计算力学(七)单元刚度矩阵

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位移

定义单元节点位移向量\(\textbf{V}_e\),已二维四边形区域为例 \[ \textbf{V}_e=[u_1,v_1,u_2,v_2\cdots u_8,v_8]^T\\ \] 单元内一点位移可以表示成 \[ \begin{bmatrix} u\\v \end{bmatrix}=\textbf{NV}_e \] 其中N为单元的插值矩阵 \[ N= \begin{bmatrix} u_1&0&u_2&0&\cdots& u_8&0\\ 0&v_1&0&v_2&\cdots& 0&v_8\\ \end{bmatrix}\\ \] 单元内位移是在局部坐标系中求得的,因此取求出该点在整体坐标系中的位置 \[ x=\sum_{i=1}^8N_ix_i\qquad y=\sum_{i=1}^8N_iy_i\\ \] 由于对位移插值和坐标变换使用了相同的插值基函数,因此称该单元为等参单元

  • 等参元(iso-parametricelement):几何形状矩阵\(N\)中的插值阶次＝位移形状矩阵\(N\)中的插值阶次
  • 超参元(super-parametricelement):几何形状矩阵\(N\)中的插值阶次>位移形状矩阵\(N\)中的插值阶次
  • 亚参元(sub-parametricelement):几何形状矩阵\(N\)中的插值阶次<位移形状矩阵\(N\)中的插值阶次

等参元、超参元以及亚参元的示意1

研究表明,对于等参元以及亚参元,位移函数可以满足完备性要求,而超参元不满足完备性要求

位移函数构造的收敛性准则

  1. 完备性要求(针对单元内部):如果在(势能)泛函中所出现位移函数的最高阶导数是\(m\)阶,则有限元解答收敛的条件之一是选区单元内的位移场函数至少是\(m\)阶完全多项式.
  2. 协调性要求(针对单元之间):如果在(势能)泛函中位移函数出现的最高阶导数是\(m\)阶,则位移函数在单元交界面上必须具有直至\(m-1\)阶的连续导数,即\(C_{m-1}\)连续性.
  • 完备性是单元收敛的必要条件,如果单元既完备,又协调,则单元一定是收敛的.
  • 完备而不协调的单元也会收敛.协调单元可以保证单调收敛.
  • 所谓收敛性,是指随着单元数目的无限增加,或单元面积趋近于零时,其解答趋近于真实解.
  • 对协调单元的不正确使用,也会出现不协调因素.

应变

定义单元中的应变向量\(\varepsilon\),在平面情况中 \[ \varepsilon=\begin{bmatrix} \varepsilon_x\\ \varepsilon_y\\ \gamma_{xy}\\ \end{bmatrix}= \begin{bmatrix} \dfrac{\partial u}{\partial x}\\ \dfrac{\partial v}{\partial y}\\ \dfrac{\partial u}{\partial y}+\dfrac{\partial v}{\partial x}\\ \end{bmatrix}\\ \]

轴对称情况

\[ \varepsilon=\begin{bmatrix} \varepsilon_x\\ \varepsilon_y\\ \gamma_{xy}\\ \varepsilon_\theta \end{bmatrix}= \begin{bmatrix} \dfrac{\partial u}{\partial x}\\ \dfrac{\partial v}{\partial y}\\ \dfrac{\partial u}{\partial y}+\dfrac{\partial v}{\partial x}\\ \dfrac{u}{x}\\ \end{bmatrix}\\ \]

\[ \begin{align*} \varepsilon&= \begin{bmatrix} \dfrac{\partial u}{\partial x}\\ \dfrac{\partial v}{\partial y}\\ \dfrac{\partial u}{\partial y}+\dfrac{\partial v}{\partial x}\\ \dfrac{u}{x}\\ \end{bmatrix}= \begin{bmatrix} \dfrac{\partial}{\partial x} & 0\\ 0 & \dfrac{\partial}{\partial y}\\ \dfrac{\partial}{\partial y}&\dfrac{\partial}{\partial x}\\ \dfrac{1}{x} & 0\\ \end{bmatrix} \begin{bmatrix} u\\ v\\ \end{bmatrix}= \begin{bmatrix} \dfrac{\partial}{\partial x} & 0\\ 0 & \dfrac{\partial}{\partial y}\\ \dfrac{\partial}{\partial y}&\dfrac{\partial}{\partial x}\\ \dfrac{1}{x} & 0\\ \end{bmatrix}\textbf{NV}_e\\ &= \begin{bmatrix} \dfrac{\partial N_1}{\partial x} & 0 & \dfrac{\partial N_2}{\partial x} & 0 & \cdots & \dfrac{\partial N_8}{\partial x} & 0\\ 0 &\dfrac{\partial N_1}{\partial y} & 0 & \dfrac{\partial N_2}{\partial y} & \cdots & 0 & \dfrac{\partial N_8}{\partial y}\\ \dfrac{\partial N_1}{\partial y} &\dfrac{\partial N_1}{\partial x} &\dfrac{\partial N_1}{\partial y} &\dfrac{\partial N_1}{\partial x} &\cdots&\dfrac{\partial N_1}{\partial y} &\dfrac{\partial N_1}{\partial x} &\\ \dfrac{N_1}{x} & 0&\dfrac{N_2}{x} & 0&\cdots &\dfrac{N_8}{x} & 0&\\ \end{bmatrix}\textbf{V}_e=\textbf{BV}_e \end{align*}\\ \]

平面情况

\[ \textbf{B}= \begin{bmatrix} \dfrac{\partial N_1}{\partial x} & 0 & \dfrac{\partial N_2}{\partial x} & 0 & \cdots & \dfrac{\partial N_8}{\partial x} & 0\\ 0 &\dfrac{\partial N_1}{\partial y} & 0 & \dfrac{\partial N_2}{\partial y} & \cdots & 0 & \dfrac{\partial N_8}{\partial y}\\ \dfrac{\partial N_1}{\partial y} &\dfrac{\partial N_1}{\partial x} &\dfrac{\partial N_1}{\partial y} &\dfrac{\partial N_1}{\partial x} &\cdots&\dfrac{\partial N_1}{\partial y} &\dfrac{\partial N_1}{\partial x} \\ \end{bmatrix}\\ \]

三维情况

\[ \varepsilon=\begin{bmatrix} \varepsilon_x\\ \varepsilon_y\\ \varepsilon_z\\ \gamma_{xy}\\ \gamma_{yz}\\ \gamma_{zx}\\ \end{bmatrix}= \begin{bmatrix} \dfrac{\partial u}{\partial x}\\ \dfrac{\partial v}{\partial y}\\ \dfrac{\partial w}{\partial z}\\ \dfrac{\partial u}{\partial y}+\dfrac{\partial v}{\partial x}\\ \dfrac{\partial v}{\partial z}+\dfrac{\partial w}{\partial y}\\ \dfrac{\partial u}{\partial z}+\dfrac{\partial w}{\partial x}\\ \end{bmatrix}= \mathbf{BV}_e\\ 其中\qquad\mathbf{B}=\begin{bmatrix} \dfrac{\partial N_1}{\partial x} & 0 & 0 & \cdots & \dfrac{\partial N_{20}}{\partial x} & 0&0\\ 0 &\dfrac{\partial N_1}{\partial y} & 0 & \cdots & 0 & \dfrac{\partial N_{20}}{\partial y}&0\\ 0&0 &\dfrac{\partial N_1}{\partial z} & \cdots &0& 0 & \dfrac{\partial N_{20}}{\partial z}\\ \dfrac{\partial N_1}{\partial y} &\dfrac{\partial N_1}{\partial x} &0&\cdots&\dfrac{\partial N_{20}}{\partial y} &\dfrac{\partial N_{20}}{\partial x}&0\\ 0&\dfrac{\partial N_1}{\partial z} &\dfrac{\partial N_1}{\partial y} &\cdots&0&\dfrac{\partial N_{20}}{\partial z} &\dfrac{\partial N_{20}}{\partial y}\\ \dfrac{\partial N_1}{\partial z} &0&\dfrac{\partial N_1}{\partial x} &\cdots&\dfrac{\partial N_{20}}{\partial z}&0 &\dfrac{\partial N_{20}}{\partial x}\\ \end{bmatrix}\\ \]

应力

定义单元中的应力向量\(\mathbf{\sigma}\) \[ \mathbf{\sigma}=[\sigma_x\quad\sigma_y\quad\tau_{xy}\quad\sigma_\theta]^T\\ \]

平面应力情况

\[ \mathbf{\sigma}= \begin{bmatrix} \sigma_x\\ \sigma_y\\ \tau_{xy}\\ \end{bmatrix}= \frac{E}{1-v^2} \begin{bmatrix} 1&\nu&0\\ \nu&1&0\\ 0&0&\dfrac{1-\nu}{2}\\ \end{bmatrix} \begin{bmatrix} \varepsilon_x\\ \varepsilon_y\\ \gamma_{xy}\\ \end{bmatrix}= \mathbf{D}\varepsilon=\mathbf{DBV}_e\\ \]

平面应变情况

\[ \mathbf{D}=\frac{E(1-\nu)}{(1+v)(1-2\nu)} \begin{bmatrix} 1&\dfrac{\nu}{1-\nu}&0\\ \dfrac{\nu}{1-\nu}&1&0\\ 0&0&\dfrac{1-2\nu}{2(1-\nu)}\\ \end{bmatrix}\\ \]

轴对称情况

\[ \mathbf{D}=\frac{E(1-\nu)}{(1+v)(1-2\nu)} \begin{bmatrix} 1&\dfrac{\nu}{1-\nu}&0&\dfrac{\nu}{1-\nu}\\ \dfrac{\nu}{1-\nu}&1&0&\dfrac{\nu}{1-\nu}\\ 0&0&\dfrac{1-2\nu}{2(1-\nu)}&0\\ \dfrac{\nu}{1-\nu}&\dfrac{\nu}{1-\nu}&0&1 \end{bmatrix}\\ \]

三维情况

\[ \mathbf{D}=\frac{E(1-\nu)}{(1+v)(1-2\nu)} \begin{bmatrix} 1&\dfrac{\nu}{1-\nu}&\dfrac{\nu}{1-\nu}&0&0&0\\ \dfrac{\nu}{1-\nu}&1&\dfrac{\nu}{1-\nu}&0&0&0\\ \dfrac{\nu}{1-\nu}&\dfrac{\nu}{1-\nu}&1&0&0&0\\ 0&0&0&\dfrac{1-2\nu}{2(1-\nu)}&0&0\\ 0&0&0&0&\dfrac{1-2\nu}{2(1-\nu)}&0\\ 0&0&0&0&0&\dfrac{1-2\nu}{2(1-\nu)}\\ \end{bmatrix}\\ \]

单元应变能

\[ \begin{align*} W^e&=\dfrac{1}{2}\iiint_{\Omega_e}\mathbf{\sigma}^T\cdot\mathbf{\varepsilon}\mathrm d\Omega=\dfrac{1}{2}\iiint_{\Omega_e}\mathbf{V}^T_e\mathbf{B}^T\mathbf{DBV}_e\mathrm d\Omega\\ &=\dfrac{1}{2}\mathbf{V}^T_e\iiint_{\Omega_e}\mathbf{B}^T\mathbf{DB}\mathrm d\Omega\;\mathbf{V}_e =\dfrac{1}{2}\mathbf{V}^T_e\mathbf{K}_e\mathbf{V}_e \end{align*} \]

单元刚度矩阵

定义单元刚度矩阵 \[ \mathbf{K}_e=\iiint_{\Omega_e}\mathbf{B}^T\mathbf{DB}\mathrm d\Omega\\ \]

\[ \begin{align*} 平面应力\qquad\mathbf{K}_e&=h\int_{-1}^1\int_{-1}^1\mathbf{B}^T\mathbf{DB}\det\mathbf{J}\mathrm d\xi\mathrm d\eta\qquad h为单元厚度\\ 平面应变\qquad\mathbf{K}_e&=\int_{-1}^1\int_{-1}^1\mathbf{B}^T\mathbf{DB}\det\mathbf{J}\mathrm d\xi\mathrm d\eta\\ 轴对称\qquad\mathbf{K}_e&=2\pi\int_{-1}^1\int_{-1}^1\mathbf{B}^T\mathbf{DB}\det\mathbf{J}\mathrm d\xi\mathrm d\eta\\ 三维情况\qquad\mathbf{K}_e&=\int_{-1}^1\int_{-1}^1\int_{-1}^1\mathbf{B}^T\mathbf{DB}\det\mathbf{J}\mathrm d\xi\mathrm d\eta d\zeta\\ \end{align*}\\ \]

好像整了很多公式但又好像啥也没写,感觉还有两三次就差不多了

#! https://zhuanlan.zhihu.com/p/695438202

  1. 曾攀.有限元分析基础教程.高等教育出版社,2009:151 ↩︎↩︎