\[
\sigma_{\text{tresca}} = \max\left\{ |\sigma_1 - \sigma_2|, |\sigma_1 - \sigma_3|, |\sigma_2 - \sigma_3| \right\}\geq\sigma_{y},
\]
最大剪切应力
设法向量为 \(\mathbf{n}\) 的平面上的应力为 \(\mathbf{t}_{\mathbf{n}}\)，正应力为 \(\sigma_{n}\)，切应力为 \(\tau_{\mathbf{n}}\)，则
\[
\tau_{\mathbf{n}}^{2} = |\mathbf{t}_{\mathbf{n}}|^{2} - \sigma_{\mathbf{n}}^{2} = \mathbf{n}^{T}\boldsymbol{\sigma}^{2}\mathbf{n} - (\mathbf{n}^{T}\boldsymbol{\sigma}\mathbf{n})^{2},
\]
将 \(\mathbf{n}\) 沿 \(\boldsymbol{\sigma}\) 的单位特征向量 \(\mathbf{n}_{1},\mathbf{n}_{2},\mathbf{n}_{3}\) （对应的特征值分别为 \(\sigma_{1},\sigma_{2},\sigma_{3}\)）分解
\[
\mathbf{n} = n_{1}\mathbf{n}_{1}+n_{2}\mathbf{n}_{3}+n_{2}\mathbf{n}_{3},
\]
其中 \(n_{1}^2+n_{2}^2+n_{3}^2=1\)， 于是
(72)\[
\tau_{\mathbf{n}}^2 = n_{1}^{2}\sigma_{1}^{2}+n_{2}^{2}\sigma_{2}^{2}+n_{3}^{2}\sigma_{3}^{2}-(n_{1}^{2}\sigma_{1}+n_{2}^{2}\sigma_{2}+n_{3}^{2}\sigma_{3})^2,
\]
于是需在约束 \(n_{1}^2+n_{2}^2+n_{3}^2=1\) 下最大化 \(\tau_{\mathbf{n}}^2\)，引入拉格朗日乘子 \(\lambda\)，构造目标函数
\[
L = \tau_{\mathbf{n}}^2 - \lambda(n_{1}^2+n_{2}^2+n_{3}^2-1)
\]
函数 \(L\) 对 \(n_{1},n_{2},n_{3}\) 的导数为
(73)\[\begin{split}
\begin{equation}
\begin{aligned}
&(\sigma_{1}^2-2\sigma_{\mathbf{n}}\sigma_{1} - \lambda)n_{1}= 0 \\
&(\sigma_{2}^2-2\sigma_{\mathbf{n}}\sigma_{2} - \lambda)n_{2}= 0 \\
&(\sigma_{3}^2-2\sigma_{\mathbf{n}}\sigma_{3} - \lambda)n_{3}= 0
\end{aligned}
\end{equation}
\end{split}\]
不失一般性
1. 若 \(n_{2}=n_{3}=0\)，此时 \(n_{1}=\pm1\)，代入到式 (72) 中，得到 \(\tau_{\mathbf{n}}=0\)，显然不是最大值
2. 若 \(n_{1}\neq0,n_{2}\neq0,n_{3}=0\)，代入到式 (72) 中，得到
\[
\tau_{\mathbf{n}}^2 = n_{1}^2\sigma_{1}^2+n_{2}^2\sigma_{2}^2 - (n_{1}^2\sigma_{1}+n_{2}^2\sigma_{2})^2,
\]
代入 \(n_{2}^2 = 1 - n_{1}^2\)，并记 \(\ell = n_{1}^2\)，于是
(74)\[\begin{split}
\begin{equation}
\begin{aligned}
\tau_{\mathbf{n}}^2 &= \ell\sigma_{1}^2+(1 - \ell)\sigma_{2}^2 - (\ell\sigma_{1}+(1 - \ell)\sigma_{2})^2\\
&=-(\sigma_{1}-\sigma_{2})^2\ell^2-(\sigma_{1}^2-\sigma_{2}^2 - 2(\sigma_{1}-\sigma_{2})\sigma_{2})\ell\\
&=-(\sigma_{1}-\sigma_{2})^2(\ell^2 - \ell),
\end{aligned}
\end{equation}
\end{split}\]
因此，当 \(\ell = n_{1}^2=\frac{1}{2}\) 时，\(\tau_{\mathbf{n}}\) 取得极值，此时 \(n_{2}^2=\frac{1}{2}\)，代入到式 (72)，得到
\[
\tau_{\mathbf{n}} = \frac{1}{2}|\sigma_{1}-\sigma_{2}|.
\]
截面法向量为
\[
(\pm\frac{\sqrt{2}}{2})\mathbf{n_{1}}+(\pm\frac{\sqrt{2}}{2})\mathbf{n}_{2},
\]
与 \(\sigma_{1}\) 截面和 \(\sigma_{2}\) 截面夹角为 \(45\) 度
3. 若 \(n_{1}\neq0,n_{2}\neq0,n_{3}\neq0\)，则此时 \(\sigma_{1},\sigma_{2},\sigma_{3}\) 都是二次方程 \(\sigma^2-2\sigma_{\mathbf{n}}\sigma - \lambda = 0\) 的根
\[
\boldsymbol{\sigma} = Q^{T}(\sigma_{1}I)Q = \sigma_{1}I \quad\Longrightarrow\quad \tau_{\mathbf{n}}\equiv0
\]
(75)\[
\tau_{\mathbf{n}}^2 = (n_{1}^{2}+n_{2}^{2})\sigma_{1}^{2}+n_{3}^{2}\sigma_{3}^{2}-((n_{1}^{2}+n_{2}^{2})\sigma_{1}+n_{3}^{2}\sigma_{3})^2,
\]
记 \(n_{1}^2+n_{2}^2 = \ell\)，于是 \(n_{3}^2 = 1-\ell\)，此时式 (75) 与式 (74) 有完全一样的形式，由此可知极值为
\[
\tau_{\mathbf{n}} = \frac{1}{2}|\sigma_{1}-\sigma_{3}|.
\]
此时 \(n_{1}^2+n_{2}^2=n_{3}^2=\frac{1}{2}\)，截面法向量为
\[
n_{1}\mathbf{n_{1}}+(\pm\sqrt{\frac{1}{2}-n_{1}^2})\mathbf{n}_{2}+(\pm\frac{\sqrt{2}}{2})\mathbf{n}_{3},
\]
与 \(\sigma_{3}\) 平面夹角为 \(45\) 度
综上，考虑到方程 (73) 的对称性，得到
\[
\max_{\mathbf{n}} \tau_{\mathbf{n}} = \frac{1}{2}\max\left\{ |\sigma_1 - \sigma_2|, |\sigma_2 - \sigma_3|, |\sigma_3 - \sigma_1| \right\}.
\]
在截面与主应力面呈 \(45\) 度夹角的平面处达到
