A Highorder extended finite element method for ...

INTERNATIONAL JOURNAL FOR NUMERICAL METHODS IN ENGINEERING Int. J. Numer. Meth. Engng (2013) Published online in Wiley Online Library (wileyonlinelibrary.com). DOI: 10.1002/nme.4584

A High-order extended finite element method for extraction of mixed-mode strain energy release rates in arbitrary crack settings based on Irwin’s integral Mengyu Lan1 , Haim Waisman1, * ,† and Isaac Harari2 1 Department

of Civil Engineering & Engineering Mechanics, Columbia University, New York, NY 10027, USA 2 Faculty of Engineering, Tel Aviv University, 69978 Ramat Aviv, Israel

SUMMARY An analytical formulation based on Irwin’s integral and combined with the extended finite element method is proposed to extract mixed-mode components of strain energy release rates in linear elastic fracture mechanics. The proposed formulation extends our previous work to cracks in arbitrary orientations and is therefore suited for crack propagation problems. In essence, the approach employs high-order enrichment functions and evaluates Irwin’s integral in closed form, once the linear system is solved and the algebraic degrees of freedom are determined. Several benchmark examples are investigated including off-center cracks, inclined cracks, and two crack growth problems. On all these problems, the method is shown to work well, giving accurate results. Moreover, because of its analytical nature, no special post-processing is required. Thus, we conclude that this method may provide a good and simple alternative to the popular J-integral method. In addition, it may circumvent some of the limitations of the J-integral in 3D modeling and in problems involving branching and coalescence of cracks. Copyright © 2013 John Wiley & Sons, Ltd. Received 30 May 2013; Revised 6 September 2013; Accepted 18 September 2013 KEY WORDS:

extended finite element method; Irwin’s integral; mixed-mode fracture; stress intensity factors; strain energy release rate; high-order asymptotic functions

1. INTRODUCTION It is important to obtain accurate estimation of stress intensity factors (SIFs) or equivalently strain energy release rates (SERRs) in order to predict accurate crack paths and the overall response of the structure. These quantities are used to predict the ‘stress intensity’ near the tip of a crack to determine its stability and direction of propagation [1–3]. Some of the most well-known and used methods are the J-integral method [4], its domain (area) variant [5–7], and the related M-integral (or interaction integral) method [8–10]; the stiffness derivative method [11, 12]; and the virtual crack closure technique [13], which is inspired by Irwin’s integral [14]. In particular, the virtual crack closure technique has been implemented in many commercial software packages because of its simplicity and effectiveness in computing the individual components of the mixed-mode energy release rates [15, 16]. An extensive review of the method can be found in [17]. The extended finite element method (XFEM) has been proposed by Belytschko and co-workers [18, 19] for fracture problems. It alleviates shortcomings associated with re-meshing. Thus one can model cracks without the need for special conforming meshes, which imply that for crack

*Correspondence to: Haim Waisman, Department of Civil Engineering & Engineering Mechanics, Columbia University, New York, NY 10027, USA. † E-mail: [email protected] Copyright © 2013 John Wiley & Sons, Ltd.

M. LAN, H. WAISMAN AND I. HARARI

propagation problems, re-meshing may completely be avoided. Enrichment functions, such as Heaviside functions behind the crack tip and four asymptotic functions (called branch functions) at the tip element, are incorporated to model the presence of cracks. The branch functions are obtained p from Williams analytical solution [20] and incorporate the r terms in the displacement field, which provides the stress singularity at the crack tip. We will review the XFEM in a later section. Although the cracks in XFEM can be modeled without re-meshing, it is still necessary to determine the stability and direction of crack propagation. Most of the XFEM literature related to linear elastic fracture mechanics has focused on computation of SIFs by the J-integral method developed by Rice [4] and its variants. A different approach that extends the Parks classical stiffness derivative method [11] to XFEM was proposed by Waisman [21]. In that work, it was shown that the stiffness derivative can be computed in closed form during the analysis and thus the virtual crack extension, and the error inherent in the finite difference scheme of the classical method can completely be avoided. In our previous work [22], we proposed a new approach to compute mixed-mode components of SERRs by direct evaluation of Irwin’s integral in the framework of XFEM. High-order enrichment functions in XFEM were employed, and closed form expressions for SERRs were obtained. We have shown that special, costly post-processing procedures may be avoided and the SERRs can be obtained directly from the algebraic degrees of freedom. Results of the numerical examples indicate that high-order enrichment functions have significant effect on the convergence. When the integral limits tend to zero, simpler SERR expressions are obtained, and high-order terms vanish. Nonetheless, these terms contribute indirectly via coefficients of first-order terms. Thus, the approach was found to be simple, efficient, and accurate. In the current paper, we generalize our previous work to compute SERRs in arbitrary crack settings. That is, the newly proposed method can handle inclined cracks whose tip can be located at any point within an element, where higher-order asymptotic fields are used to enrich the tip element in the XFEM framework. Closed form expressions are obtained by direct evaluation of Irwin’s integral, and so there is no need for special post-processing procedures. For convenience, the derivation is carried out in polar coordinates, and the integrals, obtained in closed form, are verified via numerical integration. Several benchmark examples are studied, and the results show that high accuracy can be achieved with high-order enrichment terms. High-order terms have also been studied by [23–28]. In particular, the enrichment of the approximation space by higher-order terms of asymptotic field using h-p clouds were studied by Liszka et al. [23] and Duarte et al. [24]. Liu et al. [25] replaced the branch functions with high-order asymptotic terms to retain not only the leading terms but also the associated coefficients. It was shown by the authors that SIFs can directly be extracted without any post-processing. This approach has inspired Zamani et al. [26] and Réthoré et al. [28] to achieve high accuracy by combining appropriate modifications of the enrichment scheme with an overlapping domain decomposition scheme. Potentially, our approach can circumvent some of the limitations of the J-integral when applied to 3D problems and problems involving branching and coalescence of cracks. In three dimensions, the J-integral requires the evaluation of a surface integral, which can be quite complicated [29]. In problems involving branching and coalescence of cracks, the J-integral path of a particular crack must cross other nearby cracks, and hence, the J-integral result is not valid in such cases [30]. The remainder of the paper is organized as follows. In Section 2, we present a brief introduction to the XFEM. In Section 3, we discuss high-order enrichment XFEM and our proposed approach, which includes both analytical and numerical formulations of Irwin’s integral, followed by several benchmark examples in Section 4 and conclusions.

2. MODELING CRACKS BY THE EXTENDED FINITE ELEMENT METHOD 2.1. Problem statement Consider a two-dimensional solid with an internal crack in the domain , as illustrated in Figure 1. The solid is subjected to body forces b in , traction loading tN applied on t , and displacement Copyright © 2013 John Wiley & Sons, Ltd.

Int. J. Numer. Meth. Engng (2013) DOI: 10.1002/nme

HIGH ORDER XFEM FOR EXTRACTION OF SERRs USING IRWIN’S INTEGRAL

Figure 1. A solid with a plane crack represented by red solid line.

boundary conditions u D uN on u . In cartesian coordinates, the displacement field is decomposed into its components u D ¹ux , uy º as illustrated. Additionally, the crack is defined by internal boundaries c , which are assumed to be traction free. The Galerkin approximation of the proposed problem is to seek a kinematically admissible displacement field uh 2 U h , which is a finite-dimensional subspace of the solution space U , such that Z Z Z uh W C W wh d D b wh d C (1) tN wh d 8wh 2 U0h ,

t

where and C are the standard strain and elasticity tensors. The weighting functions wh , whose values vanish on the Dirichlet boundary u , belong to the finite-dimensional subspace U0h . The aim of this work is to compute mixed-mode components of SERRs directly from Irwin’s integral, by employing the XFEM with high-order crack tip asymptotic functions. These functions allow for the evaluation of the integral quantities in closed form and would therefore result in an accurate and efficient method. 2.2. Extended finite element method overview The key idea of XFEM is to locally enrich the standard finite element approximation with local partition of unity enrichment functions, which are chosen according to the problem at hand. It follows that for crack problems, the mesh is independent of the crack orientation [31–33]. An excellent review on the XFEM can be found in [34, 35]. Similar enrichment methods for modeling cracks are based on the generalized finite element method [36, 37]. Let uh 2 U h be an extended finite element approximation to the discretized weak form of elasticity, where U h is the appropriate Sobolev space [38]. The XFEM enriches the conventional shape function space with a set of functions H.x/ and Fj .x.r, //, such that 2 3 nJ nT nF n X X X X 4NI .x/ (2) uh .x/ D NI .x/uI C NI .x/H.x/aI C Fj .x.r, // bjI 5 , I D1

I D1

I D1

j D1

where x D ¹x, yºT are the spatial cartesian coordinates. n, nJ , nT , and nF are the number of total nodal points, jump-enriched nodes, tip-enriched nodes, and number of enrichment functions, respectively. NI .x/ are the standard finite element shape functions associated with standard degrees of freedom uI , whereas aI and bjI are the degrees of freedom associated with the enriched nodes. Typically, for linear elastic fracture problems, the crackptip zone is enriched with the classical analytical solution for the near tip field [20], and only the r terms, which are given in Equation (9), Copyright © 2013 John Wiley & Sons, Ltd.



Figure 2. Enrichment visualization of cracks in extended finite element method. Crack lines are illustrated in red, blue circles are jump-enriched nodes, and green squares are tip-enriched nodes.

are employed. Note that these functions are given in polar coordinates .r, /, which is illustrated in Figure 1. Element nodes (behind the crack tip), which are fully cut by the crack, are enriched with the Heaviside function ´ C1 above cC H.x/ D , (3) 1 below c where cC andpc define the edges of the discontinuity line that splits the element into two parts. Note that the r term in the displacement field is directly built into the displacement equation, and hence, the stress singularity of p1r appears in the solution. The enriched nodes of inclined 2D cracks are illustrated in Figure 2. The reader is referred to references [33, 39] for more details on the XFEM. 2.3. High-order enrichment functions For traction-free cracks in local crack coordinates (Figure 1), the asymptotic fields for the displacement components u and v near the crack tip are given by Williams [20] as u.r, / D

² i i i r i=2 i C C .1/i cos cos KI i 2 p 2 2 2 2 2n 2 i D1 ³ i i i i C .1/i sin sin 2 C KII i 2 2 2 2

(4)

² i i r i=2 i i .1/i sin C sin KI i 2 p 2 2 2 2 2i 2 i D1 ³ i i i i C .1/i cos cos 2 , C KII i 2 2 2 2

(5)

1 X

v.r, / D

1 X

where KI i and KII i are coefficients and and are shear modulus and Kolosov constant, respectively. The Kolosov constant is defined as 8 < 3 4, plane strain D 3 : , plane stress 1C

(6)

where is Poisson’s ratio. Copyright © 2013 John Wiley & Sons, Ltd.



The expressions in Equations (4) and (5) may be simplified as u.r, / D

4 X

ui r i=2 , C O r 5=2 ,

(7)

vi r i=2 , C O r 5=2 ,

(8)

i D1

v.r, / D

4 X i D1

where the coefficients u1 , : : : , u4 and v1 , : : : , v4 in Equations (7) and (8) are given in [22]. As opposed to [25], here we incorporate high-order terms in a more traditional XFEM fashion, considering only the space spanned by these terms. See [22] for a complete derivation. Thus, the enrichment functions for different orders of r are ³ ² p p p p p (9) r W F1 D r sin , r cos , r sin sin , r sin cos 2 2 2 2 r W F2 D ¹r cos , r sin º ³ ² 3=2 3=2 3=2 3=2 3 3=2 W F D r cos , r sin , r sin sin , r sin cos r 2 2 2 2 ® ¯ r 2 W F4 D r 2 , r 2 sin 2 , r 2 cos 2 , and the full set of enrichments, with 13 terms, used in our analysis is therefore ® ¯ F .r, / D F1 , F2 , F3 , F4 ,

(10) (11) (12)

(13)

3. EXTRACTION OF MIXED-MODE COMPONENTS OF STRAIN ENERGY RELEASE RATES USING IRWIN’S INTEGRAL AND EXTENDED FINITE ELEMENT METHOD In our previous work [22], Irwin’s integral was evaluated directly in closed form and mixed-mode components of SERR were obtained directly. This approach was shown to yield accurate results by testing the method on well known benchmark examples, but its application is limited to straight crack orientation with crack tip located at the element center. In this paper, we show that the proposed direct analyticalmethod to extract mixed-mode components of SERRs from Irwin’s integral can be extended to general crack orientations. 3.1. Analytical expansion of Irwin’s integral We start from Irwin’s work [14], which describe the work required to extend a crack by an infinitesimal distance c to be equal to the work required to close the crack to its original length. Thus, the SERR for a mixed-mode state, expressed in a polar coordinate system .r, / with the origin at the crack tip, is defined by G D GI C GII ,

(14)

where G is the total energy release rate, additively decomposed into individual components GI and GII corresponding to mode I and II deformations Z c 1

. c r, 0/uN .r/dr c!C0 2 c 0 Z c 1 D lim

r . c r, 0/uN r .r/dr, c!C0 2 c 0

GI D lim GII

Copyright © 2013 John Wiley & Sons, Ltd.

(15) (16)



Figure 3. Crack sliding and opening in polar coordinate system.

where and r are the normal and shear stresses in polar coordinates, uN r and uN are relative sliding and opening displacements between corresponding points on crack surfaces, and c is the crack extension at the crack tip. See Figure 3 for an illustration. The sliding and opening displacement jumps are defined by ur .r/ D .ur .r, / ur .r, // u .r/ D .u .r, / u .r, //,

(17) (18)

Thus, under the assumption that the local crack coordinate system is aligned with the crack axis (Figure 1), the relations are simplified, and the sliding and opening displacement jumps, Equations (17)–(18), become ur .r/ D u .r, / u .r, / u .r/ D v .r, / v .r, / .

(19) (20)

Substituting Williams expansion, Equations (4) and (5), into Equations (17) and(18), retaining only the first four leading terms of r, yields ur .r/ D

1 X

1 4 X X 9 2i 1 2 ui r i=2 , ui r i=2 , D mII r C O r2 i

i D1

u .r/ D

1 X

i D1

1 4 X X 9 2i 1 vi r i=2 , vi r i=2 , D mIi r 2 C O r 2 .

i D1


i D1

(21)

i D1

(22)

i D1



More details on the expansion of the terms in Equations (21) and (22) are shown in Appendix A and [40]. Note that even order terms of r in the expansion of the displacement jump vanish from the sum as these terms are continuous across the discontinuity; hence, only odd order terms appear in the derivation in Equations (21) and (22). Also, note that we have truncated the resulting sum after the order of r 7=2 that appear in the formulation. To clarify, this is due to our choice of high-order enrichment functions (Equation (13)) and considering the XFEM formulation given in Section 3.2. In Section 3.2, shape functions are introduced (Equation (40)), which add additional r terms, and so it will become apparent that all leading terms of r up to r 7=2 contribute to the derivation of the displacement jump and must remain. The kinematic strain-displacement in polar coordinates are defined as @ur @r

(23)

ur 1 @u C r r @

(24)

rr D D r

1 D 2

1 @ur @u u C r @ @r r

,

(25)

where rr , , and r are the radial, tangential, and shear strains, respectively. Considering a plane strain state, the stresses and r are given as

D

E .rr C .1 / / .1 C /.1 2/

(26)

E r . .1 C /

(27)

r D

Combining the stresses with strains, Equations (23)–(26) and plugging in Williams solutions, Equations (4) and (5), we arrive at the definition of normal and shear stresses ahead of the crack tip 7 X E i 6

.r, 0/ D mIi r 2 C O r 2 .1 C /.1 2/ 12

(28)

i D5

7 X E i 6 2 CO r2 mII r . i 2.1 C / 12

r .r, 0/ D

(29)

i D5

Note that all orders of r in Equations (28) and (29) are preserved in the derivation of the stresses, as opposed to the derivation of displacement jump in Equations (21) and (22). However, the sum is truncated after r 3 , which again is due to our choice of high-order enrichment functions (Equation (13)) and considering the XFEM formulation given in Section 3.2 and shape functions introduced (Equation (40)). Finally, substituting the opening displacements in Equations (21)and (22) and stresses in Equations (28) and (29) into the definition of SERR, GI in Equation (32) and GII in Equation (33) and integrating, we arrive at 13

e G I . c/ D

X E i ˛iI c 2 C O c 7 , .1 C /.1 2/

(30)

i D0

and 13

e G II . c/ D

X E i ˛iII c 2 C O c 7 . 2.1 C /

(31)

i D0




Figure 4. An inclined crack with tip at O 0 .

Here, e G I and e G II have been defined as Z c 1 f G D

. c r, 0/uN .r/dr I 2 c 0 Z c 1 GII D

r . c r, 0/uN r .r/dr, 2 c 0

e

(32) (33)

so that GI D lim e G I . c/

(34)

G II . c/. GII D lim e

(35)

c!C0 c!C0

The coefficients ˛iI and ˛iII are listed in Appendix B. The goal is to find the coefficients ˛iI and ˛iII analytically, using the enrichment functions in XFEM. 3.2. Analytical formulation of Irwin’s integral To obtain expressions for the coefficients ˛iI and ˛iII , we consider a generic rectangular element with an inclined crack and crack tip at O 0 , as shown by the red line in Figure 4, where O is the center of the element and (x,y) is global Cartesian coordinate system. (u,v) defines the local Cartesian coordinate system with ˛ being the angle between the two coordinate systems. ur and u are displacements in the polar coordinate system (r,). In polar the coordinate system, any point in the element domain can be defined with respect to the global system (x,y) as xp D r cos. C ˛/ C xo0 yp D r sin. C ˛/ C yo0 . Copyright © 2013 John Wiley & Sons, Ltd.

(36) (37) Int. J. Numer. Meth. Engng (2013) DOI: 10.1002/nme


The nodal coordinates of the element are therefore .xI , yI /, xI D ˙hx =2, yI D ˙hy =2, I D 1, : : : , 4, where hx and hy are the length of the edges in x-direction and y-direction, respectively. The transformation of displacement fields from polar coordinates to cartesian coordinates can be conveniently written in matrix form as ur cos. C ˛/ sin. C ˛/ ux D . (38) sin. C ˛/ cos. C ˛/ u uy Tip elements in XFEM consider near tip asymptotic fields, and hence, the displacement field in Equation (2) may be simplified and written in polar coordinates as 0 1 nF 4 X X (39) NI @uI C Fj bjI A . u.r, / D j D1

I D1

In polar coordinates, the standard linear shape functions are ! xI .r cos. C ˛/ C xo0 / 1 yI .r sin. C ˛/ C yo0 / 1C4 NI D 1C4 4 h2x h2y

I D 1, : : : , 4,

(40) @NI @NI @u and the derivatives of the displacement field @u , and shape functions , are given in @r @ @r @ Appendix C. First, we substitute Equation (38) into Equations (17) and (18) to arrive at ur .r/ cos. C ˛/ sin. C ˛/ ux .r, / ux .r, / D uy .r, / uy .r, / u .r/ sin. C ˛/ cos. C ˛/ (41) cos.˛/ sin.˛/ ux .r, / ux .r, / D . uy .r, / uy .r, / sin.˛/ cos.˛/ Assuming a general case where all enrichment functions, including high-order functions in Equation (13), are used to enrich the solution space (i.e. nF D 13), we substitute Equation (40) into Equation (39), which leads to 1 0 4 13 X X ux .r, / ux .r, / D NI .r, / @uxI C Fj .r, /bxjI A I D1

4 X I D1

uy .r, / uy .r, / D

4 X

4 X

j D1 13 X

NI .r, / @uxI C

j D1

0

NI .r, / @uyI C

I D1

0

0

13 X

1 Fj .r, /bxjI A 1

Fj .r, /byjI A

j D1

NI .r, / @uyI C

I D1

(42)

13 X

1 Fj .r, /byjI A .

(43)

j D1

To obtain stresses .r, 0/ and r .r, 0/, the strains in Equations (23)–(25) are expressed by substituting Equation (41) to give the expressions @ux @uy cos ˛ C sin ˛ @r @r

(44)

1 @ux @uy ux cos ˛ C uy sin ˛ sin ˛ C cos ˛ r @ @

(45)

1 @ux 1 @uy @ux @uy ux uy cos ˛ C sin ˛ sin ˛ C cos ˛ C sin ˛ cos ˛ . r @ r @ @r @r r r

(46)

rr D D r D

1 2




(a)

(b)

Figure 5. Admissible integration step c chosen as the minimum self-similar distance from the crack tip to one of the element edges. (a) c is chosen as the distance to the edge ahead of the crack tip; and (b) c is chosen as the distance to the edge behind the crack tip.

Then stresses .r, 0/ and r .r, 0/ are obtained by substituting Equations (44)–(46) into Equations (26) and (27), which gives the expressions ˇ ˇ @ux ˇˇ E @uy ˇˇ

.r, 0/ D cos ˛ C sin ˛ .1 C /.1 2/ @r ˇD0 @r ˇD0 ˇ ˇ (47) 1 @ux ˇˇ 1 @uy ˇˇ E.1 / sin ˛ C cos ˛ C .1 C /.1 2/ r @ ˇD0 r @ ˇD0 ˇ ˇ ˇ ˇ E 1 @uy ˇˇ @ux ˇˇ @uy ˇˇ 1 @ux ˇˇ

r .r, 0/D cos ˛ C sin ˛ sin ˛ C cos ˛ . 2.1 C / r @ ˇD0 r @ ˇD0 @r ˇD0 @r ˇD0 (48) The next step is to expand Equations (41), (47), and (48) by substituting Equations (39), (40), and (C.1)–(C.4) at the appropriate angle. Then, the final step is to match all coefficients of leading order terms for opening displacement jumps in Equation (41) with Equations (21) and (22). Similarly, the coefficients for stresses in Equations (47) and (48) are matched with those in Equations (28) and (29). All coefficients can be found in Lan’s thesis [40]. Finally, the general solution for SERRs with finite integration limits is obtained in closed form. Remark 1 It is important to note that in the case of a horizontal crack with ˛ D 0 and crack tip positioned at .xo0 , yo0 / D .0, 0/, all the formulations and coefficients presented in this paper reduce back to the derivations given in [22]. Thus, the current work is a generalization of the method we proposed in [22]. 3.3. Numerical formulation of Irwin’s integral Strain Energy Release Rates (SERRs) can also be obtained numerically from Irwin’s integral. In this case, the integration step of c can be chosen from 0 to some admissible distance within an element, which is defined as the crack tip to the closest boundary ahead of or behind the crack tip. The admissible distance is illustrated in Figure 5. All results related to the derivation in Section 3 have been verified by numerical integration (1D Gauss quadrature rule) of Irwin’s integral, as shown in Figure 6. Stress points in front of the crack Copyright © 2013 John Wiley & Sons, Ltd.



Figure 6. Numerical integration of Irwin’s integral used to verify the analytical results in Section 3. Same color ‘x’-symbols indicate stress and displacement opening couples used in the integration.

tip and the corresponding displacement opening behind the tip have been computed numerically by the following approximation: 1 e G I . c/ D 2 c 1 e G II . c/ D 2 c

Z

c

. c r, 0/uN .r/dr 0

Z

ngp 1 X wi c ri , 0 uN ri 2 c

(49)

1 X wi r c ri , 0 uN r ri , 2 c

(50)

i D1 ngp

c

r . c r, 0/uN r .r/dr 0

i D1

where ngp is the number of Gauss quadrature points, ri are the Gauss point coordinates, and wi are the weights associated with the integration rule. The pairs of stress and displacement opening used in Equations (49) and (50) to compute SERRs are illustrated by the same color ‘x’-symbol in Figure 6. The analytical derivation in Section 3 is verified, and found to be in excellent agreement with the numerical integration. In practical XFEM realizations, the stress and strain are computed in a Cartesian coordinate system, so both quantities need to be transformed into a polar coordinate system shown in Equations (49) and (50). The transformation relation for displacements is given in Equation (41) and for stresses is given as: 2

3 2 3 32 sin2 .˛/ 2 cos.˛/ sin.˛/ cos2 .˛/

rr .r, 0/

xx .r, 0/ 4 .r, 0/ 5 D 4 sin2 .˛/ cos2 .˛/ 2 sin.˛/ cos.˛/ 5 4 yy .r, 0/ 5 .

r .r, 0/

xy .r, 0/ sin.˛/ cos.˛/ sin.˛/ cos.˛/ cos2 .˛/ sin2 .˛/ (51) Copyright © 2013 John Wiley & Sons, Ltd.



The algorithm using numerical implementation is straightforward, although time-consuming because of the singular profile of stress, which requires many quadrature points in the vicinity of the crack tip.

Algorithm: Numerical Implementation of Irwin’s Integral Preprocessor i. Initialize the system by incorporating high-order enrichment functions F. (Equations (9)–(13)) Solution i. Solve linear system Kd D f to find node displacement d. Postprocessor i. ii. iii. iv. v.

Extract node displacement de at tip element. Compute displacements ux , uy stresses xx , yy , xy at Gauss point with de . Transform displacements ux , uy into crack opening using Equation (41). Transform stresses xx , yy , xy using Equation (51). Compute the numerical approximation of Irwin’s integral using Equation (50).

4. NUMERICAL EXAMPLES Stress intensity factors are related to SERRs and can directly be obtained by [13] s KI D

e GI E

s KII D

e G II E ,

(52)

where ² D

1 2 , plane strain , 1, plane stress

(53)

G II are computed by Equations (32) and (33) and E is the Young’s modulus. Note that the e G I and e for a finite c. The proposed approach is studied on an off-center crack tip problem, an angle-cracked plate under tension problem, and a crack growth problem. All numerical examples shown in this section are plane strain problems. To eliminate the source of error related to tip element integration, we have used a trapezoidal integration rule, with 200 200 equally spaced quadrature points, for evaluating the stiffness matrix. Less quadrature points, for example, 50 50, could also provide good accuracy with higher efficiency. In all subsequent examples, the crack extension in Irwin’s integral is taken as the admissible distance (as was indicated in Figure 5). 4.1. Off-center crack tip problem In order to show the robustness of proposed approach, the first examples are pure modes I and II cases with crack tip being off the center in the tip element. The problem consists of a square plate with dimensions 10 10 units. Displacement boundary conditions with normalized SIFs KI D 1 Copyright © 2013 John Wiley & Sons, Ltd.



Figure 7. Crack tip off the center of the tip element.

and KII D 1 are applied to generate modes I and II crack openings. These essential boundary conditions (assuming a plane strain state) are given by 8 r 2.1 C / r ˆ 2 ˆ ˆ 2 2 cos KI cos < ux D E 2 2 2 Mode I W r ˆ 2.1 C / r ˆ ˆ 2 2 cos2 KI sin : uy D E 2 2 2

(54)

8 r 2.1 C / r ˆ 2 ˆ ˆ 2 2 C cos KII sin < ux D E 2 2 2 Mode II W r , ˆ r 2.1 C / ˆ ˆ 1 2 sin2 KII cos : uy D E 2 2 2

(55)

where r is the distance from the boundary to the crack tip and is the angle between the positive x-axis and boundary node. The boundary conditions are applied to all four sides of the rectangular domain. The Young’s modulus and Poisson’s ratio are chosen as E D 1 and D 0.3, respectively, and the mesh is chosen as 19 19 elements. The crack tip is varied from 2hx =5 to 2hx =5 at an increment of hx =10 within the tip element, as illustrated in Figure 7. The computed results are summarized in Table I. It can be seen from Table I that the proposed approach yields convergent results with higherorder enrichment at the tip element. The integration step c varies at different tip location, and the computation presents relatively stable results.

4.2. Angle-cracked plate under tension problem The second benchmark example is a plate with an inclined crack placed at the center under uniaxial tension by opposing tractions (Figure 8). The square plate has a side length of W D 10 units with a crack length of a D 0.5 units. Uniaxial loading is applied to the top and bottom sides of the plate. Copyright © 2013 John Wiley & Sons, Ltd.



Table I. Results for KI and KII of off-center pure mode problem. Tip location

KI

Relative error (%)

KII

Relative error (%)

Condition number

1=2

1 2 3 4 5 6 7 8 9

0.763 0.780 0.787 0.789 0.791 0.763 0.734 0.701 0.656

23.71 21.97 21.34 21.12 20.89 23.65 26.55 29.92 34.43

0.723 0.751 0.778 0.801 0.820 0.820 0.818 0.813 0.805

27.69 24.87 22.16 19.88 18.03 18.04 18.24 18.67 19.52

3.25 104 1.53 104 9.10 103 5.81 103 3.90 103 5.79 103 8.92 103 1.51 104 2.92 104

1

1 2 3 4 5 6 7 8 9

0.864 0.856 0.893 0.936 0.966 0.990 1.009 1.023 1.030

13.58 14.40 10.73 6.44 3.49 1.03 0.93 2.30 3.01

0.823 0.848 0.885 0.915 0.935 0.950 0.960 0.967 0.973

17.69 15.24 11.54 8.53 6.54 5.06 4.00 3.28 2.65

1.29 105 8.85 104 6.13 104 4.27 104 3.00 104 1.99 104 1.68 104 3.00 104 5.55 104

3=2

1 2 3 4 5 6 7 8 9

0.910 0.965 1.000 1.007 1.007 1.010 1.013 1.014 1.011

8.97 3.46 0.03 0.72 0.65 1.03 1.26 1.39 1.06

0.931 0.972 0.996 1.003 1.004 1.007 1.011 1.015 1.022

6.89 2.75 0.37 0.33 0.42 0.72 1.08 1.50 2.16

4.80 106 3.54 106 1.97 106 1.12 106 6.20 105 7.43 105 1.20 106 2.01 106 3.05 106

2

1 2 3 4 5 6 7 8 9

0.994 0.998 1.007 1.010 1.007 1.007 1.006 1.004 1.000

0.61 0.19 0.72 0.97 0.71 0.71 0.56 0.36 0.03

0.960 0.995 1.005 1.008 1.006 1.009 1.012 1.014 1.014

4.08 0.47 0.53 0.77 0.64 0.90 1.19 1.37 1.41

6.33 107 3.59 107 2.13 107 1.32 107 8.33 106 5.76 106 4.42 106 4.65 106 6.58 106

Order

In this example, the mesh is chosen as 9999, where Poisson’s ratio is chosen as 0.3, and Young’s modulus E as 1. The exact SIFs are given by p KI D a cos2 .˛/

(56)

p KII D a sin.˛/ cos.˛/,

(57)

where ˛ is the crack angle to the horizontal. The numerical results, in comparison with exact solutions, are shown in Figure 9 and Table II. In Figure 9, the solid and dashed lines represent the exact solution in Equations (56) and (57), whereas the points show the computed values. In Table II, we provide the corresponding values shown in Figure 9(d). It can be seen that order 3/2 provides satisfactory results, whereas order 2 presents more stable results. Copyright © 2013 John Wiley & Sons, Ltd.



Figure 8. Angle-cracked plate under uniaxial loading in opposite direction.

Order = 1

Order = 1/2 1.20

1.20 KI

KI

1.05

KII

0.90

0.90

0.75

0.75

KI ,K II

KI ,K II

1.05

0.60

0.60

0.45

0.45

0.30

0.30

0.15

0.15

0.00

KII

0.00 0

10

20

30

40

50

60

70

80

0

90

(a)

10

20

30

50

60

70

80

90

80

90

(b) Order = 3/2

Order = 2

1.20

1.20

KI

1.05

KI

1.05

KII

0.90

0.90

0.75

0.75

KI ,K II

KI ,K II

40

0.60

KII

0.60

0.45

0.45

0.30

0.30

0.15

0.15 0.00

0.00 0

10

20

30

40

50

60

(c)

70

80

90

0

10

20

30

40

50

60

70

(d)

Figure 9. KI and KII in a plate with an angled center crack. Dashed line and solid line represent exact stress intensity factors KI and KII.




Table II. Results for KI and KII of angle-cracked problem for order 2. Order 2

Angle (ı )

Exact KI

KI

Relative error (%)

Exact KII

KII

Relative error (%)

0 5 10 15 20 25 30 35 40 50 55 60 65 70 75 80 85 90

1.253 1.244 1.216 1.169 1.107 1.029 0.940 0.841 0.735 0.518 0.412 0.313 0.224 0.147 0.084 0.038 0.010 0.000

1.275 1.244 1.244 1.167 1.113 1.042 1.003 0.856 0.728 0.518 0.415 0.322 0.234 0.153 0.083 0.039 0.009 0.000

1.75 0.02 2.36 0.19 0.54 1.19 6.72 1.74 1.05 0.00 0.58 2.88 4.53 4.18 1.05 2.15 1.44 0.00

0.000 0.109 0.214 0.313 0.403 0.480 0.543 0.589 0.617 0.617 0.589 0.543 0.480 0.403 0.313 0.214 0.109 0.000

0.000 0.113 0.222 0.330 0.400 0.516 0.555 0.595 0.630 0.623 0.604 0.512 0.501 0.406 0.317 0.222 0.110 0.000

0.00 4.26 3.53 5.25 0.74 7.45 2.24 1.00 2.08 0.96 2.58 5.62 4.36 0.69 1.18 3.37 0.69 0.00

Figure 10. Double cantilever beam specimen under opposite load.

4.3. Double cantilever beam We consider a crack growth example in a double cantilever beam specimen [19]. The geometry is chosen as 11.8 3.94 units shown in Figure 10. The initial horizontal crack length a D 3.94 units and opposite loads P D 197 are applied on the left side. Poisson’s ratio is chosen as 0.3 and Young’s modulus E D 3 107 . The mesh is chosen as 80 32 elements. A small perturbation with length x D 0.3 units is initiated at the crack tip at an angle . To observe the different crack propagation path corresponding to different perturbations, is set equal to 0ı , 5ı , 10ı , and 15ı . We have used the maximum hoop stress [31] criteria to compute the direction c in which the crack will propagate. The angle c is computed as 0

c D 2 arctan

1 @ KI ˙ 4 KII

s

KI KII

2

1 C 8A , < c < ,

(58)

where SIFs KI and KII are computed using the proposed approach. An illustration of the crack extension scheme is illustrated in Figure 11. The crack is propagated in the following two steps: Copyright © 2013 John Wiley & Sons, Ltd.



(b)

(a)

Figure 11. Crack extension illustration. Green-colored element indicates a tip element, and blue indicates an element fully cut by the crack. (a) Step i: red dashed line shows the tentative direction and extension of crack propagation at step i C 1 determined by the proposed method. (b) Step i C 1: the kink in tip element at step i is removed by assuming a straight line segment in the element (shown in green), and the actual crack path at propagation step i C 1 is shown by the red line.

(a)

(b)

(c) Figure 12. Crack propagation illustration by setting initial perturbation D 15.

Step i: The direction of crack propagation is determined from the proposed method q by computing KI and KII , and the crack is virtually extended by a length of a D h2x C h2y . We choose this extension length to make sure that the crack propagates to the next element. In Figure 11(a), the black line indicates the crack extension at propagation step I, and the red dashed line represents the virtual crack extension. Step i+1: The kink inside the element is removed by assuming a straight line segment in the element. This is represented by the green line in Figure 11(b), which simply connects the two Copyright © 2013 John Wiley & Sons, Ltd.


M. LAN, H. WAISMAN AND I. HARARI 2.00

Y_Tip

1.75

1.50

1.25

1.00

4.0

4.2

4.4

4.6

4.8

5.0

5.2

5.4

5.6

5.8

X_Tip

Figure 13. Comparison of crack path initiated by different perturbation.

Figure 14. Geometry and initial set-up of the L-shaped plate example.

crossing points in the element. Thus, we obtain the new crack segment at the tip element (shown in solid red line in Figure 11(b) in step i+1). Figure 12 illustrates the crack growth path initiated by perturbation D 15, and Figure 13 shows a comparison of crack path with different initial perturbations. 4.4. L-shaped plate We consider an example with multiple cracks propagating in an L-shaped plate [26, 41]. Direct evaluation of Irwin’s integral allows our method to handle multiple cracks. This is in contrast to the J-integral, which is invalid for multiple crack tips that are located in the same J-integral domain. Such cases may arise in coarse meshes [30]. The geometry of the problem and its boundary conditions are shown in Figure 14. The bottom edge of the plate is fixed, whereas vertical displacement of u D 1.55 mm is applied on the lower left corner as shown since crack initiation. The material modeled is a high-strength steel 18Ni1900 with p Young’s modulus E D 190 GPa, Poisson’s ratio D 0.3, and fracture toughness Kc D 68 MPa m [42]. The initial lengths of the three cracks, from top to bottom, are l1 D 11.67 mm, l2 D 6.80 mm, Copyright © 2013 John Wiley & Sons, Ltd.



Figure 15. Computed equivalent mode I stress intensity factor KIeq . Solid red line represents the material fracture toughness in the succeeding text, which the cracks stop growing.

(a)

(b)

(c)

(d) Figure 16. Crack propagation illustration for steps 1, 7, 14, and 19.




and l3 D 5.30 mm with an in-between separation distance of h1 D 21.5 mm and h2 D 14.0 mm. The angle ˛ for the first crack is equal to 30ı . We choose a structured mesh with 3 402 elements. The problem is solved using a quasi-static approach in which at each step, the equivalent SIF KIeq is computed to determine crack propagation [43], as shown in Figure 15. The computed value is compared with the fracture toughness Kc , and crack growth occurs if KIeq > Kc .

(59)

It is interesting to observe the ‘crack shielding effect’ [41], in which, at first, all cracks propagate; however, at some point (shown in Figure 16), crack 3 stops growing (at step 7), and later (at step 14), crack 2 also stops growing, and only crack 1 proceeds. 5. CONCLUSIONS To extend our previous work, we have proposed a general analytical approach to compute mixedmode components of SERRs by direct evaluation of Irwin’s integral, in the framework of the XFEM. The newly proposed approach can be used to investigate cracks in general formulations in terms of crack tip location and crack orientation. In this method, special post-processing procedures may be avoided, and the SERRs can be obtained directly from the algebraic degrees of freedom after solving the linear system of equations. Several benchmark examples have been studied, including off-center problem, angle-cracked problem, and crack growth problems, to investigate the high-order enrichment effects and integration limits of the proposed approach. The results indicate that high-order enrichment functions have significant effect on the convergence. The admissible distance, which is defined as the crack tip to the closest boundary ahead of or behind the crack tip, provides satisfactory accuracy. Moreover, this approach can overcome difficulties in J-integral for 3D problems and problems involving branching and coalescence of cracks in which crack tips cannot be located in the same J-integral domain. The proposed method is simple, but the current quadrature rule is inefficient for the type of crack tip asymptotic functions that we use. Improving the numerical integration will be the subject of future work. With the use of dedicated quadrature rules that exploit the tip asymptotic behavior, the method has the potential of becoming competitive. APPENDIX A. Derivation of Equations (21) and (22) Starting from Equation (21), we first show that the even order terms will vanish. Assume i D 2m, where m is an integer; by substituting D into Equation (4), we will obtain ² 2m 2m r 2m=2 2m 2m 2m u2m .r, / D KI 2m C cos C.1/ ./ cos 2 ./ p 2 2 2 2 2n 2 ³ 2m 2m 2m 2m 2m sin .1/ ./ sin 2 ./ CKII 2m C 2 2 2 2 rm D p ¹KI 2m Œ. C m C 1/ cos.m/ m cos.m 2/./ 2n 2 C KII 2m Œ. C m 1/ sin.m/ m sin.m 2/./ º (A.1) rm D p ¹KI 2m Œ. C m C 1/ cos.m/ m cos.m/ 2n 2 C KII 2m Œ. C m 1/ sin.m/ C m sin.m/ º m r D p ¹KI 2m Œ. C 1/ cos.m/ . 2n 2 Copyright © 2013 John Wiley & Sons, Ltd.



We then substitute D into Equation (4) to arrive at ² 2m 2m r 2m=2 2m 2m u2m .r, / D KI 2m C C .1/2m cos ./ cos 2 ./ p 2 2 2 2 2n 2 ³ 2m 2m 2m 2m 2m sin CKII 2m C .1/ ./ sin 2 ./ 2 2 2 2 rm D p ¹KI 2m Œ. C m C 1/ cos.m/ m cos.m 2/./ 2n 2 C KII 2m Œ. C m 1/ sin.m/ m sin.m 2/./ º (A.2) rm D p ¹KI 2m Œ. C m C 1/ cos.m/ m cos.m/ 2n 2 C KII 2m Œ. C m 1/ sin.m/ m sin.m/ º m r D p ¹KI 2m Œ. C 1/ cos.m/ . 2n 2 Finally, we arrive at u2m .r, / u2m .r, / D 0.

(A.3)

Next, we substitute D into Equation (5), and we will obtain ² 2m 2m r 2m=2 2m 2m 2m KI 2m v2m .r, / D sin .1/ ./ C sin 2 ./ p 2 2 2 2 2n 2 ³ 2m 2m 2m 2m 2m cos .1/ ./ cos 2 ./ CKII 2m C 2 2 2 2 rm D p ¹KI 2m Œ. m 1/ sin.m/ C m sin.m 2/./ 2n 2 C KII 2m Œ. C m 1/ cos.m/ m cos.m 2/./ º (A.4) m r D p ¹KI 2m Œ. m 1/ sin.m/ m sin.m/ 2n 2 C KII 2m Œ. C m 1/ cos.m/ m cos.m/ º m r D p ¹KI 2m Œ. 1/ cos.m/ . 2n 2 ² 2m 2m r 2m=2 2m 2m 2m sin v2m .r, / D KI 2m .1/ ./ C sin 2 ./ p 2 2 2 2 2n 2 ³ 2m 2m 2m 2m 2m cos CKII 2m C .1/ ./ cos 2 ./ 2 2 2 2 rm D p ¹KI 2m Œ. m 1/ sin.m/ C m sin.m 2/./ 2n 2 C KII 2m Œ. C m 1/ cos.m/ m cos.m 2/./ º (A.5) m r D p ¹KI 2m Œ. m 1/ sin.m/ C m sin.m/ 2n 2 C KII 2m Œ. C m 1/ cos.m/ m cos.m/ º m r D p ¹KI 2m Œ. 1/ cos.m/ . 2n 2 Copyright © 2013 John Wiley & Sons, Ltd.



Finally, we arrive at v2m .r, / v2m .r, / D 0.

(A.6)

Next, we show that the odd terms are preserved. Assume i D 2m C 1, where m is an integer; by substituting D into Equation (4), we will obtain

u2mC1 .r, / D

D

² 2m C 1 2m C 1 r .2mC1/=2 C .1/2mC1 cos ./ KI 2mC1 C p 2 2 2n 2 2m C 1 2m C 1 cos 2 ./ 2 2 2m C 1 2m C 1 .1/2mC1 sin ./ CKII 2mC1 C 2 2 ³ 2m C 1 2m C 1 sin 2 ./ 2 2 r mC1=2 p ¹KI 2mC1 Œ. C m 1=2/ sin.m/ C .m C 1=2/ sin.m/ 2n 2 C KII 2mC1 Œ. C m C 3=2/ cos.m/ C .m C 1=2/ cos.m/ º

D

r mC1=2 p ŒKII 2mC1 Œ. 1/ cos.m/ . 2n 2 (A.7)

We then substitute D into Equation (4) to arrive at

u2mC1 .r, / D

D

² 2m C 1 2m C 1 r .2mC1/=2 KI 2mC1 C C .1/2mC1 cos ./ p 2 2 2n 2 2m C 1 2m C 1 cos 2 ./ 2 2 2m C 1 2m C 1 2mC1 sin .1/ ./ CKII 2mC1 C 2 2 ³ 2m C 1 2m C 1 sin 2 ./ 2 2 r mC1=2 p ¹KI 2mC1 Œ. C m 1=2/ sin.m/ .m C 1=2/ sin.m/ 2n 2 C KII 2mC1 Œ. C m C 3=2/ cos.m/ .m C 1=2/ cos.m/ º

D

r mC1=2 p ŒKII 2mC1 Œ. C 1/ cos.m/ . 2n 2 (A.8)

Finally, we arrive at

u2mC1 .r, / u2mC1 .r, / D


r mC1=2 p ŒKII 2mC1 Œ. C 1/.1/m . n 2

(A.9)



Next, we substitute D into Equation (5), and we will obtain ² 2m C 1 2m C 1 r .2mC1/=2 2mC1 KI 2mC1 sin .1/ ./ v2mC1 .r, / D p 2 2 2n 2 2m C 1 2m C 1 sin. 2/./ C 2 2 2m C 1 2m C 1 .1/2mC1 cos ./ CKII 2mC1 C 2 2 ³ 2m C 1 2m C 1 cos 2 ./ 2 2 r mC1=2 p ¹KI 2mC1 Œ. m C 1=2/ cos.m/ .m C 1=2/ cos.m/ 2n 2

D

C KII 2mC1 Œ. C m C 3=2/ sin.m/ C .m C 1=2/ sin.m/ º r mC1=2 p ŒKI 2mC1 Œ. 1/ cos.m/ . 2n 2

D

(A.10)

v2mC1 .r, / D

D

² 2m C 1 2m C 1 r .2mC1/=2 KI 2mC1 .1/2mC1 sin ./ p 2 2 2n 2 2m C 1 2m C 1 sin 2 ./ C 2 2 2m C 1 2m C 1 2mC1 cos .1/ ./ CKII 2mC1 C 2 2 ³ 2m C 1 2m C 1 cos 2 ./ 2 2 r mC1=2 p ¹KI 2mC1 Œ. m C 1=2/ cos.m/ C .m C 1=2/ cos.m/ 2n 2 C KII 2mC1 Œ. C m C 3=2/ sin.m/ C .m C 1=2/ sin.m/ º

D

r mC1=2 p ŒKI 2mC1 Œ. C 1/ cos.m/ . 2n 2 (A.11)

Finally, we arrive at

v2mC1 .r, / v2mC1 .r, / D

r mC1=2 p ŒKI 2mC1 Œ. C 1/.1/m . n 2

(A.12)

It can clearly be seen from Equations (A.1)–(A.12) that even order terms of r in the expansion of the displacement jump vanish and odd order terms are preserved. Copyright © 2013 John Wiley & Sons, Ltd.



B. Coefficients of Irwin’s integral expansion The coefficients of Irwin’s integral expansion in Equations (30) and (31) are 8 ˛0r D 14 mr1 mr5 ˆ ˆ ˆ ˆ ˆ ˆ ˛1r D 13 mr1 mr6 ˆ ˆ ˆ ˆ 3 1 ˆ ˛2r D 16 mr2 mr5 C 16 mr1 mr7 ˆ ˆ ˆ ˆ 2 ˆ ˆ mr1 mr8 C 15 mr2 mr6 ˛3r D 15 ˆ ˆ ˆ ˆ 1 1 5 ˆ ˛4r D 32 mr1 mr9 C 32 mr2 mr7 C 32 mr3 mr5 ˆ ˆ ˆ ˆ 8 2 ˆ mr1 mr10 C 35 mr2 mr8 C 17 mr3 mr6 ˛5r D 105 ˆ ˆ ˆ ˆ < ˛ r D 3 mr mr C 5 mr mr C 5 mr mr C 35 mr mr r D I , II 6 256 2 9 256 3 7 256 1 11 256 4 5 8 2 16 ˆ mr2 mr10 C 63 mr3 mr8 C 315 mr1 mr12 C 19 mr4 mr6 ˛7r D 315 ˆ ˆ ˆ ˆ 3 3 7 ˆ ˛8r D 512 mr2 mr11 C 512 mr3 mr9 C 512 mr4 mr7 ˆ ˆ ˆ ˆ 8 16 2 ˆ ˆ mr3 mr10 C 1155 mr2 mr12 C 99 mr4 mr8 ˛9r D 693 ˆ ˆ ˆ ˆ 5 7 r ˆ ˛10 D 2048 mr3 mr11 C 2048 mr4 mr9 ˆ ˆ ˆ ˆ 16 8 r ˆ ˆ D 3003 mr3 mr12 C 1287 mr4 mr10 ˛11 ˆ ˆ ˆ ˆ 5 r ˆ ˛12 D 4096 mr4 mr11 ˆ ˆ ˆ : r 16 mr4 mr12 . ˛13 D 6435

(B.1)

C. Derivatives of displacement field and shape function in polar coordinates The derivatives of the displacement field in Equation (39) with respect to r and are nF nF 4 4 4 X X X @u X @NI @NI X @Fj Fj bjI C NI D uI C bjI @r @r @r @r

(C.1)

nF nF 4 4 4 X X X @u X @NI @NI X @Fj Fj bjI C NI D uI C bjI , @ @ @ @

(C.2)

I D1

I D1

j D1

I D1

j D1

I D1

I D1

I D1

j D1

j D1

where the derivatives of the shape functions are ! yI .r sin. C ˛/ C yo0 / @NI xI cos. C ˛/ 1C4 D @r h2x h2y xI .r cos. C ˛/ C xo0 / yI sin. C ˛/ 1C4 C h2y h2x

(C.3)

! @NI yI .r sin. C ˛/ C yo0 / xI r sin. C ˛/ 1C4 D @ h2x h2y yI r cos. C ˛/ xI .r cos. C ˛/ C xo0 / C 1 C 4 . h2y h2x

(C.4)

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