Research and analysis of resonant and stiffness of cylindrical dielectric elastomer actuator

The fabrication process mainly incledes two parts: (i) pre-stretching and coated electrode (figure 1(b)), (ii) multilayering (figure 1(c)) and rolling (figures 1(d), (e)). Start by cutting out a piece of tape (3 M, VHB4910). On the surface of the tape, mark the areas with length and width of ${L}_{1}$ and ${L}_{2}$ respectively. The biaxial stretching of the film is carried out by a stretching mechanism, and the length and width of the calibration region in this state are ${l}_{1p}$ and ${l}_{{\rm{2}}p},$ respectively. The pre-strains in the three main directions of the membrane are defined as ${\lambda }_{{\rm{1}}p}={l}_{1p}/{L}_{1},{\lambda }_{2p}={l}_{2p}/{L}_{2},{\lambda }_{3p}={l}_{3p}/{L}_{3},$ where ${L}_{3}=1\,{\rm{mm}}.$ The compliant electrode is manually coated in the marked area on the upper surface of the film, the wire is connected at the right end. Stretch the second film with the same pre-stretch value. Similar to the sandwich structure, two membrane are superimposed on top of each other. The upper surface of the second film is coated with a compliant electrode and connected with a wire. The spring is compressed axially to length ${l}_{1p}.$ The films are then around the pre-compressed spring. Silk thread is wound at the end cover to secure the film and spring. Cut off the excess film to get a cylindrical actuator.

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Figure 1(a) shown that the size of DE membrane in the reference state are length ${L}_{1},$width ${L}_{2}$ and thickness ${L}_{{\rm{3}}}$ in the Z, X and Y directions, respectively. Figure 1(b) illustrates that the DE membrane is pre-stretched by mechanical forces ${P}_{1}$ and ${P}_{2}$ in the ${L}_{1}$ and ${L}_{2}$ directions, respectively. Define the pre-stretches as ${\lambda }_{{\rm{1}}p}={l}_{1p}/{L}_{1},{\lambda }_{2p}={l}_{2p}/{L}_{2},{\lambda }_{{\rm{3}}p}={l}_{3p}/{L}_{3}.$ Two films of the same treatment are glued together. As shown in figure 1(d), a spring with a stiffness of K, radius r, and free length of ${L}_{0}$ is axial pre-compressed to ${\lambda }_{c}={l}_{0}/{L}_{0}.$ A double layer of dielectric elastomer is rolled around the spring to form a cylindrical dielectric elastomer actuator.

When voltage ${\rm{\Phi }}$ is applied, the actuator will deform, and the stretches of the film in the three direction of length, width and thickness change to ${\lambda }_{{\rm{1}}}={l}_{1}/{L}_{1},{\lambda }_{2}={l}_{2}/{L}_{2},$ and ${\lambda }_{{\rm{3}}}={l}_{3}/{L}_{3}$ respectively. In continuum mechanics, the material coordinate (X, Y, Z) and the spatial coordinate (x, y, z) label a certain material point in the reference state and the deformable material point, respectively. For incompressible dielectric elastomer membrane(${\lambda }_{{\rm{1}}}{\lambda }_{{\rm{2}}}{\lambda }_{{\rm{3}}}=1$), the motion of the actuator can be written as:

$$\begin{eqnarray}&&x={\lambda }_{2}X,y={\lambda }_{3}Y,z={\lambda }_{1}Z\end{eqnarray} \tag{ 2-1 }$$

The dynamic governing equation of cylindrical DEA is expressed by the Euler–Lagrange equation:

$$\begin{eqnarray}&&\displaystyle \frac{\partial {\ell }}{\partial {\lambda }_{i}}-\displaystyle \frac{d}{dt}\left(\displaystyle \frac{\partial {\ell }}{\partial {\dot{\lambda }}_{i}}\right)=0,{\ell }=T-U\end{eqnarray} \tag{ 2-2 }$$

where ${\dot{\lambda }}_{i}$ is the derivative of ${\lambda }_{i}$ with respect to time. ${\ell }$ is the Lagrange, T is the kinetic energy of the system, U is the potential energy produced by the conservative forces in the system.

The kinetic energy of the system is as follows:

$$\begin{eqnarray}&&T=\displaystyle {\int }_{\Omega }\displaystyle \frac{1}{2}\rho \left({\dot{x}}^{2}+{\dot{y}}^{2}+{\dot{z}}^{2}\right)d{\rm{\Omega }}\end{eqnarray} \tag{ 2-3 }$$

$\rho$ is the density of the membrane, ${\rm{\Omega }}$ is the volume of the membrane, $\dot{x},\dot{y},$ and $\dot{z}$ are the deformation velocities of the film in three directions respectively. It is assumed that the mass of the spring is negligible.

The rheological model was used to characterize the deformation process of dielectric elastomer materials [22–24]. As shown in figure 2, the model is divided into two parallel parts. The first part contains a reversible deforming spring $\alpha ,$ and the second part contains a series spring $\beta$ and damping. The free energy of cylindrical DEA consists of three parts: membrane elastic energy, electric energy and spring elastic potential energy. Based on the assumption of uniform deformation, the elastic energy of the membrane is obtained by multiplying the free ennergy density of the menbrane by the volume.

$$\begin{eqnarray}&&\begin{array}{l}U=\displaystyle \frac{1}{2}K{\left({L}_{0}-{L}_{1}{\lambda }_{1}\right)}^{2}+2{L}_{1}{L}_{2}{L}_{3}\\ \,\,\,\,\,\times \,\left[-\displaystyle \frac{{\mu }^{\alpha }{J}^{\alpha }}{2}\,\mathrm{log}\left(1-\displaystyle \frac{{\lambda }_{1}^{2}+{\lambda }_{2}^{2}+{\lambda }_{1}^{-2}{\lambda }_{2}^{-2}-3}{{J}^{\alpha }}\right)\right.\\ \,\,\,\,\,-\,\displaystyle \frac{{\mu }^{\beta }{J}^{\beta }}{2}\,\mathrm{log}\left(1-\displaystyle \frac{{\lambda }_{1}^{2}{\zeta }_{1}^{-2}+{\lambda }_{2}^{2}{\zeta }_{2}^{-2}+{\lambda }_{1}^{-2}{\lambda }_{2}^{-2}{\zeta }_{1}^{2}{\zeta }_{2}^{2}-3}{{J}^{\beta }}\right)\\ \left.\,\,\,\,\,-\,\displaystyle \frac{{P}_{1}}{{L}_{2}{L}_{3}}{\lambda }_{1}-\displaystyle \frac{{P}_{2}}{{L}_{1}{L}_{3}}{\lambda }_{2}-\displaystyle \frac{\varepsilon }{2}{E}_{0}^{2}{\lambda }_{1}^{2}{\lambda }_{2}^{2}\right]\end{array}\end{eqnarray} \tag{ 2-4 }$$

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In which ${\mu }^{\alpha }$ and ${\mu }^{\beta }$ are the shear moduli of spring $\alpha$ and spring $\beta ,$ respectively, ${J}^{\alpha }$ and ${J}^{\beta }$ are a constant associated with the tensile limit of DE material, ${\zeta }_{1}$ and ${\zeta }_{2}$ are the damping strains in two directions of the film plane respectively. $\varepsilon$ is the permittivity of the material, ${E}_{0}={\rm{\Phi }}/{L}_{3}$ is the normal electric field. The strain of spring $\beta$ in the two directions of the film plane is ${\lambda }_{i}^{e},i=1,2,$ which is obtained from the strain of spring $\alpha$ in the corresponding direction and the strain of damping (${\lambda }_{i}^{e}={\lambda }_{i}/{\zeta }_{i}$). Damping is regarded as Newtonian fluid in this paper, and the strain rate of damping is ${\zeta }_{1}^{-1}\,\times \,d{\zeta }_{1}/dt$ and ${\zeta }_{{\rm{2}}}^{-1}\times d{\zeta }_{{\rm{2}}}/dt,$ the relation between the strain rate of damping and the strain rate of spring $\alpha$ is [25, 26]:

$$\begin{eqnarray}&&\begin{array}{l}\displaystyle \frac{d{\zeta }_{1}}{dt}=\displaystyle \frac{{\zeta }_{1}}{3\eta }\left[{\mu }^{\beta }\left({\lambda }_{1}^{2}{\zeta }_{1}^{-2}-{\zeta }_{1}^{2}{\zeta }_{2}^{2}{\lambda }_{1}^{-2}{\lambda }_{2}^{-2}\right)\right.\\ \left.\,\,\,\,\,-\,\,{\mu }^{\beta }/{\rm{2}}\left({\lambda }_{2}^{2}{\zeta }_{2}^{-2}-{\zeta }_{1}^{2}{\zeta }_{2}^{2}{\lambda }_{1}^{-2}{\lambda }_{2}^{-2}\right)\right]\end{array}\end{eqnarray} \tag{ 2-5 }$$

$$\begin{eqnarray}&&\begin{array}{l}\displaystyle \frac{d{\zeta }_{{\rm{2}}}}{dt}=\displaystyle \frac{{\zeta }_{{\rm{2}}}}{3\eta }\left[{\mu }^{\beta }\left({\lambda }_{{\rm{2}}}^{2}{\zeta }_{{\rm{2}}}^{-2}-{\zeta }_{1}^{2}{\zeta }_{2}^{2}{\lambda }_{1}^{-2}{\lambda }_{2}^{-2}\right)\right.\\ \left.\,\,\,\,\,\,-\,\,{\mu }^{\beta }/{\rm{2}}\left({\lambda }_{{\rm{1}}}^{2}{\zeta }_{{\rm{1}}}^{-2}-{\zeta }_{1}^{2}{\zeta }_{2}^{2}{\lambda }_{1}^{-2}{\lambda }_{2}^{-2}\right)\right]\end{array}\end{eqnarray} \tag{ 2-6 }$$

Where $\eta$ is the viscosity of the dashpot.

Previous research data of our research group shows that the axial deformation of the actuator is influenced by the number of winding loops of the film N. When other conditions are determined, the largest axial deformation of the actuator is obtained when N = 6. Therefor the number of winding loops of the actuator studied in this paper is N = 6. The length of the film in the first circle satisfies: ${L}_{{2}_{1}}{\lambda }_{2p}=2\pi r;$ The thickness of the film after pre-stretching is ${L}_{{\rm{3}}}{\lambda }_{1p}^{-1}{\lambda }_{2p}^{-1},$ and the double layer film is superposition during winding. The length of the film in the second round is $2\pi \left(r+2{L}_{3}{\lambda }_{1p}^{-1}{\lambda }_{2p}^{-1}\right)={L}_{{2}_{2}}{\lambda }_{2p},$ and so on, the circumferent length of the film after six rounds of winding is ${L}_{2}=3\times \left(4\pi r+20\pi {L}_{3}{\lambda }_{1p}^{-1}{\lambda }_{2p}^{-1}\right)/{\lambda }_{2p}.$ Because the film is tightly attached to the spring, within the safe deformation range of the film, the deformation of the film in the circumferential direction can be ignored, so that ${\lambda }_{2}={\lambda }_{2p}={\zeta }_{2}.$ According to the incompressibility of the material, ${\lambda }_{3}={\lambda }_{1}^{-1}{\lambda }_{2}^{-1}={\lambda }_{1}^{-1}{\lambda }_{2p}^{-1}.$ Therefore, according to equation (2–3), the kinetic energy of the system is simplified as:

$$\begin{eqnarray}&&T=\displaystyle \frac{1}{3}\rho {L}_{1}{L}_{2}{L}_{3}^{3}{\lambda }_{1}^{-4}{\dot{\lambda }}_{1}^{2}{\lambda }_{2p}^{-2}+\displaystyle \frac{1}{3}\rho {L}_{1}^{3}{L}_{2}{L}_{3}{\dot{\lambda }}_{1}^{2}\end{eqnarray} \tag{ 2-7 }$$

and the potential is obtained from equation (2–4) as:

$$\begin{eqnarray}&&\begin{array}{l}U=\displaystyle \frac{1}{2}K{\left({L}_{0}-{L}_{1}{\lambda }_{1}\right)}^{2}+2{L}_{1}{L}_{2}{L}_{3}\\ \,\,\times \,\,\left[-\displaystyle \frac{{\mu }^{\alpha }{J}^{\alpha }}{2}\,\mathrm{log}\left(1-\displaystyle \frac{{\lambda }_{1}^{2}+{\lambda }_{2p}^{2}+{\lambda }_{1}^{-2}{\lambda }_{2p}^{-2}-3}{{J}^{\alpha }}\right)\right.\\ \,\,-\,\displaystyle \frac{{\mu }^{\beta }{J}^{\beta }}{2}\,\mathrm{log}\left(1-\displaystyle \frac{{\lambda }_{1}^{2}{\zeta }_{1}^{-2}+{\lambda }_{1}^{-2}{\zeta }_{1}^{2}-2}{{J}^{\beta }}\right)\\ \,\,\,-\,\left.\displaystyle \frac{{P}_{1}}{{L}_{2}{L}_{3}}{\lambda }_{1}-\displaystyle \frac{{P}_{2}}{{L}_{1}{L}_{3}}{\lambda }_{2p}-\displaystyle \frac{\varepsilon }{2}{E}_{0}^{2}{\lambda }_{1}^{2}{\lambda }_{2p}^{2}\right]\end{array}\end{eqnarray} \tag{ 2-8 }$$

Combined with equations (2–7) and (2–8), the dynamic governing equation of the system can be obtained from the Euler–Lagrange equation:

$$\begin{eqnarray}&&\begin{array}{l}{\ddot{\lambda }}_{1}-\displaystyle \frac{2}{{\lambda }_{1}+{L}_{1}^{2}{L}_{3}^{-2}{\lambda }_{1}^{5}{\lambda }_{2p}^{2}}{\dot{\lambda }}_{1}^{2}+\displaystyle \frac{3}{\rho \left({L}_{3}^{2}{\lambda }_{1}^{-4}{\lambda }_{2p}^{-2}+{L}_{1}^{2}\right)}\\ \,\,\times \,\left[\displaystyle \frac{{\mu }^{\alpha }\left({\lambda }_{1}-{\lambda }_{1}^{-3}{\lambda }_{2p}^{-2}\right)}{1-\left({\lambda }_{1}^{2}+{\lambda }_{2p}^{2}+{\lambda }_{1}^{-2}{\lambda }_{2p}^{-2}-3\right)/{J}^{\alpha }}-\displaystyle \frac{{P}_{1}}{{L}_{2}{L}_{3}}\right.\\ \,\,+\,\left.\displaystyle \frac{{\mu }^{\beta }\left({\lambda }_{1}{\zeta }_{1}^{-2}-{\lambda }_{1}^{-3}{\zeta }_{1}^{2}\right)}{1-\left({\lambda }_{1}^{2}{\zeta }_{1}^{-2}+{\lambda }_{1}^{-2}{\zeta }_{1}^{2}-2\right)/{J}^{\beta }}-\varepsilon {E}_{0}^{2}{\lambda }_{1}{\lambda }_{2p}^{2}\right]\\ \,\,-\,\displaystyle \frac{3K\left({L}_{0}-{L}_{1}{\lambda }_{1}\right)}{2\rho {L}_{2}{L}_{3}\left({L}_{3}^{2}{\lambda }_{1}^{-4}{\lambda }_{2p}^{-2}+{L}_{1}^{2}\right)}=0\end{array}\end{eqnarray} \tag{ 2-9 }$$

The rate of deformation in the dashpot can be simplified by equations (2–5) and (2–6) as:

$$\begin{eqnarray}&&\displaystyle \frac{d{\zeta }_{1}}{dt}=\displaystyle \frac{{\zeta }_{1}}{3\eta }\left[{\mu }^{\beta }\left({\lambda }_{1}^{2}{\zeta }_{1}^{-2}-{\zeta }_{1}^{2}{\lambda }_{1}^{-2}\right)-\displaystyle \frac{{\mu }^{\beta }}{2}\left({\rm{1}}-{\zeta }_{1}^{2}{\lambda }_{1}^{-2}\right)\right]\end{eqnarray} \tag{ 2-10 }$$

$$\begin{eqnarray}&&\displaystyle \frac{d{\zeta }_{{\rm{2}}}}{dt}={\rm{0}}\end{eqnarray} \tag{ 2-11 }$$

In order to improve the axial strain capacity of the actuator and restrain the circumferential strain, the pre-stretching of the film is satisfied ${\lambda }_{1p}\leqslant {\lambda }_{2p}.$ In calculation, the following material parameters are used: $r=6\,{\rm{mm}},{L}_{0}=70\,{\rm{mm}},K=289.02\,N.{m}^{-1},{\lambda }_{c}=0.6,{L}_{1}=14\,{\rm{mm}},$ ${L}_{2}=15.2\pi \,{\rm{mm}},{L}_{3}=1\,{\rm{mm}},{\lambda }_{1p}=3,{\lambda }_{2p}=5,\rho =1.2\times {10}^{3}{\rm{kg}}\cdot {m}^{-3}$ ${\mu }^{\alpha }=55000\,Pa,{J}^{\alpha }=90,{\mu }^{\beta }=\mathrm{45000}\,{\mathrm{Pa},J}^{\beta }=70,{\tau }_{v}=\eta /{\mu }^{\beta }=50s,$ $\varepsilon =3.98\times {10}^{-11}F\cdot {m}^{-1}$ _. The material parameters are substituted into the equation (2–9), and the nonlinear stiffness term in the equation (2–9) is expanded by Taylor third order at ${\lambda }_{1}={\lambda }_{1p}=3$ to obtain a new governing equation. Before voltage is applied, the cylindrical DEA has been static treated for 3–5 h, and the film strain is considered to be completely relaxed. ${\zeta }_{1}={\lambda }_{1p}=3:$

$$\begin{eqnarray}&&{\ddot{\lambda }}_{1}-\displaystyle \frac{2}{{\lambda }_{1}+4900{\lambda }_{1}^{5}}{\dot{\lambda }}_{{\rm{1}}}^{{\rm{2}}}+{C}_{1}+{C}_{2}{\lambda }_{1}+{C}_{3}{\lambda }_{1}^{2}+{C}_{4}{\lambda }_{1}^{3}=0\end{eqnarray} \tag{ 3-1 }$$

Where ${C}_{1}=-1.0035\times {10}^{7},{C}_{3}=-9.6005\times {10}^{5},{C}_{4}=1.142\times {10}^{5}$ ${C}_{2}=4.8363\times {10}^{6}-12.4574\times {10}^{-3}{{\rm{\Phi }}}^{2}$

It is easy to understand that when the pre-stretching forces ${P}_{1},{P}_{2},$ and voltage ${\rm{\Phi }}$ are static, the axis strain of the cylindrical DEA may reach a stable state. In the equilibrium state, the governing equation equation (3–1) can be reduced to:

$$\begin{eqnarray}&&{C}_{1}+{C}_{2}{\lambda }_{1eq}+{C}_{3}{\lambda }_{1eq}^{2}+{C}_{4}{\lambda }_{1eq}^{3}=0\end{eqnarray} \tag{ 3-2 }$$

For equation (3–1), the coefficient of damping nonlinear term differs from other coefficients in the equation by more than 10⁴ order of magnitude, so this term can be ignored in the governing equation:

$$\begin{eqnarray}&&{\ddot{\lambda }}_{1}+{C}_{1}+{C}_{2}{\lambda }_{1}+{C}_{3}{\lambda }_{1}^{2}+{C}_{4}{\lambda }_{1}^{3}=0\end{eqnarray} \tag{ 3-3 }$$

With the initial conditions:

$$\begin{eqnarray}&&{\lambda }_{1}\left({\rm{0}}\right)={\lambda }_{1p}=3,{\dot{\lambda }}_{1}\left(0\right)=0\end{eqnarray} \tag{ 3-4 }$$

The linearized equation of equation (3–3) is:

$$\begin{eqnarray}&&{\ddot{\lambda }}_{1}+{C}_{1}+\left({C}_{2}+\alpha \right){\lambda }_{1}=0\end{eqnarray} \tag{ 3-5 }$$

The difference between equations (3–3) and (3–5) is:

$$\begin{eqnarray}&&e\left({\lambda }_{1}\right)=-\alpha {\lambda }_{1}+{C}_{3}{\lambda }_{1}^{2}+{C}_{4}{\lambda }_{1}^{3}\end{eqnarray} \tag{ 3-6 }$$

According to the mean square error criterion, the unknown coefficient $\alpha$ is:

$$\begin{eqnarray}&&\displaystyle \frac{\partial }{\partial \alpha }\left\langle {e}^{2}\left({\lambda }_{1}\right)\right\rangle =0\end{eqnarray} \tag{ 3-7 }$$

It yields:

$$\begin{eqnarray}&&\alpha =\displaystyle \frac{{C}_{3}\left\langle {\lambda }_{1}^{3}\right\rangle +{C}_{4}\left\langle {\lambda }_{1}^{4}\right\rangle }{\left\langle {\lambda }_{1}^{2}\right\rangle }\end{eqnarray} \tag{ 3-8 }$$

The periodic solution of equation (3–5) is:

$$\begin{eqnarray}&&{\lambda }_{1}=3\,\cos \left(\sqrt{{C}_{2}+\alpha }\right)t-\displaystyle \frac{{C}_{1}}{{C}_{2}+\alpha }=3\,\cos \,\omega t-\displaystyle \frac{{C}_{1}}{{\omega }^{2}}\end{eqnarray} \tag{ 3-9 }$$

Where $\omega =\sqrt{{C}_{2}+\alpha }.$ Based on weighted average [27]:

$$\begin{eqnarray}&&\begin{array}{l}\left\langle \cos \left(n\omega t\right)\right\rangle =\displaystyle {\int }_{0}^{\infty }{s}^{2}{\omega }^{2}t{e}^{-s\omega t}\,\cos \left(n\omega t\right)dt\\ \,\,\,\,\,\,\,=\displaystyle {\int }_{0}^{\infty }{s}^{2}\tau {e}^{-s\tau }\,\cos \left(n\tau \right)d\tau \\ \,\,\,\,\,\,\,={s}^{2}\displaystyle \frac{{s}^{2}-{n}^{2}}{{\left({s}^{2}+{n}^{2}\right)}^{2}}\end{array}\end{eqnarray} \tag{ 3-10 }$$

where s is constant.

According to the calculation criterion of equation (3–10), the average operator in equation (3–8) is:

$$\begin{eqnarray}&&\left\langle {\lambda }_{1}^{2}\right\rangle =\displaystyle \frac{9}{2}{s}^{2}\displaystyle \frac{{s}^{2}-4}{{\left({s}^{2}+4\right)}^{2}}+\displaystyle \frac{9}{2}+{\left(\displaystyle \frac{{C}_{1}}{{\omega }^{2}}\right)}^{2}-6\displaystyle \frac{{C}_{1}}{{\omega }^{2}}{s}^{2}\displaystyle \frac{{s}^{2}-1}{{\left({s}^{2}+1\right)}^{2}}\end{eqnarray} \tag{ 3-11 }$$

$$\begin{eqnarray}&&\begin{array}{l}\left\langle {\lambda }_{1}^{3}\right\rangle =\displaystyle \frac{81}{4}{s}^{2}\displaystyle \frac{{s}^{2}-1}{{\left({s}^{2}+1\right)}^{2}}+\displaystyle \frac{{\rm{27}}}{{\rm{4}}}{s}^{2}\displaystyle \frac{{s}^{2}-9}{{\left({s}^{2}+9\right)}^{2}}\\ \,\,\,\,\,-\,\displaystyle \frac{{\rm{27}}}{{\rm{2}}}\displaystyle \frac{{C}_{1}}{{\omega }^{2}}\left[{s}^{2}\displaystyle \frac{{s}^{2}-4}{{\left({s}^{2}+4\right)}^{2}}+1\right]-{\left(\displaystyle \frac{{C}_{1}}{{\omega }^{2}}\right)}^{3}\\ \,\,\,\,\,+\,9{\left(\displaystyle \frac{{C}_{1}}{{\omega }^{2}}\right)}^{2}{s}^{2}\displaystyle \frac{{s}^{2}-1}{{\left({s}^{2}+1\right)}^{2}}\end{array}\end{eqnarray} \tag{ 3-12 }$$

$$\begin{eqnarray}&&\begin{array}{l}\left\langle {\lambda }_{1}^{{\rm{4}}}\right\rangle =\displaystyle \frac{{\rm{81}}}{2}{s}^{2}\displaystyle \frac{{s}^{2}-4}{{\left({s}^{2}+4\right)}^{2}}+\displaystyle \frac{{\rm{81}}}{{\rm{8}}}{s}^{2}\displaystyle \frac{{s}^{2}-16}{{\left({s}^{2}+16\right)}^{2}}\\ \,\,\,\,\,-\,108\displaystyle \frac{{C}_{1}}{{\omega }^{2}}\left[\displaystyle \frac{{\rm{3}}}{{\rm{4}}}{s}^{2}\displaystyle \frac{{s}^{2}-1}{{\left({s}^{2}+1\right)}^{2}}+\displaystyle \frac{{\rm{1}}}{{\rm{4}}}{s}^{2}\displaystyle \frac{{s}^{2}-9}{{\left({s}^{2}+9\right)}^{2}}\right]\\ \,\,\,\,\,+\,54{\left(\displaystyle \frac{{C}_{1}}{{\omega }^{2}}\right)}^{2}\left[\displaystyle \frac{{\rm{1}}}{{\rm{2}}}{s}^{2}\displaystyle \frac{{s}^{2}-4}{{\left({s}^{2}+4\right)}^{2}}+\displaystyle \frac{{\rm{1}}}{{\rm{2}}}\right]+\displaystyle \frac{{\rm{243}}}{{\rm{8}}}\\ \,\,\,\,\,-\,12{\left(\displaystyle \frac{{C}_{1}}{{\omega }^{2}}\right)}^{3}{s}^{2}\displaystyle \frac{{s}^{2}-1}{{\left({s}^{2}+1\right)}^{2}}+{\left(\displaystyle \frac{{C}_{1}}{{\omega }^{2}}\right)}^{4}\end{array}\end{eqnarray} \tag{ 3-13 }$$

In case s = 2, substituting equations (3–11)–(3–13) into equation (3–8), and then substituting equation (3–8) into equations (3–5) and (3–9) we get the equivalent resonance frequency and solution of the governing equation as follows:

$$\begin{eqnarray}&&\alpha =7593.1689\exp \left(\displaystyle \frac{{\rm{\Phi }}}{1938.1459}\right)-1.7668\times {10}^{6}\end{eqnarray} \tag{ 3-14 }$$

$$\begin{eqnarray}&&{\lambda }_{{\rm{1}}}\left(t\right)=3\,\cos \left(\sqrt{{C}_{2}+\alpha }\right)t-\displaystyle \frac{{C}_{1}}{{C}_{2}+\alpha }\end{eqnarray} \tag{ 3-15 }$$

$$\begin{eqnarray}&&\begin{array}{l}{\omega }_{n\_eq}^{2}={C}_{2}+\alpha =3.0695\times {10}^{6}-12.4574\times {10}^{-3}{{\rm{\Phi }}}^{2}\\ \,\,\,\,\,\,\,+\,\,7593.1689\exp \left(\displaystyle \frac{{\rm{\Phi }}}{1938.1459}\right)\end{array}\end{eqnarray} \tag{ 3-16 }$$

$$\begin{eqnarray}&&{K}_{eq}={\omega }_{n\_eq}^{2}m\end{eqnarray} \tag{ 3-17 }$$

where m = 14.08 g is the mass of the cylindrical DEA.

Equation (3–16) shows that the equivalent resonant frequency of the actuator is a function of the excitation voltage when the other parameters of the actuator are given. To further explain and explore how the viscoelasticity affects the system's dynamic response, we used Matlab to simulate the voltage curve of equivalent resonant frequency and equivalent stiffness (figure 3).

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As can be seen from figure 3, with the increase of excitation voltage, the equivalent stiffness value of the cylindrical DEA decreases. The reason for this is that electrostatic pressure causes the film to deform, making the overall actuator more 'soft' [28].

In the following, we study the dynamic response of the cylindrical DEA under sinusoidal voltage excitation. It can be seen from equation (3–1) that the voltage appears in the coefficient of the governing equation.The dynamic response of cylindrical DEAs is driven by dynamic system parameters of specific frequencies. Therefore, the dynamic excitation response of the actuator is a typical parametric excitation problem. Next, we study the vibration of the actuator under the condition that the harmonic voltage and the pre-stretching of the membrane are constant:

$$\begin{eqnarray}&&{\rm{\Phi }}={{\rm{\Phi }}}_{DC}+{{\rm{\Phi }}}_{AC}\,\sin \left(2\pi ft\right)\end{eqnarray} \tag{ 3-18 }$$

where ${{\rm{\Phi }}}_{DC}$ is DC voltage amplitude, ${{\rm{\Phi }}}_{AC}$ is AC voltage amplitude and f is excitation voltage frequency. According to equation (3–2), when the DC voltage ${{\rm{\Phi }}}_{DC}=1150\,{\rm{V}}$ is applied to the cylindrical DEA, the axial strain of the actuator reaches a stable value, ${\lambda }_{eq}=3.5129.$ Similarly, when ${{\rm{\Phi }}}_{DC}=2150\,{\rm{V}},$ ${\lambda }_{eq}=3.5762;$ when ${{\rm{\Phi }}}_{DC}=3150\,{\rm{V}},$ ${\lambda }_{eq}=3.6807.$ These states of equilibrium will be substituted into the following simulation process as the initial strain value. The influence of sinusoidal voltage with different excitation frequencies on the dynamic response of cylindrical DEA will be analyzed theoretically.

Measurements of the amplitude at frequencies from 1 to 500 Hz are made for three different combinations of ${{\rm{\Phi }}}_{DC}$ and ${{\rm{\Phi }}}_{AC}$ (figure 4). The amplitude is the half the difference between the maximum vibration strain and the minimum vibration strain under the equilibrium state. The equivalent resonant frequency is obtained by third order Taylor expansion (figure 3). When the voltage is set as ${\rm{\Phi }}=1150\,{\rm{V}},$ the equivalent resonant frequency is obtained in equation (3–16) with a value of $f=278.9\,{\rm{Hz}}.$ Similarly, when ${\rm{\Phi }}={\rm{2150}}\,{\rm{V}},$ $f=277.4\,{\rm{Hz}};$ when ${\rm{\Phi }}={\rm{3150}}\,{\rm{V}},$ $f=275.1\,{\rm{Hz}}.$ Among the three simulations, all of them exhibit one resonant frequency ${f}_{r}.$ When the DC voltage is set as ${{\rm{\Phi }}}_{DC}=1150\,{\rm{V}},$ the resonant frequency is obtained in equation (2–9) with a value of ${f}_{r}=239.2\,{\rm{Hz}}.$ Similarly, when ${{\rm{\Phi }}}_{DC}={\rm{2150}}\,{\rm{V}},$ ${f}_{r}=237.7\,{\rm{Hz}};$ when ${{\rm{\Phi }}}_{DC}={\rm{3150}}\,{\rm{V}},$ ${f}_{r}=236.7\,{\rm{Hz}}.$ The difference between the two is caused by the following two reasons: first, when the governing equation is expanded by third-order Taylor, part of the higher-order terms is discarded, resulting in some errors; second, some errors will occur in the equivalent linearization of the nonlinear governing equation. Moreover, the dynamic change process of figures 3 and 4 is same: the resonant frequency decreases with the increase of the excitation voltage.

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When the DC voltage amplitude is 2150 V and AC voltage amplitude is 150 V, figure 5 illustrates time-varying curve of axial strain ${\lambda }_{1}$ at super-harmonic frequency $f=119\,{\rm{Hz}},$ harmonic frequency $f=238\,{\rm{Hz}},$ and super-harmonic frequency $f=476\,{\rm{Hz}}.$ Figure 5(b) indicates that when the sinusoidal voltage frequency is close to the resonant frequency $f=238\,{\rm{Hz}},$ the vibration of the cylindrical DEA is the strongest and the phenomenon of 'beating' occurs [29]. When the frequency is 1/2 the resonant frequency $f=119\,{\rm{Hz}}$ and twice the resonant frequency $f=476\,{\rm{Hz}},$ the phenomenon of 'beating' does not exist.

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