Introduction
A compressive member can fail in two ways. The first is via rupture due to the direct
stress and the second is by an elastic mode of failure called buckling. Short wide compressive
member tends to fail by material crushing. When buckling occurs the strut will no longer carry
any more load and it will simply continue to buckle i.e its stiffness then becomes zero and it is
useless as a structural member. Buckling, as opposed to simple strength problems such as
drawing, pressure, bending and shearing, is primarily a stability problem. Buckling problem
number among the best-known technical examples in stability theory. Buckling plays an
important role in almost every field of technology. Examples of this are:
- Columns and supports in construction and steel engineering
- Stop rods for valve actuation and connecting rods in motor construction
- Piston rods for hydraulic cylinders and
- Lifting spindles in lifting gear
A strut is a structural component designed to resist longitudinal compression. Struts
provide outwards-facing support in their lengthwise direction, which can be used to keep two
other components separate, performing the opposite function of a tie. When the cross-section
area is not large compared to the length i.e. the member is slender, and then the member will
generally fail by buckling well before the compressive yield strength is reached.
They are commonly used in architecture and engineering, and the term is particularly
frequently applied to components of automobile chassis, where they can be passive braces to
reinforce the chassis and or body, or active components of the suspension. Struts were commonly
used in early aircraft to support wings, stabilizations and landing gear.
In 18th-century, a mathematician named Leonhard Euler derived a formula that provides
the maximum axial load that a long, slender ideal column can carried without fail by buckling.
An ideal column that is perfectly straight, homogeneous, and free from initial compressive stress.
The maximum load, or known as critical load, causes the column undergoes unstable
equilibrium, that is, any increase in the loads or the introduction of the slightest lateral force will
cause the column to fail by buckling. The Euler formula for column is:
1
Pcr=π 2EI/(L2)
Where:
Cr: critical buckling load E: modulus of elasticity
I: area moment of inertia L: unsupported length of column
The notes below relate to uniform straight members made from homogeneous
engineering materials used within the elastic operating range.
Figure A: The type column and the buckling and deflection.
It is assumed that an end load is applied along the centroid of the ends. The strut will
remain straight until the end load reaches a critical value and buckling will be initiated. Any
increase in load will result in a catastrophic collapse and a reduction in load will allow the strut
to straighten. The value of the critical load depends upon the slenderness ratio and the end filing
conditions.
Theory
a) Applying the Buckling Theory:
If a rod is subjected to longitudinal forces, as implied in the sketch, it can fail in two
ways. On the one hand, it can be plasticized and flattened if its admissible compressive strain is
exceeded (see figure below). On the other hand, it is possible that it will suddenly shift to one
side and buckle before attaining the admissible compressive strain. This effect is called buckling.
2
The shape of the rod is the factor determines which of the two cases of failure will occur. A
slender, thin rod is more likely to buckle than a thick, stout rod.
Figure B: Stout and slender rod under compressive force.
b) Euler Formula:
Buckling occurs suddenly and without warning when a certain limit load is attained. It is
therefore an extremely dangerous type of failure, which must be avoided by all means. As, soon
as a rod begins to buckle, it will become deformed to the point of total destruction. This is typical
unstable behavior. Buckling is a stability problem. The critical limit load Fkrit, above which
buckling can occur, is dependent on both the slenderness of the rod, i.e. influence of length and
diameter, and the material used. In order to define slenderness the slenderness ratio l will be
introduced here.
In this case l k is the characteristic length of the rod. It takes both the actual length of the
rod and the mounting conditions into consideration.
Figure C: Euler cases of buckling.
3
For example, clamping the ends of the odds causes rigidity. The buckling length decisive
for slenderness is shorter than the actual length of the rod. Altogether, a differentiation is made
between four types of mountings, each having a different buckling length. The influence of
diameter in the slenderness ratio is expressed by the inertia radius i. It is calculated using the
minimum geometrical moment of inertia ly and the cross-sectional area A.
The influence of material is taken into consideration by the longitudinal rigidity of the
rod EA. Here, E is the modulus of elasticity of the respective material and A is the cross-
sectional area. The influence of various factors on the critical load are summarized in the so-
called “Euler formula":
or expressed in a different form:
Purpose of work:
The purpose is to determine the buckling load for a pinned ended strut. After that, a graph
of deflection versus (deflection/load) is plotted to calculate the theoretical critical buckling load
from the formula.
Objective
This laboratory test is conducted to determine the buckling load for a pinned ended strut.
4
Apparatus
Clamping Screws
Guide Columns
Dial Gauge
Steel Strut
Digital Indicator
Attachment
Sockets Spindle
5
Procedure
1. The digital indicator is switched on and warmed it up for at least 10 minutes.
2. A specimen is chosen and its length is measured. The width and thickness of the beam is
3mm and 25mm respectively.
3. The theoretical buckling load for a strut with pinned end condition is calculated. This is
to ensure that the load applied to the strut does not exceed the buckling load.
4. The grooved support is placed into the slot of the attachment for the end conditions and
the side screws are tightened. Refer appendix for proper installation of the support.
5. The top plate is moved upwards or downwards to bring the distance between the two
supports closer to the length of the strut.
6. The tare button on the digital indicator is pressed to set the reading to zero.
7. The specimen is placed in the groove of the top support.
8. While holding the specimen, the jack is adjusted so that the lower end of the specimen
just rests in the groove of the bottom support. (If the distance between the two supports is
slightly less than the length of the strut, the screw jack handle is turned in counter
clockwise. If the distance between the two supports is slightly greater than the length of
the strut, the screw jack handle is turned in clockwise.)
9. The reading on the digital indicator is noted. If the load is greater than 10N, the jack
handle is turned counter clockwise to bring it to less than 10N.
10. The position of the dial gauge is checked to ensure that it is at the mid-length of the
specimen. The dial gauge reading is set to zero.
11. The tare button is pressed to set the load indicator to zero.
12. The specimen is loaded in small increments by turning the screw jack handle slowly in
the clockwise direction.
13. For each load increment, the load and the corresponding mid-span deflection are
recorded. (Important: please ensure that the applied load is always less than 80% of the
buckling load.)
14. The specimen is unloaded by turning the jack handle in the counter clockwise direction.
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Data Collection & Recording
Length of member = 650 mm
Width of member = 25 mm
Thickness of member = 3 mm
Moment of inertia of member = 56.25 mm4
Dial gauge reading, 1 div = 0.01 mm
Load, P Mid-Span Deflection, d d/p
N div mm mm/N
0 0
11 25
21 32
31 40
41 50
52 59
60 66
70 75
80 89
90 107
100 120
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Data Analysis (Calculation & Results)
Length of member = 650 mm
Width of member = 25 mm
Thickness of member = 3 mm
Moment of inertia of member = 56.25 mm4
Dial gauge reading, 1 div = 0.01 mm
Load, P Mid-Span Deflection, d d/p
N div mm mm/N
0 0 0 0
11 25 0.25 0.023
21 32 0.32 0.015
31 40 0.40 0.013
41 50 0.50 0.012
52 59 0.59 0.011
60 66 0.66 0.011
70 75 0.75 0.011
80 89 0.89 0.011
90 107 1.07 0.012
100 120 1.20 0.012
Calculation of d/p (mm/N)
For 11N , For 21N , For 31N , For 41N ,
0.25 0.32 0.40 0.50
= 0.023 mm/N = 0.015 mm/N = 0.013 mm/N = 0.012
11 21 31 41
mm/N
For 52N , For 60N , For 70N , For 80N ,
0.59 0.66 0.75 0.89
= 0.011 mm/N = 0.011 mm/N = 0.011 mm/N = 0.011
52 60 70 80
mm/N
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For 90N , For 100N ,
1.07 1.20
= 0.012 mm/N = 0.012 mm/N
90 100
Graph Of Deflection Versus (Deflection/Load)
1.4
f(x) = − 0.82 ln(x) − 2.91
1.2
1
Deflection (mm)
f(x) = 43.93 x
0.8
0.6
0.4
0.2
0
0 0.01 0.01 0.02 0.02 0.03
Deflection/Load (mm/N)
Slope Intercept Therefore, the buckling load of
43.926 0 experimental, P is 43.926N.
Calculation of theoretical critical buckling load, Pcr
(π 2 × EI )
Formula: Pcr =
L2
b d3
Moment of inertia of member, I =
12
25× 33
I=
12
= 56.25 mm4
Modulus of Elasticity, E= 200x103 N/mm2
(π 2)( 200 x 10 3)(56.25)
Pcr =
6502
= 262.8 N
9
( Theoretical−Experimental
% Error =
Theoretical ) x 100
262.8−43.926
=( ) x 100 = 83.29%
262.8
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Discussion
The results obtained from this experiment shows definite error during the graph plotting where
most of the points do not intersect with the generated line of best fit. The slope value of the
experimental data has a big difference when compared to theoretical slope value which produces
a percentage error of 83.29%. When the experiment was conducting, we noticed a few factors
that might be the potential cause of error to occur, which are;
The screw handles are not tightened properly causing the specimen to be out of track
which leads to errors in deflection.
Measurements or readings are not taken properly which lead to parallax error. In such
cases where we have experienced is that one of the group members reads 7 from the
digital indicator but actually the reading is 9.
Systematic error which causes by the flaw of equipment used. In this case, we have
experienced inaccurate readings from the dial gauge.
Conclusion
A compressive member can fail in two ways. The first is via rupture due to the direct stress and
the second is by an elastic mode of failure called buckling. Short wide compressive member
tends to fail by material crushing. When buckling occurs the strut will no longer carry any more
load and it will simply continue to buckle i.e its stiffness then becomes zero and it is useless as a
structural member. Buckling, as opposed to simple strength problems such as drawing, pressure,
bending and shearing, is primarily a stability problem.
A strut is a structural component designed to resist longitudinal compression. Struts
provide outwards-facing support in their lengthwise direction, which can be used to keep two
other components separate, performing the opposite function of a tie. When the cross-section
area is not large compared to the length i.e. the member is slender, and then the member will
generally fail by buckling well before the compressive yield strength is reached.
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Figure D: Deflection of column
They are commonly used in architecture and engineering, and the term is particularly
frequently applied to components of automobile chassis, where they can be passive braces to
reinforce the chassis and or body, or active components of the suspension. Struts were commonly
used in early aircraft to support wings, stabilizations and landing gear.
The notes below relate to uniform straight members made from homogeneous
engineering materials used within the elastic operating range.
Figure A: The type column and the buckling and deflection.
It is assumed that an end load is applied along the centroid of the ends. The strut will
remain straight until the end load reaches a critical value and buckling will be initiated. Any
increase in load will result in a catastrophic collapse and a reduction in load will allow the strut
to straighten. The value of the critical load depends upon the slenderness ratio and the end filing
conditions.
For the prevention of steel from buckling. Let's analyse first the axial critical load
proposed by Euler for which buckling occurs.
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Fcr=π2 EI/ Leff
The variables you can change are E, I and Leff.
The first variable, the young's modulus E, is a material property. Thus, changing the
material will increase the load capacity of your column.
The second variable, the least second moment inertia I of your column, it is a geometrical
parameter. In the case of a circular column, your inertia will be I=πr4 / 4, so you can simply
increase the radius of the the cross-section.
The effective length, Leff, depends on the type constraints of the boundaries. You should
ideally decrease it, but it is not easy to be modified in a real structure.
Summarizing, the best you can do is improve your material properties (E) and increase
the dimensions of the column, increasing the moment of inertia.
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References
Nader Hoveidae, n.d. Global buckling prevention condition of all steel buckling restrained
braces. Retrieved from
https://www.researchgate.net/publication/277774250_Global_buckling_prevention_condition_of
_all_steel_buckling_restrained_braces.
Quora.com. What are the systems used to prevent buckling of columns of heights over 15m?
Retrieved from https://www.quora.com/What-are-the-systems-used-to-prevent-buckling-of-
columns-of-heights-over-15m.
StuDocu.com. Experiment 2- Buckling of Struts-2. Retrieved from https://www.studocu.com/en-
au/document/rmit/statics/practical/experiment-2-buckling-of-struts-2/1967987/view.
T. Usami, H.B. Ge and A. Kasai (October, 2008). OVERALL BUCKLING PREVENTION
CONDITION OF BUCKLINGRESTRAINED BRACES AS A STRUCTURAL CONTROL
DAMPER. Retrieved from http://www.iitk.ac.in/nicee/wcee/article/14_05-05-0128.PDF.
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Appendix
Figure 1: Fix the steel bar into the Figure 2: Record the data from the
buckling apparatus. dial gauge.
Figure 3: Strut buckling apparatus.
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