# Chapter 11: Columns¶

## Example 11.1, page no. 763¶

In :
import math

#initialisation
E = 29000                   # Modulus of elasticity in ksi
spl = 42                    # Proportional limit in ksi
L = 25                      # Total length of coloum in ft
n = 2.5                     # factor of safety
I1 = 98                     # Moment of inertia on horizontal axis
I2 = 21.7                   # Moment of inertia on vertical axis
A = 8.25                    # Area of the cross section

#calculation
Pcr2 = (4*math.pi**2*E*I2)/((L*12)**2)          # Criticle load if column buckles in the plane of paper
Pcr1 = (math.pi**2*E*I1)/((L*12)**2)            # Criticle load if column buckles in the plane of paper
Pcr = min(Pcr1,Pcr2)                            # Minimum pressure would govern the design
scr = Pcr/A                                     # Criticle stress
Pa = Pcr/n                                      # Allowable load in k
print "The allowable load is ", round(Pa), "k"


The allowable load is  110.0 k


## Example 11.2, page no. 774¶

In :
import math

#initialisation
L = 3.25                        # Length of alluminium pipe in m
d = 0.1                         # Outer diameter of alluminium pipe
P = 100000                      # Allowable compressive load in N
n =3                            # Safety factor for eular buckling
E = 72e09                       # Modulus of elasticity in Pa
l = 480e06                      # Proportional limit

#calculation
Pcr = n*P                                   # Critice load
t = (0.1-(55.6e-06)**(1.0/4.0) )/2.0        # Required thickness

tmin = t
print "The minimum required thickness of the coloumn is", round(tmin*1000,2), "mm"

The minimum required thickness of the coloumn is 6.82 mm


## Example 11.3, page no. 780¶

In :
from sympy import *

#initialisation
P = 1500                    # Load in lb
e = 0.45                    # ecentricity in inch
h = 1.2                     # Height of cross section in inch
b = 0.6                     # Width of cross section in inch
E = 16e06                   # Modulus of elasticity
my_del = 0.12               # Allowable deflection in inch

#calculation
L = mpmath.asec(1.2667)/0.06588        # Maximum allowable length possible

#Result
print "The longest permissible length of the bar is", round(L), "inch"

The longest permissible length of the bar is 10.0 inch


## Example 11.4, page no. 785¶

In :
from sympy import *
import math

#initialisation
L = 25                      # Length of coloum in ft
P1 = 320                    # Load in K
P2 = 40                     # Load in K
E = 30000                   # Modulus of elasticity of steel in Ksi
P = 360                     # Euivalent load
e = 1.5                     # Ecentricity of compressive load
A = 24.1                    # Area of the Cross section
r = 6.05                    # in inch
c = 7.155                   # in inch
sy = 42                     # Yeild stress of steel in Ksi

#calculation

smax = (P/A)*(1+(((e*c)/r**2)*mpmath.sec((L/(2*r))*math.sqrt(P/(E*A)))))           # Maximum compressive stress
print "The Maximum compressive stress in the column ", round(smax,2), "ksi"
# Bisection method method to solve for yeilding
def stress(a,b,f):
N = 100
eps = 1e-5
if((f(a)*f(b))>0):
print 'no root possible f(a)*f(b)>0'
sys.exit()
if(abs(f(a))<eps):
print 'solution at a'
sys.exit()
if(abs(f(b))<eps):
print 'solution at b'
while(N>0):
c = (a+b)/2.0
if(abs(f(c))<eps):
x = c
return x
if((f(a)*f(c))<0 ):
b = c
else:
a = c
N = N-1
print 'no convergence'
sys.exit()

def p(x):
return  x + (0.2939*x*sec(0.02916*sqrt(x))) - 1012
x = stress(710,750,p)
Py = x                  # Yeilding load in K
n = Py/P                # Factor of safety against yeilding
print "The factor of safety against yeilding is", round(n)

The Maximum compressive stress in the column  19.32 ksi
The factor of safety against yeilding is 2.0


## Example 11.5, page no. 804¶

In :
import math
import numpy

#initialisation
E = 29000                       # Modulus of elasticity in ksi
sy = 36                         # Yeilding stress in ksi
L = 20                          # Length of coloumn in ft
r = 2.57                        # radius of gyration of coloumn
K = 1                           # Effetive Length factor

#calculation
s = math.sqrt((2*math.pi**2*E)/sy)          # Criticle slenderness ratio (K*L)/r
s_ = (L*12)/r                               # Slenderness ratio

# Part(a)
n1 = (5.0/3.0)+((3.0/8.0)*(s_/s))-((1.0/8.0)*((s_**3)/(s**3)))  # Factor of safety
sallow = (sy/n1)*(1-((1.0/2.0)*((s_**2)/(s**2))))               # Allowable axial load
A = 17.6                                                        # Cross sectional area from table E1
Pallow = sallow*A                                               # Allowable axial load
print "Allowable axial load is", round(Pallow,2), "k"

# Part (b)
Pe = 200                        # Permissible load in K
L_ = 25                         # Assumed length in ft
s__ = (L_*12)/r                 # Slenderness ratio
n1_ = (5.0/3.0)+((3.0/8.0)*(s__/s))-((1.0/8.0)*((s__**3)/(s**3)))   # Factor of safety
sallow_ = (sy/n1_)*(1-((1.0/2.0)*((s__**2)/(s**2))))                # Allowable axial load
A = 17.6                                                            # Area of the cross section in**2
Pallow = sallow_*A                                                  # Allowable load
L1 = [24, 24.4, 25]
P1 = [201, 194, 190]
L_max = numpy.interp(200.0, P1, L1)
print "The maximum permissible length is", L_max, "ft"

Allowable axial load is 242.84 k
The maximum permissible length is 25.0 ft


## Example 11.6, page no. 806¶

In :
import math
import numpy

#initialisation
L = 3.6                         # Length of steel pipe coloumn
d = 0.16                        # Outer diameter in m
P = 240e03                      # Load in N
E = 200e09                      # Modulus of elasticity in Pa
sy = 259e06                     # yeilding stress in Pa
K = 2.0
Le = K*L                        # As it in fixed-free condition

#calculation
sc = math.sqrt((2*math.pi**2*E)/sy) # Critical slenderness ratio

# First trial
t = 0.007                               # Assumed thick ness in m
I = (math.pi/64)*(d**4-(d-2*t)**4)      # Moment of inertia
A = (math.pi/4)*(d**2-(d-2*t)**2)       # Area of cross section
r = math.sqrt(I/A)                      # Radius of gyration
sc_ = round((K*L)/r)                    # Slender ness ratio
n2 = 1.92                               # From equation 11.80
sa = (sy/(2*n2))*(sc**2/sc_**2)         # Allowable stress
Pa = round((sa*A)/1000)                 # Allowable axial load in N

# Interpolation
t = [7, 8, 9]
Pa = [196, 220, 243]
t_min = numpy.interp(240.0, Pa, t)
print "The minimum required thickness of the steel pipe is", round(t_min,1), "mm"

The minimum required thickness of the steel pipe is 8.9 mm


## Example 11.7, page no. 808¶

In :
import math

#initialisation

L = 16                          # Effective length in inch
P = 5                           # axial load in K

#calculation
# Bisection method for solvong the quaderatic
def stress(a,b,f):
N = 100
eps = 1e-5
if((f(a)*f(b))>0):
print 'no root possible f(a)*f(b)>0'
sys.exit()
if(abs(f(a))<eps):
print 'solution at a'
sys.exit()
if(abs(f(b))<eps):
print 'solution at b'
while(N>0):
c = (a+b)/2.0
if(abs(f(c))<eps):
x = c
return x
if((f(a)*f(c))<0 ):
b = c
else:
a = c
N = N-1
print 'no convergence'
sys.exit()
def p(x):
return  30.7*x**2 - 11.49*x -17.69
x = stress(0.9,1.1,p)
d = x                               # Diameter in inch
sl = 49.97/d                        # Slenderness ration L/r
dmin = d                            # Minimum diameter
print "The minimum required outer diameter of the tube is", round(dmin,2), "inch"

The minimum required outer diameter of the tube is 0.97 inch


## Example 11.8, page no. 810¶

In :
import math

#initialisation
Fc = 11e06                          # Compressive demath.sing stress in Pa
E = 13e09                           # Modulus of elasticity in Pa

#calculation
# Part (a)
Kce = 0.3
c = 0.8
A = 0.12*0.16                           # Area of cross section
Sl = 1.8/0.12                           # Slenderness ratio
fi = (Kce*E)/(Fc*Sl**2)                 # ratio of stresses
Cp = ((1+fi)/(2*c)) - math.sqrt(((1+fi)/(2*c))**2-(fi/c)) # Coloumn stability factor
Pa = Fc*Cp*A
print "The allowable axial load is", Pa, "N"

# Part (b)
P = 100000                              # Allowable Axial load
Cp_ = P/(Fc*A)                          # Coloumn stability factor

# Bisection method method to solve for fi
def stress(a,b,f):
N = 100
eps = 1e-5
if((f(a)*f(b))>0):
print 'no root possible f(a)*f(b)>0'
sys.exit()
if(abs(f(a))<eps):
print 'solution at a'
sys.exit()
if(abs(f(b))<eps):
print 'solution at b'
while(N>0):
c = (a+b)/2.0
if(abs(f(c))<eps):
x = c
return x
if((f(a)*f(c))<0 ):
b = c
else:
a = c
N = N-1
print 'no convergence'
sys.exit()
def p(x):
return  ((1+x)/(2.0*c)) - math.sqrt(((1+x)/(2.0*c))**2-(x/c)) - Cp_
x = stress(0.1,1.0,p)
fi_ = x
d_ = 0.12                               # Diameter in m
L_max = d_*math.sqrt((Kce*E)/(fi_*Fc))  # Maximum length in m
print "The minimum allowable length is", round(L_max,2), "m"

The allowable axial load is 173444.30361 N
The minimum allowable length is 3.02 m