# Effect of Couple Stresses on the Stress Intensity Factors for Two Parallel Cracks in an Infinite Elastic Medium under Tension

Authors: Shouetsu Itou
Journal: Mathematical Problems in Engineering
 (2012)

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## Abstract

Stresses around two parallel cracks of equal length in an infinite elastic medium are evaluated based on the linearized couple-stress theory under uniform tension normal to the cracks. Fourier transformations are used to reduce the boundary conditions with respect to the upper crack to dual integral equations. In order to solve these equations, the differences in the displacements and in the rotation at the upper crack are expanded through a series of functions that are zero valued outside the crack. The unknown coefficients in each series are solved in order to satisfy the boundary conditions inside the crack using the Schmidt method. The stresses are expressed in terms of infinite integrals, and the stress intensity factors can be determined using the characteristics of the integrands for an infinite value of the variable of integration. Numerical calculations are carried out for selected crack configurations, and the effect of the couple stresses on the stress intensity factors is revealed.


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## Body

## 1. Introduction

In the classical theory of elasticity, the differential equations of equilibrium are derived from the equilibrium of the forces for the rectangular parallelepiped elementdxdydz with respect to the rectangular coordinates (x,y,z). Since the element dxdydz is infinitesimal, the normal stresses and shearing stresses are considered to act only on the surfaces of the element. The classical theory of elasticity is valid for a homogeneous material. In contrast, for materials with microstructures, such as porous materials and discrete materials, the differential equations of equilibrium may be derived from a parallelepiped element, which, although very small, is not infinitesimal. This produces additional stresses, called couple stresses, on the surfaces of the parallelepiped element. In the linearized couple stress theory (also referred to as the Cosserat theory with constrained rotations), the couple stresses are assumed to be proportional to the curvature, which yields a new material constant l, the dimension of which is length [1].Mindlin evaluated the effect of couple stresses on the stress concentration around a circular hole in an infinite medium under tension [1], and Itou examined the effect of the couple stresses on the stress concentration around a circular hole in an infinite elastic layer under tension [2]. A similar problem has been solved for an infinite medium containing an infinite row of equally spaced holes of equal diameter under tension in the linearized couple-stress theory [3]. In these studies [1–3], the values of the stress concentration are shown to approach those for the corresponding classical solutions as l/r approaches zero, where 2r is the diameter of the holes.Sternberg and Muki solved the stress intensity factor around a finite crack in an infinite Cosserat medium under tension and revealed that the Mode I stress intensity factor is always larger than the corresponding value for the classical theory of elasticity [4]. Yoffe assumed that a crack propagates only to the right, maintaining its length 2a to be constant and solved the stress intensity factor for the propagating crack [5]. For the Yoffe model, Itou evaluated the stress intensity factor in the linearized couple-stress theory [6]. For the crack problems, it has been shown that the values of the stress intensity factors are always larger than those for the classical theory of elasticity and that the values increase as l/a approaches zero.Savin et al. determined the characteristic lengthl by measuring the velocity of the transverse ultrasonic wave for brass, bronze, duralumin, and aluminum [7], and the material constant l has been shown to be approximately one order of magnitude smaller than the mean grain size [7]. Thus, the effect of the couple stresses does not significantly affect the stress concentrations caused by the existence of circular holes, circular inclusions, and notches, whereas the effect of the couple stresses on the stress intensity factor around a crack is always larger than the corresponding value in the classical theory of elasticity.If the weight of airplanes, high-speed trains, and automobiles can be reduced, fuel consumption can be curtailed considerably. This may be accomplished by using polycarbonate honeycomb materials and metal foam materials when designing machine elements. Mora and Waas performed a compression test of honeycomb materials with a rigid circular inclusion and estimated thatl/d falls in the range 10.0 to 15.0, where d is the diameter of each cell of the honeycomb [8]. Although no experiment has been performed to determine the value of the characteristic length l for metal foam materials, l may be expected to have a value on the order of the mean average value of the diameter of the foam. The metal foam materials and the polycarbonate honeycomb materials reduce the need for landfills because these materials are reusable. As such, these materials are increasingly being used for structural components in airplanes, high-speed trains, and automobiles. As a result, the couple-stress theory has been used increasingly for the evaluation of the stresses produced in such materials.Gourgiotis and Georgiadis solved the Mode II and Mode III stress intensity factors for a crack in an infinite medium using the couple-stress theory and the distributed dislocation technique [9]. A Mode I crack problem was later solved by Gourgiotis and Georgiadis for a crack in an infinite medium [10]. Recently, Gourgiotis and Georgiadis evaluated the stress field in the vicinity of a sharp notch in an infinite medium under tension and searing stress using the couple-stress theory [11]. To the author’s knowledge, the stress intensity factors have been only evaluated for a crack in an infinite medium. In the present paper, stresses are solved for two equal parallel cracks in an infinite medium under tension using the couple-stress theory. The Fourier transform technique is used to reduce the boundary conditions with respect to the upper crack to dual integral equations. The differences in the displacements and in the rotation at the upper crack are expanded through a series of functions that vanish outside the crack. The unknown coefficients in each series are solved using the Schmidt method [12]. The stress intensity factors and the couple-stress intensity factor are calculated numerically for several crack configurations.


## 2. Fundamental Equations

In Cartesian coordinates(x,y), the upper crack is located between -a and aat y=0, and the lower crack is located between -a and a at y=-2h, as shown in Figure 1. Under plane strain conditions, the force stresses τxx,τyy,τxy,τyx and the couple stresses μx,μy are expressed as follows:(2.1)τxx=∂2ϕ∂y2-∂2ψ∂x∂y,τyy=∂2ϕ∂x2+∂2ψ∂x∂y,τxy=-∂2ϕ∂x∂y-∂2ψ∂y2,τyx=-∂2ϕ∂x∂y+∂2ψ∂x2,μx=∂ψ∂x,μy=∂ψ∂y,
where ϕ and ψ satisfy the following equations:(2.2)∇4ϕ=0,∇2ψ-l2∇4ψ=0,(2.3)∂∂x(ψ-l2∇2ψ)=-2(1-ν)l2∂∂y∇2ϕ,∂∂y(ψ-l2∇2ψ)=2(1-ν)l2∂∂x∇2ϕ,
where ∇2 is the Laplacian operator, and l is the new material constant. The rotation ωz and the strains εx and εy are given as follows:(2.4)ωz=12×(∂ν∂x-∂u∂y),2Gεy=2G∂ν∂y=(1-ν)τyy-ντxx,2Gεx=2G∂u∂x=(1-ν)τxx-ντyy,
where u and ν are the x and y components of the displacement, and G and ν are the shear modulus and Poisson’s ratio, respectively.Figure 1
Geometry and coordinate system.


## 3. Boundary Conditions

If we assume that a tensile stressp is applied perpendicular to the two cracks, the stress field is symmetric with respect to the plane y=-h, and it is sufficient to solve the problem for -h<y<0 only. For convenience, we refer to the layer -h<y<∞ as layer (1) and to the upper half-plane 0<y<∞ as half plane (2). The boundary conditions can be expressed as follows:(3.1)τyy10=-p,for|x|<a,y=0,(3.2)τyx10=0,for|x|<a,y=0,(3.3)μy10=0,for|x|<a,y=0,(3.4)u10=u20,fora<|x|,y=0,(3.5)v10=v20,fora<|x|,y=0,(3.6)ωz10=ωz20,fora<|x|,y=0,(3.7)τyy10=τyy20,for|x|<∞,y=0,(3.8)τyx10=τyx20,for|x|<∞,y=0,(3.9)μy10=μy20,for|x|<∞,y=0,(3.10)τyx1-h=0,for|x|<∞,y=-h,(3.11)v1-h=0,for|x|<∞,y=-h,(3.12)ωz1-h=0,for|x|<∞,y=-h,
where subscripts 1 and 2 indicate layer (1) and half plane (2), respectively, and superscripts 0 and -h indicate the values at y=0 and y=-h, respectively.


## 4. Analysis

In order to find the solution, the Fourier transforms are introduced as follows:(4.1)f̅(ξ)=∫-∞∞f(x)exp⁡(iξx)dx,(4.2)f(x)=12π∫-∞∞f̅(ξ)exp⁡(-iξx)dξ.Applying (4.1) to (2.2) yields the following:(4.3)d4ϕ¯dy4-2ξ2d2ϕ¯dy2+ξ4ϕ¯=0,l2d4ψ¯dy4-(2ξ2l2+1)d2ψ¯dy2+ξ2(ξ2l2+1)ψ¯=0.The solutions for (4.3) take the following forms for i=1 and 2:(4.4)ϕ̅1=A1cosh⁡(ξy)+B1ycosh⁡(ξy)+C1sinh⁡(ξy)+D1ysinh⁡(ξy),ψ̅1=E1cosh⁡(ky)+F1cosh⁡(ξy)+H1sinh⁡(ky)+I1sinh⁡(ξy),ϕ̅2=(A2+B2y)exp⁡(-|ξ|y),ψ̅2=E2exp⁡(-|ξ|y)+F2exp⁡(-ky),
where A1, B1, C1, D1, E1, F1, H1, I1, A2, B2, E2, and F2 are unknown coefficients, and k is given by:(4.5)k=(ξ2l2+1)l2.Using the Fourier transformed expressions of (2.3), the coefficients I1,F1, and E2 can be represented by the coefficients D1,B1, and B2 as follows:(4.6)I1=-4i(1-ν)l2ξD1,F1=-4i(1-ν)l2ξB1,E2=-4i(1-ν)l2ξB2.
Thus, the stresses, the displacements, and the rotation can be expressed by nine unknown coefficients in the Fourier domain.Using (3.10) through (3.12), which are valid for -∞<x<+∞, unknowns C1, D1, and E1 are given as follows:(4.7)C1=A1c11+B1c12+iH1c13,D1=A1c21+B1c22+iH1c23,iE1=A1c31+B1c32+iH1c33,
where the expressions of the known functions cij(i,j=1,2,3) are omitted. Then, the Fourier-transformed expressions of the stresses, the displacements, and the rotation in layer (1) can be expressed in terms of only three unknown coefficients, that are, A1,B1, and H1. Thus, the displacements 2Gu̅10 and 2Gν̅10 at y=0 and the rotation 4Gω̅z10 at y=0 can be expressed in terms of three unknown coefficients, that are, A1,B1, and H1. In contrast, coefficients A1,B1, and H1 can be expressed as 2Gu̅10,2Gν̅10, and 4Gω̅z10, and then the stresses, the displacements, and the rotation can be expressed in terms of 2Gu̅10,2Gv̅10, and 4Gω̅z10 in the Fourier domain. For examples, τ̅yy10,τ̅yx10, and μ̅y10 have the following forms:(4.8)τ̅yy10=(-iu̅10)k11(ξ)+ν̅10k21(ξ)+(-iω̅z10)k31(ξ),τ̅yx10=(-iu̅10)ik41(ξ)+ν̅10ik51(ξ)+(-iω̅z10)ik61(ξ),μ̅y10=(-iu̅10)ik71(ξ)+ν̅10ik81(ξ)+(-iω̅z10)ik91(ξ),
where the expressions of the known functions ki1(ξ)(i=1,2,3,…,9) are omitted.As for the upper half plane (2), the stresses, the displacements, and the rotation are shown by the three unknown coefficientsA2,B2, and F2. Thus, the displacements 2Gu̅20 and 2Gν̅20 at y=0 and the rotation 4Gω̅z20 at y=0 can be described by three unknown coefficients A2,B2, and F2. In a similar manner to the case for layer (1), the unknown coefficients A2,B2, and F2 are represented by 2Gu̅20,2Gν̅20, and 4Gω̅z20. Then, the stresses, the displacements, and the rotation can be expressed in terms of 2Gu̅20,2Gν̅20, and 4Gω̅z20 in the Fourier domain. For examples, τ̅yy20,τ̅yx20, and μ̅y20 have following forms:(4.9)τ̅yy20=(-iu̅20)k12(ξ)+ν̅20k22(ξ)+(-iω̅z20)k32(ξ),τ̅yx20=(-iu̅20)ik42(ξ)+ν̅20ik52(ξ)+(-iω̅z20)ik62(ξ),μ̅y20=(-iu̅20)ik72(ξ)+ν̅20ik82(ξ)+(-iω̅z20)ik92(ξ),
where the expressions of the known functions ki2(ξ)(i=1,2,3,…,9) are omitted.Using (3.7), (3.8) and (3.9) the following relations are obtained:(4.10)(-iu̅10)k11(ξ)+ν̅10k21(ξ)+(-iω̅z10)k31(ξ)=(-iu̅20)k12(ξ)+ν̅20k22(ξ)+(-iω̅z20)k32(ξ),(-iu̅10)ik41(ξ)+ν̅10ik51(ξ)+(-iω̅z10)ik61(ξ)=(-iu̅20)ik42(ξ)+ν̅20ik52(ξ)+(-iω̅z20)ik62(ξ),(-iu̅10)ik71(ξ)+ν̅10ik81(ξ)+(-iω̅z10)ik91(ξ)=(-iu̅20)ik72(ξ)+ν̅20ik82(ξ)+(-iω̅z20)ik92(ξ).
Equation (4.10) can be solved for (-iu̅10),ν̅10 and (-iω̅z10) as follows:(4.11)(-iu̅10)=(-iu̅20)l1(ξ)+ν̅20l2(ξ)+(-iω̅z20)l3(ξ),ν̅10=(-iu̅20)l4(ξ)+ν̅20l5(ξ)+(-iω̅z20)l6(ξ),(-iω̅10)=(-iu̅20)l7(ξ)+ν̅20l8(ξ)+(-iω̅z20)l9(ξ),
where the expressions of the known functions li(ξ)(i=1,2,3,…,9) are omitted.In order to satisfy (3.4), (3.5), and (3.6) the differences in the displacements and in the rotation at y=0 are expanded by the following series:(4.12)π(ν10-ν20)=∑n=1∞cncos⁡[(2n-1)sin-1(xa)]for|x|≤a=0fora≤|x|≤∞,(4.13)π(u10-u20)=∑n=1∞dnsin⁡[2nsin-1(xa)]for|x|≤a=0fora≤|x|≤∞,(4.14)π(ωz10-ωz20)=∑n=1∞ensin⁡[2nsin-1(xa)]for|x|≤a=0fora≤|x|≤∞,
where cn, dn, and en are the unknown coefficients that are to be determined. The Fourier transforms of (4.12) through (4.14) are(4.15)(ν̅10-ν̅20)=∑n=1∞cn(2n-1)ξJ2n-1(aξ)(u̅10-u̅20)=i∑n=1∞dn2nξJ2n(aξ)(ω̅z10-ω̅z20)=i∑n=1∞en2nξJ2n(aξ)
where Jn(ξ) is the Bessel function.Substituting (4.11) into (4.15), we obtain the following equations: (4.16)(-iu̅20)l4(ξ)+ν̅20[l5(ξ)-1]+(-iω̅z20)l6(ξ)=∑n=1∞cn(2n-1)ξJ2n-1(aξ),(-iu̅20)[l1(ξ)-1]+ν̅20l2(ξ)+(-iω̅z20)l3(ξ)=∑n=1∞dn2nξJ2n(aξ),(-iu̅20)l7(ξ)+ν̅20l8(ξ)+(-iω̅z20)[l9(ξ)-1]=∑n=1∞en2nξJ2n(aξ).
Equations (4.16) can be solved for (-iu̅20),ν̅20 and (-iω̅z20) as follows:(4.17)(-iu̅20)=∑n=1∞cnm1(ξ)(2n-1)ξJ2n-1(aξ)+∑n=1∞dnm2(ξ)×2nξJ2n(aξ)+∑n=1∞enm3(ξ)×2nξJ2n(aξ),ν̅20=∑n=1∞cnm4(ξ)(2n-1)ξJ2n-1(aξ)+∑n=1∞dnm5(ξ)×2nξJ2n(aξ)+∑n=1∞enm6(ξ)×2nξJ2n(aξ),(-iω̅z20)=∑n=1∞cnm7(ξ)(2n-1)ξJ2n-1(aξ)+∑n=1∞dnm8(ξ)×2nξJ2n(aξ)+∑n=1∞enm9(ξ)×2nξJ2n(aξ),
where the expressions of the known functions mi(ξ)(i=1,2,3,…,9) are omitted. At this stage, the stresses, the displacements, and the rotation are represented by only three unknown coefficients, that are, cn, dn, and en. For example, stresses τyy20,τyx20, and μy20 are expressed as follows:(4.18)τyy20=∑n=1∞cn(2n-1)π×∫0∞q1(ξ)ξJ2n-1(aξ)cos⁡(ξx)dξ+∑n=1∞dn2nπ×∫0∞q2(ξ)ξJ2n(aξ)cos⁡(ξx)dξ+∑n=1∞en2nπ×∫0∞q3(ξ)ξJ2n(aξ)cos⁡(ξx)dξ,(4.19)τyx20=∑n=1∞cn(2n-1)π×∫0∞q4(ξ)ξJ2n-1(aξ)sin⁡(ξx)dξ+∑n=1∞dn2nπ×∫0∞q5(ξ)ξJ2n(aξ)sin⁡(ξx)dξ+∑n=1∞en2nπ×∫0∞q6(ξ)ξJ2n(aξ)sin⁡(ξx)dξ,(4.20)μy20=∑n=1∞cn(2n-1)π×∫0∞q7(ξ)ξJ2n-1(aξ)sin⁡(ξx)dξ+∑n=1∞dn2nπ×∫0∞q8(ξ)ξJ2n(aξ)sin⁡(ξx)dξ+∑n=1∞en2nπ×∫0∞q9(ξ)ξJ2n(aξ)sin⁡(ξx)dξ,
where the known expressions q1(ξ),q2(ξ),…,q8(ξ), and q9(ξ) are omitted. Functions qi(ξ)/ξ(i=2,3,4,6,7,8) decrease rapidly as ξ increases. Functions qi(ξ)/ξ(i=1,5,9) have the following property when ξ increases:(4.21)qi(ξ)ξ⟶qiL,
where constants qiL(i=1,5,9) can be calculated as(4.22)qiL=qi(ξL)ξL,
with ξL being a large value of ξ.Finally, the remaining boundary conditions (3.1), (3.2), and (3.3) can be reduced to the following equations:(4.23)∑n=1∞cnKn(x)+∑n=1∞dnLn(x)+∑n=1∞enMn(x)=-u(x),∑n=1∞cnOn(x)+∑n=1∞dnPn(x)+∑n=1∞enQn(x)=-ν(x),∑n=1∞cnRn(x)+∑n=1∞dnSn(x)+∑n=1∞enTn(x)=-w(x)for|x|≤a,
with(4.24)Kn(x)=(2n-1)π×{q1Lcos⁡[(2n-1)sin-1(x/a)](a2-x2)1/2∫0∞[q1(ξ)ξ-q1L]J2n-1(aξ)cos⁡(ξx)dξ+q1Lcos⁡[(2n-1)sin-1(x/a)](a2-x2)1/2},Ln(x)=2nπ∫0∞q2(ξ)ξJ2n(aξ)cos⁡(ξx)dξ,Mn(x)=2nπ∫0∞q3(ξ)ξJ2n(aξ)cos⁡(ξx)dξ,On(x)=(2n-1)π∫0∞q4(ξ)ξJ2n-1(aξ)sin⁡(ξx)dξ,Pn(x)=2nπ×{q5Lsin⁡[2nsin-1(x/a)](a2-x2)1/2∫0∞[q5(ξ)ξ-q5L]J2n(aξ)sin⁡(ξx)dξ+q5Lsin⁡[2nsin-1(x/a)](a2-x2)1/2∫0∞[q9(ξ)ξ-q9L]},Qn(x)=2nπ∫0∞q6(ξ)ξJ2n(aξ)sin⁡(ξx)dξ,Rn(x)=(2n-1)π∫0∞q7(ξ)ξJ2n-1(aξ)sin⁡(ξx)dξ,Sn(x)=2nπ∫0∞q8(ξ)ξJ2n(aξ)sin⁡(ξx)dξ,Tn(x)=2nπ×{q5Lsin⁡[2nsin-1(x/a)](a2-x2)1/2∫0∞[q9(ξ)ξ-q9L]J2n(aξ)sin⁡(ξx)dξ+q9Lsin⁡[2nsin-1(x/a)](a2-x2)1/2∫0∞[q9(ξ)ξ-q9L]},(4.25)u(x)=p,ν(x)=0,w(x)=0.
The unknown coefficients cn, dn, anden in (4.23) can be solved by the Schmidt method [12].


## 5. Stress Intensity Factors

If we slightly modify the integrands in (4.18) through (4.20) and use the relations(5.1)∫0∞Jn(aξ)[cos⁡(ξx),sin⁡(ξx)]dξ={-an(x2-a2)-1/2[x+(x2-a2)-1/2]-nsin⁡(nπ2),an(x2-a2)-1/2[x+(x2-a2)-1/2]-ncos⁡(nπ2)[x+(x2-a2)-1/2]-n(x2-a2)-1/2},fora<x,
the stress intensity factors and the couple-stress intensity factor can be determined as follows:(5.2)KI=[2π(x-a)]1/2τyy20|x→a+=∑n=1∞cn(2n-1)(-1)nq1L(πa)1/2,KII=[2π(x-a)]1/2τyx20|x→a+=∑n=1∞dn(2n)(-1)nq5L(πa)1/2,M0=[2π(x-a)]1/2μy20|x→a+=∑n=1∞en(2n)(-1)nq9L(πa)1/2.


## 6. Numerical Examples

Numerical calculations are performed with quadruple precision using a Fortran program for which overflow and underflow do not occur within the range of from10-5500 to 10+5500. The stress intensity factors and the couple-stress intensity factor are calculated for h/a=5.0,0.5, and 0.1 with a Poisson’s ratio of ν=0.25.The values of the functionsqi(ξa)/(ξa)(i=2,3,4,6,7,8) are verified to decay rapidly as (ξa) increases, and the values of the functions qi(ξa)/(ξa)(i=1,5,9)are verified to rapidly approach constants qiL(i=1,5,9) as (ξa) increases. Thus, the semi-infinite integrals in (4.24) can be evaluated numerically. The Schmidt method, truncated to 12 terms for an infinite series, was then applied to solve for coefficients cn,dn, and en in (4.23). The values of the left-hand side of (4.23) are verified to coincide with those of the right-hand side of (4.23). Then, coefficients cn, dn, and en are verified to be successfully determined by the Schmidt method.The values of the Mode I stress intensityKI/(pπa) are shown for h/a=5.0 in Table 1, in which the values in parentheses are obtained from the diagram in [4]. For h/a=5.0, the distance between two cracks is 10a, and the lower crack is considered not to affect the stress field around the upper crack and vice versa. Both values in Table 1 are well coincident with each other, and the accuracy of the method described in the present paper is considered to be superior. The values of KI,KII, and M0 are plotted with respect to l/a for h/a=5.0,0.5, and 0.1, respectively, in Figures 2, 3, and 4, in which the straight broken lines indicate the corresponding value for l/a=0.0.Table 1
Values ofKI/(pπa) for h/a=5.0. (Values in parentheses were obtained from the diagram in [4]).
l/a0.10.20.51.0KI/(pπa)1.2381.2091.1221.061(1.231)(1.202)(1.120)(1.063)Figure 2
Values ofKI, KII, and M0 plotted with respect to l/a for h/a=5.0 (broken lines show the values for l/a=0.0).Figure 3
Values ofKI,KII, and M0 plotted with respect to l/a for h/a=0.5 (broken lines show the values for l/a=0.0).Figure 4
Values ofKI,KII, and M0 plotted with respect to l/a for h/a=0.1 (broken lines show the values for l/a=0.0).


## 7. Conclusion

Based on the numerical calculations described above, the following conclusions are obtained.(i)
The values ofKI/(pπa) for h/a=5.0 are considered to be approximately coincident with those for a crack in an infinite medium, and the values of KII/(pπa) are considered to be approximately equal to zero.(ii)
Asl/a approaches zero, KI/(pπa)and KII/(pπa) do not approach the corresponding values calculated using the classical theory of elasticity, whereas the values of M0/(paπa) approach zero, which is the value calculated by the classical theory of elasticity.(iii)
The values ofKI/(pπa) decrease as h/a decreases, and the same behavior is observed for the absolute values of M0/(paπa).(iv)
The new material constantl may be comparatively small, even for materials with microstructures. Therefore, the key value is KI/(pπa), even for materials with microstructures, because the values of KII/(pπa) and M0/(paπa) are considerably smaller than the value of KI/(pπa).



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