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浙江大学学报(工学版)
浙江大学学报(工学版)  2023, Vol. 57 Issue (9): 1706-1717    DOI: 10.3785/j.issn.1008-973X.2023.09.002
土木工程、水利工程     
脆塑性迭代逼近算法的改进
金俊超1,2(),景来红1,2,杨风威1,2,宋志宇1,2,尚朋阳3
1. 黄河勘测规划设计研究院有限公司，河南 郑州 450003
2. 水利部黄河流域水治理与水安全重点实验室（筹），河南 郑州 450003
3. 中水北方勘测设计研究有限责任公司，天津 300222
Improvement of multi-step brittle-plastic approach
Jun-chao JIN1,2(),Lai-hong JING1,2,Feng-wei YANG1,2,Zhi-yu SONG1,2,Peng-yang SHANG3
1. Yellow River Engineering Consulting Limited Company, Zhengzhou 450003, China
2. Key Laboratory of Water Management and Water Security for Yellow River Basin, Ministry of Water Resources (Under Construction), Zhengzhou 450003, China
3. China Water Resources Beifang Investigation, Design and Research Limited Company, Tianjin 300222, China
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摘要：

针对应力跌落计算中脆塑性迭代逼近算法存在的问题，在主应力空间，结合特征点变形和破坏特征，通过理论推导，系统分析偏应力等比例跌落方法、最小主应力不变跌落方法、塑性位势跌落方法及应力球量不变跌落方法的缺陷. 考虑脆塑性变形和破坏过程中的泊松效应，提出改进的塑性位势跌落方法；推导应力跌落计算更新过程，编写UMAT子程序将该更新过程嵌入软件Abaqus，实现岩石弹塑性应变软化过程数值求解. 采用所提改进方法替换原有的应力跌落计算方法，实现弹塑性应变软化过程的数值模拟. 引入塑性强化算法，实现岩石弹塑性变形破坏全过程的数值计算，并进行多算例验证. 对不同地质条件的Mine-by试验洞及某水电站引水隧洞辅助洞进行开挖模拟，结果表明所提改进方法能够合理模拟工程中围岩弹塑性变形破坏现象.
关键词： 应变软化;  脆塑性;  应力跌落;  数值算法;  有限元    
Abstract:

Aiming at the problem of multi-step brittle-plastic approach in stress-drop calculation, the defects of existing stress-drop calculation method based on the deviator stress dropping, the method based on the constant minor principal stress in the brittle-plastic process, the method based on the plastic potential theory and the method based on the invariant spherical stress were systematically analyzed in the principal stress space, combined with the deformation and the failure characteristics of feature points. Considering the Poisson’s effect in the brittle-plastic deformation and failure process, the method based on the plastic potential theory was improved. The corresponding stress update process was derived and embedded in the program Abaqus through the UMAT subroutine. The original stress-drop calculation method was replaced by the improved method, realizing the numerical simulation of elastic-plastic strain softening process. The calculation of full elastic-plastic deformation and failure process was further realized by introducing the plastic-strengthening algorithm, which is verified by several examples. The excavation simulation of Mine-by tunnel and auxiliary tunnels of a hydropower station diversion tunnel shows that the improved method can reasonably simulate the elastic-plastic deformation and failure phenomenon of surrounding rock.
Key words: strain softening    brittle-plastic    stress-drop    numerical algorithm    finite element
收稿日期: 2022-11-01 出版日期: 2023-10-16
CLC:  TU 45  
基金资助: 中国博士后科学基金资助项目（2022M721299）；河南省重点研发与推广专项（232102320339）
作者简介: 金俊超（1992—），男，博士后，从事岩土工程数值计算方面的研究. orcid.org/0000-0002-0408-7241.E-mail： jinjunchao@whu.edu.cn
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金俊超
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尚朋阳

引用本文:

金俊超,景来红,杨风威,宋志宇,尚朋阳. 脆塑性迭代逼近算法的改进[J]. 浙江大学学报(工学版), 2023, 57(9): 1706-1717.

Jun-chao JIN,Lai-hong JING,Feng-wei YANG,Zhi-yu SONG,Peng-yang SHANG. Improvement of multi-step brittle-plastic approach. Journal of ZheJiang University (Engineering Science), 2023, 57(9): 1706-1717.

链接本文:

https://www.zjujournals.com/eng/CN/10.3785/j.issn.1008-973X.2023.09.002    或    https://www.zjujournals.com/eng/CN/Y2023/V57/I9/1706

图 1  岩石破坏分类
破坏类型 峰值应力 残余应力 屈服函数求解 合理性
单轴拉伸破坏 $\left[ \begin{array}{c} \sigma _1^{\rm{p} }\\ \sigma _2^{\rm{p} }\\ \sigma _3^{\rm{p} }\\ \end{array} \right] = \left[ \begin{array}{c} \sigma _1^{\rm{p} }\\ 0\\ 0\\\end{array} \right]$ $\left[ \begin{gathered} \sigma _1^{\rm{r} } \\ \sigma _2^{\rm{r} } \\ \sigma _3^{\rm{r} } \\\end{gathered} \right] = \left[ \begin{gathered} \dfrac{ {\left( {1+2\beta } \right)\sigma _1^{\rm{p} } } }{3} \\ \dfrac{ {\left( {1 - \beta } \right)\sigma _1^{\rm{p} } } }{3} \\ \dfrac{ {\left( {1 - \beta } \right)\sigma _1^{\rm{p} } } }{3} \\ \end{gathered} \right] $ $\left. {\begin{array}{*{20}{l} }\;\;\;{f_{\rm{p} }^{ {\rm{M} } - {\rm{C} } }\left( { {\sigma ^{\rm{p} } } } \right) = \sigma _1^{\rm{p} } - \dfrac{ {2{c _{\rm{p} } } \cdot \cos {\varphi _{\rm{p} } } } }{ {1 + \sin {\varphi _{\rm{p} } } } } = 0},\\\begin{array}{l}f _{\rm{r} }^{M - C}\left( { {\sigma ^{\rm{r} } } } \right) = \dfrac{ {\left( {1 + 2\beta } \right)\sigma _1^{\rm{p} } } }{3} - \dfrac{ {1 - \sin {\varphi_{\rm{r} } } } }{ {1 + \sin {\varphi_{\rm{r} } } } }\dfrac{ {\left( {1 - \beta } \right)\sigma _1^{\rm{p} } } }{3} - \dfrac{ {2{c _{\rm{r} } } \cdot \cos {\varphi_{\rm{r} } } } }{ {1 + \sin {\varphi_{\rm{r} } } } } = 0.\end{array}\end{array} } \right\}$ 错误。除了主拉应力方向，横向应力方向也存在残余应力，与事实不符
单轴压缩破坏 $\left[ \begin{array}{c} \sigma _1^{\rm{p} }\\ \sigma _2^{\rm{p} }\\ \sigma _3^{\rm{p} }\\\end{array} \right] = \left[ \begin{array}{c} 0\\ 0\\ \sigma _3^{\rm{p} }\\ \end{array} \right]$ $\left[ \begin{gathered} \sigma _1^{\rm{r} } \\ \sigma _2^{\rm{r} } \\ \sigma _3^{\rm{r} } \\ \end{gathered} \right] = \left[ \begin{gathered} \dfrac{ {\left( {1 - \beta } \right)\sigma _3^{\rm{p} } } }{3} \\ \dfrac{ {\left( {1 - \beta } \right)\sigma _3^{\rm{p} } } }{3} \\ \dfrac{ {\left( {1+2\beta } \right)\sigma _3^{\rm{p} } } }{3} \\ \end{gathered} \right]$ $\left. {\begin{array}{*{20}{l} }\;\;\;{f_{\rm{p} }^{ {\rm{M} } - {\rm{C} } }\left( { {\sigma ^{\rm{p} } } } \right) = \left( {\sin {\varphi _{\rm{p} } } - 1} \right)\sigma _3^{\rm{p} } - 2{c _{\rm{p} } } \cdot \cos {\varphi _{\rm{p} } } = 0},\\\begin{array}{l}f _{\rm{r} }^{ {\rm{M} } - {\rm{C} } }\left( { {\sigma ^{\rm{r} } } } \right) = \dfrac{ {\left( {1 - \beta } \right)\sigma _3^{\rm{p} } } }{3} - \dfrac{ {1 - \sin {\varphi_{\rm{r} } } } }{ {1 + \sin {\varphi_{\rm{r} } } } }\dfrac{ {\left( {1 + 2\beta } \right)\sigma _3^{\rm{p} } } }{3} - \dfrac{ {2{c _{\rm{r} } } \cdot \cos {\varphi_{\rm{r} } } } }{ {1 + \sin {\varphi_{\rm{r} } } } } = 0.\end{array}\end{array} } \right\}$ 错误。除了主压应力方向，横向应力方向也存在残余应力，与事实不符
二向纯剪破坏 $\left[ \begin{array}{c} \sigma _1^{\rm{p} }\\ \sigma _2^{\rm{p} }\\ \sigma _3^{\rm{p} }\\\end{array} \right] = \left[ \begin{array}{c} \sigma _1^{\rm{p} }\\ 0\\ - \sigma _1^{\rm{p} }\\ \end{array} \right]$ $\left[ \begin{array}{c} \sigma _1^{\rm{r} } \\ \sigma _2^{\rm{r} } \\ \sigma _3^{\rm{r} } \\ \end{array} \right] = \left[ \begin{array}{c} \beta \sigma _1^{\rm{p} } \\ 0 \\ - \beta \sigma _1^{\rm{p} } \\\end{array} \right]$ $\left. {\begin{array}{*{20}{l} } {f_{\rm{p} }^{ {\rm{M} } - {\rm{C} } }\left( { {\sigma ^{\rm{p} } } } \right) = \sigma _1^{\rm{p} } - \dfrac{ {1 - \sin {\varphi_{\rm{r} } } }}{ {1+\sin {\varphi_{\rm{r} } } }}\left( { - \sigma _1^{\rm{p} } } \right) - \dfrac{ {2{c _{\rm{p} } } \cdot \cos {\varphi _{\rm{p} } } }}{ {1+\sin {\varphi _{\rm{p} } } }} = 0}, \\ {f _{\rm{r} }^{ {\rm{M} } - {\rm{C} } }\left( { {\sigma ^{\rm{r} } } } \right) = \beta \sigma _1^{\rm{p} } - \dfrac{ {1 - \sin {\varphi_{\rm{r} } } }}{ {1+\sin {\varphi_{\rm{r} } } }}\left( { - \beta \sigma _1^{\rm{p} } } \right) - \dfrac{ {2{c _{\rm{r} } } \cdot \cos {\varphi_{\rm{r} } } }}{ {1+\sin {\varphi_{\rm{r} } } }} = 0}.\end{array} } \right\}$ 正确。残余阶段满足二向纯剪的应力状态，与事实相符
表 1  偏应力等比例跌落方法存在的问题
破坏类型 峰值应力 残余应力 屈服函数求解 合理性
单轴拉伸破坏 $\left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ \sigma _2^{\rm{p} } \\ \sigma _3^{\rm{p} } \\ \end{array} \right] = \left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ 0 \\ 0 \\ \end{array} \right]$ $\left[ \begin{array}{c} \sigma _1^{\rm{r} } \\ \sigma _2^{\rm{r} } \\ \sigma _3^{\rm{r} } \\ \end{array} \right] = \left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ 0 \\ 0 \\ \end{array} \right]$ $\left. {\begin{array}{*{20}{l} } {f_{\rm{p} }^{ {\rm{M} } - {\rm{C} } }\left( { {\sigma ^{\rm{p} } } } \right) = \sigma _1^{\rm{p} } - \dfrac{ {2{c _{\rm{p} } } \cdot \cos {\varphi _{\rm{p} } } } }{ {1+\sin {\varphi _{\rm{p} } } } } = 0}, \\ {f _{\rm{r} }^{ {\rm{M} } - {\rm{C} } }\left( { {\sigma ^{\rm{r} } } } \right) = \sigma _1^{\rm{p} } - \dfrac{ {2{f _{\rm{r} } } \cdot \cos { {\varphi _{\rm{r} } } } } }{ {1+\sin { {\varphi _{\rm{r} } } } } } = 0}. \end{array} } \right\}$ 错误。残余强度面屈服函数无解
单轴压缩破坏 $\left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ \sigma _2^{\rm{p} } \\ \sigma _3^{\rm{p} } \\ \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ \sigma _3^{\rm{p} } \\ \end{array} \right]$ $\left[ \begin{array}{c} \sigma _1^{\rm{r} } \\ \sigma _2^{\rm{r} } \\ \sigma _3^{\rm{r} } \\ \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ \sigma _3^{\rm{r} } \\ \end{array} \right]$ $\left. {\begin{array}{*{20}{l} } {f_{\rm{p} }^{ {\rm{M} } - {\rm{C} } }\left( { {\sigma ^{\rm{p} } } } \right) = - \dfrac{ {1 - \sin { {\varphi _{\rm{r} } } } } }{ {1+\sin { {\varphi _{\rm{r} } } } } }\sigma _3^{\rm{p} } - \dfrac{ {2{c _{\rm{p} } } \cdot \cos {\varphi _{\rm{p} } } } }{ {1+\sin {\varphi _{\rm{p} } } } } = 0}, \\ {f _{\rm{r} }^{ {\rm{M} } - {\rm{C} } }\left( { {\sigma ^{\rm{r} } } } \right) = - \dfrac{ {1 - \sin { {\varphi _{\rm{r} } } } } }{ {1+\sin { {\varphi _{\rm{r} } } } } }\sigma _3^{\rm{r} } - \dfrac{ {2{f _{\rm{r} } } \cdot \cos { {\varphi _{\rm{r} } } } } }{ {1+\sin { {\varphi _{\rm{r} } } } } } = 0}. \end{array} } \right\}$ 正确。残余阶段满足单轴压缩的应力状态，与事实相符
二向纯剪破坏 $\left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ \sigma _2^{\rm{p} } \\ \sigma _3^{\rm{p} } \\\end{array} \right] = \left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ 0 \\ - \sigma _1^{\rm{p} } \\\end{array} \right]$ $\left[ \begin{array}{c} \sigma _1^{\rm{r} } \\ \sigma _2^{\rm{r} } \\ \sigma _3^{\rm{r} } \\ \end{array} \right] = \left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ 0 \\ \sigma _3^{\rm{r} } \\ \end{array} \right]$ $\left. {\begin{array}{*{20}{l} } {f_{\rm{p} }^{ {\rm{M} } - {\rm{C} } }\left( { {\sigma ^{\rm{p} } } } \right) = \sigma _1^{\rm{p} }+\dfrac{ {1 - \sin {\varphi _{\rm{p} } } } }{ {1+\sin {\varphi _{\rm{p} } } } }\sigma _1^{\rm{p} } - \dfrac{ {2{c _{\rm{p} } } \cdot \cos {\varphi _{\rm{p} } } } }{ {1+\sin {\varphi _{\rm{p} } } } } = 0}, \\ {f _{\rm{r} }^{ {\rm{M} } - {\rm{C} } }\left( { {\sigma ^{\rm{r} } } } \right) = \sigma _1^{\rm{p} } - \dfrac{ {1 - \sin { {\varphi _{\rm{r} } } } } }{ {1+\sin { {\varphi _{\rm{r} } } } } }\sigma _3^{\rm{r} } - \dfrac{ {2{f _{\rm{r} } } \cdot \cos { {\varphi _{\rm{r} } } } } }{ {1+\sin { {\varphi _{\rm{r} } } } } } = 0}.\end{array} } \right\}$ 错误。残余阶段不满足二向纯剪的应力状态，与事实不符
表 2  最小主应力不变跌落方法存在的问题
破坏类型 峰值应力 应力跌落过程的主应力增量 合理性
单轴拉伸破坏 $\left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ \sigma _2^{\rm{p} } \\ \sigma _3^{\rm{p} } \\ \end{array} \right] = \left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ 0 \\ 0 \\ \end{array} \right]$ $ \;\;\;\; \left[ \begin{array}{c} \Delta {\sigma _1} \\ \Delta {\sigma _2} \\ \Delta {\sigma _3} \\ \end{array} \right] = \left[ {\begin{array}{*{20}{c}} {\lambda +2G}&\lambda &\lambda \\ \lambda &{\lambda +2G}&\lambda \\ \lambda &\lambda &{\lambda +2G} \end{array}} \right] \left[ \begin{array}{c} - \Delta \varepsilon _1^{\rm{p}}\\0\\0\end{array} \right] = \left[ \begin{array}{c}- \left( {\lambda +2G} \right)\Delta \varepsilon _1^{\rm{p}}\\ - \lambda \Delta \varepsilon _1^{\rm{p}}\\ - \lambda \Delta \varepsilon _1^{\rm{p}}\end{array} \right]$ 错误。除了主拉伸方向应力发生改变之外，横向应力也发生改变，与事实不符
单轴压缩破坏 $\left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ \sigma _2^{\rm{p} } \\ \sigma _3^{\rm{p} } \\ \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ \sigma _3^{\rm{p} } \\\end{array} \right]$ $ \;\;\;\;\left[ \begin{array}{c} \Delta {\sigma _1} \\ \Delta {\sigma _2} \\ \Delta {\sigma _3} \\ \end{array} \right] = \left[ {\begin{array}{*{20}{c}} {\lambda +2G}&\lambda &\lambda \\ \lambda &{\lambda +2G}&\lambda \\ \lambda &\lambda &{\lambda +2G} \end{array}} \right]\left[ \begin{array}{c} \quad 0 \\\quad 0 \\ - \Delta \varepsilon _3^{\rm{p}} \\ \end{array} \right] = \left[ \begin{array}{c}\quad - \lambda \Delta \varepsilon _3^{\rm{p}} \\ \quad - \lambda \Delta \varepsilon _3^{\rm{p}} \\ - \left( {\lambda +2G} \right)\Delta \varepsilon _3^{\rm{p}} \\ \end{array} \right] $ 错误。除了主压缩方向应力发生改变之外，横向应力也发生改变，与事实不符
二向纯剪破坏 $\left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ \sigma _2^{\rm{p} } \\ \sigma _3^{\rm{p} } \\\end{array} \right] = \left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ 0 \\ - \sigma _1^{\rm{p} } \\\end{array} \right]$ $ \left[ \begin{array}{c} \Delta {\sigma _1} \\ \Delta {\sigma _2} \\ \Delta {\sigma _3} \\ \end{array} \right] = \left[ {\begin{array}{*{20}{c}} {\lambda +2G}&\lambda &\lambda \\ \lambda &{\lambda +2G}&\lambda \\ \lambda &\lambda &{\lambda +2G} \end{array}} \right]\left[ \begin{array}{c} \Delta \varepsilon _1^{\rm{p}} \\ 0 \\ - \Delta \varepsilon _1^{\rm{p}} \\ \end{array} \right] = \left[ \begin{array}{c} 2G\Delta \varepsilon _1^{\rm{p}} \\ 0 \\ - 2G\Delta \varepsilon _1^{\rm{p}} \\ \end{array} \right] $ 正确。只有1和3方向应力发生变化，与事实相符
表 3  塑性位势跌落方法存在的问题
破坏类型 峰值应力 残余应力 屈服函数求解 合理性
单轴拉伸破坏 $\left[\begin{array}{c} \sigma _1^{\rm{p} }\\ \sigma _2^{\rm{p} }\\ \sigma _3^{\rm{p} }\\\end{array} \right] = \left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ 0 \\ 0 \\ \end{array} \right]$ $ \left. {\begin{array}{*{20}{c}} {\dfrac{{\sigma _1^{\rm{r}}}}{2} = \dfrac{{\sigma _1^{\rm{p}}}}{2}}, \\ {\dfrac{{\sigma _1^{\rm{r}}}}{{\sigma _1^{\rm{r}}}} = \dfrac{{\sigma _1^{\rm{p}}}}{{\sigma _1^{\rm{p}}}}} .\end{array}} \right\} $ $ \left. {\begin{array}{*{20}{l}} {f_{\rm{p}}^{{\rm{M}} - {\rm{C}}}\left( {{\sigma^{\rm{p}}}} \right) = \sigma _1^{\rm{p}}- \dfrac{{2{c _{\rm{p}}} \cdot \cos {\varphi ^{\rm{p}}}}}{{1+\sin {\varphi ^{\rm{p}}}}} = 0}, \\ {f_{\rm{r}}^{{\rm{M}} - {\rm{C}}}\left( {{\sigma ^{\rm{r}}}} \right) = \sigma _1^{\rm{p}}- \dfrac{{2{c _{\rm{r}}} \cdot \cos {{\varphi _{\rm{r}}}}}}{{1+\sin {{\varphi _{\rm{r}}}}}} = 0}. \end{array}} \right\} $ 错误。残余强度面屈服函数无解
单轴压缩破坏 $\left[ \begin{array}{c} \sigma _1^{\rm{p} }\\ \sigma _2^{\rm{p} }\\ \sigma _3^{\rm{p} }\\\end{array} \right] = \left[ \begin{array}{c} 0 \\ 0\\ \sigma _3^{\rm{p} }\\\end{array} \right]$ $ \left. {\begin{array}{*{20}{c}} {\dfrac{{ - \sigma _3^{\rm{r}}}}{2} = \dfrac{{ - \sigma _3^{\rm{p}}}}{2}}, \\ {\dfrac{0}{{ - \sigma _3^{\rm{r}}}} = \dfrac{0}{{ - \sigma _3^{\rm{p}}}}}. \end{array}} \right\} $ $ \left. {\begin{array}{*{20}{l}} {f_{\rm{p}}^{{\rm{M}} - {\rm{C}}}\left( {{\sigma^{\rm{p}}}} \right) = - \dfrac{{1 - \sin {{\varphi _{\rm{r}}}}}}{{1+\sin {{\varphi _{\rm{r}}}}}}\sigma _3^{\rm{p}}- \dfrac{{2{c _{\rm{p}}} \cdot \cos {\varphi ^{\rm{p}}}}}{{1+\sin {\varphi ^{\rm{p}}}}} = 0}, \\ {f_{\rm{r}}^{{\rm{M}} - {\rm{C}}}\left( {{\sigma ^{\rm{r}}}} \right) = - \dfrac{{1 - \sin {{\varphi _{\rm{r}}}}}}{{1+\sin {{\varphi _{\rm{r}}}}}}\sigma _3^{\rm{p}}- \dfrac{{2{c _{\rm{r}}} \cdot \cos {{\varphi _{\rm{r}}}}}}{{1+\sin {{\varphi _{\rm{r}}}}}} = 0}. \end{array}} \right\} $ 错误。残余强度面屈服函数无解
二向纯剪破坏 $\left[ \begin{array}{c} \sigma _1^{\rm{p} }\\ \sigma _2^{\rm{p} }\\ \sigma _3^{\rm{p} }\\\end{array} \right] = \left[ \begin{array}{c} \sigma _1^{\rm{p}}\\ 0 \\ - \sigma _1^{\rm{p}}\\ \end{array} \right]$ $ \left. {\begin{array}{*{20}{c}} {\dfrac{{\sigma _1^{\rm{r}} - \sigma _3^{\rm{r}}}}{2} = \dfrac{{\sigma _1^{\rm{p}}- \left( { - \sigma _1^{\rm{p}}} \right)}}{2}}, \\ {\dfrac{{\sigma _1^{\rm{r}}}}{{\sigma _1^{\rm{r}} - \sigma _3^{\rm{r}}}} = \dfrac{{\sigma _1^{\rm{p}}}}{{\sigma _1^{\rm{p}}- \left( { - \sigma _1^{\rm{p}}} \right)}}} .\end{array}} \right\} $ $ \left. {\begin{array}{*{20}{l}} {f_{\rm{p}}^{{\rm{M}} - {\rm{C}}}\left( {{\sigma^{\rm{p}}}} \right) = \sigma _1^{\rm{p}}+\dfrac{{1 - \sin {\varphi ^{\rm{p}}}}}{{1+\sin {\varphi ^{\rm{p}}}}}\sigma _1^{\rm{p}}- \dfrac{{2{c _{\rm{p}}} \cdot \cos {\varphi ^{\rm{p}}}}}{{1+\sin {\varphi ^{\rm{p}}}}} = 0} ,\\ {f_{\rm{r}}^{{\rm{M}} - {\rm{C}}}\left( {{\sigma ^{\rm{r}}}} \right) = \sigma _1^{\rm{p}}- \dfrac{{1 - \sin {{\varphi _{\rm{r}}}}}}{{1+\sin {{\varphi _{\rm{r}}}}}}\sigma _3^{\rm{p}}- \dfrac{{2{c _{\rm{r}}} \cdot \cos {{\varphi _{\rm{r}}}}}}{{1+\sin {{\varphi _{\rm{r}}}}}} = 0}. \end{array}} \right\} $ 错误。残余强度面屈服函数无解
表 4  球量不变跌落方法存在的问题
图 2  红砂岩单轴拉伸试验结果[22]
破坏类型 峰值应力 应力跌落过程的主应力增量 合理性
单轴拉伸破坏 $\left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ \sigma _2^{\rm{p} } \\ \sigma _3^{\rm{p} } \\\end{array} \right] = \left[ \begin{array}{c} \sigma _1^{\rm{p} }\\ 0 \\ 0 \\ \end{array} \right]$ $ \left[ \begin{array}{c} \Delta {\sigma _1} \\ \Delta {\sigma _2} \\ \Delta {\sigma _3} \\ \end{array} \right] = \left[ {\begin{array}{*{20}{c}} {\lambda +2G}&\lambda &\lambda \\ \lambda &{\lambda +2G}&\lambda \\ \lambda &\lambda &{\lambda +2G} \end{array}} \right]\left[\begin{array}{c} - \Delta \varepsilon _1^{\rm{p}} \\ v\Delta \varepsilon _1^{\rm{p}} \\ v\Delta \varepsilon _1^{\rm{p}} \\ \end{array} \right] = \left[ \begin{array}{c} - E\Delta \varepsilon _1^{\rm{p}} \\ 0 \\ 0 \\ \end{array} \right] $ 正确。仅主拉伸方向应力发生改变与事实相符
单轴压缩破坏 $\left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ \sigma _2^{\rm{p} } \\ \sigma _3^{\rm{p} } \\\end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ \sigma _3^{\rm{p} } \\\end{array} \right]$ $\left[ \begin{array}{c} \Delta {\sigma _1} \\ \Delta {\sigma _2} \\ \Delta {\sigma _3} \\ \end{array} \right] = \left[ {\begin{array}{*{20}{c} } {\lambda +2G}&\lambda &\lambda \\ \lambda &{\lambda +2G}&\lambda \\ \lambda &\lambda &{\lambda +2G} \end{array} } \right] \left[\begin{array}{c} v\Delta \varepsilon _3^{\rm{p} } \\ v\Delta \varepsilon _3^{\rm{p} } \\ - \Delta \varepsilon _3^{\rm{p} } \\ \end{array} \right] = \left[ \begin{array}{c}- \lambda \Delta \varepsilon _3^{\rm{p} } \\- \lambda \Delta \varepsilon _3^{\rm{p} }\\- \left( {\lambda +2G} \right)\Delta \varepsilon _3^{\rm{p} } \end{array} \right]$ 正确。仅主压缩方向应力发生改变，与事实相符
二向纯剪破坏 $\left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ \sigma _2^{\rm{p} } \\ \sigma _3^{\rm{p} } \\ \end{array} \right] = \left[ \begin{array}{c} \sigma _1^{\rm{p} } \\ 0 \\ - \sigma _1^{\rm{p} } \\\end{array} \right]$ $ \left[ \begin{array}{c} \Delta {\sigma _1} \\ \Delta {\sigma _2} \\ \Delta {\sigma _3} \\ \end{array} \right] = \left[ {\begin{array}{*{20}{c}} {\lambda +2G}&\lambda &\lambda \\ \lambda &{\lambda +2G}&\lambda \\ \lambda &\lambda &{\lambda +2G} \end{array}} \right] \left[ \begin{array}{c} \left( {1 - v} \right)\Delta \varepsilon _1^{\rm{p}} \\ 0 \\ - \left( {1 - v} \right)\Delta \varepsilon _1^{\rm{p}} \\\end{array} \right] = \left[\begin{array}{c} 2\left( {1 - v} \right)G\Delta \varepsilon _1^{\rm{p}} \\ 0 \\ - 2\left( {1 - v} \right)G\Delta \varepsilon _1^{\rm{p}} \\ \end{array} \right] $ 正确。只有1和3方向应力发生变化，与事实相符
表 5  基于泊松效应的改进塑性位势跌落方法的合理性
图 3  红砂岩单轴拉伸应力-应变模拟曲线对比
图 4  花岗岩三轴压缩应力-应变模拟曲线对比
MPa
试验数据 本研究方法 原方法
$ {\sigma _1} $ $ {\sigma _3} $ $ {\sigma _1} $ $ {\sigma _3} $ $ {\sigma _1} $ $ {\sigma _3} $
15.86 0 15.86 0 ?5.06 ?52.32
52.76 5 52.76 5 29.81 ?57.36
89.65 10 89.65 10 64.69 ?62.41
表 6  2种应力跌落计算方法的残余应力模拟结果对比
图 5  砂岩压剪应力-应变模拟曲线对比
图 6  岩石弹塑性变形破坏全过程计算流程
图 7  Tennessee大理岩及三峡花岗岩的三轴压缩试验模拟
图 8  圆隧围岩挖掘及有限元模型
图 9  圆隧力学响应有限元模拟结果与解析解对比
图 10  Mine-by隧洞有限元模型
参数 数值 参数 数值
弹性模量E/GPa 60 峰值内摩擦角φp/(o) 0
泊松比ν 0.25 残余内摩擦角φr/(o) 48
峰值黏聚力cp/MPa 40 剪胀角ψ/(o) 30
残余黏聚力cr/MPa 5 临界塑性应变η 0.003
表 7  Mine-by隧洞围岩计算参数
图 11  模拟的隧洞塑性区和破坏区
图 12  监测的隧洞塑性区和破坏区[29]
图 13  AK10+900断面开挖松动圈[31]
图 14  辅助洞有限元模型
参数 数值 参数 数值
弹性模量E/GPa 18.9 峰值内摩擦角φp/(o) 22.4
泊松比ν 0.22 残余内摩擦角φr/(o) 42
峰值黏聚力cp/MPa 20.9 剪胀角ψ/(o) 15
残余黏聚力cr/MPa 9.1 临界塑性应变η 0.005
表 8  辅助洞围岩计算参数
图 15  开挖模拟结果
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