Espectro de respuesta con amortiguamiento

Carga rectangular

In [1]:
import numpy as np
import matplotlib.pyplot as plt

m = 0.0236205
k = 3.73
omega = np.power(k/m,1.0/2.0)
beta = 0.05
omega_damping = omega*np.power(1-beta**2,1.0/2.0)
T = (2.0*np.pi)/omega
F0 = 4.0

delta = 0.001
t = np.arange(10,3000)*delta
F = np.zeros(2990)
        
G = np.exp(-beta*omega*t)*np.sin(omega_damping*t)

Rd = np.zeros(400)
tT = np.linspace(0.01,6,400)
plt.figure(figsize=(19,8.5))
    
for j in range(tT.size):
    for i in range(t.size):
        if t[i] <= T*tT[j]:
            F[i] = F0
        else:
            F[i] = 0.0
    posicion = np.convolve(F,G)*delta/(m*omega_damping)
    posicion = posicion[:2990]
    plt.plot(t, np.abs(posicion))
    Rd[j] = max(posicion)*k/F0
    
plt.xlabel(r'$t$')
plt.ylabel(r'$u$')
plt.grid(True)
plt.show()
In [2]:
plt.figure(figsize=(19,8.5))
plt.plot(tT, Rd)
plt.xlabel(r'$\frac{t_{1}}{T}$')
plt.ylabel(r'$R_{d}$')
plt.grid(True)
plt.show()
In [3]:
print 'Rd max =', max(Rd)

Rd max = 1.847388448

Carga senoidal

In [4]:
import numpy as np
import matplotlib.pyplot as plt

m = 0.0236205
k = 3.73
omega = np.power(k/m,1.0/2.0)
beta = 0.05
omega_damping = omega*np.power(1-beta**2,1.0/2.0)
T = (2.0*np.pi)/omega
F0 = 4.0

delta = 0.001
t = np.arange(10,3000)*delta
F = np.zeros(2990)
        
G = np.exp(-beta*omega*t)*np.sin(omega_damping*t)

Rd = np.zeros(400)
tT = np.linspace(0.01,6,400)
plt.figure(figsize=(19,8.5))
    
for j in range(tT.size):
    for i in range(t.size):
        if t[i] <= T*tT[j]:
            F[i] = F0*np.sin(np.pi*t[i]/(T*tT[j]))
        else:
            F[i] = 0.0
    posicion = np.convolve(F,G)*delta/(m*omega_damping)
    posicion = posicion[:2990]
    plt.plot(t, np.abs(posicion))
    Rd[j] = max(posicion)*k/F0
    
plt.xlabel(r'$t$')
plt.ylabel(r'$u$')
plt.grid(True)
plt.show()
In [5]:
plt.figure(figsize=(19,8.5))
plt.plot(tT, Rd)
plt.xlabel(r'$\frac{t_{1}}{T}$')
plt.ylabel(r'$R_{d}$')
plt.grid(True)
plt.show()
In [6]:
print 'Rd max =', max(Rd)

Rd max = 1.64409149177

Carga triangular simétrica

In [7]:
import numpy as np
import matplotlib.pyplot as plt

m = 0.0236205
k = 3.73
omega = np.power(k/m,1.0/2.0)
beta = 0.05
omega_damping = omega*np.power(1-beta**2,1.0/2.0)
T = (2.0*np.pi)/omega
F0 = 4.0

delta = 0.001
t = np.arange(10,3000)*delta
F = np.zeros(2990)
        
G = np.exp(-beta*omega*t)*np.sin(omega_damping*t)

Rd = np.zeros(400)
tT = np.linspace(0.01,6,400)
plt.figure(figsize=(19,8.5))
    
for j in range(tT.size):
    for i in range(t.size):
        if t[i] <= T*tT[j]*0.5:
            F[i] = (F0*t[i])/(T*tT[j]*0.5)
        elif T*tT[j]*0.5 < t[i] <= T*tT[j]:
            F[i] = -(F0*(t[i] - T*tT[j]*0.5))/(T*tT[j] - T*tT[j]*0.5) + F0
        else:
            F[i] = 0
    posicion = np.convolve(F,G)*delta/(m*omega_damping)
    posicion = posicion[:2990]
    plt.plot(t, np.abs(posicion))
    Rd[j] = max(posicion)*k/F0
    
plt.xlabel(r'$t$')
plt.ylabel(r'$u$')
plt.grid(True)
plt.show()
In [8]:
plt.figure(figsize=(19,8.5))
plt.plot(tT, Rd)
plt.xlabel(r'$\frac{t_{2}}{T}$')
plt.ylabel(r'$R_{d}$')
plt.grid(True)
plt.show()
In [9]:
print 'Rd max =', max(Rd)

Rd max = 1.4147269393