# 莫爾圓

## 應力及莫爾圓

${\displaystyle {\boldsymbol {\sigma }}=\left[{\begin{matrix}\sigma _{11}&\sigma _{12}&\sigma _{13}\\\sigma _{21}&\sigma _{22}&\sigma _{23}\\\sigma _{31}&\sigma _{32}&\sigma _{33}\\\end{matrix}}\right]\equiv \left[{\begin{matrix}\sigma _{xx}&\sigma _{xy}&\sigma _{xz}\\\sigma _{yx}&\sigma _{yy}&\sigma _{yz}\\\sigma _{zx}&\sigma _{zy}&\sigma _{zz}\\\end{matrix}}\right]\equiv \left[{\begin{matrix}\sigma _{x}&\tau _{xy}&\tau _{xz}\\\tau _{yx}&\sigma _{y}&\tau _{yz}\\\tau _{zx}&\tau _{zy}&\sigma _{z}\\\end{matrix}}\right]}$

## 二維張量下的莫爾圓

${\displaystyle {\boldsymbol {\sigma }}=\left[{\begin{matrix}\sigma _{x}&\tau _{xy}&0\\\tau _{xy}&\sigma _{y}&0\\0&0&0\\\end{matrix}}\right]\equiv \left[{\begin{matrix}\sigma _{x}&\tau _{xy}\\\tau _{xy}&\sigma _{y}\\\end{matrix}}\right]}$

### 莫爾圓的方程

${\displaystyle \sigma _{\mathrm {n} }={\frac {1}{2}}(\sigma _{x}+\sigma _{y})+{\frac {1}{2}}(\sigma _{x}-\sigma _{y})\cos 2\theta +\tau _{xy}\sin 2\theta }$
${\displaystyle \tau _{\mathrm {n} }=-{\frac {1}{2}}(\sigma _{x}-\sigma _{y})\sin 2\theta +\tau _{xy}\cos 2\theta }$

{\displaystyle {\begin{aligned}\left[\sigma _{\mathrm {n} }-{\tfrac {1}{2}}(\sigma _{x}+\sigma _{y})\right]^{2}+\tau _{\mathrm {n} }^{2}&=\left[{\tfrac {1}{2}}(\sigma _{x}-\sigma _{y})\right]^{2}+\tau _{xy}^{2}\\(\sigma _{\mathrm {n} }-\sigma _{\mathrm {avg} })^{2}+\tau _{\mathrm {n} }^{2}&=R^{2}\end{aligned}}}

${\displaystyle R={\sqrt {\left[{\tfrac {1}{2}}(\sigma _{x}-\sigma _{y})\right]^{2}+\tau _{xy}^{2}}}\quad {\text{and}}\quad \sigma _{\mathrm {avg} }={\tfrac {1}{2}}(\sigma _{x}+\sigma _{y})}$

${\displaystyle (x-a)^{2}+(y-b)^{2}=r^{2}}$

${\displaystyle (\sigma _{\mathrm {n} },\tau _{\mathrm {n} })}$ 坐標系統中，其半徑${\displaystyle r=R}$ ，圓心在坐標${\displaystyle (a,b)=(\sigma _{\mathrm {avg} },0)}$ 處。

### 符號體系

#### 莫爾圓空間符號體系

1. 將正的剪應力畫在上方（圖5，符號體系#1）
2. 將正的剪應力畫在下方，也就是${\displaystyle \tau _{\mathrm {n} }}$ 軸倒置（圖5，符號體系#2）

### 繪製莫爾圓

1. 繪制笛卡爾坐標系統${\displaystyle (\sigma _{\mathrm {n} },\tau _{\mathrm {n} })}$ ，橫軸為${\displaystyle \sigma _{\mathrm {n} }}$ ，縱軸為${\displaystyle \tau _{\mathrm {n} }}$
2. ${\displaystyle (\sigma _{\mathrm {n} },\tau _{\mathrm {n} })}$ 空間中，畫出二點${\displaystyle A(\sigma _{y},\tau _{xy})}$ ${\displaystyle B(\sigma _{x},-\tau _{xy})}$ ，分別是作用在二垂直平面${\displaystyle A}$ 平面和 ${\displaystyle B}$ 平面上的應力分量（圖4及圖6），需依照選擇的符號體系。
3. 用線段${\displaystyle {\overline {AB}}}$ 連接${\displaystyle A}$ 點和${\displaystyle B}$ 點，此即為圓的直徑。
4. 繪製莫爾圓，其圓心${\displaystyle O}$ 是線段${\displaystyle {\overline {AB}}}$ 的中點，也就是此線和${\displaystyle \sigma _{\mathrm {n} }}$ 軸的交點。

### 找主要正向應力

${\displaystyle \sigma _{1}=\sigma _{\max }=\sigma _{\text{avg}}+R}$
${\displaystyle \sigma _{2}=\sigma _{\min }=\sigma _{\text{avg}}-R}$

${\displaystyle \sigma _{\text{avg}}={\tfrac {1}{2}}(\sigma _{x}+\sigma _{y})}$

${\displaystyle R={\sqrt {\left[{\tfrac {1}{2}}(\sigma _{x}-\sigma _{y})\right]^{2}+\tau _{xy}^{2}}}}$

### 找最大和最小剪應力

${\displaystyle \tau _{\max ,\min }=\pm R}$

### 找主要平面的方向

${\displaystyle \theta _{p1}}$ ${\displaystyle \theta _{p2}}$ 也可以用以下的方程取得

${\displaystyle \tan 2\theta _{\mathrm {p} }={\frac {2\tau _{xy}}{\sigma _{x}-\sigma _{y}}}}$

## 一般三維應力下的莫爾圓

{\displaystyle {\begin{aligned}\left(T^{(n)}\right)^{2}&=\sigma _{ij}\sigma _{ik}n_{j}n_{k}\\\sigma _{\mathrm {n} }^{2}+\tau _{\mathrm {n} }^{2}&=\sigma _{1}^{2}n_{1}^{2}+\sigma _{2}^{2}n_{2}^{2}+\sigma _{3}^{2}n_{3}^{2}\end{aligned}}}
${\displaystyle \sigma _{\mathrm {n} }=\sigma _{1}n_{1}^{2}+\sigma _{2}n_{2}^{2}+\sigma _{3}n_{3}^{2}.}$

{\displaystyle {\begin{aligned}n_{1}^{2}&={\frac {\tau _{\mathrm {n} }^{2}+(\sigma _{\mathrm {n} }-\sigma _{2})(\sigma _{\mathrm {n} }-\sigma _{3})}{(\sigma _{1}-\sigma _{2})(\sigma _{1}-\sigma _{3})}}\geq 0\\n_{2}^{2}&={\frac {\tau _{\mathrm {n} }^{2}+(\sigma _{\mathrm {n} }-\sigma _{3})(\sigma _{\mathrm {n} }-\sigma _{1})}{(\sigma _{2}-\sigma _{3})(\sigma _{2}-\sigma _{1})}}\geq 0\\n_{3}^{2}&={\frac {\tau _{\mathrm {n} }^{2}+(\sigma _{\mathrm {n} }-\sigma _{1})(\sigma _{\mathrm {n} }-\sigma _{2})}{(\sigma _{3}-\sigma _{1})(\sigma _{3}-\sigma _{2})}}\geq 0.\end{aligned}}}

${\displaystyle \tau _{\mathrm {n} }^{2}+(\sigma _{\mathrm {n} }-\sigma _{2})(\sigma _{\mathrm {n} }-\sigma _{3})\geq 0}$  因為其分母${\displaystyle \sigma _{1}-\sigma _{2}>0}$ 而且${\displaystyle \sigma _{1}-\sigma _{3}>0}$
${\displaystyle \tau _{\mathrm {n} }^{2}+(\sigma _{\mathrm {n} }-\sigma _{3})(\sigma _{\mathrm {n} }-\sigma _{1})\leq 0}$  因為其分母${\displaystyle \sigma _{2}-\sigma _{3}>0}$ 而且${\displaystyle \sigma _{2}-\sigma _{1}<0}$
${\displaystyle \tau _{\mathrm {n} }^{2}+(\sigma _{\mathrm {n} }-\sigma _{1})(\sigma _{\mathrm {n} }-\sigma _{2})\geq 0}$  因為其分母${\displaystyle \sigma _{3}-\sigma _{1}<0}$ 而且${\displaystyle \sigma _{3}-\sigma _{2}<0.}$

{\displaystyle {\begin{aligned}\tau _{\mathrm {n} }^{2}+\left[\sigma _{\mathrm {n} }-{\tfrac {1}{2}}(\sigma _{2}+\sigma _{3})\right]^{2}\geq \left({\tfrac {1}{2}}(\sigma _{2}-\sigma _{3})\right)^{2}\\\tau _{\mathrm {n} }^{2}+\left[\sigma _{\mathrm {n} }-{\tfrac {1}{2}}(\sigma _{1}+\sigma _{3})\right]^{2}\leq \left({\tfrac {1}{2}}(\sigma _{1}-\sigma _{3})\right)^{2}\\\tau _{\mathrm {n} }^{2}+\left[\sigma _{\mathrm {n} }-{\tfrac {1}{2}}(\sigma _{1}+\sigma _{2})\right]^{2}\geq \left({\tfrac {1}{2}}(\sigma _{1}-\sigma _{2})\right)^{2}\\\end{aligned}}}

## 腳註

1. ^ Parry, Richard Hawley Grey. Mohr circles, stress paths and geotechnics 2. Taylor & Francis. 2004: 1–30 [2018-02-05]. ISBN 0-415-27297-1. （原始内容存档于2020-08-07）.
2. ^ Russell C. Hibbeler. Mechanics Of Materials 8th Edition. 2010: 461–462. ISBN 978-0136022305 （英语）.