This document contains 4 numerical problems related to calculating the buckling or crippling load of columns using Euler's formula. Problem 1 calculates the safe load capacity of an alloy tube column with one end fixed and the other hinged. Problem 2 finds the crippling load of a T-section strut with both ends hinged. Problem 3 determines the crippling load of an I-section joist with both ends fixed. Problem 4 compares the strength ratio of a solid column section to a hollow column section with the same cross-sectional area and length, both with pinned ends. The document provides step-by-step working and uses Euler's formula to calculate the critical buckling loads.

1. Sub- Solid Mechanics
Axially and Eccentrically Loaded Columns
Lect. 3 Numerical on Euler’s formula-II
Sanjivani Rural Education Society’s
Sanjivani College of Engineering, Kopargaon-423603
An Autonomous Institute, Affiliated to Savitribai Phule Pune University, Pune
ISO 9001:2015 Certified, Approved by AICTE, Accredited by NAAC (A Grade) & NBA
Department of Civil Engineering
Prepared by:
Dr. Ghumare S. M.
Asso Professor, Civil Engg. Department

2. Methods of analysis of Buckling or Crippling load
Critical buckling Load can be calculated using Euler’s
Theory or Euler’s Formula
min
2
min
2
,
. .
E
e
e
xx yy
E I
P
L
Where L Effective length of the Column
Minimum M I Minimum of I and I
Modulus of Elasticity
I
E







3. Problem 1: An alloy tube of 25mm internal dia. and 40mm
external dia., when subjected to axial forces of 60KN
undergoes an extension of 3.84mm over it’s length of 3m.
What is the safe load carrying capacity of column, when one
end of column is fixed and other end is hinged? F.O.S.= 4.
Given, d= 25mm, D=40mm, P=60KN=60x103N,
L=3m=3000mm, dl=3.84mm, F.O.S.= 4, PE=????
3
3 2
60 10 3000
3.84 ,
765.375
61.244 10 /
, l
PL
Use
AE
X x
x E
Here
E x N mm
 



4. Problem 1 Continue……
 
 
2 2 2 2 2
4 4 4 4 3 4
2 2
2
(40 25 ) 765.375 ,
4 4
(40 25 ) 106.434 10
64 64
&
/ 2 3000 / 2 21
' :( ):
21.32
61.244 10
One end of column is fixed other hinged
End Cond.:
e
E cr
E
e
Euler s Crippling Load P
A D d mm
I D d x mm
L L mm
E I x x
P
L
P
 
 
 
    
    
  



3 3
2
3
3
3
106.434 10
,
(2121.32)
14.282 10 ,
14.282 10
3.5705 10 3.57
4
E
E
x x
P x N
P x
Safe Load x N KN
FOS

   

5. Problem 2: A T-section 150x120x20mm is used as a mild
steel strut of 4m length hinged at both ends. Calculate the
crippling load if Young’s modulus E=200Gpa
Given, T-section 150x120x20mm, L = 4m = 4000mm,
E=200Gpa=200x103mpa,Find PE=???
End Cond: Both Ends are end is hinged Le=L=4000mm
1 1 2 2
1 2
2
1
2
2
1
2
150 / 2 75
100 20 2000 ,
150 20 3000 ,
100 / 2 50 ,
100 20 /
sec @
98.8
2 0
9
11
T tion is syymetrical yy axis
y m
x mm
A y A y
find y
A A
A x mm
A x mm
y mm
y mm
m
 



 



 
  

6. 3 3
1 1 2 2
2 2
1 1 2 2
1 1
1 2
3 3
1 1 2
6
2
4
,
12 12
98.89 50 48.89 ,
98.89 1
9.512 1
. @
0 ,
5.705
10 11.11 ,
,
,
12 12
xx
yy
xx
yy
b d b d
I A h A h
h y y mm
h y y mm
Put all the values
d b d b
I
Find M I x xand y y axis
I x mm
I x
   
   
   
   
   
    
     
   
 
   
   
 

 
 

6 4
mi
6 4
n 5.705 1
10 ,
0
yy
I I x m
mm
m
 

7. 2
min
2
2 3 6
2
6
200 10 5.705 10
,
(4000)
703.83 10 ,
703.
703
8
.83
3
Using Euler's Formula,
E
e
E
E
E
E
E I
P
L
X X X X
P
P X
Crippling Lo
N
P KN
ad P KN







Problem 2 Continue……

8. Problem 3: An I section joist shown in Fig. is used as a strut
with both ends fixed. What is the crippling load acting on
the column? E = 200Gpa and L=6m
Given, L=6m=6000mm, E = 200Gpa = 200x103mpa,
Both ends fixed Le= L/2 =6/2 = 3m = 3000mm
sec @
200/ 2 100
&
I tion is syymetrical x x y y a
x
xis
y mm
  
 
6 4
3 3
3 3
,
12 12
200 400 180 360
,
1
366.82 1
@
2
0
2
.
1
,
xx
xx
xx
Find M I x xand y y axis
BD bd
I
I
X X
I
x mm
 
 
 
 
 



 
 
 

9. Problem 3 Continue……
3 3 3
m
4
n
6
3
i
400 200 360 180
2.91
, ,
12 1
1
2 1 12
0
.
2
yy
yy
DB db X
Find M I about Y Y Ax
X
I
I x m
i
I
m
s
   
   
   



   
2
min
2
2 3 6
2
3
200 10 2.91 10
,
(3000)
638.294 10 ,
638.294
Using Euler's Formula, E
e
E
E
E
E I
P
L
X X X X
P
P
Cr
X N
P ippling Load
KN





 

10. Problem 4: Compare the ratio of solid column section to
that of hollow section od same c/s area. The internal dia. Of
the hollow section of column is 3/4th of Ext. diameter. Both
columns are having same length and Pinned at both ends.
Given,
1. Solid column Section
Let, D of the solid section
2. Hollow column Section
Let, D1 be the Ext. Diameter,
d1 be the Int. Diameter
   
2 2 2
1 1
(0.75 ) ,
4 4
S H
A A
D D D
 

 
 
 
2 2
1 1
2 2
1 1
,
4
(0.75 ) ,
4
H
H
A D d
A D D


 
 
1 1 1
3
0.75
4
d D D
 
Given that area od c/s is same
for both column
 
2
4
S
A D



11. Problem 4 Continue……
   
   
2 2 2
1 1
2 2
1
1 1
2 2
1 1
(0.75 ) ,
4 4
0.5
0.6614 1.5
6 0.4375
11
S H
D D
A A
D D D
D D
D
D
R D
D
O
 


 
  

M.I. for solid section
4
4
64
0.04906 (1)
s
s
I D
I D


    
M.I. for Hollow section
 
 
4 4
1 1
4 4
1 1
4
1
( ) ,
64
(0.75 ) ,
64
0.03353 (2)
H
H
H
I D d
I D D
I D


 
 
    

12.  
2 2 4 4
2 2 2
2 2 4
1
1
2 2
4
2
2
0.0490 0.4837
( )
(0.03353 )
, 1.511
0.03353 1.511
( )
Crippling load for solid section:
To find the Ratio of Strength of solid to Hollow sectio
S
s
e e e
H
H
e e
H
e
E I E x D E D
P A
L L L
E I E x D
P Put D D
L L
E x D
P B
L
 
 

    
  
 
 
     
 
4
2
4
2 2
0.03353 1.511
0.4837
/
63
,
3.5
n
Strength of solid to Hollow sectio
Take ratio Eq.(A) and Eq.(B)
n
S
H e e
S
H
E x D
P E D
P L
P
P
L
 


 

Problem 4 Continue……

13. Next Lecture will be on
Rankine’s Formula

14. Thank You