# 如何使用 COMSOL Multiphysics 模拟残余应力

2014年 6月 5日

### 纯弯曲下的梁

#### 计算残余应力

(1)

\sigma_x=\frac{M_\mathrm{b}}
{I_z}y

(2)

M_y=\frac{\sigma_\mathrm{yield} I_z}{b/2}=\frac{\sigma_\mathrm{yield} ab^3/12}{b/2}=\frac{ab^2\sigma_\mathrm{yield}}{6}

(3)

M_\mathrm{ep}=M_\mathrm{e}+M_\mathrm{p}=\frac{2\sigma_\mathrm{yield}I_\mathrm{e}}{b-2h_\mathrm{p}}+\sigma_\mathrm{yield}ah_\mathrm{p}(b-h_\mathrm{p})

(4)

M_\mathrm{ep}=\frac{ab^2\sigma_\mathrm{yield}}{6}\left[1+\frac{2h_p}{b}\left(1-\frac{h_p}{b}\right)\right]

(5)

\sigma_\mathrm{r}=\sigma-\sigma_\mathrm{e}

(6)

\sigma_\mathrm{e}=\frac{M_\mathrm{tot}y}{I_z}
=\frac{2\sigma_\mathrm{yield}}

{b}\left[1+\frac{2h_\mathrm{p}}{b}
\left(1-\frac{h_\mathrm{p}}

{b}\right)\right]y

(7)

\sigma_\mathrm{r}=\sigma_\mathrm{yield}
-\frac{2\sigma_\mathrm{yield}}{b}\left[1+\frac{2h_\mathrm{p}}{b}
\left(1-\frac{h_\mathrm{p}}{b}\right)\right]y

(8)

\sigma_\mathrm{e}=\frac{M_\mathrm{e}y}{I_\mathrm{e}}=\frac{2y\sigma_\mathrm{yield}}{b-2h_\mathrm{p}}

(9)

\sigma_\mathrm{r}=\sigma_\mathrm{yield}\left[\frac{2}
{b-2h_\mathrm{p}}-\frac

{2} {b}\left[1+\frac{2h_\mathrm{p}}{b}
\left(1-\frac{h_\mathrm{p}}

{b}\right)\right]\right]y

RZ 平面上的表现的回弹现象。