# Chapter 9:Columns And Struts¶

## Example 9.9.1,Page No.377¶

In [1]:
import math
from math import sin, cos, tan, pi, radians

#Initilization of Variables
L=5000 #mm #Length of strut
dell=10 #mm #Deflection

#Calculations

#Central Deflection of a simply supported beam with central concentrated load is
#dell=W*L**3*(48*E*I)**-1

#Let E*I=X
X=W*L**3*(48*dell)**-1 #mm

P=pi**2*X*(L**2)**-1

#Result

Critical Load of Bar is 1028.08 N


## Example 9.9.2,Page No.377¶

In [2]:
import math
from math import sin, cos, tan, pi, radians

#Initilization of Variables

L=2000 #mm #Length of square column
E=12*10**3 #N/mm**2 #Modulus of Elasticity
sigma=12 #N/mm*2 #stress
FOS=3

#Calculations

#From Euler's Formula
#P=pi**2*E*I*(L**2)**-1  .........(1)

#W=P*(FOS)**-1

#Part-1

#At W1=95*10**3 #N
#W1=P*(3*L**2)**-1

#Let 'a' be the side of the square
#I=1*12**-1*a**4

#sub value of I in Equation 1 and further rearranging we get
a=(W1*3*12*L**2*(pi**2*E)**-1)**0.25 #mm

#From Consideration of direct crushing
#sigma*a**2=W1
#After Reaaranging the above equation we get
a2=(W1*(sigma)**-1)**0.5 #mm

#required size is 103.67*103.67 i.e a*a

#Part-2

#At W2=200*10**3 #N
#W2=P*(3*L**2)**-1
#After substituting values and further Rearranging the above equation we get
a3=(W2*3*12*L**2*(pi**2*E)**-1)**0.25 #mm

#From consideration of direct compression,size required is
a4=(W2*sigma**-1)**0.5

#required size is 129.10*129.10 i.e a4*a4

#Result
print"For W1 Load Required size is",round(a*a,2),"mm**2"
print"For W2 Load Required size is",round(a4*a4,2),"mm**2"

For W1 Load Required size is 10747.38 mm**2
For W2 Load Required size is 16666.67 mm**2


## Example 9.9.3,Page No.378¶

In [3]:
import math
from math import sin, cos, tan, pi, radians

#Initilization of Variables

#Flange
b=100 #mm #Width

D=80 #mm #Overall Depth
t=10 #mm #Thickness of web and flanges
L=3000 #mm #Length of strut
E=200*10**3 #N/mm**2 #Modulus of Elasticity

#Calculations

#Let centroid be at depth y_bar from top fibre
y_bar=(b*t*t*2**-1+(D-t)*t*((D-t)*2**-1+t))*(b*t+(D-t)*t)**-1 #mm

#M.I at x-x axis
I_x=1*12**-1*b*t**3+b*t*(y_bar-t*2**-1)**2+1*12**-1*t*((D-t))**3+t*((D-t))*((((D-t)*2**-1)+t)-y_bar)**2

#M.I at y-y axis
I_y=1*12**-1*t*b**3+1*12**-1*(D-t)*t**3 #mm**3

#Least M.I
I=I_y

#Since both ends are hinged
#Feective Length=Actual Length
L=l=3000 #mm

P=pi**2*E*I*(l**2)**-1*10**-3 #KN

#Result
print"The Buckling Load for strut of tee section",round(P,2),"KN"

The Buckling Load for strut of tee section 184.05 KN


## Example 9.9.4,Page No.379¶

In [4]:
import math
from math import sin, cos, tan, pi, radians

#Initilization of Variables

D=400 #mm #Overall Depth

#Flanges
b=300 #mm #Width
t=50 #mm #Thickness

t2=30 #mm #Web Thickness

dell=10 #mm #Deflection
FOS=1.75 #Factor of safety
E=2*10**5 #N/mm**2

#Calculations

#M.I at x-x axis
I_x=1*12**-1*(b*D**3-(b-t2)*b**3) #mm**4

#Central Deflection
#dell=5*w*L**4*(384*E*I)**-1
#After sub values in above equation and further simplifying we get
L=(dell*384*E*I_x*(5*w)**-1)**0.25

#M.I aty-y axis
I=I_y=1*12**-1*t*b**3+1*12**-1*b*t2**3+1*12**-1*t*b**3 #mm**4

#Both the Ends of column are hinged

P=pi**2*E*I*(L**2)**-1 #N

S=P*(FOS)**-1*10**-3 #N

#Result
print"Safe Load if I-section is used as column with both Ends hhinged",round(S,2),"KN"

Safe Load if I-section is used as column with both Ends hhinged 4123.29 KN


## Example 9.9.5,Page No.381¶

In [5]:
import math
from math import sin, cos, tan, pi, radians

#Initilization of Variables

D=200 #mm #External Diameter
t=20 #mm #hickness
d=200-2*t #mm #Internal Diameter
E=1*10**5 #N/mm**2
a=1*(1600)**-1 #Rankine's Constant
L=4.5 #m #Length
sigma=550 #N/mm**2 #Stress
FOS=2.5

#Calculations

#Moment of Inertia
I=pi*D**4*64**-1-pi*d**4*64**-1

#Both Ends are fixed

#Effective Length
l=1*2**-1*L*10**3 #mm

P_E=pi**2*E*I*(l**2)**-1

A=pi*4**-1*(D**2-d**2) #mm*2

k=(I*A**-1)**0.5

P_R=sigma*A*(1+a*(l*k**-1)**2)**-1

X=P_E*P_R**-1

S=P_R*(FOS)**-1*10**-3 #KN

#Result
print"Safe Load by Rankine's Formula is",round(S,2),"KN"

Safe Load by Rankine's Formula is 1404.36 KN


## Example 9.9.6,Page No.382¶

In [6]:
import math
from math import sin, cos, tan, pi, radians

#Initilization of Variables

L=3000 #mm #Length of column
a=1*1600**-1 #Rankine's constant
FOS=4 #Factor of safety
sigma=550 #N/mm**2 #stress

#Calculations

#Effective Length
l=L*2**-1 #mm

#Let d1=outer diameter & d2=inner diameter
#d1=5*8**-1*d2

#M.I
#I=pi*64**-1*(d1**4-d2**4) #mm**4

#Area of section
#A=pi4**-1*(d1**2-d2**2) #mm**2

#k=(I*A**-1)
#substituting values in above equation
#k=1*16**-1*(d1**2-d2**2)
#after simplifying further we get
#k=0.2948119.d1

#X=l*k**-1
#substituting values in above equation and after simplifying further we get
#X=5087.9898*d1**-1

P=W*FOS #N

#P2=sigma*A*(1+a*(X)**2)**-1
#substituting values in above equation and after simplifying further we get
#d1**4-12156618*d1**4-1.96691*10**8=0
#d1**2-12156618*d1-196691000=0
a=1
b=-12156.618
c=-196691000

Y=b**2-4*a*c

d1_1=((-b+Y**0.5)*(2*a)**-1)**0.5 #mm
d1_2=((-b-Y**0.5)*(2*a)**-1) #mm

d2=5*8**-1*d1_1

#Result
print"Section of cast iron hollow cylindrical column is:d1_1",round(d1_1,2),"mm"
print"                                                 :d2  ",round(d2,2),"mm"

Section of cast iron hollow cylindrical column is:d1_1 146.16 mm
:d2   91.35 mm


## Example 9.9.7,Page No.383¶

In [7]:
import math
from math import sin, cos, tan, pi, radians

#Initilization of Variables

#Let X=(P*A**-1) #Average Stress at Failure
Lamda_1=70 #Slenderness Ratio
Lamda_2=170 #Slenderness Ratio
X1=200 #N/mm**2
X2=69 #N/mm**2

#Rectangular section
b=60 #mm #width
t=20 #mm #Thickness

L=1250 #mm #Length of strut
FOS=4 #Factor of safety

#Calculations

#Slenderness ratio
#Lamda=L*k**-1

#The Rankine's Formula for strut
#P=sigma*A*(1+a*(L*k**-1)**-1

#From test result 1,
#After sub values in above equation we get and further simplifying we get
#sigma_1=200+980000*a   ...................(1)

#From test result 2,
#After sub values in above equation we get and further simplifying we get
#sigma_2=69+1994100*a   ...................(2)

#Substituting it in equation (1) we get
a=131*1014100**-1

#Substituting a in equation 1
sigma_1=200+980000*a #N/mm**2

#Effective Length
l=1*2**-1*L #mm

#Least of M.I
I=1*12**-1*b*t**3 #mm**4

#Area
A=b*t #mm**2

k=(I*A**-1)**0.5

#Slenderness ratio
Lamda=l*k**-1

#From Rankine's Ratio
P=sigma_1*A*(1+a*(Lamda)**2)**-1

S=P*(FOS)**-1*10**-3 #N

#Result
print"Constant in the Formula is:a      ",round(a,6)
print"                          :sigma_1",round(sigma_1,2)

Constant in the Formula is:a       0.000129
:sigma_1 326.6


## Example 9.9.8,Page No.385¶

In [8]:
import math
from math import sin, cos, tan, pi, radians

#Initilization of Variables

D=200 #mm #Depth
b=140 #mm #width

#Plate
b2=160 #mm #Width
t2=10 #mm #Thickness

L=l=4000 #mm #Length
FOS=4 #Factor of safety
sigma=315 #N/mm**2 #stress
a2=1*7500**-1
I_xx=26.245*10**6 #mm**4 #M.I at x-x
I_yy=3.288*10**6 #mm**4 #M.I at y-y
a=3671 #mm**2 #Area
k_x=84.6#mm
k_y=29.9 #mm

#Calculations

#Total Area
A=a+2*t2*b2 #mm**2

#M.I
I=I_yy+2*12**-1*t2*b2**3 #mm**4

k=(I*A**-1)**0.5 #mm

#Let X=L*k**-1
X=L*k**-1

#Appliying Rankine's Formula
P=sigma*A*(1+a2*(X)**2)**-1 #N

S=P*(FOS)**-1*10**-3 #KN

#Result

Safe axial Load is 220.93 KN


## Example 9.9.9,Page No.389¶

In [9]:
import math
from math import sin, cos, tan, pi, radians

#Initilization of Variables

E=200*10**3 #N/mm**2 #Modulus of elasticity
sigma=330 #N/mm**2 #Stress
a=1*7500**-1 #Rankine's constant
A=5205 #mm**2 #area of column
I_xx=59.431*10**6 #mm**4 #M.I at x-x axis
I_yy=8.575*10**6 #mm**24#M.I at y-y axis

#Calculations

#Total M.I
I=I_xx+I_yy #mm**4

#Area of compound Section
A2=2*A #mm**2

k=(I*A2**-1)**0.5 #mm

#pi**2*E*I*(L**2)**-1=sigma*A*(1+a*(L*k)**2)**-1
#After Substitt=uting values and further simplifying we get
L=(39076198*(1-0.7975432)**-1)**0.5*10**-3 #m

#Result
print"Length of column for which Rankine's formula and Euler's Formula give the same result is",round(L,2),"m"

Length of column for which Rankine's formula and Euler's Formula give the same result is 13.89 m


## Example 9.9.10,Page No.387¶

In [10]:
import math
from math import sin, cos, tan, pi, radians

#Initilization of Variables

sigma=326 #N/mm**2 #stress
E=2*10**5 #N/mm**2 #Modulus of Elasticity
FOS=2 #Factor of safety
a=1*7500**-1 #Rankine's constant
D=350 #mm #Overall Depth

#Cover plates
b1=500 #mm #width
t1=10  #mm #Thickness

d=220 #mm #Distance between two channels

L=6000 #mm #Length of column

A=5366 #mm**2 #Area of Column section
I_xx=100.08*10**6 #mm**4 #M.I of x-x axis
I_yy=4.306*10**6 #mm**4 #M.I of y-y axis
C_yy=23.6 #mm #Centroid at y-y axis

#Calculations

#Symmetric axes are the centroidal axes is

#M.I of Channel at x-x axis
I_xx_1=2*I_xx+2*(1*12**-1*b1*t1**3+b1*t1*(D*2**-1+t1*2**-1)**2)

#M.I of Channel at y-y axis
I_yy_1=2*(I_yy+A*(d*2**-1+C_yy)**2)+2*12**-1*t1*b1**3

#As I_yy<I_xx
#So
I=I_yy_1 #mm**4

A2=2*A+2*t1*b1 #Area of channel

k=(I*A2**-1)**0.5 #mm

P=sigma*A2*(1+a*(L*k**-1)**2)**-1

S=P*2**-1*10**-3 #KN

#Result

Safe Load carrying Capacity is 2717.35 KN


## Example 9.9.11,Page No.390¶

In [11]:
import math
from math import sin, cos, tan, pi, radians

#Initilization of Variables

I=4.085*10**8 #mm**4 #M.I
A=20732.0 #mm**2 #area of column
f_y=250 #N/mm**2
L=6000 #mm #Length of column

#Calculations

k=(I*A**-1)**0.5 #mm
lamda=L*k**-1 #Slenderness ratro

#From Indian standard table
lamda_1=40
sigma_a_c_1=139 #N/mm**2
lamda_2=50
sigma_a_c_2=132 #N/mm**2

#Linearly interpolating between these values for lambda=42.744

sigma_a_c_3=sigma_a_c_1-2.744*10**-1*(sigma_a_c_1-sigma_a_c_2)

#Safe Load carrying capacity of column
P=sigma_a_c_3*A*10**-3

#Result

Safe Load carrying capacity is 2841.93 KN