Modified complex variable method for displacement induced by surcharge loads and shallow tunnel excavation

Abstract

In geo-engineering, mechanical problems of surcharge loads acting on ground surface and shallow tunnel excavation are often encountered. When the complex variable method is applied, such problems turn to non-zero resultant issues and generally result in infinite displacement singularity in geomaterial at infinity, which violates the fact that displacement components at infinity should be zero in field observations. The displacement singularity at infinity caused by non-zero resultants corresponds non-zero coefficients of logarithmic items in complex potentials. To eliminate the singularity at infinity, the symmetrical method is adopted so that the far-field displacement obtained from the modified displacement solution conducted by the complex variable method would be logically identical to the reality. Two fundamental cases of single surcharge load acting on ground surface and single tunnel excavation are presented to verify the modified displacement solution, and good agreements are observed. The main advantage that the proposed solution eliminates displacement singularity at infinity and makes far-field displacement logically accord with field observation is presented by comparing to the original solution. The drawback that the proposed solution is dependent on modified depth is discussed as well.

Appendices

Expanding formulae of the components of the linear system

1.1 Expressions for unknown coefficients

The expressions of $A_k$ in Eq. (28) are

$$\begin{aligned} \left\{ \begin{array}{l} A_0=-\frac{\mathrm{i}\gamma R_0^2}{4}, \qquad A_1=-\frac{\mathrm{i}\gamma R_0^2}{4(1+\kappa )}, \qquad A_{-1}=-\frac{\mathrm{i}\kappa \gamma R_0^2}{4(1+\kappa )}, \qquad A_k=A_{-k}=0, \quad k\ge 2. \end{array} \right. \end{aligned}$$

(A.1)

The expressions of $B_k$ in Eq. (28) are

$$\begin{aligned} \left\{ \begin{array}{l} B_0=W_1(1-\alpha ^2)-W_2\alpha +W_3e_0+W_4f_1+W_5f_2+W_6f_3,\\ B_1=W_1\alpha +W_2(1-\alpha ^2)+W_4e_0+W_3e_1+W_5f_1+W_6f2,\\ B_2=W_2\alpha +W_5e_0+W_4e_1+W_3e_2+W_6f_1,\\ B_k=W_6e_{k-3}+W_5e_{k-2}+W_4e_{k-1}+W_3e_k,\qquad k\ge 3,\\ B_{-1}=-W_1\alpha +W_3f_1+W_4f_2+W_5f_3+W_6f_4,\\ B_{-k}=W_3f_k+W_4f_{k+1}+W_5f_{k+2}+W_6f_{k+3},\qquad k\ge 2, \end{array} \right. \end{aligned}$$

(A.2)

where

$$\begin{aligned} e_k=\frac{\alpha ^{k+2}}{(1-\alpha ^2)^2},\qquad k\ge 0, \quad f_k=\frac{\left[ k(1-\alpha ^2)+2\alpha ^2-1\right] a^{k-2}}{(1-\alpha ^2)^2},\qquad k\ge 1. \end{aligned}$$

(A.3)

The expressions of $W_l (l=1,2,3,\ldots ,6)$ are

$$\begin{aligned}&W_1=\frac{\mathrm{i}\kappa \gamma R_0^2}{4(1+\kappa )}, \quad W_2=\frac{\mathrm{i}\gamma R_0^2}{4(1+\kappa )\alpha }, \quad W_3=-\frac{\mathrm{i}\gamma a^2\alpha ^2(1+\alpha ^2)}{1-\alpha ^2}, \quad W_4=\frac{2\mathrm{i}\gamma a^2\alpha (1+\alpha ^2+\alpha ^4)}{1-\alpha ^2},\nonumber \\&W_5=-\frac{1}{2}\mathrm{i}k_0\gamma a^2(1-\alpha ^2)^2-\frac{\mathrm{i}\gamma a^2(1+5\alpha ^2+5\alpha ^4+\alpha ^6)}{2(1-\alpha ^2)}, \quad W_6=\frac{2\mathrm{i}\gamma a^2\alpha ^3}{1-\alpha ^2}. \end{aligned}$$

(A.4)

1.2 Expressions for components of the linear system

1.2.1 Coefficient matrix

The coefficient matrix in Eq. (35) can be expanded as

$$\begin{aligned} \varvec{U}=\left( \varvec{A}_1\quad \varvec{A}_{-N-1}\quad \varvec{A}_{-1}\quad \varvec{A}_2\quad \varvec{A}_{-2}\quad \varvec{A}_3\quad \varvec{A}_{-3}\quad \varvec{A}_4\quad \cdots \quad \varvec{A}_{-N}\quad \varvec{A}_{N+1}\right) ^\mathrm{T}, \end{aligned}$$

(A.5)

where

$$\begin{aligned} \varvec{A}_1= & {} \left( 1-\alpha ^2\quad \underbrace{0\quad \cdots \quad 0}_{N}\quad 1-\alpha ^2\quad \underbrace{0\quad \cdots \quad 0}_{N}\right) ^\mathrm{T}, \end{aligned}$$

(A.6)

$$\begin{aligned} \varvec{A}_{-N-1}= & {} \left( \underbrace{0\quad \cdots \quad 0}_{N}\quad (N+1)(1-\alpha ^2)\alpha ^{N+1}\quad \underbrace{0\quad \cdots \quad 0}_{N}\quad \alpha ^{-N-1}-\alpha ^{N+1}\right) ^\mathrm{T}, \end{aligned}$$

(A.7)

$$\begin{aligned} \varvec{A}_{-k}= & {} \left( \underbrace{0\quad \cdots \quad 0}_{k-1}\quad k(1-\alpha ^2)\alpha ^k\quad -(k+1)(1-\alpha ^2)\alpha ^k\quad \underbrace{0\quad \cdots \quad 0}_{N-1},\right. \nonumber \\&\left. \alpha ^{-k}-\alpha ^{k}\quad \alpha ^{k+2}-\alpha ^{-k}\quad \underbrace{0\quad \cdots \quad 0}_{N-k}\right) ^\mathrm{T} \end{aligned}$$

(A.8)

$$\begin{aligned} \varvec{A}_{k+1}= & {} \left( \underbrace{0\quad \cdots \quad 0}_{k-1}\quad \alpha ^{1-k}-\alpha ^{k+1}\quad -\alpha ^{-k-1}-\alpha ^{k+1}\quad \underbrace{0\quad \cdots \quad 0}_{N-1},\right. \nonumber \\&\left. k(1-\alpha ^2)\alpha ^{-k-1}\quad -(k+1)(1-\alpha ^2)\alpha ^{-k-1}\quad \underbrace{0\quad \cdots \quad 0}_{N-k}\right) ^\mathrm{T}. \end{aligned}$$

(A.9)

1.2.2 Solution vector

The solution vector in Eq. (35) can be expanded as

$$\begin{aligned} \varvec{X}=\left( \varvec{a}\quad \varvec{b}\right) ^\mathrm{T}, \end{aligned}$$

(A.10)

where

$$\begin{aligned} \varvec{a}= & {} \left( a_1\quad a_2\quad a_3\quad \cdots \quad a_{N+1}\right) , \end{aligned}$$

(A.11)

$$\begin{aligned} \varvec{a}= & {} \left( b_1\quad b_2\quad b_3\quad \cdots \quad b_{N+1}\right) . \end{aligned}$$

(A.12)

Fig. 9

Displacement components on the ground surface and along the excavation face for different values of H/h

1.2.3 Constant item vector

The constant item vector in Eq. (35) can be expanded as

$$\begin{aligned} \begin{aligned} \varvec{u}=&\left( -\overline{B}_1\alpha -A_0\alpha ^2+A_1+C\alpha ^2\quad \overline{B}_{-N-1}\quad \overline{B}_{-1}-\overline{A}_{-1}\alpha \quad B_2+A_1\right. \\&\left. \overline{B}_{-2}\quad B_3\quad \overline{B}_{-3} \quad B_4\quad \cdots \quad \overline{B}_{-N}\quad B_{N+1}\right) . \end{aligned} \end{aligned}$$

(A.13)

Discussion on symmetrically modified depth (H/h)

As is mentioned in Sects. 3 and 4.2, the value of H, the symmetrically modified depth, would affect the results of modified displacement. To be specific to the case study in Sect. 4.2, it is the value of H/h, the ratio of the symmetrically modified depth to the tunnel depth, that influences the modified displacement. In this appendix, the reason of the value $H=5.8h$ in the case study will be explained, and the influence of the value of H/h on geomaterial displacement in Sect. 4.2 would be discussed.

The procedure to find the suitable value of H/h is briefly presented below. To be clear, the suitable value of H/h for the case study that makes the modified displacement be approximate to the numerical result is not known beforehand at first. We obtained the horizontal and vertical displacement datum (scattered points) on the ground surface and along the excavation face of the numerical results. Since we already knew the exact displacement expression of the modified solution, except for H, we used the least square method to search the values within the range of $H/h\in \left[ 2,20\right]$ in several subsections to find the most suitable one for the displacement datum of the numerical results. The criterion is the minimum value of the total sums of all the squares of the difference between the numerical and analytical datum for all four displacement together. The found value of H/h is approximate to 5.836. To be simple, we took it as $H/h=5.8$.

After obtaining the value $H/h=5.8$ for the case study, we can discuss the influence of H/h. Figure 9 shows the influence of H/h on displacement components on the ground surface and along the excavation face. The thick black solid lines in Fig. 9 denote the curve of $H/h=5.8$, which are identical to the ones of “Modified-$u_x$” and “Modified-$u_y$” in Fig. 7b, d. The red and blue sets of different types of lines in Fig. 9 denote modified displacement datum with H/h larger and smaller than $H/h=5.8$, respectively. It can be seen that as the value of H/h diverges from the fitted one $H/h=5.8$, the curves of modified displacement components diverge correspondingly. Comparing Fig. 9a, c to the numerical curves of the horizontal displacement in Fig. 7, it can be found that a larger value of H/h would make the horizontal displacement of the symmetrically modified solution more approximate to the numerical results. Meanwhile, comparing Fig. 9b, d to the numerical curves of vertical displacement in Fig. 7, it can be found that the value of $H/h\approx 5.8$ makes the curves obtained from the symmetrically modified solution be more approximate to the numerical ones. Thus, Fig. 9 graphically verifies that $H/h=5.8$ is the suitable value for symmetrically modified solution. Besides, Fig. 9 also illustrates that the symmetrical modified solution is dependent on H, the symmetrically modified depth. Similar findings can be also found in the study by Wang et al. [21].