python的拱桥振型简化计算

from sympy import *
import numpy as np
import matplotlib.pyplot as plt
import math
#基本参数
g=9.8
l=60
f=2
Ac=8.15
Ic=3622000*(10e-8)
E=2.06*10e3
m=1435.2
#方程运算
H0=(m*g*(l)**2)/(8*f)
He=20502900
v1=1-(H0/He)
v2=1-H0/(4*He)
v3=1-H0/(9*He)
p=l**2/(8*f)
L=l*(1+(16/3)*(f/l)**2)
a21=8*E*Ac*l
a22=(m*((np.pi)**2)*(p)**2)*L
a2=(a21/a22)*10e4
c2=(2*(np.pi)/l)**2*sqrt((E*Ic*v2)/m)
c2=c2*1000#平衡系数
c1=((np.pi/l)**2)*sqrt((E*Ic*v1)/m)
c1=c1*1000#平衡系数
c3=((3*np.pi/l)**2)*sqrt((E*Ic*v1)/m)
c3=c3*1000
w=symbols('w')
G=solve([w**4-(c1**2+c3**2+(10/9)*a2)*w**2+(c1**2*c3**2+a2*(c1**2/9+c3**2))],[w])print(v1,v2,v3,a21,a22,a2)
print(c2,c1,c3)
print(G,c2)

以抛物线型拱为例，采用传统的数值计算方法过于繁琐（此题以李国豪版桥梁结构稳定与振动为例）

考虑拱桥的前几阶振型必须先考虑前几阶振型所对应的频率

（以三阶为例，一个为基频另外两个为对称频率）

对于此联立方程不只能求出振型比率

此外思考拱桥抛物线的矢高与拱桥长度的关系

采用最小2乘法拟合分析

根据上式代码计算出的结果

import matplotlib.pyplot as plt
import numpy as np
from scipy.optimize import leastsq
import math
plt.rcParams['font.sans-serif']=['SimHei']#显示中文
plt.rcParams['axes.unicode_minus']=False
X=np.array([2,2.5,3,3.5,4,4.5,5,5.5,6,6.5,7,7.5,8,8.5,9,9.5,10])
Y1=np.array([24.52,24.61,24.68,24.73,24.76,24.788,24.81,24.827,24.842,24.855,24.865,24.875,24.883,24.89,24.896,24.902,24.907])
Y2=np.array([14.75,17.9,21.05,24.17,27.23,30.214,33.086,35.81,38.352,40.662,42.7,44.44,45.877,47.035,47.955,48.683,49.26])def func1(params,x):A,B=paramsreturn A*np.sqrt(1-B*(1/x))def error1(params,x,y):return func1(params,x) - ydef  slovePara1():p0=[1,1]Para = leastsq(error1,p0,args=(X,Y1))return Paradef func2(params,x):k1,k2,k3,k4,k5,k6,k7,k8,k9,k10,k11,k12=paramsreturn 0.235702260395516*np.sqrt((k1+k2*x+k3*x**2+k4*x**3+(k5+k6*x+k7*x**2+k8*x**3+k9*x**4+k10*x**5+k11*x**6)**0.5)/(x*(k12*x**2+1)))def error2(params,x,y):return func2(params,x) - ydef slovePara2():p0=[1,1,1,1,1,1,1,1,1,1,1,1]Para = leastsq(error2,p0,args=(X,Y2))return Paradef solution():Para = slovePara1()A, B,  =Para[0]plt.scatter(X, Y1, color='green',label=u'已知样点',linewidth=0.3)x=np.linspace(2,10,50)y1=A*np.sqrt(1-B*(1/x))plt.plot(x,y1,color='red',label='w2',linewidth=2)plt.legend()Para = slovePara2()k1, k2, k3,k4,k5,k6,k7,k8,k9,k10,k11,k12  =Para[0]x=np.linspace(2,10,50)y2=0.235702260395516*np.sqrt((k1+k2*x+k3*x**2+k4*x**3+(k5+k6*x+k7*x**2+k8*x**3+k9*x**4+k10*x**5+k11*x**6)**0.5)/(x*(k12*x**2+1)))plt.plot(x,y2,color='blue',label='w1',linewidth=2)plt.legend()plt.scatter(X, Y2, color='green',linewidth=0.3)plt.title(u'w2与w1随f的变化情况')plt.show()solution()

所画图形如下