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# UEME 1263 Solid Mechanics I Lab 2 Report ```UNIVERSITI TUNKU ABDUL RAHMAN
LEE KONG CHIAN FACULTY OF ENGINEERING AND SCIENCE
UEME1263 SOLID MECHANICS I
LABORATORY REPORT (II)
Experiment Title: Tensile Test
Date of Experiment: 11th March 2019 (Monday)
Student’s Name
Student ID no.
Course
Khoo Yi Feng
1700666
ME
Mok Hou Yin
1706035
ME
Ng Xing Yu
1800481
ME
Teoh Kok Siang
1601997
ME
Wan Shijun
1603870
ME
Practical Group: P2
Lecturer in Charge: Ir Dr. Kwong Kok Zee
Objective
To perform tension test on steel specimen according to ASTM E8, to determine stress-strain
relationship, yield strength, tensile strength, elongation and necking, modulus of elasticity
and rupture strength.
Introduction
When an object was applied to an external tensile loading, the metal will undergo
elastic and plastic deformation. Initially, the metal will elastically deform giving a linear
relationship of stress and strain. By knowing the force applied, cross-sectional area and the
𝑃
change in length of that object, the stress, 𝜎 = 𝐴 and strain, 𝜀 =
𝐿𝑓 −𝐿𝑜
𝐿𝑜
can be determined.
Figure 1: Stress-strain graph
The yield strength is defined as the stress at which permanent deformation starts to
occur. It is also a material constant that represents the limit of its elastic behaviour. Besides,
yield point is the point where plastic deformation begins. Before the yield point the material
deforms elastically and returns to its original shape when the applied stress is removed. Once
the stress applied exceeds the yield point, some fraction of the deformation will be permanent
and it is non-reversible. Tensile strength is the ratio of tensile force per unit cross sectional
area. As the load is continuously and incrementally applied, the stress-strain curve will reach
the maximum point, which is the ultimate tensile strength. At this point, the object can
withstand the highest stress before necking takes place. Young’s modulus measures the
ability of a material to withstand changes in length when undergo tension or compression.
During elastic deformation, the engineering stress-strain relationship follows the Hooke’s
𝜎
Law and the slope of the curve indicates the Young’s modulus (E), 𝐸 = 𝜀 . After necking,
plastic deformation is not uniform anymore and stress decreases accordingly until fracture.
The fracture strength can be calculated from the load at fracture divided by the original crosssectional area, A, by using this formula, 𝜎𝑓𝑟𝑎𝑐𝑡𝑢𝑟𝑒 =
𝑃𝑓𝑟𝑎𝑐𝑡𝑢𝑟𝑒
𝐴𝑜
.
Procedures
1. The diameter of the reinforced steel bar specimen was measured using a digital Vernier
calliper. The length of the steel bar was measured using a ruler. Each specific dimension
was measured three times and the average value of the three measurements was taken.
2. The gauge length on the steel bar specimen was measured by 5 cm each for both end
and marked by a marker pen.
3. The specimen was then placed in the loading machine, which was the Instron 5582
Universal Testing Machine.
4. The extensometer was attached to the specimen.
5. The load reading was set to zero, then load was applied at a rate less than 690 kPa/min.
Unless otherwise specified, any convenient speed of testing may be used up to half of
the specified yield strength or yield point, or one quarter of the specified tensile strength,
whichever was smaller. After the yield strength or yield point had been determined, the
strain rate may be increased to a maximum of 15 mm/min of the gauge length per
minute.
6. The tensile test was started.
7. The load was applied continuously to the specimen until it breaks.
8. The tested specimen was removed from the gripping head.
9. The stress versus strain curve and the results were recorded in the computer.
10. Steps 1 to 9 were repeated using the mild steel bar specimen.
Apparatus
1. There was a testing machine capable of applying tensile load at a controlled rate of
deformation or load. These types of machine can be classified into mechanical or closed loop
electro hydraulic. A dial gauge could be equipped onto the machine to indicate the load or
could be connected to a chart recorder or computer to measure and record the load and
deformation.
2. There was a gripping device. It was used to transmit the load from the testing machine to
the test specimen. Moreover, it was used to ensure axial stress within the gauge length of the
specimen.
3. An extensometer with an LVDT or dial gauge was used to measure the deformation of the
specimen during the experiment.
4. Callipers were used to measure the dimensions of the steel specimen.
Results:
Mild Steel:
162
9.98
Length(mm)
Diameter(mm)
163
9.95
163
9.97
Average
163
9.97
Original Area, 𝐴0 = 𝜋𝑟 2
= 𝜋(
9.97 2
)
2
= 78.07 mm2
Length(mm)
Diameter(mm)
176
5.35
175
5.37
175
5.39
Average
175
5.37
Final Area, 𝐴 = 𝜋𝑟 2
= 𝜋(
5.37 2
)
2
= 22.65 mm2
Reinforced Steel:
Length(mm)
Diameter(mm)
202
8.39
Original Area, 𝐴0 = 𝜋𝑟 2
= 𝜋(
8.39 2
)
2
= 55.29 mm2
203
8.37
203
8.40
Average
203
8.39
231
5.22
Length(mm)
Diameter(mm)
230
5.21
230
5.22
Final Area, 𝐴 = 𝜋𝑟 2
= 𝜋(
5.22 2
)
2
= 21.40 mm2
Calculations:
Calculate the stress and strain for each load increment until failure
Stress, 𝜎 =
Strain, 𝜀 =
𝑃
𝐴𝑜
𝐿𝑓 −𝐿𝑜
𝐿0
=
∆𝐿
𝐿𝑜
Young Modulus, 𝐸 =
𝜎
𝜀
Average
230
5.22
Mild Steel:
σ =
ε=
2200 N
78.07 mm2
175−163
163
= 28.18 MPa
= 0.0736
𝐸𝑒𝑥𝑝 = 1.882 &times; 1011 Pa
𝐸𝑡ℎ𝑒𝑜 = 2.05 &times; 1011 Pa
Force
2.2
pulled
(kN)
Tensile
28.18
Stress
(MPa)
Tensile
0.015
Strain (%)
Force
pulled
(kN)
Tensile
Stress
(MPa)
Tensile
Strain (%)
4.4
6.6
8.8
11
13.2
15.4
17.6
19.8
56.36 84.54
112.7
140.9
169.1
197.3
225.4
253.6
0.030 0.045
0.060
0.075
0.090
0.105
0.120
0.135
22
24.2
26.4
28.6
30.8
33
35.2
281.8
310.0
338.2
366.3
394.5
422.7
Plastic
deformation
0.150
0.165
0.180
0.195
0.210
0.225
Failure begun
Reinforced Steel:
σ =
ε=
2200 N
55.29 mm2
230−203
203
= 39.79 MPa
= 0.133
𝐸𝑒𝑥𝑝 = 1.551 &times; 1011 Pa
𝐸𝑡ℎ𝑒𝑜 = 2.10 &times; 1011 Pa
Force
pulled
(kN)
2.2
Tensile
Stress
(MPa)
Tensile
Strain
(%)
4.4
6.6
8.8
11
13.2
15.4
17.6
19.8 43.6
39.79 79.5
8
119.
4
159.
2
199.
0
238.
7
278.
5
318.
3
358.
1
Plastic
deformation
0.026 0.05
1
0.07
7
0.10
3
0.12
8
1.53
9
0.18
0
0.20
5
0.23
1
Failure
begun
Stress versus strain curve
Mild steel:
Reinforced steel:
Mild steel, Py = 30225 N
Reinforced steel, Py = 23538 N
Yield Strength
Mild Steel:
Yield Strength = 385 MPa
𝜎𝑌𝑆 (experimental result from graph) = 385 MPa
𝜎𝑌𝑆 (from formula) =
𝑃𝑦
𝐴𝑜
=
30225
78.07
𝜎𝑌𝑆 (theoretical) = 370 MPa
= 387 MPa
Reinforced Steel:
Yield Strength = 310
MPa
𝜎𝑌𝑆 (experimental result from graph) = 310 MPa
𝜎𝑌𝑆 (from formula) =
𝑃𝑦
𝐴𝑜
=
23538
55.29
𝜎𝑌𝑆 (theoretical) = 460 MPa
= 426 MPa
Mild steel, Pmax = 33.354 kN
Reinforced steel, Pmax = 27.470 kN
Ultimate Tensile Strength (UTS)
For mild steel
𝜎UTS (experimental from graph) = 425.53 MPa
𝜎UTS (from formula) =
𝑃𝑚𝑎𝑥
𝐴0
33.354 kN
= 78.07 mm2
= 427.23 MPa
𝜎UTS (theoretical) = 440 MPa
For reinforced steel
𝜎UTS (experimental from graph) = 361.96 MPa
𝜎UTS (from formula) =
𝑃𝑚𝑎𝑥
𝐴0
27.470 kN
= 55.29 mm2
= 496.83 MPa
𝜎UTS (theoretical) = 595 MPa
Elongation
For mild steel:
Original Length, Lo = 163 mm
Final Length, Lf = 175 mm
Therefore, the Elongation = [(Lf – Lo) / Lo] &times; 100
= [(175 – 163) / 163] &times; 100
= 7.36%
For reinforced steel:
Original Length, Lo = 203 mm
Final Length, Lf = 230 mm
Therefore, the Elongation = [(Lf – Lo) / Lo] &times; 100
= [(230 – 203) / 203] &times; 100
= 13.3%
Modulus of Elasticity
For mild steel:
E (experimental from graph) = 188171 MPa
E (theoretical) = 205 GPa = 205000 MPa
For reinforced steel:
E (experimental from graph) = 155148 MPa
E (theoretical) = 210000 MPa
Rupture Strength
For mild steel
Rupture Strength
1 MPa
𝜎𝑟𝑢𝑝𝑡𝑢𝑟𝑒 = 1 MPa
For reinforced steel
Rupture Strength
49 MPa
𝜎𝑟𝑢𝑝𝑡𝑢𝑟𝑒 = 49 MPa
Reduction of cross-sectional area
For mild steel:
Original Area, A0 = 78.07 mm2
Final Area, As = 22.65 mm2
Percent reduction in cross-sectional area = [
(𝐴𝑜 −𝐴𝑠 )
𝐴0
=[
] &times; 100
(78.07−22.65)
] &times; 100
78.07
= 70.99 %
For reinforced steel:
Original Area, A0 = 55.29 mm2
Final Area, As = 21.40mm2
Percent reduction in cross-sectional area = [
(𝐴𝑜 −𝐴𝑠 )
=[
𝐴0
] &times; 100
(55.29−21.40)
55.29
= 61.29 %
] &times; 100
Fracture strain
Mild steel
Fracture strain = 0.43
Reinforced steel
Fracture strain = 0.40
Work hardening exponent (n)
𝜎 = 𝐾𝜀 𝑛
where
σ represents the applied stress on the material,
ε is the strain,
and K is the strength coefficient (Kmild steel = 530 MPa, Kreinforced steel = 640 MPa)
𝜎
𝐾
= 𝜀𝑛
𝜎
log 𝐾 = log(𝜀 𝑛 )
𝜎
log 𝐾 = n log 𝜀
𝑛=
𝜎
𝐾
log
log 𝜀
Mild steel
𝑛=
385.601
530
log
log 0.39286
= 0.3404 ; the yield stress and yield strain were used in this calculation
Reinforced steel
𝑛=
310.145
640
log
log 0.39306
= 0.7758 ; the yield stress and yield strain were used in this calculation
Fracture mode and fracture surface
Mild steel
Fracture mode: Mode I – Opening
Fracture surface: Nearly to cup and cone surface
Reinforced steel
Fracture mode: Mode I – Opening
Fracture surface: Nearly to cup and cone surface
Percentage error
𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 (%) =
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 − 𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙
&times; 100
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙
Percentage error of maximum tensile strength
𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 (%)𝑚𝑖𝑙𝑑 𝑠𝑡𝑒𝑒𝑙 =
440 − 425.53
&times; 100 = 3.289 %
440
𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 (%)𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝑠𝑡𝑒𝑒𝑙 =
595 − 361.96
&times; 100 = 39.17 %
595
Percentage error of Young’s Modulus
𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 (%)𝑚𝑖𝑙𝑑 𝑠𝑡𝑒𝑒𝑙 =
205 − 188.171
&times; 100 = 8.209 %
205
𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 (%)𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝑠𝑡𝑒𝑒𝑙 =
210 − 155.148
&times; 100 = 26.12 %
210
Percentage error of yield’s stress
𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 (%)𝑚𝑖𝑙𝑑 𝑠𝑡𝑒𝑒𝑙 =
|370 − 385|
&times; 100 = 4.054 %
370
𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 (%)𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝑠𝑡𝑒𝑒𝑙 =
460 − 310
&times; 100 = 32.609 %
460
Discussion
In this experiment, tensile test experiment had been performed on a mild steel bar and a
reinforced steel bar.
After conducting the experiment, the data for mild steel bar had been tabulated and
some of the key properties were calculated. The mild steel bar used experienced a percentage
of elongation about 7.36% and percentage of reduction in cross-sectional area of 70.99%.
Besides, yield stress strength, 𝜎𝑦 , of 385 MPa, Young’s Modulus, E, of 1.882 &times; 1011 Pa and
ultimate tensile strength, 𝜎𝑈𝑇𝑆 , of 425.53 MPa were obtained from the experiment table.
However, there were some percentage errors occurred between experimental values and
theoretical. Lastly, the mild steel bar had rupture strength of 1MPa.
In the reinforced steel bar, the reinforced steel bar experienced a percentage of
elongation about 13.3% and percentage of reduction in cross-sectional area at 61.29%. From
the tabulated data and graph, the yield stress, 𝜎𝑦 , Young’s Modulus E, and ultimate tensile
strength, 𝜎𝑈𝑇𝑆 obtained from the experiment were 310 MPa, 1.551&times; 1011 Pa and 361.96 MPa
respectively. However, some percentage errors were obtained between experimental values and
theoretical. The reinforced steel bar had rupture strength at 49 MPa.
Due to the specimen was a ductile material, the group were able to observe every
phase that the specimen undergone under tensile load which were the elastic deformation,
plastic deformation, necking and fracture. However, the experimental results suggested that
uncertainties existed in the experiment because there was deviation of experimental result
from the theoretical one. These deviations were caused by the imperfections of the
experimental set up. One of the main imperfections was the diameter of the specimen at
breaking end (necking region) at one end was bigger than the other one. This indicates that
the initial diameter throughout the specimen itself was not uniform, which results in different
stress concentration throughout the entire specimen. Moreover, due to the reinforced steel
specimen had spiral extruded pattern, this factor affected the stress concentration within the
specimen as well. Not only that, the clamping of the specimen might not be rigid and strong
enough and this caused the specimen displaces when tensile load was applied. Consequently,
the stress on the specimen was scattered and inconcentrated. Lastly, the strength of the
specimen that used might be affected due to the specimen has been put for a long period and
the surface of the specimen was rusted.
1) What does the area under the stress-strain curve represent?
The area under the stress-strain curve shows the toughness of a material. The definition of
toughness is the ability of a material to absorb energy and plastically deform without fracturing.
Another definition of material toughness is the amount of energy per unit volume that a material
can absorb before rupturing. It is also defined as a material's resistance to fracture the object is
under stress. The toughness in the elastic region is called the modulus of resilience. Strength
and toughness are different in describing a material. The larger the elongation and stress applied
the higher the toughness requires. Thus, the area under the graph is larger.
2) How would you distinguish between “brittle” and “ductile” modes of failure of a
specimen under tensile test?
There are three ways to differentiate between “brittle” and “ductile” modes of failure
of a specimen under tensile test. Firstly, it can be distinguished by observing the stress-strain
graph. For ductile materials, they undergo extensive plastic deformation and energy absorption
before fracture occurs. In contrast, for brittle materials, they undergo little plastic deformation
and they have much lower energy absorption ability before fracture occurs. Hence, the stressstrain curve for ductile material must be extended longer than the one for brittle materials. The
area under the stress-strain curve as known as toughness of the material for ductile material
also must be larger than the one for brittle material. (People.virginia.edu, n.d.)
Figure 2: The stress-strain graph for brittle and ductile materials
In this experiment, by observing the stress-strain graphs plotted, the lengths of the
stress-strain curve for both materials were almost the same and it was uncertain to judge which
one was longer. However, the area under the stress-strain curve for reinforced steel was larger
than the one for mild steel. This indicates that reinforced steel is more ductile and can absorb
more energy than mild steel before fracture occurs in this way of analysis.
Next, the second way to differentiate the modes of failure is by observing the fracture
surface. If the fracture surface has cup and cone characteristics (Figure 3), this indicates that
the metal has high ductility. When the stress-strain curve has passed the ultimate tensile
strength (maximum point of the curve), necking starts to occur due to the uneven plastic
deformation. The formation of micro voids occurs at the necking region. The micro voids
become larger and consequently merge together with each other when the tensile load increases.
Crack with a plane normal to the external tensile load occurs. Before the specimen breaks, a
shear plane is formed along the outer surface of the specimen with a slanted angle of 45&deg; to the
tensile axis. Then, this shear plane merges with the cracks and the cup and cone surface is
generated.
Figure 3: Cup and cone surface for ductile fracture
In contrary, for brittle fracture, since the plastic deformation for this material is little,
the crack propagates rapidly and the direction of the propagation is nearly perpendicular to the
tensile axis. Metallic materials normally form crystalline solids and they have crystallographic
planes. In brittle material, the cracking mostly propagates through these specific
crystallographic planes and therefore, the breaking of the atomic bonds is more uniform. Hence,
the fracture surface of brittle materials is flat and smooth. (People.virginia.edu, n.d.)
Figure 4: Flat surface for brittle fracture
In this experiment, based on the observation (Figure 5 and 6), both of the steel
specimens tended to appear as cup and cone surface after fracture. This indicates that the
fracture was moderately ductile and both of the materials were ductile. Comparing the flatness
of the specimens, the fracture surface of the mild steel was flatter than the one of reinforced
steel. This indicates that reinforced steel is more ductile than mild steel in this analysis. This
method of analysis agrees with the previous method used.
Figure 5: The fracture surface of mild steel
Figure 6: The fracture surface of reinforced steel
Thirdly, the results of the elongation of length and reduction of cross sectional area are
helpful to determine the mode of fracture. The summary of the calculations for these two
parameters is shown below:
Elongation of length
Mild steel
Reinforced steel
7.36%
13.3%
Reduction of cross sectional 70.99%
61.29%
area
Table 1: The summary of the calculations for the elongation of length and reduction of cross
sectional area
In term of elongation of length, the elongation of reinforced steel is larger than the one
of mild steel. This indicates that reinforced steel was easier to be deformed permanently and it
had better ductility than mild steel. In term of reduction of cross sectional area, the reduction
of cross sectional area of mild steel is larger than the one of reinforced steel. This indicates that
mild steel was easier to be deformed permanently and it had better ductility than reinforced
steel. In short, both of these parameters contradicted to each other.
In fact, the ductility of the mild steel should be greater than reinforced steel because
mild steel has lower content of carbon (0.05% - 0.25%) (Quora, 2018) compared to reinforced
steel (0.2% - 0.4%) (Concretecorrosion.net, n.d.). Most of the mild steel can be cold formed
easily which means mild steel can be machined and fabricated under room temperature due to
its great ductility. However, reinforced steel has to be hot formed in the manufacturing
processes which means it is less ductile. In this experiment, the results were not expected with
the desired result due to the errors and uncertainties discussed previously.
3) What are the reasoning to prepare the gauge length region of a specimen (region away
from the gripped end-regions) with reduced cross-sectional area?
The cross-sectional area of the gauge length region is relative smaller than the
remaining region of the specimen is because deformation and failure are expected to be
occurred at the gauge length region. The gauge length region is where meausrements are made
and it is centered within the reduced region. The ends of the gauge length region and the
shoulders of the specimen have to be further enough with the purpose to prevent constrain
deformation of the larger ends within the gauge length region (Asminternational.org, 2004).
Moreover, the gauge length should be larger than the diameter of the specimen or else a more
complex stress mode other than simple tension will be occurred.
Conclusion
In conclusion, based on the experimental result, it showed that the mild steel bar can resist
higher stress compared to the reinforced steel. However, theoretically, reinforced steel should
have higher resistance to the high stress. Hence, the results of this experiment were not
reliable. This was caused by a few of uncertainties such as the uneven stress concentration on
the specimen, the condition of the specimen and also the status of clamping. Lastly, the
objective was achieved as the stress-strain relationship, yield strength, tensile strength,
elongation and necking, modulus of elasticity and rupture strength were obtained in this
experiment.
References
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How
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Materials
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at:
https://www.quora.com/What-is-the-difference-between-mild-steel-and-TOR-steel
[Accessed 17 Mar. 2019].
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