University of the south pacific
School of engineering and physics
CV211 solid mechanics 1
Lab 2: Calculate forces in a simple cantilever truss and the modulus of elasticity of
a specimen (i.e. mild steel) using strain gauge
Lab session: Friday 5-8pm
Name: Torilofa Javin
ID: s11152656
Aim
This experiment is designed to learn about elasticity, strain gauges, modulus, and force computation.
Introduction
Trusses are structures composed of several structural parts joined together at pin joints or nodes. In
principle, pin joints are hinges since they offer no resistance to rotation. In practice, however, this is not
always the case. A truss' members are predominantly axially loaded, which is advantageous. This might
indicate that they are in tension, compression, or zero-force. Because of this, trusses are an extremely
efficient structural structure. [1]
applied Stress (σ) = applied force (F) / Area of application (A)
The amount of applied stress onto a surface area of material can be in (2) forms. The tensile stress where
the stress is a pull on the cross-sectional area of a material or referred to as positive. In addition, the
compressive stress is define as negative as the stress is a push onto an area of material. Where N/m2 is
the unit of stress
Strain is the response of a system to a stressful situation. Stress causes a material to deform when an
amount of force is applied to it. Engineering can be defined as amount of material deformation in the
direction of the applied force divided by the material's original length. This results in a value of one unit
less, which is commonly stated as meters per meter. Amount of the changes in the length divided by the
starting length is the strain in a bar stretched under tension, for example. The strain distribution of a
complicated structural element, such as stress, may or may not be uniform. [2]
Equipment


Loads
SM1009 (strain gauge trainer)
Procedure




A pin joint frame was developed using 45- and 90-degree angles in the framework. Each
framework element had force sensors glued to the surface.
An electronic load cell was employed to apply loads to the framework.
After that, the load cell was reset to zero and a 100N preload was added
The loads were then carefully applied at 100N intervals until 500N, with the strain in each
member AB, AC, and AD monitored using a force sensor and the findings calculated.
Result
Table 1: Strain.
Load (N)
0
100
200
300
400
500
Strain reading for
Members AB (µƐ)
-7
-17
-21
-29
-37
-45
Strain reading for
Members AC (µƐ)
23
48
73
97
122
147
Strain reading for
Members AD (µƐ)
-15
-31
-47
-64
-80
-97
True Strain for
Members AB (µƐ)
0
-7
-14
-22
-30
-38
Strain (True) for
Members AC (µƐ)
0
25
50
74
99
124
Strain (True) for
Members AD (µƐ)
0
-16
-32
-49
-65
-82
Table 2: Strain (True).
Load (N)
0
100.0
200.0
300.0
400.0
500.0
Table 3: Compare the theoretical and experimental forces
Members
AB
AC
AD
Experimental (N)
-267.4350
-873.6210
-576.4710
Theoretical (N)
0.0
707.1070
-500.0
GRAPH OF STRAIN VS
LOAD
600
500
500
LOAD (N)
400
400
Strain AD
300
300
Strain AC
200
200
100
100
0
82
65
49
32
16
0
0
20
40
60
80
100
120
140
STRAIN
Graph 1.0 depicts the connection or relationship between AC and AD load and true strain.
Sample calculation of the force AC (theoretical)


+↑ ∑F𝑌 = 0 = −𝑊 + 𝑅𝐴𝑌 = 𝑅𝐴𝑌 = 𝑊,
+→ ∑F𝑋 = 0 = 𝑅𝐵𝑋 + 𝑅𝐴𝑋 = −𝑅𝐴𝑋
Analysis of Pin joint at D
√2
 +↑ ∑F𝑌 = 0 = −W + 𝐹𝐴𝐶 ( 2 ) = FAC =

√𝟐
𝐖
𝟐
√2
+→ ∑F𝑋 = 0 = −𝐹𝐴𝐷 − 𝐹𝐴𝐶 ( 2 ) = 𝐹𝐴𝐷 =
√2
√2
√2
−𝐹𝐴𝐶 ( 2 ) = 𝐹𝐴𝐷 − ( 2 ) 𝑊 ( 2 ) = 𝑭𝑨𝑫 = −𝑾
Sample calculation of the force AC (experimental)
E=
σ
σ
= 210X109 =
= σ = 25.20X106
ɛ
120X10−6
σ=
𝐹
𝐹
= 25.2X106 =
= 𝑭𝑨𝑪= 𝟖𝟕𝟑
π(6.03X10−3 )2
A
4
Discussion
The goal of the lab was to investigate a mild steel specimen's modulus of elasticity and the amount of
strain possessed as a Pin Jointed Frame when a force is applied. Each member's stress were thoroughly
examined. The load versus strain graphs for the AC and AD members were created using the data
obtained throughout the experiment. The Pin Jointed Frame was put through its paces in several ways.
The difference in slope shown using the data obtained demonstrates this. Member AD had experience
larger strain than member AC. This conclusion may be derived from the slopes of the two graphs. The
theoretical and experimental force values calculated with the modulus of elasticity and the stress equation
varied somewhat. The disparity between the theoretical and actual results was attributable to errors made
during the experiment, such as incorrect strain gauge calibration or rounding off decimal digits.
Answer to question
The sort of force experienced by the member must be determined in order to establish whether it is in
tension or compression. The modulus of elasticity and stress formula were used to calculate the amount of
force experienced by the member joint. Members force AB and force AD were compressed, while
member AC possessed tension, according to the experiment. Member AB should be redundant, member
AD in compression, and member AC in tension, according to theoretical calculations.
Conclusion
Finally, the goal of this experiment was accomplished by performing a strain test on a mild steel Pin
Jointed Frame using a strain gauge. The findings of the experiment revealed that there are discrepancies
between the theoretical and actual values of forces computed using the modulus of elasticity and strain
formula. The discrepancy between these results was caused by a mistake made during the experiment.
Reference
[1 D. S. Carroll, "What is a Truss?," 5th January 2019 . [Online]. Available:
] https://www.degreetutors.com/what-is-atruss/#:~:text=A%20truss%20is%20a%20structure,members%20are%20predominantly%20axially%2
0loaded.. [Accessed 6 may 2022].
[2 i. s. university, "Engineering and True Stress and Strain," 2009. [Online]. Available:
] https://www.nde-ed.org/Physics/Materials/Mechanical/StressStrain.xhtml. [Accessed 4 may 2022].