Theoretical Study on Diaphragm Wall and Surface Deformation Due to Foundation Excavation Based on Three-Parameter Kerr Model

Abstract

To calculate the horizontal displacement of the diaphragm wall and surface settlement caused by foundation pit excavation, the three-parameter Kerr foundation model was applied to a diaphragm wall and derived the flexural differential equations of the diaphragm wall and calculated the horizontal displacement of the diaphragm wall using the finite difference calculation method. The boundary element method combined with the Mindlin displacement solution was then used to invert the additional horizontal stress near the diaphragm wall. Lastly, the Mindlin solution was used to calculate the surface settlement. The effectiveness of the proposed calculation method was verified by comparing the horizontal displacement of the diaphragm wall and the surface settlement between the theoretical calculation and the actual project. The theory proves that there is a certain connection between the horizontal displacement of the diaphragm wall and the surface settlement, and the horizontal displacement of the diaphragm wall is larger than the surface settlement. Using this theory to further analyze the foundation pit construction parameters, the greater the thickness and elasticity modulus of the diaphragm wall, and the greater the diameter and number of internal supports, the smaller the horizontal displacement of the diaphragm wall and the surface settlement. The theory can accurately predict the horizontal displacement of the diaphragm wall and surface settlement and provides guidance for the construction of foundation pit projects.

1. Introduction

Excessive deformation caused by pit excavation leads to slope instability, which affects construction safety and is not conducive to the goal of sustainable development of a project. At present, research on the horizontal displacement of diaphragm walls caused by foundation excavation mainly includes field monitoring [1], model tests [2], theoretical calculations [3], and numerical simulations [4,5]. The existing theoretical studies mainly include the limit equilibrium method, elastic foundation beam method, and variational method. The limit equilibrium method is applicable only to rigid short piles with little burial depth, and the application range is relatively narrow. The variational method is based on the principle of minimum potential energy and calculates the horizontal displacement of diaphragm walls through energy relations, but its boundary conditions are demanding, and accuracy cannot be guaranteed [6].

The elastic foundation beam method is a commonly used method to calculate the displacement of a retaining structure, which is based on the principle of treating the support pile as a Winkler elastic foundation beam placed vertically and calculating the pile bending moment, shearing force, and flexural deformation by the flexural differential equations of the beam [7]. However, the model is oversimplified and ignores the shear deformation of the soil, which cannot reflect the continuity of the soil force deformation due to pile–soil interactions. This leads to a smaller calculation result than the on-site monitoring result, resulting in a safety risk in the on-site construction. For the problems of a Winkler elastic foundation beam, Zhu et al. [8] employed the two-parameter Pasternak foundation model to enhance the precision of calculating the horizontal displacement of a retaining structure. Fu et al. [9] applied the Pasternak foundation model for tunnel force deformation analysis with good results. Zhang et al. [10] considered the rheological effects in viscoelastic soils along with the neighboring pile foundations as a Pasternak foundation model and analyzed the time-domain deformation response of the neighboring pile foundations caused by foundation excavation. Although Pasternak’s model is an improvement compared with the Winkler foundation model in considering shear forces, the shear layer and the spring are independent in this model, and the connection between them cannot be established, which leads to computational errors in the model. Feng et al. [11] assumed a tunnel to be a Timoshenko beam placed on a three-parameter Kerr foundation and calculated the tunnel deformation response resolution. The authors showed better results for the three-parameter Kerr foundation model, and the calculation error of the three-parameter Kerr foundation model was obviously smaller than that of the Winkler elastic foundation beam and Pasternak model. The rationality of the three parameters of the Kerr foundation model has been proven, but no studies have applied it to the calculation of diaphragm wall deformation.

In terms of surface subsidence prediction, empirical methods [12,13,14] and semi-empirical methods [15] are easy to use and are widely used in engineering. However, the complexity of the project itself makes empirical and semi-empirical methods sometimes not applicable [16]. As a commonly used surface settlement prediction method, the ground loss method has been continuously improved in a large number of studies. Based on the principle of the ground loss method, Li [17] treated the surface settlement surrounding a deep foundation pit as a skewed distribution curve and successfully addressed the surface settlement of the pile-anchor-supported foundation pit. Hong [18] derived an approximate analytical solution for the tunnel-induced response of a foundation system considering contact effects by using Sagaseta’s loss-of-ground settlement formulation instead of the complex solution surface of the elastic half-plane. However, the ground loss method is based on empirical formulas, and the choice of fitting curve directly affects the final settlement pattern. The error is difficult to avoid, and the theory is not strong. Qian et al. [19] and Fan et al. [20] used the separation variable method to solve the general solution of the displacement equilibrium equation of the plane strain problem and regarded the displacement of the retaining wall of the foundation pit as a known displacement boundary in calculating the surface settlement. This method is a theoretical method for solving surface settlements outside the pit at present, but it requires many soil parameters, and the derivation process is very complicated, which is difficult in practical applications. Some theories consider the unloading of pit excavation to deduce the surface settlement outside the pit but do not consider the effect of diaphragm walls, which produces a large error [10].

To summarize, most of the existing theories are limited to the study of the diaphragm wall displacement caused by pit excavation, and there is a relative lack of theories that further consider the effect of pit excavation on surface settlement. Therefore, it is necessary to establish a set of theories to calculate the displacement of the diaphragm wall and surface subsidence. Using the three-stage analysis method, the three-parameter Kerr foundation model [21] was first applied to a diaphragm wall to accurately calculate the horizontal displacement of the diaphragm wall and establish the foundation for the later calculation of surface settlement. Subsequently, the present study introduced the boundary element method and the Mindlin solution to establish the connection between the horizontal displacement of the diaphragm wall and the surface subsidence [22], and the boundary element method was utilized to invert the additional horizontal stresses. Lastly, the Mindlin solution was used to calculate the surface settlement.

2. Simplified Calculation Method of the Three-Parameter Kerr Foundation Model [11]

Computational Model and Basic Assumption

By introducing parameters G and c, the three-parameter Kerr foundation model considers the shear action of soil and the continuity of soil deformation. Compared with the Winkler model and the Pasternak model, the Kerr model has better calculation accuracy, and it has been proved in the calculation of tunnel deformation [11] and pipeline deformation [23]. The three-parameter Kerr foundation model is shown in Figure 1. G is the foundation soil shear stiffness (kPa); c and k are the foundation reaction forces of the left spring and the right spring (kPa), respectively. P(x,z) is the applied load. The basic assumptions of the model include the following:

(1) The unit-width diaphragm wall is used as a vertically placed elastic beam with thickness D and flexural stiffness EI [24,25].

(2) The subsurface diaphragm wall is in close proximity to the foundation soil, and its deformation is coordinated with the foundation deformation at the contact point.

(3) The friction between the diaphragm wall and foundation is not considered.

The differential expression for the shear layer deflection is given by [11]:

$$-\frac{EIG}{c_i}\frac{d^6{w}_2}{d{z}^6}+\frac{EI\left(c_i+k_i\right)}{c_i}\frac{d^4{w}_2}{d{z}^4}-G_{i}d\frac{d^2{w}_2}{d{z}^2}+k_{i}d{w}_2=0$$

(1)

where E is the elastic modulus of the diaphragm wall (kPa), and I is the moment of inertia of the diaphragm wall (m⁴). The horizontal displacement of the diaphragm wall is calculated by the following formula:

$$w=\left(1+\frac{k}{c}\right)w_2-\frac{G}{cd}\frac{d^2{w}_2}{d{z}^2}$$

(2)

3. Establishment and Solution of the Flexural Differential Equations of Diaphragm Wall

3.1. Establishment of Shear Layer Flexural Differential Equations

Based on the three-parameter Kerr foundation model, a computational analysis model of the internally supported diaphragm wall support structure was established, as shown in Figure 2. K_t is the stiffness coefficient of the inner support, L₁ is the length of the wall above the excavation surface, and L₂ is the length of the wall below the excavation surface.

The soil on both sides of the diaphragm wall is equated to a spring, and the soil shear force is considered through the shear layer, the soil compression force through spring k and spring c, and the support force through k_t. The calculation of soil pressure is based on static earth pressure, and it is stipulated that the positive horizontal displacement w of the diaphragm wall corresponds to the active deformation of the soil and that the negative horizontal displacement w of the diaphragm wall corresponds to the passive deformation of the soil. Therefore, when the active deformation of the soil occurs, the increment in the soil pressure is negative, and the increment in the internal support pressure is positive; when the passive deformation of the soil occurs, the increment in the soil pressure is positive, and the increment in the internal support pressure is negative.

In the three-parameter Kerr foundation model, considering the variation in the soil pressure and internal support pressure, and substituting them into the model, the shear layer flexural differential equations are established for the loaded section above the excavation surface and the embedded section below the excavation surface, respectively, with the excavation surface as the partition interface. For the loaded section, the shear flexural differential Equation (1) can be expressed as:

$$-\frac{EI{G}_i}{c_i}\frac{d^6{w}_2}{d{z}^6}+\frac{EI\left(c_i+k_i\right)}{c_i}\frac{d^4{w}_2}{d{z}^4}-G_{i}d\frac{d^2{w}_2}{d{z}^2}-\left(P_i-k_i{w}_2\right)d+k_{t_i}{w}_2+\frac{k_{\mathit{ti}}\left(c_i+k_i\right)}{c_i}{w}_2=0$$

(3)

For the embedded segment, flexural differential Equation (1) can be expressed as:

$$-\frac{EI{G}_i}{c_i}\frac{d^6{w}_2}{d{z}^6}+\frac{EI\left(c_i+k_i\right)}{c_i}\frac{d^4{w}_2}{d{z}^4}-2G_{i}d\frac{d^2{w}_2}{d{z}^2}+2k_i{w}_{2}d-\Delta P_{i}d=0$$

(4)

where P_i is the static earth pressure behind the wall above the excavation surface (kN/m), k_ti is the coefficient of the elastic modulus of the internal support, $\Delta P_i$ is the static earth pressure difference between the diaphragm wall (kN/m).

3.2. Determination of Calculation Parameters of the Three-Parameter Kerr Foundation Model

The empirical formula proposed by Tanahashi was used to calculate the shear modulus G [26], where G_i is the shear modulus of the i-th layer soil (kPa):

$$G_i=\frac{E_{ti}t}{6\left(1+v_i\right)}$$

(5)

where E_ti is the elastic modulus of the i-th layer soil (kPa), and v is the Poisson’s ratio of the i-th layer soil. t is the thickness of the shear layer (m). The thickness of the shear layer can be selected according to the literature [27]. The influence range of the soil on the pile side is 11D. D is the diameter of the diaphragm wall.

For the selection of the foundation reaction force k_i, this study used the formula proposed by Vesic [28], where the outer spring elastic modulus k_i of the three-parameter Kerr foundation model is three times the inner spring elastic modulus c_i according to the simplified elastic space method.

$$k_i=0.65{\left(\frac{E_{ti}{d}^4}{EI}\right)}^{0.083}\cdot \frac{E_{\mathit{ti}}}{{\left(1-v_i\right)}^2}$$

(6)

$$c_i=3k_i$$

(7)

The stiffness coefficient of the inner support was selected according to a previous study [24]:

$$F_h=k_t\left(w_t-w_{t0}\right)$$

(8)

$$k_t=\frac{{\alpha }_t{E}_{c}Ad}{\lambda l_{0}s}$$

(9)

where w_t is the horizontal displacement of the diaphragm wall in the inner support section (m), $w_{t0}$ is the initial horizontal displacement of the diaphragm wall in the inner support section, $\lambda$ is the adjustment coefficient of the immobile point of the support, ${\alpha }_t$ is the relaxation coefficient of the support, $E_c$ is the elasticity modulus of the support material (kPa), A is the cross-sectional area of the support (m²), l₀ is the length of the supported member under pressure (m), and s is the horizontal spacing of the support (m).

3.3. Differential Calculation of Flexural Differential Equations

The finite difference method was employed to computationally solve the flexural differential equation of the shear layer by dividing the wall into n equal sections, where each section has a length of h. In this paper, h = 0.5 is assumed in the theoretical calculations. Three virtual nodes are added at the top and bottom of the wall, and the nodes from the top of the wall to the excavation face are numbered −3, −2, −1, 0, 1, 2, 3, …, K − 3, K − 2, K − 1, K; the nodes from the excavation face to the bottom of the wall are numbered K, K + 1, K + 2, K + 3, …, N − 3, N − 2, N − 1, N, N + 1, N + 2, and N + 3, as shown in Figure 3. K is the node at the excavation face, and N is the node at the bottom of the wall.

The difference form of the flexural differential equation for the shear layer is first obtained for different order derivatives using Taylor’s formula [29]. The differentiation of the flexural differential equations for the loaded and embedded segments is performed to obtain the differential forms of the flexural differential equations, respectively.

The loaded section is expressed as:

$$A_{1,i}{\left(w_2\right)}_{i+3}+B_{1,i}{\left(w_2\right)}_{i+2}-C_{1,i}{\left(w_2\right)}_{i+1}+D_{1,i}{\left(w_2\right)}_i-C_{1,i}{\left(w_2\right)}_{i-1}+B_{1,i}{\left(w_2\right)}_{i-2}+A_{1,i}{\left(w_2\right)}_{i-3}=P_{i}d$$

(10)

The embedded section is expressed as:

$$A_{2,i}{\left(w_2\right)}_{i+3}+B_{2,i}{\left(w_2\right)}_{i+2}-C_{2,i}{\left(w_2\right)}_{i+1}+D_{2,i}{\left(w_2\right)}_i-C_{2,i}{\left(w_2\right)}_{i-1}+B_{2,i}{\left(w_2\right)}_{i-2}+A_{2,i}{\left(w_2\right)}_{i-3}=\Delta P_{i}d$$

(11)

where

$$A_{1,i}=\frac{-EI{G}_i}{c_i{h}^6},B_{1,i}=\frac{6EI{G}_i}{c_i{h}^6}+\frac{EI\left(c_i+k_i\right)}{c_i{h}^4},C_{1,i}=\frac{15EI{G}_i}{c_i{h}^6}+\frac{4EI\left(c_i+k_i\right)}{c_i{h}^4}+\frac{G_{i}d}{h^2}$$

$$D_{1,i}=\frac{20EI{G}_i}{c_i{h}^6}+\frac{6EI\left(c_i+k_i\right)}{c_i{h}^4}+\frac{2G_{i}d}{h^2}+k_{i}d+\frac{k_{ti}\left(c_i+k_i\right)}{c_i}$$

$$A_{2,i}=\frac{-EI{G}_i}{c_i{h}^6},B_{2,i}=\frac{6EI{G}_i}{c_i{h}^6}+\frac{EI\left(c_i+k_i\right)}{c_i{h}^4},C_{2,i}=\frac{15EI{G}_i}{c_i{h}^6}+\frac{4EI\left(c_i+k_i\right)}{c_i{h}^4}+\frac{2G_{i}d}{h^2}$$

$$D_{2,i}=\frac{20EI{G}_i}{c_i{h}^6}+\frac{6EI\left(c_i+k_i\right)}{c_i{h}^4}+\frac{4G_{i}d}{h^2}+2k_{i}d$$

3.4. Horizontal Displacement Solution for Shear Layer of Loaded Section

The top of the diaphragm wall is considered to be a free boundary condition with wall bending moment $M_0=0$, wall shear $Q_0=0$, shear layer bending moment $M{\prime }_0=0$, and shear layer shear $Q{\prime }_0=0$. Given that the diaphragm wall in this pit is embedded in the rock layer, the bottom of the wall is considered to be a fixed boundary condition; therefore, the wall displacement $w_N=0$, the wall rotating angle ${\theta }_N=0$, the shear layer displacement $w{\prime }_N=0$, and the shear layer rotating angle $\theta {\prime }_N=0$. In a previous study [30], the expressions of the deflection, rotating angle, bending moment, and shear force of the diaphragm wall were given, and the displacement of the shear layer of the diaphragm wall in the loading section was solved by combining the boundary conditions.

In the boundary condition of the top of the wall, according to the boundary condition of the shear layer bending moment $M{\prime }_0=0$, letting i = 0, the following can be obtained:

$${\left(w_2\right)}_{-1}=a_{-1}{\left(w_2\right)}_0-b_{-1}{\left(w_2\right)}_1+c_{-1}{\left(w_2\right)}_2+d_{-1}$$

(12)

where $a_{-1}=2$, $b_{-1}=1$, $c_{-1}=0$, and $d_{-1}=0$.

According to the boundary condition of the shear layer shear $Q{\prime }_0=0$, we obtain:

$${\left(w_2\right)}_{-2}=2{\left(w_2\right)}_{-1}-{\left(w_2\right)}_1+{\left(w_2\right)}_2$$

(13)

According to the boundary condition of the wall bending moment $M_0=0$, we obtain:

$${\left(w_2\right)}_{-2}=a_{-2}{\left(w_2\right)}_{-1}-b_{-2}{\left(w_2\right)}_0+c_{-2}{\left(w_2\right)}_1+d_{-2}$$

(14)

where $a_{-2}=\frac{\left(c_0+k_0\right)h^2+6G_0}{2G_0}$, $b_{-2}=\frac{\left(c_0+k_0\right)h^2+3G_0}{G_0}$, $c_{-2}=\frac{\left(c_0+k_0\right)h^2+2G_0}{2G_0}$, and $d_{-2}=0$.

According to the boundary condition of the wall shear $Q_0=0$, we can obtain:

$${\left(w_2\right)}_0=a_0{\left(w_2\right)}_1-b_0{\left(w_2\right)}_2+c_0{\left(w_2\right)}_3+d_0$$

(15)

where $a_0=-\frac{\left(6A_{1,0}-4B_{1,0}\right)}{D_{1,0}-2C_{1,0}+4B_{1,0}-6A_{1,0}}$, $b_0=-\frac{2B_{1,0}}{D_{1,0}-2C_{1,0}+4B_{1,0}-6A_{1,0}}$, $c_0=-\frac{-2A_{1,0}}{D_{1,0}-2C_{1,0}+4B_{1,0}-6A_{1,0}}$, and $d_0=-\frac{P_{0}d}{D_{1,0}-2C_{1,0}+4B_{1,0}-6A_{1,0}}$.

Therefore, imitating Equations (12)–(15) yields:

$${\left(w_2\right)}_i=a_i{\left(w_2\right)}_{i+1}-b_i{\left(w_2\right)}_{i+2}+c_i{\left(w_2\right)}_{i+3}+d_i$$

(16)

$${\left(w_2\right)}_{i-1}=a_{i-1}{\left(w_2\right)}_i-b_{i-1}{\left(w_2\right)}_{i+1}+c_{i-1}{\left(w_2\right)}_{i+2}+d_{i-1}$$

(17)

$${\left(w_2\right)}_{i-2}=a_{i-2}{\left(w_2\right)}_{i-1}-b_{i-2}{\left(w_2\right)}_i+c_{i-2}{\left(w_2\right)}_{i+1}+d_{i-2}$$

(18)

$${\left(w_2\right)}_{i-3}=a_{i-3}{\left(w_2\right)}_{i-2}-b_{i-3}{\left(w_2\right)}_{i-1}+c_{i-3}{\left(w_2\right)}_i+d_{i-3}$$

(19)

Substituting Equations (16)–(19) into Equation (10) obtains:

$${\left(w_2\right)}_i=a_i{\left(w_2\right)}_{i+1}-b_i{\left(w_2\right)}_{i+2}+c_i{\left(w_2\right)}_{i+3}+d_i$$

(20)

where

$$a_i=-(C_{1,i}-C_{1,i}{b}_{i-1}-B_{1,i}{b}_{i-1}{a}_{i-2}+B_{1,i}{c}_{i-2}-A_{1,i}{a}_{i-3}{a}_{i-2}{b}_{i-1}+A_{1,i}{a}_{i-3}{c}_{i-2}+A_{1,i}{b}_{i-3}{b}_{i-1})/e_i$$

$$b_i=(B_{1,i}+C_{1,i}{c}_{i-1}+B_{1,i}{c}_{i-1}{a}_{i-2}+A_{1,i}{a}_{i-3}{a}_{i-2}{c}_{i-1}-A_{1,i}{b}_{i-3}{c}_{i-1})/e_i$$

$$c_i=-A_{1,i}/e_i$$

$$d_i=-(C_{1,i}{d}_{i-1}+B_{1,i}{d}_{i-1}{a}_{i-2}+B_{1,i}{d}_{i-2}+A_{1,i}{a}_{i-3}{a}_{i-2}{d}_{i-1}+A_{1,i}{a}_{i-3}{d}_{i-2}-A_{1,i}{b}_{i-3}{d}_{i-1}+A_{1,i}{d}_{i-3}-P_{i}d)/e_i$$

$$e_i=D_{1,i}+C_{1,i}{a}_{i-1}+B_{1,i}{a}_{i-1}{a}_{i-2}-B_{1,i}{b}_{i-2}+A_{1,i}{a}_{i-1}{a}_{i-2}{a}_{i-3}-A_{1,i}{a}_{i-3}{b}_{i-2}-A_{1,i}{b}_{i-3}{b}_{i-1}+A_{1,i}{c}_{i-3}$$

The horizontal displacement of any node in the shear layer of the loaded section of the wall can be obtained by recursion through Equations (17)–(20).

3.5. Solution of Horizontal Displacement of Shear Layer in Embedded Section

The differential form of the shear layer flexural differential equation for the embedded section is given in Equation (11).

Given that the underground continuous wall in the foundation pit is embedded in the rock layer, the bottom of the wall is considered to be a fixed boundary condition. At this time, the wall displacement $w_N=0$, the wall rotation angle ${\theta }_N=0$, the shear layer displacement ${\left(w_2\right)}_N=0$, and the shear layer rotation angle ${\left({\theta }_2\right)}_N=0$. According to the above boundary conditions and letting i = N − 1, i = N − 2, and i = N − 3, respectively, the embedded differential equations ${\left(w_2\right)}_{N-1}$, ${\left(w_2\right)}_{N-2}$, and ${\left(w_2\right)}_{N-3}$ of the different nodes can be obtained.

When i = N − 1, the embedded segment differential equation can be expressed as:

$${\left(w_2\right)}_{N-1}={\left(a_2\right)}_{N-1}{\left(w_2\right)}_{N-2}-{\left(b_2\right)}_{N-1}{\left(w_2\right)}_{N-3}+{\left(c_2\right)}_{N-1}{\left(w_2\right)}_{N-4}+{\left(d_2\right)}_{N-1}$$

(21)

where ${\left(a_2\right)}_{N-1}=-\left(\frac{A_{2,\left(N-1\right)}-C_{2,\left(N-1\right)}}{B_{2,\left(N-1\right)}+D_{2,\left(N-1\right)}}\right)$,${\left(b_2\right)}_{N-1}=-\left(\frac{B_{2,\left(N-1\right)}}{B_{2,\left(N-1\right)}+D_{2,\left(N-1\right)}}\right)$

$${\left(c_2\right)}_{N-1}=-\left(\frac{A_{2,\left(N-1\right)}}{B_{2,\left(N-1\right)}+D_{2,(N-1)}}\right),{\left(d_2\right)}_{N-1}=\left(\frac{\Delta P_{\left(N-1\right)}d}{B_{2,\left(N-1\right)}+D_{2,\left(N-1\right)}}\right)$$

When i = N − 2, the embedded segment differential equation can be expressed as:

$${\left(w_2\right)}_{N-2}={\left(a_2\right)}_{N-2}{\left(w_2\right)}_{N-3}-{\left(b_2\right)}_{N-2}{\left(w_2\right)}_{N-4}+{\left(c_2\right)}_{N-2}{\left(w_2\right)}_{N-5}+{\left(d_2\right)}_{N-2}$$

(22)

where ${\left(a_2\right)}_{N-2}=-\frac{\left(A_{2,(N-2)}-C_{2(N-2)}\right){\left(b_2\right)}_{N-1}+C_{2,(N-2)}}{\left(A_{2,\left(N-2\right)}-C_{2\left(N-2\right)}\right){\left(a_2\right)}_{N-1}+D_{2,\left(N-2\right)}}$, ${\left(b_2\right)}_{N-2}=-\frac{\left(A_{2,\left(N-2\right)}-C_{2,\left(N-2\right)}\right){\left(c_2\right)}_{N-1}+B_{2,\left(N-2\right)}}{\left(A_{2,\left(N-2\right)}-C_{2,\left(N-2\right)}\right){\left(a_2\right)}_{N-1}+D_{2,\left(N-2\right)}}$, and ${\left(c_2\right)}_{N-2}=\frac{-A_{2,\left(N-2\right)}}{\left(A_{2,\left(N-2\right)}-C_{2,\left(N-2\right)}\right){\left(a_2\right)}_{N-1}+D_{2,\left(N-2\right)}}$${\left(d_2\right)}_{N-2}=-\left(\frac{\Delta P_{N-2}d-\left(A_{2,\left(N-2\right)}-C_{2,\left(N-2\right)}\right){\left(d_2\right)}_{N-1}}{\left(A_{2,\left(N-2\right)}-C_{2,\left(N-2\right)}\right){\left(a_2\right)}_{N-1}+D_{2,\left(N-2\right)}}\right)$

When i = N − 3, the embedded segment differential equation can be expressed as:

$${\left(w_2\right)}_{N-3}={\left(a_2\right)}_{N-3}{\left(w_2\right)}_{N-4}-{\left(b_2\right)}_{N-3}{\left(w_2\right)}_{N-5}+{\left(c_2\right)}_{N-3}{\left(w_2\right)}_{N-6}+{\left(d_2\right)}_{N-3}$$

(23)

where

${\left(a_2\right)}_{N-3}=-\left(\frac{B_{2,\left(N-3\right)}\left({\left(c_2\right)}_{N-1}-{\left(a_2\right)}_{N-1}{\left(b_2\right)}_{N-2}\right)+C_{2,\left(N-3\right)}\left({\left(b_2\right)}_{N-2}-1\right)}{B_{2,\left(N-3\right)}\left({\left(a_2\right)}_{N-1}{\left(a_2\right)}_{N-2}-{\left(b_2\right)}_{N-1}\right)-C_{2,\left(N-3\right)}{\left(a_2\right)}_{N-2}+D_{2,\left(N-3\right)}}\right)$, ${\left(b_2\right)}_{N-3}=\frac{B_{2,\left(N-3\right)}\left({\left(a_2\right)}_{N-1}{\left(c_2\right)}_{N-2}+1\right)-C_{2,\left(N-3\right)}{\left(c_2\right)}_{N-2}}{B_{2,\left(N-3\right)}{\left(a_2\right)}_{N-1}{\left(c_2\right)}_{N-2}-C_{2,\left(N-3\right)}{\left(c_2\right)}_{N-2}+B_{2,\left(N-3\right)}}$, ${\left(c_2\right)}_{N-3}=-\frac{A_{2,\left(N-3\right)}}{B_{2,\left(N-3\right)}\left({\left(a_2\right)}_{N-1}{\left(a_2\right)}_{N-2}-{\left(b_2\right)}_{N-1}\right)-C_{2,\left(N-3\right)}{\left(a_2\right)}_{N-2}+D_{2,\left(N-3\right)}}$, ${\left(d_2\right)}_{N-3}=-\left(\frac{B_{2,\left(N-3\right)}\left({\left(a_2\right)}_{N-1}{\left(d_2\right)}_{N-2}+{\left(d_2\right)}_{N-1}\right)-C_{2,\left(N-3\right)}{\left(d_2\right)}_{N-2}-\Delta P_{\left(N-3\right)}d}{B_{2,\left(N-3\right)}\left({\left(a_2\right)}_{N-1}{\left(a_2\right)}_{N-2}-{\left(b_2\right)}_{N-1}\right)-C_{2,\left(N-3\right)}{\left(a_2\right)}_{N-2}+D_{2,\left(N-3\right)}}\right)$

Imitating Equations (21)–(23) yields:

$${\left(w_2\right)}_i={\left(a_2\right)}_i{\left(w_2\right)}_{i-1}-{\left(b_2\right)}_i{\left(w_2\right)}_{i-2}+{\left(c_2\right)}_i{\left(w_2\right)}_{i-3}+{\left(d_2\right)}_i$$

(24)

$${\left(w_2\right)}_{i+1}={\left(a_2\right)}_{i+1}{\left(w_2\right)}_i-{\left(b_2\right)}_{i+1}{\left(w_2\right)}_{i-1}+{\left(c_2\right)}_{i+1}{\left(w_2\right)}_{i-2}+{\left(d_2\right)}_{i+1}$$

(25)

$${\left(w_2\right)}_{i+2}={\left(a_2\right)}_{i+2}{\left(w_2\right)}_{i+1}-{\left(b_2\right)}_{i+2}{\left(w_2\right)}_i+{\left(c_2\right)}_{i+2}{\left(w_2\right)}_{i-1}+{\left(d_2\right)}_{i+2}$$

(26)

$${\left(w_2\right)}_{i+3}={\left(a_2\right)}_{i+3}{\left(w_2\right)}_{i+2}-{\left(b_2\right)}_{i+3}{\left(w_2\right)}_{i+1}+{\left(c_2\right)}_{i+3}{\left(w_2\right)}_i+{\left(d_2\right)}_{i+3}$$

(27)

Substituting Equations (25)–(27) into Equation (11) and organizing them yields:

$${\left(w_2\right)}_i={\left(a_2\right)}_i{\left(w_2\right)}_{i-1}-{\left(b_2\right)}_i{\left(w_2\right)}_{i-2}+{\left(c_2\right)}_i{\left(w_2\right)}_{i-3}+{\left(d_2\right)}_i$$

(28)

where

$${\left(a_2\right)}_i=-\left(\frac{C_{2,i}{\left(b_2\right)}_{i+1}-C_{2,i}-B_{2,i}{\left(a_2\right)}_{i+2}{\left(b_2\right)}_{i+1}+B_{2,i}{\left(c_2\right)}_{i+2}-A_{2,i}{\left(a_2\right)}_{i+3}{\left(a_2\right)}_{i+2}{\left(b_2\right)}_{i+1}+A_{2,i}{\left(a_2\right)}_{i+3}{\left(c_2\right)}_{i+2}+A_{2,i}{\left(b_2\right)}_{i+3}{\left(b_2\right)}_{i+1}}{-C_{2,i}{\left(a_2\right)}_{i+1}+B_{2,i}{\left(a_2\right)}_{i+2}{\left(a_2\right)}_{i+1}-B_{2,i}{\left(b_2\right)}_{i+2}+A_{2,i}{\left(a_2\right)}_{i+3}{\left(a_2\right)}_{i+2}{\left(a_2\right)}_{i+1}-A_{2,i}{\left(a_2\right)}_{i+3}{\left(b_2\right)}_{i+2}-A_{2,i}{\left(b_2\right)}_{i+3}{\left(a_2\right)}_{i+1}+A_{2,i}{\left(c_2\right)}_{i+3}+D_{2,i}}\right)$$

$${\left(b_2\right)}_i=\frac{-C_{2,i}{\left(c_2\right)}_{i+1}+B_{2,i}{\left(a_2\right)}_{i+2}{\left(c_2\right)}_{i+1}+A_{2,i}{\left(a_2\right)}_{i+3}{\left(a_2\right)}_{i+2}{\left(c_2\right)}_{i+1}-A_{2,i}{\left(b_2\right)}_{i+3}{\left(c_2\right)}_{i+1}+B_{2,i}}{-C_{2,i}{\left(a_2\right)}_{i+1}+B_{2,i}{\left(a_2\right)}_{i+2}{\left(a_2\right)}_{i+1}-B_{2,i}{\left(b_2\right)}_{i+2}+A_{2,i}{\left(a_2\right)}_{i+3}{\left(a_2\right)}_{i+2}{\left(a_2\right)}_{i+1}-A_{2,i}{\left(a_2\right)}_{i+3}{\left(b_2\right)}_{i+2}-A_{2,i}{\left(b_2\right)}_{i+3}{\left(a_2\right)}_{i+1}+A_{2,i}{\left(c_2\right)}_{i+3}+D_{2,i}}$$

$${\left(c_2\right)}_i=\frac{-A_{2,i}}{-C_{2,i}{\left(a_2\right)}_{i+1}+B_{2,i}{\left(a_2\right)}_{i+2}{\left(a_2\right)}_{i+1}-B_{2,i}{\left(b_2\right)}_{i+2}+A_{2,i}{\left(a_2\right)}_{i+3}{\left(a_2\right)}_{i+2}{\left(a_2\right)}_{i+1}-A_{2,i}{\left(a_2\right)}_{i+3}{\left(b_2\right)}_{i+2}-A_{2,i}{\left(b_2\right)}_{i+3}{\left(a_2\right)}_{i+1}+A_{2,i}{\left(c_2\right)}_{i+3}+D_{2,i}}$$

$${\left(d_2\right)}_i=-\left(\frac{-C_{2,i}{\left(d_2\right)}_{i+1}+B_{2,i}{\left(a_2\right)}_{i+2}{\left(d_2\right)}_{i+1}+B_{2,i}{\left(d_2\right)}_{i+2}+A_{2,i}{\left(a_2\right)}_{i+3}{\left(a_2\right)}_{i+2}{\left(d_2\right)}_{i+1}+A_{2,i}{\left(a_2\right)}_{i+3}{\left(d_2\right)}_{i+2}-A_{2,i}{\left(b_2\right)}_{i+3}{\left(d_2\right)}_{i+1}+A_{2,i}{\left(d_2\right)}_{i+3}-\Delta P_{i}d}{-C_{2,i}{\left(a_2\right)}_{i+1}+B_{2,i}{\left(a_2\right)}_{i+2}{\left(a_2\right)}_{i+1}-B_{2,i}{\left(b_2\right)}_{i+2}+A_{2,i}{\left(a_2\right)}_{i+3}{\left(a_2\right)}_{i+2}{\left(a_2\right)}_{i+1}-A_{2,i}{\left(a_2\right)}_{i+3}{\left(b_2\right)}_{i+2}-A_{2,i}{\left(b_2\right)}_{i+3}{\left(a_2\right)}_{i+1}+A_{2,i}{\left(c_2\right)}_{i+3}+D_{2,i}}\right)$$

The horizontal displacement of any node in the shear layer of the embedded section can be obtained by recursion through Equations (24)–(27).

3.6. Overall Differential Calculation and Analysis of Wall Displacement

With reference to the differential diagram of the supporting structure in Figure 4, because the load section and the embedded section of the wall meet the continuous conditions of the shear layer of the boundary surface and the continuous conditions of the wall at the excavation surface K, combined with the boundary conditions of the top and bottom of the wall, the linear equations can be expressed by a matrix:

$$\left[a\right]\left[\left(w_2\right)\right]=\left[d\right]$$

(29)

where

$$\left[a\right]=\left[\begin{array}{cccccccccccccc}& & & 1& & & & & & & -1& & & \\ & & -1& & 1& & & & & 1& & -1& & \\ & & 1& -2& 1& & & & & -1& 2& -1& & \\ & -1& 2& & -2& 1& & & 1& -2& & 2& -1& \\ J_1& -L_1& M_1& & -M_1& L_1& -J_1& -J_1& L_1& -M_1& & M_1& -L_1& J_1\\ & J_2& -L_2& & L_2& -J_2& & & -J_2& L_2& & -L_2& J_2& \\ & & & 1& -a_K& b_K& -c_K& & & & & & & \\ & & 1& -a_{K-1}& b_{K-1}& -c_{K-1}& & & & & & & & \\ & 1& -a_{K-2}& b_{K-2}& -c_{K-2}& & & & & & & & & \\ 1& -a_{K-3}& b_{K-3}& -c_{K-3}& & & & & & & & & & \\ & & & & & & & -{\left(c_2\right)}_K& {\left(b_2\right)}_K& -{\left(a_2\right)}_K& 1& & & \\ & & & & & & & & -{\left(c_2\right)}_{K+1}& {\left(b_2\right)}_{K+1}& -{\left(a_2\right)}_{K+1}& 1& & \\ & & & & & & & & & -{\left(c_2\right)}_{K+2}& {\left(b_2\right)}_{K+2}& -{\left(a_2\right)}_{K+2}& 1& \\ & & & & & & & & & & -{\left(c_2\right)}_{K+3}& {\left(b_2\right)}_{K+3}& -{\left(a_2\right)}_{K+3}& 1\end{array}\right]$$

$$\left[w_2\right]={\left[\begin{array}{cccccccccccccc}{\left(w_2\right)}_{k-3}& {\left(w_2\right)}_{k-2}& {\left(w_2\right)}_{k-1}& {\left(w_2\right)}_k& {\left(w_2\right)}_{k+1}& {\left(w_2\right)}_{k+2}& {\left(w_2\right)}_{k+3}& {\left(w_2\right)}_{k-3}^{\prime }& {\left(w_2\right)}_{k-2}^{\prime }& {\left(w_2\right)}_{k-1}^{\prime }& {\left(w_2\right)}_k^{\prime }& {\left(w_2\right)}_{k+1}^{\prime }& {\left(w_2\right)}_{k+2}^{\prime }& {\left(w_2\right)}_{k+3}^{\prime }\end{array}\right]}^T$$

$$\left[d\right]={\left[\begin{array}{cccccccccccccc}0& 0& 0& 0& 0& 0& d_{K-3}& d_{K-2}& d_{K-1}& d_K& {\left(d_2\right)}_K& {\left(d_2\right)}_{K+1}& {\left(d_2\right)}_{K+2}& {\left(d_2\right)}_{K+3}\end{array}\right]}^T$$

$$J_1=\frac{G_0}{2c_0{h}^5},H_1=\frac{c_0+k_0}{2c_0{h}^3},J_2=\frac{G_0}{2c_0{h}^3},H_2=\frac{c_0+k_0}{2c_{0}h},L_1=H_1+5J_1,M_1=2H_1+7J_1,L_2=H_2+2J_2.$$

${\left(w_2\right)}_{K-3}-{\left(w_2\right)}_{K+3}$ is the displacement at point K of the excavation surface and its seven adjacent nodes calculated by the loaded section; ${\left(w_2\right)}_{K-3}^{\prime }-{\left(w_2\right)}_{K+3}^{\prime }$ is the displacement at point K of the excavation surface and its seven adjacent nodes calculated by the embedded section. The displacement at the excavation surface can be calculated by Equation (29), and then the displacement at any node of the shear layer can be solved iteratively by Equations (20) and (28). Finally, the horizontal displacement at any node of the diaphragm wall can be obtained using Equation (2). This process needs to be solved with the help of MATLAB software (MATLAB R2017a).

4. Calculation of Surface Settlement Caused by Excavation of Foundation Pit

4.1. Virtual Force Inversion

According to the boundary element method, the pit boundary generates the virtual stress field and virtual displacement field corresponding to the soil outside the pit under the influence of the virtual force. When the pit boundary conditions in the virtual displacement field are consistent with the actual pit displacement boundary conditions, it can be considered that the virtual displacement field and virtual force field are generated by the virtual force in response to the boundary effect caused by the pit excavation.

Mindlin [22] gave a displacement solution at any position when a horizontal concentrated load is applied in an elastic semi-infinite space, For a composite soil layer, weighted average parameters can be used for the soil layer, as shown in Figure 4.

The horizontal displacement of point A caused by the horizontal concentration force $P_x\left(B\right)$ at point B inside the semi-infinite space is expressed as

$$U\left(A\right)=K_x\left(A,B\right)\cdot P_x\left(B\right)$$

(30)

$$K_x\left(A,B\right)=\frac{1}{16\pi G_d\left(1-v_d\right)}\left[\begin{array}{l}\frac{3-4v_d}{R_1}+\frac{1}{R_2}+\frac{X^2}{R_1^3}+\frac{\left(3-4v_d\right)X^2}{R_2^3}+\frac{2lz}{R_2^3}\left(1-\frac{3X^2}{R_2^3}\right)\\ +\frac{4\left(1-v_d\right)\left(1-2v_d\right)}{R_2+Z_2}\left(1-\frac{X^2}{R_2\left(R_2+Z_2\right)}\right)\end{array}\right]$$

(31)

where $v_d$ is the equivalent Poisson’s ratio of the soil layer. G_d is the equivalent shear modulus of the soil layer. x, y, and z represent the coordinates of the displacement calculation point A, and u, v, and l represent the coordinates of the load point B, respectively: $X=x-u$, $Y=y-v$, $Z_1=z-l$, $Z_2=z+l$, $R_1=\sqrt{X^2+Y^2+Z_1^2}$, and $R_2=\sqrt{X^2+Y^2+Z_2^2}$.

As shown in Figure 5, assuming that a virtual force acts on the boundary where the pit enclosure is located, the following boundary integral equation can be constructed by integrating Equation (30) along this boundary:

$$U\left(A\right)={\displaystyle {\int }_L{K}_x\left(A,B\right)\cdot P_x\left(B\right)dL}$$

(32)

where L is the pit boundary, and $P_x\left(B\right)$ is the virtual force acting at point B on L.

If L is subdivided sufficiently to consider the virtual forces on each segment to be uniformly distributed, then Equation (32) can be rewritten as:

$$U\left(A\right)={\displaystyle \sum _{i=1}^m{P}_x\left(B_i\right)\cdot }{\displaystyle {\int }_{L_i}{K}_x\left(A,B_i\right)d{L}_i}$$

(33)

where L_i is the i-th segment of L; $P_x\left(B_i\right)$ is the horizontal uniform concentration force acting on the i-th segment; and i = 1~m, where m is the total number of segments of the pit boundary.

Taking the calculated displacement of the diaphragm wall as the actual displacement boundary condition, and considering that point A is located on the boundary, the following matrix expression can be obtained from Equation (34):

$$\left[\begin{array}{c}{U}_x\left(A_1\right)\\ ⋮\\ U_x\left(A_j\right)\\ ⋮\\ U_x\left(A_m\right)\end{array}\right]=\left[\begin{array}{ccccc}{G}_{11}& \cdots & G_{1j}& \cdots & G_{1m}\\ ⋮& \ddots & ⋮& ⋰& ⋮\\ G_{j1}& \cdots & G_{jj}& \cdots & G_{jm}\\ ⋮& ⋰& ⋮& \ddots & ⋮\\ G_{m1}& \cdots & G_{mj}& \cdots & G_{mm}\end{array}\right]\cdot \left[\begin{array}{c}{P}_x\left(B_1\right)\\ ⋮\\ P_x\left(B_j\right)\\ ⋮\\ P_x\left(B_m\right)\end{array}\right]$$

(34)

where $G_{ji}={\displaystyle {\int }_{L_i}{K}_x}\left(A,B_i\right)d{L}_i$

Equation (34) can be abbreviated as:

$$P_x=G^{-1}\cdot U_x$$

(35)

In this equation, the matrix G_ji and the actual boundary conditions U_x are known, and the virtual additional horizontal stress P_x on the boundary of the pit can be obtained by solving the matrix relation equation using the MATLAB software (MATLAB R2017a).

4.2. Calculation of the Surface Settlement

Pit excavation creates additional stresses, which lead to surface settlement, as shown in Figure 6.

$$w_z=\frac{PX}{16\pi G\left(1-v_d\right)}\left[\frac{Z_1}{R_1^3}+\frac{\left(3-4v_d\right)Z_1}{R_2^3}-\frac{6wz{Z}_2}{R_2^5}+\frac{4\left(1-v_d\right)\left(1-2v_d\right)}{R_2\left(R_2+Z_2\right)}\right]$$

(36)

Using the obtained additional horizontal stress and the Mindlin vertical displacement solution, the following settlement calculation matrix expression is obtained:

$$\left[\begin{array}{c}{U}_z\left(C_1\right)\\ ⋮\\ U_z\left(C_j\right)\\ ⋮\\ U_z\left(C_m\right)\end{array}\right]=\left[\begin{array}{ccccc}{K}_{11}& \cdots & K_{1j}& \cdots & K_{1m}\\ ⋮& \ddots & ⋮& ⋰& ⋮\\ K_{j1}& \cdots & K_{jj}& \cdots & K_{jm}\\ ⋮& ⋰& ⋮& \ddots & ⋮\\ K_{m1}& \cdots & K_{mj}& \cdots & K_{mm}\end{array}\right]\cdot \left[\begin{array}{c}{P}_x\left(B_1\right)\\ ⋮\\ P_x\left(B_j\right)\\ ⋮\\ P_x\left(B_m\right)\end{array}\right]$$

(37)

where $K_{ji}={\displaystyle {\int }_{L_i}{\displaystyle {\int }_v{K}_z}\left(C,B_i\right)d{L}_i}dv$.

$U_z\left(C_j\right)$ is the settlement at different locations. The surface settlement caused by the foundation excavation can finally be calculated using Equation (37).

A flowchart for calculating the displacement of the diaphragm wall and surface settlement is shown in Figure 7.

5. Engineering Example Analysis

5.1. Project Overview

For the subway station in Xiamen, the diaphragm wall is made of C30 concrete with a depth of 24.5 m and a thickness of 1000 mm. The first is a concrete support, with a horizontal spacing of 5 m and an elasticity modulus of 30 GPa; the second and third supports are $\phi$ = 609, with t = 16 mm steel support, a horizontal spacing of 5 m, and an elasticity modulus of 200 GPa. The foundation support form is shown in Figure 8.

According to the monitoring program of this project, the horizontal displacement is monitored by the inclinometer tube, which is laid along the depth direction of the whole diaphragm wall. The surface settlement is monitored by a level meter, and the monitoring points are set at 3 m, 8 m, 13 m, 18 m, and 23 m from the diaphragm wall, as shown in Figure 9 for the field measurement point layout.

5.2. Engineering Geology

The soil layers are, from top to bottom, plain fill, 1–2 residual sand cohesive soil, 1–3 residual sand cohesive soil, completely weathered granite, and scattered strongly weathered granite. The mechanical parameters are shown in

Table 1. Mechanical parameters of soil layers.

Soil Layer	Thickness /m	$\mathbf{Gravity}\mathit{\gamma }$$/(\mathbf{kN}\cdot {\mathbf{m}}^{-3}$)	Cohesion c/kPa	Internal Friction Angle φ/°	Elastic Modulus/MPa	Poisson’s Ratio/v
Plain fill	1.6	17.8	8	10	7	0.3
1–2 residual sand cohesive soil	4	18.5	20	15	5.26	0.3
1–3 residual sand cohesive soil	4.2	18.4	22	16	5.58	0.3
Completely weathered granite	7.7	19.7	28	19	8.21	0.28
Scattered strongly weathered granite	7	20.5	21	28	9.56	0.25

.

When Mindlin’s theory is used to calculate layered soils, the weighted average parameters of the soil layers are usually required. The properties of the weighted average soil layer are shown in

Table 2. Weighted average parameters of soil layer.

Soil Layer Name	Thickness /m	$\mathbf{Gravity}\mathit{\gamma }$$/(\mathbf{kN}\cdot {\mathbf{m}}^{-3}$)	Cohesion c/kPa	Internal Friction Angle φ/°	Elastic Modulus E/MPa	Poisson’s Ratio/v
Weighted average soil layer	24.5	19.39	22.36	19.82	7.58	0.28

[31,32].

5.3. Theoretical Validation

The calculation results of comparing the theoretical calculation and the two-parameter Pasternak model with the actual measured data in the field to verify the rationality of the theory of calculating the horizontal displacement are shown in Figure 10a. The results show that the theoretical calculation results based on Kerr’s three-parameter model in this paper are more accurate and closer to the field monitoring results compared with the theoretical calculation results of Pasternak’s two-parameter model. The maximum horizontal displacement calculated by the two theories is 17.13 mm and 16.99 mm, respectively, which are very close to each other. However, the development trend of the horizontal displacement varies greatly, and the theoretical calculation results of the three-parameter Kerr model are more accurate compared with the monitoring results. As for the depth of the maximum horizontal displacement, the Kerr model’s value was calculated to be 9.5 m, and the Pasternak model’s value was calculated to be 11.5 m. The Kerr model’s value is closer to the field data, which proves that the three-parameter Kerr model has a better calculation accuracy.

The theoretical calculations show that the displacement at the top is 2.4 mm, and the horizontal displacement at the position 9.5 m below the ground is the largest, which is 17.1 mm, and the bottom of the diaphragm wall, because of being embedded in the rock, has a displacement of 0. As for the on-site monitoring, the horizontal displacement at the top of the diaphragm wall is 1.69 mm, and the horizontal displacement at the position 10 m below the ground is the largest, where the horizontal displacement is 18.1 mm, and the displacement at the bottom is 0. In terms of the size and the trend of horizontal displacement, the theoretical calculation is basically the same as the field data, and the maximal error is not more than 5.5%. The theoretical calculation and the field data are basically consistent, which proves the reasonableness of the prediction of the horizontal displacement of the diaphragm wall.

As for the surface settlement, the surface settlement data were analyzed by comparing the three-parameter Kerr model, the two-parameter Pasternak model, and the field measurement results, as shown in Figure 10b. In terms of the surface settlement distribution, the settlement distribution of the three-parameter Kerr model is closer to the field monitoring data than that of the two-parameter Pasternak model. In terms of the maximum settlement, the settlements calculated by the two theories are 15.54 mm and 15.62 mm respectively, which are very close to the calculated results. As for the location of the maximum settlement, the maximum settlement of the three-parameter Kerr model is 6.5 m away from the diaphragm wall, the maximum settlement of the two-parameter Pasternak model is 13 m away from the diaphragm wall, and the maximum settlement of the on-site monitoring is 8 m away from the diaphragm wall. It is obvious that the results of the calculation of the surface settlement based on the three-parameter Kerr model are closer to the actual situation. This proves that the three-parameter Kerr model has a better calculation accuracy than the two-parameter Pasternak model.

The theoretical calculations show that the surface settlement shows a trend of “increasing first and then decreasing” with the increase in the distance from the diaphragm wall, and the size of the surface settlement ranges from 0 to 15.54 mm, and the range of the surface settlement is from 0 to 36 m. The maximum settlement is 15.54 mm at 6.5 m from the diaphragm wall. As the distance from the diaphragm wall increases, the surface settlement slowly decreases, and the displacement is 0 at the position of 36 m from the diaphragm wall. The on-site monitoring data show the same trend, with a maximum settlement of 14.9 mm, and the maximum error is not more than 4.3%. The measured values of the surface settlement and the theoretical predicted values are given in Figure 10b. This theory can accurately reflect the surface settlement of the pit on site.

5.4. Influence of Wall Thickness

This study further analyzed the effects of five variables, namely, the thickness of diaphragm wall D, the elasticity modulus of diaphragm wall E, the diameter of internal support φ, the number of internal supports S, and the geological conditions of the soil strata. When one of the variables is studied, the other four variables are consistent with the actual working conditions (D = 1.0 m, E = 30 GPa, φ = 609 mm, S = 3 sections, and the site soil layer).

Five working conditions were set: D = 0.6 m, 0.8 m, 1.0 m, 1.2 m, and 1.4 m. The distribution law of displacement of the diaphragm wall and settlement under different thicknesses of the wall was analyzed. Figure 11 shows the horizontal displacement curves and surface settlement curves with different wall thicknesses, respectively.

As shown in Figure 11a, the horizontal displacement of the wall decreases by increasing the thickness of the wall, which is due to the increased bending stiffness of the diaphragm wall. The maximum displacements of the walls all appear at a depth of about 9.5 m. This is because the soil pressure and support force are fixed, and the magnitude and location of the loads are the same; thus, changing the thickness only increases the resistance to deformation at the depth of the corresponding ground link wall. The maximum horizontal displacements are 14.55 mm, 15.87 mm, 17.13 mm, 18.38 mm, and 19.39 mm, respectively. The maximum difference is 4.84 mm, which is 28.3% of the horizontal displacement under field conditions, and the effect of increasing the wall thickness on controlling the horizontal displacement of the diaphragm wall is obvious.

As shown in Figure 11b, the wall thickness increases, and the corresponding surface settlement decreases. This is because an increase in the thickness of the diaphragm wall leads to a decrease in the horizontal displacement of the diaphragm wall and a decrease in the additional stress, so the surface settlement decreases. The maximum surface settlements are 12.89 mm, 14.26 mm, 15.54 mm, 16.79 mm, and 17.64 mm, respectively. The maximum settlement difference is 4.75 mm, which is 30.6% of the maximum settlement at the site, indicating that increasing the thickness of the diaphragm wall has a significant effect on controlling surface settlement.

5.5. Influence of Elasticity Modulus

Four working conditions are set for the elastic modulus of the diaphragm wall, with elastic moduli of E = 28 GPa, 30 GPa, 31.5 GPa, and 32.5 GPa, and the corresponding concrete grades are C25, C30, C35, and C40, respectively. Figure 12 shows the horizontal displacement curves and surface settlement curves with different elasticity moduli of the wall, respectively.

As shown in Figure 12a, the displacement decreases as the elasticity modulus increases because the flexural stiffness of the diaphragm wall increases when the elasticity modulus increases, and, therefore, the horizontal displacement of the diaphragm wall decreases.

The displacement of the diaphragm wall decreases as the elasticity modulus increases. This is due to the increase in flexural stiffness, and, hence, the horizontal displacement of the diaphragm wall decreases. The maximum displacement of the wall appears at the depth of the diaphragm wall, at about 9.5 m. The maximum horizontal displacements are 15.62 mm, 16.37 mm, 17.13 mm, and 18.05 mm, respectively. The maximum displacement difference is 2.43 mm, which is 14.2% of the horizontal displacement of the field condition, indicating that using higher-grade concrete and increasing the modulus of elasticity is effective in controlling the horizontal displacement of a diaphragm wall.

As shown in Figure 12b, the increase in the elastic modulus reduces the surface settlement, which is because the increase in the elastic modulus of the diaphragm wall leads to a reduction in the horizontal displacement of the diaphragm wall and a reduction in additional stresses, which eventually produces the reduction in the surface settlement. The maximum surface settlements are 14.11 mm, 14.83 mm, 15.54 mm, and 16.59 mm, respectively. The maximum settlement difference is 2.48 mm, which is 16.0% of the maximum settlement under field conditions, indicating that increasing the modulus of elasticity of the diaphragm wall has a significant effect on controlling the surface settlement.

5.6. Influence of Internal Support Diameter

To analyze the influence of the diameter of the internal support on the horizontal displacement of the diaphragm wall, four working conditions were set up, with the diameter of the second steel support as the representative. The diameters of the internal support were taken as φ = 500 mm, φ = 609 mm (consistent with the field), φ = 700 mm, and φ = 800 mm to analyze the influence of the change in the stiffness of the internal support on the horizontal displacement of the diaphragm wall and surface settlement with different diameters. Figure 13 shows the horizontal displacement curves and surface settlement curves with different diameters of the internal support, respectively.

As shown in Figure 13a, the displacement decreases when the diameter of the second internal support increases. The maximum horizontal displacements are 15.88 mm, 16.48 mm, 17.13 mm, and 17.65 mm, respectively. The maximum displacement difference is 1.77 mm, which is 10% of the horizontal displacement under the current working conditions, indicating that using a larger diameter internal support and increasing the compressive stiffness of the support has an obvious effect on controlling the horizontal displacement of the diaphragm wall.

As shown in Figure 13b, the diameter of the second internal support increases, the surface settlement decreases, and the maximum surface settlements are 14.19 mm, 14.92 mm, 15.54 mm, and 16.02 mm respectively. The maximum settlement difference is 1.83 mm, which is 11.8% of the maximum settlement under the site conditions, indicating that increasing the diameter of the internal support and changing the compressive stiffness of the support have obvious effects on controlling the surface settlement.

5.7. Influence of the Number of Internal Supports

Four working conditions were set for the number of internal supports, and the number of internal supports was adjusted to 2, 3, 4, and 5, as shown in

Table 3. Arrangement of internal support under different working conditions.

Construction Condition	Construction Condition 1		Construction Condition 2		Construction Condition 3		Construction Condition 4
	Support Type	Depth (m)	Support Type	Depth (m)	Support Type	Depth (m)	Support Type	Depth (m)
The first support	Concrete support	0.5	Concrete support	0.5	Concrete support	0.5	Concrete support	0.5
The second support	Steel support	4	Steel support	7.4	Concrete support	4	Concrete support	4
The third support			Steel support	12.9	Steel support	7.4	Steel support	7.4
The fourth support					Steel support	12.9	Steel support	12.9
The fifth support							Steel support	16

. Figure 14 shows the horizontal displacement curves and surface settlement curves under different numbers of internal supports, respectively.

As shown in Figure 14a, the displacement of the wall decreases as the number of internal supports increases. The maximum horizontal displacements are 14.68 mm, 15.50 mm, 17.13 mm, and 19.45 mm, respectively. The maximum displacement difference is 4.77 mm, which is 27.8% of the horizontal displacement under the current working conditions.

As shown in Figure 14b, the surface settlement decreases as the number of internal supports increases because the horizontal displacement of the diaphragm wall decreases as the number of internal supports increases, and the additional stress decreases, which eventually leads to a decrease in the surface settlement. The maximum surface settlements are 13.06 mm, 14.05 mm, 15.54 mm, and 18.42 mm, respectively. The maximum settlement difference is 5.36 mm, which is 34.5% of the maximum settlement under the site working conditions, and increasing the number of internal supports has a significant effect on controlling the surface settlement.

5.8. Influence of Soil Properties

According to the properties of the soil layer, four working conditions are set up: the site soil layer, 1–2 residual sandy clay soil, scattered strongly weathered granite, and medium coarse sand. The medium coarse sand layer weight is 17.7 $\mathrm{kN}\cdot {\mathrm{m}}^{-3}$, the cohesion is 24 kPa, the internal friction angle is 21°, the compression modulus is 7.53 MPa, and Poisson’s ratio is 0.25. Figure 15 shows the displacement of the diaphragm wall and surface settlement curves in different soil layers.

As shown in Figure 15a, the displacement of the ground link wall and the horizontal displacement of the top of the wall are 1–2 residual sand cohesive soil > the site soil layer > medium coarse sand > scattered strongly weathered granite. The maximum horizontal displacements are 11.53 mm, 15.87 mm, 17.13 mm, and 22.38 mm, respectively. The greater the strength of the soil layer, the smaller the horizontal displacement.

As shown in Figure 15b, the surface settlement near the diaphragm wall in the residual sandy clay soil is the largest, and the settlement of the remaining soil layers is equal, which is due to the large horizontal displacement of the top of the diaphragm wall in the residual sand cohesive soil and the weak soil quality, so there is a large surface settlement near the diaphragm wall. The size of the surface settlement is in the order of 1–2 residual sand cohesive soil, the site soil layer, medium coarse sand layer, and scattered strongly weathered granite. The maximum surface settlements are 11.10 mm, 14.87 mm, 15.54 mm, and 20.50 mm, respectively. The overall distribution rule is that the greater the strength of the soil layer, the smaller the soil settlement, and the location of the maximum settlement is farther from the foundation pit.

6. Conclusions

In this paper, the three-parameter Kerr foundation model was applied to a diaphragm wall structure, and a new method for calculating the horizontal displacement of the diaphragm wall was proposed. We established the connection between the horizontal displacement of the diaphragm wall and the surface settlement outside the pit by introducing the boundary element method and the Mindlin solution and inverted the additional horizontal stress. The Mindlin solution was then used to calculate the surface settlement caused by the excavation of the pit. The specific conclusions from the findings are as follows:

• The proposed method obtained the horizontal displacements of the diaphragm wall and surface settlement and compared them with the two-parameter Pasternak model as well as the measured values on site. The findings show that the horizontal displacement of the diaphragm wall and surface settlement outside the pit theoretically predicted by Kerr’s model are in good agreement with the site monitoring results, the horizontal displacement of the diaphragm wall is 17.13 mm and 18.1 mm, and the surface settlement is 15.54 mm and 14.9 mm, respectively, with the error being less than 5.4%, which proves the reasonableness of the proposed theoretical prediction method.
• There is a certain connection between the horizontal displacement of the diaphragm wall and the surface settlement outside the pit due to the unloading of the pit excavation. Under the same conditions, the maximum horizontal displacement of the diaphragm wall is greater than the value of the surface settlement outside the pit. The depth of the maximum horizontal displacement of the diaphragm wall has an effect on the position of the maximum surface settlement outside the pit, and the greater the depth of the maximum horizontal displacement, the further away from the pit the location of the maximum surface settlement.
• Analysis of the influencing factors, such as the thickness of the diaphragm wall, the elastic modulus of the diaphragm wall, the diameter of the internal support, and the number of internal supports, revealed that the greater the thickness of the diaphragm wall, the higher the elastic modulus, the larger the diameter of the internal support, the greater the number of internal supports, the higher the strength of the soil layer, and the smaller the value of the horizontal displacement of the diaphragm wall and the value of the surface settlement outside the pit. The maximum horizontal displacement differences are 4.84 mm, 2.43 mm, 1.77 mm, 4.77 mm, and 10.85 mm, and the maximum surface settlement differences are 4.75 mm, 2.48 mm, 1.83 mm, 5.36 mm, and 9.4 mm, respectively, which proves that these parameters have a good controlling effect on the horizontal displacement of the diaphragm wall and the surface settlement.