多级弹簧-质量系统瞬态分析(基于Newmark)

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匿名技术用户   2020-12-27 06:25   11   0

该程序主要用于实现多自由度下动态系统的隐式分析,计算模型参见多级弹簧-质量系统的瞬态计算(基于中心差分法)。下面是具体的Python程序。

#coding:utf-8
"""
Created on Tue Nov 20 12:41:40 2018

@author: yujin.wang
"""
import time
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import sys
sys.setrecursionlimit(10000) # 10000 is an example, try with different values
################Force###############
def CenterDiff(t,endtime,m,k,fext,u,delta,du0):
 fint = k.dot(u)
 ddu = m.I.dot((fext - fint))
 du = du0 + ddu * delta
 u = u + du*delta
 t += delta
 t_list.append(t);u1_list.append(u[0,0]);u2_list.append(u[1,0])
 if t > endtime:
  return t_list,u1_list,u2_list
 else:
  return CenterDiff(t,endtime,m,k,fext,u,delta,du)
 
def NewmarkSolver(t,endT,deltaT,TOL,u,du,ddu,k,c,m,iternum):
 t += deltaT
 a1 = m/(beta*deltaT*deltaT) + gamma *c/(beta*deltaT)
 a2 = m/(beta*deltaT) + c*(gamma/beta-1)
 a3 = m*(1./(2*beta)-1) + deltaT*c*(gamma/(2*beta)-1)

 k_ = k + a1

 f_ = f + a1*u + a2*du + a3*ddu
 un = k_.I*f_
 Rn = f_ - k*un - a1*un
# print Rn
 print '##############TIME: %f####################' %(t)
 print time.ctime()
 print 'Iter.:%d Rn:%f' %(iternum,abs(Rn).max())
#Newton-Raphson
 while abs(Rn).max() > TOL:
  iternum += 1
  k_= k_ + a1
  dun = k_.I*Rn
  un = un + dun
  Rn = f_ - k*un - a1*un
  print 'Iter.:%d Rn:%f' %(iternum,abs(Rn).max())
 dun = gamma*(un-u)/(beta*deltaT) + du*(1-gamma/beta) + deltaT*ddu*(1-gamma/(2*beta))
 ddun = (un-u)/(beta*deltaT*deltaT) - du/(beta*deltaT) - ddu*(1./(2*beta)-1)
 t_list.append(t)
 u1_list.append(un[0,0])
 u2_list.append(un[1,0])
 if t+deltaT > endT:
  return  t_list,u1_list,u2_list
 else:
  print t,un
  return NewmarkSolver(t,endT,deltaT,TOL,un,dun,ddun,k,c,m,iternum)
 
if __name__ == "__main__":
 iternum = 0
 gamma=0.5;
 beta=0.25;
 TOL =1e-3
 deltaT = 0.02
 m1=1.;m2=1.
 k1 = 100.;k2=100.;k3=0.001
 f1=0.;f2=1.
 m = np.matrix([[m1,0],[0,m2]])
 c = np.matrix([[0,0],[0,0]])
 k = np.matrix([[k1+k2,-k2],[-k2,k2+k3]])
 f = np.matrix([[f1],[f2]])

 u0 = np.matrix([[0],[0]])
 du0 = np.matrix([[0],[0]])
 u1_list = [];u2_list = []
 t_list = []
 un_list = []

 ddu0 = np.linalg.inv(m)*(f - c*du0 )
 t_list,u1_list,u2_list = NewmarkSolver(0,25.,deltaT,TOL,u0,du0,ddu0,k,c,m,iternum)
 plt.figure(1)
 plt.plot(t_list,u1_list,'red')
 plt.plot(t_list,u2_list,'blue')
 plt.legend(['X1','X2'])
 plt.grid()
 plt.show()

设置阻尼矩阵C=[[1,0],[0,1]]

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