CBE2027 Structural Analysis I
Chapter 6 Long Column
LONG (SLENDER) COLUMNS
A long (slender) column fails by elastic buckling when an axial
compressive load reaches a critical value. The Swiss mathematician
Leonard Euler (1707 1783) was the first to formulate an expression for the
critical buckling load of a column.
Long column with pin-supports
The following assumptions are made in deriving the formula for critical
buckling load (Euler Buckling Load).
1. The compressive loads are applied at the ends of the column without
eccentricity.
2. The column is perfectly straight before the loads are applied.
3. The ends of the column are frictionless pins (or hinges), which
allow the column to buckle about any axis of the cross-sections.
4. The column is made of homogeneous and isotropic material.
5. The self-weight of the column is ignored.
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CBE2027 Structural Analysis I
Chapter 6 Long Column
The critical buckling load (Euler Buckling Load) Pcr for an ideal pin-ended
column is given by:-
Pcr =
2 EI
L2
(in N)
where E = the modulus of elasticity of the material (MPa)
I = the least moment of inertia of the cross-section (mm4)
L = the length of the column from pin-end to pin-end (mm)
= constant pi (= 3.1416)
The above formula applies to long column with both ends pinned.
How to determine the buckling load of long column with other support
conditions? The concept of effective length or effective column height
will be outlined below.
EFFECTIVE LENGTH (Columns with different support conditions)
The Eulers Formula gives the buckling load for an ideal column with both
ends pinned. The formula may be extended to columns with other end
conditions through the concept of an effective length.
The effective length is the distance between points of inflection
(contraflexure) on the deflected shape of the column. These are points of
zero bending moment.
The effective length is often expressed in terms of an effective length factor
K, where
Effective Length
in which,
Le = K*L
K = effective length factor
(Its value depends on the support conditions)
L = actual unbraced column length.
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CBE2027 Structural Analysis I
Chapter 6 Long Column
The critical buckling load is then given by
Pcr =
2 EI
le
or
Pcr =
2 EI
( KL) 2
The Effective Length Factor for Different Support Conditions
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CBE2027 Structural Analysis I
Chapter 6 Long Column
SLENDERNESS RATIO
Radius of Gyration
The measure of slenderness of the cross-section of the column is its radius
of gyration, r, defined as:-
rxx =
I xx
A
and
ryy =
I yy
A
where Ixx and Iyy = moment of inertia about x-x and y-y axis.
A = area of the cross-section.
In most cases, it is required to determine the axis for which the radius of
gyration is the smallest, because that is the axis about which the column
would likely buckle.
Slenderness Ratio
A column has been described as relatively long, slender member loaded in
compression. This description is stated in relative terms and is not very
useful for analysis.
The measure of the slenderness of a column must take into account the
length, the cross-sectional shape and dimensions of the column, and the
manner of attaching the ends of the column to the structures that supply
loads and reactions to the column. The commonly used measure of
slenderness is the slenderness ratio, defined as:-
SR =
( KL) = Le
r
where L = actual length of the column between points of support or
lateral restraint.
Le = effective length, taking into account of the effects of support
conditions
K = effective length factor
r = radius of gyration of the cross-section of the column
(normally the smallest r is used when the effective lengths of
the column about both principal axes are the same.)
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CBE2027 Structural Analysis I
Chapter 6 Long Column
Euler Buckling Load in Long Columns
(c)
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(d)
(e)
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CBE2027 Structural Analysis I
Chapter 6 Long Column
In (a), the Euler buckling load of a long column with both ends pinned is
2 EI
given by Pcr =
.
L2
In (b), a curve of critical buckling load Pcr is plotted against the column
length L.
It shows that when the column length approaches to infinity, the critical
buckling load approaches to zero. This means that when you exert a very
small compressive load on a very slender column, the column buckles.
When the column length approaches to zero, the buckling load approaches
to infinity. This means that a very short column can support an extremely
large load regardless the material strength of the column. Is this make sense?
Definitely NOT. From the above discussion, we note that there are
limitations on the use of the Eulers Formula.
In (c), the moment of inertia about one axis is greater than that about the
other and the effective lengths about both axes are the same. Theoretically,
the member can potentially fail by buckling about either axis as shown in (d).
However, the load required to cause it to buckle about the stronger axis
exceeds the load which will cause buckling about the weak axis. Therefore
2 EI yy
as shown in (e).
the member will buckling at Pcry =
L2
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CBE2027 Structural Analysis I
Chapter 6 Long Column
Critical Buckling Stress for Columns
(Validity Limit of the Eulers Theory)
The critical buckling stress cr is the average stress over the cross-sectional
area A of a column at the critical load Pcr.
cr
Pcr
2E
=
=
A Le 2
r
Consider Pcr =
2 EI
Le
and cr
Pcr 2 EI 2 Er 2
2E
=
= 2 =
=
2
2
A
Le A
Le
Le
r
The critical buckling stress cr depends inversely on the square of the
slenderness ratio. The higher the slenderness ratio, the lower is the critical
stress that will cause buckling, and vice versa. The slenderness ratio (Le/r)
is an important way of thinking about columns, since it is the single
measurable parameter on which the buckling of a column depends.
As the slenderness ratio is decreased, a limit is reached when the critical
buckling stress cr equals the yield or crushing stress y of the material. For
a slenderness ratio less than this, the computed critical stress cr would be
greater than y and the Eulers formula would no longer be valid the
column no longer qualifies as a long column. It should be treated as a short
column.
A critical part of the design or analysis process for a column is to determine
whether the Eulers formula is applicable to it or not:
(a) If cr is less than y, the Eulers formula can be applied.
(b) If cr is greater than y, the Eulers formula is NOT valid. Failure of
the column is governed by y.
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CBE2027 Structural Analysis I
Chapter 6 Long Column
The above criteria may be expressed in terms of the slenderness ratio the
column can be considered a long column if the following is true:
Le
2E
Slenderness Ratio
>
y
rmin
where
y = yield stress of the column material
If the slenderness ratio Le/r is greater than the expression on the right, the
column is considered long and the Eulers formula can be applied.
If Le/r is less than the expression on the right, the Eulers formula cannot
be used. The column should be designed as a short column or
intermediate column.
Critical
Stress cr
Euler's Formula
Invalid
Euler's Formula
Valid
(Le /r)1
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Slenderness Ratio (Le /r)
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CBE2027 Structural Analysis I
Chapter 6 Long Column
SHORT COLUMNS AND LONG COLUMNS
Critical
Stress cr
yield C
Short
Columns
Long Columns
Slenderness Ratio (Le /r)
When the slenderness ratio falls in between CD, the column is treated as a
short column. When the slenderness ratio falls in between DB, the column
is treated as a long column and its behaviour is governed by Eulers
Formula.
EFFICIENT SHAPES FOR COLUMN CROSS-SECTIONS
In general, column buckling can occur in any direction. Therefore, it is
desirable to have uniform properties with respect to any axis. A hollow
circular section makes a very efficient shape for a column. Closely
approximating that is the hollow square tube. Fabricated sections made
from standard structural sections can also be used.
Building columns are often made from special wide-flange shapes called
H-sections. They have relatively wide, thick flanges as compared with the
shapes typically selected for I-beams. This makes the moment of inertia
with respect to the Y-Y axis more nearly equal to that for X-X axis. The
result is that the radii of gyration for the two axes are more nearly equal.
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CBE2027 Structural Analysis I
Chapter 6 Long Column
Some examples of efficient column shapes are shown below. (a) Hollow
circular section, (b) Hollow square section, (c) Built-up box section, (d)
Equal-leg angles with plates, (e) Channels with plates and (f) Two equal-leg
angles.
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CBE2027 Structural Analysis I
Chapter 6 Long Column
BRACING
In order to increase the compressive load carrying capacity of a column,
bracing is normally used. The addition of bracing decreases the effective
length of columns. The addition of bracing must be done with care. If it is
not done properly, no benefit is gained and even a loss is incurred due to the
additional material and effort expenditures in the bracing.
(a)
(b)
(d)
In (a) and (b), both Ixx and Iyy are the same. Bracing is added in one
direction only, the column will buckle in the other unrestrained direction as
it has a higher slenderness ratio in this direction.
Although bracing is added in (c), it is not effective as it restrains the stronger
direction. The column will buckle in the weaker direction as it has a higher
slenderness ratio in this direction.
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CBE2027 Structural Analysis I
Chapter 6 Long Column
EFFECTIVE USE OF BRACING
(a)
(b)
(c)
For symmetrical sections in (a), bracing should be added in both directions
in order to reduce the effective length of the column.
For column section with different Ixx and Iyy (see (b) and (c)), bracing should
be added in the weaker direction so as to reduce the effective length in the
weaker direction.
What is the buckling load for the column in (b)?
Buckling load = Pcrx or Pcry?
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CBE2027 Structural Analysis I
Chapter 6 Long Column
Buckling Load for Column in (b)
The column in (b) is braced in the weaker direction. The column will have a
tendency to buckle in the direction associated with the highest slenderness
ratio.
In order to determine the buckling load of this column, it is necessary to
calculate the buckling load about x-x axis, Pcrx, and the buckling load about
y-y axis, Pcry. The buckling load of the column is the smaller of Pcrx and Pcry.
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CBE2027 Structural Analysis I
Chapter 6 Long Column
Effects of Lateral Bracing on Column Buckling
Pinned
Le = L/2
Bracing a column changes its effective length and consequently its buckling
mode. The more a column is braced, the shorter its effectively length
becomes and the greater the load that is required to cause buckling. If
bracing is used, it is usually more effective when placed symmetrically.
Bracing in
both directions
Le = L/2
Pinned
(a)
No bracing
2 EI
P1 =
L2
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(b) Bracing at mid-height
2 EI 4 2 EI
P2 =
=
= 4 P1
2
2
L
L
2
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(c)
Third-point bracing
2 EI
P3 =
= 9 P1
2
L
3
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Le = L/3
Chapter 6 Long Column
Bracing in
both directions
Le =2 L/3
Bracing in
both directions
Le = L/3
Le = L/3 Le = L/3
CBE2027 Structural Analysis I
(d) Asymmetric bracing
2 EI 9 2 EI 9
P4 =
=
= P1
2
2
4
4L
2L
3
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CBE2027 Structural Analysis I
Chapter 6 Long Column
ALLOWABLE LOAD
The allowable load is calculated by dividing the critical buckling load by a
Factor of Safety, F.S.. That is,
Pa =
Pcr
F. S.
where Pa = allowable load
Pcr = critical buckling load
F.S.= Factor of Safety
ALLOWABLE STRESSES IN COMPRESSION
Similarly, the allowable stress is computed by dividing the crushing stress or
buckling stress by a F.S. depending on whether the column is a short or
slender column.
Compressive
Stress = P/A
Crushing stress
c =P/A
Buckling stress
cr
Factor of Safety
Allowable
stress
Slenderness Ratio (Le /r)
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CBE2027 Structural Analysis I
Chapter 6 Long Column
Example 1
A uniform column AB is 3 m in height and having a square hollow section
as shown in the figure. Using Eulers formula and a factor of safety of two,
determine the allowable load for the column and the corresponding normal
stress. Use E = 200 GPa and yield = 250 N/mm2.
P
Free
10 mm
120 mm
3m
120 mm
A Fixed
Solution:
Effective length = K*L = 2* 3 = 6m
Cross-sectional area = 1202 1002 = 4400 mm2
Moment of Inertia, Ixx = Iyy = (1204 1004)/12 = 8.947*106 mm4
Euler Buckling Load, Pcr =
2 EI
2
2 200 x10 3 8.947 x10 6
6000 2
le
= 490600 N = 490.6 kN
Pcr 490600
= 111.5 N / mm 2
=
A
4400
< yield = 250 N / mm 2 O.K.
Critical stress cr =
Allowable Load = Buckling Load / F.S. = 490.6/2 = 245.3 kN
And the corresponding normal stress = 245.3*103 / 4400 = 55.8 N/mm2
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CBE2027 Structural Analysis I
Chapter 6 Long Column
Example 2
An axially loaded aluminum column is 6 m long and has the cross-section as
shown in the figure. The column is pin-connected at both ends. Determine
(a) the critical Euler load for this column, (b) the critical stress for this
column. Use E = 75 GPa and yield = 150 MPa.
200 mm
Y
20 mm
X
Y
360 mm
Solution:
Effective length le = K*L = 1* 6 = 6m
Cross-sectional area = 360*200 320*160 = 20800 mm2
As Ixx < Iyy, then the column will buckle about x-x axis.
Ixx = (360*2003 320*1603) / 12 = 1.308*108 mm4
The Euler buckling load is: 2 EI xx 2 75 x10 3 1.308 x10 8
Pcr =
=
= 2689500 N = 2689.5 kN
2
2
6000
le
The critical stress is:P
2689500
cr = cr =
= 129.3 MPa < yield = 150 MPa O.K.
A
20800
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CBE2027 Structural Analysis I
Chapter 6 Long Column
Example 3
A circular hollow aluminum tube is used as an axially loaded column. The
column is fixed at the bottom and is pinned at the top. The tube is 4.5 m
long, has an outside diameter of 60 mm and an inside diameter of 40 mm.
Calculate the allowable load that the column can carry if a factor of safety of
2.5 is used. Use E = 75 GPa and yield = 150 MPa.
Solution:
Effective length le = K*L = 0.707* 4.5 = 3.1815m
Cross-sectional area =
I=
(60
64
(60
4
40 2 = 1571 mm 2
40 4 = 510509 mm 4
The Euler buckling load is: 2 EI 2 75 x10 3 510509
Pcr =
=
= 37330 N = 37.33 kN
2
2
3181
.
5
le
The critical stress is:P
37330
cr = cr =
= 23.8 MPa < yield = 150 MPa
A
1571
Euler formula can be used.
The allowable load is:Pallow = Pcr / F.S. = 37.33 / 2.5 = 14.93 kN
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CBE2027 Structural Analysis I
Chapter 6 Long Column
Example 4
A steel column with a height of 8 m is fixed at its bottom. At the top of the
column, it is braced in its weaker direction and it is free to move in its
stronger direction. The cross-section of the column is shown in the figure.
Determine the allowable load that the column can carry if a factor of safety
of 3 is used. Use E = 205 GPa and yield = 275 MPa.
Y
350 mm
30 mm
250 mm
350 mm
50 mm
50 mm
Y
Solution:
Area of the section, A = (3502 320*250) = 42500 mm2
Moment of Inertia Ixx = (3504 320*2503) / 12 = 8.339*108 mm4
Moment of Inertia Iyy = 2*50*3503 / 12 + 250*303 / 12 = 3.579*108 mm4
Effective length when buckling about x-x axis,
lex = Kx*L = 2*8 = 16 m (Fixed Free support condition)
Effective length when buckling about y-y axis,
ley = Ky*L = 0.707*8 = 5.656 m (Fixed Pinned support condition)
Buckling load about x-x axis,
2 EI xx 2 205 x10 3 8.339 x10 8
Pcrx =
=
= 6590600 N = 6590.6 kN
2
16000 2
lex
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CBE2027 Structural Analysis I
Chapter 6 Long Column
Buckling load about y-y axis,
2 EI yy 2 205 x10 3 3.579 x10 8
Pcry =
=
= 22635800 N = 22635.8 kN
2
5656 2
ley
As Pcrx < Pcry, the buckling load of the column is:Pcrx = 6590.6 kN
cr = Pcrx / A = 6590.6*103 / 42500 = 155.1 N/mm2
< y = 275 N/mm2
Eulers Formula can be used.
Pallow = Pcrx / F.S. = 6590.6 / 3 = 2196.9 kN
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CBE2027 Structural Analysis I
Chapter 6 Long Column
Tutorial 7 (Long Columns)
A rectangular aluminum (E = 70 GPa) tube of uniform thickness t
15 mm serves as a 8 m long column fixed at both ends. Calculate the
critical stress in the column.
200 mm
Q1.
15 mm
300 mm
Q2.
A 50 mm by 100 mm timber is used as a column with one end fixed
and one end pinned. Determine the minimum length at which
Eulers formula can be used if E = 10 GPa and the proportional limit
is 25 MPa. What central load can be carried with a factor of safety of
2 if the length is 2 m?
Q3.
The jib crane shown in the figure is of capacity W = 25 kN. For =
30o and a factor of safety of 3, calculate the required minimum
cross-sectional area for circular steel bar AB (E = 200 GPa).
B
W
3m
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CBE2027 Structural Analysis I
Q4.
Chapter 6 Long Column
A horizontal bar AB is supported by a pinned-end column CD as
shown in the figure. The column is a steel hollow circular section (E
= 210 GPa) having length of 4 m. The inside diameter and the
outside diameter of the section is 80 mm and 100 mm. Calculate the
allowable load Q if the factor of safety with respect to buckling of the
column is n = 2.5.
2m
1m
C
4m
Q5.
Given a factor of safety of 3, determine the largest load F that may be
applied to the structure shown in the figure. Assume that each
column is of 50 mm diameter steel bar (E = 200 GPa).
F
B
2m
C
1.5m
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1m
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CBE2027 Structural Analysis I
Q6.
Chapter 6 Long Column
Given a factor of safety of 3, determine the largest load F that may be
applied to the structure shown in the figure. Assume that each
column is of 50 x 50 mm square steel bar (E = 200 GPa).
F
C
3m
A
1m 1.5m
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