EXPLICIT SOLUTION OF STRESS COMPLEX POTENTIAL FUNCTION FOR SURROUNDING ROCK OF SHALLOW SUBSEA TUNNEL
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摘要: 利用弹性复变函数理论将浅埋海底隧道简化为半无限平面问题, 考虑围岩自重和海水压力的影响, 对隧道开挖后的围岩应力分布进行研究. 采用分式映射函数将围岩域映射为像平面圆环域, 在圆环域内将复势单值解析函数展开为Laurent级数. 利用无穷远点应力有界性对Laurent级数幂次项进行确定, 根据地表边界和孔口不均匀应力边界条件得到Laurent级数系数迭代表达式, 将已确定的Laurent级数条件代入迭代表达式中得出复势函数显式解, 从而实现复势函数系数从低次幂迭代至高次幂. 根据应力分量的复变函数表达式即可得到隧道周围各点应力分量. 研究了两个单值解析函数取不同幂次时对结果的影响, 分析了浅埋隧道埋深对环向压应力的影响. 研究结果表明: 幂级数解具有较高的可靠性, 在隧道上半部分幂级数解与有限元数值解吻合效果良好, 在隧道下半部分幂级数解最终计算结果比有限元结果相对保守; 为了保证计算结果的准确性复势函数需取足够多项; 随着隧道埋深增大, 隧道底部及两侧孔腰处环向压应力不断增大; 腰部与底部环向应力的差值也随之增大.Abstract: The elastic complex function theory is used to simplify the shallow-buried subsea tunnel into a semi-infinite plane problem. The stress distribution of surrounding rock after tunnel excavation is explored considering the effects of self-weight of surrounding rock and sea water pressure. The fractal mapping function is used to map the surrounding rock domain to a circular domain like a plane, and the complex potential single-value analytical function is expanded to a Laurent series in the circular domain. The power term of Laurent series is determined by using the stress boundedness at infinite distances. The iteration expression of Laurent series coefficient is obtained according to the surface boundary and the non-uniform stress boundary condition at the orifice. The determined Laurent series condition is substituted into the iteration expression to obtain the explicit solution of complex potential function, thus realizing the iteration of complex potential function coefficient from low power to high power. According to the complex function expression of the stress component, the stress component of all points around the tunnel can be obtained. The influence of two single-value analytical functions with different powers on the results is studied, and the influence of buried depth of shallow tunnel on the toroidal compressive stress is analyzed. The results show that the power series solution has high reliability, and it agrees well with the finite element solution in the first half of the tunnel. The final calculation results of power series solution in the second half of the tunnel are relatively conservative compared with the finite element results. Sufficient numbers of complex potential functions are required to ensure the accuracy of calculation results. As the buried depth of the tunnel increases, the circumferential compressive stress at the bottom of the tunnel and at the waist of the holes on both sides increases continuously. The difference in circumferential stress between the lumbar and the bottom increases as well.
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引 言
国内经济持续发展、城市规模不断扩大、城市之间联系愈发紧密, 客流量与货物运输量急剧增加, 为了缓解城市交通压力、节约土地资源和开发地下空间, 建设一批市政管道和跨水域隧道成为最优方案. 特别是在沿海城市, 海底隧道具有抗震性能好、不影响货船航运、隐蔽性高和不受气候影响等优势而被广泛采用. 这些隧道部分区段往往埋深较浅, 建设中不可忽视地下水压、岩体自重以及地表边界上其他复杂工况的影响. 准确得出隧道周边应力分布与变形对指导隧道设计以及安全施工具有重要的理论意义和工程价值.
20世纪初, 俄国力学家Kolosov[1]采用复应力函数法解决二维弹性静力学问题, Muskhelishvili[2]系统论述了弹性力学平面问题的复变函数法, 极大促进该方法的应用与发展. 保角映射能将复杂区域转化为简单规则区域, 使得复变函数在研究平面含孔洞问题时具有明显优势, 因此, 平面问题复变函数方法在岩土工程领域得到了广泛应用.
深埋单孔隧道属于无限单连通域问题, 研究中可忽略岩体重力梯度的影响, 不同孔型的深埋隧道其映射函数均可采用Laurent级数表示, 对深埋隧道的理论研究已趋于成熟[3-17]. 采用复变函数方法研究浅埋隧道问题起步较晚, 1997年Verruijt[9-10]提出分式映射函数, 将含圆孔的半平面岩石物理区域保角映射为像平面上同心圆环域, 获得了在不考虑岩体重力影响, 仅有孔口均布径向位移与均布径向面力边界条件时的解答; Verruijt等[11]考虑周围岩体自重应力对开挖浅埋单孔隧道的影响, 将未被开挖的初始状态、考虑岩体作用时开挖岩石等效合力状态和孔口附加等效载荷3种状态的解叠加成为最终解; Verruijt等[12]在原有复应力函数中增加了能够反映弹性体在不平衡力系作用下的影响项, 给出了二维平面问题在不平衡力系作用下复应力函数的一般形式, 并对孔口位移边值条件进行了研究. 蔚立元等[13]、王志良等[14]和陆文超等[15]得到地面任意分布载荷下的围岩应力场; 韩凯航等[16]基于复变函数法并结合柯西−黎曼方程得出浅埋隧道应力及位移函数的级数显式表达式; 宋文杰等[17]求解了不排水黏土浅埋盾构隧道地层变位和地层应力; 宋浩然等[18-19]与Fang等[20]研究围岩自重及水的作用, 将问题分解为四部分求解, 分析水对隧道围岩应力的影响; 杨公标等[21]研究在重力影响下的含空洞地层浅埋隧道围岩应力及位移解析解.
最近, 许多学者对隧洞不均匀边界条件问题展开了研究. Lu等[22-24]、Zeng等[25-26]和Cai等[27]研究了实际开挖过程引起的边界不平衡力分布, 提出不平衡力系由整个开挖边界决定, 并给出浅埋非圆孔隧道映射函数, 对地表有水平初始应力作用的浅埋隧道、水工隧道以及马蹄形非圆孔隧道等工程问题进行了研究. 申航等[28]研究矩形孔口径向位移边界条件下的应力场和位移场; 文明[29]提出在重力作用下浅埋非圆形隧道力学分析的解耦保角映射方法. Lu等[30]和Kong等[31-33]利用傅里叶级数来表示圆形浅埋隧道横截面变形的统一位移函数, 在地表水平以及边坡边界模型中考虑隧道“浮力效应”得到单孔隧道解析解, 并在此基础上采用Schwartz交替法得到双孔隧道解答; 采用傅里叶级数表示出水下浅埋隧道周边复杂应力状态的统一应力函数, 以此作为孔口应力边界条件, 给出了应力和位移的弹性解析解.
采用平面弹性复变函数方法求解浅埋隧道问题一直存在复势函数系数求解困难的问题. Verruijt等[12]在求解孔口应力边界条件问题时借助复势函数$\varphi (\zeta )$和$\psi (\zeta )$在像平面圆环域内的收敛性, 假设1阶正幂项系数为0, 利用迭代方程组经过多次迭代(比如1000或10000次)直至高幂次项系数为非零常数, 以此得到其余系数. Lu等[30]推导时取$2 N$个线性组方程求$2 N$个未知数, 并假定其余正负高幂次项为零, 其中$N$需要足够大才能保证足够的精度.
本文利用无穷远点应力有界性对解析函数的Laurent级数展开式幂次项进行了确定, 从而得到1阶正负幂次项系数, 再根据边界条件所得迭代公式解出完整的解析函数. 此方法区别于以上两种方法, 通过从低次幂系数迭代求解至高次幂系数, 复势函数每一项系数都能够用公式明确表示, 使浅埋隧道复势函数求解更方便, 更容易实现编程计算. 以浅埋海底隧道孔口受不均匀应力边界条件为例, 得出孔口应力幂级数解, 与有限元数值解对比验证幂级数解的可靠性; 研究了复势函数$\varphi (\zeta )$和$\psi (\zeta )$级数展开后取不同幂次项和不同函数系数对求解结果的影响机理; 分析了浅埋隧道埋深对环向压应力的影响. 此方法可求解半无限平面含孔洞问题, 求解结果具有较高精度, 将程序嵌入小型计算仪器中可方便工程应用.
1. 问题描述和基本理论
根据文献[11-32]提出的海底隧道力学模型可知, 所研究的浅埋单孔海底隧道位于完全饱和且均匀、各向同性的弹性含水地层中, 力学模型如图1所示. 水平地表上作用有深度为${h_w}$, 容重为${\gamma _w}$的海水, 海水处于不可压缩的稳定状态. 隧道半径为$r$, 隧道中心至地面距离为$h$, 岩石容重为$\gamma $, 侧压力系数为${k_0}$. 将该力学模型分解为两部分(图2和图3), 其关系式如下
$$ \left. \begin{split} & {{\sigma _x} = \sigma _x^0 + \sigma _x^*} \\ & {{\sigma _y} = \sigma _y^0 + \sigma _y^*} \\ & {{\sigma _{yx}} = \sigma _{xy}^0 + \sigma _{xy}^*} \end{split} \right\} $$ (1) 第1部分解答为
$$ \left.\begin{split} & \sigma _x^0 = - {\gamma _w}{h_w} + ({\gamma _w} + {k_0}\gamma )y \\ & \sigma _y^0 = - {\gamma _w}{h_w} + ({\gamma _w} + \gamma )y \\ & \sigma _{xy}^0 = 0\end{split} \right\} $$ (2) 运用平面弹性复变函数法研究第2部分模型的围岩应力. 第2部分为半无限平面含单圆孔模型, 共包含地表和孔口两个边界条件, 对于岩体任意点的应力和位移均可用复势函数$\varphi (z)$和$\psi (z)$表示. 可求得第2部分应力解
$$ \sigma _x^* + \sigma _y^* = 4{{\rm{Re}}} [\varphi '(z)] $$ (3) $$ \sigma _y^* - \sigma _x^* + 2{\rm{i}}\sigma _{xy}^* = 2[\bar z\varphi ''(z) + \psi '(z)] $$ (4) 位移解
$$ 2G(u + {\rm{i}}v) = K\varphi (z) - z\overline {\varphi '(z)} - \overline {\psi (z)} $$ (5) 式中, $ G $为剪切模量, $G = {E \mathord{\left/ {\vphantom {E {[2(1 + \mu )]}}} \right. } {[2(1 + \mu )]}}$, 对于平面应变问题$K = 3 - 4\mu $, $\mu $为泊松比, $ {k_0} = {\mu \mathord{\left/ {\vphantom {\mu {(1 - \mu )}}} \right. } {(1 - \mu )}} $.
复势函数$\varphi (z)$和$\psi (z)$是区域内的单值解析函数, 其形式如下
$$ \varphi (z) = - \frac{{{F_x} + {\rm{i}}{F_y}}}{{2\text{π} (1 + k)}}\left[ {K\ln (z - {{\bar z}_c}) + \ln (z - {z_c})} \right] + {\varphi _0}(z) $$ (6) $$ \psi (z) = \frac{{{F_x} - {\rm{i}}{F_y}}}{{2\text{π} (1 + k)}}\left[ {\ln (z - {{\bar z}_c}) + K\ln (z - {z_c})} \right] + {\psi _0}(z) $$ (7) 式中
$$\qquad\qquad\qquad {\varphi _0}(z) = \sum\limits_{k = - \infty }^{ + \infty } {a_k^*{z_k}} $$ (8) $$\qquad\qquad\qquad {\psi _0}(z) = \sum\limits_{k = - \infty }^{ + \infty } {b_k^*{z_k}} $$ (9) $$\qquad\qquad\qquad \left.\begin{aligned} &{F_x} = \oint {{X_n}{\rm{d}}s} \\ &{F_y} = \oint {{Y_n}{\rm{d}}s} \end{aligned} \right\} $$ (10) $$\qquad\qquad\qquad {z_c} = - {\rm{i}}h\frac{{1 - {\alpha ^2}}}{{1 + {\alpha ^2}}} $$ (11) 其中, ${\varphi _0}(z)$和${\psi _0}(z)$为单值解析函数, 可将其展开成Laurent级数. 在本文中考虑岩体自重和水压力, 则${F_x} = 0$, ${F_y} = \text{π} {r^2}(\gamma + {\gamma _w})$.
2. 复数形式的边界条件
2.1 应力边界条件和映射函数
采用平面弹性复变函数解法的应力边界条件如下
$$ \varphi (z) + z\overline {\varphi '(z)} + \overline {\psi (z)} = {\rm{i}}\int_A^B {({X_n} + {\rm{i}}{Y_n}){\text{d}}s} + C $$ (12) 式中
$$ \left. \begin{aligned} & {X_n} = - (l\sigma _x^0 + m\sigma _{xy}^0) = - \frac{{{\rm{d}}y}}{{{\rm{d}}s}}\sigma _x^0 \\ & {Y_n} = - (l\sigma _{xy}^0 + m\sigma _y^0) = \frac{{{\rm{d}}x}}{{{\rm{d}}s}}\sigma _y^0 \end{aligned}\right\} $$ (13) $$ \left. \begin{aligned} & l = {\rm{cos}}(n,x) = \frac{{{\rm{d}}y}}{{{\rm{d}}s}} \\ & m = {\rm{cos}}(n,y) = - \frac{{{\rm{d}}x}}{{{\rm{d}}s}}\end{aligned} \right\} $$ (14) 其中, C为积分得到的复常数, ${X_n}$和${Y_n}$分别表示边界上一点沿x轴与y轴平行的面力分量, l和m是所讨论边界点的单位外法向量的方向余弦, 如图4所示.
第2部分解中地表没有施加载荷, 则地表边界${L_1}$上有${X_n} = 0$, ${Y_n} = 0$, 此时不失计算结果的有效性, 复常数C可以假设为0, 地表边界的应力边界条件可以表示为
$$ \varphi (z) + z\overline {\varphi '(z)} + \overline {\psi (z)} = 0 $$ (15) 图2中的虚线表示未开挖的虚拟孔口, 其与图3两模型叠加得到开挖后的模型如图1, 开挖后孔口面力分量为0. 在隧道孔口边界${L_2}$上施加面力分量${X_n}$和${Y_n}$, 该面力分量与图2中虚拟孔口边界上面力分量大小相等, 方向相反, 孔口边界${L_2}$的应力边界条件表达式中复常数C不能再假设为0.
运用保角变换方法通过映射函数$z = \omega (\zeta )$将图3中Z平面中岩土区域映射到复平面$\zeta $区域上的圆环中, 圆环外径为1, 内径为$\alpha $. $\alpha $是与隧洞埋深$h$和隧洞半径$r$有关的系数, 其可通过如下确定.
$$\qquad\qquad \omega (\zeta ) = - {\rm{i}}h\frac{{1 - {\alpha ^2}}}{{1 + {\alpha ^2}}}\frac{{1 + \zeta }}{{1 - \zeta }} $$ (16) $$\qquad\qquad \frac{r}{h} = \frac{{2\alpha }}{{1 + {\alpha ^2}}} $$ (17) 将映射函数$z = \omega (\zeta )$代入${\varphi _0}(z)$和${\psi _0}(z)$中有
$$ {\varphi _0}(\zeta ) = \sum\limits_{k = - \infty }^{ + \infty } {{a_k}{\zeta _k}} $$ (18) $$ {\psi _0}(\zeta ) = \sum\limits_{k = - \infty }^{ + \infty } {{b_k}{\zeta _k}} $$ (19) 2.2 地表边界条件求解
为方便求解和表示复平面圆环上的点, 令$\zeta = \rho \sigma ,$ $\;\rho$为半径, $\sigma = {{\rm{e}}^{{\rm{i}}\theta '}}$. 图3中$z$平面地表边界$y = 0$映射到$\zeta $平面的单位圆$\left| \zeta \right| = 1$, 地表无穷远点对应点$\zeta = 1$, 则地表边界在复平面中用$\zeta = \sigma $表示. 将$ \phi (z) $和$ \psi (z) $代入地表应力边界条件得到$z$平面的表达式
$$ {\varphi _0}(z) + z\overline {{{\varphi '}_0}(z)} + \overline {{\psi _0}(z)} = \frac{{{F_x} - {\rm{i}}{F_y}}}{{2\text{π} (1 + K)}}\left( {\frac{{Kz}}{{\bar z - {z_c}}} + \frac{z}{{\bar z - {{\bar z}_c}}}} \right) $$ (20) 将映射函数$z = \omega (\zeta )$代入式(20), 其中
$$ z\overline {{{\varphi '_0}}(z)} = \frac{{\omega (\zeta )}}{{\overline {\omega '(\zeta )} }}\overline {{{\varphi '_0}}_{}(\zeta )} $$ (21) $$ \frac{{\omega (\zeta )}}{{\overline {\omega '(\zeta )} }} = - \frac{1}{2}\frac{{\left( {1 + \rho \sigma } \right){{\left( {\sigma - \rho } \right)}^2}}}{{{\sigma ^2}\left( {1 - \rho \sigma } \right)}} $$ (22) 地表边界$\zeta = \sigma $, 可将式(20)左右各转换为$\zeta $平面中有关$\sigma $的表达式, 分别如下所示
$$\begin{split} & {\varphi _0}(\sigma ) + \frac{1}{2}\left( {1 - {\sigma ^{ - 2}}} \right)\overline {{{\varphi '_0}}(\sigma )} + \overline {{\psi _0}(\sigma )} = \\ &\qquad \sum\limits_{k = - \infty }^{ + \infty } {{a_k}} {\sigma ^k} + \frac{1}{2}\sum\limits_{k = - \infty }^{ + \infty } {\left( { - k + 1} \right)\overline {{a_{ - k + 1}}} } {\sigma ^k}- \\ &\qquad \frac{1}{2}\sum\limits_{k = - \infty }^{ + \infty } {\left( { - k - 1} \right)\overline {{a_{ - k - 1}}} } {\sigma ^k} + \sum\limits_{k = - \infty }^{ + \infty } {\overline {{b_{ - k}}} } {\sigma ^k} \end{split} $$ (23) $$ \begin{split} & \frac{{{F_x} - {\rm{i}}{F_y}}}{{2\text{π} \left( {1 + k} \right)}}\left( {\frac{{kz}}{{\bar z - {z_c}}} + \frac{z}{{\bar z - {{\bar z}_c}}}} \right) = \frac{{{F_x} - {\rm{i}}{F_y}}}{{4\text{π} \left( {1 + k} \right)}}k{\sigma ^{ - 1}} +\\ &\qquad \frac{{{F_x} - {\rm{i}}{F_y}}}{{4\text{π} }} + \frac{{{F_x} - {\rm{i}}{F_y}}}{{4\text{π} \left( {1 + k} \right)}}\sigma = \sum\limits_{k = - \infty }^{ + \infty } {{A_k}} {\sigma ^k}\\[-12pt] \end{split} $$ (24) ${A_k}$可展开为
$$ {A}_{k} = \left\{\begin{split} &0\text{}\text{}\text{}\text{, }\quad k\leqslant -2 \\ &k\frac{{F}_{x}-{\rm{i}}{F}_{y}}{4\text{π} \left(1 + k\right)}\text{, }\quad k = -1 \\ &\frac{{F}_{x}-{\rm{i}}{F}_{y}}{4\text{π} }\text{, }\quad k = 0 \\ &\frac{{F}_{x}-{\rm{i}}{F}_{y}}{4\text{π} \left(1 + k\right)}\text{, }\quad k = 1 \\ &0\text{, }k\geqslant -2\end{split}\right. $$ (25) 将式(23)与式(25)相应的$k$阶幂指数对应有
$$ {a_k} + \frac{1}{2}\left( { - k + 1} \right) \overline{{ a_{ - k + 1}}} - \frac{1}{2}\left( { - k - 1} \right) \overline{{ a_{ - k - 1}}} + \overline{{ b_{ - k}}} = {A_k} $$ (26) 为确定函数${\varphi _0}(\zeta )$和${\psi _0}(\zeta )$中${a_k}$与${b_k}$的系数还需要通过孔口边界条件得到其余关系式.
2.3 孔口边界条件求解
隧道孔口${x^2} + {(y + h)^2} = {r^2}$经过映射成复平面$\zeta $上半径为$\alpha $的单位圆$\zeta = \alpha \sigma $. 在孔口边界为
$$ \varphi (z) + z\overline {\varphi '(z)} + \overline {\psi (z)} = {\rm{i}}\int_A^B {({X_n} + {\rm{i}}{Y_n}){\text{d}}s} + C $$ (27) 式中
$$ \frac{{\omega (\zeta )}}{{\overline {\omega '(\zeta )} }} = \frac{{ - \alpha \sigma - (1 - 2{\alpha ^2}) + \alpha (2 - {\alpha ^2}){\sigma ^{ - 1}} - {\alpha ^2}{\sigma ^{ - 2}}}}{{2(1 - \alpha \sigma )}} $$ (28) 将复势函数$ \phi (z) $, $ \psi (z) $和映射函数式(16)代入边界条件(27)变换为有关$\sigma $的表达式
$$ \begin{split} & \left( {1 - \alpha \sigma } \right)[{\varphi _0}(\sigma ) + \frac{{\omega (\zeta )}}{{\overline {\omega '(\zeta )} }}\overline {{{\varphi '_0}}(\sigma )} + \overline {{\psi _0}(\sigma )} ] = \\ &\qquad \left( {1 - \alpha \sigma } \right)f(\alpha \sigma ) + \left( {1 - \alpha \sigma } \right)C \end{split} $$ (29) 将式(29)左侧展开
$$ \begin{split} & \left( {1 - \alpha \sigma } \right)[{\varphi _0}(\sigma ) + \frac{{\omega (\zeta )}}{{\overline {\omega '(\zeta )} }}\overline {{{\varphi '_0}}(\sigma )} + \overline {{\psi _0}(\sigma )} ] = \\ &\qquad - \frac{1}{2}\sum\limits_{k = - \infty }^{ + \infty } {\left( { - k + 2} \right){\alpha ^{ - k + 2}} {\overline { a_{ - k + 2}}} } {\sigma ^k}- \\ &\qquad \frac{1}{2}\sum\limits_{k = - \infty }^{ + \infty } {\left( { - k + 1} \right)\left( {1 - 2{\alpha ^2}} \right){\alpha ^{ - k}} {\overline{ a_{ - k + 1}}} } {\sigma ^k} + \\ &\qquad \frac{1}{2}\sum\limits_{k = - \infty }^{ + \infty } {\left( { - k} \right)\left( {2 - {\alpha ^2}} \right){\alpha ^{ - k}} {\overline{ a_{ - k}}} } {\sigma ^k}- \\ &\qquad \frac{1}{2}\sum\limits_{k = - \infty }^{ + \infty } {\left( { - k - 1} \right){\alpha ^{ - k}} {\overline{ a_{ - k - 1}}} } {\sigma ^k} + \\ &\qquad \sum\limits_{k = - \infty }^{ + \infty } {{\alpha ^k}{a_k}} {\sigma ^k} - \sum\limits_{k = - \infty }^{ + \infty } {{\alpha ^k}{a_{k - 1}}} {\sigma ^k}+ \\ &\qquad \sum\limits_{k = - \infty }^{ + \infty } {{\alpha ^{ - k}} {\overline{ b_{ - k}}} } {\sigma ^k} - \sum\limits_{k = - \infty }^{ + \infty } {{\alpha ^{ - k + 2}} {\overline{ b_{ - k + 1}}} } {\sigma ^k} \end{split} $$ (30) 式(29)右侧用级数形式表示
$$ \left( {1 - \alpha \sigma } \right)f(\alpha \sigma ) + \left( {1 - \alpha \sigma } \right)C = \sum\limits_{k = - \infty }^{ + \infty } {{B_k}} {\sigma ^k} $$ (31) 令$\left( {1 - \alpha \sigma } \right)f(\alpha \sigma ) = \displaystyle\sum\limits_{k = - \infty }^{ + \infty } {B_k^ * } {\sigma ^k}$, $C$为积分复常数.
${B_k}$可展开为
$$ \begin{split} & - \frac{1}{2}\sum\limits_{k = - \infty }^{ + \infty } {\left( { - k + 2} \right){\alpha ^{ - k + 2}} {\overline{ a_{ - k + 2}}} } {\sigma ^k}- \\ &\qquad \frac{1}{2}\sum\limits_{k = - \infty }^{ + \infty } {\left( { - k + 1} \right)\left( {1 - 2{\alpha ^2}} \right){\alpha ^{ - k}} {\overline{ a_{ - k + 1}}} } {\sigma ^k}+ \\ &\qquad \frac{1}{2}\sum\limits_{k = - \infty }^{ + \infty } {\left( { - k} \right)\left( {2 - {\alpha ^2}} \right){\alpha ^{ - k}} {\overline{ a_{ - k}}} } {\sigma ^k}- \\ &\qquad \frac{1}{2}\sum\limits_{k = - \infty }^{ + \infty } {\left( { - k - 1} \right){\alpha ^{ - k}} {\overline{ a_{ - k - 1}}} } {\sigma ^k} + \sum\limits_{k = - \infty }^{ + \infty } {{\alpha ^k}{a_k}} {\sigma ^k}- \\ &\qquad \sum\limits_{k = - \infty }^{ + \infty } {{\alpha ^k}{a_{k - 1}}} {\sigma ^k} + \sum\limits_{k = - \infty }^{ + \infty } {{\alpha ^{ - k}} {\overline{ b_{ - k}}} } {\sigma ^k}- \\ &\qquad \sum\limits_{k = - \infty }^{ + \infty } {{\alpha ^{ - k + 2}} {\overline{ b_{ - k + 1}}} } {\sigma ^k} = \sum\limits_{k = - \infty }^{ + \infty } {{B_k}} {\sigma ^k} \end{split} $$ (32) 将式(26)代入式(32)中可以得到迭代公式如下
$$ \begin{split} &\left({\alpha }^{k}-{\alpha }^{-k}\right){a}_{k} + \left({\alpha }^{-k + 2}-{\alpha }^{k}\right){a}_{k-1}-\\ &\qquad \left(-k + 1\right)\left(1-{\alpha }^{2}\right){\alpha }^{-k}\overline{{a}_{-\left(k-1\right)}}-k\left(1-{\alpha }^{2}\right){\alpha }^{-k}\overline{{a}_{-k}}=\\ &\qquad {B}_{k}-{\alpha }^{-k}{A}_{k} + {\alpha }^{-k + 2}{A}_{k-1}\text{, }\quad k\geqslant 2\end{split} $$ (33) $$ \begin{split} &-k\left(1-{\alpha }^{2}\right){\alpha }^{k-1}{a}_{k} + \left(k-1\right)\left(1-{\alpha }^{2}\right){\alpha }^{k-1}{a}_{k-1}+\\ &\qquad \left({\alpha }^{-\left(k-1\right)}-{\alpha }^{k-1}\right)\overline{{a}_{-\left(k-1\right)}} + \left({\alpha }^{k + 1}-{\alpha }^{1-k}\right)\overline{{a}_{-k}} =\\ &\qquad \overline{{B}_{-\left(k-1\right)}}-{\alpha }^{k-1}\overline{{A}_{-\left(k-1\right)}} + {\alpha }^{k + 1}\overline{{A}_{-k}}\text{, }\quad k\geqslant 2\end{split} $$ (34) 当$k = 0$和$k = 1$时
$$ \begin{split} & ({\alpha ^2} - 1){a_{ - 1}} + ({\alpha ^2} - 1) \overline{{ a_1}} = \\ &\qquad B_0^ * - {A_0} + {\alpha ^2}{A_{ - 1}} + C = {B_0} - {A_0} + {\alpha ^2}{A_{ - 1}} \end{split} $$ (35) $$\begin{split} & ({\alpha ^2} - 1){a_{ - 1}} + ({\alpha ^2} - 1) \overline{{ a_1}} = \\ &\qquad \alpha \overline{ B_1^ * } - \overline {{ A_1}} + {\alpha ^2}\overline {{ A_0}} - {\alpha ^2}\overline C = \alpha \overline{{ B_1}} - \overline{{ A_1}} + {\alpha ^2} \overline{{ A_0}}\end{split} $$ (36) 可求得复常数$C$, 其实部与虚部为
$$\begin{split} & {{\rm{Re}}} (C) = \frac{1}{{1 + {\alpha ^2}}}{{\rm{Re}}} (\alpha \overline { B_1^ * } - \overline{{ A_1}}+\\ &\qquad {\alpha ^2} \overline{{ A_0}} - B_0^ * + {A_0} - {\alpha ^2}{A_{ - 1}}) \end{split} $$ (37) $$ \begin{split} & {{\rm{Im}}} (C) = \frac{1}{{1 - {\alpha ^2}}}{\text{Im}}(\alpha \overline{ B_1^ * } - \overline{{ A_1}}+\\ &\qquad {\alpha ^2} \overline{{ A_0}} - B_0^ * + {A_0} - {\alpha ^2}{A_{ - 1}})\end{split} $$ (38) 其中${A_k}$已由地表边界得出, ${B_k}$可根据隧道孔口边界求得. 将式(2)、式(13)和映射函数(16)代入孔口应力边界条件(27)右侧积分项中并将其展开为有关$\sigma $的多项式得到式(39), 则式(29)左侧可以展开为式(40), 式(40)通过同幂次项系数对应可得到级数系数$B_k^*$, 复常数$C$已知从而求得${B_k}$
$$ \begin{split} & \left( {1 - \alpha \sigma } \right){\rm{i}}\int_A^B {({X_n} + {\rm{i}}{Y_n}){\text{d}}s}= \\ &\qquad \left( {1 - \alpha \sigma } \right){\rm{i}}\int_A^B {\left[ {{\gamma _w}{h_w} - ({\gamma _w} + {k_0}\gamma )y} \right]{\text{d}}y} + \left( {1 - \alpha \sigma } \right)\int_A^B {\left[ {{\gamma _w}{h_w} - ({\gamma _w} + \gamma )y} \right]{\text{d}}x}= \\ &\qquad - {\rm{i}}a{\gamma _w}{h_w}\left( {1 + \alpha \sigma } \right) - \left( {1 - \alpha \sigma } \right)({\gamma _w} + {k_0}\gamma ){\rm{i}}\frac{{{a^2}}}{8}{\left( {\frac{{1 + \alpha \sigma }}{{1 - \alpha \sigma }} + \frac{{\sigma + \alpha }}{{\sigma - \alpha }}} \right)^2}- \\ &\qquad \left( {1 - \alpha \sigma } \right)({\gamma _w} + \gamma )\frac{{{\rm{i}}{a^2}}}{2}\alpha \left[ { - \frac{\alpha }{{{{(\alpha - \sigma )}^2}}} - \frac{2}{{({\alpha ^2} - 1)(\alpha - \sigma )}} + \frac{1}{{\alpha {{(\alpha \sigma - 1)}^2}}} + \frac{{2\alpha }}{{({\alpha ^2} - 1)(\alpha \sigma - 1)}}} \right]- \\ &\qquad \left( {1 - \alpha \sigma } \right)({\gamma _w} + \gamma )\frac{{{\rm{i}}{r^2}}}{2}\left[ {{\text{ln}}(\sigma - \alpha ) - {\text{ln}}(1 - \alpha \sigma )} \right] \end{split} $$ (39) $$ \begin{split} & (1 - \alpha \sigma )\left[ {{\varphi _0}(z) + z\overline {{{\varphi '}_0}(z)} + \overline {{\psi _0}(z)} } \right]= \\ &\qquad \frac{{{F_x} - {\rm{i}}{F_y}}}{{2\text{π} (1 + K)}}(1 - \alpha \sigma )\left(\frac{{zK}}{{\bar z - {z_c}}} + \frac{z}{{\bar z - {{\bar z}_c}}}\right) + (1 - \alpha \sigma ){\rm{i}}\int_A^B {({X_n} + {\rm{i}}{Y_n}){\text{d}}s} + \left( {1 - \alpha \sigma } \right)C- \\ &\qquad \frac{{{F_x} + {\rm{i}}{F_y}}}{{2\text{π} (1 + K)}}(1 - \alpha \sigma )[\ln (\bar z - {z_c}) + K\ln (\bar z - {{\bar z}_c})] + \frac{{{F_x} + {\rm{i}}{F_y}}}{{2\text{π} (1 + K)}}(1 - \alpha \sigma )\left[ {K\ln (z - {{\bar z}_c}) + \ln (z - {z_c})} \right] = \\ &\qquad \frac{{{F_x} - {\rm{i}}{F_y}}}{{2\text{π} (1 + K)}}\left[ { - \frac{1}{2}{\sigma ^2} + \frac{{(1 - K){\alpha ^2} - 1}}{{2\alpha }}\sigma + \frac{{K{\alpha ^2} + 1 - K}}{2} + \frac{{K\alpha }}{2}{\sigma ^{ - 1}}} \right] + \left( {1 - \alpha \sigma } \right)C- \\ &\qquad (1 - \alpha \sigma )\left[ {(\gamma + {\gamma _w})\frac{{{\rm{i}}{r^2}}}{2}\frac{{(K - 1)}}{{(1 + K)}}\ln \alpha - (\gamma + {\gamma _w})\frac{{\text{π} {r^2}}}{2}} \right] - {\rm{i}}a{\gamma _w}{h_w}\left( {1 + \alpha \sigma } \right) - \left( {1 - \alpha \sigma } \right)({\gamma _w} + {k_0}\gamma ){\rm{i}}\frac{{{a^2}}}{8}{\left( {\frac{{1 + \alpha \sigma }}{{1 - \alpha \sigma }} + \frac{{\sigma + \alpha }}{{\sigma - \alpha }}} \right)^2}- \\ &\qquad ({\gamma _w} + \gamma )\frac{{{\rm{i}}{a^2}}}{2}\alpha \left[ { - \frac{{\alpha \left( {1 - \alpha \sigma } \right)}}{{{{(\alpha - \sigma )}^2}}} - \frac{{2\left( {1 - \alpha \sigma } \right)}}{{({\alpha ^2} - 1)(\alpha - \sigma )}} + \frac{1}{{\alpha (1 - \alpha \sigma )}} + \frac{{2\alpha }}{{1 - {\alpha ^2}}}} \right] \end{split} $$ (40) 3. 解析函数求解与说明
3.1 解析函数求解
将复势函数$ {\varphi _0}(\zeta ) $和${\psi _0}(\zeta )$Laurent级数展开, 其中未知系数${a_k}$和${b_k}$可由迭代式(33)和式(34)求出, 但当$k = 0$和$k = 1$时得到相同的等式, 则无法直接求解${a_{ - 1}}$和${\bar a_1}$, 其余${a_k}$和${b_k}$也无法根据迭代公式求得. 为解决这一问题, 利用无穷远点应力有界性提出一种求解方法. 在图3模型中仅有隧道孔口边界${L_2}$施加载荷, 则距离边界${L_2}$无穷远的岩体应力有界. 为保证无穷远点应力有界性这一条件, 根据式(3)和式(4),则复势函数可表达为
$$ {\varphi _0}(z) = \sum\limits_{k = - \infty }^2 {a_k^*} {z^k} $$ (41) $$ {\psi _0}(z) = \sum\limits_{k = - \infty }^1 {b_k^*} {z^k} $$ (42) 将$z = w(\zeta )$代入上式, 并将映射函数中含$\zeta $分式级数展开为
$$ \frac{{1 + \zeta }}{{1 - \zeta }} = 1 + 2\left[ {\zeta + {\zeta ^2} + \cdots + {\zeta ^n} + o({\zeta ^n})} \right] $$ (43) 则有
$$ {\varphi _0}(\zeta ) = \sum\limits_{k = - 1}^\infty {{a_k}{\zeta ^k}} $$ (44) $$ {\psi _0}(\zeta ) = \sum\limits_{k = - 1}^\infty {{b_k}{\zeta ^k}} $$ (45) 从上式可知当$k \leqslant - 2$时有${a_k} = 0$和${b_k} = 0$.
由式(35)得
$$ {a_{ - 1}} + \overline{ a_1} = \frac{1}{{{\alpha ^2} - 1}}({B_0} - {A_0} + {\alpha ^2}{A_{ - 1}}) $$ (46) 式(26)中$k = 0$时, 有
$$ {a_{ - 1}} + {a_1} = 2 \overline{{ A_0}} $$ (47) 将式(46)和式(47)实部与虚部分开计算, 再采用复变函数方法能够求出幂级数解工况中${{\rm{Re}}} {(a_{ - 1})}$值为趋于0的极小值, 假定${{\rm{Re}}} {(a_{ - 1})} = 0$, 则${a_1}$实部${{\rm{Im}}} {(a_{ - 1})}$和${{\rm{Im}}} {(a_1)}$可由下式准确求得
$$ {{\rm{Im}}} {(a_{ - 1})} = \frac{1}{2}{{\rm{Im}}} \left[ {\frac{1}{{{\alpha ^2} - 1}}({B_0} - {A_0} + {\alpha ^2}{A_{ - 1}}) + 2 \overline{{ A_0}} } \right] $$ (48) $$ {{\rm{Re}}} {(a_1)} = {{\rm{Re}}} \left[ {\frac{1}{{{\alpha ^2} - 1}}({B_0} - {A_0} + {\alpha ^2}{A_{ - 1}})} \right] $$ (49) $$ {{\rm{Im}}} {(a_1)} = \frac{1}{2}{{\rm{Im}}} \left[ {2 \overline{{ A_0}} - \frac{1}{{{\alpha ^2} - 1}}({B_0} - {A_0} + {\alpha ^2}{A_{ - 1}})} \right] $$ (50) 根据迭代式(33), 有
$$\begin{split} & {a_2} = \frac{1}{{{\alpha ^2} - {\alpha ^{ - 2}}}}{B_2} - {\alpha ^{ - 2}}{A_2}+ \\ &\qquad {A_1} - \left( {1 - {\alpha ^2}} \right){a_1} - \left( {1 - {\alpha ^2}} \right){\alpha ^{ - 2}} \overline{{ a_{ - 1}}} \end{split} $$ (51) $$ \begin{split} &{a}_{k} = \frac{1}{{\alpha }^{k}-{\alpha }^{-k}}{B}_{k}-{\alpha }^{-k}{A}_{k}+\\ &\qquad{\alpha }^{-k + 2}{A}_{k-1}-\left({\alpha }^{-k + 2}-{\alpha }^{k}\right){a}_{k-1}\text{, }\quad k\geqslant 3\end{split} $$ (52) 根据地表边界条件(26)可求得系数${b_k}$
$$ \begin{split} &{b}_{k} = \overline{{ A}_{-k}}-\overline{{ a}_{-k}}-\frac{1}{2}\left(k + 1\right){a}_{k + 1}+ \\ &\qquad \frac{1}{2}\left(k-1\right){a}_{k-1}\text{, }\quad k\geqslant -1\end{split} $$ (53) 式(48) ~ 式(53)即为复势函数$ {\varphi _0}(\zeta ) $和${\psi _0}(\zeta )$ Laurent级数展开中各未知系数的求解公式.
3.2 求解方法说明
从映射函数式中可知点$\zeta = 1$为奇点, 由于这一奇点的存在给解析计算带来很大的困扰. 从3个不同的方向趋近于该奇点分别对应物理平面中岩体的3个无穷远处. 在$\zeta $平面单位圆$\zeta = {{\rm{e}}^{{\rm{i}}\theta '}}$上, 当$\theta ' \to 0$和$\theta ' \to 2\text{π}$其分别对应$z$平面地表边界无穷远点$x = + \infty $和$ x = - \infty $处, $\zeta $沿$\xi $轴$\zeta \to 1$对应无穷远点$y = - \infty $处. 同时表达式${1 \mathord{\left/ {\vphantom {1 {(1 - \zeta) }}} \right. } {(1 - \zeta )}}$在$\left| \zeta \right| < 1$区域内收敛, 可以表示为幂级数, 但其在奇点所处的单位圆上, 不能将其展开为幂级数而无法采用幂级数解法.
为保证求解精准性和计算的可行性, 我们重新定义地表边界${L^ * }$(如图5所示)为$z$平面中$y \to 0$的直线和$\zeta $平面中$\zeta = (1 - t)\sigma $, $t \in (0,0.1)$的近似单位圆. 当t足够小时, 新定义的地表边界${L^ * }$将趋近于原地表边界${L_1}$, 这不仅能够确保边界条件与原地表的边界条件相同, 还能确保岩体映射后的讨论域$\zeta $均在表达式$ {1 \mathord{\left/ {\vphantom {1 {(1 - \zeta )}}} \right. } {(1 - \zeta )}} $的收敛域内. 虽然 $\zeta $平面上边界 ${L^ * }$($\rho = 1 - t$)和边界${L_1}$($\rho = 1$)所表示的物理含义不同, 但在解析计算时可以将$\rho = 1 - t$近似为$\rho = 1$, 这并不影响最后的计算结果.
在求解计算时, 式${1 \mathord{\left/ {\vphantom {1 {(1 - \zeta )}}} \right. } {(1 - \zeta )}}$级数展开的项数一定, $\left| \zeta \right|$越小幂级数的拟合效果越好, 因此半无限平面含孔洞问题的复变函数解法在孔口周围求解效果最好; 经过映射函数映射后, 地表边界所对应的$\left| \zeta \right|$值最大, 其对级数展开后的拟合效果有较高的要求. 同等的条件下, 在最后的求解结果中隧道孔口边界的计算误差恒定小于地表边界. 从上文定义中可知新地表边界${L^ * }$上$\left| \zeta \right| \to 1$, 要使得邻近地表边界范围内的区域有较高的计算精度, 则需要幂级数展开更多的高次项.
4. 应力求解
将映射函数代入式(3)和式(4)得到迪卡尔坐标系下的应力求解公式
$$ \frac{{\sigma _x^*}}{{{k_0}}} + \sigma _y^* = 4{\rm{Re}}\left[ {\frac{{\varphi '(\zeta )}}{{\omega '(\zeta )}}} \right] $$ (54) $$ \begin{split} & \sigma _y^* - \frac{{\sigma _x^*}}{{{k_0}}} + 2{\rm{i}}\sigma _{xy}^* = \frac{2}{{\omega '(\zeta )}} \cdot \\ &\qquad \left\{ {\frac{{\overline {\omega (\zeta )} \left[ {\varphi ''(\zeta )\omega '(\zeta ) - \omega ''(\zeta )\varphi '(\zeta )} \right]}}{{{{\left[ {\omega '(\zeta )} \right]}^2}}} + \psi '(\zeta )} \right\} \end{split} $$ (55) 在极坐标中各应力分量的关系式为
$$ \sigma _\rho ^* + \sigma _\theta ^* = 4{\rm{Re}}\left[ {\frac{{\varphi '(\zeta )}}{{\omega '(\zeta )}}} \right] $$ (56) $$ \begin{split} & \sigma _\theta ^* - \sigma _\rho ^* + 2{\rm{i}}\sigma _{\rho \theta }^* = \frac{{2{\zeta ^2}}}{{{\rho ^2}\overline {\omega '(\zeta )} }} \cdot \\ &\qquad \left\{ {\frac{{\overline {\omega (\zeta )} \left[ {\varphi ''(\zeta )\omega '(\zeta ) - \omega ''(\zeta )\varphi '(\zeta )} \right]}}{{{{\left[ {\omega '(\zeta )} \right]}^2}}} + \psi '(\zeta )} \right\} \end{split} $$ (57) 式中, $\sigma _\rho ^*$, $\sigma _\theta ^*$和$\sigma _{\rho \theta }^*$分别表示径向应力、环向应力和切向应力.
将复势函数$\varphi (\zeta )$和$\psi (\zeta )$代入式(54) ~ 式(57)即可得到不同坐标系下的应力分量.
5. 不同方法对比
当本文浅埋海底隧道模型所含参数${h_w}$ = 0和${\gamma _w}$ = 0时即可简化为文献[22]中考虑重力和任意侧向应力作用下的弹性半平面含圆孔模型. 为了验证本文方法的准确性, 对侧压力系数${k_0} = 0.5$下的工况求解, 将其与Lu等[22]得出的结果和有限元结果进行对比. 复势函数${\varphi _0}(\zeta )$和${\psi _0}(\zeta )$分别计算至${a_6}$和${b_5}$. 分析孔口应力时所取计算范围与文献[22]相同即隧道右半部分, $\alpha = 0^\circ $为隧道底部, $\alpha = 180^\circ $为隧道顶部. 其余参数也与文献[22]中参数相同: $\mu = 0.3$, $E = 10.0\;{\text{MPa}}$, $\gamma = 25.0\;{\text{kN/}}{{\text{m}}^{\text{3}}}$和$h/r = 2.0$. 极坐标系中的计算结果对比见图6.
从图6可知, 本文解得的复势函数所计算的孔边环向应力与Lu等[22]得出结果和有限元结果基本相同. 与其他两种方法相比, 本文计算结果压应力偏大. 当${k_0} = 0.5$, ${h \mathord{\left/ {\vphantom {h r}} \right. } r} = 2.0$时, Lu等[22]计算复势函数需求解32元线性方程组. 本文只需根据复势函数系数的显式求解式(52)和式(53)分别求出系数${a_6}$与${b_5}$, 根据不同精度需求可确定k 的取值, k取值越高精度越高, 同时本计算过程也减少了依靠地表和孔口边界条件所求系数${A_k}$和${B_k}$的需求. 因此, 在满足工程精度需求下, 运用复势函数显式计算公式求解, 计算过程简便, 计算量小.
6. 算例分析
为了验证本文理论方法的可靠性, 将本文在迪卡尔坐标系下的理论解答与数值计算结果进行对比. 此例定义参数如下: 隧道孔洞半径r = 4 m, 埋深$h = 10\;{\text{m}}$, 海水深度${h_w} = 2\;{\text{m}}$, 海水重度${\gamma _w}$ = 10 kN/m3, 土体重度$\gamma $ = 25 kN/m3, 泊松比$\mu $ = 0.3, 弹性模量E = 200 MPa. 规定应力拉为正, 压为负. 利用ANSYS软件建立第2部分解答的有限元模型计算数值解, 模型尺寸为100 m × 60 m, 采用PLANE183单元, 共划分单元3028个, 图7为网格划分情况. 模型上边界为自由边界, 对模型左右边界施加x向约束, 下边界施加y向约束, 隧道圆形孔口施加与幂级数解边界${L_2}$相同的面力.
本文主要是对第2部分模型(图3)采用复变函数方法求解, 同样利用ANSYS对第2部分解答模型进行模拟. 通过有限元模拟得出应力分量$\sigma _x^*$, $\sigma _y^*$和$\sigma _{xy}^*$的应力云图, 如图8所示. 利用MATLAB软件编程计算得到幂级数解. 分别将幂级数解和有限元结果与第1部分解叠加得到隧道开挖后的围岩应力. 孔边应力幂级数解和数值解对比见图9.
6.1 数值解对比
从图9可以看出幂级数解(power series solutions 1)与数值解能较好吻合, 验证了本文方法的正确性. 当$ \theta \in [0^\circ ,180^\circ ] $时为隧道上半部分, 幂级数解与数值解得到的孔边应力(${\sigma _x}$, ${\sigma _y}$和${\sigma _{xy}}$)基本一致; 但当$\theta \in (180^\circ ,360^\circ )$时为隧道下半部分, 幂级数解与数值解具有一定的差异. 幂级数解与数值解产生的计算差异是由所选映射函数造成的. 与数值解相比, 幂级数解得到的${\sigma _y}$略小于数值解, 而孔口下半部分各点应力分量${\sigma _x}$与${\sigma _{xy}}$的绝对值则比数值解的绝对值大. 幂级数解最终计算结果比数值解相对保守.
6.2 解析函数系数影响分析
为了研究单值解析函数${\varphi _0}(\zeta )$与${\psi _0}(\zeta )$中系数${a_k}$与${b_k}$取不同项数对运算结果的影响, 取3组${a_k}$与${b_k}$展开不同项数时的幂级数解进行对比. 系数展开项数见表1, 幂级数解对比见图9.
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图 9 不同系数的幂级数解和数值解对比Figure 9. Comparison of power series and numerical solutions with different coefficients表 1 项数展开表Table 1. Number of terms expansion tableThe different of ${\varphi _0}(\zeta )$ and ${\psi _0}(\zeta )$ ${a_k}$ ${b_k}$ power series solutions 1 ${a_{ - 1}} \sim {a_6}$ ${b_{ - 1}} \sim {b_5}$ power series solutions 2 ${a_{ - 1}} \sim {a_2}$ ${b_{ - 1}} \sim {b_6}$ power series solutions 3 ${a_{ - 1}} \sim {a_6}$ ${b_{ - 1}} \sim {b_2}$ 从图9中可以看出power series solutions 1与power series solutions 3在3个应力分量${\sigma _x}$, ${\sigma _y}$和${\sigma _{xy}}$的计算结果都基本保持一致, power series solutions 2与其结果相比在不同位置、不同应力分量上均有一定的差距. 特别在隧道孔口上半部分$ \theta \in [0^\circ ,180^\circ ] $, power series solutions 3要优于power series solutions 2.
从中可以得出单值解析函数${\varphi _0}(\zeta )$对最终结果准确性的影响要大于${\psi _0}(\zeta )$. 求解时为了保证计算结果的准确性, ${\varphi _0}(\zeta )$需要展开足够多的项, ${\psi _0}(\zeta )$则至少要展开${b_{ - 1}} \sim {b_1}$项.
6.3 隧道埋深影响分析
为了分析浅埋隧道埋深对环向应力的影响, 隧道中心埋深h分别取8, 10, 12和14 m, 单值解析函数${\varphi _0}(\zeta )$与${\psi _0}(\zeta )$中系数${a_k}$与${b_k}$展开项与power series solutions 1相同, 其余条件与参数不变, 不同隧道埋深孔口应力情况见图10. 隧道底部($\theta $ = 270°)及两侧孔腰处($\theta $ = 200°或$\theta $ = 340°)环向压应力随着隧道埋深h增大而增大, 而孔口顶部($\theta $ = 90°)环向压应力无显著变化. 隧道埋深增加1.75倍, 腰部及底部环向压应力增加3 ~ 4倍. 隧道腰部与底部环向应力的差值也随着隧道埋深的增大而增大. 由此可见隧道埋深对隧道底部及腰部环向压应力的大小影响较大, 且对腰部的影响最为显著.
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图 10 不同埋深孔口环向应力对比Figure 10. Comparison of annular stress at orifices with different burial depth7. 结 论
本文运用平面弹性复变函数方法研究了浅埋圆孔海底隧道在围岩重力和海水静水压力作用下的围岩应力幂级数解. 采用分式映射函数将半无限平面含圆孔模型映射为像平面上的圆环域, 在圆环域内将复势单值解析函数展开为Laurent级数. 利用无穷远点应力有界性对级数展开式的幂次项进行确定, 根据级数系数迭代公式得到复势函数的显式解, 运用应力分量的复变函数表达式得到孔边各点的应力分量, 将幂级数解与有限元结果对比得出以下结论.
(1) 为保证无穷远点应力有界, 根据应力求解公式, 将复势单值解析函数${\varphi _0}(\zeta )$和${\psi _0}(\zeta )$的Laurent级数重新定义为负一项至正无穷项. 在${{\rm{Re}}} {(a_{ - 1})} = 0$条件下, 由地表与孔口边界条件得出复势解析函数系数显式解, 此解可迭代求得其余系数, 从而实现级数系数从低次幂迭代至高次幂这一过程, 使得浅埋隧道复势函数求解更方便.
(2)幂级数解与有限元数值解能很好吻合. 在隧道上半部分($\theta \in [0^\circ ,180^\circ ]$)幂级数解与数值解得到的孔边应力(${\sigma _x}$, ${\sigma _y}$, ${\sigma _{xy}}$)基本一致; 在隧道下半部分($\theta \in (180^\circ ,360^\circ )$)由于所选映射函数的影响, 幂级数解得到的${\sigma _y}$略小于数值解, ${\sigma _x}$与${\sigma _{xy}}$的绝对值则比数值解的绝对值大. 幂级数解最终计算结果较数值解偏保守.
(3) 单值解析函数${\varphi _0}(\zeta )$对最终结果准确性的影响要大于${\psi _0}(\zeta )$. 求解时为了保证计算结果的准确性, ${\varphi _0}(\zeta )$需要展开足够多项, ${\psi _0}(\zeta )$则至少要展开${b_{ - 1}} \sim {b_1}$项.
(4) 随着海底隧道埋深增大, 隧道底部及两侧孔腰处的环向压应力随之增大. 隧道埋深增加1.75倍, 腰部及底部环向压应力增加3 ~ 4倍; 而孔口顶部环向压应力无显著变化. 隧道腰部与底部环向应力的差值也随着隧道埋深的增大而增大. 隧道埋深对隧道底部及腰部环向压应力的大小影响较大, 且对腰部的影响最为显著.
数据可用性声明
支撑本研究的科学数据已在中国科学院科学数据银行(science data bank) ScienceDB平台公开发布, 访问地址为https://www.doi.org/10.57760/sciencedb. j00140.00011或http://resolve.pid21.cn/31253.11.sciencedb.j00140.00011.
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图 9 不同系数的幂级数解和数值解对比
Figure 9. Comparison of power series and numerical solutions with different coefficients
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图 10 不同埋深孔口环向应力对比
Figure 10. Comparison of annular stress at orifices with different burial depth
表 1 项数展开表
Table 1 Number of terms expansion table
The different of ${\varphi _0}(\zeta )$ and ${\psi _0}(\zeta )$ ${a_k}$ ${b_k}$ power series solutions 1 ${a_{ - 1}} \sim {a_6}$ ${b_{ - 1}} \sim {b_5}$ power series solutions 2 ${a_{ - 1}} \sim {a_2}$ ${b_{ - 1}} \sim {b_6}$ power series solutions 3 ${a_{ - 1}} \sim {a_6}$ ${b_{ - 1}} \sim {b_2}$ 
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