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Tensile Test Report A+ Lab report
Report Tensile Testing A+ Lab report Material Science
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Engineering Materials (ENGN3601)
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Title: Tensile Testing
1. Objectives:
1. To obtain the following information from mild steel, Al alloy, PE and ABS.
(a) Engineering tensile stress
(b) Strain
(c) Yield stress
(d) Ultimate tensile stress
(e) Percent elongation, %EL
(f) Percent reduction in area, %AR
2. To compare the mechanical properties between mild steel and Al alloy and between
PE and ABS.
2. Introduction:
In this experiment, the main apparatus that used is the Universal Testing Machine which is to
undergo the elongation of a specimen until it finally reaches the point where it fractures. This
test is conducted to specimens that have uniform horizontal cross section in the Universal
Testing Machine that can show the load applied for any elongation is obtained from the
gauge length. Due to the increasing tensile load where force is applied uniaxial along the axis
of the specimen, the specimen will break. Then, the result of the tensile test is recorded on a
computer as load versus displacement or force versus elongation. These load-deformation
characteristics depend on the size of specimen. Through this useful machine, we can obtain
the tensile strength at yield and other mechanical properties of a specimen. There is an
extensometer on the Universal Testing Machine which used to determine the elongation and
tensile modulus.
i.) Engineering stress,𝜎 =
𝐹 𝐴
ii.) Strain, 𝜀 =
𝐿 1 −𝐿 0 𝐿 0
iii.) Tensile Strength at Yield, 𝜎𝑦 =
𝑙𝑜𝑎𝑑 𝑎𝑡 𝑦𝑖𝑒𝑙𝑑 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
iv.) Tensile Strength at Maximum, 𝜎𝑚=
𝑙𝑜𝑎𝑑 𝑎𝑡 𝑚𝑎𝑥 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
v.) Elongation and percentage elongation =
𝐿 1 −𝐿 0 𝐿 0
× 100%
vi.) Reduction and percentage reduction in area =
𝐴 1 −𝐴 0 𝐴 0
× 100%
Where
F = strength (N)
A 0 = original cross-sectional area (mm 2 )
A 1 = instantaneous cross-sectional area (mm 2 )
L 0 = original gauge length (mm)
L 1 = instantaneous gauge length (mm)
Figure 1: The shape of Mild Steel and Aluminium alloy rod
𝐴 0
Lo
BEFORE THE EXPERIMENT (metallic element)
𝐴 1
L 1
AFTER THE EXPERIMENT (metallic element)
Figure 2
The load-displacement graph for mild steel and aluminum alloy is illustrated as below:
For mild steel:
Graph 1: load-elongation graph
Where
FIGURE 4
FIGURE 5
When tensile load which is applied uniaxial along the long axis of a specimen is
slowly increasing, plastic undergoes the same process as the metal. But, plastic can undergo
either elastic deformation or plastic deformation or both deformations. Plastic deformation
invloves chemical bond break down or planes shear. Whereas, elastic deformation involves
only stretching of chemical bond with no chemical bond breal down or planes shear.
The strength – elongation graph for plastic is illustrated as below:
GRAPH 5
Where y = Yield strength
x = Fracture Strength
Apparatus:
The instruments that are used for this experiment are Universal Testing Machine, vernier
caliper, sharp knife and ruler.
The specimens that are used for the experiment are
a) Mild steel
b) Carbon steel
c) PE
d) ABS
Load
3. Procedures:
Metals
1. The inner diameter of mild steel are measured using a vernier calipers. The original gauge
length is measured using a ruler. Three readings are taken and the average value is calculated.
The data are recorded
2. Mild steel are divided into 1cm per part and marked by knife and a ruler.
3. Mild steel is placed securely into the Universal Testing Machine, the mild steel is screwed
perpendicularly for it to experience uniaxial pull.
4. The specimen experienced increasing tensile load and is pulled into failure.
5. The results are recorded by a computer and is printed out.
6. The mild steel is taken out from the Universal Testing Machine. The diameter and the
gauge length are measured again. The data are recorded.
7. All recorded data is tabulated.
8. Step 1 to 7 is repeated by using aluminium alloy as specimen.
Plastics
1. The width and thickness of Polyethlene (PE) is measured using a vernier calipers. The
original gauge length is measured using a ruler. Three readings are taken and the average
value is calculated. The data are recorded.
Specimen
Extensometer
Figure 3
Final gauge
length (mm)
70 70 70 70.
Percentage Elongation =
####### 𝑙1− 𝑙
####### 𝑙 0
𝑥 100%
=
####### 70−50.
####### 50.
𝑥 100%
= 40%
The stress and strain of mild steel are calculated and tabulated as follows:
Load (kN) Displacement (mm) Stress (MPa) Strain
0 0 0 0.
5 2 64 0.
15 4 194 0.
20 4 259 0.
25 7 324 0.
27 12 315 0.
30 17 360 0.
32 25 390 0.
33 28 396 0.
34 40 425 0.
32 55 445 1.
32 72 422 1.
31 73 420 1.
24 74 312 1.
𝑌𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ, 𝜎𝑦 =
𝑙𝑜𝑎𝑑 𝑎𝑡 𝑦𝑖𝑒𝑙𝑑
𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
=
25𝑘𝑁
7 𝑥 10−
= 324
Ultimate strength, σm =
𝑙𝑜𝑎𝑑 𝑎𝑡 𝑚𝑎𝑥
𝑐𝑟𝑜𝑠𝑠−𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
=
34𝑘𝑁
7 𝑥 10−
= 445
Breaking Stress, σ =
𝐿𝑜𝑎𝑑 𝑎𝑡 𝑏𝑟𝑒𝑎𝑘𝑖𝑛𝑔
𝐼𝑛𝑖𝑡𝑎𝑙 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
=
24𝑘𝑁
7𝑥 10−5 𝑚 2
= 312
ii) Aluminium alloy
a) Diameter of aluminium alloy
Reading 1 2 3 Average
Original diameter,𝑑 0 (mm) 9 9 9 9.
Final diameter,𝑑 1 (mm) 7 7 6 7.
Change in diameter of aluminium alloy, Δd= |7.06-9|mm=-2
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎, 𝐴 0
0
50
100
150
200
250
300
350
400
450
500
0 0 0 0 0 1 1 1 1.
Stress (MPa)
Strain
Stress-Strain Curve for Mild Steel
𝑌𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ, 𝜎𝑦 =
𝑙𝑜𝑎𝑑 𝑎𝑡 𝑦𝑖𝑒𝑙𝑑
𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
=
14𝑘𝑁
6 𝑥 10−
= 205
Ultimate strength, σm =
𝑙𝑜𝑎𝑑 𝑎𝑡 𝑚𝑎𝑥
𝑐𝑟𝑜𝑠𝑠−𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
=
15𝑘𝑁
6 𝑥 10−
= 223
Breaking Stress, σ =
𝐿𝑜𝑎𝑑 𝑎𝑡 𝑏𝑟𝑒𝑎𝑘𝑖𝑛𝑔
𝐼𝑛𝑖𝑡𝑎𝑙 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
=
12𝑘𝑁
6𝑥 10−5 𝑚 2
= 175
b) Plastics
i) Polyethlene (PE)
Initial (mm) Final (mm)
Reading width thickness width thickness
1 12 3 6 1.
2 12 3 6 1.
3 12 3 6 1.
0
50
100
150
200
250
0 0 0 0 0 0 0 0 0 0.
Stress (MPa)
Strain
Stress-Strain Curve for Aluminium Alloy
Average 12 3 6 1.
Original cross-sectional area, A 0
= 12 (3)
= 41 2
Final cross-sectional area, Ai
= 6 (1)
= 6 2
Percentage reduction in area:
𝐴𝑖 − 𝐴𝑜
𝐴 0
× 100%
=
####### 6−41.
####### 41.
× 100%
= 83%
Gauge length
Reading 1 2 3 Average
Original gauge length(mm) 50 50 50 50.
Fracture length(mm) 267 269 268 268.
The change in length of PE, ΔL = (268.0-50) mm = 218 mm
Percent Elongation:
= 𝐿
1 − 𝐿 0
𝐿 0
× 100%
=
268−50.
50.
= ×100% = 430%
Load(kN) Displacement(mm) Stress (MPa) Strain
0 0 0 0.
0 2 2 0.
0 4 11 0.
0 6 21 0.
0 10 22 0.
0 16 21 0.
0 20 20 0.
0 25 19 0.
0 34 11 0.
0 38 11 0.
0 155 11 3.
0 242 9 4.
0 245 7 4.
0 246 0 4.
= 13 x 3 = 4 x 10-5 m 2
Final cross-sectional area, Ai
= 12 x 3 = 4 x 10-5 m 2
Percent reduction in area:
= 𝐴
1 − 𝐴 0
𝐴 0
× 100%
=
####### 4−4.
####### 4.
= ×100% = 2%
The gauge length of ABS before and after tensile testing:
Reading 1 2 3 Average
Original gauge length(cm) 50 50 50 50.
Fracture length(cm) 51 51 51 51.
𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝐸𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛
= 𝐿
1 − 𝐿 0
𝐿 0
× 100%
=
51−50.
50.
× 100% = 2%
The stress and strain values of ABS are calculated, recorded and tabulated as below:
Load (kN) Displacement (mm) Stress (MPa) Strain
0 0 0 0.
0 0 2 0.
0 0 4 0.
0 0 9 0.
0 1 11 0.
0 1 14 0.
0 1 19 0.
0 1 21 0.
1 2 35 0.
1 3 39 0.
1 3 44 0.
1 3 40 0.
1 3 35 0.
1 3 23 0.
0 3 11 0.
0 4 0 0.
𝑌𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ, 𝜎𝑦 =
𝑙𝑜𝑎𝑑 𝑎𝑡 𝑦𝑖𝑒𝑙𝑑
𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
=
1𝑘𝑁
4 × 10−5 𝑚 2
= 38
Ultimate strength, σm =
𝑙𝑜𝑎𝑑 𝑎𝑡 𝑚𝑎𝑥
𝑐𝑟𝑜𝑠𝑠−𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
=
1𝑘𝑁
4 × 10−5 𝑚 2
= 44
Breaking Stress, σ =
𝐿𝑜𝑎𝑑 𝑎𝑡 𝑏𝑟𝑒𝑎𝑘𝑖𝑛𝑔
𝐼𝑛𝑖𝑡𝑎𝑙 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
=
1𝑘𝑁
4 × 10−5 𝑚 2
= 41
5. Discussion:
Metals
From the results of this experiment, the value of the yield stress is obtained by using
the 0% strain offset method. Yield point plays the important role in the graph of load
against displacement. Yield point is defined as the maximum point on the curve which ends
the elastic deformation and the start point for plastic deformation. The yield stress of the mild
steel is higher than that of aluminium alloy. This implies that mild steel can withstand higher
stress without experiencing plastic deformation. The breaking stress of mild steel is also
higher than that of aluminium alloy. This implies that mild steel is harder than aluminium
alloy. Mild steel has a higher strain values when compared to aluminium alloy. This means
0
5
10
15
20
25
30
35
40
45
50
0 0 0 0 0 0 0 0 0 0.
Stress (MPa)
Strain
Stress-Strain Curve for ABS
2. PE plastic is more ductile compared to ABS plastic. PE has larger area under the curve
thus it can absorb more energy before fracture. Furthermore, the higher value of percentage
elongation for PE compared to ABS means that PE can undergoes more plastic deformation
than ABS.
3. Endurance of materials increases in the following order:
PE < ABS < Aluminium alloy < Mild steel
7. References
1. William F, J. H. (2006). Foundations of Materials Science and Engineering, 4th
Edition. McGraw-Hill.
2. William D. Callister, J. (2002). Material Science and Engineering - An Introduction, 6th
Edition. John Wiley & Sons Inc.
3. Tensile testing - Wikipedia, the free encyclopedia. (2012). Retrieved from Wikipedia:
en.wikipedia/wiki/Tensile_testing
4. Tensile Strength Testing: What is a Tension Test? (2012). Retrieved from Instron:
instron/wa/applications/test_types/tension/default.aspx
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Tensile Test Report A+ Lab report
Course: Engineering Materials (ENGN3601)
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Name: ***********
1
Title: Tensile Testing
1. Objectives:
1. To obtain the following information from mild steel, Al alloy, PE and ABS.
(a) Engineering tensile stress
(b) Strain
(c) Yield stress
(d) Ultimate tensile stress
(e) Percent elongation, %EL
(f) Percent reduction in area, %AR
2. To compare the mechanical properties between mild steel and Al alloy and between
PE and ABS.
2. Introduction:
In this experiment, the main apparatus that used is the Universal Testing Machine which is to
undergo the elongation of a specimen until it finally reaches the point where it fractures. This
test is conducted to specimens that have uniform horizontal cross section in the Universal
Testing Machine that can show the load applied for any elongation is obtained from the
gauge length. Due to the increasing tensile load where force is applied uniaxial along the axis
of the specimen, the specimen will break. Then, the result of the tensile test is recorded on a
computer as load versus displacement or force versus elongation. These load-deformation
characteristics depend on the size of specimen. Through this useful machine, we can obtain
the tensile strength at yield and other mechanical properties of a specimen. There is an
extensometer on the Universal Testing Machine which used to determine the elongation and
tensile modulus.
i.) Engineering stress,𝜎 = 𝐹
𝐴
ii.) Strain, 𝜀 = 𝐿1−𝐿0
𝐿0
iii.) Tensile Strength at Yield, 𝜎𝑦=𝑙𝑜𝑎𝑑 𝑎𝑡 𝑦𝑖𝑒𝑙𝑑
𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
iv.) Tensile Strength at Maximum, 𝜎𝑚=𝑙𝑜𝑎𝑑 𝑎𝑡 𝑚𝑎𝑥
𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
v.) Elongation and percentage elongation = 𝐿1−𝐿0
𝐿0
×100%
vi.) Reduction and percentage reduction in area =𝐴1−𝐴0
𝐴0
×100%
Where
F = strength (N)
A0 = original cross-sectional area (mm2)
A1 = instantaneous cross-sectional area (mm2)
L0 = original gauge length (mm)
L1 = instantaneous gauge length (mm)
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