浙江大学学报(工学版), 2020, 54(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,

Structural analysis of shield tunnel lining using theory of curved beam resting on elastic foundation

HUANG Wei-ming,, WANG Jin-chang, XU Ri-qing, YANG Zhong-xuan, XU Rong-qiao,

通讯作者: 徐荣桥,男,教授. orcid.org/0000-0002-0005-9737. E-mail: xurongqiao@zju.edu.cn

收稿日期: 2019-03-26  

Received: 2019-03-26  

作者简介 About authors

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 , E-mail:10912032@zju.edu.cn

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.

Keywords: tunnel lining ; elastic foundation ; curved beam theory ; state space method ; joint of lining segment

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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. 浙江大学学报(工学版)[J], 2020, 54(4): 787-795 doi:10.3785/j.issn.1008-973X.2020.04.018

HUANG Wei-ming, WANG Jin-chang, XU Ri-qing, YANG Zhong-xuan, XU Rong-qiao. Structural analysis of shield tunnel lining using theory of curved beam resting on elastic foundation. Journal of Zhejiang University(Engineering Science)[J], 2020, 54(4): 787-795 doi:10.3785/j.issn.1008-973X.2020.04.018

盾构隧道工法因具有经济、对地面影响小等优点而被广泛应用于城市市政交通工程建设当中. 衬砌结构分析方法多种多样,包括有限元法、有限差分法[1-3]. 这些方法能够考虑诸如施工、复杂接触关系等因素的影响,为模拟盾构隧道的性状提供了强有力的工具. 但与此同时,由于精细化建模、复杂本构模型参数的确定和计算所耗费的资源多、时间长等缺点限制了其在工程中尤其是初始设计阶段的应用. 而简化力学分析模型[4-11]具有物理概念清晰,计算效率高等优点,受到工程界的广泛欢迎.

梁弹簧模型[5, 10, 11]是一类应用广泛的简化力学分析模型. 朱合华等[5]从卡式第二定理出发,推导了梁弹簧模型的刚度矩阵. Lee等[10]通过推导衬砌受力的力法方程,求解了接头对称分布时,一种特定荷载形式下的梁弹簧模型的解析解. Do等[11]总结了已有的解析结果,指出这些解析解多只适用于衬砌接头分布对称的情况,通过共轭梁理论,将管片接头转动弹簧的影响引入单元刚度矩阵中,得到适用于任意接头分布情况的数值解,但未考虑接头处的变形的非连续性的影响. 综上,对弹性地基上的盾构隧道衬砌受力分析的梁弹簧模型通用意义上的解析研究相对欠缺,笔者认为有必要对这类应用广泛的模型展开解析研究.

状态空间法[12]通过将能量对偶的两类物理量如内力与位移作为状态变量,使得描述结构力学行为的复杂控制方程变为简洁的矩阵形式,具有计算效率高,便于编程求解的特点,被广泛的应用于工程领域[13-15],但在隧道结构分析中的应用还尚未见诸于报道.

本文基于欧拉曲梁理论,采用状态空间法,得到了弹性地基上连续曲梁的首末端状态变量间的矩阵传递关系. 根据梁弹簧模型对接头力学行为的线性化假定,得到了接头处状态变量的矩阵传递关系. 通过推导,最终将问题归结为求解关于初始截面内力和变形的6个线性代数方程. 本文的解析解具有形式简洁,计算效率高,便于编程求解的优点,并且可适用于任意的荷载和接头分布形式.

1. 问题描述

图1所示,盾构隧道衬砌在横向上是由连续管片和管片接头组成的整体结构. 主要假定如下:1)正常使用状态下,衬砌变形为小变形问题,考虑到管片的几何尺寸和形状特点,采用欧拉曲梁理论[16]描述管片的力学行为;2)考虑接头处位移的非连续性,采用径向、轴向和转动方向的集中弹簧模拟接头的力学行为;3)采用径向和切向的Winkler弹簧模拟衬砌和地层的相互作用.

图 1

图 1   衬砌结构分析模型

Fig.1   Model scheme of segmental lining


2. 盾构隧道衬砌环解析模型

2.1. 弹性地基上的曲梁模型

弹性地基上单个连续曲梁段受力如图2所示,衬砌中曲线半径为R,管片厚度为h,曲梁分段的圆心角为[θ0θ1]. 沿衬砌中曲线建立坐标轴s轴,沿半径辐射方向建立坐标轴z轴,假设沿这2个坐标方向作用在衬砌上的任意分布荷载分别为qsqz. 径向位移记为w,环向位移记为u,弯矩记为M,剪力记为Q,轴力记为N,它们的正方向如图2所示. Winkler弹簧的径向抗力系数和环向抗力系数分别记作kzks.

图 2

图 2   弹性地基上单个连续曲梁段模型

Fig.2   Model diagram of single continuous curved beam resting on elastic foundation


根据欧拉曲梁理论[16]可知,截面上的轴力和弯矩为

$N = \iint\limits_A {{\sigma _s}{\rm{d}} A = EA\varepsilon = EA\left(\frac{{{\rm{d}} u}}{{{\rm{d}} s}} + \frac{w}{R}\right)},$

$M = - \iint\limits_A {{\sigma _s}z{\rm{d}} A} = EI\kappa = EI\frac{{{\rm{d}} \varphi }}{{{\rm{d}} s}}.$

式中:AI分别为截面的面积和抗弯惯性矩. 截面的转角φ定义为

$\varphi = \frac{{{\rm{d}} w}}{{{\rm{d}} s}} - \frac{u}{R}.$

弹性地基上曲梁的平衡方程为

$\frac{{{\rm{d}} N}}{{{\rm{d}} s}}{\rm{ + }}\frac{Q}{R} = {k_s}u - {q_s}, \frac{{{\rm{d}} Q}}{{{\rm{d}} s}} - \frac{N}{R} = {k_z}w - {q_z}.$

式中:Q为剪力, $Q = - {{{\rm{d}} M}}/{{{\rm{d}} s}} $.

把式(1)~(4)整理成矩阵形式,无量纲化可得

$\frac{{{\rm{d}} {\bar{{x}}}}}{{{\rm{d}} \theta }} = {\bar{{A}}}{\bar{{x}}} + {\bar{{q}}}.$

式中: ${\bar{{x}}}$为无量纲的状态向量,由截面的3个内力和3个位移组成,

${\bar{{x}}} = {\left[ {\begin{array}{*{20}{c}} {\bar w},&{\bar u},&\varphi, &{\bar Q},&{\bar N},&{\bar M} \end{array}} \right]^{\rm{T}} };$

${\bar{{q}}}$为衬砌受到的外荷载向量,

${\bar{{q}}} = {\left[ {\begin{array}{*{20}{c}} 0,&0,&0,&{ - {{\bar q}_z}},&{ - {{\bar q}_s}},&0 \end{array}} \right]^{\rm{T}} };$

${\bar{{A}}}$为系统矩阵,

${\bar{{A}}} = \left[ {\begin{array}{*{20}{c}} {\overline {{B}} }&{\overline {{C}} } \\ {\overline {{D}} }&{ - {{\overline {{B}} }^{\rm{T}}}} \end{array}} \right],$

其中

${\bar{{B}}} = \left[ {\begin{array}{*{20}{c}} 0&1&1 \\ { - 1}&0&0 \\ 0&0&0 \end{array}} \right],$

${\bar{{C}}} = \left[ {\begin{array}{*{20}{c}} 0&0&0 \\ 0&1&0 \\ 0&0&{{1 / {\bar I}}} \end{array}} \right], \;\;{\bar{{D}}} = \left[ {\begin{array}{*{20}{c}} {{{\bar k}_z}}&0&0 \\ 0&{{{\bar k}_s}}&0 \\ 0&0&0 \end{array}} \right].$

无量纲位移、内力和坐标为

$\left. {\begin{array}{*{20}{l}} s = R\theta ,\;\;\bar I = {I / {(A{R^2})}},\;\;w = R\bar w,\;\;u = R\bar u, \\ N = EA\bar N,\;\;Q = EA\bar Q,\;\;M = EAR\bar M,{{\bar k}_z} = \dfrac{{{R^2}}}{{EA}}{k_z}, \\ {{\bar k}_s} = \dfrac{{{R^2}}}{{EA}}{k_s},\;\;{{\bar q}_z} = \dfrac{R}{{EA}}{q_z},\;\;{{\bar q}_s} = \dfrac{R}{{EA}}{q_s}. \end{array}} \right\} $

根据矩阵理论可知,式(5)的解为

${\bar{{x}}}(\theta ) = {\bar{{T}}}(\theta - {\theta _0}){{\bar{{x}}}_0} + {\bar{{f}}}(\theta - {\theta _0}).$

式中: ${{\bar{{x}}}_0}$为状态向量 ${\bar{{x}}}$θ=θ0时的值; ${\bar{{T}}}\left( {\theta - {\theta _0}} \right)$为传递矩阵, ${\bar{{f}}}\left( {\theta - {\theta _0}} \right)$为荷载积分向量,

$ {{\bar{{T}}}(\theta - {\theta _0}) = {{\rm{exp}}\;[{{\bar{{A}}}(\theta - {\theta _0})}]},} $

${{\bar{{f}}}(\theta - {\theta _0}) = \int_{{\theta _0}}^\theta {{{\rm{exp}}\; [{{\bar{{A}}}(\theta - \xi )}]}{\bar{{q}}}(\xi ){\rm{d}} \xi } .}$

传递矩阵 ${\bar{{T}}}\left( {\theta - {\theta _0}} \right)$的具体形式见附录A. 一旦外荷载 ${\bar{{q}}}$向量的形式确定,则可以由式(14)求得 ${\bar{{f}}}$,附录B给出图B.1所示荷载的积分向量 ${\bar{{f}}}$.

图 B.1

图 B.1   衬砌荷载分布

Fig.B.1   Distribution of ground pressure acting on lining


在式(11)中取θ=θ1,得到单个曲梁段两端状态向量 ${{\bar{{x}}}_0}$${{\bar{{x}}}_1}$之间的传递关系,即

${{\bar{{x}}}_1} = {\bar{{x}}}({\theta _1}) = {\bar{{T}}}({\theta _1} - {\theta _0}){{\bar{{x}}}_0} + {\bar{{f}}}({\theta _1} - {\theta _0}).$

若只有一段管片,则式(15)中两端状态向量共12个未知数,方程数为6,需要补充两端各6个边界条件进行求解.

2.2. 接头模型

由于管片间接头的存在,衬砌的整体刚度将受到削弱. 采用集中弹簧,模拟接头的抗弯、抗剪和抗压特性. 当需要考虑接头在受到不同方向内力时的刚度差异时,可以针对不同的接头受力方向采用不同的弹簧刚度,具体可以通过多次试算实现. 假定分析对象为介于第j个曲梁段和第j+1个曲梁段间的第j个接头,根据弹簧的变形关系及内力的连续性,有

$\left. {\begin{array}{*{20}{l}} Q_0^{(j{\rm{ + }}1)} = Q_1^{(j)} = k_w^{(j)}\left( {w_0^{(j{\rm{ + }}1)} - w_1^{(j)}} \right), \\ N_0^{(j{\rm{ + }}1)} = N_1^{(j)} = k_u^{(j)}\left( {u_0^{(j{\rm{ + }}1)} - u_1^{(j)}} \right), \\ M_0^{(j{\rm{ + }}1)} = M_1^{(j)} = k_\varphi ^{(j)}\left( {\varphi _0^{(j{\rm{ + }}1)} - \varphi _1^{(j)}} \right). \end{array}} \right\} $

式中: $k_w^{\left( j \right)}$$k_u^{\left( j \right)}$$k_\varphi ^{\left( j \right)}$分别为第 $j$个接头的纵向每延米的接头弹簧剪切刚度、轴向刚度和弯曲刚度, $Q_1^{\left( j \right)}$$Q_0^{\left( {j{\rm{ + }}1} \right)}$分别为第 $j$个接头前后第 $j$段曲梁末端截面和第 $j + 1$段曲梁始端截面处的剪力, $N_1^{\left( j \right)}$$N_0^{\left( {j + 1} \right)}$$M_1^{\left( j \right)}$$ M_0^{\left( {j + 1} \right)}$分别为对应的轴力和弯矩. 式(16)可以写成无量纲矩阵形式,即

${\bar{{x}}}_0^{(j + 1)} = {{\bar{{J}}}^{(j)}}{\bar{{x}}}_1^{(j)}.$

式中: ${\bar{{x}}}_1^{\left( j \right)}$${\bar{{x}}}_0^{\left( {j + 1} \right)}$分别为第 $j$段曲梁末端截面和第 $j + 1$段曲梁始端截面的无量纲状态向量. ${{\bar{{J}}}^{\left( j \right)}}$为第 $j$个接头的无量纲传递矩阵,

${{\bar{{J}}}^{(j)}} = \left[ {\begin{array}{*{20}{c}} 1&0&0&{{1 / {\bar k_w^{(j)}}}}&0&0 \\ 0&1&0&0&{{1 / {\bar k_u^{(j)}}}}&0 \\ 0&0&1&0&0&{{1 / {\bar k_\varphi ^{(j)}}}} \\ 0&0&0&1&0&0 \\ 0&0&0&0&1&0 \\ 0&0&0&0&0&1 \end{array}} \right],$

其中

$\bar k_w^{(j)} = \frac{{k_w^{(j)}R}}{{EA}},\;\;\bar k_u^{(j)} = \frac{{k_u^{(j)}R}}{{EA}},\;\;\bar k_\varphi ^{(j)} = \frac{{k_\varphi ^{(j)}}}{{EAR}}.$

若弹簧的刚度无穷大,则表示管片之间的连接是完美的,等同于连续管片截面. 由式(18)可知,此时矩阵 ${{\bar{{J}}}^{\left( j \right)}}$变为单位矩阵.

2.3. 弹性地基上的盾构隧道衬砌环曲梁模型

如果衬砌由n个接头和n段管片组成管片环,如图3所示,结合2.1和2.2节建立的2个状态变量传递关系,从第n段管片末端截面依次传递到第1个管片起始端,可以得到第n个接头两侧截面状态向量间的传递关系如下:

图 3

图 3   衬砌环状态向量传递路线

Fig.3   Transmitting route of state vector


$\begin{array}{l} {\bar{{x}}}_1^{(n)} = {{{\bar{{T}}}}^{(n)}}{\bar{{x}}}_0^{(n)} + {{{\bar{{f}}}}^{(n)}} = {{{\bar{{T}}}}^{(n)}}{{{\bar{{J}}}}^{(n - 1)}}{\bar{{x}}}_1^{(n - 1)} + \\ {{{\bar{{f}}}}^{(n)}} = \cdots = {{H}}{\bar{{x}}}_0^{(1)} + {{p}} . \\ \end{array}$

式中:

${{H}} = {{\bar{{T}}}^{(n)}}{{\bar{{J}}}^{(n - 1)}}{{\bar{{T}}}^{(n - 1)}}\cdots{{\bar{{J}}}^{(1)}}{{\bar{{T}}}^{(1)}};\quad\quad$

${{p}} \!\!=\!\! {{{\bar{{T}}}}^{(n)}}{{{\bar{{J}}}}^{(\!n - 1\!)}}{{{\bar{{T}}}}^{(\!n - 1\!)}}\!\cdots\!{{{\bar{{T}}}}^{(2)}}{{{\bar{{J}}}}^{(1)}}{{{\bar{{f}}}}^{(1)}} \!+\!\cdots\!{\rm{ + }}\! {{{\bar{{T}}}}^{\!(\!n\!)\!}}{{{\bar{{J}}}}^{(\!n \!- \!1\!)}}{{{\bar{{f}}}}^{(\!n \!-\! 1\!)}}\! + \!{{{\bar{{f}}}}^{(\!n\!)}}; $

${{\bar{{T}}}^{(j)}} = {\bar{{T}}}(\theta _1^{(j)} - \theta _0^{(j)}),\;j = 1,2,\cdots,n;$

${{\bar{{f}}}^{(j)}} = {\bar{{f}}}(\theta _1^{(j)} - \theta _0^{(j)}),\;j = 1,2,\cdots,n.$

第1段的起始端和第n段的末端通过第n个接头相连,于是有

${\bar{{x}}}_0^{(1)} = {{\bar{{J}}}^{(n)}}{\bar{{x}}}_1^{(n)}.$

把式(25)代入式(20),可得只关于第1段管片的起始端截面处状态向量的代数方程:

$({{I}} - {\bar{{H}}}){\bar{{x}}}_0^{(1)} = {\bar{{p}}}.$

式中:

${\bar{{H}}} = {{\bar{{J}}}^{(n)}}{{H}},$

${\bar{{p}}} = {{\bar{{J}}}^{(n)}}{{p}}.$

式(26)由6个关于第1段管片起始端截面6个物理量的代数方程组成,求解后可得第1段管片始端截面处的3个内力及对应的3个位移,然后可由式(12)求得第1段管片内任意截面处的内力和位移. 结合式(17),求得第2段管片的起始端截面的状态向量,重复以上过程,可以确定所有管片分段内任意截面处的内力和位移.

3. 数值算例

3.1. 本文解析结果的验证

对文献[10]的上海地铁工程实际案例进行模拟. 为了对比验证,采用统一的模型参数,分别利用Abaqus和本文的解析解进行分析. 衬砌由6块均等的管片组成,接头的具体位置为30°、90°、150°、210°、270°和330°(角度由衬砌底部开始沿顺时针方向计算). 管片的弹性模量为35 GPa,截面高度为0.3 m,

衬砌外径为2.5 m,衬砌纵向取1.0 m计算. 在数值计算时,采用欧拉伯努利梁单元B23模拟管片,沿衬砌全圆周设置径向和环向接地弹簧单元(connect points to ground). 土弹簧的径向抗力系数kz=5 000 kPa/m,环向抗力系数ku取径向抗力系数的1/3[11]. 管片间接头的行为分别由径向、轴向和转动方向的点对点弹簧单元模拟(connect two points),纵向每延米的接头弹簧转动刚度参考文献[10]的取值范围,取5 400 kN·m/(rad·m). 考虑到实际工程中其他2个方向的刚度较大,分别取纵向每延米的接头弹簧剪切刚度和轴向刚度为kw=107 kN/m2ku=1010 kN/m2. 衬砌受到的荷载如附录B中图B.1所示,各项荷载的幅值分别为q1=165.8 kPa,q2=187.9 kPa,q3=116.1 kPa,q4=60.2 kPa和q5=7.5 kPa. 图4(a)~(c)分别给出本文解析解与Abaqus数值计算径向位移、弯矩和轴力计算结果的对比. 可以看出,两者结果非常接近,验证了本文解析解的正确性.

图 4

图 4   本文解析解与Abaqus数值解计算结果对比

Fig.4   Comparison between present analytical results and numerical results of Abaqus


通过竖向和水平向位移的平均化来估算土体反力的作用,Blom[6]给出便于计算的简化解析解. 为了与Blom[6]的简化解对比,根据假定的荷载分布模式,对算例荷载稍作修改,即令衬砌受到的竖向荷载q1=165.8 kPa,水平荷载q2=146.2 kPa,其他荷载项q3~ q5为零,模型的其他参数保持不变. 图5给出本文的精确解与简化解[6]计算结果的对比,其中本文解1表示同时考虑荷载径向和水平分量作用的精确解,本文解2 表示只考虑荷载径向分量作用的精确解. 从图5可以看出,Blom[6]的简化解的计算结果总体偏小,与本文解2的结果更接近,这是由于其简化解推导过程中只考虑了荷载的径向分量的作用.

图 5

图 5   本文解析解与简化解[6]计算结果对比

Fig.5   Comparison between present analytical results and simplified analytical results[6]


3.2. 接头弹簧刚度的影响

考察接头弹簧刚度对计算结果的影响,为了便于分析,忽略了接头弹簧刚度在不同方向弯矩作用下的刚度变化. 在下面的分析中,接头的位置分别为42°、107°、172°、188°、253°和318°,假定管片的截面几何尺寸h和材料参数E以及衬砌的半径R与算例中一致,保持不变. 根据式(29)[11]可知, ${\bar k_z}$的变化只与地层的弹性模量有关.

图6(a)~(d)给出在 $ {\bar k_w}$=2.24, ${\bar k_u} $=2.24×103${\bar I} $=1.4×10−3的条件下,不同接头转动刚度下,无量纲衬砌内力和收敛位移随无量纲土弹簧径向抗力系数 ${\bar k_z}$的变化曲线. 图中, $\overline N_{\rm{max}} $为无量纲最大轴力; $\overline M_{\max} $为无量纲最大弯矩; ${\bar \delta _{\rm v}} = {\delta _{\rm v}}/R$${\bar \delta _{\rm h}} = {\delta _{\rm h}}/R$,其中δvδh分别为衬砌的竖向和水平向中心线上的收敛位移. 可以看出,随着 ${\bar k_z}$的增大,衬砌的收敛位移和内力均逐渐减小,两者的变化曲线由平缓变得陡直,而后再次变得平缓,并最终趋近于零. 根据国际隧协ITA[17]的阐述,适用盾构法的典型地层的土体弹性模量为2.5~125 MPa,如图6(a)~(d)中灰色填充区域所示(此处式(29)中土体泊松比取0.3). 在该范围内,接头转动刚度的差异对衬砌弯矩和变形的影响显著,对轴力的影响基本可以忽略不计. 当转动刚度较大时,相应的收敛位移较小而弯矩较大,体现了刚性接头对位移的限制作用.

图 6

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

Fig.6   Curves of normalized maximum internal forces and convergences varying with normalized soil reaction coefficient under different normalized rotational joint stiffness


${k_z} = \frac{{2{E_{\rm{s}}}}}{{(1 + v)R}}.$

图7(a)~(d)给出在 $ {\bar k_\varphi}$=2.19×10−4${\bar k_u} $=2.24×103${\bar I} $=1.4×10−3的条件下,不同接头剪切刚度下,无量纲衬砌内力和收敛位移随 ${\bar k_z}$的变化曲线. 在典型地层范围内,与接头的转动刚度差异的影响相比,接头剪切刚度的差异对衬砌弯矩和收敛位移的影响相对较小.

图 7

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

Fig.7   Curves of normalized maximum internal forces and convergences varying with normalized soil reaction coefficient under different normalized shear joint stiffness


图8(a)~(d)给出在 $ {\bar k_\varphi}$=2.19×10−4${\bar k_w} $=2.24, ${\bar I} $=1.4×10−3的条件下,不同接头轴向刚度下,无量纲衬砌内力和收敛位移随 ${\bar k_z}$的变化曲线. 在典型地层范围内,接头轴向刚度的变化对衬砌轴力和位移的影响显著,对弯矩的影响相对较小. 在地层荷载作用下,衬砌整体处于受压状态,结构有内缩的趋势,水平和竖向荷载的差异使得结构有竖缩而横扩(即“横鸭蛋”形变形)的趋势,这2种趋势的共同作用决定衬砌的变形形态. 衬砌的水平收敛位移为正表示衬砌在水平中心线处的变形为横扩(即传统“横鸭蛋”型变形),反之,为内缩. 不同的接头轴向刚度水平下,衬砌水平收敛位移曲线表现出不同的变化规律. 依据曲线的特征,水平收敛位移曲线可以划分为以下4类.

图 8

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

Fig.8   Curves of normalized maximum internal forces and convergences varying with normalized soil reaction coefficient under different normalized axial joint stiffness


1)当接头轴向刚度无穷大时,此时衬砌轴向刚度无削弱,内缩趋势微小,衬砌的水平变形由横扩趋势主导,衬砌水平收敛位移随 ${\bar k_z}$增大而逐渐减小并最终趋近于零,在 ${\bar k_z}$的变化域内衬砌的水平中心线处变形始终表现为外扩.

2)当无量纲接头轴向刚度大于1.2×10−2时,由于接头轴向刚度的削弱加剧了衬砌的内缩趋势,但横扩趋势占主导地位. 当 ${\bar k_z}$较小时,衬砌水平变形为横扩,且随着 ${\bar k_z}$的增大横扩量减小,衬砌变形逐渐转为内缩,到达内缩极值后内缩量开始逐渐减小并趋近于零. 这是由于土弹簧的存在对于2种趋势的发展均有限制作用,但对于横扩趋势发展的限制作用更加明显,且这种限制作用随 ${\bar k_z}$的增大而增大.

3)当无量纲化接头轴向刚度介于1.2×10−2和4.8×10−3时,接头轴向刚度的进一步削弱加剧了衬砌的内缩趋势,内缩趋势开始发挥主要作用. 当 ${\bar k_z}$较小时,衬砌水平变形为内缩,且内缩量先增大后减小并最终趋近于零.

4)当无量纲化接头轴向刚度小于4.8×10−3时,接头轴向刚度的削弱过多,导致内缩趋势对衬砌变形起控制作用,此时衬砌的水平变形始终为内缩,且随着 ${\bar k_z}$的增大内缩量逐渐减小趋近于零. 综上所述,当接头轴向的构造出现误差或者破坏而造成轴向刚度的软化时,即使荷载未发生变化,也将出现有别于“横鸭蛋”形盾构隧道变形模式的情况. 接头轴向刚度临界值为1.2×10−2,在该刚度下,衬砌的内缩趋势和横扩趋势相当,表现为当 ${\bar k_z}$较小时,衬砌的水平收敛位移接近零.

4. 结 论

(1)在盾构隧道应用的典型地层范围内,相比其他2个方向的接头刚度,接头剪切刚度对衬砌内力和位移的影响可以忽略. 接头转动刚度对衬砌弯矩的影响显著,接头轴向刚度对衬砌的轴力影响显著.

(2)衬砌的变形形态由衬砌整体受压内缩趋势以及竖向和水平向荷载差异导致的竖缩横扩趋势共同决定. 当竖缩横扩趋势起主导作用时,衬砌表现为传统的“横鸭蛋型”变形形态;反之,衬砌表现为整体内缩变形形态.

(3)接头轴向刚度的削弱,将导致内缩作用的加剧,存在临界接头轴向刚度,在该刚度下衬砌在较小的土弹簧径向抗力系数下,已表现为整体内缩变形形态.

(4)土弹簧的存在对以上2种趋势的发展起限制作用,随着土弹簧抗力系数的增大,这种限制作用愈加明显,且对竖缩横扩趋势发展的限制作用更显著,这会导致衬砌由原来的“横鸭蛋型”变形形态发展为整体内缩变形形态.

附录A

矩阵 ${\bar{{T}}}$的表达式

$\tag{A.1}\begin{split} & {\bar{{T}}}(\theta - {\theta _0}) = \sum\limits_{m = 0}^2 {\left\langle {\sum\limits_{i = 1}^3 {\left\{ {{\alpha _{m,i}}{{{\bar{{A}}}}^{2m}}\cosh \left[ {{\lambda _i}\left( {\theta - {\theta _0}} \right)} \right]} \right\}} } \right\rangle } {\rm{ + }} \\ & \sum\limits_{m = 0}^2 {\left\langle {\sum\limits_{i = 1}^3 {\left\{ {{\beta _{m,i}}{{{\bar{{A}}}}^{2m{\rm{ + }}1}}\sinh \left[ {{\lambda _i}\left( {\theta - {\theta _0}} \right)} \right]} \right\}} } \right\rangle } . \\ \end{split} $

矩阵 ${\bar{{A}}}$有3对互为相反数的特征根,式(A.1)中的λi $\left( {i = 1,2,3} \right)$为每对特征根的其中一个. 系数 ${\alpha _{m,i}}$${\beta _{m,i}}\left( {m = 0,1,2} \right)$的具体表达式如下:

$\tag{A.2}\begin{aligned} {\alpha _{0,i}} = \frac{{\sum\limits_{(j1,j2) \in {U_{i,2}}} {\left( {\lambda _{j1}^2\lambda _{j2}^2} \right)} }}{{\prod\limits_{j \in {U_{i1}}} {\left( {\lambda _i^2 - \lambda _j^2} \right)} }},\;{\beta _{0,i}}{\rm{ = }}\frac{{{\alpha _{0,i}}}}{{{\lambda _i}}},\; {i = 1,2,3}; \end{aligned} $

$\tag{A.3}\begin{aligned} {\alpha _{1,i}} = - \frac{{\sum\limits_{(j1) \in {U_{i,1}}} {\left( {\lambda _{j1}^2} \right)} }}{{\prod\limits_{j \in {U_{i1}}} {\left( {\lambda _i^2 - \lambda _j^2} \right)} }},\;\;{\beta _{1,i}}{\rm{ = }}\frac{{{\alpha _{1,i}}}}{{{\lambda _i}}},{i = 1,2,3} ; \end{aligned}$

$\tag{A.4}\begin{aligned} {\alpha _{2,i}} = \frac{1}{{\prod\limits_{j \in {U_{i1}}} {\left( {\lambda _i^2 - \lambda _j^2} \right)} }},\;\;{\beta _{2,i}}{\rm{ = }}\frac{{{\alpha _{2,i}}}}{{{\lambda _i}}},\; {i = 1,2,3}. \end{aligned}$

${{{U}}_{i,k}}$为集合 $\{1,2,3\}$去除i后,从中取k个元素形成的所有组合的集合,比如集合 ${{{U}}_{1,2}}$表示从集合 $\{1,2,3\}$中去除1后取2个元素形成的组合,故只包括1种组合,即 ${{{U}}_{1,2}} = \left\{ {\left( {2,3} \right)} \right\}$${{{U}}_{1,1}}$有2种组合,即 ${{{U}}_{1,1}} = \left\{ {\left( 2 \right),\left( 3 \right)} \right\}$.

附录B

将式(A.1)代入式(14),可以得到荷载积分向量 ${\bar{{f}}}\left( {\theta - {\theta _0}} \right)$的具体表达式:

$\tag{B.1}\begin{aligned} & {\bar{{f}}}(\theta - {\theta _0}) =\\ & \sum\limits_{m = 0}^2 {\left( {\sum\limits_{i = 1}^3 {{\alpha _{m,i}}{{{\bar{{A}}}}^{2m}}{{[ {0,0,0, - {f_{z,{\rm ch},i}}\left( \theta \right),}}}}}\right.}- {f_{s,{\rm ch},i}}\left( \theta \right),0 ]^{\rm T} \Bigg) +\\& \sum\limits_{m = 0}^2 {\left( {\sum\limits_{i = 1}^3 {{\alpha _{m,i}}{{{\bar{{A}}}}^{2m{\rm{ + }}1}}{{\left[ {0,0,0, - {f_{z,{\rm sh},i}}\left( \theta \right), - {f_{s,{\rm sh},i}}\left( \theta \right),0} \right]}^{\rm T}}} } \right)}. \\ \end{aligned} $

式中:

$\tag{B.4} \left. \begin{split} &{f_{z,\;{\rm ch},i,j}}{\rm{ = }}{\bar q_j}\frac{{{\eta _{z,\;{\rm ch},i,j}}}}{{{B_{z,\;{\rm ch},i,j}}}},\;\;{f_{s,\;{\rm ch},i,j}}{\rm{ = }}{\bar q_j}\frac{{{\eta _{s,\;{\rm ch},i,j}}}}{{{B_{s,\;{\rm ch},i,j}}}}, \\& {\eta _{z,\;{\rm ch},i,j}}\;,\;{\eta _{s,\;{\rm ch},i,j}} = {\beta _1} + {\beta _2}\cosh\;[{\lambda _i}(\theta - {\theta _0})] + \\& {\beta _3}\sinh\;[{\lambda _i}(\theta - {\theta _{\rm{0}}})]. \end{split} \right\} $

$\tag{B.5} \left. \begin{split} &{f_{z,\;{\rm sh},i,j}}{\rm{ = }}{\bar q_j}\frac{{{\eta _{z,\;{\rm sh},i,j}}}}{{{B_{z,\;{\rm sh},i,j}}}},\;\;{f_{s,\;{\rm sh},i,j}}{\rm{ = }}{\bar q_j}\frac{{{\eta _{s,\;{\rm sh},i,j}}}}{{{B_{s,\;{\rm sh},i,j}}}},\\ & {\eta _{z,\;{\rm sh},i,j}}\;,\;{\eta _{s,\;{\rm sh},i,j}} = {\beta _1} + {\beta _2}\cosh\;[{\lambda _i}(\theta - {\theta _0})]{\rm{ + }} \\ & {\beta _3}\sinh\;[{\lambda _i}(\theta - {\theta _{\rm{0}}})]. \end{split} \right\} $

图B.1给出常用的衬砌受到的地层荷载分布形式[10]. 图中,荷载方向沿垂直和水平方向. 为了计算 ${\bar{{f}}}\left( {\theta - {\theta _0}} \right)$,分别将其沿径向和环向进行分解,计算如表B.1后2栏所示的积分,将这些积分叠加得到:与第i个特征根相关的双曲余弦函数和z方向荷载乘积的积分 ${f_{z,{\rm ch},i}}\left( \theta \right)$,与第i个特征根相关的双曲正弦函数和z方向荷载乘积的积分 ${f_{z,{\rm sh},i}}\left( \theta \right)$以及其他2个意义类似的积分 ${f_{s,{\rm ch},i}}\left( \theta \right)$${f_{s,{\rm sh},i}}\left( \theta \right)$. 将这些积分代入式(B.1),可得 ${\bar{{f}}}\left( {\theta - {\theta _0}} \right)$.

表 B.1   各荷载相关积分

Tab.B.1  

${\bar q_j}$ 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} $

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