问题描述
我正在尝试数值求解伯努利的梁方程并绘制结果.该方程的一阶导数是斜率，二阶导数是挠度.我逐步解决了这个问题，首先绘制函数，然后对其进行积分，然后将积分结果绘制在同一张图上.
I'm trying to solve Bernoulli's beam equation numerically and plotting the results. First derivative of the equation is the slope and the second derivative is the deflection. I approached the problem step-by-step, first plot the function and then integrate it and plot the integration results on the same diagram.
到目前为止，我的代码如下.我怀疑问题在于，integrate.quad返回单个值，而我正试图从中获取多个值.有人知道如何解决吗?
My code so far is bellow. I suspect the problem lies in the fact that the integrate.quad returns a single value and I'm trying to get multiple values from it. Does anyone know how to approach it?
from scipy import integrate
import numpy as np
from pylab import *
# Beam parameters
L = 100
w = 10
h = 10
I = (w*h**3)/12
E = 200000
F = 100
def d2y_dx2(x):
return (-F*x)/(E*I)
a = 0.0
b = L
res, err = integrate.quad(d2y_dx2, a, b)
t = np.linspace(a,b,100)
ax = subplot(111)
ax.plot(t, d2y_dx2(t))
ax.plot(t, res(t))
show()
波纹管是用willcrack的答案修改的代码.该代码现在可以使用，但是结果不正确.在底部，我添加了使用正确的梁方程的解析解绘制结果图的代码.
Bellow is modified code with willcrack's answer. This code now works, but the results are not correct. On the bottom I added the code for plotting the results using analytical solutions of the beam equation which are correct.
from scipy import integrate
import numpy as np
import matplotlib.pyplot as plt
# Beam parameters
L = 100
w = 10
h = 10
I = (w*h**3)/12
E = 200000
F = 100
# Integration parameters
a = 0.0
b = L
# Define the beam equation
def d2y_dx2(x,y=None):
return (-F*x)/(E*I)
def something(x):
return integrate.quad(d2y_dx2)[0]
# Define the integration1 - slope
def slope(t):
slope_res = []
for x in t:
res1, err = integrate.quad(d2y_dx2, a, b)
slope_res.append(res1)
return slope_res
# Define the integration1 - deflection
def defl(t1):
defl_res = []
for t in t1:
res2, err = integrate.dblquad(d2y_dx2,a,b, lambda x: a, lambda x: b)
defl_res.append(res2)
return defl_res
# Plot
fig, (ax1, ax2, ax3) = plt.subplots(3)
t = np.linspace(a,b,100)
t1 = np.linspace(a,b,100)
ax1.plot(t, d2y_dx2(t))
ax2.plot(t, slope(t))
ax3.plot(t1, defl(t1))
plt.show()
以下是分析解决方案，代码和结果.偏转光束的形状被扭转，光束的末端在x = 0.
Analytical solution, code and results bellow. The shape of deflected beam is turned around, the end of the beam is at x = 0.
from __future__ import division #to enable normal floating division
import numpy as np
import matplotlib.pyplot as plt
# Beam parameters
w = 10 #beam cross sec width (mm)
h = 10 #beam cross sec height (mm)
I = (w*h**3)/12 #cross sec moment of inertia (mm^4)
I1 = (w*h**3)/12
E = 200000 #steel elast modul (N/mm^2)
L = 100 #beam length(mm)
F = 100 #force (N)
# Define equations
def d2y_dx2(x):
return (-F*x)/(E*I)
def dy_dx(x):
return (1/(E*I))*(-0.5*F*x**2 + 0.5*F*L**2)
def y(x):
return (1/(E*I))*(-(1/6)*F*(x**3) + (1/2)*F*(L**2)*x - (1/3)*F*(L**3))
# Plot
fig, (ax1, ax2, ax3) = plt.subplots(3)
a = 0
b = L
x = np.linspace(a,b,100)
ax1.plot(x, d2y_dx2(x))
ax2.plot(x, dy_dx(x))
ax3.plot(x, y(x))
plt.show()
推荐答案
也许您可以尝试这样的事情
Maybe you can try something like this
from scipy import integrate
import numpy as np
import matplotlib.pyplot as plt
# Beam parameters
L = 100
w = 10
h = 10
I = (w*h**3)/12
E = 200000
F = 100
# Integration parameters
a = 0.0
b = L
# Define the beam equation
def d2y_dx2(x,y=None):
return (-F*x)/(E*I)
def something(x):
return integrate.quad(d2y_dx2)[0]
# Define the integration1 - slope
def slope(t):
slope_res = []
for x in t:
res1, err = integrate.quad(d2y_dx2, a, b)
slope_res.append(res1)
return slope_res
# Define the integration1 - deflection
def defl(t1):
defl_res = []
for t in t1:
res2, err = integrate.dblquad(d2y_dx2,a,b, lambda x: a, lambda x: b)
defl_res.append(res2)
return defl_res
# Plot
fig, (ax1, ax2, ax3) = plt.subplots(3)
t = np.linspace(a,b,100)
t1 = np.linspace(a,b,100)
ax1.plot(t, d2y_dx2(t))
ax2.plot(t, slope(t))
ax3.plot(t1, defl(t1))
plt.show()
结果:
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