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Linear elastic fracture simulation directly from CAD: 2D NURBS-based implementation and role of tip enrichment

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Abstract

We propose a method for simulating linear elastic crack growth through an isogeometric boundary element method directly from a CAD model and without any mesh generation. To capture the stress singularity around the crack tip, two methods are compared: (1) a graded knot insertion near crack tip; (2) partition of unity enrichment. A well-established CAD algorithm is adopted to generate smooth crack surfaces as the crack grows. The M integral and \(J_k\) integral methods are used for the extraction of stress intensity factors (SIFs). The obtained SIFs and crack paths are compared with other numerical methods.

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Acknowledgments

The first and last authors would like to acknowledge the financial support of the Framework Programme 7 Initial Training Network Funding under grant number 289361 ‘Integrating Numerical Simulation and Geometric Design Technology’. S. P. A. Bordas also thanks partial funding for his time provided by the UK Engineering and Physical Science Research Council (EPSRC) under grant EP/G069352/1 Advanced discretization strategies for ‘atomistic’ nano CMOS simulation; the EPSRC under grant EP/G042705/1 ‘Increased Reliability for Industrially Relevant Automatic Crack Growth Simulation with the eXtended Finite Element Method’ and the European Research Council Starting Independent Research Grant (ERC Stg grant agreement No. 279578) entitled ‘Towards real time multiscale simulation of cutting in non-linear materials with applications to surgical simulation and computer guided surgery’. E. Atroshchenko was partially supported by Fondecyt grant number 11130259 entitled ‘Boundary element modeling of crack propagation in micropolar materials’.

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Authors and Affiliations

  1. Institute of Mechanics Materials and Advanced Manufacturing, Cardiff University, Cardiff, CF24 3AA, UK

    X. Peng, P. Kerfriden & S. P. A. Bordas

  2. Department of Mechanical Engineering, University of Chile, Santiago, 8370448, Chile

    E. Atroshchenko

  3. Faculté des Sciences, de la Technologie et de la Communication, Université du Luxembourg, 6 rue Richard Coudenhove-Kalergi, Research Unit in Engineering Science Campus Kirchberg, G 007, 1359, Luxembourg, Luxembourg

    S. P. A. Bordas

  4. University of Western Australia, 35 Stirling Hwy, Crawley, WA, 6009, Australia

    S. P. A. Bordas

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  1. X. Peng
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  2. E. Atroshchenko
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  3. P. Kerfriden
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Correspondence to E. Atroshchenko.

Appendices

Appendix 1

The fundamental solutions for traction BIE are:

$$\begin{aligned} K_{ij}= & {} \frac{1}{4\pi (1-\nu )r}[(1-2\nu )(\delta _{ij}r_{,k}+\delta _{jk}r_{,i} -\delta _{ik}r_{,j})\nonumber \\&+\,2r_{,i}r_{,j}r_{,k}]n_k(\mathbf {s}) \end{aligned}$$
(49)
$$\begin{aligned} S_{ij}= & {} \frac{\mu }{2\pi (1-\nu )r^2}\Big \{2\frac{\partial r}{\partial n}[(1-2\nu )\delta _{ik}r_{,j}\nonumber \\&+\,\nu (\delta _{ij}r_{,k}+\delta _{jk}r_{,i})-4r_{,i}r_{,j}r_{,k}] \nonumber \\&+\,2\nu (n_ir_{,j}r_{,k}+n_kr_{,i}r_{,j})-(1-4\nu )\delta _{ik}n_j \nonumber \\&+\,(1-2\nu )(2n_jr_{,i}r_{,k}+\delta _{ij}n_k+\delta _{jk}n_i)\Big \}n_k(\mathbf {s})\nonumber \\ \end{aligned}$$
(50)

Now we present the SST formula for the hyper-singular integral as follows. Expanding the components of distance between field and source points as Taylor series in parent space gives:

$$\begin{aligned} x_i-s_i= & {} \frac{\text {d}x_i}{\text {d}\hat{\xi }} \Big |_{\hat{\xi }=\hat{\xi _s}}(\hat{\xi }-\hat{\xi _s})+\frac{\text {d}^2x_i}{\text {d}{\hat{\xi }}^2} \Big |_{\hat{\xi }=\hat{\xi _s}}\frac{(\hat{\xi }-\hat{\xi _s})^2}{2}+\cdot \cdot \cdot \nonumber \\:= & {} A_i(\hat{\xi }-\hat{\xi _s})+B_i(\hat{\xi }-\hat{\xi _s})^2+\cdot \cdot \cdot \nonumber \\= & {} A_i\delta +B_i\delta ^2+O(\delta ^3) \end{aligned}$$
(51)

and

$$\begin{aligned} A:= & {} \left( \sum ^2_{k=1}A^2_k\right) ^{\frac{1}{2}} \nonumber \\ C:= & {} \sum ^2_{k=1}A_kB_k \end{aligned}$$
(52)

The first and second derivatives are:

$$\begin{aligned} \begin{aligned} \frac{\text {d}x_i}{\text {d}\xi }&=\frac{\text {d}N_a}{\text {d}\xi }x_i^a \\ \frac{\text {d}^2x_i}{\text {d}\xi ^2}&=\frac{\text {d}^2N_a}{\text {d}\xi ^2}x_i^a \\ \frac{\text {d}x_i}{\text {d}\hat{\xi }}&=\frac{\text {d}x_i}{\text {d}\xi }\frac{\text {d}\xi }{\text {d}\hat{\xi }}\\ \frac{\text {d}^2x_i}{\text {d}\hat{\xi }^2}&=\frac{\text {d}^2x_i}{\text {d}\xi ^2}\Big (\frac{\text {d}\xi }{\text {d}\hat{\xi }}\Big )^2 \end{aligned} \end{aligned}$$
(53)

The derivative \(r_{,i}\) can be expressed as

$$\begin{aligned} \begin{aligned} r_{,i}&=\frac{x_i-s_i}{r}=\frac{A_i}{A}+\left( {B_i}{A}-A_i\frac{A_kB_k}{A^3}\right) \delta +O(\delta ^2)\\&:=d_{i0}+d_{i1}\delta +O(\delta ^2) \end{aligned} \end{aligned}$$
(54)

The term \(1/r^2\) can be expressed as

$$\begin{aligned} \begin{aligned} \frac{1}{r^2}&=\frac{1}{A^2\delta ^2}-\frac{2C}{A^4\delta }+O(1)\\&:=\frac{S_{-2}}{\delta ^2}+\frac{S_{-1}}{\delta }+O(1) \end{aligned} \end{aligned}$$
(55)

The component of Jacobian from parametric space to physical space can be expressed as:

$$\begin{aligned} \begin{aligned} J_1(\xi )&=J_{10}(\xi _s)+J_{11}(\xi _s)(\xi -\xi _s)+O((\xi -\xi _s)^2) \\&=J_{10}(\xi _s)+\frac{\text {d}\xi }{\text {d}\hat{\xi }}\Big |_{\xi =\xi _s}J_{11}(\xi _s)\delta +O(\delta ^2) \\ J_2(\xi )&=J_{20}(\xi _s)+J_{21}(\xi _s)(\xi -\xi _s)+O((\xi -\xi _s)^2) \\&=J_{10}(\xi _s)+\frac{\text {d}\xi }{\text {d}\hat{\xi }}\Big |_{\xi =\xi _s}J_{21}(\xi _s)\delta +O(\delta ^2) \\ \text {i.e.}, \\ J_k(\xi )&:=J_{k0}(\xi _s)+\frac{\text {d}\xi }{\text {d}\hat{\xi }}\Big |_{\xi =\xi _s} J_{k1}(\xi _s)\delta +O(\delta ^2) \\ \end{aligned} \end{aligned}$$
(56)

and we note that

$$\begin{aligned} \begin{aligned} J(\xi )&=\sqrt{J_1^2(\xi )+J_2^2(\xi )}=\sqrt{\left( \frac{\text {d}y}{\text {d}\xi }\right) ^2+ \left( -\frac{\text {d} x}{\text {d}\xi }\right) ^2} \\ \mathbf {n}(\xi )&=\left[ \frac{\text {d}y}{\text {d}\xi },-\frac{\text {d}x}{\text {d}\xi }\right] \\ \text {i.e.}, \\ n_k(\xi )&=J_k(\xi )/J(\xi ) \end{aligned} \end{aligned}$$
(57)

And the NURBS basis function is also expanded as:

$$\begin{aligned} \begin{aligned} N_{a}(\hat{\xi })&=N_a(\hat{\xi }_s)+\frac{\text {d}N_a}{\text {d}\xi }\Big |_{\xi =\xi _s}(\xi -\xi _s)+\cdot \cdot \cdot \\&=N_a(\hat{\xi }_s)+\frac{\text {d}N_a}{\text {d}\xi }\Big |_{\xi =\xi _s}\frac{\text {d}\xi }{\text {d}\hat{\xi }}\Big |_{\hat{\xi }=\hat{\xi }_s}\delta +\cdot \cdot \cdot \\&:=N_{a0}(\hat{\xi }_s)+N_{a1}(\hat{\xi }_s)\frac{\text {d}\xi }{\text {d}\hat{\xi }}\Big |_{\hat{\xi }=\hat{\xi }_s}\delta +O(\delta ^2) \end{aligned} \end{aligned}$$
(58)

The detail form of hyper-singular kernel \(S_{ij}\) is (plane strain)

$$\begin{aligned} S_{ij}(\mathbf {s},\mathbf {x})= & {} \frac{\mu }{2\pi (1-\nu )r^2} \Big \{2\frac{\partial r}{\partial n}\left[ (1-\nu )\delta _{ik}r_{,j}\right. \nonumber \\&\left. +\,\nu (\delta _{ij}r_{,k}+\delta _{jk}r_{,i} -4r_{,i}r_{,j}r_{,k})\right] \nonumber \\&+\,2\nu (n_ir_{,j}r_{,k}+n_kr_{,i}r_{,j})- (1-4\nu )\delta _{ik}n_j\nonumber \\&+\,(1-2\nu )(2n_jr_{,i}r_{,k}+\delta _{ij}n_k+\delta _{jk}n_i) \Big \}n_k(\hat{\xi }_s)\nonumber \\:= & {} \frac{1}{r^2}h(\hat{\xi }) \end{aligned}$$
(59)

Noting that \(n_k(\xi )=J_k(\xi )/J(\xi )\), Use the above expansions to rewrite \(h(\xi )\) as:

$$\begin{aligned} h(\hat{\xi })= & {} \frac{h_0(\hat{\xi _s})}{J(\xi )}+\frac{h_1(\hat{\xi _s})}{J(\xi )}\delta +O(\delta ^2) \end{aligned}$$
(60)
$$\begin{aligned} h_0(\hat{\xi _s})= & {} \Big (2\nu (J_{i0}d_{j0}d_{k0}+J_{k0}d_{i0}d_{j0})\nonumber \\&+\,(1-2\nu )(2J_{j0}d_{i0}d_{k0}+\delta _{ij}J_{k0}+\delta _{jk}J_{i0})\nonumber \\&+\,(1-4\nu )\delta _{ik}J_{j0}\Big )\frac{\mu }{2\pi (1-\nu )}n_k(\hat{\xi }_s)\end{aligned}$$
(61)
$$\begin{aligned} h_1(\hat{\xi _s})= & {} \Big [2(d_{l1}J_{l0}+d_{l0}J_{l1})\Big ((1-2\nu ) \delta _{ik}d_{j0}\nonumber \\&+\,\nu (\delta _{ij}d_{k0}+\delta _{jk}d_{i0})-4d_{i0}d_{j0}d_{k0}\Big ) \nonumber \\&+\,2\nu \Big (J_{i0}(d_{j1}d_{k0}+d_{j0}d_{k1})+J_{i1}d_{j0}d_{k0}\nonumber \\&+\,J_{k0}(d_{i1}d_{j0}+d_{i0}d_{j1})+J_{k1}d_{i0}d_{j0}\Big ) \nonumber \\&+\,(1-2\nu )\Big (2(J_{j1}d_{i0}d_{k0}+J_{j0}(d_{i1}d_{k0}+d_{i0}d_{k1}))\nonumber \\&+\,\delta _{ij}J_{k1}+\delta _{jk}J_{i1}\Big ) \nonumber \\&-\,(1-4\nu )\delta _{ik}J_{j1}\Big ]\frac{\mu }{2\pi (1-\nu )}n_k(\hat{\xi }_s) \end{aligned}$$
(62)

Thus,

$$\begin{aligned}&h(\hat{\xi })N_a(\hat{\xi })J(\hat{\xi })=\Big (h_0(\hat{\xi }_s)+h_1(\hat{\xi }_s)\delta +O(\delta ^2)\Big )\nonumber \\&\quad \Big (N_{a0}(\hat{\xi }_s)+\frac{\text {d}\xi }{\text {d}\hat{\xi }}\Big |_{\hat{\xi }=\hat{\xi }_s}N_{a1}(\hat{\xi }_s)\delta +O(\delta ^2)\Big )\nonumber \\&\quad =h_0N_{a0}+\Big (h_1N_{a0}+h_0N_{a1}\frac{\text {d}\xi }{\text {d}\hat{\xi }}\Big |_{\hat{\xi }=\hat{\xi }_s}\Big )\delta +O(\delta ^2)\nonumber \\ \end{aligned}$$
(63)
$$\begin{aligned}&F(\hat{\xi _s},\hat{\xi })=\frac{1}{r^2(\hat{\xi }_s,\hat{\xi })}h(\hat{\xi })N_a(\hat{\xi })J(\hat{\xi }) \nonumber \\&\quad =\Big (\frac{S_{-2}}{\delta ^2}+\frac{S_{-1}}{\delta }+O(1)\Big )\nonumber \\&\qquad \Big (h_0N_{a0}+\Big (h_1N_{a0}+h_0N_{a1}\frac{\text {d}\xi }{\text {d}\hat{\xi }}\Big |_{\hat{\xi }=\hat{\xi }_s}\Big )\delta +O(\delta ^2)\Big )\nonumber \\&\quad =\frac{S_{-2}h_0N_{a0}}{\delta ^2}\nonumber \\&\qquad +\frac{S_{-1}h_0N_{a0}+S_{-2}\left( h_1N_{a0}+h_0N_{a1}\frac{\text {d}\xi }{\text {d}\hat{\xi }}\Big |_{\hat{\xi }=\hat{\xi }_s}\right) }{\delta }+O(1) \nonumber \\&\quad :=\frac{F_{-2}}{\delta ^2}+\frac{F_{-1}}{\delta }+O(1) \end{aligned}$$
(64)

Appendix 2

Once the \(J_1\) and \(J_2\) are evaluated properly, \(K_I\) and \(K_{II}\) can be found easily. Since

$$\begin{aligned} J_1= & {} \frac{K_I^2+K_{II}^2}{E'} \end{aligned}$$
(65a)
$$\begin{aligned} J_2= & {} -\frac{2K_IK_{II}}{E'} \end{aligned}$$
(65b)

where \(E'=E/(1-\nu ^2)\) for plane strain condition. And \(K_I\) and \(K_{II}\) can be solved as Eischen (1987):

$$\begin{aligned} K_I= & {} \pm \Big \{\frac{E'J_1}{2}\Big [1\pm \Big (1-\Big (\frac{J_2}{J_1}\Big )^2\Big )^{1/2}\Big ]\Big \}^{1/2} \end{aligned}$$
(66a)
$$\begin{aligned} K_{II}= & {} \pm \Big \{\frac{E'J_1}{2}\Big [1\mp \Big (1-\Big (\frac{J_2}{J_1}\Big )^2\Big )^{1/2}\Big ]\Big \}^{1/2} \end{aligned}$$
(66b)

The signs of \(K_I\) and \(K_{II}\) correspond to the signs of crack opening displacement \(\llbracket u_1\rrbracket \) and \(\llbracket u_2 \rrbracket \), respectively. If \(\llbracket u_1 \rrbracket >0\), \(K_I>0\). The term in brace can be determined as :

$$\begin{aligned}&\text {if}|\llbracket u_1\rrbracket |\ge |\llbracket u_2\rrbracket |, \text {take} + \end{aligned}$$
(67a)
$$\begin{aligned}&\text {if}|\llbracket u_1\rrbracket |< |\llbracket u_2\rrbracket |, \text {take} - \end{aligned}$$
(67b)

Combined with Eq. 65a, the following relationship can be obtained for the M integral,

$$\begin{aligned} M^{(1,2)}=\frac{2}{E'}\left( K_I^{(1)}K_I^{(2)}+K_{II}^{(1)}K_{II}^{(2)}\right) \end{aligned}$$
(68)

Let state 2 be the pure mode I asymptotic fields with \(K_I^{(2)}=1\), \(K_{II}^{(2)}=0\) and \(K_I\) in real state 1 can be found as

$$\begin{aligned} K_{I}^{(1)}=\frac{2}{E'}M^{(1, \text { mode }I)} \end{aligned}$$
(69)

The \(K_{II}\) can be given in a similar fashion.

The auxiliary stress field \(\sigma ^{(2)}_{ij}\) and displacement field \(u^{(2)}_{j}\) are given as:

$$\begin{aligned} \sigma _{xx}(r,\theta )= & {} \frac{K_I^{(2)}}{\sqrt{2\pi r}}\text {cos}\frac{\theta }{2}\Big (1-\text {sin}\frac{\theta }{2}\text {sin}\frac{3\theta }{2}\Big )\nonumber \\&-\frac{K_{II}^{(2)}}{\sqrt{2\pi r}}\text {sin}\frac{\theta }{2}\Big (2+\text {cos}\frac{\theta }{2}\text {cos}\frac{3\theta }{2}\Big ) \nonumber \\ \sigma _{yy}(r,\theta )= & {} \, \frac{K_{I}^{(2)}}{\sqrt{2\pi r}}\text {cos}\frac{\theta }{2}\Big (1+\text {sin}\frac{\theta }{2}\text {sin}\frac{3\theta }{2}\Big )\nonumber \\&+\frac{K_{II}^{(2)}}{\sqrt{2\pi r}}\text {sin}\frac{\theta }{2}\text {cos}\frac{\theta }{2}\text {cos}\frac{3\theta }{2}\nonumber \\ \tau _{xy}(r,\theta )= & {} \, \frac{K_{I}^{(2)}}{\sqrt{2\pi r}}\text {sin}\frac{\theta }{2}\text {cos}\frac{\theta }{2}\text {cos}\frac{3\theta }{2}\nonumber \\&+\frac{K_{II}^{(2)}}{\sqrt{2\pi r}}\text {cos}\frac{\theta }{2}\Big (1-\text {sin}\frac{\theta }{2}\text {sin}\frac{3\theta }{2}\Big )\nonumber \\ u_x(r,\theta )= & {} \,\frac{K_I}{2\mu }\sqrt{\frac{r}{2\pi }} \text {cos}\frac{\theta }{2}\left( \kappa -1+2\text {sin}^2\frac{\theta }{2}\right) \nonumber \\&+\frac{(1+\nu )K_{II}}{E}\sqrt{\frac{r}{2\pi }}\text {sin}\frac{\theta }{2}\left( \kappa +1 +2\text {cos}^2\frac{\theta }{2}\right) \nonumber \\ u_y(r,\theta )= & {} \,\frac{K_I}{2\mu }\sqrt{\frac{r}{2\pi }} \text {sin}\frac{\theta }{2}\left( \kappa +1-2\text {cos}^2\frac{\theta }{2}\right) \nonumber \\&+\frac{(1+\nu )K_{II}}{E}\sqrt{\frac{r}{2\pi }}\text {cos}\frac{\theta }{2}\left( 1-\kappa +2\text {sin}^2\frac{\theta }{2}\right) \nonumber \\ \end{aligned}$$
(70)

where \((r,\theta )\) are the crack tip polar coordinates and

$$\begin{aligned} \mu= & {} \frac{E}{2(1+\nu )} \end{aligned}$$
(71)
$$\begin{aligned} \kappa= & {} \left\{ \begin{array}{ll} 3-4\nu , &{}\quad \text {Plane strain} \\ (1-\nu )/(3+\nu ), &{}\quad \text {Plane stress} \\ \end{array}\right. \end{aligned}$$
(72)

The auxiliary strain field can be obtained by differentiating \(u_j\) with respect to the physical coordinate.

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Peng, X., Atroshchenko, E., Kerfriden, P. et al. Linear elastic fracture simulation directly from CAD: 2D NURBS-based implementation and role of tip enrichment. Int J Fract 204, 55–78 (2017). https://doi.org/10.1007/s10704-016-0153-3

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