Note_Fem边界条件的处理和numpy实现的四种方法

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 np

from typing import Dict,List,Tuple

import scipy as sc

def 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,P

def 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 1

u=

[[  0.           0.         -16.56666667  -0.248     ]]

p=

[[  50. 4980.  -50.   20.]]

solving by method 2

u=

[[  0.           0.         -16.56666667  -0.248     ]]

p=

[[  50. 4980.  -50.   20.]]

solving by method 3

u=

[[-1.04166667e-11 -3.11250000e-13 -1.65666667e+01 -2.48000000e-01]]

p=

[[  50. 4980.  -50.   20.]]

solving by method 4

u=

[[  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