Note_Fem边界条件的处理和numpy实现的四种方法

S8nHor5PPK · 3 天前

将单元刚度矩阵组装为全局刚度矩阵后,有:

此时的线性方程没有唯一解,\([K]\)是奇异矩阵,这是没有引入边界条件,消除刚体位移的原因.
边界条件分为两类:Forced and Geometric;对于力边界条件可以直接附加到节点力向量\([P]\)中,即\(P_j=P_j^{*}\),\(P_j^{*}\)是给定的节点力值.
因此我们基本只需要处理Geometric Boundary condition.下面介绍三种方法,将Bcs引入到\([K]、[P]\)
以位移边界条件为例,指定相关自由度值即:\(\Phi_j=\Phi_j^{*}\)
Method 1

将开头的\([K][\Phi]=[P]\)划分为:

\[\begin{bmatrix}[K_{11}] & [K_{12}] \\[K_{21}] & [K_{22}]\end{bmatrix}\begin{Bmatrix}\overrightarrow{\Phi}_1 \\\overrightarrow{\Phi}_2\end{Bmatrix}=\begin{Bmatrix}\overrightarrow{P}_1 \\\overrightarrow{P}_2\end{Bmatrix}\tag{1}\]

其中,\(\Phi_1\)是未知的自由节点自由度向量(free dofs);\(\Phi_2\)是已知的约束节点自由度值\(\Phi_j^{*}\)向量(specified nodal dof);\(P_1\)是已知节点力向量;\(P_2\)是未知的支反力向量
公式2进一步:

\[[K_{11}]\overrightarrow{\Phi}_1+[K_{12}]\overrightarrow{\Phi}_2=\overrightarrow{P}_1\tag{2}\]

\[[K_{21}]\overrightarrow{\Phi}_1+[K_{22}]\overrightarrow{\Phi}_2=\overrightarrow{P}_2\tag{3}\]

这时,\([K_{11}]\)是非奇异矩阵.因此自由节点自由度(未知节点位移)可求:

\[\overrightarrow{\Phi}_1=[K_{11}]^{-1}(\overrightarrow{P}_1-[K_{12}]\overrightarrow{\Phi}_2)\tag{4}\]

一旦\(\Phi_1\)求得,则未知支反力\(P_2\)可由公式3求得.
Method 2

也称划行划列法.method 1 中需要对\([K] ,[\Phi],[P]\)进行行列对调,重新排序.当出现非0位移边界时,method 1耗时长且需要记录过程,之后还需要恢复刚度矩阵.因此和method 1等效的处理方法是构建下式:

\[\begin{bmatrix}\left[K_{11}\right] & \left[0\right] \\\left[0\right] & \left[I\right]\end{bmatrix}\begin{bmatrix}\overrightarrow{\Phi}_1 \\\overrightarrow{\Phi}_2\end{bmatrix}=\begin{bmatrix}\overrightarrow{P}_1-\left[K_{12}\right]\overrightarrow{\Phi}_2\\{\overrightarrow{\Phi}_2}\end{bmatrix}\tag{5}\]

实际计算中,不需要对刚度阵重新排序.算法操作如下:

对所有的约束自由度\(\Phi_j\)重复Step 1~3即可,这种操作能够保持刚度和方程的对称性.
Method 3

该方法也称乘大数法.假设约束自由度为\(\Phi_j=\Phi_j^*\),操作如下:

该方法通用性强,适合大多数的静力学线性问题,但数值精度与大数的取值有关,太小了精度差,太大了容易出现"矩阵奇异"的现象

Method 4(对角元素置1法)

该方法的做法是,对于约束自由度\(\Phi_j=0\),把\([K]\)的j行j列置0,但对角元素Kjj=1,\([P]\)中对应元素置0.
以6x6的刚度矩阵为例子,

\[\begin{gathered}\begin{bmatrix}k_{11} & k_{12} & 0 & k_{14} & k_{15} & k_{16} \\k_{21} & k_{22} & 0 & k_{24} & k_{25} & k_{26} \\0 & 0 & 1 & 0 & 0 & 0 \\k_{41} & k_{42} & 0 & k_{44} & k_{45} & k_{46} \\k_{51} & k_{52} & 0 & k_{54} & k_{55} & k_{56} \\k_{e1} & k_{e3} & 0 & k_{eA} & k_{e5} & k_{e6}\end{bmatrix}\begin{bmatrix}\delta_1 \\\delta_2 \\\delta_3 \\\delta_4 \\\delta_5 \\\delta_6\end{bmatrix}=\begin{bmatrix}f_1 \\f_2 \\0 \\f_4 \\f_5 \\f_6\end{bmatrix}\end{gathered}\]

不引入大数,避免了数值稳定性的问题,不会影响矩阵的条件数; 但只适合\(\Phi_j=0\)这样的简单边界;可能影响系统矩阵的特性,直接替换可能改变矩阵的对称性(尤其在动力学和非线性问题中);不能处理非0的位移加载,只能处理力加载

Example

例题来自《The Finite Element Method in Engineering》的悬臂梁模型(example6.4, page227)

静力平衡方程为:

解为:

\[W=[0,0,-16.5667,-0.2480]\]

\[P=[50.0,4980.0,-50,20]\]

solve by method 1

solve by method 2

循环每个位移约束,需要注意高亮处的操作:

求解:

solve by method 3

Code Realize

四种方法进行Python+Numpy+Scipy编程实现,并与Example的解进行对比.
#-------------------------------------------------------------------------------# Name: BcsProcess# Purpose: 引入边界条件到[K]中,并返回解[U],[P]# input:# K:全局刚度矩阵,(M,M) numpy.array# BcDict:位移约束,key (int) = 自由度序号(1-based) , value (float) = 自由度约束值# LoadDict:节点力加载,key (int) = 自由度序号(1-based) , value (float) = 施加的节点力加载或者等效节点力加载## Author: Administrator## Created: 08-03-2025# Copyright: (c) Administrator 2025# Licence: <your licence>#-------------------------------------------------------------------------------import numpy as npfrom typing import Dict,List,Tupleimport scipy as scdef Method1(K:np.ndarray,BcDict:Dict[int,float],LoadDict:Dict[int,float])->Tuple: dofNum=K.shape[0] # 初始化向量 U,P=np.zeros((dofNum,1)),np.zeros((dofNum,1)) prescribedDofIndexs=np.array(list(BcDict.keys()))-1 #使用集合运算,全部自由度与约束自由度求差, 得到自由位移自由度的 freeDofIndexs=np.array(list(set(range(dofNum))-set(prescribedDofIndexs.tolist())),dtype=int) # 已知节点力加到P for label,Pval in LoadDict.items(): ind=label-1 P[ind,0]+=Pval # 已知节点位移(prescribed dof) for label,Uval in BcDict.items(): ind=label-1 U[ind,0]+=Uval U2=U[np.ix_(prescribedDofIndexs,[0])].copy() # 已知节点力(free dof) P1=P[np.ix_(freeDofIndexs,[0])].copy() # 重新划分K行列 K11=K[np.ix_(freeDofIndexs,freeDofIndexs)].copy() K12=K[np.ix_(freeDofIndexs,prescribedDofIndexs)].copy() K21=K[np.ix_(prescribedDofIndexs,freeDofIndexs)].copy() K22=K[np.ix_(prescribedDofIndexs,prescribedDofIndexs)].copy() # 计算自由节点位移值 U1=np.dot(sc.linalg.inv(K11),P1-K12.dot(U2)) # 计算支反力 P2=np.dot(K21,U1)+np.dot(K22,U2) # 合并到U,P向量 U[np.ix_(freeDofIndexs,[0])]=U1 P[np.ix_(prescribedDofIndexs,[0])]=P2 return U,Pdef Method2(K:np.ndarray,BcDict:Dict[int,float],LoadDict:Dict[int,float])->Tuple: K_origin=K.copy() dofNum=K.shape[0] # 初始化向量 U,P=np.zeros((dofNum,1)),np.zeros((dofNum,1)) # 已知节点力加到 P for label,Pval in LoadDict.items(): ind=label-1 P[ind,0]+=Pval # 循环所有的位移约束 for label,Uval in BcDict.items(): j=label-1 #Step1 for i in range(dofNum): P[i,0]=P[i,0]-K[i,j]*Uval #Step2 for i in range(dofNum): K[i,j]=0 K[j,i]=0 K[j,j]=1 #Step3 P[j,0]=Uval # 求解 K'U=P' U_=sc.linalg.solve(K,P) P_=np.dot(K_origin,U_) return U_,P_def Method3(K:np.ndarray,BcDict:Dict[int,float],LoadDict:Dict[int,float])->Tuple: C=np.max(K)*10e6 K_origin=K.copy() dofNum=K.shape[0] # 初始化向量 U,P=np.zeros((dofNum,1)),np.zeros((dofNum,1)) # 已知节点力加到 P for label,Pval in LoadDict.items(): ind=label-1 P[ind,0]+=Pval # 循环所有位移约束 for label,Uval in BcDict.items(): j=label-1 # Step1 K[j,j]=K[j,j]*C # Step2 P[j,0]=K[j,j]*Uval # 求解 K'U=P' U_=sc.linalg.solve(K,P) P_=np.dot(K_origin,U_) return U_,P_def Method4(K:np.ndarray,BcDict:Dict[int,float],LoadDict:Dict[int,float])->Tuple: if np.any(np.array(list(BcDict.values())) != 0): raise ValueError('该方法不能处理非0位移加载') K_origin=K.copy() dofNum=K.shape[0] # 初始化向量 U,P=np.zeros((dofNum,1)),np.zeros((dofNum,1)) # 已知节点力加到 P for label,Pval in LoadDict.items(): ind=label-1 P[ind,0]+=Pval # loop all nodal bcs for label, Uval in BcDict.items(): j=label-1 K[j,:]=0.0 K[:,j]=0.0 K[j,j]=1.0 P[j,0]=0 # solve K'U=P' U_=sc.linalg.solve(K,P) P_=np.dot(K_origin,U_) return U_,P_if __name__ == '__main__': K=np.array([[12,600,-12,600], [600,40000,-600,20000], [-12,-600,12,-600], [600,20000,-600,40000]]) Bcs={1:0,2:0} loads={3:-50,4:20} # 精确解 extract_U=np.array([0,0,-16.5667,-0.2480]) extract_P=np.array([50.0,4980.0,-50,20]) # 求解 u,p=Method4(K,Bcs,loads) print(f"extract U=\n{extract_U}") print(f"u=\n{u.T}") print(f"extract_P=\n{extract_P}") print(f"p=\n{p.T}")计算结果:
extract_U=[ 0. 0. -16.5667 -0.248 ]extract_P=[ 50. 4980. -50. 20.]solving by method 1u=[[ 0. 0. -16.56666667 -0.248 ]]p=[[ 50. 4980. -50. 20.]]solving by method 2u=[[ 0. 0. -16.56666667 -0.248 ]]p=[[ 50. 4980. -50. 20.]]solving by method 3u=[[-1.04166667e-11 -3.11250000e-13 -1.65666667e+01 -2.48000000e-01]]p=[[ 50. 4980. -50. 20.]]solving by method 4u=[[ 0. 0. -16.56666667 -0.248 ]]p=[[ 50. 4980. -50. 20.]]总结

列举了四种直接节点位移边界条件的处理办法,并编程实现,求解案例.对比结果发现:相比Method3存在数值误差,其他三个都更加精确.
如果需要处理多点耦合边界条件,则有罚函数法,拉格朗日乘子法等.
参考资料:

Note Completed at 2025/03/08