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Mathematical Problems in Engineering
Volume 2012 (2012), Article ID 575891, 14 pages
http://dx.doi.org/10.1155/2012/575891
Research Article

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

Department of Mechanical Engineering, Kanagawa University, Rokkakubashi, Kanagawa-ku, Yokohama 221-8686, Japan

Received 8 February 2012; Accepted 22 March 2012

Academic Editor: Oleg V. Gendelman

Copyright © 2012 Shouetsu Itou. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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.

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 element 𝑑𝑥𝑑𝑦𝑑𝑧 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 [13], 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 length 𝑙 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 that 𝑙/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=2, 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:𝜏𝑥𝑥=𝜕2𝜙𝜕𝑦2𝜕2𝜓𝜕𝑥𝜕𝑦,𝜏𝑦𝑦=𝜕2𝜙𝜕𝑥2+𝜕2𝜓,𝜏𝜕𝑥𝜕𝑦𝑥𝑦𝜕=2𝜙𝜕𝜕𝑥𝜕𝑦2𝜓𝜕𝑦2,𝜏𝑦𝑥𝜕=2𝜙+𝜕𝜕𝑥𝜕𝑦2𝜓𝜕𝑥2,𝜇𝑥=𝜕𝜓𝜕𝑥,𝜇𝑦=𝜕𝜓,𝜕𝑦(2.1) where 𝜙 and 𝜓 satisfy the following equations:4𝜙=0,2𝜓𝑙24𝜕𝜓=0,(2.2)𝜕𝑥𝜓𝑙22𝜓=2(1𝜈)𝑙2𝜕𝜕𝑦2𝜕𝜙,𝜕𝑦𝜓𝑙22𝜓=2(1𝜈)𝑙2𝜕𝜕𝑥2𝜙,(2.3) 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:𝜔𝑧=12×𝜕𝜈𝜕𝑥𝜕𝑢,𝜕𝑦2𝐺𝜀𝑦=2𝐺𝜕𝜈𝜕𝑦=(1𝜈)𝜏𝑦𝑦𝜈𝜏𝑥𝑥,2𝐺𝜀𝑥=2𝐺𝜕𝑢𝜕𝑥=(1𝜈)𝜏𝑥𝑥𝜈𝜏𝑦𝑦,(2.4) 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.

575891.fig.001
Figure 1: Geometry and coordinate system.

3. Boundary Conditions

If we assume that a tensile stress 𝑝 is applied perpendicular to the two cracks, the stress field is symmetric with respect to the plane 𝑦=, and it is sufficient to solve the problem for <𝑦<0 only. For convenience, we refer to the layer <𝑦< as layer (1) and to the upper half-plane 0<𝑦< as half plane (2). The boundary conditions can be expressed as follows:𝜏0𝑦𝑦1𝜏=𝑝,for|𝑥|<𝑎,𝑦=0,(3.1)0𝑦𝑥1𝜇=0,for|𝑥|<𝑎,𝑦=0,(3.2)0𝑦1𝑢=0,for|𝑥|<𝑎,𝑦=0,(3.3)01=𝑢02𝑣,for𝑎<|𝑥|,𝑦=0,(3.4)01=𝑣02𝜔,for𝑎<|𝑥|,𝑦=0,(3.5)0𝑧1=𝜔0𝑧2𝜏,for𝑎<|𝑥|,𝑦=0,(3.6)0𝑦𝑦1=𝜏0𝑦𝑦2𝜏,for|𝑥|<,𝑦=0,(3.7)0𝑦𝑥1=𝜏0𝑦𝑥2𝜇,for|𝑥|<,𝑦=0,(3.8)0𝑦1=𝜇0𝑦2𝜏,for|𝑥|<,𝑦=0,(3.9)𝑦𝑥1𝑣=0,for|𝑥|<,𝑦=,(3.10)1𝜔=0,for|𝑥|<,𝑦=,(3.11)𝑧1=0,for|𝑥|<,𝑦=,(3.12) where subscripts 1 and 2 indicate layer (1) and half plane (2), respectively, and superscripts 0 and indicate the values at 𝑦=0 and 𝑦=, respectively.

4. Analysis

In order to find the solution, the Fourier transforms are introduced as follows:𝑓(𝜉)=1𝑓(𝑥)exp(𝑖𝜉𝑥)𝑑𝑥,(4.1)𝑓(𝑥)=2𝜋𝑓(𝜉)exp(𝑖𝜉𝑥)𝑑𝜉.(4.2)

Applying (4.1) to (2.2) yields the following:𝑑4𝜙𝑑𝑦42𝜉2𝑑2𝜙𝑑𝑦2+𝜉4𝑙𝜙=0,2𝑑4𝜓𝑑𝑦42𝜉2𝑙2𝑑+12𝜓𝑑𝑦2+𝜉2𝜉2𝑙2+1𝜓=0.(4.3)

The solutions for (4.3) take the following forms for 𝑖=1 and 2:𝜙1=𝐴1cosh(𝜉𝑦)+𝐵1𝑦cosh(𝜉𝑦)+𝐶1sinh(𝜉𝑦)+𝐷1𝑦sinh(𝜉𝑦),𝜓1=𝐸1cosh(𝑘𝑦)+𝐹1cosh(𝜉𝑦)+𝐻1sinh(𝑘𝑦)+𝐼1sinh(𝜉𝑦),𝜙2=𝐴2+𝐵2𝑦||𝜉||𝑦,exp𝜓2=𝐸2||𝜉||𝑦exp+𝐹2exp(𝑘𝑦),(4.4) where 𝐴1, 𝐵1, 𝐶1, 𝐷1, 𝐸1, 𝐹1, 𝐻1, 𝐼1, 𝐴2, 𝐵2, 𝐸2, and 𝐹2 are unknown coefficients, and 𝑘 is given by:𝑘=𝜉2𝑙2+1𝑙2.(4.5)

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:𝐼1=4𝑖(1𝜈)𝑙2𝜉𝐷1,𝐹1=4𝑖(1𝜈)𝑙2𝜉𝐵1,𝐸2=4𝑖(1𝜈)𝑙2𝜉𝐵2.(4.6) 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:𝐶1=𝐴1𝑐11+𝐵1𝑐12+𝑖𝐻1𝑐13,𝐷1=𝐴1𝑐21+𝐵1𝑐22+𝑖𝐻1𝑐23,𝑖𝐸1=𝐴1𝑐31+𝐵1𝑐32+𝑖𝐻1𝑐33,(4.7) 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 2𝐺𝑢01 and 2𝐺𝜈01 at 𝑦=0 and the rotation 4𝐺𝜔0𝑧1 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 2𝐺𝑢01,2𝐺𝜈01, and 4𝐺𝜔0𝑧1, and then the stresses, the displacements, and the rotation can be expressed in terms of 2𝐺𝑢01,2𝐺𝑣01, and 4𝐺𝜔0𝑧1 in the Fourier domain. For examples, 𝜏0𝑦𝑦1,𝜏0𝑦𝑥1, and 𝜇0𝑦1 have the following forms:𝜏0𝑦𝑦1=𝑖𝑢01𝑘11(𝜉)+𝜈01𝑘12(𝜉)+𝑖𝜔0𝑧1𝑘13(𝜉),𝜏0𝑦𝑥1=𝑖𝑢01𝑖𝑘14(𝜉)+𝜈01𝑖𝑘15(𝜉)+𝑖𝜔0𝑧1𝑖𝑘16(𝜉),𝜇0𝑦1=𝑖𝑢01𝑖𝑘17(𝜉)+𝜈01𝑖𝑘18(𝜉)+𝑖𝜔0𝑧1𝑖𝑘19(𝜉),(4.8) where the expressions of the known functions 𝑘1𝑖(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 coefficients 𝐴2,𝐵2, and 𝐹2. Thus, the displacements 2𝐺𝑢02 and 2𝐺𝜈02 at 𝑦=0 and the rotation 4𝐺𝜔0𝑧2 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 2𝐺𝑢02,2𝐺𝜈02, and 4𝐺𝜔0𝑧2. Then, the stresses, the displacements, and the rotation can be expressed in terms of 2𝐺𝑢02,2𝐺𝜈02, and 4𝐺𝜔0𝑧2 in the Fourier domain. For examples, 𝜏0𝑦𝑦2,𝜏0𝑦𝑥2, and 𝜇0𝑦2 have following forms:𝜏0𝑦𝑦2=𝑖𝑢02𝑘21(𝜉)+𝜈02𝑘22(𝜉)+𝑖𝜔0𝑧2𝑘23(𝜉),𝜏0𝑦𝑥2=𝑖𝑢02𝑖𝑘24(𝜉)+𝜈02𝑖𝑘25(𝜉)+𝑖𝜔0𝑧2𝑖𝑘26(𝜉),𝜇0𝑦2=𝑖𝑢02𝑖𝑘27(𝜉)+𝜈02𝑖𝑘28(𝜉)+𝑖𝜔0𝑧2𝑖𝑘29(𝜉),(4.9) where the expressions of the known functions 𝑘2𝑖(𝜉)(𝑖=1,2,3,,9) are omitted.

Using (3.7), (3.8) and (3.9) the following relations are obtained:𝑖𝑢01𝑘11(𝜉)+𝜈01𝑘12(𝜉)+𝑖𝜔0𝑧1𝑘13=(𝜉)𝑖𝑢02𝑘21(𝜉)+𝜈02𝑘22(𝜉)+𝑖𝜔0𝑧2𝑘23(𝜉),𝑖𝑢01𝑖𝑘14(𝜉)+𝜈01𝑖𝑘15(𝜉)+𝑖𝜔0𝑧1𝑖𝑘16=(𝜉)𝑖𝑢02𝑖𝑘24(𝜉)+𝜈02𝑖𝑘25(𝜉)+𝑖𝜔0𝑧2𝑖𝑘26(𝜉),𝑖𝑢01𝑖𝑘17(𝜉)+𝜈01𝑖𝑘18(𝜉)+𝑖𝜔0𝑧1𝑖𝑘19=(𝜉)𝑖𝑢02𝑖𝑘27(𝜉)+𝜈02𝑖𝑘28(𝜉)+𝑖𝜔0𝑧2𝑖𝑘29(𝜉).(4.10) Equation (4.10) can be solved for (𝑖𝑢01),𝜈01 and (𝑖𝜔0𝑧1) as follows:𝑖𝑢01=𝑖𝑢02𝑙1(𝜉)+𝜈02𝑙2(𝜉)+𝑖𝜔0𝑧2𝑙3(𝜉),𝜈01=𝑖𝑢02𝑙4(𝜉)+𝜈02𝑙5(𝜉)+𝑖𝜔0𝑧2𝑙6(𝜉),𝑖𝜔01=𝑖𝑢02𝑙7(𝜉)+𝜈02𝑙8(𝜉)+𝑖𝜔0𝑧2𝑙9(𝜉),(4.11) 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:𝜋𝜈01𝜈02=𝑛=1𝑐𝑛(cos2𝑛1)sin1𝑥𝑎𝜋𝑢for|𝑥|𝑎=0for𝑎|𝑥|,(4.12)01𝑢02=𝑛=1𝑑𝑛sin2𝑛sin1𝑥𝑎𝜋𝜔for|𝑥|𝑎=0for𝑎|𝑥|,(4.13)0𝑧1𝜔0𝑧2=𝑛=1𝑒𝑛sin2𝑛sin1𝑥𝑎for|𝑥|𝑎=0for𝑎|𝑥|,(4.14) where 𝑐𝑛, 𝑑𝑛, and 𝑒𝑛 are the unknown coefficients that are to be determined. The Fourier transforms of (4.12) through (4.14) are𝜈01𝜈02=𝑛=1𝑐𝑛(2𝑛1)𝜉𝐽2𝑛1(𝑎𝜉)𝑢01𝑢02=𝑖𝑛=1𝑑𝑛2𝑛𝜉𝐽2𝑛(𝑎𝜉)𝜔0𝑧1𝜔0𝑧2=𝑖𝑛=1𝑒𝑛2𝑛𝜉𝐽2𝑛(𝑎𝜉)(4.15) where 𝐽𝑛(𝜉) is the Bessel function.

Substituting (4.11) into (4.15), we obtain the following equations: 𝑖𝑢02𝑙4(𝜉)+𝜈02𝑙5(+𝜉)1𝑖𝜔0𝑧2𝑙6(𝜉)=𝑛=1𝑐𝑛(2𝑛1)𝜉𝐽2𝑛1(𝑎𝜉),𝑖𝑢02𝑙1+(𝜉)1𝜈02𝑙2(𝜉)+𝑖𝜔0𝑧2𝑙3(𝜉)=𝑛=1𝑑𝑛2𝑛𝜉𝐽2𝑛(𝑎𝜉),𝑖𝑢02𝑙7(𝜉)+𝜈02𝑙8(𝜉)+𝑖𝜔0𝑧2𝑙9=(𝜉)1𝑛=1𝑒𝑛2𝑛𝜉𝐽2𝑛(𝑎𝜉).(4.16) Equations (4.16) can be solved for (𝑖𝑢02),𝜈02 and (𝑖𝜔0𝑧2) as follows:𝑖𝑢02=𝑛=1𝑐𝑛𝑚1(𝜉)(2𝑛1)𝜉𝐽2𝑛1(𝑎𝜉)+𝑛=1𝑑𝑛𝑚2(𝜉)×2𝑛𝜉𝐽2𝑛(+𝑎𝜉)𝑛=1𝑒𝑛𝑚3(𝜉)×2𝑛𝜉𝐽2𝑛(𝑎𝜉),𝜈02=𝑛=1𝑐𝑛𝑚4(𝜉)(2𝑛1)𝜉𝐽2𝑛1(𝑎𝜉)+𝑛=1𝑑𝑛𝑚5(𝜉)×2𝑛𝜉𝐽2𝑛+(𝑎𝜉)𝑛=1𝑒𝑛𝑚6(𝜉)×2𝑛𝜉𝐽2𝑛(𝑎𝜉),𝑖𝜔0𝑧2=𝑛=1𝑐𝑛𝑚7(𝜉)(2𝑛1)𝜉𝐽2𝑛1(𝑎𝜉)+𝑛=1𝑑𝑛𝑚8(𝜉)×2𝑛𝜉𝐽2𝑛+(𝑎𝜉)𝑛=1𝑒𝑛𝑚9(𝜉)×2𝑛𝜉𝐽2𝑛(𝑎𝜉),(4.17) 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 𝜏0𝑦𝑦2,𝜏0𝑦𝑥2, and 𝜇0𝑦2 are expressed as follows:𝜏0𝑦𝑦2=𝑛=1𝑐𝑛(2𝑛1)𝜋×0𝑞1(𝜉)𝜉𝐽2𝑛1(+𝑎𝜉)cos(𝜉𝑥)𝑑𝜉𝑛=1𝑑𝑛2𝑛𝜋×0𝑞2(𝜉)𝜉𝐽2𝑛+(𝑎𝜉)cos(𝜉𝑥)𝑑𝜉𝑛=1𝑒𝑛2𝑛𝜋×0𝑞3(𝜉)𝜉𝐽2𝑛𝜏(𝑎𝜉)cos(𝜉𝑥)𝑑𝜉,(4.18)0𝑦𝑥2=𝑛=1𝑐𝑛(2𝑛1)𝜋×0𝑞4(𝜉)𝜉𝐽2𝑛1(+𝑎𝜉)sin(𝜉𝑥)𝑑𝜉𝑛=1𝑑𝑛2𝑛𝜋×0𝑞5(𝜉)𝜉𝐽2𝑛+(𝑎𝜉)sin(𝜉𝑥)𝑑𝜉𝑛=1𝑒𝑛2𝑛𝜋×0𝑞6(𝜉)𝜉𝐽2𝑛𝜇(𝑎𝜉)sin(𝜉𝑥)𝑑𝜉,(4.19)0𝑦2=𝑛=1𝑐𝑛(2𝑛1)𝜋×0𝑞7(𝜉)𝜉𝐽2𝑛1+(𝑎𝜉)sin(𝜉𝑥)𝑑𝜉𝑛=1𝑑𝑛2𝑛𝜋×0𝑞8(𝜉)𝜉𝐽2𝑛+(𝑎𝜉)sin(𝜉𝑥)𝑑𝜉𝑛=1𝑒𝑛2𝑛𝜋×0𝑞9(𝜉)𝜉𝐽2𝑛(𝑎𝜉)sin(𝜉𝑥)𝑑𝜉,(4.20) 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:𝑞𝑖(𝜉)𝜉𝑞𝐿𝑖,(4.21) where constants 𝑞𝐿𝑖(𝑖=1,5,9) can be calculated as𝑞𝐿𝑖=𝑞𝑖𝜉𝐿𝜉𝐿,(4.22) 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:𝑛=1𝑐𝑛𝐾𝑛(𝑥)+𝑛=1𝑑𝑛𝐿𝑛(𝑥)+𝑛=1𝑒𝑛𝑀𝑛(𝑥)=𝑢(𝑥),𝑛=1𝑐𝑛𝑂𝑛(𝑥)+𝑛=1𝑑𝑛𝑃𝑛(𝑥)+𝑛=1𝑒𝑛𝑄𝑛(𝑥)=𝜈(𝑥),𝑛=1𝑐𝑛𝑅𝑛(𝑥)+𝑛=1𝑑𝑛𝑆𝑛(𝑥)+𝑛=1𝑒𝑛𝑇𝑛(𝑥)=𝑤(𝑥)for|𝑥|𝑎,(4.23) with𝐾𝑛(𝑥)=(2𝑛1)𝜋×0𝑞1(𝜉)𝜉𝑞𝐿1𝐽2𝑛1+𝑞(𝑎𝜉)cos(𝜉𝑥)𝑑𝜉𝐿1cos(2𝑛1)sin1(𝑥/𝑎)𝑎2𝑥21/2,𝐿𝑛(𝑥)=2𝑛𝜋0𝑞2(𝜉)𝜉𝐽2𝑛𝑀(𝑎𝜉)cos(𝜉𝑥)𝑑𝜉,𝑛(𝑥)=2𝑛𝜋0𝑞3(𝜉)𝜉𝐽2𝑛𝑂(𝑎𝜉)cos(𝜉𝑥)𝑑𝜉,𝑛(𝑥)=(2𝑛1)𝜋0𝑞4(𝜉)𝜉𝐽2𝑛1𝑃(𝑎𝜉)sin(𝜉𝑥)𝑑𝜉,𝑛(𝑥)=2𝑛𝜋×0𝑞5(𝜉)𝜉𝑞𝐿5𝐽2𝑛+𝑞(𝑎𝜉)sin(𝜉𝑥)𝑑𝜉𝐿5sin2𝑛sin1(𝑥/𝑎)𝑎2𝑥21/2,𝑄𝑛(𝑥)=2𝑛𝜋0𝑞6(𝜉)𝜉𝐽2𝑛𝑅(𝑎𝜉)sin(𝜉𝑥)𝑑𝜉,𝑛(𝑥)=(2𝑛1)𝜋0𝑞7(𝜉)𝜉𝐽2𝑛1𝑆(𝑎𝜉)sin(𝜉𝑥)𝑑𝜉,𝑛(𝑥)=2𝑛𝜋0𝑞8(𝜉)𝜉𝐽2𝑛𝑇(𝑎𝜉)sin(𝜉𝑥)𝑑𝜉,𝑛(𝑥)=2𝑛𝜋×0𝑞9(𝜉)𝜉𝑞𝐿9𝐽2𝑛+𝑞(𝑎𝜉)sin(𝜉𝑥)𝑑𝜉𝐿9sin2𝑛sin1(𝑥/𝑎)𝑎2𝑥21/2,(4.24)𝑢(𝑥)=𝑝,𝜈(𝑥)=0,𝑤(𝑥)=0.(4.25) 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 relations0𝐽𝑛[]=(𝑎𝜉)cos(𝜉𝑥),sin(𝜉𝑥)𝑑𝜉𝑎𝑛𝑥2𝑎21/2𝑥𝑥+2𝑎21/2𝑛sin𝑛𝜋2,𝑎𝑛𝑥2𝑎21/2𝑥𝑥+2𝑎21/2𝑛cos𝑛𝜋2,for𝑎<𝑥,(5.1) the stress intensity factors and the couple-stress intensity factor can be determined as follows:𝐾𝐼=[]2𝜋(𝑥𝑎)1/2𝜏0𝑦𝑦2||𝑥𝑎+=𝑛=1𝑐𝑛(2𝑛1)(1)𝑛𝑞𝐿1(𝜋𝑎)1/2,𝐾𝐼𝐼=[]2𝜋(𝑥𝑎)1/2𝜏0𝑦𝑥2||𝑥𝑎+=𝑛=1𝑑𝑛(2𝑛)(1)𝑛𝑞𝐿5(𝜋𝑎)1/2,𝑀0=[]2𝜋(𝑥𝑎)1/2𝜇0𝑦2||𝑥𝑎+=𝑛=1𝑒𝑛(2𝑛)(1)𝑛𝑞𝐿9(𝜋𝑎)1/2.(5.2)

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 from 105500 to 10+5500. The stress intensity factors and the couple-stress intensity factor are calculated for /𝑎=5.0,0.5, and 0.1 with a Poisson’s ratio of 𝜈=0.25.

The values of the functions 𝑞𝑖(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 𝑞𝐿𝑖(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 intensity 𝐾𝐼/(𝑝𝜋𝑎) are shown for /𝑎=5.0 in Table 1, in which the values in parentheses are obtained from the diagram in [4]. For /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 /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.

tab1
Table 1: Values of 𝐾𝐼/(𝑝𝜋𝑎) for /𝑎=5.0. (Values in parentheses were obtained from the diagram in [4]).
575891.fig.002
Figure 2: Values of 𝐾𝐼, 𝐾𝐼𝐼, and 𝑀0 plotted with respect to 𝑙/𝑎 for /𝑎=5.0 (broken lines show the values for 𝑙/𝑎=0.0).
575891.fig.003
Figure 3: Values of 𝐾𝐼,𝐾𝐼𝐼, and 𝑀0 plotted with respect to 𝑙/𝑎 for /𝑎=0.5 (broken lines show the values for 𝑙/𝑎=0.0).
575891.fig.004
Figure 4: Values of 𝐾𝐼,𝐾𝐼𝐼, and 𝑀0 plotted with respect to 𝑙/𝑎 for /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 of 𝐾𝐼/(𝑝𝜋𝑎) for /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)As 𝑙/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 of 𝐾𝐼/(𝑝𝜋𝑎) decrease as /𝑎 decreases, and the same behavior is observed for the absolute values of 𝑀0/(𝑝𝑎𝜋𝑎).(iv)The new material constant 𝑙 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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