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 哈尔滨工业大学学报  2019, Vol. 51 Issue (2): 90-99  DOI: 10.11918/j.issn.0367-6234.201712022 0

### 引用本文

TU Hongliang, QIAO Chunsheng, ZHU Ju. Anisotropic characteristic and calculation of the resistant coefficient of the jointed rock mass[J]. Journal of Harbin Institute of Technology, 2019, 51(2): 90-99. DOI: 10.11918/j.issn.0367-6234.201712022.

### 文章历史

Anisotropic characteristic and calculation of the resistant coefficient of the jointed rock mass
TU Hongliang, QIAO Chunsheng, ZHU Ju
School of Civil Engineering, Beijing Jiaotong University, Beijing 100044, China
Abstract: In load-structure model for tunnel design, the rock mass resistant coefficient (k) is an important parameter which affects the behavior of the tunnel structure prominently. The anisotropy of k appears due to the presence of joints. However, no adequate efforts have been made to research the anisotropic distribution of k in jointed rock masses. Ten influencing factors including the elastic modulus of rock, Poisson's ratio, and the properties of two sets of joints were analyzed for evaluating the anisotropic distribution of k, by using orthogonal array testing strategy (OATS) and distinct element method (DEM), with Xinggongjie Station Tunnel Project on No. 2 Line of Dalian Metro as an example. The results show that the distribution curves of k were oval-shaped. The maximum value was along the direction of the two joints angle bisector. Using the variance analysis of OATS, the significant influencing factors at the level of five percent were the elastic modulus of the rock, the normal stiffness of the joints, the spacing of the joints, and the intersection angle of two sets of joints. With the increase of the ratio of tunnel diameter to joint spacing, the anisotropy coefficient of k increased first and then started to drop, which converged to one when the ratio was equal to zero or infinite. Based on the above analysis results, the elliptic function of k was derived and was verified. The engineering example shows that the resistant coefficient of jointed rock mass is obviously anisotropic, which has little influence on the axial force of lining and remarkable influence on bending moment.
Keywords: jointed rock masses     rock mass resistant coefficient     anisotropy     orthogonal array testing strategy     variance analysis     elliptical distribution function

1 工程概况

 图 1 兴工街站地理位置 Fig. 1 Location of Xinggongjie Station
 图 2 兴工街站隧道节理分布 Fig. 2 Joints schematic diagram of Xinggongjie Station Tunnel
 图 3 兴工街站隧道地质纵剖面 Fig. 3 Profile showing geological conditions of Xinggongjie Station Tunnel

2 试验方案

 图 4 径向液压枕法测围岩抗力系数的数值计算模型 Fig. 4 Numerical model of the radial hydraulic pressure pillow testing to measure rock mass resistant coefficient

 图 5 洞壁径向压力与径向位移的关系 Fig. 5 Relation curves between radial pressure and displacement at the cavity excavation interface

 $p = \frac{C}{{1 - \tan \varphi }}.$ (1)

C=0.8 MPa，φ=35°代入上式可得p=6.67 MPa，与数值计算结果一致.

2.1 正交试验

2.2 方差分析

3 k的各向异性分布特征

 图 6 不同节理间距下围岩抗力系数的分布曲线 Fig. 6 Distributing curves of rock mass resistant coefficient under different joint spacing

 图 7 各向异性系数拟合曲线 Fig. 7 Fitted curve of the anisotropy coefficient

 $\left\{ \begin{array}{l} \mathop {\lim }\limits_{\lambda \to 0} f\left( \lambda \right) = 1,\\ \mathop {\lim }\limits_{\lambda \to \infty } f\left( \lambda \right) = 1. \end{array} \right.$ (2)

 $\xi = f\left( \lambda \right) = 0.256 \cdot \lambda \cdot {{\rm{e}}^{ - 0.157 \lambda}} + 1.$ (3)
4 k的理论计算 4.1 k分布曲线的椭圆方程

 $\frac{{{t^2}}}{{k_{\max }^2}} + \frac{{{h^2}}}{{k_{\min }^2}} = 1.$ (4)
 图 8 围岩抗力系数的椭圆分布模型 Fig. 8 Ovalshape model of rock mass resistant coefficient

 $\left( {\begin{array}{*{20}{c}} t\\ h \end{array}} \right) = \left[ {\begin{array}{*{20}{c}} {\cos \delta }&{\sin \delta }\\ { - \sin \delta }&{\cos \delta } \end{array}} \right]\left( {\begin{array}{*{20}{c}} x\\ y \end{array}} \right).$ (5)

 $\frac{{{{\left[ {x\cos \delta + y\sin \delta } \right]}^2}}}{{k_{\max }^2}} + \frac{{{{\left[ {y\cos \delta - x\sin \delta } \right]}^2}}}{{k_{\min }^2}} = 1.$ (6)

 $\frac{{{k^2}{{\cos }^2}\left( {\theta - \delta } \right)}}{{k_{\max }^2}} + \frac{{{k^2}{{\sin }^2}\left( {\theta - \delta } \right)}}{{k_{\min }^2}} = 1.$ (7)

 $\frac{1}{k} = {\left[ {\frac{{{{\cos }^2}\left( {\theta - \delta } \right)}}{{k_{\max }^2}} + \frac{{{{\sin }^2}\left( {\theta - \delta } \right)}}{{k_{\min }^2}}} \right]^{\frac{1}{2}}}.$ (8)

 $\frac{1}{k} = \frac{1}{{{k_{\max }}}}{\left[ {{{\cos }^2}\left( {\theta - \delta } \right) + {\xi ^2}{{\sin }^2}\left( {\theta - \delta } \right)} \right]^{\frac{1}{2}}}.$ (9)

ξ=1时，可得k=kmax=kmin，洞周各个方向的围岩抗力系数相等，分布曲线退化为圆形，岩体为各向同性体.由上式可以看出，围岩抗力系数由各向异性系数ξ和最大值kmax决定，ξ代表椭圆的形状，而kmax代表数值的大小.

4.2 kmax的计算 4.2.1 节理面上的作用力

 图 9 最大围岩抗力系数计算模型 Fig. 9 Model of the maximal rock mass resistant coefficient

 ${\sigma _1} = \frac{{{F_{{{\rm{N}}_{\rm{1}}}}}}}{{\left| {EH} \right|}},{\tau _1} = \frac{{{F_{{{\rm{S}}_{\rm{1}}}}}}}{{\left| {EH} \right|}}.$ (10)

x方向上力的分量为σρdθcos θy方向上力的分量为σρdθsin θ.将θα1α2积分，得到$\overset\frown{CD}$段上σx方向产生的力的分量为

 $\int_{{\alpha _1}}^{{\alpha _2}} {\sigma \cdot 1 \cdot \rho {\rm{d}}\theta } \cdot \cos \theta = \sigma r\left[ {\sin {\alpha _2} - \sin {\alpha _1}} \right].$ (11)

σy方向产生的力的分量为

 $\int_{{\alpha _1}}^{{\alpha _2}} {\sigma \cdot 1 \cdot \rho {\rm{d}}\theta } \cdot \sin \theta = \sigma r\left[ {\cos {\alpha _1} - \cos {\alpha _2}} \right].$ (12)

FH边上，x方向分量为

 $- {F_{{{\rm{N}}_{\rm{1}}}}} \cdot \sin {\alpha _2} - {F_{{{\rm{S}}_1}}} \cdot \cos {\alpha _2}.$ (13)

y方向分量为

 ${F_{{{\rm{N}}_{\rm{1}}}}} \cdot \cos {\alpha _2} - {F_{{{\rm{S}}_1}}} \cdot \sin {\alpha _2}.$ (14)

EH边上，x方向分量为

 ${F_{{{\rm{N}}_{\rm{1}}}}} \cdot \sin {\alpha _1} - {F_{{{\rm{S}}_1}}} \cdot \cos {\alpha _1}.$ (15)

y方向分量为

 $- {F_{{{\rm{N}}_{\rm{1}}}}} \cdot \cos {\alpha _1} - {F_{{{\rm{S}}_1}}} \cdot \sin {\alpha _1}.$ (16)

 $\begin{array}{*{20}{c}} {\sigma r\left[ {\sin {\alpha _2} - \sin {\alpha _1}} \right] - {F_{{{\rm{N}}_{\rm{1}}}}}\sin {\alpha _2} - }\\ {{F_{{{\rm{S}}_1}}}\cos {\alpha _2} + {F_{{{\rm{N}}_{\rm{1}}}}}\sin {\alpha _1} - {F_{{{\rm{S}}_1}}}\cos {\alpha _1} = 0.} \end{array}$ (17)

y方向平衡方程为

 $\begin{array}{*{20}{c}} {\sigma r\left[ {\cos {\alpha _1} - \cos {\alpha _2}} \right] + {F_{{{\rm{N}}_{\rm{1}}}}}\cos {\alpha _2} - }\\ {{F_{{{\rm{S}}_1}}}\sin {\alpha _2} - {F_{{{\rm{N}}_{\rm{1}}}}}\cos {\alpha _1} - {F_{{{\rm{S}}_1}}}\sin {\alpha _1} = 0.} \end{array}$ (18)

σOH方向上力的分量为

 $\begin{array}{*{20}{c}} {\int_{{\alpha _1}}^{\frac{{{\alpha _1} + {\alpha _2}}}{2}} {\sigma \rho {\rm{d}}\theta \cdot \cos \left( {\frac{{{\alpha _1} + {\alpha _2}}}{2} - \theta } \right)} + }\\ {\int_{\frac{{{\alpha _1} + {\alpha _2}}}{2}}^{{\alpha _2}} {\sigma \rho {\rm{d}}\theta \cdot \cos \left( {\theta - \frac{{{\alpha _1} + {\alpha _2}}}{2}} \right)} = 2r\sigma \sin \left( {\frac{{{\alpha _2} - {\alpha _1}}}{2}} \right).} \end{array}$ (19)

FN1OH的夹角为(π+α1-α2)/2，OH方向平衡方程为

 $\begin{array}{*{20}{c}} {2r\sigma \sin \left( {\frac{{{\alpha _2} - {\alpha _1}}}{2}} \right) - 2\left[ {{F_{{{\rm{N}}_{\rm{1}}}}}\cos \left( {\frac{{{\rm{ \mathsf{ π} }} + {\alpha _1} - {\alpha _2}}}{2}} \right) + } \right.}\\ {\left. {{F_{{{\rm{S}}_{\rm{1}}}}}\sin \left( {\frac{{{\rm{ \mathsf{ π} }} + {\alpha _1} - {\alpha _2}}}{2}} \right)} \right] = 0.} \end{array}$ (20)

 $\left\{ \begin{array}{l} {F_{{{\rm{N}}_{\rm{1}}}}} = r\sigma ,\\ {F_{{{\rm{S}}_{\rm{1}}}}} = 0. \end{array} \right.$ (21)

CDGLI块体，边IL上的作用力FN2FS2，同理可得

 $\left\{ \begin{array}{l} {F_{{{\rm{N}}_{\rm{2}}}}} = {F_{{{\rm{N}}_{\rm{1}}}}},\\ {F_{{{\rm{S}}_{\rm{2}}}}} = {F_{{{\rm{S}}_{\rm{1}}}}}. \end{array} \right.$ (22)

 $\left\{ \begin{array}{l} {F_{{{\rm{N}}_{\rm{1}}}}} = {F_{{{\rm{N}}_{\rm{2}}}}} = {F_{{{\rm{N}}_{\rm{3}}}}} = \cdots = {F_{{{\rm{N}}_n}}},\\ {F_{{{\rm{S}}_{\rm{1}}}}} = {F_{{{\rm{S}}_{\rm{2}}}}} = {F_{{{\rm{S}}_{\rm{3}}}}} = \cdots = {F_{{{\rm{S}}_n}}}. \end{array} \right.$ (23)

FHILKN节理面上作用力虽然相同，但作用面积不同，因此应力不同，节理长度$|FH|=~\frac{s}{\text{sin}({{\alpha }_{2}}-{{\alpha }_{1}})}$，应力分别为

FH

 $\left\{ \begin{array}{l} {\sigma _1} = \frac{{{F_{{{\rm{N}}_{\rm{1}}}}}}}{{\left| {FH} \right|}} = \frac{{r\sigma \cdot \sin \left( {{\alpha _2} - {\alpha _1}} \right)}}{s},\\ {\tau _1} = \frac{{{F_{{{\rm{S}}_{\rm{1}}}}}}}{{\left| {FH} \right|}} = 0. \end{array} \right.$ (24)

IL

 $\left\{ \begin{array}{l} {\sigma _2} = \frac{{{F_{{{\rm{N}}_{\rm{2}}}}}}}{{\left| {IL} \right|}} = \frac{{r\sigma \cdot \sin \left( {{\alpha _2} - {\alpha _1}} \right)}}{{2s}},\\ {\tau _2} = \frac{{{F_{{{\rm{S}}_{\rm{2}}}}}}}{{\left| {FH} \right|}} = 0. \end{array} \right.$ (25)

KN

 $\begin{array}{*{20}{c}} {\left\{ \begin{array}{l} {\sigma _3} = \frac{{{F_{{{\rm{N}}_{\rm{3}}}}}}}{{\left| {KN} \right|}} = \frac{{r\sigma \cdot \sin \left( {{\alpha _2} - {\alpha _1}} \right)}}{{3s}},\\ {\tau _3} = \frac{{{F_{{{\rm{S}}_{\rm{3}}}}}}}{{\left| {KN} \right|}} = 0. \end{array} \right.}\\ \cdots \end{array}$ (26)

n

 $\left\{ \begin{array}{l} {\sigma _n} = \frac{{{F_{{{\rm{N}}_n}}}}}{{\left| {nFN} \right|}} = \frac{{r\sigma \cdot \sin \left( {{\alpha _2} - {\alpha _1}} \right)}}{{ns}},\\ {\tau _n} = \frac{{{F_{{{\rm{S}}_n}}}}}{{\left| {nFN} \right|}} = 0. \end{array} \right.$ (27)
4.2.2 M点位移的计算

 ${u_{\rm{r}}} = \frac{{r\sigma \left( {1 + \mu } \right)}}{E}.$ (28)

M点径向方向上，节理产生的变形包括该方向上所有节理法向和切向变形在该方向上的分量之和，节理EHFH的位移为

 $\begin{array}{l} {u_{{{\rm{J}}_{\rm{1}}}}} = \frac{{{\sigma _1}}}{{{k_n}}} \cdot \cos \left( {\frac{{{\rm{ \mathsf{ π} }} + {\alpha _1} - {\alpha _2}}}{2}} \right) + \frac{{{\tau _1}}}{{{k_s}}} \cdot \sin \left( {\frac{{{\rm{ \mathsf{ π} }} + {\alpha _1} - {\alpha _2}}}{2}} \right) = \\ \;\;\;\;\;\;\;\frac{{r\sigma }}{{s{k_n}}}\sin \left( {{\alpha _2} - {\alpha _1}} \right)\sin \left( {\frac{{{\alpha _2} - {\alpha _1}}}{2}} \right). \end{array}$ (29)

 ${u_{{{\rm{J}}_{\rm{2}}}}} = \frac{{r\sigma }}{{2s{k_n}}}\sin \left( {{\alpha _2} - {\alpha _1}} \right)\sin \left( {\frac{{{\alpha _2} - {\alpha _1}}}{2}} \right).$ (30)

n条节理的位移为

 ${u_{{{\rm{J}}_n}}} = \frac{{r\sigma }}{{ns{k_n}}}\sin \left( {{\alpha _2} - {\alpha _1}} \right)\sin \left( {\frac{{{\alpha _2} - {\alpha _1}}}{2}} \right).$ (31)

 $\begin{array}{l} {u_{\rm{M}}} = {u_{\rm{r}}} + {u_{\rm{J}}} = \frac{{r\sigma \left( {1 + \mu } \right)}}{E} + \frac{{r\sigma }}{{s{k_n}}}\sin \left( {{\alpha _2} - {\alpha _1}} \right) \cdot \\ \;\;\;\;\;\;\;\;\sin \left( {\frac{{{\alpha _2} - {\alpha _1}}}{2}} \right) \cdot \left( {1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}} \right). \end{array}$ (32)

 $\begin{array}{l} 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} = \ln \left( {n + 1} \right) + a = \\ \;\;\;\;\ln \left[ {\frac{{\left( {R - r} \right) \cdot \sin \left( {{\alpha _2} - {\alpha _1}} \right)}}{s} + 1} \right] + a. \end{array}$ (33)

 $\begin{array}{l} \frac{1}{{{k_{\max }}}} = \frac{{r\left( {1 + \mu } \right)}}{E} + \frac{r}{{s{k_n}}}\sin \left( {{\alpha _2} - {\alpha _1}} \right) \cdot \sin \left( {\frac{{{\alpha _2} - {\alpha _1}}}{2}} \right) \cdot \\ \;\;\;\;\;\;\;\;\;\;\left\{ {\ln \left[ {\frac{{\left( {R - r} \right) \cdot \sin \left( {{\alpha _2} - {\alpha _1}} \right)}}{s} + 1} \right] + a} \right\}. \end{array}$ (34)

 $\frac{1}{k} = \frac{{\sqrt {{{\cos }^2}\left( {\theta - \frac{{{\alpha _1} + {\alpha _2}}}{2}} \right) + {\xi ^2}{{\sin }^2}\left( {\theta - \frac{{{\alpha _1} + {\alpha _2}}}{2}} \right)} }}{{{k_{\max }}}}.$ (35)
5 公式验证

 $k = \frac{E}{{r\left( {1 + \mu } \right)}}.$ (36)

 图 10 解析公式与数值模拟结果的对比 Fig. 10 Comparisons between the analytical calculation results and the numerical modeling results
6 工程应用

 图 11 隧道周围围岩抗力系数分布 Fig. 11 Distribution of k around the tunnel

 图 12 监测断面DK15+613的围岩压力分布(kPa) Fig. 12 Distribution ofstress on section DK15+613

 图 13 弯矩随各向异性系数变化曲线 Fig. 13 Bending moment vs anisotropy coefficient

7 结论

1) 当岩石为线弹性体、节理服从库伦滑移本构模型时，节理岩体围岩抗力系数不受围岩压力和径向液压力的影响，可称为弹性抗力系数.

2) 岩体中存在两组等间距节理时，围岩抗力系数分布曲线呈椭圆形，沿两组节理夹角角平分线方向围岩抗力系数最大，为椭圆的长轴，垂直于两组节理夹角角平分线方向围岩抗力系数最小，为椭圆的短轴.

3) 岩石弹性模量、节理法向刚度、节理间距和节理夹角对围岩抗力系数有显著影响，岩石泊松比对围岩抗力系数有一定影响，其他因素对围岩抗力系数无显著影响.

4) 隧道直径与节理间距的比值是影响围岩抗力系数尺寸效应的主要因素，比值接近于0时，可以认为围岩是均质体，比值趋于无穷大时，可以将围岩当作等效各向同性体，并用数学函数描述这一物理现象，得到了各向异性系数的表达式.

5) 推导出了围岩抗力系数的计算公式，并论证了公式的准确性，工程实例计算表明，围岩抗力系数各向异性对衬砌轴力的影响较小，对弯矩的影响显著.

6) 本文模型适用于岩石坚硬、节理填充物较少的围岩，对于软岩和节理填充物复杂的岩体工程需要采用不同的本构模型，是下一步的研究重点.

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