ABSTRACT

The buckling of column is a form of deformation as a result of axial- compression forces. This
leads to bending of the column, due to instability of the column. This mode of failure is quick
and hence dangerous. Length, strength and other factors determine how a column buckled. The
laboratory experiment was carried out to investigate the compression members with different
end restraint under axial loading and to derive the Euler buckling factor by accounting a few
columns length and end restraint condition. These result were then compared with theoretical
with experimental data. There are three specimens used which is aluminium struts with
different length. The force will be exerted on the strut with different end condition which is
pinned-pinned, pinned-fixed and fixed-fixed. From result and data, we can identify which one
is the strongest strut with some end condition to withstand high pressure.

CHAPTER 1
INTRODUCTION

1.1 General

A column is a structural member that resisting compression force must meet specific strength,
deflection and stability requirements. Buckling is a phenomenon where column often occurred
or collapsed due to excessive load exerted exceeding its critical load which then caused the
column to deflect laterally. The maximum axial load that a column can support when it is on
the verge of buckling is called critical load.
1.2 Goal and objectives
There are three main objectives in this laboratory as follows:
i. To investigate the influence of multiple column lengths and end restraints (support
condition) under axial loading
ii. To derive Euler buckling factor by accounting a few columns length and end
restraint conditions (refer table 3.1)
iii. Identify how to overcome the problem by applying appropriate solution through
application

1.3 Laboratory Scope

The laboratory that is conducted is called Buckling Struts Test. The specimen that were used
is strut with different length and is made up of aluminum. There are 5 specimens, and each of
the length of strut is 0.32m, 0.37m, 0.42mm, 0.47m and 0.52m with cross section area of 0.0192
m x 0.0017m. And the end restraint condition used is pinned-pinned end condition, pinned-
fixed end condition and fixed-fixed end condition.
The standard code of practice for this laboratory is based on BS EN ISO 6892-1:2016.
This testing is to check the maximum critical load on an aluminum strut of how much pressure
can it withstand before the strut bended or collapsed. An ideal column is initially perfectly
straight, made up of homogenous material and the load is applied through the centroid of its
cross-section. However, a pin-connected column will buckle about the principle axis of the
cross section having the least moment of inertia.
 In this experiment, we used these equations:

𝜋𝐸𝐼
𝑃𝑒 = for pinned pinned end
𝐿2

2𝜋𝐸𝐼
𝑃𝑒 = for pinned fixed end
𝐿2

4𝜋𝐸𝐼
𝑃𝑒 = for fixed fixed end
𝐿2

1.4 Significance of laboratory testing

In this laboratory testing, students are able to experience themselves on what happens when a
strut has achieved its maximum load to carry before buckle. By applying the Euler Equation,
student can also determine the theoretical value of critical load and once the experiment is
done, students can identify the difference between theoretical and experimental values
Students can also find out the maximum critical load that can withstand against axial loading
with different conditions and materials by the end of the experiment. This laboratory is very
useful as student can implement this knowledge in real life especially in the construction
works to prevent from unwanted things to occur.
1.5 Theory
Buckling of columns is a form of deformation as a result of axial-compression forces. This
leads to bending of the column due to instability of the column. Quite often the buckling of a
column can lead to sudden and dramatic failure of a structure or mechanism and as a result,
special attention must be given to the design of columns so that they can safely support their
intended loadings without buckling. Failure buckling can happen in many ways in both beams,
the most well-known and common failure mode is flexural buckling or often referred as
buckling.

Figure shows types of end moments, instead for this experiment, we only use three which is pinned pinned, fixed pinned, and fixed
fixed end .
 CHAPTER 2
LITERATURE REVIEWS

2.1 Euler Theory

Euler’s theory predicts the axial compressive force required to initiate buckling in a long thin
strut. Ideally, a perfectly straight strut, when subjected to a purely compressive load, would
compress and it buckles. Buckling is therefore the result of imperfections that prevent the load
from being applied perfectly axially; e.g. from eccentric loading and lack of initial straightness.
The combined effects of these imperfections on overall buckling behavior is predictable when
long struts are to operate under elastic conditions.

2.2 Type of buckling

Buckling is an instability that leads to structural failure. The failure modes can in simple cases
be found by simple mathematical solutions. For complex structures the failure modes are found
by numerical tools.

When a structure is subjected to compressive axial stress, buckling may occur. Buckling is
characterized by a sudden sideways deflection of a structural member. This may occur even
though the stresses that develop in the structure are well below those needed to cause failure of
the material of which the structure is composed. As an applied axial load is increased on a
member, such as a column, it will ultimately become large enough to cause the member to
become unstable and it is said to have buckled. Further loading will cause significant and
somewhat unpredictable deformations, possibly leading to complete loss of the member's load-
carrying capacity. If the deformations that occur after buckling do not cause the complete
collapse of that member, the member will continue to support the load that caused it to buckle.
If the buckled member is part of a larger assemblage of components such as a building, any
load applied to the buckled part of the structure beyond that which caused the member to buckle
will be redistributed within the structure.
There are three ways a compression member can buckle or become unstable. These are flexural
buckling, torsional buckling and flexural-torsional buckling.

i. Flexural buckling
This type of buckling can occur in any compression member that experiences a deflection
caused by bending or flexure. Flexural buckling occurs about the axis with the largest
slenderness ratio, and the smaller cross-sectional area.

Figure 3.2 Flexural buckling of a column

ii. Torsional buckling

This type of buckling only occurs in compression members that are doubly-symmetric and
have very slender cross-sectional elements. It is caused by turning about the longitudinal axis.
Torsional buckling occurs mostly in built-up sections, and almost never in rolled sections.

Figure 3.3 Lateral torsional buckling

iii. Flexural-torsional buckling
This type of buckling only occurs in compression members that have unsymmetrical cross-
section with one axis of symmetry. Flexural-torsional buckling is the simultaneous bending
and twisting of a member. This mostly occurs in channels, structural test, double-angle shapes
and equal-leg single angles.
When a beam is being bent about its major axis, flexural torsional buckling may occur. As
shown in the figure 3.2, the beam deflects downwards, but at some stage buckling occurs over
the length of the member in which the cross section moves laterally. The buckling deformations
create bending about the minor axis ad occur over the entire length of the beam and hence
sometimes it is called member buckle and the associated strength is sometimes called member
strength. Figure 3.2 shows the flexural-torsional buckling of an RHS under experimental
conditions.

Figure 3.4 Flexural-torsional buckling

CHAPTER 3
METHODOLOGY

3.1 Flow-chart of the laboratory

3.2 Type of instrumentation

- Buckling of strut testing equipment
- Aluminum struts
- Allen key
- Digital vernier caliper
- Pinned end
-Fixed end
3.3 Procedures

1. Fit the bottom chuck to the machine and remove top chuck (to give two pinned ends).
Select the aluminum strut with length of 0.32mm and measure the cross section using
the digital vernier caliper provided and then calculate the second moment of area, I, for
the strut.
2. Adjust the position of the sliding crosshead to accept the strut using the thumbnut to
lock off the slider. Ensure that there is the maximum amount of travel available on the
hand wheel threat to compress the strut. Finally tighten the locking screw.
3. Carefully back-off the hand wheel so that the strut is resting in the notch but not
transmitting any load. Re-zero the force meter using the front panel control.
4. Carefully start to load the strut. If the strut begins to buckle to the left, “flick” the strut
to right and vice versa (this reduces any error associated with the straightness of strut).
Turn the hand wheel until there is not further increase in the load (the load may peak
and then drop as it settles in the notches).
5. Then, record the reading to analyze its data later and use another aluminum strut with
length of 0.37m, 0.42mm, 0.47m and 0.52m.
Part 2
1. To study the effect of end conditions, follow the same basic procedure as in part 1 but
this time remove the bottom chuck and clamp the specimen using the cap head screw
and plate to make a pinned-fixed end condition.
2. Record the result and calculate the value of 1/𝐿& for the struts.

3. Fit the top chuck with two cap head screws and clamp both ends of the specimen to
make a pinned-pinned end condition.
 CHAPTER 4
RESULTS, DATA ANALYSIS AND DISCUSSIONS

4.1 Results

Material of strut = Aluminium

Area Of Strut = 0.0192 m x 0.0017m

Height = 0.0017m
Base = 0.0192m

𝑏𝑑 3 (0.0192)(0.0017)3
I= = = 7.8608 𝑥 10−12 𝑚4
12 12

Modulus Of Elasticity (E) = 69 x 109 N/𝑚2

EI = 5.424 x 10−1 N𝑚2

Length of strut (m) 1

( 2)𝑚−2
𝐿
0.32 1 1
= = 9.77
𝐿2 0.322
0.37 1 1
= = 7.30
𝐿2 0.372
0.42 1 1
= = 5.77
𝐿2 0.422
0.47 1 1
= = 4.53
𝐿2 0.472
0.52 1 1
= = 3.70
𝐿2 0.522
Type of end moment 1: Pinned-pinned end moment

Length = 0.32m, 0.37m, 0.42m, 0.47m, 0.52m

𝑃e =
𝜋 2 𝐸𝐼
𝐿2

EI = 5.424 x 10−1 N𝑚2

𝜋2 (5.424 x 10−1 )
0.32m= 𝑃𝑒 = = 52.28 N
0.322
𝜋2 (5.424 x 10−1 )
0.37m= 𝑃𝑒 = = 39.10 N
0.372
𝜋2 (5.424 x 10−1 )
0.42m= 𝑃𝑒 = = 30.35 N
0.422
𝜋2 (5.424 x 10−1 )
0.47m= 𝑃𝑒 = = 24.23 N
0.472
𝜋2 (5.424 x 10−1 )
0.52m= 𝑃𝑒 = = 19.80 N
0.522
Type of end moment 2: Pinned-fixed end moment

Length = 0.32m, 0.37m, 0.42m, 0.47m, 0.52m

𝑃e =
2𝜋 2 𝐸𝐼
𝐿2

EI = 5.424 x 10−1 N𝑚2

2𝜋2 (5.424 x 10−1 )

0.32m= 𝑃𝑒 = = 104.68 N
0.322
2𝜋2 (5.424 x 10−1 )
0.37m= 𝑃𝑒 = = 78.86 N
0.372
2𝜋2 (5.424 x 10−1 )
0.42m= 𝑃𝑒 = = 60.69 N
0.422
2𝜋2 (5.424 x 10−1 )
0.47m= 𝑃𝑒 = = 48.46 N
0.472
2𝜋2 (5.424 x 10−1 )
0.52m= 𝑃𝑒 = = 39.59 N
0.522
Type of end moment 3: Fixed-fixed end moment

Length = 0.32m, 0.37m, 0.42m, 0.47m, 0.52m

𝑃e =
4𝜋 2 𝐸𝐼
𝐿2

EI = 5.424 x 10−1 N𝑚2

4𝜋2 (5.424 x 10−1 )

0.32m= 𝑃𝑒 = = 209.11 N
0.322
4𝜋2 (5.424 x 10−1 )
0.37m= 𝑃𝑒 = = 156.41 N
0.372
4𝜋2 (5.424 x 10−1 )
0.42m= 𝑃𝑒 = = 121.39 N
0.422
4𝜋2 (5.424 x 10−1 )
0.47m= 𝑃𝑒 = = 96.94 N
0.472
4𝜋2 (5.424 x 10−1 )
0.52m= 𝑃𝑒 = = 79.19 N
0.522
Data Analysis
Pinned-pinned end moment

Strut no Length(m) 1 Buckling Load Buckling

( 2)m−2
L
Theory(N) Load
Experiment(N)

1 0.32 9.77 52.28 86

2 0.37 7.30 39.10 60
3 0.42 5.77 30.35 57
4 0.47 4.53 24.23 38
5 0.52 3.70 19.80 31

Fixed-pinned end moment

Strut no Length(m) 1 Buckling Load Buckling Load

( 2)𝑚−2
𝐿
Theory(N) Experiment
(N)
1 0.32 9.77 104.68 144
2 0.37 7.30 78.86 127
3 0.42 5.77 60.69 114
4 0.47 4.53 48.46 83
5 0.52 3.70 39.59 66

Fixed-Fixed end moment

Strut no Length(m) 1 Buckling Load Buckling Load

( 2)𝑚−2
𝐿
Theory(N) Experiment
(N)
1 0.32 9.77 209.11 164
2 0.37 7.30 156.41 131
3 0.42 5.77 121.39 118
4 0.47 4.53 96.94 119
5 0.52 3.70 79.19 69
Discussion

1. The relation between buckling load and Euler Buckling load is that, Euler Buckling 𝑃𝑒
shows that if the length is longer, it will have a smaller buckling value. We know that struts
will fail by buckling when their critical load is reached. This shows that the longer the
struts, the smaller the load that is needed for it to buckle.

Strut no Length(m) Buckling 𝑃𝑒 (N)

Load
Experiment(N)

1 0.32 86 52.28
2 0.37 60 39.10
3 0.42 57 30.35
4 0.47 38 24.23
5 0.52 31 19.80

Pinned-Fixed End

Relation Between Experimental Buckling and Euler

Buckling

100

86
80

60 60 57
52.28
40 39.1 38
30.35 31
24.23
20 19.8
Theoritical Buckling Load(N) Experimental Buckling Load(N)
0
0.32 m 0.37m 0.42m 0.47m 0.52m

4. Euler Formula can predict buckling load as they show the same relation on every length that
is given. But a slight error must be considered and need to give attention although it can predict
the buckling load. In this experiment, the data we get is not totally accurate as there a some
error such as human error while handling and doing the experiment. And also error from the
apparatus that gave not totally accurate reading.
CHAPTER 5
CONCLUSION
What can we conclude IS that fixed-fixed end condition is the strongest end support to
withstand axial loading from buckling especially the short-length strut which is 0.32m with
the critical load value of 209.11 N. Meanwhile, the weakest end support among these is
pinned-pinned end support with the longest strut length 0.52m with the critical load value of
19.8N. Lastly, factor that affecting buckling behaviour is the length and type of strut used.
Euler's theory for buckling factor can be used to determine the buckling load while using
various type of end supports.

REFERENCES

[1] MS Gregory - 1960

[2] R. C. Hibbler, Mechanics of Materials, vol. 10th, Pearson Education Limited, 2018.
[3] Y. Ahmad, Mekanik Bahan dan Struktur, Universiti Teknologi Malaysia, 2001.
[4] D. W. A. Rees, Basic Solid Mechanics, Macmillan Press LTD, 1