A nine nodes solid-shell finite element with enhanced pinching stress

Abstract

In this paper we present a low-order solid-shell element formulation—having only displacement degrees of freedom (DOFs), i.e., without rotational DOFs. The element has an additional middle node, that allows efficient and accurate analyses of shell structures using elements at extremely high aspect ratio. The formulation is based on the Hu–Washizu variational principle leading to a novel enhancing strain and stress tensor that renders the computation particularly efficient, with improved in-plane and out-of-plane bending behavior (Poisson thickness locking). The middle-node is endowed with only one degree of freedom, in the thickness direction, allowing the assumption of a quadratic interpolation of the transverse displacement. Unlike solid-shell finite elements reported previously in the literature and formulated under the hypothesis of plane stress or with enhanced assumed strain parameter, the new solid-shell element here mentioned uses a complete three-dimensional constitutive law and gives an enhanced pinching stress, thanks to the middle-node. Moreover, to handle the various locking problems that usually arise on solid-shell formulation, the reduced integration technique is used as well as the assumed shear strain method. Finally to assess the effectiveness and performance of this new formulation, a set of popular benchmark problems, involving geometric non-linear analysis as well as elastic-plastic behavior has been investigated.

Appendix

$$\begin{aligned} \mathbf {J}^0= & {} \frac{1}{8}\begin{bmatrix} a_1^I x_1^I &{} a_1^I x_2^I &{} a_1^I x_3^I\\ a_2^I x_1^I &{} a_2^I x_2^I &{} a_2^I x_3^I\\ a_3^I x_1^I &{} a_3^I x_2^I &{} a_3^I x_3^I \end{bmatrix} \end{aligned}$$

(57)

$$\begin{aligned} \mathbf {J}^{\xi }= & {} \frac{1}{8}\begin{bmatrix} 0 &{} 0 &{} 0\\ h_1^I x_1^I &{} h_1^I x_2^I &{} h_1^I x_3^I\\ h_3^I x_1^I &{} h_3^I x_2^I &{} h_3^I x_3^I \end{bmatrix} \end{aligned}$$

(58)

$$\begin{aligned} \mathbf {J}^{\eta }= & {} \frac{1}{8}\begin{bmatrix} h_1^I x_1^I &{} h_1^I x_2^I &{} h_1^I x_3^I\\ 0 &{} 0 &{} 0\\ h_2^I x_1^I &{} h_2^I x_2^I &{} h_2^I x_3^I \end{bmatrix} \end{aligned}$$

(59)

$$\begin{aligned} \mathbf {J}^{\zeta }= & {} \frac{1}{8}\begin{bmatrix} h_3^I x_1^I &{} h_3^I x_2^I &{} h_3^I x_3^I\\ h_2^I x_1^I &{} h_2^I x_2^I &{} h_2^I x_3^I\\ 0 &{} 0 &{} 0 \end{bmatrix} \end{aligned}$$

(60)

$$\begin{aligned} \mathbf {J}^{\xi \eta }= & {} \frac{1}{8}\begin{bmatrix} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0\\ h_4^I x_1^I &{} h_4^I x_2^I &{} h_4^I x_3^I \end{bmatrix} \end{aligned}$$

(61)

$$\begin{aligned} \mathbf {J}^{\xi \zeta }= & {} \frac{1}{8}\begin{bmatrix} 0 &{} 0 &{} 0\\ h_4^I x_1^I &{} h_4^I x_2^I &{} h_4^I x_3^I\\ 0 &{} 0 &{} 0 \end{bmatrix} \end{aligned}$$

(62)

$$\begin{aligned} \mathbf {J}^{\eta \zeta }= & {} \frac{1}{8}\begin{bmatrix} h_4^I x_1^I &{} h_4^I x_2^I &{} h_4^I x_3^I\\ 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 \end{bmatrix} \end{aligned}$$

(63)

Here are given the in-plane part of the parameter of equations (22) and (23). The normal and shear part being sorted from Sect. 5.

$$\begin{aligned} \hat{\mathbf {B}}_{mI}^0= & {} \begin{bmatrix} a_{1I}\mathbf {J}_1^{0}\\ a_{2I}\mathbf {J}_2^{0}\\ a_{1I}\mathbf {J}_2^{0}+g_{2I}\mathbf {J}_1^{0}\\ \end{bmatrix} \end{aligned}$$

(64)

$$\begin{aligned} \hat{\mathbf {B}}_{mI}^\zeta= & {} \begin{bmatrix} h_{3I}\mathbf {J}_1^{0}+a_{1I}\mathbf {J}_1^{\zeta }\\ h_{2I}\mathbf {J}_2^{0}+a_{2I}\mathbf {J}_2^{\zeta }\\ h_{2I}\mathbf {J}_1^{0}+h_{3I}\mathbf {J}_2^{0} +a_{2I} \mathbf {J}_1^{\zeta }+a_{1I}\mathbf {J}_2^{\zeta }\\ \end{bmatrix} \end{aligned}$$

(65)

$$\begin{aligned} \hat{\mathbf {B}}_{mI}^{\zeta \zeta }= & {} \begin{bmatrix} h_{3I}\mathbf {J}_1^{\zeta }\\ h_{2I}\mathbf {J}_2^{\zeta }\\ h_{2I}\mathbf {J}_1^{\zeta }+h_{3I}\mathbf {J}_2^{\zeta } \\ \end{bmatrix} \end{aligned}$$

(66)

$$\begin{aligned} \hat{\mathbf {B}}_{mI}^\xi= & {} \begin{bmatrix} 0\\ h_{1I}\mathbf {J}_2^{0}+a_{2I}\mathbf {J}_2^{\zeta }\\ h_{1I}\mathbf {J}_2^{0}+a_{1I}\mathbf {J}_2^{\zeta }\\ \end{bmatrix} \end{aligned}$$

(67)

$$\begin{aligned} \hat{\mathbf {B}}_{mI}^\eta= & {} \begin{bmatrix} h_{1I}\mathbf {J}_1^{0}+a_{1I}\mathbf {J}_1^{\eta }\\ 0\\ h_{1I}\mathbf {J}_2^{0}+a_{2I}\mathbf {J}_1^{\eta }\\ \end{bmatrix} \end{aligned}$$

(68)

$$\begin{aligned} \hat{\mathbf {B}}_{mI}^{\eta \zeta }= & {} \begin{bmatrix} h_{4I}\mathbf {J}_1^{0}+h_{3I}\mathbf {J}_1^{\eta }+h_{1I}\mathbf {J}_1^{\zeta } +a_{1I}\mathbf {J}_1^{\eta \zeta }\\ 0\\ h_{4I}\mathbf {J}_2^{0}+h_{2I}\mathbf {J}_1^{\eta }+h_{1I}\mathbf {J}_2^{\zeta } +g_{2I}\mathbf {J}_1^{\eta \zeta }\\ \end{bmatrix} \end{aligned}$$

(69)

$$\begin{aligned} \hat{\mathbf {B}}_{mI}^{\xi \zeta }= & {} \begin{bmatrix} 0\\ h_{4I}\mathbf {J}_2^{0}+h_{2I}\mathbf {J}_2^{\xi } +h_{1I}\mathbf {J}_2^{\zeta }+a_{2I}\mathbf {J}_2^{\xi \zeta }\\ h_{4I}\mathbf {J}_1^{0}+h_{3I}\mathbf {J}_2^{\xi }+h_{1I}\mathbf {J}_1^{\zeta }+a_{1I}\mathbf {J}_2^{\xi \zeta }\\ \end{bmatrix} \end{aligned}$$

(70)

$\mathbf {J}_1,\mathbf {J}_2,\mathbf {J}_3$ being the rows of the Jacobian matrix in the current configuration. Note that the rows of the jacobian matrix contrain the coordinates of the covariant $\mathbf {g}_i$ vectors.

1.1 Decomposition of T

The Inverse of the Jacobian matrix is decomposed keeping only the constant and linear terms as follows

$$\begin{aligned} \mathbf {J}^{-1} \approx \mathbf {J}^{-1}|_{\xi =0}+\sum _{i=1}^3\mathbf {J}^{-1}_{,\xi _i}|_{\xi =0}\xi _i \end{aligned}$$

(71)

The constant term being easily determined, the work will be to determine the linear terms with very limited resources. To do so, the Eq. (72) is simply derived with respect to the corresponding convective parameter and gives the Eq. (73).

$$\begin{aligned} \mathbf {J}\mathbf {J}^{-1}\approx & {} (\mathbf {J}\mathbf {J}^{-1})|_{\xi =0}+\sum _{i=1}^3(\mathbf {J}\mathbf {J}^{-1})_{,\xi _i}|_{\xi =0}\xi _i \end{aligned}$$

(72)

$$\begin{aligned} 0= & {} (\mathbf {J}\mathbf {J}^{-1})_{,\xi _i}|_{\xi = \mathbf {0}} = \mathbf {J}_{,\xi _i}|_{\xi =\mathbf {0}}\mathbf {J}^{-1}|_{\xi =\mathbf {0}} +\mathbf {J}|_{\xi =\mathbf {0}} \mathbf {J}^{-1}_{,\xi _i}|_{\xi =\mathbf {0}}\nonumber \\ \end{aligned}$$

(73)

From Eq. (73) one can easily determine the linear Jacobian terms, as follows, all terms being known or determined easily.

$$\begin{aligned} \mathbf {J}_{,\xi _i}^{-1}|_{\xi =\mathbf {0}} = -(\mathbf {J}^0)^{-1}\mathbf {J}^{\xi _i}(\mathbf {J}^0)^{-1} \end{aligned}$$

(74)

Hence, a good representation of the inverse Jacobian matrix is known and one can simply insert these terms into Eq. (19) to sort the taylor decomposition of the matrix $\mathbf {T}$

$$\begin{aligned} \mathbf {J}^{-1}\approx \mathbf {J}^{-1}|_{\xi = \mathbf {0} } - \sum _{i=1}^3(\mathbf {J}^0)^{-1}\mathbf {J}^{\xi _i}(\mathbf {J}^0)^{-1}\xi _i \end{aligned}$$

(75)