工程科学与技术   2018, Vol. 50 Issue (2): 32-41

Consolidation Solution for Sand-drained Ground with Impeded Boundaries Under Vacuum Preloading
HU Yayuan
Research Center of Coastal and Urban Geotechnical Eng.,Zhejiang Univ.,Hangzhou 310058,China
Key words: sand-drained ground    vacuum preloading    impeded boundary    consolidation settlement    analytical theory of consolidation

1 等应变条件下的固结解析解 1.1 固结方程

 图1 砂井地基单井固结计算简图 Fig. 1 Analytical sketch of consolidation for unit cell in sand-drained ground

1）等应变假定成立。地基只有竖向变形而无侧向变形，同一深度平面上任意一点的竖向变形相等。

2）有效应力原理成立。饱和土的应力–应变关系符合线弹性本构模型，地基压缩模量为 ${E_{{s}}}$

3）达西定理成立。垂直向渗透系数 ${k_{{v}}}$ 和砂井井料的渗透系数 ${k_{{w}}}$ 为常数，水平向渗透系数沿径向呈 ${k_{{h}}}f(r)$ 变化。其中： ${k_{{h}}}$ 为原状土的水平向渗透系数； $f(r)$ 为涂抹系数，反映砂井施工对水平向渗透系数的影响，为随 $r$ 变化的函数。

4）土区的径竖向固结可分别单独考虑。计算竖向固结时 ${k_{{h}}}$ 取0，计算径向固结时 ${k_{{v}}}$ 取0。综合考虑径竖向的总固结度按Carrillo建议的方法计算。

5）假定砂井地基的上下边界为半透水边界，土区顶部边界条件为：

 ${\left. {\frac{{\partial {u_{\textit{z}}}}}{{\partial {\textit{z}}}}} \right|_{{\textit{z}} = H}} = - \frac{{{R_{{U}}}}}{H}({u_{\textit{z}}} + {u_0})$ (1)

 ${\left. {\frac{{\partial {u_{\textit{z}}}}}{{\partial {\textit{z}}}}} \right|_{{\textit{z}} = 0}} = \frac{{{R_{{L}}}}}{H}{u_{\textit{z}}}$ (2)

 ${\left. {\frac{{\partial {u_{ w}}}}{{\partial {\textit{z}}}}} \right|_{{\textit{z}} = H}} = - \frac{{{R_{{{wU}}}}}}{H}({u_{{w}}} + {u_0})$ (3)
 ${\left. {\frac{{\partial {u_{ w}}}}{{\partial {\textit{z}}}}} \right|_{{\textit{z}} = 0}} = - \frac{{{R_{{{wL}}}}}}{H}{u_{{w}}}$ (4)

6）假定在 $r = {r_{{e}}}$ 处沿径向为不透水边界，根据基本假定4）有：

 ${\left. {\frac{{\partial {u_{r}}}}{{\partial {r}}}} \right|_{r = {r_{{e}}}}} = 0$ (5)

7）假定砂井中孔压沿径向的变化可忽略不计，任一深度 ${\textit{z}}$ 处砂井中向上水流的增量等于土中渗入砂井的流量，由此推导出砂井渗流连续方程为[11]

 $\frac{{\partial {}^2{u_{{w}}}}}{{\partial {{\textit{z}}^2}}} = - \frac{{2{k_{{h}}}f({r_{{w}}})}}{{{k_{{w}}}{r_{{w}}}}}{\left. {\frac{{\partial {u_r}}}{{\partial r}}} \right|_{r = {r_{{w}}}}}$ (6)

8）假定土区和砂井中初始孔压为0。根据基本假定4）有：

 ${\left. {{u_{\textit{z}}}} \right|_{t = 0}} = 0$ (7)
 ${\left. {{u_{r}}} \right|_{t = 0}} = 0$ (8)

 $\frac{{\partial {\varepsilon _{\upsilon 1}}}}{{\partial t}} = - \frac{1}{{{E_{{s}}}}}\frac{{\partial {u_{\textit{z}}}}}{{\partial t}}$ (9)
 $- \frac{{{k_{{v}}}}}{{{\gamma _{{w}}}}}\frac{{{\partial ^2}{u_{\textit{z}}}}}{{\partial {{\textit{z}}^2}}} = \frac{{\partial {\varepsilon _{\upsilon 1}}}}{{\partial t}}$ (10)

 $\frac{{\partial {\varepsilon _{\upsilon 2}}}}{{\partial t}} = - \frac{1}{{{E_{{s}}}}}\frac{{\partial {{\bar u}_r}}}{{\partial t}}$ (11)
 $- \frac{1}{r}\frac{\partial }{{\partial r}}[\frac{{{k_{{h}}}f(r)}}{{{\gamma _{{w}}}}}r\frac{{\partial {u_r}}}{{\partial r}}] = \frac{{\partial {\varepsilon _{\upsilon 2}}}}{{\partial t}}$ (12)

 ${\bar u_{r}} = \frac{1}{{{\text{π}} (r_{{e}}^2 - r_{{w}}^2)}}\int_{{r_{{w}}}}^{{r_{{{ e}}}}} {2{\text{π}} r{u_{r}}{{d }}r}$ (13)
1.2 方程求解 1.2.1 竖向固结方程的求解

 ${{\textit{z}}^*} = H - {\textit{z}}$ (14)
 $u_{\textit{z}}^* = {u_{\textit{z}}} + ({\alpha _{\textit{z}}} - {\beta _{\textit{z}}}\frac{{{{\textit{z}}^*}}}{H}){u_0}$ (15)

 ${\alpha _{\textit{z}}} = \frac{{(1 + {R_{{L}}}){R_{{U}}}}}{{(1 + {R_{{L}}}){R_{{U}}} + {R_{{L}}}}}$ (16)
 ${\beta _{\textit{z}}} = \frac{{{R_{{L}}}{R_{{U}}}}}{{({R_{{L}}} + 1){R_{{U}}} + {R_{{L}}}}}$ (17)

 $\frac{{{k_{ v}}{E_{{s}}}}}{{{\gamma _{ w}}}}\frac{{{\partial ^2}u_{\textit{z}}^*}}{{\partial {{\textit{z}}^*}^2}} = \frac{{\partial u_{\textit{z}}^*}}{{\partial t}}$ (18)

 ${\left. {\frac{{\partial u_{\textit{z}}^*}}{{\partial {{\textit{z}}^*}}}} \right|_{{{\textit{z}}^*} = 0}} = \frac{{{R_{{U}}}}}{H}u_{\textit{z}}^*$ (19)
 ${\left. {\frac{{\partial u_{\textit{z}}^*}}{{\partial {{\textit{z}}^*}}}} \right|_{{{\textit{z}}^*} = H}} = - \frac{{{R_{{L}}}}}{H}u_{\textit{z}}^*$ (20)

 ${\left. {u_{\textit{z}}^*} \right|_{t = 0}} = ({\alpha _{\textit{z}}} - \frac{{{\beta _{\textit{z}}}}}{H}{{\textit{z}}^*}){u_0}$ (21)

 $u_{\textit{z}}^* = {u_0}\sum\limits_{m = 1}^\infty {\frac{{{a_{{\textit{z}}m}}}}{{{G_{{\textit{z}}m}}}}[\frac{{{\lambda _{{\textit{z}}m}}}}{{{R_{{U}}}}}\cos (\frac{{{\lambda _{{\textit{z}}m}}}}{H}{{\textit{z}}^*}) + \sin } (\frac{{{\lambda _{{\textit{z}}m}}}}{H}{{\textit{z}}^*})] \cdot {{ e}^{ - {T_{{v}}}\lambda _{{\textit{z}}m}^2}}$ (22)

 $\tan \; \lambda _{{\textit{z}}m}^{} = \frac{{\lambda _{{\textit{z}}m}^{}({R_{{U}}} + {R_{{L}}})}}{{\lambda _{{\textit{z}}m}^2 - {R_{{U}}}{R_{{L}}}}}$ (23)

${a_{{\textit{z}}m}}$ 按式（24）计算：

 \begin{aligned} {a_{{\textit{z}}m}} = & {\alpha _{\textit{z}}}[\frac{1}{{{R_{{U}}}}}\sin\; {\lambda _{{\textit{z}}m}} + \frac{{1 - \cos\; {\lambda _{{\textit{z}}m}}}}{{{\lambda _{{\textit{z}}m}}}}] - {\beta _{\textit{z}}}[ - \frac{1}{{{\lambda _{{\textit{z}}m}}{R_{{U}}}}} + \\& (\frac{1}{{{R_{{U}}}}} + \frac{1}{{\lambda _{{\textit{z}}m}^2}})\sin\; {\lambda _{{\textit{z}}m}} + (\frac{1}{{{R_{{U}}}}} - 1)\frac{{\cos\; {\lambda _{{\textit{z}}m}}}}{{{\lambda _{{\textit{z}}m}}}}]\end{aligned} (24)

${G_{{\textit{z}}m}}$ 按式（25）计算：

 \begin{aligned}{G_{{\textit{z}}m}} = & \frac{1}{2}[{(\frac{{{\lambda _{{\textit{z}}m}}}}{{{R_{{U}}}}})^2} + 1 + \frac{1}{{{R_{{U}}}}} + \frac{1}{2}(\frac{{{\lambda _{{\textit{z}}m}}}}{{R_{{U}}^2}} - \frac{1}{{{\lambda _{{\textit{z}}m}}}})\sin (2{\lambda _{{\textit{z}}m}}) - \\& \frac{1}{{{R_{{U}}}}}\cos (2{\lambda _{{\textit{z}}m}})]\end{aligned} (25)

 \begin{aligned}{S\!_{{{v}}{\textit{z}}}}(t) = & - \int_0^H {\frac{{{u_{\textit{z}}}({\textit{z}},t)}}{{{E_{{s}}}}}{ d}{\textit{z}}} = \frac{{H{u_0}}}{{{E_{{s}}}}}({\alpha _{\textit{z}}} - \frac{{{\beta _{\textit{z}}}}}{2}) - \\& \frac{{H{u_0}}}{{{E_{{s}}}}}\sum\limits_{m = 1}^\infty {\frac{{{A_{{\textit{z}}m}}}}{{{G_{{\textit{z}}m}}}}[\frac{{\sin\; {\lambda _{{\textit{z}}m}}}}{{{R_{{U}}}}} + \frac{{1 - \cos\; {\lambda _{{\textit{z}}m}}}}{{{\lambda _{{\textit{z}}m}}}}]{{ e}^{ - {T_{{v}}}\lambda _{{\textit{z}}m}^2}}} \end{aligned} (26)

 ${S\!_{{{v}}{\textit{z}}}}(\infty ) = \frac{{H{u_0}}}{{{E_{{s}}}}}({\alpha _{\textit{z}}} - \frac{{{\beta _{\textit{z}}}}}{2})$ (27)

 \begin{aligned}[b]{{\overline U}_{\textit{z}}}(t) = & \frac{{{S\!_{{{v}}{\textit{z}}}}(t)}}{{{S\!_{{{v}}{\textit{z}}}}(\infty )}} = 1 - \frac{2}{{2{\alpha _{\textit{z}}} - {\beta _{\textit{z}}}}}\sum\limits_{m = 1}^\infty {\frac{{{A_{{\textit{z}}m}}}}{{{G_{{\textit{z}}m}}}}[\frac{{\sin \;{\lambda _{{\textit{z}}m}}}}{{{R_{{U}}}}}} + \\& \frac{{1 - \cos {\lambda _{{\textit{z}}m}}}}{{{\lambda _{{\textit{z}}m}}}}]{{ e}^{ - {T_{ v}}\lambda _{{\textit{z}}m}^2}}\end{aligned} (28)

${U_{\textit{z}}} > 30\%$ 时，式（28）可只取第1个特征值 $\lambda _{{\textit{z}}1}^{}$ 所对应的特解作为近似值，由此得：

 ${{\overline U}_{\textit{z}}}(t) = 1 - \frac{{{A_{{\textit{z}}1}}}}{{{G_{{\textit{z}}1}}}}\left(\frac{{\sin\; {\lambda _{{\textit{z}}1}}}}{{{R_{{U}}}}} + \frac{{1 - \cos \;{\lambda _{{\textit{z}}1}}}}{{{\lambda _{{\textit{z}}1}}}}\right)\frac{{2{{ e}^{ - {T_{ v}}\lambda _{{\textit{z}}1}^2}}}}{{2{\alpha _{\textit{z}}} - {\beta _{\textit{z}}}}}$ (29)
1.2.2 径向固结方程的求解

 $u_{{w}}^* = {u_{{w}}} + ({\alpha _{{w}}} - {\beta _{{w}}}\frac{{{{\textit{z}}^*}}}{H}){u_0}$ (30)
 $u_{r}^* = {u_{r}} + ({\alpha _{{w}}} - {\beta _{{w}}}\frac{{{{\textit{z}}^*}}}{H}){u_0}$ (31)

 ${\alpha _{{w}}} = \frac{{(1 + {R_{{{wL}}}}){R_{{{wU}}}}}}{{(1 + {R_{{{wL}}}}){R_{{{wU}}}} + {R_{{{wL}}}}}}$ (32)
 ${\beta _{{w}}} = \frac{{{R_{{{wL}}}}{R_{{{wU}}}}}}{{({R_{{{wL}}}} + 1){R_{{{wU}}}} + {R_{{{wL}}}}}}$ (33)

 $\frac{{\partial {}^2u_{{w}}^*}}{{\partial {{\textit{z}}^*}^2}} = - \frac{{2{k_{{h}}}f({r_{{w}}})}}{{{r_{{w}}}{k_{{w}}}}}{\left. {\frac{{\partial u_{r}^*}}{{\partial r}}} \right|_{r = {r_{{w}}}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$ (34)
 $\frac{{\partial {\varepsilon _{\upsilon 2}}}}{{\partial t}} = - \frac{1}{{{E_{{s}}}}}\frac{{\partial \bar u_{r}^*}}{{\partial t}}$ (35)
 $\quad \quad\quad\quad\quad\quad - \frac{1}{r}\frac{\partial }{{\partial r}}[\frac{{{k_{{h}}}f(r)}}{{{\gamma _{{w}}}}}r\frac{{\partial u_{r}^*}}{{\partial r}}] = \frac{{\partial {\varepsilon _{\upsilon 2}}}}{{\partial t}}$ (36)

 $\bar u_{r}^* = \frac{1}{{{\text{π}} (r_{{e}}^2 - r_{{w}}^2)}}\int_{\displaystyle{r_{{w}}}}^{\displaystyle{r_{{e}}}} {2{\text{π}} ru_{r}^*{ d}r}$ (37)

 ${\left. {\frac{{\partial u_{{w}}^*}}{{\partial {{\textit{z}}^*}}}} \right|_{{{\textit{z}}^*} = 0}} = \frac{{{R_{{{wU}}}}}}{H}u_{{w}}^*$ (38)
 ${\left. {\frac{{\partial u_{{w}}^*}}{{\partial {{\textit{z}}^*}}}} \right|_{{{\textit{z}}^*} = H}} = - \frac{{{R_{{{wL}}}}}}{H}u_{{w}}^*$ (39)
 ${\left. {u_r^*} \right|_{t = 0}} = ({\alpha _{{w}}} - \frac{{{\beta _{{w}}}}}{H}{{\textit{z}}^*}){u_0}$ (40)

 $\xi = \frac{{8{c_{{h}}}}}{{d_{{e}}^2{F_{{a}}}}} ,\; \rho = \frac{{2{k_{{h}}}({n^2} - 1)}}{{{k_{{w}}}r_{{e}}^2{F_{{a}}}}}$ (41)

 ${c_{{h}}} = \frac{{{k_{{h}}}{E_{{s}}}}}{{{\gamma _{{w}}}}} , \;n = \frac{{{r_{{e}}}}}{{{r_{{w}}}}}$ (42)
 ${F_{{a}}} = \frac{2}{{r_{{e}}^2 - r_w^2}}\int_{{r_{{w}}}}^{{r_{{e}}}} {r[\int_{{r_{{w}}}}^r {\frac{{\operatorname{d} x}}{{xf\left( x \right)}}} } - \frac{1}{{r_{{e}}^2}}\int_{{r_{{w}}}}^r {\frac{{x\operatorname{d} x}}{{f\left( x \right)}}]} { d}r$ (43)

 \begin{aligned}[b] {u_{{w}}^*} = & {{u_0}\sum\limits_{m = 1}^\infty {\frac{{{A_{{{w}}m}}}}{{{G_{{{w}}m}}}}\frac{{\xi - {\eta _m}}}{\xi } \times } }\\& {[\frac{{{\lambda _{{{w}}m}}}}{{{R_{{{wU}}}}}}\cos (\frac{{{\lambda _{{{w}}m}}}}{H}{\textit{z}^*}) + \sin (\frac{{{\lambda _{{{w}}m}}}}{H}{\textit{z}^*})]{{ e}^{ - {\eta _m}t}}}\end{aligned} (44)
 $\bar u_{r}^* = {u_0}\sum\limits_{m = 1}^\infty {\frac{{{A_{{{w}}m}}}}{{{G_{{{w}}m}}}}[\frac{{{\lambda _{{{w}}m}}}}{{{R_{{{wU}}}}}}\cos (\frac{{{\lambda _{{{w}}m}}}}{H}{{\textit{z}}^*}) + \sin (\frac{{{\lambda _{{{w}}m}}}}{H}{{\textit{z}}^*})]} {{ e}^{ - {\eta _m}t}}$ (45)

 $\tan \;\lambda _{{{w}}m}^{} = \frac{{\lambda _{{{w}}m}^{}({R_{{{wU}}}} + {R_{{{wL}}}})}}{{\lambda _{{{w}}m}^2 - {R_{{{wU}}}}{R_{{{wL}}}}}}$ (46)

${\eta _m}$ 的计算式为：

 ${\eta _m} = \frac{{\xi \lambda _{{{w}}m}^2}}{{{\rho ^2}{H^2} + \lambda _{{{w}}m}^2}} = \frac{{8{c_{{h}}}}}{{{d}_{{e}}^2}}/\left( {{F_{{a}}} + \frac{8}{{\lambda _{{{w}}m}^2}}\frac{{{n^2} - 1}}{{{n^2}}}{R_{{J}}}} \right)$ (47)

${A_{{{w}}m}}$ 按式（48）计算：

 \begin{aligned}& {A_{{{w}}m}} = {\alpha _{{w}}}[\frac{1}{{{R_{{{wU}}}}}}\sin\; {\lambda _{{{w}}m}}+ \frac{{1 - \cos \;{\lambda _{{{w}}m}}}}{{{\lambda _{{{w}}m}}}}] + {\beta _{{w}}}[\frac{1}{{{\lambda _{{{w}}m}}{R_{{{wU}}}}}} - \\& \quad \quad (\frac{1}{{{R_{{{wU}}}}}} + \frac{1}{{\lambda _{{{w}}m}^2}})\sin \;{\lambda _{wm}} - (\frac{1}{{{R_{{{wU}}}}}} - 1)\frac{{\cos\; {\lambda _{{{w}}m}}}}{{{\lambda _{{{w}}m}}}}]\end{aligned} (48)

${G_{{{w}}m}}$ 按式（49）计算：

 \begin{aligned}{G_{{{w}}m}} = & \frac{1}{2}[\frac{1}{2}(\frac{{{\lambda _{{{w}}m}}}}{{R_{{{wU}}}^2}} - \frac{1}{{{\lambda _{{{w}}m}}}})\sin (2{\lambda _{{{w}}m}}) - \frac{1}{{{R_{{{wU}}}}}}\cos (2{\lambda _{{{w}}m}})+\\ & {(\frac{{{\lambda _{{{w}}m}}}}{{{R_{{{wU}}}}}})^2} + 1 + \frac{1}{{{R_{{{wU}}}}}}]\end{aligned} (49)

 \begin{aligned}{S\!_{{{v}}{r}}}(t) = & - \int_0^H {\frac{{{{\bar u}_{r}}({\textit{z}},t)}}{{{E_{{s}}}}}{ d}{\textit{z}}} = \frac{{H{u_0}}}{{{E_{{s}}}}}({\alpha _{{w}}} - \frac{{{\beta _{{w}}}}}{2}) - \\ & \frac{{H{u_0}}}{{{E_{{s}}}}}\sum\limits_{m = 1}^\infty {\frac{{{A_{{{w}}m}}}}{{{G_{{{w}}m}}}}[\frac{{\sin \;{\lambda _{{{w}}m}}}}{{{R_{{{wU}}}}}} + \frac{{1 - \cos \;{\lambda _{{{w}}m}}}}{{{\lambda _{{{w}}m}}}}]{{ e}^{ - {\eta _m}t}}} \end{aligned} (50)

 ${S\!_{{{v}}{r}}}(\infty ) = \frac{{H{u_0}}}{{{E_{{s}}}}}({\alpha _{{w}}} - \frac{{{\beta _{{w}}}}}{2})$ (51)

 \begin{aligned}{{\overline U}_r}(t) = & \frac{{{S\!_{{{v}}{r}}}(t)}}{{{S\!_{{{v}}{r}}}(\infty )}} = 1 - \frac{2}{{2{\alpha _{{w}}} - {\beta _{{w}}}}}\sum\limits_{m = 1}^\infty {\frac{{{A_{{{w}}m}}}}{{{G_{{{w}}m}}}}[\frac{{\sin \;{\lambda _{{{w}}m}}}}{{{R_{{{wU}}}}}} + } \\& \frac{{1 - \cos \;{\lambda _{{{w}}m}}}}{{{\lambda _{{{w}}m}}}}]{{ e}^{ - {\eta _m}t}}\end{aligned} (52)

 $\frac{1}{H}\int_0^H {{{\left. {\bar u_r^*} \right|}_{t = 0}}{ d}{\textit{z}}} = {u_0}({\alpha _{{w}}} - \frac{{{\beta _{{w}}}}}{2})$ (53)

 ${\overline U_r}(t) = 1 - {{ e}^{\scriptstyle - 8{T_{{h}}}/\left( {{F_{{a}}} + \textstyle\frac{8}{{\lambda _{{{w}}1}^2}}\textstyle\frac{{{n^2} - 1}}{{{n^2}}}{\scriptstyle R_{{J}}}} \right)}}$ (54)

1.3 Carrillo法的必要条件

 ${\left. {{{\left. { - u'({\textit{z}})} \right|}_{t = \infty }} = \sigma '({\textit{z}})} \right|_{t = \infty }} = ({\alpha _{\textit{z}}} - {\beta _{\textit{z}}}\frac{{H - {\textit{z}}}}{H}){u_0}$ (55)

 ${\left. { - u'({\textit{z}})} \right|_{t = \infty }} = {\left. {\sigma '({\textit{z}})} \right|_{t = \infty }} = ({\alpha _{{w}}} - {\beta _{{w}}}\frac{{H - {\textit{z}}}}{H}){u_0}$ (56)

 ${\alpha _{\textit{z}}} = {\alpha _{{w}}}, {\beta _{\textit{z}}} = {\beta _{{w}}}$ (57)

 ${R_{{{wU}}}} = {R_{{U}}}, {R_{{{wL}}}} = {R_{{L}}}$ (58)

 $\overline U(t) = 1 - [1 - {\overline U_{r}}(t)][1 - {\overline U_{\textit{z}}}(t)]$ (59)

$\sigma {'_\infty }$ 为固结增加的最终有效应力，图2给出了 ${R_{{{wU}}}} = {R_{{U}}} = \infty$ 时， ${R_{{{wL}}}}$ ${R_{{L}}}$ ）对 $\sigma {'_\infty }/{u_0}$ 沿深度分布特性的影响；图3给出了 ${R_{{{wL}}}} = {R_{{L}}} = 1$ 时， ${R_{{{wU}}}}$ ${R_{{U}}}$ ）对 $\sigma {'_\infty }/{u_0}$ 沿深度分布特性的影响。

 图2 ${{ R}_{{\bf{wU}}}} \!=\! {{ R}_{\bf{U}}} \!=\! { \infty}$ 时 ${{ R}_{{\bf{wL}}}}$ （ ${{ R}_{\bf{L}}}$ ）对 ${\sigma {'_{ \infty}}/{ u_{ 0}}}$ 随深度 ${{z}}$ 分布的影响 Fig. 2 Effect of ${{ R}_{{\bf{wL}}}}$ ( ${{ R}_{\bf{L}}}$ ) on the distribution of ${\sigma {'_{ \infty}}/{ u_{ 0}}}$ along with depth ${{z}}$ at ${{ R}_{{\bf{wU}}}} = {{ R}_{\bf{U}}} = { \infty}$

 图3 ${{ R}_{{\bf{wL}}}} = {{ R}_{\bf{L}}} = { 1}$ 时 ${{ R}_{{\bf{wU}}}}$ （ ${{ R}_{\bf{U}}}$ ）对 $\sigma {'_{ \infty}}/{{ u}_{ 0}}$ 随深度 ${{z}}$ 分布的影响 Fig. 3 Effect of ${{ R}_{{\bf{wU}}}}$ ( ${{ R}_{\bf{U}}}$ ) on the distribution of ${\sigma} {'_{ \infty}}/{{ u}_{ 0}}$ along with depth ${{z}}$ at ${{ R}_{{\bf{wL}}}} = {{ R}_{\bf{L}}} = { 1}$

1.4 理论解的退化

${R_{{{wU}}}} \!=\! {R_{{U}}} \!=\! \infty$ ${R_{{{wL}}}} \!=\! {R_{{L}}} \!=\! 0$ 时，此时有 ${\alpha _{\textit{z}}} \!=\! {\alpha _{{w}}} \!=\! 1$ ${\beta _{\textit{z}}} = {\beta _{{w}}} = 0$ ${\lambda _{{\textit{z}}m}} = {\lambda _{{{w}}m}} = (m + 0.5){\text{π}}$ ，竖向固结的解由式（24）、（25）、（28）得：

 ${\overline U_{\textit{z}}}(t) = 1 - \frac{8}{{{{\text{π}} ^2}}}\sum\limits_{m = 1}^\infty {\frac{1}{{{{(2m + 1)}^2}}}} {{ e}^{ - {{\text{π}} ^2}{T_{ v}}{{(2m + 1)}^2}}}$ (60)

 ${\overline U_r}(t) = 1 - \sum\limits_{m = 1}^\infty {\frac{2}{{{{[(m + 0.5){\text{π}} ]}^2}}}{{ e}^{ - {\eta _m}t}}}$ (61)

2 透水系数对固结度与沉降的影响分析

 $f\left( r \right) = 1,\; {r_{{w}}} \le r \le {r_{{e}}}$ (62)

 ${F_{{a}}} = \frac{{{n^2}}}{{{n^2} - 1}}\ln n - \frac{{3{n^2} - 1}}{{4{n^2}}}$ (63)

 f\left( r \right) = \left\{ {\begin{aligned}& {\delta, \;{r_{{w}}} \le r \le {r_{{s}}}};\\& {1,\;{r_{{s}}} \le r \le {r_{{e}}}}\end{aligned}} \right. (64)

 ${F_{{a}}} = \frac{{{n^2}}}{{{n^2} - 1}}[\ln n + (\frac{1}{\delta } - 1)(\ln s + \frac{{1 - {s^2}}}{{{n^2}}} + \frac{{{s^4} - 1}}{{4{n^4}}})] - \frac{{3{n^2} - 1}}{{4{n^2}}}$ (65)

 $f\left( r \right) = \left\{ {\begin{array}{*{20}{c}}\!\!\!\!\!{\displaystyle\frac{{(r{{ - }}{r_{{w}}}) + \delta ({r_{{s}}}{{ - }}r)}}{{{r_{{s}}}{{ - }}{r_{{w}}}}},}& {{r_{{w}}} \le r \le r;}\\\!\!\!\!\!{1,}& {{r_{{s}}} \le r \le {r_{{e}}}}\end{array}} \right.$ (66)

 \begin{aligned}{F_{{a}}} = & \frac{{{n^2}}}{{{n^2} - 1}} [\frac{{s - 1}}{{\delta s - 1}}\ln (\delta s) - \frac{{{{(s - 1)}^2}}}{{{n^2}(1 - \delta )}} +\\[6pt]& \frac{{2(s - 1)(\delta s - 1)}}{{{n^2}(1 - \delta ){}^2}}\ln \frac{1}{\delta } - \frac{{2{s^3} - 3{s^2} + 1}}{{3{n^4}(1 - \delta )}}(s - 1) - \\[6pt]& \frac{{(s - 1)(\delta s - 1)}}{{{n^4}(1 - \delta ){}^2}}\!\![\frac{{{s^2} - 1}}{2} \!\!-\!\! \frac{{(s - 1)(\delta s - 1)}}{{1 - \delta }} \!\!+\!\! \frac{{{{(\delta s - 1)}^2}}}{{{{(1 - \delta )}^2}}}\ln \frac{1}{\delta }] - \\[6pt]&\frac{{({n^2} - {s^2}){{(1 - s)}^2}}}{{{n^4}(1 - \delta )}} + \ln \frac{n}{s} - \frac{3}{4} + \frac{{4{n^2}{s^2} - {s^4}}}{{4{n^4}}}]\quad\quad\;\;\;\; (\;67\;)\end{aligned} (67)
2.1 顶部透水系数恒定、底部半透水情形

 图4 ${{ R}_{\bf{J}}} = { 1}$ 时 ${{ R}_{{\bf{wL}}}}$ 对固结度的影响 Fig. 4 Effect of ${{ R}_{{\bf{wL}}}}$ on consolidation degree at ${{ R}_{\bf{J}}} = { 1}$

 图5 ${{ R}_{\bf{J}}} = { 1}$ 时 ${{ R}_{{\bf{wL}}}}$ 对无量纲沉降的影响 Fig. 5 Effect of ${{ R}_{{\bf{wL}}}}$ on dimensionless settlement at ${{ R}_{\bf{J}}} = { 1}$

 图6 ${{ R}_{\bf{J}}} = { 4}$ 时 ${{ R}_{{\bf{wL}}}}$ 对固结度的影响 Fig. 6 Effect of ${{ R}_{{\bf{wL}}}}$ on consolidation degree at ${{ R}_{\bf{J}}} = { 4}$

 图7 ${{ R}_{\bf{J}}} = { 4}$ 时 ${{ R}_{{\bf{wL}}}}$ 对无量纲沉降的影响 Fig. 7 Effect of ${{ R}_{{\bf{wL}}}}$ on dimensionless settlement at ${{ R}_{\bf{J}}} = { 4}$

2.2 底部透水系数恒定顶部半透水情形

 图8 底部不透水和 ${{ R}_{\bf{J}}} = { 4}$ 时 ${{ R}_{{\bf{wU}}}}$ 对无量纲沉降的影响 Fig. 8 Effect of ${{ R}_{{\bf{wU}}}}$ on dimensionless settlement for impermeable bottom and ${{ R}_{\bf{J}}} = { 4}$

 图9 ${{ R}_{{\bf{wL}}}} = { 1}$ 和 ${{ R}_{\bf{J}}} = { 1}$ 时 ${{ R}_{{\bf{wU}}}}$ 对固结度的影响 Fig. 9 Effect of ${{ R}_{{\bf{wU}}}}$ on consolidation degree at ${{ R}_{\bf{wL}}} = { 1}$ and ${{ R}_{\bf{J}}} = { 1}$

 图10 ${{ R}_{\bf{wL}}} { =} { 1}$ 和 ${{ R}_{\bf{J}}} = { 1}$ 时 ${{ R}_{{\bf{wU}}}}$ 对无量纲沉降的影响 Fig. 10 Effect of ${{ R}_{{\bf{wU}}}}$ on dimensionless settlement at ${{ R}_{\bf{wL}}} = { 1}$ and ${{ R}_{\bf{J}}} { =} { 1}$

2.3 渗透系数不同分布对固结度和沉降的影响

 图11 渗透系数分布对固结度影响 Fig. 11 Effect of distribution of permeability coefficient on consolidation degree

3 算　例

 图12 两种不同理论的超孔隙水压力计算值与实测值比较 Fig. 12 Comparison between the calculated excess pore pressure values by two different theories and measured data

 图13 两种不同理论的沉降计算值与实测值比较 Fig. 13 Comparison between the calculated settlement values by two different theories and measured data

4 结　论

1）当顶部透水系数恒定时，砂井地基的固结度随底部透水系数的增加而增大，但径向固结时间因子相同时地基固结沉降随底部透水系数的增大而减小。砂井地基的最终沉降也随底部透水系数的增大而减小。

2）当底部不透水时，砂井地基的固结度随顶部透水系数的增加而增大，径向固结时间因子相同时的地基固结沉降也随顶部透水系数的增大而增大，但最终沉降不随顶部透水系数的变化而变化；不管顶部透水和土层渗透系数如何变化，最终负孔压和最终有效应力不会随深度发生衰减。当底部为半透水边界且透水系数恒定时，砂井地基的固结度随顶部透水系数的增加而增大，径向固结时间因子相同时的地基固结沉降和地基的最终沉降也随顶部透水系数的增大而增大。

3）数值分析了渗透系数不同分布模式对固结度的影响规律，结果表明：涂抹区的渗透系数向原状土的渗透系数靠拢得越快，土层的固结度越大。