# Determination of the Stress State of a Piecewise Homogeneous Elastic Body with a Row of Cracks on an Interface Surface Subject to Antiplane Strains with Inclusions at the Tips

Authors: Ali Golsoorat Pahlaviani; Suren Mkhitaryan
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

The stress state of a bimaterial elastic body that has a row of cracks on an interface surface is considered. It is subjected to antiplane deformations by uniformly distributed shear forces acting on the horizontal sides of the body. The governing equations of the problem, the stress intensity factors, the deformation of the crack edges, and the shear stresses are derived. The solution of the problem via the Fourier sine series is reduced to the determination of a singular integral equation (SIE) and consequently to a system of linear equations. In the end, the problem is solved in special cases with inclusions. The results of this paper and the previously published results show that the used approach based on the Gauss-Chebyshev quadrature method can be considered as a generalized procedure to solve the collinear crack problems in mode I, II, or III loadings.


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## Body

## 1. Introduction

Stress analysis near a fracture in an elastic material is one of the most explored topics in solid mechanics. Calculation and stress analysis of engineering structures, particularly their connections and the determination of the stress and strain distribution fields of cracked bodies, have received attention from numerous investigators in recent years. The stress intensity factors, shear stresses, and crack opening displacements are major concepts that must be determined. The stress intensity factor is an important parameter that denotes the magnitude of the stress singularity. The singular order is a single real value, for example, 0.5 for a crack in a homogeneous material. The singular order of general interface corners may be real or complex. An asymptotic stress near the tip of a sharp interfacial corner is generally singular as a result of a mismatch between the materials’ elastic constants.In this paper, the problem of a piecewise homogeneous rectangular prismatic elastic body in an antiplane strain state due to antiplane forces is discussed. The plate is composed of two bonded dissimilar materials with a number of arbitrary collinear cracks on their interface surface. The aim of this work is the derivation of analytical expressions for the stress intensity factors of the cracks and the presentation of a new mathematical-numerical approach to solve singular integral equations related to the beginning of stresses at the tips of cracks in antiplane deformations; thus, a numerical method to calculate the SIFs of an interface crack between dissimilar materials is developed. Calculation of the SIFs for bimaterial plates in dimensionless form is conducted for several cases: a row of cracks, one and two cracks, and one crack with inclusions at the tips. The objectives of the present study are (i) to present a new method for calculating SIFs of interfacial cracks subject to antiplane loading and (ii) to investigate the influence of inclusion moduli on SIFs to reduce SIFs in cracked bodies and to direct us towards a method for repairs.The existence of three-dimensional effects at cracks has been known for many years, but understanding has been limited, and for some situations, it still is. Understanding improved when the existence of corner point singularities and their implications became known for straight through-the-thickness cracks [1, 2].It has been known for a long time that shear and antiplane fracture modes are coupled. This means that shear or antiplane loading of an elastic plate with a through-the-thickness crack generates a coupled three-dimensional antiplane or shear singular stress state, respectively. These singular stress states (or coupled fracture modes) are currently largely ignored in theoretical and experimental investigations as well as in standards and failure assessment codes of structural components, in which it is implicitly assumed that the intensities of these modes as well as other three-dimensional effects are negligible in comparison with the stress fields generated by the primary modes (modes I, II, and III) [3, 4].The theoretical bases of fracture stresses are discussed in the literature [5–7], and the conclusions of numerous studies and investigations on the derivation of SIFs are categorized in [8–10]. Most of these studies were performed on homogeneous plates. It is proven that, under certain circumstances, the three-dimensional governing equations of elasticity can be reduced to a system where a biharmonic equation and a harmonic equation have to be simultaneously satisfied. The former provides the solution of the corresponding plane problem, while the latter provides the solution of the corresponding out-of-plane shear problem [11]. On the other hand, a mixed fracture mode under antiplane loading may also occur. This coupled fracture mode represents one of three-dimensional phenomena that are currently largely ignored in numerical simulations and failure assessments of structural components weakened by cracks. It arises due to the boundary conditions on the plate-free surfaces, which negate the transverse shear stress components corresponding to classical mode III. Instead, a new singular stress state in addition to the well-known 3D corner singularity is generated. This singular stress state can affect or contribute significantly to the fracture initiation conditions [12, 13].Inclusions and cavities are also important in understanding the mechanical behavior of structures and are studied in several papers, for example, 2D linear elastic materials [14] and antiplane shear cracks [15]. Photoelasticity and finite element methods have also been employed to study the interaction between collinear cracks, and good agreement was found [16]. Photoelasticity is very helpful in investigating the stress state near inclusions. The results show that the singular stress field predicted by the linear elastic solution for an inclusion can be generated in reality with great accuracy [17]. The inclusions form a thin material that constituted a rigid line inclusion, embedded in a linear elastic body to produce an inhomogeneous stress state. The experiments fully validate the stress state calculated for an elastic plate [18].The mechanical behavior of thin inclusions is fundamental to the design of composite materials. It is realized that, for a given geometry and boundary condition,KIII depends on the gradation of both the modulus of elasticity and Poisson’s ratio [19]. The cracked sandwich plate twist specimen is viable to characterize mode III fracture [20].The stress intensity factors can be calculated by a path-independent h-integral and through the virtual crack closure-integral method (VCCM) for numerical implementation [21]. The present study is aimed at investigating the stress state of a piecewise homogeneous elastic body which has a row of collinear cracks in mode III. The numerical procedure based on the loading Gauss-Chebyshev quadrature method is applied. This approach can be used for other multicrack problems or more complicated types of loading [22, 23].At the end of this paper, we discuss the influence of the bimaterial nature of the body, the distance and geometry of the cracks, and the presence of inclusions at the tips on the characteristics of antiplane shear stresses and deformations.


## 2. Derivation of the Singular Integral Equation for the General Form of the Problem

A piecewise homogeneous rectangular body in the Cartesian coordinate systemOxyz is considered as shown in Figure 1.Figure 1
Rectangular piecewise homogeneous elastic body with several collinear cracks.In the equations below, subscript 1 denotes rectangular plateD1 and subscript 2 denotes D2.The top rectangular plateD1=0≤x≤l;0≤y≤h1 has a rigidity modulus G1, length l, and height h1, and the bottom rectangular plate D2=0≤x≤l;-h2≤y≤0 has a rigidity modulus G2, length l, and height h2. On the bonding surface of the two materials in the interval 0≤x≤l, there are N arbitrary collinear cracks with total length L and individual lengths Lk; that is,(1)Lk=ak,bkk=1,N¯a1≥0,bN≤l,L=⋃k=1NLk=⋃k=1Nak,bk,ak<bkk=1,N¯,bk<ak+1k=1,N-1¯.The upper and lower boundaries of these cracks are subjected to antiplane shear stresses τ±(0)x, which act as varying internal loadings of openings: (2)τyz1y=+0=τ+0xx∈L+;τyz2y=-0=τ-0xx∈L-,L±=⋃k=1NLk±.The edgesx=0 and x=l of rectangles Dj (j=1,2) are restricted, and the edges of the rectangles y=h1 and y=-h2 are subjected to antiplane shear stresses: τj(x) (j=1,2), (3)τyz1y=h1-0=τ1x,τyz2y=-h2+0=τ2x0<x<l,where the notation τyzjj=1,2 for shearing stress components in plates Dj is used. Because the pair of vertical sides of the rectangle is rigidly fixed, the boundary value problem for the complex potential of state can be transformed to the problem in the half-plane, and its solution can be obtained in quadrature using conformal mapping [24, 25].Through the above assumptions, the body is under the antiplane strain condition, so the displacements of the crack edges are along theOz-axis with the base on theOxy surface.It is required to calculate the dislocation density of the crack boundaries, the related stress intensity factors, and the distribution of shearing stresses in the regions that are outside of the crack system, with total lengthL′=[0,l]∖L.For the development of the above-mentioned functions, the plate is divided into an upper rectangle (D1) and a lower rectangle (D2) on theOx-axis with 0≤x≤l, and we introduce a function wj(x,y) (j=1,2) as the only nonzero displacement component along theOz-axis for both Dj (j=1,2) rectangular plates and separately investigate their individual elastic equilibria. In this manner, for rectangle (D1), according to Hooke’s law, the below boundary value problem is found:(4)Δw1=∂2w1∂x2+∂2w1∂y2=00<x<l,0<y<h1,w1x,yx=0=w1x,yx=l=00<y<h1,τyz1y=+0=G1∂w1∂yy=+0=τ+x,τyz1y=h1=G1∂w1∂yy=h1=τ1x0<x<l.Suppose the notation(5)τyz1y=+0=τ+x=τ+0xx∈L+τxx∈L′in which the function τ+0x, as mentioned before, is the shear stress loading on the cracks’ top boundaries L+ and τ(x) is the unknown tearing shear stress that acts on the noncracked interface surface system L′.According to [26], to solve (4) by means of the Fourier sine series, it is obtained that (6)w-1n,y=∫0lw1x,ysinπnxldxn=1,2,….Hence from the inverse Fourier series we deduce (7)w1x,y=2l∑n=1∞w-1n,ysinπnxl0≤x≤l.Equation (7) is the Fourier sine series with coefficients given by (6). The reason for using the Fourier sine series is that boundary value problem (4) with the Laplacian and Dirichlet boundary conditions can be solved by separation of variables, and because the boundary conditions in the x variable are the Dirichlet boundary conditions, this generates a Fourier sine series with eigenfunctions sin⁡(nπx/l) and eigenvalues (nπx/l)2, n=1,2,….Considering (6) and (7), it follows that by taking the two parts of the differential equation (4), multiplying by sin⁡(πnx/l), integrating it over the interval (0,l), and taking into consideration the boundary condition (4), we obtain the below boundary value problem:(8)d2w-1dx2-π2n2l2w-1=00<y<h1,G1dw-1dyy=+0=τ-+n,G1dw-1dyy=h1=τ-1nwith(9)τ-+n=∫0lτ+xsinπnxldx,τ-1n=∫0lτ1xsinπnxldx.Boundary value problem (8) can be solved by the following formula:(10)w-1n,y=lπnG1sinπnh1/l×πny-h1lτ-1nchπnyl000000-τ-+nchπny-h1l00000000000000000000≤y≤h1.For (n=1,2,…), we can conclude(11)w-1n,0=lτ-1n-τ-+nchπnh1/lπnG1shπnh1/l.Using a similar approach, the boundary value problem related to the rectangular plate(D2) is obtained: (12)w-2n,0=lτ--nchπnh2/l-τ-2nπnG2shπnh2/l,w-2n,y,τ--n,τ-2n=∫0lw2x,y,τ-x,τ2x0000000·sinπnxldx,τ-x=τyz2y=-0=τ-0xx∈L-τxx∈L′.Then, considering thew1(x,0) and w2(x,0) displacements and the following functions:(13)Φx=w1x,0-w2x,02=2l∑n=1∞Φnsinπnxl,Ψx=w1x,0+w2x,02=2l∑n=1∞Ψnsinπnxl,Ωx=τ+x+τ-x2=2l∑n=1∞Ωnsinπnxl,Xx=τ+x-τ-x2=2l∑n=1∞Xnsinπnxliiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii0<x<l,where Φn, Ψn, Ωn, and Xn are the Fourier sine series coefficients of the above functions that are similar to those introduced in (7) and (9), we obtained(14)Φn=12w-1n,0-w-2n,0,Ψn=12w-1n,0+w-2n,0,Ωn=12τ-+n+τ--n,Xn=12τ-+n-τ--n.Substituting coefficients (14) into (11) and (12), a system of two linear equations with respect to Ωn and Ψn is obtained. Solving the system, coefficients Ωn and Ψn can be expressed by Φn and Xn and, in turn, all of the functions in (13) can be expressed by the Φn and Xn coefficients. In this case, as we expect, the required function (13) after some simple transforms is found in dimensionless form as follows (see Appendix):(15)Ω0ξ=-12π∫L0ctgξ-η2+ctgξ+η2φ0ηdη+1π∫L0K∗ξ-η+K∗ξ+ηφ0ηdη+μ-1μ+1τ~+ξ-τ~-ξ+2μπ1+μ×∫L0L∗ξ-η-L∗ξ+η0000000·τ~+η-τ~-ηdη+1π∫0πL1∗ξ-η-L1∗ξ+ητ-1ηdη+1π∫0πL2∗ξ-η-L2∗ξ+ητ-2ηdη,where the first part in the first integral when η=ξ is the main quantity in the Cauchy formulation.The other parameters are(16)K∗ξ=∑n=1∞sinnξΔnμ,h+,h-K∗ξ=×μshnh+e-nh-+shnh-e-nh+,L∗ξ=∑n=1∞cos⁡nξΔnμ,h+,h-shnh+-h-,L1∗ξ=∑n=1∞cos⁡nξΔnμ,h+,h-shnh-0<ξ<π,L2∗ξ=∑n=1∞cos⁡nξΔnμ,h+,h-shnh+and Δnμ,h+,h-=μsh(nh+)ch(nh-)+ch(nh+)sh(nh-) and the corresponding dimensionless parameters are(17)ξ=πxl;η=πsl;h+=πh1l;h-=πh2l;μ=G1G2,τ-jξ=G∗τjlξπ;τ~±ξ=G∗τ±0lξπ,τ-ξ=G∗τlξπ;G∗=G1+G22G1G2,αk=πakl;βk=πbkl;k=1,N-,L0=⋃k=1Nαk,βk;L0′=0,π∖L0;0<ξ,η<π.In this manner the system of lengths (L) converts to system (L0), and, through function (13), the dislocation density is imposed at the boundaries of the crack edges:(18)ϕ′x=φ′xx∈L;0x∈L′;=2π∑n=1∞φncos⁡nξ,φn=πnlϕn=lπ∫L0φ0ξcos⁡nξdξ,φ0ξ=φ′lξπ.Then, by applying key equation (15) to the system (L0) and introducing the new variables(19)t=cos⁡ξ,u=cos⁡ηregarding the unknown dislocation density φ0(ξ) from (18), the singular integral equation governing the problem is obtained as follows:(20)1π∫Λ0ω0uduu-t+1π∫Λ0u+tK0t,u1-u2-2Kt,uiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii×ω0udu=ftt∈Λ0,ft=-12h~0+t+μ-1μ+1h~0-tft=+4μπ1+μ∫Λ0Lt,uh~0-uduft=+2π∫-11L1t,uω-1uduft=+2μπ∫-11L2t,uω-2udu,h~0±t=τ~+arccost±τ~-arccost,ω-jt=τ-jarccostj=1,2,…,ω0t=φ0arccost,K0t,u=1-t2+1-u2-1,Λ0=⋃k=1Nδk,γk,δk=cos⁡βN+1-k=cos⁡πbN+1-kl,γk=cos⁡αN+1-k=cos⁡πaN+1-kl.Based on (16), the following influencing factors are derived:(21)Kt,u=1-t21-u2∑n=1∞TnuUn-1tΔnμ,h+,h-Kt,u=×μshnh+e-nh+shnh-e-nh,Lt,u=1-t2∑n=1∞Un-1uUn-1tΔnμ,h+,h-shnh+-h-,L1t,u=1-t2∑n=1∞Un-1uUn-1tΔnμ,h+,h-shnh-,L2t,u=1-t2∑n=1∞Un-1uUn-1tΔnμ,h+,h-shnh+,where Tn(u) and Un-1(t) are Chebyshev polynomials of the first kind and the second kind, respectively.The singular integral equations (20) and (21) in a condition that explains the continuity of displacements at the crack tips are expressed as (22)∫akbkφ′sds=0k=1,N¯which, based on the previously discussed dimensionless variables, converts to (23)∫δkγkω0udu1-u2=0k=1,N¯.Hence after solving the singular integral equations (20) and (21) by considering condition (23) and verifying key equation (15) in the region outside of the crack system L0′, the antiplane shear stress τ-ξ formula in the dimensionless form is obtained.


## 3. Derivation of the Linear System of Equations

To solve the singular integral equations (20) and (21) under condition (23), the numerical solution for singular integral equations [7, 27, 28], which is based on the Gauss quadratic solutions for ordinary and singular integrals, is used. First, each section of δk,γkk=1,N¯ from system Λ0, by means of the technique of transitioning to new variables r,ρ-1≤r,ρ≤1(24)t=γk-δk2r+γk+δk2;u=γk-δk2ρ+γk+δk2k=1,N¯can be converted to the section [-1,1]. The singular integral equations (20) and (21) can therefore be written as follows:(25)1π∫-11ωk0ρdρρ-r+1π∑n=1(n≠k)N∫-11Lknr,ρωn0ρdρ+12π∑n=1N∫-11Kknr,ρωn0ρdρ=fkr,fkr=fγk-δk2r+γk+δk2,Lknr,ρ=ρ-γk-δkγn-δnr+γn+δnγn-δn-γk+δkγn-δn-1,Kknr,ρ=γn-δn×γn-δnρ+γk-δkr+γn+δn+γk+δk4-γn-δnρ+γn+δn2000000×K0γk-δk2r+γk+δk2,γn-δn2ρ+γn+δn2000000-2Kγk-δk2r+γk+δk2,γn-δn2ρ+γn+δn2γn-δnρ+γk-δkr+γn+δn+γk+δk4-γn-δnρ+γn+δn2,ωk(0)(r)=ω0γk-δk2r+γk+δk2k=1,N¯,-1<r<1,-1<r,ρ<1.The functionf(t) and kernel functions K0(t,u) and K(t,u) are expressed by relations (20) and (21). In this case, condition equation (23) may be expressed in the following form:(26)∫-11ωk0ρdρ4-γk-δkρ+γk+δk2=0k=1,N¯.Using the above approach, the singular integral equation (25) through condition equation (26) is reduced to a system of linear algebraic equations:(27)1M∑p=1Mψkρpρp-rm+∑n=1(n≠k)N∫-11Lknrm,ρpψnρp0000i00+12∑n=1(n≠k)NKknrm,ρpψnρp=fkrm0000000000000000000m=1,M-1¯,k=1,N¯,1M∑p=1Mψkρp4-γk-δkρp+γk-δk2=0.In (25), (26), it is supposed that(28)ωk0ρ=ψkρ1-ρ2k=1,N¯,where M is an arbitrary natural number. Consider (29)rm=cos⁡πmMm=1,M-1,ρp=cos⁡2p-12Mπp=1,M¯are the roots of the Chebyshev polynomials of the second kind Um-1(r) and of the first kind TM(ρ).The opening displacement of each crackLk=[ak,bk] is given by the equations:(30)Φx=∫akxφ′sds,Φx=-∫xbkφ′sdsak≤x≤bk,k=1,N¯which can be expressed by means of the dimensionless variables as (31)ΦN+1-k0r=-γN+1-k-δN+1-k2π×∫-11sgnr-ρωN+1-k0ρdρ4-γN+1-k-δN+1-kρ+γN+1-k+δN+1-k2iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii-1≤r≤1,k=1,N¯,where (32)ΦN+1-k0r=Φlξπl-1=l-1Φlπarccost.It must be noted that the system of equations (27) can be solved by the Gaussian technique, and (31) also can be solved with high accuracy by means of the Gauss quadratic formula using (28).At the end, the stress intensity factors, SIFs, for the antiplane mode of fracture at the crack tips[ak,bk] are obtained according to [24] k=1,N¯: (33)KIIIak=G1G2G1+G2limx→ak+02πx-akφ′(x),KIIIbk=G1G2G1+G2limx→bk+02πbk-xφ′(x)which can be converted to the dimensionless form and be expressed through ψk(ρ) using formula (28).Because of the large number of variable parameters involved in the problem, only three different cases are considered.CaseA: a plate with one symmetric crack: the crack length is assumed arbitrary, and antiplane shear loading is two point loads. The stress and dislocation states are studied for different crack sizes and shear modulus ratios.CaseB: a plate with two cracks: the cracks are located symmetrically and have arbitrary length and distance. The stress and dislocation states are studied for different crack sizes, distances between adjacent crack tips, and shear modulus ratios.CaseC: it is a plate with thin-walled inclusions at the crack tips, in which the sensitivity of SIFs to the size of the inclusion domain with respect to crack length and the effect of the shear modulus ratio of the inclusion material to the plate on the stress state are studied.


## 4. Case A: Plate with One Crack

A rectangular plate in the Cartesian coordinate systemOxyz is considered, which is composed of an upper rectangle D1 with modulus of rigidity G1, length l, and height h and a lower rectangle D2 with modulus of rigidity G2, length the same as l, and height h. On the interface surface line of the two rectangular plates in the interval 0≤x≤l, there is a central crack with length equal to 2a (Figure 2): (34)L=a1,b1,a1=l2-a,b1=l2+a0<a<l2.Figure 2
Plate with one crack.The boundaries of the crack indeed have not any loadings. Furthermore, the vertical edgesx=0 and x=l of rectangles Dj (j=1,2) are restrained, and the upper and lower horizontal edges y=±h at the midpoints (l/2,±h) are stressed by antiplane shear forces equal to P(35)τyz1y=h-0=τyz2y=-h+0=Pδx-l20<x<lin which τyzjj=1,2 are antiplane shear stresses on the top and bottom boundaries of plates Dj and δ(x) is the Dirac delta function.It is necessary to determine the dislocation field around the crack boundaries, the stress intensity factors, SIFs, and the shearing stresses on the interface surfaceL′=[0,l]∖L outside of the crack region.The singular integral equation (SIE) governing the current problem can be found in the previous section, particularly by observing the dislocation field around the crack boundaries as follows:(36)ϕ′x=φ′xx∈a1,b1=L;0x∈L′;ϕx=12w1x,0-w2x,00≤x≤lin which wj(x,0) are the displacement components of region y=±0,0≤x≤l of rectangles Dj (j=1,2) in the direction ofOz, so the singular integral equation of this boundary value problem from [29] is written as follows:(37)1π∫Λ0ω0uduu-t+1πu+tK0t,u1-u2-2Kt,u×ω0udu=ftt∈Λ0,ω0t=φ′1πarccost,K0t,u=1-t2+1-u2-1,Kt,u=1-t21-u2∑n=1∞TnuUn-1tchnh0e-nh,Λ0=δ1,γ1;δ1=-sinα;γ1=sinα,α=πal;h0=πhl,ft=2Q01-t2∑n=1∞-1n-1U2n-2tch2n-1h0,Q0=G1+G2P2G1G2l,where Tn(t) and Un-1(t) are Chebyshev polynomials of the first kind and the second kind, respectively.To solve the singular integral equation (37), the Gaussian quadratic solution for ordinary and singular Cauchy integrals can be used. By substitution,(38)ω10ρ=ψ1ρ1-ρ2-1<ρ<1.Using the approach shown here and also in [7, 27, 28], solving the singular integral equation (37) leads to a system of linear algebraic equations as follows:(39)∑p=1M1M1ρp-rm+12K11rm,ρpψ1ρp=f1rmm=1,M-1¯,∑p=1M1Mψ1ρp1-ρp2sin2α=0,K11rm,ρp=2sin2α(rm+ρp)1-ρp2sin2α11-rm2sin2α+1-ρp2sin2α000000000000000×11-rm2sin2α+1-ρp2sin2αK11rm,ρp=-4sinα1-rm2sin2α1-ρp2sin2αK11rm,ρp=×∑n=1∞e-nh0chnh0TnρpsinαUn-1rmsinα,f1rm=2Q01-rm2sin2αf1rm=×∑n=1∞-1n-1U2n-2rmsinαch2n-1h0,rm=cos⁡πmMm=1,M-1¯,ρp=cos⁡2p-12Mπp=1,M¯,where M is an arbitrary natural number and rm and ρp are the roots of Chebyshev polynomials of the second kind UM-1(r) and the first kind TM(ρ), respectively. We can substitute the below parameters to express the system of equations (39) in a simpler form: (40)Xp=ψ1ρpp=1,M¯;am=f1rmm=1,M-1¯;0m=M,Kmp=1M1ρp-rm+12K11rm,ρp0000p=1,M¯;m=1,M-1¯1M11-ρp2sin2α0000p=1,M;m=1,M-1.Hence, the system of equations (39) is reduced to(41)∑p=1MKmpXp=amm=1,M¯.By assumption of the below expression for the right hand side of (41)(42)am(0)=1-rm2sin2α∑n=1∞-1n-1U2n-2(rmsinα)ch2n-1h0000000000000000000000m=1,M-1¯0000000000000000000000m=M.X
p
0
p
=
1
,
M
¯ is found. The solution of system (26) is thereby obtained as follows: (43)Xp=2Q0Xp0p=1,M¯.For the derivation of SIF, the general formula from [7, 30] may be used; that is,(44)KIII=limx→b1+02πx-b1τyzx,0=limx→b1+02πx-b1τx.The above formula can be converted to a dimensionless form as follows:(45)KIII=2G1G2G1+G2×limx→1+02larccos-rsinα-π2+ατ0t.


## 5. Case B: Plate with Two Cracks

In this section, a rectangular plate in the Cartesian coordinate systemOxyz with two cracks is considered. On the interface surface of the two segments in the interval 0≤x≤l there are two central cracks located symmetrically at L=∪k≃12ak,bk that have equal lengths (Figure 3):(46)a1=l2-a,b1=l2-b,a2=l2+b,b2=l2+a0000000<a<l2;b<a.The boundaries of the cracks have no traction; furthermore, the vertical edges of plates Dj (j=1,2) at x=0 and x=l are clamped, and the upper and lower horizontal edges y=±h are loaded by antiplane distributed shear loading T(x), so that(47)τyz1y=h-0=τyz2y=-h+0=Tx0<x<l,in which τyzjj=1,2 are the antiplane shear stresses on the top and bottom boundaries of segment Dj.Figure 3
Plate with two cracks.The displacement of the crack boundariesL, the stress intensity factors SIFs, and the shear stresses at the interface surface L′=[0,l]∖L outside of the cracks are determined. Moreover, it is necessary to investigate the influence of adjacent crack tips a2 and b1.Suppose that the upper and lower edges of the plate are stressed by antiplane shear forcesP so that T(x)=Pδ(x-l/2) in which δ(x) is the known Dirac delta function. By this assumption, the function f(t) is calculated as follows:(48)f(t)=Q01-t2∑n=1∞-1n-1ch2n-1h0U2n-2t000000000000000000000000000-1<t<1,Q0=PG1+G2G1G2l.To calculate the shear stress, making use of the variables shown before and variableρ and taking into consideration (38), the shear stress is concluded:(49)τ0t=-12π∑k=12γk-δkψk(0)(ρ)dρ1-ρ2γk-δk/2ρ+γk+δk/2-t00000×∫-11ψk(0)(ρ)dρ1-ρ2γk-δk/2ρ+γk+δk/2-t-12π∑k=12γk-δkγk-δkρ+γk+δk+2t4-γk-δkρ+γk+δk20000000×∫-11γk-δkρ+γk+δk+2t4-γk-δkρ+γk+δk2000000000000000×K0t,γk-δk2ρ+γk+δk2000000000000000γk-δkρ+γk+δk+2t4-γk-δkρ+γk+δk2-2Kt,γk-δk2ρ+γk+δk2×ψk0ρdρ1-ρ2+ftt∈-1,1Λ0Λ0=-sinα,-sinβ∪sinβ,sinα.From the symmetry of the problem due to axisx=l/2 and taking into consideration only the right hand crack with end points at x=a2 and x=b2, the stress intensity factors are defined as follows [3, 4, 7, 26]:(50)KIIIa2=limx→a2-02πa2-xτyz=limx→a2-02πa2-xτx,KIIIb2=limx→b2+02πx-b2τyz=limx→b2+02πx-b2τx,where τx are the shear stresses according to (31).


## 6. Case C: Plate with Inclusions at the Crack Tips

In this section, attention is specifically paid to the effect of inclusions at the crack tips on the stress intensity factors (SIFs).A prismatic elastic bodyΩ with a rectangular cross section in Cartesian coordinatesOxyz occupying an area Ω=0≤x≤l;-h≤y≤h;-∞<z<∞ and possessing a shear modulus G is considered. The elastic body is rigidly clamped at x=0 and x=l and loaded by shear forces equal to T(x) acting both in the positive and in negative directions of theOz-axis by the horizontal y=±h. Furthermore, on the symmetry plane y=0, it has a through-the-thickness crack in the shape of the strip ω={y=0;l/2-a<x<l/2+a; -∞<z<∞}, a<l/2. Shear forces of equal intensities T0(x) are acting in opposite directions along theOz-axis on the upper (+) and lower (−) areas of the crack edges(51)ω±=y=±0;l2-b<x<l2+b;-∞<z<∞b<aof the crack. Additionally, at the tips (52)ω0±=y=±0;x∈l2-a;l2-b=y=±0;x∈0∪l2+b;l2+a;-∞<z<∞the edges of the crack are joined by thin-walled inclusions with shear modulus G0 deforming by the Winkler model (1867) that act as linear elastic springs: p=kω(Figure 4) [31].Figure 4
Plate with inclusions at crack tips.The prismatic bodyΩ subjected to the above-mentioned shear forces is in a state of antiplane deformation in the direction of theOz-axis on the basic planeOxy. The main rectangle D=0≤x≤l;-h≤y≤h with the central crack ω0=y=0;l/2-a<x<l/2+a0<a<l/2 is a cross section of the body Ω at the planeOxy.It is necessary to determine the dislocation density on the crack edges, the SIFs, the shear contact stresses on the edges of the inclusion, and the shear stresses outside the crack along the surface of its location.This is formally similar to the Dugdale-Barenblatt model for a central crack containing yielding as confined and localized narrow plastic zones, which shows the effect of yielding on the crack length [32, 33], but, in this problem, which is a linear elastic fracture mechanics problem, we investigate the effect of inclusions at the end areas of crack on the mechanical behavior of crack tips, decreasing the dislocation density and antiplane SIFs.The componentuz=w(x,y) in the direction of theOx-axis is the only nonzero component of the displacement, and τxz and τyz are the only nonzero stress components. Therefore, the problem can be mathematically stated as a boundary value problem in the following way: (53)∂2w∂x2+∂2w∂y2=0x,y∈D∖ω0,wx,yx=0=wx,yx=l=0-h<y<h;τyzy=h-0=τyzy=-h+0=G∂w∂yy=±h∓0=Tx00000000000000000000000000000<x<l;τyzy=±0=G∂w∂yy=±0=T0x00x∈l2-b;l2+b;b<a;τyzy=±0=±G0wx,yy=±0=±kGwx,yy=±0000x∈l2-a;l2-b∪l2+b;l2+a;k=G0G.Let us again consider the following functions based on [18, 34]:(54)Φx=w+x,0-w-x,02=2l∑n=1∞Φnsinπnxl,Ψx=w+x,0+w-x,02=2l∑n=1∞Ψnsinπnxl,Ωx=τ+x+τ-x2=2l∑n=1∞Ωnsinπnxliiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii0<x<l,Xx=τ+x-τ-x2=2l∑n=1∞Xnsinπnxl,where Φn, Ψn, Ωn, and Xn are sine Fourier series coefficients as before. Inserting these functions into the displacements and stresses on the edges of the crack at the interval [0,l] of rectangle D2 then according to (4), the following equations are obtained:(55)Φx=w+x,0;Ψx≡0;Ωx=τ+x,Xx≡0;Φn=w¯+n,0;Ψn=0,Ωn=τ-+n;Xn=0n=1,2,….After some simple transformations and calculations, the following equations can be derived:(56)1π∫-11ω0ρdρρ-r+1π∫-11K0r,ρω0ρdρ-1π∫-11Kr,ρω0ρdρ=fr-grr∈-c;cfr-λ0ψ0rr∈-1,-c∪c,10000000000000000000000000000c=sinβsinαwith(57)ω0r=φ0arccosr·sinα,fr=f~arccosr·sinα,gr=T~0arccosr·sinα,ψ0r=φ~0arccosr·sinα,fr=2π1-sin2α·r2×∫-11∑n=1∞Un-1r·sinαUn-1uchnh0T~arccosuduK0r,ρ=r+ρsin2α1-sin2α·ρ2×11-sin2α·r2+1-sin2α·ρ2000000000000000000000-1<r,ρ<1.Transforming the above equation, the following equation is derived:(58)τ~0t=τ0arccostτ~0t=-sinαπ∫-11ω0ρdρsinα·ρ-t-sinαπτ~0t=×∫-11t+ρsinαω0ρdρ1-ρ2·sin2α1-t2+1-ρ2·sin2ατ~0t=+2sinαπ1-t2τ~0t=·∫-11∑n=1∞e-nh0chnh0Un-1tTnρ·sinα×ω0ρdρ1-ρ2sin2α+2π1-t2×∫-11∑n=1∞Un-1tUn-1uchnh0T~arccosudu00000000000000000t∈-1;-sinα∪sinα;1.As mentioned above, the determinative singular integral equation (SIE) can be reduced to a system of linear equations. For this purpose, applying the Heaviside function(59)Hr=1r>00r<0this equation can be expressed in the following way:(60)1π∫-111ρ-r+K0r,ρ-Kr,ρHc+r-Hc-rsignr1-ρ2·sin2α000000+λsinα2Hc+r-Hc-rsignr1-ρ2·sin2αsignr-ρ×ω0ρdρ=fr-Hr+c-Hr-cgr0000000000000000000000-1<r<1;λ=πλ0=klwith the notations explained above.Supposing(61)ω0ρ=X0ρ1-ρ2-1<ρ<1the determinative SIE (39) with conditions of (25) and (40) is reduced to the system of linear equations (62)∑p=1MKmpXp=amm=1,M¯,Kmp=1M1ρp-rm+K0rm,ρp-Krm,ρp+λsinαsignrm21-ρp2sin2αHc+rm-Hc-rm×signrm-ρpm=1,M-1¯;p=1,M¯11-ρp2·sin2αm=M;p=1,M¯,am=frm-Hrm+c-Hrm-cgrm000000000m=1,M-1¯0,0m=M,Xp=X0ρpp=1,M¯;ρp=cos⁡2p-12Mπp=1,M¯;rm=cos⁡πmMm=1,M-1¯.For this case, the crack edges are free of shearing forces, and concentrated shearing forces are acting on the horizontal sides of the rectangular plate; that is,(63)T0x≡0,Tx=Pδx-l2,where δ(x) is a certain Dirac delta function. Additionally, the following equation can be obtained:(64)T-n=∫0lPδx-l2sinπnxldxT-n=Psinπn2=0,npun=2q;-1q+1P,npun=2q-100000000000000000000000000000000q=1,2,….Taking into consideration the above-mentioned definition, the function f~ξ can be expressed in the following way:(65)f~ξ=2P0∑n=1∞-1n+1sin2n-1ξch2n-1h0,P0=PlG0<ξ<πand the function f(r) from (57) is obtained:(66)fr=f~arccosr·sinα=2P01-r2·sin2α·∑n=1∞-1n+1U2n-2r·sinαch2n-1h0000000000000000-1<r<1.It is obvious that the function ω0(r) with respect to the symmetry of line x=l/2 in this special case and, consequently, the function X0(r) according to (61) are odd functions. Therefore, the components of the second integrals in (56) and (58) containing polynomial r or t in arguments tend to zero, so that the above-mentioned equations and the kernel-matrix Kmp of the system of equations (62) are simplified.It must be emphasized that the expressions of functionsf~ξ and f(r) from (65), (66) on various intervals are used in the equations.The numerical analysis of the main characteristics of the stated problem can be carried out for the considered special case.


## 7. Numerical Results and Discussions

To solve the system of equations (40) that is summarized in system (41) with the parameters shown in (42), M=10,20,30,40,50, factor α (which represents the cracking length) equal to π/4, h0=0.1π, and Q0=0.01 can be used. By means of Chebyshev polynomials, the deformation parameter Xp is obtained from (43). For example, M=10 leads to solving (10∗10) matrix equation (26).The dimensionless SIFKIII(0) according to formula (50) using the vector Xp, M=10 and considering the assumptions α=π/3, π/4, π/8, π/16, and π/32 for different crack lengths is calculated. The results are shown in Table 1, indicating that longer cracks lead to larger SIFs.Table 1
Dimensionless SIFKIII(0) as a function of M and α.
α
=
π
/
32
α
=
π
/
16
α
=
π
/
8
α
=
π
/
4
α
=
π
/
3
M
=
10
1.12782
1.69511
2.44264
3.80191
5.00603
M
=
20
1.42096
2.01939
2.91422
4.52924
5.96371
M
=
30
1.67019
2.37358
3.42536
5.32364
7.00971
M
=
40
2.29343
3.25929
4.70355
7.31018
9.62542
M
=
50
2.30110
3.26362
4.70979
7.31988
9.63819The shear stressτ0(t) from (49) in dependence of t and μ taking the rigidity ratios of the two materials μ=G1/G2=0.1,0.3,0.5,0.8,1.0,2,5,10 is obtained and the graphs are shown in Figure 5. The dimensionless crack dislocations that depend on parameters α and r are calculated, and the curves are represented in Figure 6. This figure shows that the maximum dislocation of a crack is at midpoint of its length, and it has a zero value at the crack tips, the points where maximum shear occurs.Figure 5
τ
0
(
t
) curve for α=π/4, h0=0.1π, and Q0=0.01.Figure 6
Crack dislocation(ϕ1(0)(r)) for M=50, h0=0.1π, and Q0=0.01.To solve the system of linear equations (27) for the problem with two cracks we use iteration M=10,15,20,30,… in order that the results converge to the order 10-4. Calculation is based on function (27) fk(r), Lkn(r,ρ), Kkn(r,ρ), and δk, γk and (25), (37), and (39).Parametersα and β are supposed to be equal to α=π/3 and β=π/4;π/8;π/16;π/32;π/64 in order to determine the influence of the distance between the two adjacent cracks on their fracture characteristics. The shear stress τ0(t) is calculated according to (49), in which function f(t) is obtained from (37), supposing the quantity Q0=0.01. Making use of the above calculated stresses, the variation curves of τ0(t) can be drawn for μ=0.1, 0.3, 0.5, and 1.0, as shown in Figure 7.Figure 7
Variation of antiplane shear stressτ0(t) for α=π/3, β=π/4.The crack dislocation density as a function of the edges’ opening displacementsΨk0r may be found through formula (31). The Ψk0r curves are shown in Figure 8 again for varying μ and β. At the end, using (50), the stress intensity factors in the dimensionless form can easily be calculated, as shown in Figure 9 and Tables 2 and 3, for the above parameters. The variation of SIF KIII based on parameter β and μ = 0.3, 0.5, and 1.0 is shown in Figure 10 to recognize the state of the SIFs under the change of the lengths of cracks.Table 2
Variation ofKIII for α=π/3 based on β, μ.
μ
β
π
/
4
π
/
8
π
/
16
π
/
32
π
/
64
0.1
7.70
7.35
7.11
6.69
6.41
0.3
5.36
5.54
5.73
6.17
6.35
0.5
4.88
5.12
5.34
5.87
6.13
1
4.21
4.35
4.63
5.41
5.92Table 3
Variation ofKIII(0) for α=π/3 based on β, λ.
λ
β
0
π
/
64
π
/
32
π
/
16
π
/
8
π
/
4
0.0
3.92
3.96
4.01
4.12
4.19
4.26
0.1
2.87
3.23
3.69
3.80
3.98
4.13
0.5
2.79
2.95
3.40
3.64
3.82
3.94
1.0
2.54
2.72
3.08
3.37
3.69
3.82
5.0
2.42
2.60
2.81
3.10
3.46
3.60
20
2.17
2.34
2.63
2.88
3.18
3.45Figure 8
Dislocation density curves of cracks boundariesψk(0)rm for α=π/3, β=π/4.Figure 9
Variation ofKIII for α=π/3 based on β, μ.Figure 10
Variation ofKIII based on β for α=π/3 and μ=1,0.5,0.3.For the case ofμ=1.0, the cracks approaching each other (that means a smaller value of β) lead us to construct curves as shown in Figure 11 for τ0(t) and Figure 12 for Ψk0r.Figure 11
Variation of antiplane shear stressτ0(t) for α=π/3, μ=1.Figure 12
Dislocation density curves of cracks boundariesψk(0)rm for α=π/3, μ=1.The main characteristics of the crack problem with inclusions in the mentioned special case can be calculated through numerical analysis. For calculation it is considered as some engineering practical ratios for the shear modulus quantitiesk=G0/G that are k=0.0 (crack without repair), k=0.5,1.0 (repair with the same material), and k = 2.0, 5.0, and 10.0 (for very high rigid repair materials). By solving system of linear equations (62) which leads to solving a matrix (10 × 10) and obtaining vectors Xp(10×1), the shear stress τ0t can be calculated from (58) and the results are shown in Figure 13.Figure 13
Variation of antiplane shear stressτ~0t, τ-1rm for α=π/3, β=π/64.Dislocation functionsψ~0r=2φ0r were investigated using relation (31), as shown in Figure 14 in which the maximum displacement of crack boundaries occurred at midpoint and obviously with zero value at crack tips.Figure 14
Dislocation curve of crack boundariesψ~0rm for α=π/3, β=π/64.The stress intensity factors SIFsKIII(0) calculated through (50) are shown in Figure 17, which presents the decreasing trend of the KIII curve when the ratio of k=G0/G increases. It also shows that the crack tip repairing by this method can reduce the SIF by approximately 50 percent, which means that this approach is very effective to control the crack propagations in cracked plates and it is also a treatment for singularities at the crack tips, defects, and holes. Figures 15 and 16 show that the reduction of the magnitude KIII is not high when we use a very rigid material for tip repair. For example, for k=2.0 and k=10.0 (ratio = 5), the reduction of KIII is only 18 percent (1.98/2.42).Figure 15
Variation ofKIII0 for α=π/3 based on β, λ.Figure 16
Curve of decreasingKIII0.Figure 17
Curve of decreasingKIII0 for various inclusion materials on base metal steel with G=840000 Kg/cm2.The above results coincide with experimental observations of high concentrations and singularities in the stress fields within elastic materials [17, 18, 35, 36].


## 8. Conclusion

In the present paper, the singular integral equations governing the piecewise homogeneous elastic plate problem subject to uniform remote antiplane shear loading have been considered. To determine the antiplane shear stresses, the crack edge dislocation densities, and the mode III stress intensity factors, a new mathematical-numerical calculation is developed. To perform elastic analyses and investigations on the state of stresses at the crack tips, the dimensions of the cracks, the distances between adjacent tips, and the influence of shear modulus ratios between the two materials have been varied and studied. The governing singular integral equation of a problem of stress-strain state of a piecewise homogeneous elastic prismatic body of a rectangular cross section, when there is a system of an arbitrary finite number of collinear cracks on the interface line of dissimilar materials, is obtained. For problem solving, the method of mechanical quadrature is used. It allowed conducting a detailed analysis of the main mechanical characteristics of the stated problem and revealed their dependence on the mechanical and geometrical parameters of the problem. For the case of cracks coming together it has been shown that the stress intensity factorsKIII grow based on two parameters, the total distance between their far tips and the closeness of their near tips. The technique presented in this paper can be used to solve a class of problems associated with cracking in bimaterial interfaces.For a crack with inclusions at the tips, it is observed that the insertion of a material at the crack tips avoids singularities, reduces the antiplane SIFKIII, and strengthens and controls the crack propagation in the region at the tips. This is an effective method and does not require using a material with a very high shear rigidity value G0. Meanwhile, the crack opening displacement (COD) and the shear stresses at the crack tips decrease. This suggests a method to repair cracked plates and members.Practice shows that the method of mechanical quadrature is a very effective method and results in a very good convergence. The data of Table2 confirms this. The accuracy of the method in predicting the intensity factor may be verified by a comparison with experimental measurements, carried out by a photoelasticity method, and by commercial finite element software. The drawn conclusions provide meaningful reference for the analysis of SIFs in mode III. Numerical results show the influence of ratios of shear moduli on the stress state.The results of this paper and the previously published results show that the used approach based on the Gauss-Chebyshev quadrature method can be considered as a generalized procedure to solve the collinear crack problems in mode I, II, or III loadings.



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