Effect of Couple Stresses on the Stress Intensity Factors ...downloads.hindawi.com/journals/mpe/2012/575891.pdfthe stress intensity factor in the linearized couple-stress theory 6

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2012, Article ID 575891, 14 pagesdoi:10.1155/2012/575891

Research ArticleEffect of Couple Stresses on theStress Intensity Factors for Two Parallel Cracks inan Infinite Elastic Medium under Tension

Shouetsu Itou

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

Correspondence should be addressed to Shouetsu Itou, [email protected]

Received 8 February 2012; Accepted 22 March 2012

Academic Editor: Oleg V. Gendelman

Copyright q 2012 Shouetsu Itou. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

Stresses around two parallel cracks of equal length in an infinite elastic medium are evaluatedbased on the linearized couple-stress theory under uniform tension normal to the cracks. Fouriertransformations are used to reduce the boundary conditions with respect to the upper crack to dualintegral equations. In order to solve these equations, the differences in the displacements and in therotation at the upper crack are expanded through a series of functions that are zero valued outsidethe crack. The unknown coefficients in each series are solved in order to satisfy the boundary con-ditions inside the crack using the Schmidt method. The stresses are expressed in terms of infiniteintegrals, and the stress intensity factors can be determined using the characteristics of the inte-grands for an infinite value of the variable of integration. Numerical calculations are carried outfor selected crack configurations, and the effect of the couple stresses on the stress intensity factorsis revealed.

1. Introduction

In the classical theory of elasticity, the differential equations of equilibrium are derivedfrom the equilibrium of the forces for the rectangular parallelepiped element dx dy dz withrespect to the rectangular coordinates (x, y, z). Since the element dx dy dz is infinitesimal,the normal stresses and shearing stresses are considered to act only on the surfaces of theelement. The classical theory of elasticity is valid for a homogeneous material. In contrast,for materials with microstructures, such as porous materials and discrete materials, thedifferential equations of equilibrium may be derived from a parallelepiped element, which,although very small, is not infinitesimal. This produces additional stresses, called couplestresses, on the surfaces of the parallelepiped element. In the linearized couple stress theory

2 Mathematical Problems in Engineering

(also referred to as the Cosserat theory with constrained rotations), the couple stresses areassumed to be proportional to the curvature, which yields a new material constant l, thedimension of which is length [1].

Mindlin evaluated the effect of couple stresses on the stress concentration around acircular hole in an infinite medium under tension [1], and Itou examined the effect of thecouple stresses on the stress concentration around a circular hole in an infinite elastic layerunder tension [2]. A similar problem has been solved for an infinite medium containing aninfinite 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 toapproach those for the corresponding classical solutions as l/r approaches zero, where 2r isthe diameter of the holes.

Sternberg andMuki solved the stress intensity factor around a finite crack in an infiniteCosserat medium under tension and revealed that the Mode I stress intensity factor is alwayslarger than the corresponding value for the classical theory of elasticity [4]. Yoffe assumedthat a crack propagates only to the right, maintaining its length 2a to be constant and solvedthe stress intensity factor for the propagating crack [5]. For the Yoffe model, Itou evaluatedthe stress intensity factor in the linearized couple-stress theory [6]. For the crack problems, ithas been shown that the values of the stress intensity factors are always larger than those forthe 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 thetransverse ultrasonic wave for brass, bronze, duralumin, and aluminum [7], and the materialconstant l has been shown to be approximately one order of magnitude smaller than themeangrain size [7]. Thus, the effect of the couple stresses does not significantly affect the stress con-centrations caused by the existence of circular holes, circular inclusions, and notches, whereasthe effect of the couple stresses on the stress intensity factor around a crack is always largerthan the corresponding value in the classical theory of elasticity.

If the weight of airplanes, high-speed trains, and automobiles can be reduced, fuelconsumption can be curtailed considerably. This may be accomplished by using polycar-bonate honeycomb materials and metal foam materials when designing machine elements.Mora and Waas performed a compression test of honeycomb materials with a rigid circularinclusion and estimated that l/d falls in the range 10.0 to 15.0, where d is the diameter ofeach cell of the honeycomb [8]. Although no experiment has been performed to determinethe value of the characteristic length l for metal foam materials, l may be expected to have avalue on the order of the mean average value of the diameter of the foam. The metal foammaterials and the polycarbonate honeycomb materials reduce the need for landfills becausethese materials are reusable. As such, these materials are increasingly being used for struc-tural 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 suchmaterials.

Gourgiotis and Georgiadis solved the Mode II and Mode III stress intensity factors fora crack in an infinite medium using the couple-stress theory and the distributed dislocationtechnique [9]. A Mode I crack problem was later solved by Gourgiotis and Georgiadis fora crack in an infinite medium [10]. Recently, Gourgiotis and Georgiadis evaluated the stressfield in the vicinity of a sharp notch in an infinite medium under tension and searing stressusing the couple-stress theory [11]. To the author’s knowledge, the stress intensity factorshave been only evaluated for a crack in an infinite medium. In the present paper, stresses aresolved 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

Mathematical Problems in Engineering 3

a

a

y

−a

−a

h

h

(1)

(2)

x

Figure 1: Geometry and coordinate system.

respect to the upper crack to dual integral equations. The differences in the displacementsand in the rotation at the upper crack are expanded through a series of functions that vanishoutside the crack. The unknown coefficients in each series are solved using the Schmidtmethod [12]. The stress intensity factors and the couple-stress intensity factor are calculatednumerically for several crack configurations.

2. Fundamental Equations

In Cartesian coordinates (x, y), the upper crack is located between −a and a at y = 0, and thelower crack is located between −a and a at y = −2h, as shown in Figure 1. Under plane strainconditions, the force stresses τxx, τyy, τxy, τyx and the couple stresses μx, μy are expressedas follows:

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,

(2.1)

where φ and ψ satisfy the following equations:

∇4φ = 0, ∇2ψ − l2∇4ψ = 0, (2.2)

∂

∂x

(ψ − l2∇2ψ

)= −2(1 − ν)l2 ∂

∂y∇2φ,

∂

∂y

(ψ − l2∇2ψ

)= 2(1 − ν)l2 ∂

∂x∇2φ, (2.3)


where ∇2 is the Laplacian operator, and l is the new material constant. The rotation ωz andthe strains εx and εy are given as follows:

ωz =12×(∂ν

∂x− ∂u

∂y

),

2Gεy = 2G∂ν

∂y= (1 − ν)τyy − ντxx, 2Gεx = 2G

∂u

∂x= (1 − ν)τxx − ντyy,

(2.4)

where u and ν are the x and y components of the displacement, and G and ν are the shearmodulus and Poisson’s ratio, respectively.

3. Boundary Conditions

If we assume that a tensile stress p is applied perpendicular to the two cracks, the stress fieldis symmetric with respect to the plane y = −h, and it is sufficient to solve the problem for−h < y < ∞ only. For convenience, we refer to the layer −h < y < 0 as layer (1) and to theupper half-plane 0 < y < ∞ as half plane (2). The boundary conditions can be expressed asfollows:

τ0yy1 = −p, for |x| < a, y = 0, (3.1)

τ0yx1 = 0, for |x| < a, y = 0, (3.2)

μ0y1 = 0, for |x| < a, y = 0, (3.3)

u01 = u02, for a < |x|, y = 0, (3.4)

v01 = v

02 , for a < |x|, y = 0, (3.5)

ω0z1 = ω

0z2, for a < |x|, y = 0, (3.6)

τ0yy1 = τ0yy2, for |x| <∞, y = 0, (3.7)

τ0yx1 = τ0yx2, for |x| <∞, y = 0, (3.8)

μ0y1 = μ

0y2, for |x| <∞, y = 0, (3.9)

τ−hyx1 = 0, for |x| <∞, y = −h, (3.10)

v−h1 = 0, for |x| <∞, y = −h, (3.11)

ω−hz1 = 0, for |x| <∞, y = −h, (3.12)

where subscripts 1 and 2 indicate layer (1) and half plane (2), respectively, and superscripts0 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:

f(ξ) =∫∞

−∞f(x) exp(iξx)dx, (4.1)

f(x) =12π

∫∞

−∞f(ξ) exp(−iξx)dξ. (4.2)

Applying (4.1) to (2.2) yields the following:

d4φ

dy4− 2ξ2

d2φ

dy2+ ξ4φ = 0,

l2d4ψ

dy4−(2ξ2l2 + 1

)d2ψ

dy2+ ξ2

(ξ2l2 + 1

)ψ = 0.

(4.3)

The solutions for (4.3) take the following forms for i = 1 and 2:

φ1 = A1 cosh(ξy

)+ B1y cosh

(ξy

)+ C1 sinh

(ξy

)+D1y sinh

(ξy

),

ψ1 = E1 cosh(ky

)+ F1 cosh

(ξy

)+H1 sinh

(ky

)+ I1 sinh

(ξy

),

φ2 =(A2 + B2y

)exp

(−|ξ|y),ψ2 = E2 exp

(−|ξ|y) + F2 exp(−ky),

(4.4)

whereA1, B1,C1,D1, E1, F1,H1, I1,A2, B2, E2, and F2 are unknown coefficients, and k is givenby:

k =

√(ξ2l2 + 1

)

l2. (4.5)

Using the Fourier transformed expressions of (2.3), the coefficients I1, F1, and E2 canbe represented by the coefficients D1, B1, and B2 as follows:

I1 = −4i(1 − ν)l2ξD1, F1 = −4i(1 − ν)l2ξB1, E2 = −4i(1 − ν)l2ξB2. (4.6)

Thus, the stresses, the displacements, and the rotation can be expressed by nine unknowncoefficients in the Fourier domain.

Using (3.10) through (3.12), which are valid for −∞ < x < +∞, unknowns C1, D1, andE1 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) canbe expressed in terms of only three unknown coefficients, that are, A1, B1, andH1. Thus, thedisplacements 2Gu01 and 2Gν01 at y = 0 and the rotation 4Gω0

z1 at y = 0 can be expressed interms of three unknown coefficients, that are, A1, B1, andH1. In contrast, coefficients A1, B1,andH1 can be expressed as 2Gu01, 2Gν

01, and 4Gω0

z1, and then the stresses, the displacements,and the rotation can be expressed in terms of 2Gu01, 2Gv

01, and 4Gω0

z1 in the Fourier domain.For examples, τ0yy1, τ

0yx1, and μ

0y1 have the following forms:

τ0yy1 =(−iu01

)k11(ξ) + ν

01k

12(ξ) +

(−iω0

z1

)k13(ξ),

τ0yx1 =(−iu01

)ik14(ξ) + ν

01ik

15(ξ) +

(−iω0

z1

)ik16(ξ),

μ0y1 =

(−iu01

)ik17(ξ) + ν

01ik

18(ξ) +

(−iω0

z1

)ik19(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 and2Gν02 at y = 0 and the rotation 4Gω0

z2 at y = 0 can be described by three unknown coefficientsA2, B2, and F2. In a similar manner to the case for layer (1), the unknown coefficients A2, B2,and F2 are represented by 2Gu02, 2Gν

02, and 4Gω0

z2. Then, the stresses, the displacements, andthe rotation can be expressed in terms of 2Gu02, 2Gν

02, and 4Gω0

z2 in the Fourier domain. Forexamples, τ0yy2, τ

0yx2, and μ

0y2 have following forms:

τ0yy2 =(−iu02

)k21(ξ) + ν

02k

22(ξ) +

(−iω0

z2

)k23(ξ),

τ0yx2 =(−iu02

)ik24(ξ) + ν

02ik

25(ξ) +

(−iω0

z2

)ik26(ξ),

μ0y2 =

(−iu02

)ik27(ξ) + ν

02ik

28(ξ) +

(−iω0

z2

)ik29(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:

(−iu01

)k11(ξ) + ν

01k

12(ξ) +

(−iω0

z1

)k13(ξ)

=(−iu02

)k21(ξ) + ν

02k

22(ξ) +

(−iω0

z2

)k23(ξ),

(−iu01

)ik14(ξ) + ν

01ik

15(ξ) +

(−iω0

z1

)ik16(ξ)

=(−iu02

)ik24(ξ) + ν

02ik

25(ξ) +

(−iω0

z2

)ik26(ξ),

(−iu01

)ik17(ξ) + ν

01ik

18(ξ) +

(−iω0

z1

)ik19(ξ)

=(−iu02

)ik27(ξ) + ν

02ik

28(ξ) +

(−iω0

z2

)ik29(ξ).

(4.10)


Equation (4.10) can be solved for (−iu 01), ν

01 and (−iω 0

z1) as follows:

(−iu01

)=(−iu02

)l1(ξ) + ν

02l2(ξ) +

(−iω0

z2

)l3(ξ),

ν01 =(−iu02

)l4(ξ) + ν

02l5(ξ) +

(−iω0

z2

)l6(ξ),

(−iω0

1

)=(−iu02

)l7(ξ) + ν

02l8(ξ) +

(−iω0

z2

)l9(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:

π(ν01 − ν02

)=

∞∑n=1

cn cos[(2n − 1)sin−1

(xa

)]for |x| ≤ a

= 0 for a ≤ |x| ≤ ∞,

(4.12)

π(u01 − u02

)=

∞∑n=1

dn sin[2n sin−1

(xa

)]for |x| ≤ a

= 0 for a ≤ |x| ≤ ∞,

(4.13)

π(ω0z1 −ω0

z2

)=

∞∑n=1

en sin[2n sin−1

(xa

)]for |x| ≤ a

= 0 for a ≤ |x| ≤ ∞,

(4.14)

where cn, dn, and en are the unknown coefficients that are to be determined. The Fouriertransforms of (4.12) through (4.14) are

(ν01 − ν02

)=

∞∑n=1

cn(2n − 1)

ξJ2n−1(aξ)

(u01 − u02

)= i

∞∑n=1

dn2nξJ2n(aξ)

(ω0z1 −ω0

z2

)= i

∞∑n=1

en2nξJ2n(aξ)

(4.15)

where Jn(ξ) is the Bessel function.


Substituting (4.11) into (4.15), we obtain the following equations:

(−iu02

)l4(ξ) + ν

02[l5(ξ) − 1] +

(−iω0

z2

)l6(ξ) =

∞∑n=1

cn(2n − 1)

ξJ2n−1(aξ),

(−iu02

)[l1(ξ) − 1] + ν02l2(ξ) +

(−iω0

z2

)l3(ξ) =

∞∑n=1

dn2nξJ2n(aξ),

(−iu02

)l7(ξ) + ν

02l8(ξ) +

(−iω0

z2

)[l9(ξ) − 1] =

∞∑n=1

en2nξJ2n(aξ).

(4.16)

Equations (4.16) can be solved for (−iu02), ν02 and (−iω0z2) as follows:

(−iu02

)=

∞∑n=1

cnm1(ξ)(2n − 1)

ξJ2n−1(aξ) +

∞∑n=1

dnm2(ξ) × 2n

ξJ2n(aξ)

+∞∑n=1

enm3(ξ) × 2n

ξJ2n(aξ),

ν02 =∞∑n=1

cnm4(ξ)(2n − 1)

ξJ2n−1(aξ) +

∞∑n=1

dnm5(ξ) × 2n

ξJ2n(aξ)

+∞∑n=1

enm6(ξ) × 2n

ξJ2n(aξ),

(−iω0

z2

)=

∞∑n=1

cnm7(ξ)(2n − 1)

ξJ2n−1(aξ) +

∞∑n=1

dnm8(ξ) × 2n

ξJ2n(aξ)

+∞∑n=1

enm9(ξ) × 2n

ξJ2n(aξ),

(4.17)

where the expressions of the known functions mi(ξ) (i = 1, 2, 3, . . . , 9) are omitted. At thisstage, the stresses, the displacements, and the rotation are represented by only three unknowncoefficients, that are, cn, dn, and en. For example, stresses τ0yy2, τ

0yx2, and μ

0y2 are expressed as

follows:

τ0yy2 =∞∑n=1

cn(2n − 1)

π×∫∞

0

q1(ξ)ξ

J2n−1(aξ) cos(ξx)dξ

+∞∑n=1

dn2nπ

×∫∞

0

q2(ξ)ξ

J2n(aξ) cos(ξx)dξ

+∞∑n=1

en2nπ

×∫∞

0

q3(ξ)ξ

J2n(aξ) cos(ξx)dξ,

(4.18)


τ0yx2 =∞∑n=1

cn(2n − 1)

π×∫∞

0

q4(ξ)ξ

J2n−1(aξ) sin(ξx)dξ

+∞∑n=1

dn2nπ

×∫∞

0

q5(ξ)ξ

J2n(aξ) sin(ξx)dξ

+∞∑n=1

en2nπ

×∫∞

0

q6(ξ)ξ

J2n(aξ) sin(ξx)dξ,

(4.19)

μ0y2 =

∞∑n=1

cn(2n − 1)

π×∫∞

0

q7(ξ)ξ

J2n−1(aξ) sin(ξx)dξ

+∞∑n=1

dn2nπ

×∫∞

0

q8(ξ)ξ

J2n(aξ) sin(ξx)dξ

+∞∑n=1

en2nπ

×∫∞

0

q9(ξ)ξ


(4.20)

where the known expressions q1(ξ), q2(ξ), . . . , q8(ξ), and q9(ξ) are omitted. Functions q i(ξ)/ξ (i = 2, 3, 4, 6, 7, 8) decrease rapidly as ξ increases. Functions qi(ξ)/ξ (i = 1, 5, 9) have thefollowing property when ξ increases:

qi(ξ)ξ

−→ qLi , (4.21)

where constants qLi (i = 1, 5, 9) can be calculated as

qLi =qi(ξL)ξL

, (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=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,

(4.23)


with

Kn(x) =(2n − 1)

π×{∫∞

0

[q1(ξ)ξ

− qL1]J2n−1(aξ) cos(ξx)dξ

+qL1 cos

[(2n − 1)sin−1(x/a)

]

(a2 − x2)1/2

},

Ln(x) =2nπ

∫∞

0

q2(ξ)ξ

J2n(a ξ) cos(ξ x)dξ,

Mn(x) =2nπ

∫∞

0

q3(ξ)ξ

J2n(aξ) cos(ξx)dξ,

On(x) =(2n − 1)

π

∫∞

0

q4(ξ)ξ

J2n−1(aξ) sin(ξx)dξ,

Pn(x) =2nπ

×{∫∞

0

[q5(ξ)ξ

− qL5]J2n(aξ) sin(ξx)dξ

+qL5 sin

[2n sin−1(x/a)

]

(a2 − x2)1/2

},

Qn(x) =2nπ

∫∞

0

q6(ξ)ξ


Rn(x) =(2n − 1)

π

∫∞

0

q7(ξ)ξ

J2n−1(aξ) sin(ξx)dξ,

Sn(x) =2nπ

∫∞

0

q8(ξ)ξ


Tn(x) =2nπ

×{∫∞

0

[q9(ξ)ξ

− qL9]J2n(aξ) sin(ξx)dξ

+qL9 sin

[2n sin−1(x/a)

]

(a2 − x2)1/2

},

(4.24)

u(x) = p, ν(x) = 0, w(x) = 0. (4.25)

The unknown coefficients cn, dn, and en in (4.23) can be solved by the Schmidt method [12].


KI

KII

M0

0 0.2 0.4 0.6 0.8 1

l/a

−0.5

0

0.5

1

1.5

(KI,K

II)/

(p√ π

a),M

0/(pa√ π

a)

Figure 2: Values ofKI ,KII , andM0 plotted with respect to l/a for h/a = 5.0 (broken lines show the valuesfor l/a = 0.0).

5. Stress Intensity Factors

If we slightly modify the integrands in (4.18) through (4.20) and use the relations

∫∞

0Jn(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

)}, for a < x,

(5.1)

the stress intensity factors and the couple-stress intensity factor can be determined as follows:

KI = [2π(x − a)]1/2τ0yy2∣∣∣x→a+

=∞∑n=1

cn(2n − 1)(−1)nqL1

(πa)1/2,

KII = [2π(x − a)]1/2τ0yx2∣∣∣x→a+

=∞∑n=1

dn(2n)(−1)nqL5(πa)1/2

,

M0 = [2π(x − a)]1/2μ0y2

∣∣∣x→a+

=∞∑n=1

en(2n)(−1)nqL9(πa)1/2

.

(5.2)


Table 1: Values of KI/(p√πa) for h/a = 5.0. (Values in parentheses were obtained from the diagram in

[4]).

l/a 0.1 0.2 0.5 1.0

KI/(p√πa ) 1.238 1.209 1.122 1.061

(1.231) (1.202) (1.120) (1.063)

KI

KII

M0

0 0.2 0.4 0.6 0.8 1l/a

−0.5

0

0.5

1

1.5(K

I,K

II)/

(p√ π

a),M

0/(pa√ π

a)

Figure 3:Values ofKI, KII , andM0 plotted with respect to l/a for h/a = 0.5 (broken lines show the valuesfor l/a = 0.0).

6. Numerical Examples

Numerical calculations are performed with quadruple precision using a Fortran program forwhich overflow and underflow do not occur within the range of from 10−5500 to 10+5500. Thestress 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 functions qi(ξa)/(ξa) (i = 2, 3, 4, 6, 7, 8) are verified to decay rapidlyas (ξa) increases, and the values of the functions qi(ξa)/(ξa) (i = 1, 5, 9) are verified torapidly approach constants qLi (i = 1, 5, 9) as (ξ a) increases. Thus, the semi-infinite integralsin (4.24) can be evaluated numerically. The Schmidt method, truncated to 12 terms for aninfinite series, was then applied to solve for coefficients cn, dn, and en in (4.23). The values ofthe 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 Schmidtmethod.

The values of the Mode I stress intensity KI/(p√πa) are shown for h/a = 5.0

in Table 1, in which the values in parentheses are obtained from the diagram in [4]. Forh/a = 5.0, the distance between two cracks is 10a, and the lower crack is considered notto affect the stress field around the upper crack and vice versa. Both values in Table 1 are wellcoincident with each other, and the accuracy of the method described in the present paper is


KI

KII

M0

0 0.2 0.4 0.6 0.8 1

l/a

−0.5

0

0.5

1

1.5

(KI,K

II)/

(p√ π

a),M

0/(pa√ π

a)

Figure 4:Values ofKI, KII , andM0 plotted with respect to l/a for h/a = 0.1 (broken lines show the valuesfor l/a = 0.0).

considered to be superior. The values of KI, KII , and M0 are plotted with respect to l/a forh/a = 5.0, 0.5, and 0.1, respectively, in Figures 2, 3, and 4, in which the straight broken linesindicate the corresponding value for l/a = 0.0.

7. Conclusion

Based on the numerical calculations described above, the following conclusions are obtained.

(i) The values of KI/(p√πa) for h/a = 5.0 are considered to be approximately coin-

cident with those for a crack in an infinite medium, and the values of KII/(p√πa)

are considered to be approximately equal to zero.

(ii) As l/a approaches zero, KI/(p√πa)and KII/(p

√πa) do not approach the cor-

responding values calculated using the classical theory of elasticity, whereas thevalues ofM0/(pa

√πa) approach zero, which is the value calculated by the classical

theory of elasticity.

(iii) The values of KI/(p√πa) decrease as h/a decreases, and the same behavior is

observed for the absolute values ofM0/(pa√πa).

(iv) The new material constant l may be comparatively small, even for materials withmicrostructures. Therefore, the key value is KI/(p

√πa), even for materials with

microstructures, because the values of KII/(p√πa) and M0/(pa

√πa) are consid-

erably smaller than the value of KI/(p√πa).


References

[1] R. D. Mindlin, “Influence of couple-stresses on stress concentrations—main features of Cosserattheory are reviewed by lecturer and some recent solutions of the equations, for cases of stress con-centration around small holes in elastic solids, are described,” Experimental Mechanics, vol. 3, no. 1,pp. 1–7, 1963.

[2] S. Itou, “The effect of couple-stresses on stress concentrations around a circular hole in a strip undertension,” International Journal of Engineering Science, vol. 14, no. 9, pp. 861–867, 1976.

[3] S. Itou, “The effect of couple-stresses on stress concentrations in a plate containing an infinite row ofholes,” Letters in Applied and Engineering Sciences, vol. 5, pp. 351–358, 1977.

[4] E. Sternberg and R. Muki, “The effect of couple-stresses on the stress concentration around a crack,”International Journal of Solids and Structures, vol. 3, no. 1, pp. 69–95, 1967.

[5] E. H. Yoffe, “The moving Griffith crack,” Philosophical Magazine, vol. 42, pp. 739–750, 1951.[6] S. Itou, “The effect of couple-stresses on the stress concentration around amoving crack,” International

Journal of Mathematics and Mathematical Sciences, vol. 4, no. 1, pp. 165–180, 1981.[7] G. N. Savin, A. A. Lukasev, E. M. Lysko, S. V. Veremejenko, and G. G. Agasjev, “Elastic wave propa-

gation in a Cosserat continuum with constrained particle rotation,” Prikladnaya Mekhanika, vol. 6, pp.37–41, 1970.

[8] R. Mora and A. M. Waas, “Measurement of the Cosserat constant of circular-cell polycarbonate hon-eycomb,” Philosophical Magazine A, vol. 80, no. 7, pp. 1699–1713, 2000.

[9] P. A. Gourgiotis and H. G. Georgiadis, “Distributed dislocation approach for cracks in couple-stresselasticity: shear modes,” International Journal of Fracture, vol. 147, no. 1–4, pp. 83–102, 2007.

[10] P. A. Gourgiotis and H. G. Georgiadis, “An approach based on distributed dislocations and disclina-tions for crack problems in couple-stress elasticity,” International Journal of Solids and Structures, vol.45, no. 21, pp. 5521–5539, 2008.

[11] P. A. Gourgiotis and H. G. Georgiadis, “The problem of sharp notch in couple-stress elasticity,” Inter-national Journal of Solids and Structures, vol. 48, no. 19, pp. 2630–2641, 2011.

[12] S. Itou, “Indentations of an elastic Cosserat layer by moving punches,” Zeitschrift fur AngewandteMathematik und Mechanik, vol. 52, no. 2, pp. 93–99, 1972.

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