user2882635
user2882635

Reputation: 163

Python - Integrating a function and plotting results

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.

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()

EDIT: 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()

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()

enter image description here

Upvotes: 1

Views: 1759

Answers (2)

willcrack
willcrack

Reputation: 1852

I think I found the solution for the slope. I'll try the other one later. Here's the update.

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()

New Result:

NEW

Still struggling with the last one...

Upvotes: 1

willcrack
willcrack

Reputation: 1852

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()

Result:
Result

Upvotes: 1

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