# A Crack between Orthotropic Materials with a Shear Yield Zone at the Crack Tip

Authors: V. Loboda; I. Gergel; T. Khodanen; O. Mykhail
Journal: Mathematical Problems in Engineering
 (2019)

---

## Abstract

A plane problem for a crack between two anisotropic semi-infinite spaces under remote tensile-shear loading is considered. In the framework of the assumption that the crack faces are free of stresses an exact analytical solution of the problem is given on basis of the complex potentials approach. This solution possesses oscillating square root singularities in stresses and in the derivatives of the displacement jumps at the crack tips. To remove these singularities a new model founded on the introduction of the shear yield zones at the crack tips is suggested. This model is appropriate for the cases where interface adhesive layer is softer than the surrounding matrixes. Under this assumption the problem is reduced to the nonhomogeneous combined Dirichlet-Riemann boundary value problem with the conditions at infinity. An exact analytical solution of this problem is presented for the case of a single yield zone. The length of this zone is found from the finiteness of the shear stress at the end point of the zone. Due to such simulation the shear stress becomes finite at any point and the normal stress possesses only square root singularity at the crack tip. Therefore, the conventional stress intensity factor of the normal stress at the crack tip is used. The numerical illustration of the obtained solution is given.


---

## Body

## 1. Introduction

Interface cracks in many cases are the main reason of composite constructions failure. Therefore, much attention has been paid to the investigation of interface fracture in the framework of the open crack model. Such model is connected with oscillating singularity, which was the most clearly investigated in the paper [1] for a crack between two isotropic materials. The contact zone model, assuming a partial closing of the crack faces and eliminating a physically unreal oscillation, has been suggested in [2] and analytically developed in the papers [3–5].The problem of an interface crack between two different anisotropic materials is much more complicated and therefore less studied compared to an isotropic case. An analytical analysis of this problem in the framework of an open interface crack model was carried out in [6] and continued in [7–11]. The phenomenon of oscillation was confirmed in these papers for an anisotropic case. Accounting of the crack faces contact for anisotropic bimaterials has been done in papers [12, 13] in an analytically numerical way and in [14, 15] analytically.In many cases the interface is much softer than the adhered matrixes; therefore, thin yield zones develop at the crack tips along the interface. Accounting of such zones was performed in [16] with use of Coulomb law of dry friction and in [17, 18] by means of applying Dugdale [19] model for the simulation of the mechanical fields in these zones. However to the author’s knowledge a thin yield zone based upon the Treska’s-Saint-Venant’s theory has been never used for an interface crack in neither orthotropic nor isotropic bimaterial.In this paper an interface crack with a yield zone based upon maximum shear stress criterion is studied. An exact analytical solution for the associated mathematical model is derived. This solution is free from an oscillation and, therefore, the conventional form of stress fracture criterion can be used.


## 2. Formulation of the Basic Relations

The constitutive relations of elasticity for a linear anisotropic material in the absence of body forces in a fixed rectangular coordinate systemxi can be presented in the form [20](1)σij=Cijkluk,l,(2)σij,j=0,where σij, uk are stress and displacements components; Cijkl are the elastic moduli, for which Cijkl=Cklij=Cjikl=Cijlk hold true and Einstein’s summation convention from 1 to 3 for Latin suffixes has been used.Substituting (1) in (2) one gets the following Lame equations:(3)Cijkluk,lj=0.Assuming that all fields are independent of the coordinatex3, the solution of (3) according to the method suggested in [21] and developed in [6] can be presented in the form:(4)uk=akfx1+px2,where f is an arbitrary analytic function of the complex variable z=x1+px2; p anda1,a2,a3 are eigenvalue and eigenvector components of the following system:(5)Q+pR+RT+p2Ta=0.The elements of the 3×3 matricesQ,R andT are defined as(6)Q=Qik,R=Rik,T=TikwithQik=Ci1k1,Rik=Ci1k2,Tik=Ci2k2i,k=1,2,3.Here and afterwards the superscript T stands for the transposed matrix. A nontrivial solution of (5) exists if p is a root of the equation(7)det⁡Q+pR+RT+p2T=0.Since (7) has no real roots [9] we denote the roots of this equation with positive imaginary parts as pα and the associated eigenvectors of (5) as aα (subscript α here and afterwards takes the numerals 1, 2, 3). The most general real solution of (3) can be presented as [9](8)U=Afz+A-f-z-,where U=u1,u2,u3T, A=[a1,a2,a3] is a matrix composed of eigenvectors, fz=f1z1,f2z2,f3(z3)T is an arbitrary vector function, zα=x1+pαx2, and the overbar stands for the complex conjugate. Introducing the vector(9)t=σ12,σ22,σ32Tand using (2), this vector can be presented in the form(10)t=Bf′z+B-f-′z-,where the components of 3×3 matrixB are defined as(11)Biα=Ci2k1+pαCi2k2akαnotsummedoverindexα.Further a bimaterial composed of two different anisotropic semi-infinite spacesx2>0 and x2<0 with mechanical properties defined by the matrices Cijkl(1) and Cijkl(2), respectively, is considered. We assume that the vectort is continuous across the whole bimaterial interface and the part L={-∞,c1∪b1,c2∪...bn,∞} of the interface -∞<x1<∞, x2=0 are bounded. That is, the boundary conditions at the interface x2=0 are as follows:(12)t1x1,0=t2x1,0forx1∈-∞,∞,(13)U1x1,0=U2x1,0forx1∈L.Satisfying (12), using for each material the presentations (8), (10), and applying the method written in detail in the paper [14] the following expressions at the interface are obtained:(14)U′x1=W+x1-W-x1,(15)t1x1,0=GW+x1-G-W-x1,where U′(x1)=U′(1)(x1,0)-U′(2)(x1,0) is the jump of the function U′(x1) across the material interface, G=B(1)D-1, D=A(1)-L-B(1), L=A(2)(B(2))-1.It should be noted that the vector functionW(z)= W1(z),W2(z),W3(z)T is analytic in the whole plane with a cut along (-∞,∞)∖L. The relations (14), (15) play an important role for the formulation of various problems of linear relationship for anisotropic bimaterials with cuts at the material interfaces under the action of mechanical loadings.The main attention in the following will be paid to the consideration of orthotropic materials with the axis of the material symmetry parallel tox3 as the most important class of anisotropic materials. In this case the matrixG has the following structure:(16)G=G11G12G13G21G22G23G31G32G33=ig11g120g21ig22000ig33,where all gij are real. It is clear that the plane and out of plane problems can be decoupled. Because of the simplicity of the out of plane problem solution our attention will be focused on the plane problem for the displacement components (u1,u2). In this case similarly to the contracted notations of the anisotropic elasticity [20] the following relations for the elastic coefficients related to the (x1,x2)-plane can be introduced: C1111=C11,C1122=C12,C2222=C22,C1212=C44 and the combination of each of the first two equations of (14), (15) can be presented in the form (see for details [14])(17)σ221x1,0+im1σ121x1,0=q1Ω1+x1+γ1Ω1-x1,j=1,2,(18)u1′x1+iS1u2′x1=Ω1+x1-Ω1-x1,where(19)Ω1z=W1z+iS1W2z,(20)γ1=-g21+m1g11q1,S1=g22+m1g12g21-m1g11,q1=g21-m1g11,m1=--g21g22g11g12.It is worthy to note that the functionΩ1(z) is analytic in the whole plane with a cut along (-∞,∞)∖L.


## 3. A Crack at the Interface of Two Materials

Let us assume further that a crack takes place at the section[c,a] of the material interface. The half-spaces are subjected to uniformly distributed normal stress σ22∞ and shear stress σ12∞ at infinity, which do not depend on the coordinate x3. The crack faces are free of loading. This kind of external fields initiates plane deformation state; therefore, only the cross-section orthogonal to x3 (Figure 1) can be considered and the relations (17), (18) with n=1 are valid. We do not pay attention for a while to the section (a,b) of the interface, which will need consideration later.Figure 1
A crack between two anisotropic materials with a yield zone.The boundary conditions for the formulated problem are as follows:(21)σ211=σ212=0,σ221=σ222=0forc<x1<a,(22)σ21=0,σ22=0,u1′=0,u2′=0forx1∉c,a.Satisfying conditions (21) and (22) with the use of (17), (18) provides the continuity of the function Ω1z over the segments x1∉c,a of the material interface and also leads to the following equation:(23)Ω1+x1+γ1Ω1-x1=0forc<x1<a.The conditions at infinity follow from (17) and can be written in the form(24)Ω1zz→∞=σ~22+im1σ~21,where σ~22=σ22∞/r1, σ~21=σ21∞/r1, r1=1+γ1q1.The solution of (23) under the condition at infinity (24) has been obtained with use of [22] in the form(25)Ω1z=σ~22+im1σ~21z-a+c/2-iεl0z-cz-az-cz-aiεwhere ε=1/2πln⁡γ1,. l0=a-c.The stresses at the interface are found from (17), (25) as follows:(26)σ221x1,0+im1σ211x1,0=σ22∞+im1σ21∞x1-a+c/2-iεl0x1-cx1-ax1-cx1-aiεforx1>a.The derivative of the displacement jumps is obtained by use of the formula (18) in the form(27)u1′x1,0+iS1u2′x1,0=-σ22∞i-m1σ21∞q1γ1x1-a+c/2-iεl0x1-ca-x1x1-ca-x1iεforc<x1<a.After integrating the last relation, we obtain(28)u1x1,0+iS1u2x1,0=x1-ca-x1σ22∞i-m1σ21∞q1γ1x1-ca-x1iεforc<x1<a.


## 4. Formulation of the Problem and Development of the Interface Crack Model Free from Oscillation

It is clearly seen from (26), (27) that the mechanical stresses and the derivative of displacement jumps are singular at the crack tips. Moreover this singularity is oscillating. To remove this oscillation different models were suggested, for example, [2, 17, 18]. In the present paper the model based upon the introduction of the shear yield zones is suggested. As it will be shown later, either both of these zones are very short or one zone is substantially shorter than the other one; therefore their mutual influence can be neglected. Taking into account this circumstance we will consider only the longer zone (a,b) for simplicity, assuming that it occurs at the right crack tip (Figure 1). If the longer zone occurs at another crack tip then this zone can be considered by simple transformation of half-spaces.Thus the boundary conditions for the considered model can be written as follows:(29)σ211=σ212=0,σ221=σ222=0forc<x1<a,(30)σ211=σ212=τS,σ22=0,u2′=0fora<x1<b,(31)σ21=0,σ22=0,u1′=0,u2′=0forx1∉c,b.where τS is the shear yield limit of the adhesive layer.Satisfying the interface conditions (29) and using (17) one gets (23). Additionally the first and third conditions (30) lead to the equations(32)ImΩ1+x1+γ1Ω1-x1=-m1τsq1,ImΩ1+x1-Ω1-x1=0fora<x1<b.Satisfaction of the boundary conditions (31) provides the analyticity of the function Ω1z outside of the interval c,b and the last relations lead to the equation(33)ImΩ1±x1=τfora<x1<b,where τ=-m1τS/r1.Equations (23) and (33) present the nonhomogeneous combined Dirichlet-Riemann boundary value problem. The conditions at infinity (24) are valid for this problem also. By using the results of the paper by Nahmein and Nuller [23] the general solution of the homogeneous problem corresponding to (23), (33) can be presented in the form(34)Ω1hz=PzE1z+QzE2z,where(35)Pz=C1z+C2,Qz=D1z+D2.The functions(36)E1z=ieiφzz-cz-b,E2z=eiφzz-cz-apresent the canonical solutions of the homogeneous problem (23), (33), where φz=2εln⁡b-az-c/lz-a+a-cz-b, l=b-c, and C1,C2,D1,D2 are arbitrary real coefficients.A particular solution of the nonhomogeneous combined Dirichlet-Riemann boundary value problem for certain right sides of (23) and (33) was analyzed in [24]. In the present case of the problem (23), (33) its particular solution can be found in the form(37)Ω1pz=ΦzE1z,where Φz is assumed to be analytic in the whole complex plane with a cut [a,b] along the x1-axis. It should be mentioned that Ω1pz satisfies (33). Substituting (37) into (33) and taking into account that ImE1±x1=0 on (a,b) one has the following equation:(38)ImΦ±x1=ψ±x1fora<x1<b,where ψ(x1)=τ/E1x1.A solution of the Dirichlet problem (38) has the following form [25, formula (46.25)]:(39)Φz=Yz2π∫abψ+t+ψ-tY+tt-zdt+12π∫abψ+t-ψ-tt-zdt,where Y(z)=(z-a)(z-b) and 0≤arg⁡(z-a)≤2π, 0≤arg⁡(z-b)≤2π.Using(40)ψ+t+ψ-t=-2τt-cb-tsinh⁡φ0t,ψ+t-ψ-t=2τt-cb-tcosh⁡φ0t,φ0x1=2εtan-1⁡a-cb-x1b-cx1-a,Y+t=-it-ab-tata,b,the formula (39) takes the form(41)Φz=τπ-iYzL1z+L2z,where L1(z)=∫abt-c/t-asinh⁡φ0(t)/t-zdt, L2z=∫ab(t-c)(b-t)cosh⁡φ0(t)/t-zdt.The general solution of the problem (23), (33) can be found by summing the solutions (34) and (37). Arbitrary constants C1,C2,D1,D2 can be found from the condition at infinity (24) together with condition of the displacement uniqueness which due to (18) can be written in the form(42)∫cbΩ1+x1-Ω1-x1dx1=0.Considering that for the validity of last equation the coefficient beforez-1 in the expansion of Ω1z at infinity should be equal to zero [24] and also (43)E1zz→∞=iz-2eiβz+iβ1+c+b2+oz-3,E2zz→∞=z-2eiβz+iβ1+c+a2+oz-3,Φzz→∞=-iR+oz-1,R=τπ∫abt-ct-asinh⁡φ0tdt,one gets the following expressions for the unknown coefficients(44)C1=-σ~23sin⁡β-E~1cos⁡β,D1=σ~23cos⁡β-E~1sin⁡β,C2=-c+b2C1-β1D1,D2=β1C1-c+a2D1-R,where β=εln⁡1-1-λ/1+1-λ, β1=ε(a-c)(b-c), (45)λ=b-ab-c.Using the obtained solutions the general solution of the nonhomogeneous combined Dirichlet-Riemann boundary value problem (23), (33) can be presented in the form(46)Ω1z=Pz+ΦzE1z+QzE2z.From this solution all required quantities at the material interface can be found.


## 5. Stress Intensity Factor and Yield Zone Length

According to (17) the stress field on the right side from the yield zone can be presented in the form:(47)σ221x1,0+im1σ211x1,0=r1Px1+Φx1E1x1+Qx1E2x1.For an arbitrary position of the pointb, which defines the yield zone length, the right hand side of (47) is singular for x1→b+0 and besides Pb+Φ(b) is real and E1x1x1→b+0=i/b-cx1-b is pure imaginary. Therefore, for any b the normal stress σ22(1)(x1,0) is finite for x1→b+0 whilst σ12(1)(x1,0) is singular. For removing of this singularity the equation(48)Pb+Φb=0should be valid. After some transformation this equation can be written as follows:(49)m1δcos⁡β-sin⁡β-2ε1-λcos⁡β+m1δsin⁡β-2m1πb-cτSσ22∞∫abt-cb-tcosh⁡φ0tdt=0.where δ=σ21∞/σ22∞.This equation should be solved with respect toλ and after that the position of the point b can be found from (45). Usually (49) can be solved numerically and that allows finding the largest root of this equation from the interval (0,1) which we denote λ0.The normal stress at the interval(a,b) according to (17), (46) can be found in the form(50)σ221x1,0=q1Ω1+x1+γ1Ω1-x1-im1τS.Substituting the formula (46), taking into account that [24](51)E1±x1=±e±ϕ0x1x1-cb-x1,E2±x1=e±ϕ0x1x1-cx1-aforx1∈a,band using Plemeli formulas [22] one gets the following expression:(52)q1-1σ221x1,0=Px1eφ0x1-γ1e-φ0x1x1-cb-x1+Qx1eφ0x1+γ1e-φ0x1x1-cx1-a+τπ-x1-ax1-ceφ0x1+γ1e-φ0x1L1x1+eφ0x1-γ1e-φ0x1x1-cb-x1L2x1,where the integrals L1(x1) and L2(x1) should be considered here in sense of principal value on Cauchy [25].Consider further the stress intensity factor (SIF) of the normal stress at the pointa(53)K1=limx1→a+0⁡2πx1-aσ221x1,0.Considering thatL1(x1) has a square root singularity for x1→a+0 and L2(x1) has the logarithmic singularity at this point we get the following formula:(54)K1=2q12πγ1Qaa-c.This formula after some transformations takes the form(55)K1=2q12πγ1a-cl21-λ2εC1+1-λD1-R.The derivative of the crack faces displacement jump at the interval(c,a) (crack opening) can be found due to (18) in the form(56)S1u2′x1,0=ImΩ1+x1-Ω1-x1.Substituting the expression (46) one gets(57)u2′x1,0=γ1+1S1γ1Px1+Φx1b-x1-iQx1a-x1exp⁡iφ∗x1x1-cforc<x1<a,where φ∗x1=2εln⁡b-ax1-c/la-x1+a-cb-x1.The crack opening can be found by using the following formula:(58)u2x1,0=∫cx1u2′t,0dt.


## 6. Numerical Results and Discussion

The calculations were performed for a bimaterial composed of boron-epoxy orthotropic material (upper one) withC11(1)=26.9C66(1), C22(1)=3.6C66(1), C12(1)=3.15C66(1), C66(1)=4.78×1010Pa and isotropic material (lower one) having the following characteristics: μ(2)=0.478×1010Pa, ν(2)=0.345. τS=107Pa, c=-10mm, b=10mm were chosen and different values of the external mechanical loadings were considered.In Table1 the yield zone lengths and the SIF K1 are presented for σ12∞=0 and different values of σ22∞. It can be seen from these results that growing of σ22∞ leads to increasing of λ0 and K1. Such conclusion completely agrees with the finding which follows from simple physical inferences.Table 1
Yield zone length and SIFK1 for σ12∞=0 and different values of σ22∞.
10 - 6 σ 22 ∞  [ P a ] 1 2 3 4 5  100 λ 0 0.1318 0.3700 0.6581 0.9750 1.313  10 - 5 K 1  [ P a  /  m 3 / 2 ] 1.644 3.396 5.161 6.931 8.704The dependence of the same variables as in Table1 with respect to the intensity of the shear stress at infinity is given in Table 2. It can be seen that the yield zone length essentially grows with growing of the mentioned stress; however the SIF remain slightly dependent on this parameter. It is also explainable that the applied shear stress does not greatly influence the SIF of the normal stress.Table 2
Yield zone length and SIFK1 for σ22∞=3×106Pa and different values of σ12∞.
10 - 6 σ 12 ∞  [ P a ] 0 1 2 3 4 5 6  100 λ 0 0.6581 1.902 4.039 7.100 11.04 15.77 21.19  10 - 5 K 1  P a  /  m 3 / 2 5.161 5.049 5.107 5.290 5.576 5.954 6.425The crack openingu2x1,0 for c=-10mm,  b=10mm,  σ22∞=3×106Pa is presented in Figure 2 for different values of σ12∞. Lines I, II, III, and IV correspond to σ12∞=0, 2×106Pa, 4×106Pa and 6×106Pa, respectively. It follows from these results that the crack opening is almost symmetrical for σ12∞=0 because the yield zone length is extremely small in this case and the problem is almost symmetrical. However, increasing of σ12∞ leads to some contortion of the curve and its deviation from the symmetrical state.Figure 2
Crack opening106u2x1,0m for σ22∞=3×106Pa and different values of σ12∞.The variation of the shear stressσ21(1)(x1,0) in the right hand side of the yield area for σ22∞=3×106Pa, the same crack geometry, mechanical loading and σ12∞ equals to 0 (line I), 2×106Pa (II), 4×106Pa (III) and 6×106Pa (IV) is presented in Figure 3. The values of b for the lines I, II, III, and IV are found with use of Table 1 and are equal to 10.13mm, 10.84mm, 12.48mm, and 15.38mm, respectively. As it can be seen the considered yield model eliminated the singularities of the shear stress and the oscillation of the normal stress at the right crack tip. Thereby such approach allows getting shear stress field at the point b free from singularities and transforming the oscillating singularity of the normal stress into conventional square root singularity at the point a. The last circumstance gives the possibility to introduce the SIF of the normal stress in a commonly used form.Figure 3
Variation of the shear stressσ211(x1,0)MPa in the right hand side of the yield area for σ22∞=3×106Pa and different values of σ12∞.


## 7. Conclusion

An interface crack in a bimaterial anisotropic space under tensile-shear loading at infinity is considered. Plane strain conditions are assumed and the crack faces are free of stresses. Due to complex potentials approach an exact analytical solution (25) of the formulated problem is given. The stresses and the derivatives of the displacement jumps obtained from this solution have the oscillating square root singularities at the crack tips. The new model based upon the introduction of the shear yield zones at the crack tips is suggested for removing of these singularities. This model is used under the assumption that the interface adhesive layer is softer than the surrounding matrixes. The problem is reduced to the nonhomogeneous combined Dirichlet-Riemann boundary value problem (23), (33) with the conditions at infinity (24). An exact analytical solution of this problem is presented for the case of a single yield zone. The assumption concerning consideration of a single yield zone is approved by the fact that another zone is extremely short and does not influence the longer zone. The length of yield zone is found from the finiteness of the shear stress at its end. This gives the simple transcendental equation (49) for the determination of the yield zone length.As a result of the performed modeling the shear stress becomes restricted in the whole region and the normal stress has only square root singularity at the crack tip. It means that the stress intensity factor of the normal stress can be introduced and calculated in the conventional way. The dependencies of the mentioned stress intensity factor, yield zone length, crack opening, and the shear stress on the applied loading are illustrated in the tables and graphical forms. It is particularly shown that the stress intensity factor of the normal stress essentially depends on the normal applied stress and moderately on the external shear stress. Finally, it is worthy to mention that the results of the paper are obtained in a simple analytical form which is convenient for the engineering applications.



---

*Source: 9723089-2019-06-25.xml*