# High-Precision Elastoplastic Four-Node Quadrilateral Shell Element
## 2 Introduction to DOF of Rotation around Element Normal  
The method proposed by [PERSON] [17] is employed to interpolate the secondary displacement of the nodes located at the edges of the quadrilateral shell element, incorporating a degree of freedom for rotation about the element's normal. As illustrated in Figure 1, the displacement of node \(K\) on the \(J\) side can be expressed as follows:  
\[\begin{array}{l}u_{K}=\frac{u_{I}+u_{I}}{2}+\frac{v_{I}-y_{I}}{8}(\omega_{Z }-\omega_{ZI})\\ v_{K}=\frac{v_{I}+v_{I}}{2}-\frac{x_{I}-x_{I}}{8}(\omega_{Z}-\omega_{ZI})\end{array} \tag{1}\]  
In a quadrilateral element with middle nodes, the interpolation function of displacement \(u\) and \(v\) is  
\[u=\widetilde{N}_{I}u_{I}v=\widetilde{N}_{I}v_{I} \tag{2}\]  
where the uppercase subscripts \(I\), \(J\), and \(K\) represent the node number, the repeated uppercase subscripts signify the conventional summation, and \(\widetilde{N}_{I}\) is the standard eight-node serendipity interpolation basis function.  
\[\begin{array}{l}\widetilde{N}_{I}=0.25(1+\xi\xi_{I})(1+\eta\eta_{I})(\xi\xi_{I}+ \eta\eta_{I}-1)(I=1,2,3,4)\\ \widetilde{N}_{I}=0.5(1-\xi^{2})(1+\eta\eta_{I})(I=5,7)\\ \widetilde{N}_{I}=0.5(1+\xi\xi_{I})\big{(}1-\eta^{2}\big{)}(I=6,8)\end{array} \tag{3}\]  
where \(\xi_{I}=\big{[}-1\quad 0\quad 1\quad 1\quad 1\quad 0\quad-1\quad-1\big{]}\), \(\eta_{I}=\big{[}-1\quad-1\quad-1\quad 0\quad 1\quad 1\quad 1\quad 0\big{]}\).  
In quadrilateral elements with intermediate nodes, each edge of the element is consistent with a three-node straight line, and the intermediate nodes on each edge of the element can be expressed according to Equation (1). All middle nodes of a quadrilateral shell element are acquired by introducing them into Equation (2) as per Equation (1).  
\[u=N_{I}u_{I}+N_{I}^{u}\omega_{ZI}v=N_{I}v_{I}+N_{I}^{v}\omega_{ZI} \tag{4}\]  
where \(N_{I}\) is the interpolation basis function of four standard nodes,  
\[N_{I}=0.25(1+\xi\xi_{I})(1+\eta\eta_{I})(I=1,2,3,4) \tag{5}\]  
\(N_{I}^{u}\) and \(N_{I}^{v}\) are \(\omega_{I}\) corresponding to the interpolation basis function of \(u\) and \(v\).  
\[\begin{array}{l}N_{1}^{u}=0.125\Big{(}y_{14}\widetilde{N}_{8}+y_{12} \widetilde{N}_{5}\Big{)}N_{2}^{u}=0.125\Big{(}y_{21}\widetilde{N}_{5}+y_{23} \widetilde{N}_{6}\Big{)}\\ N_{3}^{u}=0.125\Big{(}y_{32}\widetilde{N}_{6}+y_{34}\widetilde{N}_{7}\Big{)}N_ {4}^{u}=0.125\Big{(}y_{43}\widetilde{N}_{7}+y_{41}\widetilde{N}_{8}\Big{)} \end{array} \tag{6}\]  
\[\begin{array}{l}N_{1}^{v}=-0.125\Big{(}x_{14}\widetilde{N}_{8}+x_{12} \widetilde{N}_{5}\Big{)}N_{2}^{v}=-0.125\Big{(}x_{21}\widetilde{N}_{8}+x_{23} \widetilde{N}_{5}\Big{)}\\ N_{3}^{v}=-0.125\Big{(}x_{32}\widetilde{N}_{8}+x_{34}\widetilde{N}_{5}\Big{)}N_ {4}^{v}=-0.125\Big{(}x_{43}\widetilde{N}_{8}+x_{41}\widetilde{N}_{5}\Big{)} \end{array} \tag{7}\]  
where \(y_{IJ}=y_{I}-y_{J}\), \(x_{IJ}=x_{I}-x_{J}\).  
According to the Mindlin-Reissner formula, the complete three-dimensional velocity field of the shell element is  
\[V_{i}(\xi,\eta,\zeta)=N_{I}^{3D}(\xi,\eta,\zeta)V_{il} \tag{8}\]  
where \(N_{I}^{3D}\) denotes the full three-dimensional isoparametric shape functions, \(V_{i}=\big{[}u\quad v\quad w\big{]}\). The symbols in Equation (8) and Equation (4) have different meanings. Equation (4) represents the two-dimensional velocity field, while Equation (8) represents the complete three-dimensional velocity field. Since the normals remain straight, we can write  
\[V_{i}(\xi,\eta,\zeta)=N_{I}\Big{[}V_{il}+\frac{a}{2}\zeta(-\delta_{i2}\omega_{ xI}+\delta_{i1}\omega_{yI})\Big{]} \tag{9}\]  
where \(a\) is the thickness, Equation (4) is brought into Equation (9), where \(u\) and \(v\) are expressed as  
\[\begin{array}{l}u=N_{I}u_{I}+N_{I}^{u}\omega_{zI}+N_{I}\frac{a}{2}\zeta \omega_{yI}\\ v=N_{I}v_{I}+N_{I}^{v}\omega_{zI}-N_{I}\frac{a}{2}\zeta\omega_{xI}\end{array} \tag{10}\]  
Figure 1: The DOF of a three-node straight line.