46308 由矩陣特徵值證明力學莫爾圓性質並應用於實際工程問題

### 2. 土壤力學與材料力學應力分析

#### 2.2. 應力轉換公式

\begin{align} &\left\{\begin{array}{l} \sigma_\theta =\dfrac{\sigma_x+\sigma_z}{2}+\dfrac{\sigma_x-\sigma_z}{2}\cos 2\theta +\tau_{xz}\sin 2\theta, \\[7pt] \tau_\theta =-\dfrac{\sigma_x-\sigma_z}{2}\sin 2\theta +\tau_{x z}\cos 2\theta, \end{array}\right.\label{1}\\[7pt] &\left\{\begin{array}{l} \sigma_\theta =\dfrac{\sigma_x+\sigma_z}{2}+\dfrac{\sigma_x-\sigma_z}{2}\cos 2\theta -\tau_{xz}\sin 2\theta,\\[7pt] \tau_\theta =\dfrac{\sigma_x-\sigma_z}{2}\sin 2\theta +\tau_{x z}\cos 2\theta. \end{array}\right.\label{2} \end{align}

#### 2.3. 應力莫爾圓

\begin{align} \Big(\sigma_\theta -\frac{\sigma_x+\sigma_z}{2}\Big)^2+\tau_\theta^2=\Big(\frac{\sigma_x-\sigma_z}{2}\Big)^2+\tau_{xz}^2.\label{3} \end{align}

\begin{align} \left\{\begin{array}{l} \sigma_1=\dfrac{\sigma_x+\sigma_z}{2}+\sqrt{\Big(\dfrac{\sigma_x-\sigma_z}{2}\Big)^2+\tau_{xz}^2}, \\[7pt] \sigma_3=\dfrac{\sigma_x+\sigma_z}{2}-\sqrt{\Big(\dfrac{\sigma_x-\sigma_z}{2}\Big)^2+\tau_{xz}^2}.\end{array}\right.\label{4} \end{align}

### 3. 工程數學觀點

#### 3.1. 應力矩陣

\begin{align} \mathbf{\tau} =\left[\begin{array}{ccc}\sigma_x&~&\tau_{xz}\\ \tau_{zx}&&\sigma_z \end{array}\right]. \label{5} \end{align}

\begin{align} \tau_{xz}=\tau_{zx}. \label{6} \end{align}

#### 3.2. 旋轉矩陣

\begin{align} \left[\begin{array}{c} x'\\ y'\end{array}\right]={\mathbf R}_{xy} \left[\begin{array}{c} x\\ y\end{array}\right]=\left[\begin{array}{ccc}\cos\theta_p&~&-\sin\theta_p\\ \sin\theta_p&&\cos\theta_p\end{array}\right]\left[\begin{array}{c} x\\ y\end{array}\right].\label{7} \end{align}

\begin{align} \left[\begin{array}{c}\overline x'\\ \overline y'\end{array}\right] =\overline {\mathbf R}_{xy} \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{ccc}\cos\theta_O&~&\sin\theta_O\\ -\sin\theta_O&&\cos\theta_O\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right].\label{8} \end{align}

\begin{align} \left[\begin{array}{c} x'\\ z'\end{array}\right]={\mathbf R}_{xz} \left[\begin{array}{c}x\\ z\end{array}\right]=\left[\begin{array}{ccc}\cos\theta_p&~&\sin\theta_p\\ -\sin\theta_p&&\cos\theta_p\end{array}\right] \left[\begin{array}{c}x\\ z\end{array}\right].\label{9} \end{align}

\begin{align} \left[\begin{array}{c}\overline x'\\ \overline z'\end{array}\right]={\overline{\mathbf R}}_{xz} \left[\begin{array}{c}x\\ z\end{array}\right]=\left[\begin{array}{ccc} \cos\theta_O&~&-\sin\theta_O\\ \sin\theta_O&~&\cos\theta_O\end{array}\right]\left[\begin{array}{c}x\\ z\end{array}\right].\label{10} \end{align}

#### 3.3. 利用應力矩陣與旋轉矩陣獲得應力轉換公式

\begin{align} \mathbf{\tau} =\left[\begin{array}{ccc}\sigma_x&~&\tau_{xz}\\ {\tau}_{xz}&~&\sigma_z\end{array}\right],\label{11} \end{align}

\begin{align} \mathbf{\tau}_\theta =\,&\left[\begin{array}{ccc}\tau_{\theta_{11}}&~&\tau_{\theta_{12}}\\ {\tau}_{\theta_{21}}&~&\tau_{\theta_{22}}\end{array}\right] =\mathbf{R}_{xz} \mathbf{\tau} \mathbf{R}_{xz}^T,\label{12}\\ {\hbox{其中：}} \tau_{\theta_{11}}=\,&\frac{(\sigma_x+\sigma_z)}{2}+\frac{(\sigma_x-\sigma_z )}{2}\cos 2\theta_p+\tau_{xz}\sin 2\theta_p,\label{13}\\ \tau_{\theta_{12}}=\,&\tau_{\theta_{21}}=-\frac{(\sigma_x-\sigma_z )}{2}\sin 2\theta_p+\tau_{xz}\cos 2\theta_p,\label{14}\\ \tau_{\theta_{22}}=\,&\frac{(\sigma_x+\sigma_z )}{2}-\frac{(\sigma_x-\sigma_z )}{2}\cos 2\theta_p-\tau_{xz}\sin 2\theta_p.\hskip 2cm~\label{15} \end{align}

\begin{align} \mathbf{\tau}_\theta' =\,&\left[\begin{array}{ccc}\tau_{\theta_{11}}'&~&\tau_{\theta_{12}}'\\ {\tau}_{\theta_{21}}'&~&\tau_{\theta_{22}}'\end{array}\right] ={\overline{\mathbf{R}}_{xz}} \mathbf{\tau} \overline{\mathbf{R}}_{xz}^T,\label{16}\\ {\hbox{其中：}} \tau_{\theta_{11}}'=\,&\frac{(\sigma_x+\sigma_z)}{2}+\frac{(\sigma_x-\sigma_z )}{2}\cos 2\theta_O-\tau_{xz}\sin 2\theta_O,\label{17}\\ \tau_{\theta_{12}}'=\,&\tau_{\theta_{21}}'=\frac{(\sigma_x-\sigma_z )}{2}\sin 2\theta_O+\tau_{xz}\cos 2\theta_O,\label{18}\\ \tau_{\theta_{22}}'=\,&\frac{(\sigma_x+\sigma_z )}{2}-\frac{(\sigma_x-\sigma_z )}{2}\cos 2\theta_O+\tau_{xz}\sin 2\theta_O.\hskip 2cm~\label{19} \end{align}

#### 3.4. 應力分量的變化軌跡即為應力莫爾圓

\begin{align} r=\,&\sqrt{\Big(\frac{\sigma_x-\sigma_z}{2}\Big)^2+\tau_{xz}^2},\label{20}\\ {\hbox{與}} &\left\{\begin{array}{l}\dfrac{\sigma_x-\sigma_z}{2}=r\cos\alpha\\ \tau_{xz}=r\sin\alpha \end{array}\right.,\label{21} \end{align}

\begin{align} {\mathbf\tau}_\theta =\,&\left[\begin{array}{ccc}\frac{(\sigma_x+\sigma_z )}{2}+r\cos(\alpha -2\theta_p )&r\sin(\alpha -2\theta_p) \\ r\sin(\alpha -2\theta_p )&\frac{(\sigma_x+\sigma_z )}{2}-r\cos(\alpha -2\theta_p )\end{array}\right]\nonumber\\ =\,&\left[\begin{array}{ccc}\sigma_c+r\cos t&~&r\sin t\\ r\sin t&~&\sigma_c-r\cos t\end{array}\right],\label{22}\\ {\hbox{及}} {\mathbf\tau}_\theta'=\,&\left[\begin{array}{ccc}\frac{(\sigma_x+\sigma_z )}{2}+r\cos(\alpha +2\theta_O )&~&r\sin(\alpha +2\theta_O)\\ r\sin(\alpha +2\theta_O)&~&\frac{(\sigma_x+\sigma_z )}{2}-r\cos(\alpha +2\theta_O )\end{array}\right]\nonumber\\ =\,&\left[\begin{array}{ccc}\sigma_c+r\cos t'&~&r\sin t'\\ r\sin t'&~&\sigma_c-r\cos t'\end{array}\right],\label{23} \end{align}

\begin{align} &\left\{\begin{array}{l}\sigma =\sigma_c+r\cos\overline t,\\ \tau =r\sin \overline t.\end{array}\right.\label{24}\\ {\hbox{並可得：}} &(\sigma -\sigma_c)^2+\tau^2=r^2.\label{25} \end{align}

#### 3.5. 應力矩陣之特徵值即為主應力大小

\begin{align} \left|\begin{array}{ccc}\sigma_x- \sigma &~&\tau_{xz}\\ \tau_{xz}&~&\sigma_z- \sigma\end{array}\right|=0, \label{26} \end{align}

\begin{align} \sigma =\frac{(\sigma_x+\sigma_z )\pm \sqrt{(\sigma_x+\sigma_z )^2-4(\sigma_x \sigma_z-\tau_{xz}^2 )}}{2},\label{27} \end{align}

\begin{align} \sigma =\sigma_c\pm r.\label{28} \end{align}

\begin{align} \left\{\begin{array}{l}\sigma_1=\sigma_c+r,\\ \sigma_3=\sigma_c-r.\end{array}\right.\label{29} \end{align}

#### 3.6. 相同應力莫爾圓上有著相同特徵值

\begin{align} {\mathbf\tau}_\theta =\left[\begin{array}{ccc} \sigma_c+r\cos(\overline t)&~&r\sin(\overline t)\\ r\sin(\overline t)&~&\sigma_c-r\cos(\overline t)\end{array}\right].\label{30} \end{align}

\begin{align} \left|\begin{array}{ccc}\sigma_c+r\cos(\overline t)- \sigma &~&r\sin(\overline t)\\ r\sin(\overline t)&~&\sigma_c-r\cos(\overline t)- \sigma\end{array}\right|=0,\label{31} \end{align}

\begin{aligned} \sigma =\,&\frac{2\sigma_c\pm \sqrt{4\sigma_c^2-4(\sigma_c^2-r^2 )}}{2}\\ =\,&\sigma_c\pm r. \label{32} \end{aligned}

#### 3.7. 特徵方向所旋轉的角度恰為應力莫爾圓上旋轉角之半

\begin{align} \left[\begin{array}{ccc}\sigma_c+r\cos(\overline t)-(\sigma_c+r)&~&r\sin(\overline t)\\ r\sin(\overline t)&~&\sigma_c-r\cos(\overline t)-(\sigma_c+r)\end{array}\right]\left[\begin{array}{c} x_{11}\\ x_{21}\end{array}\right]={\mathbf 0},\label{33}\\[6pt] \left[\begin{array}{ccc}\sigma_c+r\cos(\overline t)-(\sigma_c-r)&~&r\sin(\overline t)\\ r\sin(\overline t)&~&\sigma_c-r\cos(\overline t)-(\sigma_c-r)\end{array}\right] \left[\begin{array}{c}x_{13}\\ x_{23}\end{array}\right]={\mathbf 0},\label{34} \end{align}

\begin{align} {\mathbf u}_1=\,&\left[\begin{array}{c}u_{11}\\ u_{21}\end{array}\right]=\left[\begin{array}{c}\cos\Big(\frac{\overline t}2\Big)\\[5pt] \sin\Big(\frac{\overline t}2\Big)\end{array}\right],\label{35}\\ {\hbox{及}} {\mathbf u}_3=\,&\left[\begin{array}{c}u_{13}\\ u_{23}\end{array}\right]=\left[\begin{array}{c}-\sin\Big(\frac{\overline t}2\Big)\\[5pt] \cos\Big(\frac{\overline t}2\Big)\end{array}\right].\label{36} \end{align}

#### 3.8. 具備相同特徵向量之莫爾圓性質

\begin{align} {\mathbf\tau}_A=\,&\left[\begin{array}{ccc} \sigma_{cA}+r_A\cos ({\overline t}_A)&~&r_A\sin ({\overline t}_A)\\ r_A\sin ({\overline t}_A)&~&\sigma_{cA}-r_A\cos ({\overline t}_A)\end{array}\right],\label{37}\\ {\mathbf\tau}_B=\,&\left[\begin{array}{ccc}\sigma_{cB}+r_B\cos ({\overline t}_B)&~&r_B\sin ({\overline t}_B)\\ r_B\sin ({\overline t}_B)&~&\sigma_{cB}-r_B\cos ({\overline t}_B)\end{array}\right],\label{38} \end{align}

\begin{align} \overline t={\overline t}_A={\overline t}_B.\label{39} \end{align}

\begin{align} \left[\begin{array}{ccc} {\mathbf u}_{1\overline t}&~&{\mathbf u}_{3\overline t} \end{array}\right]= \left[\begin{array}{ccc} \cos\left(\frac{\overline t}2\right)&~&-\sin\left(\frac{\overline t}2\right)\\[5pt] \sin\left(\frac{\overline t}2\right)&~&\cos\left(\frac{\overline t}2\right)\end{array}\right].\label{40} \end{align}

\begin{align} \left[\begin{array}{ccc} {\mathbf u}_{1\overline t}&~&{\mathbf u}_{3\overline t}\end{array}\right]^{-1} {\mathbf \tau}_A \left[\begin{array}{ccc} {\mathbf u}_{1\overline t}&~&{\mathbf u}_{3\overline t} \end{array}\right]=\,&\left[\begin{array}{ccc} \sigma_{1A}&~&0\\ 0&~&\sigma_{3A}\end{array}\right],\label{41}\\ {\hbox{與}} \left[\begin{array}{ccc}{\mathbf u}_{1\overline t}&~&{\mathbf u}_{3\overline t}\end{array}\right]^{-1} {\mathbf\tau}_B \left[\begin{array}{ccc} {\mathbf u}_{1\overline t}&~&{\mathbf u}_{3\overline t}\end{array}\right]=\,&\left[\begin{array}{ccc}\sigma_{1B}&0\\ 0&\sigma_{3B}\end{array}\right].\label{42} \end{align}

### 4. 論例

\begin{align} &{\mathbf\tau} =\left[\begin{array}{ccc} -4&~&-2\\ -2&~&6\end{array}\right],\label{43}\\ {\hbox{並可得}} &\left\{\begin{array}{l}\sigma_c=1.0\\ r=\sqrt{29}\end{array}\right.,\label{44}\\ {\hbox{與}} &\quad \overline t=201.8^\circ.\label{45} \end{align}

\begin{align} &\left\{\begin{array}{l}\sigma_1=1+\sqrt{29}\\ \sigma_3=1-\sqrt{29}\end{array}\right. ,\label{46}\\ {\hbox{而其所對應的特徵向量分別為式 \eqref{47} 與式 \eqref{48}。}} {\mathbf u}_1=\,&\left[\begin{array}{c} \cos\Big(\frac{\overline t}2\Big)\\[7pt] \sin\Big(\frac{\overline t}2\Big)\end{array}\right]=\left[\begin{array}{c}-0.189\\ 0.982\end{array}\right],\label{47}\\[5pt] {\mathbf u}_3=\,&\left[\begin{array}{c} -\sin\Big(\frac{\overline t}2\Big)\\[7pt] \cos\Big(\frac{\overline t}2\Big)\end{array}\right]=\left[\begin{array}{c} -0.982\\ -0.189\end{array}\right].\label{48} \end{align}

### 5. 基礎工程

#### 5.1. 平面應變

\begin{align} {\mathbf \tau} =\,&\left[\begin{array}{ccccc}\sigma_x&~&0&~&\tau_{xz}\\ 0&~&\sigma_y&~&0\\ \tau_{xz}&~&0&~&\sigma_z\end{array}\right]=\left[\begin{array}{ccccc}\sigma_x&~&0&~&\tau_{xz}\\ 0&~&\nu(\sigma_x+\sigma_z )&~&0\\ \tau_{xz}&~&0&~&\sigma_z\end{array}\right],\label{49} \end{align}

\begin{align} \left\{\begin{array}{l} \sigma_2=\nu (\sigma_x+\sigma_z) \\[4pt] {\mathbf u}_2=\left[\begin{array}{c}0\\[-3pt] 1\\[-3pt] 0\end{array}\right]\end{array}\right..\label{50} \end{align}

#### 5.2. 平面應變下的加載問題

\begin{align} \Delta \sigma_x=\,&\frac q\pi [\alpha -\sin\alpha\cos(\alpha +2\delta )],\label{51}\\ \Delta \sigma_z=\,&\frac q\pi [\alpha +\sin\alpha\cos(\alpha +2\delta )],\label{52}\\ {\hbox{與}} \Delta \tau_{xz}=\,&\frac q\pi [\sin\alpha \sin(\alpha +2\delta )].\hskip 4cm ~\label{53} \end{align}

### 6. 結論與建議

I. 利用應力矩陣與點位旋轉型式的旋轉矩陣可獲得式 \eqref{1} 之應力轉換公式, 而此結果與基於力平衡條件所得公式一致。 另一方面, 若採用座標旋轉型式的旋轉矩陣則可獲得式 \eqref{2}, 且此亦與力平衡條件所得公式相同。 此外, 若兩者顯示於 $\sigma$-$\tau$ 平面, 則其軌跡皆為圓形, 而此即為應力莫爾圓。

II. 微小受力單元所對應的應力矩陣, 其特徵值即為主應力大小。 再者, 對於一固定的應力莫爾圓, 其所對應的應力矩陣皆有著相同的特徵值大小。

III. 若將應力矩陣所對應的單位特徵向量依特徵值由大而小排列, 則此即為旋轉矩陣。

IV. 若不同莫爾圓具有相同的旋轉角則具備相同的特徵向量, 且特徵向量的旋轉角恰為應力莫爾圓旋轉角之半。 此外, 若矩陣擁有相同的特徵方向, 則可進行同步對角化。