Tension Test on Metallic Materials
Experiment #2
Performed: 2007.10.16
Submitted: 2007.10.23
Morgan Kelly Gilmour
1
Objective
The purpose of this experiment was to obtain a stress-strain curve and determine basic
mechanical properties for four metallic materials, Aluminum 2024, Aluminum 7075, and
Steel 1018.
Procedure
Each of the three, dog-boned shaped, metallic materials, Aluminum 2024, Aluminum
7075, and Steel 7075, is measured with a caliper for thickness and width. They are
individually placed in a Q-test machine and subjected to tensile load until fracture occurs,
or until the machine can no longer exert force. The experimenter observes the
deformation of each material, and refers to the data for stress and strain information. The
experimenter then uses this data to make conclusions about the materials’ mechanical
properties.
Below are the three materials’ cross-sectional areas:
Material = (width)(thickness) = area in^2
Width (in)
Thickness
(in)
Area (in^2)
Aluminum
2024
0.513
Aluminum
7075
0.5095
Steel
1018
0.515
0.1275
0.0654
0.1255
0.0639
0.1265
0.0651
Table 1. Materials and their areas based on width and thickness
Specimens are 9 in x ½ in x ½ in
The experimenter must make a stress vs. strain curve for each material. Using the force
and elongation data gathered by the Q-test machine, the experimenter divides the force by
the cross sectional area of the material to obtain stress (units = lb/in^2 = psi), and divides
the distance the material was stretched by the initial distance (change in length / initial
length) to obtain strain. The initial distance between the grips of the machine was 6.1 in.
Results and Discussion
Data from the Q-test machine was collected and inputted into an Excel document,
producing three separate stress-strain curves for each of the materials. Samples of the
stress and strain calculations, resulting graphs, and analysis of the graphs are shown
below.
2
Aluminum 2024 Stress vs Strain
80000
70000
Stress (psi)
60000
50000
40000
30000
20000
10000
0
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
Strain (%)
Figure 1. Aluminum 2024 Stress-strain curve.
See Appendix Calculations for sample calculations.
The Al 2024 stress-strain curve shows us that relatively brittle. We can see a clear elastic
state, where the deformation is linear from about 0-0.02 Strain, and from 0 to about
50000 psi stress. Then the curve arcs, indicating it has entered an inelastic state, and
eventually breaks off. The transition point is very obvious, indicating clearly when the
material starts to fail. We observed no necking at the break point, but did observe a nearperfect 45º angle in the break point, indicating a brittle material. The break point was
smooth and straight, with no material curving into another plane. These observations
suggest a brittle material, but the fact that it has entered an inelastic state allows us to
reject the option of it being completely brittle. This metal is slightly ductile, but
relatively brittle.
3
Aluminum 7075 Stress vs. strain
90000
80000
Stress (psi)
70000
60000
50000
40000
30000
20000
10000
0
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
Strain (%)
Figure 2. Aluminum 7075 Stress-strain curve
The machine was unable to stretch Aluminum 7075 to a point where it broke. Therefore,
the Aluminum had no break point. However, when compressed, the material buckled and
bent easily. Clearly, the material is either more ductile or just stronger than Aluminum
2024, or perhaps both.
Steel 1018 Stress vs. Strain
60000
Stress (psi)
50000
40000
30000
20000
10000
0
0
0.05
0.1
0.15
Strain (%)
Figure 3. Steel 1018 Stress-Strain curve
4
0.2
0.25
Steel 1018 is very ductile. There was necking, observable plastic deformation, and a
rough break at the break point. It also took a considerable amount of strain to reach its
break point, and though it has a clear transition between elastic and inelastic regions, the
prominent arc in the inelastic region suggests ductility.
Below are the results we gather from the graphs. For explanations as to how to derive
these numbers, and for sample calculations, see Appendix Calculations.
E (psi) experimental
E (psi) accepted
E % error
Yield Strength (psi) experimental
Yield Strength (psi) accepted
Yield Strength % error
Tensile Strength (psi)
experimental
Tensile Strength (psi) accepted
Tensile Strength % error
Rupture Strain (%) experimental
Rupture Strain (%) accepted
Rupture Strain % error
Aluminum
2024
2,500,000
10,587,754.871
-76.39%
53000
470004
12.76%
68000
680007
0%
0.17
0.1810
-5.56%
Aluminum
7075
3,000,000
10,732,792.592
-72.04%
72000
716305
.52%
Steel 1018
2,400,000
30,457,7703
-92.1%
38000
559706
-32.1%
80000
830008
-3.6%
X
.1111
X
49000
725009
-32.2%
0.225
0.1812
25%
Table 2. The values of E, Yield Strength, Tensile Strength, and Rupture strain for each of
the three metals.
We see from these graphs and their analyses that Steel is far more ductile than either
Aluminum sample. We see that its tensile strength is in between the two other values, but
that its yield strength was much lower than Al 2024 and Al 7075’s values were. This is
not necessarily always the case, as we saw from the accepted values. Generally,
Young’s Modulus Al 2024. http://edboyden.org/constants.html.
Young’s Modulus Al 7075. http://www.freepatentsonline.com/5047092.html.
3
Young’s Modulus Steel 1018. http://en.wikipedia.org/wiki/Carbon_steel
4
Yield Strength Al 2024. http://asm.matweb.com/search/SpecificMaterial.asp?bassnum=MA2024T4.
5
Yield Strength Al 7075. http://www.springerlink.com/index/E3Q808137N237R1X.pdf.
6
Yield Strength Steel 1018.
http://www.efunda.com/Materials/alloys/carbon_steels/show_carbon.cfm?ID=AISI_1018&prop=all&Page
_Title=AISI%201018.
7
Tensile Strength Al 2024. http://asm.matweb.com/search/SpecificMaterial.asp?bassnum=MA2024T4.
8
Tensile Strength Al 7075. http://www.finishing.com/70/11.shtml.
9
Tensile Strength Steel 1018. . http://en.wikipedia.org/wiki/Carbon_steel.
10
Rupture Strain Al 2024. http://www.matweb.com/search/SpecificMaterial.asp?bassnum=MA2024AT3.
11
Rupture Strain Al 7075. http://www.matweb.com/search/SpecificMaterial.asp?bassnum=MA7075T6.
12
Rupture Strain Steel 1018. http://www.matweb.com/search/SpecificMaterial.asp?bassnum=M1018E.
1
2
5
Aluminum 2024 has the lowest yield strength. In our test, our samples could have been
flawed, causing a lower yield stress than normal.
The Aluminum 7075 could not be broken in the machine, proving that it is stronger than
Aluminum 2024. The buckling that occurred when it was compressed does not mean that
it would fail differently than 2024 or Steel, because these samples would have buckled
had we compressed them, as well.
Clearly, Aluminum 7075 was more resistant to failure than Aluminum 2024, proving it to
be the stronger aluminum of the two.
Steel weighs much more than Aluminum, making it a cumbersome material, and perhaps
not as good to build with than Aluminum 7075. Aluminum 7075, all in all, is the best of
the three materials.
There is significant error involved in all three of the Young’s Modulus values. This can
be attributed to the fact that this report was written with the incorrect data. Because the
data was incorrect, we can also suspect that the unusually high error for Steel 1018’s
numbers was due to this data problem. However, there might also have been material
flaws that caused the Steel to break more easily, and this source of error might have also
caused the small but acceptable error found in most of the other numbers.
There was no rupture strain for Al 7075 because it was so strong that the machine was
unable to break it. The machine ceased the tensile load before the Al 7075 broke, leaving
us with no rupture strain value. However, we know that the rupture strain was supposed
to be at .11, which is curiously lower than the other two materials’ numbers.
Conclusion
After careful analysis of the Young’s Modulus, Tensile Strength, Yield Strength, and
Rupture Strength of Aluminum 2024, Aluminum 7075, and Steel 1018 under tension, we
can conclude that Aluminum 7075 is the superior material because of its light weight,
large tensile strength and yield strength, and superior strength to the other two materials.
Aluminum 7075 was the only material that did not fracture in the machine, and although
Steel is more ductile than Aluminum 7075, Steel did fracture under less stress.
6
Appendix
References
Conversion Factors, Material Properties, and Physical Constants. 2007.
http://edboyden.org/constants.html. Visited 2007.10.20
Young’s Modulus Al 7075. 2007. http://www.freepatentsonline.com/5047092.html.
Visited 2007.10.20
2
3
Carbon Steel. 2007. http://en.wikipedia.org/wiki/Carbon_steel. Visited 2007.10.20
4
ASM Aerospace Specification Metals. 2007.
http://asm.matweb.com/search/SpecificMaterial.asp?bassnum=MA2024T4. Visited
2007.10.20
5
The Effect of Joint Design on Mechanical Properties of AL7075. 2005.
http://www.springerlink.com/index/E3Q808137N237R1X.pdf. Visited 2007.10.20
6
AISI 1018. 2007.
http://www.efunda.com/Materials/alloys/carbon_steels/show_carbon.cfm?ID=AISI_1018
&prop=all&Page_Title=AISI%201018. Visited 2007.10.20
7
ASM Aerospace Specification Metals. 2007.
http://asm.matweb.com/search/SpecificMaterial.asp?bassnum=MA2024T4. Visited
2007.10.20
8
Seeking Suppliers of Alloys for Air Force. 2007.
http://www.finishing.com/70/11.shtml. Visited 2007.10.20
9
Carbon Steel. 2007. http://en.wikipedia.org/wiki/Carbon_steel. Visited 2007.10.20
10
Al 2024, 2007.
http://www.matweb.com/search/SpecificMaterial.asp?bassnum=MA2024AT3. Visited
2007.10.20
11
Al 7075, 2007.
http://www.matweb.com/search/SpecificMaterial.asp?bassnum=MA7075T6. Visited
2007.10.20
12
Steel 1018, 2007.
http://www.matweb.com/search/SpecificMaterial.asp?bassnum=M1018E. Visited
2007.10.20
7
Sample Calculations
Stress Sample Calculation:
Force = 4271.415 lbs
Cross sectional area of Nylon = .0654 in^2
4271.415 /
.0654 = 65312.16 psi = stress
Strain Sample Calculation:
Distance = 1.053 in
Initial grip separation = 6.1 in
1.053 in
/ 6.1 in = 0.172623 = strain
Aluminum 2024 Stress vs Strain
80000
70000
Stress (psi)
60000
50000
40000
30000
20000
10000
0
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
Strain (%)
We find by looking at the graph that Al 2024’s Young’s Modulus, E, is equal to the slope
of the elastic region, 2,500,000 psi.
E = delta y / delta x = [(50000) psi]/[(.02-0)] = 2500000 psi
8
The internet suggests that Young’s Modulus should be about 7.310E+10 Pa, which
translates to 10,587,754.87 psi.13 The error is extremely large, and the procedure for
calculating percent error is below:
[(experimental – accepted value) / (accepted value)] *100 = percent error
The tensile strength is calculated by observing where the stress-strain curve reaches its
peak. In Al 2024’s graph, we see that the tensile strength is about 68000 psi.
The yield strength is calculated by following the slope of the elastic region of the stressstrain curve after observing an offset strain of .2%. We can estimate this to be about
53000 psi after looking at the graph.
Finally, the rupture strain is the last point on the graph. We estimate this point to be at
about .17%.
13
Young’s Modulus for Al 2024. http://edboyden.org/constants.html.
9