THREE-DIMENSIONAL ANALYTICAL METHOD FOR CALCULATING VIBRATIONS OF A COUPLED TUNNEL-SOIL-FLUID SYSTEM
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摘要: 轨道交通隧道行车会引发地表或水中环境的振动, 从而对沿线居民或珍惜鱼类产生不良影响. 对此, 提出一种双洞隧道−地基−流体耦合系统动力响应的三维解析方法. 隧道和地基模拟为弹性介质, 空气或水模拟为理想流体介质, 利用时间和纵向坐标的双重傅里叶变换, 将时间−空间域内的三维动力问题转化为频域−波数域内求解. 利用波面转换和平移公式, 满足隧道、地基、流体及其交界面的边界条件, 实现动力耦合求解, 获得简谐载荷作用下双洞隧道−地基−流体系统动力响应的解析解. 通过与地基−流体耦合系统的动力基本解以及有限元−边界元耦合模型进行对比, 验证本文所提出方法的正确性. 对比分析单洞和双洞隧道内作用动力点源引起的地基、水以及地表空气的响应. 结果表明, 邻近隧道的存在会改变地基土中波动能量的分布, 进而改变水中或者空气中振动传播规律, 且邻近隧道的影响与载荷频率、观察点位置以及隧道间的相对位置关系密切相关. 当双洞隧道间距小于4倍隧道直径时, 隧道间的动力相互作用不可忽略. 研究成果可为隧道内行车引发的空气噪声以及水下振动传播特性研究和预测评估分析提供理论支撑.Abstract: Train-induced vibrations from underground railway tunnels transmit to the ground surface or water, which may disturb adjacent residents or rare fish. This study presents a three-dimensional analytical method for calculating the dynamic response of the coupled tunnel-soil-fluid system. The twin tunnels and soils are simulated as the elastic solid, while the air or water is simulated as the ideal fluid medium. The three-dimensional dynamic problem in the time-space domain is transformed into the frequency-wavenumber domain by using the double Fourier transforms of time and longitudinal coordinates. The boundary conditions on the fluid-soil interface and the tunnel-soil interface are satisfied by introducing the transformation between the cylindrical waves and the plane waves and the transform between different cylindrical waves. The solution for dynamic response of the coupled tunnel-soil-fluid system under point loads is therefore derived. The accuracy of the proposed method is verified by comparing with the Green’s function and FE-BE model for a coupled soil-fluid system. The dynamic responses of soil, water, and air induced by point loads in a single tunnel and twin tunnels are compared and analyzed via two numerical cases. The results demonstrate that the existence of the adjacent tunnel can change the energy distribution of train-induced waves in the soil, thus affecting the propagation characteristics of waves in the water or in the air. The influence of the adjacent tunnel is highly dependent of the loading frequency, the observation position and the relative position between twin tunnels. When the distance between the two adjacent tunnels is less than four times of the diameter of the tunnel, the dynamic interaction between the twin tunnels play a relevant role in the response of the water or the air. This study can provide benefit for the evaluation of vibrations and radiated noise in air or water from underground railway tunnels.
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Keywords:
- railway tunnel /
- environmental vibration /
- dynamic interaction /
- analytical solution
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引 言
随着城市轨道交通建设的快速发展, 地下列车运行时引起的环境振动和噪声问题也愈发突出. 一方面, 地下列车行车振动会通过隧道和地层传播至地表环境, 引发空气噪声, 从而对沿线周边的居民生活和工作产生不良影响[1-2]. 另一方面, 跨江河甚至跨海隧道工程愈发普遍[3], 而水下隧道内行车振动通过土层传播到水中, 会对水中的生物和生态环境产生不良影响, 尤其是位于育种区的珍惜鱼类[4-5]. 因此, 有必要建立可靠的轨道交通隧道车致振动在空气或水中传播的计算方法, 理清振动的传播特性, 从而采取有效的减隔振措施.
国内外学者首先针对隧道−地基系统动力响应问题开展了诸多研究, 提出系列数值和解析/半解析计算方法. 常见的数值法包括有限元(FE)法和有限元−边界元(FE-BE)耦合法, 相关研究人员通过建立FE模型分析了地铁列车运行引起地表环境的响应[6-10]. 有限元法的优点在于可以模拟复杂的几何结构, 但存在计算效率低和人工截断边界易产生误差等问题. 为了提高三维数值计算的效率, 通过假设模型沿纵向保持不变, 相关学者提出2.5维的数值计算方法. 高广运等[11]和边学成[12]运用2.5维FE方法计算了地铁运行引起的地基动力响应. Sheng等[13]和François等[14]提出2.5维FE-BE耦合法, 并计算了列车运行引起的地基响应. 何超等[15]在此基础上, 建立饱和地基中隧道车致振动响应的2.5维有限元−边界元耦合模型. 利用该模型, 进一步研究分块盾构隧道引起的动力响应特性[16]. 此外, Yang等[17]还提出隧道−地基系统动力响应的2.5维有限元/无限元法, 利用无限单元模拟波在半无限远场的衰减. 基于2.5维有限元−完美匹配层法, 朱志辉等[18]建立轨道−隧道−地基系统的车致振动计算模型. 通过假设结构在纵向的周期性分布, Degrande等[19]提出周期性FE-BE方法. 狄宏规等[20]将自由波传播理论引入2.5维有限元−完美匹配层法, 建立了隧道−地基系统的多周期性有限元模型.
相比于数值法, 解析/半解析法具有计算效率高和操作简便的优势, 便于工程人员直接应用. Metrikine等[21]将地基中的隧道简化为欧拉梁, 分析不同载荷下地表的动力响应. Hunt等[22-23]将土体和隧道视为两个同心圆柱管, 提出隧道−地基耦合系统动力响应的PiP模型. Yuan等[24]计算了移动简谐载荷作用下半空间中隧道动力响应的基本解. He等[25-26]借助传递矩阵法和波面转换理论, 提出水平分层地基中隧道动力响应的波函数解析法. 而考虑轨道交通线路普遍采用双洞隧道形式, He等[27-28]进一步引入Graf加法定理, 建立了均质和水平分层地基中双洞隧道车致振动响应的三维解析模型. 郭慧吉等[29]将其扩展至非饱和土−隧道动力相互作用问题.
上述研究仅针对隧道−地基耦合系统的动力响应分析, 并未考虑其与流体的动力耦合作用, 无法用于隧道内行车引发的空气噪声以及水中振动评估分析. 对此, Romero等[30]在隧道−地基系统的2.5维FE-BE模型基础上, 将空气视为理想流体并采用2.5维BE模拟, 分析了隧道车致振动引发的空气噪声问题. Song等[5]采用三维FE模型分析了水下隧道内行车引起的水中振动问题.
综上所述, 针对隧道−地基系统车致动力响应问题的研究已较为成熟, 建立了系列数值和解析方法. 然而, 隧道行车引发的空气噪声以及水下振动涉及隧道−地基−流体的动力相互作用, 目前仅见少量基于数值方法的研究分析, 尚无隧道−地基−流体耦合系统动力响应的理论解析方法. 此外, 尽管已有部分学者针对流体−地基耦合系统的动力响应开展了研究, 并推导了相应的基本解. 但也都未考虑地基−流体系统与埋置隧道结构的动力相互作用, 因此也无法应用于隧道内行车引发的空气噪声以及水下振动预测评估.
对此, 本文将建立一种计算隧道−地基−流体耦合系统动力响应的高效解析方法. 采用连续介质力学和流体力学分别建立隧道、地基和流体的振动控制方程, 通过双重傅里叶变换, 将时间−空间域内的三维动力问题转化为频域−波数域内求解; 引入波面转换和平移公式, 满足流体、地基、隧道及其交界面的边界条件, 推导双洞隧道−地基−流体系统动力响应的解析解. 通过与既有方法对比验证其正确性. 最后, 利用该方法对隧道内作用动力点源引起的空气和水下振动进行了分析. 研究成果可为隧道内行车引发的空气噪声以及水下振动传播特性研究和预测评估分析提供理论支撑.
1. 控制方程及其通解
隧道−地基−流体耦合系统计算模型如图1所示, 其中隧道和地基土体均视为弹性介质, 而地基以上的空气或水视为理想流体. 考虑到轨道交通通常采用双线形式以实现双向运输乘客, 本文采用双洞隧道的设计形式. 其中, 右线隧道(隧道1)的埋深为 ht, 两隧道中心的距离为 dt, 左线隧道(隧道2)相对右侧隧道的角度为 θt, 两隧道的内径和外径分别为$ {R}_{i}^{\mathrm{1,2}} $和$ {R}_{o}^{\mathrm{1,2}} $. 以两个隧道中心为原点, 分别建立两个局部直角坐标系(x1, y1, z), (x2, y2, z)和对应的圆柱坐标系(r1, θ1, z), (r2, θ2, z). 此外, 以隧道1中心在流体−地基交界面上的投影点为原点建立描述流体的局部直角坐标系(xf, yf, z).
由于车致振动引起的应变水平低, 可假定车致振动响应为弹性变形[31]. 此外, 假定隧道−地基−流体系统沿隧道纵向无限长, 且材料特性和几何形状沿隧道纵向保持不变. 为此, 便可引入时间t和纵向坐标z的傅里叶变换, 从而将时间−空间域动力问题转化至频域−波数域内求解. 其中, 对应的傅里叶变换如下所示
$$ \left.\begin{aligned} & \hat f(\omega ) = \int_{ - \infty }^{ + \infty } {f(t){{\rm{e}}^{ - {\rm{i}}wt}}{\text{d}}t} \\ & f(t) = \frac{1}{{2{\text{π }}}}\int_{ - \infty }^{ + \infty } {\hat f(\omega ){{\rm{e}}^{{\rm{i}}\omega t}}{\text{d}}\omega } \end{aligned} \right\}\tag{1a} $$ $$ \left.\begin{aligned} & \tilde f({k_z}) = \int_{ - \infty }^{ + \infty } {f(z){{\rm{e}}^{ - {\rm{i}}{k_z}z}}{\text{d}}z} \\ & {\text{ }}f(z) = \frac{1}{{2{\text{π }}}}\int_{ - \infty }^{ + \infty } {\tilde f({k_z}){{\rm{e}}^{{\rm{i}}{k_z}z}}{\text{d}}{k_z}} \end{aligned} \right\}\tag{1b} $$ 式中, 上标“ˆ”和“˜”分别代表对时间t和坐标z的傅里叶变换, ω为角频率, kz为z向(隧道纵向)波数.
1.1 地基波动方程及其通解
地基土体和隧道均视为各向同性的弹性连续介质, 其波动方程可以表示为[32]
$$ \left( {\lambda + 2\mu } \right)\nabla \nabla \cdot {\boldsymbol{u}} - \mu \nabla \times (\nabla \times {\boldsymbol{u}}) = \rho \ddot {\boldsymbol{u}} $$ (2) 式中, ${\boldsymbol{u}}$为介质的位移向量; λ和μ为拉梅常数, $ \rho $为介质密度; 上标“··”代表对时间的2阶求导.
式(2)可通过势函数分解法求解, 对于三维问题, 位移可通过势函数表示为
$$ {\boldsymbol{u}} = \nabla \times ({{{\boldsymbol{e}}}_z}{\varphi _{{\rm{SH}}}}) + \nabla \times \nabla \times ({{{\boldsymbol{e}}}_z}{\varphi _{{\rm{SV}}}}) + \nabla {\varphi _{\rm{P}}} $$ (3) 式中, φSH, φSV和φP分别代表介质中SH波、SV波和P波的势函数.
将式(3)代入式(2), 并进行式(1)所示的双重傅里叶变换, 可获得3个亥姆霍兹方程
$$ \left.\begin{aligned} & {{\tilde \nabla }^2}{{\hat {\tilde {\varphi}} }_{{\rm{SH}}}} + k_{rs}^2{{\hat {\tilde {\varphi}} }_{{\rm{SH}}}} = 0 \\ & {{\tilde \nabla }^2}{{\hat {\tilde {\varphi}} }_{{\rm{SV}}}} + k_{rs}^2{{\hat {\tilde {\varphi}} }_{{\rm{SV}}}} = 0 \\ & {{\tilde \nabla }^2}{{\hat {\tilde {\varphi}} }_{\rm{P}}} + k_{rp}^2{{\hat {\tilde \varphi} }_{\rm{P}}} = 0 \end{aligned} \right\} $$ (4) 式中, $ {k}_{rs,p} = {\left({k}_{s,p}^{2}-{k}_{z}^{2}\right)}^{1/2} $; $ {k}_{s,p} = \omega /{C}_{s,p} $为S波和P波的波数; $ {C}_{s} = {(\mu /\rho )}^{1/2} $, $ {C}_{p} = {\left[(2\mu + \lambda )/\rho \right]}^{1/2} $ 分别为介质中S波和P波的波速.
根据波的多重散射定律[33], 地基中存在由隧道−土体界面激发的柱面波和土体−流体界面激发的平面波, 因此, 地基土体的位移和应力的通解可表示为
$$ \begin{split} & {{\hat {\tilde {\boldsymbol{u}}}}_s} = \frac{1}{{2{\text{π }}}}\int\limits_{ - \infty }^{ + \infty } {\sum\limits_{j = 1}^3 {{{A}_{dj}}{{\hat {\bar {\tilde {\boldsymbol{\phi}}}} }_{dj}}} } {\text{d}}{k_y}+ \\ &\qquad \sum\limits_{m = - M}^M {\sum\limits_{j = 1}^3 {\left( {A_{mj}^1\hat {\tilde { {\textit{χ}}}} _{mj}^1 + A_{mj}^2\hat {\tilde {{ {\textit{χ}}}}} _{mj}^2} \right)} } \end{split} $$ (5) $$ \begin{split} & {{\hat {\tilde {\boldsymbol{\sigma}}} }_s} = \frac{1}{{2{\text{π }}}}\int\limits_{ - \infty }^{ + \infty } {\sum\limits_{j = 1}^3 {{A_{dj}}{{\hat {\bar {\tilde {\boldsymbol{\psi}}}} }_{dj}}} } {\text{d}}{k_y}+ \\ &\qquad \sum\limits_{m = - M}^M {\sum\limits_{j = 1}^3 {\left( {A_{mj}^1\hat {\tilde {\boldsymbol{\eta}}} _{mj}^1 + A_{mj}^2\hat {\tilde {\boldsymbol{\eta}}} _{mj}^2} \right)} } \end{split} $$ (6) 式中, ${\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{dj}}$和${\hat {\bar {\tilde {\boldsymbol{\psi}}}} _{dj}}$分别表示对应地基土体位移和应力的下行平面波函数, j = 1, 2, 3分别代表SH波、SV波和P波; $\hat {\tilde {{ {\textit{χ}}}}} _{mj}^{1,2}$和$\hat {\tilde {\boldsymbol{\eta}}} _{mj}^{1,2}$分别为对应地基土体位移和应力的外行柱面波函数, 上标1和2对应隧道1-土体界面和隧道2-土体界面激发的散射波, M为环向模态的阶数, 一般取M = 10便可获得足够的收敛. 上述波函数的具体表达式可见附录A. $ {A_{dj}} $和$ A_{mj}^{1,2} $分别为对应的未知系数.
1.2 隧道波动方程及其通解
由于隧道中存在着由其内、外界面激发的柱面波, 因此, 隧道的位移和应力通解为
$$ \left.\begin{split} &{\hat{\tilde{{\boldsymbol{u}}}}}_{t}^{1\text{, }2} = {\displaystyle \sum _{m = -M}^{M}{\displaystyle \sum _{j = 1}^{3}\left({B}_{mj}^{1\text{, }2}{\hat{\tilde{{{ {\textit{χ}}}} }}}_{mj}^{t1,t2} + {C}_{mj}^{1\text{, }2}{}^{{\rm{Re}}}\hat{\tilde{{{ {\textit{χ}}}} }}{}_{mj}^{t1,t2}\right)}}\\ &{\hat{\tilde{{\boldsymbol{\sigma}} }}}_{t}^{1,2} = {\displaystyle \sum _{m = -M}^{M}{\displaystyle \sum _{j = 1}^{3}\left({B}_{mj}^{1,2}{\hat{\tilde{{\boldsymbol{\eta}} }}}_{mj}^{t1,t2} + {C}_{mj}^{1\text{, }2}{}^{{\rm{Re}}}\hat{\tilde{{\boldsymbol{\eta}} }}{}_{mj}^{t1,t2}\right)}}\end{split}\right\} $$ (7) 式中, $\hat {\tilde {{ {\textit{χ}}}}} _{mj}^{t1,t2}$和$\hat {\tilde {\boldsymbol{\eta}}} _{mj}^{t1,t2}$分别对应于隧道1, 2位移和应力的外行柱面波(向外传播)函数; ${}^{{\rm{Re}}}\hat {\tilde {{ {\textit{χ}}}}} _{mj}^{t1,t2}$和${}^{{\rm{Re}}}\hat {\tilde{\boldsymbol{ \eta}}} _{mj}^{t1,t2}$分别对应于隧道1, 2位移和应力的规则柱面波(向内传播)函数, 表达式与地基中的柱面波函数一致. $ B_{mj}^{1,2} $和$ C_{mj}^{1,2} $分别为对应的未知系数.
1.3 流体波动方程及其通解
由于理想流体内仅存在P波, 因此其波动方程为[34]
$$ {K_f}\nabla \nabla {{\boldsymbol{u}}_f} = {\rho _f}{\ddot {\boldsymbol{u}}_f} $$ (8) $$ {{\boldsymbol{u}}_f} = \frac{1}{{{\rho _f}{\omega ^2}}}\nabla {p_f} $$ (9) 式中, ${{\boldsymbol{u}}_f}$为流体的位移向量, $ {p_f} $为流体压强, $ {K_f} $为体积模量, $ {\rho _f} $为流体密度.
同理, 采用类似式(3)和式(4)所示的势函数分解法, 可得流体位移和压强的通解为
$$ {\hat {\tilde {\boldsymbol{u}}}_f} = \frac{1}{{2{\text{π }}}}\int\limits_{ - \infty }^{ + \infty } {({A_{fd}}{{\hat {\bar {\tilde {\boldsymbol{\phi}}}} }_{fd}} + {A_{fu}}{{\hat {\bar {\tilde {\boldsymbol{\phi}}}} }_{fu}}){{\rm{e}}^{{\rm{i}}{k_y}y}}} {\text{d}}{k_y} $$ (10) $$ {\hat {\tilde p}_f} = \frac{1}{{2{\text{π }}}}\int\limits_{ - \infty }^{ + \infty } {({A_{fd}}{{\hat {\bar {\tilde \psi}} }_{fd}} + {A_{fu}}{{\hat {\bar {\tilde \psi}} }_{fu}}){{\rm{e}}^{{\rm{i}}{k_y}y}}} {\text{d}}{k_y} $$ (11) 式中, ${\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{fd}}$和${\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{fu}}$分别为对应流体位移的下行和上行平面波函数, $ {\hat {\bar {\tilde \psi}} _{fd}} $和$ {\hat {\bar {\tilde \psi}} _{fu}} $分别为对应流体压强的下行和上行平面波函数, 表达式见附录A; $ {A_{fd}} $和$ {A_{fu}} $为对应的未知系数.
2. 边界条件
通过上述推导, 已获得隧道、地基和流体动力响应的通解, 其中包含的未知系数$ {A_{dj}} $, $ A_{mj}^{1,2} $, $ B_{mj}^{1,2} $, $ C_{mj}^{1,2} $, $ {A_{fd}} $和$ {A_{fu}} $可通过对应的边界条件来确定.
对于流体, 主要考虑其无限深(对应空气)和有限深(对应水)两种情况.
当流体为无限深(hf = ∞)时, 根据辐射条件, 此时, 流体中无下行平面波, 因此, 对应的下行波未知系数为零, 即
$$ {A_{fd}} = 0 $$ (12) 当考虑流体为有限深度(hf ≠ ∞)时, 由流体表面的自由应力条件可得
$$ {\hat {\tilde p}_f}({x_f} = {h_f}) = 0 $$ (13) 将式(13)带入式(11)中, 可得
$$ {A_{fd}} = - {{\rm{e}}^{2{\rm{i}}{k_{xf}}{h_f}}}{A_{fu}} $$ (14) 此外, 对于土体−流体分界面, 假设其竖向位移和竖向应力满足连续条件, 从而可得
$$ \left. \begin{split} & {{\hat {\tilde u}}_{sx}}({x_1} = {h_t}) = {{\hat {\tilde u}}_{fx}}({x_f} = 0) \\ & {{\hat {\tilde \sigma} }_{sxx}}({x_1} = {h_t}) = - {{\hat {\tilde p}}_f}({x_f} = 0) \\ & {{\hat {\tilde \sigma} }_{sxy}}({x_1} = {h_t}) = 0 \\ & {{\hat {\tilde \sigma} }_{sxz}}({x_1} = {h_t}) = 0\end{split} \right\} $$ (15) 为了满足式(15)中的边界条件, 地基土体的位移和应力通解式(5)和式(6)需统一由直接坐标系描述. 如附录A所示, 平面波函数和柱面波函数对应的基函数分别为直角坐标系下的指数函数和圆柱坐标系下的柱函数. 对此, 可引入柱函数向指数函数的转化公式来实现波面转换[35]
$$ H_m^{(1)}({k_r}r){{\rm{e}}^{{\text{i}}m\theta }} = \frac{1}{{\text{π }}}\int\limits_{ - \infty }^{ + \infty } {\frac{{{{\text{i}}^{ - m}}{{\rm{e}}^{{\text{i}}m\beta }}}}{{{k_x}}}{{\rm{e}}^{{\text{i}}{k_x}x + {\text{i}}{k_y}y}}{\text{d}}{k_y},{\text{ }}x > 0} $$ (16) 利用式(16)可实现地基中柱面波函数向平面波函数的转化, 进一步联立式(5)、式(10)、式(12)、式(14)和式(15)可得
$$\begin{split} & {A_{fu}} = {E_{fu}}\sum\limits_{j = 1}^3 {\Bigg[ {{A_{dj}}{{\hat {\bar {\tilde \phi}} }_{dj1}}(x = {h_t})} } + {{\hat {\bar {\tilde \phi}} }_{uj1}}(x = {h_t}) \cdot \\ &\qquad \frac{2}{{{k_{xj}}}}\sum\limits_{m = - M}^M {\left( {T_{mj}^uA_{mj}^1 + T_{mj}^uT_j^{12}A_{mj}^2} \right)} \Bigg] \end{split} $$ (17) 式中
$$ E_{fu}=\left\{\begin{array}{l} \dfrac{1}{\hat{\tilde{\tilde{\phi}}}_{ful}\left(x_f=0\right)}, \quad h_f=\infty \\ \dfrac{1}{\hat{\tilde{\tilde{\phi}}}_{ful}\left(x_f=0\right)-{\rm{e}}^{2 {\rm{i}} k_{xf} h_f} \hat{\tilde{\tilde{\phi}}}_{fdl}\left(x_f=0\right)}, h_f \neq \infty \end{array}\right. $$ (18) $$ T_{mj}^u =\left\{\begin{array}{l} {{\text{i}}^{ - m}} {{{\rm{e}}^{{\text{i}}m{\beta _s}}}{\text{ }}{\beta _s} = \arcsin ({k_y}/{k_{rs}}),{\text{ }}j = 1,2} \\ {{\text{i}}^{ - m}} {{{\rm{e}}^{{\text{i}}m{\beta _p}}}{\text{ }}{\beta _p} = \arcsin ({k_y}/{k_{rp}}),{\text{ }}j = 3{\text{ }}} \end{array}\right. $$ (19) 其中, $ {\hat {\bar {\tilde \phi}} _{fd1}} $, $ {\hat {\bar {\tilde \phi}} _{fu1}} $, $ {\hat {\bar {\tilde \phi}} _{dj1}} $和$ {\hat {\bar {\tilde \phi}} _{uj1}} $分别为${\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{fd}}$, ${\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{fu}}$, ${\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{dj}}$和${\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{uj}}$中的第1个元素; 转换系数$T_j^{12} = {\text{exp[i}}( - {k_{xj}}{d_t}\cos {\theta _t} + {k_y}{d_t}\sin {\theta _t}{\text{]}}$; 当j = 1, 2时, $ {k}_{xj} = {k}_{xs} $, 当j = 3时, $ {k}_{xj} = {k}_{xp} $.
将式(17)代入式(15), 并结合式(6)和式(11), 可得到地基中下行平面波未知系数$ {A_{dj}} $与外行柱面波未知系数$ A_{mj}^{1,2} $的关系式
$$ {A_{dj}} = \sum\limits_{m' = - M}^M {\sum\limits_{j' = 1}^3 {\left( {K_{jm'j'}^{d1}A_{m'j'}^1 + K_{jm'j'}^{d2}A_{m'j'}^2} \right)} } $$ (20) 式中, $ K_{jm'j'}^{d1} $和$ K_{jm'j'}^{d2} $的表达式见附录B.
同理, 假设隧道−土体界面($ {r}_{\mathrm{1,2}} = {R}_{o}^{\mathrm{1,2}} $)也满足位移和应力连续条件, 即
$$ \left. \begin{split} & {{\hat {\tilde {\boldsymbol{u}}}}_s}({r_{1,2}} = R_o^{1,2}) = \hat {\tilde {\boldsymbol{u}}}_t^{1,2}({r_{1,2}} = R_o^{1,2}) \\ & {{\hat {\tilde {\boldsymbol{\sigma}}} }_s}({r_{1,2}} = R_o^{1,2}) = \hat {\tilde {\boldsymbol{\sigma}}} _t^{1,2}({r_{1,2}} = R_o^{1,2}) \end{split} \right\} $$ (21) 为了满足式(21)中的边界条件, 土体的位移和应力通解需统一由柱坐标系描述. 对此, 可引入指数函数向柱函数的转化公式来实现波面转换[35]
$$ {{\text{e}}^{ - {\text{i}}{k_x}x + {\text{i}}{k_y}y}} = \sum\limits_{m = - M}^M {{{\text{i}}^{ - m}}{{\text{e}}^{{\rm{i}}m\beta }}{J_m}({k_r}r){{\text{e}}^{{\text{i}}m\theta }}} $$ (22) 此外, 隧道1和隧道2与土体的界面所激发的柱面波函数$\hat {\tilde {{ {\textit{χ}}}}} _{nj}^1$和$\hat {\tilde { {\textit{χ}}}} _{nj}^2$分别在对应的圆柱坐标系(r1, θ1, z)和(r2, θ2, z)描述. 对此, 可引入Graf加法定理[36-37], 实现两隧道激发的柱面波之间的相互平移
$$ \left.\begin{split} & \hat {\tilde {{ {\textit{χ}}}}} _{nj}^1({r_1},{\theta _1}) = \sum\limits_{m = - M}^M {G_{nmj}^{12}{}^{{\rm{Re}}}\hat {\tilde {{ {\textit{χ}}}}} _{mj}^2({r_2},{\theta _2})} \\ & \hat {\tilde {{ {\textit{χ}}}}} _{nj}^2({r_2},{\theta _2}) = \sum\limits_{m = - M}^M {G_{nmj}^{21}{}^{{\rm{Re}}}\hat {\tilde {{ {\textit{χ}}}}} _{mj}^1({r_1},{\theta _1})} \end{split} \right\} $$ (23) 式中
$$ \left. \begin{split} & G_{nmj}^{12} = {{\rm{e}}^{ - {\text{i}}(n - m){\theta _t}}}H_{n - m}^{(1)}({k_{rj}}{d_t}) \\ & G_{nmj}^{21} = {{\rm{e}}^{{\text{i}}(n - m)(\text{π} - {\theta _t})}}H_{n - m}^{(1)}({k_{rj}}{d_t})\\ &{k_{rj}} = \left\{ \begin{split} & {k_{rs}},j = 1,2 \\ & {k_{rp}},j = 3 \end{split} \right.\end{split} \right\} $$ (24) 利用式(23)和式(24), 可将式(5)和式(6)中的地基位移和应力通解统一由圆柱坐标(r1, θ1, z)或(r2, θ2, z)进行描述. 而对于隧道内表面, 考虑频率为f0的简谐载荷作用在角度为γ1,2处, 则根据应力平衡条件可得
$$ \hat {\tilde {\boldsymbol{\sigma}}} _t^{1,2}({r_{1,2}} = R_i^{1,2}) = {\left[ {\begin{array}{*{20}{c}} {{{\text{e}}^{ - {\text{i}}m{\gamma _{1,2}}}}/\left(2{\text{π }}R_i^{1,2}\right)},&{0,}&0 \end{array}} \right]^{\text{T}}} $$ (25) 联立式(21) ~ 式(25)并结合式(5) ~ 式(7)可得地基和隧道中柱面波的未知系数$ {A}_{mj}^{\mathrm{1,2}} $, $ {B}_{mj}^{\mathrm{1,2}} $和$ {C}_{mj}^{\mathrm{1,2}} $, 然后由式(19)可得地基中平面波的未知系数$ {A}_{dj} $, 再式(14)和式(17)可得流体中平面波的未知系数$ {A}_{fd} $和$ {A}_{fu} $. 而一旦所有未知系数确定后, 则可根据式(5) ~ 式(7)和式(10) ~ 式(11)确定双洞隧道−地基−流体耦合系统内任意位置在频域−波数域的动力响应. 而利用傅里叶逆变换则可得三维时间−空间域内的结果.
需注意的是, 本文所提出的方法仅适用于圆形隧道, 尚无法应用于矩形和马蹄形等异性断面隧道. 此外, 本文推导了简谐载荷作用下隧道−地基−流体耦合系统动力响应计算方法, 以获得其频响函数. 而轨道交通行车受轮轨不平顺等激扰, 激振力更为复杂, 包括的频率成分多. 例如: 环境振动问题关注的频率范围为1 ~ 80 Hz, 而噪声关注的频率范围为4 ~ 250 Hz. 对此, 可将本文提出的计算方法与车辆−轨道动力学模型进行耦合计算, 以获得行车激扰下的动力响应, 其推导过程与文献[26]一致, 不再赘述.
3. 模型验证
通过与现有的地基−流体耦合系统的动力响应[30,38]进行对比, 可验证本文所提出的三维解析计算方法的正确性. 为此, 将隧道的力学参数退化为地基土体的力学参数, 并将隧道的内径取足够小的值(本算例取$ {R}_{i}^{\mathrm{1,2}} $ = 0.02 m). 此外, 隧道外径$ {R}_{o}^{\mathrm{1,2}} $ = 1 m, 两隧道的埋深ht = 1.98 m, 两隧道的距离 dt = 10 m, 土体和空气材料参考文献[38].
图2给出了右侧隧道仰拱S0 (−0.02, 0, 0) 上作用单位简谐载荷时, 地基中点S1 (−3.02, 2, 0) 的位移和空气中点S2 (2.98, 2, 0) 的压强响应(单位: m). 载荷频率范围为2.5 ~ 320 Hz. 由图可知, 本文方法结果与两种既有方法的计算结果一致, 从而验证了本文所提出方法的正确性.
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图 2 点S0 (−0.02, 0, 0)上作用单位简谐载荷引起的地基中点S1 (−3.02, 2, 0) 的位移和空气中S2 (2.98, 2, 0)的压强响应 (单位: m)Figure 2. Displacements of S1 (−3.02, 2, 0) in soil and pressures of S2 (2.98, 2, 0) in air caused by unit harmonic load on S0 (−0.02, 0, 0) (unit: m)4. 算例分析
本节将通过2个算例分析隧道内作用简谐载荷引起的空气噪声和水下振动响应, 并对比单洞隧道和双洞隧道引起的振动响应情况, 以研究车致振动在隧道−地基−流体系统中的传播特性.
4.1 隧道−地基−水系统动力响应
首先考虑水下隧道的情况. 如图3所示, 外径为3 m, 内径为2.75 m的双洞隧道平行埋置于水底的土层中, 隧道中心间隔为dt = 10 m, 到水底的埋深为 ht = 15 m. 水深hf = 30 m. 隧道、土体和水的材料参数如表1所示. 其中, 介质阻尼可通过引入复数形式的拉梅常数 λ* = λ(1 + 2iβ) 和 μ* = μ(1 + 2iβ) 或弹性模量 E* = E(1 + 2iβ) 进行考虑, 其中β为阻尼系数, 隧道、土体和水的阻尼系数分别取0.015, 0.025和0.01.
表 1 隧道、土体和水的材料参数Table 1. Material parameters of tunnel, soil and waterMedium Parameter Value water Kf/kPa 2.25 × 109 ρf/(kg·m−3) 1000 βf 0.01 soil λ 3.84 × 108 μ 4.22 × 107 $ {\rho }_{s} $/(kg·m−3) 1875 βs 0.025 tunnel Et/Pa 5.0 × 1010 υt 0.3 ρt/(kg·m−3) 2500 βt 0.015 采用插入增益(insertion gain (IG))表示双洞隧道相比单洞隧道情况下的位移幅值差异, 单位为分贝(dB), 其计算公式如下
$$ {{IG}}({\boldsymbol{x}},\omega ){\text{ = 20l}}{{\text{g}}}\left( {\frac{{\left| {{{\hat u}_{t2}}({\boldsymbol{x}},\omega )} \right|}}{{\left| {{{\hat u}_{t1}}({\boldsymbol{x}},\omega )} \right|}}} \right) $$ (26) 式中, $ {\hat u_{t1}} $和$ {\hat u_{t2}} $分别对应单洞隧道和双洞隧道条件下的位移.
图4对比了单洞隧道和双洞隧道条件下, 频率为40 Hz的简谐载荷作用下引起的土体竖向位移场. 由图可知, 双洞隧道明显改变了地基中的竖向位移场分布规律. 当仅考虑单洞隧道时, 地基土中的位移场关于y轴对称, 隧道仰拱产生的振动会随着向四周传播而逐渐衰减. 当考虑双洞隧道时, 隧道2的存在起到了类似隔振屏障的作用, 进而显著改变地基中波的传播规律. 当隧道1产生的振动波传播至隧道2时, 会被其反射回去, 从而导致隧道2后方区域的位移幅值减小了近20 dB, 隧道上方受邻近隧道2反射波叠加的影响, 并使隧道上方部分区域内位移增大了近5 ~ 10 dB.
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图 4 频率f = 40 Hz的单位简谐载荷作用于隧道1仰拱处引起的地基竖向位移场Figure 4. Vertical displacements in the soil caused by a unit harmonic load with frequency f = 40 Hz acting on the invert of tunnel 1图5给出了频率为40 Hz的简谐载荷作用下引起的水压分布. 当仅考虑单洞隧道时, 水压也关于y轴对称, 最大值出现距离隧道一定距离的侧上方水−土界面处, 并向隧道两侧逐渐衰减. 当考虑双洞隧道时, 由于邻近隧道对波的反射作用, 隧道上方大部分水域的响应增强.
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图 5 频率f = 40 Hz的单位简谐载荷作用于隧道1仰拱处引起的水压分布Figure 5. Water pressures distribution caused by a unit harmonic load with frequency f = 40 Hz acting on the invert of tunnel 1进一步分析不同频率载荷作用下引起的水压响应. 如图3所示, 选取了水中距离水底5 m的3个观测点, 其中点A1 (20, −10, 0)位于隧道2的正上方, A2 (20, 0, 0)位于隧道1的正上方, 而A3 (20, 10, 0)位于隧道1的右侧(单位: m). 载荷频率为1 ~ 250 Hz, 结果如图6所示. 由图可知, 当频率较低时(f < 30 Hz), 单洞和双洞隧道条件下, 引起的压强随频率变化趋势一致, 隧道2的存在引起的水压变化很小. 这是因为, 低频时地基中弹性波的波长要大于隧道的尺寸, 导致其能够绕过隧道继续传播. 而随着载荷频率的增加, 地基中弹性波的波长逐渐减小, 更多的波动能量会受到隧道2的影响, 导致单洞和双洞隧道条件下水压的差异增大. 而由于受到隧道−土以及土−流体相互作用的影响, 差异会随着频率的变化而呈现一定的波动. 此外, 位于隧道2正上方的观测点A1的差异最为显著, 并随着观测点向右偏移, 差异会逐渐减小.
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图 6 水中不同观察点处的压强随载荷频率变化图Figure 6. Pressures at different observation points in water varying with frequencies4.2 隧道−地基−空气系统动力响应
接下来分析地下隧道振动引起的空气响应. 其中隧道和地基土体的参数与4.1节一致. 空气的体积模量Kf = 1.41 × 105 kPa, 密度ρf = 1.22 kg/m3.
图7显示了f = 40 Hz的单位简谐载荷作用于隧道1仰拱时, 引起的地表空气压强响应. 同理, 单洞隧道条件下的地表气压关于y轴和z轴对称, 压强场向两侧扩散并先增大后逐渐减小. 当隧道2存在时, 其对地基土中的振动波产生了显著散射效应, 进而导致地表空气压强分布发生显著变化, 尤其是距离隧道较近的地表, 部分区域的空气压强会出现接近 ± 20 dB的差异.
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图 7 频率f = 40 Hz的单位简谐载荷作用于隧道1仰拱处引起的地表空气压强分布Figure 7. Air pressures distribution caused by a unit harmonic load with frequency f = 40 Hz acting on the invert of tunnel 1随后分析双洞隧道不同相对位置对空气压强的影响. 如图8所示, 保持隧道1(载荷作用隧道)的位置不变, 隧道2分别考虑位于隧道1的正上方(case 1)、左侧(case 2)和正下方(case 3) 3种情况. 并考虑位于地表的3个观测点: B1 (15, −10, 0), B2 (15, 0, 0) 和B3 (15, 10, 0)(单位: m).
图9给出了双洞隧道不同相对位置时, 隧道1的仰拱处作用简谐载荷引起的3个观察点的气压随载荷频率的变化. 由图可知, 当载荷频率较低时, 地基中的振动波可绕过隧道而传播, 因此, 不同工况下的气压变化趋势一致, 且趋近单洞隧道下的响应. 而随着载荷频率的增加, 不同工况下的气压变化显著. 当隧道2位于隧道1的正上方时(case 1), 隧道直接位于振源和观察点的传播路径上. 因此, 此时隧道2的屏障效应最为显著, 导致的响应差异也最大; 而当隧道2位于隧道1的正下方时(case 3), 此时, 由隧道1激发的振动波向下传播后经隧道2反射回去的几乎已衰减了. 因此, 对于该工况下, 双洞隧道引起的气压响应与单洞隧道的基本吻合. 而当隧道2位于隧道的左侧时, 此时, 单洞隧道相对于双洞隧道引起的气压差值位于其他两种工况之间. 此外, 隧道载荷频率的进一步升高. 此时对于隧道两侧的观察点B1和B3, case 2下的气压与单洞隧道的差异比case 1的更大. 这是因为此时的振动波的波长很短, 对于case 1的情况, 隧道1激发的振动波可绕过隧道2而传播至观察点B1和B3; 而对于case 2的情况, 此时B1和B3的响应除了直接从隧道1传播过来的, 还有一部分是经过隧道2反射后传播过来的.
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图 9 双洞隧道不同相对位置下地表压强随载荷频率的变化图Figure 9. Air pressures on the ground surface varying with frequencies under different relative positions of twin tunnels进一步分析双洞隧道不同间距对空气压强的影响. 考虑水平并行的双洞隧道, 隧道2位于隧道1左侧, 保持隧道1位置不变, 移动隧道2的位置, 使得两隧道间距分别为2D(D = 6 m为隧道直径)、3D和4D.
图10给出了双洞隧道不同间距时, 隧道1的仰拱处作用简谐载荷引起的观察点B1和B2的气压随载荷频率的变化. 由图可知, 当双洞隧道间距为2倍隧道直径时, 隧道2的散射效应十分明显, 不仅改变了地表观察点的气压响应水平, 且改变了地表气压在高频段随频率的变化趋势. 随着两隧道之间的距离增大, 隧道2的散射效应逐渐减小. 当双洞隧道间距为4倍隧道直径时, 双洞隧道引起的观察点B1和B2的气压响应基本趋近单洞隧道的情况, 隧道2的影响基本可忽略. 然而, 在工程实际中, 地铁隧道的间距通常小于4倍隧道直径. 因此, 在进行地铁车致环境振动和噪声评估时, 应当考虑双洞隧道间相互作用的影响.
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图 10 双洞隧道不同间距下地表压强随载荷频率的变化图Figure 10. Air pressures on the ground surface varying with frequencies under different distance between twin tunnels5. 结 论
本文提出了一种双洞隧道−地基−流体耦合系统动力响应的三维解析计算方法, 可实现隧道车致振动在空气或水中传播的快速计算分析. 通过与既有方法进行对比, 验证了本文所提出方法的正确性. 通过水下和地下埋置双洞隧道的数值算例, 对隧道车致振动在水下和空气中的传播规律进行了研究, 并对比分析了双洞隧道和单洞隧道引起的振动响应差异, 得到以下结论.
(1)邻近隧道的存在起到了类似隔振屏障的作用, 改变了地基中波动能量的分布, 导致其后方区域的振动水平显著降低, 隧道上方受邻近隧道反射波叠加的影响, 导致地表部分区域的振动水平增加, 并进一步改变水中或者空气中的振动传播规律.
(2)邻近隧道存在对空气或水中振动响应的影响与载荷频率密切相关. 当载荷频率较低时, 地基中的振动波长大于隧道尺寸, 邻近隧道的存在影响小; 当载荷频率较高时, 邻近隧道的散射效应大, 显著改变了空气或水中振动响应.
(3)双洞隧道间距对上方空气或水中振动响应的影响显著. 当双洞隧道间距小于4倍隧道直径时, 临近隧道的存在会改变振动水平的量值和分布规律. 因此, 在进行地铁车致环境振动和噪声评估时, 应当考虑双洞隧道间动力相互作用的影响.
附录A
1. 对应地基土体位移和应力的下行平面波函数${\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{dj}}$和${\hat {\bar {\tilde {\boldsymbol{\psi}}}} _{dj}}$可表示为[27]
$$ \left. \begin{split} & {{\hat {\bar {\tilde {\boldsymbol{\phi}}}} }_{d1}} = \nabla \times \left( {{{\boldsymbol{e}}_z}{\varphi _{SH}}} \right) = {{\text{e}}^{ - {\text{i}}{k_{xs}}{x_1} + {\text{i}}{k_y}{y_1}}}{\left[ {{\text{i}}{k_y},{\text{i}}{k_{xs}},0} \right]^{\text{T}}} \\ & {{\hat {\bar {\tilde {\boldsymbol{\phi}}}} }_{d2}} = \nabla \times \nabla \times \left( {{{\boldsymbol{e}}_z}{\varphi _{SV}}} \right)= \\ & \qquad {{\text{e}}^{ - {\text{i}}{k_{xs}}{x_1} + {\text{i}}{k_y}{y_1}}}{\left[ {{k_z}{k_{xs}}, - {k_z}{k_y},(k_y^2 + k_{xs}^2)} \right]^{\text{T}}} \\ & {{\hat {\bar {\tilde {\boldsymbol{\phi}}}} }_{d3}} = \nabla {{\varphi _P}} = {{\text{e}}^{ - {\text{i}}{k_{xp}}{x_1} + {\text{i}}{k_y}{y_1}}}{\left[ { - {\rm{i}}{k_{xp}},{\rm{i}}{k_y},{\rm{i}}{k_z}} \right]^{\text{T}}} \end{split} \right\}\tag{A1} $$ $$ \left. \begin{split} & {{\hat {\bar {\tilde {\boldsymbol{\psi}}}} }_{d1}} = \mu {{\text{e}}^{ - {\text{i}}{k_{xs}}{x_1} + {\text{i}}{k_y}{y_1}}}{\left[ {2{k_y}{k_{xs}},k_{xs}^2 - k_y^2, - {k_z}{k_y}} \right]^{\text{T}}} \\ & {{\hat {\bar {\tilde {\boldsymbol{\psi}}}} }_{d2}} = \mu {{\text{e}}^{ - {\text{i}}{k_{xs}}{x_1} + {\text{i}}{k_y}{y_1}}}\cdot \\ &\qquad {\left[ { - 2{\text{i}}{k_z}k_{xs}^2,2{\text{i}}{k_z}{k_y}{k_{xs}}, - {\text{i}}{k_{xs}}\left( {k_y^2 + k_{xs}^2 - k_z^2} \right)} \right]^{\text{T}}} \\ & {{\hat {\bar {\tilde {\boldsymbol{\psi}}}} }_{d3}} = \mu {{\text{e}}^{ - {\text{i}}{k_{xp}}{x_1} + {\text{i}}{k_y}{y_1}}} \cdot \\ &\qquad {\left[ {2k_p^2 - k_s^2 - 2k_{xp}^2,2{k_y}{k_{xp}},2{k_z}{k_{xp}}} \right]^{\text{T}}} \end{split} \right\} \tag{A2}$$ 其中, ${k_{xs,xp}} = {\left( {k_{s,p}^2 - k_z^2 - k_y^2} \right)^{1/2}}$为x向的波数; ${k_{s,p}} = \omega /{C_{s,p}}$为S波和P波的波数; ${C_s} = \mu /\rho $, ${C_p} = \left( {2\mu + \lambda } \right)/\rho $分别为S波和P波的波速; 此外, 下行平面波函数${\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{uj}}$和${\hat {\bar {\tilde {\boldsymbol{\psi}}}} _{uj}}$的表达式只需式(A1)和(A2)中的$ {k_{xs,p}} $替换为$ - {k_{xs,p}} $即可.
2. 对应地基土体位移和应力的外行柱面波波函数$\hat {\tilde {{ {\textit{χ}}}}} _{mj}^{1,2}$和$\hat {\tilde {\boldsymbol{\eta}}} _{mj}^{1,2}$可表示为[27]
$$ \left. \begin{split} & \hat {\tilde {{ {\textit{χ}}}}} _{m1}^{1,2} = \nabla \times \left( {{{\boldsymbol{e}}_z}{{\hat {\tilde \varphi} }_{SH}}} \right) = {{\text{e}}^{{\text{i}}m{\theta _{1,2}}}}\cdot \\ & \qquad {\left[ {\frac{{{\text{i}}m}}{{{r_{1,2}}}}{\rm{H}}_m^{(1)}({k_{rs}}{r_{1,2}}), - {k_{rs}}{\rm{H}}{{_m^{(1)}}^\prime }({k_{rs}}{r_{1,2}}),0} \right]^{\text{T}}} \\ & \hat {\tilde {{ {\textit{χ}}}}} _{m2}^{1,2} = \nabla \times \nabla \times \left( {{{\boldsymbol{e}}_z}{{\hat {\tilde \varphi} }_{SV}}} \right) = {{\text{e}}^{{\text{i}}m{\theta _{1,2}}}}\Bigg[ {{\text{i}}{k_z}{k_{rs}}{\rm{H}}{{_m^{(1)}}^\prime }({k_{rs}}{r_{1,2}})}, \\ & \qquad{ {\frac{{ - m{k_z}}}{{{r_{R,L}}}}{\rm{H}}_m^{(1)}({k_{rs}}{r_{1,2}}),k_{rs}^2{\rm{H}}_m^{(1)}({k_{rs}}{r_{1,2}})} \Bigg]^{\text{T}}} \\ & \hat {\tilde {{ {\textit{χ}}}}} _{m3}^{1,2} = \tilde \nabla \left( {{{\hat {\tilde \varphi }}_P}} \right) = {{\text{e}}^{{\text{i}}m{\theta _{1,2}}}}\Bigg[ {{k_{rp}}{\rm{H}}{{_m^{(1)}}^\prime }({k_{rp}}{r_{1,2}}),} \\ &\qquad { {\frac{{{\text{i}}m}}{{{r_{1,2}}}}{\rm{H}}_m^{(1)}({k_{rp}}{r_{1,2}}),{\text{i}}{k_z}{\rm{H}}_m^{(1)}({k_{rp}}{r_{1,2}})} \Bigg]^{\text{T}}} \end{split} \right\}\tag{A3} $$ $$ \left. \begin{split} & \hat {\tilde {\boldsymbol{\eta}}} _{m1}^{1,2} = \mu {{\text{e}}^{{\text{i}}m{\theta _{1,2}}}}\Biggr\{ {\frac{{2{\text{i}}m\left[ {{k_{rs}}{r_{1,2}}{\rm{H}}{{_m^{(1)}}^\prime }({k_{rs}}{r_{1,2}}) - {\rm{H}}_m^{(1)}({k_{rs}}{r_{1,2}})} \right]}}{{r_{1,2}^2}},} \\ & \qquad { { - k_{rs}^2\left[ {2{\rm{H}}{{_m^{(1)}}^\prime }^\prime ({k_{rs}}{r_{1,2}}) + {\rm{H}}_m^{(1)}({k_{rs}}{r_{1,2}})} \right],\frac{{ - m{k_z}}}{{{r_{R,L}}}}{\rm{H}}_m^{(1)}({k_{rs}}{r_{1,2}})} \Biggr\}^{\text{T}}} \\ & \hat {\tilde {\boldsymbol{\eta}}} _{m2}^{1,2} = \mu {{\text{e}}^{{\text{i}}m{\theta _{1,2}}}}\Biggr\{ {2{\text{i}}{k_z}k_{rs}^2{\rm{H}}{{_m^{(1)}}^\prime }^\prime ({k_{rs}}{r_{1,2}}),} \\ & \qquad \frac{{2m{k_z}\left[ {{\rm{H}}_m^{(1)}({k_{rs}}{r_{1,2}}) - {k_{rs}}{r_{1,2}}{\rm{H}}{{_m^{(1)}}^\prime }({k_{rs}}{r_{1,2}})} \right]}}{{r_{1,2}^2}},\\ &\qquad {k_{rs}}(k_s^2 - 2k_z^2){\rm{H}}{{_m^{(1)}}^\prime }({k_{rs}}{r_{1,2}}) \Biggr\}^{\text{T}} \\ & \hat {\tilde {\boldsymbol{\eta}}} _{m3}^{1,2} = \mu {{\text{e}}^{{\text{i}}m{\theta _{1,2}}}}\Biggr\{ {\left( {2k_p^2 - k_s^2} \right){\rm{H}}_m^{(1)}({k_{rp}}{r_{1,2}}) + 2k_{rp}^2{\rm{H}}{{_m^{(1)}}^{\prime \prime }}({k_{rp}}{r_{1,2}}),} \\ &\qquad { {\frac{{ - 2{\text{i}}m\left[ {{\rm{H}}_m^{(1)}({k_{rp}}{r_{1,2}}) - {k_{rp}}{r_{1,2}}{\rm{H}}{{_m^{(1)}}^\prime }({k_{rp}}{r_{1,2}})} \right]}}{{r_{1,2}^2}},2{\text{i}}{k_z}{k_{rp}}{\rm{H}}{{_m^{(1)}}^\prime }({k_{rp}}{r_{1,2}})} \Biggr\}^{\text{T}}} \end{split} \right\} \tag{A4}$$ 其中, ${k_{rs,rp}} = {\left( {k_{s,p}^2 - k_z^2} \right)^{1/2}}$为r向的波数; ${{\rm{H}}}_{m}^{\left(1\right)}(\cdot )$为第一类汉克尔函数. 规则柱面波函数${}^{Re}\hat {\tilde {{ {\textit{χ}}}}} _{mj}^{1,2}$和${}^{Re}\hat {\tilde {\boldsymbol{\eta}}} _{mj}^{1,2}$的表达式只需式(A3)和(A4)中的汉克尔${{\rm{H}}}_{m}^{\left(1\right)}(\cdot )$替换为贝塞尔函数${{\rm{J}}}_{m}^{\left(1\right)}(\cdot )$即可.
3. 对应流体位移和压强的下行平面波函数${\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{fd}}$和${\hat {\bar {\tilde {\boldsymbol{\psi}}}} _{fd}}$可表示为
$$ \left. \begin{split} & {{\hat {\bar {\tilde {\boldsymbol{\phi}}}} }_{fd}} = \nabla {\varphi _f} = {{\text{e}}^{ - {\text{i}}{k_{xf}}{x_f} + {\text{i}}{k_y}{y_1}}}{\left[ { - {\text{i}}{k_{xf}},{\text{i}}{k_y},{\text{i}}{k_z}} \right]^{\text{T}}} \\ & {{\hat {\bar {\tilde {\boldsymbol{\psi}}}} }_{fd}} = {\rho _f}{\omega ^2}{{\text{e}}^{ - {\text{i}}{k_{xf}}{x_f} + {\text{i}}{k_y}{y_1}}} \end{split} \right\}\tag{A5} $$ 其中, ${k_{xf}} = {\left( {k_f^2 - k_z^2 - k_y^2} \right)^{1/2}}$为x向的波数; ${k_f} = \omega /{C_f}$为流体的纵波波数; ${C_f} = {K_f}/{\rho _f}$为流体的纵波波速.
附录B
${\boldsymbol{K}}_{jm'j'}^{d1}$和${\boldsymbol{K}}_{jm'j'}^{d2}$可表示为如下矩阵形式
$$ \left. \begin{split} & {\boldsymbol{K}}_{jm'j'}^{d1} = - {({{\boldsymbol{S}}_{de}} - {T_f}{\boldsymbol{D}}_{de}^{fs})^{ - 1}}({\boldsymbol{S}}_{tue}^1 - {T_f}{\boldsymbol{D}}_{tue}^{fs1}) \\ & {{\boldsymbol{K}}_{jm'j'}^{d2}} = - {({{\boldsymbol{S}}_{de}} - {T_f}{\boldsymbol{D}}_{de}^{fs})^{ - 1}}({\boldsymbol{S}}_{tue}^2 - {T_f}{\boldsymbol{D}}_{tue}^{fs2}) \end{split} \right\} \tag{B1}$$ 其中, 系数Tf表示为
$$ {T_f} = \left\{ \begin{split} & - \frac{{{{\hat {\bar {\tilde \psi}} }_{fu}}({x_f} = 0)}}{{{{\hat {\bar {\tilde \phi}} }_{fu1}}({x_f} = 0)}}\begin{array}{*{20}{c}} ,&{{h_f} = \infty } \end{array} \\ & - \frac{{{{\hat {\bar {\tilde \psi}} }_{fu}}({x_f} = 0) - {{\rm{e}}^{{\rm{i}}2{k_{xf}}{h_f}}}{{\hat {\bar {\tilde \psi}} }_{fd}}({x_f} = 0)}}{{{{\hat {\bar {\tilde \phi}} }_{fu1}}({x_f} = 0) - {{\rm{e}}^{{\rm{i}}2{k_{xf}}{h_f}}}{{\hat {\bar {\tilde \phi }}}_{fd1}}({x_f} = 0)}}\begin{array}{*{20}{c}} ,&{{h_f} \ne \infty } \end{array} \end{split} \right. \tag{B2}$$ 3 × 3系数矩阵 ${{\boldsymbol{S}}_{de}}$ 和 ${\boldsymbol{D}}_{de}^{fs}$ 表示为
$$ \left. \begin{aligned} & {{\boldsymbol{S}}_{de}} = {{\boldsymbol{S}}_d}{{\boldsymbol{E}}_d} \\ & {{\boldsymbol{S}}_d} = \left[ {{{\hat {\bar {\tilde {\boldsymbol{\psi}}}} }_{d1}},{{\hat {\bar {\tilde {\boldsymbol{\psi}}}} }_{d2}},{{\hat {\bar {\tilde {\boldsymbol{\psi}}}} }_{d3}}} \right] \\ & {{\boldsymbol{E}}_d}{\text{ = diag}}\left[ {{{\rm{e}}^{ - {\rm{i}}{k_{xs}}{h_t}}},{{\rm{e}}^{ - {\rm{i}}{k_{xs}}{h_t}}},{{\rm{e}}^{ - {\rm{i}}{k_{xp}}{h_t}}}} \right] \end{aligned} \right\} \tag{B3}$$ $$ \left.\begin{aligned} & {\boldsymbol{D}}_{de}^{fs}{\text{ = }}{\boldsymbol{D}}_d^{fs}{{\boldsymbol{E}}_d} \\ & {\boldsymbol{D}}_d^{fs}{\text{ = }}\left[ {\begin{array}{*{20}{c}} {\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{d1}^{fs},}&{\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{d2}^{fs},}&{\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{d3}^{fs}} \end{array}} \right] \\ & \hat{ \bar {\tilde {\boldsymbol{\phi}}}} _{dj}^{fs} = {\left[ {\begin{array}{*{20}{c}} {{{\hat {\bar {\tilde \phi}} }_{dj1}},}&{0,}&0 \end{array}} \right]^{\text{T}}} \end{aligned} \right\}\tag{B4} $$ 3(2M + 1) × 3(2M + 1)系数矩阵${\boldsymbol{S}}_{tu}^{1,2}$ 和 ${\boldsymbol{D}}_{tu}^{fs1,2}$ 表示为
$$ \begin{split} & {\boldsymbol{S}}_{tu}^{1,2} = 2\left[ {\frac{{T_{ - M1}^{1,2}{{\text{e}}^{{\text{i}}{k_{xs}}{h_t}}}}}{{{k_{xs}}}}{{\hat {\bar {\tilde {\boldsymbol{\psi}}}} }_{u1}},\frac{{T_{ - M2}^{1,2}{{\text{e}}^{{\text{i}}{k_{xs}}{h_t}}}}}{{{k_{xs}}}}{{\hat {\bar {\tilde {\boldsymbol{\psi}}}} }_{u2}},\frac{{T_{ - M3}^{1,2}{{\text{e}}^{{\text{i}}{k_{xp}}{h_t}}}}}{{{k_{xp}}}}{{\hat {\bar {\tilde {\boldsymbol{\psi}} }}}_{u3}},} . ..,\right. \\ &\qquad \left. {\frac{{T_{M1}^{R,L}{{\text{e}}^{{\text{i}}{k_{xs}}{h_t}}}}}{{{k_{xs}}}}{{\hat {\bar {\tilde {\boldsymbol{\psi}}}} }_{u1}},\frac{{T_{M2}^{R,L}{{\text{e}}^{{\text{i}}{k_{xs}}{h_t}}}}}{{{k_{xs}}}}{{\hat {\bar {\tilde {\boldsymbol{\psi}}}} }_{u2}},\frac{{T_{M3}^{R,L}{{\text{e}}^{{\text{i}}{k_{xp}}{h_t}}}}}{{{k_{xp}}}}{{\hat {\bar {\tilde {\boldsymbol{\psi}} }}}_{u3}}} \right] \end{split}\tag{B5} $$ $$ \begin{split} & {\boldsymbol{D}}_{tu}^{fs1,2} = 2\left[ {\frac{{T_{ - M1}^{1,2}{{\text{e}}^{{\text{i}}{k_{xs}}{h_t}}}}}{{{k_{xs}}}}\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{u1}^{fs},\frac{{T_{ - M2}^{1,2}{{\text{e}}^{{\text{i}}{k_{xs}}{h_t}}}}}{{{k_{xs}}}}\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{u2}^{fs},\frac{{T_{ - M3}^{1,2}{{\text{e}}^{{\text{i}}{k_{xp}}{h_t}}}}}{{{k_{xp}}}}\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{u3}^{fs},}. .., \right. \\ &\qquad \left. {\frac{{T_{M1}^{1,2}{{\text{e}}^{{\text{i}}{k_{xs}}{h_t}}}}}{{{k_{xs}}}}\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{u1}^{fs},\frac{{T_{M2}^{1,2}{{\text{e}}^{{\text{i}}{k_{xs}}{h_t}}}}}{{{k_{xs}}}}\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{u2}^{fs},\frac{{T_{M3}^{1,2}{{\text{e}}^{{\text{i}}{k_{xp}}{h_t}}}}}{{{k_{xp}}}}\hat {\bar {\tilde {\boldsymbol{\phi}}}} _{u3}^{fs}} \right] \end{split} \tag{B6}$$ $$ \hat {\bar {\tilde {\boldsymbol{\phi}}}} _{uj}^{fs} = {\left[ {{{\hat {\bar{ \tilde \phi}} }_{uj1}},0,0} \right]^{\rm{T}}}\tag{B7} $$ -
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图 2 点S0 (−0.02, 0, 0)上作用单位简谐载荷引起的地基中点S1 (−3.02, 2, 0) 的位移和空气中S2 (2.98, 2, 0)的压强响应 (单位: m)
Figure 2. Displacements of S1 (−3.02, 2, 0) in soil and pressures of S2 (2.98, 2, 0) in air caused by unit harmonic load on S0 (−0.02, 0, 0) (unit: m)
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图 4 频率f = 40 Hz的单位简谐载荷作用于隧道1仰拱处引起的地基竖向位移场
Figure 4. Vertical displacements in the soil caused by a unit harmonic load with frequency f = 40 Hz acting on the invert of tunnel 1
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图 5 频率f = 40 Hz的单位简谐载荷作用于隧道1仰拱处引起的水压分布
Figure 5. Water pressures distribution caused by a unit harmonic load with frequency f = 40 Hz acting on the invert of tunnel 1
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图 6 水中不同观察点处的压强随载荷频率变化图
Figure 6. Pressures at different observation points in water varying with frequencies
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图 7 频率f = 40 Hz的单位简谐载荷作用于隧道1仰拱处引起的地表空气压强分布
Figure 7. Air pressures distribution caused by a unit harmonic load with frequency f = 40 Hz acting on the invert of tunnel 1
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图 9 双洞隧道不同相对位置下地表压强随载荷频率的变化图
Figure 9. Air pressures on the ground surface varying with frequencies under different relative positions of twin tunnels
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图 10 双洞隧道不同间距下地表压强随载荷频率的变化图
Figure 10. Air pressures on the ground surface varying with frequencies under different distance between twin tunnels
表 1 隧道、土体和水的材料参数
Table 1 Material parameters of tunnel, soil and water
Medium Parameter Value water Kf/kPa 2.25 × 109 ρf/(kg·m−3) 1000 βf 0.01 soil λ 3.84 × 108 μ 4.22 × 107 $ {\rho }_{s} $/(kg·m−3) 1875 βs 0.025 tunnel Et/Pa 5.0 × 1010 υt 0.3 ρt/(kg·m−3) 2500 βt 0.015 
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