Investigation of electrically actuated geometrically nonlinear clamped circular nanoplate

Abstract

In this paper, oscillations of the electrically actuated nonlinear clamped circular nanoplate are considered. The pull-in phenomenon, i.e., the transition of the oscillatory regime to the attraction one, is studied. Original PDE takes into account Coulomb, Casimir and van der Waals forces and geometrical nonlinearity of plate. A simple and physically clear algorithm for determining the DC voltage values at which the system suddenly collapses is proposed. This algorithm is based on the criterion consisting in the merging of points of a stable equilibrium (center) and an unstable equilibrium (saddle). Comparisons with the results obtained by other authors show sufficient accuracy of the proposed algorithm. The influence of geometric nonlinearity on pull-in values is studied. It is shown that neglect of this factor can lead to significant errors. Influence of Casimir and van der Waals forces on the pull-in value is estimated. It is shown that the dependence of the pull-in value on the initial displacements in the radial direction is almost linear.

Appendices

Appendix 1

Berger [21] proposed an effective technique for simplifying the equations of geometrically nonlinear plates. The method is based upon the assumption of neglecting the plane inertia and strain energy associated with the second invariant of the strains in the plate middle surface. For a circular plate in the case of axial symmetry, this means

$$\frac{{\partial^{2} U}}{{\partial t^{2} }} \approx 0,\quad \varepsilon_{r} \varepsilon_{\theta } \approx 0,$$

(A1)

where $\varepsilon_{r} = \frac{\partial U}{{\partial \overline{r}}} + \frac{1}{2}\left( {\frac{\partial W}{{\partial \overline{r}}}} \right)^{2} ,\varepsilon_{\theta } = \frac{U}{{\overline{r}}}$, $U(t)$ is the radial displacement.

Then from equation for in-plane deformation, one obtains

$$\frac{\partial U}{{\partial \overline{r}}} + \frac{U}{{\overline{r}}} + \frac{1}{2}\left( {\frac{\partial W}{{\partial \overline{r}}}} \right)^{2} = \frac{TD}{B},$$

(A2)

so T is the function of t only.

Let the following conditions for displacements in the plane be specified at the edge of the plate:

$$U = U_{e} (t)\quad {\text{at}}\;\quad \overline{r} = R.$$

(A3)

Multiplying both sides of equation (A2) by $\overline{r}$, integrating on $\overline{r}$ over 0 to R and taking into account the boundary conditions (A3), as well as the symmetry conditions at $\overline{r} = 0$, we find

$$T(t) = \frac{B}{{DR^{2} }}\left[RU_{e} (t) + \int\limits_{0}^{R} {\left( {\frac{\partial W}{{\partial \overline{r}}}} \right)^{2} \overline{r}d\overline{r}}\right ].$$

(A4)

The equation of motion in the normal direction can be written as (T is the function on t only)

$$\rho \delta \frac{{\partial^{2} W}}{{\partial t^{2} }} + D\left[ {\frac{{\partial^{4} W}}{{\partial \overline{r}^{4} }} + \frac{2}{{\overline{r}}}\frac{{\partial^{3} W}}{{\partial \overline{r}^{3} }} - \frac{1}{{\overline{r}^{2} }}\frac{{\partial^{2} W}}{{\partial \overline{r}^{2} }} + \frac{1}{{\overline{r}^{3} }}\frac{\partial W}{{\partial \overline{r}}}} \right] - DT\frac{\partial }{{\overline{r}\partial \overline{r}}}\left( {\overline{r}\frac{\partial W}{{\partial \overline{r}}}} \right) = 0.$$

(A5)

Substituting expression (A4) into equation (A5), one obtains

$$\rho \delta \frac{{\partial^{2} W}}{{\partial t^{2} }} + D\left[ {\frac{{\partial^{4} W}}{{\partial \overline{r}^{4} }} + \frac{2}{{\overline{r}}}\frac{{\partial^{3} W}}{{\partial \overline{r}^{3} }} - \frac{1}{{\overline{r}^{2} }}\frac{{\partial^{2} W}}{{\partial \overline{r}^{2} }} + \frac{1}{{\overline{r}^{3} }}\frac{\partial W}{{\partial \overline{r}}}} \right] - \frac{B}{{R^{2} }}\left[RU(R,t) + \int\limits_{0}^{R} {\left( {\frac{\partial W}{{\partial \overline{r}}}} \right)^{2} \overline{r}d\overline{r}} \right]\frac{\partial }{{\overline{r}\partial \overline{r}}}\left( {\overline{r}\frac{\partial W}{{\partial \overline{r}}}} \right) = 0.$$

(A6)

For the case of piezoelectric thin plates, Berger’s approximation is generalized in the paper [22].

Comparisons of the results with known solutions indicate that for a broad range of problems, Berger’s approach yields sufficiently accurate results.

Appendix 2

Integrals:

$$I_{1} {\text{ }} = \int\limits_{0}^{1} {\frac{{\left( {1 - r^{2} } \right)^{2} }}{{\left[ {1 - u\, \left( {1 - r^{2} } \right)^{2} } \right]^{2} }}} \, dr{\text{ }} = \frac{1}{{4u\left( {1 - u} \right)}} - \frac{1}{8}\frac{{\left( {1 - 2u^{{1/2}} } \right)\arctan \left( {\frac{{u^{{1/4}} }}{{\sqrt {1 - u^{{1/2}} } }}} \right)}}{{u^{{5/4}} \left( {1 - u^{{1/2}} } \right)^{{3/2}} }} - \frac{1}{{16}}\frac{{\left( {1 + 2u^{{1/2}} } \right)\ln \left( {\frac{{\left( {1 + u^{{1/2}} } \right)^{{1/2}} + u^{{1/4}} }}{{\left( {1 + u^{{1/2}} } \right)^{{1/2}} - u^{{1/4}} }}} \right)}}{{u^{{5/4}} \left( {1 + u^{{1/2}} } \right)^{{3/2}} }};$$

$$I_{2} = \int\limits_{0}^{1} {\frac{{\left( {1 - r^{2} } \right)^{2} }}{{\left[ {1 - u\, \left( {1 - r^{2} } \right)^{2} } \right]^{3} }}} \, dr = \frac{3}{{32}}\frac{{1 + u}}{{u\left( {1 - u} \right)^{2} }} + \frac{1}{{64}}\frac{{\left( {10u^{{1/2}} - 4u - 3} \right)\arctan \left( {\frac{{u^{{1/4}} }}{{\sqrt {1 - u^{{1/2}} } }}} \right)}}{{u^{{5/4}} \left( {1 - u^{{1/2}} } \right)^{{5/2}} }}\quad - \frac{1}{{128}}\frac{{\left( {4u + 10u^{{1/2}} + 3} \right)\ln \left( {\frac{{\left( {1 + u^{{1/2}} } \right)^{{1/2}} + u^{{1/4}} }}{{\left( {1 + u^{{1/2}} } \right)^{{1/2}} - u^{{1/4}} }}} \right)}}{{u^{{5/4}} \left( {1 + u^{{1/2}} } \right)^{{5/2}} }};$$

$$I_{3} = \int\limits_{0}^{1} {\frac{{\left( {1 - r^{2} } \right)^{2} }}{{\left[ {1 - u\, \left( {1 - r^{2} } \right)^{2} } \right]^{4} }}} \, dr = - \frac{1}{{512}}\frac{1}{{u^{{5/4}} \left( {1 + u^{{1/2}} } \right)^{{7/2}} }}\quad \ln \left( {\frac{{\left( {1 + u^{{1/2}} } \right)^{{1/2}} - u^{{1/4}} }}{{\left( {1 + u^{{1/2}} } \right)^{{1/2}} + u^{{1/4}} }}} \right)\left( {40u^{{3/2}} + 140u + 172u^{{1/2}} + 77} \right)\quad + \frac{1}{{512}}\frac{1}{{u^{{5/4}} \left( { - 1 + u^{{1/2}} } \right)^{{7/2}} }}\ln \left( {\frac{{\left( { - 1 + u^{{1/2}} } \right)^{{1/2}} - u^{{1/4}} }}{{\left( { - 1 + u^{{1/2}} } \right)^{{1/2}} + u^{{1/4}} }}} \right)\left( {40u^{{3/2}} - 140u + 172u^{{1/2}} - 77} \right)\quad - \frac{1}{{384}}\frac{{14u^{2} - 53u - 21}}{{u\left( {1 - u} \right)^{{3}} }} + \frac{3}{{64}}\frac{{\arctan \left( {\frac{{u^{{1/4}} }}{{\sqrt {1 - u^{{1/2}} } }}} \right)\left( { - 4u + 10u^{{1/2}} - 7} \right)}}{{u^{{5/4}} \left( {1 - u^{{1/2}} } \right)^{{5/2}} }}\quad - \frac{3}{{128}}\frac{{\ln \left( {\frac{{\left( {1 + u^{{1/2}} } \right)^{{1/2}} + u^{{1/4}} }}{{\left( {1 + u^{{1/2}} } \right)^{{1/2}} - u^{{1/4}} }}} \right)\left( {4u + 10u^{{1/2}} + 7} \right)}}{{u^{{5/4}} \left( {1 + u^{{1/2}} } \right)^{{5/2}} }}.$$