预氧化—稳定化—固化联合修复As(Ⅲ)污染土

张文杰; 余海生; 蒋墨翰

doi:10.11779/CJGE20220279

预氧化—稳定化—固化联合修复As(Ⅲ)污染土 English Version

上海大学力学与工程科学学院, 上海 200444

基金项目:

国家自然科学基金项目 41772300

详细信息

作者简介:
张文杰(1978—)，男，博士，教授，主要从事环境岩土工程方面的研究工作。E-mail: zhwjlyl@163.com

中图分类号: TU411
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出版历程
- 收稿日期: 2022-03-15
- 网络出版日期: 2023-02-15
- 刊出日期: 2023-05-31

Combined remediation of As(Ⅲ)-contaminated soils by pre-oxidation, stabilization and solidification

School of Mechanics and Engineering Science, Shanghai University, Shanghai 200444, China

摘要

摘要: 固化/稳定化是重金属污染土治理的最常用方法，但因为As(Ⅲ)的迁移性强，固化处理As(Ⅲ)污染土的效果欠佳。提出了As(Ⅲ)污染土的预氧化—稳定化—固化联合修复方法，首先利用Fenton试剂将土中As(Ⅲ)氧化为迁移性弱的As(Ⅴ)，再用FeCl₃稳定As(Ⅴ)，最后通过水泥固化进一步固定As(Ⅴ)。通过无侧限抗压强度试验、毒性浸出试验（TCLP）、合成沉降浸出试验（SPLP）、pH测试、连续萃取试验和光谱分析研究了联合修复的效果和机理。结果表明，Fenton预氧化能够有效提高修复效率，当Fe与As的摩尔比为1︰1、添加10%水泥时，TCLP和SPLP得到As的浸出毒性分别低至2.51，1.33 mg/L，修复效率分别高达97.46%，98.53%；FeCl₃能促进水泥的水化并改善固化体的孔隙结构，提高固化体的强度；修复方法可将As转化为更稳定的形态，有效降低As的潜在环境风险；大多数As可以与铁的水合氧化物结合而被固定，但随着水泥用量增加pH增加，Fe-As的结合减弱，导致部分As释放；光谱分析结果表明，联合修复可以将土中92.5%的As(Ⅲ)转化为As(Ⅴ)，并通过水化硅酸钙的固封和钙矾石的离子交换将As固定。该研究为As(Ⅲ)污染土的高效修复提供了新的视角。
- As(Ⅲ)污染土 /
- 预氧化 /
- 稳定化 /
- 固化 /
- 浸出毒性
Abstract: The solidification/stabilization (S/S) is the most popular method for treatment of heavy metal-contaminated soils, however, the S/S treatment of As(Ⅲ)-contaminated soils is not effective due to the high mobility of As(Ⅲ). A combined remediation technique is proposed, in which As(Ⅲ) is first oxidized to As(Ⅴ) by the Fenton reagent, then stabilized by FeCl₃ and finally stabilized by cement. The unconfined compressive strength tests, toxicity characteristic leaching procedure (TCLP), synthetic precipitation leaching procedure (SPLP), pH measurements, sequential extraction procedure and spectroscopic investigations are carried out to investigate the effects and mechanism of the proposed technique. The results show that the Fenton pre-oxidation process significantly improves the remediation efficiency. Under an Fe-to-As molar ration of 1:1 and a cement dosage of 10%, the leaching toxicity of As in TCLP and SPLP is as low as 2.51 and 1.33 mg/L, and the immobilization efficiency reaches 97.46% and 98.53%, respectively. The hydration degree of the cement and the pore structure of the curing body are improved by FeCl₃ and therefore the strength increases. The combined remediation can transform As to more stable phases and effectively reduce the potential environmental risk. The majority of As is bound to hydrous oxides of Fe, but an increase in pH due to the increasing cement dosage will affect the Fe-As binding and cause potential release of As. The spectroscopic investigations show that the proposed remediation can transform 92.5% of As(Ⅲ)to As(Ⅴ) and immobilize As by the encapsulation of calcium silicate hydrate and the ion exchange of ettringite. This study provides a new insight into the effective remediation of As(Ⅲ)-contaminated soils.
- As(Ⅲ)-contaminated soil /
- pre-oxidation /
- stabilization /
- solidification /
- leaching toxicity

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0. 引言

瑞利波传播过程中主要引起水平方向振动，这种振动与建筑物的自然频率相近时，会引起共振，导致更大的破坏。为进一步明确瑞利波传播特性和衰减规律，大量学者对其进行理论分析^[1]、数值模拟^[2]和试验研究^[3]。瑞利波在介质中的传播受多种因素影响，主要包括介质的弹性参数、密度、重力以及分层结构等^[4]。

桩基在瑞利波作用下的动力响应的研究成为抗震过程中的重要组成部分。Yang等^[5]在考虑桩顶柔性约束的情况下，研究了非饱和土-桩体系在瑞利波作用下的动力响应。此外Cai等^[6]采用刚性排桩作为隔离瑞利波的方法，考虑了软土地质所产生的影响。

通过以上讨论发现：考虑竖向荷载影响的瑞利波作用下桩水平动力响应分析比较复杂。且对于软土地基中桩在瑞利波作用下的动力响应的研究也相对较少。因此，本文建立考虑竖向荷载对瑞利波作用下饱和软土地基中桩的水平动力响应影响的计算模型。研究结果为瑞利波作用下桩基动力分析和设计提供理论指导。

1. 计算模型

1.1 问题描述

在瑞利波的作用下饱和土-桩体系的数学模型如图 1所示，其中桩长为L，半径为r_a，截面积为A_p，密度为ρ_p，杨氏模量为E_p，剪切模量为G_p。桩体模型简化为桩端为固定边界条件的Timoshenko梁，桩顶视为质量为M₀的刚性块体。桩周饱和土视为线弹性材料，泊松比、阻尼比和剪切模量分别取为 ${\nu _0}$ ， ${\varepsilon _0}$ ，s和G_s。其中瑞利波的作用方式是以波的形式穿过土体传播到桩体上。

图 1 瑞利波作用下桩在饱和土中的计算模型

Figure 1. Computational model for pile in saturated soils under Rayleigh waves

下载: 全尺寸图片幻灯片

1.2 基本方程

根据Biot’s理论，运动方程可以表示为

${\sigma _{ij, j}} = {\rho _{\text{s}}}{\ddot u_i} + {\rho _{\text{f}}}{\ddot w_i} \text{, }\;$

(1)

$- {p_{{\text{f}}, i}} = {\rho _{\text{f}}}{\ddot u_i} + {m_1}{\ddot w_i} + {r_1}{\dot w_i} 。$

(2)

式中： ${u_i}$ 为土体位移； ${w_i}$ 为流体位移； $\lambda$ ，μ为拉梅常数；m₁= ${\rho _{\text{f}}}$ /n为孔隙中流体密度与孔隙率的比值；r₁= ${\rho _{\text{f}}}$ g/k_d，k_d为土壤达西渗透系数， ${p_{\text{f}}}$ 为孔隙中流体压力；α和M分别为两相材料的Biot’s参数。

基于Timoshenko理论，建立桩段控制微分方程^[7]：

$\begin{array}{l} {A_{\text{p}}}{G_{\text{p}}}{k_{\text{0}}}\left( {\frac{{{\partial ^2}{u_{\text{p}}}}}{{\partial {z^2}}} - \frac{{\partial {\theta _{\text{p}}}}}{{\partial z}}} \right) + {\rho _{\text{p}}}{A_{\text{p}}}\frac{{{\partial ^2}{u_{\text{p}}}}}{{\partial {t^2}}} +\\ \;\;\;\;\;\;\;\;\;\;\;\;({k_{\text{h}}} + {\text{i}}{c_{\text{h}}})({u_{\text{p}}} - {u_{\text{r}}}\cos \theta ) = 0 \text{, }\; \end{array}$

(3)

$\begin{array}{l} {A_{\text{p}}}{G_{\text{p}}}{k_{\text{0}}}\left( {\frac{{\partial {u_{\text{p}}}}}{{\partial z}} - {\theta _{\text{p}}}} \right) - {\rho _{\text{p}}}{I_{\text{p}}}\frac{{{\partial ^2}{u_{\text{p}}}}}{{\partial {t^2}}} - \\ \;\;\;\;\;\;\;\;\;\;\;\;{E_{\text{p}}}{I_{\text{p}}}\frac{{{\partial ^2}{\theta _{\text{p}}}}}{{\partial {{\text{z}}^2}}} + {A_{\text{p}}}P\frac{{\partial {u_{\text{p}}}}}{{\partial z}} = 0 。 \end{array}$

(4)

式中： ${u_{\text{p}}}(z, t) = {\bar u_{\text{p}}}(z){{\text{e}}^{{\text{i}}\omega t}}$ ， ${\theta _{\text{p}}}(z, t) = {\bar \theta _{\text{p}}}(z){{\text{e}}^{{\text{i}}\omega t}}$ 分别为桩的水平位移和转角；k₀为Timoshenko梁理论中的修正剪切因子；P为由上部结构质量引起的竖向荷载。

2. 方程的求解

由式（2）得：

${w_{\text{r}}} = \frac{{{\omega ^2}{\rho _{\text{f}}}{u_{\text{r}}}}}{\vartheta } - \frac{1}{\vartheta }\frac{{\partial {p_{\text{f}}}}}{{\partial r}} \text{, }\; {w_\theta } = \frac{{{\omega ^2}{\rho _{\text{f}}}{u_\theta }}}{\vartheta } - \frac{1}{\vartheta }\frac{{\partial {p_{\text{f}}}}}{{r\partial \theta }} 。$

(5)

式中： $\eta _{\text{a}}^2 = 2/(1 - \nu )$ ， $\vartheta = {\text{i}}\omega {\rho _{\text{f}}}g/{k_{\text{d}}} - {m_1}{\omega ^2}$ 。

将式（5）代入式（1），（2）结合本构方程得

$\begin{array}{c} \eta _{\text{b}}^{\text{2}}\frac{\partial }{{\partial r}}\left[ {\frac{1}{r}\left( {\frac{{\partial (r{u_r})}}{{\partial r}} + \frac{{\partial {u_\theta }}}{{\partial \theta }}} \right)} \right] - \frac{1}{{{r^2}}}\frac{\partial }{{\partial \theta }}\left[ {\frac{{\partial (r{u_\theta })}}{{\partial r}} - \frac{{\partial {u_r}}}{{\partial \theta }}} \right] +\\ \frac{{{\partial ^2}{u_r}}}{{\partial {z^2}}} + {c_{\text{1}}}\frac{{{\rho _{\text{s}}}{\omega ^2}}}{{{G_s}}}{u_r} - {c_{\text{2}}}\frac{{\partial {p_{\text{f}}}}}{{\partial r}} = 0 \text{, }\; \end{array}$

(6)

$\begin{array}{c} \frac{\partial }{{\partial r}}\left[ {\frac{1}{r}\left( {\frac{{\partial (r{u_\theta })}}{{\partial r}} - \frac{{\partial {u_{\text{r}}}}}{{\partial \theta }}} \right)} \right] + \eta _{\text{b}}^{\text{2}}\frac{1}{{{r^2}}}\frac{\partial }{{\partial \theta }}\left[ {\frac{{\partial (r{u_r})}}{{\partial r}} + \frac{{\partial {u_\theta }}}{{\partial \theta }}} \right] +\\ \frac{{{\partial ^2}{u_\theta }}}{{\partial {z^2}}} + {c_{\text{1}}}\frac{{{\rho _{\text{s}}}{\omega ^2}}}{{{G_{\text{s}}}}}{u_\theta } - {c_{\text{2}}}\frac{{\partial {p_{\text{f}}}}}{{r\partial \theta }} = 0 。 \end{array}$

(7)

式中： ${c_{\text{1}}} = 1 + \frac{{{\omega ^2}\rho _{\text{f}}^2}}{{{\rho _{\text{s}}}\vartheta }};{c_{\text{2}}} = \frac{\alpha }{{2\mu }}\frac{{1 - 2\nu }}{{1 - \nu }} + \frac{{{\omega ^2}{\rho _{\text{f}}}}}{{{\mu _{\text{s}}}\vartheta }};$ $\eta _{\text{b}}^{\text{2}} = \frac{{2 - \nu }}{{1 - \nu }}$

忽略流体的垂直位移：

$\begin{array}{l} \left( {\frac{{{\omega ^{\text{2}}}{\rho _{\text{f}}}}}{\vartheta } - \alpha \frac{{1 - 2{\nu _{\text{0}}}}}{{1 - {\nu _{\text{0}}}}}} \right)\left( {\frac{{\partial {u_r}}}{{\partial r}} + \frac{{{u_r}}}{r} + \frac{1}{r}\frac{{\partial {u_\theta }}}{{\partial \theta }}} \right) \\ \;\;\;\;\;\;\;\;\;\;\;\; = \frac{1}{\vartheta }{\nabla ^2}{p_{\text{f}}} + \left( {\frac{{{\alpha ^2}}}{\lambda }\frac{{{\nu _{\text{0}}}}}{{1 - {\nu _{\text{0}}}}} + \frac{1}{M}} \right){p_{\text{f}}} 。 \end{array}$

(8)

水平位移和切向位移用两个势函数Φ和Ψ表示为

${u_r} = \frac{{\partial \phi }}{{\partial r}} + \frac{1}{r}\frac{{\partial \psi }}{{\partial \theta }} \text{, }\; {u_\theta } = \frac{1}{r}\frac{{\partial \phi }}{{\partial \theta }} - \frac{{\partial \psi }}{{\partial {\text{r}}}} 。$

(9)

将式（9）代入式（6）~（8），联立得

$\eta _{\text{b}}^2{\nabla ^2}\phi + \frac{{{\partial ^2}\phi }}{{\partial {z^2}}} + {c_{\text{1}}}\frac{{{\rho _{\text{s}}}}}{{{G_{\text{s}}}}}{\omega ^2}\phi = {c_{\text{2}}}{p_{\text{f}}} \text{, }\;$

(10)

${\nabla ^2}\psi + \frac{{{\partial ^2}\psi }}{{\partial {z^2}}} + {c_{\text{1}}}\frac{{{\rho _{\text{s}}}}}{{{G_{\text{s}}}}}{\omega ^2}\psi = 0 \text{, }\;$

(11)

${\nabla }^{2}{p}_{\text{f}}+\left(\frac{\vartheta {\alpha }^{2}}{\lambda }\frac{{\nu }_{\text{0}}}{1-{\nu }_{\text{0}}}+\frac{\vartheta }{M}\right){p}_{\text{f}}=\left({\omega }^{\text{2}}{\rho }_{\text{f}}-\alpha \vartheta \frac{1-2{\nu }_{\text{0}}}{1-{\nu }_{\text{0}}}\right){\nabla }^{2}\phi 。$

(12)

采用分离变量法，设 $\phi (r, \theta , z) = {\phi _1}(r, \theta )Z(z),$ 假设 $\varphi (r, \theta , z) = {\varphi _1}(r, \theta )Z(z)$ ， $\frac{{{{\text{d}}^2}Z}}{{{\text{d}}{z^2}}} + {a^2}Z = 0$ ，将Z(z)表示为

$Z(z)=B_0 \cos (a z)+B_1 \sin (a z)。$

(13)

联立式（10）~（12）并将式（13）代入得

${\nabla ^2}{\psi _{\text{1}}} - \left[ {{a^2} - {c_{\text{1}}}\frac{{{\rho _{\text{s}}}}}{{{G_{\text{s}}}}}{\omega ^2}} \right]{\psi _{\text{1}}} = 0 。$

(14)

$({\nabla ^{\text{2}}} - \zeta _{{\text{11}}}^{\text{2}})({\nabla ^{\text{2}}} - \zeta _{{\text{12}}}^{\text{2}}){\phi _{\text{1}}} = {\text{0}}$

(15)

这里设： ${d_{11}} = \frac{{{c_1}}}{{\eta _{\text{b}}^{\text{2}}}}\frac{{{\rho _{\text{s}}}}}{{{G_{\text{s}}}}}{\omega ^2} + \frac{\vartheta }{M} - \frac{{{a^2}}}{{\eta _{\text{b}}^{\text{2}}}} + \frac{{{c_2}\alpha \vartheta }}{{\eta _{\text{b}}^{\text{2}}}}\frac{{1 - 2{\nu _{\text{0}}}}}{{1 - {\nu _{\text{0}}}}} - \frac{{{c_2}{\omega ^{\text{2}}}{\rho _{\text{f}}}}}{{\eta _{\text{b}}^{\text{2}}}}$ + $\frac{\vartheta {\alpha }^{2}}{\lambda }\frac{{\nu }_{\text{0}}}{1-{\nu }_{\text{0}}}$ ， ${d_{{\text{12}}}} = \frac{\vartheta }{{\eta _{\text{b}}^{\text{2}}}}\left[ {{c_1}\frac{{{\rho _{\text{s}}}}}{{{G_{\text{s}}}}}{\omega ^2}\left( {\frac{{{\alpha ^2}}}{\lambda }\frac{{{\nu _{\text{0}}}}}{{1 - {\nu _{\text{0}}}}} + \frac{{\text{1}}}{M}} \right) - } \right.$ $\left. {\frac{{{a^2}{\alpha ^2}}}{\lambda }\frac{{{\nu _{\text{0}}}}}{{1 - {\nu _{\text{0}}}}} - \frac{{{a^2}}}{M}} \right] \text{, }\; \zeta _{{\text{11}}}^{\text{2}} = \frac{{ - {d_{11}} + \sqrt {d_{11}^2 - 4{d_{12}}} }}{2}$ , $\zeta _{{\text{12}}}^{\text{2}} =\frac{{ - {d_{11}} - \sqrt {d_{11}^2 - 4{d_{12}}} }}{2}$ ， ${v_{\text{s}}} = \sqrt {\frac{{{G_{\text{s}}}}}{{{\rho _{\text{s}}}}}}$ ，从而得到

$({\nabla ^{\text{2}}} + k_{{\text{a1}}}^2)({\nabla ^{\text{2}}} + k_{{\text{a2}}}^2){\phi _{\text{1}}} = {\text{0}} \text{, }\;$

(16)

$({\nabla ^2} + k_{{\text{b1}}}^2){\psi _{\text{1}}} = 0 。$

(17)

通过算子分解理论和分离变量法得：

$\begin{array}{l} {\phi _{\text{1}}} = {A_{{\text{s1}}}}\exp ( - {s_1}r\sin \theta - {\text{i}}{k_{\text{R}}}r\cos \theta ) +\\ \;\;\;\;{A_{{\text{s2}}}}\exp ( - {s_2}r\sin \theta - {\text{i}}{k_{\text{R}}}r\cos \theta ) \text{, }\; \end{array}$

(18)

${\psi _{\text{1}}} = {B_{\text{s}}}\exp ( - \gamma r\sin \theta - {\text{i}}{k_{\text{R}}}r\cos \theta ) 。$

(19)

式中：A_s1，A_s2，B_s分别为与边界条件有关的待定常数；k_R，V_R=ω/k_R分别为瑞利波的复波速和相速度； ${s_i} =$ $\sqrt {k_{\text{R}}^2 - k_{{\text{a}}i}^2}$ (i=1, 2)和 $\gamma = \sqrt {k_{\text{R}}^2 - k_{{\text{b1}}}^2}$ ，分别为压缩波和剪切波对应的衰减指数； ${k}_{{a}_{1}}^{2}=-{\zeta }_{11}^{2}, {k}_{{a}_{2}}^{2}=-{\zeta }_{12}^{2}，$ $k_{{b_{\text{1}}}}^2 =$ $- {a^2} + {\kappa _{\text{1}}}\frac{{{\rho _{\text{s}}}}}{{{G_{\text{s}}}}}{\omega ^2}$ 。

因此由式（13），结合边界条件即 ${\left. {{\tau _{rz}}} \right|_{z = 0}} = 0$ ， ${\left. {{u_r}} \right|_{z = L}} = {\left. {{u_\theta }} \right|_{z = L}} = 0$ 可知

$\left. \begin{array}{l} {B}_{0}={B}_{1}=0且\mathrm{cos}(aL)=0\text{ }\text{, }\;\\ {a}_{m}=\frac{(2m-1){\rm{ \mathsf{ π}}}}{2L}, m=1, 2, 3\text{ }。 \end{array} \right\}$

(20)

$\begin{array}{l} \varphi = [{A_{{\text{s1}}}}\exp ( - {s_1}r\sin \theta - {\text{i}}{k_{\text{R}}}r\cos \theta ) +\\ \;\;{A_{{\text{s2}}}}\exp ( - {s_2}r\sin \theta - {\text{i}}{k_{\text{R}}}r\cos \theta )]\sin ({a_m}z) \text{, }\; \end{array}$

(21)

$\psi = {B_{\text{s}}}\exp ( - \gamma r\sin \theta - {\text{i}}{k_{\text{R}}}r\cos \theta )\sin ({a_m}z) 。$

(22)

将式（21）代入式（10）得

$\begin{array}{l} \;\;\;\;{p_{\text{f}}} = \left[ { - \frac{{a_m^2}}{{{c_2}}} + \frac{{{c_1}}}{{{c_2}}}{{\left( {\frac{\omega }{{{V_{\text{s}}}}}} \right)}^2}} \right]\left[ {{A_{{\text{s1}}}}\exp \left( \begin{array}{l} - {s_1}r\sin \theta \hfill \\ - {\text{i}}{k_R}r\cos \theta \hfill \end{array} \right) + } \right. \hfill \\ \;\;\;\;\left. {{A_{{\text{s2}}}}\exp \left( \begin{array}{l} - {s_2}r\sin \theta \hfill \\ - {\text{i}}{k_{\text{R}}}r\cos \theta \hfill \end{array} \right)} \right]\sin ({a_m}z) + \hfill \\ \frac{{\eta _{\text{b}}^{\text{2}}}}{{{c_2}}}(s_1^2 - k_{\text{R}}^{\text{2}}){A_{{\text{s1}}}}\exp ( - {s_1}r\sin \theta - {\text{i}}{k_{\text{R}}}r\cos \theta )\sin ({a_m}z) + \\ \frac{{\eta _{\text{b}}^{\text{2}}}}{{{c_2}}}(s_2^2 - k_{\text{R}}^{\text{2}}){A_{{\text{s2}}}}\exp ( - {s_2}r\sin \theta - {\text{i}}{k_{\text{R}}}r\cos \theta )\sin ({a_m}z) 。\end{array}$

(23)

通过式（9）结合本构方程及式（21），（22）并代入边界条件 ${\left. {{\tau _{rz}}} \right|_{z = 0}} = 0, {\left. {{u_\theta }} \right|_{z = L}} = 0$ 可知

${A_{{\text{s2}}}} = {d_1}{A_{{\text{s}}1}}, {B_{\text{s}}} = {d_2}{A_{{\text{s1}}}} 。$

(24)

联立式（9）即可求得位移转角u_r、u_θ表达式。

因此自由场饱和土在瑞利波作用下位移由体积平均原理得

${\bar u_r} = (1 - n){u_r} + n{w_r} \text{, }\;$

(25)

${\bar u_\theta } = (1 - n){u_\theta } + n{w_\theta } 。$

(26)

3. 瑞利波作用下桩的动力响应

3.1 水平阻力

桩在瑞利波作用下的动力响应由桩周土体决定，采用动力Winkler模型来描述饱和土桩体系的水平动力响应，进而计算单位桩长上的水平阻力^[8-9]。因此设单位桩长上的水平阻力为

${q_h} = ({k_h} + {i} {c_h}){u_{{\bar{\text p}}}} 。$

(27)

式中：k_h和c_h分别为Winkler模型中弹簧刚度和阻尼系数。

3.2 考虑竖向荷载影响的桩的动力响应

将式（25），（27）代入式（3），（4）并省略因子e^iωt得

$\frac{{{{\text{d}}^{\text{4}}}{{\bar u}_{\text{p}}}}}{{{\text{d}}{z^4}}} + W\frac{{{{\text{d}}^2}{{\bar u}_{\text{p}}}}}{{{\text{d}}{z^2}}} + J{\bar u_{\text{p}}} = {H_1}{A_{{\text{s1}}}}\sin ({a_m}z) \text{, }\;$

(28)

$\frac{{{{\text{d}}^4}{{\bar \theta }_{\text{p}}}}}{{{\text{d}}{z^4}}} + W\frac{{{{\text{d}}^2}{{\bar \theta }_{\text{p}}}}}{{{\text{d}}{z^2}}} + J{\bar \theta _{\text{p}}} = {H_2}{A_{{\text{s1}}}}\cos ({a_m}z) 。$

(29)

其中， $W = \frac{{{\rho _{\text{P}}}{\omega ^2}}}{{{E_{\text{p}}}}} + \frac{{{\rho _{\text{P}}}{\omega ^2}}}{{{k_{\text{0}}}{G_{\text{p}}}}} + \frac{{{A_{\text{p}}}P}}{{{E_{\text{p}}}{I_{\text{p}}}}} - \frac{{({k_{\text{h}}} + {\text{i}}{c_{\text{h}}})}}{{{k_{\text{0}}}{A_{\text{p}}}{G_{\text{p}}}}}$ ， $J =$ $\frac{{\rho _{\text{p}}^{\text{2}}{\omega ^4}}}{{{k_{\text{0}}}{G_{\text{p}}}{E_{\text{p}}}}} - \frac{{{\rho _{\text{P}}}{A_{\text{p}}}{\omega ^2}}}{{{E_{\text{p}}}{I_{\text{p}}}}} - \left( {\frac{{{\rho _{\text{P}}}{\omega ^2}}}{{{k_{\text{0}}}{A_{\text{p}}}{G_{\text{p}}}{E_{\text{p}}}}} - \frac{1}{{{E_{\text{p}}}{I_{\text{p}}}}}} \right)({k_{\text{h}}} + {\text{i}}{c_{\text{h}}})$

方程（28）、（29）对应的解设为

$\begin{array}{c} {\bar u_{\text{p}}} = {M_{\text{1}}}\cos ({\lambda _{\text{1}}}z) + {M_{\text{2}}}\sin ({\lambda _{\text{1}}}z) + {M_{\text{3}}}ch({\lambda _{\text{2}}}z) +\\ {M_{\text{4}}}sh({\lambda _{\text{2}}}z) + {b_{\text{1}}}{A_{{\text{s1}}}}\sin ({a_m}z) \text{, }\; \end{array}$

(30)

$\begin{array}{c} {\bar \theta _{\text{p}}} = {M_{\text{1}}}{\chi _{\text{1}}}\sin ({\lambda _{\text{1}}}z) + {M_{\text{2}}}{\chi _{\text{2}}}\cos ({\lambda _{\text{1}}}z) + {M_{\text{3}}}{\chi _{\text{3}}}sh({\lambda _{\text{2}}}z) +\\ {M_{\text{4}}}{\chi _{\text{4}}}ch({\lambda _{\text{2}}}z) + {b_{\text{2}}}{A_{{\text{s1}}}}\cos ({a_m}z) \text{, }\; \end{array}$

(31)

式中： ${\lambda }_{1}=\sqrt{\frac{W+\sqrt{{W}^{2}-4J}}{2}}, {\lambda }_{2}=\sqrt{\frac{-W+\sqrt{{W}^{2}-4J}}{2}}，$ M_i(1，2，3，4)为与桩的边界条件有关的待定系数。其中 ${\chi }_{\text{1}}=-{\chi }_{\text{2}}=-\frac{{k}_{0}{A}_{\text{p}}{G}_{\text{p}}{\lambda }_{\text{1}}+{A}_{\text{p}}P{\lambda }_{\text{1}}}{{k}_{0}{A}_{\text{p}}{G}_{\text{p}}+{\rho }_{\text{p}}{I}_{\text{p}}{\omega }^{2}+{E}_{\text{p}}{I}_{\text{p}}{\lambda }_{\text{1}}^{2}}，$ ${\chi _{\text{3}}} =$ ${\chi _{\text{4}}} = \frac{{{k_0}{A_{\text{p}}}{G_{\text{p}}}{\lambda _{\text{2}}} + {A_{\text{p}}}P{\lambda _{\text{2}}}}}{{{k_0}{A_{\text{p}}}{G_{\text{p}}} + {\rho _{\text{p}}}{I_{\text{p}}}{\omega ^2} - {E_p}{I_{\text{p}}}\lambda _{\text{1}}^2}},$ ${b_{\text{1}}} = \frac{{{H_{\text{1}}}}}{{a_m^4 - Wa_m^2 + J}}$ ， ${b_{\text{2}}} = \frac{{{H_2}}}{{a_m^4 - Wa_m^2 + J}}$ 。

沿桩身的弯矩和剪力可由弹性力学推导为

$\begin{array}{l} {\bar M_{\text{p}}} = {E_{\text{P}}}{I_P}\left[ {{M_1}{\chi _1}{\lambda _1}\cos ({\lambda _1}z) - {M_2}{\chi _2}{\lambda _1}\sin ({\lambda _1}z)} \right. +\\ {M}_{3}{\chi }_{3}{\lambda }_{2}ch({\lambda }_{2}z)+{M}_{4}{\chi }_{4}{\lambda }_{2}sh({\lambda }_{2}z)-{b}_{\text{2}}{a}_{m}{A}_{\text{s1}}\mathrm{sin}({a}_{m}z)]。 \end{array}$

(32)

$\begin{array}{l} {{\bar Q}_{\text{p}}} = {k_0}{A_{\text{P}}}{G_{\text{P}}}\left[ { - {M_1}({\lambda _1} + {\chi _1})\sin ({\lambda _1}z) + {M_2}({\lambda _1} - {\chi _2})} \right. \cdot \hfill \\ \cos ({\lambda _1}z) + {M_3}({\lambda _2} - {\chi _3})sh({\lambda _2}z) + {M_4}({\lambda _2} - {\chi _4})ch({\lambda _2}z) + \hfill \end{array} \left. \\ \;\;\;\;\;\;\;\;\;\;{({b_{\text{1}}}{a_m} - {b_{\text{2}}}){A_{s{\text{1}}}}\cos ({a_m}z)} \right] 。$

(33)

桩顶柔性状态下，桩端处于固定状态，桩的边界条件为：

$\left. \begin{array}{l} {\overline{M}}_{\text{p}}(z)={K}_{1}{\overline{\theta }}_{c}+{K}_{2}\left[{\overline{\theta }}_{\text{p}}(0)-{\overline{\theta }}_{c}\right]\text{, }\;z=0\text{ }\text{, }\;\\ {\overline{Q}}_{\text{p}}(z)-{A}_{\text{p}}P{\overline{\theta }}_{\text{p}}(z)+{\omega }^{2}{M}_{\text{0}}{\overline{u}}_{\text{p}}(z)=0, z=0\text{ }\text{, }\;\\ {\overline{u}}_{\text{p}}(z)=0\text{, }\;{\overline{\theta }}_{\text{p}}(z)=0, z=L\text{ }。 \end{array} \right\}$

(34)

将边界条件代入桩体水平位移、旋转角度沿桩身的弯矩和剪力表达式，由此可以导出所有未确定的待定参数M_i（i=1, 2, 3, 4），A_s1。

4. 数值分析及讨论

通过数值算例验证，分析竖向荷载及桩顶柔性约束下各参数对于桩身位移、转角和弯矩的影响。各相关参数属性： ${\rho _{\text{s}}}$ =2.7×10³ kg/m³， ${\rho _{\text{f}}}$ =2.2×10³ kg/m³，E_a=2×10⁹ Pa， ${\nu _{\text{0}}}$ =0.4，n=0.4，α=0.9，k_d=1×10^-8 m/s，K_s=3.6×10¹⁰ Pa，K_f=2×10⁹ Pa，G_s=2.5×10⁶ Pa，d=1 m，r_a=0.5 m，L=20 m， ${\rho _{\text{p}}}$ =2.5×10³ kg/m³，E_p= 2.5×10¹⁰ Pa，k₀=0.75，I_p=π/64，ν_p=0.2，A_p=0.25π，M₀=1×10⁵ kg，P=10×10⁶ Pa。引入无量纲频率a₀=ωL_a/v_s。

4.1 验证

为验证模型的准确性，将Makris^[10]解与模型进行比较。在相同条件下，对本文的计算模型进行验证。图 2将水平位移与Makris^[10]解的结果进行比较。可以看出两种解法有较高的一致性。

图 2 本文解与Makris解的比较

Figure 2. Comparison between authors' and Makris' solutions

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4.2 参数分析

图 3，4给出了瑞利波作用下，竖向荷载对单桩的水平动力响应的影响，其研究了桩长、竖向荷载大小和无量纲频率对桩基水平振动的影响。从图 3可以看出，随着竖向荷载的增加，位移、转角和弯矩都在增大，且增加的趋势也越来越大。随着深度的增加，竖向荷载改变所引起的变化不再明显，最终趋于稳定。

图 3 竖向荷载对单桩横向响应沿桩长变化的影响

Figure 3. Influences of vertical load on lateral response of a single pile along pile length

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图 4 竖向荷载对单桩水平响应随无因次频率变化的影响

Figure 4. Influences of vertical load on horizontal response of a single pile with dimensionless frequency

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图 4显示了垂直载荷对位移、转角和弯矩的影响。当频率等于固有频率时，发生共振。在共振区，当竖向荷载较小时，随竖向荷载的增大，桩顶位移和桩端弯矩整体呈增大趋势。随无量纲频率的增大，竖向荷载的改变引起水平位移和转角的变化将不再明显。

5. 结论

考虑竖向荷载对瑞利波作用下饱和软土中单桩结构水平动力响应的影响，建立瑞利波作用下饱和软土中桩顶柔性约束下单桩水平动力响应的计算模型。通过数值计算结果，得出以下3点结论。

（1）随深度的增加，位移和转角先减小后增大；弯矩整体呈减小趋势；最终接近桩端固结端时桩的水平动力响应趋于稳定。

（2）随无量纲频率的增加，位移、转角和弯矩在共振区发生共振后最终趋于稳定。

（3）随竖向荷载的增加，位移、转角和弯矩都在增大。但是这种改变所引起的变化随着深度的增加不再明显，最终趋于稳定。

图 1 修复后土样的pH

Figure 1. pH of treated soil

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图 2 修复后土样的无侧限抗压强度

Figure 2. UCS of treated soil

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图 3 仅用水泥固化时As的浸出浓度

Figure 3. Leached As concentrations after cement treatment

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图 4 联合修复后As的浸出浓度

Figure 4. Leached As concentrations after combined remediation

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图 5 不同方法修复后As的形态分布

Figure 5. Speciation of As after different treatments

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图 6 土样的XRD图谱

Figure 6. XRD patterns of specimens

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图 7 修复前后土样的XPS图谱

Figure 7. XPS spectra before and after treatment

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表 1 试验用土的金属含量

Table 1 Metal contents in test soil (单位：mg/kg)

金属元素	As	Mn	Al	Fe	Ca	Mg
含量	8.9	679	67700	13100	9650	2470

下载: 导出CSV

表 2 连续萃取试验步骤

Table 2 Sequential extraction procedure

形态	萃取剂	萃取条件	固液比	洗涤步骤
F1	(NH₄)₂SO₄ (0.05 mol/L)	震荡4 h，20 ℃	1︰25
F2	(NH₄)H₂PO₄ (0.05 mol/L)	震荡16 h，20 ℃	1︰25
F3	草酸铵缓冲液（0.2 mol/L）；pH 3.25	暗处震荡4 h，20 ℃	1︰25	草酸铵溶液(0.2 mol/L)；pH 3.25；固液比1︰12.5；暗处震荡10 min
F4	草酸铵缓冲液(0.2 mol/L)；+抗坏血酸(0.1 mol/L) pH 3.25	开盖和非避光条件下，96 ℃水浴处理30 min	1︰25	草酸铵溶液(0.2 mol/L)；pH 3.25；固液比1︰12.5；暗处震荡10 min
F5	HNO₃/H₂O₂	微波消解	1︰50

下载: 导出CSV

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0. 引言
1. 计算模型
1.1 问题描述
1.2 基本方程
2. 方程的求解
3. 瑞利波作用下桩的动力响应
3.1 水平阻力
3.2 考虑竖向荷载影响的桩的动力响应
4. 数值分析及讨论
4.1 验证
4.2 参数分析
5. 结论

预氧化—稳定化—固化联合修复As(Ⅲ)污染土 English Version

作者简介: 张文杰(1978—)，男，博士，教授，主要从事环境岩土工程方面的研究工作。E-mail: zhwjlyl@163.com

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