Buckling of Columns

Introduction
• Load-carrying structures may fail in a variety of ways
depending upon the type of structure, support conditions,
type of loading, material, etc.

• Buckling is a common type of failure in structures in particular

long and slender structural members loaded axially in
COMPRESSION.
Crushing failure – 1985
Mexico earthquake.

Buckling failure
Stable and Unstable
Equilibrium
• The concept of the stability of equilibrium can be
demonstrated easily by considering the equilibrium on three
different surfaces as illustrated in the Figure 1 below.
• In all three situations the ball is in an equilibrium position i.e.
It satisfies 𝐹𝑥 = 0, 𝐹𝑦 = 0 and 𝑀 = 0.

Figure 1. Stability of Equilibrium

(a) Stable equilibrium
- If the ball is moved to a new position, it will always return to
the lowest point.

(b) Neutral equilibrium

-The ball will remain wherever it is placed (no unbalanced force).

(c) Unstable equilibrium

-On the top, it is in equilibrium but unstable. If it is moved to a
different position it will roll away.
 Consider a simply-supported long slender column subjected to a
compressive axial load P as shown in Figure 2 below.

- For small values of P, the column will remain essentially

P straight and the stresses can be determined easily (i.e. ).
𝛿

Stable eqm.
- As P increases, the column bends and there is small
lateral deflection. Stresses now not only depend on
L
axial load but also on the deflection.

Unstable eqm.
- As P is increased further, the load will reach to a value
where there is sudden large deflection and buckling
P without warning.
Figure 2. Pin-
ended column
This failure is not due to the failure of column material but
due to buckling. The transition between stable and unstable
equilibrium occurs at a special value of the axial force known
as the critical buckling load Pcr.
So, generally:

• P < Pcr – column is in stable equilibrium in the straight

position.

• P = Pcr - column is in neutral equilibrium. At this load,

the column may undergo small lateral deflection
with no change in axial load. When load is
removed, the bent shape does not disappear.

• P > Pcr - column is in unstable equilibrium and may

collapse by buckling under the slightest
disturbance.
Euler Buckling Load
– Pin-Ended Column
• To investigate the stability of real columns, we begin by
considering the ideal pin-ended column, as shown in Figure 3
below:

Figure 3. Buckling of a pin-ended column

Main assumptions:

 The column is initially perfectly straight, and it is made of linearly

elastic material.

 The column is free to rotate, at its ends, about frictionless pins;

that is, it is restrained like a simply supported beam. Each pin
passes through the centroid of the cross section.

 The column is symmetric about the xy plane, and any lateral

deflection of the column takes place in the xy plane.

 The column is loaded by an axial compressive force P.

Using moment-curvature relationship, the bending-
moment equation is given by:

(1)

where M is the bending moment at any cross-section,

v is the lateral deflection in the y-direction, and EI is the
flexural rigidity for bending in the xy plane.

From FBD,
(2)

where v is the deflection at the cross-section.

Eqn (1) now becomes

(3) – Homogeneous, linear, second-order DE

Solving eqn (3) will give the magnitude of the critical load and
the deflected shape of the buckled column.
Solution to DE:
(3) – Homogeneous, linear, second-order DE

Let
(4)

Rewriting eqn (3)

(5)

The general solution for eqn (5) is

(6)

where C1 and C2 are constants of integration.

 (6)

Boundary conditions
(7)

First condition gives C2 = 0, therefore

(8)

Second condition gives

(9)

Therefore, either C1 = 0 or sin kL = 0.

C1 = 0 is a trivial solution. It gives the behaviour of an ideal
column that is in equilibrium with no deflection under the action
of axial compressive load P.

Second possibility: - known as buckling equation

Therefore,
- n = 0 is omitted since it means
that k = 0 and hence P = 0.
Or from eqn (4)

(10)

Eqn (10) gives the critical loads for this column.

The deflection curve is given by:

(11)
From eqn (10) the lowest critical load (or Euler buckling load) for
a column with pinned ends is obtained when n = 1:

(12)

The corresponding buckled shape (or mode shape) is

(13)

The constant C1 represents the deflection at the mid-point of

the column and may have any small value, positive or negative.

Higher values of n would give an infinite number of critical loads

and corresponding mode shapes.
For example, the mode shape for n = 2 has two half-waves. The
corresponding load is four times larger than the critical load for
n = 1.

Figure 4. Buckled shapes for an ideal column with pinned ends

Euler Critical Stress
The critical load for a pin-ended column can be determined
using eqn (12). The corresponding critical stress is obtained
simply by dividing the load by its cross-sectional area:

(14)

where I is the second moment of area for the principal axis

about which buckling occurs.

𝐼
Eqn (14) is simplified further by using the notation 𝑟 =
𝐴
where r is the radius of gyration of the cross-section in the plane
of bending.
Eqn (14) becomes

(15)

Where L/r is a non-dimensional ratio known as the slenderness

ratio.
The critical stress can be plotted as a function of slenderness
ratio to obtain the Euler’s Curve as shown in Figure 5.

Figure 5. Euler’s curve for structural steel

Example 1
A 7.2-m long A-36 steel tube having the
x-section shown is to be used a pin-ended
column. Determine the maximum allowable
axial load the column can support so that
it does not buckle.
Est = 200 GPa
 2 EI
Pcr 
L2
2
   6 1
4
2

 200 10 kN/m   704 1 m / 1000 mm 4

7.2 m 2
 228.2 kN
This force creates an average compressive stress in the
column of
Pcr 228.2 kN 1000 N/kN 
 cr  
 
A  752   702 mm2
 100.2 N/mm 2  100 MPa
Since cr < Y = 250 MPa, application of Euler’s eqn is
appropriate.
Example 2
The A-36 steel W20046 member shown is to be used as
a pin-connected column. Determine the largest axial load it
can support before it either begins to buckle
or the steel yields.

A = 5890 mm2,
Ix = 45.5106 mm4,and
Iy = 15.3106 mm4
By inspection, buckling will occur about the y-y axis.

 2 EI
Pcr 
L2


      
 2 200 106 kN/m 2 15.3 104 mm4 1 m / 1000 mm 4
4 m 2
 1887.6 kN
When fully loaded, average compressive stress in column is

Pcr 1887.6 kN 1000 N/kN 

 cr  
A 5890 mm2
 320.5 N/mm2
Since this stress exceeds yield stress (250 N/mm2), the load P is
determined from simple compression:

P
250 N/mm  2
5890 mm2
P  1472.5 kN
Other Support Conditions
The critical loads for columns with various support conditions
can be related to the critical load of a pinned-end column
through the concept of an effective length.

Consider the deflected shape of a column

fixed at the base and free at the top as
shown in Figure 6.

The deflection curve is a quarter of a

complete sine wave. Extending the curve
becomes half of a complete sine wave.

Figure 6. Deflection curves showing the effective length

for a fixed-free column
The effective length Le for any column is the length of an
equivalent pinned-end column i.e. it is the length of a pinned-
end column having a deflection curve that exactly matches all or
part of the deflection curve of the original column.

It can also be expressed as the distance between points of

inflection in its deflection curve assuming that the curve is
extended (if necessary) until points of inflection are reached.

For a fixed-free column, the effective

length is:
The formula for critical loads in Eqn (12) can be rewritten in
terms of effective length since effective length is the length of
an equivalent pinned-end column:

(16)

The effective length is often expressed in terms of an effective

length factor K i.e. Le = KL. Therefore the critical load in eqn (16)
becomes

(17)

K = 2 for fixed-free column and K = 1 for pinned-end column.

For other support conditions:

Figure 7. Effective lengths for columns with different end conditions

 Example 3
A W15024 steel column is 8 m
long and is fixed at its ends as
shown. Its load-carrying capacity
is increased by bracing it about
the y-y axis using struts that are
assumed to be pin-connected
to its mid-height. Determine the
load it can support sp that the
column does not buckle nor
material exceed the yield stress.
Take Est = 200 GPa and Y = 410 MPa.

A = 3060 mm2
Ix = 13.4106 mm4 ; Iy = 1.83106 mm4
rx = 66.2 mm ; ry = 24.5 mm
Columns with Eccentric
Loading
• We apply a load P to column at a short eccentric distance
e from centroid of x-section.
• This is equivalent to applying
a load P and moment M’ = Pe.

Figure 8. Column subjected to eccentric load

From free-body diagram, internal moment in column is

M   Pe  v  18
Thus, the general solution for the differential eqn of the
deflection curve is

v  C1 sin
P
x  C2 cos
P
xe 19
EI EI

Applying boundary conditions to determine the constants,

deflection curve is written as

  P L  P   P  
v  e  tan  sin 
  x   cos x   1 20
  EI 2   EI   EI  
Due to symmetry of loading, both maximum deflection and
maximum stress occur at column’s midpoint. Therefore,
when x = L/2,  = max, so

  P L 
vmax  e sec   1
 21
  EI 2  
Maximum stress in column occur when maximum moment
occurs at the column’s midpoint. Using eqns (18) and (21)

M  Pe  vmax 
 P L
M  Pe sec 
 22
 EI 2 
Maximum stress is compressive and therefore

P Mc
 max  
A I

P Pec  P L 
 max   sec 
 23
A I  EI 2 
Since radius of gyration r2 = I/A,

P  ec  L P 
 max  1  2 sec  24
A  r  2r EA 

Eqn (24) is known as the Secant Formula for columns.

The secant formula gives the maximum compressive stress in

the column as a function of the average compressive stress P/A,
the modulus of elasticity E, and two non-dimensional ratios –
𝑒𝑐
the slenderness ratio L/r and the eccentricity ratio 2 .
𝑟
max = maximum elastic stress in column, at inner concave side of
midpoint (compressive).
P = vertical load applied to the column. P < Pcr unless e = 0, then P
= Pcr (Eqn 13-5)
e = eccentricity of load P, measured from the neutral axis of
column’s x-sectional area to line of action of P.
c = distance from neutral axis to outer fiber of column where
maximum compressive stress max occurs.
A = x-sectional area of column
L = unsupported length of column in plane of bending. For non
pin-supported columns, Le should be used.
E = modulus of elasticity of material.
r = radius of gyration, r = √(I/A), where I is computed about the
neutral or bending axis.
Graph of the secant formula can be plotted as shown in Figure 9.

Figure 9. Graph of secant formula

Further reading
• Gere, J.M. and Goodno, B.J., Mechanics of Materials, 8th Ed, pp
900 – 940.

Tutorial Questions
• Gere, J.M. and Goodno, B.J., Mechanics of Materials, 8th Ed, pp
953-965.