# Natural Frequency Response of FG-CNT Coupled Curved Beams in Thermal Conditions
## 3 Theory and Formulation  
The theoretical background of the selected in-plane vibration problem stems from the FSDT curved beam assumptions, while neglecting the deflections and rotations, and assuming a distribution of linear elastic normal springs at the interfaces. Thus, the kinematic field of the reference layer for each curved beam is defined as  
\[ \begin{array}{rcl}D_{1}^{i}\,(x_{i},z,t)&=&d_{1}^{i}\,(x_{i},t)+z\,\Phi_{i} \,(x_{i},t)\\ D_{2}^{i}\,(x_{i},z,t)&=&d_{2}^{i}\,(x_{i},t)\end{array}  \tag{10}\]  
where \(D_{1}^{i}\) and \(D_{2}^{i}\) define the displacement components of an arbitrary point in the longitudinal \((x_{i})\) and thickness \((z)\) directions, respectively, whereas \(d_{1}^{i}\) and \(d_{2}^{i}\) refer to the displacement field of the neutral axis in the longitudinal \((x_{i})\) and thickness \((z)\) directions, respectively. For details about the geometrical definition of this system in its bottom and upper parts can be found in [1]. The relation between the strain components of an arbitrary point and the reference neutral strain field is defined as  
\[ \begin{array}{c}\left\{\begin{matrix}\varepsilon_{x}^{i}\\ \gamma_{xz}^{i}\end{matrix}\right\}=\left[\begin{matrix}\phi_{1}^{j}&\phi_{2}^ {j}&0\\ 0&0&\phi_{3}^{j}\end{matrix}\right]\begin{cases}\varepsilon_{x}^{i0}\\ \eta_{xz}^{i0}\\ \gamma_{xz}^{i0}\end{cases}\end{array}  \tag{11}\]  
with  
\[\phi_{1}^{i}=\frac{1}{1+Y_{i}}\,\,\phi_{2}^{i}=\frac{z}{1+Y_{i}} \tag{12}\]  
where \(\varepsilon_{x}^{i}\) and \(\gamma_{xz}^{i}\) are normal and shear strain components, \(\varepsilon_{x}^{i0}\) and \(\gamma_{xz}^{i0}\) are the normal and shear neutral strain components, and \(\eta_{x}^{i}\) represent the curvature of beams on the reference surface, with \(Y_{i}=\frac{z}{K_{i}}\). Based on the FDST assumptions for cylindrical panels, the kinematic relations are defined for each \(i\)th beam as  
\[ \begin{array}{c}\left\{\begin{matrix}\varepsilon_{x}^{i0}\\ \eta_{x}^{i}\\ \gamma_{xz}^{i0}\end{matrix}\right\}=\left\{\begin{matrix}\frac{\partial}{ \partial x_{i}}&\frac{1}{K_{i}}&0\\ -\frac{\partial}{K_{i}}&\frac{\partial}{\partial x_{i}}&1\\ 0&0&\frac{\partial}{\partial x_{i}}\end{matrix}\right\}\left\{\begin{matrix}d_ {1}^{i}\\ \Phi_{i}\\ d_{2}^{i}\end{matrix}\right\}\end{array}  \tag{13}\]For an FG material, we consider the following constitutive relations:  
\[\left\{ \begin{array}{c}\sigma_{x}^{i}\\ \tau_{xx}^{i}\end{array} \right\}=\left[ \begin{array}{cc}E(z)&0\\ 0&\frac{E(z)}{2\left(1+\
u\left(z\right)\right)}\end{array} \right]\left\{ \begin{array}{c}\varepsilon_{x}^{i}\\ \tau_{xx}^{j}\end{array} \right\} \tag{14}\]  
where \(\sigma_{x}^{i}\) is the normal stress, \(\tau_{xx}^{i}\) is the shear stress, and \(E\) and \(\
u\) are elastic modulus and Poisson's ratio of the beam, respectively. Therefore, forces and moments can be calculated by employing the integration of stresses in the \(z\)-direction. \[\left[ \begin{array}{c}N_{x}^{i}\\ Q_{x}^{i}\\ M_{x}^{i}\end{array} \right]=b_{i}\int\limits_{ \begin{array}{c}-\frac{b_{i}} {2}\\ -\frac{b_{i}}{2}\end{array} }^{\frac{b_{i}}{2}}\left\{ \begin{array}{c}\sigma_ {x}^{i}\\ \tau_{xx}^{i}\end{array} \right\}dz \tag{15}\]  
Forces and moments can be, thus, defined as  
\[ \begin{array}{c}N_{x}^{i}=\varepsilon_{x0}^{i0}A_{11}^{i}+\ \eta_{x}^{i}\ B_{11}^{i}\\ Q_{x}^{i}=\gamma_{xx}^{i0}A_{55}^{i0}\\ M_{x}^{i}=\varepsilon_{x}^{i0}B_{11}^{i}+\eta_{x}^{i}\ D_{11}^{i}\end{array}  \tag{16}\]  
in which  
\[ \begin{array}{c}A_{11}^{i}=\ b_{i}\int\limits_{\begin{array}{c}\frac{b_{i} }{2}\\ \end{array}}^{\frac{b_{i}}{2}}\phi_{1}^{i}E(z)dz,\ B_{11}^{i}=\ b_{i}\int \limits_{\begin{array}{c}\frac{b_{i}}{2}\\ -\frac{b_{i}}{2}\end{array}}^{\frac{b_{i}}{2}}\phi_{2}^{i}E(z)dz\\ D_{11}^{i}=\ b_{i}\int\limits_{\begin{array}{c}\frac{b_{i}}{2}\\ -\frac{b_{i}}{2}\end{array}}^{\frac{b_{i}}{2}}\phi_{3}^{i}E(z)dz,\ A_{55}^{i}= \kappa\ b_{i}\int\limits_{\begin{array}{c}\frac{b_{i}}{2}\\ -\frac{b_{i}}{2}\end{array}}^{\frac{b_{i}}{2}}\phi_{1}^{i}\frac{E(z)}{2(1+\
u )}\,dz\end{array}