# 3-Node Triangle without Rotational DOF

The need of simple and efficient element in nonlinear analysis of shells undergoing large rotations is evident in crash and sheet metal forming simulations. The constant-moment plate elements fit this need. One of the famous concepts in this field is that of Batoz et al. 1 known under DKT elements where DKT stands for Discrete Kirchhoff Triangle. The DKT12 element 1, 2 has a total of 12 DOFs. The discrete Kirchhoff plate conditions are imposed at three mid-point of each side. The element makes use of rotational DOF. at each edge to take into account the bending effects. A simplified three-node element without rotational DOF is presented in 3. The rotational DOF is computed with the help of out-of-plane translational DOF in the neighbor elements. This attractive approach is used in Radioss in the development of element SH3N6 which based on DKT12.

## Strain Computation

Consider two adjacent coplanar elements with a common edge i-j as shown in 図 2. Due to out-of-plane displacements of nodes $m$ and $k$ , the elements rotate around the side i-j. The angles between final and initial positions of the elements are respectively ${\alpha }_{m}$ and ${\alpha }_{k}$ for corresponding opposite nodes $m$ and $k$ . Assuming, a constant curvature for both of elements, the rotation angles ${\theta }_{m}$ and ${\theta }_{k}$ related to the bending of each element around the common side are obtained by:(1)
${\theta }_{k}=\frac{{h}_{k}}{2R}\text{ }\text{ }\text{and}\text{ }\text{ }{\theta }_{m}=\frac{{h}_{m}}{2R}$
However, for total rotation you have:(2)
${\theta }_{k}+{\theta }_{m}={\alpha }_{k}+{\alpha }_{m}$
${\theta }_{k}=\frac{\left({\alpha }_{k}+{\alpha }_{m}\right){h}_{k}}{\left({h}_{k}+{h}_{m}\right)}\text{ }\text{ }\text{and}\text{ }\text{ }{\theta }_{m}=\frac{\left({\alpha }_{k}+{\alpha }_{m}\right){h}_{m}}{\left({h}_{k}+{h}_{m}\right)}$
Consider the triangle element in 図 2. The outward normal vectors at the three sides are defined and denoted $n1$ , $n2$ and $n3$ . The normal component strain due to the bending around the element side is obtained using plate assumption:(4)
$\left\{\begin{array}{c}{\epsilon }_{n1}\\ {\epsilon }_{n2}\\ {\gamma }_{n3}\end{array}\right\}=\left[\begin{array}{cccccc}\frac{2}{\left({h}_{1}+{h}_{5}\right)}& 0& 0& 0& \frac{2}{\left({h}_{1}+{h}_{5}\right)}& 0\\ 0& \frac{2}{\left({h}_{2}+{h}_{6}\right)}& 0& 0& 0& \frac{2}{\left({h}_{2}+{h}_{6}\right)}\\ 0& 0& \frac{2}{\left({h}_{3}+{h}_{4}\right)}& \frac{2}{\left({h}_{3}+{h}_{4}\right)}& 0& 0\end{array}\right]\left\{\begin{array}{c}{\text{α}}_{\text{1}}\\ {\text{α}}_{\text{2}}\\ {\text{α}}_{\text{3}}\\ {\text{α}}_{\text{4}}\\ {\text{α}}_{\text{5}}\\ {\text{α}}_{\text{6}}\end{array}\right\}$
The six mid-side rotations ${\alpha }_{i}$ are related to the out-of-plane displacements of the six apex nodes as shown in 図 3 by the following relation:(5)
$\left\{\begin{array}{c}{\text{α}}_{\text{1}}\\ {\text{α}}_{\text{2}}\\ {\text{α}}_{\text{3}}\\ {\text{α}}_{\text{4}}\\ {\text{α}}_{\text{5}}\\ {\text{α}}_{\text{6}}\end{array}\right\}=\left[\begin{array}{cccccc}\frac{1}{{h}_{1}}& -\frac{\mathrm{cos}{\beta }_{3}}{{h}_{2}}& -\frac{\mathrm{cos}{\beta }_{2}}{{h}_{3}}& 0& 0& 0\\ -\frac{\mathrm{cos}{\beta }_{3}}{{h}_{1}}& \frac{1}{{h}_{2}}& -\frac{\mathrm{cos}{\beta }_{1}}{{h}_{3}}& 0& 0& 0\\ -\frac{\mathrm{cos}{\beta }_{2}}{{h}_{1}}& -\frac{\mathrm{cos}{\beta }_{1}}{{h}_{2}}& \frac{1}{{h}_{3}}& 0& 0& 0\\ -\frac{\mathrm{cos}{\gamma }_{2}}{{q}_{1}}& -\frac{\mathrm{cos}{\gamma }_{1}}{{q}_{2}}& 0& \frac{1}{{h}_{4}}& 0& 0\\ 0& -\frac{\mathrm{cos}{\psi }_{3}}{{r}_{2}}& -\frac{\mathrm{cos}{\psi }_{2}}{{r}_{3}}& 0& \frac{1}{{h}_{5}}& 0\\ -\frac{\mathrm{cos}{\phi }_{2}}{{s}_{1}}& 0& -\frac{\mathrm{cos}{\phi }_{1}}{{s}_{3}}& 0& 0& \frac{1}{{h}_{6}}\end{array}\right]\left\{\begin{array}{c}{\text{w}}_{\text{1}}\\ {\text{w}}_{\text{2}}\\ {\text{w}}_{\text{3}}\\ {\text{w}}_{\text{4}}\\ {\text{w}}_{\text{5}}\\ {\text{w}}_{\text{6}}\end{array}\right\}$
Where, $\left(\begin{array}{ccc}{h}_{1}& {h}_{2}& {h}_{3}\end{array}\right)$ , $\left(\begin{array}{ccc}{q}_{1}& {h}_{4}& {q}_{2}\end{array}\right)$ , $\left(\begin{array}{ccc}{r}_{2}& {h}_{5}& {r}_{3}\end{array}\right)$ and $\left(\begin{array}{ccc}{s}_{3}& {h}_{6}& {s}_{1}\end{array}\right)$ are respectively the heights of the triangles (1,2,3), (1,4,2), (2,5,3) and (3,6,1).
The non-null components of strain tensor in the local element reference are related to the normal components of strain by the following relation: 1 3(6)
$\left\{\begin{array}{c}{\epsilon }_{xx}\\ {\epsilon }_{yy}\\ {\gamma }_{xy}\end{array}\right\}=\left[\begin{array}{ccc}{\left(\frac{{y}_{3}-{y}_{2}}{{l}_{1}}\right)}^{2}& {\left(\frac{{y}_{1}-{y}_{3}}{{l}_{2}}\right)}^{2}& {\left(\frac{{y}_{2}-{y}_{1}}{{l}_{3}}\right)}^{2}\\ {\left(\frac{{x}_{2}-{x}_{3}}{{l}_{1}}\right)}^{2}& {\left(\frac{{x}_{3}-{x}_{1}}{{l}_{2}}\right)}^{2}& {\left(\frac{{x}_{1}-{x}_{2}}{{l}_{3}}\right)}^{2}\\ 2\left(\frac{{y}_{3}-{y}_{2}}{{l}_{1}}\right)\left(\frac{{x}_{2}-{x}_{3}}{{l}_{1}}\right)& 2\left(\frac{{y}_{1}-{y}_{3}}{{l}_{2}}\right)\left(\frac{{x}_{3}-{x}_{1}}{{l}_{2}}\right)& 2\left(\frac{{y}_{2}-{y}_{1}}{{l}_{3}}\right)\left(\frac{{x}_{1}-{x}_{2}}{{l}_{3}}\right)\end{array}\right]\left\{\begin{array}{c}{\epsilon }_{n1}\\ {\epsilon }_{n2}\\ {\gamma }_{n3}\end{array}\right\}$

## Boundary Conditions Application

As the side rotation of the element is computed using the out-of-plane displacement of the neighbor elements, the application of clamped or free boundary conditions needs a particular attention. It is natural to consider the boundary conditions on the edges by introducing a virtual and symmetric element outside of the edge as described in 図 4. In the case of free rotation at the edge, the normal strain ${\epsilon }_{nk}$ is vanished. From 式 4, this leads to:(7)
${\alpha }_{k}=-{\alpha }_{m}$
In 式 5 the fourth row of the matrix is then changed to:(8)
$\left[\begin{array}{cccccc}\frac{\mathrm{cos}{\beta }_{2}}{{h}_{1}}& \frac{\mathrm{cos}{\beta }_{1}}{{h}_{2}}& -\frac{1}{{h}_{3}}& 0& 0& 0\end{array}\right]$
The clamped condition is introduced by the symmetry in out-of-plane displacement, that is, ${w}_{m}={w}_{k}$ . This implies ${\alpha }_{k}={\alpha }_{m}$ . The fourth row of the matrix in 式 5 is then changed to:(9)
$\left[\begin{array}{cccccc}-\frac{\mathrm{cos}{\beta }_{2}}{{h}_{1}}& -\frac{\mathrm{cos}{\beta }_{1}}{{h}_{2}}& -\frac{1}{{h}_{3}}& 0& 0& 0\end{array}\right]$
1 Batoz J.L. and Dhatt G., 「Modeling of Structures by finite element」, volume 3, Hermes, 1992.
2

Batoz J.L., Guo Y.Q., Shakourzadeh H., 「Nonlinear Analysis of thin shells with elasto-plastic element DKT12」, Revue Europénne des Eléments Finis, Vol. 7, N° 1-2-3, pp. 223-239.1998.

3

Sabourin F. and Brunet M., 「Analysis of plates and shells with a simplified three-node triangle element」, Thin-walled Structures, Vol. 21, pp. 209-223, Elsevier, 1995.