School of Science and Engineering
EGR 2412L Mechanics of Materials Laboratory
Laboratory Experiment No. 1: Determination of the
Stress Withstood by a Specimen
Fall 2026
Team Members
Asmae El HARTI
Imane Abdoun
Ilyas Ourrai
Anwar Zizoune
Wari HIBATALLAH
Submission Date: September 22, 2026
Supervised by Ms. Noura Lasfar
Mechanics Laboratory
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Experiment 1
Objective
The objective of this experiment is to determine and compare the tensile stress withstood by
pure aluminum and H111 aluminum specimens using a Mechanical Tensile Testing Unit (MTTU)
(See Figure 1). For each applied load, the deformation is obtained from the difference between the
initial and final reading on the ruler that measure the elongation of the specimen. The corresponding tensile force is then calculated by multiplying this deformation by the spring constant of the
testing unit. Finally, the stress is determined by dividing the calculated force by the specimen’s
cross-sectional area.
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Theoretical Background
The specimen undergoes tensile deformation when an axial pulling force is applied at one end
while the opposite end is held fixed. In this experiment, the deformation, ∆L, is determined from
the difference between the initial ruler reading, RC0 , and the reading recorded after the load is
applied, RC:
∆L = RC0 − RC
Using Hooke’s law and the operating principle of the testing device, the corresponding tensile
force is calculated as follows:
F = K∆L,
K = 39 kN/mm
where K is the equivalent spring constant of the device. The normal tensile stress developed
in the specimen is then calculated by dividing the applied force by the cross-sectional area:
F
A
The cross-sectional area at the narrowest section of the specimen is determined from its width,
a, and thickness, e:
σ=
A = ae
When F is expressed in kilonewtons and A in mm2 , the resulting stress is expressed in kN/mm2 ,
which is equivalent to GPa.
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Apparatus and Experimental Procedure
3.1
Apparatus
The experiment was conducted using a manual tensile testing apparatus equipped with a springloading system and a ruler for measuring displacement. A digital caliper was used to measure the
width and thickness of the pure aluminum and H111 aluminum specimens.
3.2
Experimental Setup
Figure 1 shows an apparatus similar to the one used in the experiment. The specimen is secured
between a fixed grip and a movable grip. The load is applied manually by rotating the handle,
causing the springs to deform and producing a tensile force on the specimen. The corresponding
displacement is measured using the ruler attached to the apparatus.
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Figure 1: An experimental setup similar to the tensile testing apparatus used.
3.3
Experimental Procedure
Each specimen was placed in the MTTU and secured tightly using the grips. The tensile load
was applied by rotating the lever by one-third of a turn at each iteration. After every iteration, the
ruler reading was recorded and used to determine the deformation relative to the initial position.
This procedure was performed for 10 iterations for the pure aluminum specimen and 6 iterations
for the H111 aluminum specimen.
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Force Diagram
Figure 2 shows the free-body diagram of the specimen. The applied tensile force F acts at the
right end, while an equal reaction force R acts at the fixed left end.
F
Loaded end
A = ae
Narrow section
W = mg
Fixed end
R
Figure 2: Free-body diagram of the specimen subjected to tensile loading and its own weight.
The specimen’s weight is shown for completeness but neglected in the calculations because it
is insignificant compared with the applied force.
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Mechanics Laboratory
Experiment 1
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Measurements and Results
5.1
Pure Aluminum Specimen
A = 48.3mm2
RC0 = 87mm
Table 1: Measurements and calculated results for the pure aluminum specimen.
∆L = RC0 − RC
(mm)
-13
-14
-15
-24
-33
-43
-53
-63
-73
-83
RC
(mm)
100
101
102
111
120
130
140
150
160
170
5.2
|F |
(kN)
507
546
585
936
1287
1677
2067
2457
2847
3237
|σ|
(GPa)
10.5
11.3
12.1
19.4
26.6
34.7
42.8
50.9
58.9
67.0
H111 Specimen
A = 48.3mm2
RC0 = 90mm
Table 2: Measurements and calculated results for the H111 aluminum specimen.
∆L = RC0 − RC
(mm)
-10
-15
-20
-30
-40
-50
RC
(mm)
100
105
110
120
130
140
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|F |
(kN)
390
585
780
1170
1560
1950
|σ|
(GPa)
8.07
12.11
16.15
24.22
32.30
40.37
Sample Calculations
The cross-sectional area of each specimen is calculated using its width a and thickness e:
A = ea
A = 48.3 mm2
⇐⇒
The following calculations use the spring constant of the testing device:
K = 39 kN/mm
6.1
Pure Aluminum
For the first measurement of the pure aluminum specimen, the displacement is calculated as:
∆L = RC0 − RC
⇐⇒
∆L = 87 − 100 = −13 mm
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Mechanics Laboratory
Experiment 1
The negative sign indicates that the displacement occurs in the direction opposite to the selected
reference direction. Therefore, its magnitude is
|∆L| = 13 mm
The corresponding tensile force is:
|F | = K|∆L|
⇐⇒
|F | = (39 kN/mm)(13 mm) = 507 kN
Finally, the tensile stress is calculated as:
|σ| =
6.2
|F |
A
⇐⇒
|σ| =
507 kN
= 10.5 kN/mm2 = 10.5 GPa
48.3 mm2
H111 Aluminum
For the first measurement of the H111 aluminum specimen, the displacement is:
∆L = RC0 − RC
⇐⇒
∆L = 90 − 100 = −10 mm
Therefore, the magnitude of the displacement is
|∆L| = 10 mm
The corresponding tensile force is:
|F | = K|∆L|
⇐⇒
|F | = (39 kN/mm)(10 mm) = 390 kN
Finally, the tensile stress is:
|σ| =
7
|F |
A
⇐⇒
|σ| =
390 kN
= 8.07 kN/mm2 = 8.07 GPa
48.3 mm2
Plots
(a) Pure aluminum
(b) H111 aluminum
Figure 3: Tensile stress versus displacement for the pure aluminum and H111 aluminum specimens.
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Mechanics Laboratory
Experiment 1
(a) Pure aluminum
(b) H111 aluminum
Figure 4: Stress–strain curves for the pure aluminum and H111 aluminum specimens.
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Discussion
While the stress–displacement curves provide a clear idea of how the specimens deform under
the applied load, displacement depends on the initial geometry of the specimen. Therefore, it
cannot be used directly to generalize the material behavior. To obtain a more general measure of
deformation, we calculate the engineering strain by normalizing the elongation with respect to the
initial length:
∆L
.
L0
ε=
For a specimen subjected to an axial tensile force, the strain can also be expressed as
ε=
F
σ
∆L
=
= .
L0
AE
E
Based on the stress–strain curves, an approximately linear relationship between stress and
strain is observed for both specimens:
σ ∝ ε.
This linear relationship corresponds to Hooke’s law:
σ = Eε,
where E is Young’s modulus. Therefore, the slope of each stress–strain curve represents the
Young’s modulus of the tested material:
∆σ
.
∆ε
From the slope of the pure aluminum stress–strain curve in Figure 4a, the experimental Young’s
modulus is
E=
EAl = 70.232 GPa.
Similarly, the slope of the H111 aluminum stress–strain curve in Figure 4b gives
EH111 = 72.675 GPa.
The experimental Young’s modulus values are compared with the reference values as follows:
EAl,exp = 70.232 GPa,
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EAl,ref ≈ 70 GPa,
Mechanics Laboratory
Experiment 1
and
EH111,ref ≈ 69.2 GPa.
EH111,exp = 72.675 GPa,
The experimental values are therefore close to the expected values, although the H111 result
is slightly higher. The H111 specimen also has an experimental modulus approximately 3.5%
greater than that of pure aluminum, indicating a slightly steeper stress–strain response. The linear
behavior of both curves indicates elastic deformation, meaning that the specimens should return to
their original dimensions after unloading if their elastic limits were not exceeded. Small differences
from the reference values may result from ruler-reading uncertainty, dimensional measurements,
specimen alignment, grip slippage, or variations in the loading increments.
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The Handout Questions
Question 1: Draw a force diagram of the specimen undergoing tensile testing and specify all the
forces acting on it.
The required free-body diagram is presented in Figure 2. It includes the applied
tensile force, the reaction force at the fixed end, and the specimen’s weight. The
weight is neglected in the calculations because it is insignificant compared with the
applied tensile force.
Question 2: The test performed in this laboratory is called tensile testing. Why do you think it
is called so?
It is called tensile testing because the specimen is subjected to an axial pulling
force that produces tension and causes it to elongate. The test is used to study the
behavior of a material under uniaxial tensile loading.
Question 3: H111 is an aluminum derivative. What is the composition of H111?
H111 does not refer to a specific aluminum alloy or chemical composition. It is a
temper designation used for aluminum alloys that have undergone slight strain hardening after annealing.1 Therefore, the elements and their percentages depend on the
base alloy associated with the H111 temper. Common industrial examples include
aluminum 5754-H111, frequently used for flooring and treadplates, and aluminum
5083-H111, widely used in marine and chemical environments.
As an example, the approximate chemical composition of aluminum 5083-H111 is
presented in Table 3. The complete alloy reference is therefore required to determine
the exact chemical composition.
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Annealing is a heat-treatment process in which a material is heated and cooled under controlled conditions to
reduce internal stresses, soften the material, and improve its ductility.
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Mechanics Laboratory
Experiment 1
Table 3: Approximate chemical composition of aluminum 5083-H111 by mass.
Element
Magnesium (Mg)
Manganese (Mn)
Chromium (Cr)
Silicon (Si)
Iron (Fe)
Copper (Cu)
Zinc (Zn)
Titanium (Ti)
Aluminum (Al)
Composition by mass (%)
4.00–4.90
0.40–1.00
0.05–0.25
≤ 0.40
≤ 0.40
≤ 0.10
≤ 0.25
≤ 0.15
Remaining
Question 4: Which material appears to be stronger? Explain your answer using the experimental
results.
Based on the experimental stress–strain curves, H111 aluminum appears to be
slightly stiffer than pure aluminum because it produced a greater Young’s modulus:
EH111 = 72.675 GPa > EAl = 70.232 GPa.
This means that H111 aluminum undergoes slightly less elastic deformation under
the same applied stress. However, Young’s modulus measures stiffness rather than
material strength. A definitive comparison of strength would require determining
the yield strength or ultimate tensile strength by continuing the test until yielding
or failure.
Question 5: Why is stress measured at the narrowest section of the specimen, which is also its
middle section?
Normal tensile stress is defined as
σ=
F
.
A
For the same applied force, stress is greatest where the cross-sectional area is smallest. The narrow middle section is therefore designed to concentrate deformation
within a uniform and measurable region. It also encourages yielding or failure to
occur away from the grips, where the stress distribution may be affected by clamping.
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Conclusion
This experiment served as an introduction to the indirect measurement of tensile stress using
other measurable physical quantities. The displacement of the testing device, its spring constant,
and the cross-sectional area of each specimen were used to calculate the applied force and the
resulting tensile stress. This method was applied to pure aluminum and H111 aluminum specimens,
whose stress–strain curves showed an approximately linear response, indicating elastic behavior
over the tested range.
The experimental Young’s moduli were 70.232 GPa for pure aluminum and 72.675 GPa for
H111 aluminum. These values are close to the expected values, with H111 aluminum exhibiting
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Mechanics Laboratory
Experiment 1
slightly greater stiffness. However, since the specimens were not loaded until yielding or failure,
their ultimate strengths could not be directly compared to clearly established which of the two
material is stronger. The results may have been affected by measurement uncertainty, specimen
alignment, grip slippage, variations in the loading increments, and cyclic loading effects caused by
the repeated use of the same specimens.
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