Three-Dimensional Volume Integral Equation Method for Solving

Article ThreeThree-Dimensional-Dimensional Volume IntegralIntegral EquationEquation Method for Solving Isotropic/Anisotropic Isotropic/Anisotropic Inhomogeneity Problems Jungki Lee * and Mingu Han Jungki Lee * and Mingu Han Department of Mechanical and Design Engineering, Hongik University, Sejong City 30016, Korea; [email protected] of Mechanical and Design Engineering, Hongik University, Sejong City 30016, Korea; [email protected]* Correspondence: [email protected]; Tel.: +82-44-860-2619 * Correspondence: [email protected]; Tel.: +82-44-860-2619 Received: 18 18 September September 2020 2020;; Accepted: Accepted: 22 22 October October 2020 2020;; Published: Published: 26 26 October October 2020 2020

Abstract: InIn this this paper, paper, the volume integral equation method (VIEM) is is introduced for the analysis of an an unbounded unbounded isotropic isotropic solid solid composed composed of of multiple multiple isotropic/anisotropic isotropic/anisotropic inhomogeneities. inhomogeneities. A comprehensiveA comprehensive examination examination of a a three-dimensional three-dimensional elastostatic elastostatic VIEM VIEM is introduced is introduced for the analysis for the analysisof an unbounded of an unbounded isotropic isotropic solid composed solid composed of isotropic of /anisotropicisotropic/anisotropic inhomogeneity inhomogeneity of arbitrary of arbitraryshape. The shape authors. The hope authors that the hope volume that integralthe volume equation integral method equation can be method used to compute can be used critical to computevalues of critical practical values interest of inpractical realistic interest models in of realistic composites models composed of composites of strong comp anisotropicosed of andstrong/or anisotropicheterogeneous and/or inhomogeneities heterogeneous of inhomogeneities arbitrary shapes. of arbitrary shapes.

Keywords: numericalnumerical modelling; modelling; multiple isotropic multiple/anisotropic isotropic/anisotropic inhomogeneities; three-dimensional inhomogeneities; threevolume-dimensional integral equation volume method; integral equationﬁnite element method; method; finite boundary element method; integral boundary equation method; integral equationelastostatics method;/elastodynamics elastostatics/elastodynamics

1. Introduction Introduction The fibers ﬁbers and matrix in composites are usually composed of isotropic material. However, to satisfy advanced advanced composites, composites, a a portion portion of of the the constituents constituents can can be be anisotropic. anisotropic. For For example, example, in in SiC/Ti SiC/Ti (silicon carbide/titanium) carbide/titanium) metal matrix composites, the matrix is is almost isotropic, isotropic, whereas whereas the the SiC fiberﬁber has strong anisotropy and consist consistss of an interphase and a core. A A transverse cross section of a SiC/TiSiC/Ti-15-3-15-3 composite is shown in FigureFigure1 1..

FigureFigure 1. 1. TransverseTransverse cross cross-section-section of of a SiC/Ti SiC/Ti-15-3-15-3 composite.

Mathematics 2020, 8, x; doi: FOR PEER REVIEW www.mdpi.com/journal/mathematics Mathematics 2020, 8, 1866; doi:10.3390/math8111866 www.mdpi.com/journal/mathematics Mathematics 2020, 8, 1866 2 of 26

Various analytical methods are accessible for solving inhomogeneity problems when the geometry of the inhomogeneities is straightforward (e.g., ellipsoidal, cylindrical and spherical) or when the inhomogeneities are separated well from one another [1–3]. We cannot apply these methods to realistic models for a general problem when the inhomogeneities or voids are of arbitrary shape and the density of the voids or inhomogeneities is high. Therefore, stress analysis of heterogeneous solids frequently requires the utilization of numerical techniques in view of the finite element method (FEM) or the boundary integral equation method (BIEM or BEM). These techniques experience challenges in dealing with problems which include infinite media or multiple anisotropic inhomogeneities. In response to this concern, it has been shown that the volume integral equation method (VIEM) can eliminate both of these limitations in heterogeneous problems which include infinite media [4,5]. By contrast with the BIEM, the VIEM does not require utilization of Green’s function for anisotropic inhomogeneities and is not sensitive to the geometry of the inhomogeneities. Moreover, as opposed to the FEM, where the whole domain needs to be discretized, the VIEM needs discretization of the multiple inhomogeneities only. Problems associated with multiple inhomogeneities have been examined by several authors [6–8]. In this paper, the stress field in an unbounded isotropic elastic matrix, composed of multiple isotropic inhomogeneities and whose number and shape are arbitrary under uniform loading at infinity could be evaluated. The efficiency, accuracy and capability of the volume integral equation method will also be examined using these results. The VIEM is applied to a class of three-dimensional elastostatic inhomogeneity problems. The details of the numerical treatment, especially the evaluation of singular integrals, for resolving three-dimensional problems in view of the VIEM are introduced in this paper. The accuracy of the VIEM is examined by comparing the results obtained from analytical solutions. The purpose of this paper is to introduce the volume integral equation method (VIEM) as an efficient numerical method for solving multiple anisotropic inhomogeneity problems.

2. Volume Integral Equation Method (VIEM) The geometry of the overall elastodynamic problem considered herein is shown in Figure2a. It is assumed that an unbounded isotropic elastic solid, containing a number of isotropic or anisotropic inhomogeneities of arbitrary shape, is subjected to prescribed dynamic loading at inﬁnity. In Figure2a, V and S show the volume and surface of the inhomogeneity and the symbol n indicates the outward (1) (1) unit normalMathematics to 20 S.20 The, 8, x FOR symbols PEER REVIEWρ and cijkl denote the density and fourth-order elasticity3 of 26 tensors of (2) (2) the inhomogeneity whereas ρ and cijkl denote the density and the fourth-order elasticity tensors 6). The value gim(ξ,x) denotes the elastodynamic Green’s function for the unbounded homogeneous of the unbounded matrix material. The unbounded matrix material is presumed to be isotropic and matrix material. The value gim(ξ,x) stands for the ith displacement component at the point ξ as a (2) (1) homogeneous, so that c is a constant-iωt isotropic tensor, whereas c can be arbitrary. Therefore, result of unit concentratedijkl force, eme , at the point x in the mth direction.ijkl It should be noted that the the inhomogeneitiessymbol em represents may, a unit in general, vector in bethe anisotropicmth direction. and uk,l(ξ inhomogeneous.) represents the strain The field interfaces inside the between the inhomogeneitiesinhomogeneities. asA detailed well as expression the unbounded of three- matrixdimensional material elastostatic are assumed gi,jm(ξ,x) will to be be perfectlyin a later bonded, ensuringsection. continuity of the stress and displacement vectors.

(a) (b)

FigureFigure 2. Geometry 2. Geometry of of the the generalgeneral (a (a) )elastodynamic elastodynamic and and(b) elastostatic (b) elastostatic problem. problem.

The geometry of the general elastostatic problem is shown in Figure 2b. Lee and Mal [4] showed that the elastostatic displacement, um(x), within the composite, fulfills the volume integral equation:

u (xx ) = uom ( ) - δc g (ξ ,x )  u (ξ )  dξ m mR ijkl i, j k,l (2)

where the integral is over entire space R and δcijkl = cijkl(1) − cijkl(2). The value gim(ξ,x) is the elastostatic Green’s function (or Kelvin’s solution) for the unbounded homogeneous matrix material. The value gim(ξ,x) stands for the ith displacement component at the point ξ as a result of unit concentrated force at the point x in the mth direction. The elastodynamic volume integral Equation (1) requires gim(ξ,x) and gi,jm(ξ,x) while the elastostatic volume integral Equation (2) requires only gi,jm(ξ,x). This is in contrast to the BEM. In Equations (1) and (2), the summation convention and comma notation have been utilized and we differentiate them with respect to the integration variable ξi. It is evident that the integrand is non-zero within the inhomogeneities only, since δcijkl = 0, outside the inhomogeneities. In Equations (1) and (2), x represents a field point while ξ represents a point inside and on the boundary of the inhomogeneities (see Figure 3). Here, the field point is an exterior point or a point inside and on the boundary of the inhomogeneities. The field point (x) and the point inside and on the boundary of the inhomogeneities (ξ) are independent of each other.

Figure 3. The field point and the interior point (including the boundary point).

Mathematics 2020, 8, x FOR PEER REVIEW 3 of 26

6). The value gim(ξ,x) denotes the elastodynamic Green’s function for the unbounded homogeneous matrix material. The value gim(ξ,x) stands for the ith displacement component at the point ξ as a Mathematics 2020, 8, 1866 3 of 26 result of unit concentrated force, eme-iωt, at the point x in the mth direction. It should be noted that the symbol em represents a unit vector in the mth direction. uk,l(ξ) represents the strain field inside the o iωt m inhomogeneities.Let um (x, ω A)e detailed− signify expression the mth of component three-dimensional of the displacement elastostatic gi vector,j (ξ,x) aswill a be result in a oflater the iωt section.incident field at x in the absence of the inhomogeneities. Let um(x, ω)e− signify the same within the inhomogeneities, where ω is the circular frequency of the waves. In the following example, iωt we suppress the common time factor e− and the explicit dependence of ω for all field quantities. Mal and Knopoff [9] showed that the elastodynamic displacement um(x), within the composite, fulfills the volume integral equation: Z o δρ 2 m ξ ξ δ m ξ ξ ξ um(x) = um(x) + [ ω gi ( , x) ui( ) cijkl gi,j ( , x) uk,l( )] d (1) R · · · − · · · where the integral is over entire space R, δρ and δcijkl show differences in the densities and the fourth-order elasticity tensors of the inhomogeneities and the unbounded matrix material: (1) (2) (1) (2) δρ = ρ ρ and δcijkl = cijkl cijkl . Since cijkl = cjikl = cijlk = cklij (i, j, k, l = 1, 2, 3), cijkl can − − m be reduced as cαβ (α, β = 1, 6). The value gi (ξ,x) denotes the elastodynamic Green’s function for the unbounded homogeneous matrix material. The value g m(ξ,x) stands for the ith displacement i iωt component at the point ξ as a( resulta) of unit concentrated force, eme(−b) , at the point x in the mth direction. It should be noted that the symbol em represents a unit vector in the mth direction. uk,l(ξ) Figure 2. Geometry of the general (a) elastodynamic and (b) elastostatic problem. represents the strain field inside the inhomogeneities. A detailed expression of three-dimensional elastostatic g m(ξ,x) will be in a later section. The geometryi,j of the general elastostatic problem is shown in Figure 2b. Lee and Mal [4] The geometry of the general elastostatic problem is shown in Figure2b. Lee and Mal [ 4] showed showed that the elastostatic displacement, um(x), within the composite, fulfills the volume integral that the elastostatic displacement, u (x), within the composite, fulfills the volume integral equation: equation: m Z ( ) = omo ( ) δ m(ξ ) (ξ)  ξ uum (xxx ) = uum (x ) - R δccijkl ggi,j (ξ, , x ) uuk,l (ξ ) dξ (2(2)) m m− R ijkl· i, j· k,l · wherewhere the integral isis overover entireentire space space R R and andδc δcijkl= =c cijkl(1)(1) − cijkl(2)(2). .The The value value g gimm(ξ(ξ,x,)x )is is the the elasto elastostaticstatic ijkl ijkl − ijkl i Green’sGreen’s function function (or (or Kelvin’s Kelvin’s solution) solution) for for the the unbounded unbounded homogeneous homogeneous matrix matrix material material.. The The value value gimm(ξ,x) stands for the ith displacement component at the point ξ as a result of unit concentrated gi (ξ,x) stands for the ith displacement component at the point ξ as a result of unit concentrated force force at the point x in the mth direction. The elastodynamic volume integral Equation (1) requiresm at the point x in the mth direction. The elastodynamic volume integral Equation (1) requires gi (ξ,x) gim(ξ,x) mand gi,jm(ξ,x) while the elastostatic volume integral Equation (2) requires onlym gi,jm(ξ,x). This and gi,j (ξ,x) while the elastostatic volume integral Equation (2) requires only gi,j (ξ,x). This is in iscontrast in contrast to the to BEM.the BEM. InIn Equation Equationss (1) andand (2), (2), the the summation summation convention convention and and comma comma notation notation have have been been utilized utilized and and we differentiate them with respect to the integration variable ξi. It is evident that the integrand we differentiate them with respect to the integration variable ξi. It is evident that the integrand is is non-zero within the inhomogeneities only, since δcijkl = 0, outside the inhomogeneities. non-zero within the inhomogeneities only, since δcijkl = 0, outside the inhomogeneities. InIn Equation Equationss (1) (1) and and (2) (2),, xx representsrepresents a a field field point point while while ξξ representsrepresents a a point point inside inside and and on on the the boundaryboundary of of the the inhomogeneities inhomogeneities (see (see Figure Figure 3)3).. Here, Here, the the field field point is an exterior point or a point insideinside and and on the boundaryboundary ofof thethe inhomogeneities. inhomogeneities. The The field field point point (x ()x and) and the the point point inside inside and and on on the theboundary boundary of theof the inhomogeneities inhomogeneities (ξ) ( areξ) are independent independent of eachof each other. other.

FigureFigure 3. 3. TheThe field field point point and and the the interior interior point point (including (including the the boundary boundary point) point).. If x lies in the domain occupied by the inhomogeneities, then Equations (1) and (2) are integro-differential equations for the unknown displacement vector u(x) within the inhomogeneities, which can, in principle, be resolved through the solution of Equations (1) and (2). It is extremely Mathematics 2020, 8, 1866 4 of 26 challenging and sometimes impossible to solve systems of linear equations analytically even within the presence of a single inhomogeneity of arbitrary shape. An algorithm based on the discretization of Equations (1) and (2) was developed by Lee and Mal [4,5] to compute numerically the unknown displacement vector u(x) by discretizing the inhomogeneities utilizing standard finite elements. Once u(x) inside the inhomogeneities is determined, the displacement field outside the inhomogeneities can be acquired from Equations (1) and (2) by assessing the integral. The stress field inside and outside the inhomogeneities can also be resolved without difficulty. Details of the numerical treatment of Equations (1) and (2) can be seen in references [5,10,11] for plane elastodynamic problems and in Lee and Mal [4] for plane elastostatic problems. Further mathematical detailing of the elastostatic VIEM can also be seen in Section 4.3 from the book “Volume Integral Equation Method” by Buryachenko [12]. A general description of the volume integral equation method can be viewed in Chapter 4 entitled “Volume Integral Equation Method (VIEM)” written by the first author of this paper, contained in the book “Advances in Computers and Information in Engineering Research, Vol. 2”, edited by Michopoulos et al. (eds.) [13]. It should be noted that Lee and his co-workers e.g., [4,5,7,10,11,14] have been developing the VIEM based on numerical modeling and analysis, while Buryachenko e.g., [12,15,16] has been performing research more mathematically. In Section 3.1 of the reference [16], Buryachenko pointed out that “the VIEM was quite time-consuming and no optimized commercial software existed for its application.” Standard parallel programming, such as MPI (message passing interface), has been utilized to speed up computation in the VIEM. The parallel volume integral equation method (PVIEM) enabled us to investigate more complicated multiple anisotropic inhomogeneity problems elastostatically or elastodynamically. Since there is no commercial software for the VIEM, we are developing a VIEMAP (volume integral equation method application program). This VIEM modeling software includes a pre-processor, a solver and a post-processor, which are adapted to solve multiple isotropic/anisotropic inhomogeneity problems more easily and efficiently. The authors intend to help university researchers and college students in undergraduate degree programs make VIEM models using the VIEMAP more conveniently than the finite element method and solve multiple isotropic/anisotropic inhomogeneity problems in an unbounded isotropic media more accurately and easily than the boundary element method.

3. Numerical Analysis Based On the Three-Dimensional Volume Integral Equation Method

3.1. A Single Cubic Inhomogeneity Problem In this section, in order to introduce the volume integral equation method for three-dimensional elastostatic problems, we consider a single isotropic or orthotropic cubic inhomogeneity in an o unbounded isotropic matrix subject to uniform remote tensile loading, σ xx, as shown in Figure4 (see also Figure2b). We first consider a single isotropic cubic inhomogeneity ( 4 mm x, y, z 4 mm) in the − ≤ ≤ unbounded isotropic matrix. Four models with a different number of quadratic hexahedral elements for the isotropic inhomogeneity are used for the convergence test; Model_8 8 8 (512 elements), × × Model_10 10 10 (1000 elements), Model_14 14 14 (2744 elements) and Model_16 16 16 × × × × × × (4096 elements). Figure4 shows typical discretized models used in the VIEM [17]. Next we consider a single orthotropic cubic inhomogeneity ( 4 mm x, y, z 4 mm) in the − ≤ ≤ unbounded isotropic matrix. The same four models for the isotropic inhomogeneity are used for the orthotropic inhomogeneity. The elastic constants for the isotropic matrix, the isotropic inhomogeneity and three different kinds of orthotropic inhomogeneity are listed in Tables1 and2. In order to distinguish di fferent material properties easily, we assign a distinct material name (mat_01, mat_02, mat_03, —) to each material. Mathematics 2020, 8, 1866 5 of 26 Mathematics 2020, 8, x FOR PEER REVIEW 5 of 26

(a) (b)

FigureFigure 4. TypicalTypical discretized cubic models (−4 m mmm ≤ x,x, y, y, z ≤ 44 m mm)m) in in the the volume volume integral equation − ≤ ≤ methodmethod (VIEM) (VIEM).. ( (aa)) A A single single cube cube (Model_1 (Model_100 × 1010 × 110);0); (b (b) )A A single single cube cube (Model_1 (Model_166 × 1616 × 16).16). × × × × Table 1. Material properties of the isotropic matrix and the isotropic inhomogeneity. The elastic constants for the isotropic matrix, the isotropic inhomogeneity and three different kinds of orthotropic(Unit: GPa) inhomogeneity Isotropic are Matrix listed (mat_01) in Tables Isotropic 1 and 2. Inhomogeneity In order to distinguish (mat_01) different material propertiesλ easily, we assign a 67.34distinct material name (mat_01, 176.06 mat_02, mat_03, ---) to each material. µ 37.88 176.06 Table 1. Material properties of the isotropic matrix and the isotropic inhomogeneity.

Table 2. Material(Unit: properties GPa) Isotropic of the isotropicMatrix (mat_01) matrix and Isotropic the isotropic Inhomogeneity and orthotropic (mat_01) cubic inhomogeneity.

λ 67.34 Orthotropic 176.06Orthotropic Orthotropic Isotropic Isotropicμ Matrix 37.88 Inhomogeneity176.06Inhomogeneity Inhomogeneity Inhomogeneity (mat_01, 02, 03, 04) #1 #2 #3 (mat_01) Table 2. Material properties of the isotropic matrix(mat_02) and the isotropic(mat_03) and orthotropic(mat_04) cubic

c11in[GPa]homogeneity. 143.10 528.18 139.54 279.08 418.61 c12 [GPa] 67.34 176.06 3.90 7.80 11.70 Isotropic Orthotropic Orthotropic Orthotropic c13 [GPa]Isotropic Matrix 67.34 176.06 3.90 7.80 11.70 Inhomogeneity Inhomogeneity #1 Inhomogeneity #2 Inhomogeneity #3 c [GPa](mat_01, 02, 03, 143.10 04) 528.18 15.28 30.56 45.83 22 (mat_01) (mat_02) (mat_03) (mat_04) c23 [GPa] 67.34 176.06 3.29 6.59 9.88 c11 [GPa] 143.10 528.18 139.54 279.08 418.61 c33 [GPa] 143.10 528.18 15.28 30.56 45.83 c12 [GPa] 67.34 176.06 3.90 7.80 11.70 c44 [GPa] 37.88 176.06 5.90 11.80 17.70 c13 [GPa] 67.34 176.06 3.90 7.80 11.70 c55 [GPa] 37.88 176.06 5.90 11.80 17.70 c22 [GPa] 143.10 528.18 15.28 30.56 45.83 c [GPa] 37.88 176.06 5.90 11.80 17.70 c23 [GPa66 ] 67.34 176.06 3.29 6.59 9.88 c33 [GPa] 143.10 528.18 15.28 30.56 45.83 c3.1.1.44 [GPa] Volume Integral37.88 Equation176.06 for an Unbounded5.90Isotropic Matrix Containing11.80 a Single Isotropic17.70 cCubic55 [GPa] Inhomogeneity37.88 176.06 5.90 11.80 17.70 c66 [GPa] 37.88 176.06 5.90 11.80 17.70 The coordinate axes are x1, x2, x3. The stress–strain relationships (σα = cαβεβ (α, β = 1, 6)) for 3.1.1.the isotropic Volume inhomogeneities Integral Equation can for be an expressed Unbounded in the Isotropic form: Matrix Containing a Single Isotropic Cubic Inhomogeneity       σ1   λ + 2µ λ λ 0 0 0  ε1  The coordinate axes are x1, x2, x3. The stress–strain relationships (σα = cαβεβ (α, β = 1, 6)) for the  σ   λ λ + 2µ λ 0 0 0  ε   2    2  isotropic inhomogeneities can be expressed in the form:    σ3   λ λ λ + 2µ 0 0 0  ε3    =   . (3)  σ4σ  λ +20μ λ 0λ 0 000µ 0 0 ε ε4    11      µ  ε   σ5 σ  0λ λ +2 0μ λ 0000 0 0 ε 5    22       σ6 0 0 0 0 0 µ  ε6  σ 33 λ λ λ +2μ 000 ε   =   . (3) It should be noted thatσ the44  stress–strain000 relationships areμ also00 referred ε  to as constitutive equations.     In Equation (3), the stiﬀσness55 constants0 c 0are related 0 to the0 μ engineering0 ε elastic constants through  ij   λ µ λ µ c11 = c22 = c33 = ( + 2 ), c12σ=66 c13= c023 = , and 0 c44 = 0c55 = 0c66 0= μ. ε 

It should be noted that the stress–strain relationships are also referred to as constitutive equations.

Mathematics 2020, 8, 1866 6 of 26

The displacement components in the volume integral Equation (2) for multiple isotropic inhomogeneities can be expressed in the form: R ( ) = o( ) δ(λ + µ) 1 + δλ( 1 + 1 ) + δλ( 1 + 1 ) u1 x u1 x 2 g1,1u1,1 g1,1u2,2 g2,2u1,1 g1,1u3,3 g3,3u1,1 − V { 1 1 1 1 +δ(λ + 2µ)g u2,2 + δλ(g u3,3 + g u2,2) + δ(λ + 2µ)g u3,3 h 2,2 2,2 3,3 i 3,3 +δµ 1 ( + ) + 1 ( + ) (4) g2,3 u2,3 u3,2 g3,2 u2,3 u3,2 h 1 1 i +δµ g (u1,3 + u3,1) + g (u1,3 + u3,1) h 1,3 3,1 io +δµ 1 ( + ) + 1 ( + ) ξ ξ ξ g1,2 u1,2 u2,1 g2,1 u1,2 u2,1 d 1d 2d 3 R ( ) = o( ) δ(λ + µ) 2 + δλ( 2 + 2 ) + δλ( 2 + 2 ) u2 x u2 x 2 g1,1u1,1 g1,1u2,2 g2,2u1,1 g1,1u3,3 g3,3u1,1 − V { 2 2 2 2 +δ(λ + 2µ)g u2,2 + δλ(g u3,3 + g u2,2) + δ(λ + 2µ)g u3,3 h 2,2 2,2 3,3 i 3,3 +δµ 2 ( + ) + 2 ( + ) (5) g2,3 u2,3 u3,2 g3,2 u2,3 u3,2 h 2 2 i +δµ g (u1,3 + u3,1) + g (u1,3 + u3,1) h 1,3 3,1 io +δµ 2 ( + ) + 2 ( + ) ξ ξ ξ g1,2 u1,2 u2,1 g2,1 u1,2 u2,1 d 1d 2d 3 R ( ) = o( ) δ(λ + µ) 3 + δλ( 3 + 3 ) + δλ( 3 + 3 ) u3 x u3 x 2 g1,1u1,1 g1,1u2,2 g2,2u1,1 g1,1u3,3 g3,3u1,1 − V { 3 3 3 3 +δ(λ + 2µ)g u2,2 + δλ(g u3,3 + g u2,2) + δ(λ + 2µ)g u3,3 h 2,2 2,2 3,3 i 3,3 +δµ 3 ( + ) + 3 ( + ) (6) g2,3 u2,3 u3,2 g3,2 u2,3 u3,2 h 3 3 i +δµ g (u1,3 + u3,1) + g (u1,3 + u3,1) h 1,3 3,1 io +δµ 3 ( + ) + 3 ( + ) ξ ξ ξ g1,2 u1,2 u2,1 g2,1 u1,2 u2,1 d 1d 2d 3 (1) where u1(x), u2(x) and u3(x) are the three-dimensional displacement components, δcαβ = cαβ (2) (1) − cαβ (α, β = 1, 6), where cαβ represent the elastic stiﬀness constants of the isotropic inhomogeneities (2) while cαβ denote those for the isotropic matrix material; δc = δc = δc = (λ + 2µ ) (λ + 2µ ), 11 22 33 1 1 − 2 2 δc12 = δc13 = δc23 = λ1 λ2, and δc44 = δc55 = δc66 = µ1 µ2. − m − In Equations (4)–(6), gi (ξ,x) shows the Green’s function for the unbounded isotropic matrix material and is presented by Banerjee [18] and Pao and Varatharajulu [19] as:

2 1 (x1 ξ1) g1 = [ − + (3 4ν)] 1 16π(1 ν)µr r2 − −

1 (x1 ξ1)(x2 ξ2) g1 = g2 = [ − − ] 2 1 16π(1 ν)µr r2 − 1 (x1 ξ1)(x3 ξ3) g1 = g3 = [ − − ] 3 1 16π(1 ν)µr r2 − 2 1 (x2 ξ2) g2 = [ − + (3 4ν)] (7) 2 16π(1 ν)µr r2 − − 1 (x2 ξ2)(x3 ξ3) g2 = g3 = [ − − ] 3 2 16π(1 ν)µr r2 − 2 1 (x3 ξ3) g3 = [ − + (3 4ν)] 3 16π(1 ν)µr r2 − − q where r = | x ξ | = (x ξ )2 + (x ξ )2 + (x ξ )2, ν is Poisson’s ratio, and µ is the shear modulus. − 1 − 1 2 − 2 3 − 3 Mathematics 2020, 8, 1866 7 of 26

m Also, gi,j (ξ,x) is presented as:    ( ξ )3 ( ν)( ξ )  1 1 3 x1 1 1 4 x1 1  g =  − + − −  1.1 16π(1 ν)µr2  r3 r  −    ( ξ )2( ξ ) ( ν)( ξ )  1 1 3 x1 1 x2 2 3 4 x2 2  g =  − − + − −  1.2 16π(1 ν)µr2  r3 r  −    ( ξ )2( ξ ) ( ν)( ξ )  1 1 3 x1 1 x3 3 3 4 x3 3  g =  − − + − −  1.3 16π(1 ν)µr2  r3 r  −    ( ξ )2( ξ ) ( ξ )  1 2 1 3 x1 1 x2 2 x2 2  g = g =  − − −  2.1 1,1 16π(1 ν)µr2  r3 − r  −    ( ξ )( ξ )2 ( ξ )  1 2 1 3 x1 1 x2 2 x1 1  g = g =  − − −  2.2 1,2 16π(1 ν)µr2  r3 − r  − ( ) 1 3(x1 ξ1)(x2 ξ2)(x3 ξ3) g1 = g2 = − − − 2.3 1,3 16π(1 ν)µr2 r3 −    ( ξ )2( ξ ) ( ξ )  1 3 1 3 x1 1 x3 3 x3 3  g = g =  − − −  3.1 1,1 16π(1 ν)µr2  r3 − r  − ( ) 1 3(x1 ξ1)(x2 ξ2)(x3 ξ3) g1 = g3 = − − − 3.2 1,2 16π(1 ν)µr2 r3 −    ( ξ )( ξ )2 ( ξ )  1 3 1 3 x1 1 x3 3 x1 1  g = g =  − − −  3.3 1,3 16π(1 ν)µr2  r3 − r  −    ( ξ )( ξ )2 ( ν)( ξ )  2 1 3 x1 1 x2 2 3 4 x1 1  g =  − − + − −  2.1 16π(1 ν)µr2  r3 r  −    ( ξ )3 ( ν)( ξ )  2 1 3 x2 2 1 4 x2 2  g =  − + − −  2.2 16π(1 ν)µr2  r3 r  −    ( ξ )2( ξ ) ( ν)( ξ )  2 1 3 x2 2 x3 3 3 4 x3 3  g =  − − + − −  2.3 16π(1 ν)µr2  r3 r  − ( ) 1 3(x1 ξ1)(x2 ξ2)(x3 ξ3) g2 = g3 = − − − 3,1 2,1 16π(1 ν)µr2 r3 −    ( ξ )2( ξ ) ( ξ )  2 3 1 3 x2 2 x3 3 x3 3  g = g =  − − −  3,2 2,2 16π(1 ν)µr2  r3 − r  −    ( ξ )( ξ )2 ( ξ )  2 3 1 3 x2 2 x3 3 x2 2  g = g =  − − −  3,3 2,3 16π(1 ν)µr2  r3 − r  −    ( ξ )( ξ )2 ( ν)( ξ )  3 1 3 x1 1 x3 3 3 4 x1 1  g =  − − + − −  3.1 16π(1 ν)µr2  r3 r  −    ( ξ )( ξ )2 ( ν)( ξ )  3 1 3 x2 2 x3 3 3 4 x2 2  g =  − − + − −  3.2 16π(1 ν)µr2  r3 r  − Mathematics 2020, 8, 1866 8 of 26

   ( ξ )3 ( ν)( ξ )  3 1 3 x3 3 1 4 x3 3  g =  − + − −  (8) 3.3 16π(1 ν)µr2  r3 r  − It should be noted that for the applied uniformly remote stress, the displacement vector u◦ is of the form: o = o = o = u1 c1x, u2 c2y, u3 c3z (9) where the constants C1–C3 are related to the tensile and shear components of the applied uniformly remote stress.

3.1.2. Volume Integral Equation for an Unbounded Isotropic Matrix Containing a Single Orthotropic Cubic Inhomogeneity

Let the coordinate axes x1, x2, x3 be taken parallel to the symmetry axes of the orthotropic material. The constitutive equations for the orthotropic inhomogeneities can be expressed in the form:       σ1   c11 c12 c13 0 0 0  ε1        σ   c c c 0 0 0  ε   2   12 22 23  2        σ3   c13 c23 c33 0 0 0  ε3    =   . (10)  σ4   0 0 0 c44 0 0  ε4           ε   σ5   0 0 0 0 c55 0  5        σ6  0 0 0 0 0 c66  ε6 

The stiﬀness constants cij are related to the engineering elastic constants through:

1 ν ν ν +ν ν ν +ν ν 1 ν ν c = − 23 32 , c = 21 31 23 = c , c = 31 21 32 = c , c = − 31 13 , 11 E2E3∆ 12 E2E3∆ 21 13 E2E3∆ 31 22 E3E1∆ ν +ν ν 1 ν ν (11) c = 32 31 12 = c , c = − 12 12 , c = µ , c = µ , c = µ , 23 E3E1∆ 32 33 E1E2∆ 44 23 55 31 66 12

where, 1 ν ν ν ν ν ν 2ν ν ν ∆ = − 12 21 − 23 32 − 31 13 − 12 23 31 . (12) E1E2E3

In the above equations, E1,E2,E3 are Young’s moduli in 1, 2 and 3 directions, respectively, µ23, µ31 and µ12 are the shear moduli in the 2-3, 3-1 and 1-2 planes, respectively. νij is Poisson’s ratio for transverse strain in the j-direction when stressed in the i-direction. It should be noted that the elastic moduli satisfy the reciprocal νij/Ei =νji/Ej. The displacement components in the volume integral Equation (2) for multiple orthotropic inhomogeneities can be expressed in the form: R ( ) = o( ) δ 1 + δ ( 1 + 1 ) + δ ( 1 + 1 ) u1 x u1 x c11g1,1u1,1 c12 g1,1u2,2 g2,2u1,1 c13 g1,1u3,3 g3,3u1,1 − V { 1 1 1 1 +δc22g u2,2 + δc23(g u3,3 + g u2,2) + δc33g u3,3 h 2,2 2,2 3,3 i 3,3 +δ 1 ( + ) + 1 ( + ) (13) c44 g2,3 u2,3 u3,2 g3,2 u2,3 u3,2 h 1 1 i +δc55 g (u1,3 + u3,1) + g (u1,3 + u3,1) h 1,3 3,1 io +δ 1 ( + ) + 1 ( + ) ξ ξ ξ c66 g1,2 u1,2 u2,1 g2,1 u1,2 u2,1 d 1d 2d 3 R ( ) = o( ) δ 2 + δ ( 2 + 2 ) + δ ( 2 + 2 ) u2 x u2 x c11g1,1u1,1 c12 g1,1u2,2 g2,2u1,1 c13 g1,1u3,3 g3,3u1,1 − V { 2 2 2 2 +δc22g u2,2 + δc23(g u3,3 + g u2,2) + δc33g u3,3 h 2,2 2,2 3,3 i 3,3 +δ 2 ( + ) + 2 ( + ) (14) c44 g2,3 u2,3 u3,2 g3,2 u2,3 u3,2 h 2 2 i +δc55 g (u1,3 + u3,1) + g (u1,3 + u3,1) h 1,3 3,1 io +δ 2 ( + ) + 2 ( + ) ξ ξ ξ c66 g1,2 u1,2 u2,1 g2,1 u1,2 u2,1 d 1d 2d 3 Mathematics 2020, 8, 1866 9 of 26

and, R ( ) = o( ) δ 3 + δ ( 3 + 3 ) + δ ( 3 + 3 ) u3 x u3 x c11g1,1u1,1 c12 g1,1u2,2 g2,2u1,1 c13 g1,1u3,3 g3,3u1,1 − V { 3 3 3 3 +δc22g u2,2 + δc23(g u3,3 + g u2,2) + δc33g u3,3 h 2,2 2,2 3,3 i 3,3 +δ 3 ( + ) + 3 ( + ) (15) c44 g2,3 u2,3 u3,2 g3,2 u2,3 u3,2 h 3 3 i +δc55 g (u1,3 + u3,1) + g (u1,3 + u3,1) h 1,3 3,1 io +δ 3 ( + ) + 3 ( + ) ξ ξ ξ c66 g1,2 u1,2 u2,1 g2,1 u1,2 u2,1 d 1d 2d 3

(1) (2) where u1(x), u2(x) and u3(x) are the three-dimensional displacement components, δcαβ = cαβ cαβ (1) − (α, β = 1, 6), where cαβ represent the elastic stiﬀness constants of the orthotropic inhomogeneities (2) while cαβ denote those for the isotropic matrix material; δc = c (λ + 2µ ), δc = c (λ + 2µ ), 11 11 − 2 2 22 22 − 2 2 δc33 = c33 (λ2 + 2µ2), δc12 = c12 λ2, δc13 = c13 λ2, δc23 = c23 λ2, and δc44 = c44 µ2, δc55 = c55 µ2, − − − − − − δc66 = c66 µ2. − m In Equations (13)–(15), gi (ξ,x) is Green’s function for the unbounded isotropic matrix material. Thus, the VIEM does not need to use Green’s function for the orthotropic material of the inhomogeneity.

3.1.3. Numerical Formulation The integrands in the volume integral Equations (4)–(6) and Equations (13)–(15) contain singularities with diﬀerent orders due to the singular characteristics of Green’s function at x = ξ m (i.e., r = 0), and the evaluation of the singular integrals requires special attention. In general, gi (ξ,x) behaves as 1/r while its derivatives behave as 1/r2 as r 0. It ought to be noted that the VIEM involves m → only gi (ξ,x) for the isotropic matrix and its derivatives. By contrast, the BEM involves Green’s function for these and for the anisotropic inhomogeneities and their derivatives. The singularities in the VIEM are weaker (integrable) when compared with those in the BEM. Therefore, we have utilized the direct integration scheme introduced by Li et al. [20] after reasonable adjustments to address these singularities in the integrands. The order of singularity of the singular elements decreases by one degree with tetrahedron polar co-ordinates [20] in the VIEM. This strategy also converts weakly singular integrals into integrals over smooth functions. The tetrahedron polar co-ordinates are applied to quadratic, hexahedral, isoparametric singular elements as follows: (1) In Figure5a, Ωf stands for the quadratic hexahedral element indicating the interior of the considered body. Ωf is associated with a global three-dimensional Cartesian coordinate system. Ωf is mapped onto a cube Ωf0 of side-length 2, characterized by the local three-dimensional Cartesian coordinates, ξ1, ξ2 and ξ3. (2) If the singular point is located at a corner f f f node, Ω 0 is divided into two triangular prisms, Ω1 0 and Ω2 0 , as shown in Figure5b. If the singular f f f f point is located at a midside node, then Ω 0 is divided into three triangular prisms, Ω1 0 , Ω2 0 and Ω3 0 , as shown in Figure5b. (3) A subdivision of each triangular prism is divided into three tetrahedral subcells as shown in Figure5c. (4) Each subcell maps onto a unit cube, applying the tetrahedron f f f f polar coordinates [20]; the subcell Ω11 0 maps onto Ω11 00 . Subsequently, Ω11 00 maps onto Ω11 000 in Figure5d. The last mapping stands for a linear transformation carrying out numerical integrations in a standard way. Figure6 shows a typical hexahedral element used in the VIEM. Standard 20-node quadratic hexahedral elements were used in ﬁnite element discretization of the VIEM models. Figure7 shows a detailed explanation for a subdivision of each triangular prism into three tetrahedral subcells after a f f f f f f subdivision of Ω 0 into two (Ω1 0 and Ω2 0 ) or three triangular prisms (Ω1 0 , Ω2 0 and Ω3 0 ) (see Figure5b) for the hexahedral element used in the VIEM (Figure6). Mathematics 2020, 8, x FOR PEER REVIEW 10 of 26 hexahedral, isoparametric singular elements as follows: (1) In Figure 5a, Ωf stands for the quadratic hexahedral element indicating the interior of the considered body. Ωf is associated with a global three-dimensional Cartesian coordinate system. Ωf is mapped onto a cube Ωf′ of side-length 2, characterizedMathematics 2020 ,by8, 1866the local three-dimensional Cartesian coordinates, ξ1, ξ2 and ξ3. (2) If the singular10 of 26 ′ ′ ′ point is located at a corner node, Ωf is divided into two triangular prisms, Ω1f and Ω2f , as shown in

ξ 2 ξ 3

ξ 3 P1 ξ P10 2 P＇(-1,1,1) P＇ (0,1,1) P＇(1,1,1) P2 2 10 1 P17 P ＇＇ 9 P11(-1,0,1) P (1,0,1) f ＇ 9 P11 Ω ξ 1 P (0,-1,1) P＇(-1,-1,1) 12 ＇(1,-1,1) P18 3 P4 P4 P 5 ＇＇ (1,1,0) P18(-1,1,0) P17 P12 P3 ξ1 P P6 14 ＇ (-1,-1,0) f＇＇ (1,-1,0) P P19 Ω P20 20 P 13 P＇(-1,1,-1) ＇(1,1,-1) P 6 ＇ P 5 19 P15 P14(0,1,-1) ＇ P＇ (1,0,-1) P15(-1,0,-1) 13 x3 P8 P＇(-1,-1,-1) P＇ (0,-1-1) P＇(1,-1,-1) P 7 16 8 7 P16 x2 x 1 (a) 3 (-1,1,1) 2 (1,1,1) 2 (-1,1,1) 1 (1,1,1) 1 (1,1,1) 2 (-1,1,1) 1 (1,0,1)

3(-1,-1,1) 1 (1,0,1) 1 (1,0,1) f＇ Ω 1 3(-1,-1,1) 2 (-1,-1,1) 3 (1,-1,1)

f＇ 2(-1,-1,1) 3 (1,-1,1) Ω 2 ＇f f＇ Ω 2 or 6(-1,1,-1) Ω 1 5 (1,1,-1) 5 (-1,1,-1) 4 (1,1,-1) 4 (1,1,-1) 5 (-1,1,-1) f＇ Ω 3 4 (1,0,-1) 4 (1,0,-1)

6 (-1,-1,-1) 5 (-1,-1,-1) 6 (1,-1,-1) 6 (-1,-1,-1) 4 (1,0,-1) 5 (-1,-1,-1) 6 (1,-1,-1) (b)

2 (-1,1,1) 1 (1,1,1) 2 (-1,1,1) 1 (1,1,1) 1 (1,1,1) 1 (1,1,1)

2 (-1,1,1) 3 (-1,-1,1) f＇ 3 (-1,-1,1) f＇ f＇ Ω 1 Ω 11 Ω 12

f＇ Ω 13 4 (-1,1,-1) (-1,1,-1) (1,1,-1) (-1,1,-1) 5 4 4 (-1,1,-1) 3 4 (1,1,-1)

6 (-1,-1,-1) 3 (-1,-1,-1) 2 (-1,-1,-1)

(c)

η 3

2 (-1,1,1) 1 (1,1,1) (-1,1,1) (1,1,1) 8 7 η 2 ρ 3 8 (0,1,1) 7 (1,1,1) (-1,-1,1) 6(1,-1,1) 3(-1,-1,1) 5 f＇ 5 (0,0,1) Ω 11 f＂ 6 (1,0,1) Ω 11 ρ2 η (0,1,0) f＂ 1 4 3 (1,1,0) Ω 11 ρ 1 4(-1,1,-1) 3(1,1,-1) 1 (0,0,0) 2 (1,0,0) 4 (-1,1,-1)

1 (-1,-1,-1) 2 (1,-1,-1) (d)

FigureFigure 5 5.. ApplicationApplication of of the the tetrahedron polar co co-ordinates-ordinates to quadratic, hexahedral, isoparametric singularsingular elements elements [20]. [20]. (a ()a )Mapping Mapping of of ΩΩf ontof onto ΩfΩ′; f(0b;() bsubdivision) subdivision of Ω off′ Ωatf 0theat thesingular singular point point 1; (c 1;) ′ f0 ′ f0 ′′ f00 ′′ s(ubdivisionc) subdivision of triangular of triangular prism prism Ω1fΩ into1 into three three tetrahedral; tetrahedral; (d) m (dapping) mapping of Ω of11fΩ onto11 onto Ω11f Ω and11 ofand Ω11 off f00 ′′′ f000 ontoΩ11 Ωonto11f . Ω11 .

Tables3 and4 show a detailed explanation for a subdivision of each triangular prism into three tetrahedral subcells after a subdivision of Ωf0 into two or three triangular prisms (see Figure5b,c), respectively, for the hexahedral element used in the VIEM (Figure6). Table3 shows a subdivision of each triangular prism into three tetrahedral subcells after a subdivision of Ωf0 into two triangular Mathematics 2020, 8, x FOR PEER REVIEW 11 of 26 Mathematics 2020, 8, 1866 11 of 26 Figure 5b. If the singular point is located at a midside node, then Ωf′ is divided into three triangular ′ ′ ′ prisms, (seeΩ1f , FigureΩ2f and5b,c). Ω3f This, as shownsubdivision in Figure technique 5b. (3) is A applied subdivision to all the of each corner triangular nodes, 1, prism 3, 5, 7, is 13,divided 15, 17 into and three 19 of tetrahedral the hexahedral subcells element as shown shown in in Figure Figure 5c.6. Thus, (4) Each the sub totalcell number maps ofonto cases a unit in Table3 increases to 48 (8 nodes 6 cases). Table4 shows a subdivision of eachf ′ triangular prismf ″ cube, applying the tetrahedron× polar coordinates [20]; the subcell Ω11 maps onto Ω11 . f0 into three tetrahedral″ subcells after a subdivision‴ of Ω into three triangular prisms (see Figure5b,c). Subsequently, Ω11f maps onto Ω11f in Figure 5d. The last mapping stands for a linear The same subdivision technique is applied to all the midside nodes, 2, 4, 6, 8, 9, 10, 11, 12, 14, 16, transformation carrying out numerical integrations in a standard way. 18 and 20 of the hexahedral element shown in Figure6. Thus, the total number of cases in Table4 Figure 6 shows a typical hexahedral element used in the VIEM. Standard 20-node quadratic rises to 108 (12 nodes 9 cases). It should be noted that, in Tables3 and4, 1, 2 or 1, 2 and 3 in the hexahedral elements were× used in finite element discretization of the VIEM models. Figure 7 shows triangular prism column and 1, 2 and 3 in the tetrahedron column are arbitrarily selected. There is no a detailed explanation for a subdivision of each triangular prism into three tetrahedral subcells after particular order in the numbers, 1, 2 or 1, 2 and 3. Therefore, in order to compute singular integrals a subdivision of Ωf′ into two (Ω1f′ and Ω2f′) or three triangular prisms (Ω1f′, Ω2f′ and Ω3f′) (see Figure accurately, it is necessary to investigate 156 total cases (48 cases in Table3 and 108 cases in Table4) per 5b) for the hexahedral element used in the VIEM (Figure 6). hexahedral element.

Figure 6. A typical quadratic hexahedral element used in the VIEM.

Table 3. A subdivision of each triangular prism into three tetrahedral subcells at each singular point

after a subdivision of Ωf0 into two triangular prisms (see Figure5b,c). The singular point is located at the corner point of the hexahedral element shown in Figure6.

Singular Point Triangular Prism Tetrahedron Node 1 Node 2 Node 3 Node 4 1 1 5 7 19 1 2 1 17 5 19 3 1 13 17 19 1 1 1 3 5 17 2 2 1 15 3 17 3 1 13 15 17

Mathematics 2020, 8, 1866 12 of 26

Table 3. Cont.

Singular Point Triangular Prism Tetrahedron Node 1 Node 2 Node 3 Node 4 1 3 7 1 13 1 2 3 19 7 13 3 3 15 19 13 3 1 3 5 7 19 2 2 3 17 5 19 3 3 15 17 19 1 5 7 1 19 1 2 5 1 13 19 3 5 13 17 19 5 1 5 1 3 13 2 2 5 3 15 13 3 5 15 17 13 1 7 1 3 13 1 2 7 3 15 13 3 7 15 19 13 7 1 7 3 5 15 2 2 7 5 17 15 3 7 17 19 15 1 13 17 15 3 1 2 13 5 17 3 3 13 1 5 3 13 1 13 19 17 5 2 2 13 7 19 5 3 13 1 7 5 1 15 19 17 5 1 2 15 7 19 5 3 15 3 7 5 15 1 15 13 19 7 2 2 15 1 13 7 3 15 3 1 7 1 17 15 13 3 1 2 17 13 1 3 3 17 1 5 3 17 1 17 13 19 1 2 2 17 19 7 1 3 17 7 5 1 1 19 17 15 5 1 2 19 15 3 5 3 19 3 7 5 19 1 19 15 13 3 2 2 19 13 1 3 3 19 1 7 3 Mathematics 2020, 8, x FOR PEER REVIEW 12 of 26 Mathematics 2020, 8, 1866 13 of 26

(a)

(b)

FigureFigure 7. 7A. A subdivision subdivision of of each each triangular triangular prism prism into into three three tetrahedral tetrahedral subcells subcells after after a subdivisiona subdivision of of f′ ΩfΩ0 into into two two or or three three triangular triangular prisms prisms (see (see Figure Figure5b,c). 5b,c ().a ()a A) A subdivision subdivision of of each each triangular triangular prism prism f′ intointo three three tetrahedral tetrahedral subcells subcells at at the the singular singular point point 15 15 after afte ar subdivision a subdivision of Ωoff 0Ωinto into two two triangular triangular f′ f′ prisms (fΩ0 1 and f0 Ω2 ) (see Figure 5b,c); (b) a subdivision of each triangular prism into three prisms (Ω1 and Ω2 ) (see Figure5b,c); ( b) a subdivision of each triangular prism into three tetrahedral f′ tetrahedral subcells at the singular point 14 after a subdivisionf0 of Ω into three triangularf0 f 0prisms subcells at the singular point 14 after a subdivision of Ω into three triangular prisms (Ω1 , Ω2 and f0 1f′ 2f′ 3f′ Ω3(Ω) (see, Ω Figure and Ω5b,c).) (see Figure 5b,c).

Tables 3 and 4 show a detailed explanation for a subdivision of each triangular prism into f′ three tetrahedral subcells after a subdivision of Ω into two or three triangular prisms (see Figure 5b,c), respectively, for the hexahedral element used in the VIEM (Figure 6). Table 3 shows a f′ subdivision of each triangular prism into three tetrahedral subcells after a subdivision of Ω into two triangular prisms (see Figure 5b,c). This subdivision technique is applied to all the corner nodes, 1, 3, 5, 7, 13, 15, 17 and 19 of the hexahedral element shown in Figure 6. Thus, the total number of

Mathematics 2020, 8, 1866 14 of 26

Table 4. A subdivision of each triangular prism into three tetrahedral subcells at each singular point after a subdivision of Ωf0 into three triangular prisms (see Figure5b,c). The singular point is located at the midside of the hexahedral element shown in Figure6.

Singular Point Triangular Prism Tetrahedron Node 1 Node 2 Node 3 Node 4 1 2 3 5 15 1 2 2 5 17 15 3 2 17 14 15 1 2 5 7 19 2 2 2 2 17 5 19 3 2 14 17 19 1 2 7 1 13 3 2 2 19 7 13 3 2 14 19 13 1 4 1 3 13 1 2 4 3 15 13 3 4 15 16 13 1 4 7 1 13 4 2 2 4 19 7 13 3 4 16 19 13 1 4 5 7 19 3 2 4 17 5 19 3 4 16 17 19 1 6 3 5 15 1 2 6 5 17 15 3 6 17 18 15 1 6 1 3 15 6 2 2 6 13 1 15 3 6 18 13 15 1 6 7 1 13 3 2 6 19 7 13 3 6 18 19 13 1 8 1 3 15 1 2 8 13 1 15 3 8 20 13 15 1 8 3 5 15 8 2 2 8 5 17 15 3 8 17 20 15 1 8 5 7 17 3 2 8 7 19 17 3 8 19 20 17 Mathematics 2020, 8, 1866 15 of 26

Table 4. Cont.

Singular Point Triangular Prism Tetrahedron Node 1 Node 2 Node 3 Node 4 1 9 13 15 17 1 2 9 19 13 17 3 9 12 19 17 1 9 15 3 17 9 2 2 9 3 5 17 3 9 5 12 17 1 9 3 1 5 3 2 9 1 7 5 3 9 7 12 5 1 10 13 15 19 1 2 10 15 17 19 3 10 17 11 19 1 10 1 13 19 10 2 2 10 7 1 19 3 10 11 7 19 1 10 3 1 7 3 2 10 5 3 7 3 10 11 5 7 1 11 17 19 13 1 2 11 15 17 13 3 11 10 15 13 1 11 19 7 13 11 2 2 11 7 1 13 3 11 1 10 13 1 11 7 5 1 3 2 11 5 3 1 3 11 3 10 1 1 12 17 19 15 1 2 12 19 13 15 3 12 13 9 15 1 12 5 17 15 12 2 2 12 3 5 15 3 12 9 3 15 1 12 7 5 3 3 2 12 1 7 3 3 12 9 1 3 Mathematics 2020, 8, 1866 16 of 26

Table 4. Cont.

Singular Point Triangular Prism Tetrahedron Node 1 Node 2 Node 3 Node 4 1 14 17 15 5 1 2 14 15 3 5 3 14 3 2 5 1 14 19 17 5 14 2 2 14 7 19 5 3 14 2 7 5 1 14 13 19 7 3 2 14 1 13 7 3 14 2 1 7 1 16 19 17 7 1 2 16 17 5 7 3 16 5 4 7 1 16 13 19 7 2 16 2 16 1 13 7 3 16 4 1 7 1 16 15 13 1 3 2 16 3 15 1 3 16 4 3 1 1 18 17 15 3 1 2 18 5 17 3 3 18 6 5 3 1 18 15 13 3 18 2 2 18 13 1 3 3 18 1 6 3 1 18 13 19 1 3 2 18 19 7 1 3 18 7 6 1 1 20 15 13 3 1 2 20 13 1 3 3 20 1 8 3 1 20 17 15 3 20 2 2 20 5 17 3 3 20 8 5 3 1 20 19 17 5 3 2 20 7 19 5 3 20 8 7 5

3.1.4. Numerical Results In the VIEM, the displacements and stresses inside the cubic inhomogeneity are ﬁrst calculated using the models in Figure4. We calculate the displacements and stresses outside the cubic inhomogeneity using Equation (2), by adding standard ﬁnite elements in appropriate locations. Mathematics 2020, 8, 1866 17 of 26

Figure8 shows typical VIEM models for the single cubic inhomogeneity ( 4 mm x, XX 4 mm) Mathematics 2020, 8, x FOR PEER REVIEW − ≤ ≤ 17 of 26 (between two green lines) including the outside of the cubic inhomogeneity ( 8 mm x 4 mm and − ≤ ≤ − ≤4 mm4 mm)x and8 outside mm). Since the isotropicx , ξ outside cubic the inhomogeneity inhomogeneity, (−8 singularities mm ≤ XX ≤ do −4 not mm exist and while 4 mm calculating ≤ XX ≤ 8 ≤ ≤ mthem displacements), respectively. andThe stressesresults using outside the the VIEM inhomogeneity for the different in the models VIEM. Therefore,converged thevery displacement well within thisand range stress of ﬁelds hexahedral inside and elements. outside the inhomogeneities can be solved without diﬃculty.

(a) (b)

Figure 8. A typical single cubic model (−4 m mmm ≤ xx,, XX ≤ 44 m mm)m) (between (between two two green lines) including − ≤ ≤ the outside of thethe cubecube ((−88 mmmm ≤ x ≤ −44 m mmm and and 4 4 m mmm ≤ xx ≤ 88 m mm)m) in in the the VIEM. VIEM. ( (aa)) A single cubic − ≤ ≤ − ≤ ≤ model (−44 mm mm ≤ xx,, XX XX ≤ 44 mm mm)) with 1000 elements (Model_10(Model_10 × 1010 × 1010)) (between (between two two green green lines) − ≤ ≤ × × including the the outside outside of of the the cube cube (−8 ( 8 mm mm ≤ x x≤ −4 mm4 mm and and 4 mm 4 mm ≤ x ≤ 8x mm8) mm).. 1000 1000elements elements are used are − ≤ ≤ − ≤ ≤ forused the for discretization the discretization of the of theoutside outside of the of the cube cube;; (b) ( ba) single a single cubic cubic model model ( (−44 mm mm ≤ xx, ,XXXX ≤ 44 mm mm)) − ≤ ≤ with 4096 elements (Model_16 × 16 × 16)16) (between (between two two green green lines) lines) including the outside of the cube × × ((−88 mm ≤ x ≤ −44 mm and 4 mm ≤ x ≤ 88 mmmm).). 4096 elements are used for the discretization of the − ≤ ≤ − ≤ ≤ outside of the cube.

o Figure9 shows normalized tensile stress component ( σxx/σ xx) along the x-axis and the XX-axis ( 8 mm x, XX 8 mm) for Model_8 8 8 (512 elements), Model_10 10 10 (1000 elements), − ≤ ≤ × × × × Model_14 14 14 (2744 elements) and Model_16 16 16 (4096 elements) under uniform remote × × × × tensile loading for the isotropic cubic inhomogeneity. The remote applied load was assumed to be o σ xx = 143.1 GPa. Standard 20-node quadratic hexahedral elements in Figure6 were used in the VIEM models. In the Model_nxnxn, the symbol n represents the number of line segments split in each side of a cube. A cubic inhomogeneity ( 4 mm x, y, z 4 mm) is divided into n3 hexahedral elements in the Model_nxnxn. − ≤ ≤ For example, in the Model_14 14 14, a cubic inhomogeneity ( 4 mm x, y, z 4 mm) is divided × × − ≤ ≤ into 2744 (=143) elements. o Figure9a shows the normalized tensile stress component ( σxx/σ xx) along the x-axis inside the isotropic cubic inhomogeneity ( 4 mm x 4 mm) and outside the isotropic cubic inhomogeneity − ≤ ≤ ( 8 mm x 4 mm and 4 mm x 8 mm), respectively. Figure9b shows the normalized − ≤ ≤ − o ≤ ≤ tensile stress component (σxx/σ xx) along the XX-axis inside the isotropic cubic inhomogeneity ( 4 mm XX 4 mm) and outside the isotropic cubic inhomogeneity ( 8 mm XX 4 mm and − ≤ ≤ − ≤ ≤ − 4 mm XX 8 mm), respectively. The results using the VIEM for the diﬀerent models converged very ≤ ≤ well within this range of hexahedral elements. o Figures 10–12 show the normalized tensile stress component (σxx/σ xx) along the x-axis and the XX-axis ( 8 mm x, XX 8 mm) for Model_8 8 8, Model_10 10 10, Model_14 14 14 − ≤ ≤ × × × × × × and Model_16 16 16 under uniform remote tensile loading for the three diﬀerent orthotropic × × (a) inhomogeneities (mat_01, mat_02 and mat_03), respectively. The remote applied load was assumed to o be σ xx = 143.1 GPa.

Mathematics 2020, 8, x FOR PEER REVIEW 17 of 26

≤ 4 mm) and outside the isotropic cubic inhomogeneity (−8 mm ≤ XX ≤ −4 mm and 4 mm ≤ XX ≤ 8 mm), respectively. The results using the VIEM for the different models converged very well within this range of hexahedral elements.

(a) (b)

Figure 8. A typical single cubic model (−4 mm ≤ x, XX ≤ 4 mm) (between two green lines) including the outside of the cube (−8 mm ≤ x ≤ −4 mm and 4 mm ≤ x ≤ 8 mm) in the VIEM. (a) A single cubic model (−4 mm ≤ x, XX ≤ 4 mm) with 1000 elements (Model_10 × 10 × 10) (between two green lines) including the outside of the cube (−8 mm ≤ x ≤ −4 mm and 4 mm ≤ x ≤ 8 mm). 1000 elements are used for the discretization of the outside of the cube; (b) a single cubic model (−4 mm ≤ x, XX ≤ 4 mm) with 4096 elements (Model_16 × 16 × 16) (between two green lines) including the outside of the cube (−8 mm ≤ x ≤ −4 mm and 4 mm ≤ x ≤ 8 mm). 4096 elements are used for the discretization of the Mathematicsoutside2020, 8 of, 1866 the cube. 18 of 26

Mathematics 2020, 8, x FOR PEER REVIEW 18 of 26

(a)

(b)

FigureFigure 9. Numerical 9. Numerical results results using using the the volume volume integral integral equation equation method method (VIEM) (VIEM) for for the the normalized normalized o o tensiletensile stress stress component component (σxx (σ/σxx/σxx)xx along) along the thex-axis x-axis and and the the XX-axis XX-axis ( 8 (−8 mm mm x, ≤ XX x, XX 8 ≤ mm)8 mm) for for − ≤ ≤ Model_8Model_88 × 88, × Model_108, Model_1010 × 1010, × 1 Model_140, Model_1414 × 1414 × and14 and Model_16 Model_1616 × 1616 × under16 under uniform uniform × × × × × × × × remoteremote tensile tensile loading. loading. (a) ( x-axis;a) x-axis; (b) ( XX-axis.b) XX-axis.

o FigureFigures 10 a 10 shows–12 show the normalizedthe normalized tensile tensile stress stress component component (σxx (σ/σxx/σxxo)xx along) along the the x-axis x-axis inside and the theXX orthotropic-axis (−8 m cubicm ≤ x, inhomogeneity XX ≤ 8 mm) for #1 M (mat_02odel_8 in× 8 Table × 8, 2M)(odel_14 mm0 × 10x × 10,4 mm)Model_1 and4 outside× 14 × 1 the4 and − ≤ ≤ cubicModel_1 inhomogeneity6 × 16 × 1( 6 8 under mm uniformx 4 mm remote and 4 tensile mm x loading8 mm), for respectively. the three different Figure 10 orthotropicb shows − ≤ ≤ − o ≤ ≤ theinhomogeneities normalized tensile (mat_01, stress mat_02 component and mat_03) (σxx/σ xx, respectively) along the. XX-axis The remote inside applied the orthotropic load was assumed cubic inhomogeneityto be σoxx = 143.1 #1 ( GPa.4 mm XX 4 mm) and outside the cubic inhomogeneity ( 8 mm XX 4 mm − ≤ ≤ − ≤ ≤ − Figure 10 a shows the normalized tensile stress component (σxx/σoxx) along the x-axis inside the orthotropic cubic inhomogeneity #1 (mat_02 in Table 2) (−4 mm ≤ x ≤ 4 mm) and outside the cubic inhomogeneity (−8 mm ≤ x ≤ −4 mm and 4 mm ≤ x ≤ 8 mm), respectively. Figure 10b shows the normalized tensile stress component (σxx/σoxx) along the XX-axis inside the orthotropic cubic inhomogeneity #1 (−4 mm ≤ XX ≤ 4 mm) and outside the cubic inhomogeneity (−8 mm ≤ XX ≤ −4 mm and 4 mm ≤ x ≤ 8 mm), respectively. The VIEM solutions using the different models converged very well within this range of hexahedral elements. Figures 11a and 12a show the same as Figure 10a for the orthotropic cubic inhomogeneity #2 and #3 (mat_03 and mat_04 in Table 2) (−8 mm ≤ x ≤ 8 mm), respectively. Figures 11b and 12b show the same as Figure 10b for the orthotropic cubic inhomogeneity #2 and #3 (−8 mm ≤ XX ≤ 8 mm), respectively. The results using the VIEM for the different models converged very well within this range of hexahedral elements. The normalized tensile stress component (σxx/σoxx) for the isotropic cubic inhomogeneity in Figure 9 appears to be considerably different from those of the orthotropic cubic inhomogeneities in Figures 10–12.

Mathematics 2020, 8, 1866 19 of 26 and 4 mm x 8 mm), respectively. The VIEM solutions using the diﬀerent models converged very ≤ ≤ wellMathematics within 20 this20, 8, rangex FOR PEER of hexahedral REVIEW elements. 19 of 26

(a)

(b)

Figure 10.10. Numerical results using the volume integral equation method (VIEM) for the normalizednormalized xx oxx tensiletensile stress stress component component ( (σσxx/σ/σ xx) ) along along the the x x-axis-axis and and the XXXX-axis-axis (−88 mmmm ≤ x, XX ≤ 8 mm)mm) for − ≤ ≤ Model_8 × 8 × 8, Model_10 × 10 × 10, Model_14 × 14 × 14 and Model_16 × 16 × 16 under uniformuniform × × × × × × × × remoteremote tensiletensile loading.loading. (a) x-axis;x-axis; (b)) XX-axis.XX-axis.

Mathematics 2020, 8, 1866 20 of 26 Mathematics 2020, 8, x FOR PEER REVIEW 20 of 26

(a)

(b)

Figure 11.11. Numerical results using the volume integral equation method (VIEM) for the normalizednormalized o tensiletensile stress stress component component ( (σσxxxx/σ/σ xxxx) ) along along the the x x-axis-axis and and the XXXX-axis-axis (−88 mmmm ≤ x, XX ≤ 8 mm)mm) for − ≤ ≤ Model_8Model_8 × 8 × 8,8, Model_10Model_10 × 10 × 10,10, Model_14Model_14 × 14 × 1144 and Model_16Model_16 × 16 × 1616 underunder uniformuniform × × × × × × × × remoteremote tensiletensile loading.loading. (a) x-axis;x-axis; (b)) XX-axis.XX-axis.

Figures 11a and 12a show the same as Figure 10a for the orthotropic cubic inhomogeneity #2 and #3 (mat_03 and mat_04 in Table2)( 8 mm x 8 mm), respectively. Figures 11b and 12b show − ≤ ≤ the same as Figure 10b for the orthotropic cubic inhomogeneity #2 and #3 ( 8 mm XX 8 mm), − ≤ ≤ respectively. The results using the VIEM for the diﬀerent models converged very well within this range of hexahedral elements.

Mathematics 2020, 8, 1866 21 of 26 Mathematics 2020, 8, x FOR PEER REVIEW 21 of 26

(a)

(b)

FigureFigure 12. 12.Numerical Numerical results results using using the the volume volume integral integral equation equation method method (VIEM) (VIEM) for for the the normalized normalized o o tensiletensile stress stress component component (σxx (σ/σxx/σxx)xx along) along the the x-axis x-axis and and the the XX-axis XX-axis ( 8 (−8 mm mm x, ≤ XX x, XX8 ≤ mm)8 mm) for for − ≤ ≤ Model_8Model_88 × 88, × Model_108, Model_1010 × 1010, × 1 Model_140, Model_1414 × 1414 × and14 and Model_16 Model_1616 × 1616 × under16 under uniform uniform × × × × × × × × remoteremote tensile tensile loading. loading. (a) x-axis;(a) x-axis; (b) ( XX-axis.b) XX-axis.

o 3.2.The A single normalized Spherical tensile Inhomogeneity stress component Problem (σxx/σ xx) for the isotropic cubic inhomogeneity in Figure9 appears to be considerably di ﬀerent from those of the orthotropic cubic inhomogeneities in Figures3.2.1. 10 A– S12ingle. Isotropic Spherical Inhomogeneity In order to examine the accuracy of the numerical results using the VIEM, the numerical results using the VIEM for a single isotropic spherical inhomogeneity were first compared to the analytical solutions [21]. We considered a single isotropic spherical inhomogeneity with a radius of 7 mm in an unbounded isotropic matrix subject to uniform remote tensile loading, σxxo, as shown in

Mathematics 2020, 8, 1866 22 of 26

3.2. A single Spherical Inhomogeneity Problem

3.2.1. A Single Isotropic Spherical Inhomogeneity In order to examine the accuracy of the numerical results using the VIEM, the numerical results using the VIEM for a single isotropic spherical inhomogeneity were ﬁrst compared to the analytical solutions [21]. We considered a single isotropic spherical inhomogeneity with a radius of 7 mm in an Mathematics 2020, 8, x FOR PEER REVIEW o 22 of 26 unbounded isotropic matrix subject to uniform remote tensile loading, σxx , as shown in Figure 13 (also see Figure2b). In Figure 13, standard 20-node quadratic hexahedral elements in Figure6 were Figure 13 (also see Figure 2b). In Figure 13, standard 20-node quadratic hexahedral elements in used in the discretization [17]. The number of hexahedral elements was 4320. In order to investigate Figure 6 were used in the discretization [17]. The number of hexahedral elements was 4320. In order the accuracy of the hexahedral elements, only hexahedral elements without tetrahedral elements were to investigate the accuracy of the hexahedral elements, only hexahedral elements without used in Figure 13. The elastic properties of the isotropic matrix and the isotropic inhomogeneity for the tetrahedral elements were used in Figure 13. The elastic properties of the isotropic matrix and the calculations are given in Table5. isotropic inhomogeneity for the calculations are given in Table 5.

(a) (b)

Figure 13. AA typical typical discretized discretized spherical spherical model model in inthe the volume volume integral integral equation equation method method (VIEM) (VIEM).. (a) (Aa )spherical A spherical model model;; (b) (abn) aninside inside view view of a of spherical a spherical model. model.

TableTable 5. MaterialMaterial properties properties of the isotropic matrix and the isotropic spherical inhomogeneity.inhomogeneity.

IsotropicIsotropic M Matrixatrix IsotropicIsotropic Inhomogeneity Inhomogeneity IsotropicIsotropic Matrix Matrix IsotropicIsotropic Inhomogeneity Inhomogeneity (mat_05) (mat_05) (mat_06) (mat_06) (mat_05) (mat_05) (mat_06) (mat_06) λλ [GPa][GPa] 75.0 75.0 150.0 150.0 75.0 75.0 375.0 375.0 µ [GPa] 37.5 75.0 37.5 187.5 μ [GPa] 37.5 75.0 37.5 187.5

Figure 14 14 shows shows a comparison a comparison between between the numericalthe numerical results results using the using volume the integral volume equation integral o equationmethod (VIEM) method and (VIEM) the analytical and the solutions analytical [21] for solutions the normalized [21] for tensile the stressnormalized component tensile (σxx stress/σ xx) componentalong the x-axis (σxx/σ insideoxx) along the isotropicthe x-axis spherical inside the inhomogeneities isotropic spherical with inhomogeneities a radius of 7 mm with under a radius uniform of o 7remote mm un tensileder uniform loading. remote The remote tensile applied loading. load The was remote assumed applied to be σ loadxx = was150.0 assumed GPa. to be σoxx = o 150.0It GPa. should be noted that the normalized tensile stress component (σxx/σ xx) inside the isotropic spherical inhomogeneities was found to be constant [21,22]. Eshelby [22] solved the inhomogeneity problem of an ellipsoidal inclusion using the equivalent inclusion method. The stresses inside the inclusion are determined by:

i (2) c (2) σ = c (ε ε∗ ) = c (S ε∗ ε∗ ) (16) ij ijkl kl − kl ijkl klpq pq − kl

(2) (c) where cijkl represents the fourth-order elasticity tensors of the matrix and εkl stands for the (c) constraint strain (εkl = Sklpqεpq*). εkl* represents the eigenstrain of the equivalent inclusion and Sklpq represents Eshelby’s fourth-order tensor. It should be noted that the Eshelby’s tensor satisﬁes minor symmetries, Sijkl = Sjikl = Sijlk.

Figure 14. Comparison of the numerical results using the volume integral equation method (VIEM)

and the analytical solutions for the normalized tensile stress component (σxx/σoxx) along the x-axis inside the isotropic spherical inclusions with a radius of 7 mm under uniform remote tensile loading.

Mathematics 2020, 8, x FOR PEER REVIEW 22 of 26

Figure 13 (also see Figure 2b). In Figure 13, standard 20-node quadratic hexahedral elements in Figure 6 were used in the discretization [17]. The number of hexahedral elements was 4320. In order to investigate the accuracy of the hexahedral elements, only hexahedral elements without tetrahedral elements were used in Figure 13. The elastic properties of the isotropic matrix and the isotropic inhomogeneity for the calculations are given in Table 5.

Mathematics 2020, 8, 1866 (a) (b) 23 of 26

Figure 13. A typical discretized spherical model in the volume integral equation method (VIEM). (a) AIn spherical order to model obtain; ( theb) an analytical inside view solution, of a spherical we used model. Equations (5) and (6) in Shibata and Ono [21] and the expressions in Equation (17): Table 5. Material properties of the isotropic matrix and the isotropic spherical inhomogeneity. (7 5v) = = = − Isotropic Matrix IsotropicS1111 InhomogeneityS2222 S3333 Isotropic15(1 vM) ,atrix Isotropic Inhomogeneity − (5v 1) ======− (mat_05)S 1122 S2233 (mat_05)S3311 S1133 S3322(mat_06)S2211 15(1 v) , (mat_06) (17) (4 5v) − λ [GPa] 75.0 150.0= = = 75.0− 375.0 S1212 S2323 S3131 15(1 v) . μ [GPa] 37.5 75.0 37.5− 187.5 In Equation (17), ν is Poisson’s ratio for the matrix material. FigureFor the isotropic14 shows spherical a comparison inhomogeneity between (mat_05), the numerical the analytical results solution using was the 1.309091.volume integral For the equationisotropic spherical method (VIEM) inhomogeneity and the (mat_06), analytical the solutions analytical [21] solution for the was normalized 1.617336. tensile An excellent stress componentagreement was(σxx/σ foundoxx) along between the x the-axis analytical inside the and isotropic numerical spherical solutions inhomogeneities using the VIEM with for a both radius cases of 7considered. mm under Figure uniform 14 showsremote that tensile the percentageloading. The diﬀ erences remote for applied the two load sets was of results assumed are lessto be than σoxx 1% = 150.0in both GPa. cases.

Figure 14. Comparison of the numerical results using the volume integral equation method (VIEM) oo and the analyticalanalytical solutionssolutions forfor thethe normalizednormalized tensiletensile stressstress componentcomponent (σ(σxxxx//σσ xx)) along along the x-axisx-axis inside the isotropic spherical inclusions with a radius of 7 m mmm under uniform remote tensile loading.

3.2.2. A Single Orthotropic Spherical Inhomogeneity We consider a single orthotropic spherical inhomogeneity in an unbounded isotropic matrix under o uniform remote tensile loading, σxx , as shown in Figure 13 (also see Figure2b). The elastic properties of the isotropic matrix and orthotropic inhomogeneities used in these calculations are given in Table2. Figure 15 shows the numerical results using the volume integral equation method (VIEM) for o the normalized tensile stress component (σxx/σ xx) along the x-axis inside the orthotropic spherical inhomogeneities (the orthotropic inhomogeneity #1 and #2; mat_02 and mat_03 in Table2) with a radius of 7 mm under uniform remote tensile loading. The remote applied load was assumed to be o o σ xx = 143.1 GPa. It should be noted that the normalized tensile stress component (σxx/σ xx) inside the orthotropic spherical inhomogeneities was found to be constant [23]. Mathematics 2020, 8, x FOR PEER REVIEW 24 of 26

inhomogeneities was also found to be constant for the single orthotropic spherical inhomogeneity

Mathematicsproblems.2020 , 8, 1866 24 of 26

Figure 15. Numerical results using the volume integral equation method (VIEM) for the normalized Figure 15. Numerical results using the volume integral equation method (VIEM) for the normalized tensile stress component (σxx/σoxx) along the x-axis inside the orthotropic spherical inclusions with a tensile stress component (σxx/σoxx) along the x-axis inside the orthotropic spherical inclusions with radius of 7 mm under uniform remote tensile loading. a radius of 7 mm under uniform remote tensile loading. 4. Conclusions The major merit of the volume integral equation method (VIEM), compared to the finite elementThe method volume (FEM), integral is equation that it needs method discretization was applied of to the a class inhomogeneities of three-dimensional only as opposed elastostatic to inhomogeneitydiscretization of problems. the whole First, domain. we introduced In elastodynamic the details problems of the numerical, involving treatment, multiple especially anisotropic the evaluationinhomogeneities, of singular numerical integrals, treatment for solving of three-dimensional the BIEM is very problems difficult using to achieve the VIEM., since We closed considered-form a single isotropic or orthotropic cubic inhomogeneity ( 4 mm x, y, z 4 mm) in the unbounded analytical expressions of the elastodynamic Green’s functions− in≤ anisotropic≤ media are currently not isotropicavailable. matrix By contrast, under uniform the VIEM remote does tensile not require loading. the use of Green’s functions for anisotropic inhomogeneities.It was determined Since thatthe VIEM the VIEM is designed solutions, to using use standar the differentd finite models, elements, converged it is simpler verywell and formore all cases.advantageous Next, we for considered dealing with a single multiple isotropic non-/smoothorthotropic anisotropic spherical inhomogeneities. inhomogeneity Thus, in an unboundedthe VIEM is isotropicnow generally matrix moreunder applicable uniform remote and executabletensile loading. than An the excellent boundary agreement element was or found finite between element themethods. analytical Moreover, and numerical the volume solutions integral using equation the VIEM method for single(VIEM) isotropic can be sphericalused to compute inhomogeneity critical o problems.values of practical The normalized interest in tensile realistic stress models component of composites (σxx/σ withxx) inside strong the anisotropic spherical inhomogeneitiesinhomogeneities wasof arbitrary also found shapes. to be constant for the single orthotropic spherical inhomogeneity problems. The major merit of the volume integral equation method (VIEM), compared to the finite element methodAuthor Contributions: (FEM), is that Conceptualization, it needs discretization J.L.; VIEM of the Analysis, inhomogeneities J.L.; Investigation, only as opposed J.L. and toM.H.; discretization Validation, ofM.H. the and whole J.L.; domain. Writing-original In elastodynamic draft, J.L. All problems, authors have involving read and multiple agreed anisotropic to the published inhomogeneities, version of the numericalmanuscript.treatment of the BIEM is very difficult to achieve, since closed-form analytical expressions ofFunding: the elastodynamic This research Green’s was supported functions by in the anisotropic Ministry of media Trade, are Industry currently and not Energy available. (MOTIE), By contrast, KOREA, thethrough VIEM the does Education not require Program the for use Creative of Green’s and Industrialfunctions Convergence for anisotropic (No. inhomogeneities. N0000717) and 2019 Since Hongik the VIEMUniversity is designed Research to Fund use standard. This work finite was elements, also supported it is simpler by the and National more advantageous Supercomputing for dealing Center with with multiplesupercomputing non-smooth resources anisotropic including inhomogeneities. technical support Thus,(No. KSC the-2020 VIEM-CRE is now-0107) generally and the National more applicable Research andFoundation executable of Korea than (NRF) the boundary grant funded element by the or Korea finite government element methods. (MSIT) (No. Moreover, 2019R1H1A2039647) the volume. integral equationAcknowledgments: method (VIEM) The authors can bewould used like to computeto thank all critical reviewers values and of journal practical editors interest for their in realisticconstructive models and ofmotivating composites comments. with strong The authors anisotropic also thank inhomogeneities Jason P. Buschman of arbitrary at Hongik shapes. University (Sejong Campus) for his thorough proofreading of the revised manuscript. Author Contributions: Conceptualization, J.L.; VIEM Analysis, J.L.; Investigation, J.L. and M.H.; Validation, M.H. andConflicts J.L.; Writing-original of Interest: The authors draft, J.L. declare All authors no conflict have of read interest. and agreed to the published version of the manuscript.

Mathematics 2020, 8, 1866 25 of 26

Funding: This research was supported by the Ministry of Trade, Industry and Energy (MOTIE), KOREA, through the Education Program for Creative and Industrial Convergence (No. N0000717) and 2019 Hongik University Research Fund. This work was also supported by the National Supercomputing Center with supercomputing resources including technical support (No. KSC-2020-CRE-0107) and the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (No. 2019R1H1A2039647). Acknowledgments: The authors would like to thank all reviewers and journal editors for their constructive and motivating comments. The authors also thank Jason P. Buschman at Hongik University (Sejong Campus) for his thorough proofreading of the revised manuscript. Conﬂicts of Interest: The authors declare no conﬂict of interest.

References

1. Bose, S.K.; Mal, A.K. Elastic waves in a fiber-reinforced composite. J. Mech. Phys. Solids 1974, 22, 217–229. [CrossRef] 2. Maslov, B.P. Stress concentration in an isotropic matrix bonded by anisotropic fibers. Sov. Appl. Mech. 1987, 23, 963–969. [CrossRef] 3. Yang, R.B.; Mal, A.K. The effective transverse moduli of a composite with degraded fiber-matrix interfaces. Int. J. Eng. Sci. 1995, 33, 1623–1632. [CrossRef] 4. Lee, J.K.; Mal, A.K. A volume integral equation technique for multiple inclusion and crack interaction problems. J. Appl. Mech. 1997, 64, 23–31. [CrossRef] 5. Lee, J.K.; Mal, A.K. A volume integral equation technique for multiple scattering problems in elastodynamics. Appl. Math. Comput. 1995, 67, 135–159. [CrossRef] 6. Nimmer, R.P.; Bankert, R.J.; Russel, E.S.; Smith, G.A.; Wright, P.K. Micromechanical modeling of fiber/matrix interface effects in transversely loaded SiC/Ti-6-4 metal matrix composites. J. Compos. Tech. Res. 1991, 13, 3–13. 7. Lee, J.; Mal, A. Characterization of matrix damage in metal matrix composites under transverse loads. Comput. Mech. 1998, 21, 339–346. [CrossRef] 8. Aghdam, M.M.; Falahatgar, S.R. Micromechanical modeling of interface damage of metal matrix composites subjected to transverse loading. Compos. Struct. 2004, 66, 415–420. [CrossRef] 9. Mal, A.K.; Knopoff, L. Elastic wave velocities in two component systems. J. Inst. Math. Appl. 1967, 3, 376–387. [CrossRef] 10. Lee, J.K.; Lee, H.C.; Jeong, H.G. Multiple scattering using parallel volume integral equation method: Interaction of SH waves with multiple multilayered anisotropic elliptical inclusions. Math. Probl. Eng. 2015. [CrossRef] 11. Lee, J.K.; Lee, H.M.; Mal, A. A mixed volume and boundary integral equation technique for elastic wave field calculations in heterogeneous materials. Wave Motion 2004, 39, 1–19. [CrossRef] 12. Buryachenko, V.A. Micromechanics of Heterogeneous Materials; Springer: New York, NY, USA, 2007. 13. Advances in Computers and Information in Engineering Research; Michopoulos, J.G.; Rosen, D.W.; Paredis, C.J.; Vance, J.M. (Eds.) ASME Press: New York, NY, USA, 2020; Volume 2, in press. 14. Lee, J.K.; Oh, S.M.; Mal, A. Calculation of interfacial stresses in composites containing elliptical inclusions of various types. Eur. J. Mech. A Solid. 2014, 44, 17–40. [CrossRef] 15. Buryachenko, V.A.; Bechel, V.T. A series solution of the volume integral equation for multiple-inclusion interaction problems. Compos. Sci. Technol. 2000, 60, 2465–2469. [CrossRef] 16. Buryachenko, V.A. Solution of general integral equations of micromechanics of heterogeneous materials. Int. J. Solids Struct. 2014, 51, 3823–3843. [CrossRef] 17. PATRAN User’s Manual, version 2008 r2; MSC Sotware: Newport Beach, CA, USA, 2008. 18. Banerjee, P.K. The Boundary Element Methods in Engineering; McGraw-Hill: London, UK, 1993. 19. Pao, Y.-H.; Varatharajulu, V. Huygens’ principle, radiation conditions, and integral formulas for the scattering of elastic waves. J. Acoust. Soc. Am. 1976, 59, 1361–1369. [CrossRef] 20. Li, H.B.; Han, G.M.; Mang, H.A. A new method for evaluating singular integrals in stress analysis of solids by the direct boundary element method. Int. J. Numer. Methods Eng. 1985, 21, 2071–2098. [CrossRef] 21. Shibata, M.; Ono, K. Stress concentration due to a prolate spheroidal inclusion. J. Compos. Mater. 1978, 12, 132–138. [CrossRef] Mathematics 2020, 8, 1866 26 of 26

22. Eshelby, J.D. The determination of the elastic ﬁeld of an ellipsoidal inclusion and related problems. Proc. Royal Soc. A 1957, 241, 376–396. 23. Ma, H. Solutions of Eshelby-Type Inclusion Problems and a Related Homogenization Method Based on a Simpliﬁed Strain Gradient Elasticity Theory. Ph.D. Thesis, Department of Mechanical Engineering, Texas A&M University, College Station, TX, USA, 2010.

Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional aﬃliations.

© 2020 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).