Numerical analysis of a main crack interactions with micro-defects/inhomogeneities using two-scale generalized/extended finite element method

Abstract

Generalized or extended finite element method (G/XFEM) models the crack by enriching functions of partition of unity type with discontinuous functions that represent well the physical behavior of the problem. However, this enrichment functions are not available for all problem types. Thus, one can use numerically-built (global-local) enrichment functions to have a better approximate procedure. This paper investigates the effects of micro-defects/inhomogeneities on a main crack behavior by modeling the micro-defects/inhomogeneities in the local problem using a two-scale G/XFEM. The global-local enrichment functions are influenced by the micro-defects/inhomogeneities from the local problem and thus change the approximate solution of the global problem with the main crack. This approach is presented in detail by solving three different linear elastic fracture mechanics problems for different cases: two plane stress and a Reissner–Mindlin plate problems. The numerical results obtained with the two-scale G/XFEM are compared with the reference solutions from the analytical, numerical solution using standard G/XFEM method and ABAQUS as well, and from the literature.

Appendix A: Intensity factors for Reissner–Mindlin plate problem

The J-integral contour for a Reissner–Mindlin plate problem is defined as [16]:

$$\begin{aligned} J = \oint _{\Gamma }^{} \left\{ W \delta _{1\beta } - \left[ M_{\alpha \beta } \psi _{\alpha ,1} + Q_{\beta } w_{,1} \right] \right\} n_{\beta } d\Gamma \end{aligned}$$

(A.1)

Table 6 Auxiliary states for the displacement fields of Reissner–Mindlin plate

where $M_{\alpha \beta }$ is the bending moment, $Q_{\beta }$ is the shear, w is the transverse displacement and $\psi _{\alpha }$ is section rotation about the $x_{\alpha }$ axes that define the middle plane of the plate, in which $\alpha \text { and } \beta$ ranging over the values 1,2.

Referring to Fig. 4a and following similar mathematical to the Sect. 4, one can reach following interaction energy integral equation for the Reissner–Mindlin plate, and for a crack with traction free faces as:

$$\begin{aligned} I= & {} \oint _{C}^{} \left\{ -W \delta _{1\beta } + \left[ M_{\alpha \beta }^{(1)} \psi _{\alpha ,1}^{(2)} + M_{\alpha \beta }^{(2)} \psi _{\alpha ,1}^{(1)} \right. \right. \nonumber \\&\quad \left. \left. +\,Q_{\beta }^{(1)} w_{,1}^{(2)} + Q_{\beta }^{(2)} w_{,1}^{(1)} \right] \right\} m_{\beta } \, q \, dC \end{aligned}$$

(A.2)

in which, similar to the Sect. 4, State (1) represents the current state and State (2) is an auxiliary state. The interaction strain energy, W, is defined by:

$$\begin{aligned} W^{(1, 2)}= & {} M^{(1)} : \epsilon _b^{(2)} + Q^{(1)} \cdot \epsilon _s^{(2)}\nonumber \\= & {} M^{(2)} : \epsilon _b^{(1)} + Q^{(2)} \cdot \epsilon _s^{(1)} \end{aligned}$$

(A.3)

where $\epsilon _b$ and $\epsilon _s$ are bending and shear strains, respectively. Applying the divergence theorem to the integral over A, we obtain:

$$\begin{aligned} I= & {} \int _{A}^{} \left\{ \left[ M_{\alpha \beta }^{(1)} \psi _{\alpha ,1}^{(2)} + M_{\alpha \beta }^{(2)} \psi _{\alpha ,1}^{(1)} + Q_{\beta }^{(1)} w_{,1}^{(2)} + Q_{\beta }^{(2)} w_{,1}^{(1)} \right] \right. \nonumber \\&\quad \left. - W \delta _{1\beta } \right\} q_{\beta } \, dA \end{aligned}$$

(A.4)

The above integral can be reduced depending on whether the quantity of interest is $K_I$, $K_{II}$, or $K_{III}$, as certain terms in the auxiliary fields vanish for each case. For example, for $K_I$ and $K_{II}$ the integral takes the form:

$$\begin{aligned} I = \int _{A}^{} \left\{ \left[ Q_{\beta }^{(1)} w_{,1}^{(2)} + Q_{\beta }^{(2)} w_{,1}^{(1)} \right] - W \delta _{1\beta } \right\} q_{\beta } \, dA \end{aligned}$$

(A.5)

whereas for $K_{III}$ the integral is:

$$\begin{aligned} I = \int _{A}^{} \left\{ \left[ M_{\alpha \beta }^{(1)} \psi _{\alpha ,1}^{(2)} + M_{\alpha \beta }^{(2)} \psi _{\alpha ,1}^{(1)} \right] - W \delta _{1\beta } \right\} q_{\beta } \, dA \end{aligned}$$

(A.6)

The auxiliary state for the displacement fields in Reissner–Mindlin plate theory can be found in Sosa [46] as a power series in $\sqrt{r}$ and are shown in the following Table 6, in which $C_{\theta }$ and $S_{\theta }$ represent $\cos \theta$ and $\sin \theta$ functions, respectively. Moreover, the auxiliary bending moments and shear are as follows:

$$\begin{aligned} M_{11}= & {} \frac{K_1}{\sqrt{2r}} \cos \frac{\theta }{2} \left( 1 - \sin \frac{\theta }{2} \sin \frac{3\theta }{2} \right) \nonumber \\&-\,\frac{K_2}{\sqrt{2r}} \sin \frac{\theta }{2} \left( 2 + \cos \frac{\theta }{2} \cos \frac{3\theta }{2} \right) \end{aligned}$$

(A.7)

$$\begin{aligned} M_{22}= & {} \frac{K_1}{\sqrt{2r}} \cos \frac{\theta }{2} \left( 1 - \sin \frac{\theta }{2} \sin \frac{3\theta }{2} \right) \nonumber \\&-\,\frac{K_2}{\sqrt{2r}} \sin \frac{\theta }{2} \cos \frac{\theta }{2} \cos \frac{3\theta }{2} \end{aligned}$$

(A.8)

$$\begin{aligned} M_{12}= & {} \frac{K_1}{\sqrt{2r}} \sin \frac{\theta }{2} \cos \frac{\theta }{2} \cos \frac{3\theta }{2}\nonumber \\&+\,\frac{K_2}{\sqrt{2r}} \cos \frac{\theta }{2} \left( 1 - \sin \frac{\theta }{2} \sin \frac{3\theta }{2} \right) \end{aligned}$$

(A.9)

$$\begin{aligned} Q_1= & {} -\frac{K_3}{\sqrt{2r}} \sin \frac{\theta }{2} \end{aligned}$$

(A.10)

$$\begin{aligned} Q_2= & {} \frac{K_3}{\sqrt{2r}} \cos \frac{\theta }{2} \end{aligned}$$

(A.11)

As an example, $K_2$ and $K_3$ must be set equal to zero in all equation in order to calculate auxiliary moment intensity factor of mode-I. The process of evaluating the mixed-mode intensity factors must be carried out with a judicious choice of the auxiliary moment and shear force intensity factors to evaluating the interaction energy integral. From the Eq. (A.1) and the energy release rate formulation, one can obtain the following expression:

$$\begin{aligned} I = \frac{24 \pi }{E t^3} \left[ K_I K_I^{(2)} + K_{II} K_{II}^{(2)} \right] + \frac{12 \pi }{10 \mu t} K_{III} K_{III}^{(2)} \end{aligned}$$

(A.12)

where, to extract $K_I$, the following values is chosen $K_i^{(2)} = 1 , K_{II}^{(2)} = 0$, and $K_{III}^{(2)} = 0$. Then, the moment intensity factor $K_I$ can be calculated as:

$$\begin{aligned} K_I = \frac{E t^3}{24 \pi } I \end{aligned}$$

(A.13)