personal.us.es · 720 m. p. ariza and j. dominguez are based on the classical displacement integral...

INTERNATIONAL JOURNAL FOR NUMERICAL METHODS IN ENGINEERINGInt. J. Numer. Meth. Engng 2004; 60:719–753 (DOI: 10.1002/nme.984)

Boundary element formulation for 3D transverselyisotropic cracked bodies

M. P. Ariza and J. Dominguez∗,†

Escuela Superior de Ingenieros, Universidad de Sevilla, Camino de los Descubrimientos,s.n. 41092-Sevilla, Spain

SUMMARY

The boundary traction integral representation is obtained in elasticity when the classical displacementrepresentation is differentiated and combined according to Hooke’s law. The use of both traction anddisplacement integral representations leads to a mixed (or dual) formulation of the BEM where thediscretization effort for crack problems is much smaller than in the classical formulation. A boundaryelement analysis of three-dimensional fracture mechanics problems of transversely isotropic solidsbased on the mixed formulation is presented in this paper. The hypersingular and strongly singularkernels appearing in the formulation are regularized by using two terms of the displacement seriesexpansion and one term of the traction expansion, at the collocation point. All the remaining integralsare analytically evaluated or transformed by means of Stokes’ theorem into regular or weakly singularintegrals, which are numerically computed. The method is general and can be used for elements ofany shape including quarter-point crack front elements. No change of co-ordinates is required for theintegration. The formulation as presented in this paper is something as clear, general and easy tohandle as the classical BE formulation. It is used in combination with three-dimensional quadraticand quarter-point elements to obtain accurate results for several different crack problems. Cracks inboundless and finite transversely isotropic domains are studied. The meshes are simple and includeonly discretization of the crack and the external boundary. The obtained results are in good agreementwith those existing in the literature. Copyright � 2004 John Wiley & Sons, Ltd.

KEY WORDS: boundary elements; transversely isotropic materials; fracture mechanics

1. INTRODUCTION

There is a significant number of papers dedicated to the boundary element analysis of three-dimensional fracture mechanics problems (see the review paper by Aliabadi [1]). Some of them

∗Correspondence to: J. Dominguez, Escuela Superior de Ingenieros, Universidad de Sevilla, Camino de losDescubrimientos, s.n., 41092-Sevilla, Spain.

†E-mail: [email protected]

Contract/grant sponsor: Comision Interministerial de Ciencia y Tecnologia of Spain; contract/grant number:DPI2000-1217-C02-01 and DPI2001-2377-C02-01

Received 3 September 2002Revised 13 June 2003

Copyright � 2004 John Wiley & Sons, Ltd. Accepted 21 July 2003

720 M. P. ARIZA AND J. DOMINGUEZ

are based on the classical displacement integral representation and a subdomain technique [2, 3].Others, on the traction integral representation proposed by Ioakimidis [4] and Sladek and Sladek[5], for isotropic and anisotropic bodies, respectively, 20 years ago. Mi and Aliabadi [6] pre-sented the so-called Dual BEM for the general solution of three-dimensional crack problems infinite or infinite domains. This method is based on a combination of displacement and tractionintegral representation for external boundaries and crack surface. Discontinuous elements areused and no subregions or internal boundaries are required. The hypersingular integrals wereevaluated, by the authors, following the scheme developed by Guiggiani et al. [7]. More recently,Dominguez et al. [8, 9] presented another mixed (or dual) BE approach where the hypersin-gular and strongly singular integrals appearing in the formulation, are analytically transformedinto line and surface integrals, which are at most weakly singular and consequently can beanalytically or numerically evaluated without difficulty. The formulation is general and is usedwith different continuous elements including quadratic quarter-point elements at the crack front.

An analytical transformation of the hypersingular and strongly singular integrals also basedon Stokes’ theorem was described by Young [10] who used modified interpolation functionsto include unique displacement gradients at the source point. Different approaches for theintegration of strongly singular and hypersingular kernels are described on the recent book byAliabadi [11].

The aforementioned mixed BE approaches are restricted to isotropic solids. Only recentlyPan and Yuang [12] have presented a BE approach for anisotropic fracture mechanics problems,based on the combined use of displacement and traction integral representation for externalboundaries and crack surface, respectively. A general fundamental solution given in terms ofintegrals is used by those authors who obtain the required fundamental solution derivativesby a finite difference numerical scheme [13]. The hypersingular kernels are integrated usingKutt’s numerical quadrature formulae. Nine different types of quadrilateral elements, and thecorresponding set of shape functions, are introduced to handle the possible geometric andboundary conditions.

A boundary element analysis of three-dimensional crack problems in transversely isotropicsolids is presented in the present paper. The explicit fundamental solution presented by Pan andChou [14] is used. For the first time, for the authors’ knowledge, explicit expressions for thefundamental solution traction derivatives appearing in the traction integral equation are given.The hypersingular and strongly singular terms of the traction equation kernels are analyticallytransformed into weakly singular terms following a similar procedure to that used by Dominguezet al. [8] for the isotropic case. Thus, all integrals are carried out either analytically or bysimple Gauss quadrature. General quadrilateral and triangular continuous quadratic elements areused. Quarter-point elements are located at the crack front. The numerical treatment of theseelements is the same as others with the only specific characteristic being the location of themid nodal line.

The BE approach presented is general and easy to handle. It does not require special elementsor hypersingular integrals. Crack problems in finite and boundless domains are studied. Thestress intensity factor (SIF) is obtained from the quarter-point nodal opening displacement. Theresults presented show the accuracy and robustness of the approach. This paper is intended toshow a general numerical approach and to present results for some practical crack problemsin transversely isotropic solids, since the number of problems of this kind previously studiedis very small, most of them being restricted to infinite domains (see for instance the book byFabrikant [15]).

Copyright � 2004 John Wiley & Sons, Ltd. Int. J. Numer. Meth. Engng 2004; 60:719–753

BOUNDARY ELEMENT FORMULATION FOR 3D TRANSVERSELY ISOTROPIC CRACKED BODIES 721

2. TRACTION BIE FORMULATION

The well known classical displacement integral representation for an internal point y of anisotropic or anisotropic elastic body � bounded by a regular surface � with unit outwardnormal n(x) under zero body forces conditions can be written as

ul(y) +∫

�p∗

lk(x, y)uk(x) d� −∫

�u∗

lk(x, y)pk(x) d� = 0 (1)

for l, k = 1, 2, 3; where uk and pk stand for the k component of displacement and tractionvectors, respectively, and u∗

lk , p∗lk are the fundamental solution displacement and traction tensors,

respectively. Explicit expressions of u∗lk and the stress tensor for three-dimensional transversely

isotropic materials can be found in Pan and Chou’s paper [14]. Tractions p∗lk are immediately

obtained by projection over any surface defined by its normal.Internal stresses can be obtained by differentiation of displacement at point y and introduction

of the corresponding strains into the stress–strain relationship. The traction vector componentsat an internal point y over a surface with unit outward normal N is

pl(y) = �lm(y)Nm(y) (2)

and the integral representation for the traction components

pl(y) +∫

�s∗lmk(x, y)Nm(y)uk(x) d� −

∫�

d∗lmk(x, y)Nm(y)pk(x) d� = 0 (3)

where d∗lmk and s∗

lmk are linear combinations of derivatives of u∗lk and p∗

lk , respectively. Theexpressions for d∗

lmk coincide, except for one change of sign, with the stress tensor components.The change of sign is due to the different differentiation point: integration point for thestress tensor and collocation point for d∗

lmk . Explicit expressions of d∗lmk for the case of

transversely isotropic solids were given by Pan and Chou [14]. Explicit expressions for s∗lmk

for 3-D transversely isotropic materials are given in Appendix A. They have been obtainedby differentiation and combination of Pan and Chou’s solution. The process has been carriedout by hand in the present paper and validated by checking their analytical convergence to theisotropic solution when the material properties are set for that case. They have been validatedalso by comparison with the static part of the s∗

lmk terms obtained from the dynamic fundamentalsolution presented by Wang and Achenbach [16]. Values computed in both cases for differentmaterials match each other completely. The kernel d∗

lmk has a strong singularity of order r−2

whereas s∗lmk is hypersingular of order r−3, as r → 0.

A limit to the boundary process is carried out to obtain the integral representation for asmooth boundary point y. Using the modified geometry of Figure 1, the following expressioncan be written as

pl(y) + lim�→0+

{∫(�−e�)+��

[s∗lmk(x, y)Nm(y)uk(x) − d∗

lmk(x, y)Nm(y)pk(x)] d�}

= 0 (4)

Several aspects related to the above traction representation can be pointed out: (1) the existenceof this limit to the boundary to obtain an integral equation establishes a continuity requirementof the type uk ∈ C1,�, at y; i.e. uk must be differentiable at y with its first derivative satisfyinga Hölder condition and � being a positive constant as has been stated by Martin and Rizzo



Figure 1. Vanishing contour around a boundary point.

[17] and Martin et al. [18]; (2) At non-smooth boundary points it is not possible to obtain anintegral representation of the tractions only in terms of boundary displacements and tractionsfollowing the limit to the boundary process. However, such a process can be followed for thedisplacement derivatives and after inverting the c matrix obtain a representation of the tractionsby linear combination of the displacement derivatives representation; (3) The surface �� usedto remove point y from the boundary may have any shape provided it is a regular surface.A spherical surface has been selected because of its simplicity.

Taking into account that uk ∈ C1,�, at y, it can be expanded as

uk(x) = uk(y) + uk,h(y)(xh − yh) + O(r1+�) (5)

and the tractions

pk(x) = �kh(x)nh(x) = �kh(y)nh(x) + O(r�) (6)

Notice that this particular expansion of pk(x) around y with n(x) as a variable, is required forthe surface ��, where n(x) remains variable as � → 0, but not for � − e� where n(x) tends toa constant value N(y), as � → 0.

By using expansions (5) and (6), Equation (4) can be written as

pl(y) + lim�→0+

{∫�−e�

{s∗lmkNm[uk(x) − uk(y) − uk,h(y)(xh − yh)]

− d∗lmkNm[pk(x) − pk(y)]} d� +

∫��


− d∗lmkNm[pk(x) − �kh(y)nh(x)]} d� + uk(y)

∫�−e�

s∗lmkNm d� + uk(y)

∫��

s∗lmkNm d�

+ uk,h(y)∫

�−e�

s∗lmkNm(xh − yh) d� + uk,h(y)

∫��

s∗lmkNm(xh − yh) d�

− pk(y)∫

�−e�

d∗lmkNm d� − �kh(y)

∫��

d∗lmkNmnh d�

}= 0 (7)

or

pl(y) + T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 = 0 (8)

where each Ti term is the limit for � → 0 of one of the above integrals.



The gradient components of uk at y, i.e. uk,h(y), should be understood as components ofthe tangential gradient. These tangential gradient components will eventually be represented byderivatives of the displacements representation over the boundary elements.

Since s∗lmk is hypersingular of order r−3, and d∗

lmk is singular of order r−2, as r → 0, thekernels in the integrals T1 and T2 are weakly singular. Thus, T1 can be evaluated numerically,and T2 = 0.

The integrals over the spherical surface �� (T4, T6 and T8) can be evaluated analyticallywithout much difficulty using a local spherical system of co-ordinates centred at y. The inte-gration process is similar as for the isotropic case (see, Dominguez et al. [8]) leading to thefollowing limit values:

T6 + T8 = − 12 pl(y) (9)

By using Stokes’ theorem and some simple algebra [19] the integrals in T3, T5 and T7 canbe transformed in a summation of regular and weakly singular surface integrals, line integralsover the contour of the boundary and an unbounded term which cancels out T4.

After substitution of terms T1–T8, the general traction BIE in terms only of regular andweakly singular integrals becomes

1

2pl(y) +

∫�{s∗

lmkNm[uk(x) − uk(y) − uk,h(y)(xh − yh)]

− d∗lmkNm[pk(x) − pk(y)]} d� + [uk(y)Ilk + uk,h(y)Jlhk + pk(y)Klk] = 0 (10)

where the terms Ilk , Jlhk and Klk , are the limit values for � → 0 of the result of applica-tion of Stokes’ theorem to

∫�−e�

s∗lmkNm d� + ∫

��s∗lmkNm d�; ∫

�−e�s∗lmkNm(xh − yh) d�; and∫

�−e�d∗lmkNm d�, respectively. They contain only regular and weakly singular integrals. Equa-

tion (10) is valid for general anisotropic solids. Explicit expressions of the terms Ilk , Jlhk andKlk have been obtained in the present work for transversely isotropic materials. Their valuesfor cracks perpendicular to the axis of symmetry of the material are given in Appendix B.

3. BOUNDARY ELEMENTS

Once the BIE has been transformed and it does not contain any strongly singular or hypersin-gular integral, the boundary element implementation is greatly simplified. Nevertheless, thereare some aspects of the boundary element discretization that should be addressed.

When dealing with infinite domain crack problems only one crack surface is discretized intoquadratic BE. The traction BIE is written for collocation points which may be assumed tobe on any one of the two crack surfaces (in both cases the obtained equation would be thesame). To avoid the need of writing another equation for points on the crack surface, the crackopening displacement (COD) is chosen as the basic variable on the crack. This is all that oneusually needs in Fracture Mechanics.

The elements used on the crack surface are standard nine- or six-node quadratic elements ex-cept for those with one side at the crack front which are quarter-point elements. These elementsare quadrilateral with its nine nodes on a plane and two straight line sides perpendicular tothe crack front. The mid-nodes 4, 8 and 9 (Figure 2) are located at one quarter of the element



Figure 2. Nine node quarter-point quadratic element.

width. The displacement interpolation in these quarter-point elements is able to reproduce theactual behaviour in the vicinity of the crack front [3].

In problems where in addition to the crack, external boundaries and/or other internal bound-aries exist, these boundaries are discretized into quadratic BE and the classical BIE is writtenon them. Displacements and tractions are the boundary variables for the regular boundaries,whereas the COD is the unknown for the crack nodes.

In special cases where surface cracks exist or where the actual displacement componentsat the crack surfaces are required one may write both the tractions BIE and the standarddisplacement BIE for the crack surface and use the displacement at both crack surfaces asboundary variables.

It is worth to mention that there is no sense in talking about writing the displacement BIEfor one of the crack surfaces and the traction BIE for the other since the BIE equations arewritten for points which actually are on both crack surfaces and include in any case the sameterms for the displacements of the upper and the lower surface.

The two facts which make the discretization of the tractions BIE written for collocation pointson the crack surface, different to the discretization of the classical BIE are: (1) the boundarydisplacement uk satisfies the Hölder continuity condition uk ∈ C1,�, at y; and (2) derivatives,at collocation point y, of the displacement components exist in the boundary integral equation.As a consequence of the first of these two characteristics, one should not use collocation atnodal points located at the contour of the element. Due to the second, one cannot use constantelements. Following the idea of Gallego and Dominguez [20] for two dimensions, the boundaryintegral equation is discretized using quadratic surface elements with six or nine nodes locatedat their usual position to represent the geometry and the boundary variables. However, thecollocation is not done at the contour nodes; i.e. at �1, �2 = ±1, but at certain points close tothe nodes, inside the element. Results for isotropic and transversely isotropic materials showthat �1, �2 = ±0.75 produce accurate results.

Notice that elements of this type are continuous C0 since the boundary variables are writtenas usual in terms of their values at nodes located at the contour of the element. Only thecollocation points are shifted to the interior of the element. Tractions and displacements ateach collocation point are expressed in terms of nodal values and shape function values atcollocation point. The displacement derivatives (tangential gradient) are expressed in terms ofnodal displacements and the shape function derivatives at collocation point.



As a consequence of the collocation strategy, one may have two or more equations for eachnodal component, obtained by collocation at as many points as elements contain the node.These equations are added up to yield only one per nodal component. It is worth to noticethat this multiple collocation approach (MCA) is only used for the nodes on the contour of theelements on one crack surface but neither for those at the crack front, where no collocationat all is required (�uk = 0), nor for external boundary nodes, where only C0 continuity isrequired and the usual BE nodal collocation is used.

It is also important to mention that since the regularization process leads to expressionsof the boundary integrals with some more terms than the original hypersingular and stronglysingular integrals, the regular expressions of the integral may be used only over a part ofthe surface � close to the collocation point whereas the original expressions of the integralsare used in the rest of the boundary where they are non-singular. Thus, the expression to bediscretized into boundary elements is

1

2pl(y) +

∫�0


− d∗lmkNm[pk(x) − pk(y)]} d� + [uk(y)I 0lk + uk,h(y)J 0

lhk + pk(y)K0lk]

+∫

�−�0

{s∗lmkNmuk(x) − d∗

lmkNmpk(x)} d� = 0 (11)

where �0 is a part of the boundary close to the collocation point, i.e. the elements containingthe node which equation is being written by multiple collocation, and � − �0 is the rest ofthe boundary.

4. STRESS INTENSITY FACTORS COMPUTATION

According to the work of Kassir and Sih [21] there is an asymptotic relation between theCOD near the crack front and the SIFs, for cracked transversely isotropic media. Usingthe leading terms of those relations, the SIFs can be evaluated from the three componentsof the COD at a distance r from the crack front. Assuming that z is the material axis ofsymmetry perpendicular to the crack plane and the t-axis is tangent to the crack front line atthe point where a node is located (Figure 3), the SIF components at this point can be writtenin terms of the cracks opening displacements at the quarter-point node as

KI =√

�

2L�uz

�1m1

1 + m1− m2

1 + m2

KII =√

�

2L�un

C44(√

n2 − √n1 )(1 + m1)(1 + m2)

(m2 − m1)√

n1n2(12)

KIII =√

�

2L�ut

C44√n3



Figure 3. Local co-ordinates at crack boundary point P.

where L is the quarter-point element width, n1 and n2 are the two solutions of the quadraticequation of the material properties

C11C44n2 + [C13(C13 + 2C44) − C11C33]n + C33C44 = 0 (13)

n3 = 2C44

C11 − C12(14)

mj = (C13 + C44)nj

C33 − C44nj

, j = 1, 2 (15)

and

�1 = C44(√

n1 − √n2)

(16)

5. NUMERICAL RESULTS FOR CRACK PROBLEMS

To show the generality and robustness of the approach several different 3-D fracture mechanicsproblems in transversely isotropic media are studied in this section: cracks in an infinite domain(5.1 and 5.2), internal crack in finite domain (5.3) and surface crack in finite domain (5.3, 5.4and 5.5). In all cases the discretization used is simple and the obtained results accurate.

5.1. Penny shaped crack in infinite domain

This is a classical problem for which there is analytical solution. The material is a transverselyisotropic graphite-epoxy with the following properties: C11 = 13.92 GPa, C33 = 160.7 GPa,C12 = 6.92 GPa, C13 = 6.44 GPa and C44 = 7.07 GPa. Two loading states are assumed: oneis a uniform traction which produces a mode-I SIF, and the second is a shear traction intwo opposite directions over the crack faces which yields mode-II and mode-III SIFs. TheBE mesh is shown in Figure 4. All the elements are quadrilateral nine-node elements withthe mid-nodes of the external elements at one quarter of their width. The material axis ofsymmetry is perpendicular to the crack surface.



Figure 4. Boundary element discretization of a penny shaped crack in infinite domain.

Figure 5. Elliptical crack in infinite domain: (a) geometry; and (b) boundary element discretization.

The analytic solutions for mode-I [21], mode-II and mode-III [22], have been recovered inall cases within a 1% accuracy by the present BE approach using the COD at the quarter-pointnodes to calculate the SIF values.

5.2. Elliptical crack in infinite domain

The analytical solution for the SIFs of elliptical cracks (Figure 5(a)) in transversely isotropicboundless domains can be found in the work of Kassir and Sih [22]. In the present work theproblem is analysed for the same material as in the previous example using the mesh shownin Figure 5(b). The computed values reproduce the analytic solution along the crack front withan error �1% at any point.

5.3. Penny shaped crack in cylindrical bar

A penny shaped crack is assumed to be at the centre of a cylinder as shown in Figure 6(a).The length of the cylinder is twice its diameter. The material axis of symmetry coincides withthe axis of the cylinder. Two different materials are considered: one is the same graphite-epoxy composite as for the previous examples; the other, is a laminate composite with verydifferent properties: C11 = 5.37 GPa, C33 = 251.168 GPa, C12 = 1.34 GPa, C13 = 3.35 GPaand C44 = 5 GPa. The cylinder is subject to uniform traction at the two opposite faces. TheBE discretization of the external surface is shown in Figure 6(b), where all the elements are



Figure 6. (a) Penny shaped crack in cylindrical bar; and (b) external surface discretization.

Figure 7. Effect of the ratio between the cylinder radius R and the crack radius a on the SIF.

standard quadratic nine nodes quadrilateral. The mesh for the crack is the same as for the firstexample. The classical displacement integral representation is written for the external surfaceelements whereas the traction integral representation is used for the crack surface.

An analysis of the KI stress intensity factor variation with the size of the internal crack hasbeen carried out. Results for the two materials and values of the ratio between the cylinderradius R and the crack radius a from 10 to 2 are shown in Figure 7. The KI values arenormalized by its value for the same crack in a boundless domain. It can be noticed from thefigure that the effect of the crack size with respect to the cylinder size is much more importantfor the two transversely isotropic materials than for an isotropic material with Poisson’s ratio� = 0.3. The material with the greatest difference between C11 and C33 modulae is the onethat shows a greater influence of the crack size.



Figure 8. Geometry and BE discretization of rectangular bar with edge crack.

Figure 9. SIF mode-I for edge cracked rectangular bar.

5.4. Edge cracked rectangular bar under uniform traction

The geometry, boundary conditions and the BE mesh for this problem are shown in Figure 8.The plate geometry is defined by H/a = 1.75, W/a = 3 and t/a = 2, where a is the crackdepth. The same three materials of the previous example are analysed in this case. The cracksurface is discretized into 48 nine nodes quadratic elements, 12 of them being quarter pointelements along the crack front. Elements are concentrated at both ends of the crack wherestress changes across the front are expected. The study of this problem is a better test ofthe technique than the previous one since the SIF varies across the crack front and there isa more significant crack opening. Normalized values of the SIF across the crack front arerepresented for an isotropic material, graphite-epoxy and composite laminate. Results for theisotropic case obtained by Li et al. [23] using a symmetric integral equation method are invery good agreement with the present results as shown in Figure 9. To the authors’ knowledge,no previous results of this problem for transversely isotropic material exist.



Figure 10. Rectangular bar with a semi-circular surface crack: (a) geometry; and (b) SIF mode-I.

Figure 11. BE discretization of one half of a rectangular bar with a semi-circular surface crack.

5.5. Rectangular bar with a semi-circular surface crack under uniform traction

The prismatic bar shown in Figure 10(a) contains a surface semi-circular crack of radius a. Thebar geometry is shown in the figure where H = W and a/W = 0.25. A uniform traction � isprescribed on the ends of the bar. This problem has been studied considering the same threedifferent materials of the previous cases. Using the symmetry conditions with respect to thevertical plane, only one half of the problem is discretized with the mesh shown in Figure 11.All the elements are quadratic with quarter-point elements for the crack front. The distributionof KI along the crack front as a function of the angle � measured from the external frontfrom the external face is shown in Figure 10(b). Once again the upper curve corresponds tothe composite laminate, the intermediate curve to graphite-epoxy and the lower curve to the� = 0.3 isotropic material. Finite element results given by Newman and Raju [24] for the



Figure 12. Rectangular bar with a semi-circular surface crack: (a) eeometry; and (b) SIF mode-I.

isotropic case are also shown in Figure 10(b). The agreement with the present results for thismaterial is very good. As in the previous problem, the SIF variation across the crack front isquite similar for the three materials.

5.6. Square bar with a corner quarter-circle crack under uniform traction

The problem geometry for this case is shown in Figure 12(a). The BE mesh used is the sameas in the previous example (Figure 11) with free traction over all lateral faces. A uniformtraction � is prescribed on the ends of the bar. The distribution of KI along the crack front asa function of � is shown in Figure 12(b) for the same three materials studied in the previousexamples. Once more the KI variation with � is quite similar for the three analysed materials.These values are smaller as the material behaviour is more isotropic.

The results for the isotropic case are compared with those obtained by Li et al. [23] usinga symmetric integral equation formulation. The agreement is very good except for a boundarylayer effect near the free boundary which exist in Li et al. results and not in the present.It is worth to mention that such effect was also obtained with the present approach when afiner mesh for the crack was used. Li et al. also reported that only with very small elementsnear the crack ends the boundary layer effect is obtained. This effect is due to the fact thatthe stress field around the vertex does not posses a −(1/2) singularity but rather the order ofthe singularity depends upon Poisson’s ratio and other factors [25]. If the standard definitionof the SIF is retained its value may go to zero or unbounded depending on the actual orderof the singularity. Therefore, there is not much sense in computing a SIF value based on the−(1/2) singularity at the external boundary. The effect of the vertex is usually restricted toa very thin boundary layer and its careful treatment may not be essential [21, 23]. Thus, thiseffect has not been considered in depth in the present paper.

6. CONCLUSION

A mixed single-domain BE formulation for the analysis of fracture problems in three-dimen-sional transversely isotropic solids has been presented in this paper. Closed-form expressions of



the fundamental solution traction derivatives required for the traction boundary integral equationhave been derived. The hypersingular and strongly singular terms in the integral equation areregularized by subtracting and adding back terms of the Taylor series expansion of the densities.The new hypersingular and strongly singular integrals appearing in the added back terms areanalytically transformed to yield regular or at most weakly singular integrals which are evaluatedusing a Gauss quadrature. Explicit expressions of those integrals for cracks perpendicular tothe axis of symmetry of the material are given.

The numerical approach presented and implemented in this paper is as general, clear andeasy to handle as the classical BE formulation. No integration of hypersingular or stronglysingular terms is required. The boundary approximation is done as usual in BE techniques bycontinuous quadratic isoparametric elements with the crack front elements having the mid-nodeline at one quarter of the element width. The displacement and traction continuity over theboundaries is retained, contrary to non-conforming elements. The collocation points for thecrack surface are always inside the elements with multiple collocation for nodes between twoor more elements.

Solution to several three-dimensional crack problems in transversely isotropic boundless andfinite solids have been presented. Stress intensity factors have been computed from the crackopening displacement at the quarter-point nodes. Numerical solutions obtained are shown to bein very good agreement with those in the literature when available. The results obtained arevery accurate even with a relative small number of elements.

The effect of the crack size in relation to that of the domain have been studied for a pennyshaped crack inside a cylinder. The obtained results show important influence of the crack sizeon the SIF. This influence is much more significant for transversely isotropic solids than forisotropic ones. Other crack problems of practical importance have been studied in order to testthe numerical tool. Problems including flexural behaviour and where the stress intensity factorvaries across the crack front have been analysed. The computed values for isotropic materialare in good agreement with those existing in the literature. In most problems analysed, noprevious results for non-isotropic materials existed. The present results show that SIF valuesincrease with the material anisotropy in greater extent as crack opening and bending effectincrease.

APPENDIX A

A.1. Transversely isotropic solids

There exist five independent constants characterizing the material elastic behaviour: C11, C12,C13, C33 and C44. P13 = √

C11C33.Expressions for the slmk kernels when P13 − C13 − 2C44 �= 0 and

�1A1 = −�2A2 = C13 + C44

4�C33C44(�22 − �21), Bi = −Ai

A′1 = B ′

1 = �21C33 − C44

8�C33C44(�21 − �22)�21

, A′2 = B ′

2 = C44 − �22C33

8�C33C44(�21 − �22)�22



s∗111 = n1

{(C11 + C12)

(D71

R31

+ D72

R32

)− 3X1

(D71

R51

+ D72

R52

)

+ 4C66[A′1�1[−R11(C11 + 3C12) + R31(C11R

2 + 6C12x22 ) − 3R51x

22X1]

+ A′2�2[−R12(C11 + 3C12) + R32(C11R

2 + 6C12x22 ) − 3R52x

22X1]]

+ 2C66D(C12 − C11)(−R13 + R33R2 − 3R53x

21x

22 ) − C13

[D41A1�21

R31

(1 − 3z21

R21

)

+ D42A2�22R32

(1 − 3z22

R22

)]− 2C66C13

[A1�21R31

(1 − 3x2

2

R21

)+ A2�22

R32

(1 − 3x2

2

R22

)]}

+ 3C66x1x2n2

{− 4A′

1�1[R31(C11 + C12) − R51X1] − 4A′2�2[R32(C11 + C12)

−R52X1] + D(x22 − x2

1 )(C11 − C12)R53 − 2C13

(A1�21R51

+ A2�22R32

)}

+ x1n3

{2C44A3′

1D31[−R21(3C11+C12)+R41X1]+2C44A′2D32[−R22(3C11+C12)

+ R42X1] + C12 − C11

4�(R23 − R43x

22 ) − 3C44C13x3

(D31A1�21

R51

+ D32A2�22R52

)}

(A1)

s∗121 = 6C66x1x2n1

{4[�1A′

1(R31 − R51x22 ) + �2A

′2(R32 − R52x

22 )]

− 2

(D71

R51

+ D72

R52

)+ DR53(x

22 − x2

1 )

}

+ C66n2{D[2R13 − 4R33x22 + 3R53x

21 (x

21 − x2

2 )]− 8[�1A′

1(−R11 + R2R31 − 3R51x21x

22 ) + �2A

′2(−R12 + R2R32 − 3R52x

21x

22 )]}

+ C66x2n3

{1

4�[−2R23 + R43(x

22 − x2

1 )]

+ 4C44[A′1D31(−R21 + R41x

21 ) + A′

2D32(−R22 + R42x21 )]}

(A2)



s∗131 = C44x1n1

{3x3

[�21R51

(A1D41 − D71) + �22R52

(A2D42 − D72)

]

+ 2C66[�1D61(R21 − R41x22 ) + �2D62(R22 − R42x

22 ) − �3D(R23 − R43x

22 )]}

+ C44C66x2n2{−2[�1D61(R21 − R41x21 ) + �2D62(R22 − R42x

21 )]

− �3D[2R23 − R43(x21 − x2

2 )]}

− C44n3

{−C44

[D61D31

R31

(1 − 3x2

1

R21

)+ D62D32

R32

(1 − 3x2

1

R22

)]− �3

4�R33

(1 − 3x2

2

R23

)}

(A3)

s∗112 = 3C66x1x2n1

{−4�1A

′1[R31(C11 + C12) − R51X1] − 4�2A

′2[R32(C11 + C12)

− R52X1] + DR53(C11 − C12)(x22 − x2

1 ) − 2C13

(A1�21R51

+ A2�22R52

)}

+ n2

{(C11 + C12)

(D71

R31

+ D72

R32

)− 3X1

(D71

R51

+ D72

R52

)

+ 4C66[A′1�1[−R11(3C11 + C12) + R31(6C11x

21 + C12R

2) − 3R51x21X1]

+ A′2�2[−R12(3C11 + C12) + R32(6C11x

21 + C12R

2) − 3R52x21X1]]

+ 2C66D(C11 − C12)(−R13 + R33R2 − 3R53x

21x

22 ) − C13

[D41A1�21

R31

(1 − 3z21

R21

)

+ D42A2�22R32

(1 − 3z22

R22

)]− 2C66C13

[A1�21R31

(1 − 3x2

1

R21

)+ A2�22

R32

(1 − 3x2

1

R22

)]}

+ x2n3

{2C44A

′1D31[−R21(C11 + 3C12) + R41X1] + 2C44A

′2D32[−R22(C11 + 3C12)

+ R42X1] + C11 − C12

4�(R23 − R43x

21 ) − 3C44C13x3

(D31A1�21

R51

+ D32A2�22R52

)}(A4)



s∗122 = C2

66n1{8[�1A′1(R11 − R31R

2 + R51x21x

22 ) + �2A

′2(R12 − R32R

2 + R52x21x

22 )]

+ D[2R13 − 4R33x22 + 3R53x

21 (x

22 − x2

1 )]} + 6C66n2

{C66[4�1A′

1(R31 − R51x21 )

+ 4�2A′2(R32 − R52x

21 ) + DR53x

21x

22 (x

21 − x2

2 )] − x21x

22

(D71

R51

+ D72

R52

)}

+ C266x1n3

{4C44[A′

1D31(−R21 + R41x22 ) + A′

2D32(−R22 + R42x22 )]

+ 1

4�[−2R23 + R43(x

21 − x2

2 )]}

(A5)

s∗132 = C44C66x2n1{2[�1D61(−R21 + R41x

21 ) + �2D62(−R22 + R42x

21 )] + �3D[−2R23

+ R43(x22 − x2

1 )]} + C44x1n2

{3x3

[�21R51

(D41A1 − D71) + �22R52

(D42A2 − D72)

]

+ 2C66[�1D61(3R21 − R41x21 ) + �2D62(3R22 − R42x

21 ) + �3D(R23 − R43x

22 )]}

+ 3C44x1x2n3

{C44

(D41D31

R51

+ D42D32

R52

)− �3

4�R53

}(A6)

s∗113 = x1n1

{2C44[A′

1D31[−R21(3C11 + C12) + R41X1] − A′2D32[R22(3C11 + C12)

− R42X1]] + C12 − C11

4�(R23 − R43x

22 ) − 3C44C13x3

(D31A1�21

R51

+ D32A2�22R52

)}

+ x2n2

{2C44[A′

1D31[−R21(C11 + 3C12) + R41X1] + A′2D32[−R22(C11 + 3C12)

+ R42X1]] + C11 − C12

4�(R23 − R43x

21 ) − 3C44C13x3

(D31A1�21

R51

+ D32A2�22R52

)}

+ n3

{2�1A′

1D51

R31

(C11 + C12 − 3

X1

R21

)+ 2�2A′

2D52

R32

(C11 + C12 − 3

X1

R22

)



− C13

[D51A1�21

R31

(1 − 3z21

R21

)+ D52A2�22

R32

(1 − 3z22

R22

)]}(A7)

s∗123 = C66x2n1

{4C44[A′

1D31(−R21 + x21R41) + A′

2D32(−R22 + x21R42)]

+ 1

4�[−2R23 + R43(x

22 − x2

1 )]}

+ C66x1n2

{4C44[A′

1D31(−R21 + x22R41) + A′

2D32(−R22 + x22R42)]

+ 1

4�[−2R23 + R43(x

21 − x2

2 )]}

− 12C66x1x2n3

{D51B

′1�1

R51

+ D52B′2�2

R52

}(A8)

s∗133 = C44n1

{C44

[D31D61

R31

(−1 + 3x2

1

R21

)+ D32D62

R32

(−1 + 3x2

1

R22

)]

+ �34�R3

3

(−1 + 3x2

2

R23

)}+ 3C44x1x2n2

{− �34�R5

3

+ C44

[D31D61

R51

+ D32D62

R52

]}

− 3C44x1x3n3

{D51D61�21

R51

+ D52D62�22R52

}(A9)

s∗231 = C44x2n1

{3x3

[�21

A1D41 − D71

R51

+ �22A2D42 − D72

R52

]

+ 2C66[�1D61(3R21 − R41x22 ) + �2D62(3R22 − R42x

22 ) + �3D(R23 − R43x

21 )]}

− C44C66x1n2{2�1D61(R21 − R41x22 ) + 2�2D62(R22 − R42x

22 )

+ �3D[2R23 + R43(x22 − x2

1 )]}

+ 3C44x1x2n3

{C44

(D61D31

R51

+ D62D32

R52

)− �3

4�R53

}(A10)

s∗221 = n1

{(C11 + C12)

(D71

R31

+ D72

R32

)− 3X2

(D71

R51

+ D72

R52

)



+ 4C66[A′1�1[−R11(3C11 + C12) + R31(C12R

2 + 6C11x22 ) − 3R51x

22X2]

+ A′2�2[−R12(3C11 + C12) + R32(C12R

2 + 6C11x22 ) − 3R52x

22X2]]

+ 2C66D(C11 − C12)(−R13 + R33R2 − 3R53x

21x

22 ) − C13

[D41A1�21

R31

(1 − 3z21

R21

)

+ D42A2�22R32

(1 − 3z22

R22

)]− 2C66C13

[A1�21R31

(1 − 3x2

2

R21

)+ A2�22

R32

(1 − 3x2

2

R22

)]}

+ 3C66x1x2n2

{−4A′

1�1[R31(C11 + C12) − R51X2] − 4A′2�2[R32(C11 + C12) − R52X2]

+ D(x22 − x2

1 )(C12 − C11)R53 − 2C13

(A1�21R51

+ A2�22R32

)}

+ x1n3

{2C44A

′1D31[−R21(C11 + 3C12) + R41X2] + 2C44A

′2D32[−R22(C11 + 3C12)

+ R42X2] + C11 − C12

4�(R23 − R43x

22 ) − 3C44C13x3

(D31A1�21

R51

+ D32A2�22R52

)}(A11)

s∗222 = 3C66x1x2n1

{−4A′

1�1[R31(C11 + C12) − R51X2] − 4A′2�2[R32(C11 + C12)

− R52X2] + D(x22 − x2

1 )(C12 − C11)R53 − 2C13

(A1�21R51

+ A2�22R32

)}

+ n2

{(C11 + C12)

(D71

R31

+ D72

R32

)− 3X2

(D71

R51

+ D72

R52

)

+ 4C66[A′1�1[−R11(C11 + 3C12) + R31(C11R

2 + 6C12x21 ) − 3R51x

21X2]

+ A′2�2[−R12(C11 + 3C12) + R32(C11R

2 + 6C12x21 ) − 3R52x

21X2]]

+ 2C66D(C12 − C11)(−R13 + R2R33 − 3x21x

22R53) − C13

[D41A1�21

R31

(1 − 3z21

R21

)

+ D42A2�22R32

(1 − 3z22

R22

)]− 2C66C13

[A1�21R31

(1 − 3x2

1

R21

)+ A2�22

R32

(1 − 3x2

1

R22

)]}



+ x2n3

{2C44A

′iD31[−R21(3C11 + C12) + R41X2] + 2C44A

′2D32[−R22(3C11 + C12)

+ R42X2] + C12 − C11

4�(R23 − x2

1R43) − 3C44C13x3

(D31A1�21

R51

+ D32A2�22R52

)}(A12)

s∗232 = C44C66x1n1{2[�1D61(−R21 + R41x

22 ) + �2D62(−R22 + R42x

22 )]

+ �3D[−2R23(x21 − x2

2 )]} + C44x2n2

{3x3

[�21R51

(D41A1 − D71) + �22R52

(D42A2 − D72)

]

+2C66[�1D61(R21 − R41x21 ) + �2D62(R22 − R42x

21 ) + �3D(R23 − R43x

22 )]}

+ C44n3

{−C44

[D61D31

R31

(1 − 3x2

2

R21

)+ D62D32

R32

(1 − 3x2

2

R22

)]− �3

4�R33

(1 − 3x2

1

R23

)}

(A13)

s∗223 = x1n1

{2C44[A′

1D31[−R21(C11 + 3C12) + R41X2] + A′2D32[−R22(C11 + 3C12)

+ R42X2]] + C11 − C12

4�(R23 − R43x

22 ) − 3C44C13x3

(D31A1�21

R51

+ D32A2�22R52

)}

+ x2n2

{2C44[A′

1D31[−R21(3C11 + C12) + R41X2] + A′2D32[−R22(3C11 + C12)

+ R42X2]] + C12 − C11

4�(R23 − R43x

21 ) − 3C44C13x3

(D31A1�21

R51

+ D32A2�22R52

)}

+ n3

{2�1A′

1D51

R31

(C11 + C12 − 3

X2

R21

)+ 2�2A′

2D52

R32

(C11 + C12 − 3

X2

R22

)

− C13

[D51A1�21

R31

(1 − 3z21

R21

)+ D52A2�22

R32

(1 − 3z22

R22

)]}(A14)



s∗233 = C44x1x2n1

{3C44

(D61D31

R51

+ D61D31

R51

)− 3

�34�R5

3

}

− C44n2

{C44

[D61D31

R31

(1 − 3x2

2

R21

)+ D62D32

R32

(1 − 3x2

2

R22

)]− �3

4�R33

(1 − 3x2

1

R23

)}

− 3C44x2x3n3

{D61D51�21

R51

+ D62D52�22R52

}(A15)

s∗331 = n1

{C13

[D81

R31

(2 − 3

R2

R21

)+ D82

R32

(2 − 3

R2

R22

)]+ 4C66C13[A′

1�1[−4R11

+ R31(x21 + 7x2

2 ) − 3R51x22R

2] + A′2�2[−4R12 + R32(x

21 + 7x2

2 ) − 3R52x22R

2]]

+ C33

[−2C66

[�21A1

R31

(1 − 3

x22

R21

)+ �22A2

R32

(1 − 3

x22

R22

)]

− �21A1D41

R31

(1 − 3

z21

R21

)− �22A2D42

R32

(1 − 3

z22

R22

)]}

− 6C66x1x2n2

{2C13[�1A′

1(2R31 − R51R2) + �2A

′2(2R32 − R52R

2)]

−C33

(�21A1

R51

+ �22A2

R52

)}+ C44x1n3

{2C13[D31A

′1(−4R21 + R41R

2)

+D32A′2(−4R22 + R42R

2)] − 3x3C33

(�21A1D31

R51

+ �22A2D32

R52

)}(A16)

s∗332 = −6C66x1x2n1

{2C13[�1A′

1(2R31 − R51R2) + �2A

′2(2R32 − R52R

2)]

− C33

(�21A1

R51

+ �22A2

R52

)}+ n2

{C13

[D81

R31

(2 − 3

R2

R21

)+ D82

R32

(2 − 3

R2

R22

)]

+ 4C66C13[A′1�1[−4R11 + R31(7x

21 + x2

2 ) − 3R51x21R

2]

+ A′2�2[−4R12 + R32(7x

21 + x2

2 ) − 3R52x21R

2]] + C33

[−2C66

[�21A1

R31

(1 − 3

x21

R21

)



+ �22A2

R32

(1 − 3

x21

R22

)]− �21A1D41

R31

(1 − 3

z21

R21

)− �22A2D42

R32

(1 − 3

z22

R22

)]}

+ C44x2n3

{2C13[D31A

′1(−4R21 + R41R

2) + D32A′2(−4R22 + R42R

2)]

− 3x3C33

(�21A1D31

R51

+ �22A2D32

R52

)}(A17)

s∗333 = C44x1n1

{2C13[D31A

′1(−4R21 + R41R

2) + D32A′2(−4R22 + R42R

2)]

− 3C33x3

(�21A1D31

R51

+ �22A2D32

R52

)}+ C44x2n2

{−3C33x3

(�21A1D31

R51

+ �22A2D32

R52

)+ 2C13[D31A

′1(−4R21 + R41R

2) + D32A′2(−4R22 + R42R

2)]}

× n3

{−2C13

[�1A′

1D51

R31

(1 − 3

z21

R21

)+ �2A′

2D52

R32

(1 − 3

z22

R22

)]

− C33

[�21A1D51

R31

(1 − 3

z21

R21

)+ �22A2D52

R32

(1 − 3

z22

R22

)]}(A18)

When P13 − C13 − 2C44 = 0 and

A1 = A2 = 0, B1 = B2 = − C13 + C44

16�C11C44, A′

1 = A′2 = 1

16�C11, B ′

1 = B ′2 = 1

16�C44�21

s∗111 = n1

{�32�

(C11 − C12)(R13 − R33R2 + 3R53x

21x

22 ) + �1

2�

1

R31

(C11 + C12 − 3

X1

R21

)

+ �232��1

C11(−R11 + R31R2 − 3R51x

21x

22 ) + 3

�232��1

C12(−R11 + 2R31x22 − R51x

42)

+ �23�1

D11

R31

(C11 + 3C12 − 3

x21C11 + 3x2

2C12

R21

)− �1D2

1

R31

(C11 + C12 − 3

X1

R21

)

− 3�23�1

D1x22

R51

(C11 + 3C12 − 5

X1

R21

)+ C13

(C13�214�P13

+ (�21 + �23)D1

)1

R31

(1 − 3

z21

R21

)



− 3�1D1z21

R51

(C11+C12 − 5

X1

R21

)−3C13D1

1

R51

[�21z

21

(3−5

z21

R21

)+�23x

22

(1−5

z21

R21

)]}

+ x1x2n2

{3

�23�1

D11

R51

(3(C11 + C12) − 5

X1

R21

)+ 3

�232��1

(−R31(C11 + C12) + R51X1)

+ 3�34�

(C11 − C12)R53(x22 − x2

1 ) + 3�23D1C131

R51

(1 − 5

z21

R21

)}

+ x1n3

{3D1

z1

R51

(3C11 + C12 − 5

X1

R21

)+ 1

4�(C11 + C12)(−R23 + x2

2R43)

+ 1

4�(R21(−3C11 + C12) + R41(C11x

21 − C12x

22 ))

+3�1C13z1

R51

[−D2 + D1

(2 − 5

z21

R21

)]}(A19)

s∗121 = 3C66x1x2n1

{−�1

�

1

R51

+ �23��1

(R31 − R51x22 ) + �3

2�(2R33 − R2R53)

+ 2

R51

(−3

�23�1

D1 + �1D2 + 5�1D1z21

R21

+ 5�23�1

D1x22

R21

)}

+ C66n2

{2

�23�1

D11

R31

(−1 + 3

R2

R21

− 15x21x

22

R41

)+ �3

�(R13−R33R

2) + 3�34�

R53(x21−x2

2 )

× �23��1

(R11 − R31R2 + 3R51x

21x

22 )

}

+ C66x2n3

{6D1

z21

R51

(1 − 5

x21

R21

)+ 1

4�R53(x

21 − x2

2 )2

}

(A20)

s∗131 = C44x1n1

{3�21

z1

R51

(−1

2�+ D2

)+ �23

2�(R21 − R41x

22 ) − �23

2�(R23 − R43x

22 )

+ 3D1z1

R51

(−�23 − 2�21 + 5

�23x22 + �21z

21

R21

)



− 3

(C13�214�P13

+ (�21 + �23)D1

)x3

R51

+ 15D1x3

R71

(�21z21 + �23x

22 )

}

+ C44x2n2

{3�23(1 + �1)D1

x3

R51

(1 − 5

x21

R21

)+ �23

4�[−2R23 + R43(x

22 − x2

1 )]

− �232�

(R21 − R41x21 )

}+ C44n3

{(�1D1 + �1

4�− D2

) 1

R31

(−1 + 3

x21

R21

)

+ 3D1(1 + �1)z21

R51

(1 − 5

x21

R21

)+ �3

4�

1

R33

(−1 + 3

x22

R23

)}(A21)

s∗112 = x1x2n1

{3�23C13D1

1

R51

(1 − 5

z21

R21

)+ 3

�34�

(C11 − C12)R53(x22 − x2

1 )

+ 3�23�1

D11

R51

(3(C11 + C12) − 5

X1

R21

)− 3

�232��1

(R31(C11 + C12) − R51X1)

}

+ n2

{�12�

1

R31

(C11 + C12 − 3

X1

R21

)+ �23

2��1C12(−R11 + R31R

2 − 3R51x21x

22 )

+ 3�23

2��1C11(−R11 + 2R31x

21 − R51x

41) + �3

2�(C12 − C11)(R13 − R33R

2 + 3R53x21x

22 )

+ �23�1

D11

R31

(3C11 + C12 − 3

3x21C11 + x2

2C12

R21

)− �1D2

1

R31

(C11 + C12 − 3

X1

R21

)

− 3�1D1z21

R51

(C11 + C12 − 5

X1

R21

)− 3�23

�1D1

x21

R51

(3C11 + C12 − 5

X1

R21

)+ C13

(C13�214�P13

+ (�21 + �23)D1

)1

R31

(1−3

z21

R21

)−3C13D1

1

R51

[�21z

21

(3−5

z21

R21

)+ �23x

21

(1−5

z21

R21

)]}

+ x2n3

{3D1

z1

R51

(C11 + 3C12 − 5

X1

R21

)+ 1

4�(R21(C11−3C12)+R41(−C11x

21+C12x

22 ))

+ 1

4�(C11 + C12)(−R23 + R43x

21 ) + 3�1C13

z1

R51

[−D2 + D1

(2 − 5

z21

R21

)]}(A22)



s∗122 = C66n1

{2

�23D1

�1

1

R31

(−1 + 3

R2

R21

− 15x21x

22

R41

)

+ �3�

(R13 − R33R2) + 3

�34�

R53(x21 − x2

2 )2 + �23

�1�(R11 − R31R

2 + 3R51x21x

22 )

}

+ C66x1x2n2

{3

R51

(2�1D2 − �1

�− 6

�23D1

�1

)+ 3

�23�1�

(R31 − R51x21 ) + 3

�32�

R53(x21 − x2

2 )

+ 30D1

R71

(�1z

21 + �23

�1x21

)}+ C66x1n3

{6�1D1

z1

R51

(1−5

x22

R21

)+ 1

4�(−4R23+R43R

2)

}

(A23)

s∗132 = C44x2n1

{3D1�

23(1 + �1)

x3

R51

(1 − 5

x21

R21

)�234�

[−2R23 + R43(x22 − x2

1 )]

− �232�

(R21 − R41x21 )

}+ C44x1n2

{3�21

z1

R51

(−1

2�+ D2

)+ �23

2�(3R21 − R41x

21 )

+ �232�

(R23 − R43x22 ) + 3�23D1

z1

R51

(−3 + 5

x21

R21

)− 3

(C13�214�P13

+ (�21 + �23)D1

)x3

R51

+ 15D1�21

x3z21

R71

− 3D1�23

x3

R51

(2 − 5

x21

R21

)− 3D1�

21

z1

R51

(2 − 5

z21

R21

)}

+ C44x1x2n3

{− 3

R51

(D2 + �1

4�

)+ 3

�34�

1

R53

− 15D1z21

R71

+ 3�1D11

R51

(1 − 5

z21

R21

)}

(A24)

s∗113 = x1n1

{3D1

z1

R51

(3C11 + C12 − 5

X1

R21

)− 1

4�(C11 + C12)(R23 − x2

2R43)

+ 1

4�(R21(−3C11 + C12) + R41(C11x

21−C12x

22 ))−3�1C13

z1

R51

[D2−D1

(2−5

z21

R21

)]}

+ x2n2

{3D1

z1

R51

(C11 + 3C12 − 5

X1

R21

)− 1

4�(R21(C11−3C12)+R41(−C11x

21+C12x

22 ))



− 1

4�(C11 + C12)(R23 − x2

1R43) − 3�1C13z1

R51

[D2 − D1

(2 − 5

z21

R21

)]}

+ n3

{3

D1

�1

z21

R51

(C11 + C12 − 5

X1

R21

)− D2

�1

1

R31

(C11 + C12 − 3

X1

R21

)

+ D2C131

R31

(1 − 3

z21

R51

)+ 3D1C13

z21

R51

(3 − 5

z21

R51

)}(A25)

s∗123 = C66x2n1

{6D1

z1

R51

(1 − 5

x21

R21

)+ 1

4�(−4R23 + R43R

2)

}

+ C66x1n2

{6D1

z1

R51

(1 − 5

x22

R21

)+ 1

4�(−4R23 + R43R

2)

}

+ 6C66x1x2n3

{D2

�1

1

R51

− 5D1

�1

z21

R71

}(A26)

s∗133 = C44n1

{1

R31

(�1D1 + �1

4�+ D2

)(−1 + 3

x21

R21

)

+ 3(�1 + 1)D1z21

R51

(1 − 5

x21

R21

)− �3

4�

1

R33

(1 − 3

x22

R23

)}

+ C44x1x2n2

{− 3

R51

(D2 + �1

4�

)+ 3

�34�

1

R53

− 15D1z21

R71

+ 3�1D11

R51

(1 − 5

z21

R21

)}

+ C44x1n3

{3D2(�1 − 1)

x3

R51

− 15D1x3z

21

R71

+ 3D1z1

R51

(2 − 5

z21

R21

)}(A27)

s∗221 = n1

{�12�

1

R31

(C11 + C12 − 3

X2

R21

)+ �23

2��1C12(−R11 + R31R

2 − 3R51x21x

22 )

+ 3�23

2��1C11(−R11 + 2R31x

22 − R51x

42) + �3

2�(C12 − C11)(R13 − R33R

2 + 3R53x21x

22 )

+ �23�1

D11

R31

(3C11 + C12 − 3

x21C12 + 3x2

2C11

R21

)− �1D2

1

R31

(C11 + C12 − 3

X2

R21

)



− 3�23�1

D1x22

R51

(3C11 + C12 − 5

X2

R21

)+ C13

(C13�214�P13

+ (�21 + �23)D1

)1

R31

(1 − 3

z21

R21

)

− 3�1D1z21

R51

(C11+C12−5

X2

R21

)−3C13D1

1

R51

[�21z

21

(3−5

z21

R21

)+�23x

22

(1−5

z21

R21

)]}

+ x1x2n2

{3�23C13D1

1

R51

(1 − 5

z21

R21

)+ 3

�34�

(C11 − C12)R53(x21 − x2

2 )

+ 3�23�1

D11

R51

(3(C11 + C12) − 5

X2

R21

)− 3

�232��1

(R31(C11 + C12) − R51X2)

}

+ x1n3

{3D1

z1

R51

(C11 + 3C12 − 5

X2

R21

)+ 1

4�(R21(C11 − 3C12) + R41(x

21C12 − x2

2C11))

+ 1

4�(C11 + C12)(−R23 + R43x

22 ) + 3�1C13

z1

R51

[−D2 + D1

(2 − 5

z21

R21

)]}(A28)

s∗222 = x1x2n1

{3

�23�1

D11

R51

(3(C11 + C12) − 5

X2

R21

)+ 3

�34�

(C11 − C12)R53(x21 − x2

2 )

+ 3�23

2��1(−R31(C11 + C12) + R51X2) + 3�23D1C13

1

R51

(1 − 5

z21

R21

)}

+ n2

{�32�

(C11 − C12)(R13 − R33R2 + 3R53x

21x

22 ) + �1

2�

1

R31

(C11 + C12 − 3

X2

R21

)

+ �232��1

C11(−R11 + R31R2 − 3R51x

21x

22 ) + 3

�232��1

C12(−R11 + 2R31x21 − R51x

41)

+ �23�1

D11

R31

(C11 + 3C12 − 3

x21C12 + 3x2

2C11

R21

)− 3�1D1

z21

R51

(C11 + C12 − 5

X2

R21

)

− 3�23�1

D1x21

R51

(C11 + 3C12 − 5

X2

R21

)+ C13

(C13�214�P13

+ (�21 + �23)D1

)1

R31

(1 − 3

z21

R21

)

− �1D21

R31

(C11+C12−3

X2

R21

)−3C13D1

1

R51

[�21z

21

(3−5

z21

R21

)+�23x

21

(1−5

z21

R21

)]}



+ x2n3

{3D1

z1

R51

(3C11 + C12−5

X2

R21

)+ 1

4�(R21(−3C11+C12)+R41(C11x

22−C12x

21 ))

+ 1

4�(C11 + C12)(−R23 + x2

1R43) + 3�1C13z1

R51

[−D2 + D1

(2 − 5

z21

R21

)]}(A29)

s∗231 = C44x2n1

{3�21

z1

R51

(−1

2�+ D2

)+ �23

2�(3R21 − R41x

22 ) + �23

2�(R23 − R43x

21 )

+ 3�23D1z1

R51

(−3 + 5

x22

R21

)− 3

(C13�214�P13

+ (�21 + �23)D1

)x3

R51

+ 15D1�21

x3z21

R71

− 3D1�23

x3

R51

(2 − 5

x22

R21

)− 3D1�

21

z1

R51

(2 − 5

z21

R21

)}+ C44x1n2

{− �23

2�(R21

− R41x22 ) + 3D1�

23(1 + �1)

x3

R51

(1 − 5

x22

R21

)+ �23

4�[−2R23 + R43(x

21 − x2

2 )]}

+ C44x1x2n3

{− 3

R51

(D2 + �1

4�

)+ 3

�34�

1

R53

− 15D1z21

R71

+ 3�1D11

R51

(1 − 5

z21

R21

)}

(A30)

s∗232 = C44x1n1

{3�23(1 + �1)D1

x3

R51

(1 − 5

x22

R21

)− �23

2�(R21 − R41x

22 )

+ �234�

[−2R23 + R43(x21 − x2

2 )]}

+ C44x2n2

{3�21

z1

R51

(−1

2�+ D2

)+ �23

2�(R21 − R41x

21 )

− �232�

(R23 − R43x21 ) + 3

x3

R51

(C13�214�P13

+ (�21 + �23)D1

)+ 15D1

x3

R71

(�21z21 + �23x

21 )

+ 3D1z1

R51

(−�23 − 2�21 + 5

�23x21 + �21z

21

R21

)}+ C44n3

{3D1(1 + �1)

z21

R51

(1 − 5

x22

R21

)

+(�1D1 + �1

4�− D2

) 1

R31

(−1 + 3

x22

R21

)+ �3

4�

1

R33

(−1 + 3

x21

R23

)}(A31)

s∗223 = x1n1

{3D1

z1

R51

(C11 + 3C12 − 5

X2

R21

)− 1

4�(C11 + C12)(R23 − x2

2R43)



+ 1

4�(R21(C11 − 3C12) + R41(x

21C12 − x2

2C11)) − 3�1C13z1

R51

[D2 − D1

(2 − 5

z21

R21

)]}

+ x2n2

{3D1

z1

R51

(3C11 + C12 − 5

X2

R21

)+ 1

4�(R21(C12 − 3C11) + R41(x

22C11 − x2

1C12))

− 1

4�(C11 + C12)(R23 − x2

1R43) − 3�1C13z1

R51

[D2 − D1

(2 − 5

z21

R21

)]}

+ n3

{3

D1

�1

z21

R51

(C11 + C12 − 5

X2

R21

)− D2

�1

1

R31

(C11 + C12 − 3

X2

R21

)

+ D2C131

R31

(1 − 3

z21

R51

)+ 3D1C13

z21

R51

(3 − 5

z21

R51

)}(A32)

s∗233 = C44x1x2n1

{− 3

R51

(D2 + �1

4�

)+ 3

�34�

1

R53

− 15D1z21

R71

+ 3�1D11

R51

(1 − 5

z21

R21

)}

+ C44n2

{1

R31

(�1D1 + �1

4�+ D2

)(−1 + 3

x22

R21

)+ 3(�1 + 1)D1

z21

R51

(1 − 5

x22

R21

)

− �34�

1

R33

(1 − 3

x21

R23

)}+ C44x2n3

{3D2(�1 − 1)

x3

R51

− 15D1x3z

21

R71

+ 3D1z1

R51

(2 − 5

z21

R21

)}(A33)

s∗331 = n1

{�23

2��1C13(−R11 + R31R

2 − 3R51x21x

22 ) + 3�23

2��1C13(−R11 + 2R31x

22 − R51x

42)

+ �1C13

(1

2�− D2

)1

R31

(2 − 3

R2

R21

)+ �23

�1C13D1

1

R31

(4 − 3

3x22 + x2

1

R21

)

− 3D1C131

R51

[�1z

21

(2 − 5

R2

R21

)− �23

�1x22

(4 − 5

R2

R21

)]+ C33

(C13�214�P13

+ (�21 + �23)D1

)

× 1

R31

(1 − 3

z21

R21

)− 3�21D1C33

z21

R51

(3 − 5

z21

R21

)− 3�23D1C33

x22

R51

(1 − 5

z21

R21

)}



+ x1x2n2

{3�23C33D1

1

R51

(1 − 5

z21

R21

)+ 3

�23�1

C13D11

R51

(6 − 5

R2

R21

)+ 3

�232��1

C13

× (−2R31 + R51R2)

}+ x1n3

{3C13D1

z1

R51

(2 − 5

R2

R21

)+ C13

4�[−2R21 + R41(x

21 − x2

2 )]

+ C13

2�(−R23 + R43x

22 ) − 3�1C33D2

z1

R51

+ 3�1C33D1z1

R51

(2 − 5

z21

R21

)}(A34)

s∗332 = x1x2n1

{3�23C33D1

1

R51

(1 − 5

z21

R21

)+ 3

�23�1

C13D11

R51

(6 − 5

R2

R21

)

+ 3�23

2��1C13(−2R31 + R51R

2)

}+ n2

{�23

2��1C13(−R11 + R31R

2 − 3R51x21x

22 )

+ 3�232��1

C13(−R11 + 2R31x21 − R51x

41) + �1C13

(1

2�− D2

)1

R31

(2 − 3

R2

R21

)

+ �23�1

C13D11

R31

(4 − 3

x22 + 3x2

1

R21

)−3D1C13

1

R51

[�1z

21

(2−5

R2

R21

)−�23

�1x21

(4−5

R2

R21

)]

+ C33

(C13�214�P13

+ (�21 + �23)D1

)1

R31

(1 − 3

z21

R21

)− 3�21D1C33

z21

R51

(3 − 5

z21

R21

)

− 3�23D1C33x21

R51

(1 − 5

z21

R21

)}+ x2n3

{3C13D1

z1

R51

(2 − 5

R2

R21

)

+C13

4�[−2R21 + R41(x

22 − x2

1 )] + C13

2�(−R23 + R43x

21 ) − 3�1C33D2

z1

R51

+ 3�1C33D1z1

R51

(2 − 5

z21

R21

)}(A35)

s∗333 = x1n1

{3D1C13

z1

R51

(4 − 5

R2

R21

)+ C13

4�[−2R21 + R41(x

21 − x2

2 )]

+ C13

2�(−R23 + R43x

22 ) − 3�1C33D2

z1

R51

+ 3�1C33D1z1

R51

(2 − 5

z21

R21

)}

+ x2n2

{3D1C13

z1

R51

(4 − 5

R2

R21

)+ C13

4�[−2R21 + R41(x

22 − x2

1 )]



+ C13

2�(−R23 + R43x

21 ) − 3�1C33D2

z1

R51

+ 3�1C33D1z1

R51

(2 − 5

z21

R21

)}

+ n3

{3

C13D1

�1

z21

R51

(2 − 5

R2

R21

)− C13D2

�1

1

R31

(2 − 3

R2

R21

)

+ C33D21

R31

(1 − 3

z21

R21

)+ 3C33D1

z21

R51

(3 − 5

z21

R21

)}(A36)

In any case

s∗211 = s∗

121, s∗212 = s∗

122, s∗213 = s∗

123, s∗311 = s∗

131, s∗321 = s∗

231

s∗312 = s∗

132, s∗322 = s∗

232, s∗313 = s∗

133, s∗323 = s∗

233

and

R2 = x21 + x2

2 , R2i = R2 + z2i , R∗

i = Ri + zi, zi = �ix3

R1i = 1

RiR∗2i

, R2i = 1

R3i R

∗i

+ 1

R2i R

∗2i

, R3i = 1

R3i R

∗2i

+ 2

R2i R

∗3i

R4i = 3

R5i R

∗i

+ 3

R4i R

∗2i

+ 2

R3i R

∗3i

, R5i = 1

R5i R

∗2i

+ 2

R4i R

∗3i

+ 2

R3i R

∗4i

D1 = C13 + C44

4�P13, D2 = C44

4�P13, D3i = C11 + C13�2i

C13 + C44, D4i = C11 − C13�

2i li

D5i = C13 − C33�2i li , D6i = 2A′

i�i − Ai, D7i = 2C44�3i (1 + li)A

′i

li = C11/�2i − C44

C13 + C44, X1 = C11x

21 + C12x

22 , X2 = C12x

21 + C11x

22

APPENDIX B

B.1 Transversely isotropic solids

There exists five independent constants characterizing the material elastic behaviour: C11, C12,C13, C33 and C44. P13 = √

C11C33.Expressions for the Ilk terms are

I11 = −C44

( �34�

+ C44(D61D31 + D62D32)) ∮

��

r × Nr3

dl



+ 3C244(D61D31 + D62D32)

∮��

r2,1r × N

r3dl + 3

C44�34�

∮��

r2,2r × N

r3dl (B1)

I22 = −C44

( �34�

+ C44(D61D31 + D62D32)) ∮

��

r × Nr3

dl

+ 3C244(D61D31 + D62D32)

∮��

r2,2r × N

r3dl + 3

C44�34�

∮��

r2,1r × N

r3dl (B2)

I33 = −(2C13(�1A′1D51 + �2A

′2D52) − C33(�

21A1D51 + �22A2D52))

∮��

r × Nr3

dl (B3)

I12 = 3C44

(C44(D41D31 + D42D32) − �3

4�

) ∮��

r,1r,2r × N

r3dl (B4)

I21 = 3C44

(C44(D61D51 + D62D52) − �3

4�

) ∮��

r,1r,2r × N

r3dl (B5)

when P13 − C13 − 2C44 �= 0, and

I11 = −C44

(�3 + �14�

+ �1D1 − D2

)∮��

r × Nr3

dl

+ 3C44

(�1D1 − D2 + �1

4�

) ∮��

r2,1r × N

r3dl + 3

C44�34�

∮��

r2,2r × N

r3dl (B6)

I22 = −C44

(�3 + �14�

+ �1D1 − D2

)∮��

r × Nr3

dl

+ 3C44

(�1D1 − D2 + �1

4�

) ∮��

r2,2r × N

r3dl + 3

C44�34�

∮��

r2,1r × N

r3dl (B7)

I33 = D2

(C13

�1+ C33

)∮��

r × Nr3

dl (B8)

I12 = 3C44

(�1D1 − D2 + �3 − �1

4�

)∮��

r,1r,2r × N

r3dl (B9)

I21 = I12

when P13 − C13 − 2C44 = 0. In any case

I13 = I23 = I31 = I32 = 0



Expressions for the Jlhk terms are

J1h1 = −C44

( �34�

+ C44(D61D31 + D62D32)) ∮

��

eh × Nr

dl

+ C244(D61D31 + D62D32)

[∮��

r2,1eh × N

rdl + 2h1

∮��

eh × Nr

dl]

+ C44�34�

[∮��

r2,2eh × N

rdl + 2h2

∮��

eh × Nr

dl]

(B10)

J2h2 = −C44

( �34�

+ C44(D61D31 + D62D32)) ∮

��

eh × Nr

dl

+ C244(D61D31 + D62D32)

[∮��

r2,2eh × N

rdl + 2h2

∮��

eh × Nr

dl]

+C44�34�

[∮��

r2,1eh × N

rdl + h1

∮��

2eh × Nr

dl]

(B11)

J3h3 = −(2C13(�1A′1D51+�2A

′2D52)−C33(�

21A1D51+�22A2D52))

∮��

eh × Nr

dl (B12)

J1h2 = C44

(C44(D41D31 + D42D32) − �3

4�

)

×[h1

∮��

r2,1e2 × N

rdl + h2

∮��

r2,2e1 × N

rdl]

(B13)

J2h1 = C44

(C44(D61D51 + D62D52) − �3

4�

)

×[h1

∮��

r2,1e2 × N

rdl + h2

∮��

r2,2e1 × N

rdl]

(B14)

when P13 − C13 − 2C44 �= 0, and

J1h1 = C44

{(�1D1 − D2 + �1

4�

) [∮��

r2,1eh × N

rdl + 2h1

∮��

eh × Nr

dl]

−(

�3 + �14�

+ �1D1 − D2

)∮��

eh × Nr

dl

+ �34�

[∮��

r2,2eh × N

rdl + 2h2

∮��

e1 × Nr

dl]}

(B15)



J2h2 = C44

{(�1D1 − D2 + �1

4�

) [∮��

r2,1eh × N

rdl + 2h1

∮��

eh × Nr

dl]

−(

�3 + �14�

+ �1D1 − D2

)∮��

eh × Nr

dl

+ �34�

[∮��

r2,2eh × N

rdl + 2h2

∮��

eh × Nr

dl]}

(B16)

J3h3 = D2

(C13

�1+ C33

)∮��

eh × Nr

dl (B17)

J1h2 = C44

(�1D1 − D2 + �3 − �1

4�

)[h1

∮��

r2,1e2 × N

rdl + h2

∮��

r2,2e1 × N

rdl](B18)

J2h1 = J1h2

when P13 − C13 − 2C44 = 0.In any case

J1h3 = J2h3 = J3h1 = J3h2 = 0

The Klk terms are zero except for

Kl3 = C44(D31A1 + D32A2)(1 − l3)N3

∮��

el × Nr

dl (B19)

when P13 − C13 − 2C44 �= 0, and

Kl3 = C11 − C13�218�C11

(1 − l3)N3

∮��

el × Nr

dl (B20)

when P13 − C13 − 2C44 = 0.

ACKNOWLEDGEMENTS

This work was supported by the Ministerio de Ciencia y Tecnologia of Spain. (DPI2000-1217-C02-01and DPI2001-2377-C02-01). The financial support is gratefully acknowledged.

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