【小呆的力学笔记】有限元专题之循环对称结构有限元原理

1. 循环对称问题的提出

许多工程结构都是其中某一扇面的n次周向重复，也就是是周期循环对称结构。如果弹性体的几何形状、约束情况以及所受的外部载荷都是对称于某一轴，则所有的应力、应变和位移也就对称于对称轴，那么这就是循环对称问题。典型的有发动机轮盘受离心力载荷下的应力分析，轮盘结构如下图1所示。观察轮盘结构，不难发现轮盘是扇形段重复多次的结构，那么离心力是周期循环对称的，并假设轮盘温度场是沿周向均布的，那么轮盘的应力应变应该也是周期循环对称的。

对于循环对称问题，事实上可以通过仅对某一扇面进行有限元模型就能获得正确的应力、应变和位移分析结果，当然需要在有限元算法中引入特殊的条件。

2. 循环对称条件

2.1 节点位移的循环对称关系

在循环对称问题中，需要引入柱坐标系，来给定循环对称条件。如下图，其中$xyz$是笛卡尔坐标系，$r\theta z$是柱坐标系，结构是典型轮盘的某一扇段。

在该循环对称问题中，扇面的面A的节点$i$和面B的对应节点$j$在柱坐标系$r\theta z$应该具有相同的坐标，同时应该也具备相同的位移变量。假设节点$i$和节点$j$分别属于面A和面B的一对对应节点，见下面示意图，那么其柱坐标下的位移变量应该满足下式关系：
$$\begin{gathered} u_{ri}=u_{rj} \\ {u}_{\theta i}=u_{\theta j} \\ {u}_{zi}=u_{zj} \end{gathered}$$

节点$i$在柱坐标系下的位移与在笛卡尔坐标系下的位移进行变换，具体的变换关系如下

$$\begin{gathered} -u_{ri}\sin \alpha -u_{\theta i}\cos \alpha =u_{xi} \\ {u}_{ri}\cos \alpha -u_{\theta i}\sin \alpha =u_{yi} \\ {u}_{zi}=u_{zi} \end{gathered}$$
写成矩阵形式
$$\left[\begin{array}{c}{u}_{xi}\\ u_{yi}\\ u_{zi}\end{array}\right]=\left[\begin{array}{ccc}-\sin \alpha & -\cos \alpha & 0\\ \cos \alpha & -\sin \alpha & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{ri}\\ u_{\theta i}\\ u_{zi}\end{array}\right]$$
节点$j$在柱坐标系下的位移与在笛卡尔坐标系下的位移进行变换，具体的变换关系如下
$$\begin{gathered} u_{rj}\sin \beta -u_{\theta j}\cos \beta =u_{xj} \\ {u}_{rj}\cos \beta +u_{\theta j}\sin \beta =u_{yj} \\ {u}_{zj}=u_{zj} \end{gathered}$$
写成矩阵形式
$$\left[\begin{array}{c}{u}_{xj}\\ u_{yj}\\ u_{zj}\end{array}\right]=\left[\begin{array}{ccc}\sin \beta & -\cos \beta & 0\\ \cos \alpha & \sin \beta & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{rj}\\ u_{\theta j}\\ u_{zj}\end{array}\right]$$
那么
$$\left[\begin{array}{c}{u}_{rj}\\ u_{\theta j}\\ u_{zj}\end{array}\right]=\left[\begin{array}{ccc}\sin \beta & \cos \beta & 0\\ -\cos \alpha & \sin \beta & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{xj}\\ u_{yj}\\ u_{zj}\end{array}\right]$$
由于
$$\left[\begin{array}{c}{u}_{ri}\\ u_{\theta i}\\ u_{zi}\end{array}\right]=\left[\begin{array}{c}{u}_{rj}\\ u_{\theta j}\\ u_{zj}\end{array}\right]$$
那么
$$\begin{gathered} \left[\begin{array}{c}{u}_{xi}\\ u_{yi}\\ u_{zi}\end{array}\right]=\left[\begin{array}{ccc}-\sin \alpha & -\cos \alpha & 0\\ \cos \alpha & -\sin \alpha & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{ri}\\ u_{\theta i}\\ u_{zi}\end{array}\right]=\left[\begin{array}{ccc}-\sin \alpha & -\cos \alpha & 0\\ \cos \alpha & -\sin \alpha & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{rj}\\ u_{\theta j}\\ u_{zj}\end{array}\right] \\ =\left[\begin{array}{ccc}-\sin \alpha & -\cos \alpha & 0\\ \cos \alpha & -\sin \alpha & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}\sin \beta & \cos \beta & 0\\ -\cos \alpha & \sin \beta & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{xj}\\ u_{yj}\\ u_{zj}\end{array}\right] \\ =\left[\begin{array}{ccc}-\sin \alpha \sin \beta +\cos \alpha \cos \beta & -\sin \alpha \cos \beta -\cos \alpha \sin \beta & 0\\ \cos \alpha \sin \beta +\sin \alpha \cos \beta & \cos \alpha \cos \beta -\sin \alpha \sin \beta & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{xj}\\ u_{yj}\\ u_{zj}\end{array}\right] \\ =\left[\begin{array}{ccc}\cos (\alpha +\beta )& -\sin (\alpha +\beta )& 0\\ \sin (\alpha +\beta )& \cos (\alpha +\beta )& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{xj}\\ u_{yj}\\ u_{zj}\end{array}\right] \\ =\left[\begin{array}{ccc}\cos (\theta )& -\sin (\theta )& 0\\ \sin (\theta )& \cos (\theta )& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{u}_{xj}\\ u_{yj}\\ u_{zj}\end{array}\right]=\left[\begin{array}{c}{\theta }_1\end{array}\right]\left[\begin{array}{c}{u}_{xj}\\ u_{yj}\\ u_{zj}\end{array}\right] \end{gathered}$$

2.2 节点内力的循环对称关系

扇形段I除了节点位移存在循环对称关系，剩余扇形对扇形段I的节点力也存在循环对称关系。典型的扇形段相互作用关系见下图，其中扇形段I是分析对象，扇形段II和扇形段III对扇形段I有相互作用。

其中扇形段I、II、III是重复扇形段，$i$、$i^{{}^{\prime }}$、$i^{{}^{\prime \prime }}$是一组对应周期循环节点，$j$、$j^{{}^{\prime }}$、$j^{{}^{\prime \prime }}$是一组对应周期循环节点。
其中$j^{{}^{\prime }}$对$i$的作用力为$f_{ri}$、$f_{\theta i}$、$f_{zi}$，$j$对$i^{{}^{\prime \prime }}$的作用力为$f_{r{i}^{{}^{\prime \prime }}}$、$f_{\theta i^{{}^{\prime \prime }}}$、$f_{z{i}^{{}^{\prime \prime }}}$，从周期循环对称特征定义，可知
$$\begin{gathered} f_{ri}=f_{r{i}^{{}^{\prime \prime }}} \\ {f}_{\theta i}=f_{\theta i^{{}^{\prime \prime }}} \\ {f}_{zi}=f_{z{i}^{{}^{\prime \prime }}} \end{gathered}$$
那么，$i^{{}^{\prime \prime }}$对$j$的作用力$f_{rj}$、$f_{\theta j}$、$f_{zj}$，存在如下关系式
$$\begin{gathered} f_{ri}=-f_{rj} \\ {f}_{\theta i}=-f_{\theta j} \\ {f}_{zi}=-f_{zj} \end{gathered}$$
注：上述节点力均在柱坐标系下。
参照上节节点位移的转换关系推导过程，不难推得在上述节点力关系式在笛卡尔坐标系下的表达式
$$\left[\begin{array}{c}{f}_{xi}\\ f_{yi}\\ f_{zi}\end{array}\right]=-\left[\begin{array}{c}{\theta }_1\end{array}\right]\left[\begin{array}{c}{f}_{xj}\\ f_{yj}\\ f_{zj}\end{array}\right]$$

3. 在平衡方程中引入循环对称条件

若某循环结构包含一对循环对称节点$i$、$j$，不失一般性，平衡方程可以写成下式

$$\left[\begin{array}{cccccccc}{k}_{11}& k_{12}& \cdots & k_{1i}& \cdots & k_{1j}& \cdots & k_{1n}\\ k_{21}& k_{22}& \cdots & k_{2i}& \cdots & k_{2j}& \cdots & k_{2n}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ k_{i1}& k_{i2}& \cdots & k_{ii}& \cdots & k_{ij}& \cdots & k_{in}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ k_{j1}& k_{j2}& \cdots & k_{ji}& \cdots & k_{jj}& \cdots & k_{jn}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ k_{n1}& k_{n2}& \cdots & k_{ni}& \cdots & k_{nj}& \cdots & k_{nn}\end{array}\right]\left[\begin{array}{c}{u}_1\\ u_2\\ ⋮\\ u_i\\ ⋮\\ u_j\\ ⋮\\ u_n\end{array}\right]=\left[\begin{array}{c}{R}_1+F_1\\ F_2\\ ⋮\\ F_i+f_i\\ ⋮\\ F_j+f_j\\ ⋮\\ F_n\end{array}\right]$$
式中$u_1$为模型的位移约束，有$u_1={\overline{u}}_1$，$R_1$为支反力；$F_i，i=1,\cdots$为节点外载荷，$f_i、f_j$为其他扇形段对扇形段I的作用力，这里引入循环对称条件，

$$\left[\begin{array}{c}{f}_i\end{array}\right]=-\left[\begin{array}{c}{\theta }_1\end{array}\right]\left[\begin{array}{c}{f}_j\end{array}\right]$$
上面平衡方程变成如下形式
$$\left[\begin{array}{cccccccc}{k}_{11}& k_{12}& \cdots & k_{1i}& \cdots & k_{1j}& \cdots & k_{1n}\\ k_{21}& k_{22}& \cdots & k_{2i}& \cdots & k_{2j}& \cdots & k_{2n}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ k_{i1}& k_{i2}& \cdots & k_{ii}& \cdots & k_{ij}& \cdots & k_{in}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ k_{j1}& k_{j2}& \cdots & k_{ji}& \cdots & k_{jj}& \cdots & k_{jn}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ k_{n1}& k_{n2}& \cdots & k_{ni}& \cdots & k_{nj}& \cdots & k_{nn}\end{array}\right]\left[\begin{array}{c}{\overline{u}}_1\\ u_2\\ ⋮\\ u_i\\ ⋮\\ u_j\\ ⋮\\ u_n\end{array}\right]=\left[\begin{array}{c}{R}_1+F_1\\ F_2\\ ⋮\\ F_i-\theta f_j\\ ⋮\\ F_j+f_j\\ ⋮\\ F_n\end{array}\right]$$
进一步，用${\theta }^T$左乘第$i$行，则
$$\left[\begin{array}{cccccccc}{k}_{11}& k_{12}& \cdots & k_{1i}& \cdots & k_{1j}& \cdots & k_{1n}\\ k_{21}& k_{22}& \cdots & k_{2i}& \cdots & k_{2j}& \cdots & k_{2n}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ {\theta }^T{k}_{i1}& {\theta }^T{k}_{i2}& \cdots & {\theta }^T{k}_{ii}& \cdots & {\theta }^T{k}_{ij}& \cdots & {\theta }^T{k}_{in}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ k_{j1}& k_{j2}& \cdots & k_{ji}& \cdots & k_{jj}& \cdots & k_{jn}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ k_{n1}& k_{n2}& \cdots & k_{ni}& \cdots & k_{nj}& \cdots & k_{nn}\end{array}\right]\left[\begin{array}{c}{\overline{u}}_1\\ u_2\\ ⋮\\ u_i\\ ⋮\\ u_j\\ ⋮\\ u_n\end{array}\right]=\left[\begin{array}{c}{R}_1+F_1\\ F_2\\ ⋮\\ {\theta }^T{F}_i-f_j\\ ⋮\\ F_j+f_j\\ ⋮\\ F_n\end{array}\right]$$
将第$i$行加到第$j$行，上式进一步变换为
$$\left[\begin{array}{cccccccc}{k}_{11}& k_{12}& \cdots & k_{1i}& \cdots & k_{1j}& \cdots & k_{1n}\\ k_{21}& k_{22}& \cdots & k_{2i}& \cdots & k_{2j}& \cdots & k_{2n}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ {\theta }^T{k}_{i1}& {\theta }^T{k}_{i2}& \cdots & {\theta }^T{k}_{ii}& \cdots & {\theta }^T{k}_{ij}& \cdots & {\theta }^T{k}_{in}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ {\theta }^T{k}_{i1}+k_{j1}& {\theta }^T{k}_{i2}+k_{j2}& \cdots & {\theta }^T{k}_{ii}+k_{ji}& \cdots & {\theta }^T{k}_{ij}+k_{jj}& \cdots & {\theta }^T{k}_{in}+k_{jn}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ k_{n1}& k_{n2}& \cdots & k_{ni}& \cdots & k_{nj}& \cdots & k_{nn}\end{array}\right]\left[\begin{array}{c}{\overline{u}}_1\\ u_2\\ ⋮\\ u_i\\ ⋮\\ u_j\\ ⋮\\ u_n\end{array}\right]=\left[\begin{array}{c}{R}_1+F_1\\ F_2\\ ⋮\\ {\theta }^T{F}_i-f_j\\ ⋮\\ F_j+{\theta }^T{F}_i\\ ⋮\\ F_n\end{array}\right]$$
将位移循环对称条件引入上式中
$$\left[\begin{array}{c}{u}_i\end{array}\right]=\left[\begin{array}{c}{\theta }_1\end{array}\right]\left[\begin{array}{c}{u}_j\end{array}\right]$$
那么平衡方程变换为
$$\left[\begin{array}{cccccccc}{k}_{11}& k_{12}& \cdots & k_{1i}& \cdots & k_{1j}& \cdots & k_{1n}\\ k_{21}& k_{22}& \cdots & k_{2i}& \cdots & k_{2j}& \cdots & k_{2n}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ {\theta }^T{k}_{i1}& {\theta }^T{k}_{i2}& \cdots & {\theta }^T{k}_{ii}& \cdots & {\theta }^T{k}_{ij}& \cdots & {\theta }^T{k}_{in}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ {\theta }^T{k}_{i1}+k_{j1}& {\theta }^T{k}_{i2}+k_{j2}& \cdots & {\theta }^T{k}_{ii}+k_{ji}& \cdots & {\theta }^T{k}_{ij}+k_{jj}& \cdots & {\theta }^T{k}_{in}+k_{jn}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ k_{n1}& k_{n2}& \cdots & k_{ni}& \cdots & k_{nj}& \cdots & k_{nn}\end{array}\right]\left[\begin{array}{c}{\overline{u}}_1\\ u_2\\ ⋮\\ \theta u_j\\ ⋮\\ u_j\\ ⋮\\ u_n\end{array}\right]=\left[\begin{array}{c}{R}_1+F_1\\ F_2\\ ⋮\\ {\theta }^T{F}_i-f_j\\ ⋮\\ F_j+{\theta }^T{F}_i\\ ⋮\\ F_n\end{array}\right]$$
将$\theta$提出来，右乘到第$i$列，那么上式变为
$$\left[\begin{array}{cccccccc}{k}_{11}& k_{12}& \cdots & k_{1i}\theta & \cdots & k_{1j}& \cdots & k_{1n}\\ k_{21}& k_{22}& \cdots & k_{2i}\theta & \cdots & k_{2j}& \cdots & k_{2n}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ {\theta }^T{k}_{i1}& {\theta }^T{k}_{i2}& \cdots & {\theta }^T{k}_{ii}\theta & \cdots & {\theta }^T{k}_{ij}& \cdots & {\theta }^T{k}_{in}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ {\theta }^T{k}_{i1}+k_{j1}& {\theta }^T{k}_{i2}+k_{j2}& \cdots & {\theta }^T{k}_{ii}\theta +k_{ji}\theta & \cdots & {\theta }^T{k}_{ij}+k_{jj}& \cdots & {\theta }^T{k}_{in}+k_{jn}\\ ⋮& ⋮& & ⋮& & ⋮& & ⋮\\ k_{n1}& k_{n2}& \cdots & k_{ni}\theta & \cdots & k_{nj}& \cdots & k_{nn}\end{array}\right]\left[\begin{array}{c}{\overline{u}}_1\\ u_2\\ ⋮\\ u_j\\ ⋮\\ u_j\\ ⋮\\ u_n\end{array}\right]=\left[\begin{array}{c}{R}_1+F_1\\ F_2\\ ⋮\\ {\theta }^T{F}_i-f_j\\ ⋮\\ F_j+{\theta }^T{F}_i\\ ⋮\\ F_n\end{array}\right]$$

在上式中用缩减节点的位移列阵替换全节点位移列阵，即用$\left[\begin{array}{c}{\overline{u}}_1,u_2,\cdots ,u_{i-1},u_{i+1},\cdots ,u_j,\cdots ,u_n\end{array}\right]$替换$\left[\begin{array}{c}{\overline{u}}_1,u_2,\cdots ,u_{i-1},u_j，u_{i+1},\cdots ,u_j,\cdots ,u_n\end{array}\right]$
那么相应的要将位移列阵中第$i$行归属$u_j$合并到第$j$列，那么平衡方程变换为

$${\left[\begin{array}{ccccccccc}{k}_{11}& k_{12}& \cdots & k_{1i-1}& k_{1i+1}& \cdots & k_{1i}\theta +k_{1j}& \cdots & k_{1n}\\ k_{21}& k_{22}& \cdots & k_{2i-1}& k_{2i+1}& \cdots & k_{2i}\theta +k_{2j}& \cdots & k_{2n}\\ ⋮& ⋮& & ⋮& ⋮& & ⋮& & ⋮\\ {\theta }^T{k}_{i1}& {\theta }^T{k}_{i2}& \cdots & {\theta }^T{k}_{ii-1}& {\theta }^T{k}_{ii+1}& \cdots & {\theta }^T{k}_{ii}\theta +{\theta }^T{k}_{ij}& \cdots & {\theta }^T{k}_{in}\\ ⋮& ⋮& & ⋮& ⋮& & ⋮& & ⋮\\ {\theta }^T{k}_{i1}+k_{j1}& {\theta }^T{k}_{i2}+k_{j2}& \cdots & {\theta }^T{k}_{ii-1}+k_{ji-1}& {\theta }^T{k}_{ii+1}+k_{ji+1}& \cdots & {\theta }^T{k}_{ii}\theta +k_{ji}\theta +{\theta }^T{k}_{ij}+k_{jj}& \cdots & {\theta }^T{k}_{in}+k_{jn}\\ ⋮& ⋮& & ⋮& ⋮& & ⋮& & ⋮\\ k_{n1}& k_{n2}& \cdots & k_{ni-1}& k_{ni+1}& \cdots & k_{ni}\theta +k_{nj}& \cdots & k_{nn}\end{array}\right]}_{n\times (n-1)}{\left[\begin{array}{c}{\overline{u}}_1\\ u_2\\ ⋮\\ u_{i-1}\\ u_{i+1}\\ ⋮\\ u_j\\ ⋮\\ u_n\end{array}\right]}_{(n-1)\times 1}=\left[\begin{array}{c}{R}_1+F_1\\ F_2\\ ⋮\\ {\theta }^T{F}_i-f_j\\ ⋮\\ F_j+{\theta }^T{F}_i\\ ⋮\\ F_n\end{array}\right]$$
事实上如果位移列阵自由度为$(n-1)$，那么相应的方程也只需要$(n-1)$个，因此我们去掉第$i$方程，那么平衡方程变成
$${\left[\begin{array}{ccccccccc}{k}_{11}& k_{12}& \cdots & k_{1,i-1}& k_{1,i+1}& \cdots & k_{1i}\theta +k_{1j}& \cdots & k_{1n}\\ k_{21}& k_{22}& \cdots & k_{2,i-1}& k_{2,i+1}& \cdots & k_{2i}\theta +k_{2j}& \cdots & k_{2n}\\ ⋮& ⋮& & ⋮& ⋮& & ⋮& & ⋮\\ k_{i-1,1}& k_{i-1,2}& \cdots & k_{i-1,i-1}& k_{i-1,i+1}& \cdots & k_{i-1,i}\theta +k_{i-1,j}& \cdots & k_{i-1,n}\\ k_{i+1,1}& k_{i+1,2}& \cdots & k_{i+1,i-1}& k_{i+1,i+1}& \cdots & k_{i+1,i}\theta +k_{i+1,j}& \cdots & k_{i+1,n}\\ ⋮& ⋮& & ⋮& ⋮& & ⋮& & ⋮\\ {\theta }^T{k}_{i1}+k_{j1}& {\theta }^T{k}_{i2}+k_{j2}& \cdots & {\theta }^T{k}_{i,i-1}+k_{j,i-1}& {\theta }^T{k}_{i,i+1}+k_{j,i+1}& \cdots & {\theta }^T{k}_{ii}\theta +k_{ji}\theta +{\theta }^T{k}_{ij}+k_{jj}& \cdots & {\theta }^T{k}_{in}+k_{jn}\\ ⋮& ⋮& & ⋮& ⋮& & ⋮& & ⋮\\ k_{n1}& k_{n2}& \cdots & k_{ni-1}& k_{ni+1}& \cdots & k_{ni}\theta +k_{nj}& \cdots & k_{nn}\end{array}\right]}_{(n-1)\times (n-1)}{\left[\begin{array}{c}{\overline{u}}_1\\ u_2\\ ⋮\\ u_{i-1}\\ u_{i+1}\\ ⋮\\ u_j\\ ⋮\\ u_n\end{array}\right]}_{(n-1)\times 1}=\left[\begin{array}{c}{R}_1+F_1\\ F_2\\ ⋮\\ F_{i-1}\\ F_{i+1}\\ ⋮\\ F_j+{\theta }^T{F}_i\\ ⋮\\ F_n\end{array}\right]$$
将上式写成分块矩阵形式
$$\left[\begin{array}{cc}{k}_{11}& K_{12}\\ K_{21}& K_{22}\end{array}\right]\left[\begin{array}{c}{\overline{u}}_1\\ U_2\end{array}\right]=\left[\begin{array}{c}{R}_1+F_1\\ \hat{F}\end{array}\right]$$
将其展开
$$\begin{gathered} k_{11}{\overline{u}}_1+K_{12}{U}_2=R_1+F_1 \\ {K}_{21}{\overline{u}}_1+K_{22}{U}_2=\hat{F} \end{gathered}$$
那么$U_2$可以从下式求解
$$U_2=K_{22}^{-1}(\hat{F}-K_{21}{\overline{u}}_1)$$
那么，有
$$R_1=k_{11}{\overline{u}}_1+K_{12}{U}_2-F_1$$
同时，在确定$u_j$后，将其回代入下式
$$\left[\begin{array}{c}{u}_i\end{array}\right]=\left[\begin{array}{c}{\theta }_1\end{array}\right]\left[\begin{array}{c}{u}_j\end{array}\right]$$
可以确定$u_i$，那么就确定全部节点位移，带入平衡方程可以得到$f_i、f_j$，解得所有未知量。