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

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.

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.