This document discusses the use of Euler's formula to calculate the buckling or critical load of columns. It provides the following key points: 1. Euler's formula can be used to calculate the critical buckling load of long columns based on the column's effective length, which depends on its end conditions, and its minimum moment of inertia. 2. Several examples are provided to demonstrate calculating the slenderness ratio and critical load of columns with different cross-sections, lengths, materials and end conditions using Euler's formula. The examples analyze columns that are circular, hollow circular, hollow rectangular, and tubular. 3. Parameters that must be provided to use Euler's formula include the column's dimensions, length

1. Sub- Solid Mechanics
Axially and Eccentrically Loaded Columns
Lect. 2 Numerical on Euler’s formula
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
(Euler’s Formula is used only for Long Columns)
Limitations: Not used for short columns
2
2
min
2 2
min
, .
. .
E
e e
e
xx yy
Where Effective length of the Columnbased on End contions
Minimum M I Minimum of I and I
Modulus of Elasti
E I
E I
city
P
L L
L
I
E





 


3. .
.
Effective Length (Le)
Definition: It is the length between the two points
of zero Bending moments

4. Effective Length (Le)
Sr.
No.
End Condition Eff. Length Euler’s Formula
1 Both Ends Fixed
Le = L/2
2 One end is Fixed
and Other End
Hinged
Le = L/
3 Both Ends Hinged
Le= L
4 One end is Fixed
and Other is free Le= 2L
2
min
2
4
E
E I
P
L


2
min
2
2
E
E I
P
L


2
min
2
E
E I
P
L


2
min
2
4
E
E I
P
L



5. Numerical on
Euler’s Formula
2
min
2
E
e
E I
P
L



6. Problem 1: Determine the slenderness ratio of circular
column with 50mm diameter and 6m length if 1) both ends
of the is fixed ii) Both ends of column is hinged.
Take E=200Gpa
Given, D=50mm, L=6m=6000mm, E=200Gpa=200x103mpa
2 2 2
4 4 3 4
min
3
(50) 1962.5 ,
4 4
(50) 30
1.
6.64 10
64 64
/ 2 6000 / 2 3000
, .
306.64 10
156.24 12.50
196
( )
2.5
xx yy
e
e
A D mm
I I I D x mm
L L mm
Slenderness Ratio
Slenderness R
Both end
L I
k Rad of Gyration r
k A
a
I x
k r mm
A
tio
s fixed
 
 


  
    
  
  
    
3000
240
12.5
e
L
k
  

7. 3
2 2 3 3
min
2 2
6000
306.64 10
, 12.50
1962.5
6000
480
12.5
200 10 306.64 10
(6000)
1
(
2.
( )
6
)
.79
:
e
e
e
E
e
E
E
Slenderness Ratio
Slendernes
L L mm
L I x
where k r mm
k A
L
k
E I
s Rat
x
Both ends Hinged
EulersCrippli
x x
P
L
ng
P N
Load P
io





 
    
  
 

Problem 1 Continue……

8. Problem 2: A solid tubular section of brass with diameter
40mm is used as a column with one end is fixed and other
end is hinged. Determine the minimum length for which
Euler’s formula can be used. Take E=100Gpa & Proportional
limit=150mpa
Given, Brass column, D=40mm,L=??, E=100Gpa=100x103mpa
& Proportional limit=6E=150mpa
End Cond: One end is fixed and other end is hinged
2
min
2
,
/ 2
( )
. )
&
(
e
E
e
E
E E E
L L
E I
P A
L
One ends fixed oth
P
P x A Put in Eq A
A
er End Hinged

 

         
 

9.  
2
min
2
2 2
2 2
2 3 4
2
2
11
2
sin ' , ( )
2
,
/ 2
2 100 10 (40)
64
150 (40) ,
4
2.48 10
1315680.86
188495.56
1147.03 1.147
E
e
E
E I
U g Euler s Formula P A
L
E I E I
x A
L
L
X X
x
L
x
L
L mm m

 




    
 

 
 
Problem 2 Continue……

10. Problem 3: Calculate the compressive load on Hollow cast
iron column with one end fixed and other end hinged. The
external dia. of column is 150mm and internal dia. 100mm.
Length of column is 10m.Factor of Safety = 5, E = 95Gpa
Given, Hollow cast iron column D=150mm, d=100mm,
L=10m=10,000mm, E = 95Gpa = 95x103mpa, FOS=5
   
4 4 4 4 6 4
min min
2 2 3 6
min
2 2
3
, 150 100 19.94 10
64 64
/ 2 10000 / 2 7071.06 ,
95 10 19.94 10
(7071.06)
373.95 10
3 95
&
73.
xx yy
e
E
e
E
E
One ends fixed other En
I I I D d I x mm
L L mm
E I x x x
d Hi
P
L
P x N
P KN Crippling L d
n d
oa
ge
 
 
      
  
 

    

11. Safe Compressive load
( )
. . 5
373.95
,
5
74.79
E
Safe Compressive load P
P
Crippling Load
P
F O S
P
P KN
 


Problem 3 Continue……

12. Problem 4: Determine the Crippling load for Hollow
rectangular column of outer dimension 100 x 80 mm and
thickness 10mm. The actual length of column is 6m, with
both ends fixed. Take E=120Gpa.
Given, Hollow Rectangular column= 100 x 80 mm
L=6m=6000mm, E =120= 120x103mpa, To find PE
6 4
min
3 3 3 3
6 4
3 3 3 3
6 4
100 80 80 60
12 12 12 12
2.827 10
80 100 60 80
12 12 12 12
4.106
2.827 1
1
0
6 0
xx
xx
xx
yy
yy
BD bd x x
I
I x mm
DB db
I I x m
x x
I
I x mm
m
 
   

   


13. 6 4
min
2 2 3 6
min
2 2
3
2.827 10
6 ,
/ 2 3 3000 ,
120 10 2.827 10
,
(3000)
372 10 ,
' ( ):
372
xx
e
E
e
E
E
E
Both ends of column is fixed
Euler s Load P
I I x mm
L m
L L m mm
E I x x x x
P
L
P X N
about x x axi
N s
P K
 
 


  
 


Problem 4 Continue……

14. Problem 5: A Steel rod 6m length and 72mm diameter is
used as a column which is fixed at one end and free at other
end. Find the crippling load using Euler’s formula.
Take E=200Gpa
Given, D= 72mm, L=6m=6000mm, E =200Gpa= 200x103mpa,
To find PE
   
4 4 6 4
min
2 2 3 6
min
2 2
3
72 1.3184 10
64 64
2 2 6 12 12000 ,
200 10 1.3184 10
(12000)
18.055 10
18 5
&
.05
xx yy
e
E
e
E
E
I I I D x mm
L L X
One
Crippling
ends fixed other is free
Loa
m mm
E I x x x
P
L
P x N
d
P KN
 
 
    
   
 

    

15. Problem 6: An Aluminium tube of length 8m is used as a
simply supported column with two ends hinged carrying
1.4KN axial load. If outer dia.is 50mm, find the inner dia. of
the tube that would be provide factor of safety 2. E =70Gpa
Given, D= 50mm, d=????, L=8m=8000mm, FOS=2
E =70Gpa= 70x103mpa, P = 1.4KN
   
3
4 4 4 4
min
1.4 2,
2.8 2.8 10
50
64 64
8 800 ,
,
0
'
E
E
xx yy
e
Euler s Load Ax
P X
P KN X N
I I I D d d
ial load x FOS
Column is hinged at both e
L L mm
nds
m
 

 
     
  


16.  
2 2 3
3
min min
2 2
3 4
min
4 4 3
min
3
4 4
4 6 4
4 3 4 3
,
5
70 10
2.8 10
(8000)
259.645 10 ,
259.645 10 ,
64
259.645 10 64
50 ,
5.29 10 50 ,
957.86 10 , 957
0 ,
31.
.86 10 ,
55 .
E
e
E I x x I
P X
L
Solving
D
I x mm
I D d x
x x
d
d x
d x d x
mm
d mm Internal dia
 


  

  
 
  
  





Problem 6 Continue……

17. Thank You