= 0. It Itis iseasily easily seen seen that that K KII0II0/K/KII00 II00decreasesdecreases with with an increase an increase of Ec of for Ec forε =ε 0.01.= 0.01. Figure5a,b display K II0/KII00 versus ε. It is easily seen that the KII0/KII00 decreases (witha) an increase of Ec and ε, respectively). (b)

Figure 3. (a) Qc0/Q00 versus ε with x/a = 0.25, 0.5, 0.75, 0.9 for Ec = 1. The orthotropic material listed in Table 1; (b) Qc0/Q00 versus ε with x/a =0, 0.25, 0.5, 0.75 for Ec = 1. The isotropic material listed in Table 2.

Figure 4a,b display KII0/KII00 versus Ec, where KII0 represents KI00 when Qc0 = 0 and ε =0. It is easily seen that KII0/KII00 decreases with an increase of Ec for ε = 0.01.

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(a) (b)

Figure 4. (a) KII0/KI00 versus Ec with x / a = 0.25, 0.5, 0.75, 0.9 forε = 0.01. The orthotropic material listed in Table 1; (b) KII0/KI00 versus Ec with x(a/) a = 0.25, 0.5, 0.75, 0.9 forε = 0.01. The isotropic material listed in(b Table) 2. = ε FigureFigure 4. 4. ((aa))K KII0II0/K/KI00I00 versusversus EcEc with withx /xa/a =0.25,0.25, 0.5, 0.5, 0.75, 0.75, 0.9 0.9 for for ε== 0.01. 0.01. Theε The orthotropic orthotropic material material listed listed in in Table Table 1;1;( (bb)) = Figure 5a,b displayε KII0/KII00 versus . It is easily seen that the KII0/KII00 decreases KKII0II0/K/KI00I00 versusversus Ec Ecwith withx /xa/awith=0.25,0.25, an 0.5, 0.5, increase 0.75, 0.75, 0.9 0.9 of for for Ecε =and= 0.01. 0.01. ε The The, respectively). isotropic isotropic material material listed listed in in Table Table 22..

Figure 5a,b display KII0/KII00 versus ε . It is easily seen that the KII0/KII00 decreases with an increase of Ec and ε , respectively).

(a) (b)

Figure 5. (a) KII0/KII00 versus ε with x/a = 0.25, 0.5, 0.75, 0.9 for Ec = 1 The orthotropic material listed in Table 1; (b) KII0/KII00 Figure 5. (a)KII0/KII00 versus ε with x/a = 0.25, 0.5, 0.75, 0.9 for Ec = 1 The orthotropic material listed in Table1;( b) versus ε with x/a = 0.25, 0.5, 0.75, 0.9 for Ec = 1. The isotropic material listed in Table 2. KII0/KII00 versus ε with x/a(a=) 0.25, 0.5, 0.75, 0.9 for Ec = 1. The isotropic material listed in Table(b2).

Figure 5. (a) KII0/KII00 versus ε with x/a = 0.25, 0.5, 0.75, 0.9 for Ec = 1 The orthotropic material listed in Table 1; (b) KII0/KII00 FiguresFigures6 –6–88 show showG G0,0, as as computed computed using using Equation Equation (89), (89), for for the the extended extended partially partially versus ε with x/a = 0.25, 0.5, 0.75, 0.9 for Ec = 1. The isotropicε material listed in Table 2. insulatedinsulated crack crack (E (Ecc= = 0,0,ε = 0) = with0) with avariety a variety of of thermal thermal flux flux and and crack crack lengths. lengths. With With the the increaseincrease of of crack crack length, length, the the thermal thermal flux flux around around the the crack crack surface surface increases. increases. It It means means that that Figures 6–8 show G0, as computed using Equation (89), for the extended partially thethe value value of of energy energy release release rate rate increases increases with with the the increase increase of of the the shearing shearing stress stress and and shear shear insulated crack (Ec = 0, ε = 0) with a variety of thermal flux and crack lengths. With the displacementdisplacement induced induced by by the the thermal thermal flux. flux. When When thermal thermal flux flux outweighs outweighs remote remote tensile tensile increase of crack length, the thermal flux around the crack surface increases. It means that stress,stress, the the value value of ofG G0 0may may turn turn negative. negative. the value of energy release rate increases with the increase of the shearing stress and shear displacement induced by the thermal flux. When thermal flux outweighs remote tensile stress, the value of G0 may turn negative.

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(a) (b) (a) (b) Figure 6. (a) G0 versus tension stress with σ0 = 0, 0.5, 1, 2 MPa for Ec = 0, ε = 0, 2a = 1 mm. The orthotropic material listed FigureFigure 6. 6.a (aG) G0 versus tension stress withσ σ0 = 0, 0.5, 1, 2 MPa for c ε, ε = 0, 2a = 1 mm. The orthotropic material listed in Table (1;) (b0) versusG0 versus tension tension stress stress with with0 = σ 0,0 = 0.5, 0, 0.5, 1, 2 1, MPa 2 MPa for EforcE= E = 0,c 0= =0, 0, ε 2a= =0, 1 2a mm. = 1 Themm. orthotropic The isotropic material material listed listed in ε Tableinin Table Table1;( b 2. )1;G (0bversus) G0 versus tension tension stress stress with withσ0 = σ 0,0 = 0.5, 0, 0.5, 1, 2 1, MPa 2 MPa for Eforc = E 0,c =ε =0,0, 2a= =0, 12a mm. = 1 mm. The isotropicThe isotropic material material listed listed in Tablein Table2. 2.

(a) (b) (a) (b) Figure 7. (a) G0 versus tension stress with σ0 = 0, 0.5, 1, 2 MPa for Ec = 0, ε = 0. (2a = 2 mm). The orthotropic material listed Figure 7. (a)G0 versus tension stress with σ0 = 0, 0.5, 1, 2 MPa for Ec = 0, εε= 0. (2a = 2 mm). The orthotropic material listed inFigure Table 7. 1. ( (ab) )G G0 0versus versus tension tension stress stress with with σ σ0 0= = 0, 0, 0.5, 0.5, 1, 1, 2 2MPa MPa for for E Ec =c =0, 0, ε = =0. 0. (2a (2a = =2 mm).2 mm). The The orthotropic isotropic material material listed listed in Table1.( b) G0 versus tension stress with σ0 = 0, 0.5, 1, 2 MPa for Ec = 0, ε = 0. (2a = 2 mm). The isotropic material listed in Table 1. (b) G0 versus tension stress with σ0 = 0, 0.5, 1, 2 MPa for Ec = 0, ε = 0. (2a = 2 mm). The isotropic material listed inin Table Table2. 2. in Table 2.

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(a()a ) (b(b) ) ε FigureFigure 8.8. 8. ((a a())a G)G 0G0 versusversus0 versus tensiontension tensionstress stress stress with with withσ 0σ 0σ= =0 0, =0, 0.5,0, 0.5, 0.5, 1, 1, 21, 2 MPa 2MPa MPa for for for Ec E= Ec 0,=c =ε0 ,=0 , 0.ε (2a = =0. =0. (2a 4 (2a mm). = =4 4mm).The mm). orthotropic The The orthotropic orthotropic material material material listed ε inlistedlisted Table in in1 Table.( Tableb) G1. 01. (versusb ()b G) G0 versus0 tensionversus tension tension stress stress with stressσ with0 with= 0, σ 0 0.5,σ =0 =0, 1,0, 0.5, 20.5, MPa 1, 1, 2 2MPa for MPa E forc for= E 0, Ec ε=c ==0 0.,0 ,ε (2a = = =0. 40. (2a mm). (2a = =4 The4mm). mm). isotropic The The isotropic isotropic material material material listed inlistedlisted Table in in2 Table. Table 2. 2.

22 ε FiguresFigures 99–11– 9–1111 showshow show SS 0S0/0/((/σ(0σ20ππaπ)a versusa) )versus versus E Ec Ecfor cforfor the the the extended extended extended partially partially partially insulated insulated insulated crack crack crack ( ε( (ε = 0.01) with a variety of thermal ﬂux. It is found that S /(2 σ 2πa), as computed using = =0.01) 0.01) with with a avariety variety of of thermal thermal flux. flux. It It is is found found that that S 0S/0(/σ(0σ020ππa),a0), as as computed computed using using Equa- Equa- Equation (92), decreases with E . tiontion (92), (92), decreases decreases with with E Ec. c. c

(a()a ) (b(b) )

0 2022 c 00 2 2 2 ε 0 FigureFigure 9. 9. ((a a()a)S S) 0S0/(/(/(σ0σπ0 πa)πa) a)versusversus versus E Ec Ewith cwithwith Q Q00Q =00 =20, =20, 20,40, 40, 40,80 80 80J/(m J/(m J/(m·s)·s) for ·fors) forε = ε =0.01.= 0.01. 0.01. (2 (2a (2a= a=2mm, =2mm, 2 mm, x/a x/a = x/a =0.99, 0.99,= 0.99,σ 0σ = =0σ 0MPa).0 MPa).= 0 MPa).The The ortho- ortho- The 2 2 2 2 ε orthotropictropictropic material material material listed listed listed in in Table Table in Table 1. 1. (b (1)b .(S) 0S/(b0/(σ)S0σ2π0 /(πa)a)σ of0 of versusπ versusa) of versusE Ec withc with E cQ withQ00 00= =20, 20,Q 0040, 40,= 80 20, 80 J/(m 40, J/(m 802·s)·s) J/(m for for ε ·s) = for =0.01. 0.01.ε = (2a 0.01. (2a =2 =2(2a mm, mm, = 2x/a mm,x/a = =0.99, 0.99,x/a σ0 = 0 MPa). The isotropic material listed in Table 2. =σ0 0.99, = 0 MPa).σ0 = 0 The MPa). isotropic The isotropic material material listed in listed Table in 2. Table 2.

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(a) (b) (a) (b)

Figure 10. (a) S0/(σ02πa) versus Ec with Q00= 20, 40, 80 J/(m2·s) for ε = 0.01. (2a = 2 mm, x/a = 0.99, σ0 = 2 MPa). The Figure 10. (a) S0/(σ02π2a) versus Ec with Q00= 20, 40, 80 J/(m2·s)2 for ε = 0.01. (2a = 2 mm, x/a = 0.99, σ0 = 2 MPa). The Figure 10. (a)S0/(σ0 πa) versus Ec with Q00 2= 20, 40, 80 J/(m ·s) for ε = 0.01. (2a = 22 mm, x/a = 0.99, σ0 = 2 MPa). The orthotropic material listed in Table 1. (b) S0/(σ0 πa) versus Ec with Q00 = 20, 40, 80 J/(m ·s) for ε = 0.01 (2a =2 mm, x/a = orthotropic material listed in Table 1. (b) S0/(σ02π2a) versus Ec with Q00 = 20, 40, 80 J/(m2·s)2 for ε = 0.01 (2a =2 mm, x/a = orthotropic material listed in Table1.( b)S0/(σ0 πa) versus Ec with Q00 = 20, 40, 80 J/(m ·s) for ε = 0.01 (2a = 2 mm, x/a = 0.99, σ0 =1 MPa). The isotropic material listed in Table 2. 0.99, σ0 =1 MPa). The isotropic material listed in Table 2. 0.99, σ0 = 1 MPa). The isotropic material listed in Table2.

(a) (b) (a) (b) Figure 11. (a) S0/(σ02πa)2 versus Ec with Q00 = 20, 40, 80 J/(m2·s) for2 ε = 0.01. (2a = 2 mm, x/a = 0.99, σ0 = 2 MPa). The ortho- FigureFigure 11. 11. (a()a )SS0/(0σ/(02σπ0a)π versusa) versus Ec with Ec with Q00 =Q 20,00 = 40, 20, 80 40, J/(m 802·s) J/(m for ·s)ε for = 0.01.ε = 0.01.(2a = (2a2 mm, = 2 x/a mm, = 0.99,x/a = σ 0.99,0 = 2 MPa).σ0 = 2 The MPa). ortho- The tropic material listed in Table 1. (b) S0/(σ02πa) versus2 Ec with Q00 = 20, 40, 80 J/(m2·s) for ε 2 = 0.01. (2a = 2 mm, x/a = 0.99, σ0 tropicorthotropic material material listed in listed Table in 1. Table (b) 1S.(0/(bσ0)S2π0a)/( versusσ0 πa) E versusc with EQc00with = 20, Q40,00 80= 20,J/(m 40,2·s) 80 for J/(m ε · =s) 0.01. for ε(2a=0.01. = 2 mm, (2a =x/a 2 mm,= 0.99,x/a σ0= = 2 MPa). The isotropic. =0.99, 2 MPa).σ0 = The 2 MPa). isotropic. The isotropic.

Figures 12–14 show S0/(σ02πa2) for the extended partially insulated crack (EC= 1) with FiguresFigures 12–14 12–14 show show S0 S/(0σ/(02σπ0a)π fora) forthe theextended extended partially partially insulated insulated crack crack (EC= (E1) Cwith= 1) a variety of thermal flux. From Equation (92), S0/(σ02πa) consist 2of the mode-I stress inten- awith variety a variety of thermal of thermal flux. From flux. Equation From Equation(92), S0/(σ0 (92),2πa) consist S0/(σ0 ofπ athe) consist mode-I of stress the mode-Iinten- sity factor induced by thermal flux and the mode-II stress intensity factor created by me- sitystress factor intensity induced factor by thermal induced flux by and thermal the mode-II flux and stress the intensity mode-II factor stress created intensity by factor mechanical loading. When thermal flux and mechanical loading increase, the corresponding chanicalcreated loading. by mechanical When loading.thermal flux When and thermal mechanical flux andloading mechanical increase, loading the corresponding increase, the intensity factors and S0/(σ02πa) increase. It isσ revealed2π that crack face boundary conditions intensitycorresponding factors and intensity S0/(σ02 factorsπa) increase. and S 0It/( is revealed0 a) increase. that crack It is face revealed boundary that conditions crack face have a great effect on strain energy density factor on the crack line based on numerical haveboundary a great conditions effect on strain have aenergy great effect density on strainfactor energyon the crack density line factor based on on the numerical crack line results. results.based on numerical results.

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(a(a) ) (b(b) )

22 2 εε 22 2 FigureFigure 12. 12.12. ( a((a) ))SS S0/(00/(σ/(σ00σππ0aa)π )versus aversus) versus ε withwith Q QQ000000 = = 20, =20, 20,40, 40, 40,80 80 J/(m 80J/(m J/(m·s)·s) for for·s) E Ec for c= = 1. 1. E(2 c(2a=a = = 1.2 2 mm, (2mm,a = x/a x/a 2 mm,= = 0.99, 0.99,x/a σ σ0 0= = 00.99, 0 MPa). MPa).σ0 The =The 0 orthotropic MPa).orthotropic The material listed in Table 1. (b) S0/(σ02π2 a) versus 2εε with Q00 = 20, 40, 80 J/(m2·s)2 for Ec = 1.2 (2a = 2 mm, x/a = 0.99, σ0 = 0 MPa). orthotropicmaterial listed material in Table listed 1. in(b) Table S0/(σ10.(πba))S versus0/(σ0 πa )with versus Q00ε =with 20, 40,Q00 80= J/(m 20, 40,·s)80 for J/(m Ec = 1.·s) (2 fora = E2c mm,= 1. (2x/aa == 20.99, mm, σ0x/a = 0= MPa). 0.99, The isotropic material listed in Table 2. σThe0 = 0isotropic MPa). The material isotropic listed material in Table listed 2. in Table2.

(a(a) ) (b(b) )

Figure 13. (a) S0/(σ02π2 a) versus εε with Q00 = 20, 40, 80 J/(m2·s)2 for Ec = 1. (2a = 2 mm, x/a = 0.99, σ0 = 1 MPa). The orthotropic FigureFigure 13.13. ((aa))S S0/(/(σ0σπ2aπ) aversus) versus ε with QQ00 = 20,= 20, 40, 40, 80 J/(m 80 J/(m·s) for2·s) Ec for = 1. Ec (2 =a 1.= 2 (2mm,a = x/a 2 mm, = 0.99,x/a σ=0 =0.99, 1 MPa).σ =The 1 MPa).orthotropic The 0 0 2 00 ε 2 0 materialmaterial listed listed in in Table Table 1. 1. ( b(b) )S S0/0(/σ(σ00π2πaa) )versus versus 2ε withwith Q Q0000 = = 20, 20, 40, 40, 80 80 J/(m J/(m·s)2·s) for for E Ec c= = 1. 21. (2 (2aa = = 2 2 mm, mm, x/a x/a = = 0.99, 0.99, σ σ0 0= = 1 1 MPa). MPa). orthotropic material listed in Table1.( b) S0/(σ0 πa) versus ε with Q00 = 20, 40, 80 J/(m ·s) for Ec = 1. (2a = 2 mm, x/a = 0.99, TheThe isotropic isotropic material material listed listed in in Table Table 2. 2. σ0 = 1 MPa). The isotropic material listed in Table2.

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(a) (b)

0 02 2 ε 00 2 2 c 0 FigureFigure 14. 14. (a()a S)S/(0σ/(πσ0a) πversusa) versus εwithwith QQ=0020,= 40, 20, 80 40, J/(m 80 J/(m·s) for· s)E for= 1. E(2ca= = 1.2 mm, (2a =x/a 2 = mm, 0.99,x/a σ ==2 0.99,MPa).σ 0The= 2 orthotropic MPa). The material listed in Table 1. (b) S0/(σ02πa) versus ε2 with Q00 = 20, 40, 80 J/(m2·s) for Ec = 1.2 (2a = 2 mm, x/a = 0.99, σ0 = 2 MPa). orthotropic material listed in Table1.( b)S0/(σ0 πa) versus ε with Q00 = 20, 40, 80 J/(m ·s) for Ec = 1. (2a = 2 mm, x/a = 0.99, The isotropic material listed in Table 2. σ0 = 2 MPa). The isotropic material listed in Table2. 5. Conclusions 5. ConclusionsThis paper presents a detailed study into the fracture analysis for a crack in ortho- tropicThis material paper subjected presents a to detailed combined study 2 intoi-order the fracturesymmetrical analysis thermal for a crackflux and in orthotropic 2j-order symmetricalmaterial subjected mechanical to combined loading. 2i -orderAs the symmetrical closed form thermal solutions flux of and main 2j-order fracture symmetrical parame- tersmechanical have been loading. obtained, As theit provides closed form a quick solutions and convenient of main fracture way to parameters solve predicting have been the remainingobtained, itlife provides of above a quickproblems. and convenient way to solve predicting the remaining life of aboveThe problems. analysis presented in this paper has also revealed that, for a given mechanical load, Thethe presence analysis presentedof thermal inloads this can paper induce has also the revealedenergy release that, for rate a givento become mechanical either negativeload, the or presence positive, of depending thermal loads on the can rela inducetive magnitude the energy of release the thermal rate to loads become and either the cracknegative surface or positive,conditions depending of impermeability. on the relative In contrast, magnitude the strain of the energy thermal density loads ahead and theof thecrack crack surface is always conditions positive. of impermeability.This finding should In contrast, not be taken the strainto infer energy that energy density release ahead rateof theapproaches crack is alwaysshould be positive. avoided This when finding assessing should structural not be integrity, taken to merely infer that that energy only therelease positive rate excursion approaches of shouldG should be be avoided used to when assess assessing delamination structural growth. integrity, This recom- merely mendationthat only the mimics positive the excursionapproach commonly of G should used be used to assess to assess crack delamination growth in metals, growth. where This onlyrecommendation the positive excursion mimics the of K approach is used to commonly compute usedcrack togrowth. assess crack growth in metals, where only the positive excursion of K is used to compute crack growth. Author Contributions: Conceptualization, B.W. and D.P.; Methodology, B.W. and D.P.; Formal Analysis,Author Contributions:B.W.; Writing—OriginalConceptualization, Draft Preparation, B.W. and B.W.; D.P.; Writ Methodology,ing—Review B.W. & Editing, and D.P.; R.J. Formaland D.P.Analysis, All authors B.W.; Writing—Originalhave read and agreed Draft to Preparation,the published B.W.; version Writing—Review of the manuscript. & Editing, R.J. and D.P. All authors have read and agreed to the published version of the manuscript. Funding: This research received no external funding. Funding: This research received no external funding. Conflicts of Interest: The authors declare no conflict of interest. Conflicts of Interest: The authors declare no conflict of interest. References References 1. Nowacki, W. Thermoelasticity; Pergamon Press: New York, NY, USA, 1962. 1. Nowacki, W. Thermoelasticity; Pergamon Press: New York, NY, USA, 1962. 2. Sih, G.C.; Michaopoloulos, J.; Chou, S.C. 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