## 3 Solution Procedure
### Electromechanical Responses  
When the base excitation is applied (\(w_{b}=W_{b}\cos(\omega t)\)), the vibration response is assumed to be of the following form:  
\[w=W(t)\sin\Big{(}\frac{\pi x}{L}\Big{)} \tag{48}\]By substituting Equation (48) in Equations (29) and (30), and applying the [PERSON] method, one obtains:  
\[\alpha^{2}W^{3}+\alpha_{V}VW+\omega_{n}^{2}W+\overline{c}_{v}\dot{W}+\ddot{W}= \overline{W}_{b}\omega^{2}\cos\omega t \tag{49}\]  
\[\overline{c}_{p}\dot{V}+V-\overline{e}_{311}W\dot{W}-\overline{\mu}_{3113}\dot {W}=0 \tag{50}\]  
where \(\omega_{n}^{2}\) is defined in Equation (42), and:  
\[\overline{c}_{v}=\frac{\pi c_{v}(\pi^{2}(\omega_{0}a)^{2}+L^{2})}{\pi l_{0}( \pi^{2}(\omega_{0}a)^{2}+L^{2})+4 Lm_{s}},\alpha_{V}=\frac{2\pi^{3}b_{\rm e11}} {\pi l_{0}(\pi^{2}(\omega_{0}a)^{2}+L^{2})+4 Lm_{s}}\]  
\[\overline{W}_{b}=\frac{4L(l_{0}L+m_{s})}{\pi l_{0}(\pi^{2}(\omega_{0}a)^{2}+L ^{2})+4 Lm_{s}}W_{b},\overline{c}_{p}=\frac{\alpha_{3}bL}{2h_{p}} \tag{51}\]  
\[\overline{\mu}_{3113}=\frac{\pi bR}{L}\mu_{3113},\overline{e}_{311}=\frac{ \pi^{2}bR}{4L}e_{311}\]  
The harmonic balance method is used to solve Equations (49) and (50). Setting:  
\[W(t)=W_{s}+h_{t}(a_{1}\cos\omega t+b_{1}\sin\omega t) \tag{52}\]  
\[V(t)=c_{1}\cos\omega t+d_{1}\sin\omega t+c_{2}\cos 2\omega t+d_{2}\sin 2\omega t \tag{53}\]  
where \(W_{s}\) is the static displacement by buckling. When the beam is in the state of prebuckling (\(P\)\(<\)\(P_{cr}\)), \(W_{s}\) is zero, and when the beam is in the state of postbuckling (\(P\)\(\geq\)\(P_{cr}\)), \(W_{s}\) is calculated by Equation (37). Substituting Equations (52) and (53) into Equations (49) and (50), the coefficients for \(\cos\omega t\) and \(\sin\omega t\) should be balance. Then the differential equations are converted to a series of nonlinear algebraic equations such as:  
\[\begin{array}{l}\frac{3}{4}\alpha^{2}b_{1}h_{t}^{3}a_{1}^{2}+\frac{1}{2}h_{ t}\alpha_{v}a_{1}d_{2}+\frac{3}{4}h_{t}^{3}\alpha^{2}b_{1}{}^{3}-\overline{c}_{ v}h_{t}\omega a_{1}\\ -\frac{1}{2}h_{t}\alpha_{v}c_{2}b_{1}+h_{t}\big{(}\omega_{n}{}^{2}-\omega^{2} \big{)}b_{1}+3\alpha^{2}h_{t}W_{s}^{2}b_{1}+W_{s}\alpha_{v}d_{1}=0\end{array} \tag{54}\]  
\[\begin{array}{l}\frac{3}{4}\alpha^{2}h_{t}^{2}a_{1}^{3}+h_{t}\big{(}\frac{3} {4}\alpha^{2}h_{t}^{2}b_{1}^{2}+\omega_{n}^{2}-\omega^{2}\big{)}a_{1}+ \overline{c}_{v}\omega b_{1}-\overline{W}_{b}\omega\\ +\frac{1}{2}c_{2}h_{t}\alpha_{v}a_{1}+3\alpha^{2}h_{t}W_{s}^{2}a_{1}+\frac{1} {2}h_{t}\alpha_{v}b_{1}d_{2}+W_{s}\alpha_{v}c_{1}=0\end{array} \tag{55}\]  
\[\begin{array}{l}h_{t}\omega\overline{\mu}_{3113}a_{1}+\overline{e}_{311}h_{ t}\omega W_{s}a_{1}-\omega\overline{c}_{p}c_{1}+d_{1}=0\end{array} \tag{56}\]  
\[\begin{array}{l}-\overline{e}_{311}h_{t}\omega W_{s}b_{1}-h_{t}\overline{ \mu}_{3113}\omega b_{1}+c_{1}+\overline{c}_{p}\omega d_{1}=0\end{array} \tag{57}\]  
\[\begin{array}{l}\frac{1}{2}\overline{e}_{311}h_{t}{}^{2}\omega\Big{(}a_{1}^ {2}-b_{1}^{2}\Big{)}-2\overline{c}_{p}\omega c_{2}+d_{2}=0\end{array} \tag{58}\]  
\[\begin{array}{l}-\overline{e}_{311}h_{t}{}^{2}\omega a_{1}b_{1}+c_{2}+2 \overline{c}_{p}\omega d_{2}=0\end{array} \tag{59}\]  
in which, \(a_{1}\), \(b_{1}\), \(c_{1}\), \(d_{1}\), \(c_{2}\), and \(d_{2}\) can be solved numerically.  
The generated voltage is the alternating voltage, which is described as a function of time. In order to evaluate the energy capture efficiency conveniently, the root mean square (RMS) voltage \(V_{rms}\) is introduced as follows:  
\[V_{rms}=\sqrt{\frac{1}{T_{V}}{\int_{0}^{T_{V}}{[V(t)]}^{2}\mathrm{d}t}} \tag{60}\]  
where \(T_{V}\) is the period of the alternating voltage. The RMS voltage is the square root of the mean over one cycle of the square of the instantaneous voltage. In fact, the RMS voltage is equal to the value of the direct current that would produce the same average power dissipation in a resistive load [69].