## 3 Mechanical Model of the Existing Pile under Adjacent Pile Driving
### Response of Existing Pile Based on Winkler Model
\begin{cases}\omega_{-1}-2\omega_{0}+\omega_{1}=0\\ -\omega_{-2}+2\omega_{-1}-2\omega_{1}+\omega_{2}=0\\ \omega_{n-1}-2\omega_{n}+\omega_{n+1}=0\\ \omega_{n-2}-2\omega_{n-1}+2\omega_{n+1}-\omega_{n+2}=0\end{cases}  \tag{16}\]  
(i). For \(i=0\),\[ \begin{cases}\frac{1}{H^{4}}\omega_{-2}-\frac{4}{H^{4}}\omega_{-1}+\left(\frac{6}{H ^{4}}+\frac{E_{\text{s}}D_{0}}{E_{\text{p}}P_{\text{p}}}\right)\omega_{0}-\frac {4}{H^{4}}\omega_{1}+\frac{1}{H^{4}}\omega_{2}=\frac{p_{0}D_{0}}{E_{\text{p}}P_ {\text{p}}}\\ \omega_{-1}-2\omega_{0}+\omega_{1}=0\\ -\omega_{-2}+2\omega_{-1}-2\omega_{1}+\omega_{2}=0\end{cases}  \tag{17}\]  
Eliminating \(\omega_{-2}\) and \(\omega_{-1}\), Equation (17) can be written as:  
\[(4\alpha+2\beta+\gamma)\omega_{0}-4\alpha\omega_{1}+2\alpha\omega_{2}=0 \tag{18}\]  
(ii). For \(i=1\),  
\[ \begin{cases}\frac{1}{H^{4}}\omega_{-1}-\frac{4}{H^{4}}\omega_{0}+\left(\frac{ 6}{H^{4}}+\frac{E_{\text{s}}D_{0}}{E_{\text{p}}P_{\text{p}}}\right)\omega_{1 }-\frac{4}{H^{4}}\omega_{2}+\frac{1}{H^{4}}\omega_{3}=\frac{p_{1}D_{0}}{E_{ \text{p}}P_{\text{p}}}\\ \omega_{-1}-2\omega_{0}+\omega_{1}=0\end{cases}  \tag{19}\]  
Eliminating \(\omega_{-1}\), Equation (19) can be written as:  
\[(2\alpha+\beta)\omega_{0}+(\gamma-\alpha)\omega_{1}+\beta\omega_{2}+\alpha \omega_{3}=0 \tag{20}\]  
(iii). For \(i=n-1\),  
\[ \begin{cases}\frac{1}{H^{4}}\omega_{n-3}-\frac{4}{H^{4}}\omega_{n-2}+\left( \frac{6}{H^{4}}+\frac{E_{\text{s}}D_{0}}{E_{\text{p}}P_{\text{p}}}\right) \omega_{n-1}-\frac{4}{H^{4}}\omega_{n}+\frac{1}{H^{4}}\omega_{n+1}=\frac{p_{n- 1}D_{0}}{E_{\text{p}}P_{\text{p}}}\\ \omega_{n-1}-2\omega_{n}+\omega_{n+1}=0\end{cases}  \tag{21}\]  
Eliminating \(\omega_{n+1}\), Equation (21) can be written as:  
\[ \begin{cases}\frac{1}{H^{4}}\omega_{n-2}-\frac{4}{H^{4}}\omega_{n-1}+\left( \frac{6}{H^{4}}+\frac{E_{\text{s}}D_{0}}{E_{\text{p}}P_{\text{p}}}\right) \omega_{n}-\frac{4}{H^{4}}\omega_{n+1}+\frac{1}{H^{4}}\omega_{n+2}=\frac{p_{0} D_{0}}{E_{\text{p}}P_{\text{p}}}\\ \omega_{n-1}-2\omega_{n}+\omega_{n+1}=0\\ -\omega_{n-2}+2\omega_{n-1}-2\omega_{n+1}+\omega_{n+2}=0\end{cases}  \tag{22}\]  
Eliminating \(\omega_{n+1}\) and \(\omega_{n+2}\), Equation (23) can be written as:  
\[2\alpha\omega_{n-2}-4\alpha\omega_{n-1}+(4\alpha+2\beta+\gamma)\omega_{n}=0 \tag{24}\]  
Substituting Equations (16)-(24), the stiffness matrix can be expressed as:  
\[\{K\}= \begin{bmatrix}4\alpha+2\beta+\gamma&-4\alpha&2\alpha\\ 2\alpha+\beta&\gamma-\alpha&\beta&\alpha\\ \alpha&\beta&\gamma&\beta&\alpha\\ &\ddots&\ddots&\ddots&\ddots&\\ &&\alpha&\beta&\gamma&\beta&\alpha\\ &&&\alpha&\beta&\gamma-\alpha&2\alpha+\beta\\ &&&2\alpha&-4\alpha&4\alpha+2\beta+\gamma\end{bmatrix}  \tag{25}\]  
Case 2:  
If the existing pile is the end-bearing pile, the pile top is free, and the pile tip is fixed. The boundary can be expressed as:  
\[ \begin{cases}\omega_{-1}-2\omega_{0}+\omega_{1}=0\\ -\omega_{-2}+2\omega_{-1}-2\omega_{1}+\omega_{2}=0\\ \omega_{n+1}=0\\ \omega_{n+2}-\omega_{n}=0\end{cases}  \tag{26}\](i). For \(i=0\),  
\[\left\{ \begin{matrix}\frac{1}{h^{4}}\omega_{-2}-\frac{4}{h^{4}}\omega_{-1}+\Big{(} \frac{6}{h^{4}}+\frac{E_{\text{s}}D_{0}}{E_{\text{p}}l_{\text{p}}}\Big{)}\omega_ {0}-\frac{4}{h^{4}}\omega_{1}+\frac{1}{h^{4}}\omega_{2}=\frac{p_{0}D_{0}}{E_{ \text{p}}l_{\text{p}}}\\ \omega_{-1}-2\omega_{0}+\omega_{1}=0\\ -\omega_{-2}+2\omega_{-1}-2\omega_{1}+\omega_{2}=0\end{matrix} \right. \tag{27}\]  
Eliminating \(\omega_{-2}\) and \(\omega_{-1}\), Equation (27) can be written as:  
\[(4\alpha+2\beta+\gamma)\omega_{0}-4\alpha\omega_{1}+2\alpha\omega_{2}=0 \tag{28}\]  
(ii). For \(i=1\),  
\[\left\{ \begin{matrix}\frac{1}{h^{4}}\omega_{-1}-\frac{4}{h^{4}}\omega_{0}+ \Big{(}\frac{6}{h^{4}}+\frac{E_{\text{s}}D_{0}}{E_{\text{p}}l_{\text{p}}} \Big{)}\omega_{1}-\frac{4}{h^{4}}\omega_{2}+\frac{1}{h^{4}}\omega_{3}=\frac{p _{1}D_{0}}{E_{\text{p}}l_{\text{p}}}\\ \omega_{-1}-2\omega_{0}+\omega_{1}=0\end{matrix} \right.