## 2 Theory
\begin{array}{l}a_{1}=1,a_{2}=\frac{4(k-1)}{(k+1)}a_{1},a_{3}=\frac{4k}{(k+2 )}a_{2}-\frac{6(k+1)(k-2)+\overline{D}_{5}}{(k+1)(k+2)}a_{1},\\ a_{4}=\frac{4(k+1)}{(k+3)}a_{3}-\frac{6(k+2)(k-1)+\overline{D}_{5}}{(k+2)(k+3) }a_{2}\\ -\frac{(\overline{D}_{4}-4(k+3)(k-2))(k-1)-\overline{D}_{6}}{(k+1)(k+2)(k+3) }a_{1},\\ a_{5}=\frac{4(k+2)}{(k+4)}a_{4}-\frac{6k(k+3)+\overline{D}_{5}}{(k+3)(k+4)(k+ 4)}-\frac{\left(\overline{D}_{4}-4(k+4)(k-1)\right)k-\overline{D}_{6}}{(k+2)(k +3)(k+4)}a_{2}\\ -\frac{k\left\{\overline{D}_{4}(k+5)(k-2)+\overline{D}_{5}\right\}+(k-1)+ \overline{D}_{6}}{(k+1)(k+2)(k+3)(k+4)}a_{1},\\ a_{6}=\frac{4(k+3)}{(k+5)}a_{5}-\frac{6(k+4)(k+1)+\overline{D}_{5}}{(k+4)(k+5) }a_{4}-\frac{\left(\overline{D}_{4}-4(k+5)\right)(k+1)-\overline{D}_{6}}{(k+3) (k+4)(k+5)}a_{3}\\ -\frac{(k+1)\left[\left\{(k+6)(k-1)+\overline{D}_{3}\right\}+3\overline{D}_{6} \right]+\overline{D}_{7}}{(k+2)(k+3)(k+4)(k+5)}a_{2}\\ -\frac{k\left\{\overline{D}_{2}(k-1)-3\overline{D}_{6}\right\}-2\overline{D}_ {7}}{(k+2)(k+3)(k+4)(k+5)}a_{1}.\end{array}  \tag{23}\]  
The bending moment and shear force obtained by substituting Equations (11) and (15) into Equations (8) and (10) can be rewritten as follows:  
\[M(\zeta)=N_{1}\Big{\{}(1-\zeta)^{4}W^{\prime\prime}+\overline{D}_{8}W^{ \prime}\Big{\}}, \tag{24}\]  
\[V(\zeta)=-N_{2}\Big{[}(1-\zeta)^{4}W^{\prime\prime\prime}-4(1-\zeta)^{3}W^{ \prime\prime}-\Big{\{}\overline{D}_{9}-\overline{D}_{1}(1-\zeta)^{4}\Big{\}}W ^{\prime}+\big{(}\overline{D}_{10}-\overline{D}_{11}\big{)}W\Big{]}, \tag{25}\]  
\[\text{where }\overline{D}_{8}=\frac{K_{80}L}{E_{10}c^{2}},\overline{D}_{9}=\frac{F_{ 0}L^{2}}{E_{10}c^{2}},\overline{D}_{10}=\frac{K_{W2}L^{3}}{E_{10}c^{3}}, \overline{D}_{11}=\frac{m_{l2}\omega^{2}L^{3}}{E_{10}c^{3}}.\]  
The bending displacement with respect to the four roots of the initial equation can be defined as  
\[W(\zeta)=A_{1}f(\zeta,0)+A_{2}f(\zeta,1)+A_{3}f(\zeta,2)+A_{4}f(\zeta,3), \tag{26}\]  
where  
\[f(\zeta,k)=\sum_{i=0}^{\infty}a_{i+1}(k)\zeta^{k+i}. \tag{27}\]  
The slope of the deformation curve can be obtained by differentiating the bending displacement:  
\[\Phi(\zeta)=\frac{c}{L}\frac{\text{d}W(\zeta)}{\text{d}\zeta}=\frac{c}{L}(A_{1} f(\zeta,0)+A_{2}f(\zeta,1)+A_{3}f(\zeta,2)+A_{3}f(\zeta,3)). \tag{28}\]  
The expressions for the bending displacement \(W(\zeta)\) and the slope \(\Phi(\zeta)\) defined in Equations (26) and (28) can be utilized for \(n=1\) and \(n=2\), respectively. The nondimensional length coordinates \(\overline{x}\) of the beam are \(\overline{x}=0\) and \(\overline{x}=1\) at the left- and right-hand endpoints, respectively. The state vectors at \(\overline{x}=0\) and \(\overline{x}=1\) can be determined by substituting \(\overline{x}=0\) and \(\overline{x}=1\) into the displacements and forces for \(n=1\) and \(n=2\), respectively, and the state vectors \(\mathbf{Z}_{0}\) at \(\overline{x}=0\) for \(n=1\) and \(n=2\) can be expressed in matrix form as follows:  
\[\mathbf{Z}_{0}=\mathbf{Q}\mathbf{A} \tag{29}\]  
where \(\mathbf{Z}_{0}=[W,\ \Phi,\ M,V]_{\overline{x}=0}^{T}\), \(\mathbf{A}=[A_{1},A_{2},A_{3},A_{4}]^{T}\), and the components of \(\mathbf{Q}\) matrix for \(n=1\) can be defined as  
\[ \begin{array}{c}Q_{1j}=\sum\limits_{j=1}^{4}f(0,j-1),\ Q_{2j}=\xi\sum \limits_{j=1}^{4}f^{\prime}(0,j-1),\\ Q_{3j}=N_{1}\sum\limits_{j=1}^{4}\big{\{}f^{\prime\prime}(0,j-1)-\overline{D}_ {7}f^{\prime}(0,j-1)\big{\}},\\ Q_{4j}=-N_{2}\sum\limits_{j=1}^{4}\big{\{}f^{\prime\prime}(0,j-1)-3f^{\prime \prime}(0,j-1)-\big{\{}\overline{D}_{8}-\overline{D}_{1}\big{\}}f^{\prime}(0,j -1)-\big{(}\overline{D}_{9}-\overline{D}_{10}\big{)}f(0,j-1)\big{\}},\\ \end{array}  \tag{30}\]  
and for \(n=2\),  
\[ \begin{array}{c}Q_{3j}=N_{1}\sum\limits_{j=1}^{4}\big{\{}f^{\prime\prime}(0, j-1)-\overline{D}_{8}f^{\prime}(0,j-1)\big{\}},\\ Q_{4j}=-N_{2}\sum\limits_{j=1}^{4}\big{\{}f^{\prime\prime\prime}(0,j-1)-4f^{ \prime\prime}(0,j-1)-\big{\{}\overline{D}_{9}-\overline{D}_{1}\big{\}}f^{ \prime}(0,j-1)-\big{(}\overline{D}_{10}-\overline{D}_{11}\big{)}f(0,j-1)\big{\}}.\\ \end{array}