Chapter 13: Statically Indeterminate Problems

Example 13.2 page number 693

In [4]:
#Given 
#First we will solve without the reaction at middle
#Given
%matplotlib inline 
import numpy
import matplotlib.pyplot as plt
import numpy as np
l_ab = 1.0   #2L in - The length of the beam
F_D = 1.0    #W lb/in - The force distribution 
F = F_D*l_ab #WL - The force applied
#Beause of symmetry the moment caliculations can be neglected
#F_Y = 0
R_A = F/2 #wl - The reactive force at A
R_B = F/2 #wl - The reactive force at B
#EI - The flxure rigidity is constant and 1/EI =1 # k

#part - A
#section 1--1
l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance 
M_1 = [0,0,0,0,0,0,0,0,0,0,0]
v = [0,0,0,0,0,0,0,0,0,0,0]
for i in range(10):
    v[i] = R_A - F_D*l_1[i]  
    M_1[i] = R_A*l_1[i] - F_D*(l_1[i]**2)/2
# (EI)y'' = M_1[i] we will integrate M_1[i] twice where variable is l_1[i]
#(EI)y'- 

M_1_intg1 = R_A*(l_1[i]**2)/4 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**3)*l_1[i]/24 #integration of x**n = x**n+1/n+1
#(EI)y- Using end conditions for caliculating constants 

M_1_intg2 = R_A*(l_1[i]**3)/12.0 - F_D*(l_1[i]**4)/24.0 + F_D*(l_ab**3)*l_1[i]/24.0 
#Equations 

l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance 
M_1_intg2 = [0,0,0,0,0,0,0,0,0,0,0]
Y = [0,0,0,0,0,0,0,0,0,0,0]
for i in range(10):
    M_1_intg2[i] = (l_1[i]**3)/12.0 - (l_1[i]**4)/24.0 - l_1[i]/24.0   # discluding every term for ruling out float values
    Y[i] = M_1_intg2[i] #W(l**4)/EI  k = 1/EI
Y_min = 16*min(Y)
print "a) The maximum displacement in y direction is",16*min(Y),"W(l**4)/EI "
print "a) The maximum deflection occured at",2*l_1[Y.index(min(Y))],"L"
f_bb = 2**3/48.0 #l**3/EI - flexibility coefficient
Reac = - Y_min/f_bb #WL , The reaction at the mid of the bar
print "The reaction at the mid of the bar",Reac ,"WL"

#Graphs 
Y.extend(Y) #Because of symmetry
values = Y 
y = np.array(values)
t = np.linspace(0,1,22)
poly_coeff = np.polyfit(t, y, 2)

plt.plot(t, y, 'o')
plt.plot(t, np.poly1d(poly_coeff)(t), '-')
plt.show()
print "b)The above graph is beam displacement graph"
print "b)The minimum occures in the middle from the above graph  "
a) The maximum displacement in y direction is -0.208333333333 W(l**4)/EI 
a) The maximum deflection occured at 1.0 L
The reaction at the mid of the bar 1.25 WL
b)The above graph is beam displacement graph
b)The minimum occures in the middle from the above graph  

Example 13.3 page number 694

In [5]:
#Given 
#First we will solve without the reaction at middle
#Given
import numpy as np
l_ab = 1.0   #2L in - The length of the beam
F_D = 1.0    #W lb/in - The force distribution 
F = F_D*l_ab #WL - The force applied
#Beause of symmetry the moment caliculations can be neglected
#F_Y = 0
R_A = F/2 #wl - The reactive force at A
R_B = F/2 #wl - The reactive force at B
#EI - The flxure rigidity is constant and 1/EI =1 # k

#part - A
#section 1--1
l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance 
M_1 = [0,0,0,0,0,0,0,0,0,0,0]
v = [0,0,0,0,0,0,0,0,0,0,0]
for i in range(10):
    v[i] = R_A - F_D*l_1[i]  
    M_1[i] = R_A*l_1[i] - F_D*(l_1[i]**2)/2
# (EI)y'' = M_1[i] we will integrate M_1[i] twice where variable is l_1[i]
#(EI)y'- 

M_1_intg1 = R_A*(l_1[i]**2)/4 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**3)*l_1[i]/24 #integration of x**n = x**n+1/n+1
#(EI)y- Using end conditions for caliculating constants 

M_1_intg2 = R_A*(l_1[i]**3)/12.0 - F_D*(l_1[i]**4)/24.0 + F_D*(l_ab**3)*l_1[i]/24.0 
#Equations 

l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance 
M_1_intg2 = [0,0,0,0,0,0,0,0,0,0,0]
Y = [0,0,0,0,0,0,0,0,0,0,0]
for i in range(10):
    M_1_intg2[i] = (l_1[i]**3)/12.0 - (l_1[i]**4)/24.0 - l_1[i]/24.0   # discluding every term for ruling out float values
    Y[i] = M_1_intg2[i] #W(l**4)/EI  k = 1/EI
e_1 = 16*min(Y)         #WL4/EI - The maximum defection 
e_2 = - F_D*((2*l_ab)**3)/24.0 #WL3/EI - The maximum angle
#Caliculating for momentum and force
f_ab = ((2*l_ab)**2)/16.0 #L2/EI 
f_bb = ((2*l_ab)**3)/48.0 #L3/EI 
f_aa = 2*l_ab/3.0 #L/EI
f_ba = ((l_ab)**2)/4.0 #L2/EI
#F*X = e - Matrix multiplication 
#Solving for X
a = np.array([[f_aa,f_ba], [f_ba,f_bb]])
b = np.array([e_2,e_1])
x = np.linalg.solve(a, b)
print "The reactive moment at A i.e M_A",x[0],"WL**2"
print "The reactive force at A i.e R_A",x[1],"WL"
The reactive moment at A i.e M_A -0.0714285714286 WL**2
The reactive force at A i.e R_A -1.14285714286 WL
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