## 3 Pile Deformation Induced by Capsule Grouting  
Considering the pile as an Euler-Bernoulli beam placed on a Winkler foundation, the differential control equations unfold as follows:  
\[E_{\text{p}}I_{\text{p}}\frac{\text{d}^{4}w(y)}{\text{d}z^{4}}+kD_{\text{p}}w(y )-q(y)D_{\text{p}}=0 \tag{9}\]  
\[q(y)=k\Delta(y) \tag{10}\]  
where \(w(y)\) is the lateral deformation of the pile foundation, \(D_{\text{p}}\) is the diameter of the pile, \(E_{\text{p}}\) and \(I_{\text{p}}\) are the elasticity modulus and inertia moment of the pile foundation, respectively, \(k\) is the spring stiffness of the foundation, and \(q(y)\) is the additional load induced by the expansion of the capsule. [PERSON] [36] obtained the spring stiffness of the foundation based on the surface long beam test with a comprehensive consideration of formation properties and beam stiffness. Additionally, [PERSON] et al. [37] derived the calculation method for the foundation spring stiffness, taking into account the effect of the soil burial depth:  
\[k=\frac{3.08E_{\text{s}}}{\lambda D_{\text{p}}(1-v^{2})}\sqrt{\frac{E_{\text{ s}}D_{\text{p}}^{4}}{E_{\text{p}}I_{\text{p}}}} \tag{11}\]  
Figure 5: Profile of the capsule and the pile. \[\lambda=\left\{ \begin{array}{ll}2.18&z_{1}/D_{\rm p}\leq 0.5\\ 1+\frac{1}{1.7z_{1}/D_{\rm p}}&z_{1}/D_{\rm p}>0.5\end{array} \right. \tag{12}\]  
where \(E_{\rm s}\) is the modulus of elasticity of the foundation soil, \(\
u\) is the Poisson ratio of the foundation soil, and \(\lambda\) is the spring stiffness depth coefficient. The finite difference method is then employed to solve Equation (9), and the nodal unit is illustrated in Figure 6. The pile shaft is equally divided into \(n\) parts, each with length \(l\). Two imaginary nodes are added at each end of pile, resulting in \((n+5)\) nodes totally. The finite difference format of Equation (9) is as follows:  
\[E_{\rm p}I_{\rm p}\frac{6w_{i}-4(w_{i+1}+w_{i-1})+(w_{i+2}+w_{i-2})}{l^{4}}+k_{ i}D_{\rm p}w_{i}=D_{\rm p}q_{i} \tag{13}\]  
When both ends of the pile are free, the bending moment \(M\) and shear force \(Q\) at both ends are 0:  
\[\left\{ \begin{array}{ll}M_{0}=-E_{\rm p}I_{\rm p}\frac{w_{1}-2w_{0}+w_{-1}}{ l^{2}}=0\\ M_{n}=-E_{\rm p}I_{\rm p}\frac{w_{n+1}-2w_{n}+w_{n-1}}{l^{2}}=0\\ Q_{0}=-E_{\rm p}I_{\rm p}\frac{w_{-2}w_{1}+2w_{-1}-w_{-2}}{2l^{3}}=0\\ Q_{n}=-E_{\rm p}I_{\rm p}\frac{w_{n+2}-2w_{n+1}+2w_{n-1}-w_{n-2}}{2l^{3}}=0 \end{array} \right. \tag{14}\]  
The expressions for \(w_{-2}\), \(w_{-1}\), \(w_{n+1}\), \(w_{n+2}\) can be obtained by solving Equation (14), and the expressions are then substituted into Equation (13) and written in matrix form as  
\[(K_{1}+K_{2})w=P \tag{15}\]  
where \(K_{1}\) is the deformation stiffness matrix of the pile, \(K_{2}\) is the stiffness matrix of the foundation, \(w\) is the lateral displacement column vector of the pile, \(P\) is the additional load column vector, and each matrix expression unfolds as follows:  
Figure 6: Discretization of the pile. \[K_{1}=\frac{E_{\text{p}}I_{\text{p}}}{l^{4}} \begin{bmatrix}2&-4&2\\ -2&5&-4&1&&\textbf{0}\\ 1&-4&6&-4&1\\ &\ddots&\ddots&\ddots&\ddots&\ddots\\ &&1&-4&6&-4&1\\ \textbf{0}&&1&-4&5&-2\\ &&&2&-4&2\end{bmatrix} _{(n+1)(n+1)} \tag{16}\]  
\[K_{2}=D_{\text{p}} \begin{bmatrix}k_{0}\\ k_{1}\\ &k_{2}\\ &\ddots&\ddots&\ddots&\ddots\\ &k_{n-2}\\ \textbf{0}\end{bmatrix} _{(n+1)(n+1)} \tag{17}\]  
\[P=D_{\text{p}} \begin{bmatrix}q_{0}&q_{1}&q_{2}&\cdots&q_{n-2}&q_{n-1}&q_{n} \end{bmatrix} ^{\text{T}}_{(n+1)} \tag{18}\]  
When both ends of the pile are fixed constraints, i.e., the angle of rotation \(\theta\) and displacement \(w\) are 0 at both ends, the following is applied:  
\[\left\{ \begin{array}{l}\theta_{-1}=\frac{w_{0}-w_{-2}}{2l}=0\\ \theta_{n+1}=\frac{w_{n+2}-w_{n}}{2l}=0\\ w_{-1}=0\\ w_{n+1}=0\end{array} \right.