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650:350 Mechanical Measurements
Section 1B
10/23/2023
Lazarus
TA: Shai
Sabaroche
Strain Gauge
Measurements
Morgan Lazarus
, ‘25
Department of Mechanical and Aerospace Engineering,
Rutgers University, Piscataway New Jersey 08854
Finding the axial and transverse strain for a constant
stress beam can help in finding the stresses
and the beam’s
material properties as well. Using a rectangular strain
gauge rosette for the aluminum cantilever beam allows for
the flexural stress to be calculated. For Part 1: The
Poisson ratios for aluminum, brass, and steel are 0.26,
0.27, and 0.24 respectively. The
Young’s
moduli for
aluminum, brass, and steel are 63.0 GPa, 47.1 GPa, and
148.4 GPa respectively. To check the validity of the strain
values found from the Wheatstone bridge voltagemethod,
a strain gauge indicator was used. For Part 2: The
experimental flexural stress values for the cantilever beam
are 0 psi, 1755 psi, 3496 psi, 5251 psi,and 7012 psi for each
of the given deflection values, which are very close to the
theoretical valuescalculated via the given formulas.
Introduction
Engineers often use strain measurements to gain
an understanding of how certain materials will deform
given an applied load or force. These measurements aid
us in designing objects as well as engineering tools. To
obtain such measurements, we use a strain gauge. A
strain gauge is a device that measures the strain on an
object by the change in its resistance. Strain gauges are
often used in conjunction with a Wheatstone bridge
circuit when analyzing the strain while changing the
voltage. A Wheatstone
bridge
circuit
consists
of
four resistances, an input voltage, as well as an output
voltage.
The following lab experiment uses two different
beams to see the effects of a given load on stress and
strain. The first type of beam in the experiment is known
as a constant stress beam, on which the stress is constant
along the axial direction due to the changing width of
the beam. The second type of beam used is known as a
cantilever beam. A cantilever beam has a load applied
on the opposite side of the clamped end. The load
amount on the end of the beam can be varied by the
equipment for analysis purposes.
During the experiment we used two different
types of strain gauge sensors. For the constant stress
beam, there are two single axis gauges that measure
axial
and
transverse
strain.
However,
when
measuring strain in different directions it is best to
use a rectangular strain gauge rosette. A rectangular
strain gauge consists of three strain gauges, each of
which are 45 degrees with respect to one another.
The rectangular strain gauge rosette is shown below
in Figure 1.
Figure 1. Image of a rectangular strain gauge
rosette
Figure 2. Image of a rectangular strain gauge
rosette
Since the electrical resistance of a wire in strain
gauges is changed when subjected to strain, an
equation relating the gauge factor, change in
resistance, and strain can be formulated. This is
shown in Equation 1.
ε =
1
??
ΔR
𝑅
[1]
When
a
Wheatstone
bridge
is
used
in
combination with a strain gauge, the output v, and
input voltages also come into the equation along
650:350 Mechanical Measurements
Section 1B
10/23/2023
Lazarus
TA: Shai
Sabaroche
with the change of resistance. This exact equation
is shown in Equation 2.
∆?
𝑂??
?
𝑖𝑛
=
∆𝑅/𝑅
4+2∆𝑅/𝑅
=
??
4+2???
≈
???
4
[2]
To
calculate young’s
modulus for a constant stress
beam, we must calculate the axial stress as well as the
axial strain on the beam. Using equation 2, we can find
the Axial strain on the beam. However, the axial stress
for a constant stress beam requires knowing the
magnitude of the applied load, P, the distance from the
strain gauge location to the load, x, the base width at the
strain gauge location, b, as well as the thickness at the
strain gauge location, h. The equation relating these
variables to axial stress is shown below in Equation 3.
𝜎
𝑥
=
6𝑃𝑥
?ℎ
2
[3]
Once axial stress and axial strain are calculated, they can
be related to each other to find young’s
modulus. the
materials from the voltage method and the axial
modulus, which would be the slope of the corresponding
graph. This relationship is shown in equation 4 below.
?
?
=
𝜎
?
[4]
The strain gauge rosette that was used for the cantilever
beam produced three different strains. Each of the
strains can then be used to find the two maximum and
minimum principal strains. The relationship between
these maximum and minimum strains is shown in
Equation 5.
?
(?
?
+?
?
)
2
± √
(?
?
−?
?
)
2
+(?
?
−?
?
)
2
2
??𝑥,?𝑖?
[5]
Upon finding the principal strains for the cantilever
beam, the stresses can be calculated
using young’s
modulus, E, and
Poisson’s ratio,
. The exact
relationship between these values is shown in Equation
6. These calculated stresses can then be compared to the
theoretical values.
𝜎
1
=
?
1−𝜈
2
(ℰ
1
+ 𝜈ℰ
2
), 𝜎
2
=
?
1−𝜈
2
(ℰ
2
+ 𝜈ℰ
1
)
[6]
The theoretical values for the cantilever beam stress
can be calculated via the load formula, Equation 7,
and the flexural stress formula for a cantilever beam,
Equation 8.
? =
3?𝐼?
𝐿
3
[7]
𝜎
𝑥
=
?𝑥?
𝐼
[8]
Results and Discussion
The first part of the lab involving the
constant stress beams required finding the axial
and transverse strain values via the voltage
method (VM) and the strain gauge indicator
(SGI) for each of the beam materials (aluminum,
brass, and steel.) The voltage method involved
the use of a Wheatstone bridge circuit given
certain input and output voltage values, which
could then be used to calculate the strain via
Equation 2. The strain gauge outputted values for
the axial strain of steel. The axial strains for each
of the materials from the voltage method and the
axial strain from the strain gauge indicator are
shown in Figure 3.
There is a small discrepancy between the axial
strain values for steel from the two different
methods. The main source of error that affects
this is zero drift because one of the strain values
when unloading the beam at 0 is much higher than
its counterpart. The percent error between the
experimental voltage method (VM) and the
theoretical SGI is 11.9%, which was found via the
difference in slope values of the plots and the
percent error formula.
650:350 Mechanical Measurements
Section 1B
10/23/2023
Lazarus
TA: Shai
Sabaroche
Figure 3: Comparison of Axial Strain values for
Steel’s constant stress beam from the VM and SGI
To calculate the Poisson ratios for all the
constant stress beams, the axial and transverse strains
must be plotted against each other. The resulting graph
depicts a negative slope equivalent to the
Poisson’s ratio
of the material. The following graph is depicted in
Figure 4. The Poisson ratios for aluminum, brass, and
steel from the plot are 0.2552, 0.2746, and 0.2361
respectively. There is a huge discrepancy between these
values and the theoretical Poisson ratios. This is mainly
due to an instrumental systematic offset error in the
output voltages, which systematically affects the
calculated strain values. This explains why all the
Poisson ratio values are offset by 0.06-0.08. Also, we
observe the existence of the hysteresis error, which is
the behavior of a system where the output variable is
dependent on the input variable as well as the previous
state of the output variable. In such a system, when the
input increases versus when the input decreases, the
output variable will assume differing values. The
hysteresis error for the aluminum values is noticeable
because the data points when unloading and loading are
not very close to each other. The hysteresis error is the
behavior of a system which This heavily impacts the
Poisson ratio for Aluminum. The percent errors for the
Poisson ratios for Aluminum, Brass, and Steel are
22.7%, 19.2%, and 17.2% respectively.
Figure 4: Plot of Axial Strain vs Transverse
Strain to find Poisson’s ratio for Aluminu
m,
Brass, and Steel
To calculate Young’s Modulus for
each of the
materials, it is necessary to plot axial strain and axial
stress against each other. The resulting slope of this
plot is Young’s Modulus for the corresponding
material. This plot is shown in Figure 4 and is
modeled in Equation 4.
The experimental Young’s
Moduli for Aluminum, Brass, and Steel from the
slopes were found to be 9,131,000 psi (63.0 GPa),
6,825,000 psi (47.1 GPa), and 21,520,000 psi (148.4
GPa) respectively. These values are much lower than
the theoretical values.
The percent errors Aluminum, Brass, and Steel when
compared to the theoretical ones are 10%, 52.9%,
and 25.8% respectively, which is larger than
expected. The main source of these large errors is
the existence of an experimental systematic error.
Which could have possibly occurred if the
clamp’s
was not positioned all the way on the edge of the
table. Additionally, the clamp needed to be at the
point where the base meets the triangular section of
the beam. The setup error is believed to be the main
contributor to the percent error between the
experimental and theoretical values. Such can only
be confirmed however by replicating the experiment.
Commented [ML1]:
Reword
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I need answer within 20 minutes please please with my best wishes
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-6
- 195 × 10 (oriented in the postive x direction)
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a) The normal strain and shear strain on a plane inclined at 29° clockwise from
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gauge B are 0.3 microstrain and 301.1 microstrain, respectively.
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d) The normal strain and shear strain on a plane inclined at 29° clockwise from
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7:49
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Expert Q&A
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show all work step by step
A cube of a material with a yield strength of 100x10° N/m*is subjectedtothe loads shown onthe
right. Will plastic/permanent deformation occur as a result of this loading? Consider Both the
Tresca (Maximum Shear Stress) and Von-Mises (Distortion Energy) criteria
F,| = 45x10'N
F= 30x10'N
FJ= 60x10'N
Cube with side length = 3x10°m
Would this change if the directions of E, and E were reversed? Numerical Values Required.
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7:Y. O %Y. O
36 ll ll
く ロ
quiz 2
elearning.uowa.edu.iq
Marked out of 1.00
P Flag question
quiz 2:
flexural stress which fluctuates between +
250 MN/m2 and - 115 MN/m² applied on a
machine component. Calculate the ultimate
strength value according to Goodman's
relation. Take yield strength = 0.55 Ultimate
%3D
strength; Endurance strength = 0.5 Ultimate
strength; and factor of safety = 1.5.
A-
В
I
D
II
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Answer the problem with detailed step by step solution refer to the table of equations when answering the problem.
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The stress-strain diagram for a steel alloy having an
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Figure
(ks)
70
40
30
40
30
20
10
0 004 00
-
612 636 630 6.34 0.38
BLS GURE LUNGS GORD
1 of 1
(in/in)
X Incorrect; Try Again; 2 attempts remaining
Part B
Determine the load on the specimen that causes yielding
Express your answer to three significant figures and include appropriate units.
Py = 74
Submit
kips
Previous Answers Request Answer
?
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gauges system mounted at a point on a structural member that is subjected to
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Ex = 6.686 x104 ;&,
3.0 x10-4; & = - 6.857 x10-5; Yxy = 2.580 x10-3;
%3D
%3D
Determine the stress components if the member is made of an isotropic material with linear
elastic behavior having the following elastic properties:
Young's modulus E = 70 Gpa and Poisson's ratio v = 0.29;
Select one:
Ox = 70 Mpa; Oy = 50 Mpa; o,
3 30 Мра; Тҳу
35 Mра;
%3D
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%3D
Ох %3D 140 Мра; бу
%3D
Ox = 90 Mpa; oy
= 50 Mpa; o, = 60 Mpa; Ty = 35 Mpa%3;
Ox = 70 Mpa; oy = 50 Mpa; 0z = 30 Mpa; Txy = 70 Mpa;
= 70 Mpa3;
= 100 Mpa; Oy = 50 Mpa; 0z = 80 Mpa; Txy
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Problem 2:
A specimen of steel having an initial diameter of 0.503 in was tested in tension using a gauge length of 2
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Elastic State
Load P Ibf
Elongation in
1 000
0.0004
2 000
0.0006
3 000
0.0010
4 000
0.0013
7 000
0.0023
8 400
0.0028
8 800
0.0036
9 200
0.0089
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Element 1
Ук
**
Answer in MPa.
2 cm
Element 2
Assume the reaction forces at the origin were
solved to be Rf = [325, 100, 0] N and
=
Rm [10, -2, 5] Nm. We will consider a
hypothetical cut at the connection of the dolly to
the axle. Z is positive out of the page.
What is the value of normal stress in the
x-direction Ox for element 2?
Answer options:
a) Ox < -300
b) -300 ≤ Ox < -100
c) -100 ≤ Ox < -200
d) 0 ≤ Ox < 100
e) 100 ≤ Ox < 300
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Tuield =130 mPA
Joltimate=185 mPA
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B) If we have a
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3-0 block of Al-5456 with the
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Fy=1000N
) Is Our block Safes (consider only plane stes)
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Uniaxial strain (%)
Lateral strain (%)
(MPa)
0.0
0.0000
0.0000
5.0
0.0319
-0.0150
10.0
0.0720
-0.0275
15.0
0.1025
-0.0425
20.0
0.1450
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25.0
0.1755
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35.0
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0.3125
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52.0
0.3515
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(f)
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Ex = 3.8 x104; &y = 5.5 x10-5; Ez = - 7.5 x10-5; Yy = 9.1 x104;
%3D
Determine the stress components if the member is made of an isotropic material with linear elastic behavior
having the following elastic properties:
Young's modulus E = 200 Gpa and Poisson's ratio v = 0.3;
Select one:
Ох %3D -100 Мра; оу %3D -50 Мра; о, %3D 30 Мра; Тху 3D 70 Мра;
Ox = -75 Mpa; o, = -50 Mpa; 0z = 40 Mpa; Ty = 80 Mpa3;
Oy = 300 Mpa; oy = -150 Mpa; 0z
3 100 Мра;B тyу
= 100 Mpa;
TOSHIBA
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Load (N)
Elongation (mm)
0
0
24372
0,183
23008
0,315
28357
5,777
35517
12,315
27555
17,978
23750
23,865
Based on the information above; draw stress-strain diagram of the material and answer the following questions.
- Determine the true stress (in MPa) at yield point.
- Determine the true stress (in MPa) at point of ultimate strength.
- Determine the true stress (in MPa) at fracture point.
- Determine the true strain (in mm/mm) at yield point. (Use at least five decimal units)
- Determine the true strain (in mm/mm) at point of ultimate strength. (Use at least five decimal units)
- Determine the true strain (in mm/mm) at fracture point. (Use at least five decimal units)
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Find Fac for Fy=285 MPa and l/r =37
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Shear Modulus; 7,69 x 10¹⁰ Pa
TB-60 Nm
dac-78 mm
dcp-54 mm
dD34mm
TE-29 Nm
B
E
0.7 m
C
0.45 m
0.5 m
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