﻿ 一维六方准晶中带双裂纹的椭圆孔口问题的塑性模拟

# 一维六方准晶中带双裂纹的椭圆孔口问题的塑性模拟Plastic Simulation of Elliptical Orifice with Double Cracks in One-Dimensional Hexagonal Quasi-Crystals

Abstract: In this paper, the problem of elliptical hole with double cracks in one-dimensional hexagonal quasi-crystals is simulated. By using conformal mapping method, the physical plane problem that is difficult to deal with in quasi-crystals is transformed into a familiar and regular mathematical plane problem. Based on the existing theory, the size of the cohesive force area at the tip of the double cracks in the elliptical hole is obtained. In the case of limit, it is the same as the existing result.

1. 引言

2. 一维六方准晶中带双裂纹椭圆孔口问题的塑性模拟

2.1. 一维六方准晶中的基本公式

$\left\{\begin{array}{l}{\sigma }_{11}={C}_{11}{\epsilon }_{11}+{C}_{12}{\epsilon }_{22}+{C}_{13}{\epsilon }_{33}+{R}_{1}{w}_{3},{\sigma }_{22}={C}_{12}{\epsilon }_{11}+{C}_{11}{\epsilon }_{22}+{C}_{13}{\epsilon }_{33}+{R}_{1}{w}_{3}\hfill \\ {\sigma }_{33}={C}_{13}{\epsilon }_{11}+{C}_{13}{\epsilon }_{22}+{C}_{33}{\epsilon }_{33}+{R}_{2}{w}_{3},{\sigma }_{23}={\sigma }_{32}=2{C}_{44}{\epsilon }_{23}+{R}_{3}{w}_{2}\hfill \\ {\sigma }_{31}={\sigma }_{13}=2{C}_{44}{\epsilon }_{31}+{R}_{3}{w}_{1},{\sigma }_{12}={\sigma }_{21}=2{C}_{66}{\epsilon }_{12},{H}_{1}=2{R}_{3}{\epsilon }_{31}+{K}_{2}{w}_{1}\hfill \\ {H}_{2}=2{R}_{3}{\epsilon }_{23}+{K}_{2}{w}_{2},{H}_{3}={R}_{1}\left({\epsilon }_{11}+{\epsilon }_{22}\right)+{R}_{2}{\epsilon }_{33}+{K}_{1}{w}_{3}\hfill \end{array}$ (1)

${\epsilon }_{ij}=\left({\partial }_{j}{u}_{i}+{\partial }_{i}{u}_{j}\right)/2,{w}_{j}={\partial }_{j}v$ (2)

$\left\{\begin{array}{l}{\partial }_{1}{\sigma }_{11}+{\partial }_{2}{\sigma }_{12}+{\partial }_{3}{\sigma }_{13}=0,{\partial }_{1}{\sigma }_{21}+{\partial }_{2}{\sigma }_{22}+{\partial }_{3}{\sigma }_{23}=0\hfill \\ {\partial }_{1}{\sigma }_{31}+{\partial }_{2}{\sigma }_{32}+{\partial }_{3}{\sigma }_{33}=0,{\partial }_{1}{H}_{1}+{\partial }_{2}{H}_{2}+{\partial }_{3}{H}_{3}=0\hfill \end{array}$ (3)

${\partial }_{3}{u}_{i}=0,{\partial }_{3}v=0,{\partial }_{3}{\sigma }_{ij}=0,{\partial }_{3}{H}_{j}=0,i,j=1,2,3$ (4)

${\sigma }_{23}={\sigma }_{32}=2{C}_{44}{\epsilon }_{23}+{R}_{3}{w}_{2},{\sigma }_{31}={\sigma }_{13}=2{C}_{44}{\epsilon }_{31}+{R}_{3}{w}_{1}$ (5)

${H}_{1}=2{R}_{3}{\epsilon }_{31}+{K}_{2}{w}_{1},{H}_{2}=2{R}_{3}{\epsilon }_{23}+{K}_{2}{w}_{2}$ (6)

${\partial }_{1}{\sigma }_{31}+{\partial }_{2}{\sigma }_{32}=0,{\partial }_{1}{H}_{1}+{\partial }_{2}{H}_{2}=0$ (7)

${\epsilon }_{3j}={\epsilon }_{j3}={\partial }_{j}{u}_{3}/2,{w}_{j}={\partial }_{j}v,j=1,2$ (8)

${C}_{44}{\nabla }^{2}{u}_{3}+{R}_{3}{\nabla }^{2}v=0$${R}_{3}{\nabla }^{2}{u}_{3}+{K}_{2}{\nabla }^{2}v=0$，其中， ${\nabla }^{2}={\partial }^{2}/\partial {x}_{1}^{2}+{\partial }^{2}/\partial {x}_{2}^{2}$

${C}_{44}{K}_{2}-{R}_{3}^{2}\ne 0$ 时，有

${\nabla }^{2}{u}_{3}=0,{\nabla }^{2}v=0$ (9)

${u}_{3}=\mathrm{Re}{\varphi }_{1}\left(z\right),v=\mathrm{Re}{\psi }_{1}\left(z\right)$ (10)

2.2. 一维六方准晶中带双裂纹椭圆孔口问题的塑性模拟

${\partial }_{1}{\sigma }_{11}+{\partial }_{2}{\sigma }_{12}=0,{\partial }_{1}{\sigma }_{21}+{\partial }_{2}{\sigma }_{22}=0,{\partial }_{1}{\sigma }_{31}+{\partial }_{2}{\sigma }_{32}=0,{\partial }_{1}{H}_{1}+{\partial }_{2}{H}_{2}=0.$ (11)

$\left\{\begin{array}{l}\sqrt{{x}_{1}^{2}+{x}_{2}^{2}}\to \infty :{\sigma }_{32}=\tau ,{H}_{3}={\tau }_{0},{\sigma }_{31}={H}_{1}={H}_{2}=0\hfill \\ \frac{{x}_{1}^{2}}{{a}^{2}}+\frac{{x}_{2}^{2}}{{b}^{2}}=1:{\sigma }_{32}=0,{H}_{2}=0\hfill \\ a<|{x}_{1}| (12)

$\left\{\begin{array}{l}\sqrt{{x}_{1}^{2}+{x}_{2}^{2}}\to \infty :{\sigma }_{32}=\tau ,{H}_{3}={\tau }_{0},{\sigma }_{31}={H}_{1}={H}_{2}=0\hfill \\ \frac{{x}_{1}^{2}}{{a}^{2}}+\frac{{x}_{2}^{2}}{{b}^{2}}=1:{\sigma }_{32}=0,{H}_{2}=0\hfill \\ a<|{x}_{1}| (13a)

$\left\{\begin{array}{l}\sqrt{{x}_{1}^{2}+{x}_{2}^{2}}\to \infty :{\sigma }_{32}={\sigma }_{31}={H}_{1}={H}_{2}=0\hfill \\ \frac{{x}_{1}^{2}}{{a}^{2}}+\frac{{x}_{2}^{2}}{{b}^{2}}=1:{\sigma }_{32}=0,{H}_{2}=0\hfill \\ a<|{x}_{1}| (13b)

Figure 1. Plastic simulation of elliptical orifice with double cracks in one-dimensional hexagonal quasi-crystals

$\left\{\begin{array}{l}\sqrt{{x}_{1}^{2}+{x}_{2}^{2}}\to \infty :{\sigma }_{32}={\sigma }_{31}={H}_{1}={H}_{2}=0\hfill \\ \frac{{x}_{1}^{2}}{{a}^{2}}+\frac{{x}_{2}^{2}}{{b}^{2}}=1:{\sigma }_{32}=-\tau ,{H}_{2}=-{\tau }_{0}\hfill \\ a<|{x}_{1}| (14)

$z={x}_{1}+i{x}_{2},i=\sqrt{-1}$，若 $f\left(z\right)$ 为解析函数，则有

$\frac{\partial f}{\partial {x}_{1}}=\frac{\text{d}f}{\text{d}z},\frac{\partial f}{\partial {x}_{2}}=i\frac{\text{d}f}{\text{d}z}$ (15)

$f\left(z\right)=P\left({x}_{1},{x}_{2}\right)+iQ\left({x}_{1},{x}_{2}\right)=\mathrm{Re}f\left(z\right)+i\mathrm{Im}f\left(z\right)$ (16)

$\frac{\partial P}{\partial {x}_{1}}=\frac{\partial Q}{\partial {x}_{2}},\frac{\partial P}{\partial {x}_{2}}=-\frac{\partial Q}{\partial {x}_{1}}$ (17)

$\left\{\begin{array}{l}{\sigma }_{23}={\sigma }_{32}={C}_{44}\frac{\partial }{\partial {x}_{2}}\mathrm{Re}{\varphi }_{1}+{R}_{3}\frac{\partial }{\partial {x}_{2}}\mathrm{Re}{\psi }_{1}\hfill \\ {\sigma }_{13}={\sigma }_{31}={C}_{44}\frac{\partial }{\partial {x}_{1}}\mathrm{Re}{\varphi }_{1}+{R}_{3}\frac{\partial }{\partial {x}_{1}}\mathrm{Re}{\psi }_{1}\hfill \\ {H}_{1}={K}_{2}\frac{\partial }{\partial {x}_{1}}\mathrm{Re}{\psi }_{1}+{R}_{3}\frac{\partial }{\partial {x}_{1}}\mathrm{Re}{\varphi }_{1},{H}_{2}={K}_{2}\frac{\partial }{\partial {x}_{2}}\mathrm{Re}{\psi }_{1}+{R}_{3}\frac{\partial }{\partial {x}_{2}}\mathrm{Re}{\varphi }_{1}\hfill \end{array}$ (18)

${\sigma }_{31}-i{\sigma }_{32}={C}_{44}{{\varphi }^{\prime }}_{1}+{R}_{3}{{\psi }^{\prime }}_{1},{H}_{1}-i{H}_{2}={K}_{2}{{\psi }^{\prime }}_{1}+{R}_{3}{{\varphi }^{\prime }}_{1}$ (19)

$\left\{\begin{array}{l}{\sigma }_{32}={\sigma }_{23}=-\frac{1}{2i}\left[{C}_{44}\left({{\varphi }^{\prime }}_{1}-\stackrel{¯}{{{\varphi }^{\prime }}_{1}}\right)+{R}_{3}\left({{\psi }^{\prime }}_{1}-\stackrel{¯}{{{\psi }^{\prime }}_{1}}\right)\right]+\tau \hfill \\ {H}_{2}=-\frac{1}{2i}\left[{K}_{2}\left({{\psi }^{\prime }}_{1}-\stackrel{¯}{{{\psi }^{\prime }}_{1}}\right)+{R}_{3}\left({{\varphi }^{\prime }}_{1}-\stackrel{¯}{{{\varphi }^{\prime }}_{1}}\right)\right]+{\tau }_{0}\hfill \end{array}$ (20)

$\left\{\begin{array}{l}{C}_{44}\left({{\varphi }^{\prime }}_{1}-\stackrel{¯}{{{\varphi }^{\prime }}_{1}}\right)+{R}_{3}\left({\psi }_{1}-\stackrel{¯}{{{\psi }^{\prime }}_{1}}\right)=2\tau i\hfill \\ {K}_{2}\left({{\psi }^{\prime }}_{1}-\stackrel{¯}{{{\psi }^{\prime }}_{1}}\right)+{R}_{3}\left({{\varphi }^{\prime }}_{1}-\stackrel{¯}{{{\varphi }^{\prime }}_{1}}\right)=2{\tau }_{0}i\hfill \end{array}$ $z\in L$ (21)

$\begin{array}{l}z=\omega \left(\zeta \right)=\frac{a+b}{2}\frac{2\left({d}^{2}+1\right)\left(1+{\zeta }^{2}\right)+\left({d}^{2}-1\right)\sqrt{{\left(1+\zeta \right)}^{4}+2k{\left(1-{\zeta }^{2}\right)}^{2}+{\left(1-\zeta \right)}^{4}}}{8d\zeta }\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{a-b}{2}\frac{8d\zeta }{2\left({d}^{2}+1\right)\left(1+{\zeta }^{2}\right)+\left({d}^{2}-1\right)\sqrt{{\left(1+\zeta \right)}^{4}+2k{\left(1-{\zeta }^{2}\right)}^{2}+{\left(1-\zeta \right)}^{4}}}\end{array}$ (22)

$d=\frac{\left(c+R\right)+\sqrt{{\left(c+R\right)}^{2}-{a}^{2}+{b}^{2}}}{a+b},k=\frac{{d}^{4}+6{d}^{2}+1}{{\left({d}^{2}-1\right)}^{2}}$ (23)

$\varphi \left(\zeta \right)={\varphi }_{1}\left(z\right)={\varphi }_{1}\left(\omega \left(\zeta \right)\right),\psi \left(\zeta \right)={\psi }_{1}\left(z\right)={\psi }_{1}\left(\omega \left(\zeta \right)\right)$，则有

${{\varphi }^{\prime }}_{1}\left(z\right)={\varphi }^{\prime }\left(\zeta \right)/{\omega }^{\prime }\left(\zeta \right),{{\psi }^{\prime }}_{1}\left(z\right)={\psi }^{\prime }\left(\zeta \right)/{\omega }^{\prime }\left(\zeta \right)$ (24)

$\begin{array}{l}{\varphi }^{\prime }\left(\sigma \right)-\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\stackrel{¯}{{\varphi }^{\prime }\left(\sigma \right)}+\frac{{R}_{3}}{{C}_{44}}\left[{\psi }^{\prime }\left(\sigma \right)-\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\stackrel{¯}{{\psi }^{\prime }\left(\sigma \right)}\right]=\frac{2\tau i}{{C}_{44}}{\omega }^{\prime }\left(\sigma \right),\\ {\psi }^{\prime }\left(\sigma \right)-\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\stackrel{¯}{{\psi }^{\prime }\left(\sigma \right)}+\frac{{R}_{3}}{{K}_{2}}\left[{\varphi }^{\prime }\left(\sigma \right)-\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\stackrel{¯}{{\varphi }^{\prime }\left(\sigma \right)}\right]=\frac{2{\tau }_{0}i}{{K}_{2}}{\omega }^{\prime }\left( \sigma \right)\end{array}$

$\begin{array}{l}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\varphi }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma -\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\frac{\stackrel{¯}{{\varphi }^{\prime }\left(\sigma \right)}}{\sigma -\zeta }\text{d}\sigma +\frac{{R}_{3}}{{C}_{44}}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\psi }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma \\ -\frac{{R}_{3}}{{C}_{44}}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\frac{\stackrel{¯}{{\psi }^{\prime }\left(\sigma \right)}}{\sigma -\zeta }\text{d}\sigma =\frac{2\tau i}{{C}_{44}}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma \end{array}$ (25)

$\begin{array}{l}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\psi }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma -\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\frac{\stackrel{¯}{{\psi }^{\prime }\left(\sigma \right)}}{\sigma -\zeta }\text{d}\sigma +\frac{{R}_{3}}{{K}_{2}}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\varphi }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma \\ -\frac{{R}_{3}}{{K}_{2}}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\frac{\stackrel{¯}{{\varphi }^{\prime }\left(\sigma \right)}}{\sigma -\zeta }\text{d}\sigma =\frac{2{\tau }_{0}i}{{K}_{2}}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma \end{array}$ (26)

$\frac{1}{2\pi i}{\int }_{\tau }\frac{{\varphi }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma ={\varphi }^{\prime }\left(\zeta \right),\frac{1}{2\pi i}{\int }_{\tau }\frac{{\psi }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma ={\psi }^{\prime }\left(\zeta \right),|\zeta |<1$ (27)

$\begin{array}{c}{\omega }^{\prime }\left(\zeta \right)=\left(1-{\zeta }^{2}\right)\left[\left({d}^{2}+1\right)+\frac{\left({d}^{2}-1\right)\left(1+k\right)\left(1+{\zeta }^{2}\right)}{\sqrt{{\left(1+\zeta \right)}^{4}+2k{\left(1-{\zeta }^{2}\right)}^{2}+{\left(1-\zeta \right)}^{4}}}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left\{-\frac{a+b}{8d{\zeta }^{2}}+8d\left(a-b\right)/\left\{{\left[2\left({d}^{2}+1\right)\left(1+{\zeta }^{2}\right)+\left({d}^{2}-1\right)\sqrt{{\left(1+\zeta \right)}^{4}+2k{\left(1-{\zeta }^{2}\right)}^{2}+{\left(1-\zeta \right)}^{4}}\right]}^{2}\right\}\right\}\end{array}$ (28)

${\varphi }^{\prime }\left(\zeta \right)$${\psi }^{\prime }\left(\zeta \right)$ 可以展成下面的Taylor级数，因为它们都是 $|\zeta |<1$ 内的解析函数，

${\varphi }^{\prime }\left(\zeta \right)=\underset{n=0}{\overset{\infty }{\sum }}{a}_{n}{\zeta }^{n},{\psi }^{\prime }\left(\zeta \right)=\underset{n=0}{\overset{\infty }{\sum }}{b}_{n}{\zeta }^{n},|\zeta |<1$

$\stackrel{¯}{{\varphi }^{\prime }\left(\sigma \right)}$$\stackrel{¯}{{\psi }^{\prime }\left(\sigma \right)}$ 可看作 $\stackrel{¯}{{\varphi }^{\prime }}\left(\frac{1}{\zeta }\right)=\underset{n=0}{\overset{\infty }{\sum }}\stackrel{¯}{{a}_{n}}{\left(\frac{1}{\zeta }\right)}^{n}$$\stackrel{¯}{{\psi }^{\prime }}\left(\frac{1}{\zeta }\right)=\underset{n=0}{\overset{\infty }{\sum }}\stackrel{¯}{{b}_{n}}{\left(\frac{1}{\zeta }\right)}^{n}$ 在单位圆周 $\tau$ 上的边值，所以(25)与(26)中的 $\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\stackrel{¯}{{\varphi }^{\prime }\left(\sigma \right)}$$\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\stackrel{¯}{{\psi }^{\prime }\left(\sigma \right)}$$|\zeta |>1$ 内的函数 $-\frac{1}{{\zeta }^{2}}\stackrel{¯}{{\varphi }^{\prime }}\left(\frac{1}{\zeta }\right)$$-\frac{1}{{\zeta }^{2}}\stackrel{¯}{{\psi }^{\prime }}\left(\frac{1}{\zeta }\right)$ 的边值。利用文献 [7] 中所提到的无穷远处的Cauchy积分公式，即， $|\zeta |<1$，有

$\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\frac{\stackrel{¯}{{\varphi }^{\prime }\left(\sigma \right)}}{\sigma -\zeta }\text{d}\sigma =\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\frac{\stackrel{¯}{{\psi }^{\prime }\left(\sigma \right)}}{\sigma -\zeta }\text{d}\sigma =0$ (29)

$\left\{\begin{array}{l}{\varphi }^{\prime }\left(\zeta \right)+\frac{{R}_{3}}{{C}_{44}}{\psi }^{\prime }\left(\zeta \right)=\frac{2\tau i}{{C}_{44}}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma \hfill \\ {\psi }^{\prime }\left(\zeta \right)+\frac{{R}_{3}}{{K}_{2}}{\varphi }^{\prime }\left(\zeta \right)=\frac{2{\tau }_{0}i}{{K}_{2}}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma \hfill \end{array}$ (30)

$\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma ={\omega }^{\prime }\left(\infty \right)=\frac{\left({d}^{2}+1\right)\left(a+b\right)}{4d}=F\left(\zeta \right)$ (31)

${\varphi }^{\prime }\left(\zeta \right)+\frac{{R}_{3}}{{C}_{44}}{\psi }^{\prime }\left(\zeta \right)=\frac{2\tau i}{{C}_{44}}F\left(\zeta \right),{\psi }^{\prime }\left(\zeta \right)+\frac{{R}_{3}}{{K}_{2}}{\varphi }^{\prime }\left(\zeta \right)=\frac{2{\tau }_{0}i}{{K}_{2}}F\left(\zeta \right)$ (32)

${\varphi }^{\prime }\left(\zeta \right)=2i\frac{\tau {K}_{2}-{\tau }_{0}{R}_{3}}{{C}_{44}{K}_{2}-{R}_{3}^{2}}F\left(\zeta \right),{\psi }^{\prime }\left(\zeta \right)=2i\frac{{\tau }_{0}{C}_{44}-\tau {R}_{3}}{{C}_{44}{K}_{2}-{R}_{3}^{2}}F\left(\zeta \right)$ (33)

$\left\{\begin{array}{l}{{\varphi }^{\prime }}_{1}\left(z\right)=\frac{{\varphi }^{\prime }\left(\zeta \right)}{{\omega }^{\prime }\left(\zeta \right)}=\frac{2i\left(\tau {K}_{2}-{\tau }_{0}{R}_{3}\right)}{{C}_{44}{K}_{2}-{R}_{3}^{2}}\frac{F\left(\zeta \right)}{{\omega }^{\prime }\left(\zeta \right)}\hfill \\ {{\psi }^{\prime }}_{1}\left(z\right)=\frac{{\psi }^{\prime }\left(\zeta \right)}{{\omega }^{\prime }\left(\zeta \right)}=\frac{2i\left({\tau }_{0}{C}_{44}-\tau {R}_{3}\right)}{{C}_{44}{K}_{2}-{R}_{3}^{2}}\frac{F\left(\zeta \right)}{{\omega }^{\prime }\left(\zeta \right)}\hfill \end{array}$ (34)

${\sigma }_{31}-i{\sigma }_{32}=2\tau i\frac{F\left(\zeta \right)}{{\omega }^{\prime }\left(\zeta \right)},{H}_{1}-i{H}_{2}=2{\tau }_{0}i\frac{F\left(\zeta \right)}{{\omega }^{\prime }\left(\zeta \right)}$ (35)

$\left\{\begin{array}{l}{C}_{44}\left({{\varphi }^{\prime }}_{1}-\stackrel{¯}{{{\varphi }^{\prime }}_{1}}\right)+{R}_{3}\left({{\psi }^{\prime }}_{1}-\stackrel{¯}{{{\psi }^{\prime }}_{1}}\right)=-2{\sigma }_{Y}i,\hfill \\ {K}_{2}\left({{\psi }^{\prime }}_{1}-\stackrel{¯}{{{\psi }^{\prime }}_{1}}\right)+{R}_{3}\left({{\varphi }^{\prime }}_{1}-\stackrel{¯}{{{\varphi }^{\prime }}_{1}}\right)=0.\hfill \end{array}$ (36)

$\left\{\begin{array}{l}{\varphi }^{\prime }\left(\sigma \right)-\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\stackrel{¯}{{\varphi }^{\prime }\left(\sigma \right)}+\frac{{R}_{3}}{{C}_{44}}\left[{\psi }^{\prime }\left(\sigma \right)-\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\stackrel{¯}{{\psi }^{\prime }\left(\sigma \right)}\right]=-\frac{2{\sigma }_{Y}i}{{C}_{44}}{\omega }^{\prime }\left(\sigma \right),\hfill \\ {\psi }^{\prime }\left(\sigma \right)-\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\stackrel{¯}{{\psi }^{\prime }\left(\sigma \right)}+\frac{{R}_{3}}{{K}_{2}}\left[{\varphi }^{\prime }\left(\sigma \right)-\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\stackrel{¯}{{\varphi }^{\prime }\left(\sigma \right)}\right]=0.\hfill \end{array}$

$\begin{array}{l}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\varphi }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma -\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\frac{\stackrel{¯}{{\varphi }^{\prime }\left(\sigma \right)}}{\sigma -\zeta }\text{d}\sigma +\frac{{R}_{3}}{{C}_{44}}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\psi }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma \\ -\frac{{R}_{3}}{{C}_{44}}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\frac{\stackrel{¯}{{\psi }^{\prime }\left(\sigma \right)}}{\sigma -\zeta }\text{d}\sigma =-\frac{2{\sigma }_{Y}i}{{C}_{44}}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma \end{array}$

$\begin{array}{l}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\psi }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma -\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\frac{\stackrel{¯}{{\psi }^{\prime }\left(\sigma \right)}}{\sigma -\zeta }\text{d}\sigma +\frac{{R}_{3}}{{K}_{2}}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\varphi }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma \\ -\frac{{R}_{3}}{{K}_{2}}\frac{1}{2\pi i}{\int }_{\tau }\frac{{\omega }^{\prime }\left(\sigma \right)}{\stackrel{¯}{{\omega }^{\prime }\left(\sigma \right)}}\frac{\stackrel{¯}{{\varphi }^{\prime }\left(\sigma \right)}}{\sigma -\zeta }\text{d}\sigma =\text{0}\end{array}$

$\frac{1}{2\pi i}{\int }_{\tau }\frac{{\varphi }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma ={\varphi }^{\prime }\left(\zeta \right),\frac{1}{2\pi i}{\int }_{\tau }\frac{{\psi }^{\prime }\left(\sigma \right)}{\sigma -\zeta }\text{d}\sigma ={\psi }^{\prime }\left(\zeta \right),|\zeta |<1$

${\varphi }^{\prime }\left(\zeta \right)+\frac{{R}_{3}}{{C}_{44}}{\psi }^{\prime }\left(\zeta \right)=-\frac{2{\sigma }_{Y}i}{{C}_{44}}F\left(\zeta \right),{\psi }^{\prime }\left(\zeta \right)+\frac{{R}_{3}}{{K}_{2}}{\varphi }^{\prime }\left(\zeta \right)=0$

${\varphi }^{\prime }\left(\zeta \right)=\frac{2i{K}_{2}{\sigma }_{Y}}{{R}_{3}^{2}-{K}_{2}{C}_{44}}F\left(\zeta \right),{\psi }^{\prime }\left(\zeta \right)=\frac{2i{R}_{3}{\sigma }_{Y}}{{K}_{2}{C}_{44}-{R}_{3}^{2}}F\left(\zeta \right)$ (37)

$\left\{\begin{array}{l}{{\varphi }^{\prime }}_{1}\left(z\right)=\frac{{\varphi }^{\prime }\left(\zeta \right)}{{\omega }^{\prime }\left(\zeta \right)}=\frac{2i{K}_{2}{\sigma }_{Y}}{{R}_{3}^{2}-{C}_{44}{K}_{2}}\frac{F\left(\zeta \right)}{{\omega }^{\prime }\left(\zeta \right)}\\ {{\psi }^{\prime }}_{1}\left(z\right)=\frac{{\psi }^{\prime }\left(\zeta \right)}{{\omega }^{\prime }\left(\zeta \right)}=\frac{2i{R}_{3}{\sigma }_{Y}}{{C}_{44}{K}_{2}-{R}_{3}^{2}}\frac{F\left(\zeta \right)}{{\omega }^{\prime }\left(\zeta \right)}\end{array}$ (38)

${\sigma }_{31}-i{\sigma }_{32}=-2{\sigma }_{Y}i\frac{F\left(\zeta \right)}{{\omega }^{\prime }\left(\zeta \right)},{H}_{1}-i{H}_{2}=0$

3. 应力强度因子的计算

${K}_{\text{Ш}}^{‖}=\underset{\zeta \to {\zeta }_{1}}{\mathrm{lim}}2\sqrt{\pi }\tau \frac{F\left(\zeta \right)}{\sqrt{{\omega }^{″}\left(\zeta \right)}},{K}_{\text{Ш}}^{\perp }=\underset{\zeta \to {\zeta }_{1}}{\mathrm{lim}}2\sqrt{\pi }{\tau }_{0}\frac{F\left(\zeta \right)}{\sqrt{{\omega }^{″}\left(\zeta \right)}}$ (39)

$\sqrt{{\omega }^{″}\left(\zeta \right)}\to \frac{\sqrt{\left({d}^{2}+1\right)\left[\left({d}^{2}-1\right)a+\left({d}^{2}+1\right)b\right]}}{\sqrt{2d\left({d}^{2}-1\right)}},\zeta \to 1$ (40)

$F\left(\zeta \right)\to \frac{\left({d}^{2}+1\right)\left(a+b\right)}{4d},\zeta \to 1$ (41)

$\frac{F\left(\zeta \right)}{\sqrt{{\omega }^{″}\left(\zeta \right)}}\to \frac{\left(a+b\right)\sqrt{2\left({d}^{4}-1\right)}}{4\sqrt{d\left[\left({d}^{2}-1\right)a+\left({d}^{2}+1\right)b\right]}},\zeta \to 1$ (42)

$\begin{array}{l}{K}_{\text{Ш}}^{‖}=\underset{\zeta \to 1}{\mathrm{lim}}2\sqrt{\pi }\tau \frac{F\left(\zeta \right)}{\sqrt{{\omega }^{″}\left(\zeta \right)}}=\frac{\tau \left(a+b\right)\sqrt{\pi \left({d}^{4}-1\right)}}{\sqrt{2d\left[\left({d}^{2}-1\right)a+\left({d}^{2}+1\right)b\right]}}\\ {K}_{\text{Ш}}^{\perp }=\underset{\zeta \to 1}{\mathrm{lim}}2\sqrt{\pi }{\tau }_{0}\frac{F\left(\zeta \right)}{\sqrt{{\omega }^{″}\left(\zeta \right)}}=\frac{{\tau }_{0}\left(a+b\right)\sqrt{\pi \left({d}^{4}-1\right)}}{\sqrt{2d\left[\left({d}^{2}-1\right)a+\left({d}^{2}+1\right)b\right]}}\end{array}$ (43)

${K}_{\text{Ш}}^{{\sigma }_{Y}}=\underset{\zeta \to 1}{\mathrm{lim}}2\sqrt{\pi }\left(-{\sigma }_{Y}\right)\frac{F\left(\zeta \right)}{\sqrt{{\omega }^{″}\left(\zeta \right)}}=-\frac{{\sigma }_{Y}\left(a+b\right)\sqrt{\pi \left({d}^{4}-1\right)}}{\sqrt{2d\left[\left({d}^{2}-1\right)a+\left({d}^{2}+1\right)b\right]}}$ (44)

4. 裂纹尖端塑性区尺寸的计算

${K}_{\text{Ш}}^{‖}+{K}_{\text{Ш}}^{{\sigma }_{Y}}=\frac{\tau \left(a+b\right)\sqrt{\pi \left({d}^{4}-1\right)}}{\sqrt{2d\left[\left({d}^{2}-1\right)a+\left({d}^{2}+1\right)b\right]}}-\frac{{\sigma }_{Y}\left(a+b\right)\sqrt{\pi \left({d}^{4}-1\right)}}{\sqrt{2d\left[\left({d}^{2}-1\right)a+\left({d}^{2}+1\right)b\right]}}=0$

$R=a\left[\mathrm{sec}\left(\frac{\pi }{2}\frac{\tau }{{\sigma }_{Y}}\right)-1\right]$

$R=c\left(\frac{{\tau }^{2}}{{\sigma }_{Y}^{2}}-1\right)$

$b\to 0$ 时， $d\to \frac{c+\sqrt{{c}^{2}-{a}^{2}}}{a}$，则 ${d}^{2}=\frac{2c\left(c+\sqrt{{c}^{2}-{a}^{2}}\right)}{{a}^{2}}-1$，且有 ${d}^{4}=\frac{8{c}^{4}-8{a}^{2}{c}^{2}+4c\left(2{c}^{2}-{a}^{2}\right)\sqrt{{c}^{2}-{a}^{2}}}{{a}^{4}}+1$，则对于式 ${K}_{\text{І}}^{\tau }+{K}_{\text{І}}^{{\sigma }_{Y}}=0$${K}_{\text{І}}^{{\sigma }_{Y}}$ 式中有 ${d}_{\sigma }\to \frac{c+R+\sqrt{{\left(c+R\right)}^{2}-{a}^{2}}}{a}$，经过化解，得 $\frac{{\tau }^{2}}{{\sigma }_{Y}^{2}}=\frac{\left({d}_{\sigma }^{4}-1\right)d\left({d}^{2}-1\right)}{\left({d}^{4}-1\right){d}_{\sigma }\left({d}_{\sigma }^{2}-1\right)}$，再令 $a\to 0$，有 $\frac{8{c}^{3}\cdot 16{\left(c+R\right)}^{4}}{16{c}^{4}\cdot 8{\left(c+R\right)}^{3}}=\frac{c+R}{c}=\frac{{\tau }^{2}}{{\sigma }_{Y}^{2}}$

5. 结语

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