WEIGHT FUNCTIONS OF A CRACK IN A TWO ...yantipov/qjmam2012.pdfWEIGHT FUNCTIONS OF A CRACK IN A...

WEIGHT FUNCTIONS OF A CRACK IN ATWO-DIMENSIONAL MICROPOLAR SOLID

by Y. A. ANTIPOV†

(Department of Mathematics, Louisiana State University, Baton Rouge, LA 70803, USA)

[Received 10 September 2011. Revised 20 November 2011]

Summary

Two models for a crack in an unbounded two-dimensional micropolar solid are analysed. Thefirst one concerns plane-strain deformation, while the second problem assumes the conditions ofantiplane-strain loading. In both cases, one of three modes is uncoupled, while the other two arecoupled. For a semi-infinite crack, the models reduce to a scalar and an order-2 vector Riemann–Hilbert problems with a Hermitian matrix coefficient. In the plane-strain case, the problem issolved exactly, and the weight functions are derived by quadratures. In the antiplane-strain case,the matrix coefficient cannot be factorized by the methods available in the literature. For a finiteantiplane-strain crack, an approximate solution is obtained by the method of orthogonal polyno-mials. The weight functions are found in a series form through the solution of an infinite systemof linear algebraic equations. In both cases, the plane-strain and antiplane-strain micropolar the-ory, it is shown that if a certain micropolar parameter (the same for both theories) tends to zero,then the solution and the weight functions tend to those of classical elasticity. Numerical resultsfor the weight functions are reported.

1. Introduction

The theory of couple–stress elasticity (also known as asymmetric elasticity) was initiated by E.Cosserat and F. Cosserat (1). According to this theory, an infinitesimal surface element is subjectnot only to the classical stresses limδs→0 δp/δs but also to the couple–stresses limδs→0 δm/δs.Here, δp and δm are the force and the moment, respectively, acting on the surface element δswith an external normal n. Another difference between the classical elasticity and the Cosserattheory is that the deformation of the body is described by two vectors, the displacement u and therotation ϕϕϕ. The modern theory of the Cosserat medium including its linearized two-dimensional(2D) and three-dimensional (3D) versions was developed in (2 to 6). In the constrained couple–stress theory, the rotation vector is not free, ϕϕϕ = 1

2 curl u, while in the unconstrained couple–stresstheory, the rotation and displacement vectors are independent. To distinguish the two cases, Eringen(6) suggested to call the unconstrained theory as micropolar elasticity. A review of the governingequations and derivation of two conservation laws of micropolar elasticity were given in Lubardaand Markenscoff (7).

In the framework of the first theory, the effect of couple–stresses on the stress concentration at thetip of a finite crack under conditions of plane strain subject to uniform uniaxial tension was studiedby Sternberg and Muki (8). By using the method of Fredholm integral equations, they found thatwhen the couple–stress parameter l → 0, the stress intensity factor does not approach that obtainedfor the elastic case (l = 0). Both cases, the constrained couple–stress theory and micropolar elas-ticity, were analysed by Atkinson and Leppington (9) for a semi-infinite and a finite crack under

†⟨[email protected]⟩c⃝ The author 2012. Published by Oxford University Press; all rights reserved.

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2 of 33 Y. A. ANTIPOV

conditions of plane strain. In the case of a semi-infinite crack, the boundary conditions were chosento be σy = −τ0ex/a , τyx = 0, m yz = 0, x < 0, y = 0±, and the problem was reduced to a systemof two Wiener–Hopf equations. The system was decoupled, and the solution was found by quadra-tures. The problem of a finite crack was solved by the method of matched asymptotic expansions.For both theories, they showed that when a certain parameter tends to zero, then the energy releaserate from a crack tip tends to the classical elastic value, while the stress intensity factor does not.That parameter was taken as the couple–stress parameter l (also used in Sternberg and Muki (8))for the constrained couple–stress theory, and the parameter

s1 = γ (κ + µ)

κ(κ + 2µ)(1.1)

for the micropolar case. Here, γ , κ and µ are material constants. Another interesting limiting cases → ∞ (s = l in the the constrained couple–stress theory and s = s1 in micropolar elastic-ity) is hard to implement if the solutions derived in Sternberg and Muki (8) and Atkinson andLeppington (9) are used, and it was discussed neither in Sternberg and Muki (8), nor in Atkinson andLeppington (9).

Asymptotics of the stresses, couple–stresses, displacements and rotations at the tip of a crackwas studied in (10, 11). The plane-strain problem of micropolar elasticity for a finite crack wasreduced to a system of singular integral equations and solved numerically in Li and Lee (12). TheDirichlet, Neumann and mixed problems of plane-strain micropolar elasticity were analysed by theboundary integral equation method in Schiavone (13). Existence and uniqueness of a weak solutionin a Sobolev space for a smooth curved crack in a micropolar medium were studied in Shmoylovaand Potapenko (14).

Our goal of the paper was to derive the weight functions of a micropolar crack under condi-tions of plane strain and antiplane shear. The concept of weight functions in fracture was intro-duced by Bueckner, who found the weight functions for a semi-infinite and a penny-shaped cracksin a homogeneous elastic medium (15, 16). For the elastic case, exact and approximate expres-sions for the weight functions are available for a variety of models including the exact formulasfor 3D problems of a dynamic semi-infinite crack (17) and a static interfacial crack (18, 19). Theweight functions for a semi-infinite crack in a 2D poroelastic medium were derived in Craster andAtkinson (20).

In section 2, we analyse the problem for a semi-infinite crack subject to loading τyx = p1(x),σy = p2(x), m yz = p3(x). In contrast to (8, 9), in our derivation, we do not employ the formulationin terms of the generalized Airy stress function (21). Instead, we apply the Fourier transform directlyto the governing equilibrium equations of micropolar elasticity. This ultimately gives a representa-tion of the solution that is easy to analyse when s1 → ∞. In this section, we reduce the problem toa vector Riemann–Hilbert problem (RHP). The mode-II is decoupled, while the other two equationsare coupled. The coefficient of the RHP is a Hermitian matrix G = ||gmn|| (m, n = 1, 2), g12 = g21,that can be represented as G = b0 Q0 + b1 Q1 (b0 and b1 are Holder functions and Q0 and Q1 arepolynomial matrices). Such a matrix G always admits a closed-form factorization (for example,see (22)). The approach we use for the solution of the RHP is similar to the method proposed inAtkinson and Leppington (9). The only one difference is that we diagonalize the matrix coefficientand represent it in the form G(ζ ) = T0(ζ )3(ζ )T1(ζ ) (Tj (ζ ) are rational 2 × 2 matrices and 3(ζ)is a diagonal matrix), while the method (9) triangularizes the coupled problem.

In section 2, we also derive exact formulas for the stress intensity factors and the four weightfunctions, the functions WI, j and WVI, j ( j = I,VI) associated with the modes I (σy) and VI (m yz),



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WEIGHT FUNCTIONS OF A CRACK IN A MICROPOLAR SOLID 3 of 33

Fig. 1 The mode-IV, V and VI cracks

respectively (for the standard mode-I, II (τyx ) and III (τyz) cracks, see the diagram (23, p. 59); forthe other modes, see Fig. 1). We show that if the elastic parameter κ → 0+, and γ remains finite andnon-zero (in this case, s1 → +∞), then the equilibrium equations, the stress-strain relations andthe solution of the RHP reduce to the classical equations and formulas of elasticity. As κ → 0+, theweight functions WI,I, WII and WVI,VI tend to −

√2(πξ)−1, that is, the mode-I and mode-II weight

function of elasticity. The other two functions WI,VI and WVI,I vanish. If γ → 0+ and κ/γ → 0+,then WVI,VI → 0, and the only non-zero weight functions are WI = WI,I and WII. The asymptoticanalysis is illustrated by numerical results. Section 3 contains the analysis of the antiplane-strainproblem on a crack subject to the boundary conditions

τyz = p1(x), m yx = p2(x), m yy = p3(x), 0 < x < a, y = 0±. (1.2)

We have proved that the m yx -mode (mode-V) is decoupled, while the τyz- and m yy-modes (modesIII and IV) are coupled. Therefore, the antiplane-strain problem for a crack in a plane subject to theboundary conditions (1.2) is equivalent to two boundary value problems for the upper half-plane.The first one,

τyz = p1(x), m yx = 0, m yy = p3(x), 0 < x < a, y = 0,

uz = 0, m yx = 0, ϕy = 0, x < 0, x > a y = 0, (1.3)



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is associated with the modes III and IV, while the second problem,

m yx = p2(x), 0 < x < a; ϕx = 0, x < 0, x > a, y = 0,

τyz = 0, m yy = 0, −∞ < x < ∞, y = 0, (1.4)

recovers the mode-V solution. Notice that the numerical solution (24) relates to the antiplane-strainequations in the upper half-plane subject to the boundary conditions,

τyz = p(x), m yx = 0, m yy = 0, |x | < a, y = 0,

uz = 0, ϕx = 0, ϕy = 0, |x | > a, y = 0, (1.5)

different from (1.3). Therefore, this boundary value problem does not model the antiplane-strainproblem of micropolar elasticity for a crack, and that is why in the solution (24), all three modes,III, IV and V, are coupled. This also explains why in the case of loading (1.5), the mode-V stressintensity factor does not vanish (24), although it must identically be zero.

For a semi-infinite crack, the antiplane-strain problem reduces to a scalar RHP and an order-2vector RHP. We solve the mode-V scalar RHP in closed form and find the mode-V weight functionby two quadratures. Although, as in the plane-strain case, the matrix coefficient of the RHP for theother two modes is a Hermitian matrix, its structure is different, G = b0 Q0 + b1 Q1 + b2 Q2, wherebj are Holder functions and Q j are 2 × 2 polynomial matrices ( j = 0, 1, 2). Such a matrix cannotbe factorized in closed form by methods available in the literature. However, we have shown thatthe eigenvalues of the matrix are positive and it is positive definite. According to Shmul’yan (25)(see also (26)), the partial indices of a Hermitian positive definite matrix vanish, and the indicesand the solution of the vector RHP are stable (27). We do not derive an approximate solution ofthe RHP associated with a semi-infinite crack and focus on the case of a finite crack. The problemis equivalent to a system of two singular integral equations with Cauchy and logarithmic kernels.The system is solved by the method of orthogonal polynomials. We derive the weight functions in aseries form by using the generating function of the Chebyshev polynomials Un(x). The coefficientsof the series solve a certain infinite system of linear algebraic equations. Again, as in the plane-strain case, if κ → 0+, then the mode-III and IV weight functions associated with the crack tipsx = 0 and x = a tend to the classical elastic weight functions −

√2(a − ξ)(πaξ)−1 (for x = 0)

and −√

2ξ [πa(a − ξ)]−1 (for x = a).

2. Plane-strain problem of micropolar elasticity

2.1 Formulation

The problem to be analysed is that of a semi-infinite crack {0 < x < ∞ y = 0±} in an otherwiseunbounded micropolar solid. The crack faces are subjected to plane-strain loading

τyx = p1(x), σy = p2(x), m yz = p3(x), 0 < x < ∞, y = 0±. (2.1)

If −∞ < x 6 0, the displacements ux and uy and the microrotation ϕz are continuous through theline y = 0,



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ux (x, 0+)− ux (x, 0−) = 0, uy(x, 0+)− uy(x, 0−) = 0,

ϕz(x, 0+)− ϕz(x, 0−) = 0, −∞ < x 6 0, (2.2)

and they are discontinuous if 0 < x < ∞. The stresses (force–stresses) σx , σy , τxy and τyx and thecouple–stresses mxz and m yz satisfy the equilibrium equations

∂σx

∂x+ τyx

∂y= 0,

∂τxy

∂x+ ∂σy

∂y= 0,

∂mxz

∂x+ ∂m yz

∂y+ τxy − τyx = 0. (2.3)

The force– and couple–stresses are expressed through the displacements ux and uy and the micro-rotation ϕz as follows:

σx = (κ + λ+ 2µ)∂ux

∂x+ λ

∂uy

∂y, σy = (κ + λ+ 2µ)

∂u y

∂y+ λ

∂ux

∂x,

τxy = (κ + µ)∂u y

∂x+ µ

∂ux

∂y− κϕz, τyx = (κ + µ)

∂ux

∂y+ µ

∂uy

∂x+ κϕz,

mxz = γ∂ϕz

∂x, m yz = γ

∂ϕz

∂y. (2.4)

Here, λ, µ, γ and κ are material constants such that γ > 0, κ > 0, κ+2µ > 0 and κ+2µ+3λ > 0(28, 29). As γ → 0+ and κ → 0+, the parameters λ and µ tend to the Lame constants, and thegoverning equations become those of classical elasticity.

Equations (2.3) and (2.4) can be rearranged to give

(κ + λ+ 2µ)∂2ux

∂x2 + (λ+ µ)∂2uy

∂x∂y+ (κ + µ)

∂2ux

∂y2 + κ∂ϕz

∂y= 0,

(κ + λ+ 2µ)∂2uy

∂y2 + (λ+ µ)∂2ux

∂x∂y+ (κ + µ)

∂2u y

∂x2 − κ∂ϕz

∂x= 0,

γ

(∂2ϕz

∂x2 + ∂2ϕz

∂y2

)+ κ

(∂u y

∂x− ∂ux

∂y

)− 2κϕz = 0. (2.5)

2.2 Vector RHP

On applying the Fourier transform to (2.5) uxuy

ϕz

(ζ ; y) =∫ ∞

−∞

uxuyϕz

(x, y)eiζ x dx, (2.6)

we obtain the following system of ordinary differential equations



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(κ + µ)u′′x − ζ 2(κ + λ+ 2µ)ux − iζ(λ+ µ)u′

y + κϕ′z = 0,

−iζ(λ+ µ)u′x + (κ + λ+ 2µ)u′′

y − ζ 2(κ + µ)uy + iζκϕz = 0,

− κ u′x − iζκ uy + γ ϕ′′

z − (ζ 2γ + 2κ)ϕz = 0. (2.7)

Suppose first that y > 0. The general solution of the system is derived in the form

ux =[

iC1 sgn ζ −(

3κ2 − λ

ζ(2κ2 − κ1)− y sgn ζ

)iC2

]e−|ζ |y + γ r

κ1C3e−ry,

uy = (C1 + C2 y)e−|ζ |y − iζγκ1

C3e−r y, ϕz = −2iκ2 sgn ζ2κ2 − κ1

C2e−|ζ |y + C3e−r y, (2.8)

for 0 < y < ∞. Here, C1, C2 and C3 are free constants to be determined,

r =√ζ 2 + ρ2, ρ =

√κ(κ + 2µ)γ (κ + µ)

> 0, κ1 = κ + 2µ > 0, κ2 = κ + λ+ 2µ > 0. (2.9)

The next stage is to extend the boundary conditions (2.1) for y = 0+ to the whole real axis andapply the Fourier transform. With the aid of (2.4), we find p1

p2p3

(ζ ) =κ + µ 0 0

0 κ2 00 0 γ

u′x

u′yϕ′

z

(ζ ; 0+)+ 0 −iζµ κ

−iζλ 0 00 0 0

uxu y

ϕz

(ζ ; 0+), (2.10)

where

pj (ζ ) = 8−j (ζ )+ p+

j (ζ ), p+j (ζ ) =

∫ ∞

0pj (x)eiζ x dx, j = 1, 2, 3,

8−1 (ζ ) =

∫ 0

−∞τyx (x, 0)eiζ x dx, 8−

2 (ζ ) =∫ 0

−∞σy(x, 0)eiζ x dx,

8−3 (ζ ) =

∫ 0

−∞m yz(x, 0)eiζ x dx . (2.11)

On substituting the expressions (2.8) into the boundary conditions (2.10) and solving the system forthe constants C1, C2 and C3, we obtain

C1 = iκ1(2κ2 − κ1)ζ1(ζ )

[(−κ2

1 + 2γ κ2ζ2)r p1(ζ )− 2iκ2 sgn ζ(|ζ |3γ − κ1r) p2(ζ )

+ κ1|ζ |(κ1|ζ | − 2κ2r) p3(ζ )],

C2 = iζ1(ζ )

[− |ζ |r p1(ζ )+ iζr p2(ζ )+ (|ζ | − r)ζ 2 p3(ζ )],

C3 = − iγ (2κ2 − κ1)1(ζ )

[2iγ κ2|ζ | p1(ζ )+ 2γ κ2ζ p2(ζ )− iκ1(2κ2 − κ1) p3(ζ )

],



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where

1(ζ) = κ1r − 2γ κ2ζ2

2κ2 − κ1(|ζ | − r). (2.12)

This gives, with reference to (2.8), the Fourier images of the two displacements and the microrota-tion in terms of the functions pj (ζ ), j = 1, 2, 3. On putting y = 0, we get

ux (ζ ; 0+) = iκ1(2κ2 − κ1)ζ1(ζ )

{2i sgn ζκ1κ2r p1(ζ )+ [κ2

1r − 2γ κ2ζ2(r − |ζ |)] p2(ζ )

+ 2iκ1κ2ζ(r − |ζ |) p3(ζ )},

u y(ζ ; 0+) = iκ1(2κ2 − κ1)ζ1(ζ )

{[2γ κ2ζ

2(r − |ζ |)− κ21r ] p1(ζ )+ 2i sgn ζκ1κ2r p2(ζ )

− 2κ1κ2|ζ |(r − |ζ |) p3(ζ )},

ϕz(ζ ; 0+) = iγ (2κ2 − κ1)1(ζ )

{[2iγ κ2(r − |ζ |) p1(ζ )+ 2γ κ2 sgn ζ(r − |ζ |) p2(ζ )

+ i[κ1(2κ2 − κ1)+ 2γ κ2|ζ |(r − |ζ |)] p3(ζ )}. (2.13)

The expressions for the functions ux (ζ ; 0−), u y(ζ ; 0−) and ϕz(ζ ; 0−) can be obtained from (2.13)by replacing |ζ | and r by −|ζ | and −r , respectively.

To satisfy the conditions of continuity (2.2) for the displacements ux and u y and the rotationϕz for negative x , we subtract the expressions ux (ζ ; 0−), uy(ζ ; 0−) and ϕz(ζ ; 0−) from the onesin (2.13). The new relations may be written in the form of the following vector RHP on the realaxis L:

888+(ζ ) = d(ζ )G(ζ )[888−(ζ )+ p+(ζ )], ζ ∈ L , (2.14)

where

G(ζ ) =r/|ζ | 0 0

0 r/|ζ | −iζ (1 − r/|ζ |)0 iζ (1 − r/|ζ |) κ2

0 − ζ 2 (1 − r/|ζ |)

,

888+(ζ ) =∫ ∞

0

ux (x, 0+)− ux (x, 0−)uy(x, 0+)− uy(x, 0−)ϕz(x, 0+)− ϕz(x, 0−)

eiζ x dx,

κ0 =√κ1(2κ2 − κ1)

2γ κ2, d(ζ ) = − 4κ2

(2κ2 − κ1)1(ζ ), (2.15)

the parameter κ0 is real and positive (κ1 > 0, κ2 > 0, γ > 0, 2κ2 − κ1 > 0), and the components ofthe vectors888−(ζ ) and p+(ζ ) are given by (2.11).

2.3 Mode-II crack

The structure of the matrix G(ζ ) indicates that the τyx -mode (mode-II) is uncoupled, while the σy-mode (mode-I) and the m yz-mode (mode-VI) are coupled. The scalar RHP for the determination of



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the Fourier transform 8+1 (ζ ) of the function [ux ](x) = ux (x, 0+)− ux (x, 0−),

8+1 (ζ ) = d(ζ )r

|ζ | [8−1 (ζ )+ p+

1 (ζ )], ζ ∈ L , (2.16)

can be solved in a standard way. Split the coefficient of the problem as

d(ζ )r|ζ | = − γ0

|ζ |δ0(ζ ), (2.17)

where

γ0 = 4γ (2κ2

0 + ρ2), δ0(ζ ) =

(1 + ρ2

2κ20

)−1 [1 + ζ 2

κ20

(1 − |ζ |

r

)]. (2.18)

The function δ0(ζ ) possesses the following properties: δ0(ζ ) = 1 + O(ζ−2) as ζ → ±∞ and

δ0(ζ ) =(

1 + 12ρ

2/κ20

)−1 + O(ζ 2) as ζ → 0. Since it is real and even on the real axis, we factorizeit as

δ0(ζ ) = χ+0 (ζ )

χ−0 (ζ )

, ζ ∈ L , (2.19)

where χ±0 (ζ ) = χ0(ζ ± i0), ζ ∈ L , and

χ0(ζ ) = exp{

12π i

∫ ∞

−∞ln δ0(ζ0)dζ0

ζ0 − ζ

}= exp

{ζ

π i

∫ ∞

0

ln δ0(ζ0)dζ0

ζ 20 − ζ 2

}, ζ ∈ C±, (2.20)

C+ and C− are the upper and lower half-planes, respectively, and ln δ0(ζ ) is the branch fixed by thecondition ln 1 = 0.

Factorize next the function |ζ | as |ζ | = s+(ζ )s−(ζ ), ζ ∈ L , and introduce the Cauchy integral

9±0 (ζ ) = 1

2π i

∫ ∞

−∞χ−

0 (ζ0) p+1 (ζ0)dζ0

s−(ζ0)(ζ0 − ζ ), ζ ∈ C± \ L , (2.21)

whose boundary values 9±0 (ζ ) = 90(ζ ± i0), ζ ∈ L , due to the Sohkotski–Plemelj formulas, have

the property

9+0 (ζ )−9−

0 (ζ ) = χ−0 (ζ ) p

+1 (ζ )

s−(ζ ), ζ ∈ L . (2.22)

Here, s±(ζ ) = √|ζ | exp{iθ±/2}, 0 < θ+ < π , and −π < θ− < 0. Referring to the continuityprinciple and the Liouville theorem, we deduce the formulas

8+1 (ζ ) = − γ09

+0 (ζ )

s+(ζ )χ+0 (ζ )

, ζ ∈ C+ and 8−1 (ζ ) = s−(ζ )9−

0 (ζ )

χ−0 (ζ )

, ζ ∈ C−. (2.23)



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2.4 Modes I and VI

In order to determine the other two components of the vectors888±(ζ ), 8±2 (ζ ) and 8±

3 (ζ ), we needto solve the following vector RHP:

ϕϕϕ+(ζ ) = d(ζ )A(ζ )[ϕϕϕ−(ζ )+ p+(ζ )], ζ ∈ L , (2.24)

where

A(ζ ) =(

r/|ζ | −iζ (1 − r/|ζ |)iζ (1 − r/|ζ |) κ2

0 − ζ 2 (1 − r/|ζ |)), (2.25)

and

ϕϕϕ±(ζ ) =(8±

2 (ζ )

8±3 (ζ )

), p+(ζ ) =

(p+

2 (ζ )

p+3 (ζ )

). (2.26)

To factorize the matrix A(ζ ), it is advisable to split it first as A(ζ ) = R(ζ )A◦(ζ ), where

R(ζ ) =(

0 −iζiζ κ2

0 − ζ 2

), A◦(ζ ) = I2 − r

|ζ |ζ 2

(κ2

0 iκ20 ζ

−iζ ζ 2

)(2.27)

and I2 = diag{1, 1}. The next stage is to find the eigenvalues of the matrix A◦(ζ ),

λ1 = 1, λ2(ζ ) = 1 − (ζ 2 + κ20 )r

|ζ |ζ 2 , (2.28)

and diagonalize the matrix A◦(ζ ). Finally, we obtain the following splitting of the matrix coefficientA(ζ ) of the RHP (2.24):

A(ζ ) = 1ζ 2 + κ2

0

(−iζ −ζ 2

κ20 iζ 3

)(1 00 λ2(ζ )

)(iζ κ2

01 iζ

). (2.29)

This representation, after rearrangement, transforms the boundary condition (2.24) as((κ2

0 + 12ρ

2)

iζ κ20 + 1

2ρ2

κ20 iζ

)ϕϕϕ+(ζ ) = − 2

γ

(1

rδ0(ζ )0

0 δ1(ζ )|ζ |δ0(ζ )

)(iζ κ2

01 iζ

)[ϕϕϕ−(ζ )+ p+(ζ )], (2.30)

for ζ ∈ L , where δ0(ζ ) was given in (2.18) and

δ1(ζ ) = (ζ 2 + κ20 )r − ζ 2|ζ |(

κ20 + ρ2/2

)r

. (2.31)

Since the function δ1(ζ ) is even, δ1(ζ ) = 1 + O(ζ−2) as ζ → ±∞ and δ1(ζ ) ∼(

1 + 12ρ

2/κ20

)−1

as ζ → 0, it can be factorized as



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δ1(ζ ) = χ+1 (ζ )

χ−1 (ζ )

, ζ ∈ L , (2.32)

where χ±1 (ζ ) = χ1(ζ ± i0), ζ ∈ L , and

χ1(ζ ) = exp{

12π i

∫ ∞

−∞ln δ1(ζ0)dζ0

ζ0 − ζ

}, ζ ∈ C±. (2.33)

The function r = r(ζ ) is factorized explicitly in the form r(ζ ) = r+(ζ )r−(ζ ), ζ ∈ L , wherer±(ζ ) = (ζ ± iρ)1/2 are analytic functions in the half-planes C±, and arg(ζ + iρ) ∈ [0, π ],arg(ζ − iρ) ∈ [−π, 0]. We introduce next the vector function

999(ζ) = − 1γπ i

∫ ∞

−∞

iζ0r−(ζ0)

κ20

r−(ζ0)1

s−(ζ0)χ−1 (ζ0)

iζ0s−(ζ0)χ

−1 (ζ0)

χ−0 (ζ0)p+(ζ0)dζ0

ζ0 − ζ(2.34)

and rewrite the boundary condition (2.30) in the form

χ+0 (ζ )

(κ2

0 + ρ2

2

)iζr+(ζ )

(κ2

0 + ρ2

2

)r+(ζ )

κ20 s+(ζ )χ+

1 (ζ )

iζ s+(ζ )χ+

1 (ζ )

ϕϕϕ+(ζ )−999+(ζ )

= − 2γχ−

0 (ζ )

iζr−(ζ )

κ20

r−(ζ )1

s−(ζ )χ−1 (ζ )

iζs−(ζ )χ−

1 (ζ )

ϕϕϕ−(ζ )−999−(ζ ), ζ ∈ L . (2.35)

On applying the Liouville theorem, we obtain that the left- and right-hand sides of (2.35) equal aconstant vector D = (D1, D2)

⊤. Thus, the vector functions ϕϕϕ±(ζ ) are defined by

ϕϕϕ+(ζ ) = − 2χ+1 (ζ )

(2κ20 + ρ2)χ+

0 (ζ )r+(ζ )s+(ζ )(ζ 2 + κ2

0 )

× iζ s+(ζ )

χ+1 (ζ )

−(κ2

0 + ρ2

2

)r+(ζ )

− κ20 s+(ζ )χ+

1 (ζ )

(κ2

0 + ρ2

2

)iζr+(ζ )

[D +999+(ζ )], ζ ∈ C+,

ϕϕϕ−(ζ ) = γχ−1 (ζ )r

−(ζ )s−(ζ )2χ−

0 (ζ )(ζ2 + κ2

0 )

iζχ−

1 (ζ )s−(ζ ) − κ2

0r−(ζ )

− 1s−(ζ )χ−

1 (ζ )

iζr−(ζ )

[D +999−(ζ )], ζ ∈ C−. (2.36)

Due to the polynomial ζ 2 + κ20 in the denominator, ϕϕϕ+(ζ ) and ϕϕϕ−(ζ ) in (2.36) have inadmissible

poles at the points ζ = iκ0 ∈ C+ and ζ = −iκ0 ∈ C−, respectively. These points becomeremovable singularities if and only if the constants D1 and D2 are chosen to be

D1 = 1a+ − a−

{a−9+

1 (iκ0)− a+9−1 (−iκ0)+ a+a−[9+

2 (iκ0)−9−2 (−iκ0)]

},

D2 = − 1a+ − a− [9+

1 (iκ0)−9−1 (−iκ0)+ a+9+

2 (iκ0)− a−9−2 (−iκ0)], (2.37)

where

a+ = (2κ20 + ρ2)χ+

1 (iκ0)r+(iκ0)

2κ0s+(iκ0), a− = −κ0s−(−iκ0)χ

−1 (−iκ0)

r−(−iκ0). (2.38)



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2.5 Weight functions

Analysis of the solution found indicates that the stresses σy and τyx and the couple–stress m yz havethe square root singularities at the tip of the crack

σy ∼ KI√2π(−x)−1/2, τyx ∼ KII√

2π(−x)−1/2, m yz ∼ KVI√

2π(−x)−1/2, y = 0, x → 0−,

where KI, KII and KVI are the stress intensity factors. By using the abelian theorem for the Laplacetransform, we obtain

8−1 (ζ ) =

∫ 0

−∞τyx (x, 0)eiζ x dx ∼ KII√

2e−iπ/4ζ−1/2,

8−2 (ζ ) =

∫ 0

−∞σy(x, 0)eiζ x dx ∼ KI√

2e−iπ/4ζ−1/2,

8−3 (ζ ) =

∫ 0

−∞m yz(x, 0)eiζ x dx ∼ KVI√

2e−iπ/4ζ−1/2, ζ → ∞, ζ ∈ C−. (2.39)

On the other hand, from (2.23) and (2.36), analysing the asymptotics as ζ → ∞, ζ ∈ C−, of thesolution of the RHP, we derive

8−1 (ζ ) ∼ −9◦

0ζ−1/2, 8−

2 (ζ ) ∼ iγ D1

2ζ−1/2, 8−

3 (ζ ) ∼ iγ D2

2ζ−1/2,

ζ → ∞, ζ ∈ C−, (2.40)

where

9◦0 = 1

2π i

∫ ∞

−∞χ−

0 (ζ ) p+1 (ζ )

s−(ζ )dζ. (2.41)

By comparing formulas (2.39) and (2.40), we determine the stress intensity factors

KI = γ√2

e3iπ/4 D1, KII = −√

2eiπ/49◦0 , KVI = γ√

2e3iπ/4 D2. (2.42)

These formulas can be rewritten in terms of the weight functions. Remembering (2.11), we substi-tute the expression (2.41) into (2.42). This gives

KII =∫ ∞

0WII(ξ)p1(ξ)dξ, (2.43)

where WII(ξ) is the mode-II weight function,

WII(ξ) = − eiπ/4√

2π i

∫ ∞

−∞χ−

0 (ζ )eiζ ξ

s−(ζ )dζ. (2.44)



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In order to determine the weight functions associated with the modes I and VI, we transform theexpressions (2.42) for the stress intensity factors KI and KVI and write them in the form(

KIKVI

)= γ e3π i/4

√2(a+ − a−)

(a−9+

1 (iκ0)− a+9−1 (−iκ0)+ a+a−[9+

2 (iκ0)−9−2 (−iκ0)]

−9+1 (iκ0)+9−

1 (−iκ0)− a+9+2 (iκ0)+ a−9−

2 (−iκ0)

),

(2.45)

where (91(ζ )

92(ζ )

)= − 1

π iγ

∫ ∞

0

(vI,I(ζ, ξ) vI,VI(ζ, ξ)

vVI,I(ζ, ξ) vVI,VI(ζ, ξ)

)(p2(ξ)

p3(ξ)

)dξ (2.46)

and

vI,I(ζ, ξ) = i∫ ∞

−∞ζ0χ

−0 (ζ0)eiζ0ξdζ0

r−(ζ0)(ζ0 − ζ ), vI,VI(ζ, ξ) = κ2

0

∫ ∞

−∞χ−

0 (ζ0)eiζ0ξdζ0

r−(ζ0)(ζ0 − ζ ),

vVI,I(ζ, ξ) =∫ ∞

−∞χ−

0 (ζ0)eiζ0ξdζ0

s−(ζ0)χ−1 (ζ0)(ζ0 − ζ )

, vVI,VI(ζ, ξ) = i∫ ∞

−∞ζ0χ

−0 (ζ0)eiζ0ξdζ0

s−(ζ0)χ−1 (ζ0)(ζ0 − ζ )

.

(2.47)

Denote by {WI,I(ξ),WI,VI(ξ)} and {WVI,I(ξ),WVI,VI(ξ)} the weight functions for the modes I andVI, respectively. Then,

KI =∫ ∞

0WI,I(ξ)p2(ξ)dξ +

∫ ∞

0WI,VI(ξ)p3(ξ)dξ,

KVI =∫ ∞

0WVI,I(ξ)p2(ξ)dξ +

∫ ∞

0WVI,VI(ξ)p3(ξ)dξ. (2.48)

On substituting (2.46) into (2.45) and comparing the result with (2.48), we obtain

WI, j (ξ)= eiπ/4√

2π(a− − a+){a−vI, j (iκ0, ξ)− a+vI, j (−iκ0, ξ)

+ a+a−[vVI, j (iκ0, ξ)− vVI, j (−iκ0, ξ)]},

WVI, j (ξ)= eiπ/4√

2π(a− − a+)[vI, j (−iκ0, ξ)− vI, j (iκ0, ξ)+ a−vVI, j (−iκ0, ξ)

− a+vVI, j (iκ0, ξ)], (2.49)

where j = I,VI.

2.6 Evaluation of the weight functions

The problem of finding the numerical values of the weight functions involves the evaluation of theintegrals (2.44), (2.47) and the limiting value of the singular integrals (2.20) and (2.33) defined bythe Sokhotski–Plemelj formulas as

χ−j (ζ ) = exp

{−1

2ln δj (ζ )+ 1

2π i

∫ ∞

−∞ln δj (ζ0)dζ0

ζ0 − ζ

}, ζ ∈ L , j = 0, 1. (2.50)



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Since the functions δj (ζ ) are even, the principal value of the integrals can be evaluated by convertingthem into the integrals over the interval (−1, 1) as

χ−j (ζ ) = exp

{−1

2ln δj (ζ )+ 2ζ

π i

∫ 1

−1ln δj

(1 + η

1 − η

)dη

(1 + η)2 − (1 − η)2ζ 2

}, ζ ∈ L .

In order to compute the weight functions WII(ξ) and Wm, j (ξ) (m, j = I,VI), we represent theintegrals (2.44) and (2.49) in the form

W (ξ) = a∗{∫ ∞

0

cos ζ ξ√ζ

[3(ζ)+ i3(−ζ )]dζ + i∫ ∞

0

sin ζ ξ√ζ

[3(ζ)− i3(−ζ )]dζ}. (2.51)

Here, for the mode-II weight function, W (ξ) = WII(ξ),

a∗ = i − 12π

, 3(ζ ) = χ−0 (ζ ), (2.52)

and for the other modes, W (ξ) = Wm, j (ξ),

a∗ = 1 + i2π(a− − a+)

, 3(ζ ) = 3mj (ζ ), m, j = I, V I, (2.53)

where

3I,I(ζ ) = iχ−0 (ζ )

ζ 2 + κ20

{ζ s−(ζ )r−(ζ )

[(ζ + iκ0)a− − (ζ − iκ0)a+]+ 2κ0a+a−

χ−1 (ζ )

},

3I,VI(ζ ) = χ−0 (ζ )

ζ 2 + κ20

{κ2

0 s−(ζ )r−(ζ )

[(ζ + iκ0)a− − (ζ − iκ0)a+]− 2κ0ζa+a−

χ−1 (ζ )

},

3VI,I(ζ ) = χ−0 (ζ )

ζ 2 + κ20

{2κ0ζ s−(ζ )

r−(ζ )+ (ζ − iκ0)a− − (ζ + iκ0)a+

χ−1 (ζ )

},

3VI,VI(ζ ) = iχ−0 (ζ )

ζ 2 + κ20

{−2κ3

0 s−(ζ )r−(ζ )

+ ζ

χ−1 (ζ )

[(ζ − iκ0)a− − (ζ + iκ0)a+]} .

For computational purposes, it is desirable to improve the rate of convergence of the integral (2.51).Notice that 3(ζ) = 3◦ + O(ζ ) as ζ → 0 and 3(ζ) ∼ 3∗ +3∞/ζ as ζ → ∞, where for mode II,

3◦ =√

1 + ρ2/(2κ20 ), 3∗ = 1, 3∞ = χ∞

0 , (2.54)

and for the other modes,

3◦ = 3◦mj , 3∗ = 3∗

mj , 3∞ = 3∞mj ,

3◦I,I = 2ia+a−/κ0, 3◦

I,VI = 3◦VI,VI = 0, 3◦

VI,I = −i(a+ + a−)/κ0,

3∗I,I = 3∗

VI,VI = i(a− − a+), 3∗I,VI = 3∗

VI,I = 0,

3∞I,I = i(χ∞

0 + iρ/2)(a− − a+)− κ0(a− + a+), 3∞I,VI = κ2

0 (a− − a+)− 2κ0a+a−,

3∞VI,I = 2κ0 + a− − a+, 3∞

VI,VI = i(χ∞0 − χ∞

1 )(a− − a+)+ κ0(a− + a+).



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Here,

χ∞j = − 1

π i

∫ ∞

0ln δj (ζ )dζ = − 2

π i

∫ 1

−1ln δj

(1 + η

1 − η

)dη

(1 − η)2, (2.55)

and ln δj ((1 + η)/(1 − η)) = O((1 − η)2), η → 1, j = 0, 1. Next, we use the following integrals:∫ 1

0xa−1eiξ x dx = S(a; ξ), Re a > 0,

∫ ∞

1xa−1eiξ x dx = Ɣ(a)

ξa eiπa/2 − S(a; ξ), Re a < 1, (2.56)

where

S(a; ξ) =∞∑

j=0

(iξ) j

j!(a + j). (2.57)

These formulas furnish the resulting representation for the weight functions used for thecomputations

W (ξ)

a∗ =∫ ∞

0F(ζ, ξ)dζ + (1 + i)3∗

√2πξ

+ (1 + i)(3◦ −3∗)[

Re S(

12; ξ)

+ Im S(

12; ξ)]

− (1 − i)3∞[

2√

2πξ + Re S(

−12; ξ)

− Im S(

−12; ξ)]. (2.58)

Here,

F(ζ, ξ) = cos ζ ξ√ζ

{3(ζ)+ i3(−ζ )− (1 + i)3◦ψ−(ζ )−

[(1 + i)3∗ + (1 − i)

3∞

ζ

]ψ+(ζ )

}+ i

sin ζ ξ√ζ

{3(ζ)− i3(−ζ )− (1 − i)3◦ψ−(ζ )−

[(1 − i)3∗ + (1 + i)

3∞

ζ

]ψ+(ζ )

},

ψ−(ζ ) ={

1, 0 < ζ < 1,0, 1 < ζ < ∞,

ψ+(ζ ) ={

0, 0 < ζ < 1,1, 1 < ζ < ∞.

The function F(ζ, ξ) is bounded as ζ → 0, and F(ζ, ξ) = O(ζ−5/2) as ζ → ∞.Figure 2 shows the weight functions WI,I(ξ), WI,VI(ξ), WVI,I(ξ), WVI,VI(ξ) and WII(ξ) for the

parameters κ/µ = 1, λ/µ = 1 and γ /µ = 1. It turns out that the weight functions WI,I(ξ) andWVI,VI(ξ) tend to −∞ as ξ → 0+, while the off-diagonal functions WI,VI(ξ) and WVI,I(ξ) arebounded. The function WI,I is very close to the classical elastic weight function for modes I and II,W0 = −√

2/(πξ). The differences between WII and W0 are substantial. If the parameters ξ , κ/µand λ/µ are fixed, and the parameter γ /µ approaches zero, then the weight functions WI,VI, WVI,VIand WII change their values drastically, while the changes in the other weight functions, WI,I andWVI,I, are insignificant (Fig. 3).

2.7 Limiting case κ → 0+

We now analyse the limit case κ → 0+ assuming that γ > 0. The third equation in (2.5) isseparated from the other two, and the microrotation ϕz is a harmonic function. The vector RHP(2.14) is uncoupled and has the form

888+(ζ ) = −|ζ |−1 diag{e0, e0, 2/γ }[888−(ζ )+ p+(ζ )], ζ ∈ L , (2.59)



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Fig. 2 The weight functions WI,I(ξ), WI,VI(ξ), WVI,I(ξ), WVI,VI(ξ) and WII(ξ) for the parameters κ/µ = 1,λ/µ = 1 and γ /µ = 1. W0(ξ) is the classical elastic mode-I weight function

where e0 = (λ+ 2µ)/{µ(λ+ µ)}. The solution is immediately derived as

8+j (ζ ) = −

ejψ+j (ζ )

s+(ζ ), 8−

j (ζ ) = s−(ζ )ψ−j (ζ ), j = 1, 2, 3. (2.60)

Here, e1 = e2 = e0, e3 = 2/γ , and

ψj (ζ ) = 12π i

∫ ∞

−∞

p+j (ζ0)dζ0

s−(ζ0)(ζ0 − ζ ). (2.61)

By evaluating the integral ∫ ∞

−∞eiζ ξdζs−(ζ )

= (1 + i)

√2πξ, ξ > 0, (2.62)

we obtain that the modes I, II and VI share the same weight function

W0(ξ) = −√

2/(πξ). (2.63)



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Fig. 3 The weight functions WI,I, WI,VI, WVI,I, WVI,VI and WII versus the parameter γ /µ when ξ = 0.1,κ/µ = 1 and λ/µ = 1

It is also possible to pass to the limit κ → 0+ in the final formulas (2.44) and (2.49) for the weightfunctions obtained for the case κ > 0. As κ → 0+, the functions δj (ζ ) → 1 ( j = 0, 1) andr(ζ ) → |ζ |. Therefore, χ(ζ )± → 1 and r±(ζ ) → s±(ζ ). Formula (2.44) becomes identical to(2.63). Set now κ = 0 in (2.38) and (2.47). This gives a± = ±κ0 and

(vI,I vI,VIvVI,I vVI,VI

)(ζ, ξ) =

∫ ∞

−∞

(iζ0 κ2

01 iζ0

)eiζ0ξdζ0

s−(ζ0)(ζ0 − ζ ). (2.64)

By substituting these formulas into (2.49), we derive ultimately

WI,I = WVI,VI = −√

2/(πξ), WI,VI = WVI,I = 0. (2.65)

Figure 4 shows that if κ/µ → 0+, then the weight functions WI,VI and WVI,I vanish, while theother three functions tend to −√

2/(πξ) that is approximately equal to −2.5231 at ξ = 0.1.



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Fig. 4 The weight functions WI,I, WI,VI, WVI,I, WVI,VI and WII versus the parameter κ/µ when ξ = 0.1,γ /µ = 1 and λ/µ = 1

3. Antiplane-strain problem of micropolar elasticity

3.1 Derivation of the governing vector RHP

The balance laws of micropolar elasticity in the antiplane-strain case reduce to

∂τxz

∂x+ ∂τyz

∂y= 0,

∂mxx

∂x+ ∂m yx

∂y+ τyz − τzy = 0,

∂mxy

∂x+ ∂m yy

∂y− τxz + τzx = 0,

where τxz , τzx , τyz and τzy are the stress tensor components, and mxx , m yy , mxy and m yx are thecouple–stress tensor components. The constitutive and kinematic equations when combined lead tothe relations

τxz = (µ+ κ)∂uz

∂x+ κϕy, τzx = µ

∂uz

∂x− κϕy,

τyz = (µ+ κ)∂uz

∂y− κϕx , τzy = µ

∂uz

∂y+ κϕx ,



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mxx = δ∂ϕx

∂x+ α

∂ϕy

∂y, m yy = δ

∂ϕy

∂y+ α

∂ϕx

∂x,

mxy = β∂ϕx

∂y+ γ

∂ϕy

∂x, m yx = β

∂ϕy

∂x+ γ

∂ϕx

∂y, (3.1)

where δ = α + β + γ , and α, β, γ , κ and µ are material constants such that (28, 29)

γ > |β|, κ > 0, κ + 2µ > 0, 3α + β + γ > 0. (3.2)

The problem to be solved here consists in finding a solution {w,ϕx , ϕy} of the system of differentialequations

(µ+ κ)1uz + κ∂ϕy

∂x− κ

∂ϕx

∂y= 0,

δ∂2ϕx

∂x2 + γ∂2ϕx

∂y2 + (α + β)∂2ϕy

∂x∂y− 2κϕx + κ

∂uz

∂y= 0,

γ∂2ϕy

∂x2 + δ∂2ϕy

∂y2 + (α + β)∂2ϕx

∂x∂y− 2κϕy − κ

∂uz

∂x= 0, (3.3)

satisfying the boundary conditions

τyz = p1(x), m yx = p2(x), m yy = p3(x), 0 < x < ∞, y = 0±, (3.4)

and the continuity conditions

uz(x, 0+)− uz(x, 0−) = 0, ϕx (x, 0+)− ϕx (x, 0−) = 0,

ϕy(x, 0+)− ϕy(x, 0−) = 0, −∞ < x 6 0. (3.5)

The solution can be represented in terms of the Fourier integrals uzϕxϕy

(x, y) = 12π

∫ ∞

−∞

uz

ϕx

ϕy

(ζ ; y)e−iζ x dx, (3.6)

where the functions uz , ϕx and ϕy yield the solution to the system of ordinary differential equations

(µ+ κ)(u′′z − ζ 2uz)− κ(iζ ϕy + ϕ′

x ) = 0,

−δζ 2ϕx + γ ϕ′′x − iζ(α + β)ϕ′

y − 2κϕx + κ u′z = 0,

δϕ′′y − γ ζ 2ϕy − iζ(α + β)ϕ′

x − 2κϕy + iκζ uz = 0, y ∈ R1 \ {0}, (3.7)



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subject to the boundary conditions

(µ+ κ)u′z − κϕx = p1, −iζβϕy + γ ϕ′

x = p2, δϕ′y − iαζ ϕx = p3, y = 0±, (3.8)

where

pj (ζ ) = 8−j (ζ )+ p+

j (ζ ), p+j (ζ ) =

∫ ∞

0pj (x)eiζ x dx, j = 1, 2, 3,

8−1 (ζ ) =

∫ 0

−∞τyz(x, 0)eiζ x dx, 8−

2 (ζ ) =∫ 0

−∞m yx (x, 0)eiζ x dx,

8−3 (ζ ) =

∫ 0

−∞m yy(x, 0)eiζ x dx . (3.9)

This boundary value problem, similarly to the plane-strain case, can be reduced to a vector RHP.The general solution of the system (3.7) in the case 0 < y < ∞ is given by

uz = −2iC1

ζe−|ζ |y − iκC2

(κ + µ)ζe−r0 y,

ϕx = i |ζ |C1

ζe−|ζ |y + ir0C2

ζe−r0 y + iδζr1C3

2κ + δζ 2 e−r1 y,

ϕy = C1e−|ζ |y + C2e−r0 y + C3e−r1 y, (3.10)

where C1, C2 and C3 are arbitrary constants, and

rj =√ζ 2 + ρ2

j , j = 0, 1, ρ0 =√κ(κ + 2µ)γ (κ + µ)

> 0, ρ1 =√

2κδ> 0.

By employing the boundary conditions (3.8), we express the constants C j through pj ( j = 1, 2, 3).Then, we substitute them into (3.10) and fix y = 0. This gives us expressions for uz(ζ ; 0+),ϕx (ζ ; 0+) and ϕy(ζ ; 0+). After that we obtain similar expressions for the functions uz(ζ ; 0−),ϕx (ζ ; 0−) and ϕy(ζ ; 0−). This can be done by replacing |ζ |, r0 and r1 in the solution on the liney = 0+ by −|ζ |, −r0 and −r1, respectively. We then introduce the vector function

888+(ζ ) =∫ ∞

0

uz(x, 0+)− uz(x, 0−)ϕx (x, 0+)− ϕx (x, 0−)ϕy(x, 0+)− ϕy(x, 0−)

eiζ x dx, (3.11)

and, as in the plane-strain case, derive the following vector RHP:

888+(ζ ) = [h(ζ )]−1G(ζ )[888−(ζ )+ p+(ζ )], ζ ∈ L , (3.12)

where L is the real axis,

h(ζ ) = (β + γ )2r0

[(κ + 2µ)|ζ | + κζ 2

r1

]− 2[κ(κ + 2µ)+ (β + γ )(κ + µ)ζ 2]2

(κ + µ)|ζ |r1, (3.13)



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G(ζ ) = G11(ζ ) 0 iG13(ζ )

0 G22(ζ ) 0−iG13(ζ ) 0 G33(ζ )

,

G11(ζ ) = 2(β + γ )2(κ|ζ |κ + µ

− 2r0

)+ 2(κ + 2µ)[2κ + (β + γ )ζ 2]2

(κ + µ)ζ 2r1,

G13(ζ ) = 2κ(κ + 2µ)(κ + µ)ζ

[(β + γ )|ζ | − 2κ + (β + γ )ζ 2

r1

],

G22(ζ ) = 4κ(κ + 2µ)r0

|ζ |r1, G33(ζ ) = 2κ(κ + 2µ)

κ + µ

(κ + 2µ

|ζ | + κ

r1

).

As in the plane-strain problem, one of the equations (the second one) is separated from the others.Therefore, the m yx -mode (mode-V) is uncoupled, while the τyz- and m yy-modes, modes III and IV,respectively, are coupled.

3.2 Mode-V

The scalar RHP associated with mode-V has the form

8+2 (ζ ) = G22(ζ )

h(ζ )[8−

2 (ζ )+ p+2 (ζ )], ζ ∈ L . (3.14)

The analysis of the behaviour of the function h(ζ ) at infinity shows that

h(ζ ) = −κ(κ + 2µ)h1 + O(ζ−2), ζ → ∞, (3.15)

where

h1 = (β + γ )

(3α + 2β + 2γα + β + γ

− β

γ

). (3.16)

Because of the conditions (3.2), the parameter h1 is always positive. It will be convenient to intro-duce a new function

G◦22(ζ ) = −h1r0(ζ )G22(ζ )

4h(ζ ). (3.17)

This function is even, positive at ζ = 0 and at infinity,

G◦22(ζ ) ∼ h1

2γ> 0, ζ → 0, G◦

22(ζ ) = 1 + O(ζ−2), ζ → ∞, (3.18)

and does not have poles or zeros in the real axis. Therefore, it can be factorized in terms of theCauchy integral as follows:

G◦22(ζ ) = χ+(ζ )

χ−(ζ ), ζ ∈ L , (3.19)



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where

χ(ζ ) = exp{

12π i

∫ ∞

−∞ln G◦

22(ζ0)dζ0

ζ0 − ζ

}, ζ ∈ C±. (3.20)

Hence, the solution to the problem exists and is unique in the class of functions whose asymptoticsat infinity are

8+2 (ζ ) = O(ζ−3/2), 8−

2 (ζ ) = O(ζ−1/2), ζ → ∞. (3.21)

The solution to the RHP is given by

8+2 (ζ ) = χ+(ζ )

r+0 (ζ )

9+(ζ ), 8−2 (ζ ) = −h1

4r−

0 (ζ )χ−(ζ )9−(ζ ), (3.22)

where

9(ζ) = − 2π ih1

∫L

p+2 (ζ0)dζ0

r−0 (ζ0)χ−(ζ0)(ζ0 − ζ )

, (3.23)

and r±0 (ζ ) = (ζ ± iρ0)

1/2 are analytic functions in the half-planes C±, and arg(ζ + iρ0) ∈ [0, π ],arg(ζ − iρ0) ∈ [−π, 0]. As in the mode-II case, we derive expressions for the stress intensity factorKV and the weight function. They are

KV =√

2π limx→0−

√−xm yx (x, 0) =∫ ∞

0WV(ξ)p2(ξ)dξ,

WV(ξ) = i − 12π

∫ ∞

−∞eiζ ξdζ

r−0 (ζ )χ

−(ζ ), (3.24)

where p2(ξ) = m yx (ξ, 0±), 0 < ξ < ∞.

3.3 System of singular integral equations for modes III and IV: a finite crack

To find the functions 8±1 (ζ ) and 8±

3 (ζ ), we need to solve the following RHP:

ϕϕϕ+(ζ ) = [h(ζ )]−1 A(ζ )[ϕϕϕ−(ζ )+ p+(ζ )], ζ ∈ L , (3.25)

where A(ζ ) is the 2 × 2 matrix

A(ζ ) =(

G11(ζ ) iG13(ζ )

−iG13(ζ ) G33(ζ )

)= 2κ(κ + µ)|ζ | Q0(ζ )+ 2(κ + 2µ)

(κ + µ)ζ 2r1Q1(ζ )+ 4(β + γ )2r0 Q2,

Q j are the polynomial matrices

Q0(ζ ) =(

(β + γ )2ζ 2 i(κ + 2µ)(β + γ )ζ

−i(κ + 2µ)(β + γ )ζ (κ + 2µ)2

),

Q1(ζ ) =(

[2κ + (β + γ )ζ 2]2 −iκζ [2κ + (β + γ )ζ 2]iκζ [2κ + (β + γ )ζ 2] κ2ζ 2

), Q2 =

(−1 00 0

), (3.26)



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and

ϕϕϕ±(ζ ) =(8±

1 (ζ )

8±3 (ζ )

), p+(ζ ) =

(p+

1 (ζ )

p+3 (ζ )

). (3.27)

Unfortunately, the structure of the matrix A(ζ ) does not allow for exact factorization by methodsavailable in the literature, and therefore, modes III and IV cannot be decoupled. However, it ispossible to determine the partial indices of the RHP. The eigenvalues of the matrix A(ζ ) have theform

λ1 = G11 + G33

2+ 1

2

√10, λ2 = G11 + G33

2− 1

2

√10, (3.28)

where 10 = (G11 − G33)2 + 4G2

13. It can be directly verified that both eigenvalues are positive onthe real axis, and therefore, the Hermitian matrix A(ζ ) is positive definite. The partial indices, ν1 andν2, are equal to zero (25). This implies that the difference between the number of arbitrary constantsin the solution of the problem (3.25) and the number of additional conditions equals zero, and thesolution of the RHP is stable (27). This guarantees that if |A(ζ )− A(ζ )| < ε, −∞ < ζ < +∞, andA(ζ ) admits a factorization A(ζ ) = X+(ζ )[X−(ζ )]−1, then the factors X±(ζ ) approach the exactfactors X±(ζ ) of the matrix A(ζ ) as ε → 0.

By focusing on the case of a finite crack {0 < x < a, y = ±0}, we transform the associatedvector RHP into a system of singular integral equations. This system admits an efficient approximatesolution. On using the convolution theorem and integration by parts, we have(

p1(x)p3(x)

)= 1π

∫ a

0K (x − τ)

(χ1(τ )

χ3(τ )

)dτ, 0 < x < a, (3.29)

where

K (x) = − 12i

∫ ∞

−∞h(ζ )[A(ζ )]−1e−iζ x dζ

ζ,

[A(ζ )]−1 = 11(ζ)

(G33(ζ ) −iG13(ζ )

iG13(Gz) G11(ζ )

), 1(ζ ) = G11(ζ )G33(ζ )− G2

13(ζ ),

χ1(x) = ddx

[uz(x, 0+)− uz(x, 0−)

], χ3(x) = d

dx

[ϕy(x, 0+)− ϕy(x, 0−)

]. (3.30)

To understand the structure of the kernel, it is crucial to determine the asymptotics of the matrixh(ζ )[A(ζ )]−1 as ζ → 0 and ζ → ∞. We have

h(ζ )[A(ζ )]−1 ∼ −14

((κ + 2µ)|ζ | −i(β + γ )|ζ |2 sgn ζ

i(β + γ )|ζ |2 sgn ζ 4κρ1

), ζ → 0. (3.31)

Analysis of the entries of the matrix A(ζ ) yields

Gmm(ζ ) = G0mm

|ζ | + G1mm

|ζ |3 + O(|ζ |−5), m = 1, 3, G13(ζ ) = G013 sgn ζ|ζ |2 + O(|ζ |−4), ζ → ∞,



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where

G011 = 2κ(κ + 2µ)h1

κ + µ, G1

11 = κ + 2µκ + µ

[8κ2 − 4κ(β + γ )ρ2

1 + 34ρ4

1(β + γ )2]

+ ρ40

2(β + γ )2,

G033 = 4κ(κ + 2µ), G1

33 = −κ2(κ + 2µ)ρ2

1κ + µ

, G013 = κ(κ + 2µ)

κ + µ

[(β + γ )ρ2

1 − 4κ],

and therefore, as ζ → ∞,

h(ζ )[A(ζ )]−1 = 14

(−2(κ + µ)|ζ | + c0|ζ |−1 i (β+γ )ρ2

1−4κ2 sgn ζ + ic1(ζ |ζ |)−1

−i (β+γ )ρ21−4κ

2 sgn ζ − ic1(ζ |ζ |)−1 −h1|ζ | + c2|ζ |−1

),

where cj , ( j = 0, 1, 2) are real constants. Ultimately, we arrive at the following system of singularintegral equations:

1π

∫ a

0

[ν0

τ − x+ k11(x − τ)

]χ1(τ )dτ

+ 1π

∫ a

0

[ν1 ln |τ − x | + k13(x − τ)

]χ3(τ )dτ = p1(x), 0 < x < a,

− 1π

∫ a

0

[ν1 ln |τ − x | + k13(x − τ)

]χ1(τ )dτ

+ 1π

∫ a

0

[ν2

τ − x+ k33(x − τ)

]χ3(τ )dτ = p3(x), 0 < x < a, (3.32)

where ν0 = 12 (κ + µ), ν1 = 1

8 {(β + γ )ρ21 − 4κ}, ν2 = 1

4 h1, kmj (m, j = 1, 3) are regular kernelsgiven by

kmm(t)=∫ ∞

0Lmm(ζ ) sin ζ tdζ + ν∗

m−13(t), m = 1, 3,

k13(t)=∫ ∞

0L13(ζ ) cos ζ tdζ + ν1C + ν1

∞∑k=1

(−1)k t2k

2k(2k)!,

L11(ζ )= h(ζ )G33(ζ )

ζ1(ζ )+ ν0 + ν∗

0ψ+(ζ )ζ 2 , L33(ζ ) = h(ζ )G11(ζ )

ζ1(ζ )+ ν2 + ν∗

2ψ+(ζ )ζ 2 ,

L13(ζ )= h(ζ )G13(ζ )

ζ1(ζ )+ ν1ψ+(ζ )

ζ, ψ+(ζ ) =

{0, 0 < ζ < 1,1, 1 < ζ < ∞.

3(t)= t (C + ln |t | − 1)+∞∑

k=1

(−1)k t2k+1

2k(2k + 1)!, ν∗

0 = ν2G133 − δ∗G0

11, ν∗

2 = ν0G111 − δ∗G0

33,

δ∗ = δ2 − δ1

(G1

11

G011

+ G133

G033

− (G013)

2

G011G0

33

), δ1 = −κ(κ + 2µ)h1,

δ2 = (β + γ )2

4

[32ρ4

1 − κρ20ρ

21 − ρ4

0(κ + µ)

]− 2κ + µ

[κ2(κ + 2µ)2 − ρ2

1κ(κ + µ)(κ + 2µ)(β + γ )+ 38ρ4

1(κ + µ)2(β + γ )2].



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Here, the series representation of the cosine integral (30)

ci(t) = C + ln |t | +∞∑

k=1

(−1)k t2k

2k(2k)!(3.33)

(C is the Euler constant) was used.In order to solve the system of integral equations (3.32), we employ the method of orthogonal

polynomials. As a matter of utility, it will be desirable to have the system (3.32) in the interval(−1, 1) with respect to the functions χm(x) = χm(2x/a − 1). We expand the unknown functionsχm(t) in terms of the Chebyshev polynomials of the first kind Tn(x) as

χm(t) = 1√1 − t2

∞∑j=0

a(m)j Tj (t), m = 1, 3, (3.34)

where the coefficients a(m)j are to be determined. From the natural condition (it follows from thedefinition (3.30)) ∫ 1

−1χm(t)dt = 0, (3.35)

we immediately obtain that a(m)0 = 0, m = 1, 3. The next step of the method is the use of thespectral relations

1π

∫ 1

−1

Tj (t)dt√1 − t2(t − η)

={

0, j = 0,Uj−1(η), j = 1, 2, . . . , − 1 < η < 1,

1π

∫ 1

−1

Tj (t)√1 − t2

ln |t − η|dt ={ − ln 2, j = 0,

j−1Tj (η), j = 1, 2, . . . ,− 1 < η < 1,

the relation

∫ 1

−1Tj (t)Un(t)

√1 − t2dt =

π/2, j = n = 0,π/4, j = n = 0,

−π/4, j = n + 2,(3.36)

and the orthogonality condition for the Chebyshev polynomials of the second kind Un(t). Ulti-mately, the system of integral equations reduces to the following infinite system of linear algebraicequations:

πν0

2a(1)n+1 +

∞∑j=1

c(1,1)nj a(1)j − ν1[σna(3)n − σn+2a(3)n+2] +∞∑

j=1

c(1,3)nj a(3)j = b(1)n , n = 0, 1, . . . ,

πν2

2a(3)n+1 +

∞∑j=1

c(3,3)nj a(3)j + ν1[σna(1)n − σn+2a(1)n+2] −∞∑

j=1

c(1,3)nj a(1)j = b(3)n , n = 0, 1, . . . .

(3.37)



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Here,

σ0a(1)0 = σ0a(3)0 = 0, σn = π

4n(n = 1, 2 . . .),

c(m,l)nj = 1π

∫ 1

−1

∫ 1

−1

Tj (t)√1 − t2

√1 − η2Un(η)kml(η − t)dηdt,

b(m)n =∫ 1

−1

√1 − η2Un(η) pm(η)dη, pm(t) = pm

(a2(t + 1)

),

kmm(t)= a2

kmm

(at2

), k13(t) = a

2

[ν1 ln

a2

+ k13

(at2

)].

By using the properties of the kernels and the Chebyshev polynomials, we find that

c(11)nj = c(33)

nj = 0 if n + j = 2, 4, 6, . . . and c(13)nj = 0 if n + j = 1, 3, 5, . . . .

It can be verified (31) that the non-zero coefficients decay as n or j grow (c(ml)nj = o(n−1) as n → ∞

when j is fixed, and c(ml)nj = o( j−1) as j → ∞ when n is fixed, m, l = 1, 3), and the infinite system

(3.37) can be solved approximately by the reduction method.

3.4 Weight functions for modes III and IV

Determine now the stress intensity factors and the weight functions pertaining to them. At the tipsof the crack, the stress τyz and the couple–stress m yy have the square root singularity:

τyz(x, 0) ∼ K (0)III√−2πx

, m yy(x, 0) ∼ K (0)IV√−2πx

, x → 0−,

τyz(x, 0) ∼ K (a)III√

2π(x − a), m yy(x, 0) ∼ K (a)

IV√2π(x − a)

, x → a+. (3.38)

On the other hand, as x → 0−, we have

τyz(x, 0) ∼ ν0

π

∫ a

0

χ1(τ )dττ − x

, m yy(x, 0) ∼ ν2

π

∫ a

0

χ3(τ )dττ − x

, x → 0−. (3.39)

On replacing the functions χm(τ ) by χm(t) = χ(a(t + 1)/2), using (3.34) and the asymptoticformula for the singular integral

I (x ′) = 1π

∫ 1

0

Tj (2τ ′ − 1)dτ ′√τ ′(1 − τ ′)(τ ′ − x ′)

∼ (−1) j (−x ′)−1/2, x ′ → 0−, (3.40)

we obtain

τyz ∼ ν0

2

(− x

a

)−1/2 ∞∑j=1

(−1) j a(1)j , m yy ∼ ν2

2

(− x

a

)−1/2 ∞∑j=1

(−1) j a(3)j , x → 0−. (3.41)

By comparing formulas (3.39) and (3.41), we find that

K (0)III = ν0

2

√2πa

∞∑j=1

(−1) j a(1)j , K (0)IV = ν2

2

√2πa

∞∑j=1

(−1) j a(3)j . (3.42)



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Similarly,

K (a)III = −ν0

2

√2πa

∞∑j=1

a(1)j , K (a)IV = −ν2

2

√2πa

∞∑j=1

a(3)j . (3.43)

Having determined formulas for the stress intensity factors K (c)III and K (c)

IV (c = 0, a), we nowintroduce the associated weight functions W (0)

j,m and W (a)j,m , j,m = III, IV, such that(

K (c)III

K (c)IV

)=∫ a

0

(W (c)

III,III(ξ) W (c)III,IV(ξ)

W (c)IV,III(ξ) W (c)

IV,IV(ξ)

)(p1(ξ)

p3(ξ)

)dξ, c = 0, a. (3.44)

By choosing the functions p1(x) and p3(x) as p1(x) = δ(x − ξ) and p3(x) = 0 (δ(x) is thegeneralized δ-function), we find from (3.44)

W (c)III,III(ξ) = K (c,1)

III , W (c)IV,III(ξ) = K (c,1)

IV , c = 0, a, (3.45)

where K (c,1)III = K (c)

III and K (c,1)IV = K (c)

IV are expressed through the solution a(m)n (n = 1, 2, . . . ; m =1, 3) of the infinite system (3.37) by (3.42) and (3.43). This system has to be solved with the specialright-hand side

b(1)n = qn, b(3)n = 0, qn = (4/a2)√ξ(a − ξ)Un(2[ξ/a] − 1). (3.46)

By solving next the infinite system (3.37) with the right-hand side

b(1)n = 0, b(3)n = qn, (3.47)

we determine the associated coefficients a(m)n (n = 1, 2, . . . ,m = 1, 3). By means of (3.42) and(3.43), the coefficients K (c,3)

III = K (c)III and K (c,3)

IV = K (c)IV may be computed and identified as the

other weight functions

W (c)III,IV(ξ) = K (c,3)

III , W (c)IV,IV(ξ) = K (c,3)

IV , c = 0, a. (3.48)

Analyse now the series representations of the the weight functions. Consider first W (c)III,III and

W (c)IV,III, c = 0, a. In this case, b(1)n = qn and b(3)n = 0. From the infinite system (3.37),

a(1)n = 2πν0

qn−1 + a(1)n , a(3)n = a(3)n + a(3)n ,

a(3)n = − 4ν1

π2ν0ν2(σn−1qn−2 − σn+1qn), a(m)n = o(n−1), m = 1, 3, n → ∞. (3.49)

Here, σ0q−1 = 0, a(m)0 = a(m)0 = 0, m = 1, 3. Introduce next the power series

F(z) = ν0

√πa2

∞∑j=1

(−1) j a(1)j z j , (3.50)

which is an analytic function of z in the open disc |z| < 1. On substituting a(1)n = 2(πν0)−1qn−1 +

a(1)n along with (3.46) into (3.50) and employing the generating function for the Chebyshev polyno-mials Un(x)



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∞∑j=0

Uj (x)z j = 11 − 2xz + z2 , |z| < 1, (3.51)

we obtain

F(z) = ν0

√πa2

[− 8zπaν0

√ξ(a − ξ)

a + 2(2ξ − a)z + az2 +∞∑

j=1

(−1) j a(1)j z j]. (3.52)

By evaluating the limit of F(z) as z → 1−, we determine the factor K (0)III for the loading functions

p1(x) = δ(x − ξ), p3(x) = 0. That factor is the weight function W (0)III,III given by

W (0)III,III = −

√2(a − ξ)

πaξ+ ν0

√πa2

∞∑j=1

(−1) j a(1)j , (3.53)

and a(1)j = o( j−1) as j → ∞. Similarly,

W (a)III,III = −

√2ξ

πa(a − ξ)− ν0

√πa2

∞∑j=1

a(1)j . (3.54)

We wish to evaluate next the weight functions W (c)IV,III or, equivalently, the factors K (c)

IV , c = 0, a,as p1(x) = δ(x − ξ), p3(x) = 0. Since

∞∑j=1

(σj−1qj−2 − σj+1qj ) = σ1q0, (3.55)

from (3.42) and (3.49),

W (0)IV,III = −2ν1

ν0a

√2ξ(a − ξ)

πa+ ν2

√πa2

∞∑j=1

(−1) j a(3)j ,

W (a)IV,III = 2ν1

ν0a

√2ξ(a − ξ)

πa− ν2

√πa2

∞∑j=1

a(3)j .

By solving the system (3.37) with b(1)n = 0 and b(3)n = qn , we determine a new set of the coefficientsa(1)n and a(3)n . These coefficients admit the representations

a(3)n = 2πν2

qn−1 + a(3)n , a(1)n = a(1)n + a(1)n ,

a(1)n = 4ν1

π2ν0ν2(σn−1qn−2 − σn+1qn), a(m)n = o(n−1), m = 1, 3, n → ∞.



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Similarly to the previous case,

W (0)III,IV = 2ν1

ν2a

√2ξ(a − ξ)

πa+ ν0

√πa2

∞∑j=1

(−1) j a(1)j ,

W (a)III,IV = −2ν1

ν2a

√2ξ(a − ξ)

πa− ν0

√πa2

∞∑j=1

a(1)j ,

W (0)IV,IV = −

√2(a − ξ)

πaξ+ ν2

√πa2

∞∑j=1

(−1) j a(3)j ,

W (a)IV,IV = −

√2ξ

πa(a − ξ)− ν2

√πa2

∞∑j=1

a(3)j . (3.56)

It is seen that the weight functions are expressed in terms of the new coefficients a(1)n and a(3)n . Todetermine these coefficients associated with the weight functions W (c)

III,III and W (c)IV,III (c = 0, a), one

needs to solve the infinite system (3.37) with the right-hand side b(1)n = b(1)n and b(3)n = b(3)n , where

b(1)n = − 4ν21

π2ν0ν2

[σn(σn−1qn−2 − σn+1qn)− σn+2(σn+1qn − σn+3qn+2)

]− 2πν0

∞∑j=1

c(1,1)nj qj−1 + 4ν1

π2ν0ν2

∞∑j=1

c(1,3)nj (σj−1qj−2 − σj+1qj ),

b(3)n = 2πν0

∞∑j=1

c(1,3)nj qj−1 + 4ν1

π2ν0ν2

∞∑j=1

c(3,3)nj (σj−1qj−2 − σj+1qj ), (3.57)

where σ0 = σ−1 = 0. To make computations of the weight functions W (c)III,IV and W (c)

IV,IV (c = 0, a)by means of (3.56), we have to replace the right-hand side in (3.37) by b(1)n and b(3)n given by

b(1)n = − 4ν1

π2ν0ν2

∞∑j=1

c(1,1)nj (σj−1qj−2 − σj+1qj )− 2πν2

∞∑j=1

c(1,3)nj qj−1,

b(3)n = − 4ν21

π2ν0ν2

[σn(σn−1qn−2 − σn+1qn)− σn+2(σn+1qn − σn+3qn+2)

]+ 4ν1

π2ν0ν2

∞∑j=1

c(1,3)nj (σj−1qj−2 − σj+1qj )− 2πν2

∞∑j=1

c(3,3)nj qj−1 (3.58)

and solve the infinite system for the coefficients a(1)n = a(1)n and a(3)n = a(3)n . It can be directlyverified that in both cases, (3.57) and (3.58), the coefficients b(1)n and b(3)n decay as n → ∞, andb(m)n = o(n−1), n → ∞, m = 1, 3.

The diagonal weight functions W (c)III,III(ξ) and W (c)

IV,IV(ξ) (c = 0, a) as functions of ξ when α/µ =1, β/µ = 1, γ /µ = 2 and κ/µ = 1 are plotted in Fig. 5. The off-diagonal functions are presented inFig. 6. It is seen that when ξ → c, then the absolute values of the diagonal functions W (c)

III,III(ξ) andW (c)

IV,IV(ξ) grow to infinity, while the absolute values of the other two diagonal functions become



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Fig. 5 The weight functions W (0)III,III(ξ), W (a)

III,III(ξ), W (0)IV,IV(ξ) and W (a)

IV,IV(ξ), for the parameters a = 1,α/µ = 1, β/µ = 1, γ /µ = 2 and κ/µ = 1

small. The absolute values of the four off-diagonal functions are small as ξ approach both tips ofthe crack. Figure 7 shows the effect on the weight functions of changing the parameter κ whenξ = 0.5a and therefore, W (0)

mj = W (a)mj , m, j = III, IV. Regardless of ξ , as κ → 0+, the diagonal

functions W (c)III,III and W (c)

IV,IV tend to the classical elastic mode-I weight functions for a finite crack,while the off-diagonal functions W (c)

III,IV and W (c)IV,III vanish.

4. Conclusions

We have studied two 2D problems of a crack in an unbounded solid modelled in the frameworkof the unconstrained couple–stress theory (micropolar elasticity). In the first problem, the crack issemi-infinite, and the conditions of plane strain are assumed. That problem has been reduced to ascalar RHP for mode-II (the shear τyx -mode) and an order-2 vector RHP for modes I and VI, the σy-and m yz-modes, respectively. These two problems have been solved by quadratures. We have foundthe stress intensity factors and the five associated weight functions, one for the uncoupled mode andtwo for each of the coupled modes. It has been shown that if the micropolar parameter κ → 0+



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Fig. 6 The weight functions W (0)III,IV(ξ), W (a)

III,IV(ξ), W (0)IV,III(ξ) and W (a)

IV,III(ξ), for the parameters a = 1,α/µ = 1, β/µ = 1, γ /µ = 2 and κ/µ = 1

and γ = 0, then the stress intensity factors and the diagonal weight functions WI,I, WII and WVI,VItend to the function W0(ξ) = −√

2/(πξ), the weight function for the modes I and II in the classicalelasticity, while the off-diagonal functions WI,VI and WVI,I vanish. If γ → 0+ and κ = o(γ ), thenthe couple–stress m yz → 0, WI,I and WII tend to W0(ξ), and the other weight functions vanish.

The second model problem concerns the antiplane-strain problem of micropolar elasticity for acrack subject to loading τyz = p1(x), m yx = p2(x), m yy = p3(x). We have shown that the modeV (the m yx -mode) is uncoupled and found the associated weight function WV by quadratures. Theother two modes, III and IV (the τyz- and m yy-modes), are coupled. It turned out that to find theweight functions for these modes, we needed to solve a vector RHP with the coefficient admittedthe representation G = b0 Q0 +b1 Q1 +b2 Q2 (bj are Holder functions and Q j are 2×2 polynomialmatrices). We were not able to factorize this matrix. However, we have managed to determine thatit has zero partial indices. Therefore, they are stable, and the number of arbitrary constants in thesolution of the RHP coincides with the number of solvability conditions.

For a finite antiplane-strain crack, we have derived a system of singular integral equations andsolved it approximately by the method of orthogonal polynomials. The weight functions have beenfound in a series form in terms of the solution of an infinite system of linear algebraic equations. Asin the plane-strain case, if the micropolar parameter κ → 0+, while the other parameters meet the



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Fig. 7 The weight functions W (0)III,III(ξ), W (0)

III,IV(ξ), W (0)IV,III(ξ) and W (0)

IV,IV(ξ) versus κ/µ when ξ = 0.5a,a = 1, α/µ = 1, β/µ = 1 and γ /µ = 2

physical conditions γ > |β|, µ > 0, 3α + β + γ > 0, then the weight functions WIII,III and WIV,IVassociated with the tips of the crack coincide with the classical elastic weight functions for a finitemode-III crack. The other weight functions, WIII,IV and WIV,III, tend to zero.

Acknowledgements

This work was partly funded by National Scientific Foundation through grant DMS0707724. Theauthor is thankful to P. Schiavone and I. M. Spitkovsky for discussions of the paper.

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