## 3 Statement of the Problem
Equation (13), when it is inserted into Equation (11), yields  
\[EI\frac{\partial^{4}w}{\partial x^{4}}+\rho A\frac{\partial^{2}w}{\partial t^ {2}}+\big{(}\sigma_{0}A-K_{p}\big{)}\frac{\partial^{2}w}{\partial x^{2}}+K_{ S}w+K_{C}\frac{\partial w}{\partial t}+\alpha_{T}EI\frac{\partial^{2}M_{T}}{ \partial x^{2}}=0 \tag{14}\]  
The following is a generalized form of the Moore-Gibson-Thompson heat transfer Equation (6), which may also be written as [25; 27]:  
\[\bigg{(}1+\tau_{0}\frac{\partial}{\partial t}\bigg{)}\bigg{[}\rho C_{E}\frac {\partial^{2}\theta}{\partial t^{2}}-z\alpha_{T}T_{0}E\frac{\partial^{2}}{ \partial t^{2}}\bigg{(}\frac{\partial^{2}w}{\partial x^{2}}\bigg{)}-\rho \frac{\partial Q}{\partial t}\bigg{]}=\bigg{(}K^{*}+K\frac{\partial}{\partial t }\bigg{)}\bigg{(}\frac{\partial^{2}\theta}{\partial x^{2}}+\frac{\partial^{2} \theta}{\partial z^{2}}\bigg{)} \tag{15}\]  
The temperature increase for the present microbeam is supposed to vary in the following form:  
\[e_{xx}=e=-z\frac{\partial^{2}w}{\partial x^{2}} \tag{8}\]0  
Consequently, we obtain this by inserting Equation (16) into Equations (13) and (14):  
\[e_{xx}=e=-z\frac{\partial^{2}w}{\partial x^{2}} \tag{8}\]1  
\[e_{xx}=e=-z\frac{\partial^{2}w}{\partial x^{2}} \tag{8}\]2After multiplying by \(12z/h^{3}\), the result of doing an integration concerning z along the microbeam thickness using Equation (15) is shown below:  
\[ \begin{array}{c}\left(1+\tau_{0}\frac{\partial}{\partial t}\right)\left[\frac{ \rho C_{E}}{K}\frac{\partial^{2}\Theta}{\partial t^{2}}-\frac{\pi_{T}h\pi^{2}T _{0}E}{24K\eta_{0}}\frac{\partial^{2}}{\partial t^{2}}\left(\frac{\partial^{2 }w}{\partial x^{2}}\right)\right]=\left(K^{*}+K\frac{\partial}{\partial t} \right)\left(\frac{\partial^{2}}{\partial x^{2}}-\frac{\pi^{2}}{h^{2}}\right) \Theta\\ +\frac{\pi^{2}}{2K\eta^{2}}\left(1+\tau_{0}\frac{\partial}{\partial t}\right) \int_{-h/2}^{h/2}z\frac{\partial\Omega}{\partial t}dz\end{array}  \tag{19}\]  
We now offer the non-dimensional variables that are stated as follows. \[ \begin{array}{c}\left\{x^{\prime},z^{\prime},u^{\prime},w^{\prime},L^{ \prime},h^{\prime}\right\}=c_{0}\eta_{0}\{x,z,u,w,L,h\},\{t^{\prime},\tau_{ 0}^{\prime},\tau_{d}^{\prime}\}=c_{0}^{2}\eta_{0}\{t,\tau_{0},\tau_{d}\}\\ \Theta^{\prime}=\frac{\Theta}{\tau_{0}^{\prime}},\sigma_{x}^{\prime}=\frac{ \sigma_{x}}{L},Q^{\prime}=\frac{Q}{c_{0}^{2}\eta_{0}KT_{0}},M^{\prime}=\frac{ M}{c_{0}\eta_{0}EI},c_{0}=\sqrt{\frac{E}{\rho}},\eta_{0}=\frac{\rho C_{E}}{K}\end{array}  \tag{20}\]  
When it comes to these non-dimensional values (after removing the primes), the equations that control the situation consist of the following:  
\[ \begin{array}{c}\frac{\partial^{4}w}{\partial x^{4}}+\frac{12}{h^{2}}\frac{ \partial^{2}w}{\partial t^{2}}+\frac{1}{h^{2}}(\sigma_{0}-K_{2})\frac{ \partial^{2}w}{\partial x^{2}}+\frac{K_{1}}{h^{2}}w+\frac{K_{0}}{h^{2}}\frac{ \partial w}{\partial t}+\frac{24\pi_{T}}{h\pi^{2}}\frac{\partial^{2}\Theta}{ \partial x^{2}}=0\\ \\ \left(1+\tau_{0}\frac{\partial}{\partial t}\right)\left[\frac{\partial^{2} \Theta}{\partial t^{2}}-\frac{\pi_{T}h\pi^{2}E}{24K\eta_{0}}\frac{\partial^{2 }}{\partial t^{2}}\left(\frac{\partial^{2}w}{\partial x^{2}}\right)\right]= \left(\frac{K^{*}}{c_{0}^{2}\eta_{0}K}+\frac{\partial}{\partial t}\right) \left(\frac{\partial^{2}}{\partial x^{2}}-\frac{\pi^{2}}{h^{2}}\right)\Theta \\ +\frac{\pi^{2}}{2}\left(1+\tau_{0}\frac{\partial}{\partial t}\right)\int_{-h/2 }^{h/2}z\frac{\partial\Omega}{\partial t}dz\\ \\ M(x,t)=-\left[\frac{\partial^{2}w}{\partial x^{2}}+\frac{24T_{0}a\tau_{T}}{h \pi^{2}}\Theta\right]\\ \\ \sigma_{x}=-\left[z\frac{\partial^{2}w}{\partial x^{2}}+T_{0}a\tau_{T}\Theta \right]\end{array}  \tag{22}\]  
where \(K_{0}\), \(K_{1}\), and \(K_{2}\) each stand for the amount of time it takes the elastic substrate to mechanically relax due to the viscoelasticity, the Winkler model, and the shear stiffness of the elastic substrate, respectively.