A structural member subjected to axial compressive force is called strut.When strut is vertical it is known as column.

1. COLUMN & STRUTS
Presenting by
Prof. Brijesh Raychanda
M.Tech (Structural Design)

2. STRUTS
• A structural member subjected to axial compressive
force is called strut.
• Strut may be vertical, horizontal or inclined.
• The cross – sectional dimensions of strut are small.
• Normally, struts carry smaller compressive loads.
• Struts are used in roof truss and bridge trusses.

3. COLUMN
• When strut is vertical it is known as column.
• The cross – sectional dimensions of column are large.
• Normally, columns carry heavy compressive loads.
• Columns are used in concrete and steel buildings.

4. STRUT COLUMN

5. • Radius of Gyration ( K ) :
The distance from the given axis at which if all
small elements of lamina placed, the M.I. Of the lamina
about the axis does not changed. This distance is called
radius of gyration.
k = √(I/A) or I = AK²
K=radius of gyration
I = Moment of Inertia (mm4)
A = Area of Section (mm2)

6. • Slenderness ratio ( λ ) :
Slenderness Ratio =
effective length of column/Minimum radius of gyration
λ = le/kmin
If λ is more , its load carrying capacity will be less.

7. Long Column :-
• When length of column is more as compared to its c/s
dimension, it is called long column.
Le/kmin > 50
For mild steel λ > 80 is called long column.
Short Column:-
• When length of column is less as compared to its c/s
dimension, it is called Short column.
Le/kmin <50

8. Crushing Load :
The load at which, short column fails by crushing is called
crushing load.
Crippling Load and Buckling Load :
The load at which, long column starts buckling(bending) is
called buckling load or crippling load.
• It depends upon the following factors.
1. Amount of load.
2. Length of column
3. End condition of column
4. C/s dimensions of column
5. Material of column.

9. Column End Condition And Effective Length
1.Both end hinged.
2.Both end fixed.
3.One end fixed and other hinged.
4.One end fixed and other free.

10. COLUMNS HAVING VARIOUS TYPES OF SUPPORTS
Effective length
10

11. Euler’s Formula
Euler’s Crippling Load,
PE = ∏²EI /le²
Where, E = is Modulus of Elasticity (Mpa)
I = is MOI or 2nd Moment of area (mm4)
Le =is Effective length (mm)
Also known as Critical Buckling Load

12. Euler’s Formula for both end of the columns are
hinged
• The Bending Moment at the Section is given by
• General solution is
• Since x = 0 at y = 0, then C2 = 0.
Since x = l at y = 0, then
 

EI
 y 0
dx2
d 2
y
  P


   
 
x
EI 
P
P x  C sin
EI  2 
y  C1cos
12


 
L   0
E I 
P
C 2 sin

13. IDEAL COLUMN WITH HINGED SUPPORTS
for C1 = 0, we get
• Which is satisfied if
• or
L2
2
EI
P 
EI
P
L  
13

 
 EI
sin P
L  0

14. Assumption of Euler’s formula :
 The material is elastic, homogeneous, isotropic.
 The column is long.
 The load is truely axial.
 Failure is due to buckling.
 The cross section is uniform throughout.
 The Hook’s law is valid.
 The column is straight before application of load.

15. • Pcr
= (π 2 EI) / Le
2
But I =Ak2
∴ Pcr/A= π 2E/(Le/K)2
σcr = π2E/(Le/K)2
Where σcr is crippling stress or critical stress or stress
at failure
Limitation Of Euler’s Formula

16. Le/K= √ (π2E / σc)
For steel σc = 320N/mm2
and E =2 x 105 N/mm2
Limiting value (Le/K) is given by
Hence, the limiting value of crippling stress is the crushing
stress. The corresponding slenderness ratio may be found by the
relation
σcr = σc
∴ σc = π 2E/(Le/K)2
(Le/K)lim =√ (π2E / σc) = √ π2 × 2 × 105/320) = 78.54
Hence if Le /k < (Le /k)lim Euler's formula will not be
valid.

17. We know that, Euler’s formula for calculating crippling load is
valid only for long columns.
But the real problem arises for intermediate columns which fails
due to the combination of buckling and direct stress.
The Rankine suggested an empirical formula which is valid for
all types of columns. The Rankine’s formula is given by,
1/ PR= 1/ Pc + 1/ PE
PR = crippling load by Rankine’s
formula
Pc = crushing load = σc .A
PE = buckling load= PE= (π 2 EI) / Le
2
Rankine Formula

18. A
substituting
R
R
PE
E
C
C
R
2
2
2
2
2
2
Le2
Le2
 E
c
Le2
 c A
 E
 c A
 c (Le / K ) 2
 E K2
 c A
 c Le 2
 E ( AK 2
)
 EI
 c A
 EI
where a 
1 a (Le / K )2
P 
1 

1 

1 

1
c
P 
A and PE 
 c A
 c A
PC c
PC
1
PC

P  P
P 
PR PC PE
1

PC  PE
PR PC  PE
P  PE
1

1

1

19. / π 2EI
where a = Rankine’s constant =σc
and λ = slenderness ratio = Le/ k
PR = σcA / (1+a.λ2 )

24. Eccentric Loading
• Short Column
σmax = P/A + P.e/Z = P/A (1 + eyc/k2)
Z = Ak2/ yc
• Long Column
– Rankine’s Formula σc= P/A (1 + eyc/k2) (
– Euler’s Formula
– σmax = P/A + Pe v/Z
–σmin = P/A – Pe v /Z
v = sec {(le/2) /√[
1 + αle/k)
P/(EI)]}

27. THANK YOU