An efficient quadrature method for vibration analysis of thin elliptical plates with continuous and discontinuous edge conditions

Abstract

Mapping an irregular domain into a square one is a common technique in analyzing problems of plates and shells with irregular shapes. For the irregular shape without four corners such as the elliptical shape, the difficulty arises that the Jacobian determinant is zero at the corner points. An efficient quadrature method is presented to analyze the transverse vibration of thin plates with an elliptical shape. To circumvent the above-mentioned difficulty, Gauss quadrature is used in numerical integration. Besides, derivative degrees of freedom are not used, and a boundary point is modeled by two nodes separated by a very small distance. Since the nodes are not coinciding with integration points, a way indirectly using the differential quadrature law is employed to derive the explicit formulas to ease the programming. A convergence study is performed. Free vibration of elliptical plates with continuous and discontinuous edge conditions is analyzed to demonstrate the efficiency of the developed rotation-free weak-form quadrature method.

Appendix

The general shape functions of a 16-node Serendipity element with nodes of any type are given below:

$$f_{1} = \frac{1}{4}(1 - \xi )(1 - \eta ) - \beta (f_{5} + f_{16} ) - \frac{1}{2}(f_{15} + f_{6} ) - \gamma (f_{7} + f_{14} ),$$

$$\begin{gathered} f_{2} = \frac{1}{4}(1 + \xi )(1 - \eta ) - \beta (f_{7} + f_{8} ) - \frac{1}{2}(f_{9} + f_{6} ) - \gamma (f_{5} + f_{10} ), \hfill \\ f_{3} = \frac{1}{4}(1 + \xi )(1 + \eta ) - \beta (f_{10} + f_{11} ) - \frac{1}{2}(f_{12} + f_{9} ) - \gamma (f_{8} + f_{13} ), \hfill \\ f_{4} = \frac{1}{4}(1 - \xi )(1 + \eta ) - \beta (f_{13} + f_{14} ) - \frac{1}{2}(f_{15} + f_{12} ) - \gamma (f_{11} + f_{16} ), \hfill \\ \end{gathered}$$

$$\begin{gathered} f_{5} = (\xi - x_{p} )(\xi - \xi^{3} )(1 - \eta )/\alpha, \hfill \\ f_{6} = (\xi^{2} - x_{p}^{2} )(1 - \xi^{2} )(1 - \eta )/\tau, \hfill \\ f_{7} = (\xi + x_{p} )(\xi - \xi^{3} )(1 - \eta )/\alpha, \hfill \\ \end{gathered}$$

$$\begin{gathered} f_{13} = (\xi - x_{p} )(\xi - \xi^{3} )(1 + \eta )/\alpha, \hfill \\ f_{12} = (\xi^{2} - x_{p}^{2} )(1 - \xi^{2} )(1 + \eta )/\tau, \hfill \\ f_{11} = (\xi + x_{p} )(\xi - \xi^{3} )(1 + \eta )/\alpha, \hfill \\ \end{gathered}$$

$$\begin{gathered} f_{8} = (\eta - x_{p} )(\eta - \eta^{3} )(1 + \xi )/\alpha, \hfill \\ f_{9} = (\eta^{2} - x_{p}^{2} )(1 - \eta^{2} )(1 + \xi )/\tau, \hfill \\ f_{10} = (\eta + x_{p} )(\eta - \eta^{3} )(1 + \xi )/\alpha, \hfill \\ \end{gathered}$$

$$\begin{gathered} f_{14} = (\eta + x_{p} )(\eta - \eta^{3} )(1 - \xi )/\alpha, \hfill \\ f_{15} = (\eta^{2} - x_{p}^{2} )(1 - \eta^{2} )(1 - \xi )/\tau, \hfill \\ f_{16} = (\eta - x_{p} )(\eta - \eta^{3} )(1 - \xi )/\alpha \hfill \\ \end{gathered}$$

where $x_{p}$($0 < x_{p} < 1$) is shown in Fig. 1c, $\tau = - 2x_{p}^{2} , \, \alpha = 4(x_{p}^{2} - x_{p}^{4} ), \, \beta = (1 + x_{p} )/2$, and $\, \gamma = (1 - x_{p} )/2$.