# Dual Boundary Element Method Applied to Antiplane Crack Problems

Authors: Wei-Liang Wu
Journal: Mathematical Problems in Engineering
 (2009)

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## Abstract

This paper is concerned with an efficient dual boundary element
method for 2d crack problems under antiplane shear loading. The dual equations are the displacement and the traction boundary integral equations. When the displacement equation is applied on the outer boundary and the traction equation on one of the crack surfaces, general crack problems with anti-plane shear loading can be solved with a single region formulation. The outer boundary is discretised with continuous quadratic elements; however, only one of the crack surfaces needs to be discretised with discontinuous quadratic elements. Highly accurate results are obtained, when the stress intensity factor is evaluated with the discontinuous quarter point element method. Numerical examples are provided to demonstrate the accuracy and efficiency of the present formulation.


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## Body

## 1. Introduction

The problem of a cracked body subjected to an antiplane shear loading had been studied extensively. Sih [1] provided analytical solutions for mode III cracks in infinite regions by using Westergaard stress functions and Muskhelishvili's method. Chiang [2] presented analytical solutions for slightly curved cracks in antiplane strain in infinite regions using perturbation procedures similar to those carried out for in-plane loading cases by Cotterell and Rice [3]. Zhang [4, 5] and Ma and Zhang [6] gave analytical solutions for a mode III stress intensity factor considering a finite region with an eccentric straight crack. Ma [7] provided analytical solutions for mode III straight cracks in finite regions using Fourier transforms and Fourier series. Smith [8] studied the elastic stress distribution in the immediate vicinity of a blunt notch. However, their solutions were concerned with specified geometries or boundary conditions. To deal with the complexities of general geometries and boundary conditions, an accurate and efficient numerical method is essential [9–12].Several numerical solutions had been devised for antiplane crack problems. Wallentin et al. [13] investigated the railway wheel crack problem numerically, based on Betti's reciprocity theorem. Guagliano and Vergani [14] described the experimental and numerical analysis of internal cracks in wheels under Hertzian loads. Paulino et al. [15] provided numerical solutions for a curved crack subjected to an antiplane shear loading in finite regions by using the boundary integral equation method. Ting et al. [16] provided numerical solutions for mode III crack problems by using the boundary element alternating method. Liu and Altiero [17] provided numerical solutions for mode III crack problems using the boundary integral equation with linear approximation on displacements and stresses. Barlow and Chandra [18] discussed the computational fatigue crack growth rate by using the crack opening displacement approach to calculate the stress intensity factors. Mews and Kuhn [19] provided numerical solutions for the traction free central crack problem by using Green's function, instead of the usual fundamental solution. Mir-Mohamad-Sadegh and Altiero [20] used the indirect boundary integral equation method to solve traction problems, using displacement-based formulations. Sun et al. [21] derived a new boundary integral equation to analyse cracked anisotropic bodies under antiplane shear. Also, for the further study, the crack front plastic deformation in a ductile material was introduced to apply the effective Dugdale strip yield model [22–24]. In general, the boundary element method (BEM) is a well-established numerical technique for the analysis of linear fracture mechanics problems. However, the solution of general crack problems cannot be achieved with the direct application of the BEM, because the coincidence of the crack surfaces gives rise to a singular system of algebraic equations.To overcome this shortcoming, we provide an efficient numerical procedure, based on the dual boundary element method (DBEM), for antiplane shear loading problems. The dual boundary element method seems to have certain apparent advantages for in-plane loading problems with a single region formulation. This method incorporates two independent boundary integral equations, the displacement and traction equations. Portela et al. [25] considered the effective numerical implementation of the two-dimensional DBEM for solving general in-plane fracture mechanics problems. W. H. Chen and T. C. Chen [26] proposed a different DBEM formulation for in-plane crack problems. Chen and Chen suggested the use of the displacement integral equation applied only on the outer boundary and the traction integral equation on one of the crack surfaces. In Chen and Chen's formulation, relative displacement of crack surfaces was used instead of the displacement. This reduces the degrees of freedom and hence the computational effort. This study uses an integral equation formulation that combines with the crack modelling strategy of quadratic boundary elements for antiplane crack problems. The stress intensity factor is calculated based on the near tip displacement method. More accurate results are obtained by placing discontinuous quarter point elements at crack tips [27], which correctly model the behaviour of the crack tip displacement. This is a similar technique to that used for continuous quarter point elements [28]. Numerical examples are provided to demonstrate the accuracy and efficiency of the present formulation.


## 2. The Dual Boundary Integral Equation for Antiplane Problems

Consider a finite domain subjected to an arbitrary antiplane shear loading, where the only nonzero displacement componentuz in the z direction may be specified as follows [7]:
(2.1)∇2uz=0.
The Laplace equation (2.1) can be transformed into a boundary integral equation, as is typical with the BEM. The boundary integral formulation of the displacement component, uz, at an internal point I, is given by [29]
(2.2)uz(I)+∫ΓH(I,x)uz(x)dΓ(x)=∫ΓG(I,x)tz(x)dΓ(x),
where tz represents the traction component at a boundary point x. H(I,x) and G(I,x) represent the fundamental traction and displacement solutions, respectively, which are given as
(2.3)H(I,x)=-12πr∂r∂n,G(I,x)=12πμln(1r),
where μ is the shear modulus, r is the distance between I and x, and n denotes the outward normal unit vector at the point x on the boundary Γ. If we consider a finite body with L cracks, (2.2) can be written as
(2.4)uz(I)+∫ΓSH(I,x)uz(x)dΓ(x)+∑l=1L∫Γl+H(I,x+)uz(x+)dΓ(x)+∑l=1L∫Γl-H(I,x-)uz(x-)dΓ(x)=∫ΓSG(I,x)tz(x)dΓ(x)+∑l=1L∫Γl+G(I,x+)tz(x+)dΓ(x)+∑l=1L∫Γl-G(I,x-)tz(x-)dΓ(x),
where x+ and x- are the field points located on upper and lower crack surfaces, respectively. Note that ΓS denotes the outer boundary of the body, Γl+ the lth upper crack boundary, Γl- the lth lower crack boundary, and Γ=ΓS+∑l=1L(Γl++Γl-). Using the fact that H(I,x+)∣Γ+=H(I,x-)∣-Γ- and G(I,x+)∣Γ+=G(I,x-)∣-Γ-, (2.4) can be simplified to
(2.5)uz(I)+∫ΓSH(I,x)uz(x)dΓ(x)+∑l=1L∫Γl+H(I,x+)Δuz(x)dΓ(x)=∫ΓSG(I,x)tz(x)dΓ(x)+∑l=1L∫Γl+G(I,x+)Δtz(x)dΓ(x),
where Δuz=uz(x+)-uz(x-) and Δtz=tz(x+)-tz(x-), however Δtz is always zero on the crack faces. As the internal point approaches the outer boundary, that is, as I→B, the displacement equation becomes
(2.6)c(B)uz(B)+⨍ΓSH(B,x)uz(x)dΓ(x)+∑l=1L∫Γl+H(B,x+)Δuz(x)dΓ(x)=∫ΓSG(B,x)tz(x)dΓ(x),
where ⨍ represents the Cauchy principle value integral and c(B)=1/2, given a smooth boundary at the point B.The stress componentsσiz are obtained from differentiation of equation (2.5), followed by the application of Hooke's law. At an internal point I, these components are given by
(2.7)σiz(I)+∫ΓSSi(I,x)uz(x)dΓ(x)+∑l=1L∫Γl+Si(I,x)Δuz(x)dΓ(x)=∫ΓSDi(I,x)tz(x)dΓ(x),
where Si(I,x) and Di(I,x) contain derivatives of H(I,x) and G(I,x) in the i direction, respectively, which are given as
(2.8)Si(I,x)=μ2πr2[∂r∂xi∂r∂n-(δij-∂r∂xj∂r∂xi)nj],Di(I,x)=-12πr∂r∂xi,
where ni denotes the ith component of the outward normal to the boundary at point x, and δij is the Kronecker delta. Again, by moving the source point I to the upper crack boundary B, and using tz=σizni, we obtain the traction integral equation
(2.9)12tz(B)+∫ΓSni(B)Si(B,x)uz(x)dΓ(x)+∑l=1L⨎Γl+ni(B)Si(B,x)Δuz(x)dΓ(x)=∫ΓSni(B)Di(B,x)tz(x)dΓ(x),
where ⨎ represents the Hadamard principal value integral. Both Cauchy and Hadamard principal-value integrals in (2.6) and (2.9) are finite parts of improper integrals. To solve the finite part integrals, we can follow the method mentioned in Portela et al. [25].The displacement integral equation (2.6) and the traction integral equation (2.9) are the two main equations to solve for the displacement of the outer boundary and the relative displacement of the crack faces. Equation (2.6) is applied for collocation on the outer boundary where continuous quadratic elements are used, and (2.9) is applied on the upper crack faces which are modelled by discontinuous quadratic elements. By taking all the discretised nodes on the outer boundary ΓS and upper crack surfaces ∑l=1LΓl+ at the source point B, the system of (2.6) and (2.9) for the multiple cracks problem can be written in a matrix form as
(2.10)[H1H20S1S2I][uz,SΔuz,ctz,c+]=[G1D1][tz,S],
where H1, H2, G1 and S1, S2, D1 are the corresponding assembled matrices from (2.6) and (2.9), respectively. The uz,S and tz,S are the displacement and traction vectors on the outer boundary ΓS, respectively. Δuz,c and tz,c+ are the relative displacement vector and the traction vector on the upper crack faces.


## 3. Calculation of the Mode III Stress Intensity Factor

Near tip displacement extrapolation is used to evaluate the numerical values of the stress intensity factor. The relative displacements of the crack surfaces are calculated using the DBEM and are used in the near crack tip stress field equations to obtain the stress intensity factor. Due to the singular behaviour of the stress around the crack tip, it is reasonable to expect a better approximation by replacing the normal discontinuous quadratic element with a transition element possessing the same order of singularity at the crack tip. The discontinuous quarter point element method is used in the present formulation [27, 30]. The mode III stress intensity factor is evaluated as
(3.1)KIII=μ42πrΔuz(r),
where r is the distance from the crack tip to the nearest node on the upper crack face, and Δuz(r) denotes the relative displacement in the antiplane direction.


## 4. Numerical Examples

In order to demonstrate the accuracy and efficiency of the technique previously described, and to illustrate possible applications, we now consider several examples. In all the numerical tests, the outer boundary is modelled by 24 continuous quadratic elements, and each crack discretization is carried out with three different meshes of 6, 8, and 10 discontinuous quadratic elements, respectively. The best accuracy is achieved with 6 elements, in which the crack discretization is graded, towards the tip, with ratios 0.25, 0.15, and 0.1. The plate is subjected to a uniform antiplane shear loadingτ, and the stress intensity factor is normalised with respect to
(4.1)K0=τπa,
where a defines the half length of the crack. All computations are carried out under the condition of plane strain.


### 4.1. A Rectangular Plate Containing a Central Slant Crack

Firstly, consider a rectangular plate containing a central slant crack as shown in Figure1. The crack has length 2a and makes an angle θ with the horizontal direction. For a horizontal crack (θ=0°), the normalised mode III stress intensity factor is calculated for various ratios of a/h and a/w and compared to those given in [17, 31] (see Table 1). The largest difference between these does not exceed 1.65 per cent. Further, the normalised mode III stress intensity factor is calculated for h/w=2, while the crack slanted an angle θ with the various ratios of a/w. Three cases are considered, where θ=30°, 45°, and 60°, respectively. The results obtained are presented in Figure 2. As it can be seen, when the ratio of a/w increases, the stress intensity factor increases due to edge effect.Table 1
Normalised mode III stress intensity factor for a straight central crack.
a:h1:0.251:0.51:11:21:4a:wPresent1.9091.7241.6891.6881.6611:1.2Reference [17]1.8971.7231.6891.6861.686Reference [31]1.9001.7251.6911.6891.689a:wPresent1.7961.4671.3711.3611.3611:1.4Reference [17]1.7801.4601.3691.3591.358Reference [31]1.7821.4631.3701.3611.360a:wPresent1.7841.4051.2571.2361.2361:1.6Reference [17]1.7711.3991.2541.2331.233Reference [31]1.7731.4011.2561.2351.235a:wPresent1.7921.3841.1791.1311.1291:2.0Reference [17]1.7701.3771.1761.1271.126Reference [31]1.7721.3791.1781.1301.128Figure 1
Rectangular plate with a central slant crack.Figure 2
The rectangular plate with a central slant crack at (a)θ=30°, (b) θ=45°, and (c) θ=60°.For the case wherea/w=1/50, which could be considered as the case of infinite geometry since a≪w, we compare the results with the analytical results for the latter as given in [32]. The results are plotted in Figure 3. Excellent agreement is observed; the maximum error is around 0.02 per cent.Figure 3
The infinite plate with a central slant crack: (a) the analytical solutions and (b) the present method.


### 4.2. A Rectangular Plate Containing Two Identical Collinear Cracks

As shown in Figure4, the second example is a rectangular plate containing two identical collinear cracks. 2a is the length of the inclined crack and 2d is the distance between the centre of the cracks. The geometric parameters are h/w=2 and a/w=1/50. Figure 4 displays the variations of normalised mode III stress intensity factors at tip A and tip B versus different ratios of a/d. Due to the interaction between the two cracks, the computed normalised mode III stress intensity factor at tip A is always larger than that at tip B. Hence, as the crack centre distance d decreases, the difference of stress intensity factor increases. There is excellent correlation between the computed results using the present method and those from analytical solutions; the difference between these results does not exceed 0.03 per cent at tip A or 0.09 per cent at tip B.The rectangular plate with two identical collinear cracks. Normalised mode III stress intensity factors versusa/d for tip A: (a) the analytical results and (b) the present method, and for tip B: (c) the analytical solutions and (d) the present method.
(a)(b)


### 4.3. An Infinite Plate Containing Two Parallel Cracks

The third example is an infinite plate (h/w=2, a/w=1/50) containing two parallel cracks, as shown in Figure 5. 2a is the length of the two identical cracks and 2d is the distance between the cracks. The computed results are compared with the published results in [32]. The results of normalised mode III stress intensity factor for different s are plotted in Figure 5, where s=a/(a+d). The effect of the interaction of cracks on the mode III stress intensity factor is observed. The largest difference between the present and the published results does not exceed 0.65 per cent.The rectangular plate with two parallel cracks. Normalised mode III stress intensity factors versuss for (a) [32] and (b) the present method.
(a)(b)


## 4.1. A Rectangular Plate Containing a Central Slant Crack

Firstly, consider a rectangular plate containing a central slant crack as shown in Figure1. The crack has length 2a and makes an angle θ with the horizontal direction. For a horizontal crack (θ=0°), the normalised mode III stress intensity factor is calculated for various ratios of a/h and a/w and compared to those given in [17, 31] (see Table 1). The largest difference between these does not exceed 1.65 per cent. Further, the normalised mode III stress intensity factor is calculated for h/w=2, while the crack slanted an angle θ with the various ratios of a/w. Three cases are considered, where θ=30°, 45°, and 60°, respectively. The results obtained are presented in Figure 2. As it can be seen, when the ratio of a/w increases, the stress intensity factor increases due to edge effect.Table 1
Normalised mode III stress intensity factor for a straight central crack.
a:h1:0.251:0.51:11:21:4a:wPresent1.9091.7241.6891.6881.6611:1.2Reference [17]1.8971.7231.6891.6861.686Reference [31]1.9001.7251.6911.6891.689a:wPresent1.7961.4671.3711.3611.3611:1.4Reference [17]1.7801.4601.3691.3591.358Reference [31]1.7821.4631.3701.3611.360a:wPresent1.7841.4051.2571.2361.2361:1.6Reference [17]1.7711.3991.2541.2331.233Reference [31]1.7731.4011.2561.2351.235a:wPresent1.7921.3841.1791.1311.1291:2.0Reference [17]1.7701.3771.1761.1271.126Reference [31]1.7721.3791.1781.1301.128Figure 1
Rectangular plate with a central slant crack.Figure 2
The rectangular plate with a central slant crack at (a)θ=30°, (b) θ=45°, and (c) θ=60°.For the case wherea/w=1/50, which could be considered as the case of infinite geometry since a≪w, we compare the results with the analytical results for the latter as given in [32]. The results are plotted in Figure 3. Excellent agreement is observed; the maximum error is around 0.02 per cent.Figure 3
The infinite plate with a central slant crack: (a) the analytical solutions and (b) the present method.


## 4.2. A Rectangular Plate Containing Two Identical Collinear Cracks

As shown in Figure4, the second example is a rectangular plate containing two identical collinear cracks. 2a is the length of the inclined crack and 2d is the distance between the centre of the cracks. The geometric parameters are h/w=2 and a/w=1/50. Figure 4 displays the variations of normalised mode III stress intensity factors at tip A and tip B versus different ratios of a/d. Due to the interaction between the two cracks, the computed normalised mode III stress intensity factor at tip A is always larger than that at tip B. Hence, as the crack centre distance d decreases, the difference of stress intensity factor increases. There is excellent correlation between the computed results using the present method and those from analytical solutions; the difference between these results does not exceed 0.03 per cent at tip A or 0.09 per cent at tip B.The rectangular plate with two identical collinear cracks. Normalised mode III stress intensity factors versusa/d for tip A: (a) the analytical results and (b) the present method, and for tip B: (c) the analytical solutions and (d) the present method.
(a)(b)


## 4.3. An Infinite Plate Containing Two Parallel Cracks

The third example is an infinite plate (h/w=2, a/w=1/50) containing two parallel cracks, as shown in Figure 5. 2a is the length of the two identical cracks and 2d is the distance between the cracks. The computed results are compared with the published results in [32]. The results of normalised mode III stress intensity factor for different s are plotted in Figure 5, where s=a/(a+d). The effect of the interaction of cracks on the mode III stress intensity factor is observed. The largest difference between the present and the published results does not exceed 0.65 per cent.The rectangular plate with two parallel cracks. Normalised mode III stress intensity factors versuss for (a) [32] and (b) the present method.
(a)(b)


## 5. Conclusions

An efficient and accurate dual boundary element technique has been successfully developed for the analysis of two dimensional cracks subjected to an antiplane shear loading. The dual boundary equations are the usual displacement boundary integral equation and the traction boundary integral equation. When the displacement equation is applied on the outer boundary and the traction equation is applied on one of the crack surfaces, a general crack problem can be solved in a single region formulation. The discontinuous quarter point elements are used for evaluating the mode III stress intensity factor, which correctly describes ther1/2 behaviour of the near tip displacements. This, therefore, allows accurate results for mode III stress intensity factors to be calculated.



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*Source: 132980-2009-11-05.xml*