Uneingeschränkter Zugang

Axisymmetric Torsion of an Elastic Layer Sandwiched between Two Elastic Half-Spaces with Two Interfaced Cracks


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Introduction

The class of problems related to the behavior of rigid disc inclusions embedded in bonded contact with an elastic medium, has been a subject of much interest in geomechanics, civil engineering, and applied mechanics. This work is motivated by both theoretical and practical interests in the problems of turbines disks, some pipes, and many industrial applications. It may give a better understanding of the behavior of foundations under external loads. In structure-medium interaction problems arising in foundation engineering, the foundation is usually modeled using a rigid or flexible inclusion having circular, strip, rectangular, or arbitrary shape. Nowadays, composites play a very important role in geomechanics engineering.

It is common knowledge that all existing structural materials contain different inter- and intra-component defects (cracks, delaminations, etc.) [1]. The problem of the torsion of an infinite elastic medium by a rigid inclusion (deeply embedded) was considered by Selvadurai [2, 3]. His results depend on the rotational and translational stiffnesses of the embedded rigid circular disc. The problem of the torsion of an elastic half- space was considered, at first, by Reissner and Sagoci [4]. They studied the static interaction of a rigid disc and an elastic isotropic half-space for which they obtained the solution by means of the spheroidal coordinates. The same problem was solved by Sneddon [5] using a different method. He used the Hankel transforms method for reducing the problem to a pair of dual integral equations. Collins [6] treated the torsional problem of an elastic half-space by assuming the displacement at any point in the half- space to be due to a distribution of wave sources over the part of the free surface in contact with the disc. The solution of the forced vibration problem of elastic layer of finite thickness when the lower face is either stress free or rigidly clamped was given by Gladwell [7]. Pak and Saphores [8] provided an analytical formulation for the general torsional problem of a rigid disc embedded in an isotropic half-space. Besides, Bacci and Bennati [9] considered the torsional of circular rigid disc adhered to the upper surface of an elastic layer fixed to an undefonnable support. More recently, Singh et al. [10] studied the torsional of a non-homogeneous, isotropic, half- space by rotating a circular part of its boundary surface. Cai and Zue [11] discussed the torsional vibration of a rigid disc bonded to a poroelastic multilayered medium.. Yu [12] studied the forced torsional oscillations inside the multilayered solid. The elastodynamic Green’s function of the center of rotation and a point load method were used to solve the problem. Pal and Mandal [13] considered the forced torsional oscillations of a transversely isotropic elastic half- space under the action of an inside rigid disc. A similar problem with the rocking rotation was solved later on by Ahmadi and Eskandari [14]. They used an appropriate Green’s function to write the mixed boundary -value problem posed as a dual integral equation. All these problems are based on the theory of Hankel integral transformation in order to bring the mixed boundary -value problem into a system of dual integral equations. Then, the corresponding solution is sought from an integral equation of Fredholm type.

The torsional of elastic layers with a penny -shaped crack was considered by some researchers. Sih and Chen [15] studied the problem of a penny-shaped crack in layered composite under a uniform torsional stress. The displacement and stress fields throughout the composite were obtained by solving a standard Fredholm integral equation of the second kind. Low [16] investigated a problem of the effects of embedded flaws in the form of an inclusion or a crack in an elastic half- space subjected to torsional deformations. The corresponding Fredholm integral equations were solved numerically by quadrature approach. The same method was used by Dhawan [17] for solving the problem of a rigid disc attached to an elastic half-space with an internal crack. By using Hankel and Laplace transforms and taking numerical inversion of Laplace transform, Basu and Mandal [18] treated the torsional load on a penny-shaped crack in an elastic layer sandwiched between two elastic half-spaces. The purpose of this article is to study an axisymmetric torsion of an embedded circular rigid disc in bonded contact with an isotropic elastic layer sandwiched between two elastic half-spaces with two penny-shaped cracks symmetrically located at each of the two interfaces between the layer and the half-spaces. A similar method was used in arecently published work by Madani and Kebli [20], dealing with the case of a penny-shaped crack problem in the interior of a homogeneous elastic material at the symmetry plane, under an axisymmetric torsion by two circular rigid discs symmetrically located in the elastic medium.

Basic Equations of the Problem

In view of the axial symmetry of the problem, it is natural to consider (r, θ, z) the cylindrical polar co-ordinates. Here we consider an axisymmetric torsion of an embedded circular rigid disc with a radius b at the symmetry plane z = 0 in an isotropic and homogeneous layer of thickness 2h and with one material sandwiched between two half-spaces of the second material. Two penny -shaped cracks with a radius a were symmetrically located at the interface between the layer and the half- spaces z = ±h. The faces of the cracks are assumed to be stress free while the discs rotate with an equal angle ω about the z- axis passing through their center as shown in Figure .1.

Figure 1

Geometry and coordinate system.

Owing to symmetry about the z = 0 plane, it is sufficient to consider the problem in the upper half- space where z ≥ 0. In this case of an axisymmetric torsion problem, the displacement vector assumes the form (0, uθ, 0) in the cylindrical polar coordinate system (r, θ, z). It is convenient to identify a layer region (superscript (1)) occupying the region r ∈ (0, ∞); z ∈ (0, h) and a layer region (superscript (2)) occupying the region r ∈ (0, ∞); z ∈ (h, ∞). The signs h+ and h denote the variables in the upper surface and the lower surface of plane z = h, respectively.

The only non-zero components of stress are given by

τiθz=Giuiθz,τiθr=Girr(uiθr),i=1,2$${{\tau }^{i}}_{\theta z}={{G}_{i}}\frac{\partial {{u}^{i}}_{\theta }}{\partial z},\,\,\,\,\,\,{{\tau }^{i}}_{\theta r}={{G}_{i}}r\frac{\partial }{\partial r}\left( \frac{{{u}^{i}}_{\theta }}{r} \right),\,\,\,\,i=1,2$$

where uθ = uθ (r, z) and Gi is the shear modulus of the material. As the torsion of the homogeneous material is static, the displacement uθ (r, z) must satisfy

2uθr2+uθrruθr2+2uθz2=0$$\frac{{{\partial }^{2}}{{u}_{\theta }}}{\partial {{r}^{2}}}+\frac{\partial {{u}_{\theta }}}{r\partial r}-\frac{{{u}_{\theta }}}{{{r}^{2}}}+\frac{{{\partial }^{2}}{{u}_{\theta }}}{\partial {{z}^{2}}}=0$$

By means of Hankel’s transformation, integral and its inverse given in [19],

F(λ,z)=0f(r,z)rJ1(λr)dr$$F\left( \lambda ,z \right)=\int_{0}^{\infty }{f\left( r,z \right)r\,{{J}_{1}}\left( \lambda r \right)dr}$$

And

f(r,z)=0F(λ,z)λJ1(λr)dλ$$f\left( r,z \right)=\int_{0}^{\infty }{F\left( \lambda ,z \right)}\,\lambda {{J}_{1}}\left( \lambda r \right)d\lambda $$

The solution of equation (2) for the regions I (0 ≤ zh) and I I (zh) is expressed as

uθ(i)(r,z)=0[Ai(λ)eλz+Bi(λ)eλz]J1(λr)dλi=,2$$u_{\theta }^{\left( i \right)}\left( r,z \right)=\int_{0}^{\infty }{\left[ {{A}_{i}}\left( \lambda \right){{e}^{-\lambda z}}+{{B}_{i}}\left( \lambda \right){{e}^{\lambda z}} \right]}{{J}_{1}}\left( \lambda r \right)d\lambda \,i=,2$$

where λ is the transform variable, J1 is the Bessel function of the first kind of order one, and Ai and Bi are unknown functions

Boundary and Continuity Conditions

We consider the regularity conditions at infinity, the symmetry plane condition at z = 0, and the boundary and continuity conditions at the bonded interfaces z=h+$z={{h}^{+}}$and h${{h}^{-}}$. Therefore, we find the following conditions

limr,zuθi(r,z)=0,limr,zτθzi(r,z)=0$$\underset{r,z\to \infty }{\mathop{\lim }}\,u_{\theta }^{i}\left( r,z \right)=0,\,\,\,\,\,\,\,\underset{r,z\to \infty }{\mathop{\lim }}\,\tau _{\theta z}^{i}\left( r,z \right)=0$$τθz(1)(r,0)=0,r>b$${{\tau }_{\theta z}}^{\left( 1 \right)}\left( r,0 \right)=0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,r>b$$τθz(1)(r,h)τθz(2)(r,h+)=0,ra$${{\tau }_{\theta z}}^{\left( 1 \right)}\left( r,{{h}^{-}} \right)-{{\tau }_{\theta z}}^{\left( 2 \right)}\left( r,{{h}^{+}} \right)=0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,r\ge a$$uθ(1)(r,h)uθ(2)(r,h+)=0,ra$${{u}_{\theta }}^{\left( 1 \right)}\left( r,{{h}^{-}} \right)-{{u}_{\theta }}^{\left( 2 \right)}\left( r,{{h}^{+}} \right)=0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,r\ge a$$τθz(2)(r,h+)=τθz(1)(r,h)=0r<a$${{\tau }_{\theta z}}^{\left( 2 \right)}\left( r,{{h}^{+}} \right)={{\tau }_{\theta z}}^{\left( 1 \right)}\left( r,{{h}^{-}} \right)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,r<a$$uθ(1)(r,0)=ωrrb$${{u}_{\theta }}^{\left( 1 \right)}\left( r,0 \right)=\omega r\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,r\le b$$

Applying the regularity conditions at infinity given in Eq. (4a), we obtain

uθ(1)(r,z)=0[A1(λ)eλz+B1(λ)eλz]J1(λr)dλ$${{u}_{\theta }}^{\left( 1 \right)}\left( r,z \right)=\int_{0}^{\infty }{\left[ {{A}_{1}}\left( \lambda \right){{e}^{-\lambda z}}+{{B}_{1}}\left( \lambda \right){{e}^{\lambda z}} \right]}{{J}_{1}}\left( \lambda r \right)d\lambda $$τθz(1)(r,z)=G10λ[A1(λ)eλz+B1(λ)eλz]J1(λr)dλ$${{\tau }_{\theta z}}^{\left( 1 \right)}\left( r,z \right)={{G}_{1}}\int_{0}^{\infty }{\lambda \left[ -{{A}_{1}}\left( \lambda \right){{e}^{-\lambda z}}+{{B}_{1}}\left( \lambda \right){{e}^{\lambda z}} \right]}{{J}_{1}}\left( \lambda r \right)d\lambda $$uθ(2)(r,z)=0[A2(λ)eλz]J1(λr)dλ$${{u}_{\theta }}^{\left( 2 \right)}\left( r,z \right)=\int_{0}^{\infty }{\left[ {{A}_{2}}\left( \lambda \right){{e}^{-\lambda z}} \right]}{{J}_{1}}\left( \lambda r \right)d\lambda $$τθz(2)(r,z)=G20λ[A2(λ)eλz]J1(λr)dλ$${{\tau }_{\theta z}}^{\left( 2 \right)}\left( r,z \right)=-{{G}_{2}}\int_{0}^{\infty }{\lambda \left[ -{{A}_{2}}\left( \lambda \right){{e}^{-\lambda z}} \right]}{{J}_{1}}\left( \lambda r \right)d\lambda $$

where A1(λ), B1(λ), and A2(λ) are arbitrary functions that need to be determined by satisfying the boundary and continuity conditions.

The boundary and the continuity conditions in Eqs. (4c) and (4e) lead to

τθz(2)(r,h+)τθz(1)(r,h)=0$${{\tau }_{\theta z}}^{\left( 2 \right)}\left( r,{{h}^{+}} \right)-{{\tau }_{\theta z}}^{\left( 1 \right)}\left( r,{{h}^{-}} \right)=0$$

The above equation implies

A2(λ)=γ(A1(λ)B1(λ)e2λh)$${{A}_{2}}\left( \lambda \right)=\gamma \left( {{A}_{1}}\left( \lambda \right)-{{B}_{1}}\left( \lambda \right){{e}^{2\lambda h}} \right)$$

where γ=G1G2.$\gamma =\frac{{{G}_{1}}}{{{G}_{2}}}.$

The mixed boundary conditions Eqs. (4e), (4d), (4f), and (4b) are satisfied if A1 and B1 are solutions of the following dual integral equations:

0λ[B1(λ)eλhA1(λ)eλh]J1(λr)dλ=00r<a$$\int_0^\infty\lambda{\lbrack B_1{(\lambda)}e^{\lambda h}-A_1{(\lambda)}e^{-\lambda h}\rbrack}J_1{(\lambda r)}d\lambda=0\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0\leq r<a$$0[(1γ)eλhA1(λ)+(1+γ)eλhB1(λ)]J1(λr)dλ=0ra$$\int_{0}^{\infty }{\left[ \left( 1-\gamma \right){{e}^{-\lambda h}}{{A}_{1}}\left( \lambda \right)+\left( 1+\gamma \right){{e}^{\lambda h}}{{B}_{1}}\left( \lambda \right) \right]}{{J}_{1}}\left( \lambda r \right)d\lambda =0\,\,\,\,\,r\ge a$$0[A1(λ)+B1(λ)]J1(λr)dλ=ωr0rb$$\int_{0}^{\infty }{\left[ {{A}_{1}}\left( \lambda \right)+{{B}_{1}}\left( \lambda \right) \right]}{{J}_{1}}\left( \lambda r \right)d\lambda =\omega r\,\,\,\,\,\,\,\,\,\,\,\,\,0\le r\le b$$0λ[B1(λ)A1(λ)]J1(λr)dλ=0r>b$$\int_{0}^{\infty }{\lambda \left[ {{B}_{1}}\left( \lambda \right)-{{A}_{1}}\left( \lambda \right) \right]}{{J}_{1}}\left( \lambda r \right)d\lambda =0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,r>b$$
Limiting Cases

Let’s take the limit a → 0 and the nonhomogeneity parameter γ = 1, one can obtain the closed-form solution pertinent to the torsional rotation of a rigid disc embedded in a homogeneous elastic full-space. Owing to the symmetry of the full-space case with respect to the plane of the disc, it can be deduced that τθz is zero for r > a at the disc plane. This situation corresponds exactly to the torsion of a homogeneous elastic half-space by a circular rigid disc (0 < r < a, z = 0) bonded to the surface. This is adapted to the problem concerning isotropic half-space considered by Reissner and Sagoci [4].

By taking the parameter γ = 1, the problem is simplified to the torsional rotation of a rigid circular inclusion attached to an elastic half-space with internal crack and the dual integral equations become

0λ[B1(λ)eλhA1(λ)eλh]J1(λr)dλ=00r<a$$\int_{0}^{\infty }{\lambda \left[ {{B}_{1}}\left( \lambda \right){{e}^{\lambda h}}-{{A}_{1}}\left( \lambda \right){{e}^{-\lambda h}} \right]}{{J}_{1}}\left( \lambda r \right)d\lambda =0\,\,\,\,\,\,\,\,\,\,\,\,0\le r<a$$0[2eλhB1(λ)]J1(λr)dλ=0ra$$\int_{0}^{\infty }{\left[ 2{{e}^{\lambda h}}{{B}_{1}}\left( \lambda \right) \right]}\,{{J}_{1}}\left( \lambda r \right)d\lambda =0\,\,\,\,\,\,\,\,\,\,\,\,r\ge a$$0[A1(λ)+B1(λ)]J1(λr)dλ=ωr0rb$$\int_{0}^{\infty }{\left[ {{A}_{1}}\left( \lambda \right)+{{B}_{1}}\left( \lambda \right) \right]}{{J}_{1}}\left( \lambda r \right)d\lambda =\omega r\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\le r\le b$$0λ[B1(λ)A1(λ)]J1(λr)dλ=0r>b$$\int_{0}^{\infty }{\lambda \left[ {{B}_{1}}\left( \lambda \right)-{{A}_{1}}\left( \lambda \right) \right]}{{J}_{1}}\left( \lambda r \right)d\lambda =0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,r>b$$

This system of dual integral equations has the same meaning as Eqs. (13), (14), (15), and (16) in Dhawan’s paper [17].

Reduction of the dual integral equations

Equations 8b and 8d are identically satisfied if we introduce the following representation:

1γeλhA1λ+1+γeλhB1λ=λ0atϕtJ32λtdt$$\left[ {\left( 1-\gamma \right){{e}^{-\lambda h}}{{A}_{1}}\left( \lambda \right)+\left( 1+\gamma \right){{e}^{\lambda h}}{{B}_{1}}\left( \lambda \right)=\sqrt{\lambda }\int_{0}^{a}{\sqrt{t}\phi \left( t \right){{J}_{\frac{3}{2}}}\left( \lambda t \right)dt}} \right.$$B1(λ)A1(λ)=λ0btψ(t)J12(λt)dt$${{B}_{1}}\left( \lambda \right)-{{A}_{1}}\left( \lambda \right)=\sqrt{\lambda }\int_{0}^{b}{\sqrt{t}}\psi \left( t \right){{J}_{\frac{1}{2}}}\left( \lambda t \right)dt$$

where J12${{J}_{\frac{1}{2}}}$and J32${{J}_{\frac{3}{2}}}$are the Bessel functions of the first kind of order 12$\frac{1}{2}$and 32$\frac{3}{2}$, respectively. The unknown functions are given by the following equation:

A1(λ)=eλhp(λ)(1+γ)λ0atϕ(t)J32(λt)dtλp(λ)0btψ(t)J12(λt)dt$${{A}_{1}}\left( \lambda \right)=\frac{{{e}^{-\lambda h}}}{p\left( \lambda \right)\left( 1+\gamma \right)}\sqrt{\lambda }\int_{0}^{a}{\sqrt{t}}\phi \left( t \right){{J}_{\frac{3}{2}}}\left( \lambda t \right)dt-\frac{\sqrt{\lambda }}{p\left( \lambda \right)}\int_{0}^{b}{\sqrt{t}}\psi \left( t \right){{J}_{\frac{1}{2}}}\left( \lambda t \right)dt$$B1(λ)=eλhp(λ)(1+γ)λ0atϕ(t)J32(λt)dt+λ(1λp(λ))0btψ(t)J12(λt)dt$${{B}_{1}}\left( \lambda \right)=\frac{{{e}^{-\lambda h}}}{p\left( \lambda \right)\left( 1+\gamma \right)}\sqrt{\lambda }\int_{0}^{a}{\sqrt{t}}\phi \left( t \right){{J}_{\frac{3}{2}}}\left( \lambda t \right)dt+\sqrt{\lambda }\left( 1-\frac{\sqrt{\lambda }}{p\left( \lambda \right)} \right)\int_{0}^{b}{\sqrt{t}\psi }\left( t \right){{J}_{\frac{1}{2}}}\left( \lambda t \right)dt$$

where p(λ)=(1γ1+γ)e2λh+1andϕ(t)andψ(t)$p\left( \lambda \right)=\left( \frac{1-\gamma }{1+\gamma } \right){{e}^{-2\lambda h}}+1\,\,\,\,\,\text{and}\,\,\,\,\phi \left( t \right)\,\,\,\text{and}\,\,\,\psi \left( t \right)$are continuous unknown functions of t defined over two intervals 0t<a$0\le t<a$and 0t<b$0\le t<b$, respectively. Substituting A1(λ) and B1(λ) in Eqs. (8a) and (8c), we get

0atφ(t)dt0λ32f11(λ)J32(λt)J1(λr)dλ+0btψ(t)dt0λ32f12(λ)J12(λr)dλ=0r<a$$\begin{align}& \int_{0}^{a}{\sqrt{t}}\varphi \left( t \right)dt\int_{0}^{\infty }{{{\lambda }^{\frac{3}{2}}}}{{f}_{11}}\left( \lambda \right){{J}_{\frac{3}{2}}}\left( \lambda t \right){{J}_{1}}\left( \lambda r \right)d\lambda +\int_{0}^{b}{\sqrt{t}\psi \left( t \right)dt}\int_{0}^{\infty }{{{\lambda }^{\frac{3}{2}}}}{{f}_{12}}\left( \lambda \right){{J}_{\frac{1}{2}}}\left( \lambda r \right)d\lambda \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=0\,r<a \\ \end{align}$$0atφ(t)dt0λf21(λ)J32(λr)J1(λr)dλ+0btψ(t)dt0λf22(λ)J12(λt)J1(λr)dλ=ωrr<b$$\begin{align}& \int_{0}^{a}{\sqrt{t\varphi }}\left( t \right)dt\int_{0}^{\infty }{\sqrt{\lambda }}{{f}_{21}}\left( \lambda \right){{J}_{\frac{3}{2}}}\left( \lambda r \right){{J}_{1}}\left( \lambda r \right)d\lambda +\int_{0}^{b}{\sqrt{t}\psi \left( t \right)dt}\int_{0}^{\infty }{\sqrt{\lambda }{{f}_{22}}\left( \lambda \right)}{{J}_{\frac{1}{2}}}\left( \lambda t \right){{J}_{1}}\left( \lambda r \right)d\lambda \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\omega \,r\,r<b \\ \end{align}$$

where

f11(λ)=1p(λ)(1e2λh)f12(λ)=(1+γ)((11p(λ))eλh+1p(λ)eλh)f21(λ)=2eλhp(λ)(1+γ)f22(λ)=(12p(λ))$$\begin{align}& \,\,\,\,\,\,\,\,{{f}_{11}}\left( \lambda \right)=\frac{1}{p\left( \lambda \right)}\left( 1-{{e}^{-2\lambda h}} \right) \\ & {{f}_{12}}\left( \lambda \right)=\left( 1+\gamma \right)\left( \left( 1-\frac{1}{p\left( \lambda \right)} \right){{e}^{\lambda h}}+\frac{1}{p\left( \lambda \right)}{{e}^{-\lambda h}} \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{f}_{21}}\left( \lambda \right)=\frac{2{{e}^{-\lambda h}}}{p\left( \lambda \right)\left( 1+\gamma \right)} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{f}_{22}}\left( \lambda \right)=\left( 1-\frac{2}{p\left( \lambda \right)} \right) \\ \end{align}$$

The expression for p(λ) approaches to 1 for large values of λ. Equation 14 can be converted to the Abel integral equation by means of the relation λJ1(λr)=1r2ddr[r2J2(λr)],$\lambda {{J}_{1}}\left( \lambda r \right)=\frac{1}{{{r}^{2}}}\frac{d}{dr}\left[ {{r}^{2}}{{J}_{2}}\left( \lambda r \right) \right],$and then, by taking into account the integral formula

0λJ32(λt)J2(λr)dλ={2πt32r2(r2t2)t<r0t>r$$\int_{0}^{\infty }{\sqrt{\lambda }}{{J}_{\frac{3}{2}}}\left( \lambda t \right){{J}_{2}}\left( \lambda r \right)d\lambda =\left\{ \begin{matrix}\sqrt{\frac{2}{\pi }\frac{{{t}^{\frac{3}{2}}}}{{{r}^{2}}\sqrt{\left( {{r}^{2}}-{{t}^{2}} \right)}}} & t<r \\0 & t>r \\\end{matrix} \right.$$

we obtain the Abel equation corresponding to Eq. (14)

2π0rt2φ(t)(r2t2)dt+r20atφ(t)dt0λ12(f11(λ)1)J32(λt)J2(λr)dλ+r20btψ(t)dt0λ12f12(λ)I12(λr)J2(λr)dλ=0,r<a$$\begin{array}{l}\sqrt{\frac2\pi}\int_0^r\frac{t^2\varphi{(t)}}{\sqrt{(r^2-t^2)}}dt+r^2\int_0^a\sqrt{t\varphi}{(t)}dt\int_0^\infty\lambda^\frac12{(f_{11}{(\lambda)}-1)}J_\frac32{(\lambda t)}J_2{(\lambda r)}d\lambda\\\text{ }\text{ }\text{ }\text{ }\text{ }+r^2\int_0^b\sqrt t\psi{(t)}dt\int_0^\infty\lambda^\frac12f_{12}{(\lambda)}I_\frac12{(\lambda r)}J_2{(\lambda r)}d\lambda=0,\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }r<a\end{array}$$

Next, we invert the last equation by applying the Abel transform formula

0rftr2t2dt=grthenft=2πddt0trgrt2r2dr¯$$\int_{0}^{r}{\frac{f\left( t \right)}{\sqrt{\left( {{r}^{2}}-{{t}^{2}} \right)}}d}t=g\left( r \right)\text{then}f\left( t \right)=\frac{2}{\pi }\frac{d}{dt}\int_{0}^{t}{\frac{rg\left( r \right)}{\sqrt{\left( {{t}^{2}}-{{r}^{2}} \right)}}}d{{r}^{\overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}$$

to obtain

t2φ(t)=2πddt0tr3(t2r2)[0aδφ(δ)dδ0λ12(f11(λ)1)J32(λδ)J2(λr)dλ0bδψ(δ)dδ0λ12f12(λ)J12(λt)J2(λr)dλ]dr=0,r<a$$\begin{align}& {{t}^{2}}\varphi \left( t \right)=\sqrt{\frac{2}{\pi }\frac{d}{dt}}\int_{0}^{t}{\frac{{{r}^{3}}}{\sqrt{\left( {{t}^{2}}-{{r}^{2}} \right)}}\left[ -\int_{0}^{a}{\sqrt{\delta }\varphi \left( \delta \right)d\delta \int_{0}^{\infty }{{{\lambda }^{\frac{1}{2}}}\left( {{f}_{11}}\left( \lambda \right)-1 \right){{J}_{\frac{3}{2}}}\left( \lambda \delta \right){{J}_{2}}\left( \lambda r \right)}} \right.}d\lambda \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. -\int_{0}^{b}{\sqrt{\delta }\psi \left( \delta \right)d\delta }\int_{0}^{\infty }{{{\lambda }^{\frac{1}{2}}}{{f}_{12}}}\left( \lambda \right){{J}_{\frac{1}{2}}}\left( \lambda t \right){{J}_{2}}\left( \lambda r \right)d\lambda \right]dr=0,\,\,\,\,\,\,\,r<a \\ \end{align}$$

For the left-hand side of the above equation, the integral is further simplified by using the following relationship:

2πddt0tr3t2r2J2(λr)dr=λt52J32(λt)$$\sqrt{\frac{2}{\pi }\frac{d}{dt}}\int_{0}^{t}{\frac{{{r}^{3}}}{\sqrt{{{t}^{2}}-{{r}^{2}}}}}{{J}_{2}}\left( \lambda r \right)dr=\sqrt{\lambda }{{t}^{\frac{5}{2}}}{{J}_{\frac{3}{2}}}\left( \lambda t \right)$$

we obtain the first Fredholm integral equation of the second kind

φ(t)+t0aδφ(δ)K(t,δ)dδ+r0bδψ(δ)L(t,δ)dδ=0,0<t<a$$\begin{align}& \varphi \left( t \right)+\sqrt{t}\int_{0}^{a}{\sqrt{\delta }\varphi }\left( \delta \right)K\left( t,\delta \right)d\delta +\sqrt{r}\int_{0}^{b}{\sqrt{\delta }\psi \left( \delta \right)}L\left( t,\delta \right)d\delta =0, \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0<t<a \\ \end{align}$$

where

K(t,δ)=0λ(f11(λ)1)J32(λt)J32(λδ)L(t,δ)=0λf12(λ)J32(λt)J12(λδ)dλ$$\begin{align}& K\left( t,\delta \right)=\int_{0}^{\infty }{\lambda \left( {{f}_{11}}\left( \lambda \right)-1 \right){{J}_{\frac{3}{2}}}}\left( \lambda t \right){{J}_{\frac{3}{2}}}\left( \lambda \delta \right) \\ & \,\,\,L\left( t,\delta \right)=\int_{0}^{\infty }{\lambda {{f}_{12}}\left( \lambda \right){{J}_{\frac{3}{2}}}}\left( \lambda t \right){{J}_{\frac{1}{2}}}\left( \lambda \delta \right)d\lambda \\ \end{align}$$

Following the similar procedure as before, Eq. (15) is reduced to the second Fredholm integral equation. Using the following formula

0λJ12(λt)J1(λr)dλ={2tπ1r(r2t2)t<r0t>r$$\int_{0}^{\infty }{\sqrt{\lambda }{{J}_{\frac{1}{2}}}}\left( \lambda t \right){{J}_{1}}\left( \lambda r \right)d\lambda =\left\{ \begin{matrix}\sqrt{\frac{2t}{\pi }}\frac{1}{r\sqrt{\left( {{r}^{2}}-{{t}^{2}} \right)}} & t<r \\0 & t>r \\\end{matrix} \right.$$

we obtain the following Abel-type equation:

1r2π0rtψtr2t2dt+0atφtdt0λf21λJ32λtJ1λrdλ+0btψtdt0λf22λ1λJ12λtJ1λrdλ=ωr,r<b$$ \begin{align} & \frac{1}{r}\sqrt{\frac{2}{\pi }}\int_{0}^{r}{\frac{t\psi \left( t \right)}{\sqrt{\left( {{r}^{2}}-{{t}^{2}} \right)}}}dt+\int_{0}^{a}{\sqrt{t}\varphi \left( t \right)dt\int_{0}^{\infty }{\sqrt{\lambda }{{f}_{21}}\left( \lambda \right){{J}_{\frac{3}{2}}}\left( \lambda t \right){{J}_{1}}\left( \lambda r \right)d\lambda }} \\ & +\int_{0}^{b}{\sqrt{t}\psi \left( t \right)dt}\int_{0}^{\infty }{\sqrt{\lambda }}\left( {{f}_{22}}\left( \lambda \right)-1 \right)\left( \lambda \right){{J}_{\frac{1}{2}}}\left( \lambda t \right){{J}_{1}}\left( \lambda r \right)d\lambda =\omega r, \\ & \,\,\,\,r<b \\ \end{align}$$

Now, we invert the above equation by applying the Abel transform formula to get

tψ(t)=2πddt0tr2(t2r2)[ωr0aδφ(δ)dδ0λf21(λ)J32(λδ)J1(λr)dλ0bδψ(δ)dδ0λ(f22(λ)1)J12(λδ)J1(λr)dλ]dr,r<b$$\begin{align}& t\psi \left( t \right)=\sqrt{\frac{2}{\pi }\frac{d}{dt}}\int_{0}^{t}{\frac{{{r}^{2}}}{\sqrt{\left( {{t}^{2}}-{{r}^{2}} \right)}}}\left[ \omega r-\int_{0}^{a}{\sqrt{\delta \varphi }}\left( \delta \right)d\delta \right.\int_{0}^{\infty }{\sqrt{\lambda }{{f}_{21}}\left( \lambda \right){{J}_{\frac{3}{2}}}}\left( \lambda \delta \right){{J}_{1}}\left( \lambda r \right)d\lambda \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. -\int_{0}^{b}{\sqrt{\delta }\psi \left( \delta \right)d\delta }\int_{0}^{\infty }{\sqrt{\lambda }\left( {{f}_{22}}\left( \lambda \right)-1 \right){{J}_{\frac{1}{2}}}\left( \lambda \delta \right){{J}_{1}}\left( \lambda r \right)d\lambda } \right]dr,r<b \\ \end{align}$$

Using the following relationships

ddt0tr3t2r2dr=2t22πddt0tr2J1(λr)t2r2dr=tλtJ12(λt)$$\begin{align}& \frac{d}{dt}\int_{0}^{t}{\frac{{{r}^{3}}}{\sqrt{{{t}^{2}}-{{r}^{2}}}}}dr=2{{t}^{2}} \\ & \sqrt{\frac{2}{\pi }\frac{d}{dt}}\int_{0}^{t}{\frac{{{r}^{2}}{{J}_{1}}\left( \lambda r \right)}{\sqrt{{{t}^{2}}-{{r}^{2}}}}}dr=t\sqrt{\lambda t}{{J}_{\frac{1}{2}}}\left( \lambda t \right) \\ \end{align}$$

we get the second Fredholm integral equation of the second kind

ψ(t)+t0qδφ(δ)M(t,δ)+t0bδψ(δ)N(t,δ)dδ=4ω2πt,0<t<b$$\begin{align}& \psi \left( t \right)+\sqrt{t}\int_{0}^{q}{\sqrt{\delta }\varphi }\left( \delta \right)M\left( t,\delta \right)+\sqrt{t}\int_{0}^{b}{\sqrt{\delta }\psi }\left( \delta \right)\,N\left( t,\delta \right)d\delta =\frac{4\omega }{\sqrt{2\pi }}t, \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0<t<b \\ \end{align}$$

with the kernel

M(t,δ)=0λf21(λ)J12(λt)J32(λδ)dλN(t,δ)=0λ(f22(λ)1)J12(λt)J12(λδ)dλ$$\begin{align}& \,\,\,M\left( t,\delta \right)=\int_{0}^{\infty }{\lambda {{f}_{21}}}\left( \lambda \right){{J}_{\frac{1}{2}}}\left( \lambda t \right){{J}_{\frac{3}{2}}}\left( \lambda \delta \right)d\lambda \\ & N\left( t,\delta \right)=\int_{0}^{\infty }{\lambda \left( {{f}_{22}}\left( \lambda \right)-1 \right)}{{J}_{\frac{1}{2}}}\left( \lambda t \right){{J}_{\frac{1}{2}}}\left( \lambda \delta \right)d\lambda \\ \end{align}$$

To get a non-dimensionalized equation, from Eqs. (18) to (22), let us change the variables as follows

δ={as0<δ<abs0<δ<b,t={au0<t<abu0<t<b,$$\delta =\left\{ \begin{matrix}as & 0<\delta <a \\bs & 0<\delta <b \\\end{matrix} \right.,\,\,\,\,\,t=\left\{ \begin{matrix}au & 0<t<a \\bu & 0<t<b, \\\end{matrix} \right.\,$$

Next, we multiply the above two equations of the system by 2π4aωφ(au)and2π4bωψ(bu),$\frac{\sqrt{2\pi }}{4a\omega }\varphi \left( au \right)\,\text{and}\,\frac{\sqrt{2\pi }}{4b\omega }\psi \left( bu \right),$respectively, and using the following substitutions

{Φ(u)=2π4aωφ(au)Ψ(u)=2π4bωψ(bu)c=abλ=xbH=hb$$\left\{ \begin{matrix}\Phi \left( u \right)=\frac{\sqrt{2\pi }}{4a\omega }\varphi \left( au \right) & \Psi \left( u \right)=\frac{\sqrt{2\pi }}{4b\omega }\psi \left( bu \right) \\c=\frac{a}{b}\,\,\lambda =\frac{x}{b} & H=\frac{h}{b} \\\end{matrix} \right.$$

we obtain the following equations:

Φ(u)+c2u01sΦ(s)K(u,s)ds+1cu0bsΨ(s)L(u,s)ds=0,u<1$$\Phi \left( u \right)+{{c}^{2}}\sqrt{u}\int_{0}^{1}{\sqrt{s}\Phi }\left( s \right)\,K\left( u,s \right)ds+\frac{1}{\sqrt{c}}\sqrt{u}\int_{0}^{b}{\sqrt{s}}\Psi \left( s \right)L\left( u,s \right)ds=0,\,\,\,\,\,\,u<1$$Ψ(u)+c2cu01sΦ(s)M(u,s)ds+u0bsΨ(s)N(u,s)ds=0,u<1$$\Psi \left( u \right)+{{c}^{2}}\sqrt{c}\sqrt{u}\int_{0}^{1}{\sqrt{s}}\Phi \left( s \right)\,M\left( u,s \right)ds+\sqrt{u}\int_{0}^{b}{\sqrt{s}}\Psi \left( s \right)\,N\left( u,s \right)ds=0,\,\,\,\,u<1$$

where

K(u,s)=0x(f11(x)1)J32(xcu)J32(xcs)dxL(t,δ)=0xf12(x)J32(xcu)J12(xs)dxM(t,δ)=0xf21(x)J12(xu)J32(xcs)dxN(t,δ)=0x(f22(x)1)J12(xu)J12(xs)dx$$\begin{align}& K\left( u,s \right)=\int_{0}^{\infty }{x\left( {{f}_{11}}\left( x \right)-1 \right)}{{J}_{\frac{3}{2}}}\left( xcu \right){{J}_{\frac{3}{2}}}\left( xcs \right)dx \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,L\left( t,\delta \right)=\int_{0}^{\infty }{x}{{f}_{12}}\left( x \right){{J}_{\frac{3}{2}}}\left( xcu \right){{J}_{\frac{1}{2}}}\left( xs \right)dx \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,M\left( t,\delta \right)=\int_{0}^{\infty }{x{{f}_{21}}\left( x \right){{J}_{\frac{1}{2}}}\left( xu \right){{J}_{\frac{3}{2}}}\left( xcs \right)dx} \\ & N\left( t,\delta \right)=\int_{0}^{\infty }{x}\left( {{f}_{22}}\left( x \right)-1 \right){{J}_{\frac{1}{2}}}\left( xu \right){{J}_{\frac{1}{2}}}\left( xs \right)dx \\ \end{align}$$
Numerical Results and Discussion

As the kernels K, L, M, and N are continuous on the interval [1], the system of Fredholm integral equations can be solved by direct or iterative techniques [21]. The midpoint quadrature [22] is used to find the numerical solution for the system given by Eqs. (25a) and (25b). By dividing the interval [1] into N equal subintervals, so that the midpoints are u=um=2m12s=un=2n12m,n=1,2...N.,$u={{u}_{m}}=\frac{2m-1}{2}\,\,\,\,\,s={{u}_{n}}=\frac{2n-1}{2}\,\,\,\,m,n=1,\,\,\,2...N.,$and by introducing the following notations

Φ(um)=Φmψ(um)=Ψm$$\Phi \left( {{u}_{m}} \right)={{\Phi }_{m}}\,\,\,\,\,\,\,\,\,\,\,\,\psi \left( {{u}_{m}} \right)={{\Psi }_{m}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$K(um,un)=KmmL(um,un)=Lmn$$K\left( {{u}_{m}},{{u}_{n}} \right)={{K}_{mm}}\,\,\,\,\,L\left( {{u}_{m}},{{u}_{n}} \right)={{L}_{mn}}$$

We evaluate numerically the infinite integral K, L, M, and N using the Simpson rule. After solving the above system, the unknown coefficients can be obtained.

Stress intensity factors

The stress intensity factors at the edge of the crack and at the rim of the disc are defined, respectively, by the following equations:

KIIIa=limra+2π(ra)τzθ1(r,z)|z=H$$K_{III}^{a}=\underset{r\to {{a}^{+}}}{\mathop{\lim }}\,\sqrt{2\pi \left( r-a \right)}\tau _{z\theta }^{1}\left( r,z \right){{|}_{z=H}}$$KIIIb=limrb2π(rb)τzθ1(r,z)|z=0$$K_{III}^{b}=\underset{r\to {{b}^{-}}}{\mathop{\lim }}\,\sqrt{2\pi \left( r-b \right)}\tau _{z\theta }^{1}\left( r,z \right){{|}_{z=0}}$$

On the planes z = h for ra and z = 0, the expressions of stress are given by

τzθ1(r,θ)=G10λ32[1e2λhp(λ)(1+γ)0atϕ(t)J32(λt)dt+(eλh(11p(λ))eλhp(λ))0btψ(t)J12(λt)dt]J1(λr)dλ$$\begin{align}& \tau _{z\theta }^{1}\left( r,\theta \right)={{G}_{1}}\int_{0}^{\infty }{{{\lambda }^{\frac{3}{2}}}}\left[ \frac{1-{{e}^{-2\lambda h}}}{p\left( \lambda \right)\left( 1+\gamma \right)} \right.\int_{0}^{a}{\sqrt{t}\phi \left( t \right){{J}_{\frac{3}{2}}}}\left( \lambda t \right)dt+ \\ & \left. \left( {{e}^{\lambda h}}\left( 1-\frac{1}{p\left( \lambda \right)} \right)-\frac{{{e}^{\lambda h}}}{p\left( \lambda \right)} \right)\int_{0}^{b}{\sqrt{t}\psi \left( t \right){{J}_{\frac{1}{2}}}\left( \lambda t \right)dt} \right]{{J}_{1}}\left( \lambda r \right)d\lambda \\ \end{align}$$τzθ1(r,h)=G10λ320b[tψ(t)J12(λt)dt]J1(λr)dλ$$\tau _{z\theta }^{1}\left( r,h \right)={{G}_{1}}\int_{0}^{\infty }{{{\lambda }^{\frac{3}{2}}}}\int_{0}^{b}{\left[ \sqrt{t}\psi \left( t \right){{J}_{\frac{1}{2}}}\left( \lambda t \right)dt \right]}{{J}_{1}}\left( \lambda r \right)d\lambda $$

The second part of the integrals in Eq. (29) converge quickly as their limits ra and rb automatically vanishes; however, the limits of the other two integrals analyzed asymptotically as follows. Using the relation J1(λR)=1RddRJ0(λR),${{J}_{1}}\left( \lambda \,R \right)=-\frac{1}{R}\frac{d}{dR}{{J}_{0}}\left( \lambda \,R \right),$we obtain

τzθ1(r,θ)=G10atϕ(t)dt0λ12p(λ)(1+γ)J32(λt)J0(λr)dλG10λ32[e2λhp(λ)(1+γ)0atϕ(t)J32(λt)dt+(eλh(11p(λ))eλhp(λ))0btψ(t)J12(λt)dt]J1(λr)dλ$$\begin{align}& \tau _{z\theta }^{1}\left( r,\theta \right)=-{{G}_{1}}\int_{0}^{a}{\sqrt{t}\phi \left( t \right)dt\int_{0}^{\infty }{\frac{{{\lambda }^{\frac{1}{2}}}}{p\left( \lambda \right)\left( 1+\gamma \right)}{{J}_{\frac{3}{2}}}\left( \lambda t \right){{J}_{0}}\left( \lambda r \right)d\lambda -{{G}_{1}}\int_{0}^{\infty }{{{\lambda }^{\frac{3}{2}}}\left[ \frac{{{e}^{-2\lambda h}}}{p\left( \lambda \right)\left( 1+\gamma \right)} \right.}}} \\ & \left. \int_{0}^{a}{\sqrt{t}\phi }\left( t \right){{J}_{\frac{3}{2}}}\left( \lambda t \right)dt+\left( {{e}^{\lambda h}}\left( 1-\frac{1}{p\left( \lambda \right)} \right)-\frac{{{e}^{\lambda h}}}{p\left( \lambda \right)} \right)\int_{0}^{b}{\sqrt{t}\psi \left( t \right){{J}_{\frac{1}{2}}}\left( \lambda t \right)dt} \right]{{J}_{1}}\left( \lambda r \right)d\lambda \\ \end{align}$$τzθ1(r,h)=G10btψ(t)dt0λ12J12(λt)J0(λr)dλ$$\tau _{z\theta }^{1}\left( r,h \right)=-{{G}_{1}}\int_{0}^{b}{\sqrt{t}}\psi \left( t \right)dt\int_{0}^{\infty }{{{\lambda }^{\frac{1}{2}}}}{{J}_{\frac{1}{2}}}\left( \lambda t \right){{J}_{0}}\left( \lambda r \right)d\lambda $$

For large values of λ, we use the following asymptotic behavior of the Bessel function of the first Jv(λ)2λπcos(λvπ2π4)${{J}_{v}}\left( \lambda \right)\cong \sqrt{\frac{2}{\lambda \pi }}\cos \left( \lambda -v\frac{\pi }{2}-\frac{\pi }{4} \right)$and using the following integral formulas for the first infinite integral in the right part of the Eqs. (31) and (32), respectively,

0cos(λt)J0(λr)dλ={1(r2t2)t<r0t>rand0sin(λt)J0(λr)dλ={0t<r1(t2r2)t>r$$\begin{align}& \int_{0}^{\infty }{\cos }\left( \lambda t \right){{J}_{0}}\left( \lambda r \right)d\lambda =\left\{ \begin{matrix}\frac{1}{\sqrt{\left( {{r}^{2}}-{{t}^{2}} \right)}} & t<r \\0 & t>r \\\end{matrix} \right.\text{and} \\ & \,\,\,\,\int_{0}^{\infty }{sin}\left( \lambda t \right){{J}_{0}}\left( \lambda r \right)d\lambda =\left\{ \begin{matrix}0 & t<r \\\frac{1}{\sqrt{\left( {{t}^{2}}-{{r}^{2}} \right)}} & t>r \\\end{matrix} \right. \\ \end{align}$$

As p(λ) → 1 as λ → ∞, Eqs. (31) and (32) become

τzθ1(r,θ)=G1γ+12πddr0aϕ(t)(r2t2)dt+R1(r)$$\tau _{z\theta }^{1}\left( r,\theta \right)=\frac{{{G}_{1}}}{\gamma +1}\sqrt{\frac{2}{\pi }\frac{d}{dr}}\int_{0}^{a}{\frac{\phi \left( t \right)}{\sqrt{\left( {{r}^{2}}-{{t}^{2}} \right)}}}dt+{{R}_{1}}\left( r \right)$$τzθ1(r,h)=G12πddr0bψ(t)(t2r2)dt$$\tau _{z\theta }^{1}\left( r,h \right)=-{{G}_{1}}\sqrt{\frac{2}{\pi }\frac{d}{dr}}\int_{0}^{b}{\frac{\psi \left( t \right)}{\sqrt{\left( {{t}^{2}}-{{r}^{2}} \right)}}}dt$$

where

R1(r)=G10λ32[e2λhp(λ)(1+γ)0atϕ(t)J32(λt)dt+(eλh(11p(λ))eλhp(λ))0btψ(t)J12dt]J1(λr)dλ$$\begin{align}& {{R}_{1}}\left( r \right)=-{{G}_{1}}\int_{0}^{\infty }{{{\lambda }^{\frac{3}{2}}}}\left[ \frac{{{e}^{-2\lambda h}}}{p\left( \lambda \right)\left( 1+\gamma \right)}\int_{0}^{a}{\sqrt{t}\phi \left( t \right){{J}_{\frac{3}{2}}}\left( \lambda t \right)dt}+ \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \left( {{e}^{\lambda h}}\left( 1-\frac{1}{p\left( \lambda \right)} \right)-\frac{{{e}^{\lambda h}}}{p\left( \lambda \right)} \right)\int_{0}^{b}{\sqrt{t}\psi \left( t \right){{J}_{\frac{1}{2}}}dt} \right]{{J}_{1}}\left( \lambda r \right)d\lambda \\ \end{align}$$

Now integrating by parts, we get

τzθ1(r,0)=G1γ+12π[aϕ(a)r(r2a2)0atϕ(t)(r2a2)dt]+R1(r)$$\tau _{z\theta }^{1}\left( r,0 \right)=\frac{{{G}_{1}}}{\gamma +1}\sqrt{\frac{2}{\pi }}\left[ \frac{a\phi \left( a \right)}{r\sqrt{\left( {{r}^{2}}-{{a}^{2}} \right)}}\int_{0}^{a}{\frac{t{\phi }'\left( t \right)}{\sqrt{\left( {{r}^{2}}-{{a}^{2}} \right)}}dt} \right]+{{R}_{1}}\left( r \right)$$

We note that the infinite integrals in the preceding expressions are convergent throughout the medium except at the singular points ra+, which occupy the crack boundary.

τzθ1(r,h)=G12π[bψ(b)r(b2r2)rbtψ(t)(r2t2)dt]$$\tau _{z\theta }^{1}\left( r,h \right)={{G}_{1}}\sqrt{\frac{2}{\pi }}\left[ \frac{b\psi \left( b \right)}{r\sqrt{\left( {{b}^{2}}-{{r}^{2}} \right)}}-\int_{r}^{b}{\frac{t{\psi }'\left( t \right)}{\sqrt{\left( {{r}^{2}}-{{t}^{2}} \right)}}dt} \right]$$

In this case, the integral in the above relation converges quickly and the integral is bonded as rb. As a result, we obtain a square root singularity at r = b and the constant ψ(b) is the measure of the strength of singularity at the vicinity of the rigid inclusion. By using the following transformations ϕ(a)=4aω2πΦN,$\phi \left( a \right)=\frac{4a\omega }{\sqrt{2\pi }}{{\Phi }_{N}},$ψ(b)=4bω2πΨN.$\psi \left( b \right)=\frac{4b\omega }{\sqrt{2\pi }}{{\Psi }_{N}}.$.

we obtain the stress intensity factor at the edge of the crack and at the rim of the disc

KIIIa=4G1ωa(1+γ)πΦN$$K_{III}^{a}=\frac{-4{{G}_{1}}\omega \sqrt{a}}{\left( 1+\gamma \right)\sqrt{\pi }}{{\Phi }_{N}}$$KIIIb=4G1ωb(1+γ)πΨN$$K_{III}^{b}=\frac{4{{G}_{1}}\omega \sqrt{b}}{\left( 1+\gamma \right)\sqrt{\pi }}{{\Psi }_{N}}$$

Figure 2 shows the variation of the normalized stress intensity factor KIIIa$K_{III}^{a}$at the edge of the crack defined by Eq. (37a) against c for various values of the layer thickness H = 1, 0.75, 0.5, 0.25 calculated using the shear modulus ratio γ = 1. It is observed that the values of stress intensity factor increase and attain its maximum values at c = 1, and with the increase in the value of c, the stress intensity factor decreases. In addition, the effect of the axial distance between the crack and the disc H on the stress intensity factor is also shown in this figure. The increase in the distance H induces the decrease in stress intensity factor for all the values of parameter c. Figure 3 illustrates the variation of the stress intensity factor at the crack because of variations in shear modulus ratio γ for different values: the normalized crack size c = 0.25, 0.5, 0.75, 1 and the layer thickness H = 1. We observe from the figure that as the shear parameter γ increases, the stress intensity factor decreases for all values of c.

Figure 2

Variation of the normalized stress intensity factor at the edge of the crack KIIIa$K_{III}^{a}$with a/b.

Figure 3

Variation of the normalized stress intensity factor at the edge of the crack KIIIa$K_{III}^{a}$with G1 ⁄ G2.

Figure 4 illustrates the variation of the normalized stress intensity factor KIIIb$K_{III}^{b}$at the edge of the rigid inclusion defined by Eq. (37b) versus c for H = 1, 0.75, 0.5, and 0.25. Relatively, small variation for smaller values of c and considerable variation for larger values of c are observed. Also, the interaction between the cracks and the rigid disc is greater when the cracks are closer to the disc. In addition to the interaction, the stress intensity factor KIIIb$K_{III}^{b}$KIIIb$K_{III}^{b}$increases as the crack radius increases. From the

Figure 4

Variation of the normalized stress intensity factor at the edge of the rigid disc KIIIb$K_{III}^{b}$with a/b.

formulation and the presented figures, conclusions may be deduced:

Singularity at the edge of the internal crack and the internal rigid inclusion is observed (The results seem to agree with the previous works (Low [16], Dhawan [17], and Madani and Kebli [20]).

The Mode III stress intensity factor KIIIb$K_{III}^{b}$at the edge of the crack is negative, and it decreases with increasing γ.

There is considerable interaction between the cracks and the rigid inclusion when a/b is large and the cracks are close to the rigid inclusion.

The moment required to produce rotation of the disc

The torque required to sustain the rotation of the disc can be computed using the following equation:

T=2π0br2τzθ1(r,0)dr$$T=2\pi \int_{0}^{b}{{{r}^{2}}\tau _{z\theta }^{1}}\left( r,0 \right)dr$$

Using the relation 0br2J1(λr)dr=b2λJ2(λb),$\int_{0}^{b}{{{r}^{2}}{{J}_{1}}}\left( \lambda r \right)dr=\frac{{{b}^{2}}}{\lambda }{{J}_{2}}\left( \lambda b \right),$we get

T=2πb2G10[A1(λ)+B1(λ)]J2(λb)dλ$$T=2\pi {{b}^{2}}{{G}_{1}}\int_{0}^{\infty }{\left[ {{A}_{1}}\left( \lambda \right)+{{B}_{1}}\left( \lambda \right) \right]}{{J}_{2}}\left( \lambda b \right)d\lambda $$

Here, as the moment is applied only to the rigid inclusion, the integrand is expressed in terms of ψ(t). Substituting the values of A1(λ) and B1(λ) from Eqs. (12) and (13) into Eq. (39) and using the asymptotic behavior of the Bessel function of the first kind J12${{J}_{\frac{1}{2}}}$, we find that

T=22πb2G10bψ(t)dt0sin(λt)J2(λb)dλ$$T=2\sqrt{2}\pi {{b}^{2}}{{G}_{1}}\int_{0}^{b}{\psi \left( t \right)dt}\int_{0}^{\infty }{\text{sin}\left( \lambda t \right)}{{J}_{2}}\left( \lambda b \right)d\lambda $$

Taking into account the relation 0sin(λt)J2(λb)dλ=2tb2,$\int_{0}^{\infty }{\text{sin}\left( \lambda t \right){{J}_{2}}}\left( \lambda b \right)d\lambda =\frac{2t}{{{b}^{2}}},$ we obtain the moment applied to the inclusion

T=42πG10bψ(t)dt$$T=4\sqrt{2\pi }{{G}_{1}}\int_{0}^{b}{\psi \left( t \right)\,\,}dt$$

By using the following transformations t = bu and ψ(bu)=4bω2πψu.$\psi \left( bu \right)=\frac{4b\omega }{\sqrt{2\pi }}{{\psi }_{\text{u}}}.$

T=16ωb3G10buΨ(u)du$$T=16\omega {{b}^{3}}{{G}_{1}}\int_{0}^{b}{u\Psi \left( u \right)\,}du$$

The moment required to effect the rotation ω, when the medium contain no crack, can be formulated as T0=16ωb3G13.${{T}_{0}}=\frac{16\omega {{b}^{3}}{{G}_{1}}}{3}.$Equation 42 can be expressed as

TT0=30buΨ(u)du$$\frac{T}{{{T}_{0}}}=3\int_{0}^{b}{u\Psi \left( u \right)}\,du$$

The problem of inclusion–crack interaction has a remarkable use in the design of composite anchoring systems with flat disc inclusions, in deep foundations in a geological medium, in in situ load tests at the base of a borehole, in injection anchoring regions in soft rock masses with sealing materials, or when penetrating single-propelled anchors in steep soil masses, such as over-consolidated clays.

Displacement and stress fields

The results for the variation of the normalized displacement u(i)(ρ, ξ)/ωa and stress τ(i)(ρ, ξ)/Giωa with ρ = r/b are shown graphically in Figures 58 for the different values of the dimensionless axial distances ξ = z/b. For each region, five different axial distances are selected as I (ξ =0; H/5; 2H/5; 3H/5; H) and I I (ξ = H ; 6H/5; 7H/5; 8H/5; 2H), with the particular values of the height H = 1, the dimensionless

Figure 5

Tangential displacement uθ1$\text{u}_{\text{ }\!\!\theta\!\!\text{ }}^{1}$versus ρ for various ξ, 0≤z≤h.

crack size c = 1 and the shear parameter γ = 1 and γ = 2. The variation of the normalized displacements is shown in Figures 5 and 6. We notice that the displacements in the two regions increase at first, reach maximum values at ρ = c, and then decrease out of the disc band with increasing ρ. The distribution of the shear stresses in the elastic medium are also discussed and shown in Figures 7 and 8. The stresses initially increase, attain its maximum values, and then with the increase in the value of ρ, the stresses go on decreasing.

Figure 6

Tangential displacement uθ2$\text{u}_{\theta }^{2}$versus ρ for various ξ, z≥h.

Figure 7

Shear stress τθz1${{\text{ }\!\!\tau\!\!\text{ }}_{\text{ }\!\!\theta\!\!\text{ z}}}^{\text{1}}$versus ρ for various ξ, 0≤z≤h.

Figure 8

Shear stress τθz2${{\text{ }\!\!\tau\!\!\text{ }}_{\text{ }\!\!\theta\!\!\text{ z}}}^{\text{2}}$versus ρ for various ξ, z≥h.

Conclusion

In this article, an axisymmetric torsion of a rigid disc embedded in the interior of a homogeneous elastic layer sandwiched between two half-spaces containing two interface cracks is analytically addressed. By using the Hankel integral transformation and its inverse, the mixed boundary value problem is reduced to a system of dual integral equations, which are reduced to a Fredholm integral equation system of the second kind. The numerically computed results of the displacements, the stresses, and the stress intensity factors are presented graphically for some dimensionless parameters. The numerical results show that the discontinuities around the crack and the inclusion cause a large increase in the stresses that decay with distance from the disc loaded. Furthermore, the dependence of the stress intensity factor on the disc size, the distance between the crack and the disc and the nonhomogeneity parameter is observed.

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