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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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650:350 Mechanical Measurements
Section 1B
10/23/2023
Lazarus
TA: Shai
Sabaroche
Figure 5: Plot of Axial Strain and Axial Stress to find
Young’s Moduli for Aluminum, Brass, and Steel
The second part of the lab involved the aluminum
cantilever beam along with the rectangular strain gauge
rosette, which measures three different strain values.
The theoretical values were found by using equations 5
and 6. The calculated values were found using
Equations 7 and 8. The experimental and theoretical
values for the stress in the cantilever beam are very close
to each other. For each of the deflections from 0.1 to 0.4,
the percent errors are 4.0%, 3.6%, 3.8%, and 3.9%
respectively. This small error can be attributed to the
resolution error of the tape measure used to measure the
dimensions of the cantilever beam.
Table 1: Calculated and Measured strains and
stresses for the cantilever beam
?
?
??𝑥
𝜎
1
𝜎
2
?
?𝑖?
?
???𝑡𝑖?𝑒𝑣𝑒𝑟
in
in/in
psi
psi
in/in
psi
meas.
calc
calc
meas.
calc
0
0
0
0.1
0.2
0.3
0.4
Conclusion
The primary objective of the lab is to help gain a better
understanding of how strain gauges work via the use of
constant stress and cantilever beams. As mentioned
prior, the strain gauge indicator for the constant
stress beam gives the theoretical value of the axial
stress, which were compared to the experimental
value of it using a Wheatstone bridge. Also, there
were some large discrepancies in the young’s moduli
and the Poisson ratios for the given materials for the
constant stress beams. This was majorly due to the
systematic error in the equipment and setup, and
minorly due to the hysteresis and drift error.
However, the experimental and theoretical stress
values for the cantilever beam were very close to
each other with only a minor percent error mainly
due to the resolution uncertainty in the measuring
equipment.
Collected Data Tables
Table 2: ???
Clamped
To Load
Application
(in)
Bottom
of Strain
Gauge
to Load
App (in)
Width
(in)
Thickness
(in)
Steel
11-10/16
5-11/16
1-14/16
0.329
Aluminum
11-4/16
7-10/16
2-8/16
0.310
Brass
10-8/16
7-10/16
2-8/16
0.322
Cantilever
Table 3: ??
650:350 Mechanical Measurements
Section 1B
10/23/2023
Lazarus
TA: Shai
Sabaroche
Weight
(lbs)
Aluminum
Brass
Steel
Axial (mV)
Transverse
(mV)
Axial (mV)
Transverse
(mV)
Axial (mV)
Transverse
(mV)
SI-Axial
(10^-6)
0
0.004
0.002
0.001
0.000
-0.005
0.000
-1
1
0.061
-0.018
-0.033
-0.011
-0.016
-0.006
-7
2
0.117
-0.038
-0.067
-0.022
-0.030
-0.011
-14
3
0.173
-0.057
-0.100
-0.033
-0.045
-0.016
-20
4
0.229
-0.076
-0.134
-0.044
-0.061
-0.021
-27
5
0.284
-0.095
-0.168
-0.055
-0.078
-0.026
-33
6
0.338
-0.113
-0.202
-0.066
-0.093
-0.031
-40
7
0.394
-0.131
-0.235
-0.077
-0.110
-0.036
-46
6
0.338
-0.113
-0.201
-0.066
-0.091
-0.031
-40
5
0.283
-0.094
-0.168
-0.055
-0.074
-0.026
-33
4
0.229
-0.075
-0.134
-0.045
-0.057
-0.021
-27
3
0.174
-0.056
-0.101
-0.034
-0.040
-0.016
-20
2
0.120
-0.036
-0.067
-0.024
-0.023
-0.011
-15
1
0.065
-0.016
-0.033
-0.013
-0.006
-0.006
-8
0
0.010
0.004
0.001
-0.002
0.011
-0.001
-1
Table 4: ??
References
[1] Figliola, R. S., & Beasley, D. E. (2019).
Theory and
design for mechanical measurements. Wiley.
[2] Guo, Z. (2022). (tech.).
MAE 350 Lab_F22
.
[3] Guo, Z. (2022). (tech).
Sample Lab Report 1.
[4] Hysteris error CITATION!!
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- The strain rosette shown has angles 0, = 0° (aligned with the x-axis), %3D 0, = 60°, and 0̟ = 120°, We attach it to a body and then load the body. The strain rosette reports the following values: ɛ, = 500µ , ɛ, = 1200µ- and E, = 100µ- %3D m Recall for a delta rosette: Ex = Ea (2ɛ, +2ɛ, - Eg) 2 Y xy %3D a Sketch Mohr's Circle for Strain. Indicate the following Y Max on your sketch the following 20,, ɛ,, ɛ,, and InPlane 2 p Determine the following: E1. E2. 0p. YMax InPlanearrow_forwardA tensile specimen made of hot-rolled AISI 1020 steel is loaded to point corresponding to a strain of 40%. 60 Su = 66 ksi Stress σ (ksi) 40 20 Sy= = 39 ksi Se = 36 ksi Hot-rolled 1020 steel F | G | H 0 10 20 30 40 50 60 0 70 80 90 100 110 120 130 140 150 160 Strain € (%) ☐ T 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 Area ratio R 0.1 0.2 0.3 0.4 0.5 Area reduction A, What value of Sy is applicable to this location? 0.6arrow_forwardThe thin-walled open cross section shown is transmitting torque 7. The angle of twist ₁ per unit length of each leg can be determined separately using the equation 01 = 3Ti GLIC 3 where G is the shear modulus, ₁ is the angle of twist per unit length, T is torque, and L is the length of the median line. In this case, i = 1, 2, 3, and T; represents the torque in leg i. Assuming that the angle of twist per unit length for each leg is the same, show that T= Lic³ and Tmaz = G01 Cmax Consider a steel section with Tallow = 12.40 kpsi. C1 2 mm L1 20 mm C2 3 mm L2 30 mm C3 2 mm L3 25 mm Determine the torque transmitted by each leg and the torque transmitted by the entire section. The torque transmitted by the first leg is | N-m. The torque transmitted by the second leg is N-m. The torque transmitted by the third leg is N-m. The torque transmitted by the entire section is N-m.arrow_forward
- A tensile specimen made of hot-rolled AISI 1020 steel is loaded to point corresponding to a strain of 40%. 60 Su = 66 ksi Stress σ (ksi) S₁ = 39 ksi 40 Se = 36 ksi Hot-rolled 1020 steel 20 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 Strain € (%) 0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 Area ratio R 0.1 0.2 0.3 0.4 0.5 Area reduction A, What value of area ratio is applicable to this location? 0.6arrow_forward2arrow_forwardA tensile specimen made of hot-rolled AISI 1020 steel is loaded to point corresponding to a strain of 49%. 60 Su = 66 ksi Stress σ (ksi) Sy = 39 ksi 400B Se = 36 ksi Hot-rolled 1020 steel 20 F 0 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 Strain € (%) 0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 Area ratio R 0.1 0.2 0.3 0.4 0.5 Area reduction A, What value of Su is applicable to this location? 0.6arrow_forward
- A 45° degree strain gauge rosette positioned on the surface of a structure under stress recorded the following values of normal strain: Gauge A: Gauge B: (oriented at 45° anti-clockwise to gauge A) Gauge C: -275×10 (oriented at 90° anti-clockwise to gauge A) What is the value of normal strain and shear strain on a plane inclined at 29° clockwise from gauge B? -6 - 195 × 10 (oriented in the postive x direction) 145 x 10-6 a) The normal strain and shear strain on a plane inclined at 29° clockwise from gauge B are 0.3 microstrain and 602.1 microstrain, respectively. b) The normal strain and shear strain on a plane inclined at 29° clockwise from gauge B are 0.3 microstrain and 301.1 microstrain, respectively. c) The normal strain and shear strain on a plane inclined at 29° clockwise from gauge B are 108.5 microstrain and 334.9 microstrain, respectively. d) The normal strain and shear strain on a plane inclined at 29° clockwise from gauge B are 108.5 microstrain and 167.4 microstrain,…arrow_forward7:49 1 Search Expert Q&A Done 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.arrow_forward400 350 300 250 200 150 100 50 0.002 0.004 0.006 0.008 0.01 Strain Stress (MPa)arrow_forward
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