﻿ 考虑不同中主应力影响的柱孔卸荷缩孔效应分析
 哈尔滨工业大学学报  2020, Vol. 52 Issue (11): 63-70  DOI: 10.11918/201903189 0

### 引用本文

ZHAO Chunfeng, WANG Youbao, WU Yue, FEI Yi, GONG Xin. Analysis of cylindrical cavity unloading-contraction under different degrees of intermediate principal stress[J]. Journal of Harbin Institute of Technology, 2020, 52(11): 63-70. DOI: 10.11918/201903189.

### 文章历史

1. 岩土与地下工程教育部重点实验室(同济大学)，上海 200092;
2. 同济大学 地下建筑与工程系，上海 200092

Analysis of cylindrical cavity unloading-contraction under different degrees of intermediate principal stress
ZHAO Chunfeng1,2, WANG Youbao2, WU Yue2, FEI Yi2, GONG Xin2
1. Key Laboratory of Geotechnical and Underground Engineering(Tongji University), Ministry of Education, Shanghai 200092, China;
2. Department of Geotechnical Engineering, Tongji University, Shanghai 200092, China
Keywords: unified strength theory    cavity contraction    unloading    intermediate principal stress    soil stiffness

1 问题的提出及力学模型、屈服准则的建立 1.1 问题的提出及力学模型的建立

 图 1 土体开挖卸荷的圆孔收缩示意 Fig. 1 Schematic diagram of cavity contraction under soil excavation unloading

1) 土体初始应力场是均匀的、各向同性的；

2) 孔壁压力逐步卸荷过程中，远场应力一直保持为初始应力；

3) 土体假定为理想弹塑性体，并满足统一强度理论屈服准则；

4) 在整个卸荷缩孔过程中，出平面方向始终保持为中主应力方向；

5) 弹性区的应变很小且塑性区的弹性应变可以忽略.

 ${{\lambda }_{\text{rel}}}=\frac{p}{{{p}_{0}}},$ (1)
 ${{\lambda }_{\text{con}}}=\frac{a}{{{a}_{0}}}.$ (2)

 $\frac{\partial {{\sigma }_{r}}}{\partial r}+\frac{{{\sigma }_{r}}-{{\sigma }_{\theta }}}{r}=0,$ (3)
 ${{\varepsilon }_{r}}=\frac{1-{{\nu }^{2}}}{E}\left( {{\sigma }_{r}}-\frac{\nu }{1-\nu }{{\sigma }_{\theta }} \right),$ (4a)
 ${{\varepsilon }_{\theta }}=\frac{1-{{\nu }^{2}}}{E}\left( {{\sigma }_{\theta }}-\frac{\nu }{1-\nu }{{\sigma }_{r}} \right),$ (4b)
 ${{\varepsilon }_{r}}=-\frac{\text{d}{{u}_{r}}}{\text{d}r},{{\varepsilon }_{\theta }}=-\frac{{{u}_{r}}}{r}.$ (5)

 ${{\sigma }_{r(r=a)}}=p,$ (6a)
 ${{\sigma }_{r(r=\infty )}}={{p}_{0}}.$ (6b)
1.2 统一屈服准则的建立

 $\begin{matrix} F={{\sigma }_{1}}-\frac{\alpha }{1+b}(b{{\sigma }_{2}}+{{\sigma }_{3}})={{\sigma }_{\text{t}}}, \\ {{\sigma }_{2}}\le \frac{{{\sigma }_{1}}+\alpha {{\sigma }_{3}}}{1+\alpha }, \\ \end{matrix}$ (7a)
 $\begin{matrix} {{F}^{\prime }}=\frac{1}{1+b}({{\sigma }_{1}}+b{{\sigma }_{2}})-\alpha {{\sigma }_{3}}={{\sigma }_{\text{t}}}, \\ {{\sigma }_{2}}\ge \frac{{{\sigma }_{1}}+\alpha {{\sigma }_{3}}}{1+\alpha }. \\ \end{matrix}$ (7b)

 $\begin{matrix} {{F}^{\prime }}=\alpha {{\sigma }_{1}}-\frac{1}{1+b}(b{{\sigma }_{2}}+{{\sigma }_{3}})={{\sigma }_{\text{t}}}, \\ {{\sigma }_{2}}\le \frac{{{\sigma }_{3}}+\alpha {{\sigma }_{1}}}{1+\alpha }, \\ \end{matrix}$ (8a)
 $\begin{matrix} F=\frac{\alpha }{1+b}({{\sigma }_{1}}+b{{\sigma }_{2}})-{{\sigma }_{3}}={{\sigma }_{\text{t}}}, \\ {{\sigma }_{2}}\ge \frac{{{\sigma }_{3}}+\alpha {{\sigma }_{1}}}{1+\alpha }, \\ \end{matrix}$ (8b)

 ${{\sigma }_{2}}=\frac{m}{2}({{\sigma }_{1}}+{{\sigma }_{3}}).$ (9)

 $\begin{array}{*{35}{l}} {{\sigma }_{2}}=\frac{m}{2}({{\sigma }_{1}}+{{\sigma }_{3}})\approx \frac{1}{2}({{\sigma }_{1}}+{{\sigma }_{3}})\ge \\ \frac{1}{2}({{\sigma }_{1}}+{{\sigma }_{3}})+\frac{\text{sin}{{\varphi }_{0}}}{2}({{\sigma }_{3}}-{{\sigma }_{1}}). \\ \end{array}$ (10)

 $\frac{{{\sigma _3} + \alpha {\sigma _1}}}{{1 + \alpha }} = \frac{1}{2}({\sigma _1} + {\sigma _3}) + \frac{{{\rm{sin}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _0}}}{2}({\sigma _3} - {\sigma _1}) \le {\sigma _2}.$ (11)

 $\begin{array}{*{20}{l}} {F = {\sigma _1} - \frac{{2(1 + b)(1 + {\rm{sin}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _0}) - bm(1 - {\rm{sin}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _0})}}{{(2 + bm)(1 - {\rm{sin}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _0})}}{\sigma _3} - }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{{4(1 + b)c{\rm{cos}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _0}}}{{(2 + bm)(1 - {\rm{sin}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _0})}} = 0.} \end{array}$ (12)

 $F = {\sigma _\theta } - \zeta {\sigma _r} - {\sigma _0} = 0.$ (13)

 $\left\{ \begin{array}{l} \begin{array}{*{20}{l}} {\zeta = \frac{{1 + {\rm{sin}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _{\rm{t}}}}}{{1 - {\rm{sin}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _{\rm{t}}}}},}\\ {{\sigma _0} = \frac{{2{c_{\rm{t}}}{\rm{cos}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _{\rm{t}}}}}{{1 - {\rm{sin}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _{\rm{t}}}}},} \end{array}\\ {\rm{sin}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _{\rm{t}}} = \frac{{2(1 + b){\rm{sin}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _0}}}{{2(1 + b) - b(1 - {\rm{sin}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _0})}},\\ {c_{\rm{t}}} = \frac{{2(1 + b){c_0}{\rm{cos}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _0}}}{{2(1 + b) - b(1 - {\rm{sin}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _0})}}\frac{1}{{{\rm{cos}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\varphi _t}}}. \end{array} \right.$ (14)
2 钻孔卸荷收缩的解析解 2.1 弹性响应与初始屈服解

 $\left\{ \begin{array}{l} {\sigma _r} = {p_0} + (p - {p_0}){\left( {\frac{a}{r}} \right)^2},\\ {\sigma _\theta } = {p_0} - (p - {p_0}){\left( {\frac{a}{r}} \right)^2}. \end{array} \right.$ (15)
 ${u_r} = \frac{{p - {p_0}}}{{2G}}{\left( {\frac{a}{r}} \right)^2}r.$ (16)

 ${p_{\rm{y}}} = {\sigma _r}{|_{r = a}} = \frac{{2{p_0} - {\sigma _0}}}{{1 + \zeta }}.$ (17)
2.2 弹塑性分析

 $\left\{ {\begin{array}{*{20}{l}} {{\sigma _r} = {p_0} + ({p_{\rm{y}}} - {p_0}){{\left( {\frac{R}{r}} \right)}^2},}\\ {{\sigma _\theta } = {p_0} - ({p_{\rm{y}}} - {p_0}){{\left( {\frac{R}{r}} \right)}^2},} \end{array}} \right.r \ge R,$ (18)
 ${u_r} = \frac{{{p_{\rm{y}}} - {p_0}}}{{2G}}{\left( {\frac{R}{r}} \right)^2}r.$ (19)

 ${\frac{{\partial {\sigma _r}}}{{\partial r}} + \frac{{{\sigma _r} - (\zeta {\sigma _r} + {\sigma _0})}}{r} = 0.}$ (20)

 ${{\sigma _r} = \frac{{{{(Kr)}^{\zeta - 1}} - {\sigma _0}}}{{\zeta - 1}}.}$ (21)

K为积分常数，将式(6a)、(17)分别代入式(21), 得

 $K = \frac{{{{[p(\zeta - 1) + {\sigma _0}]}^{\frac{1}{{\zeta - 1}}}}}}{a} = \frac{{{{[{p_{\rm{y}}}(\zeta - 1) + {\sigma _0}]}^{\frac{1}{{\zeta - 1}}}}}}{R}.$ (22)

 $\left\{ {\begin{array}{*{20}{l}} {{\sigma _r} = {{\left( {\frac{r}{R}} \right)}^{\zeta - 1}}\frac{{[{p_{\rm{y}}}(\zeta - 1) + {\sigma _0}]}}{{\zeta - 1}} - \frac{{{\sigma _0}}}{{\zeta - 1}},}\\ {{\sigma _\theta } = \frac{\zeta }{{\zeta - 1}}{{\left( {\frac{r}{R}} \right)}^{\zeta - 1}}\frac{{[{p_{\rm{y}}}(\zeta - 1) + {\sigma _0}]}}{{\zeta - 1}} - \frac{{{\sigma _0}}}{{\zeta - 1}},} \end{array}\quad a \le r < R,} \right.$ (23a)

 $\left\{ {\begin{array}{*{20}{l}} {{\sigma _r} = {{\left( {\frac{r}{a}} \right)}^{\zeta - 1}} \cdot \frac{{[p(\zeta - 1) + {\sigma _0}]}}{{\zeta - 1}} - \frac{{{\sigma _0}}}{{\zeta - 1}},}\\ {{\sigma _\theta } = \frac{\zeta }{{\zeta - 1}}{{\left( {\frac{r}{a}} \right)}^{\zeta - 1}} \cdot \frac{{[p(\zeta - 1) + {\sigma _0}]}}{{\zeta - 1}} - \frac{{{\sigma _0}}}{{\zeta - 1}},} \end{array}\quad a \le r < R.} \right.$ (23b)

Mair等[33]在预测隧道周围地层运动时给出了轴对称条件下卸载柱形孔周半径为r的塑性区超孔隙水压力公式为

 $\Delta u = {s_{\rm{u}}}\left( {1 - \frac{{{p_0} - p}}{{{s_{\rm{u}}}}} + 2{\rm{ln}}\frac{r}{a}} \right),a \le r \le R.$ (24a)

 $\left\{ {\begin{array}{*{20}{l}} {{\sigma ^\prime }_r = {{\left( {\frac{r}{R}} \right)}^{\zeta - 1}} \cdot \frac{{[{p_{\rm{y}}}(\zeta - 1) + {\sigma _0}]}}{{\zeta - 1}} - \frac{{{\sigma _0}}}{{\zeta - 1}} - \Delta u,}\\ {{\sigma ^\prime }_\theta = \frac{\zeta }{{\zeta - 1}}{{\left( {\frac{r}{R}} \right)}^{\zeta - 1}} \cdot \frac{{[{p_{\rm{y}}}(\zeta - 1) + {\sigma _0}]}}{{\zeta - 1}} - \frac{{{\sigma _0}}}{{\zeta - 1}} - \Delta u.} \end{array}} \right.$ (24b)

 $\frac{R}{a} = \frac{{{{[{p_{\rm{y}}}(\zeta - 1) + {\sigma _0}]}^{\frac{1}{{\zeta - 1}}}}}}{{{{[p(\zeta - 1) + {\sigma _0}]}^{\frac{1}{{\zeta - 1}}}}}}.$ (25)

 $\frac{R}{a} = {[\frac{{{C_1}}}{{p(\zeta - 1) + {\sigma _0}}}]^{\frac{1}{{\zeta - 1}}}}.$ (26)

 ${u_R} = {u_r}{|_{r = R}} = R - {R_0} = \frac{{(1 - \zeta ){p_0} - {\sigma _0}}}{{2G(1 + \zeta )}}R.$ (27)

 $r_0^2 - {r^2} = R_0^2 - {R^2}.$ (28)

 ${u_r} = r - {r_0}.$ (29)

 ${u_r} = r - \sqrt {{r^2} + u_R^2 - 2{u_R}R} .$ (30)

 ${\left( {\frac{{{r_0}}}{r}} \right)^2} = 1 + {C_2}{[\frac{{{C_1}}}{{p(\zeta - 1) + {\sigma _0}}}]^{\frac{2}{{\zeta - 1}}}}\frac{{{a^2}}}{{{r^2}}}.$ (31)

 ${\left( {\frac{{{a_0}}}{a}} \right)^2} = 1 + {C_2}{[\frac{{{C_1}}}{{p(\zeta - 1) + {\sigma _0}}}]^{\frac{2}{{\zeta - 1}}}}.$ (32)

 $p = \frac{{{C_3}}}{{{{\left[ {{{\left( {\frac{{{a_0}}}{a}} \right)}^2} - 1} \right]}^{\frac{{\zeta - 1}}{2}}}}} - \frac{{{\sigma _0}}}{{\zeta - 1}}.$ (33)
 $a = \frac{{{a_0}}}{{\sqrt {1 + {{\left( {\frac{{{C_3}}}{{p(\zeta - 1) + {\sigma _0}}}} \right)}^{\frac{2}{{\zeta - 1}}}}} }}.$ (34)

 $\begin{array}{*{20}{l}} {{C_3} = C_2^{\frac{{\zeta - 1}}{2}}{C_1} = {{\left[ {{{\left( {1 - \frac{{(1 - \zeta ){p_0} - {\sigma _0}}}{{2G(1 + \zeta )}}} \right)}^2} - 1} \right]}^{\frac{{\zeta - 1}}{2}}} \times }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left[ {\frac{{2{p_0} - {\sigma _0}}}{{\zeta + 1}}(\zeta - 1) + {\sigma _0}} \right].} \end{array}$ (35)

 ${{\delta _1} = \frac{{{\sigma _0}}}{{{p_0}}},}$ (36a)
 ${{\delta _2} = \frac{G}{{{p_0}}}.}$ (36b)

 ${\lambda _{{\rm{rel}}}} = \frac{{{C^\prime }_3}}{{{{\left[ {{{\left( {\frac{1}{{{\lambda _{{\rm{con}}}}}}} \right)}^2} - 1} \right]}^{\frac{{\zeta - 1}}{2}}}}} - \frac{{{\delta _1}}}{{(\zeta - 1)}},$ (37a)
 ${\lambda _{{\rm{con}}}} = \frac{1}{{\sqrt {1 + {{\left( {\frac{{{C^\prime }_3}}{{{\lambda _{{\rm{rel}}}}(\zeta - 1) + {\delta _1}}}} \right)}^{\frac{2}{{\zeta - 1}}}}} }}.$ (37b)

 $\begin{array}{l} \begin{array}{*{20}{l}} {{C^\prime }_3 = C_2^{\frac{{\zeta - 1}}{2}}{C^\prime }_1 = }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {{\left[ {{{\left( {1 - \frac{{1 - \zeta }}{{2(1 + \zeta ){\delta _2}}} + \frac{{{\delta _1}}}{{2(1 + \zeta ){\delta _2}}}} \right)}^2} - 1} \right]}^{\frac{{\zeta - 1}}{2}}} \times } \end{array}\\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left[ {\frac{{2(\zeta - 1 + {\delta _1})}}{{\zeta + 1}}} \right]. \end{array}$ (38)
3 算例研究及参数化分析

3.1 中间主应力影响参数b对卸荷因子与缩孔系数的影响

3.2 中间主应力影响参数b对卸荷引起孔周位移、应力的影响

 图 4 不同b值时的孔周位移场变化 Fig. 4 Displacement field around cavity variation with different b
 图 5 不同b值时孔周应力场变化 Fig. 5 Stress field around cavity variation with different b

Yu解受黏聚力变化影响较大，黏聚力越小时，其塑性区迅速向左移动，表明塑性区迅速增大，而本文解变化较小，可以理解为未考虑中间主应力影响的Yu解更为保守.

3.3 土体刚度指数对缩孔系数的影响

Yu等[17]引入了式(36b)的土体刚度指数δ2将土体剪切刚度正则化，研究土体刚度对缩孔的影响.表 1~3列出了孔壁完全卸荷时土体刚度指数与缩孔系数的对应关系.当δ2=10时，土体刚度较小，黏聚力为0.334p0b由0.1增大为1时，缩孔系数维持在0.946，基本保持不变；而当b=0.1, 黏聚力保持为0.334p0，而δ2由10增大到50时，缩孔系数由0.945增大到0.988.而且，δ2=10，随着黏聚力减小，缩孔系数由0.946减小至0.795，变化幅度为16%；δ2=50时，缩孔系数由0.988减小为0.947，变化幅度仅为4.1%.这表明刚度越大，中间主应力效应越大，缩孔效应越小；刚度较大时，中间主应力效应对缩孔的影响较小，土体刚度对缩孔有至关重要的减弱效应，此时，b的影响较土体刚度的影响很小.

3.4 c0, φ0对卸荷缩孔关系和孔周位移的影响

 图 6 不同φ0时的孔壁卸荷缩孔曲线 Fig. 6 Unloading-contraction curves of cavity wall under different φ0
 图 7 不同c0时的孔壁卸荷缩孔曲线 Fig. 7 Unloading-contraction curves of cavity wall under different c0
 图 8 不同φ0时的孔周位移 Fig. 8 Displacement field around cavity under different φ0
 图 9 不同c0值时的孔周位移 Fig. 9 Displacement field around cavity under different c0
4 工程案例

 图 10 不同b时的孔壁位移随卸荷变化关系 Fig. 10 Relation between cavity wall displacement and unloading with different b
5 结论

1) 推导得到了基于统一强度理论的考虑中主应力影响系数b的无量纲化的柱孔卸荷收缩公式(33)、(34)或(37)、(38)，通过该式可对指定卸荷程度下的柱孔缩孔系数进行定量预测.

2) 采用考虑不同中间主应力效应得到的柱孔收缩解较Mohr-Coulomb准则大，说明不考虑中主应力影响的解答偏于保守，其本质为：b的增大，减小了初始屈服卸荷压力，推迟了临界弹塑性区峰值环向应力的出现，降低了孔周塑性区扩展.

3) 中间主应力影响参数b越大，卸荷缩孔效应越小；中间主应力影响参数对孔周径向位移的影响不可忽略；孔周径向应力受其影响较小而环向应力影响较大，b越大，推动峰值环向应力逐渐向孔壁移动，表明中主应力有利于减小孔周塑性区.

4) 土体刚度对卸荷与缩孔关系的影响很大，而且当刚度较大时，中间主应力影响系数b对卸荷缩孔的影响相对减小.

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