# Chapter 08: Columns¶

## Example 8.3 Pg.No.280¶

In [9]:
from __future__ import division
import math

E=75000         #youngs modulus(N/mm^2)
G=21000          #shear modulus (N/mm^2)
L=2              #length of column (m)
l1=75            #flange length (mm)
l2=37.5            #total length(mm)
t=2.5            #thickness(mm)

A=t*(2*l2+l1)         #observed from Fig 8.17

#chapter 16  Ixx=bd^3/12+Ab^2
Ixx=2*l2*t*l2**2+t*l1**3/12
Iyy=2*t*l2**3/12

I0=Ixx+Iyy

#eqn 18.11 J=SUM((s*t^3)/3)
J=2*l2*t**3/3+l1*t**3/3

Gama=t*l2**3*l1**2/24
Iyy=0.22*10**5
L=2*10**3
P_CRxx=math.pi**2*E*Ixx/L**2
P_CRyy=math.pi**2*E*Iyy/L**2
P_CRo=A/I0*(G*J+math.pi**2*E*Gama/L**2)

print "P_CRxx = %3.2e N\n"%(P_CRxx)
print "P_CRyy = %0.1e N\n"%(P_CRyy)
print "P_CRO = %0.2e N\n"%(P_CRo)

print "therefore buckling in the column due to axial load=%0.1e N\n"%(P_CRyy)

P_CRxx = 6.51e+04 N

P_CRyy = 4.1e+03 N

P_CRO = 2.22e+04 N

therefore buckling in the column due to axial load=4.1e+03 N



## Example 8.4 Pg.No.282¶

In [29]:
from __future__ import division
import math
import numpy as np
from sympy import solve, symbols, pprint
from sympy import diff
P=symbols('P')

E=70000        #youngs modulus (N/mm^2)
G=30000         #shear modulus (N/mm^2)
x_S=-76.2        #position of shear center(mm)
l1=l2=100        #lengths (mm)
t=2              #thickness(mm)

x_bar=2*t*l1*50/(3*l1)/2

A=600         #area (mm^2)
Ixx=1.17*10**6    #second moment of area (mm^4)
Iyy=0.67*10**6    #second moment of area (mm^4)
I0=5.32*10**6     # total second moment of area (mm^4)
J=800              #torsion constant (mm^4)
Gama=2488*10**6     #(mm^6)
L=10**3             #(mm)

P_CRxx=math.pi**2*E*Ixx/L**2
P_CRyy=math.pi**2*E*Iyy/L**2
P_CRo=A/I0*(G*J+math.pi**2*E*Gama/L**2)

fun=P**2*(1-A*x_S**2/I0)-P*(P_CRxx+P_CRo)+P_CRxx*P_CRo
solution = solve(fun, P)
print "lowest value of critical load = %1.2e N\n"%(min(solution))

lowest value of critical load = 1.68e+05 N


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