## 2 Theory
and the two differential equations, Equations (6) and (7), the bending moment and axial and shear forces can be obtained as follows [26; 27]:  
\[E_{f}A_{f}\frac{\partial^{2}u(x,t)}{\partial x^{2}}-\frac{2k_{c}A_{c}G_{c}}{h_{c }^{2}}u(x,t)-\frac{k_{c}A_{c}G_{c}}{h_{c}}\bigg{(}1+\frac{h_{f}}{h_{c}}\bigg{)} \frac{\partial w(x,t)}{\partial x}-m_{f}\frac{\partial^{2}u(x,t)}{\partial t ^{2}}=0, \tag{6}\]  
and  
\[E_{f}I_{f}\frac{\partial^{4}w(x,t)}{\partial x^{4}}-\frac{k_{c}A_{c}G_{c}}{2} \bigg{(}1+\frac{h_{f}}{h_{c}}\bigg{)}^{2}\frac{\partial^{2}w(x,t)}{\partial x ^{2}}-\frac{k_{c}A_{c}G_{c}}{h_{c}}\bigg{(}1+\frac{h_{f}}{h_{c}}\bigg{)}\frac{ \partial u(x,t)}{\partial x}+m_{f}\frac{\partial^{2}w(x,t)}{\partial t^{2}}=0. \tag{7}\]  
_Axial force_,  
\[F(x)=-2E_{f}A_{f}\frac{\partial u(x,t)}{\partial x}. \tag{8}\]  
_Bending moment_,  
\[M(x)=-2E_{f}I_{f}\frac{\partial^{2}w(x,t)}{\partial x^{2}}. \tag{9}\]  
_Shear force_,  
\[\mathcal{U}_{c}=\frac{1}{2}k_{c}A_{c}G_{c}\int_{0}^{L}\bigg{\{}\frac{2}{h_{c} }u(x,t)+\bigg{(}1+\frac{h_{f}}{h_{c}}\bigg{)}\frac{\partial w(x,t)}{\partial x }\bigg{\}}^{2}dx, \tag{2}\]0  
By assuming harmonic vibration with angular frequency \((\omega)\), the following is obtained  
\[\mathcal{U}_{c}=\frac{1}{2}k_{c}A_{c}G_{c}\int_{0}^{L}\bigg{\{}\frac{2}{h_{c} }u(x,t)+\bigg{(}1+\frac{h_{f}}{h_{c}}\bigg{)}\frac{\partial w(x,t)}{\partial x }\bigg{\}}^{2}dx, \tag{2}\]1  
By substituting Equation (11) into Equations (6) and (7), the differential equations can be rewritten as  
\[\mathcal{U}_{c}=\frac{1}{2}k_{c}A_{c}G_{c}\int_{0}^{L}\bigg{\{}\frac{2}{h_{c} }u(x,t)+\bigg{(}1+\frac{h_{f}}{h_{c}}\bigg{)}\frac{\partial w(x,t)}{\partial x }\bigg{\}}^{2}dx, \tag{2}\]2  
\[\mathcal{U}_{c}=\frac{1}{2}k_{c}A_{c}G_{c}\int_{0}^{L}\bigg{\{}\frac{2}{h_{c} }u(x,t)+\bigg{(}1+\frac{h_{f}}{h_{c}}\bigg{)}\frac{\partial w(x,t)}{\partial x }\bigg{\}}^{2}dx, \tag{2}\]3  
where  
\[\mathcal{U}_{c}=\frac{1}{2}k_{c}A_{c}G_{c}\int_{0}^{L}\bigg{\{}\frac{2}{h_{c} }u(x,t)+\bigg{(}1+\frac{h_{f}}{h_{c}}\bigg{)}\frac{\partial w(x,t)}{\partial x }\bigg{\}}^{2}dx, \tag{2}\]4  
The two differential equations can be expressed in a sixth-order differential equation:  
\[\mathcal{U}_{c}=\frac{1}{2}k_{c}A_{c}G_{c}\int_{0}^{L}\bigg{\{}\frac{2}{h_{c} }u(x,t)+\bigg{(}1+\frac{h_{f}}{h_{c}}\bigg{)}\frac{\partial w(x,t)}{\partial x }\bigg{\}}^{2}dx, \tag{2}\]5  
where  
\[\mathcal{U}_{c}=\frac{1}{2}k_{c}A_{c}G_{c}\int_{0}^{L}\bigg{\{}\frac{2}{h_{c} }u(x,t)+\bigg{(}1+\frac{h_{f}}{h_{c}}\bigg{)}\frac{\partial w(x,t)}{\partial x }\bigg{\}}^{2}dx, \tag{2}\]6  
and \(\Psi(x)=U(x)\)_or_\(W(x)\). If the solution of Equation (14) is assumed to be of the form  
\[\mathcal{U}_{c}=\frac{1}{2}k_{c}A_{c}G_{c}\int_{0}^{L}\bigg{\{}\frac{2}{h_{c} }u(x,t)+\bigg{(}1+\frac{h_{f}}{h_{c}}\bigg{)}\frac{\partial w(x,t)}{\partial x }\bigg{\}}^{2}dx, \tag{2}\]7  
then, a sixth-order polynomial in \(\lambda\) can be obtained by substituting Equation (15) into Equation (14):  
\[\mathcal{U}_{c}=\frac{1}{2}k_{c}A_{c}G_{c}\int_{0}^{L}\bigg{\{}\frac{2}{h_{c} }u(x,t)+\bigg{(}1+\frac{h_{f}}{h_{c}}\bigg{)}\frac{\partial w(x,t)}{\partial x }\bigg{\}}^{2}dx, \tag{2}\]8To determine the solution of Equation (16), this polynomial can be transformed into a third-order polynomial as follows:  
\[\mathcal{U}_{c}=\frac{1}{2}k_{c}A_{c}G_{c}\int_{0}^{L}\bigg{\{}\frac{2}{h_{c} }u(x,t)+\bigg{(}1+\frac{h_{f}}{h_{c}}\bigg{)}\frac{\partial w(x,t)}{\partial x }\bigg{\}}^{2}dx, \tag{2}\]9  
where  
\[\mathcal{U}=\mathcal{U}_{f}+\mathcal{U}_{c} \tag{3}\]0  
Thus, the displacements \(U(x)\) and \(W(x)\) can be defined by  
\[\mathcal{U}=\mathcal{U}_{f}+\mathcal{U}_{c} \tag{3}\]1  
and  
\[\mathcal{U}=\mathcal{U}_{f}+\mathcal{U}_{c} \tag{3}\]2  
where \(H_{j}\) and \(Q_{j}\) are the different constants.