Python集成函数并绘制结果

Question

我试图用数值方法求解伯努利的光束方程，并绘制结果。方程的一阶导数是斜率，二阶导数是挠度。我一步一步地解决这个问题，首先绘制函数，然后对其进行积分，并在同一个图表上绘制积分结果

到目前为止，我的代码如下。我怀疑问题在于integrate.quad返回一个值，而我正试图从中获取多个值。有人知道怎么做吗

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()

编辑：Bellow是用willcrack的答案修改的代码。此代码现在可以工作，但结果不正确。在底部，我添加了使用正确的梁方程解析解绘制结果的代码

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()

解析解、代码和结果如下。偏转梁的形状会发生改变，梁的末端位于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()

Answer 1

也许你可以试试这样的

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：

Answer 2

我想我找到了斜坡的解决办法。 我稍后再试试另一个。 这是最新消息

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)

    
# Define the integration1 - slope
def slope(x):
    slope_res = np.zeros_like(x)
    for i,val in enumerate(x):
        y,err = integrate.quad(f,a,val)
        slope_res[i]=y
    return slope_res

# Define the integration1 - deflection
def defl(x):
    
    defl_res = np.zeros_like(x)
    for i,val in enumerate(x):
        y, err = integrate.dblquad(d2y_dx2,0,val, lambda x: 0, lambda x: val)
        defl_res[i]=y
    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()

新结果：

还在和最后一个挣扎