Integral Identities for a Semi-infinite Interfacial Crack in 2D and 3D Elasticity

Abstract

The paper is concerned with the problem of a semi-infinite crack at the interface between two dissimilar elastic half-spaces, loaded by a general asymmetrical system of forces distributed along the crack faces. On the basis of the weight function approach and the fundamental reciprocal identity (Betti formula), we formulate the elasticity problem in terms of singular integral equations relating the applied loading and the resulting crack opening. Such formulation is fundamental in the theory of elasticity and extensively used to solve several problems in linear elastic fracture mechanics (for instance various classic crack problems in homogeneous and heterogeneous media).

Appendices

Appendix A: Inversion of the 2D Singular Operator S ^(s)

The inversion formula for the singular integral equation

$$ \psi(x_1) = S^{(s)}\varphi(x_1) = \frac{1}{\pi} \int_{-\infty}^{0} \frac{\varphi(\xi)}{x_1 - \xi} d\xi, \quad x_1 < 0, $$

(A.1)

strongly depends on the properties of the known function ψ(x ₁). Indeed, let us first assume that the function ψ(x ₁) has compact support −a≤x ₁≤−b, where a and b are positive constants, and belongs to a Hölder class, as it usually takes place in the classical 2D elasticity (Muskhelishvili, 1946 [16]). Then the inversion formula can be found in Muskhelishvili (1946) [16] (see also Rice, 1968 [22])

$$ \varphi(x_1) = \bigl(S^{(s)} \bigr)_1^{-1}\psi(x_1) = -\frac {1}{\pi} \int_{-\infty}^{0} \sqrt{\frac{\xi}{x_1}} \frac{\psi (\xi)}{x_1 - \xi} d\xi, \quad x_1 < 0. $$

(A.2)

Note that under such assumptions

$$ \varphi(x_1) \sim\frac{K_0}{\sqrt{-x_1}}, \quad x_1 \to0^-, \quad\quad \varphi(x_1) \sim \frac{K_\infty}{(-x_1)^{3/2}} , \quad x_1 \to -\infty. $$

(A.3)

Here the constants K ₀ and K _∞ (K ₀ is the so-called stress intensity factor) are determined by the formulae

$$ K_0 = -\frac{1}{\pi} \int _{-\infty}^{0} \frac{\psi(\xi)}{\sqrt {-\xi}} d\xi, \quad\quad K_\infty= \frac{1}{\pi} \int_{-\infty}^{0} \psi(\xi)\sqrt{-\xi } d\xi. $$

(A.4)

Let us now consider the case where the function ψ(x ₁) extends over the whole negative semi-axis, having the following behavior at zero and infinity:

$$ \psi(x_1) \sim\frac{\psi_0}{(-x_1)^{\alpha_0}}, \quad x_1 \to0^-, \quad\quad \psi(x_1) \sim\frac{\psi_\infty}{(-x_1)^{\alpha_\infty}}, \quad x_1 \to-\infty. $$

(A.5)

If we assume that α ₀<1/2 and α _∞>3/2 then the inversion formula (A.2) is still valid and leads to the asymptotics (A.3). In the case α ₀<1/2 and 1/2<α _∞<3/2 the inversion formula (A.2) remains true, the behavior of the function ψ(x ₁) near zero is the same as in (A.3)₁, while the behavior at infinity (A.3)₂ changes to:

$$ \varphi(x_1) \sim\frac{-\pi\psi_\infty\sin(\pi\alpha_\infty )}{(-x_1)^{\alpha_\infty}}, \quad x_1 \to-\infty. $$

(A.6)

However, there are situations (such as some problems arising in hydraulic fracturing, see Garagash and Detournay, 2000 [9]) where the behavior of the function ψ(x ₁) at infinity is worst, so that α _∞ in (A.5) is in the range 0<α _∞<1/2. In such cases, the classic inversion formula (A.2) is not any longer applicable and another inversion formula should be used instead. In Garagash and Detournay (2000) [9] the formula

$$ \varphi(x_1) = \bigl(S^{(s)} \bigr)_2^{-1}\psi(x_1) = -\frac{C_0}{(-x_1)^{1/2}} - \frac{1}{\pi} \int_{-\infty}^{0} \sqrt { \frac{x_1}{\xi}} \frac{\psi(\xi)}{x_1 - \xi} d\xi, \quad x_1 < 0, $$

(A.7)

was derived by easy and elegant arguments leaving, however, the constant C ₀ unknown.

Here we present an inversion formula accurately derived under slightly more restricted conditions on the behavior of the function ψ(x ₁) at infinity. Namely we assume that

$$ \psi(x_1) = \psi_\infty(x_1) + O \biggl[ \frac{1}{(-x_1)^{\beta }} \biggr], \quad x_1 \to-\infty, $$

(A.8)

where

$$ \psi_\infty(x_1) = \sum_{j = 1}^N \frac{\psi^{(j)}_\infty }{(-x_1)^{\alpha^{(j)}_\infty}}, $$

(A.9)

and $0 < \alpha^{(j)}_{\infty}< 1/2$ and β>1/2. The alternative formula reads

(A.10)

Note that this formula can be rewritten in the equivalent form (A.7), where the constant C ₀ can be determined by

$$ C_0 = \frac{1}{\pi} \int_{-\infty}^{0} \frac{\psi(\xi) - \psi_\infty(\xi)}{(-\xi)^{1/2}} d\xi. $$

(A.11)

Appendix B: Reduction of the Identities From 3D to 2D Case

The integral identities for 3D elasticity can be reduced to the 2D case by assuming that all the mechanical fields are independent of the x ₃ coordinate. We start by showing that if the fields u, σ ₂, p are independent of x ₃ then the convolution with respect to x ₁ and x ₃, denoted by ⊛, reduces to the convolution with respect to the x ₁ coordinate only, denoted by ∗.

In particular, we have the following formulae (proved in Appendix D, Theorem 1). If φ is a function of x ₁ only, then

(B.1)

(B.2)

We also make use of the following identities (proved in Appendix D, Theorems 2 and 3):

(B.3)

(B.4)

If φ is a function of x ₁ only then the identities (B.3) and (B.4) yield

(B.5)

(B.6)

Using the formulae (B.1), (B.2), (B.5) and (B.6), the integral identity (84) for 3D elasticity reduces to

(B.7)

(B.8)

This formula is fully consistent with the formulation for 2D elasticity given in Sects. 3 and 4. In a similar way, also the formula (85) can be shown to reduce to the 2D formulation.

Appendix C: Illustrative Example in 2D Case: Point Forces Applied at the Crack Faces

In this section, we illustrate the use of the singular integral identities in the 2D case, when the loading is given in terms of point forces applied at the crack faces. We analyse first the antiplane problem and then the vector problem.

Mode III. Symmetrical Point Forces

Assume that the loading is given as two symmetrical point forces applied on the crack faces, at a distance a from the crack tip, and directed along the x ₃-axis,

$$ \langle p_3 \rangle(x_1) = -F \delta(x_1 + a), \quad\quad [\![p_3 ]\!](x_1) = 0, $$

(C.1)

where δ is the Dirac delta function.

The singular integral equation relating the applied loading with the resulting crack opening is given by (32) and, by means of the inversion formula (A.2) we obtain

$$ \frac{\partial [\![u_3 ]\!]^{(-)}}{\partial x_1} = -\frac{b + e}{\pi} F \int_{-\infty}^0 \sqrt{\frac{\xi}{x_1}} \frac{\delta (\xi+ a)}{x_1 - \xi} d\xi= - \frac{b + e}{\pi} F \sqrt{-\frac{a}{x_1}} \frac{1}{x_1 + a}. $$

(C.2)

By integration of (C.2) and using the conditions that the crack opening vanishes at zero and at infinity, we get

$$ [\![u_3 ]\!](x_1) = \left \{ \begin{array}{l@{\quad}l} \frac{2F}{\pi}(b + e) \operatorname{arctanh}\sqrt {-\frac{x_1}{a}}, &-a < x_1 < 0, \\[8pt] \frac{2F}{\pi}(b + e) \operatorname{arctanh}\sqrt {-\frac{a}{x_1}}, & x_1 < -a. \end{array} \right . $$

(C.3)

From (33) we can now obtain the corresponding tractions ahead of the crack tip, namely

$$ \langle\sigma_{23} \rangle^{(+)}(x_1) = - \frac{1}{\pi(b + e)} \int_{-\infty}^0 \frac{1}{x_1 - \xi} \frac{\partial [\![u_3 ]\!]^{(-)}}{\partial\xi } d\xi= \frac{F}{\pi} \sqrt{\frac{a}{x_1}} \frac{1}{x_1 + a}. $$

(C.4)

Note that the stress intensity factor in this case is

$$ K_\mathrm{III}= \lim_{x_1 \to0} \sqrt{2\pi x_1} \langle \sigma_{23} \rangle^{(+)}(x_1) = \sqrt{ \frac{2}{\pi a}} F. $$

(C.5)

Mode III. Skew-Symmetrical Point Forces

We assume now that the loading is given by two skew-symmetrical point forces again applied at a distance a behind crack tip and directed along the x ₃-axis,

$$ \langle p_3 \rangle(x_1) = 0, \quad\quad [\![p_3 ]\!](x_1) = -2F \delta(x_1 + a). $$

(C.6)

It is easy to show that all the formulae derived above still apply, however, they contain now the factor η, so that

(C.7)

(C.8)

(C.9)

and the stress intensity factor is given by

$$ K_\mathrm{III}= \eta\sqrt{\frac{2}{\pi a}} F. $$

(C.10)

Mode I and II. Symmetrical Point Forces

We consider now the vector 2D case and suppose that the loading is given as symmetrical point forces in both x ₁ and x ₂ directions, namely

(C.11)

(C.12)

For simplicity we assume that the Dundurs parameter d is equal to zero, so that the system of singular equations (51) decouples

$$ -\frac{1}{b} {\mathcal{S}}^{(s)} \frac{\partial [\![u_j ]\!]^{(-)}}{\partial x_1} = \langle p_j \rangle(x_1), \quad j = 1,2 $$

(C.13)

and it is possible to obtain results in closed form. Indeed, the inversion formula (A.2) leads to

$$ \frac{\partial [\![u_j ]\!]^{(-)}}{\partial x_1} = - \frac{b F_j}{\pi} \sqrt{-\frac{a}{x_1}} \frac{1}{x_1 + a}, $$

(C.14)

which, after integration, gives

$$ [\![u_j ]\!](x_1) = \left \{ \begin{array}{l@{\quad}l} \frac{2b F_j}{\pi} \operatorname{arctanh}\sqrt {-\frac{x_1}{a}}, &-a < x_1 < 0, \\[8pt] \frac{2b F_j}{\pi} \operatorname{arctanh}\sqrt {-\frac{a}{x_1}}, & x_1 < -a. \end{array} \right . $$

(C.15)

Finally, the tractions ahead of the crack tip are calculated from (52)

$$ \langle\sigma_{2j} \rangle^{(+)}(x_1) = \frac{F_j}{\pi} \sqrt {\frac{a}{x_1}} \frac{1}{x_1 + a}. $$

(C.16)

The stress intensity factors are then obtained as

$$ K_\mathrm{I} = \sqrt{\frac{2}{\pi a}} F_2, \quad\quad K_\mathrm{II}= \sqrt{\frac{2}{\pi a}} F_1. $$

(C.17)

Mode I and II. Skew-Symmetrical Point Forces

In this case the point forces, applied at a distance a behind the crack tip, have the same direction

(C.18)

(C.19)

Assuming again d=0, the solution of the system of singular integral equations (51) leads to

(C.20)

(C.21)

which after integration gives

(C.22)

The tractions ahead of the crack tip are obtained from (52) and read

(C.23)

(C.24)

The stress intensity factors are then obtained as

$$ K_\mathrm{I}= \alpha\sqrt{\frac{2}{\pi a}} F_2,\quad \quad K_\mathrm{II}= \alpha \sqrt{\frac{2}{\pi a}} F_1. $$

(C.25)

The reader may be surprised by the coupling constant terms in the right-hand side of expressions (C.22), showing a discontinuity of the crack opening at the point of application of the forces. However, these terms can be easily explained by making recourse to the Flamant solution for a half-plane loaded by concentrated forces F ₁ and F ₂ at its surface (Barber, 2002 [2]).

Assuming a reference system centred at the point of application of the forces, the surface displacement are determined, up to an arbitrary constant, by

(C.26)

where μ and ν are the shear modulus and the Poisson’s ratio of the half-plane, respectively.

We now apply this solution to the two half-planes making up the cracked bimaterial plane. Of course, we do not expect this solution to match exactly with the correct solution obtained above, since the boundary conditions for the crack problem differ from the Flamant problem along the interface joining the two materials. Nonetheless, let us for the time being make this analogy and calculate from the solution (C.26) what the crack opening would be.

We consider only the coupling (diagonal) terms in the system (C.26), which can explain the observed discontinuity in the crack opening. We also adjust the arbitrary constant in the Flamant solution in order to have the same displacement for the two crack faces at the crack tip. As a result, we obtain

(C.27)

(C.28)

which is in perfect agreement with our solution (C.22).

Appendix D: Identities Involving Convolution Integrals

Theorem 1

Let φ(x ₁) be a function defined on ℝ and possess the derivatives of order 1,…,n. Assume also that the function φ(x ₁) is vanishing with all its derivatives as x ₁→±∞. Then

$$ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{\sqrt {(x_1 - \xi_1)^2 + (x_3 - \xi_3)^2}} \frac{\partial^n \varphi }{\partial\xi_1^n} d\xi_1 d\xi_3 = -2 \int_{-\infty}^{\infty} \frac{1}{x_1 - \xi_1} \frac{\partial^{n-1} \varphi}{\partial x_1^{n-1}} d\xi_1. $$

(D.1)

Proof

Consider the integral

$$ \int_{-R}^{R} \int_{-\infty}^{\infty} \frac{1}{\sqrt{(x_1 - \xi_1)^2 + (x_3 - \xi_3)^2}} \frac{\partial^n \varphi}{\partial\xi_1^n} d\xi_1 d\xi_3, $$

(D.2)

and integrate by parts the inner integral, to obtain

(D.3)

The first term in (D.3) is zero and the second term can be rewritten changing the order of integration in the form

$$ -\int_{-\infty}^{\infty} \biggl( \int _{-R}^{R} \frac{1}{[(x_1 - \xi_1)^2 + (x_3 - \xi_3)^2]^{3/2}} d\xi_3 \biggr) (x_1 - \xi_1) \frac{\partial^{n-1} \varphi}{\partial\xi_1^{n-1}} d \xi_1. $$

(D.4)

Taking the limit as R→∞, the inner integral in (D.4) gives

$$ \lim_{R \to\infty} \int_{-R}^{R} \frac{1}{[(x_1 - \xi_1)^2 + (x_3 - \xi_3)^2]^{3/2}} d\xi_3 = \frac{2}{(x_1 - \xi_1)^2}, $$

(D.5)

which concludes the proof. □

Theorem 2

Let φ(x ₁,x ₃) be a function defined on ℝ² and possess the Fourier transform with respect to x ₁ and x ₃, denoted by $\bar{\varphi}(\beta,\lambda)$. Then

(D.6)

Proof

Consider the Fourier transform $\bar{\varphi}(\beta,\lambda)$ and write

$$ \bar{\varphi}(\beta,\lambda) = \frac{\beta^2}{\beta^2 + \lambda^2} \cdot \bar{\varphi}(\beta,\lambda) + \frac{\lambda^2}{\beta^2 + \lambda^2} \cdot\bar{\varphi }( \beta,\lambda). $$

(D.7)

To conclude the proof, apply the inversion Fourier transform to both side of (D.7) and use the first two formulae given in Table 1.

Table 1 Double Fourier inversion formulae

 □

Theorem 3

Let φ(x ₁,x ₃) be a function defined on ℝ² and possess the Fourier transform with respect x ₁ and x ₃, denoted by $\bar{\varphi}(\beta,\lambda)$. Then

(D.8)

Proof

Consider the Fourier transform $\bar{\varphi}(\beta,\lambda)$ and write

$$ \frac{\beta}{\beta^2 + \lambda^2} \cdot\lambda\bar{\varphi }(\beta,\lambda) = \frac{\lambda}{\beta^2 + \lambda^2} \cdot \beta\bar{\varphi}(\beta,\lambda). $$

(D.9)

To conclude the proof, apply the inversion Fourier transform to both side of (D.7) and use the third and fourth formulae in Table 1. □

Appendix E: Double Fourier Transform Inversion for the 3D Case

The following four integrals will be used in the analysis (Gradshteyn and Ryzhik, 1965 [11]):

(E.1)

(E.2)

(E.3)

(E.4)

where K ₀(z) is the modified Bessel function of the second kind of zero order, satisfying the differential equation

$$ z^2 y'' + z y' - z^2 y = 0. $$

(E.5)

In some sense, the last two integrals are the Fourier inversion formulae for the first two.

We also shall make use of the fact that the Fourier transform of the convolution is given by the product of the Fourier transforms, which leads to

$$ {\mathcal{F}}^{-1}_{\beta}{\mathcal{F}}^{-1}_\lambda \bigl[ \bar{g}(\beta,\lambda) \cdot\bar{f}(\beta,\lambda) \bigr] = \int _{-\infty}^\infty\int_{-\infty}^\infty g(x_1 - \xi_1,x_3 - \xi_3) f( \xi_1,\xi_3) d\xi_1 d\xi_3. $$

(E.6)

It is evident what to do with the factors β and λ. Namely, these factors arise when taking the Fourier transform of the derivatives instead of the function itself:

$$ \beta\bar{f}(\beta,\lambda) = i \overline{\frac{\partial f}{\partial x_1}}(\beta,\lambda), \quad\quad \lambda\bar{f}(\beta,\lambda) = i \overline{\frac{\partial f}{\partial x_3}}(\beta, \lambda). $$

(E.7)

Analogous identities can be written for the factors β ², λ ² and βλ involving higher order derivatives:

(E.8)

Taking into account the following identity:

$$ \sqrt{\beta^2 + \lambda^2} = \frac{\beta^2 + \lambda^2}{\sqrt {\beta^2 + \lambda^2}}, $$

(E.9)

we need to evaluate only the following three basic integrals:

(E.10)

(E.11)

(E.12)

Let us consider the first integral (E.10). By (E.1) we obtain

$$ \int_{-\infty}^{\infty} e^{-ix_3\lambda} \frac{d\lambda}{\sqrt {\beta^2 + \lambda^2}} = 2 \int_0^{\infty} \frac{\cos(|x_3|\lambda)}{\sqrt{\beta^2 + \lambda^2}} d\lambda= 2 \left \{ \begin{array}{l@{\quad}l} K_0(|x_3|\beta), & \operatorname{Re}\beta> 0, \\[4pt] K_0(-|x_3|\beta), & -\operatorname{Re}\beta> 0. \end{array} \right . $$

(E.13)

Making recourse to (E.3), we conclude that

(E.14)

We note the function transforms into itself (multiplied by a constant factor) after double Fourier transform.

Let us now consider the second integral (E.11). By means of (E.2) we write

(E.15)

We can now use the formula (E.4) to obtain

(E.16)

Again, the function transforms into itself (multiplied by a constant factor) after double Fourier transform.

The last integral (E.12) can be evaluated in a similar way and we have

$$ {\mathcal{F}}^{-1}_{\beta}{\mathcal{F}}^{-1}_\lambda \biggl[\frac {\beta}{\beta^2 + \lambda^2} \biggr] = \frac{-i}{2\pi} \frac{x_1}{x_1^2 + x_3^2}. $$

(E.17)

Equations (E.14), (E.16) and (E.17) are the basic integrals we use to perform the double Fourier inversion for the 3D case in Sect. 5. The complete list of inversion formulae is given in Table 1.