#### 期刊菜单

Solution Method of Elastic Mechanics for Anisotropic Plate Based on Multiple Complex Functions
DOI: 10.12677/IJM.2021.103019, PDF, HTML, XML, 下载: 195  浏览: 359

Abstract: The anisotropic plate is the basic construction shape of composite materials. The solution method of elastic mechanics for its stress boundary problems must become the basis of loading analysis to engineering structures. By using the concept of multiple complex variable functions and selecting the typical anisotropic plate to be loaded on the local edge, the basic equations and multiple com-plex function methods have been established for solving the elastic mechanical problems. By means of introducing the multiple complex variables and the coordinate replace method, the stress fields of the anisotropic plate have been determined. The train of thought and the solution method in this paper may be an aid to improve the basic theory and research method in the composite mechanics.

1. 引言

2. 复合材料弹性力学的基本理论

$\frac{\partial {\sigma }_{x}}{\partial x}+\frac{\partial {\tau }_{xy}}{\partial y}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\partial {\tau }_{xy}}{\partial x}+\frac{\partial {\sigma }_{y}}{\partial y}=0$ (1)

${\sigma }_{x}=\frac{{\partial }^{2}F}{\partial {y}^{2}}={F}_{yy},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{y}=\frac{{\partial }^{2}F}{\partial {x}^{2}}={F}_{xx},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\tau }_{xy}=-\frac{{\partial }^{2}F}{\partial x\partial y}=-{F}_{xy}$ (2)

$\begin{array}{l}{\epsilon }_{x}={a}_{11}{\sigma }_{x}+{a}_{12}{\sigma }_{y}+{a}_{16}{\tau }_{xy}\\ {\epsilon }_{y}={a}_{12}{\sigma }_{x}+{a}_{22}{\sigma }_{y}+{a}_{26}{\tau }_{xy}\\ {\gamma }_{xy}={a}_{16}{\sigma }_{x}+{a}_{26}{\sigma }_{y}+{a}_{66}{\tau }_{xy}\end{array}\right\}$ (3)

$\frac{{\partial }^{2}{\epsilon }_{x}}{\partial {y}^{2}}+\frac{{\partial }^{2}{\epsilon }_{y}}{\partial {x}^{2}}=\frac{{\partial }^{2}{\gamma }_{xy}}{\partial x\partial y}$ (4)

$\frac{{\partial }^{4}F}{\partial {y}^{4}}+{A}_{1}\frac{{\partial }^{4}F}{\partial x\partial {y}^{3}}+{A}_{2}\frac{{\partial }^{4}F}{\partial {x}^{2}\partial {y}^{2}}+{A}_{3}\frac{{\partial }^{4}F}{\partial {x}^{3}\partial y}+{A}_{4}\frac{{\partial }^{4}F}{\partial {x}^{4}}=0$ (5)

Figure 1. Partial edge of anisotropic plate subjected to distributing pressure and coordinates

$\begin{array}{l}{\sigma }_{y}=-p,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\tau }_{xy}=0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(aa+b,\text{\hspace{0.17em}}y=0\right)\end{array}\right\}$ (6)

3. 各向异性板应力边值问题的多复变函数解法

$w=x+qy=x+gy+ihy,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{¯}{w}=x+\stackrel{¯}{q}y=x+gy-ihy$ (7)

$q+\stackrel{¯}{q}=2g,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}q-\stackrel{¯}{q}=2hi,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}q\stackrel{¯}{q}={g}^{2}+{h}^{2}$

$w\stackrel{¯}{w}=\left(x+qy\right)\left(x+\stackrel{¯}{q}y\right)={x}^{2}+2gxy+\left({g}^{2}+{h}^{2}\right){y}^{2}={\left(x+gy\right)}^{2}+{h}^{2}{y}^{2}$

$w=X+iY,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{¯}{w}=X-iY,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}w\stackrel{¯}{w}={X}^{2}+{Y}^{2}$

$L=|w|=|\stackrel{¯}{w}|=\sqrt{w\stackrel{¯}{w}}=\sqrt{{\left(x+gy\right)}^{2}+{h}^{2}{y}^{2}}=\sqrt{{X}^{2}+{Y}^{2}}$ (8)

$\begin{array}{l}{F}_{x}=\frac{\partial F}{\partial x}=\frac{\partial F}{\partial w}\frac{\partial w}{\partial x}+\frac{\partial F}{\partial \stackrel{¯}{w}}\frac{\partial \stackrel{¯}{w}}{\partial x}=\frac{\partial F}{\partial w}+\frac{\partial F}{\partial \stackrel{¯}{w}}\\ {F}_{y}=\frac{\partial F}{\partial y}=\frac{\partial F}{\partial w}\frac{\partial w}{\partial y}+\frac{\partial F}{\partial \stackrel{¯}{w}}\frac{\partial \stackrel{¯}{w}}{\partial y}=q\frac{\partial F}{\partial w}+\stackrel{¯}{q}\frac{\partial F}{\partial \stackrel{¯}{w}}\end{array}\right\}$ (9)

$\begin{array}{l}{F}_{xx}=\frac{{\partial }^{2}F}{\partial {x}^{2}}=\frac{{\partial }^{2}F}{\partial {w}^{2}}+2\frac{{\partial }^{2}F}{\partial w\partial \stackrel{¯}{w}}+\frac{{\partial }^{2}F}{\partial {\stackrel{¯}{w}}^{2}}\\ {F}_{yy}=\frac{{\partial }^{2}F}{\partial {y}^{2}}={q}^{2}\frac{{\partial }^{2}F}{\partial {w}^{2}}+2q\stackrel{¯}{q}\frac{{\partial }^{2}F}{\partial w\partial \stackrel{¯}{w}}+{\stackrel{¯}{q}}^{2}\frac{{\partial }^{2}F}{\partial {\stackrel{¯}{w}}^{2}}\\ {F}_{xy}=\frac{{\partial }^{2}F}{\partial x\partial y}=q\frac{{\partial }^{2}F}{\partial {w}^{2}}+2g\frac{{\partial }^{2}F}{\partial w\partial \stackrel{¯}{w}}+\stackrel{¯}{q}\frac{{\partial }^{2}F}{\partial {\stackrel{¯}{w}}^{2}}\end{array}\right\}$ (10)

$F=C\Psi \left(w\right)+\stackrel{¯}{C}\stackrel{¯}{\Psi }\left(\stackrel{¯}{w}\right)$ (11)

${F}_{x}=C{\Psi }^{\prime }+\stackrel{¯}{C}\stackrel{¯}{{\Psi }^{\prime }},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{F}_{y}=qC{\Psi }^{\prime }+\stackrel{¯}{q}\stackrel{¯}{C}\stackrel{¯}{{\Psi }^{\prime }}$

$\begin{array}{l}{F}_{xx}=C{\Psi }^{″}+\stackrel{¯}{C}{\stackrel{¯}{\Psi }}^{″}=C\Phi +\stackrel{¯}{C}\stackrel{¯}{\Phi }\\ {F}_{yy}={q}^{2}C{\Psi }^{″}+{\stackrel{¯}{q}}^{2}\stackrel{¯}{C}{\stackrel{¯}{\Psi }}^{″}={q}^{2}C\Phi +{\stackrel{¯}{q}}^{2}\stackrel{¯}{C}\stackrel{¯}{\Phi }\\ {F}_{xy}=qC{\Psi }^{″}+\stackrel{¯}{q}\stackrel{¯}{C}{\stackrel{¯}{\Psi }}^{″}=qC\Phi +\stackrel{¯}{q}\stackrel{¯}{C}\stackrel{¯}{\Phi }\end{array}$

$\begin{array}{l}{\sigma }_{x}={q}^{2}C\Phi +{\stackrel{¯}{q}}^{2}\stackrel{¯}{C}\stackrel{¯}{\Phi }=2\mathrm{Re}\left({q}^{2}C\Phi \right)\\ {\sigma }_{y}=C\Phi +\stackrel{¯}{C}\stackrel{¯}{\Phi }=2\mathrm{Re}\left(C\Phi \right)\\ {\tau }_{xy}=-qC\Phi -\stackrel{¯}{q}\stackrel{¯}{C}\stackrel{¯}{\Phi }=-2\mathrm{Re}\left(qC\Phi \right)\end{array}\right\}$ (12)

$\left({q}^{4}+{A}_{1}{q}^{3}+{A}_{2}{q}^{2}+{A}_{3}q+{A}_{4}\right)C{\Phi }^{″}+\left({\stackrel{¯}{q}}^{4}+{A}_{1}{\stackrel{¯}{q}}^{3}+{A}_{2}{\stackrel{¯}{q}}^{2}+{A}_{3}\stackrel{¯}{q}+{A}_{4}\right)\stackrel{¯}{C}{\stackrel{¯}{\Phi }}^{″}=0$

$\begin{array}{l}{q}^{4}+{A}_{1}{q}^{3}+{A}_{2}{q}^{2}+{A}_{3}q+{A}_{4}=0\\ {\stackrel{¯}{q}}^{4}+{A}_{1}{\stackrel{¯}{q}}^{3}+{A}_{2}{\stackrel{¯}{q}}^{2}+{A}_{3}\stackrel{¯}{q}+{A}_{4}=0\end{array}\right\}$ (13)

${x}_{1}=x-a={r}_{1}\mathrm{cos}{\theta }_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}_{2}=x-a-b={r}_{2}\mathrm{cos}{\theta }_{2}$

$y={y}_{1}={y}_{2}={r}_{1}sin{\theta }_{1}={r}_{2}\mathrm{sin}{\theta }_{2}$

$\begin{array}{l}{w}_{1}={x}_{1}+{q}_{1}{y}_{1}={x}_{1}+{g}_{1}{y}_{1}+i{h}_{1}{y}_{1}={r}_{1}\left(\mathrm{cos}{\theta }_{1}+{g}_{1}\mathrm{sin}{\theta }_{1}+i{h}_{1}\mathrm{sin}{\theta }_{1}\right)\\ {w}_{2}={x}_{1}+{q}_{2}{y}_{1}={x}_{1}+{g}_{2}{y}_{1}+i{h}_{2}{y}_{1}={r}_{1}\left(\mathrm{cos}{\theta }_{1}+{g}_{2}\mathrm{sin}{\theta }_{1}+i{h}_{2}\mathrm{sin}{\theta }_{1}\right)\\ {w}_{3}={x}_{2}+{q}_{1}{y}_{2}={x}_{2}+{g}_{1}{y}_{2}+i{h}_{1}{y}_{2}={r}_{2}\left(\mathrm{cos}{\theta }_{2}+{g}_{1}\mathrm{sin}{\theta }_{2}+i{h}_{1}\mathrm{sin}{\theta }_{2}\right)\\ {w}_{4}={x}_{2}+{q}_{2}{y}_{2}={x}_{2}+{g}_{2}{y}_{2}+i{h}_{2}{y}_{2}={r}_{2}\left(\mathrm{cos}{\theta }_{2}+{g}_{2}\mathrm{sin}{\theta }_{2}+i{h}_{2}\mathrm{sin}{\theta }_{2}\right)\end{array}\right\}$ (14)

${\Phi }_{1}=\mathrm{ln}{w}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\Phi }_{2}=\mathrm{ln}{w}_{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\Phi }_{3}=\mathrm{ln}{w}_{3},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\Phi }_{4}=\mathrm{ln}{w}_{4}$

$\begin{array}{l}{\sigma }_{x}=2\mathrm{Re}\left({q}_{1}^{2}{C}_{1}\mathrm{ln}{w}_{1}+{q}_{2}^{2}{C}_{2}\mathrm{ln}{w}_{2}+{q}_{1}^{2}{C}_{3}\mathrm{ln}{w}_{3}+{q}_{2}^{2}{C}_{4}\mathrm{ln}{w}_{4}\right)\\ {\sigma }_{y}=2\mathrm{Re}\left({C}_{1}\mathrm{ln}{w}_{1}+{C}_{2}\mathrm{ln}{w}_{2}+{C}_{3}\mathrm{ln}{w}_{3}+{C}_{4}\mathrm{ln}{w}_{4}\right)\\ {\tau }_{xy}=-2\mathrm{Re}\left({q}_{1}{C}_{1}\mathrm{ln}{w}_{1}+{q}_{2}{C}_{2}\mathrm{ln}{w}_{2}+{q}_{1}{C}_{3}\mathrm{ln}{w}_{3}+{q}_{2}{C}_{4}\mathrm{ln}{w}_{4}\right)\end{array}\right\}$ (15)

$\begin{array}{l}\mathrm{cos}{\theta }_{1}+{g}_{1}\mathrm{sin}{\theta }_{1}={J}_{1}cos{\beta }_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{h}_{1}\mathrm{sin}{\theta }_{1}={J}_{1}sin{\beta }_{1}\\ \mathrm{cos}{\theta }_{1}+{g}_{2}\mathrm{sin}{\theta }_{1}={J}_{2}cos{\beta }_{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{h}_{2}\mathrm{sin}{\theta }_{1}={J}_{2}sin{\beta }_{2}\\ \mathrm{cos}{\theta }_{2}+{g}_{1}\mathrm{sin}{\theta }_{2}={J}_{3}cos{\beta }_{3},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{h}_{1}\mathrm{sin}{\theta }_{2}={J}_{3}sin{\beta }_{3}\\ \mathrm{cos}{\theta }_{2}+{g}_{2}\mathrm{sin}{\theta }_{2}={J}_{4}cos{\beta }_{4},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{h}_{2}\mathrm{sin}{\theta }_{2}={J}_{4}sin{\beta }_{4}\end{array}\right\}$ (16)

$\begin{array}{l}\mathrm{tan}{\beta }_{1}=\frac{{h}_{1}\mathrm{sin}{\theta }_{1}}{\mathrm{cos}{\theta }_{1}+{g}_{1}\mathrm{sin}{\theta }_{1}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{J}_{1}=\sqrt{{\left(cos{\theta }_{1}+{g}_{1}sin{\theta }_{1}\right)}^{2}+{\left({h}_{1}\mathrm{sin}{\theta }_{1}\right)}^{2}}\\ \mathrm{tan}{\beta }_{2}=\frac{{h}_{2}\mathrm{sin}{\theta }_{1}}{\mathrm{cos}{\theta }_{1}+{g}_{2}\mathrm{sin}{\theta }_{1}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{J}_{2}=\sqrt{{\left(cos{\theta }_{1}+{g}_{2}sin{\theta }_{1}\right)}^{2}+{\left({h}_{2}\mathrm{sin}{\theta }_{1}\right)}^{2}}\\ \mathrm{tan}{\beta }_{3}=\frac{{h}_{1}\mathrm{sin}{\theta }_{2}}{\mathrm{cos}{\theta }_{2}+{g}_{1}\mathrm{sin}{\theta }_{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{J}_{3}=\sqrt{{\left(cos{\theta }_{2}+{g}_{1}sin{\theta }_{2}\right)}^{2}+{\left({h}_{1}\mathrm{sin}{\theta }_{2}\right)}^{2}}\\ \mathrm{tan}{\beta }_{4}=\frac{{h}_{2}\mathrm{sin}{\theta }_{2}}{\mathrm{cos}{\theta }_{2}+{g}_{2}\mathrm{sin}{\theta }_{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{J}_{4}=\sqrt{{\left(cos{\theta }_{2}+{g}_{2}sin{\theta }_{2}\right)}^{2}+{\left({h}_{2}\mathrm{sin}{\theta }_{2}\right)}^{2}}\end{array}\right\}$ (17)

${w}_{1}={r}_{1}{J}_{1}{\text{e}}^{i{\beta }_{1}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{w}_{2}={r}_{1}{J}_{2}{\text{e}}^{i{\beta }_{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{w}_{3}={r}_{2}{J}_{3}{\text{e}}^{i{\beta }_{3}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{w}_{4}={r}_{2}{J}_{4}{\text{e}}^{i{\beta }_{4}}$ (18)

$\begin{array}{l}{\sigma }_{x}=2\mathrm{Re}\left[{q}_{1}^{2}{C}_{1}\mathrm{ln}\left({r}_{1}{J}_{1}\right)+{q}_{2}^{2}{C}_{2}\mathrm{ln}\left({r}_{1}{J}_{2}\right)+{q}_{1}^{2}{C}_{3}\mathrm{ln}\left({r}_{2}{J}_{3}\right)+{q}_{2}^{2}{C}_{4}\mathrm{ln}\left({r}_{2}{J}_{4}\right)\right]\\ \begin{array}{cc}& \end{array}-2\mathrm{Im}\left({\beta }_{1}{q}_{1}^{2}{C}_{1}+{\beta }_{2}{q}_{2}^{2}{C}_{2}+{\beta }_{3}{q}_{1}^{2}{C}_{3}+{\beta }_{4}{q}_{2}^{2}{C}_{4}\right)\\ {\sigma }_{y}=2\mathrm{Re}\left[{C}_{1}\mathrm{ln}\left({r}_{1}{J}_{1}\right)+{C}_{2}\mathrm{ln}\left({r}_{1}{J}_{2}\right)+{C}_{3}\mathrm{ln}\left({r}_{2}{J}_{3}\right)+{C}_{4}\mathrm{ln}\left({r}_{2}{J}_{4}\right)\right]\\ \begin{array}{cc}& \end{array}-2\mathrm{Im}\left({\beta }_{1}{C}_{1}+{\beta }_{2}{C}_{2}+{\beta }_{3}{C}_{3}+{\beta }_{4}{C}_{4}\right)\\ {\tau }_{xy}=-2\mathrm{Re}\left[{q}_{1}{C}_{1}\mathrm{ln}\left({r}_{1}{J}_{1}\right)+{q}_{2}{C}_{2}\mathrm{ln}\left({r}_{1}{J}_{2}\right)+{q}_{1}{C}_{3}\mathrm{ln}\left({r}_{2}{J}_{3}\right)+{q}_{2}{C}_{4}\mathrm{ln}\left({r}_{2}{J}_{4}\right)\right]\\ \begin{array}{cc}& \end{array}+2\mathrm{Im}\left({\beta }_{1}{q}_{1}{C}_{1}+{\beta }_{2}{q}_{2}{C}_{2}+{\beta }_{3}{q}_{1}{C}_{3}+{\beta }_{4}{q}_{2}{C}_{4}\right)\end{array}\right\}$ (19)

① 对于图1右边自由边界 $\left(x>a+b,y=0\right)$${\theta }_{1}=0,\text{\hspace{0.17em}}{\theta }_{2}=0$，则有：

${\beta }_{1}={\beta }_{2}={\beta }_{3}={\beta }_{4}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{J}_{1}={J}_{2}={J}_{3}={J}_{4}=1$

${\sigma }_{y}=2\mathrm{Re}\left[\left({C}_{1}+{C}_{2}\right)\mathrm{ln}{r}_{1}+\left({C}_{3}+{C}_{4}\right)\mathrm{ln}{r}_{2}\right]=0$

${\tau }_{xy}=-2\mathrm{Re}\left[\left({q}_{1}{C}_{1}+{q}_{2}{C}_{2}\right)\mathrm{ln}{r}_{1}+\left({q}_{1}{C}_{3}+{q}_{2}{C}_{4}\right)\mathrm{ln}{r}_{2}\right]=0$

$\begin{array}{l}\mathrm{Re}\left({C}_{1}+{C}_{2}\right)=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Re\left({C}_{3}+{C}_{4}\right)=0\\ \mathrm{Re}\left({q}_{1}{C}_{1}+{q}_{2}{C}_{2}\right)=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Re\left({q}_{1}{C}_{3}+{q}_{2}{C}_{4}\right)=0\end{array}\right\}$ (20)

② 对于图1左边自由边界 $\left(x${\theta }_{1}=\pi ,\text{\hspace{0.17em}}{\theta }_{2}=\pi$，则有：

${\beta }_{1}={\beta }_{2}={\beta }_{3}={\beta }_{4}=\pi ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{J}_{1}={J}_{2}={J}_{3}={J}_{4}=1$

${\sigma }_{y}=2\mathrm{Re}\left[\left({C}_{1}+{C}_{2}\right)\mathrm{ln}{r}_{1}+\left({C}_{3}+{C}_{4}\right)\mathrm{ln}{r}_{2}\right]-2\pi \mathrm{Im}\left({C}_{1}+{C}_{2}+{C}_{3}+{C}_{4}\right)=0$

$\begin{array}{l}{\tau }_{xy}=-2\mathrm{Re}\left[\left({q}_{1}{C}_{1}+{q}_{2}{C}_{2}\right)\mathrm{ln}{r}_{1}+\left({q}_{1}{C}_{3}+{q}_{2}{C}_{4}\right)\mathrm{ln}{r}_{2}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+2\pi \mathrm{Im}\left({q}_{1}{C}_{1}+{q}_{2}{C}_{2}+{q}_{1}{C}_{3}+{q}_{2}{C}_{4}\right)=0\end{array}$

$\begin{array}{l}\mathrm{Im}\left({C}_{1}+{C}_{2}+{C}_{3}+{C}_{4}\right)=0\\ \mathrm{Im}\left({q}_{1}{C}_{1}+{q}_{2}{C}_{2}+{q}_{1}{C}_{3}+{q}_{2}{C}_{4}\right)=0\end{array}\right\}$ (21)

③ 对于图1中间受压边界 $\left(a${\theta }_{1}=0,\text{\hspace{0.17em}}{\theta }_{2}=\pi$，则有：

${\beta }_{1}={\beta }_{2}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\beta }_{3}={\beta }_{4}=\pi ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{J}_{1}={J}_{2}={J}_{3}={J}_{4}=1$

${\sigma }_{y}=2\mathrm{Re}\left[\left({C}_{1}+{C}_{2}\right)\mathrm{ln}{r}_{1}+\left({C}_{3}+{C}_{4}\right)\mathrm{ln}{r}_{2}\right]-2\pi \mathrm{Im}\left({C}_{3}+{C}_{4}\right)=-p$

${\tau }_{xy}=-2\mathrm{Re}\left[\left({q}_{1}{C}_{1}+{q}_{2}{C}_{2}\right)\mathrm{ln}{r}_{1}+\left({q}_{1}{C}_{3}+{q}_{2}{C}_{4}\right)\mathrm{ln}{r}_{2}\right]+2\pi \mathrm{Im}\left({q}_{1}{C}_{3}+{q}_{2}{C}_{4}\right)=0$

$2\pi \mathrm{Im}\left({C}_{3}+{C}_{4}\right)=p,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{Im}\left({q}_{1}{C}_{3}+{q}_{2}{C}_{4}\right)=0$ (22)

$\begin{array}{l}{C}_{1}=\frac{p}{2\pi }\frac{{D}_{3}-i{D}_{2}}{{D}_{12}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{C}_{2}=-\frac{p}{2\pi }\frac{{D}_{3}+i{D}_{1}}{{D}_{1}{}_{2}}\\ {C}_{3}=-\frac{p}{2\pi }\frac{{D}_{3}-i{D}_{2}}{{D}_{12}}=-{C}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{C}_{4}=\frac{p}{2\pi }\frac{{D}_{3}+i{D}_{1}}{{D}_{1}{}_{2}}=-{C}_{2}\end{array}\right\}$ (23)

$\begin{array}{l}{D}_{1}={g}_{1}\left({g}_{1}-{g}_{2}\right)+{h}_{1}\left({h}_{1}-{h}_{2}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{D}_{2}={g}_{2}\left({g}_{2}-{g}_{1}\right)+{h}_{2}\left({h}_{2}-{h}_{1}\right)\\ {D}_{12}={D}_{1}+{D}_{2}={\left({g}_{1}-{g}_{2}\right)}^{2}+{\left({h}_{1}-{h}_{2}\right)}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{D}_{3}={g}_{2}{h}_{1}-{g}_{1}{h}_{2}\end{array}\right\}$ (24)

$\begin{array}{l}{\sigma }_{x}=\frac{p}{\pi }\mathrm{Re}\left[\frac{{D}_{3}-i{D}_{2}}{{D}_{12}}{q}_{1}^{2}\mathrm{ln}\frac{{r}_{1}{J}_{1}}{{r}_{2}{J}_{3}}-\frac{{D}_{3}+i{D}_{1}}{{D}_{1}{}_{2}}{q}_{2}^{2}\mathrm{ln}\frac{{r}_{1}{J}_{2}}{{r}_{2}{J}_{4}}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }\text{\hspace{0.17em}}-\frac{p}{\pi }\mathrm{Im}\left[\frac{{D}_{3}-i{D}_{2}}{{D}_{12}}{q}_{1}^{2}\left({\beta }_{1}-{\beta }_{3}\right)-\frac{{D}_{3}+i{D}_{1}}{{D}_{1}{}_{2}}{q}_{2}^{2}\left({\beta }_{2}-{\beta }_{4}\right)\right]\end{array}$

$\begin{array}{l}{\sigma }_{y}=\frac{p}{\pi }\mathrm{Re}\left[\frac{{D}_{3}-i{D}_{2}}{{D}_{12}}\mathrm{ln}\frac{{r}_{1}{J}_{1}}{{r}_{2}{J}_{3}}-\frac{{D}_{3}+i{D}_{1}}{{D}_{1}{}_{2}}\mathrm{ln}\frac{{r}_{1}{J}_{2}}{{r}_{2}{J}_{4}}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\frac{p}{\pi }\mathrm{Im}\left[\left({\beta }_{1}-{\beta }_{3}\right)\frac{{D}_{3}-i{D}_{2}}{{D}_{12}}-\left({\beta }_{2}-{\beta }_{4}\right)\frac{{D}_{3}+i{D}_{1}}{{D}_{1}{}_{2}}\right]\end{array}$

$\begin{array}{l}{\tau }_{xy}=-\frac{p}{\pi }\mathrm{Re}\left[\frac{{D}_{3}-i{D}_{2}}{{D}_{12}}{q}_{1}\mathrm{ln}\frac{{r}_{1}{J}_{1}}{{r}_{2}{J}_{3}}-\frac{{D}_{3}+i{D}_{1}}{{D}_{1}{}_{2}}{q}_{2}\mathrm{ln}\frac{{r}_{1}{J}_{2}}{{r}_{2}{J}_{4}}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{p}{\pi }\mathrm{Im}\left[\frac{{D}_{3}-i{D}_{2}}{{D}_{12}}{q}_{1}\left({\beta }_{1}-{\beta }_{3}\right)-\frac{{D}_{3}+i{D}_{1}}{{D}_{1}{}_{2}}{q}_{2}\left({\beta }_{2}-{\beta }_{4}\right)\right]\end{array}$

$\begin{array}{l}{h}_{1}{D}_{3}-{g}_{1}{D}_{2}={h}_{2}{D}_{3}+{g}_{2}{D}_{1}={g}_{2}\left({g}_{1}^{2}+{h}_{1}^{2}\right)-{g}_{1}\left({g}_{2}^{2}+{h}_{2}^{2}\right)={D}_{4}\\ {g}_{1}{D}_{3}+{h}_{1}{D}_{2}={g}_{2}{D}_{3}-{h}_{2}{D}_{1}={h}_{1}\left({g}_{2}^{2}+{h}_{2}^{2}\right)-{h}_{2}\left({g}_{1}^{2}+{h}_{1}^{2}\right)={D}_{5}\\ \left({D}_{3}-i{D}_{2}\right){q}_{1}=\left({D}_{3}+i{D}_{1}\right){q}_{2}={D}_{5}+i{D}_{4}\\ {D}_{6}={g}_{2}{D}_{4}+{h}_{2}{D}_{5},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{D}_{7}={g}_{1}{D}_{4}+{h}_{1}{D}_{5}\\ {D}_{8}={g}_{1}{D}_{5}-{h}_{1}{D}_{4},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{D}_{9}={g}_{2}{D}_{5}-{h}_{2}{D}_{4}\end{array}\right\}$ (25)

$\begin{array}{l}{\sigma }_{x}=-\frac{p}{\pi }\left[\frac{{D}_{6}}{{D}_{1}{}_{2}}\left({\beta }_{4}-{\beta }_{2}\right)-\frac{{D}_{7}}{{D}_{12}}\left({\beta }_{3}-{\beta }_{1}\right)-\frac{{D}_{8}}{{D}_{12}}\mathrm{ln}\frac{{r}_{1}{J}_{1}}{{r}_{2}{J}_{3}}+\frac{{D}_{9}}{{D}_{1}{}_{2}}\mathrm{ln}\frac{{r}_{1}{J}_{2}}{{r}_{2}{J}_{4}}\right]\\ {\sigma }_{y}=-\frac{p}{\pi }\left[\frac{{D}_{1}}{{D}_{12}}\left({\beta }_{4}-{\beta }_{2}\right)+\frac{{D}_{2}}{{D}_{12}}\left({\beta }_{3}-{\beta }_{1}\right)-\frac{{D}_{3}}{{D}_{12}}\mathrm{ln}\frac{{J}_{1}{J}_{4}}{{J}_{2}{J}_{3}}\right]\\ {\tau }_{xy}=\frac{p}{\pi }\left[\frac{{D}_{4}}{{D}_{1}{}_{2}}\left({\beta }_{1}-{\beta }_{2}-{\beta }_{3}+{\beta }_{4}\right)-\frac{{D}_{5}}{{D}_{12}}\mathrm{ln}\frac{{J}_{1}{J}_{4}}{{J}_{2}{J}_{3}}\right]\end{array}\right\}$ (26)

4. 计算举例

$\begin{array}{l}{a}_{11}=60{\left(TPa\right)}^{-1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{a}_{22}=120{\left(TPa\right)}^{-1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{a}_{66}=160{\left(TPa\right)}^{-1}\\ {a}_{12}=-20{\left(TPa\right)}^{-1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{a}_{16}={a}_{26}=-30{\left(TPa\right)}^{-1}\end{array}$

${A}_{1}=-\frac{2{a}_{16}}{{a}_{11}}=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{2}=\frac{{a}_{66}+2{a}_{12}}{{a}_{11}}=2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{3}=-\frac{2{a}_{26}}{{a}_{11}}=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{4}=\frac{{a}_{22}}{{a}_{11}}=2$

${q}^{4}+{A}_{1}{q}^{3}+{A}_{2}{q}^{2}+{A}_{3}q+{A}_{4}={q}^{4}+{q}^{3}+2{q}^{2}+q+2=0$

${q}_{1}=0.2762+1.07223i,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{q}_{2}=-0.7762+1.01447i$

${q}_{3}=0.2762-1.07223i,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{q}_{4}=-0.7762-1.01447i$

$\mathrm{tan}{\beta }_{1}=\frac{1.07223\mathrm{sin}{\theta }_{1}}{\mathrm{cos}{\theta }_{1}+0.2762\mathrm{sin}{\theta }_{1}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{J}_{1}=\sqrt{{\left(cos{\theta }_{1}+0.2762sin{\theta }_{1}\right)}^{2}+{\left(1.07223\mathrm{sin}{\theta }_{1}\right)}^{2}}$

$\mathrm{tan}{\beta }_{2}=\frac{1.01447\mathrm{sin}{\theta }_{1}}{\mathrm{cos}{\theta }_{1}-0.7762\mathrm{sin}{\theta }_{1}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{J}_{2}=\sqrt{{\left(cos{\theta }_{1}-0.7762sin{\theta }_{1}\right)}^{2}+{\left(1.01447\mathrm{sin}{\theta }_{1}\right)}^{2}}$

$\begin{array}{l}{D}_{1}=0.3526,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{D}_{2}=0.7584,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{D}_{12}=1.111\\ {D}_{3}=-1.1125,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{D}_{4}=-1.4023,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{D}_{5}=0.5058\\ {D}_{6}=1.602,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{D}_{7}=0.155,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{D}_{8}=1.643,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{D}_{9}=1.03\end{array}$

$\begin{array}{l}{\sigma }_{x}=-p\left[1.442\frac{{\beta }_{4}-{\beta }_{2}}{\pi }-0.14\frac{{\beta }_{3}-{\beta }_{1}}{\pi }-0.471\mathrm{ln}\frac{{r}_{1}{J}_{1}}{{r}_{2}{J}_{3}}+0.295\mathrm{ln}\frac{{r}_{1}{J}_{2}}{{r}_{2}{J}_{4}}\right]\\ {\sigma }_{y}=-p\left[0.3174\frac{{\beta }_{4}-{\beta }_{2}}{\pi }+0.6826\frac{{\beta }_{3}-{\beta }_{1}}{\pi }+0.319\mathrm{ln}\frac{{J}_{1}{J}_{4}}{{J}_{2}{J}_{3}}\right]\\ {\tau }_{xy}=-p\left[1.2622\frac{{\beta }_{1}-{\beta }_{2}-{\beta }_{3}+{\beta }_{4}}{\pi }+0.145\mathrm{ln}\frac{{J}_{1}{J}_{4}}{{J}_{2}{J}_{3}}\right]\end{array}\right\}$ (27)

Figure 2. Relationship curves of parameter variations in coordinate replace

5. 结论

 [1] 史济怀. 多复变函数论基础[M]. 北京: 高等教育出版社, 2014. [2] 陆启铿, 殷慰萍. 多复变在中国的研究与发展[M]. 北京: 科学出版社, 2009. [3] 孟晓任, 曹廷彬. 多复变整函数涉及全导数的唯一性定理[J]. 数学年刊A辑, 2014, 35(2): 203-210. [4] 赵娜, 李娜. 基于多复变函数论的不定积分计算方法[J]. 科技通报, 2018, 34(12): 12-15. [5] 卞星明, 文远芳, 黄斐然. 基于泛复变函数求解Maxwell方程的方法[J]. 高电压技术, 2006, 32(4): 34-36. [6] 熊锡金. 泛系函数论的理法扩变——泛复变函数论: 理念•方法论•定理[J]. 计算机与数字工程, 2013, 41(9): 1391-1394. [7] 高健, 刘官厅. 含共线双半无限裂纹的正交异性复合材料板平面弹性问题的解析解[J]. 应用力学学报, 2016, 33(1): 1-6. [8] 贾普荣. 复合材料平面应力问题的弹性力学解法[J]. 力学研究, 2020, 9(3): 95-102. https://doi.org/10.12677/IJM.2020.93011 [9] 贾普荣. 基于泛复函的各向异性板裂纹尖端应力场解法[J]. 力学研究, 2021, 10(2): 90-98. https://doi.org/10.12677/IJM.2021.102009