Viscoelastic crack analysis using symplectic analytical singular element combining with precise time-domain algorithm

Abstract

In this study, the crack problem in linear viscoelastic material is investigated numerically. The time dependent two-dimensional (2D) viscoelastic crack problem is treated by the precise time-domain expanding algorithm (PTDEA), such that the original problem is transformed into a series of quasi-elastic crack problems. The relationships among these quasi-elastic problems are expressed in terms of the time-domain expanding coefficients of displacement and stress in an improved recursive manner. Then a symplectic analytical singular element (SASE) which has been demonstrated to be effective and efficient for 2D elastic fracture problem is applied to solve the quasi-elastic crack problems obtained above. The SASE is constructed by using the symplectic eigen solutions with higher order expanding terms. An improved convergence criterion employing both displacement and stress for PTDEA is proposed. Taking advantage of the SASE, the stress intensity factors, crack opening and sliding displacements (COD and CSD) and strain energy release rate of the studied problem can be solved directly without any post-processing. Numerical examples show that the results of the present method can be solved accurately and effectively.

Appendix A: Calculation process of SERRs

While the export nodal quasi-elastic displacements vector $\varvec{{\tilde{d}}}_m^{\mathrm{N}}$ is solved out, the coefficients ${\tilde{a}}_{n,\left( {m,k} \right) } \left( {n\ge 1;m\ge K;k\ge 0} \right)$ can be obtained by Eq. (55) or (45). And then according to the relationship of $\varvec{u}_{m,k}$ and $\varvec{{\tilde{u}}}_{m,k}$ in the third section, $a_{n,\left( {m,k} \right) } \left( {n\ge 1;m\ge 0;k\ge 0} \right)$ in Eq. (52) can be acquired by the following recursive formulations:

• (1) When $k=0$, for $0\le m<K\,\left( {\hbox {if }K>0} \right)$ via Eq. (16) gives

  $$\begin{aligned} a_{n,\left( {m,0} \right) } =0, \end{aligned}$$

  (A.1)

  and for $K\le m<I$ via Eqs. (22) and (19) gives

  $$\begin{aligned} a_{n,\left( {m,0} \right) } =\left[ {{\tilde{a}}_{n,\left( {m,0} \right) } -e_{n,\left( {m,0} \right) } } \right] /b_{m,0} , \end{aligned}$$

  (A.2)

  where

  $$\begin{aligned} \left\{ {{\begin{array}{ll} {e_{n,\left( {K,0} \right) } =0,} &{} \\ {\begin{array}{l} e_{n,\left( {m,0} \right) } =\sum \limits _{i=I-m}^{I-1} {\left[ {T_0^{J-i} \frac{q_i }{q_I }\frac{(m-I+i)!}{(m-K)!}a_{n,\left( {m-I+i,0} \right) } } \right] } \\ -\sum \limits _{j=I-m}^{J-1} {\left[ {T_0^{J-j} \frac{p_j }{p_J }\frac{(m-I+j)!}{(m-K)!}\left( {b_{m-J+j,0} a_{n,\left( {m-J+j,0} \right) } +e_{n,\left( {m-J+j,0} \right) } } \right) } \right] ,} \\ \end{array}} &{} {\left( {K<m<I} \right) .} \\ \end{array} }} \right. \end{aligned}$$

  (A.3)

• (2) When $k>0$, for $0\le m<I$ via Eq. (24) gives

  $$\begin{aligned}&a_{n,\left( {m,k} \right) } =\left( {T_k /T_{k-1} } \right) ^{m}\nonumber \\&\quad \times \sum _{j=m}^{M_{k-1} } {\left[ {j!/m!/(j-m)!} \right] a_{n,\left( {j,k-1} \right) }}, \end{aligned}$$

  (A.4)

  and for $K\le m<I$ via Eq. (28) gives

  $$\begin{aligned}&e_{n,\left( {m,k} \right) } =\left( {T_k /T_{k-1} } \right) ^{m-K}\sum _{j=m}^{M_{k-1} }\nonumber \\&\quad {\left[ {\left( {j-K} \right) !/\left( {m-K} \right) !/(j-m)!} \right] e_{n,\left( {j,k-1} \right) }}.\nonumber \\ \end{aligned}$$

  (A.5)

• (3) When $k\ge 0$, for $m\ge I$ via Eqs. (31) and (30) gives

  $$\begin{aligned} a_{n,\left( {m,k} \right) } =\left[ {{\tilde{a}}_{n,\left( {m,k} \right) } -e_{n,\left( {m,k} \right) } } \right] /b_{m,k}, \end{aligned}$$

  (A.6)

  where

  $$\begin{aligned}&e_{n,\left( {m,k} \right) } =\sum _{i=0}^{I-1} {\left[ {T^{J-i}\frac{q_i }{q_I }\frac{(m-I+i)!}{(m-K)!}a_{n,\left( {m-I+i,k} \right) } } \right] } \nonumber \\&\quad -\sum _{j=0}^{J-1} \left[ T^{J-j}\frac{p_j }{p_J }\frac{(m-I+j)!}{(m-K)!}\right. \nonumber \\&\qquad \left. \left( b_{m-J+j,k}a_{n,\left( {m-J+j,k} \right) } +e_{n,\left( {m-J+j,k} \right) } \right) \right] .\nonumber \\ \end{aligned}$$

  (A.7)