# Analysis of Lateral Displacement of Pile Foundation Caused by Large-Diameter Shield Tunneling
## 2 Lateral Displacement of Front Pile Caused by Using a Shield Machine
### Horizontal Displacement of Soil Caused by Shield Tunneling  
The Mindlin solution is a useful tool for calculating the stress and displacement fields in a semi-infinite elastic body caused by vertical or horizontal concentrated forces in a semi-infinite space. Since the additional thrust \(q\) of the shield and the horizontal distribution of the shield friction \(f\) along the shield shell surface are the uniform forces distributed on the incision surface and the shield shell surface, respectively, it is necessary to perform double integral calculations and appropriate coordinate transformation on them. The micro-element area of the circular section of the additional thrust of the shield is taken as \(A=rd\theta dr\), as shown in Figure 2, and the coordinate transformation is as follows:  
\[u=\left\{ \begin{matrix}y^{\prime}=y-r\cos\theta\\ x^{\prime}=x\\ z^{\prime}=z\\ c=H-r\sin\theta\end{matrix} \right. \tag{1}\]  
By substituting it into the Mindlin solution and integrating it, the displacement formula of the soil in the \(x\)-direction at any point in space caused by the additional thrust of the shield can be obtained:  
Figure 1: Mechanical model of large-diameter slurry shield tunneling. Figure 2: Schematic diagram of the additional thrust integral of the shield. \[ \begin{array}{l}u_{h}^{q}=\int_{0}^{R}\int_{0}^{2\pi}\frac{q\prime(1+\mu)d\theta d \mu}{8\pi E(1-\mu)}\{\frac{(3-4\mu)}{R_{1}}+\frac{1}{R_{2}}+\frac{x^{2}}{R_{1} ^{3}}\\ \qquad\qquad+\frac{(3-4\mu)x^{2}}{R_{3}^{2}}+\frac{2z(H-r\text{sin}\theta)}{R_ {2}^{3}}(1-\frac{3x^{2}}{R_{2}^{2}})\\ \qquad\qquad+\frac{4(1-\mu)(1-2\mu)}{R_{2}+z+H-r\text{sin}\theta}[1-\frac{x^{2 }}{R_{2}(R_{2}+z+H-r\text{sin}\theta)}]\}\end{array}  \tag{2}\]  
In the formula, \(q\) represents the shield machine face thrust; \(R\) represents the radius of the tunnel; \(H\) represents the buried depth of the tunnel; and \(R_{1}\) and \(R_{2}\) are, respectively, denoted by:  
\[R_{1}=\sqrt{x^{2}+(y-r\text{cos}\theta)^{2}+(z-H+r\text{sin}\theta)^{2}} \tag{3}\]  
\[R_{2}=\sqrt{x^{2}+(y-r\text{cos}\theta)^{2}+(z+H-r\text{sin}\theta)^{2}} \tag{4}\]  
The micro-element area of the cylindrical surface of the friction force acting on the shield shell is taken as \(dA=Rd\theta dl\), as shown in Figure 3, and the coordinate transformation is as follows:  
\[u=\left\{ \begin{array}{l}y^{\prime}=y-r\cos\theta\\ x^{\prime}=x-l\\ z^{\prime}=z\\ c=H-R\text{sin}\theta\end{array} \right. \tag{5}\]  
By substituting it into the Mindlin solution and integrating it, the displacement formula of soil in the \(x\)-direction at any point in space caused by the shield shell friction can be obtained:  
\[ \begin{array}{l}u_{h}^{f}=\int_{0}^{L}\int_{0}^{2\pi}\frac{fRd\theta dl}{16 \pi G(1-\mu)}\bigg{\{}\frac{(3-4\mu)}{R_{1}}\,+\frac{1}{R_{2}}+\frac{(x-l)^{2 }}{R_{1}^{3}}+\frac{(3-4\mu)(x-l)^{2}}{R_{2}^{3}}+\frac{2(H-R\text{sin}\theta )\text{z}}{R_{2}^{3}}\left(1-\frac{3(x-l)^{2}}{R_{1}^{2}}\right)+\\ \qquad\qquad\qquad\qquad\qquad\frac{4(1-\mu)(1-2\mu)}{R_{2}+z+H-R\text{sin} \theta}\bigg{[}1-\frac{(x-l)^{2}}{R_{2}(R_{2}+z+H-R\text{sin}\theta)}\bigg{]} \bigg{\}}\end{array}  \tag{6}\]  
In the formula, \(L\) is the shield shell length; \(R_{1}\) and \(R_{2}\) are, respectively, denoted by:  
\[R_{1}=\sqrt{\left(x-l\right)^{2}+\left(y-R\text{cos}\theta\right)^{2}+\left(z -H+R\text{sin}\theta\right)^{2}} \tag{7}\]  
\[R_{2}=\sqrt{\left(x-l\right)^{2}+\left(y-R\text{cos}\theta\right)^{2}+\left(z +H-R\text{sin}\theta\right)^{2}} \tag{8}\]  
The friction force \(f\) acting between the shield shell and the surrounding soil at any angle \(\theta\) on the shield shell can be calculated using the formula for the shield shell friction,  
Figure 3: Thrust integral diagram of shield machine. which considers soil softening and unevenness. This formula is developed by referring to [PERSON]'s method [4].