Al Maaref University
Faculty of Engineering
Department of Mechanical Eng
Course code: MEE322L
Course Title: Mechanics of Materials I Lab
Experiment Title: Tensile testing on metallic materials
Instructor :
Hussein Mouzanar
Group No.:
A
Students Name:
Mohammad Ali Nasreddine
Ahmad Fakih
Jawad Rachid
Yousef Kassir
Date of Testing:
17-6-2020
Date of Submission: 21-6-2020
1- Objectives
This tensile test was made on steel to:
•
Determine the modulus of elasticity of steel in addition to other properties
•
Learn how to use the Universal Testing Machine to test specimens in axial tension.
2- Theoretical Background
Tensile test is one of the most common tests for steel. The test involves straining a test piece by
tensile force, generally to fracture, for the purpose of determining tensile strength, yield strength,
ductility and reduction of area.
a) Modulus of elasticity:
The modulus of elasticity E can be determined by finding the slope of the elastic part of
the stress strain diagram; this part is linear and has the following relation:
𝜎 = 𝐸𝜀
Where:
[𝑁⁄𝑚2 ]
(1)
𝜎 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 [𝑃𝑎]
{ 𝜀 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 [𝑚𝑚/𝑚𝑚]
𝐸 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑐𝑖𝑡𝑦 [𝑃𝑎]
b) Yield Stress and Strain:
To find the yield stress and yield strain, the 0.2% offset method is used. Draw a line
parallel to the elastic part after moving 0.2% on the strain axis; the intersection of this
line with the curve will give the yielding point having the yield stress and the yield strain.
c) Ultimate Stress and Strain:
This will correspond to the stress when the loading force has its highest value.
d) Failure Stress and Strain:
•
To determine the failure stress, look at the stress reading just before the specimen
broke.
2
•
To determine the failure strain, two ways can be used:
1. Measure the final length (L) and initail length (L0) of the specimen, then use the
formula:
𝑑𝑢𝑐𝑡𝑖𝑙𝑖𝑡𝑦 =
𝐿−𝐿0
𝐿0
(2)
2. Use the graphicl method: At the failure point, draw a line parallel to the elastic
part of curve until it intersect the strain axis. The strain of the intersection point is
the permanent strain, also called ductility of the material.
3- Test Procedure
3-1 Test Setup
a) First we measured the dimesnions of the specimen using caliper meter
b) Place the specimen in the tensile machine using the grips
c) Ajust the extensometer on the specimen
d) Attach the machine to a given software on the computer to give the stess-strain graph
e) Adjust the testing machine to have suitable speed
f) The test is applied until the specimen breaks and the resulting graph is saved
3-2 Materials and Equipments
a) A cylindrical dogbone shaped steel bar is needed:
Figure 1. A cylindrical dogbone shaped steel bar
3
b) Note that in our case we weren’t able to get this shape so we used a normal steel
bar instead:
Figure 2. Circular cross sectional steel bar
c) A ruler and a caliper meter is needed to measure the dimensions properly:
Figure 3.. Caliper meter
d) A testing machine known as tensile machine:
Figure 4. Tensile machine
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e) An extensometer (of gage lenth=50mm) to get more accurate results in the elastic
zone of test:
Figure 5. Extensometer
f) A computer that is attached to the machine to visualize the graphs and variation of
the load with respect to displacement:
Figure 6. Computer
3-3 Tasks
First we measured the length using the ruler and the diameter using caliper meter; and
wrote down the measurements and the error values of the length, force and displacement.
Then we placed the steel bar in the machine using the grips in a way that it won’t slide.
We then placed the extensometer on the specimen to get more accurate values.
Afterwards, we closed the machine door to make sure that no one gets hurt during the
experiment. We then opened the software that is responsible to display the loaddisplacement graph, and entered the values of the dimensions and the speed upon which
the machine will run. Finally, we started the test and waited until the steel bar broke into
pieces.
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4- Results
1.
Dimensions of steel bar:
The steel bar had the following dimensions: {
Stress-Strain curve:
To get the stress-strain curve, you have to divide each force value by the original cross
sectional area of the specimen and divide the extension by the initial length of the bar.
800
700
600
500
Stress (MPa)
2.
𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 10𝑚𝑚
𝑔𝑎𝑔𝑒 𝑙𝑒𝑛𝑔𝑡ℎ = 219𝑚𝑚
400
300
200
100
0
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
Strain(mm/mm)
Figure 7. Stress-Strain diagram
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3.
Modulus of Elasticity:
Using equation (1), we can calculate Young’s Modulus. But first we have to zoom in into
the elastic zone of the curve as shown in figure 8.
700
600
Stress (MPa)
500
400
300
200
100
0
0
0.001
0.002
0.003
0.004
0.005
0.006
0.007
Strain
Figure 8. Elastic part of stress- strain diagram of steel
The young’s modulus is then:
𝐸=
4.
∆𝜎
400 − 200
=
= 200𝐺𝑃𝑎
∆𝜖 0.002 − 0.001
Yield Stress and Strain:
Using figure 8 and the 0.2% offset method discussed before, we can calculate the Yield
Stress and Strain:
𝜎𝑦 = 580 𝑀𝑃𝑎
{
𝜖𝑦 = 0.004
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5.
Ultimate stress:
Using figure 7:
𝜎
= 678 𝑀𝑃𝑎
{ 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒
𝜖𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 = 0.175
6.
Failure stress:
Using figure 7:
𝜎𝑓𝑎𝑖𝑙𝑢𝑟𝑒 = 380 𝑀𝑃𝑎
Ductility:
Using the graphical method -mentioned before in the theoretical background- we can
calculate ductility by drawing in figure 7 a line parallel to the elastic part from the rupture
point. The intersection with strain axis gives the ductility:
800
700
600
500
400
Stress (MPa)
7.
300
200
100
0
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
Strain(mm/mm)
𝐷𝑢𝑐𝑡𝑖𝑙𝑖𝑡𝑦 = 0.17 = 17%
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5- Discussion
1.
Steel Properties: the bar showed a great resistance to tension under the tensile loading of
speed 10mm/min. This explains the relatively high value of E. Thus steel is a hard and
strong material and have high stiffness. Comparing the values with the internet values,
the modulus of elasticity is exactly equal to that known of steel (200GPa).
2.
Errors: In any experiment there is an error percentage due to inaccuracy in
measurements and devices used.
a) Errors in length measuring that is found by dividing smallest unit on ruler over 2:
𝑤𝐿 = ±0.05𝑚𝑚
b) Errors in force measuring that is given by the tensile machine:
𝑤𝐹 = ±0.5𝑁
c) Errors in displacement that is given by the tensile machine:
𝑤𝛿 = ±0.001𝑚𝑚
d) Errors in area measurement:
𝜕𝐴
𝜋𝑑
𝑤𝐴 = √( )2 × 𝑤𝐿 2 = √( )2 × 0.052 × 10−6 = ±7.85 × 10−7 𝑚2
𝜕𝑑
2
Where:
𝑑 = 10𝑚𝑚
e) Errors in stress:
𝑤𝜎 = √(
𝜕𝜎 2
𝜕𝜎
1
−𝐹
) × 𝑤𝐹 2 + ( )2 × 𝑤𝐴 2 = √( )2 × 0.52 + ( 2 )2 × 𝑤𝐴 2
𝜕𝐹
𝜕𝐴
𝐴
𝐴
Where:
𝜋𝑑 2
𝐴=
= 7.85 × 10−5 𝑚2
4
{
𝐹
𝜎=
𝐴
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1. For Yield stress:
𝑤𝜎 = ±5.8 𝑀𝑃𝑎
2. For Ultimate stress:
𝑤𝜎 = ±6.78 𝑀𝑃𝑎
f) Errors in strain:
𝑤𝜖 = √(
𝜕𝜖 2
𝜕𝜀
−𝛿
1
) × 𝑤𝐿 2 + ( )2 × 𝑤𝛿 2 = √( 2 )2 × 𝑤𝐿 2 + ( )2 × 𝑤𝛿 2
𝜕𝐿
𝜕𝛿
𝐿
𝐿
Where:
𝐿 = 0.219 𝑚 𝑖𝑠 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑏𝑎𝑟
𝛿 𝑖𝑠 𝑡ℎ𝑒 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 (𝑚)
{
𝛿
𝜀=
𝐿
1. Yield strain:
𝑤𝜖 = 4.656 × 10−6
2. Ultimate Strain:
𝑤𝜖 = 4.021 × 10−5
g) Errors in modulus of elasticity:
𝑤𝐸 = √(
𝜕𝐸 2
𝜕𝐸
1
−𝜎
) × 𝑤𝜎 2 + ( )2 × 𝑤𝜖 2 = √( )2 × 𝑤𝜎 2 + ( 2 )2 × 𝑤𝜀 2 = ±0.9𝐺𝑃𝑎
𝜕𝜎
𝜕𝜖
𝜀
𝜀
Where:
𝐸 𝑖𝑠 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 (𝑃𝑎)
{𝜎 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 (𝑃𝑎) 𝑎𝑡 𝑎𝑛𝑦 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝑝𝑎𝑟𝑡
𝜀 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑡𝑟𝑎𝑖𝑛 𝑜𝑓 𝑡ℎ𝑎𝑡 𝑝𝑜𝑖𝑛𝑡
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3.
Reasons for errors:
i.
Sliding of the bar during the test because of rust and the deformed shape
ii.
Inaccuracy of measurments of devices used for measurement and the T.M.
6- Conclusions
In this experiment, we were able to calculate the modulus of elasticity of a circular cross section
steel bar upon doing the tensile test. We were also able to find other properties as yielding,
ultimate and failure stress. The results were reasonable and the error values were calculated for
each.
7- Appendix
. The following figure shows the steel bar during the test:
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