﻿ 正交异性材料弹性力学通解 General Solution of Elastic Mechanics for Orthotropic Materials

International Journal of Mechanics Research
Vol. 11  No. 03 ( 2022 ), Article ID: 55555 , 9 pages
10.12677/IJM.2022.113007

General Solution of Elastic Mechanics for Orthotropic Materials

Purong Jia

School of Mechanics and Civil Engineering & Architecture, Northwestern Polytechnical University, Xi’an Shaanxi

Received: Jun. 13th, 2022; accepted: Jul. 1st, 2022; published: Sep. 6th, 2022

ABSTRACT

The investigation of the mechanic property for orthotropic materials must be of great significance for the design procedure and engineering application of composite constructions. According to the constitutive relations of the orthotropic materials, the basic equations of elastic mechanics have been established to solve the stress boundary problem of the orthotropic plate. By the method of the coordinate transition and real variable functional analysis, the partial derivation equation of the orthotropic plate has been fully solved. To take the wedge plate supported a concentrated force for the typical example, and by selecting the reasonable harmonious functions, the general solutions of stress fields are derived for the orthotropic plate.

Keywords:Orthotropic Materials, Elastic Mechanics, Coordinate Transition, Harmonious Function, Stress Field

1. 引言

2. 复合材料弹性力学基本方程

${\epsilon }_{x}=\frac{{\sigma }_{x}}{{E}_{1}}-\frac{{\mu }_{12}{\sigma }_{y}}{{E}_{1}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\epsilon }_{y}=\frac{{\sigma }_{y}}{{E}_{2}}-\frac{{\mu }_{12}{\sigma }_{x}}{{E}_{1}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\gamma }_{xy}=\frac{{\tau }_{xy}}{{G}_{12}}$ (1)

$\begin{array}{l}{\epsilon }_{x}=\frac{1-{\mu }_{13}{\mu }_{31}}{{E}_{1}}{\sigma }_{x}-\frac{{\mu }_{12}+{\mu }_{13}{\mu }_{32}}{{E}_{1}}{\sigma }_{y}\\ {\epsilon }_{y}=\frac{1-{\mu }_{23}{\mu }_{32}}{{E}_{2}}{\sigma }_{y}-\frac{{\mu }_{12}+{\mu }_{13}{\mu }_{32}}{{E}_{1}}{\sigma }_{x}\\ {\gamma }_{xy}=\frac{{\tau }_{xy}}{{G}_{12}}\end{array}\right\}$ (2)

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

$\frac{{\partial }^{4}F}{\partial {y}^{4}}+2B\frac{{\partial }^{4}F}{\partial {x}^{2}\partial {y}^{2}}+C\frac{{\partial }^{4}F}{\partial {x}^{4}}=0$ (5)

$B=\frac{{E}_{1}}{2{G}_{12}}-{\mu }_{12},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}C=\frac{{E}_{1}}{{E}_{2}}$ (平面应力)

$B=\frac{{E}_{1}-2{G}_{12}\left({\mu }_{12}+{\mu }_{13}{\mu }_{32}\right)}{2{G}_{12}\left(1-{\mu }_{13}{\mu }_{31}\right)},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}C=\frac{{E}_{1}\left(1-{\mu }_{23}{\mu }_{32}\right)}{{E}_{2}\left(1-{\mu }_{13}{\mu }_{31}\right)}$ (平面应变)

$\left[1\right]\text{\hspace{0.17em}}\text{ }B=\sqrt{C},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[2\right]\text{\hspace{0.17em}}\text{ }B>\sqrt{C},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[3\right]\text{\hspace{0.17em}}\text{ }B<\sqrt{C}$

${\sigma }_{\theta }=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\tau }_{r\theta }=0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }\left(\theta =±\alpha \right)$ (6)

Figure 1. Orthotropic material plate subjected to a concentrated force

$\begin{array}{l}{\sigma }_{r}={\sigma }_{x}{\mathrm{cos}}^{2}\theta +{\sigma }_{y}{\mathrm{sin}}^{2}\theta +{\tau }_{xy}\mathrm{sin}2\theta \\ {\sigma }_{\theta }={\sigma }_{x}si{n}^{2}\theta +{\sigma }_{y}co{s}^{2}\theta -{\tau }_{xy}\mathrm{sin}2\theta \\ {\tau }_{r\theta }=\left({\sigma }_{y}-{\sigma }_{x}\right)sin\theta cos\theta +{\tau }_{xy}\mathrm{cos}2\theta \end{array}\right\}$ (7)

3. 基于坐标变换的应力场解法

3.1. 含参数的坐标变换法

$\begin{array}{l}X=x,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Y=hy\\ x=r\mathrm{cos}\theta ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=rsin\theta \\ X=L\mathrm{cos}\beta ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Y=Lsin\beta \end{array}\right\}$ (8)

$L=r\lambda ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\lambda =\sqrt{{\mathrm{cos}}^{2}\theta +{h}^{2}{\mathrm{sin}}^{2}\theta }$ (9)

$\mathrm{cos}\beta =\frac{\mathrm{cos}\theta }{\lambda },\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{sin}\beta =\frac{h\mathrm{sin}\theta }{\lambda },\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{tan}\beta =h\mathrm{tan}\theta$ (10)

$F=F\left(x,y\right)=U\left(X,Y\right)=U$

$\frac{\partial F}{\partial x}=\frac{\partial U}{\partial X}\frac{\partial X}{\partial x}+\frac{\partial U}{\partial Y}\frac{\partial Y}{\partial x}=\frac{\partial U}{\partial X},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\partial F}{\partial y}=\frac{\partial U}{\partial X}\frac{\partial X}{\partial y}+\frac{\partial U}{\partial Y}\frac{\partial Y}{\partial y}=h\frac{\partial U}{\partial Y}$

$\frac{{\partial }^{2}F}{\partial {x}^{2}}=\frac{{\partial }^{2}U}{\partial {X}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{\partial }^{2}F}{\partial {y}^{2}}={h}^{2}\frac{{\partial }^{2}U}{\partial {Y}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{\partial }^{2}F}{\partial x\partial y}=h\frac{{\partial }^{2}U}{\partial X\partial Y}$ (11)

${h}^{4}\frac{{\partial }^{4}U}{\partial {Y}^{4}}+2B{h}^{2}\frac{{\partial }^{4}U}{\partial {X}^{2}\partial {Y}^{2}}+C\frac{{\partial }^{4}U}{\partial {X}^{4}}=0$ (12)

3.2. 特殊情况的应力函数解析( ${B}^{2}=C$ )

$\frac{{\partial }^{4}U}{\partial {Y}^{4}}+2\frac{B}{{h}^{2}}\frac{{\partial }^{4}U}{\partial {X}^{2}\partial {Y}^{2}}+\frac{{B}^{2}}{{h}^{4}}\frac{{\partial }^{4}U}{\partial {X}^{4}}=0$

$h=\sqrt{B}=\sqrt[4]{C}$ (13)

$\frac{{\partial }^{4}U}{\partial {Y}^{4}}+2\frac{{\partial }^{4}U}{\partial {X}^{2}\partial {Y}^{2}}+\frac{{\partial }^{4}U}{\partial {X}^{4}}=\left(\frac{{\partial }^{2}}{\partial {Y}^{2}}+\frac{{\partial }^{2}}{\partial {X}^{2}}\right)\left(\frac{{\partial }^{2}U}{\partial {Y}^{2}}+\frac{{\partial }^{2}U}{\partial {X}^{2}}\right)=0$ (14)

${\sigma }_{X}=\frac{{\partial }^{2}U}{\partial {Y}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{Y}=\frac{{\partial }^{2}U}{\partial {X}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\tau }_{XY}=-\frac{{\partial }^{2}U}{\partial X\partial Y}$ (15)

${\sigma }_{x}={h}^{2}\frac{{\partial }^{2}U}{\partial {Y}^{2}}={h}^{2}{\sigma }_{X},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{y}=\frac{{\partial }^{2}U}{\partial {X}^{2}}={\sigma }_{Y},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\tau }_{xy}=-h\frac{{\partial }^{2}U}{\partial X\partial Y}=h{\tau }_{XY}$

${\sigma }_{x}={h}^{2}{{\sigma }_{X}|}_{Y\to hy}^{X\to x},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{y}={{\sigma }_{Y}|}_{Y\to hy}^{X\to x},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\tau }_{xy}=h{{\tau }_{XY}|}_{Y\to hy}^{X\to x}$ (16)

《实例分析》如图1所示，楔形板在顶端承受压力P作用，两斜边自由( $\theta =±\alpha$ )。选定材料主方向(1,2)与坐标轴( $x,y$ )取向平行。在 $X-Y$ 平面内选取函数U为：

$U={A}_{1}Y\mathrm{arctan}\frac{Y}{X}$ (17)

$\frac{\partial U}{\partial X}=-\frac{{A}_{1}{Y}^{2}}{{X}^{2}+{Y}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\partial U}{\partial Y}={A}_{1}\left(\mathrm{arctan}\frac{Y}{X}+\frac{XY}{{X}^{2}+{Y}^{2}}\right)$

$\frac{{\partial }^{2}U}{\partial {X}^{2}}+\frac{{\partial }^{2}U}{\partial {Y}^{2}}=\frac{2{A}_{1}X}{{X}^{2}+{Y}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\frac{{\partial }^{2}}{\partial {X}^{2}}+\frac{{\partial }^{2}}{\partial {Y}^{2}}\right)\left(\frac{{\partial }^{2}U}{\partial {X}^{2}}+\frac{{\partial }^{2}U}{\partial {Y}^{2}}\right)=0$

${\sigma }_{X}=\frac{2{A}_{1}{X}^{3}}{{\left({X}^{2}+{Y}^{2}\right)}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{Y}=\frac{2{A}_{1}X{Y}^{2}}{{\left({X}^{2}+{Y}^{2}\right)}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\tau }_{XY}=\frac{2{A}_{1}{X}^{2}Y}{{\left({X}^{2}+{Y}^{2}\right)}^{2}}$ (18)

${\sigma }_{x}=\frac{2{A}_{1}{h}^{2}{x}^{3}}{{\left({x}^{2}+{h}^{2}{y}^{2}\right)}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{y}=\frac{2{A}_{1}{h}^{2}x{y}^{2}}{{\left({x}^{2}+{h}^{2}{y}^{2}\right)}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\tau }_{xy}=\frac{2{A}_{1}{h}^{2}{x}^{2}y}{{\left({x}^{2}+{h}^{2}{y}^{2}\right)}^{2}}$ (19)

${\sigma }_{x}=\frac{2{A}_{1}{h}^{2}{\mathrm{cos}}^{3}\theta }{r{\lambda }^{4}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{y}=\frac{2{A}_{1}{h}^{2}{\mathrm{sin}}^{2}\theta \mathrm{cos}\theta }{r{\lambda }^{4}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\tau }_{xy}=\frac{2{A}_{1}{h}^{2}\mathrm{sin}\theta {\mathrm{cos}}^{2}\theta }{r{\lambda }^{4}}$

${\sigma }_{r}=\frac{2{A}_{1}{h}^{2}\mathrm{cos}\theta }{r{\lambda }^{4}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{\theta }={\tau }_{r\theta }=0$ (20)

$P+{\int }_{-x\mathrm{tan}\alpha }^{x\mathrm{tan}\alpha }{\sigma }_{x}\text{d}y=P+{\int }_{-x\mathrm{tan}\alpha }^{x\mathrm{tan}\alpha }\frac{2{A}_{1}{h}^{2}{x}^{3}}{{\left({x}^{2}+{h}^{2}{y}^{2}\right)}^{2}}\text{d}y=0$ (21)

$2{A}_{1}h\left[\omega +\frac{\mathrm{tan}\omega }{1+{\mathrm{tan}}^{2}\omega }\right]=2{A}_{1}h\left[\omega +\mathrm{sin}\omega \mathrm{cos}\omega \right]=-P$

$2{A}_{1}h=\frac{-P}{\omega +\mathrm{sin}\omega \mathrm{cos}\omega }=\frac{-P}{\phi }$

${\sigma }_{r}=\frac{-Ph\mathrm{cos}\theta }{\phi r{\left({\mathrm{cos}}^{2}\theta +{h}^{2}{\mathrm{sin}}^{2}\theta \right)}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{\theta }={\tau }_{r\theta }=0$ (22)

${\sigma }_{x}=\frac{-Ph{x}^{3}}{\phi {\left({x}^{2}+{h}^{2}{y}^{2}\right)}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\sigma }_{y}=\frac{-Phx{y}^{2}}{\phi {\left({x}^{2}+{h}^{2}{y}^{2}\right)}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\tau }_{xy}=\frac{-Ph{x}^{2}y}{\phi {\left({x}^{2}+{h}^{2}{y}^{2}\right)}^{2}}$ (23)

3.3. 正交异性材料平面应力场通解( $B>\sqrt{C}$ )

$\frac{{\partial }^{2}U}{\partial {X}^{2}}+\frac{{\partial }^{2}U}{\partial {Y}^{2}}=0$ (24)

$\left({h}^{4}-2B{h}^{2}+C\right)\frac{{\partial }^{4}U}{\partial {X}^{4}}=0$

${h}_{1}=\sqrt{B+\sqrt{{B}^{2}-C}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{h}_{2}=\sqrt{B-\sqrt{{B}^{2}-C}}$ (25)

$F={F}_{1}\left(x,{h}_{1}y\right)+{F}_{2}\left(x,{h}_{2}y\right)={U}_{1}\left(X,{Y}_{1}\right)+{U}_{2}\left(X,{Y}_{2}\right)={U}_{1}+{U}_{2}$ (26)

$\frac{{\partial }^{2}{U}_{1}}{\partial {X}^{2}}+\frac{{\partial }^{2}{U}_{1}}{\partial {Y}_{1}^{2}}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{\partial }^{2}{U}_{2}}{\partial {X}^{2}}+\frac{{\partial }^{2}{U}_{2}}{\partial {Y}_{2}^{2}}=0$ (27)

《实例解析》再以图1所示的楔形板顶端受压力P为例，且设正交异性材料弹性常数满足： $B>\sqrt{C}$，两斜边自由( $\theta =±\alpha$ )。选定材料主方向(1,2)与坐标轴( $x,y$ )取向平行。为了便于分析，先选择 $X-Y$ 面内的函数 ${U}_{0}$，并确定为调和函数：

${U}_{0}=X\mathrm{ln}\sqrt{{X}^{2}+{Y}^{2}}-Y\mathrm{arctan}\frac{Y}{X}$ (28)

$\frac{\partial {U}_{0}}{\partial X}=\mathrm{ln}\sqrt{{X}^{2}+{Y}^{2}}+1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\partial {U}_{0}}{\partial Y}=-\mathrm{arctan}\frac{Y}{X}$

$\frac{{\partial }^{2}{U}_{0}}{\partial {X}^{2}}=\frac{X}{{X}^{2}+{Y}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{\partial }^{2}{U}_{0}}{\partial {Y}^{2}}=-\frac{X}{{X}^{2}+{Y}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{\partial }^{2}{U}_{0}}{\partial {X}^{2}}+\frac{{\partial }^{2}{U}_{0}}{\partial {Y}^{2}}=0$

$\begin{array}{l}F={A}_{1}\left(X\mathrm{ln}\sqrt{{X}^{2}+{Y}_{1}^{2}}-{Y}_{1}\mathrm{arctan}\frac{{Y}_{1}}{X}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+{A}_{2}\left(X\mathrm{ln}\sqrt{{X}^{2}+{Y}_{2}^{2}}-{Y}_{2}\mathrm{arctan}\frac{{Y}_{2}}{X}\right)\end{array}$ (29)

${\sigma }_{x}=\frac{{\partial }^{2}F}{\partial {y}^{2}}=-\frac{{A}_{1}{h}_{1}^{2}X}{{X}^{2}+{Y}_{1}^{2}}-\frac{{A}_{2}{h}_{2}^{2}X}{{X}^{2}+{Y}_{2}^{2}}=-\frac{{A}_{1}{h}_{1}^{2}x}{{x}^{2}+{h}_{1}^{2}{y}^{2}}-\frac{{A}_{2}{h}_{2}^{2}x}{{x}^{2}+{h}_{2}^{2}{y}^{2}}$ (30-1)

${\sigma }_{y}=\frac{{\partial }^{2}F}{\partial {x}^{2}}=\frac{{A}_{1}X}{{X}^{2}+{Y}_{1}^{2}}+\frac{{A}_{2}X}{{X}^{2}+{Y}_{2}^{2}}=\frac{{A}_{1}x}{{x}^{2}+{h}_{1}^{2}{y}^{2}}+\frac{{A}_{2}x}{{x}^{2}+{h}_{2}^{2}{y}^{2}}$ (30-2)

${\tau }_{xy}=-\frac{{\partial }^{2}F}{\partial x\partial y}=-\frac{{A}_{1}{h}_{1}{Y}_{1}}{{X}^{2}+{Y}_{1}^{2}}-\frac{{A}_{2}{h}_{2}{Y}_{2}}{{X}^{2}+{Y}_{2}^{2}}=-\frac{{A}_{1}{h}_{1}^{2}y}{{x}^{2}+{h}_{1}^{2}{y}^{2}}-\frac{{A}_{2}{h}_{2}^{2}y}{{x}^{2}+{h}_{2}^{2}{y}^{2}}$ (30-3)

$\begin{array}{l}{\sigma }_{r}=-\frac{\mathrm{cos}\theta }{r}\left[{A}_{1}\frac{{h}_{1}^{2}+\left({h}_{1}^{2}-1\right){\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta +{h}_{1}^{2}{\mathrm{sin}}^{2}\theta }+{A}_{2}\frac{{h}_{2}^{2}+\left({h}_{2}^{2}-1\right){\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta +{h}_{2}^{2}{\mathrm{sin}}^{2}\theta }\right]\\ {\sigma }_{\theta }=\frac{\mathrm{cos}\theta }{r}\left({A}_{1}+{A}_{2}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\tau }_{r\theta }=\frac{\mathrm{sin}\theta }{r}\left({A}_{1}+{A}_{2}\right)\end{array}$ (31)

$P+{\int }_{{y}_{1}}^{{y}_{2}}{\sigma }_{x}\text{d}y=P+{\int }_{-x\mathrm{tan}\alpha }^{x\mathrm{tan}\alpha }\left[-\frac{{A}_{1}{h}_{1}^{2}x}{{x}^{2}+{h}_{1}^{2}{y}^{2}}+\frac{{A}_{1}{h}_{2}^{2}x}{{x}^{2}+{h}_{2}^{2}{y}^{2}}\right]\text{d}y=0$ (32)

${A}_{1}{|{h}_{1}\mathrm{arctan}\frac{{h}_{1}y}{x}-{h}_{2}\mathrm{arctan}\frac{{h}_{2}y}{x}|}_{y=-x\mathrm{tan}\alpha }^{y=x\mathrm{tan}\alpha }=P$

$2{A}_{1}\left[{h}_{1}\mathrm{arctan}\left({h}_{1}\mathrm{tan}\alpha \right)-{h}_{2}\mathrm{arctan}\left({h}_{2}\mathrm{tan}\alpha \right)\right]=P$

${A}_{1}=\frac{P}{2\left({h}_{1}{\omega }_{1}-{h}_{2}{\omega }_{2}\right)}=\frac{P}{2\delta }$ (33)

$\begin{array}{l}{\sigma }_{r}=-\frac{P\mathrm{cos}\theta }{2\delta r}\left[\frac{{h}_{1}^{2}+\left({h}_{1}^{2}-1\right){\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta +{h}_{1}^{2}{\mathrm{sin}}^{2}\theta }-\frac{{h}_{2}^{2}+\left({h}_{2}^{2}-1\right){\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta +{h}_{2}^{2}{\mathrm{sin}}^{2}\theta }\right]\\ {\sigma }_{\theta }={\tau }_{r\theta }=0\end{array}$ (34)

$\begin{array}{l}{\sigma }_{x}=-\frac{P}{2\delta }\left(\frac{{h}_{1}^{2}x}{{x}^{2}+{h}_{1}^{2}{y}^{2}}-\frac{{h}_{2}^{2}x}{{x}^{2}+{h}_{2}^{2}{y}^{2}}\right)=-\frac{P\mathrm{cos}\theta }{2\delta r}\left(\frac{{h}_{1}^{2}}{{\lambda }_{1}^{2}}-\frac{{h}_{2}^{2}}{{\lambda }_{2}^{2}}\right)\\ {\sigma }_{y}=-\frac{P}{2\delta }\left(\frac{x}{{x}^{2}+{h}_{2}^{2}{y}^{2}}-\frac{x}{{x}^{2}+{h}_{1}^{2}{y}^{2}}\right)=-\frac{P\mathrm{cos}\theta }{2\delta r}\left(\frac{1}{{\lambda }_{2}^{2}}-\frac{1}{{\lambda }_{1}^{2}}\right)\\ {\tau }_{xy}=-\frac{P}{2\delta }\left(\frac{{h}_{1}^{2}y}{{x}^{2}+{h}_{1}^{2}{y}^{2}}-\frac{{h}_{2}^{2}y}{{x}^{2}+{h}_{2}^{2}{y}^{2}}\right)=-\frac{P\mathrm{sin}\theta }{2\delta r}\left(\frac{{h}_{1}^{2}}{{\lambda }_{1}^{2}}-\frac{{h}_{2}^{2}}{{\lambda }_{2}^{2}}\right)\end{array}$ (35)

《算例》为了说明上列式中各个参数的确定方法，选择正交异性板处于平面应力状态，工程弹性常数设为：

${E}_{1}=84000\text{\hspace{0.17em}}\text{MPa},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{E}_{2}=12000\text{\hspace{0.17em}}\text{MPa},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{G}_{12}=6000\text{\hspace{0.17em}}\text{MPa},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\mu }_{12}=0.25$

$B=\frac{{E}_{1}}{2{G}_{12}}-{\mu }_{12}=\frac{84}{2×6}-0.25=6.75,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}C=\frac{{E}_{1}}{{E}_{2}}=\frac{84}{12}=7$

${h}_{1}=\sqrt{B+\sqrt{{B}^{2}-C}}=3.6,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{h}_{2}=\sqrt{B-\sqrt{{B}^{2}-C}}=0.735$

${\omega }_{1}={74.5}^{\circ }=1.3\left(\text{rad}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\omega }_{2}={36.3}^{\circ }=0.634\left( rad \right)$

$\delta ={h}_{1}{\omega }_{1}-{h}_{2}{\omega }_{2}=4.214$

$\begin{array}{l}{\sigma }_{x}=-\frac{P}{8.428}\left(\frac{12.96x}{{x}^{2}+12.96{y}^{2}}-\frac{0.54x}{{x}^{2}+0.54{y}^{2}}\right)\\ {\sigma }_{y}=-\frac{P}{8.428}\left(\frac{x}{{x}^{2}+0.54{y}^{2}}-\frac{x}{{x}^{2}+12.96{y}^{2}}\right)\\ {\tau }_{xy}=-\frac{P}{8.428}\left(\frac{12.96y}{{x}^{2}+12.96{y}^{2}}-\frac{0.54y}{{x}^{2}+0.54{y}^{2}}\right)\end{array}$

4. 结论

General Solution of Elastic Mechanics for Orthotropic Materials[J]. 力学研究, 2022, 11(03): 54-62. https://doi.org/10.12677/IJM.2022.113007

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