# Size-Dependent Free Vibration of Axially Moving Nanobeams Using [PERSON]'s Two-Phase Integral Model
## 2 Model Equations
(**a**) full graph, (**b**) enlargement partial of full graph. Figure 2: The critical velocity versus volume fraction \(\xi_{1}\) when \(p=0.04,\alpha=40,k_{s}=0.01\). Black dotted line, red dotted line and blue dotted line denote the results for \(k=3,2,1\) mode, respectively. (**a**) full graph, (**b**) enlargement partial of full graph. We can determine the critical velocity for the fixed boundary by a similar procedure. The critical velocity in this case can be determined by the following equation:  
\[2\cos(\lambda_{1})+\lambda_{1}\sin(\lambda_{1})-2=0. \tag{21}\]  
The effects of parameters \(\alpha,\xi_{1}\) and \(k_{s}\) on the critical velocity are obtained by the numerical method. The results are plotted in the following Figures 4-6. From these figures, we find that the difference of critical velocity between the results of simply supported boundary condition and the fixed boundary condition is very small. In other words, it implies that the critical velocity is almost independent of the boundary conditions for the axially moving beam. Figure 4: The critical velocity versus geometric parameter \(\alpha\) when \(p=0.04,k_{s}=0.01,\xi_{1}=0.5\) under fixed boundary condition. Black dotted line, red dotted line and blue dotted line denote the results for \(k=3,2,1\) mode, respectively. (**a**) full graph, (**b**) enlargement partial of full graph. Figure 3: The critical velocity versus geometric parameter \(k_{s}\) when \(p=0.04,\alpha=40,\xi_{1}=0.5\). Black dotted line, red dotted line and blue dotted line denote the results for \(k=3,2,1\) mode, respectively. In (**a**), the bottom of the three dotted lines denote the critical velocity of the classical results. (**b**) enlargement partial of full graph for nonlocal model. where \(C_{ni}\) are constants to be determined and \(\beta_{nj}(j=1,2,3,4)\) are the roots of the following equation  
\[(x\gamma+\omega_{n})^{2}-x^{2}(p+k_{s}(x^{2}\xi_{1}+2\alpha^{2}(1-\xi_{1})))=0. \tag{26}\]  
Firstly, we consider the natural frequencies of beam under simply supported boundary condition. To this end, after substituting Equation (25) into boundary condition Equation (12), a linear algebraic system for \(C_{nj}\) can be obtained as  
\[ \begin{cases}C_{n1}+C_{n2}+C_{n3}+C_{n4}=0,\\ C_{n1}e^{\beta_{n1}}+C_{n2}e^{\beta_{n2}}+C_{n3}e^{\beta_{n3}}+C_{n4}e^{\beta_{ n4}}=0,\\ C_{n1}\frac{e^{-\alpha}\beta_{n1}^{2}(\alpha\xi_{2}^{\prime}(e^{-\beta_{n1}})-2 e^{\alpha}\xi_{1}(-\beta_{n1}))}{2(4-\beta_{n1})}+C_{n2}\frac{e^{-\alpha}\beta_{n2 }^{2}\left(\alpha\xi_{2}^{\prime}\left(e^{-\beta_{n2}}\right)-2e^{\alpha}\xi_ {1}(-\beta_{n2})\right)}{2(4-\beta_{n2})}\\ +C_{n3}\frac{e^{-\alpha}\beta_{n3}^{2}(\alpha\xi_{2}^{\prime}(e^{-\beta_{n3}} )-2e^{\alpha}\xi_{1}(-\beta_{n3}))}{2(4-\beta_{n3})}+C_{n4}\frac{e^{-\alpha} \beta_{n4}^{2}\left(\alpha\xi_{2}^{\prime}(e^{-\beta_{n4}})-2e^{\alpha}\xi_{ 1}(-\beta_{n4})\right)}{2(4-\beta_{n4})}=0,\\ C_{n1}\frac{\left(e^{-\alpha}\beta_{n1}^{2}\left(2\xi_{1}e^{+\beta_{n1}}(x+ \beta_{n1})-\alpha\xi_{2}^{\prime}\left(e^{+\beta_{n1}}-1\right)\right) \right)}{2(4+\beta_{n1})}+C_{n2}\frac{\left(e^{-\alpha}\beta_{n2}^{2}\left(2 \xi_{1}e^{+\beta_{n2}}(x+\beta_{n2})-\alpha\xi_{2}^{\prime}\left(e^{+\beta_{n2 }}-1\right)\right)\right)}{2(4+\beta_{n2})}\\ +C_{n3}\frac{\left(e^{-\alpha}\beta_{n3}^{2}\left(2\xi_{1}e^{+\beta_{n3}}(x+ \beta_{n3})-4\xi_{2}^{\prime}\left(e^{+\beta_{n3}}-1\right)\right)\right)}{2( 4+\beta_{n3})}+C_{n4}\frac{\left(e^{-\alpha}\beta_{n4}^{2}\left(2\xi_{1}e^{+ \beta_{n4}}(x+\beta_{n4})-\alpha\xi_{2}^{\prime}\left(e^{+\beta_{n4}}-1\right) \right)\right)}{2(4+\beta_{n4})}=0.\end{cases}  \tag{27}\]  
For non-trivial solutions, the determinant \(D\) of the coefficient matrix of Equation (27) must be zero. Here, the long expression of \(D\) have been omitted for brevity. The natural frequencies \(\omega_{n}\) and \(\beta_{ni}\) can be solved numerically from Equation (26) and \(Det[D]=0\).