Size-dependent analysis of a functionally graded piezoelectric micro-cylinder based on the strain gradient theory with the consideration of flexoelectric effect: plane strain problem

Abstract

In this study, a size-dependent analysis of functionally graded piezoelectric (FGP) micro-rotating cylinder is presented based on the plane strain condition and strain gradient theory, which is a non-classical theory capable of capturing the size effect in microscaled structures. The present model is used to analyze the FGP micro-rotating cylinder with the consideration of flexoelectric effects exposed to a symmetric magneto-electro-mechanical loading. All mechanical and electrical properties are assumed to be graded in the thickness direction according to a power-law distribution. With respect to the fifth-order strain gradient coefficient and electromechanical coupling, the constitutive equations are obtained from electric Gibbs free energy density, which is a function of strain, second-order deformation gradient and electric field. By substituting the constitutive equations in electric and mechanical equilibrium equations, two coupled electromechanical governing differential equations in terms of radial displacement and electric potential are derived considering centrifugal force and Lorentz magnetic force obtained from Maxwell’s relations. The generalized differential quadrature method is proposed to solve the coupled governing differential equations. Numerical results attained from the strain gradient elasticity reveal the effects of flexoelectric, microstructural length scale, non-homogeneity constant, rotation and magnetic field on the response of the FGP micro-rotating cylinder.

Appendix

It is shown here that the total stress tensor has the following expression (see Eq. 13):

$$ \begin{aligned}\varvec{\sigma}& = \left\{ {\left( {1 - \frac{{\ell_{r} }}{r}} \right)A_{rr} - L^{2} \left[ {A_{rr}^{{{\prime \prime }}} + \frac{{A_{rr}^{{\prime }} }}{r} - \frac{2}{{r^{2} }}\left( {A_{rr} - A_{\theta \theta } } \right)} \right] - e_{11} E_{r} + \left[ {f_{11} E_{r}^{{\prime }} + \left( {f_{11} - 2f_{46} } \right)\frac{{E_{r} }}{r}} \right]} \right\}e_{r} \otimes e_{r} \\ & \quad + \;\left\{ {\left( {1 - \frac{{\ell_{r} }}{r}} \right)A_{\theta \theta } - L^{2} \left[ {A_{\theta \theta }^{{{\prime \prime }}} + \frac{{A_{\theta \theta }^{{\prime }} }}{r} + \frac{2}{{r^{2} }}\left( {A_{rr} - A_{\theta \theta } } \right)} \right] - e_{21} E_{r} + \left[ {f_{15} E_{r}^{{\prime }} + \left( {f_{15} + 2f_{46} } \right)\frac{{E_{r} }}{r}} \right]} \right\}e_{\theta } \otimes e_{\theta } . \\ \end{aligned} $$

(37)

Proof

A new second-order tensor $\varvec{A}$ is defined for the sake of simplicity as:

$$ A_{ij} = C_{ijkl} \varepsilon_{kl} = A_{rr} e_{r} \otimes e_{r} + A_{\theta \theta } e_{\theta } \otimes e_{\theta } . $$

(38)

Substituting Eq. (38) into Eq. (6d), the total stress tensor is expressed in terms of components of the second-order tensor $\varvec{A}$ as follows.

$$ \sigma_{ij} = A_{ij} - L^{2} A_{ij,pp} - l_{p,p} A_{ij} - e_{kij} E_{k} + \left( {f_{lijp} E_{l} } \right)_{,p} $$

(39)

Therefore, it follows from Eq. (39) that:

$$ \begin{aligned} A_{ij,p} & = \nabla \varvec{A} = \varvec{A} \otimes \nabla = \left( {A_{rr} e_{r} \otimes e_{r} + A_{\theta \theta } e_{\theta } \otimes e_{\theta } } \right) \otimes \left( {\frac{\partial }{\partial r}e_{r} + \frac{1}{r}\frac{\partial }{\partial \theta }e_{\theta } } \right) \\ & = A_{rr}^{{\prime }} e_{r} \otimes e_{r} \otimes e_{r} + \left( {\frac{{A_{rr} - A_{\theta \theta } }}{r}} \right)e_{r} \otimes e_{\theta } \otimes e_{\theta } + \left( {\frac{{A_{rr} - A_{\theta \theta } }}{r}} \right)e_{\theta } \otimes e_{r} \otimes e_{\theta } + A_{\theta \theta }^{{\prime }} e_{\theta } \otimes e_{\theta } \otimes e_{r} . \\ \end{aligned} $$

(40)

This gives the gradient of the second-order tensor $\varvec{A}$ in the cylindrical coordinate system. Using Eq. (40), the Laplacian of the second-order tensor $\varvec{A}$ is calculated as:

$$ \begin{aligned} A_{ij,pp} = \nabla^{2} \varvec{A} = \nabla \cdot \nabla \varvec{A} &= \left[ {A_{rr}^{{{\prime \prime }}} + \frac{{A_{rr}^{{\prime }} }}{r} - \frac{2}{{r^{2} }}\left( {A_{rr} - A_{\theta \theta } } \right)} \right]e_{r} \otimes e_{r} \\ & \quad + \left[ {A_{\theta \theta }^{{{\prime \prime }}} + \frac{{A_{\theta \theta }^{{\prime }} }}{r} + \frac{2}{{r^{2} }}\left( {A_{rr} - A_{\theta \theta } } \right)} \right]e_{\theta } \otimes e_{\theta } .\end{aligned} $$

(41)

Also, the divergence of vector $\varvec{l}$ is written as:

$$ l_{p,p} = \nabla \cdot \varvec{l} = \left( {\frac{\partial }{\partial r}e_{r} + \frac{1}{r}\frac{\partial }{\partial \theta }e_{\theta } } \right) \cdot \left( {l_{r} e_{r} + l_{\theta } e_{\theta } } \right) = \frac{{l_{r} }}{r}. $$

(42)

Moreover, the last term of Eq. (39) can be calculated using tensor notation in the curvilinear coordinates as follows:

$$ \left( {f_{lijp} E_{l} } \right)_{,p} = \left[ {f_{1111} E_{r}^{{\prime }} + \left( {f_{1111} - f_{1221} - f_{1212} } \right)\frac{{E_{r} }}{r}} \right]e_{r} \otimes e_{r} + \left[ {f_{1122} E_{r}^{{\prime }} + \left( {f_{1122} + f_{1221} + f_{1212} } \right)\frac{{E_{r} }}{r}} \right]e_{\theta } \otimes e_{\theta } . $$

(43)

It is important to be mentioned that there are four independent flexoelectric coefficients (regarding only two coordinates $1 = r$ and $2 = \theta$), which can be written as $f_{1111} = f_{11}$, $f_{1122} = f_{2211} = f_{15}$, $f_{1221} = f_{1212} = f_{46}$ according to [86]. Thus, Eq. (43) is simplified as:

$$ \left( {f_{lijp} E_{l} } \right)_{,p} = \left[ {f_{11} E_{r}^{{\prime }} + \left( {f_{11} - 2f_{46} } \right)\frac{{E_{r} }}{r}} \right]e_{r} \otimes e_{r} + \left[ {f_{15} E_{r}^{{\prime }} + \left( {f_{15} + 2f_{46} } \right)\frac{{E_{r} }}{r}} \right]e_{\theta } \otimes e_{\theta } $$

(44)

where superscript “′” denotes first derivative with respect to radial coordinate $r$. Substituting Eqs. (38), (41), (42) and (44) into Eq. (39) will give Eq. (37).

Similarly, the couple stress tensor in terms of components of the second-order tensor $\varvec{A}$ can be rewritten using Eq. (6c) as:

$$ \mu_{ijp} = L^{2} A_{ij,p} + l_{p} A_{ij} - f_{lijp} E_{l} $$

(45)

Substituting Eqs. (38) and (40) into Eq. (45), the third-order tensor $\varvec{\mu}$ can be determined as follows:

$$ \begin{aligned}\varvec{\mu}& = \left( {L^{2} A_{rr}^{{\prime }} + \ell_{r} A_{rr} - f_{11} E_{r} } \right)e_{r} \otimes e_{r} \otimes e_{r} + \left[ {L^{2} \left( {\frac{{A_{rr} - A_{\theta \theta } }}{r}} \right) - f_{46} E_{r} } \right]e_{r} \otimes e_{\theta } \otimes e_{\theta } \\ & \quad + \;\left[ {L^{2} \left( {\frac{{A_{rr} - A_{\theta \theta } }}{r}} \right) - f_{46} E_{r} } \right]e_{\theta } \otimes e_{r} \otimes e_{\theta } + \left( {L^{2} A_{\theta \theta }^{{\prime }} + \ell_{r} A_{\theta \theta } - f_{15} E_{r} } \right)e_{\theta } \otimes e_{\theta } \otimes e_{r} . \\ \end{aligned} $$

(46)

Also, the divergence of the third-order tensor $\varvec{ \mu }$ can be calculated as below.

$$ \mu_{ijp,p} = \nabla \cdot\varvec{\mu}= \left[ {\frac{{\partial \mu_{rrr} }}{\partial r} + \frac{{\mu_{rrr} }}{r} - \frac{{\mu_{r\theta \theta } }}{r} - \frac{{\mu_{\theta r\theta } }}{r}} \right]e_{r} \otimes e_{r} + \left[ {\frac{{\partial \mu_{\theta \theta r} }}{\partial r} + \frac{{\mu_{r\theta \theta } }}{r} + \frac{{\mu_{\theta \theta r} }}{r} + \frac{{\mu_{\theta r\theta } }}{r}} \right]e_{\theta } \otimes e_{\theta } $$

(47)

Therefore, the stress components are obtained using Eqs. (6) and (47):

$$ \begin{aligned} \sigma_{ij} & = \tau_{ij} - \mu_{ijp,p} = \left[ {\tau_{rr} - \left( {\frac{{\partial \mu_{rrr} }}{\partial r} + \frac{{\mu_{rrr} }}{r} - \frac{{\mu_{r\theta \theta } }}{r} - \frac{{\mu_{\theta r\theta } }}{r}} \right)} \right]e_{r} \otimes e_{r} \\ & \quad + \;\left[ {\tau_{\theta \theta } - \left( {\frac{{\partial \mu_{\theta \theta r} }}{\partial r} + \frac{{\mu_{r\theta \theta } }}{r} + \frac{{\mu_{\theta \theta r} }}{r} + \frac{{\mu_{\theta r\theta } }}{r}} \right)} \right]e_{\theta } \otimes e_{\theta } \\ \end{aligned} $$

(48)

and the mechanical boundary conditions utilizing Eqs. (26) and (47).

$$ \begin{aligned} \left[ {\mu_{rrr} } \right]_{r = a, b} & = 0 \\ \left\{ {\tau_{rr} + \frac{{\mu_{r\theta \theta } }}{r} + \frac{{\mu_{\theta \theta r} }}{r} + \frac{{\mu_{\theta r\theta } }}{r} - \frac{{\mu_{rrr} }}{r} - \frac{{\partial \mu_{rrr} }}{\partial r}} \right\}_{r = a, b} & = - P_{i} , P_{o} \\ \end{aligned} $$

(49)

It is worthwhile to express that Eqs. (48) and (49) are another form of Eqs. (12) and (27) in order to derive stress components and mechanical boundary conditions.