脆塑性迭代逼近算法的改进

金俊超,景来红,杨风威,宋志宇,尚朋阳

Improvement of multi-step brittle-plastic approach

Jun-chao JIN,Lai-hong JING,Feng-wei YANG,Zhi-yu SONG,Peng-yang SHANG

表 2 最小主应力不变跌落方法存在的问题

Tab.2 Defects of existing calculation method based on constant minor principal stress in brittle-plastic process

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