孔隙率各向异性下饱和多孔弹性地基动力响应$^{\bf 1)}$

图1计算模型示意图

Fig. 1Schematic diagram of calculation model

1.3 基本控制方程

几何方程

(1)\begin{eqnarray}{\varepsilon}_{{ij}} =\dfrac{1}{2}\left( {{u}_{{i},{j}} +{u}_{{j},{i}} } \right)\end{eqnarray}

式中, ${\varepsilon }_{{ij}} $表示应变张量, ${u}_{i} $表示位移张量, ","表示对于坐标求导.

本构方程

(2)\begin{eqnarray}{\sigma }_{{ij}} =2G{\varepsilon }_{{ij}} +\left( {\lambda e-\beta _1 \theta -P} \right)\delta _{ij}\end{eqnarray}

式中, ${\sigma }_{{ij}} $表示应力张量, $\lambda ,G$表示拉梅常数, $e$表示体积应变, $\theta $表示温度变化, $\theta =T-T_0 $, $T$表示绝对温度, $T_0 $表示初始温度, $\delta _{ij} $表示Kronecker符号, $\beta _1 $表示体积应变, $\beta _1 =\left( {3\lambda +2G} \right)\alpha _{\rm s} $, $\alpha _{\rm s} $表示土颗粒线性热膨胀系数, $P$表示超孔隙水压力.

运动方程

(3)\begin{eqnarray}{\sigma }_{{ij},{j}} +{F}_{i} =\rho _\xi {\ddot{{u}}}_{i}\end{eqnarray}

式中, ${F}_{i} $表示体力张量, $\rho _\xi $表示不同方向的密度, $\rho_\xi =n_\xi \rho _{\rm w} +\left( {1-n_\xi }\right)\rho _{\rm s} $, $\rho _{\rm w}$表示孔隙水密度, $\rho _{\rm s} $表示土颗粒密度, 字母上方"."表示对于时间$t$的求导.

热传导方程

(4)\begin{eqnarray}K{\theta }_{,{ii}} =\rho _\xi C_E \left( {{\dot{{\theta }}}+\tau {\ddot{{\theta }}}} \right)+\beta _1 T_0 \left( {{\dot{{e}}}+\tau {\ddot{{e}}}} \right)\end{eqnarray}

式中, $K$表示热传导系数, $C_E $表示单位质量介质的比热容, $\tau $表示弛豫时间.

不计体力的运动平衡方程

(5)\begin{eqnarray}G{u}_{{i},{jj}} +\left( {\lambda +G} \right){u}_{{j},{ij}} -\beta _1 {\theta }_{,{i}} -{P}_{,{i}} =\rho _\xi {\ddot{{u}}}_{i}\end{eqnarray}

能量方程

(6)\begin{eqnarray}m\left( {{\dot{{\theta }}}+\tau {\ddot{{\theta }}}} \right)+\beta _1 T_0 \left( {{\dot{{e}}}+\tau {\ddot{{e}}}} \right)=K{\theta }_{,{ii}}\end{eqnarray}

式中, $m$表示介质比热容, $c_{\rm w} $表示孔隙水比热容, $c_{\rm s} $表示土颗粒比热容, $m=n_\xi \rho _{\rm w} c_{\rm w} +\left( {1-n_\xi } \right)\rho _{\rm s} c_{\rm s} $.

结合达西定律的流体平衡方程

(7)\begin{eqnarray}b\left( {\alpha _u {\dot{{\theta }}}-{\dot{{e}}}} \right)+\rho _{\rm w} {\ddot{{e}}}+{P}_{,{ii}} =0\end{eqnarray}

式中, $\alpha _u $表示介质线膨胀系数, $\alpha _{\rm w} $表示孔隙水线性热膨胀系数, $\alpha _{\rm s} $表示土颗粒线性热膨胀系数, 其中, $\alpha _u =n_\xi \alpha _{\rm w} +\left( {1-n_\xi } \right)\alpha _{\rm s} $, $k_{\rm d} $表示介质的渗透系数, $g$表示重力加速度, $b={\rho _{\rm w} g}/{k_{\rm d} }$.

基于二维半无限大弹性地基假设, 同时为使后续公式推导过程更为简化, 引入笛卡尔坐标系$\left( {x,y,z} \right)$, 将位移分量表示为$u_i =\left({u,0,w} \right)$.

为使计算结果具有一定的普适性, 引入下列无量纲量, 该组无量纲量同参考文献^[38]

(8)$\begin{eqnarray} \left.\begin{array}{l} \left( {'x,'u} \right)=V_x \eta \left( {x,u} \right), \ \ '\theta =\dfrac{\beta _1 \theta }{\lambda +2G}\\ \left( {'z,'w} \right)=V_z \eta \left( {z,w} \right), \ \ V_x =\sqrt {\dfrac{\lambda +2G}{\rho _{\xi x} }} \left( {'t,'\tau } \right)=V_x V_z \eta \\ \left( {t,\tau } \right), \ \ V_z =\sqrt {\dfrac{\lambda +2G}{\rho _{\xi z} }}\\ 'P=\dfrac{P}{\lambda +2G}, \ \ '\sigma =\dfrac{\sigma }{G}, \ \ \eta =\dfrac{m}{K}\\ \end{array}\right\} \end{eqnarray}$

根据方程(8), 方程(1)$\sim$(7)可简化为(为书写简便, 省略点号)

(9)$\sigma _{xx} =\beta ^2\dfrac{\partial u}{\partial x}+\left( {\beta ^2-2} \right)\dfrac{\partial w}{\partial z}-\beta ^2\theta -\beta ^2P$

(10)$\sigma _{zz} =\left( {\beta ^2-2} \right)\dfrac{\partial u}{\partial x}+\beta ^2\dfrac{\partial w}{\partial z}-\beta ^2\theta -\beta ^2P$

(11)$\tau _{xz} =\dfrac{\partial u}{\partial z}+\dfrac{\partial w}{\partial x}$

(12)$\beta ^2\dfrac{\partial ^2u}{\partial x^2}-\beta ^2\dfrac{\partial \theta }{\partial x}+\varphi _0 \dfrac{\partial ^2u}{\partial z^2}-\beta ^2\dfrac{\partial P}{\partial x} + \left( {\beta ^2-1} \right)\dfrac{\partial ^2w}{\partial x\partial z}=\beta ^2\varphi _0 \dfrac{\partial ^2u}{\partial t^2}$

(13)$\beta ^2\dfrac{\partial ^2w}{\partial z^2}+\left( {\beta ^2-1} \right)\dfrac{\partial ^2u}{\partial x\partial z}-\beta ^2\dfrac{\partial \theta }{\partial z}-\beta ^2\dfrac{\partial P}{\partial z}+\dfrac{\partial ^2w}{\varphi _0 \partial x^2}=\beta ^2\dfrac{\partial ^2w}{\varphi _0 \partial t^2}$

(14)$\left( {1+\tau \dfrac{\partial }{\partial t}} \right)\left[ {\dfrac{\partial \theta }{\partial t}+\varphi _2 \left( {\dfrac{\partial ^2u}{\partial x\partial t}+\dfrac{\partial ^2w}{\partial z\partial t}} \right)} \right] =\varphi _1 \dfrac{\partial ^2\theta }{\partial x^2}+\dfrac{\partial ^2\theta }{\varphi _1 \partial z^2}$

(15)$\varphi _3 \dfrac{\partial \theta }{\partial t}+\varphi _6 \left( {\dfrac{\partial ^3u}{\partial x\partial t^2}+\dfrac{\partial ^3w}{\partial z\partial t^2}} \right)+\varphi _1 \dfrac{\partial ^2P}{\partial x^2} -\varphi _4 \dfrac{\partial ^2u}{\partial x\partial t}-\varphi _5 \dfrac{\partial ^2w}{\partial z\partial t}+\dfrac{\partial ^2P}{\varphi _1 \partial z^2}=0$

式中

$\begin{eqnarray*} &&\beta ^2=\dfrac{\lambda +2G}{G},\ \ \varphi _0 =\dfrac{\rho _{\xi x} }{\rho _{\xi z} },\ \ \varphi _1 =\sqrt {\dfrac{\rho _{\xi z} }{\rho _{\xi x} }} \end{eqnarray*}$

$\varphi _2 =\dfrac{\beta _1^2 T_0 }{m\left( {\lambda +2G}\right)},\ \ \varphi _3 =\dfrac{ba_u }{\eta \beta _1 },\ \ \varphi _4=\dfrac{n_{\xi x} b}{\eta \left( {\lambda +2G} \right)}$

$\varphi _5=\dfrac{n_{\xi z} b}{\eta \left( {\lambda +2G} \right)},\ \ \varphi _6 =\dfrac{\rho _{\rm w} }{\sqrt {\rho _{\xi x} \rho _{\xi z}}}$

1.4 边界条件

本文讨论地基上表面受到温度载荷、机械载荷作用时的热-水-力耦合动力响应问题, 其中边界条件如下.

(1)地基上表面受温度载荷作用

① 地基上表面应力自由

(16)\begin{eqnarray}\sigma _{xz} =0,\ \ \sigma _{zz} =0\end{eqnarray}

② 地基上表面受到温度载荷作用

(17)$\theta =Q\psi \left( {x,t} \right)$

(18)$\psi \left( {x,t} \right)=\psi ^* \left( {a,\omega } \right)\exp \left( {\omega t+{\rm i}ax} \right)$

式中, $Q$表示温度载荷, $\omega $表示频率(复时间常数), $a$表示$x$轴方向的波数, $\psi ^* \left( {a,\omega } \right)$表示载荷振幅, i为虚数单位, i $=\sqrt {-1} $.

③ 地基上表面可透水

(19)\begin{eqnarray}P=0\end{eqnarray}

(2)地基上表面受机械载荷作用

① 地基上表面受机械载荷作用

(20)\begin{eqnarray}\sigma _{xz} =0,\ \ \sigma _{zz} =-q\psi \left( {x,t} \right)\end{eqnarray}

式中, $q$表示机械载荷.

② 地基上表面不考虑温度载荷

(21)\begin{eqnarray}\theta =0\end{eqnarray}

③ 地基上表面可透水

(22)\begin{eqnarray}P=0\end{eqnarray}

2 二维热-水-力耦合问题的正则模态分析

所考虑的物理量的解均可以按照以下正则模态的形式进行分解

(23)\begin{eqnarray}&&\left[ {u,w,\sigma _{ij} ,\theta ,P} \right]\left( {x,z,t} \right) =\left[ {u^* \left( z \right),w^* \left( z \right),\sigma _{ij} ^* \left( z \right),\theta ^* \left( z \right),P^* \left( z \right)} \right]\cdot \exp \left( {\omega t+{\rm i}ax} \right)\end{eqnarray}

式中, $u^* \left( z \right)$, $w^* \left( z \right)$, $\sigma _{ij}^* \left( z \right)$, $\theta ^* \left( z \right)$, 和$P^* \left( z\right)$分别表示所考虑的物理场的振幅.

对方程(9)$\sim$(15)进行正则模态分解可得如下方程

(24)$\sigma _{xx}^* =\beta ^2u^* {\rm i}a+\left( {\beta ^2-2} \right)\dfrac{\partial w^* }{\partial z}-\beta ^2\theta ^* -\beta ^2P^*$

(25)$\sigma _{zz}^* =\left( {\beta ^2-2} \right)u^* {\rm i}a+\beta ^2\dfrac{\partial w^* }{\partial z}-\beta ^2\theta ^* -\beta ^2P^*$

(26)$\tau _{xz}^* =\dfrac{\partial u^* }{\partial z}+{\rm i}aw^*$

(27)$\left( {\dfrac{\partial ^2}{\partial z^2}-\varphi _0 } \right)u^* +\varphi _1 \dfrac{\partial w^* }{\partial z}-\varphi _2 \theta ^* -\varphi _2 P^* =0$

(28)$\left( {\dfrac{\partial ^2}{\partial z^2}-\varphi _3 } \right)w^* +\varphi _4 \dfrac{\partial u^* }{\partial z}-\dfrac{\partial \theta ^* }{\partial z}-\dfrac{\partial P^* }{\partial z}=0$

(29)$\left( {\dfrac{\partial ^2}{\partial z^2}-\varphi _5 } \right)\theta ^* =\varphi _6 u^* +\varphi _7 \dfrac{\partial w^* }{\partial z}$

(30)$\left( {\dfrac{\partial ^2}{\partial z^2}-\varphi _8 } \right)P^* -\varphi _9 u^* -\varphi _{10} \dfrac{\partial w^* }{\partial z}+\varphi _{11} \theta ^* =0$

式中

$\phi _0 =\beta ^2\left( {\omega ^2+\dfrac{a^2}{\varphi _0 }} \right),\ \ \phi _1 =\dfrac{\left( {\beta ^2-1} \right){\rm i}a}{\varphi _0 }$

$\phi _2 =\dfrac{\beta ^2{\rm i}a}{\varphi _0 },\ \ \phi _3 =\dfrac{a^2+\beta ^2w^2}{\varphi _0 \beta ^2},\ \ \phi _4 =\dfrac{\left( {\beta ^2-1} \right){\rm i}a}{\beta ^2}$

$\phi _5 =\varphi _1^2 a^2+\varphi _1 \omega +\varphi _1 \tau \omega ^2,\ \ \phi _6 =\varphi _1 \varphi _2 {\rm i}a\omega +\varphi _1 \varphi _2 \tau \omega ^2{\rm i}a$

$\phi _7 =\varphi _1 \varphi _2 \omega +\varphi _1 \varphi _2 \tau \omega ^2,\ \ \phi _9 =\varphi _1 \varphi _4 {\rm i}a\omega -\varphi _1 \varphi _6 {\rm i}a\omega ^2$

$\phi _8 =\varphi _1^2 a^2,\ \ \phi _{10} =\varphi _1 \varphi _5 \omega -\varphi _1 \varphi _6 \omega ^2,\ \ \phi _{11} =\varphi _1 \varphi _3 \omega$

令$\wp ={\partial }/{\partial z}$, 对分解后的方程进行简化整理

(31)$\left( {\wp ^2-\phi _0 } \right)u^* +\phi _1 \wp w^* -\phi _2 \theta ^* -\phi _2 P^* =0$

(32)$\left( {\wp ^2-\phi _3 } \right)w^* +\phi _4 \wp u^* -\theta ^* -P^* =0$

(33)$\left( {\wp ^2-\phi _5 } \right)\theta ^* =\phi _6 u^* +\phi _7 \wp w^*$

(34)$\left( {\wp ^2-\phi _8 } \right)P^* -\phi _9 u^* -\phi _{10} \wp w^* +\phi _{11} \theta ^* =0$

消去$w^* (z)$, $\theta ^* (z)$和$P^* \left( z \right)$, 可以得到一个关于$u^* \left( z \right)$的8阶偏微分方程

(35)$\begin{eqnarray}\left( {\wp ^{8}-B_1 \wp ^6+B_2 \wp ^4-B_3 \wp ^2+B_4 } \right)u^* \left( z \right)=0\end{eqnarray}$

将方程(35)因式分解后可得

(36)$\begin{eqnarray}\left( {\wp ^2-k_1^2 } \right)\left( {\wp ^2-k_2^2 } \right)\left( {\wp ^2-k_3^2 } \right)\left( {\wp ^2-k_4^2 } \right)u^* \left( z \right)=0\end{eqnarray}$

式中, $k_i^2 \left( {i=1,2,3,4} \right)$是如下八阶方程的特征根

(37)$\begin{eqnarray}k^{8}-B_1 k^6+B_2 k^4-B_3 k^2+B_4 =0\end{eqnarray}$

其中

$B_1 =k_1^2 +k_2^2 +k_3^2 +k_4^2 $

$B_2 =k_1^2 k_2^2 +k_1^2 k_3^2 +k_1^2 k_4^2 +k_2^2 k_3^2 +k_2^2 k_4^2 +k_3^2 k_4^2$

$B_3 =k_1^2 k_2^2 k_3^2 +k_1^2 k_2^2 k_4^2 +k_1^2 k_3^2 k_4^2 +k_2^2 k_3^2 k_4^2 $

$B_4 =k_1^2 k_2^2 k_3^2 k_4^2$

方程的解可表示为如下形式

(38)$u^* \left( z \right)=\sum\limits_{n=1}^4 {M_{1n} \left( {a,\omega } \right)} e^{-k_n z}$

(39)$w^* \left( z \right)=\sum\limits_{n=1}^4 {M_{1n} '\left( {a,\omega } \right)} e^{-k_n z}$

(40)$\theta ^* \left( z \right)=\sum\limits_{n=1}^4 {M_{1n} "\left( {a,\omega } \right)} e^{-k_n z}$

(41)$P^* \left( z \right)=\sum\limits_{n=1}^4 {M_{1n} "'\left( {a,\omega } \right)} e^{-k_n z}$

其中

$M_{1n} '\left( {a,\omega } \right)=H_{1n} M_{1n} \left( {a,\omega } \right),\ M_{1n} "\left( {a,\omega } \right)=H_{2n} M_{1n} \left( {a,\omega } \right)$

$M_{1n} "'\left( {a,\omega } \right)=H_{3n} M_{1n} \left( {a,\omega } \right),\ H_{1n} =\dfrac{m_4 \phi _2 -m_1 m_5 }{m_2 m_5 -m_3 \phi _2 }$

$H_{2n} =\dfrac{\phi _6 }{m_6 }+\dfrac{m_7 H_{1n} }{m_6 },\ H_{3n} =\dfrac{\phi _9 }{m_8 }+\dfrac{m_9 H_{1n} }{m_8 }-\dfrac{\phi _{11} H_{2n} }{m_8 }$

代入应力分量表达式中得

(42)$\sigma _{xx}^* =\sum\limits_{n=1}^4 {H_{4n} M_{1n} \left( {a,\omega } \right)} e^{-k_n z}$

(43)$\sigma _{zz}^* =\sum\limits_{n=1}^4 {H_{5n} M_{1n} \left( {a,\omega } \right)} e^{-k_n z}$

(44)$\tau _{xz}^* =\sum\limits_{n=1}^4 {H_{6n} M_{1n} \left( {a,\omega } \right)} e^{-k_n z}$

式中

$H_{4n} ={\rm i}a\beta ^2-k_n \left( {\beta ^2-2} \right)H_{1n} -\beta ^2H_{2n} -\beta ^2H_{3n}$

$H_{5n} ={\rm i}a\left( {\beta ^2-2} \right)-k_n \beta ^2H_{1n} -\beta ^2H_{2n} -\beta ^2H_{3n}$

$H_{6n} =-k_n +{\rm i}aH_{1n}$

3 算例分析及结果讨论

以饱和多孔半无限大弹性地基为例, 分别在地基的上表面施加温度载荷和机械载荷, 研究孔隙率各向异性参数变化对地基深度增加方向和波的传播方向上竖向位移、超孔隙水压力、竖向应力及温度的影响. 根据求得的解析解表达式, 利用Maple软件, 结合边界条件得到了各物理的分布规律图.

3.1 计算模型验证

本文是考虑孔隙率各向异性参数的影响而建立的控制方程, 当孔隙率各向异性参数$k=1.0$时方程可退化为各向同性耦合动力响应方程, 其结果可与参考文献^[38]进行对比, 从而验证模型的合理性.

研究中所需要的基本参数取值同参考文献^[38]一致, $\psi ^* =1$, $a=1.2$. 其他参数详见表1.

表1饱和多孔弹性地基热-水-力多物理场耦合基本参数

Table 1 Basic parameters of themo-hydro-mechanical coupling of saturated porous elastic foundation

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3.2 地基深度增加方向各物理量变化规律

在地基上表面分别施加温度载荷和机械载荷作用, 研究地基深度增加方向上孔隙率各向异性参数变化对各物理量的影响. 以下8幅图为$x=1.0$平面上$t=0.5$时各物理量的变化规律图. 其中, 退化模型曲线与参考文献结果曲线吻合较好, 进而验证了模型的合理性.

图2$\sim\!$图5为温度载荷下竖向位移、超孔隙水压力、竖向应力和温度变化规律图, 整体来说, 孔隙率各向异性参数变化对竖向位移和温度影响较小, 对超孔隙水压力和竖向应力影响较大. 竖向位移变化趋势都是由压缩状态逐渐转为膨胀最后趋于平缓衰减至零; 温度则是逐渐衰减至零. 超孔隙水压力曲线和竖向应力曲线峰值均逐渐减小, 且峰值出现位置向地基深度减小方向移动, 超孔隙水压力减幅为52.72%和61.06%, 竖向应力减幅为16.92%和26.13%. 这主要是由于孔隙水对温度较为敏感, 随着孔隙率各向异性参数增大, 两个方向孔隙分布更不均匀影响地基排水速率, 进而使得超孔隙水压力变小, 而超孔隙水压力变化影响了竖向应力的变化.

图2

图2温度载荷作用时无量纲竖向位移分布规律图

Fig. 2The distribution of non-dimension vertical displacement under thermal load

图3

图3温度载荷作用时无量纲超孔隙水压力分布规律图

Fig. 3The distribution of non-dimension excess pore water pressure under thermal load

图4

图4温度载荷作用时无量纲竖向应力分布规律图

Fig. 4The distribution of non-dimension vertical stress under thermal load

图5

图5温度载荷作用时无量纲温度分布规律图

Fig. 5The distribution of non-dimension temperature under thermal load

图6$\sim\!$图9为机械载荷下竖向位移、超孔隙水压力、竖向应力和温度变化规律图, 整体来说, 孔隙率各向异性对竖向位移和竖向应力影响较小, 对超孔隙水压力和温度影响较大.图6和图8中, 竖向位移和竖向应力均逐渐衰减至零, 但应力衰减速率比位移衰减速率略快.图7超孔隙水压力曲线随各向异性参数的增加逐渐减小, 尤其是峰值处, 有较为明显变化, 峰值减幅为8.89%和9.82%;图9中随着各向异性参数的增大, 温度的峰值逐渐减小, 减幅依次为18.46%和19.74%.

图6

图6机械载荷作用时无量纲竖向位移分布规律图

Fig. 6The distribution of non-dimension vertical displacement under mechanical load

图7

图7机械载荷作用时无量纲超孔隙水压力分布规律图

Fig. 7The distribution of non-dimension excess pore water pressure under mechanical load

图8

图8机械载荷作用时无量纲竖向应力分布规律图

Fig. 8The distribution of non-dimension vertical stress under mechanical load

图9

图9机械载荷作用时无量纲温度分布规律图

Fig. 9The distribution of non-dimension temperature under mechanical load

从图6$\sim\!$图9可以看出, 温度曲线量级相对较小, 说明单纯的机械载荷作用产生的热量不多, 可忽略不计. 孔隙率各向异性参数变化对超孔隙水压力和温度的影响趋势相同, 但对温度的影响更明显, 这是由于随着各向异性参数增大, 使得地基热量耗散更为迅速, 引起明显的温度变化, 而孔隙率各向异性参数的变化影响了地基中孔隙的分布, 参数增大, 孔隙分布更为不均匀, 导致超孔隙水压力峰值减小.

3.3 波的传播方向上各物理量变化规律

本节从波的传播方向上研究了地基上表面分别施加温度载荷和机械载荷作用时孔隙率各向异性参数变化对所考虑物理量的影响.图10$\sim\!$图17为$z=1.0$平面上$t=0.5$时各物理量的变化规律图.

图10

图10温度载荷作用时无量纲竖向位移分布规律图

Fig. 10The distribution of non-dimension vertical displacement under thermal

图11

图11温度载荷作用时无量纲超孔隙水压力分布规律图

Fig. 11The distribution of non-dimension excess pore water pressure under thermal load

图12

图12温度载荷作用时无量纲竖向应力分布规律图

Fig. 12The distribution of non-dimension vertical stress under thermal load

图13

图13温度载荷作用时无量纲温度分布规律图

Fig. 13The distribution of non-dimension temperature under thermal load

图14

图14机械载荷作用时无量纲竖向位移分布规律图

Fig. 14The distribution of non-dimension vertical displacement under mechanical load

图15

图15机械载荷作用时无量纲超孔隙水压力分布规律图

Fig. 15The distribution of non-dimension excess pore water pressure under mechanical load

图16

图16机械载荷作用时无量纲竖向应力分布规律图

Fig. 16The distribution of non-dimension vertical stress under mechanical load

图17

图17机械载荷作用时无量纲温度分布规律图

Fig. 17The distribution of non-dimension temperature under mechanical load

图10$\sim\!$图13为温度载荷下竖向位移、超孔隙水压力、竖向应力和温度变化规律图. 整体来说, 随着孔隙率各向异性参数的增大, 物理量峰值逐渐减小. 可以明显看出, 孔隙率各向异性参数变化对超孔隙水压力和竖向应力的影响远超于竖向位移和温度. 其中, 超孔隙水压力减幅依次为53.50%和66.15%, 竖向应力减幅依次为25.26%和33.77%, 而竖向位移减幅在4%附近, 温度减幅稳定在2%左右. 这是因为孔隙率的变化影响到孔隙水, 随着各向异性参数增大, 孔隙分布更为不均匀, 孔隙水排出速率受到影响, 导致超孔隙水压力减幅明显, 而超孔隙水压力变化又会影响竖向应力, 使得竖向应力变化.

图14$\sim\!$图17为机械载荷下竖向位移、超孔隙水压力、竖向应力和温度的变化规律图. 孔隙率各向异性对竖向位移和竖向应力变化不太明显, 减幅相对较小; 对超孔隙水压力和温度变化明显, 其中温度减幅更大, 为18.46%和19.74%. 单纯的机械载荷作用产生于地基中的温度很小, 孔隙率的变化使得地基散热更为迅速, 所以对温度的影响更加明显.

4 结论

本文在饱和多孔弹性地基基础上引入孔隙率各向异性参数, 并结合Lord-Shulman广义热弹性理论, 建立了考虑孔隙率各向异性的热-水-力耦合动力响应模型并推导其基本控制方程, 采用正则模态法对简化方程进行求解, 得到了各物理量的解析表达, 分析了孔隙率各向异性参数变化对相关物理量的影响. 主要结论如下:

(1) 考虑孔隙率各向异性参数的饱和多孔弹性地基热-水-力耦合动力响应模型可退化为各向同性饱和多孔弹性地基热-水-力耦合动力响应模型, 退化后物理量变化规律与参考文献^[38]曲线分布规律一致, 验证了模型的合理性;

(2) 地基深度增加方向上, 地基上表面受温度载荷作用时, 孔隙率各向异性参数变化对超孔隙水压力和竖向应力影响明显, 随着各向异性参数增大, 峰值逐渐减小, 峰值所在位置向靠近地基上表面移动; 地基上表面受机械载荷作用时, 孔隙率各向异性参数变化对超孔隙水压力和温度影响明显, 随着各向异性参数增大, 峰值逐渐减小, 峰值所在位置向靠近地基上表面移动;

(3) 波的传播方向上, 孔隙率各向异性参数变化对各物理量均有影响, 使得峰值逐渐减小, 地基上表面受温度载荷作用时, 对超孔隙水压力和竖向应力影响明显; 地基上表面受机械载荷作用时, 对超孔隙水压力和温度影响明显.

附录

${\varepsilon }_{{ij}} $ 应变张量 ${u}_{i} $ 位移张量

${\sigma }_{{ij}} $ 应力张量 $\lambda ,G$ 拉梅常数

${e}$ 体积应变 $\theta $ 温度变化

$T$ 绝对温度 $T_0 $ 初始温度

$\delta _{ij} $ Kronecker符号 $\beta _1 $ 体积应变

$P$ 超孔隙水压力 ${F}_{i} $ 体力张量

$\rho _\xi $ 不同方向的密度 $\rho _{\xi x} $ 水平方向的密度

$\rho _{\xi z} $ 竖直方向的密度 $\rho _{\rm w} $ 孔隙水密度

$\rho _{\rm s} $ 土颗粒密度 $n_\xi $ 不同方向的孔隙率

$n_{\xi x} $ 水平方向的孔隙率 $n_{\xi z} $ 竖直方向的孔隙率

$K$ 热传导系数 $C_E $ 单位质量介质比热容

$\tau $ 热弛豫时间 $m$ 介质比热容

$c_{\rm w} $ 孔隙水比热容 $c_{\rm s} $ 土颗粒比热容

$\alpha _u $ 介质线膨胀系数 $k_{\rm d} $ 介质的渗透系数

$\alpha _{\rm w} $ 孔隙水线性热膨胀系数 $\alpha _{\rm s} $ 土颗粒线性热膨胀系数

$g$ 重力加速度 $Q$ 温度载荷

$\omega $ 频率(复时间常数) $a$ $x$轴方向的波数

$\psi ^* \left( {a,\omega } \right)$ 载荷振幅 $q$ 机械载荷

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