The simplest 3-, 6- and 8-noded fully-parameterized ANCF plate elements using only transverse slopes

Abstract

In this paper, the simplest kinematical models of triangular plate finite elements using the absolute nodal coordinate formulation (ANCF) are considered. The ANCF is the finite element approach for simulating large displacements and rotations, in which the nodal position vectors and their derivatives are described in the inertial frame only. This leads to linear kinematics of elements, constant mass matrix and simple expressions for inertia terms in the equations of motion. The elastic forces appear in the ANCF in highly nonlinear form due to using the Green–Lagrange strain tensor. This fact compels researchers to find possibilities of reducing the computational complexity in using the ANCF. One of the ways in this direction is to use simplest fully-parameterized plate elements employing transverse slopes only, without using longitudinal slopes.

Appendices

Appendix A

1.1 A.1 Elements of equations of motion for element 4323

The shape functions for the triangular element 4323 are defined by Eq. (7) for the number of nodes n=3. The shape functions’ matrix S ⁴³²³ from Eq. (1) can be formulated as follows:

where I is the 3×3 identity matrix, and the shape functions $s_{i,0}^{231}(x,y),\ i = 1, \ldots,3$ are described by Eq. (2).

The element mass matrix M ⁴³²³ and generalized vector of gravity forces $\mathbf{Q}_{4323}^{\mathrm{g}}$, which are constants in the ANCF, are calculated by using Eq. (11) and can be written explicitly as

$$\begin{aligned} &\mathbf{M}^{4323} = \frac{\rho\Delta h}{144} \begin{bmatrix} 2\mathbf{D} & \mathbf{D} & \mathbf{D} \\ \mathbf{D} & 2\mathbf{D} & \mathbf{D} \\ \mathbf{D} & \mathbf{D} & 2\mathbf{D} \end{bmatrix} \quad \mbox{and } \end{aligned}$$

(20)

$$\begin{aligned} & \mathbf{Q}_{4323}^{\mathrm{g}} = \frac{\rho g\Delta h}{3}\{ 0,0,\ - 1,0,0,0,0,0,\ - 1,0,0,0,0,0,\ - 1,0,0,0\}^{\mathrm{T}}, \end{aligned}$$

(21)

where D is the 6×6 diagonal matrix in the form D=diag[12,12,12,h ²,h ²,h ²], ρ is the material density, Δ is the element area, and h is the element thickness. For convenience, the mass matrix can be rewritten in a compact form as

$$\mathbf{M}^{4323} = \frac{\rho\Delta h}{144}\mathbf{T}[2,1] \otimes \mathbf{D}, $$

where T is a symmetric Toeplitz matrix constructed by using diagonal values 2 and 1, and the symbol ⊗ represents the Kronecker block matrix product of matrices T and D.

1.2 A.2 Elements of equations of motion for element 4423

The shape functions for the rectangular element 4423 are also defined by Eq. (7) for the number of nodes n=4. The shape functions’ matrix S ⁴⁴²³ can be written explicitly as

where the bilinear shape functions $s_{i,0}^{241}(x,y), i = 1, \ldots,4$ are determined by Eq. (3).

The mass matrix and the generalized vector of gravity force are

$$\begin{aligned} &\mathbf{M}^{4423} = \frac{\rho abh}{432} \begin{bmatrix} 4\mathbf{D} & 2\mathbf{D} & \mathbf{D} & 2\mathbf{D} \\ 2\mathbf{D} & 4\mathbf{D} & 2\mathbf{D} & \mathbf{D} \\ \mathbf{D} & 2\mathbf{D} & 4\mathbf{D} & 2\mathbf{D} \\ 2\mathbf{D} & \mathbf{D} & 2\mathbf{D} & 4\mathbf{D} \end{bmatrix} \quad \mbox{and } \\ & \mathbf{Q}_{4423}^{\mathrm{g}} = \frac{\rho gabh}{4}\{ 0,0, - 1,0,0,0,0,0, - 1,0,0,0, \ldots,0,0, - 1,0,0,0\}^{\mathrm{T}}. \end{aligned}$$

Again, the mass matrix can be reformulated using a symmetric Toeplitz matrix with diagonal values 4, 2, and 1, and the same matrix D=diag[12,12,12,h ²,h ²,h ²]:

$$\mathbf{M}^{4423} = \frac{\rho abh}{432}\mathbf{T}[4,2,1] \otimes \mathbf{D}. $$

1.3 A.3 Elements of equations of motion for element 6623

The shape functions for the triangular element 6623 are defined by Eq. (7), where n equals 3, and the matrix S ⁶⁶²³ of shape functions from Eq. (1) in explicit form can be formulated as follows:

where I is the 3×3 unit matrix, and $s_{i,0}^{261}(x,y), \ i = 1, \ldots,6$ are the triangular shape functions.

The constant mass matrix and constant generalized gravity vector are calculated using Eq. (11):

$$\begin{aligned} \mathbf{M}^{6623}& = \frac{\rho\Delta h}{4320} \begin{bmatrix} 6\mathbf{D} & - \mathbf{D} & - \mathbf{D} & \mathbf{O} & - 4\mathbf{D} & \mathbf{O} \\ & 6\mathbf{D} & - \mathbf{D} & \mathbf{O} & \mathbf{O} & - 4\mathbf{D} \\ & & 6\mathbf{D} & - 4\mathbf{D} & \mathbf{O} & \mathbf{O} \\ & & & 32\mathbf{D} & 16\mathbf{D} & 16\mathbf{D} \\ & \mathrm{sym.} & & & 32\mathbf{D} & 16\mathbf{D} \\ & & & & & 32\mathbf{D} \end{bmatrix} , \\ \mathbf{Q}_{6623}^{\mathrm{g}}& = \rho g\Delta h \biggl\{ 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, - \frac{1}{3},0,0,0,0,0, \\ &\quad- \frac{1}{3},0,0,0,0,0, - \frac{1}{3},0,0,0\biggr\} ^{\mathrm{T}}, \end{aligned}$$

where D is the 6×6 diagonal matrix of the form shown above, and O is the 6×6 zero matrix.

1.4 A.4 Elements of equations of motion for element 8823

The shape functions for the triangular element 8823 are defined by Eq. (7), where n equals 8, and the matrix S ⁸⁸²³ of shape functions from Eq. (1) in explicit form can be formulated as follows:

where I is the 3×3 unit matrix, and $s_{i,0}^{281}(x,y), i = 1, \ldots,8$ are the shape functions. The constant mass matrix and constant generalized gravity vector are calculated using Eq. (11):

$$\begin{aligned} &\mathbf{M}^{8823} = \frac{\rho abh}{2160} \begin{bmatrix} 6\mathbf{D} & 2\mathbf{D} & 3\mathbf{D} & 2\mathbf{D} & - 6\mathbf{D} & - 8\mathbf{D} & - 8\mathbf{D} & - 6\mathbf{D} \\ & 6\mathbf{D} & 2\mathbf{D} & 3\mathbf{D} & - 6\mathbf{D} & - 6\mathbf{D} & - 8\mathbf{D} & - 8\mathbf{D} \\ & & 6\mathbf{D} & 2\mathbf{D} & - 8\mathbf{D} & - 6\mathbf{D} & - 6\mathbf{D} & - 8\mathbf{D} \\ & & & 6\mathbf{D} & - 8\mathbf{D} & - 8\mathbf{D} & - 6\mathbf{D} & - 6\mathbf{D} \\ & & & & 32\mathbf{D} & 20\mathbf{D} & 16\mathbf{D} & 20\mathbf{D} \\ & \mathrm{sym}. & & & & 32\mathbf{D} & 20\mathbf{D} & 16\mathbf{D} \\ & & & & & & 32\mathbf{D} & 20\mathbf{D} \\ & & & & & & & 32\mathbf{D} \end{bmatrix} , \\ & \mathbf{Q}_{8823}^{\mathrm{g}} = \rho g\Delta h \biggl\{ \underbrace{0,0,\frac{ - 1}{12},0,0,0, \ldots,}_{4\ \text{times}}\underbrace{0,0, \frac{ - 1}{3},0,0,0, \ldots}_{4\ \text{times}}\biggr\} ^{\mathrm{T}}. \end{aligned}$$

Appendix B: Cantilever beam subjected to large deflection

Tables 3 and 4 show the relative error in calculation of the vertical and horizontal deflection components, respectively, for the benchmark static problem described in Sect. 4.1. In these tables, p=Pl ²/EI is the force factor, the column ‘E.P.’ contains solution of the Euler’s Elastica problem, N is the number of elements along the beam length.

Table 3 Vertical deflection of the cantilever beam’s free end for different element types

Table 4 Horizontal deflection of the cantilever beam’s free end for different element types

Appendix C: Cantilever beam subjected the action of a bending moment

Tables 5 and 6 show the values of deflection of the free-end of a cantilever beam calculated using elements 6623 and 4423. The results are compared to the analytical solution. The relative error is presented in Tables 7 and 8.

Table 5 Vertical displacements and rotations of the plate’s points for the element 6623

Table 6 Vertical displacements and rotations of the plate’s points for the element 4423

Table 7 Error in vertical displacements and rotations of the plate’s points for the element 4423

Table 8 Error in vertical displacements and rotations of the plate’s points for the element 6623