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Executive Summary Department of Mechanical Engineering Aug.24, 2021, Statesboro, Georgia, USA 1 Copyright © 2023 by ASME TENSILE TESTING ON BRASS Lawrence Almeter Group 5 Objective
The objective of this experiment is to find the stress-strain curve, ultimate tensile strength, Young’s Modulus, yield strength, and percent elongation of a brass tensile testing coupon. These values can be solved using the load and displacement over time data given by the tensile testing machine as well as the dimensions of our sample. Brief Introduction/Theory The modern UTM or universal testing machine was not formally developed until the 19
th
century. However, Robert Hooke, who discovered Hooke’s law or the law of Elasticity, put forth many efforts during the 17
th
century for testing the properties of materials. [1] Stress is equal to force over the cross-sectional area. Strain is equal to the length of deformation over the original length of the deformed material. The stress-strain curve is used to graphically represent the overall performance that the material exhibits while under load, and it allows many different mechanical properties of materials to be found easily. [2] Method/Procedure Start by turning on the material testing machine and starting the data collection software. Select the Gauge + ESM 303 with Travel configuration and open the setup file MENG 3331 Tensile. After setting up the machine and software, prepare the sample. Draw two lines centered on the sample 2” apart along the width of the sample. Measure the distance again three times and record the average and standard deviation for better data. Measure the width and thickness three times and record the average and standard deviation. With the sample prepared, it can be put in the testing machine. Place the sample in the lower grip and then adjust the upper grip till it is at the height of the two lines drawn previously. Adjust the upper grip as close to zero travel as possible. Click start on the software and wait till the sample fails. Data& Results The sample dimensions were 0.160in or 4.056mm in width, 0.003in or 0.076mm in thickness, 2.567in or 65.21mm in gauge length, and the cross-sectional area was 0.12in
2
or 0.309mm
2
. These values are found in Table 1. The data given by the machine over 108.4 seconds showed the max load applied to the sample was 31.7lbf, and the max displacement in the brass was 0.901in or 22.8854mm. The max force calculated was 141.00N. The max stress was 456.269MPa, and the max strain was 0.3509mm/mm. These values are found in Table 2. Discussion Most of our results compared neatly to the known values. Our calculated cross-sectional area of 0.309mm
2
was close to the known 0.303mm
2
. The known tensile max tensile strength of brass is 430 MPa and our value is close at 458 MPa. Our estimation of elongation percent was 35% which is off from the known value of 25%. This is partially due to calculating it by converting the max strain into a percent which is not highly accurate, but it is a rough estimation. Our calculated Modulus of Elasticity was 41.18 GPa which was far off from the known 117 GPa. This could be due to defects in the material, improper placement in the testing machine, or even a bad estimation of the slope [Figure 2]. Performing more tests could give more consistent data and remove many outliers. Conclusion These are the calculated values for the brass testing material. The Young’s Modulus is 41.14 GPa. The Yield Strength is 300 MPa. The Ultimate Tensile Strength is 458 MPa. The percent elongation is 35%. These values are found in Table 4. Lab Questions 1.
The upper yield strength of the 1004 steel is approximately 380 MPa and the lower yield strength is approximately 360 MPa. 2.
Copper has a higher ductility since it has a larger strain value. 3.
The 1020 steel has a higher toughness since the area under its curve is larger than the area under the 1004 steel curve.
Executive Summary Department of Mechanical Engineering Aug.24, 2021, Statesboro, Georgia, USA 2 Copyright © 2023 by ASME Support Materials Data Table 1 summarizes the measurements taken on the brass testing material. Table 2 shows the data given by the testing machine as well as converted and calculated values used to find our resulting values. Due to the vast number of data points, only the first and last ten data points are shown in Table 2. Table 3 contains two data points that are used to solve the slope which is the Young’s Modulus. Table 4 contains the results and values of the experiment. Table 1. Measurements T [in] T [mm] W [in] W [mm] A [in
2
] A [mm
2
] GL [in] GL [mm] 0.003 0.0762 0.161 4.0894 0.0123 0.3116 2.5 63.5 0.003 0.0762 0.16 4.064 0.0122 0.3097 2.622 66.5988 0.003 0.0762 0.158 4.0132 0.0120 0.3058 2.58 65.532 Avg 0.003 0.076 0.160 4.056 0.012 0.309 2.567 65.210 StdD 0 0 0.0015 0.0388 0.0001 0.0030 0.0620 1.5743 Table 2. Testing Data and Calculations Load [lbF] Disp [in] Disp [mm] t [s] F [N] σ [MPa]
ε [mm/mm]
0.6000 0.0010 0.0254 0.4000 2.6688 8.6360 0.0004 3.0000 0.0030 0.0762 0.6000 13.3440 43.1800 0.0012 4.9000 0.0040 0.1016 0.8000 21.7952 70.5274 0.0016 6.9000 0.0060 0.1524 1.0000 30.6912 99.3141 0.0023 8.9000 0.0070 0.1778 1.2000 39.5872 128.1008 0.0027 10.5000 0.0090 0.2286 1.4000 46.7040 151.1302 0.0035 12.3000 0.0110 0.2794 1.6000 54.7104 177.0382 0.0043 14.2000 0.0120 0.3048 1.8000 63.1616 204.3855 0.0047 15.8000 0.0140 0.3556 2.0000 70.2784 227.4149 0.0055 17.3000 0.0160 0.4064 2.2000 76.9504 249.0049 0.0062 31.7000 0.8860 22.5044 106.6000 141.0016 456.2691 0.3451 31.7000 0.8870 22.5298 106.8000 141.0016 456.2691 0.3455 31.7000 0.8890 22.5806 107.0000 141.0016 456.2691 0.3463 31.8000 0.8900 22.6060 107.2000 141.4464 457.7085 0.3467 31.8000 0.8920 22.6568 107.4000 141.4464 457.7085 0.3474 31.8000 0.8940 22.7076 107.6000 141.4464 457.7085 0.3482 31.8000 0.8960 22.7584 107.8000 141.4464 457.7085 0.3490 31.8000 0.8970 22.7838 108.0000 141.4464 457.7085 0.3494 31.7000 0.8990 22.8346 108.2000 141.0016 456.2691 0.3502 0.0000 0.9010 22.8854 108.4000 0.0000 0.0000 0.3509 Detail Results Table 3. Calculation of Slope (Young’s Modulus) x y Point 1 0.0012 43.1800 Point 2 0.0066 267.7163 Slope 41175.668 Table 4. Calculations UTS [MPa] 458 %EL 35 Young's Modulus [MPa] 41176 Young's Modulus [GPa] 41.18 Yield Strength [MPa] 300 Stress is defined as force applied over a cross-sectional area (see equation 1). Strain is defined as the ratio of the length of deformation and original length (see equation 2). Stress and strain are both calculated in Table 2. Figure 1 describes the relationship between the calculated stress and strain. Figure 2 shows a closer view of the elastic region and the linear approximation of the slop (modulus of elasticity) as well as the yield strength. Figure 1. Stress over Strain for Brass
Figure 2. Yield Strength and Young’s Modulus 𝜎 =
ி
=
ி[ே]
[
మ
]
= 𝑆𝑡𝑟𝑒𝑠𝑠 [𝑀𝑃𝑎]
- Equation 1 𝜀 =
௧
ை ௧
=
௱
బ
= 𝑆𝑡𝑟𝑎𝑖𝑛
- Equation 2 0
100
200
300
400
500
0.00
0.10
0.20
0.30
0.40
Stress [MPa]
Strain [mm/mm]
Brass…
0
100
200
300
400
500
0.000
0.005
0.010
0.015
0.020
Stress [MPa]
Strain [mm/mm]
Brass…
Yield Strength
Slope (Young’s Modulus)
0.2% off set
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Related Questions
Select one or more:
a. 28.6
Ob. 22.8
O c. 3.7
Od. No
Oe. 4.9
Of. Yes
0
8
9
10
11
12
13
14
15
g.
18.5
295
293
280
288
263
290
298
275
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Figure 1 shows the tensile testing results for different materials. All specimens have an initial
diameter of 12 mm and an initial gauge length of 50 mm.
300
250
Low carbon steel
Network polymer
200
Crystalline polymer
150
Amorphous polymer
100
50
5
10
15
20
25
30
Strain (%)
Figure 1: Stress-strain curve
b. Determine the following parameters for each material:
• the tensile strength
the 0.2% offset yield strength
the modulus of elasticity
• the ductility
Stress (MPa)
LO
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1. A tensile test was conducted on a metal "505" specimen and the following stress-strain curves
were generated, both curves generated from the same set of data. Use the graphs to fill in the
mechanical properties of the material tested in the box below. Don't forget units!
Stress vs Strain
Stress, psi
Stress, psi
80000
70000
60000
50000
40000
30000
20000
10000
0
0.00
80000
70000
60000
50000
40000
30000
20000
10000
0.02
0
0.000 0.002
0.04
0.004
0.06
0.006
0.08
0.10
Strain
Stress vs Strain
0.008
0.12
Elastic Modulus, E:
0.2% Offset Yield Strength, oo:
Tensile Strength, ou:
Breaking Strength, of:
% Elongation:
0.14
0.010 0.012 0.014
Strain
0.16
0.18
0.016 0.018
0.20
0.020
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I need the answer as soon as possible
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You have been given the following test sample data following mechanical testing of 15
test pieces of a modified Alumina.
What is the Weibull modulus of this material?
Would you advise the use of this material over one with a Weibull Modulus of 19.6 and a
mean failure stress of 270 MPa, if you anticipate that the peak stress on the material
could be 255 MPa?
Sample
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Select one or more:
a. 185
b. No
Yes
□d. 49
□e. 28.6
3.7
Failure Stress (MPa)
297
293
270
300
g. 22.8
260
296
265
295
280
288
263
290
298
275
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6. State your answers to the following questions.Strain Gauge represents the deformation of a material through a change in resistance. If so, explain how temperature will affect the strain gauge in the experimental environment.①:In this experiment, the Strain Gauge measures the strain in micro units. Explain one possible error factor when applying a load by hanging a weight on the material with the strain gauge attached. (Hint: It is easy to shake by hanging the weight using a thread)①:
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A tensile test was performed on a metal specimen with a diameter of
1/2 inch and a gage length (the length over which the elongation is meas-
ured) of 4 inches. The data were plotted on a load-displacement graph,
P vs. AL. A best-fit line was drawn through the points, and the slope of
the straight-line portion was calculated to be P/AL = 1392 kips/in. What
is the modulus of elasticity?
BI
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The First half of this question has been answered. (First 3) Needing help with the last part.
Answered: 1) Solid / Ductile Material 2) Spring Constant: K = 05 = 250 N/m 3) Time Threshold: 5 seconds
Unanswered / Need: Viscoscity, Charasteristic Frequency & What real world model charasteristic does this describe?
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i need the answer quickly
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m2
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Please show the complete solution. Thank you
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Recording Help
Stress (MPa)
600
500-
400-
300
200
100-
0
0.00
Tell me what you
Exercise 2
Stress (MPa)
0.04
500
400
300
200-
100
0.000 0.002 0.004 0.006
Strain
0.08
Strain
0.12
0.16
0.20
Consider a cylindrical specimen of a steel alloy 10.0
mm
in
tension. Determine, its elongation when a load of
20,000 N is applied.
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The following data were recorded during the tensile test of a test specimen with a
diameter of 12.8 mm. The gage length is 50.8 mm. The given data are as follow;
Given:
Test specimen material: Aluminum
Diameter of test specimen: Do = 12.8 mm
Length of test specimen: Lo = 50.8 mm
Force and Elongation data for test specimen
Assumptions:
1. The given data is accurate and the material is isotropic
2. The direction of applied force is parallel to the length of the cylinder
Requirement:
To interpret and plot the stress Vs Strain curve for a given specimen
DATA:
LENGTH, mm
STRESS, MPa
LOAD, N
0
50.8
7 330
50.851
15 100
50.902
23 100
50.952
30 400
51.003
34 400
51.054
38 400
51.308
41 300
51.816
44 800
52.832
46 200
53.848
47 300
54.864
47 500
55.88
36 100
56.896
44 800
57.658
42 600
58.42
36 400
59.182
STRAIN
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6. The following engineering stress-strain data were obtained for 0.2% C plain
carbon steel.
(a) Plot the engineering stress-strain curve
(b) Determine the ultimate tensile strength for the alloy
(c) Determine the percent elongation at fracture
(d) Plot the true stress-strain curve
Engineering strain, in./in.
Engineering stress, ksi
30
0.001
55
0.002
60
0.005
68
0.01
72
0.02
74
0.04
75
0.06
76
0.08
75
0.1
73
0.12
69
0.14
65
0.16
56
0.18
51
0.19(fracture)
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Question 2
In designing prosthetic sockets, the latter will need to be experimentally tested for their
structural integrity. Figure 2 shows one such design of a prosthetic socket which is made of
carbon fibre composite. Strain gauges are installed to record the strains at various locations of
the legs during walking and the readings are recorded using a telemetry system to detemine
the critical stressed area. At a particular strain gauge location indicated in Figure 2, the
readings recorded by one of the 45° strain gauge rosettes are:
Ea = 2500 x 10*, es = 1500 x 10°, & = -950 x 10*
Using Mohr's Cicle or otherwise, detemine:
(a) the principal strains and the direction of the maximum principal strain relative to
the gauge "a".
(b) the corresponding principal stresses and sketch the results on a properly oriented
element.
You may assume that the prosthetic socket is made of polypropylene whose Young's
modulus of 1.0 GPa and Poisson ratio of 0.3.
Figure 2
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The following data were collected from a 12
mm diameter test specimen of Magnesium.
LOAD (N)
0
5000
10000
15000
20000
25000
26500
27000
26500
25000
GAUEGE LENGTH (mm)
30.000
30.0296
30.0592
30.0888
30.15
30.51
30.90
31.50 (maximum load)
32.10
32.79 (fracture)
After the fracture, the gauge length is 32.61 mm and
the diameter is 11.74 mm.
a) What is the elastic modulus?
b) Percent elongation at fracture?
c) Percent elongation after fracture?
d) What is the Poisson's ratio?
e)Draw the engineering stress-strain diagram
corresponding to the values in the table. Call this plot
I. Now consider this experiment is repeated at a
higher temperature with an identical sample. Draw
the new engineering stress-strain diagram, call it plot
Il and highlight the differences (on the same graph)
between I and II.
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The stress-strain data from a tensile test on a cast-iron specimen are
€ (10-3)
0
0.20
0.44
0.80
1.0
1.5
2.0
26
32
40
46
49
54
NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.
2.8
3.4
4.0
5.0
a (kpsi)
0
5
10
16
19
Determine the tangent modulus, E, at a value of a=0 psi.
106 psl.
The tangent modulus, Eis
X
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I just can't find The percent elongation at fracture.
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1. Plot the engineering stress & strain diagram of
an alloy having a tensile test result found in
Table 1. The tensile test specimen has a
diameter of 12.5mm and a gage length of
50.0mm.
The given alloy is used to make a 30.0mm diameter
cylinder, which is placed inside a hardened circular
steel casement with a 30.01mm inner diameter.
Table 1: Tensile test results of an alloy
Change In Length
(mm)
Change In Diameter
(mm)
Load (kN)
0.000
0.0000
0.0000
4.364
0.0254
0.0019
-0.0057
13.092
0.0762
21.819
0.1270
0.0095
30.547
32.729
0.1778
0.7620
34.911
3.0480
30.01 mm O
F
Rigid Plate
Cylindrical Alloy
-Steel casement
Figure k Section view of the steel casement
encapsulating the cylindrical alloy
2. Determine the required minimum value of F
such that the cylindrical alloy would touch the
walls of the steel casement.
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