单元型函数

节点线性单元

image-20241230140207887

对于杆单元,为了满足完备性要求,至少要选取一次线性多项式,即

\[\phi^e(x)=\alpha_0+\alpha_1 x \]

在1和2节点分别满足

\[\phi^e(x_1^e)=\phi_1^e x \quad \text{与} \quad \phi^e(x_2^e)=\phi_2^e x \]

此时得到两个线性方程,恰好对应(4.2)式中同样数量的待定系数\(\alpha_0\)\(\alpha_1\),因此这些待定系数可由(4.3)式解得。

写成矩阵形式

\[\phi^e(x)=\left[1 \quad x\right]\left[\begin{array}{l}\alpha_0 \\ \alpha_1\end{array}\right]=p(x) \alpha^e \]

采用节点值表示待定系数

\[\begin{array}{l} \phi^e(x_1^e)=\phi_1^e x=\alpha_0+\alpha_1 x_1^e \rightarrow\left[\begin{array}{l} 1 \\ x_1^e \end{array}\right]\left[\begin{array}{l} \alpha_0 \\ \alpha_1 \end{array}\right] \\ \phi^e(x_2^e)=\phi_2^e x=\alpha_0+\alpha_1 x_2^e \rightarrow\left[\begin{array}{l} 1 \\ x_2^e \end{array}\right]\left[\begin{array}{l} \alpha_0 \\ \alpha_1 \end{array}\right] \end{array} \]

由(4.5)式解出

\[\alpha^e=(A^e)^{-1} \phi^e \]

其中,\((A^e)^{-1}\)\(A^e\)的逆矩阵,它可按下式计算得到

\[(A^e)^{-1}=\frac{1}{\left|A^e\right|}\left(A^e\right)^* \]

\(\left|A^e\right|\)\(A^e\)的行列式。

\((A^e)^*\)\(A^e\)的伴随矩阵,它的元素\((A_{ij}^e)^*\)\(A^e\)的元素\(A_{ij}^e\)的代数余子式。那么,对于(4.5)式中的\(A^e\),其逆矩阵为

\[(A^e)^{-1}=\frac{1}{l^e}\left[\begin{array}{cc} x_2^e & -x_1^e \\ -1 & 1 \end{array}\right]=\frac{1}{l^e}\left[\begin{array}{cc} l^e & -x_1^e \\ -1 & 1 \end{array}\right] \]

其中,\(l^e\)为杆单元长度。

将(4.6)式代入(4.4)式,得

\[\phi^e(x)=N^e(x) \phi^e \]

文中首先定义了单元形函数矩阵\(N^e(x)=[N_1^e(x) \quad N_2^e(x)] = p(x)(A^e)^{-1}\),并通过将\((4.8)\)式代入\((4.9)\)式,得到:

\[N^e = [N_1^e(x) \quad N_2^e(x)] = p(x)(A^e)^{-1} = \left[1 \quad x\right]\left[\begin{array}{cc} \frac{x_2^e}{l^e} & -\frac{x_1^e}{l^e} \\ -\frac{1}{l^e} & \frac{1}{l^e} \end{array}\right] = \left[\frac{l^e - x}{l^e} \quad \frac{x}{l^e}\right] \]

\((4.10)\)式中,\(N_1^e(x)\)\(N_2^e(x)\)分别为单元在1节点与2节点的形函数,且这些形函数是线性函数,具有以下属性:

\[N_1^e(x_1^e) = 1, \quad N_1^e(x_2^e) = 0 \]

\[N_2^e(x_1^e) = 0, \quad N_2^e(x_2^e) = 1 \]

简记为:

\[N_i^e(x_j^e) = \delta_{ij} \]

其中,\(\delta_{ij}\)为Kronecker符号,即:

\[\delta_{ij} = \begin{cases} 1, & i = j \\ 0, & i \neq j \end{cases} \]

在第3章中的“弱”形式中,需要求导函数与场函数的导数,如下:

\[\frac{d\phi^e}{dx} = \frac{d(N^e \phi^e)}{dx} = \frac{dN^e}{dx} \phi^e = \frac{dN_1^e}{dx} \phi_1^e + \frac{dN_2^e}{dx} \phi_2^e \]

写成矩阵形式:

\[\frac{d\phi^e}{dx} = \left[\begin{array}{ll} \frac{dN_1^e}{dx} & \frac{dN_2^e}{dx} \end{array}\right]\left[\begin{array}{l} \phi_1^e \\ \phi_2^e \end{array}\right] = B^e \phi^e \]

其中,\(B^e\)称为单元应变矩阵,即:

\[B^e = \left[\begin{array}{ll} \frac{dN_1^e}{dx} & \frac{dN_2^e}{dx} \end{array}\right] = \left[\begin{array}{ll} -\frac{1}{l^e} & \frac{1}{l^e} \end{array}\right] \]

posted @ 2024-12-30 15:42  redufa  阅读(17)  评论(0)    收藏  举报