# Chapter 13:Buckling of Columns¶

## Example 13.1 Page NO 665¶

In [3]:
#Given
l = 12           #ft, length
E = 29*10**3         #GPa, stress
ro = 75         #mm, outside radius
ri = 70         #mm, inside  radius
sigma_y = 250   #MPa, stress

#Calculations
import math
Ix=110
Iy=37
A = 9.13
Pcr = (math.pi**2*(E*10**3)*Iy)/((l*12)**2) #Pcr = (math.pi**2*EI)/(l**2)
sigma_cr = (Pcr*1000)/A
p=36*A

print"The maximum allowable axial load that the column can support    = ",round(p,0),"kip"

The maximum allowable axial load that the column can support    =  329.0 kip


## Example 13.2 Page NO 668¶

In [9]:
#Given
E = 29*10**3   #GPa, stress
lx= 144      #inch, length
ly=100.8     #inch
A =4.43     #inch**2, area
sigma_y = 60  #ksi, stress

#Calculations
import math
Ix=29
Iy=9.32

Pcrx = ((math.pi**2)*E*Ix)/(lx**2) #Pcr = (math.pi**2*EI)/(l**2)
Pcry = ((math.pi**2)*E*Iy)/(ly**2) #Pcr = (math.pi**2*EI)/(l**2)
sigma_cr = (Pcr*1000)/A
sigmacr = Pcry/A #in kN
if sigmacr<sigma_y:
print"Buckling will occue before the material yield. So Psr=",round(Pcry,0),"kip"
else:
print"n"


Buckling will occue before the material yield. So Psr= 263.0 kip


## Example 13.3 Page No 669¶

In [53]:
#Given
E = 70          #GPa
Ix = 61.3*10**-6   #Moment of inertia along x-axis
Iy = 23.2*10**-6    ##Moment of inertia along y-axis
l = 5
KLx = 2*l       #m
KLy = 0.7*(l)   #m
FS = 3          #Factor of safety
sigma_y = 215   #MPa

#Calculation
import math
Pcrx = (math.pi**2*E*10**6*Ix)/(KLx**2) #Pcr = (math.pi**2*EI)/(l**2)
Pcry = (math.pi**2*E*10**6*Iy)/(KLy**2) #Pcr = (math.pi**2*EI)/(l**2)
Pcr = min(Pcrx,Pcry)
A = 7.5*10**-3 #mm**2
P_allow = Pcr/FS
sigma_cr = (Pcr*10**-3)/A

if(sigma_cr<sigma_y):
print"The largest allowable load that the column can support     = ",round(P_allow,0),"kN"
else:
print"n"

The largest allowable load that the column can support     =  141.0 kN


## Example 13.4 Page NO 683¶

In [54]:
#Given
z1 = 4*1000    #mm, length
e = 200       #mm, elongation
KLy = 100.8   #inch**2
Iy = 49.1     #inch**4
E = 29*10**3  #ksi, stress
sigma_y =421.2  #MPa

#Calculation
#y-y Axis Buckling
import math
Pcry = (math.pi**2*E*10**6*Iy)/(KLy**2) #Pcr = (math.pi**2*EI)/(l**2)
Pcry = Pcry/1000
#x-x Axis Yielding
Kx= 2
KLx = 288   #inch
A=11.7
c = (8.25)/2.0
rx = 3.53   #inch

#Solved by applying the Secant Formula and then finding Px by trial and error
trial_Px = 88.4 #kN
A = 7850#mm**2
sigma = (trial_Px*1000)/(A)

if(Pcry>trial_Px and sigma<sigma_y):
print'The maximum eccentric load that the column can support = ',trial_Px
print'Failure will occur about the x-x axis.'

else:
print"n"

The maximum eccentric load that the column can support =  88.4
Failure will occur about the x-x axis.


## Example 13.5 Page NO 686¶

In [4]:
#Given
d = 30  #mm, diameter
L = 600 #mm
sigma_pl = 150#MPa, stress

#Calculations
import math
I = (math.pi/4)*(r**4)
A = math.pi*r**2
r_gyr = sqrt(I/A)
K = 1
sl_ratio = (K*L)/(r_gyr)
flag1 = 0

#Assuming the critical stress is elastic
E = 150/0.001
sigma_cr1 = (math.pi**2*E)/(sl_ratio**2) #Pcr = (math.pi**2*EI)/(l**2)

if(sigma_cr1 > sigma_pl):
Et = (270 - 150)/(0.002 - 0.001)
sigma_cr2 = (math.pi**2*Et)/(sl_ratio**2) #Pcr = (math.pi**2*EI)/(l**2)

if(sigma_cr2>150 and sigma_cr2<270):
Pcr = sigma_cr2*A
Pcr = Pcr/1000.0 #in kN
print'The critical load when used as a pin supported column = ',round(Pcr,0),"kN"

else:
print""

The critical load when used as a pin supported column =  131.0 kN


## Example 13.6 page No 696¶

In [24]:
#given
l=16      #16 ft
A=29.4    #in**2, area
rx=4.60   #in
ry=2.65   #in
k=1
E=29*10**3   #Stress
sigmay=36.0  #ksi
#calculation
import math
x1=k*l*12/ry
x=math.sqrt(2*math.pi**2*E/sigmay)
a=(1-(x1**2/(2.0*x**2)))*((sigmay))
b=(5/3.0)+((3/8.0)*(x1)/(x))-((x1**3)/(8.0*(x**3)))
sigmaallow=a/b
P=sigmaallow*A

#result

 The largest load is 476.0 kip


## Example 13.7 Page NO 697¶

In [52]:
#Given
E = 29*10**3 #ksi, stress
sigma_y = 50 #ksi, shear stress
l = 15      #ft, length
k =0.5      #shape factor

#Calculations
import math
I_by_d = (1/4.0)*(math.pi)*(d/2.0)**4
A_by_d = (1/4.0)*(math.pi)*d**2
r_by_d = math.sqrt(I_by_d/A_by_d)
sl_ratio_c = math.sqrt((2*math.pi**2*E)/(sigma_y))

a1=math.sqrt(2*(math.pi)**2*E/(sigma_y))

d_=((18*4*16*23*(k**2)*(l**2)*12**2)/(12*math.pi**3*E))**(1/4.0)
print "The smallest diameter is ",round(d_,2),"inch. So use d=2.25 inch"
d=2.25
a1=k*l*12/(d/4.0)
if a1<200:
print"Use of equation is appropriate"

The smallest diameter is  2.11 inch. So use d=2.25 inch
Use of equation is appropriate


## Example 13.8 Page NO 698¶

In [46]:
#Given
L = 30    #inch
P = 12    #kip
sigma =28.0 #ksi
K = 1

#Calculations
import math
b2 = (P)/(2*sigma)
b_ = math.sqrt(b2)
A = 2*b_*b_
Iy = (1/12.0)*(2*b_*b_**3)
ry = sqrt(Iy/A)
sl_ratio = (K*L)/(ry)
if(sl_ratio>12):
b4 = (P*103.9**2)/(2*54000) #Eqn 13.26
b = b4**(1/4.0)

sl_ratio_ = (2598.1)/(b)
w = 2*b
else:
print"j"

if(sl_ratio>55):
print'The thickness of the bar  = ',round(b,2),"inch"

else:
print"h"

The thickness of the bar  =  1.05 inch


## Example 13.9 Page NO 699¶

In [47]:
#Given
y1 = 5.5     #inch, length
x1 = 1.5     #inch
A = (x1*y1)  #area
d = 1.5     #inch
K = 1

#Eqn 13.29
L2 = (540*A*d**2)/(P)
L = sqrt(L2)
KL_d = (K*L)/(d)

if(KL_d>26 and KL_d<=50):
print'The greatest allowable length L as specified by the NFPA  = ',round(L,1),"inch"
else:
print"j"

The greatest allowable length L as specified by the NFPA  =  44.8 inch


## Example 13.10 Page NO 705¶

In [48]:
#Given
#the given dimansion are
L = 80.0 #inch
K = 2.0
l = 4.0    #inch
b = 2.0    #inch
e = 1    #inch
c = 2    #inch

#Calculations
I1 = (1/12.0)*(l*b**3)
A = l*b
r = sqrt(I1/A)
sl_ratio = (K*L)/(r)

#Eqn 13.26
sigma_allow = (54000)/(sl_ratio**2)
I2 = (1/12.0)*(b*l**3)
coefficient = (1/A) + (e*c)/I2

sigma_max = sigma_allow
P = sigma_max/coefficient

#Display
print'The load that can be supported if the column is fixed at its base  ',P,"kip"

The load that can be supported if the column is fixed at its base   2.25 kip


## Example 13.11 Page No 706¶

In [49]:
#Given
import math
sigmaB_allow = 22 #ksi, allowable stress
E = 29*10**3      #ksi, stress
sigma_y = 36   #ksi, shear stress
K= 1           #shape factor
A = 5.87      #inch**2, area
Ix = 41.4    #inch**4, moment of inertia
ry = 1.5     #inch
d = 6.2      #inch
c= d/2.0
e = 30       #inch
L = 15       #ft

sl_ratio = (K*L*12)/(ry)
sl_ratio_c = math.sqrt((2*math.pi**2*E)/(sigma_y))

if(sl_ratio<sl_ratio_c):
num = (1 - (sl_ratio**2/(2*sl_ratio_c**2)))*sigma_y
denom1 = (5/3.0) + ((3/8.0)*sl_ratio/sl_ratio_c)
denom2 = (sl_ratio**3)/(8*sl_ratio_c**3)
sigmaA_allow = num/(denom1 - denom2)

coeffP = 1/(sigmaA_allow*A) + (e*c)/(Ix*sigmaB_allow)
P = 1/coeffP

sigA = (P/A)/(sigmaA_allow)
else:
print"k"

if(sigA < 0.15):
print'The maximum allowable value of eccentric load  = ',round(P,2),"kN"
else:
print"h"


The maximum allowable value of eccentric load  =  8.43 kN


## Example 13.12 Page NO 707¶

In [2]:
#Given
K = 2       #shape factor
d= 3.0      #inch, diameter
L = 60    #inch, length
e = 4    #inch
c = d
l = 3.0    #inch
b =6.0     #inch
A = l*b  #inch**2

#Calculations
sl_ratio = (K*L)/(d)

if(sl_ratio>26 and sl_ratio<50):
sigma_allow = (540)/(sl_ratio**2)
sigma_max = sigma_allow

I = (1/12.0)*(l*b**3)
coeffP = (1/A) + (e*c)/(I)
P = sigma_max/coeffP
print'The eccentric load that can be supported = ',round(P,2),"kip"
else:
print"no"

The eccentric load that can be supported =  1.22 kip