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浙江大学学报(工学版)
浙江大学学报(工学版)  2020, Vol. 54 Issue (4): 787-795    DOI: 10.3785/j.issn.1008-973X.2020.04.018
土木工程、交通工程     
基于弹性地基曲梁理论的盾构隧道管片分析方法
黄伟明(),王金昌,徐日庆,杨仲轩,徐荣桥*()
浙江大学 土木工程学系，浙江 杭州 310058
Structural analysis of shield tunnel lining using theory of curved beam resting on elastic foundation
Wei-ming HUANG(),Jin-chang WANG,Ri-qing XU,Zhong-xuan YANG,Rong-qiao XU*()
Department of Civil Engineering, Zhejiang University, Hangzhou 310058, China
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摘要：

基于弹性地基曲梁理论，考虑管片接头的抗弯、抗压和抗剪性能，采用状态空间法，求得圆形盾构隧道管片环在周围土压力作用下变形和内力的解析解. 通过与有限元和简化解结果的对比，验证了解析结果. 参数分析表明，在盾构工法的典型地层范围内，接头剪切刚度的变化对衬砌内力和变形的影响明显小于其他2个方向刚度的变化. 衬砌的水平变形由受压内缩趋势和荷载差异导致的横扩趋势共同决定. 接头轴向刚度的削弱加剧内缩趋势，土弹簧的存在对上述2种趋势的发展起限制作用. 当接头轴向刚度较小时，内缩趋势占主导地位，将导致衬砌向内的水平变形.
关键词： 隧道衬砌;  弹性地基;  曲梁理论;  状态空间法;  管片接头    
Abstract:

The state space method was used to find the exact solution for the deformations and internal forces of a lining subjected to surrounding ground pressure based on the theory of curved beam resting on elastic foundation and considering the rotation-resisting, compression-resisting and shear-resisting capacities of joints. The solution was validated by a comparison with the results of the finite element method and simplified analytical solution. The parameter analysis showed that the variation of shearing stiffness of the joints shows a less influence on the internal forces and deformation of the lining than that of the other two stiffnesses of joints in the typical range of soil deposits for the application of shield tunneling. The horizontal deformation of lining is codetermined by the contracting trend of the entire lining due to compressional force and the horizontal expansion trend caused by the difference between vertical and horizontal pressure acting on the lining. The degradation of axial stiffness of joint may increase the contracting trend of lining, and the existence of soil springs can limit the development of both trends. When the compressional stiffness of joints is small, the contracting trend would be dominant and lead to an inward horizontal deformation.
Key words: tunnel lining    elastic foundation    curved beam theory    state space method    joint of lining segment
收稿日期: 2019-03-26 出版日期: 2020-04-05
CLC:  U 451  
基金资助: 国家“973”重点基础研究发展规划资助项目（2015CB057801）；国家自然科学基金资助项目（51578499，51761130078）
通讯作者: 徐荣桥     E-mail: 10912032@zju.edu.cn;xurongqiao@zju.edu.cn
作者简介: 黄伟明（1986—），男，博士生，从事盾构隧道结构分析研究. orcid.org/0000-0002-9684-6574. E-mail： 10912032@zju.edu.cn
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引用本文:

黄伟明,王金昌,徐日庆,杨仲轩,徐荣桥. 基于弹性地基曲梁理论的盾构隧道管片分析方法[J]. 浙江大学学报(工学版), 2020, 54(4): 787-795.

Wei-ming HUANG,Jin-chang WANG,Ri-qing XU,Zhong-xuan YANG,Rong-qiao XU. Structural analysis of shield tunnel lining using theory of curved beam resting on elastic foundation. Journal of ZheJiang University (Engineering Science), 2020, 54(4): 787-795.

链接本文:

http://www.zjujournals.com/eng/CN/10.3785/j.issn.1008-973X.2020.04.018    或    http://www.zjujournals.com/eng/CN/Y2020/V54/I4/787

图 1  衬砌结构分析模型
图 2  弹性地基上单个连续曲梁段模型
图 B.1  衬砌荷载分布
图 3  衬砌环状态向量传递路线
图 4  本文解析解与Abaqus数值解计算结果对比
图 5  本文解析解与简化解[6]计算结果对比
图 6  不同接头转动弹簧刚度下，无量纲衬砌内力和收敛位移最大值随无量纲土弹簧径向抗力系数的变化曲线
图 7  不同接头剪切弹簧刚度下，无量纲衬砌内力和收敛位移最大值随无量纲土弹簧径向抗力系数的变化曲线
图 8  不同接头轴向弹簧刚度下，无量纲化衬砌内力和收敛位移最大值随土弹簧径向抗力系数的变化曲线
${\bar q_j}$ 关于双曲余弦函数的积分系数 关于双曲正弦函数的积分系数
$\begin{array}{l} \;\;\;\;\;\;\;\;\;\;\;{{\bar q}_1} \\ ({90^{\circ}} \leqslant \xi \leqslant {270^{\circ}}) \\ \end{array} $ $\begin{array}{l} {B_{z,\;{\rm ch},i,1}}{\rm{ = }}{\lambda _i}(\lambda _i^2 + 4),\;{\beta _1} = - {\lambda _i}\sin \;(2\theta ), \\ {\beta _2} = {\lambda _i}\sin \;(2{\theta _0}),\;{\beta _3} = - \left[ {2 + \lambda _i^2{{\cos }^2}{\theta _0}} \right] \\ \end{array} $ $\begin{array}{l} {B_{z,\;{\rm sh},i,1}} = {\lambda _i}(\lambda _i^2 + 4),\;{\beta _1} = 2 + \lambda _i^2{\cos ^2}\;\theta , \\ {\beta _2} = - \left[ {2 + \lambda _i^2{{\cos }^2}\;{\theta _0}} \right],\;{\beta _3} = {\lambda _i}\sin \;(2{\theta _0}), \\ \end{array} $
$\begin{array}{l} {B_{s,\;{\rm ch},i,1}}{\rm{ = }}2(\lambda _i^2 + 4),\;{\beta _1} = - 2\cos \;(2\theta ), \\ {\beta _2} = {\rm{2}}\cos \;(2{\theta _0}),\;{\beta _3} = {\lambda _i}\sin \;(2{\theta _0}) \\ \end{array} $ $\begin{array}{l} {B_{s,\;{\rm sh},i,1}} = 2(\lambda _i^2 + 4),\;{\beta _1} = - {\lambda _i}\sin \;(2\theta ), \\ {\beta _2} = {\lambda _i}{\rm{sin}}\;(2{\theta _0}),\;{\beta _3} = 2\cos \;(2{\theta _0}) \\ \end{array} $
$\begin{array}{l} \;\;\;\;\;\;\;\;\;\;\;{{\bar q}_2} \\ ({{\rm{0}}^{\circ}} \leqslant \xi \leqslant {\rm{9}}{{\rm{0}}^{\circ}}, \\ {\rm{ 27}}{{\rm{0}}^{\circ}} \leqslant \xi \leqslant {\rm{36}}{{\rm{0}}^{\circ}} ) \\ \end{array} $ ${\eta _{z,\;{\rm ch},i,2}} = {\eta _{z,\;{\rm ch},i,1}},\;{B_{z,\;{\rm ch},i,2}} = {B_{z,\;{\rm ch},i,1}}$ ${\eta _{z,\;{\rm sh},i,2}} = {\eta _{z,\;{\rm sh},i,1}},\;{B_{z,\;{\rm sh},i,2}} = {B_{z,\;{\rm sh},i,1}}$
${\eta _{s,\;{\rm ch},i,2}} = {\eta _{s,\;{\rm ch},i,1}},\;{B_{s,\;{\rm ch},i,2}} = {B_{s,\;{\rm ch},i,1}}$ ${\eta _{s,\;{\rm sh},i,2}} = {\eta _{s,\;{\rm sh},i,1}},\;{B_{s,\;{\rm sh},i,2}} = {B_{s,\;{\rm sh},i,1}}$
$\begin{array}{l} \;\;\;\;\;\;\;\;{{\bar q}_3} \\ ({0^{\circ}} \leqslant \xi \leqslant {360^{\circ}}) \\ \end{array} $ $\begin{array}{l} {B_{z,\;{\rm ch},i,3}} = {\lambda _i}(\lambda _i^2 + 4),\;{\beta _1} = {\lambda _i}\sin \;(2\theta ), \\ {\beta _2} = - {\lambda _i}\sin \;(2{\theta _0}),\;{\beta _3} = - [2 + \lambda _i^2{\sin ^2} \;{{\theta _0}}] \\ \end{array} $ $\begin{array}{l} {B_{z,\;{\rm sh},i,3}} = {\lambda _i}(\lambda _i^2 + 4),\;{\beta _1} = 2 + \lambda _i^2{\sin ^2}\;\theta , \\ {\beta _2} = - \left[ {2 + \lambda _i^2{{\sin }^2}\;{\theta _0}} \right],\;{\beta _3} = - {\lambda _i}\sin \;(2{\theta _0}) \\ \end{array} $
${\eta _{s,\;{\rm ch},i,3}} = - {\eta _{s,\;{\rm ch},i,1}},\;{B_{s,\;{\rm ch},i,3}} = {B_{s,\;{\rm ch},i,1}}$ ${\eta _{s,\;{\rm sh},i,3}} = - {\eta _{s,\;{\rm sh},i,1}},\;{B_{s,\;{\rm sh},i,3}} = {B_{s,\;{\rm sh},i,1}}$
$\begin{array}{l} \;\;\;\;\;\;\;{{\bar q}_4} \\ ({0^{\circ}} \leqslant \xi \leqslant {360^{\circ}} ) \\ \end{array} $ $\begin{array}{l} {f_{z,\;{\rm ch},i,4}} = {{\bar q}_4}\sum\limits_{k = 1}^3 {{{{\eta _{z,\;{\rm ch},i,4,k}}}}/{{{B_{z,\;{\rm ch},i,4,k}}}}} \\ {\eta _{{\rm{z}},\;{\rm ch},i,4,k}}{\rm{ = }}{\beta _{k,1}} + {\beta _{k,2}}\cosh\;[{\lambda _i}(\theta - {\theta _0})] +\\ \quad\quad\quad\quad {\beta _{k,3}}\sinh\;[{\lambda _i}(\theta - {\theta _0})]\;;\\ {B_{z,\;{\rm ch},i,4,1}} = 4{\lambda _i}(\lambda _i^2 + 4),\;{B_{z,\;{\rm ch},i,4,2}} = 8(\lambda _i^2 + 1),\\ {B_{z,\;{\rm ch},i,4,3}} = 8(\lambda _i^2 + 9),\;{\beta _{1,1}} = {\rm{2}}{\lambda _i}\sin \;(2\theta ),\\ {\beta _{1,2}} = - 2{\lambda _i}\sin \;(2{\theta _0}),\\ {\beta _{1,3}} = - \left[ {2\lambda _i^2{{\sin }^2}{\theta _0} + 4} \right],\\ {\beta _{2,1}} = - \sin \;\theta ,\;{\beta _{2,2}} = \sin \;{\theta _0},\\ {\beta _{2,3}} = - {\lambda _i}\cos \;{\theta _0},\;{\beta _{3,1}} = {\rm{3sin\;(3}}\theta ),\\ {\beta _{3,2}} = - {\rm{3sin\;(3}}{\theta _0}),\;{\beta _{3,3}} = {\lambda _i}\cos \;\left( {3{\theta _0}} \right) \end{array}$ $\begin{array}{l} {f_{z,\;{\rm sh},i,4}} = {{\bar q}_4}\sum\limits_{k = 1}^2 {{{{\eta _{z,\;{\rm {sh}},i,4,k}}}}/{{{B_{z.{\rm{sh}},i,4,k}}}}} \\ {\eta _{z,\;{\rm {sh}},i,4,k}} = {\beta _{k,1}} + {\beta _{k,2}}\cosh\;[{\lambda _i}(\theta - {\theta _0})] +\\ \quad\quad\quad\quad\;{\beta _{k,3}}\sinh\;[{\lambda _i}(\theta - {\theta _0})];\\ {B_{z,\;{\rm sh},i,4,1}} = 4{\lambda _i}(\lambda _i^2 + 4),\;{B_{z,\;{\rm sh},i,4,2}} = 2(\lambda _i^2 + 1)(\lambda _i^2 + 9)\\ {\beta _{1,1}} = 4 + 2\lambda _i^2{\sin ^2}\theta ,\;{\beta _{1,2}} = - [4 + 2\lambda _i^2{\sin ^2}{\theta _0}],\\ {\beta _{1,3}} = - 2{\lambda _i}\sin \;(2{\theta _0}),\\ {\beta _{2,1}} = 2{\lambda _i}{\cos ^3}\; \theta + {\lambda _i}\left( {\lambda _i^2 + 3} \right)\cos \; \theta {\sin ^2}\; \theta ,\\ {\beta _{2,2}} = - \left[ {2{\lambda _i}{{\cos }^3} \;{{\theta _0}} + {\lambda _i}\left( {\lambda _i^2 + 3} \right)\cos \;{{\theta _0}}{{\sin }^2} \;{{\theta _0}}} \right],\\ {\beta _{2,3}} = \left[ {\left( {\lambda _i^2 + 3} \right){{\sin }^3} \;{{\theta _0}} - 2\lambda _i^2\sin \;{{\theta _0}}{{\cos }^2} \;{{\theta _0}}} \right] \end{array}$
$\begin{array}{l} {f_{s,\;{\rm ch},i,4}} = {{\bar q}_4}\sum\limits_{k = 1}^3 {{{{\eta _{s,\;{\rm ch},i,4,k}}}}/{{{B_{s,\;{\rm ch},i,4,k}}}}} \\ {\eta _{s,\;{\rm ch},i,4,k}} = {\beta _{k,1}} + {\beta _{k,2}}\cosh\; \left[ {{\lambda _i}\left( {\theta - {\theta _0}} \right)} \right] + \\ \quad\quad\quad\quad \;\; {\beta _{k,3}}\sinh \left[ {{\lambda _i}\left( {\theta - {\theta _0}} \right)} \right]; \\ {B_{s,\;{\rm ch},i,4,1}} = 4(\lambda _i^2 + 4),\;{B_{s,\;{\rm ch},i,4,2}}{\rm{ = }}8(\lambda _i^2 + 1), \\ {B_{s,\;{\rm ch},i,4,3}} = 8(\lambda _i^2 + 9),\;{\beta _{1,1}} = 2\cos \left( {2\theta } \right), \\ {\beta _{1,2}} = - 2\cos \left( {2{\theta _0}} \right),\;{\beta _{1,3}} = - {\lambda _i}\sin \left( {2{\theta _0}} \right), \\ {\beta _{2,1}} = \cos \; \theta ,\;{\beta _{2,2}} = - \cos \;{{\theta _0}}, \\ {\beta _{2,3}} = - {\lambda _i}\sin \;{{\theta _0}},\;{\beta _{3,1}} = 3\cos \left( {3\theta } \right), \\ {\beta _{3,2}} = - 3\cos \left( {3{\theta _0}} \right),\;{\beta _{3,3}} = - {\lambda _i}\sin \left( {3{\theta _0}} \right) \\ \end{array} $ $\begin{array}{l} {f_{s,\;{\rm sh},i,4}} = {{\bar q}_4}\sum\limits_{k = 1}^3 {{{{\eta _{s,\;{\rm sh},i,4,k}}}}/{{{B_{s,\;{\rm sh},i,4,k}}}}} \\ {\eta _{s,\;{\rm sh},i,4,k}}{\rm{ = }}{\beta _{k,1}} + {\beta _{k,2}}\cosh\;[{\lambda _i}(\theta - {\theta _0})] + \\ \quad\quad\quad\quad {\beta _{k,3}}\sinh\;[{\lambda _i}(\theta - {\theta _0})]; \\ {B_{s,\;{\rm sh},i,4,1}} = 4(\lambda _i^2 + 4),\;{B_{s,\;{\rm sh},i,4,2}} = 8(\lambda _i^2 + 1), \\ {B_{s,\;{\rm sh},i,4,3}}{\rm{ = }}8(\lambda _i^2 + 9),\;{\beta _{1,1}} = {\lambda _i}\sin \;(2\theta ), \\ {\beta _{1,2}} = - {\lambda _i}\sin \;(2{\theta _0}),\;{\beta _{1,3}} = - 2\cos \;(2{\theta _0}), \\ {\beta _{2,1}} = {\lambda _i}\sin \;\theta ,\;{\beta _{2,2}} = - {\lambda _i}\sin \;{\theta _0}, \\ {\beta _{2,3}} = - \cos \;{\theta _0},\;{\beta _{3,1}} = {\lambda _i}\sin \left( {3\theta } \right), \\ {\beta _{3,2}} = - {\lambda _i}\sin \left( {3{\theta _0}} \right),\;{\beta _{3,3}} = - 3\cos \left( {3{\theta _0}} \right) \\ \end{array} $
$\begin{array}{l} \;\;\;\;\;\;\;{{\bar q}_5} \\ ({0^{\circ}} \leqslant \xi \leqslant {360^{\circ}}) \\ \end{array} $ $\begin{array}{l} {B_{z,\;{\rm ch},i,5}} = \lambda _i^2 + 1,\;{\beta _1} = - \sin \;\theta , \\ {\beta _2} = \sin \;{\theta _0},\;{\beta _3} = - {\lambda _i}\cos \;{\theta _0} \\ \end{array} $ $\begin{array}{l} {B_{z,\;{\rm sh},i,5}}{\rm{ = }}\lambda _i^2 + 1,\;{\beta _1} = - {\lambda _i}\cos \;\theta , \\ {\beta _2} = {\lambda _i}\cos \;{\theta _0},\;{\beta _3} = - \sin \;{\theta _0} \\ \end{array} $
$\begin{array}{l} {B_{s,\;{\rm ch},i,5}}{\rm{ = }}\lambda _i^2 + 1,\;{\beta _1} = \cos \;\theta , \\ {\beta _2} = - \cos \;{\theta _0},\;{\beta _3} = - {\lambda _i}\sin \;{\theta _0} \\ \end{array} $ $\begin{array}{l} {B_{s,\;{\rm sh},i,5}}{\rm{ = }}\lambda _i^2 + 1,\;{\beta _1} = {\lambda _i}\sin \;\theta , \\ {\beta _2} = - {\lambda _i}\sin \;{\theta _0},\;{\beta _3} = - \cos \;{\theta _0} \\ \end{array} $
表 B.1  各荷载相关积分
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