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. 2024 Jun 22;29(11):2224–2235. doi: 10.1177/10812865241257534

A partially debonded rigid elliptical inclusion with a liquid slit inclusion occupying the debonded portion

Xu Wang 1, Peter Schiavone 2,
PMCID: PMC11530348  PMID: 39493868

Abstract

We derive a closed-form solution to the plane strain problem of a partially debonded rigid elliptical inclusion in which the debonded portion is filled with a liquid slit inclusion when the infinite isotropic elastic matrix is subjected to uniform remote in-plane stresses. The original boundary value problem is reduced to a Riemann–Hilbert problem with discontinuous coefficients, and its analytical solution is derived. By imposing the incompressibility condition of the liquid slit inclusion and balance of moment on a circular disk of infinite radius, we obtain a set of two coupled linear algebraic equations for the two unknowns characterizing the internal uniform hydrostatic tension within the liquid slit inclusion and the rigid body rotation of the rigid elliptical inclusion. As a result, these two unknowns can be uniquely determined revealing the elastic field in the matrix.

Keywords: Rigid elliptical inclusion, liquid slit inclusion, internal uniform hydrostatic tension, rigid body rotation, Riemann–Hilbert problem, incompressibility, balance of moment

1. Introduction

Analytical solutions to interface crack problems in laminate and fibrous composites have been well recorded in the literature with one of the main objectives being the study of the effect of interface fracture on the mechanical behavior of composites (see, e.g., [114]). In a recent study, Wang and Schiavone [15] analyzed the plane strain problem of a partially debonded circular elastic inhomogeneity and observed that when the debonded portion is filled with an incompressible liquid slit inclusion, interface crack growth can be completely suppressed as if the debonded portion of the circular interface doesn’t even exist when the elastic matrix is subjected to a hydrostatic far-field load. This fact suggests that liquid inclusions could be used quite effectively to repair interface cracks in composites. Although the micromechanics of liquid inclusions has been studied extensively in the last decade (see, e.g., [1626]), the mechanism by which liquid inclusions play a role in repairing cracks has not been thoroughly investigated.

In this paper, we continue our recent study and analyze the plane strain problem of a partially debonded rigid elliptical inclusion with the debonded portion being occupied by a liquid slit inclusion when the surrounding infinite isotropic elastic matrix is subjected to uniform remote in-plane stresses. Since the initial area of the liquid slit inclusion is zero, the inclusion always behaves as if it is incompressible. The liquid slit inclusion admits an internal uniform hydrostatic stress field and remains perfectly bonded to the surrounding media. The original boundary value problem is reduced to a standard Riemann–Hilbert problem with discontinuous coefficients. An analytical solution to this Riemann–Hilbert problem can then be derived quite conveniently. However, this analytical solution contains the two unknowns characterizing the internal uniform hydrostatic tension within the liquid slit inclusion and the rigid body rotation of the rigid elliptical inclusion. By imposing the incompressibility condition of the liquid slit inclusion and the balance of moment on a circular disk of infinite radius, the two unknowns can be uniquely determined by solving a set of two coupled linear algebraic equations. In particular, when the two tips of the debonded portion (or those of the liquid slit inclusion) are symmetric with respect to the major or minor axis of the elliptical interface, the two unknowns can be determined quite simply from two separate linear algebraic equations. Our detailed numerical results demonstrate that the shape of the elliptical interface, the positions of the two tips of the debonded portion, and the Poisson’s ratio of the matrix all exert a significant influence on the internal uniform hydrostatic tension characterizing the liquid inclusion and rigid body rotation characterizing the rigid inclusion. In particular, when the remote loading satisfies the harmonic shape condition for a (perfectly bonded) rigid elliptical inclusion [27,28], the rigid elliptical inclusion is still perfectly bonded to the surrounding matrix as if the debonded portion doesn’t even exist. This fact implies that the insertion of the liquid slit inclusion into the debonded portion will also suppress crack growth along the elliptical interface.

2. Complex variable formulation

We first establish a Cartesian coordinate system {xi}(i=1,2,3) . For plane strain deformation of an isotropic elastic material, the three in-plane stresses (σ11,σ22,σ12) , two in-plane displacements (u1,u2) , and two stress functions (φ1,φ2) are given in terms of two analytic functions ϕ(z) and ψ(z) of the complex variable z=x1+ix2 as [29]

σ11+σ22=2[ϕ(z)+ϕ(z)¯],σ22σ11+2iσ12=2[z¯ϕ(z)+ψ(z)], (1)

and

2μ(u1+iu2)=κϕ(z)zϕ(z)¯ψ(z)¯,φ1+iφ2=i[ϕ(z)+zϕ(z)¯+ψ(z)¯], (2)

where κ=34ν , and µ and ν(0ν1/2) are the shear modulus and Poisson’s ratio, respectively. In addition, the stresses are related to the two stress functions through [9]

σ11=φ1,2,σ12=φ1,1,σ21=φ2,2,σ22=φ2,1. (3)

3. Closed-form solution

As shown in Figure 1, a rigid elliptical inclusion is partially bonded to an infinite isotropic elastic matrix subjected to uniform remote in-plane stresses (σ11,σ22,σ12) . The rigid inclusion remains perfectly bonded to the matrix along the portion Lb of the elliptical interface L, and is debonded from the matrix along the remaining portion Lc of the elliptical interface L. Thus, L=LbLc . In addition, the debonded portion Lc is filled with a liquid slit inclusion, which admits an internal uniform hydrostatic stress field and which is still perfectly bonded to its surrounding media.

Figure 1.

Figure 1.

A rigid elliptical inclusion partially bonded to an infinite isotropic elastic matrix subjected to uniform remote in-plane stresses. The debonded portion Lc of the elliptical interface is occupied by a liquid slit inclusion.

According to the above statement and the complex variable formulation in Section 2, the boundary value problem takes the following form in the physical z-plane:

κϕ(z)zϕ(z)¯ψ(z)¯=2iμϖ21z,zLb;ϕ(z)+zϕ(z)¯+ψ(z)¯=σ0z,zLc;ϕ(z)Az+O(1),ψ(z)Bz+O(1),|z|, (4)

where ϖ21=12(u2,1u1,2) is the unknown rigid body rotation of the rigid elliptical inclusion to be determined, σ0 is the unknown internal uniform hydrostatic tension within the liquid slit inclusion to be determined, and

A=σ11+σ224,B=σ22σ11+2iσ122. (5)

We now introduce the following conformal mapping function for the matrix:

z=ω(ξ)=R(ξ+mξ),ξ=ω1(z)=z+z24mR22R,R>0,|m|1,|ξ|1, (6)

where m is a complex number.

As shown in Figure 2, using the mapping function in equation (6), the matrix is mapped onto |ξ|1 , the portion Lb in the z-plane is mapped onto the arc lb of the unit circle in the ξ-plane, and the portion Lc in the z-plane is mapped onto the remaining arc lc of the unit circle in the ξ-plane. Furthermore, the center of the arc lb lies on ξ=1 and the central angle subtended by the arc lb is 2θ0 . The two tips of the arc lc are located at ξ=a=eiθ0 and ξ=a¯=eiθ0 . The choice of a complex number m is to account for the non-symmetry of the two tips of the debonded portion Lc with respect to the principal axes of the ellipse L.

Figure 2.

Figure 2.

The image ξ-plane.

The boundary value problem takes the following form in the image ξ-plane:

κϕ(ξ)ω(ξ)ϕ(ξ)¯ω(ξ)¯ψ(ξ)¯=2iμϖ21ω(ξ),ξlb;ϕ(ξ)+ω(ξ)ϕ(ξ)¯ω(ξ)¯+ψ(ξ)¯=σ0ω(ξ),ξlc;ϕ(ξ)ARξ+O(1),ψ(ξ)BRξ+O(1),|ξ|, (7)

where, for convenience, we write ϕ(ξ)=ϕ(ω(ξ)),ψ(ξ)=ψ(ω(ξ)) .

Considering equation (7)1, we introduce an auxiliary function Y(ξ) defined as follows:

Y(ξ)={ω(ξ)ϕ¯(1ξ)ω¯(1ξ)+ψ¯(1ξ)+R(2iμϖ21κA)ξR(B¯+Am)1ξ,|ξ|1,κϕ(ξ)κARξR(2iμmϖ21+B¯+Am)1ξ,|ξ|1. (8)

It is seen from equation (8) that Y(ξ) is analytic and single valued everywhere in |ξ|1 , and is analytic and single valued everywhere in |ξ|1 including the point at infinity. Furthermore, Y(ξ) is continuous across the arc lb , i.e.,

Y+(ξ)Y(ξ)=0,ξlb, (9)

where the superscripts “+” and “−” indicate the values when approaching the unit circle |ξ|=1 from inside and outside, respectively.

Using the definition of Y(ξ) in equation (8), the interface condition in equation (7)2 becomes

Y+(ξ)+e2πεY(ξ)=R[σ0+2iμϖ21(κ+1)A]ξ+Rκ[κmσ02iμmϖ21(κ+1)(B¯+Am)]1ξ,ξlc, (10)

where

ε=lnκ2π. (11)

Equations (9) and (10) constitute a standard Riemann–Hilbert problem with discontinuous coefficients. An analytical solution to the Riemann–Hilbert problem in equations (9) and (10) can be derived as

Y(ξ)=Rκ[σ0+2iμϖ21(κ+1)A]κ+1[ξχ(ξ)]+R[κmσ02iμmϖ21(κ+1)(B¯+Am)]κ+1[1ξχ(ξ)ξχ(0)], (12)

where the Plemelj function χ(ξ) is defined by

χ(ξ)=(ξa)1/2+iε(ξa¯)1/2iε, (13)

and

χ(0)=e2ε(πθ0). (14)

The brunch cut for the Plemelj function χ(ξ) is taken along lc so that χ(ξ)ξ as ξ . The auxiliary function Y(ξ) given by equation (12) contains the two unknowns σ0 and ϖ21 . Since the initial area of the liquid slit inclusion is zero, the change in the area of the liquid inclusion due to internal uniform hydrostatic stresses is zero. Thus, the liquid slit inclusion always behaves as if it is incompressible. The incompressibility of the liquid inclusion requires that [16,18,30,31]

Lc(un+un)ds=0, (15)

where un is the normal component of the displacement along Lc . By considering equation (8), the incompressibility condition in equation (15) can be further written in the form

Im{1ξ2ω¯(1ξ)Y(ξ)dξ}=RIm{(1ξ2m¯)Y(ξ)dξ}=0, (16)

where the integral contour is taken around lc .

Using the residue theorem, the incompressibility condition in equation (16) is equivalent to

2πRe{Y(0)}Im{|ξ|(1ξ2m¯)Y(ξ)dξ}=2πRe{Y(0)}+Im{m¯|ξ|Y(ξ)dξ}=0, (17)

where the integral is taken in the counterclockwise direction.

It is derived from equation (12) that

Y(0)=Rκ[σ0+2iμϖ21(κ+1)A]κ+1[1χ(0)]R[κmσ02iμmϖ21(κ+1)(B¯+Am)]κ+1χ(0)2χ(0), (18)

and

Im{m¯|ξ|Y(ξ)dξ}=2πRκ+1[1+cosθ02εsinθ0χ(0)]{κ|m|2σ0(κ+1)[Re{mB}+A|m|2]}πRκ(1+4ε2)sin2θ0κ+1[mσ0+2mμϖ21mA(κ+1)], (19)

where m and m are, respectively, the real and imaginary parts of m, and

χ(0)=e2ε(πθ0)(cosθ0+2εsinθ0),χ(0)=e2ε(πθ0)sin2θ0(1+4ε2). (20)

By substituting the results in equations (18) and (19) into equation (17), we obtain the following linear algebraic equation for the two unknowns σ0 and ϖ21 :

2κ{1+e2ε(πθ0)(cosθ0+2εsinθ0)m(1+4ε2)sin2θ0+|m|2[1+cosθ02εsinθ0e2ε(πθ0)]}σ0μ2m(κ+1)(1+4ε2)sin2θ0ϖ21=(κ+1){2κ[1+e2ε(πθ0)(cosθ0+2εsinθ0)]m(κ+1)(1+4ε2)sin2θ0+2|m|2[1+cosθ02εsinθ0e2ε(πθ0)]}σ11+σ224μ+(κ+1){2m[1+cosθ02εsinθ0e2ε(πθ0)](1+4ε2)sin2θ0}σ22σ112μ2m(κ+1)[1+cosθ02εsinθ0e2ε(πθ0)]σ12μ. (21)

In addition, the remote asymptotic behavior of ψ(ξ) is

ψ(ξ)BRξ+c+iRIm{Bm}ξ+O(1ξ2),|ξ|, (22)

where c is an arbitrary real number (i.e., Im{c}=0 ) to satisfy the condition that the resultant moment about the origin on the disk |z|= is zero.

It is obtained from equation (8) that the original pair of analytic functions ϕ(ξ) and ψ(ξ) can be expressed in terms of the auxiliary function Y(ξ) as follows:

ϕ(ξ)=1κY(ξ)+ARξ+Rκ(2iμmϖ21+B¯+Am)1ξ,ψ(ξ)=Y¯(1ξ)+ξ(1+m¯ξ2)mξ2ϕ(ξ)+R(2iμϖ21+κA)1ξ+R(B+Am¯)ξ. (23)

Using equation (23) to impose the condition in equation (22), we obtain another linear algebraic equation for the two unknowns σ0 and ϖ21 as follows:

12m(κ+1)(1+4ε2)sin2θ0σ0μ+2{1e2εθ0(cosθ0+2εsinθ0)m(1+4ε2)sin2θ0+|m|2[1e2εθ0(cosθ02εsinθ0)]}ϖ21=m(κ+1)(1+4ε2)sin2θ0σ11+σ224μ+m(κ+1)[1e2εθ0(cosθ02εsinθ0)]σ22σ112μ+(κ+1){m[1e2εθ0(cosθ02εsinθ0)]12(1+4ε2)sin2θ0}σ12μ. (24)

As a result, σ0 and ϖ21 can be uniquely determined by solving the two coupled linear algebraic equations in equations (21) and (24) as follows:

σ0μ=c22δ1c12δ2c11c22c12c21,ϖ21=c21δ1+c11δ2c11c22c12c21, (25)

where the six coefficients c11,c12,c21,c22,δ1 , and δ2 are defined by

c11=2κ{1+e2ε(πθ0)(cosθ0+2εsinθ0)m(1+4ε2)sin2θ0+|m|2[1+cosθ02εsinθ0e2ε(πθ0)]},c12=4c21=2m(κ+1)(1+4ε2)sin2θ0,c22=2{1e2εθ0(cosθ0+2εsinθ0)m(1+4ε2)sin2θ0+|m|2[1e2εθ0(cosθ02εsinθ0)]}, (26a)
δ1=(κ+1){2κ[1+e2ε(πθ0)(cosθ0+2εsinθ0)]m(κ+1)(1+4ε2)sin2θ0+2|m|2[1+cosθ02εsinθ0e2ε(πθ0)]}σ11+σ224μ+(κ+1){2m[1+cosθ02εsinθ0e2ε(πθ0)](1+4ε2)sin2θ0}σ22σ112μ2m(κ+1)[1+cosθ02εsinθ0e2ε(πθ0)]σ12μ,δ2=m(κ+1)(1+4ε2)sin2θ0σ11+σ224μ+m(κ+1)[1e2εθ0(cosθ02εsinθ0)]σ22σ112μ+(κ+1){m[1e2εθ0(cosθ02εsinθ0)]12(1+4ε2)sin2θ0}σ12μ. (26b)

In particular, when m is real valued ( 1m1 ), the two tips of Lc are symmetric with respect to the major or minor axis of the ellipse L. In this case, equations (21) and (24) become decoupled, and σ0 and ϖ21 can be determined in a decoupled manner as

σ0=(κ+1)({2κ[1+e2ε(πθ0)(cosθ0+2εsinθ0)]m(κ+1)(1+4ε2)sin2θ0+2m2[1+cosθ02εsinθ0e2ε(πθ0)]}(σ11+σ22)2{(1+4ε2)sin2θ02m[1+cosθ02εsinθ0e2ε(πθ0)]}(σ22σ11))8κ{1+e2ε(πθ0)(cosθ0+2εsinθ0)m(1+4ε2)sin2θ0+m2[1+cosθ02εsinθ0e2ε(πθ0)]}, (27)
ϖ21=ε12(κ+1){m[1e2εθ0(cosθ02εsinθ0)]12(1+4ε2)sin2θ0}1e2εθ0(cosθ0+2εsinθ0)m(1+4ε2)sin2θ0+m2[1e2εθ0(cosθ02εsinθ0)], (28)

where ε12 is the remote shear strain defined by

ε12=σ122μ. (29)

We can see from equations (27) and (28) that σ0 is unaffected by the remote shear stress σ12 and ϖ21 is unaffected by the remote normal stresses σ11 and σ22 when the two tips of Lc are symmetric with respect to major or minor axis of the ellipse L. We illustrate in Figures 38 the variations of σ0 and ϖ21 as functions of θ0 for different values of the two parameters m and κ. We can see from these figures that all three parameters θ0,m , and κ exert a significant influence on σ0 and ϖ21 . In addition, it is seen from Figures 3 and 4 that when m is negative (e.g., m=0.8 ), the value of σ0 is insensitive to θ0 . In the following, we further discuss several typical cases when m is a real number to demonstrate the analytical results in equations (27) and (28).

Figure 3.

Figure 3.

Variations of σ0 as a function of θ0 for different values of m with κ=2,σ22=0 .

Figure 4.

Figure 4.

Variations of σ0 as a function of θ0 for different values of m with κ=2,σ11=0 .

Figure 5.

Figure 5.

Variations of σ0 as a function of θ0 for different values of κ with m=0.5,σ22=0 .

Figure 6.

Figure 6.

Variations of σ0 as a function of θ0 for different values of κ with m=0.5,σ11=0 .

Figure 7.

Figure 7.

Variations of ϖ21 as a function of θ0 for different values of m with κ=2 .

Figure 8.

Figure 8.

Variations of ϖ21 as a function of θ0 for different values of κ with m=0.5 .

When the rigid inclusion is completely debonded from the matrix with θ0=0 , equation (27) becomes

σ0=(κ+1)[(1m)2σ11+(1+m)2σ22]4(1+m2κ). (30)

When the rigid inclusion is circular with m=0 , equation (27) becomes

σ0=(κ+1)(σ11+σ22)4+(κ+1)(1+4ε2)sin2θ0(σ11σ22)4κ[1+e2ε(πθ0)(cosθ0+2εsinθ0)], (31)

which is consistent with our recent result for a partially debonded circular elastic inclusion in which the debonded part is filled with an incompressible liquid inclusion [15].

When the remote loading satisfies the following harmonic shape condition for a perfectly bonded rigid elliptical inclusion [27,28]:

σ22σ11σ22+σ11=m(κ1)2,σ12=0, (32)

Equation (27) reduces to

σ0=(κ+1)(σ11+σ22)4=(κ+1)[(1m)2σ11+(1+m)2σ22]4(1+m2κ), (33)

which means that the rigid elliptical inclusion is still perfectly bonded to the matrix or that the whole elliptical interface L becomes a liquid–solid interface. This fact implies that the insertion of the liquid inclusion into the debonded portion Lc will suppress crack growth.

When the rigid elliptical inclusion is perfectly bonded to the matrix with θ0=π , equation (28) reduces to

ϖ21=m(κ+1)ε12κ+m2, (34)

which is consistent with our previous studies [32,33].

When the rigid inclusion is circular with m=0 , equation (28) becomes

ϖ21=ε12(κ+1)(1+4ε2)sin2θ02[1e2εθ0(cosθ0+2εsinθ0)], (35)

which also recovers our recent result for a partially debonded circular elastic inclusion in which the debonded portion is filled with an incompressible liquid inclusion [15].

Now, the auxiliary function Y(ξ) has been completely determined. The original pair of analytic functions ϕ(ξ) and ψ(ξ) can then be obtained using equation (23). Consequently, the elastic field in the matrix is known. E.g., the displacement jumps across the debonded portion Lc of the elliptical interface are given by

(u1+iu2)(u1+iu2)+=R2μ{κ[σ0+2iμϖ21(κ+1)A]+κmσ02iμmϖ21(κ+1)(B¯+Am)ξχ(0)}χ+(ξ),ξlc. (36)

When m is real valued and the remote loading satisfies equation (32), it is verified using equations (28) and (36) that ϖ21=0 and there is no displacement jump across the debonded portion Lc with the matrix remaining perfectly bonded to the rigid elliptical inclusion.

4. Conclusions

We have derived a closed-form solution to the problem of a rigid elliptical inclusion partially bonded to an infinite elastic matrix subjected to uniform remote in-plane stresses. The debonded portion of the elliptical interface is filled with a liquid slit inclusion. By enforcing the interface conditions along the elliptical interface, a Riemann–Hilbert problem with discontinuous coefficients is derived in equations (9) and (10), and its solution is given by equation (12) which contains two unknowns σ0 and ϖ21 . By imposing the incompressibility condition of the liquid slit inclusion and balance of moment on a circular disk of infinite radius, a set of two coupled linear algebraic equations is derived in equations (21) and (24) from which σ0 and ϖ21 are uniquely determined in equation (25). When the two tips of Lc are symmetric with respect to the major or minor axis of the ellipse L, σ0 and ϖ21 are obtained in an uncoupled manner in equations (27) and (28). Thus, the auxiliary function Y(ξ) and the original pair of analytic functions ϕ(ξ) and ψ(ξ) characterizing the elastic field in the matrix have been completely determined. Once again, we demonstrate that the insertion of a liquid slit inclusion into the debonded portion will suppress crack growth along the elliptical interface.

Footnotes

The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.

Funding: The author(s) disclosed receipt of the following financial support for the research, authorship, and/or publication of this article: This work is supported by a Discovery Grant from the Natural Sciences and Engineering Research Council of Canada (grant no. RGPIN-2023-03227 Schiavo).

ORCID iD: Peter Schiavone Inline graphic https://orcid.org/0000-0003-4741-0165

Contributor Information

Xu Wang, School of Mechanical and Power Engineering, East China University of Science and Technology, Shanghai, China.

Peter Schiavone, Department of Mechanical Engineering, University of Alberta, Edmonton, AB, Canada.

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