09.yasmine material lab report
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Department of Applied Chemistry
Division of Science and Engineering
SCHOOL OF ENGINEERING
ENGINEERING MATERIAL 100
Experiments 4 and 6
Mechanical Testing and Applications of Non-Metals
Name: Yasmin Ousam Shafei Mahmoud Khalil
Student ID: 7E0A7588 / 14982477
Group: B2
Due Date: May 25, 2010
Lecturer: Dr. Zeya
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Lab Assignment IIExperiments 4 and 6
By: Yasmin Khalil
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Table of Contents
0.0. Abstract 4
1.0. Introduction ....... 4
Experiment 4: Mechanical Testing 4 - 10
2.0. Aim ..4
3.0. Introduction...4
4.0. Objective..4
5.0. Theory 5 - 6
5.1. Proof stress 5
5.2.
Stress. 5
5.3. Strain. 5
5.4. Youngs Modulus of Elasticity5
5.5. Tensile Strength. 5
5.6. Yield Strength.. 5
5.7. Yield Point.. 5
5.8. 0.2% Proof Stress.. 6
5.9. Ductility6
5.10.Elastic Deformation...6
5.11.Plastic Deformation6
6.0. Apparatus 7
7.0.Procedure. 7
8.0. Observations and Results. 7 - 10
8.1.Tensile test results..7
8.2.Elongation results... 8
8.3.
Fracture Appearance. 88.4.Data from the curve..9
8.5.Calculated Values.9
8.6.Calculations.. 9
8.7.Youngs Modulus .10
9.0.Graphs..10
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Lab Assignment IIExperiments 4 and 6
By: Yasmin Khalil
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5.0.Theory5.1.Proof stressis a stress that causes a system to be in a specified small, permanent
deformation and may result in the extension of a tensile test piece or material. It can
also be defined as the stress used to demonstrate the materials ability to persist
service loads. The proof stress will produce 0.2 % extension for steel quoted in N/mm2.This value is close to the yield stress in materials which dont exhibit a definite yield
point.
5.2.Stress is the measure of the average force that acts perpendicularly to the surface ofthe body per unit area in a body. It has the SI unit of Pascal (symbol Pa or N/m
2).
5.3.
Strain is the deformation unit under a load. In other words, it is the measure of thechange in length of a body divided by its original length when a there is a force acting
on the body which causes the changes. It has no units.
5.4. YoungsModulus of Elasticity is equal to stress ( ) over strain ( ). It is the measure of
the stiffness of slope of the graph strain versus stress. Youngs Modulus ( ) has the
unit of Pascal (symbol Pa or N/m2). A low modulus means that the specimen or
structure will be flexible while stiff and inflexible will occurred in a high modulus.
5.5. Tensile Strength is the maximum force of load applied to the specimen before it
fractures divided by the original cross sectional area. The ultimate tensile strength of a
material represents the maximum stress that a material can withstand before its
deformation when a force is applied on it.
5.6. Yield Strength is the stress at the yield point which is defined as the stress required to
start a particular amount of plastic deformation when the material is loaded. The
material is elastic below the yield strength and is viscous above the yield strength.
5.7.Yield Point is the point indicated when the permanent deformation of a stressed
material will occur; it is the end of the elastic region. Yield Point is also known as the
elastic limit from the stress-strain curve which can represent the initial progress from
linearity of the curve.
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Lab Assignment IIExperiments 4 and 6
By: Yasmin Khalil
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6.0.Apparatus
Carbon Steel and Caliper Computer Universal Testing Machine
7.0.Procedure1-Use the Caliper to measure the diameter of carbon steel.
2-Measure the gauge length.
3-
Insert the carbon steel rod in the Universal Testing Machine.4-Tension the cross arm of the machine and reset the machine settings in a way that it
will be equal to zero.
5-Observe and record the elongation data from the graph in the computer screen and
take 10 points from elongation data and graph them in a graph paper.
8.0.Observations and Results
8.1. Tensile test results
SampleMeasurements Carbon Steel Bar
Gauge Length Diameter (mm) 12.5
Minimum Diameter , after testing (mm) 0.75
Gauge Length, Lo(mm) 300200 = 100
Gauge Length, Ll (mm) 337- 200 = 137
Table 1: The Tensile test results for Carbon Steel
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Lab Assignment IIExperiments 4 and 6
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8.2.Elongation Result for Carbon Steel
8.3.
Fracture Appearance
Sample Picture Sketch Description
Carbon Steel Rod The carbon steel rod has
a relatively strong
resistance to the tensile
force applied to it. It
broke two parts, where
the upper section formsa cup at the breaking
region and the bottom
section forms a cone.
Table 4: Fracture Appearance of carbon steel
Load (kN)
Elongation
(mm)
3.3 0.24.2 0.5
4.7 0.6
15.4 2.9
24.0 4.4
25.2 5.0
25.0 5.5
24.9 6.3
29.9 11.1
35.0 19.637.2 34.9
37.1 37.5
35.8 40.0
30.0 42.5
Table 2: Elongation result for
carbon steel
Stress (MPa) Strain
26890.82 0.00234146.34 0.005
38211.38 0.006
125203.25 0.029
195121.95 0.044
204878.05 0.050
203252.03 0.055
202439.02 0.063
243089.43 0.111
284552.85 0.196302439.02 0.349
301626.02 0.375
291056.91 0.400
243902.44 0.420
Table 3: Stress and Strain for
the Elongation of carbon steel
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8.4.Data from the curve
Yield Point Load (kN) 24.9
0.2% Proof Load (kN) 25
Maximum Load (kN) 37.2
Table 5: Data of Carbon steels curve
8.5. Calculated values
0.2% proof stress or Y.P Stress (MPa) 2.02*10 14 MPa
Tensile Strength (MPa) 3.02*10^14 MPa
% elongation 99.64 %
% R of Area 37 %
Table 6: Calculated values of Carbon steel
8.6. Calculations
Cross Sectional Area (A0)=
= (.0125/2) ^2 * Pi
= 1.23*10 ^-4 m^2
Final Cross Sectional Area (Af) =
= ((7.5*10^-4)/2)^2 * Pi
= 4.42*10^-7 m^2
0.2% Proof Stress or Y.P. Stress (MPa)=
=(24.9 *10^3) N/ (1.23*10 ^-4) m^2
=2.02*10 ^14 MPa
Tensile Strength (MPa) =
= (37.2 *10^3) N/ (1.23*10 ^-4) m^2
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= 3.02*10^14 MPa
% Elongation= 100 %
=((137100)/100) * 100
= 37 %
% R of A = 100 %
= [(1.23*10 ^-4 m^2)(4.42*10^-7 m^2)/ (1.23*10 ^-4 m^2)]* 100
=99.64 %
Stress=
= 24.9*10^3 / 1.23*10 ^-4
= 202439.02 MPa
Strain=
= (6.3*10^-3 / 0.1)
= 0.063
Youngs Modulus=
= 202439.02 MPa/0.063
= 3213.32 GPa
8.7. Youngs Modulus
Load at Yield Point (kN) 24.9
Extension at Yield Point (mm) 6.3
Stress (MPa) 202439.02 MPa
Strain (m/m) 0.063
Youngs Modulus (GPa) 3213.32 GPa
Table 7: Carbon Steels Youngs Modulus
9.0.GraphsPlease refer to the attached graph papers. [Graphs 1 and 2]. Thank you.
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Lab Assignment IIExperiments 4 and 6
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Tensile Test
14.0.Apparatus
15.0.
16.0.
17.0.
18.0.
19.0.
15.0. Procedure1- Use the Caliper to measure the length, thickness and width of the Nylon piece.
2- Insert the Nylon piece in the Universal Testing Machine.
4-Tension the cross arm of the machine and reset the machine settings in a way that it
will be equal to zero.
5-Observe and record the elongation data from the graph in the computer screen and
take 10 points from elongation data and graph them in a graph paper.
16.0. Observations and Results16.1. Tensile Test Results
Sample
Measurements Nylon
Gauge Length (mm) 134.92
Original Length (mm) 179.88
Final Length (mm) 209.81
Gauge Thickness (mm) 3.04
Final Thickness (mm) 1.1
Gauge Width (mm) 11.48
Final Width (mm) 7.4
Cantilever and Nylon Computer Screen Universal testing Machine
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Reduction in Width (mm) 11.487.4 = 4.08
Reduction in Thickness (mm) 3.041.1 = 1.94
Table 8: The Tensile test results for Nylon
16.2. Elongation Data for Nylon
16.3. Failure Appearance
Sample Picture Sketch Description
Nylon The nylon stick broke
into two parts in a
ductile mode. There is anecking part at the
breaking of each part,
with curves in upward
and downward
directions.
Table 10: Fracture Appearance of Nylon
Load (N)
Machine
Extension (mm)
300 0.1
302 1.2
601 2.9
838 6.9
900 11.4
902 11.5
902 12.8
902 13.6
902 14.4
904 15.6
898 15.4
828 28.5
733 40.2
578 47.1
535 52.5
255 58.1
Table 9: Elongation result for Nylon
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Lab Assignment IIExperiments 4 and 6
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16.4. Data from Curve
Yield Point Load (kN) 838
Tensile strength (MPa) 34.9
Maximum Load (kN) 11.4
Table 11: Data of Carbon steels curve
16.5. Calculated Values
% elongation 16.64
% R of Area 76.68
Table 12: Calculated values of Carbon steel
16.6. Calculations
Tensile Strength = Width x Height
= 0.01148 * 0.00304
= 3.45 x 10^-5 m
= 34.8992 = 34.9 MPa
% Elongation =
x 100 %
= ((209.81 - 179.88)/ 179.88) x 100
= 16.64 %
% R of A =
x 100 %
= ((3.04 x 11.48)(1.1 x 7.4)/(3.04 x 11.48)) x 100
= 76.68 %
Stress =
= (838/(0.0034 x 0.01148)
= 21.47 MPa
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Strain =
= (6.9/179.88)
= 0.0384
Youngs Modulus =
= (21.47 MPa /0.0384)
= 0.56 GPa
16.7. Youngs modulusLoad at Yield Point (kN) 838
Extension at Yield Point (mm) 6.9
Stress (MPa) 21.47
Strain (mm/mm) 0.0384
Youngs Modulus (GPa) 0.56
Table 13: Carbon Steels Youngs Modulus
17.0. GraphsPlease refer to the graph papers attached. [Graph 3].Thank you.
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Lab Assignment IIExperiments 4 and 6
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Cantilever Bend Test
18.0. Apparatus
19.0. Procedure
1- Measure the thickness, width and length of the cantilever beam.
2- Apply a 10g load to the free end of the cantilever beam.
3- Measure the bending/deflection by the digital meter.
4-
Add 10g loads until you have noted down 15 loads vs. deflection readings.
20.0.Observations and Results
20.1.Sample and MeasurementsSample and Measurements
Load (g) Deflection
(mm)
Load (g) Deflection
(mm)
Load (g) Deflection
(mm)
10 0.0 60 1.11 110 1.85
20 0.37 70 1.19 120 2.18
30 0.54 80 1.30 130 2.35
40 0.74 90 1.50 140 2.55
50 0.78 100 1.63 150 2.71
Table 14: Cantilever beam load vs. Deflection points
Loads Cantilever beam test machine
Aluminum Bar
Hanger with loads
Digital meter
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Lab Assignment IIExperiments 4 and 6
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20.2.
Measurements (m)
Measurement (m)
Thickness of Specimen 0.006
Width of Specimen 0.019
Effective Length of specimen 0.600
Table 15: Cantilever beam specification
The formula below is used to find the youngs modulus of a material.
d
M
wt
gt
dwt
MgtE
3
3
3
344
Here d
M
is the inverse of the deflection versus load graph
Slope = (1.741.18)/ (10070) = (.56/30) = .0187
d
M
= 1/.0187 =53.476
E = [(4 x 9.81 x .006^3)/ (0.019 x .0600^3)] x 53.476 = GPa
21.0.
Graphs
Please refer to the attached graph papers. [ GRAPH 4 ]. Thank you.
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Lab Assignment IIExperiments 4 and 6
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22.0.
Discussions1. Why does carbon steel has formed cup and cone shape after fracture?
I have observed that the carbon steel has formed a cup and a cone shape after fracture.
Which is due to the time when the carbon steel rod started to neck, small cavities formed in
the middle of the cross section. And then those cavities grew in a direction perpendicular tothe applied load as the deformation continued. Fracture was caused by the rapid creation of
a crack around the outer perimeter of the neck, at an angle of around 45o. Therefore, the
two broken pieces formed were in a shape of a cup and a cone.
2. Explain why carbon steel has broken after Ultimate tensile strength.
After ultimate tensile strength, carbon steel broke as a result of the external force applied
on it. Carbon steel has exceeded its ability to withstand force and its maximum resistance to
fracture.
3.
Why does 0.2% proof stress is important for your material selection?The metal will deform as soon as forces are applied on it. This deformation is known aselastic deformation when it is up until the 0.2% proof stress. 0.2% proof stress determines
the point of permanent damage. When the forces are applied on the material is more than
the 0.2% proof stress, it will be permanently deformed and practically unusable. Therefore,
0.2% proof stress is important for material selection in order to know how much force a
material can withstand.
4. Compare the stress and strain graphs for non metal (Acetal, Nylon,
polycarbonate)In general, fibers have the highest tensile moduli, and elastomers have the lowest, and
plastics have tensile moduli somewhere in between fibers and elastomers. Acetalhas an
UTS of 60, Elongation of 45% and Tensile modulus of 2.7, while Nylonhas an UTS of 60,
Elongation of 90% and Tensile modulus of 1.8 and PVChas an UTS of 70, Elongation of 100%
and Tensile modulus of 2.6.
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Lab Assignment IIExperiments 4 and 6
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23.0. ConclusionFinally, I conclude that the tensile test provides information on: proof stress, yield point,
tensile strength parameter, elongation and reduction of area. Other than that, the graphs
obtained can be used to identify the elastic properties, plastic deformation, fracture points,
and deflection of the materials with reference to the applied load.
Even if there were some errors done during the experiment; the experimental value of
youngs modulus proved that the aluminumbar is tough and strong. Aluminums
withstand to a high deflection at the end of the beam proves that it is strong. And
according to the graph and youngs modulus value of carbon steel, I can wrap up by saying
that it has a higher youngs modulus than aluminum and can able to withstand a large
amount of forces before fracture, as shown in the graph. In addition, the curve of graph
sketched is large which symbolizes the strength of a material.
24.0. ReferencesCallister, W.D. 2007. Mechanical Properties. In Materials Science and Engineering: An
Introduction, 7th
Edition, 132-160. John Wiley & Sons ,Inc., NY.
25.0. DeclarationI, Yasmin Khalil hereby declare that this report is the result of my own efforts and that all
the photos were taken by me on Tuesday (May 18 10). The calculations and graphs are
based on the results and data gathered by group B2.