【小呆的力学笔记】弹塑性力学的初步认知二：应力分析(1)

1.1 一点的应力状态

物体在受到外力或者自身不均匀的温度场等作用时，在其内部会产生内力，物体的内力与方向和截面都有关系。假设有一个受到外力作用的变形体，被一个平面截成A、B两个部分，B部分对A部分施加有作用力，在该截面上的dS微小面积上，作用力为dP，那么我们成dP与dS的比例极限为应力，如下式
$$\mathbf{\sigma }=\underset{\Delta S\to 0}{\lim }\frac{d\mathbf{P}}{dS}\text{\hspace{1em}(1)}$$
上式黑体表明是方向相关量。

应力是有方向，我们规定垂直于截面的分量成为正应力，平行于截面的分量为剪应力。
为了分析一点的应力状态，从物体内任一点取一微小的四面体单元，如下图。其中${\sigma }_{xx}$、${\sigma }_{yy}$、${\sigma }_{zz}$为正应力，${\tau }_{xy}$、${\tau }_{xz}$、${\tau }_{yz}$、${\tau }_{yx}$、${\tau }_{zx}$、${\tau }_{zy}$为剪应力，在很多时候我们借鉴矩阵的应用，将这些分量放在一起来表示一点的应力状态，比如如下形式。（实际上，更方便的表示一点的应力状态就是应力张量，后面在专门学习张量的时候引入张量的概念和计算，应力张量是二阶张量，具有矩阵的显像化形式）
写成矩阵形式，如下式
$$\left[\begin{array}{ccc}{\sigma }_{xx}& {\tau }_{xy}& {\tau }_{xz}\\ {\tau }_{yx}& {\sigma }_{yy}& {\tau }_{yz}\\ {\tau }_{zx}& {\tau }_{zy}& {\sigma }_{zz}\end{array}\right]$$

1.2 一点主应力状态

假设该斜面上只有正应力分量，没有剪应力，即四面体的斜面上${\mathbf{\sigma }}_v$为正应力，其中${\mathbf{\sigma }}_v$的应力方向余弦为$(l,m,n)$，根据力的平衡，可得下式。
$$\begin{gathered} {\sigma }_{xx}?S_{\Delta BOC}+{\tau }_{zx}?S_{\Delta AOB}+{\tau }_{yx}?S_{\Delta AOC}={\sigma }_{vx}{S}_{\Delta ABC} \\ {\sigma }_{yy}?S_{\Delta AOC}+{\tau }_{zy}?S_{\Delta AOB}+{\tau }_{xy}?S_{\Delta BOC}={\sigma }_{vy}{S}_{\Delta ABC} \\ {\sigma }_{zz}?S_{\Delta AOB}+{\tau }_{yz}?S_{\Delta AOC}+{\tau }_{xz}?S_{\Delta BOC}={\sigma }_{vz}{S}_{\Delta ABC}\text{\hspace{1em}(2)} \end{gathered}$$
其中
$$\begin{gathered} S_{\Delta AOC}=\frac{1}{2}dxdz=S_{\Delta ABC}?m \\ {S}_{\Delta AOB}=\frac{1}{2}dxdy=S_{\Delta ABC}?n \\ {S}_{\Delta BOC}=\frac{1}{2}dydz=S_{\Delta ABC}?l\text{\hspace{1em}(3)} \end{gathered}$$
那么力平衡方程改为
$$\begin{gathered} {\sigma }_{xx}?l+{\tau }_{zx}?n+{\tau }_{yx}?m={\sigma }_{vx}={\sigma }_{v}l \\ {\sigma }_{yy}?m+{\tau }_{zy}?n+{\tau }_{xy}?l={\sigma }_{vy}={\sigma }_{v}m \\ {\sigma }_{zz}?n+{\tau }_{yz}?m+{\tau }_{xz}?l={\sigma }_{vz}={\sigma }_{v}n\text{\hspace{1em}(4)} \end{gathered}$$
合并同类相，将其改为
$$\begin{gathered} ({\sigma }_{xx}?{\sigma }_v)?l+{\tau }_{yx}?m+{\tau }_{zx}?n=0 \\ {\tau }_{xy}?l+({\sigma }_{yy}?{\sigma }_v)?m+{\tau }_{zy}?n=0 \\ {\tau }_{xz}?l+{\tau }_{yz}?m+({\sigma }_{zz}?{\sigma }_v)?n=0\text{\hspace{1em}(5)} \end{gathered}$$
写成矩阵形式，如下式
$$\left[\begin{array}{ccc}{\sigma }_{xx}?{\sigma }_v& {\tau }_{yx}& {\tau }_{zx}\\ {\tau }_{xy}& {\sigma }_{yy}?{\sigma }_v& {\tau }_{zy}\\ {\tau }_{xz}& {\tau }_{yz}& {\sigma }_{zz}?{\sigma }_v\end{array}\right]\left[\begin{array}{c}l\\ m\\ n\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\text{\hspace{1em}(6)}$$
方向余弦存在非零解，那么系数行列式需为零，如下所示。
$$\left|\begin{array}{ccc}{\sigma }_{xx}?{\sigma }_v& {\tau }_{yx}& {\tau }_{zx}\\ {\tau }_{xy}& {\sigma }_{yy}?{\sigma }_v& {\tau }_{zy}\\ {\tau }_{xz}& {\tau }_{yz}& {\sigma }_{zz}?{\sigma }_v\end{array}\right|=0\text{\hspace{1em}(7)}$$
将其展开，如下式，并利用剪应力互等关系。
$$\begin{array}{rl}\left|\begin{array}{ccc}{\sigma }_{xx}?{\sigma }_v& {\tau }_{yx}& {\tau }_{zx}\\ {\tau }_{xy}& {\sigma }_{yy}?{\sigma }_v& {\tau }_{zy}\\ {\tau }_{xz}& {\tau }_{yz}& {\sigma }_{zz}?{\sigma }_v\end{array}\right|=& ({\sigma }_{xx}?{\sigma }_v)({\sigma }_{yy}?{\sigma }_v)({\sigma }_{zz}?{\sigma }_v)+{\tau }_{yx}{\tau }_{zy}{\tau }_{xz}+{\tau }_{zx}{\tau }_{xy}{\tau }_{yz}\\ & ?({\sigma }_{yy}?{\sigma }_v){\tau }_{zx}{\tau }_{xz}?({\sigma }_{zz}?{\sigma }_v){\tau }_{yx}{\tau }_{xy}?({\sigma }_{xx}?{\sigma }_v){\tau }_{zy}{\tau }_{yz}\\ =& {\sigma }_{xx}{\sigma }_{yy}{\sigma }_{zz}?({\sigma }_{xx}{\sigma }_{yy}+{\sigma }_{xx}{\sigma }_{zz}+{\sigma }_{yy}{\sigma }_{zz}){\sigma }_v+({\sigma }_{xx}+{\sigma }_{yy}+{\sigma }_{zz}){\sigma }_v^2\\ & ?{\sigma }_v^3+2{\tau }_{xz}{\tau }_{xy}{\tau }_{yz}+({\tau }_{xz}^2+{\tau }_{xy}^2+{\tau }_{yz}^2){\sigma }_v?{\sigma }_{yy}{\tau }_{xz}^2?{\sigma }_{zz}{\tau }_{xy}^2?{\sigma }_{xx}{\tau }_{yz}^2\\ =& ?{\sigma }_v^3+({\sigma }_{xx}+{\sigma }_{yy}+{\sigma }_{zz}){\sigma }_v^2+\\ & ({\tau }_{xz}^2+{\tau }_{xy}^2+{\tau }_{yz}^2?{\sigma }_{xx}{\sigma }_{yy}?{\sigma }_{xx}{\sigma }_{zz}?{\sigma }_{yy}{\sigma }_{zz}){\sigma }_v+\\ & {\sigma }_{xx}{\sigma }_{yy}{\sigma }_{zz}+2{\tau }_{xz}{\tau }_{xy}{\tau }_{yz}?{\sigma }_{yy}{\tau }_{xz}^2?{\sigma }_{zz}{\tau }_{xy}^2?{\sigma }_{xx}{\tau }_{yz}^2\\ =& ?{\sigma }_v^3+I_1{\sigma }_v^2+I_2{\sigma }_v+I_3\end{array}\text{\hspace{1em}(8)}$$

其中$I_1$、$I_2$、$I_3$成为应力不变量，如下所示。
$$\begin{gathered} I_1={\sigma }_{xx}+{\sigma }_{yy}+{\sigma }_{zz}=tr[\mathbf{\sigma }] \\ {I}_2={\tau }_{xz}^2+{\tau }_{xy}^2+{\tau }_{yz}^2?{\sigma }_{xx}{\sigma }_{yy}?{\sigma }_{xx}{\sigma }_{zz}?{\sigma }_{yy}{\sigma }_{zz} \\ {I}_3={\sigma }_{xx}{\sigma }_{yy}{\sigma }_{zz}+2{\tau }_{xz}{\tau }_{xy}{\tau }_{yz}?{\sigma }_{yy}{\tau }_{xz}^2?{\sigma }_{zz}{\tau }_{xy}^2?{\sigma }_{xx}{\tau }_{yz}^2=|\mathbf{\sigma }|\text{\hspace{1em}(9)} \end{gathered}$$

难么，公式（7）就变为
$$\begin{array}{r}\left|\begin{array}{ccc}{\sigma }_{xx}?{\sigma }_v& {\tau }_{yx}& {\tau }_{zx}\\ {\tau }_{xy}& {\sigma }_{yy}?{\sigma }_v& {\tau }_{zy}\\ {\tau }_{xz}& {\tau }_{yz}& {\sigma }_{zz}?{\sigma }_v\end{array}\right|=?{\sigma }_v^3+I_1{\sigma }_v^2+I_2{\sigma }_v+I_3=0\end{array}\text{\hspace{1em}(10)}$$

那么${\mathbf{\sigma }}_v$可以通过公式（10）来求解，回代公式（6），可以解的$(l,m,n)$。
根据三次方程的韦达定理，有三个方程的根（也就是主应力）的和等于下式。
$${\sigma }_1+{\sigma }_2+{\sigma }_3=?\frac{I_1}{?1}=I_1\text{\hspace{1em}(11)}$$

其实观察上面的计算，不难发现，正应力${\mathbf{\sigma }}_v$和方向余弦$(l,m,n)$为应力矩阵的特征值和特征向量。

1.3 应力偏张量、球张量、应力不变量

下图为能反映一点的应力状态的六面体，众多的金属实验表明（在常见的应力范围内）当六面体各个面上只有相等正应力无切应力时，物体不发生塑性变形和形状变化。由此，定义了这么一种应力状态，即
$$[{\mathbf{\sigma }}_m]=\left[\begin{array}{ccc}{\sigma }_m& 0& 0\\ 0& {\sigma }_m& 0\\ 0& 0& {\sigma }_m\end{array}\right]\text{\hspace{1em}(12)}$$
$${\sigma }_m={\sigma }_1={\sigma }_2={\sigma }_3=\frac{1}{3}({\sigma }_1+{\sigma }_2+{\sigma }_3)=\frac{1}{3}{I}_1\text{\hspace{1em}(13)}$$

那么将应力张量减去应力应力球张量，可得应力偏张量（同时可以确定的是[s]也是对称矩阵，由于剪应力互等）
$$[\mathbf{s}]=\left[\begin{array}{ccc}{\sigma }_{xx}?{\sigma }_m& {\tau }_{xy}& {\tau }_{xz}\\ {\tau }_{yx}& {\sigma }_{yy}?{\sigma }_m& {\tau }_{yz}\\ {\tau }_{zx}& {\tau }_{zy}& {\sigma }_{zz}?{\sigma }_m\end{array}\right]=\left[\begin{array}{ccc}{s}_{xx}& s_{xy}& s_{xz}\\ s_{yx}& s_{yy}& s_{yz}\\ s_{zx}& s_{zy}& s_{zz}\end{array}\right]\text{\hspace{1em}(14)}$$

同样，按照1.2的过程，可以得到应力偏张量的主应力的公式如下所示。
$$\begin{array}{rl}\left|\begin{array}{ccc}{s}_{xx}?s_v& s_{yx}& s_{zx}\\ s_{xy}& s_{yy}?s_v& s_{zy}\\ s_{xz}& s_{yz}& s_{zz}?s_v\end{array}\right|& =\left|\begin{array}{ccc}{s}_{xx}?s_v& s_{xy}& s_{xz}\\ s_{yx}& s_{yy}?s_v& s_{yz}\\ s_{zx}& s_{zy}& s_{zz}?s_v\end{array}\right|\\ & =?s_v^3+{\hat{I}}_1{s}_v^2+{\hat{I}}_2{s}_v+{\hat{I}}_3=0\end{array}\text{\hspace{1em}(15)}$$
其中，有应力偏张量的不变量如下所示。
$$\begin{gathered} {\hat{I}}_1=s_{xx}+s_{yy}+s_{zz}={\sigma }_{xx}?{\sigma }_m+{\sigma }_{yy}?{\sigma }_m+{\sigma }_{zz}?{\sigma }_m=0 \\ {\hat{I}}_2=s_{xz}^2+s_{xy}^2+s_{yz}^2?s_{xx}{s}_{yy}?s_{xx}{s}_{zz}?s_{yy}{s}_{zz} \\ {\hat{I}}_3=s_{xx}{s}_{yy}{s}_{zz}+2s_{xz}{s}_{xy}{s}_{yz}?s_{yy}{s}_{xz}^2?s_{zz}{s}_{xy}^2?s_{xx}{s}_{yz}^2\text{\hspace{1em}(16)} \end{gathered}$$
对公式（15）进行展开合并同类项等过程，如下所示，
$$\begin{array}{rl}?s_v^3+{\hat{I}}_1{s}_v^2+{\hat{I}}_2{s}_v+{\hat{I}}_3& =?s_v^3+{\hat{I}}_2{s}_v+{\hat{I}}_3\\ & =?s_v^3+(s_{xz}^2+s_{xy}^2+s_{yz}^2?s_{xx}{s}_{yy}?s_{xx}{s}_{zz}?s_{yy}{s}_{zz})s_v\\ & +s_{xx}{s}_{yy}{s}_{zz}+2s_{xz}{s}_{xy}{s}_{yz}?s_{yy}{s}_{xz}^2?s_{zz}{s}_{xy}^2?s_{xx}{s}_{yz}^2\\ & =?s_v^3+({\tau }_{xz}^2+{\tau }_{xy}^2+{\tau }_{yz}^2)s_v\\ & ?[({\sigma }_{xx}?{\sigma }_m)({\sigma }_{yy}?{\sigma }_m)+({\sigma }_{xx}?{\sigma }_m)({\sigma }_{zz}?{\sigma }_m)+({\sigma }_{yy}?{\sigma }_m)({\sigma }_{zz}?{\sigma }_m)]s_v\\ & +({\sigma }_{xx}?{\sigma }_m)({\sigma }_{yy}?{\sigma }_m)({\sigma }_{zz}?{\sigma }_m)+2{\tau }_{xz}{\tau }_{xy}{\tau }_{yz}\\ & ?({\sigma }_{yy}?{\sigma }_m){\tau }_{xz}^2?({\sigma }_{zz}?{\sigma }_m){\tau }_{xy}^2?({\sigma }_{xx}?{\sigma }_m){\tau }_{yz}^2\\ & =?s_v^3+({\tau }_{xz}^2+{\tau }_{xy}^2+{\tau }_{yz}^2)s_v?({\sigma }_{xx}{\sigma }_{yy}+{\sigma }_{xx}{\sigma }_{zz}+{\sigma }_{yy}{\sigma }_{zz})s_v\\ & +({\sigma }_{xx}+{\sigma }_{yy}+{\sigma }_{xx}+{\sigma }_{zz}+{\sigma }_{yy}+{\sigma }_{zz}){\sigma }_m{s}_v?3{\sigma }_m^2{s}_v+{\sigma }_{xx}{\sigma }_{yy}{\sigma }_{zz}\\ & ?({\sigma }_{xx}{\sigma }_{yy}+{\sigma }_{xx}{\sigma }_{zz}+{\sigma }_{yy}{\sigma }_{zz}){\sigma }_m+({\sigma }_{xx}+{\sigma }_{yy}+{\sigma }_{zz}){\sigma }_m^2?{\sigma }_m^3+2{\tau }_{xz}{\tau }_{xy}{\tau }_{yz}\\ & ?({\sigma }_{yy}{\tau }_{xz}^2+{\sigma }_{zz}{\tau }_{xy}^2+{\sigma }_{xx}{\tau }_{yz}^2)+({\tau }_{xz}^2+{\tau }_{xy}^2+{\tau }_{yz}^2){\sigma }_m\\ & =?s_v^3+({\tau }_{xz}^2+{\tau }_{xy}^2+{\tau }_{yz}^2)(s_v+{\sigma }_m)?({\sigma }_{xx}{\sigma }_{yy}+{\sigma }_{xx}{\sigma }_{zz}+{\sigma }_{yy}{\sigma }_{zz})(s_v+{\sigma }_m)\\ & +6{\sigma }_m^2{s}_v?3{\sigma }_m^2{s}_v+{\sigma }_{xx}{\sigma }_{yy}{\sigma }_{zz}+3{\sigma }_m^3?{\sigma }_m^3+2{\tau }_{xz}{\tau }_{xy}{\tau }_{yz}\\ & ?({\sigma }_{yy}{\tau }_{xz}^2+{\sigma }_{zz}{\tau }_{xy}^2+{\sigma }_{xx}{\tau }_{yz}^2)\\ & =?s_v^3+3{\sigma }_m^2{s}_v+2{\sigma }_m^3+({\tau }_{xz}^2+{\tau }_{xy}^2+{\tau }_{yz}^2?{\sigma }_{xx}{\sigma }_{yy}?{\sigma }_{xx}{\sigma }_{zz}?{\sigma }_{yy}{\sigma }_{zz})(s_v+{\sigma }_m)\\ & +{\sigma }_{xx}{\sigma }_{yy}{\sigma }_{zz}+2{\tau }_{xz}{\tau }_{xy}{\tau }_{yz}?({\sigma }_{yy}{\tau }_{xz}^2+{\sigma }_{zz}{\tau }_{xy}^2+{\sigma }_{xx}{\tau }_{yz}^2)\\ & =?s_v^3?{\sigma }_m^3+3{\sigma }_m^2{s}_v+3{\sigma }_m^3+I_2(s_v+{\sigma }_m)+I_3\\ & =?(s_v+{\sigma }_m)(s_v^2?s_v{\sigma }_m+{\sigma }_m^2?3{\sigma }_m^2)+I_2(s_v+{\sigma }_m)+I_3\\ & =?(s_v+{\sigma }_m)(s_v+{\sigma }_m)(s_v?2{\sigma }_m)+I_2(s_v+{\sigma }_m)+I_3\\ & =?(s_v+{\sigma }_m{)}^2(s_v+{\sigma }_m?3{\sigma }_m)+I_2(s_v+{\sigma }_m)+I_3\\ & =?(s_v+{\sigma }_m{)}^3+3{\sigma }_m(s_v+{\sigma }_m{)}^2+I_2(s_v+{\sigma }_m)+I_3\\ & =?(s_v+{\sigma }_m{)}^3+I_1(s_v+{\sigma }_m{)}^2+I_2(s_v+{\sigma }_m)+I_3=0\end{array}\text{\hspace{1em}(17)}$$
对比公式（17）和公式（10），如下所示。
$$\begin{gathered} ?(s_v+{\sigma }_m{)}^3+I_1(s_v+{\sigma }_m{)}^2+I_2(s_v+{\sigma }_m)+I_3=0 \\ ?{\sigma }_v^3+I_1{\sigma }_v^2+I_2{\sigma }_v+I_3=0 \end{gathered}$$
不难发现，$s_v+{\sigma }_m={\sigma }_v$，即应力偏张量主值和应力张量主值存在以上转换关系。
同时，从公式（16）的${\hat{I}}_1$
$${\hat{I}}_1=s_{xx}+s_{yy}+s_{zz}=0$$
那么可以得到
$$(s_{xx}+s_{yy}+s_{zz}{)}^2=s_{xx}^2+s_{yy}^2+s_{zz}^2+2(s_{xx}{s}_{yy}+s_{xx}{s}_{zz}+s_{yy}{s}_{zz})=0\text{\hspace{1em}(18)}$$
于是
$$\begin{array}{rl}?6(s_{xx}{s}_{yy}+s_{xx}{s}_{zz}+s_{yy}{s}_{zz})& =2s_{xx}^2+2s_{yy}^2+2s_{zz}^2?2(s_{xx}{s}_{yy}+s_{xx}{s}_{zz}+s_{yy}{s}_{zz})\\ & =(s_{xx}?s_{yy}{)}^2+(s_{xx}?s_{zz}{)}^2+(s_{yy}?s_{zz}{)}^2\\ & =({\sigma }_{xx}?{\sigma }_{yy}{)}^2+({\sigma }_{xx}?{\sigma }_{zz}{)}^2+({\sigma }_{yy}?{\sigma }_{zz}{)}^2\end{array}\text{\hspace{1em}(19)}$$
把公式（16）的${\hat{I}}_2$，那么
$$\begin{array}{rl}?6(s_{xx}{s}_{yy}+s_{xx}{s}_{zz}+s_{yy}{s}_{zz})& =6[{\hat{I}}_2?({\tau }_{xz}^2+{\tau }_{xy}^2+{\tau }_{yz}^2)]\\ & =({\sigma }_{xx}?{\sigma }_{yy}{)}^2+({\sigma }_{xx}?{\sigma }_{zz}{)}^2+({\sigma }_{yy}?{\sigma }_{zz}{)}^2\end{array}\text{\hspace{1em}(20)}$$

$${\hat{I}}_2=\frac{1}{6}[({\sigma }_{xx}?{\sigma }_{yy}{)}^2+({\sigma }_{xx}?{\sigma }_{zz}{)}^2+({\sigma }_{yy}?{\sigma }_{zz}{)}^2+6({\tau }_{xz}^2+{\tau }_{xy}^2+{\tau }_{yz}^2)]\text{\hspace{1em}(21)}$$

应力偏张量的第二不变量更常用的还有一个形式如下。
$${\hat{I}}_2=\frac{1}{6}[({\sigma }_1?{\sigma }_2{)}^2+({\sigma }_1?{\sigma }_3{)}^2+({\sigma }_2?{\sigma }_3{)}^2]\text{\hspace{1em}(22)}$$
看公式（21）（22）是不是特别的眼熟，其实他就是等效应力的来源，定义应力强度${\sigma }_e$为
$$\begin{gathered} {\sigma }_e=\sqrt{3I_2}=\sqrt{\frac{1}{2}[({\sigma }_1?{\sigma }_2{)}^2+({\sigma }_1?{\sigma }_3{)}^2+({\sigma }_2?{\sigma }_3{)}^2]} \\ =\sqrt{\frac{1}{2}[({\sigma }_{xx}?{\sigma }_{yy}{)}^2+({\sigma }_{xx}?{\sigma }_{zz}{)}^2+({\sigma }_{yy}?{\sigma }_{zz}{)}^2+6({\tau }_{xz}^2+{\tau }_{xy}^2+{\tau }_{yz}^2)]}\text{\hspace{1em}(22)} \end{gathered}$$