Shivendra Nandan Galgotias University

1. Stabilit
y
of
columnS

2. Columns and sturts: Structural members subjected to
compression and which are relatively long compared to
their lateral dimensions are called columns or Struts.
Generally, the term column is used to denote vertical
members and the term strut denotes inclined members
Examples: strut in a truss, Piston rods, side links in
forging machines, connecting rods etc.

3. Stable, Neutral and unstable Equilibrium
Stable equilibrium: A stable equilibrium is one in which a
body in static equilibrium on being displaced slightly,
returns to its original position and continues to remain in
equilibrium.

4. Neutral equilibrium: A neutral equilibrium is one in
which a body in equilibrium, on being displaced does
not returns to its original position, but its motion stops
and resumes its equilibrium state in its new position.

5. Unstable equilibrium: An unstable equilibrium is one in
which a body in equilibrium on being slightly disturbed,
moves away from its equilibrium position and loses its
state of equilibrium.

6. Buckling Load: The maximum load which a column can support
before becoming unstable is known as buckling load or crippling
load or critical load
The buckling takes place about the axis having
minimum radius of gyration or least moment of
inertia.
At this stage, the maximum stress in the column will be less than the
yield stress (crushing stress) of the material.

7. Safe load: It is the load to which a column is subjected
to and is well below the buckling load. It is obtained by
dividing the buckling load by a suitable factor of safety.
safe load = buckling load / factor of safety
Stability factor: The ratio of critical load to the
allowable load on a column is called stability Factor

8. MODES OF FAILURE OF THE COLUMNS
P
P
The load carrying capacity of a short column depends
only on its cross sectional area(A) and the crushing stress
of the material(σcu). The crushing load Pu for axially
loaded short column is given by Pcu =σcu × A .
The safe load on the column is obtained by dividing the crushing
load by suitable factor of safety. i.e., Psafe =Pcu/ FS
The mode of failure of columns depends upon their lengths and
depending on the mode of failure columns are classified as
a. Short columns b. Long columns
Short Columns: A short column buckles under compression as
shown in figure and fails by crushing. The load causing failure is
called crushing load.

9. Long columns: Long columns, which are also called
slender columns, when subjected to compression,
deflects or bends in a lateral direction as shown in the
figure. The lateral deflection of the long column is
called buckling.
The load carrying capacity of long column depends upon several
factors like the length of the column, M.I of its cross–section,
Modulus of elasticity of the material, nature of its support, in
addition to area of cross section and the crushing strength of the
material.
Critical load denotes the maximum load carrying
capacity of the long column.
The long column fails when there is excessive
buckling .ie when the load on the column exceeds critical
load.

10. SC – 10
Short columns fails by crushing or yielding of the material under
the load P1
Long column fails by buckling at a substantially smaller load P2
(less than P1).
P
1
P
2
The buckling load is less than the crushing
load for a long column
The value of buckling load for long column is
low whereas for short column the value of
buckling load is relatively high.

11. sc-11
Failure of long columns(contd)
Stress due to buckling
σb = ( M.ymax)/ I
= {(P.e). ymax}/I
= ( P.e) / ZWhere e = maximum bending of the column
at the centre
Consider a long column of uniform cross sectional area A
throughout its length L subjected to an axial compressive
load P. The load at which the column just buckles is known
as buckling load or crippling load.
Stress due to axial load σc = P/A
P
L e

12. Sc-12Failure of long columns(contd)
σmax = σc + σb
Extreme stress at centre of column will be the sum of direct
compressive stress and buckling stress
In case of long columns, the direct compressive
stresses are negligible when compared to buckling
stress. So always long columns fail due to buckling.

13. Modes of failures (contd.)
Intermediate Columns: These are columns which have
moderate length, length lesser than that of long columns
and greater than that of short columns.
sc-13
In these columns both bulging and buckling effects are
predominant. They show the behavior of both long
columns and short columns when loaded.

14. Euler’s Theory (For long columns)
Assumptions:
1. The column is initially straight and of uniform
lateral dimension
2. The material of the column is homogeneous,
isotropic, obeys Hookes law
3. The stresses are within elastic limit
4. The compressive load is axial and passes
through the centroid of the section
5. The self weight of the column itself is neglected.
6. The column fails by buckling alone

15. Euler’s Theory (For long columns)
A Bending moment which bends the
column as to present convexity
towards the initial centre line of the
member will be regarded as positive
Bending moment which bends the
column as to present concavity
towards the initial centre line of the
member will be regarded as negative
Sign convention for Bending Moments

16. Euler’s Formula for Pin-Ended Beams
(both ends hinged)
Consider an axially loaded long column AB of length L. Its both
ends A and B are hinged. Due to axial compressive load P, let the
deflection at distance x from A be y.
d2
y
dx2
d2
y
dx2 +
Py
EI
= 0
L
y x
P
PA
BThe bending moment at the section is given
by
EI
= - P y
-ve sign on right hand side, since as
x increases curvature decreases

17. At x =0, y =0,we get c1=0 (from eq.1)
Also at x=L, y =0 we get
c2 .sin [L√P/(EI)] =0
If c2 = 0, then y at any section is zero, which means there is no
lateral deflection which is not true
Therefore sin [L√P/(EI)] =0
This is the linear differential equation, whose solution is
Y = c1.cos [x√P/(EI)] + c2.sin[x √P/(EI)] …(1)
Where c1 and c2 are the constants of integration. They can be
found using the boundary conditions.

18. sin [L√P/(EI)] =0
=> [L√P/(EI)] = 0, π, 2 π ,……n π
Taking least non zero value we get
[L√P/(EI)] = π
Squaring both sides and simplifying
PE =
π2
E I
L2
This load is called critical or buckling
load or crippling load

19. case End condition Equivalent
length(Le)
Euler’s Buckling load
1 Both ends hinged Le=L PE= (π 2
E I) / Le
2
2 One end fixed, other
end free
Le=2L PE= (π 2
EI) / 4L2
3 One end fixed, other
end pin jointed
Le=L / √2 PE= 2(π 2
EI) / L2
4 Both ends fixed Le=L/2 PE= 4(π 2
EI) / L2
Note: L is the actual length of respective column and Le is to be
considered in calculating Euler's buckling load

20. Extension of Euler’s formula

21. Slenderness ratio: It is the Ratio of the effective length of the
column to the least radius of gyration of the cross sectional ends of
the column.
The Effective length: of a column with given end conditions is the
length of an equivalent column with both ends hinged, made up of
same material having same cross section, subjected to same
crippling load (buckling load) as that of given column.
Slenderness ratio, λ =Le /k
Least radius of gyration, k= √ Imin/A
Imin is the least of I xx and I yy
L =actual length of the column
Le=effective length of the column

22. Based on slenderness ratio ,columns are classified as
short ,long and intermediate columns.
Generally the slenderness ratio of short column is less
than 32 ,and that of long column is greater than 120,
Intermediate columns have slenderness ratio greater than
32 and less than 120.

23. Limitation of Euler's theory
Pcr
=
(π 2
EI) / Le
2
But I =Ak2
∴ Pcr/A= π 2
E/(Le/K)2
σcr = π2
E/(Le/K)2
Where σcr is crippling stress or critical stress or stress at failure
The validity of Euler’s theory is subjected to condition that
failure is due to buckling. The Euler’s formula for crippling is
The term Le/K is called slenderness ratio. As slenderness ratio
increases critical load/stress reduces. The variation of critical stress
with respect to slenderness ratio is shown in figure 1. As Le/K
approaches to zero the critical stress tends to infinity. But this
cannot happen. Before this stage the material will get crushed.

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

25. Empirical formula or Rankine - Gordon
formula
PR = crippling load by Rankine’s formula
Pc = crushing load = σc .A
PE = buckling load= PE= (π 2
EI) / Le
2
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
PE
1
PC
= +

26. For short columns: The effective length will be small and hence the
value of PE =(π2
EI) / Le
2
will be very large.
Hence 1/ PE is very small and can be neglected.
therefore 1/ PR= 1/ Pc or PR =Pc
For long column: we neglect the effect direct compression or
crushing and hence the term 1/ Pc can be neglected.
therefore 1/ PR= 1/ PE or PR =PE
Hence Rankines formula,
1/ PR= 1/ Pc + 1/ PE is satisfactory for all types of
columns

27. E
awhere
KLea
A
P
E
KLe
A
KE
Le
A
Le
AKE
A
A
Le
EI
A
A
P
Le
EI
PandAPngsubstituti
P
P
P
PP
PP
P
PP
PP
P
PPP
c
c
R
c
c
c
c
c
c
c
c
R
EcC
E
C
C
EC
EC
R
EC
EC
R
ECR
2
2222
2
2
2
2
2
2
2
2
2
)/(1
)/(
11
)(
11
1
1
111
π
σ
σ
π
σ
σ
π
σ
σ
π
σ
σ
π
σ
σ
π
σ
=
+
=
+
=
+
=
+
=
+
=
==
+
=
+
×
=
×
+
=
+=
(I=AK2
)

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

29. ILLUSTRATIVE NUMERICAL EXAMPLES
Euler’s crippling load =PE= (π2
EI) / Le 2
= [π 2
× 200 × 109
× π × (0.06)4
/64] / (1.7682
)
= 401.4 ×103
N =401.4 kN
Safe compressive load = PE /3 =133.9kN
1. A solid round bar 60mm in diameter and 2.5m long is
used as a strut. One end of the strut is fixed, while its other
end is hinged. Find the safe compressive load, for this strut,
using Euler’s formula. Assume E=200GN/m2
and factor of
safety =3.
end condition: one end hinged, other end fixed
effective length Le = L /(√ 2)= 2.5/ (√ 2)= 1.768m
Solution:

30. outside diameter of the column =D =50mm
=0.05m; E=70 × 10 9
N/m2
Inside diameter = ?
2.A slender pin ended aluminium column 1.8m long and of
circular cross-section is to have an outside diameter of
50mm. Calculate the necessary internal diameter to prevent
failure by buckling if the actual load applied is 13.6kN and
the critical load applied is twice the actual load. Take
Ea = 70GN/m2
.
Solution:

31. End condition: pin-ended ( hinged)
Le =L =1.8m
Euler’s crippling load =PE= π 2
(EI) / Le
2
d = 0.0437m =
2
44
3
8.1
64
)05.0(
1070
102.27
92
d−
×××
=×
π
π
Critical load =PE = 2 × safe load (given condition)
= 2 × 13.6=27.2kN
I= π (D4
-d4
) /64 = π (0.054
-d4
) /64

32. 3. A built up beam shown in the figure is simply supported at its
ends. Compute its length, given that when it subjected to a load of
40kN per metre length. It deflects by 1cm. Find the safe load, if this
beam is used as a column with both ends fixed. Assume a factor of
safety of 4. use Euler’s formula. Take E = 210GN/m2
.
300
mm
50 mm
1000 mm20 mm

33. L = 14.15m
Moment of inertia of section about X-X axis,
Load =40kN/m , length of the beam =?
= 994166 × 104
mm4.= 99.41×10-4
m
Using the relation, δ =
5 wL4
384EI
0.01 =
5 × 40 × 10 3
× L 4
( 384 × 210 × 109
× 99.41 × 10-4
12
100020
525)50300(
12
50300
2
33
2
×
+





×+
×
=xxI

34. Safe load, the beam can carry as
column:
End condition: Both ends fixed
PE = π 2
(EIyy) / Le
2
= (π 2
× 210 × 109
× 2.25 × 10 -4
) / (7.07)2
= 9.33 × 10 6
N = 9.33 × 10 3
kN
Safe load = Pe /F.S = 9.33 × 10 3
/ 4 = 2.333 × 10 3
kN
Le = L/2 = 14.15/2 = 7.07m
Iyy = 2[ (50 × 300 3
) /12] + (1000 × 20 3
) /12
= 22567 × 10 4
mm4
= 2.25 × 10 -4
m4

35. 4.From the test on steel struts with ends fixed in position and fixed
in direction the following results are obtained.
Assuming the values in agreement with Rankine’s formula ,find the
two constants

36. Rankine’s critical load = PR = σcA / (1+a.λ2
)
Rankine’s critical stress = PR / A
200= σc / [1+a.(70 2
) ] …. (1)
69= σc / [1+a.(170 2
) ] …(2)
(1) / (2) gives
constant a = 1.29 × 10 -4
substituting ‘a ‘ in (1) or (2)
[ ] 22
2
2
)170(1)70(18986.2
)70(1
)170(1
69
200
aa
a
a
+=+
+
+
=

37. 5.Find the Euler’s crushing load for a hollow cylindrical cast iron
column, 15cm external diameter and 2cm thick, if it is 6m long and
hinged at both ends. E = 80GPa. Compare this load with the
crushing load as given by the Rankine’s formula, using
σc= 550MPa and a =1/600. For what length of the strut of this
cross-section does the Euler’s formula ceases to apply ?
Solution: Internal diameter = 15 – 2 × 2 = 11 cm
A = π/4[ 0.152
- 0.112
] = 81.7 × 10-4
m2
.
I = π [ 0.154
- 0.114
] /64
=17.66 × 10-6
m4
.
A
I
K min
= = 0.0465 m

38. Euler’s critical load is given by
PE = π 2
(EI) / Le 2
= (π 2
× 80 × 10 9
× 17.66 × 10 -6
) / 62
= 387327.14 N (higher)
Rankine’s critical load, PR = (σc A) / [1+ a (Le / K)2
]
= (550 × 10 6
× 81.7 × 10 -4
)/ [1+1/600 (6/0.0465)2
]
= 156301.78
To calculate limiting length : σc = 550 MPa =550N/mm2
= PE / A
550 = π 2
(EI) / A Le
2
therefore Le = 1.761m

39. 6. The built up column shown in the figure consisting of
150mm × 100mm RSJ with 120mm wide plate riveted to each
flange. Calculate the safe load, the column can carry , if it is
4m long having one end fixed and other end hinged with a factor
of safety 3.5. Take the properties of the joist as
A = 21.67 × 102
mm2
; Ixx = 839.1 × 104
mm4
; Iyy = 94.8 × 104
mm4
.
Assume Rankine’s constant as 315N/mm2
and a =1/7500

40. 120mm
100mm
150mm
12m
m

41. Solution:
Ixx = 839.1 × 104
+ 2[ (120 × 123
)/12 + 120 × 12 ×( 75 +6)2
]
= 2732.1 × 104
mm4
= 2732.1 × 10-8
m4
Similarly, Iyy = 94.8 × 10 4
+ 2[ (12 × 1203
)/12 ] mm4
= 440.4 × 104
mm4
= 440.4 × 10-8
m4
(Iyy is the lower value, column will tend to buckle in
YY direction, Iyy has to be considered)
A = 21.67×102
+ 2 (120×12) = 5047 mm2
=50.47×10-4
m2

42. Exercise problems
1. Calculate the safe compressive load on a hollow cast iron
column one end fixed and other end hinged of 150mm external
diameter,100mm internal diameter and 10m length. Use Euler's
formula with a factor of safety of 5 and E=95GN/m2
Ans: 74.8kN
2. Bar of length 4m when used as a simply supported beam and
subjected to a u.d.l of 30kN/m over the whole span., deflects
15mm at the centre. Determine the crippling loads when it is
used as a column with the following end conditions:
(i) Both ends pin jointed (ii) one end fixed and other end
hinged (iii) Both ends fixed
Ans: (i) 4108 kN (ii) 8207kN (iii) 16432 kN

43. sc - 42
Exercise problems (contd)
3.Determine the ratio of the buckling strengths of two
columns of circular cross-section one hollow and other solid
when both are made of the same material, have the same
length, cross sectional area and end conditions. The internal
diameter of the hollow column is half of its external diameter
Ans: 1.66
4. Calculate the critical load of a strut 5m long which is
made of a bar circular in section and pin jointed at both ends.
The same bar when freely supported gives mid span
deflection of 10mm with a load of 80N at the centre.
Ans: 8.22kN

44. sc - 43
Exercise problems (contd)
5. A hollow C.I column whose outside diameter is 200mm
has a thickness of 20mm. It is 4.5m long and is fixed at both
ends. Calculate the safe load by Rankine’s formula using a
factor of safety of 4. Take σc =550MN/m2
, a=1/1600
Ans: 0.877 MN
6. A hollow cylindrical cast iron column is 4m long with both
ends fixed. Determine the minimum diameter of the column,
if it has to carry a safe load of 250kN with a factor of safety
of 5. Take the internal diameter as 0.8 times the external
diameter.
σC =550MN /m 2
a= 1/1600
Ans: D= 136mm d= 108.8mm