Lab 1 Tension Test
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ME 2356 Laboratory Template
1
Department of Mechanical and Industrial Engineering
ME 2356: Laboratory for Mechanics of Materials
Tension Testing
Submitted by
Sydney Rodriguez
Date Submitted:10/21/2023
Date Performed: 10/06/2023
Lab Section: 12
Lab TA: Lindsey Young
Course Instructor: Marguerite Matherne
ME 2356 Laboratory Template
2
1.
Introduction
The objective of this phase of the experiment is to gather data concerning the strength of the
supplied sample and to examine material of Steel’s attributes in depth.
2.
Methods and results
2.1
Instron
Procedure
In this experiment, an Instron 5582 apparatus is employed to assess compression and tension in a
given sample. The machine operates by applying force to the material, recording parameters such
as time, displacement, and force. Additionally, an extensometer is utilized, a compact device
attached to the sample to precisely calculate minor degrees of strain. Data from both these
sources are subsequently employed to generate graphs, thereby providing valuable insights into
the characteristics of the sample.
Results
Sample material and measurements table
Table 1. “Steel” sample.
Parameter
Initial value
Final value
Unit
Length
101.63
N/A
mm
Diameter
8.960
5.940
mm
Area
63.05
27.71
mm^2
Load vs. displacement plot
ME 2356 Laboratory Template
3
Stress vs. strain plot
Mechanical properties table
Table 2. “Steel” mechanical properties.
Parameter
Value
Unit
Young’s modulus
2.08e+09
N/m^2
Proportional limit
5.79e+08
Pa
Yield strength
1.73e+07
Pa
Ultimate strength
6.43e+08
Pa
True stress
1.14e+09
Pa
Reduction of area
45.9
%
Elastic energy density
7.19e+06
J/m^3
Discussion
6. On a stress-strain chart, the region beneath the curve at a specific point represents the strain
energy density necessary to subject the material to that point on the curve. As the modulus of
resilience encompasses the area under the curve until the yield point, it essentially signifies the
strain energy density needed to subject the material to its yield strength.
7. The calculated values closely approximate the reference values; however, it's crucial to
acknowledge potential sources of error in this experiment. Firstly, discrepancies may
inadvertently arise in the procedure and setup due to human error. Additionally, human error can
affect the values calculated, as the selection of data points may vary among individuals.
Nonetheless, in general, anyone conducting this experiment should achieve comparable results if
using the same material and procedure.
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ME 2356 Laboratory Template
4
2.2
Manual tester
Procedure
This phase of the experiment closely resembles the initial part, employing a tensile test.
However, in this case, the operation is manual rather than being performed with the Instron
machine. The sample is securely positioned in the apparatus, and the PASCOCapstone software
is configured to capture data as the user applies and removes the load from the sample.
Results
Sample material and measurements table
Table 3.
“Aluminum” sample.
Parameter
Initial value
Unit
Length
3.2
cm
Diameter
3.33
mm
Load vs. displacement plot
ME 2356 Laboratory Template
5
Stress vs. strain plot
Mechanical properties table
Table 4.
“Aluminum” mechanical properties.
Parameter
Value
Unit
Loading Young’s
modulus
1.29e+10
N/m^2
Reloading Young’s
modulus
1.30e+09
N/m^2
Yield strength
3.64e+07
Pa
Ultimate strength
1.89e+07
Pa
Discussion
4.
The values in this segment do not closely approximate the reference values, considering
there is a higher degree of error compared to the first part of the experiment. Naturally,
manual operation of the tensile test introduces some inconsistencies, leading to
irregularities in the graph. Similar errors in calculations might also manifest in this phase,
as in the first one. With repeated experiments by others they should experience different
results closer to the reference values online.
ME 2356 Laboratory Template
6
a.
Polariscope
Procedure
In the final stage of the experiment, two samples of the identical material but with distinct shapes
are employed. These samples are situated beneath a loading mechanism that can be adjusted to
augment the applied load. Additionally, a polarized light source and polarizer are incorporated,
enabling the user to visually assess the stress at various locations within the material and observe
disparities between the two samples.
Results
3 photos of the first sample
3 photos of the second sample
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ME 2356 Laboratory Template
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Discussion
3.
In the second images of both samples, the regions experiencing the most significant stress
are highlighted. In the first sample, it's apparent that the point of contact with the load
and the immediate vicinity display the highest stress concentrations. Additionally, there
are observable stress wave patterns throughout the sample, a consequence of the
concentrated nature of the applied load. For the second sample featuring a hole, it's
evident that the areas surrounding the hole exhibit the highest stress levels. This is
attributed to the varying cross-sectional area at these specific locations, which differs
from the rest of the sample.
4.
The polariscope method offers numerous potential applications owing to its
straightforward approach to stress observation. For those seeking to assess relative stress
and make material comparisons without the need for precise data, this method proves
highly effective. A company engaged in product design can readily identify both the
strongest and weakest areas within the material by simply peering through the
polariscope, facilitating informed decisions on necessary reinforcements.
ME 2356 Laboratory Template
8
3.
Appendix:
Part 1:
%clear file
clearvars;
clc;
%read data
filename =
'Lab1_InstronData_Steel.csv'
;
T = readtable(filename);
%sample properties
di = 8.960e-3;
ri = di/2;
li = 101.63e-3;
df = 5.940e-3;
rf = df/2;
Af = pi*rf^2;
Ai = pi*ri^2;
%conversions
Force = T.Force*9.80665;
%Load in N
Displace = T.Displacement*10^-3;
%Extension in m
%stress and strain
Stress = Force/Ai;
Strain = Displace/li;
eStrain = T.Strain1;
%youngs modulus
F = polyfit(eStrain(5:100), Stress(5:100), 1);
E = F(1)
%yield strength
P = polyfit(Strain(20:170), Stress(20:170), 1);
oStress = P(1)*Strain(1:200);
oStrain = Strain(1:200)+0.002;
%Elastic energy density
U = 0.5*Stress(870)*Strain(870)
%ultimate strength
StrengthUlt = Stress(4050)
%true stress at failure
TrueStress = Force(5000)/Af
%proportional limit
PL = [Strain(1100), Stress(1100)];
%plot load vs displacement
figure(1)
ME 2356 Laboratory Template
9
plot(Displace*10^3, Force)
title(
'Load vs. displacement (steel)'
);
xlabel(
'Extension (mm)'
);
ylabel(
'Load (N)'
);
%plot stress vs strain
figure(2)
plot(Strain, Stress)
hold
on
plot(eStrain, Stress)
hold
on
plot(oStrain, oStress)
hold
on
plot(PL(1), PL(2),
'r*'
)
title(
'Stress vs Strain (steel)'
);
xlabel(
'Strain (-)'
);
ylabel(
'Stress (Pa)'
);
legend(
'Crosshead displacement strain'
,
'Extensometer strain'
,
'0.2%
Offset'
,
'Proportional limit'
,
'Location'
,
'southeast'
);
Part 2:
clearvars;
clc;
filename =
'lab 1 thur 03 (1).xlsx'
;
T = readtable(filename);
di = 3.33e-3;
ri = di/2;
li = 3.2e-2;
df = 6.35e-3;
rf = df/2;
Af = pi*rf^2;
Ai = pi*ri^2;
Force = T.Force_N_Run_2;
Displace = T.Position_m_Run_2;
Stress = Force/Ai;
Strain = Displace/li;
F = polyfit(Strain(15:300), Stress(15:300), 1);
E = F(1);
F2 = polyfit(Strain(581:640), Stress(581:640), 1);
E2 = F2(1);
P = polyfit(Strain(15:300), Stress(15:300), 1);
oStress = P(1)*Strain(1:425);
oStrain = Strain(1:425)+0.002;
StrengthUlt = Stress(431)
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figure(1)
plot(Displace, Force)
title(
'Load vs. displacement (Aluminum)'
);
xlabel(
'Extension (m)'
);
ylabel(
'Load (N)'
);
figure(2)
plot(Strain, Stress)
hold
on
plot(oStrain, oStress)
title(
'Stress vs. Strain'
);
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2m
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The rigid beam ABC is supported by pin A and wires BD and CE. If the load P on the beam
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Case Study
Determine Properties from Tensile Test Data
A cylindrical specimen of stainless steel having an initial diameter of 12.8mm and initial length
of 50.8mm is pulled in tension. Use the data provided below to
a) Plot the data as engineering stress versus engineering strain using excel or similar
software.
b) Compute the modulus of elasticity.
c) Determine the yield strength at a strain offset of 0.002.
d) Determine the tensile strength of this alloy.
e) What is the approximate ductility, in percent elongation?
f) Compute the modulus of resilience.
Load (N)
Length (mm)
50.800
12,700
25,400
38,100
50,800
76,200
50.825
50.851
50.876
50.902
50.952
89,100
92,700
51.003
51.054
102,500
107,800
119,400
128,300
149,700
159,000
160,400
159,500
151,500
124,700
51.181
51.308
51.562
51.816
52.832
53.848
54.356
54.864
55.880
56.642
Fracture
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Academic)
The following tensile data were collected from a standard 12.827 mm diameter test specimen of
copper alloy. The specimen was tested under tensile load until rupture.
After fracture, the sample length is 76.708 mm and the diameter is 9.5 mm.
Load (kN) Sample Length (mm)
50.8
13.4
50.842
26.7
50.897
33.4
50.957
40.0
51.0286
46.7
51.816
53.4
57.404
55.2
63.5
50.7
76.708
Calculate:
1. Ultimate Engineering Tensile strength:
2. Engineering stress at fracture:
3. True stress at fracture:
4. Percent elongation of the tested copper alloy specimen:
5. Percent reduction of area of the tested copper alloy specimen:
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Plane-Stress Transformation
Learning Goal:
The state of in-plane stress at a point on an element of material is shown. Let |ox| = 320 MPa, |oy| = 120 MPa, and |Txy| = 75.0 MPa. Use this information to represent the state of stress of the same point that is rotated through an angle
of 0 = 25.0°.
b
Txy
σx
Part A - Normal and shear stress on element sectioned at plane a-a
Using the element sectioned at plane a-a and the rotated coordinate system shown, determine the normal and shear stresses, σ and Ta'y', respectively, acting on plane a-a.
90-0
-90-8
X
Express your answers in MPa to three significant figures separated by a comma.
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