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http://mms.sagepub.com/ Mathematics and Mechanics of Solids http://mms.sagepub.com/content/early/2012/12/20/1081286512462182 The online version of this article can be found at: DOI: 10.1177/1081286512462182 published online 20 December 2012 Mathematics and Mechanics of Solids YA Antipov and AV Smirnov Subsonic propagation of a crack parallel to the boundary of a half-plane Published by: http://www.sagepublications.com can be found at: Mathematics and Mechanics of Solids Additional services and information for http://mms.sagepub.com/cgi/alerts Email Alerts: http://mms.sagepub.com/subscriptions Subscriptions: http://www.sagepub.com/journalsReprints.nav Reprints: http://www.sagepub.com/journalsPermissions.nav Permissions: What is This? - Dec 20, 2012 OnlineFirst Version of Record >> at LOUISIANA STATE UNIV on January 16, 2013 mms.sagepub.com Downloaded from
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http://mms.sagepub.com/Mathematics and Mechanics of Solids

http://mms.sagepub.com/content/early/2012/12/20/1081286512462182The online version of this article can be found at:

 DOI: 10.1177/1081286512462182

published online 20 December 2012Mathematics and Mechanics of SolidsYA Antipov and AV Smirnov

Subsonic propagation of a crack parallel to the boundary of a half-plane  

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Article

Subsonic propagation of a crackparallel to the boundary of ahalf-plane

Mathematics and Mechanics of Solids1–15©The Author(s) 2012Reprints and permission:sagepub.co.uk/journalsPermissions.navDOI: 10.1177/1081286512462182mms.sagepub.com

YA Antipov and AV SmirnovDepartment of Mathematics, Louisiana State University, USA

Received 19 June 2012; accepted 20 June 2012

AbstractA two-dimensional steady-state model problem on a half-plane with a semi-infinite crack propagating at constant speedparallel to the boundary of the half-plane is considered. The crack faces are subjected to normal and tangential loads,while the boundary of the half-plane is free of traction. The problem is formulated as an order-2 vector Riemann–Hilbertproblem and then reduced to a system of singular integral equations in a semi-infinite segment with respect to thederivatives of the displacement jumps. The solution to the system of integral equations is represented in a series form interms of an orthonormal basis of the associated Hilbert space, the orthonormal Jacobi polynomials. The coefficients ofthe expansions solve an infinite system of linear algebraic equations of the second kind. The stress intensity factors andthe weight functions are determined and computed. The Griffith energy criterion is applied to derive a crack growthcriterion in terms of the KI - and KII -stress intensity factors, the crack speed, the shear and dilatational waves speedsand the Griffith material constant.

KeywordsGriffith energy criterion, orthogonal polynomials, subsonic crack propagation, systems of singular integral equations

1. IntroductionThe main motivation of this work is to study the boundary effects on a crack propagating parallel to the boundaryof a solid. The static problem for a semi-infinite crack parallel to the boundary of a half-plane was analyzed byZlatin and Khrapkov [1]. They transformed the model problem into an order-2 vector Riemann–Hilbert (RH)problem and managed to derive a closed-form solution by exact factorization of the matrix coefficient of theRH problem. The steady-state problem for a plane with a semi-infinite crack −∞ < x1 < 0, x2 = 0 driven bymoving normal and tangential loads applied to the crack faces was considered by Craggs [2]. Because of thesymmetry, the Craggs problem admits decoupling and can be solved in closed form by a variety of methodsincluding the factorization method for a scalar RH problem, the Mellin transform method, which bypasses theRH problem, and the method of orthogonal polynomials. Many researchers analyzed different aspects of theCraggs model problem and considered its generalizations. Surveys of the results were given by Freund [3]and Broberg [4]. To our knowledge, no analytical solution for the steady-state problem on a coupled mode-I,IIsemi-infinite crack propagating parallel to the boundary of a half-plane is available in the literature.

In Section 2, we state the model problem for a half-plane {−∞ < x1 < ∞, −∞ < x2 < δ} (δ > 0) with acrack extending from x1 = −∞, x2 = 0 to x1 = Vt, x2 = 0 and driven by normal and tangential componentsof traction applied to the crack faces. It is assumed that the loads move at the same speed V . By employing themethod of integral transformations, we map the boundary-value problem for the governing system of partial

Corresponding author:YA Antipov, Department of Mathematics, Louisiana State University, Baton Rouge, LA 70803, USA.Email: [email protected]

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2 Mathematics and Mechanics of Solids

differential equations to a vector RH problem with the matrix coefficient G(p) = a(p)I + b(p)S(p), where a(p),b(p) are Hölder functions, I = diag{1, 1},

S(p) =(

1 −iα1 tanhα0piα2 tanhα0p −1

), (1)

and α0,α1 and α2 are real nonzero constants. At this stage, all our efforts availed us no results in factorizing thematrix G(p).

In Section 3, we transform the RH problem into a system of two singular integral equations first on thesemi-infinite segment (−∞, 0) and then on the finite segment (−1, 1). Its solution is found in the Hilbert spaceL2,ρ(−1, 1), ρ = (1 + ζ )1/2(1 − ζ )−1/2. By expanding the unknown functions in terms of an orthonormal basisof this space, we reduce the system of integral equations to an infinite system of linear algebraic equations ofthe second kind. Its solution is found by the reduction method.

In Section 4, we derive the stress intensity factors (SIFs) and the weight functions. In addition, we determinethe energy released when the crack extends from x to x + r, and r is small. Then we apply the Griffith criterionand derive a Willis-type formula for a mode-I,II semi-infinite crack propagating along the boundary of a half-plane. The criterion of steady-state propagation of a semi-infinite crack in a plane was derived by Willis [5]. Wealso discuss some numerical results obtained for the SIFs, the weight functions and the Griffith criterion.

2. Formulation and a vector RH problemA semi-infinite crack that occupies, at time t, the surface S2(t) = {x : −∞ < x1 < Vt, x2 = 0, |x3| < ∞}propagates in the direction parallel to the boundary of an elastic half-space R3

+ = {x : |x1| < ∞, −∞ < x2 <δ, |x3| < ∞} (δ > 0). The boundary of the half-space is assumed to be free of traction. The speed V is constantand V < cR, where cR is the Rayleigh wave speed for the elastic solid whose density and the Lame constantsare ρ, λ and μ, respectively. The banks of the crack are subjected to plane-strain loading

σj2 = σ ◦j2(x1 − Vt), −∞ < x1 − Vt < 0, x2 = 0±, j = 1, 2. (2)

The system of equations for two-dimensional dynamic elasticity is the following

c2l�φ − ∂2φ

∂t2= 0, c2

s�ψ − ∂2ψ

∂t2= 0, x ∈ R2

+ \ S1(t), (3)

where cl and cs are the longitudinal and shear wave speeds

cl =√λ+ 2μ

ρ, cs =

√μ

ρ, (4)

S1(t) = {x : −∞ < x1 < Vt, x2 = 0}, R2+ = {x : |x1| < ∞, −∞ < x2 < δ}, and φ and ψ are dynamic

potentials such that the displacements u1, u2 and the stresses σ11, σ12 are

u1 = ∂φ

∂x1+ ∂ψ

∂x2, u2 = ∂φ

∂x2− ∂ψ

∂x1,

σ12 = μ

(2∂2φ

∂x1∂x2− ∂2ψ

∂x21

+ ∂2ψ

∂x22

),

σ22 = λ�φ + 2μ

(∂2φ

∂x22

− ∂2ψ

∂x1∂x2

). (5)

The problem is steady state, and the advantage of introducing the moving coordinates x = x1 − Vt, y = x2 isevident. The governing equations (3) are simplified as

α2 ∂2φ

∂x2+ ∂2φ

∂y2= 0, β2 ∂

∂x2+ ∂2ψ

∂y2= 0, (6)

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Antipov and Smirnov 3

where α =√

1 − V 2/c2l , β = √

1 − V 2/c2s . On applying the Fourier transform

φ(p, y) =∫ ∞

−∞φ(x, y)eipxdx, ψ(p, y) =

∫ ∞

−∞ψ(x, y)eipxdx (7)

to the differential equations (6), we find for y < 0,

φ(p, y) = C0(p)eα|p|y, ψ(p, y) = D0(p)eβ|p|y. (8)

In the case 0 < y < δ

φ(p, y) = C1(p) coshαpy + C2(p) sinhαpy,

ψ(p, y) = D1(p) coshβpy + D2(p) sinhβpy. (9)

Here, the functions Cj(p) and Dj(p) (j = 0, 1, 2) are to be determined from the boundary conditions

σxy = σyy = 0, |x| < ∞, y = δ,

σxy = σ ◦12(x), σyy = σ ◦

22(x), x < 0, y = 0±,

∂u

∂x(x, 0+) − ∂u

∂x(x, 0−) = χ1(x),

∂v

∂x(x, 0+) − ∂v

∂x(x, 0−) = χ2(x), |x| < ∞, (10)

u(x1 − Vt, x2) = u1(x1, x2, t), v(x1 − Vt, x2) = u2(x1, x2, t), σxy(x1 − Vt, x2) = σ12(x1, x2, t), and σyy(x1 − Vt, x2) =σ22(x1, x2, t). The functions χ1(x), χ2(x) vanish for x > 0 and are unknown otherwise. The Fourier-transformedboundary conditions read

d2ψ

dy2+ p2ψ − 2ip

dy= 0,

c2l

c2s

d2φ

dy2+ p2

(2 − c2

l

c2s

)φ + 2ip

dy= 0, y = δ,

d2ψ

dy2+ p2ψ − 2ip

dy= �+

1 (p) +�−1 (p), y = 0±,

c2l

c2s

d2φ

dy2+ p2

(2 − c2

l

c2s

)φ + 2ip

dy= �+

2 (p) +�−2 (p), y = 0±,

(dψ

dy− ipφ

)y=0+

−(

dy− ipφ

)y=0−

= X −1 (p)

p,

(dφ

dy+ ipψ

)y=0+

−(

dy+ ipψ

)y=0−

= X −2 (p)

p, (11)

where the functions �−1 (p), �−

2 (p) are known,

�−j (p) =

∫ 0

−∞σ ◦

j2(x, 0)eipxdx, j = 1, 2, (12)

while the other functions

�+1 (p) = 1

μ

∫ ∞

0σxy(x, 0)eipxdx, �+

2 (p) = 1

μ

∫ ∞

0σyy(x, 0)eipxdx,

X −j (p) = i

∫ 0

−∞χj(x)eipxdx, j = 1, 2, (13)

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4 Mathematics and Mechanics of Solids

are to be determined. Six out of the eight equations in (11) determine the unknown functions Cj(p), Dj(p)(j = 0, 1, 2). The other two equations constitute the following vector RH problem(

�+1 (p)

�+2 (p)

)= G(p)

(X −

1 (p)X −

2 (p)

)−(�−

1 (p)�−

2 (p)

), −∞ < p < +∞, (14)

where G(p) is a Hermitian matrix

G(p) =(

g11(p) ig12(p)−ig12(p) g22(p)

), (15)

whose entries are given by

g11(p) = e−(α+β)δ|p|

2β(1 − β2)

[R1 sinh(α + β)δp − R2 sinh(α − β)δp + 2d(p) sgn p

R1

],

g12(p) = 4R2(1 + β2)

R1(1 − β2)e−(α+β)δ|p| sinh2 (α − β)δp

2,

g22(p) = e−(α+β)δ|p|

2α(1 − β2)

[R1 sinh(α + β)δp + R2 sinh(α − β)δp + 2d(p) sgn p

R1

],

d(p) = R21 sinh2 (α + β)δp

2− R2

2 sinh2 (α − β)δp

2,

R1 = (1 + β2)2 − 4αβ, R2 = (1 + β2)2 + 4αβ. (16)

It can be shown that the eigenvalues of the matrix G(p), λ1(p) and λ2(p), do not satisfy the conditionλ1(p)λ2(p) > 0 for all p ∈ (−∞, ∞), and the matrix G(p) cannot be factorized such that G(p) = g0(p)G0(p),where g0(p) is a scalar, and the matrix G0(p) is positive definite for all real p. The vector RH problem may berewritten as (

�+1 (p)

�+2 (p)

)= G(p)

(β−1X −

1 (p)α−1X −

2 (p)

)−(�−

1 (p)�−

2 (p)

), −∞ < p < +∞, (17)

where the new matrix coefficient has the structure

G(p) = a(p)I + b(p)S(p),

a(p) = e−(α+β)δ|p|

2(1 − β2)

[R1 sinh(α + β)δp + 2d(p) sgn p

R1

],

b(p) = − R2

2(1 − β2)e−(α+β)δ|p| sinh(α − β)δp,

S(p) =(

1 − 4iαR1

(1 + β2) tanh 12 (α − β)δp

4iβR1

(1 + β2) tanh 12 (α − β)δp −1

), (18)

and I is the unit 2 × 2 matrix. The matrix G(p) does not admit a Wiener–Hopf factorization by the methodscurrently available in the literature.

3. System of integral equationsIn this section we aim to map the RH problem into a system of singular integral equations on a semi-infinitesegment and develop an efficient numerical scheme for its solution. By applying the inverse Fourier transform tothe RH problem, using the convolution theorem and evaluating the integrals, we find the following expressionsvalid for −∞ < x < ∞

σxy(x, 0) = − γ

πβ

{∫ 0

−∞

[1

ξ − x+ k11(ξ − x)

]χ1(ξ )dξ +

∫ 0

−∞k12(ξ − x)χ2(ξ )dξ

},

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Antipov and Smirnov 5

σyy(x, 0) = − γ

πα

{∫ 0

−∞k21(ξ − x)χ1(ξ )dξ +

∫ 0

−∞

[1

ξ − x+ k22(ξ − x)

]χ2(ξ )dξ

}, (19)

where

k11(x) = R(1 − R)x

2[x2 + 4α2δ2]− R(1 + R)x

2[x2 + 4β2δ2]+ (R2 − 1)x

x2 + (α + β)2δ2,

k12(x) = 4R(1 + β2)β

R1

x2 + 4α2δ2+ β

x2 + 4β2δ2− α + β

x2 + (α + β)2δ2

),

k21(x) = −αβ

k12(x),

k22(x) = − R(1 + R)x

2[x2 + 4α2δ2]+ R(1 − R)x

2[x2 + 4β2δ2]+ (R2 − 1)x

x2 + (α + β)2δ2,

R = R2

R1, γ = μR1

2(1 − β2). (20)

For −∞ < x < 0, the traction vector components σxy(x, 0) and σyy(x, 0) are known (see (10)). Slightlyrearranged, the expressions (19) give the resulting system of singular integral equations

∫ 0

−∞

[1

ξ − x+ k11(ξ − x)

]χ1(ξ )dξ +

∫ 0

−∞k12(ξ − x)χ2(ξ )dξ = −π f1(x), −∞ < x < 0,

∫ 0

−∞k21(ξ − x)χ1(ξ )dξ +

∫ 0

−∞

[1

ξ − x+ k22(ξ − x)

]χ2(ξ )dξ = −π f2(x), −∞ < x < 0, (21)

where

f1(x) = β

γσ ◦

12(x), f2(x) = α

γσ ◦

22(x). (22)

For δ = ∞ (the half-plane becomes a plane), the kernels kij(x) (i, j = 1, 2) vanish, the system decouples, and itsclosed-form solution is

χj(x) = 1

π√−x

∫ 0

−∞

√−ξ fj(ξ )dξ

ξ − x, −∞ < x < 0, j = 1, 2. (23)

In the case of finite values of δ we propose an approximate scheme based on the method of orthogonal poly-nomials. First, we transform the system into another one on the segment (−1, 1). Introduce new variables andfunctions

ξ = η + 1

η − 1, x = ζ + 1

ζ − 1,

(1 − ζ )χj(ζ ) = χj(x), (1 − ζ )fj(ζ ) = fj(x), j = 1, 2. (24)

This brings us to the system ∫ 1

−1

[− 1

η − ζ+ 2kjj(ξ − x)

(1 − η)(1 − ζ )

]χj(η)dη

+∫ 1

−1

2kj3−j(ξ − x)

(1 − η)(1 − ζ )χ3−j(η)dη = −π fj(ζ ), −1 < ζ < 1, j = 1, 2. (25)

Let now L2,ρ(μ,ν)(a, b) be the Hilbert space of functions with the norm defined by

||χ (x)|| =(∫ b

a|χ (x)|2ρ(μ, ν)dx

)1/2

, ρ(μ, ν) = (x − a)μ(b − x)ν . (26)

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6 Mathematics and Mechanics of Solids

As in the case δ = ∞, we seek the solution, the functions χ1(x) and χ2(x), in the space L2,ρ(−1/2,−3/2)(0, ∞). Atzero and infinity, the functions behave as

χj(x) ∼ C0jx−1/2, x → 0; χj(x) ∼ C1jx

−3/2, x → ∞ (27)

(C0j and C1j are nonzero constants). When written for the new functions χ1(ζ ) and χ2(ζ ), this results

χj(ζ ) ∼ C0j(1 + ζ )−1/2, ζ → −1; χj(ζ ) ∼ C1j(1 − ζ )1/2, ζ → 1, (28)

and the functions χ1(ζ ) and χ2(ζ ) belong to the space L2,ρ(−1/2,1/2)(−1, 1). The functions (1 − ζ )1/2

(1 + ζ )−1/2P1/2,−1/2n (ζ ) form an orthonormal basis of the space L2,ρ(−1/2,1/2)(−1, 1). Here, P1/2,−1/2

n (ζ ) are theorthonormal Jacobi polynomials. Therefore, the functions χ1(ζ ) and χ2(ζ ) can be expanded in terms of the basisfunctions as

χj(ζ ) = (1 − ζ )1/2(1 + ζ )−1/2∞∑

m=0

a(j)m P1/2,−1/2

m (ζ ), j = 1, 2, −1 < ζ < 1, (29)

where the coefficients a(j)n are to be determined. The original functions χ1(x) and χ2(x) may be put into the form

χj(x) = 2

(1 − x)√−x

∞∑m=0

a(j)m P1/2,−1/2

m

(x + 1

x − 1

), j = 1, 2, −∞ < x < 0. (30)

By employing next the spectral relation

∫ 1

−1

(1 − η

1 + η

)1/2 P1/2,−1/2m (η)dη

η − ζ= −π P−1/2,1/2

m (ζ ), −1 < ζ < 1, (31)

and the orthonormality of the polynomials,

∫ 1

−1

(1 + ζ

1 − ζ

)1/2

P−1/2,1/2m (ζ )P−1/2,1/2

n (ζ )dζ = δmn, n, m = 0, 1, . . . (32)

(δmn is the Kronecker symbol), we transform the system of integral equations into an infinite algebraic system

a(j)n +

∞∑m=0

[c(j,j)nm a(j)

m + c(j,3−j)nm a(3−j)

m ] = b(j)n , n = 0, 1, . . . ; j = 1, 2. (33)

Here,

c(j,l)nm = 2

π

∫ 1

−1

∫ 1

−1

√1 + ζ

(1 − ζ )3

P−1/2,1/2n (ζ )P1/2,−1/2

m (η)√1 − η2

kjl(ξ − x)dηdζ ,

b(j)n = −

∫ 1

−1

√1 + ζ

(1 − ζ )3P−1/2,1/2

n (ζ )fj(x)dζ . (34)

In order to compute the coefficients c(j,l)nm we rearrange the integrands as√

1 + ζ

(1 − ζ )3

kjj(ξ − x)√1 − η2

=√

1 + ζ

1 − ζ

√1 − η

1 + ηkjj(η, ζ ), j = 1, 2,

√1 + ζ

(1 − ζ )3

k12(ξ − x)√1 − η2

=√

1 − ζ 2

√1 − η

1 + ηk12(η, ζ ), (35)

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Antipov and Smirnov 7

where

kjj(η, ζ ) = (R2 − 1)�1(η, ζ ;α + β) + (−1)j[rj�1(η, ζ ; 2β) − r3−j�1(η, ζ ; 2α)], j = 1, 2,

k12(η, ζ ) = 2R(1 + β2)β

R1[�0(η, ζ ; 2α) +�0(η, ζ ; 2β) − 2�0(η, ζ ;α + β)],

k21(η, ζ ) = −αβ

k12(η, ζ ),

r1 = R

2(1 + R), r2 = R

2(1 − R), �0(η, ζ ; ε) = ε(1 − η)

4(η − ζ )2 + [εδ(1 − η)(1 − ζ )]2,

�1(η, ζ ; ε) = 2(ζ − η)

4(η − ζ )2 + [εδ(1 − η)(1 − ζ )]2, (36)

and apply the Gaussian type quadrature formulas [6]

∫ 1

−1

√1 + ζ

1 − ζf (ζ )dζ = 4π

2M + 1

M∑j=1

cos2 φj f (cos 2φj),

∫ 1

−1

√1 − ζ 2f (ζ )dζ = π

M + 1

M∑j=1

sin2 2ψj f (cos 2ψj), (37)

where

φj = (2j − 1)π

2(2M + 1), ψj = jπ

2(M + 1)(38)

and M is the number of abscissas. Using the connection between the Chebyshev and the orthonormal Jacobipolynomials

P−1/2,1/2n (ζ ) =

√2

π(ζ + 1)T2n+1

(√ζ + 1

2

),

P1/2,−1/2n (ζ ) = 1√

πU2n

(√ζ + 1

2

), (39)

we derive ultimately

c(j,j)nm = 32(−1)m

(2M + 1)2

M∑j=1

cosφj cos(2n + 1)φj

M∑s=1

cosφs cos(2m + 1)φskjj(− cos 2φs, cos 2φj),

c(1,2)nm = 32(−1)m

(2M + 1)(M + 1)

M∑j=1

sin2 ψj cosψj cos(2n + 1)ψj

×M∑

s=1

cosφs cos(2m + 1)φsk12(− cos 2φs, cos 2ψj). c(2,1)nm = −α

βc(1,2)

nm . (40)

The integrals b(j)n can be written in the form

b(j)n = − 1√

π

∫ 0

−∞

1√1 − x

T2n+1

(√ −x

1 − x

)fj(x)dx. (41)

Their convergence is guaranteed if fj(x) ∈ L1(0, A) for any finite A > 0, and fj(x) = o(|x|−1/2), x → −∞.

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8 Mathematics and Mechanics of Solids

Show finally that if δ → ∞, then the solution of the infinite system tends to the closed-form solution (23)for the whole plane. When δ → ∞, then c(j,l)

nm → 0 and a(j)m → b(j)

m , that is if δ = ∞, then

χj(ζ ) = −√

1 − ζ

1 + ζ

∫ 1

−1fj(ξ )

√1 + η

(1 − η)3�(η, ζ )dη, (42)

where

�(η, ζ ) =∞∑

m=0

P−1/2,1/2m (η)P1/2,−1/2

m (ζ ). (43)

To summarize this series, we employ the relations (39) and also the formula

limq→1−

∞∑m=0

q2m+1 sin(2m + 1)x = 1

2 sin x. (44)

This gives us

�(η, ζ ) = 1

π(η − ζ ), (45)

and therefore,

χj(x) = − 1

π

√(1 − ζ )3

1 + ζ

∫ 1

−1fj(ξ )

√1 + η

(1 − η)3

η − ζ, (46)

where x = (ζ + 1)/(ζ − 1), ξ = (η + 1)/(η − 1). This formula, when rearranged, coincides with (23).

4. Near-tip field

4.1. SIFs

To evaluate the SIFsKI = lim

x→0+

√2πxσyy(x, 0), KII = lim

x→0+

√2πxσxy(x, 0), (47)

we analyze the asymptotics of the stresses σxy(x, 0), σyy(x, 0) as x → 0+. From (19), we have

σyy(x, 0) ∼ − γ

πα

∫ 0

−∞

χ2(ξ )dξ

ξ − x, x → 0+. (48)

On making the substitutions (24) from (29), we derive as x → 0+ (ζ → −1−)

σyy(x, 0) ∼ γ (1 − ζ )

πα

∞∑m=0

a(2)m

∫ 1

−1

√1 − η

1 + ηP1/2,−1/2

m (η)dη

η − ζ. (49)

We now employ the following relation for the Jacobi polynomials∫ 1

0τα1 (1 − τ )α2Pα1,α2

n (1 − 2τ )dτ

τ − t= �(α1)�(α2 + n + 1)

�(α1 + α2 + n + 1)F(1 + n, −α1 − α2 − n, 1 − α1; t)

+�(−α1)�(1 + α1 + n)

n!(−t)α1F(1 + α1 + n, −α2 − n, 1 + α1; t), t /∈ (0, 1). (50)

For the orthonormal Jacobi polynomials P1/2,−1/2n (η) this relation can be simplified, and we obtain

σyy(x, 0) ∼ 2γ x−1/2

√πα

∞∑m=0

(−1)ma(2)m , x → 0+. (51)

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Antipov and Smirnov 9

2 4 6 8 10-6

-5

-4

-3

-2

-1

0

KI(1)

KII(2)

KI(2)

KII(1)

Figure 1. The SIFs versus δ for V/cR = 0.5: K(1)I and K

(1)II are the SIFs for the case σ ◦

22(x) = 1, −1 < x < 0, σ ◦22(x) = 0, x < −1,

and σ ◦12(x) = 0, x < 0; K

(2)I and K

(2)II are the SIFs for the case σ ◦

12(x) = 1, −1 < x < 0, σ ◦12(x) = 0, x < −1, and σ ◦

22(x) = 0, x < 0.

On comparing this asymptotic relation with the definition of the SIF KI , we derive the final formula for KI ,

KI = 2√

α

∞∑m=0

(−1)ma(2)m . (52)

Similarly,

KII = 2√

β

∞∑m=0

(−1)ma(1)m . (53)

For computations, we take the Poisson ratio ν = 0.3. The Rayleigh speed cR is defined explicitly [7] as

cR = cs√

s∗, s∗ = 1

3(8 − R+ − R−), R± =

(45κ0

2− 404 ± 3

√3R∗2

)1/3

,

R∗ = −14656 + 2768κ0 − 181κ20 + 4κ3

0 , κ0 = 8(2 − ν)/(1 − ν) ∈ (16, 24). (54)

The way in which the SIFs approach their asymptotic values as δ → ∞ is seen in Figure 1: the factors K(1)I

and K(2)II tend to K◦

I and K◦II , respectively, while the other two factors, K(2)

I and K(1)II tend to zero. Here,

K◦I = 2

√2γ

α

∞∑m=0

(−1)mb(2)m , K◦

II = 2√

β

∞∑m=0

(−1)mb(1)m . (55)

In Figure 1, the crack propagation speed V is chosen to be V = 0.5cR. The factors K(1)I and K(1)

II denote theSIFs for the case when σ ◦

22(x) = 1 for −1 < x < 0, and σ ◦22(x) = 0 otherwise, and σ ◦

12(x) = 0 for all x < 0.

The factors K(2)I and K(2)

II are the SIFs for the case σ ◦12(x) = 1 for −1 < x < 0, and σ ◦

12(x) = 0, x < −1, and

σ ◦22(x) = 0 for all x < 0. As δ → 0, the absolute values of all the factors except for K(2)

II are growing. The SIF

K(2)II approaches zero as the distance between the crack and the boundary of the half-plane tends to zero.

When the crack is close to the surface (δ = 2) and the crack propagation speed V is growing toward theRayleigh speed cR, the magnitudes of all the SIFs apart from K(1)

I are growing. The factor K(1)I is decreasing as

V/cR approaches 1. This is shown in Figure 2.

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10 Mathematics and Mechanics of Solids

0.2 0.3 0.4 0.5 0.6 0.7 0.8

-3.0

-2.5

-2.0

-1.5

-1.0

-0.5

0.0

KI(1)

KII(1)

KII(2)

KI(2)

Figure 2. The SIFs versus V/cR for δ = 2: K(1)I and K

(1)II are the SIFs for the case σ ◦

22(x) = 1, −1 < x < 0, σ ◦22(x) = 0, x < −1, and

σ ◦12(x) = 0, x < 0; K

(2)I and K

(2)II are the SIFs for the case σ ◦

12(x) = 1, −1 < x < 0, σ ◦12(x) = 0, x < −1, and σ ◦

22(x) = 0, x < 0.

4.2. Weight functions

Introduce next the weight functions WI ,I (x), WI ,II (x), WII ,I (x) and WII ,II (x) as follows

KI =∫ 0

−∞WI ,I (ξ )σ ◦

22(ξ )dξ +∫ 0

−∞WI ,II (ξ )σ ◦

12(ξ )dξ ,

KII =∫ 0

−∞WII ,I (ξ )σ ◦

22(ξ )dξ +∫ 0

−∞WII ,II (ξ )σ ◦

12(ξ )dξ . (56)

To determine the weight functions, we use the method introduced in Antipov [8]. The pair WI ,I (x), WII ,I (x) canbe found if loading

σ12(ξ , 0) = 0, σ22(ξ , 0) = δ(ξ − x), −∞ < ξ < 0, (57)

is applied. Here, −∞ < x < 0 and δ(x) is the Dirac delta function. For this case, the coefficients b(1)n , b(2)

n canexplicitly be computed

b(1)n = 0, b(2)

n = −αγ

√−x

1 − xP−1/2,1/2

n

(x + 1

x − 1

). (58)

It is useful to express the coefficients b(2)n through the Chebyshev polynomials. From (39)

b(2)n = − α

γ√π(1 − x)

T2n+1

(√x

x − 1

). (59)

With this relation, we now split a(j)n as

a(j)n = b(j)

n + a(j)n , j = 1, 2, (60)

and therefore the weight function WI ,I can be represented in the form

WI ,I (x) = W ◦I ,I (x) + WI ,I (x), (61)

where

W ◦I ,I (x) = 2

√2γ

α

∞∑m=0

(−1)mb(2)m , WI ,I (x) = 2

√2γ

α

∞∑m=0

(−1)ma(2)m , (62)

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Antipov and Smirnov 11

The first term can be computed explicitly

W ◦I ,I (x) = −

√2

−πx, (63)

while the elevation of the function WI ,I (x) and also the function

WII ,I (x) = 2√

β

∞∑m=0

(−1)ma(1)m , (64)

requires the solution of the infinite system

a(j)n +

∞∑m=0

[c(j,j)nm a(j)

m + c(j,3−j)nm a(3−j)

m ] = b(j)n , n = 0, 1, . . . ; j = 1, 2, (65)

where

b(1)n = 0, b(2)

n = −∞∑

m=0

c(2,2)nm b(2)

m . (66)

Similarly, the pair WI ,II (x), WII ,II (x) may be written as

WI ,II (x) = 2√

α

∞∑m=0

(−1)ma(2)m ,

WII ,II (x) = −√

2

−πx+ 2

√2γ

β

∞∑m=0

(−1)ma(1)m , (67)

where the coefficients a(1)m , a(2)

m form the solution of the infinite system (65) with the right-hand side given by

b(1)n = −

∞∑m=0

c(1,1)nm b(1)

m , b(2)n = 0. (68)

For computations of the weight functions, we choose ν = 0.3 and x = −1. When V is fixed (V < cR) andδ → ∞, the weight functions WI ,I and WII ,II tend to −√

2/π , while the other two functions, WI ,II and WII ,I ,vanish, see Figure 3. The magnitudes of all the weight functions grow as δ → 0 and V is fixed (V = 0.5cR inFigure 3).

The weight function curves in Figure 4 for δ = 1, ν = 0.3, x = −1 show that the weight function WI ,Idecreases as V → cR. The other functions may also decrease when the normalized speed V/cR is close to 1.Our numerical scheme becomes less reliable when V approaches the critical speed cR.

4.3. Griffith criterion

Following Willis [5] consider the potential energy δU released when the crack S1(t) = {x : −∞ < x1 <Vt, x2 = 0} extends to S1(t) + δS1(t) = {x : −∞ < x1 < Vt + r, x2 = 0}, where r is small. The energy δU maybe expressed as

δU = 1

2

∫ r

0{σxy(x, 0)δ[u](x) + σyy(x, 0)δ[v](x)}dx. (69)

Here, [u] + δ[u], [v] + δ[v] are the displacement jumps related to the extended crack, and

σxy ∼ KII√2πx

, σyy ∼ KI√2πx

, x ∈ (0, r), r → 0+. (70)

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12 Mathematics and Mechanics of Solids

0 2 4 6 8 10-6

-5

-4

-3

-2

-1

0

W II,I

W I,I

W I,II

W II,II

Figure 3. The weight functions Wj,l(x) (j, l = I , II) versus δ for V/cR = 0.5, x = −1.

0.0 0.2 0.4 0.6 0.8-4

-3

-2

-1

0

1

W I,II

W II,I

W II,II

W I,I

Figure 4. The weight functions Wj,l(x) (j, l = I , II) versus V/cR for δ = 1, x = −1.

To find asymptotic expansions for δ[u], δ[v], we integrate (30) and fix the constant of integration by assuringthat the displacement jumps vanish at the crack tip. For the displacement jumps [u] and [v], we have

(u(x, 0+) − u(x, 0−)v(x, 0+) − v(x, 0−)

)= 2

∞∑m=0

(a(1)

ma(2)

m

)∫ x

0P1/2,−1/2

m

(ξ + 1

ξ − 1

)dξ

(1 − ξ )√−ξ , x < 0. (71)

The displacement jumps vanish at the point x = 0 and tend to the values

−2∞∑

m=0

(a(1)

ma(2)

m

)∫ 0

−∞P1/2,−1/2

m

(ξ + 1

ξ − 1

)dξ

(1 − ξ )√−ξ (72)

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Antipov and Smirnov 13

as x → −∞. These values are finite and, in general, nonzero.Since P1/2,−1/2

m (−1) = π−1/2(−1)m, for small negative x,

[v](x) = −4

√−x

π

∞∑m=0

(−1)ma(2)m + O(|x|3/2), x → 0−. (73)

When the crack extends to x = r, due to the relation (52),

[v](x) = −√

2(r − x)

π

αKI

γ+ O{(r − x)3/2}, x → r. (74)

Similarly,

[u](x) = −√

2(r − x)

π

βKII

γ+ O{(r − x)3/2}, x → r. (75)

Now, on substituting formulas (70), (74) and (75) into (69), we obtain

δU ∼ − 1

4γ(αK2

I + βK2II )r, r → 0+. (76)

This formula can be written in terms of the Rayleigh function

D(V ) = 4

√(1 − V 2

c2l

)(1 − V 2

c2s

)−(

2 − V 2

c2s

)2

(77)

as

δU ∼ V 2r

2csμD(V )(αK2

I + βK2II ), r → 0+. (78)

According to the Griffith criterion, the crack starts propagating if the energy δU is equal to or greater than theincrease in the surface energy 2Tr, δU ≥ 2Tr, where T is the Griffith material constant. This criterion may berepresented in terms of the SIFs in the form√

1 − V 2

c2l

K2I +

√1 − V 2

c2s

K2II ≥ 4Tc2

sμD(V )

V 2. (79)

Notice, that if δ = ∞, and σ ◦12(x) = 0 for all x < 0, then the inequality (79) coincides with the criterion

K2I ≥ 4Tc2

sμD(V )

V 2√

1 − V 2/c2l

(80)

obtained by Willis [5]. For finite values of δ, even when the tangential component of loading vanishes, the SIFKII = 0, and both factors, KI and KII , are involved in the Griffith crack propagation criterion.

Another way to represent the crack propagation criterion is to rewrite inequality (79) as

H(KI , KII , V/cs, V/cl) ≥ μT , (81)

where

H = αK2I + βK2

II

4(cs/V )2D(V ). (82)

Figure 5 shows the results of calculations of the function H versus δ for V/cR = 0.5 and some loads. It isseen that H rapidly advances as the distance between the crack and the half-plane boundary decreases.

The dependence of H on the normalized crack speed V/cR when δ = 2 is plotted in Figure 6. The functionH → ∞ as V/cR → 0 and it grows as the crack speed V approaches the Rayleigh speed. The curves in Figure6 are reminiscent of the graph of modulus of cohesion kc(V ) versus V/cs [9] in the Barenblatt-type criterion forintersonic shear crack propagation.

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14 Mathematics and Mechanics of Solids

0 2 4 6 8 100

20

40

60

80

H (3)

H (2)

H (1)

Figure 5. The function H versus δ for V/cR = 0.5, x = −1 and loading σ ◦22 = δ(ξ − x), σ ◦

12 = 0: H (1); σ ◦22 = 0, σ ◦

12 = δ(ξ − x):

H (2); σ ◦22 = δ(ξ − x), σ ◦

12 = δ(ξ − x): H (3).

0.2 0.4 0.6 0.80

50

100

150

200

H (3)

H (1)

H (2)

Figure 6. The function H versus V/cR for δ = 2, x = −1 and loading σ ◦22 = δ(ξ − x), σ ◦

12 = 0: H (1); σ ◦22 = 0, σ ◦

12 = δ(ξ − x): H (2);

σ ◦22 = δ(ξ − x), σ ◦

12 = δ(ξ − x): H (3).

5. ConclusionsWe have analyzed a two-dimensional steady-state problem on propagation of a semi-infinite crack in a half-plane. The crack is subjected to normal and tangential loads applied to its faces, and it propagates at speedV along the half-plane boundary free of traction. The boundary of the half-plane breaks the symmetry of theproblem, and, in contrast to the problem for a plane, the modes I and II are coupled. We have deduced an order-2vector RH problem associated with the model. The coefficient is a Hermitian matrix that cannot be factorizedin a closed form. We have reduced the problem to a system of two singular integral equations with respect tothe derivatives of the displacement jumps. The method of orthogonal polynomials has been employed for its

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Antipov and Smirnov 15

solution. The unknown functions have been expanded in terms of the orthonormal Jacobi polynomials. Thecoefficients of the expansions have been determined from an infinite system of linear algebraic equations of thesecond kind.

We have derived formulas for the SIFs KI and KII and the weight functions WI ,I , WI ,II , WII ,I and WII ,II . Bydetermining the energy δU released when the crack extends to a small distance, we applied the Griffith criterionand established that the crack starts propagating when H ≥ μT , where

H =√

1 − (V/cl)2)K2I +

√1 − (V/cs)2)K2

II

4(cs/V )2D(V ),

D(V ) is the Rayleigh function, cs, cl are the shear and longitudinal waves speeds, μ is the shear modulus andT is the Griffith material constant. We have computed the SIFs, the weight functions and the function H fordifferent V/cR and δ (cR is the Rayleigh speed, and δ is the distance between the half-plane boundary and thecrack). It has been found that H grows to infinity when the distance δ between the crack and the half-planeboundary decreases while the crack speed does not vary. The function H monotonically decreases as δ grows.When the distance δ is fixed, H , as a function of V/cR, attains its minimum in the interval (0, 1) and grows asV/cR approaches the points 0 and 1.

Funding

This research received no specific grant from any funding agency in the public, commercial, or not-for-profit sectors.

Conflict of interest

None declared.

References

[1] Zlatin, AN, and Khrapkov, AA. A semi-infinite crack that is parallel to the boundary of an elastic half-plane. Dokl Akad NaukSSSR 1986; 291(4): 810–813.

[2] Craggs, JW. On the propagation of a crack in an elastic-brittle material. J Mech Phys Solids 1960; 8: 66–75.[3] Freund, LB. Dynamic fracture mechanics. Cambridge: Cambridge University Press, 1990.[4] Broberg, KB. Cracks and fracture. San Diego: Academic Press, 1999.[5] Willis, JR. A comparison of the fracture criteria of Griffith and Barenblatt. J Mech Phys Solids 1967; 15: 151–162.[6] Abramowitz, M, and Stegun, IA (eds). Handbook of mathematical functions. Washington: Dover Publications, 1965.[7] Antipov, YA. Subsonic semi-infinite crack with a finite friction zone in a bimaterial. J Mech Phys Solids 2009; 57: 1934–1957.[8] Antipov, YA. Weight functions of a crack in a two-dimensional micropolar solid. Quart J Mech Appl Math 2012; 65: 239–271.[9] Antipov, YA, Obrezanova, O, and Willis, JR. A fracture criterion of “Barenblatt” type for a transonic shear crack. Mathematics

and Mechanics of Solids 2004; 9: 271–283.

at LOUISIANA STATE UNIV on January 16, 2013mms.sagepub.comDownloaded from


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