A Continuum Model for Circular Graphene Membranes Under Uniform Lateral Pressure

Abstract

Despite the numerous applications of pressurized graphene membranes in new technologies, there is still a lack of accurate mechanical models. In this work we develop a continuum model for circular graphene membranes subjected to uniform lateral pressure. We adopt a semi-inverse method by defining a simplified kinematics of deformation and we describe the material behavior with a stored energy function that takes into account both nonlinearity and anisotropy of graphene. An expression of the applied pressure as a function of the deflection of the membrane is obtained from an approximate solution of the equilibrium. The simplifying hypotheses of the analytical model are verified by a finite element (FE) analysis in nonlinear elasticity. In addition, a numerical solution of the differential equilibrium equations of the exact theory is presented. The pressure-deflection response from FE and numerical solutions agree well with the prediction of the analytical formula, demonstrating its accuracy. The analytical solution is then employed for the response of a two-layered composite membrane made of graphene deposited onto a soft substrate. This application is of great interest since new nanotechnologies make use of layered nanocomposites. Differently from our entirely nonlinear approach, most continuum models in the literature are based on the assumption of linear elastic material, which is suitable only when deformations are small. The present work gives a comprehensive description of the mechanics of pressurized graphene membranes.

Appendices

Appendix A: Analytical Solution with Transverse Contraction

After deformation, thickness $t$ transforms into $t' = \lambda _{Z} t$ and the deformation gradient reads [46]

$$ \left [\mathbf{F}\right ] = \left [ \textstyle\begin{array}{c@{\quad}c@{\quad}c} \rho \dfrac{\partial \psi}{\partial R} \cos \psi & 0 & \lambda _{Z} \sin \psi \\ 0 & \dfrac{\rho \sin \psi}{R} & 0 \\ -\rho \dfrac{\partial \psi}{\partial R} \sin \psi & 0 & \lambda _{Z} \cos \psi \end{array}\displaystyle \right ]. $$

Rotation tensor $\mathbf{R}$ remains unchanged and right stretch tensor $\mathbf{U}$ becomes

$$ \left [\mathbf{U}\right ] = \left [ \textstyle\begin{array}{c@{\quad}c@{\quad}c} \rho \dfrac{\partial \psi}{\partial R} & 0 & 0 \\ 0 & \dfrac{\rho \sin \psi}{R} & 0 \\ 0 & 0 & \lambda _{Z} \\ \end{array}\displaystyle \right ]. $$

The right Cauchy-Green deformation tensor and the Green-Lagrange strain tensor expressed in cylindrical coordinates assume the form

$$ \left [\mathbf{C}\right ] = \left [ \textstyle\begin{array}{c@{\quad}c@{\quad}c} \lambda _{R}^{2} & 0 & 0 \\ 0 & \lambda _{\Theta}^{2} & 0 \\ 0 & 0 & \lambda _{Z}^{2} \\ \end{array}\displaystyle \right ], \qquad \left [\mathbf{E}\right ] = \frac{1}{2} \left [ \textstyle\begin{array}{c@{\quad}c@{\quad}c} \lambda _{R}^{2}-1 & 0 & 0 \\ 0 & \lambda _{\Theta}^{2}-1 & 0 \\ 0 & 0 & \lambda _{Z}^{2}-1 \\ \end{array}\displaystyle \right ], $$

where $\lambda _{R}$ and $\lambda _{\Theta}$ are given by (2).

Strain invariants $I_{1}$ and $I_{2}$ become

$$ \begin{aligned} & I_{1} = \frac{1}{2} \left ( \lambda _{R}^{2} + \lambda _{\Theta}^{2} + \lambda _{Z}^{2} - 3 \right ), \\ & I_{2} = \frac{1}{4} \left ( \lambda _{R}^{2} -1 \right ) \left ( \lambda _{\Theta}^{2} -1 \right ) + \frac{1}{4} \left ( \lambda _{R}^{2} -1 \right ) \left ( \lambda _{Z}^{2} -1 \right ) + \frac{1}{4} \left ( \lambda _{\Theta}^{2} -1 \right ) \left ( \lambda _{Z}^{2} -1 \right ), \end{aligned} $$

(27)

while invariant $I_{3}$ has the same expression as in (5). The second Piola-Kirchhoff stress tensor is computed using (7), obtaining

$$ \begin{aligned} \Sigma _{\mathit{RR}} & = \frac{\beta _{1}}{t}+\frac{\beta _{2}}{2 t} \left ( \lambda _{R}^{2}-1\right )+\frac{3 \beta _{3}}{4 t} \left (\lambda _{R}^{2}- \lambda _{\Theta }^{2}\right )^{2} \cos (6 \phi ), \\ \Sigma _{\Theta \Theta} & = \frac{\beta _{1}}{t}+ \frac{\beta _{2}}{2 t} \left (\lambda _{\Theta }^{2}-1\right ) - \frac{3 \beta _{3}}{4 t} \left (\lambda _{R}^{2}-\lambda _{\Theta }^{2} \right )^{2} \cos (6 \phi ), \\ \Sigma _{\mathit{ZZ}} & = \frac{\beta _{1}}{t}+ \frac{\beta _{2}}{2 t} \left ( \lambda _{Z}^{2}-1\right ), \\ \Sigma _{R \Theta} & = - \frac{3 \beta _{3}}{4 t} \left (\lambda _{R}^{2}- \lambda _{\Theta }^{2}\right )^{2} \sin (6 \phi ), \quad \Sigma _{R Z} = \Sigma _{\Theta Z} = 0, \end{aligned} $$

(28)

where $\beta _{1}$, $\beta _{2}$ and $\beta _{3}$ are given in (8). The first Piola-Kirchhoff stress tensor is thus derived

$$ \left [ \mathbf{T}_{R} \right ] = \left [ \textstyle\begin{array}{c@{\quad}c@{\quad}c} \lambda _{R} \Sigma _{R R} \cos \psi & \lambda _{R} \Sigma _{R \Theta} \cos \psi & \lambda _{Z} \Sigma _{\mathit{ZZ}} \sin \psi \\ \lambda _{\Theta } \Sigma _{R \Theta} & \lambda _{\Theta } \Sigma _{ \Theta \Theta } & 0 \\ - \lambda _{R} \Sigma _{R R} \sin \psi & - \lambda _{R} \Sigma _{R \Theta} \sin \psi & \lambda _{Z} \Sigma _{\mathit{ZZ}} \cos \psi \\ \end{array}\displaystyle \right ]. $$

Condition $\mathbf{T}_{R} \mathbf{n}_{Z} = \mathbf{0}$ for the membrane stress state requires that

$$ \lambda _{Z} \Sigma _{\mathit{ZZ}} \sin \psi = 0 \quad \text{and} \quad \lambda _{Z} \Sigma _{\mathit{ZZ}} \cos \psi = 0, \quad \forall \psi \in \left [0,\pi /6\right ], $$

(29)

which are satisfied only if $\Sigma _{\mathit{ZZ}} = 0$. From this condition and recalling (28), we derive the following implicit expression of stretch $\lambda _{Z}$:

$$ \lambda _{Z} = \sqrt{1 - 2 \frac{\beta _{1}}{\beta _{2}}}. $$

(30)

Finally, the components of the Cauchy stress tensor read

$$ \textstyle\begin{array}{lll} T_{R R} = \dfrac{\lambda _{R}}{\lambda _{\Theta } \lambda _{Z}} \Sigma _{R R} \cos ^{2}\psi , \quad &T_{\Theta \Theta} = \dfrac{\lambda _{\Theta }}{\lambda _{R} \lambda _{Z}} \Sigma _{\Theta \Theta}, \quad &T_{Z Z} = \dfrac{\lambda _{R}}{\lambda _{\Theta} \lambda _{Z}} \Sigma _{R R} \sin ^{2}\psi , \\ T_{R \Theta} = \dfrac{1}{\lambda _{Z}} \Sigma _{R \Theta} \cos \psi , \quad & T_{R Z} = - \dfrac{\lambda _{R}}{\lambda _{\Theta} \lambda _{Z}} \Sigma _{R R} \cos \psi \sin \psi , \quad & T_{\Theta Z} = - \dfrac{1}{\lambda _{Z}} \Sigma _{R \Theta} \sin \psi , \end{array} $$

with $\lambda _{Z}$ expressed by (30).

As we did in Sect. 2, we now write the equilibrium in the neighborhood of the central point of the membrane. With this aim, we firstly compute the Cauchy stress tensor for the limit case of $R \rightarrow 0$. Radial and circumferential stretches are equal to each other ($\lambda _{R} = \left .\lambda _{\Theta}\right |_{R \rightarrow 0} = \lambda$) and third invariant $I_{3}$ goes to zero. The only non-zero components of the Cauchy stress tensor are

$$ \left .T_{R R}\right |_{{R \rightarrow 0}} = \left .T_{\Theta \Theta} \right |_{R \rightarrow 0} = T_{0} = \frac{\psi _{0}^{2} \csc \psi _{0}^{2} - \left .\lambda _{Z}^{2}\right |_{{R \rightarrow 0}}}{2 t \left .\lambda _{Z}\right |_{{R \rightarrow 0}}} \left .\beta _{2}\right |_{{R \rightarrow 0}}. $$

(31)

Hereinafter, for the sake of simplicity, we indicate $\left .\lambda _{Z}\right |_{{R \rightarrow 0}}$ simply with $\lambda _{Z}$. Equilibrium equation (13) becomes

$$ p = \frac{2 T_{0} \lambda _{Z} t}{\rho}, $$

(32)

from which, using (31), (8) and (27), we derive the following expression of the applied pressure:

$$ p = \frac{\sin \psi _{0}}{8 a} \left ( \psi _{0}^{2} \csc ^{2} \psi _{0} - \lambda _{Z}^{2} \right ) \sum _{j=0}^{3} \zeta _{j} \psi _{0}^{2 j} \csc ^{2 j} \psi _{0}, $$

(32)

where

$$ \begin{aligned} \zeta _{0} = & \; \lambda _{Z}^{2} \bigl[\lambda _{Z}^{2} \left (5 c_{10} \lambda _{Z}^{2}-c_{11} \lambda _{Z}^{2}-25 c_{10}-3 c_{11}+4 c_{12}\right )+8 c_{7} \left (\lambda _{Z}^{2}-4\right )-2 c_{9} \left (\lambda _{Z}^{2}+2\right ) \\ &{}-4 c_{6}+8 c_{8}+9 \left (5 c_{10}+3 c_{11}-2 c_{12}\right )\bigr] \\ & + 12 c_{5} \left (\lambda _{Z}^{2}-3\right )-8 c_{2}+16 c_{4}+3 \left (4 c_{6}+16 c_{7}-4 c_{8}+2 c_{9}-15 c_{10}-9 c_{11}+6 c_{12}\right ), \\ \zeta _{1} = & \; 2 \lambda _{Z}^{2} \left [3 c_{11} \left (\lambda _{Z}^{2}-8\right )-2 c_{12} \left (\lambda _{Z}^{2}-7\right )+5 c_{10} \left (\lambda _{Z}^{2}-4\right )-4 c_{8}+4 c_{9}\right ]+16 c_{7} \left (\lambda _{Z}^{2}-4\right ) \\ & + 24 c_{5}-8 c_{6}+16 c_{8}-8 c_{9}+90 c_{10}+54 c_{11}-36 c_{12}, \\ \zeta _{2} = & \; 2 \left [c_{12} \left (11-5 \lambda _{Z}^{2}\right )+5 c_{10} \left (\lambda _{Z}^{2}-7\right )+3 c_{11} \left (3 \lambda _{Z}^{2}-5\right )+12 c_{7}-2 c_{8}\right ], \\ \zeta _{3} = & \; 4 \left (5 c_{10}+c_{11}-c_{12}\right ). \end{aligned} $$

An explicit expression of stretch $\lambda _{Z}$ is derived by satisfying condition (29) for the limit case of $R \rightarrow 0$, for which $\lambda _{R} = \left .\lambda _{\Theta}\right |_{R \rightarrow 0} = \lambda$. In this case, substitution of (8) and (27) into (30) gives

$$ \sqrt{\eta} - \sqrt{\frac{\Delta _{n}}{\Delta _{d}}} = 0, $$

(33)

with $\eta = \lambda _{Z}^{2}$ and

$$\begin{aligned} \Delta _{n} = & -8 c_{2} \left (\eta +2 \lambda ^{2}-2\right ) +16 c_{4} +12 c_{5} \left [-\eta +2 \left (\eta -1\right ) \lambda ^{2}+ \lambda ^{4}\right ] \\ & -4 c_{6} \left [\eta ^{2}-7 \eta +2 \left (3 \eta -7\right ) \lambda ^{2}+5 \lambda ^{4}+9\right ] \\ & +8 c_{7} \left [-\eta ^{2}+5 \eta +\left (5 \eta -8\right ) \lambda ^{4}+2 \left (\eta -5\right ) \left (\eta -1\right ) \lambda ^{2}+2 \lambda ^{6}-3\right ] \\ & -4 c_{8} \left (\lambda -1\right ) \left ( \lambda +1\right ) \left (2 \eta +\lambda ^{2}-3\right ) \left (\eta +2 \lambda ^{2}-2\right ) \\ & -2 c_{9} \left [\eta ^{3}-4 \eta ^{2}+11 \eta +2 \left (\eta -7 \right ) \lambda ^{4}+2 \left (\eta ^{2} -6 \eta +11 \right ) \lambda ^{2}+4 \lambda ^{6}-12\right ] \\ & +5 c_{10} \bigl\{ -\eta ^{3}+6 \eta ^{2}-15 \eta +8 \left (\eta -2 \right ) \lambda ^{6}+\left [\eta \left (5 \eta -32\right )+33\right ] \lambda ^{4} \\ &+2 \left (\eta -1\right ) \left [\left (\eta -6\right ) \eta +15\right ] \lambda ^{2}+3 \lambda ^{8}+9\bigr\} \\ & -c_{11} \bigl\{ \eta \left [\left (\eta -5\right ) \eta ^{2}+9 \right ]+8 \left (1-2 \eta \right ) \lambda ^{6}-3 \left [5 \left ( \eta -4\right ) \eta +9\right ] \lambda ^{4} \\ &+2 \left [\eta \left ( \eta ^{2}+9 \eta -27\right )+9\right ] \lambda ^{2}+\lambda ^{8} \bigr\} \\ & -c_{12} \left (\lambda -1\right ) \left (\lambda +1\right ) \left (2 \eta +\lambda ^{2}-3\right ) \left [2 \left (\eta -6\right ) \eta +10 \eta \lambda ^{2}+9 \lambda ^{4}-24 \lambda ^{2}+15\right ], \\ \Delta _{d} = & -8 c_{2}+16 c_{4} +12 c_{5} \left (\eta +2 \lambda ^{2}-3 \right )-4 c_{6} \left (\eta +2 \lambda ^{2}-3\right ) \\ & + 8 c_{7} \left [\eta ^{2}+2 (\eta -4) \lambda ^{2}-4 \eta +3 \lambda ^{4}+6 \right ] \\ & -4 c_{8} \left (\lambda ^{2}-1\right ) \left (2 \eta +\lambda ^{2}-3 \right )-2 c_{9} \left (\eta -1\right ) \left (\eta -4 \lambda ^{2}+3 \right ) \\ & +5 c_{10} \left (\eta +2 \lambda ^{2}-3\right ) \left [ \left (\eta -2\right ) \eta +2 \lambda ^{4}-4 \lambda ^{2}+3\right ] \\ & -c_{11} \left (\eta +2 \lambda ^{2}-3\right ) \left [\eta ^{2}+4 (3-2 \eta ) \lambda ^{2}+6 \eta -2 \lambda ^{4}-9\right ] \\ & -2 c_{12} \left ( \lambda -1\right ) \left (\lambda +1\right ) \left (2 \eta +\lambda ^{2}-3 \right ) \left (\eta +2 \lambda ^{2}-3\right ). \end{aligned}$$

Equation (33) admits four solutions in $\eta$. Two solutions are not real and another one does not respect condition $\eta = 1$ when $\lambda = 1$. The remaining solution is the correct one. We do not report this solution due to its very long mathematical expression. Having obtained an explicit expression for $\eta$, we compute $\lambda _{Z} = \sqrt{\eta}$ and by substitution into (32) we finally derive the pressure-deflection equation. We recall that relation $\psi _{0} = 2 \arctan \bar{\delta}$ allows us to obtain a direct expression of pressure as a function of deflection.

Appendix B: Fichter’s Model

Fichter’s model is based on the assumption that the material is linearly elastic. In this case, the equilibrium equations are

$$ \begin{aligned} & N^{2} \left ( \bar{R}^{2} \frac{d^{2} N}{d \bar{R}^{2}} + 3 \bar{R} \frac{d N}{d \bar{R}} \right ) - \frac{1}{2} \bar{R}^{3} \frac{d N}{d \bar{R}} + \frac{1}{2} \left ( 3 + \nu \right ) \bar{R}^{2} N + \frac{1}{4} \frac{\bar{R}^{2} E H}{p L} = 0, \\ & N \frac{d \bar{\delta}}{d \bar{R}} + \frac{1}{2} \bar{R} = 0, \end{aligned} $$

(34)

where $\bar{R} = R/a$, $E$ is the Young’s modulus, $\nu$ is the Poisson’s ratio and $N = N_{R}/(p a)$, with $N_{R}$ indicating the radial stress resultant. Young’s modulus and Poisson’s ratio of graphene are computed using (18), obtaining $E = 1042.9$ GPa and $\nu = 0.146$. The solution for both stress resultant and deflection is found in the form of a power series

$$ \begin{aligned} & N \left ( \bar{R} \right ) = \sum _{0}^{\infty }n_{2 m} \bar{R}^{2 m}, \\ & \bar{\delta} \left ( \bar{R} \right ) = \sum _{0}^{\infty }w_{2 n} \left ( 1 - \bar{R}^{2 n + 2} \right ). \end{aligned} $$

(35)

Substituting (35)₁ into (34)₁ and equating coefficients of like powers of $\bar{R}$ we obtain a system of equations that allows to derive the expressions of coefficients $n_{2 m}$ as functions of $n_{0}$. Likewise, substituting (35)₂ into (34)₂ and equating coefficients of like powers of $\bar{R}$ we derive the expressions of coefficients $w_{2 n}$ as functions of $n_{0}$. Finally, $n_{0}$ is evaluated by imposing the following boundary condition on radial displacement:

$$ \left .\left \{ \bar{R} \left [ \frac{d}{d \bar{R}} \left ( \bar{R} N \right ) - \nu N - \bar{R} \frac{d \bar{\delta}}{d \bar{R}} \right ] \right \}\right |_{\bar{R}=1} = 0. $$

(36)

This procedure was implemented in software Wolfram Mathematica. A vector of increasing pressure values was defined and, for each value, the solution was obtained by considering twelve terms in the power series ($m = 12$ and $n = 12$). More terms did not cause sensible variations in the solutions and only increased the computational burden.