| 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 |