Maryam Jahanzad
KEM110702
TITLE: TENSILE TESTING
OBJECTIVE:
This experiment is carried out to determine mechanical properties of the specimens (mild
steel, aluminum alloy, polyethylene (PE) and Acrylonitrile Butadiene Styrene (ABS)). The
mechanical properties will be obtained from the experiment, are as follow:
- Engineering tensile stress.
- Strain.
- Yield stress.
- Ultimate tensile stress.
- Percentage of elongation.
- Percentage of reduction in area.
INTRODUCTION:
The tensile experiment is the most common mechanical test that reveals several important
mechanical properties. The material to be tested is formed into a shape suitable for gripping
in the testing machine, and then pulled at constant rate until it fractures.
The tensile instrument elongates the specimen at a
constant rate and has devices to continuously
measure and record the applied load and elongation
of the specimen. During the stretching of the
specimen, changes occur in its physical dimensions
and its mechanical properties. The ability to predict
the loads that will cause a part to fail depends upon
both material properties and the part geometry. This
experiment involves testing to determine the relative
properties.
From that information (stated before) we can determine following values;
Engineering tensile stress, 𝜎 =
𝐹
𝐴0
F = force
A0 = original length
pg. 1
(𝑁⁄𝑚𝑚2 )
Maryam Jahanzad
(i) Strain, 𝜀 =
KEM110702
𝐿1 − 𝐿0
𝐿0
L1 = length after strain
L0 = original length
(ii) Yield strength, 𝜎𝑦 =
𝑌𝑖𝑒𝑙𝑑 𝑤𝑒𝑖𝑔ℎ𝑡
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
(iii)Ultimate/maximum tensile stress, 𝜎𝑚 =
(iv) Percentage of elongation =
𝐿1 − 𝐿0
𝐿0
(v) Percentage of reduction of area =
(𝑁⁄𝑚𝑚2 )
𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑙𝑜𝑎𝑑
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
× 100%
𝐴0 − 𝐴1
𝐴0
× 100%
A0 = original area
A1 = area after stress
(a) Tensile Test for metal
d
A0
L0
A0
a
L1
pg. 2
(𝑁⁄𝑚𝑚2 )
Maryam Jahanzad
KEM110702
For tensile load and maximum load, it can be obtained from the graph of load (N) versus
elongation:
(i) Mild Steel
Fm
Fp
Fy
Load
Elongation
Fy = Yield load (N)
Fm = maximum load (N)
Fp = Tensile strength (N)
(ii) For aluminum alloy or any element which do not have fixed yield strength, the yield
strength is obtained using the “0.2% strain offset method”. This is also known as
“Tegasan pruf”. The graph is shown below.
Fm
Fp
F0.2%
Load
Elongation
pg. 3
Maryam Jahanzad
KEM110702
(b)Tensile Test for plastic
Figure shows the plastic specimen before and after deformation
A0
L0
before testing
A1
L1
Graph of stress-elongation for plastics is illustrated as follow:
σy
σb
MN/m
2
E
σy = yield strength
σb = Tensile strength
pg. 4
Maryam Jahanzad
KEM110702
F
F
A
L0
F
L
F
APPARATUS:

Specimens:
o mild steel
o aluminum alloy
o polyethylene (PE)
o Acrylnitrile Butadiene Styrene (ABS)

Universal testing machine

marker

vernier calipers

ruler
PROCEDURES:
(a) Tensile stress test for metals
1. The diameter, width and original length of the mild steel are measured.
2. 10 lines are marked with 10mm each part by using knife and ruler.
3. The mild steel is clamped in the machine.
4. The machine is then operated automatically.
pg. 5
Maryam Jahanzad
KEM110702
5. After the mild steel is fractured, its final diameter, width and length are measured.
6. Step 1-5 is repeated using aluminum alloy
7. The way of specimen break is observed.
(b) Tensile stress test for plastics
1. The width, breadth, length of the polyethylene (PE) is measured
2. 10 lines are marked with 10mm each part by using marker and ruler.
3. The polyethylene (PE) is clamped in the machine.
4. The machine is then operated automatically.
5. After the polyethylene (PE) is fractured, its final breadth, width and length are
measured.
6. Step 1-5 is repeated using Acrylnitrile Butadiene Styrene (ABS)
7. The way of specimen break is observed.
PRECAUTIONS:
a) Reading of the diameter and length is repeated several times to get the mean. This is
to reduce and minimize the parallax error.
b) Make sure the specimen is not being interrupt when the load is applied on specimen.
c) The specimens are clip tightly on the machine and make sure it is in the correct
position in order not to affect the reading.
d) Make sure the reading entered in the computer is correct.
e) Read the manual on how to use the Universal testing machine before use.
pg. 6
Maryam Jahanzad
KEM110702
RESULTS AND CALCULATIONS:
Specimen
Reading
First Reading
Initial
Mild Steel
Second
Third
Average
Reading
Reading
9.90
9.65
10.00
9.85
5.95
6.35
5.70
6.00
60.00
60.00
60.00
60.00
81.05
79.85
81.05
80.65
9.30
9.40
9.40
9.37
6.70
6.50
6.20
6.47
50.00
50.00
50.00
50.00
64.65
64.40
64.60
64.55
diameter, D0
(mm)
Final
diameter, D1
(mm)
Initial length,
L0 (mm)
Final length,
L1 (mm)
Aluminum
Initial
alloy
diameter, D0
(mm)
Final
diameter, D1
(mm)
Initial length,
L0 (mm)
Final length,
L1 (mm)
(a) Mild steel
From graph, we obtained:

Load = 50.00kN

Load at 0.2% strain = 26.049 kN

Maximum load = 34.290 kN

Fracture load = 24.290kN
Original area, A0 =
𝜋𝑑0 2
4
=7.62 × 10−5 𝑚2
pg. 7
Maryam Jahanzad
KEM110702
Engineering tensile stress, 𝜎 =
50×103
7.62×10−5
= 6.56 × 108 𝑁⁄𝑚𝑚2
Strain, 𝜀 =
80.65− 60
60
= 0.34
34.29×103
Ultimate stress, σm = 7.62×10−5
=450.0 MPa
Percentage of elongation =
80.65− 60
60
× 100%
= 0.34 × 100%
= 34%
Final area, A1 =
𝜋𝑑1 2
=
4
𝜋(6×10−3 )2
4
=2.83 × 10−5 𝑚2
Percentage decrease in area=
7.62×10−5 − 2.83×10−5
7.62×10−5
= 62.86%
pg. 8
× 100%
Maryam Jahanzad
KEM110702
From the data computed by computer, mild steel has:
Ultimate
Yield strength, σy/ MPa
stress/
tensile
strength, σm/ MPa
341.848
450.0
Fracture Strength, σp/ MPa
318.70
Data transferred from the Load – Displacement Graph obtained from computer: mild steel
Displacement (mm)
Load (kN)
Stress (MPa)
Strain
0.00
0.00
0.00
0.00
5.00
27.50
360.89
0.08
10.00
28.67
376.25
0.17
15.00
27.00
354.33
0.25
20.00
30.00
393.70
0.33
25.00
31.50
413.39
0.42
30.00
32.55
427.17
0.50
35.00
33.60
440.94
0.58
40.00
33.80
443.57
0.67
45.00
33.90
444.88
0.75
50.00
34.00
446.19
0.83
55.00
33.97
445.80
0.92
68.07
0.00
0.00
1.13
pg. 9
Maryam Jahanzad
KEM110702
Graph of stress against strain
600
500
400
stress
300
(MPa)
Ряд1
200
100
0
0
0,2
0,4
0,6
strain
pg. 10
0,8
1
1,2
Maryam Jahanzad
KEM110702
(b) Aluminum alloy
From graph, we obtained:

Load = 50.00kN

Load at 0.2% strain = 13.215 kN

Maximum load = 15.710 kN

Fracture load = 11.80 kN
Original area, A0 =
𝜋𝑑0 2
4
=6.89 × 10−5 𝑚2
Engineering tensile stress, 𝜎 =
50×103
6.76×10−5
= 7.26 × 108 𝑁⁄𝑚𝑚2
Strain, 𝜀 =
64.55− 50
50
= 0.291
Ultimate stress, σm =
15.71×103
6.9×10−5
=227.68 MPa
Percentage of elongation =
64.55− 50
50
× 100%
= 0.291 × 100%
= 29.1%
Final area, A1 =
=
𝜋𝑑1 2
4
𝜋7.082
4
=3.29 × 10−5 𝑚2
pg. 11
Maryam Jahanzad
KEM110702
Percentage decrease in area=
7.26×10−5 − 3.29×10−5
7.26×10−5
× 100%
= 56.82%
From the data computed by computer, aluminum alloy has:
Yield strength, σy/
Ultimate stress/ tensile strength, σm/
Fracture Strength, σp/
MPa
MPa
MPa
195.23
227.68
171.18
pg. 12
Maryam Jahanzad
KEM110702
Data transferred from the Load – Displacement Graph obtained from computer: aluminum
alloy
Displacement (mm)
Load (kN)
Stress (MPa)
Strain
0.00
0.00
0.00
0.00
2.50
10.13
146.81
0.05
5.00
14.38
208.41
0.10
7.50
14.75
213.77
0.15
10.00
15.00
217.39
0.20
12.50
15.10
218.84
0.25
15.00
15.25
221.01
0.30
17.50
15.30
221.74
0.35
20.00
15.60
226.09
0.40
22.50
15.65
226.81
0.45
25.00
15.71
227.68
0.50
27.50
15.50
224.64
0.55
31.43
0.00
0.00
0.63
pg. 13
Maryam Jahanzad
KEM110702
Graph of strees against strain
300
250
200
stress
150
(MPa)
Ряд1
100
50
0
0
0,1
0,2
0,3
0,4
strain
pg. 14
0,5
0,6
0,7
Maryam Jahanzad
KEM110702
(b) Tensile test for plastics
Specimen
Reading
Second
Third
Reading
Reading
10.28
10.28
10.27
10.28
3.20
3.15
3.20
3.18
50.00
50.00
50.00
50.00
191.00
190.00
188.00
189.7
3.3
3.25
3.25
3.27
1.40
1.55
1.62
1.52
10.33
10.34
10.33
10.33
10.31
10.20
10.20
10.24
49.80
49.80
49.80
49.80
L1 (mm)
49.70
49.90
50.00
49.87
Initial thick,
3.30
3.25
3.25
3.3
3.20
3.15
3.20
3.18
Polyethylene
Initial width,
(PE)
W0 (mm)
Final width,
First Reading
Average
W1 (mm)
Initial length,
L0 (mm)
Final length,
L1 (mm)
Initial thick,
T0 (mm)
Final thick, T1
(mm)
Acrylontrile
Initial width,
Butadiene
W0 (mm)
Styrene
(ABS)
Final width,
W1 (mm)
Initial length,
L0 (mm)
Final length,
T0 (mm)
Final thick, T1
(mm)
pg. 15
Maryam Jahanzad
KEM110702
Polyethylene (PE)
From graph, we obtained:

Load = 50.00kN

Load at 0.2% strain = 0.498 kN

Maximum load = 0.929 kN

Fracture load = 0.091 kN
Original area, A0 = 𝑡ℎ𝑖𝑐𝑘 × 𝑤𝑖𝑑𝑡ℎ
=3.36 × 10−5 𝑚2
50×103
Engineering tensile stress, 𝜎 =
3.36×10−6
= 14.87 × 108 𝑁⁄𝑚2
Strain, 𝜀 =
189.7 − 50
50
= 2.794
Percentage of elongation =
189.7 − 50
50
× 100%
= 2.794 × 100%
= 279.40%
Final area, A1 = 𝑡ℎ𝑖𝑐𝑘 × 𝑤𝑖𝑑𝑡ℎ
=4.834 × 10−6 𝑚2
Percentage decrease in area=
3.36 ×10−5 − 4.834×10−6
3.36 ×10−5
= 85.61%
pg. 16
× 100%
Maryam Jahanzad
KEM110702
From the data computed by computer, polyethylene has:
Yield strength, σy/ MPa
14.818
Ultimate stress/ tensile
strength, σm/ MPa
27.638
Fracture Strength, σp/ MPa
2.716
Data transferred from the Load – Displacement Graph obtained from computer: PE
Displacement (mm)
Load (kN)
Stress (MPa)
Strain
0.00
0.00
0.00
0.00
10.00
0.92
27.38
0.20
20.00
0.85
25.30
0.40
30.00
0.52
15.48
0.60
50.00
0.50
14.88
1.00
70.00
0.52
15.48
1.40
90.00
0.50
14.88
1.80
110.00
0.50
14.88
2.20
120.00
0.48
14.29
2.40
130.00
0.41
12.20
2.60
140.00
0.34
10.12
2.80
150.00
0.04
1.19
3.00
160.00
0.02
0.60
3.20
pg. 17
Maryam Jahanzad
KEM110702
Graph of Stress againts Strain
35
30
25
20
Stress (MPa)
15
10
5
0
0
0,5
1
1,5
2
Strain
pg. 18
2,5
3
3,5
Maryam Jahanzad
KEM110702
Acrylontrile Butadiene Styrene (ABS)
From graph, we obtained:

Load = 50.00kN

Load at 0.2% strain = 0.965 kN

Maximum load = 0.966 kN

Fracture load = 0.761 kN
Original area, A0 = 𝑡ℎ𝑖𝑐𝑘 × 𝑤𝑖𝑑𝑡ℎ
=3.41 × 10−5 𝑚2
Engineering tensile stress, 𝜎 =
50×103
3.41×10−5
= 14.67 × 108 𝑁⁄𝑚𝑚2
Strain, 𝜀 =
49.87 – 49.8
49.8
= 1.4 × 10−3
Percentage of elongation =
49.87 – 49.8
49.8
× 100%
= 1.4 × 10−3 × 100%
= 0.14%
Final area, A1 = 𝑡ℎ𝑖𝑐𝑘 × 𝑤𝑖𝑑𝑡ℎ
=3.26 × 10−5 𝑚𝑚2
Percentage decrease in area=
3.41 ×10−5 − 3.26×10−5
3.41 ×10−5
× 100% = 4.4%
From the data computed by computer, Acrylontrile Butadiene Styrene (ABS) has:
Yield strength, σy/ MPa
28.475
pg. 19
Ultimate stress/ tensile
strength, σm/ MPa
28.514
Fracture Strength, σp/ MPa
22.456
Maryam Jahanzad
KEM110702
Data transferred from the Load – Displacement Graph obtained from computer: ABS
Displacement (mm)
Load (kN)
Stress (MPa)
Strain
0.00
0.00
0.00
0.00
0.25
0.10
2.93
0.005
0.50
0.25
7.33
0.010
0.75
0.40
11.73
0.015
1.00
0.55
16.13
0.020
1.25
0.74
21.70
0.025
1.50
0.83
24.34
0.030
1.75
0.94
27.57
0.035
2.00
0.93
27.27
0.040
2.25
0.83
24.34
0.045
2.50
0.04
1.17
0.050
3.00
0.04
1.17
0.060
9.75
0.04
1.17
0.196
pg. 20
Maryam Jahanzad
KEM110702
Graph of stress against strain
30
25
20
15
stress
(MPa)
Ряд1
10
5
0
0
-5
pg. 21
0,05
0,1
0,15
strain
0,2
0,25
Maryam Jahanzad
KEM110702
DISCUSSION:
a) Tensile test for metals
Mechanical behaviour
Mild Steel
Aluminium alloy
Yields stress (MPa)
341.848
195.23
Ultimate
tensile
strength 450.00
227.68
(MPa)
Fracture stress (MPa)
318.70
171.18
% Elongation
34%
29.1%
1. Graph stress- strain for mild steel shown its experience a yield load. Whereas, aluminium
alloy did not experience a specify yield load. And therefore, “0.2% offset method” is
used.
2. From the result, the value of tensile stress, fracture stress and yield load for mild steel is
higher than that of aluminium alloy. This means that more force is required to elongate
the mild steel. This phenomenon is caused by the composition of the carbon and iron in
the mild steel which makes it harder.
3. As mild steel is an alloy that consists of carbon and iron. The atom size of carbon and
iron is different. Because of different size of atom, the atom in the mild steel is packed
more closely. Besides that, carbon atom (filler atom) is act as resistance for the
dislocation. Therefore the sliding of the atomic layer can be minimizing. Grain size atom
fills vacancy and interstitial sites so that the iron atoms cannot slide through each other
easily.
4. The area under the graph stress- strain represented the energy absorb by the metal to
deform. Since the area under graph for mild steel is greater than aluminium alloy,
therefore mild steel can absorb more energy.
pg. 22
Maryam Jahanzad
KEM110702
b) Tensile test for plastics
Acrylonitrile
Mechanical behaviour
Polyethylene (PE)
Yields stress (MPa)
14.818
28.475
strength 27.638
28.514
Ultimate
tensile
Butadiene
Styrene (ABS)
(MPa)
Fracture stress (MPa)
2.716
22.456
% Elongation
279.40%
0.14%
1. From the experiment, the mechanical properties of PE and ABS can be observed. PE
experiences large plastic deformation before it breaks. Whereas, ABS experience little
plastic deformation before it break. This can be said that PE is more ductile than ABS.
2. From the result obtained, yields stress for ABS is higher than PE. It shows that ABS
can afford a higher stress compare to PE before undergoing plastic deformations.
Besides that, stress at fracture for ABS also higher than PE. Therefore, ABS is
stronger, harder and stiffer compare to PE.
CONCLUSION:
a) Tensile test for metal

In conclusion, mild steel is stiffer, stronger, harder, more ductile and able to undergo
high stress before fracture than Aluminum alloy.
b) Tensile test for plastic

Polyethylene plastic is harder, stiffer and stronger than Acrylontrile Butadiene
Styrene (ABS).

But Butadiene Styrene (ABS) has higher ductility than Polyethylene plastic (PE).
pg. 23
Maryam Jahanzad
KEM110702
REFERENCE:
1. Rollason : Metallurgy for Engineering
2. William F.Smith & Javad Hashemi: Foundations of Materials Science and
Engineering 4th Edition.
3. Materials Science and Engineering An Introduction, 6th Edition; William D.
Callister, Jr. John Wiley & Sons, Inc. Chapter 6
pg. 24