Lecture 6

BETM3583 Vibration and Analysis Monitoring

MODAL ANALYSIS USING PIEZO FILM SENSORS

1.1 INTRODUCTION

Today, modal analysis has become a widespread means of finding the modes of vibration of a
machine or structure. Modal analysis is the study of the dynamic properties of structures under
vibration excitation. Modal analysis used to determine a structure’s vibration characteristics
such as natural frequencies and mode shapes. It is the most fundamental of all dynamic
analysis types and is generally the starting point for other, more detailed dynamic analysis. The
benefit of modal analysis is to allow the design to avoid resonant vibrations or to vibrate at a
specified frequency (speakers, for example). Modal analysis also gives the engineers an idea of
how the design will respond to different types of dynamic loads. It is also helps in calculating
solution controls (time steps, etc.) for other dynamic analyses.

During the design and engineering applications, the characteristics of the material are
regarded as one of the most significant aspects of the process. This is to prevent any
component or structure failure. Modes are inherent properties of a structure, and are
determined by the material properties (mass, damping, and stiffness), and boundary conditions
of the structure. Each mode is defined by a natural (modal or resonant) frequency, modal
damping, and a mode shape (i.e. the so-called “modal parameters”). If either the material
properties or the boundary conditions of a structure change, its modes will change.

1.2 PROBLEM STATEMENT

The majority of structures can be made to resonate, i.e. to vibrate with excessive oscillatory
motion. Resonant vibration is mainly caused by an interaction between the inertial and elastic
properties of the materials within a structure. Resonance is often the cause of, or at least a
contributing factor to many of the vibration and noise related problems that occur in structures
and operating machinery, i.e. that small forces can result in important deformation, and
possibly, damage can be induced in the structure. To better understand any structural vibration
1
problem, the resonant frequencies of a structure need to be identified and quantified. A
common and useful way of doing this is to define its modes of vibration. Each mode is defined
by a modal frequency and a mode shape. Today, modal analysis has become a widespread
means of finding the modes of vibration of a machine or structure. In every development of a
new or improved mechanical product, structural dynamics testing on product prototypes is used

to assess its real dynamic behavior. Vibration tests as well as numerical simulations are used to
get some prior insight into the problem.

1.3 OBJECTIVES

a) To obtained the vibration signal of circular plate brass under impact.

b) To identify the natural frequency and mode shape of circular plate brass using
piezofilm sensor.
c) To determine materials properties of Poisson ratio, Young Modulus and Shear Modulus
of brass.
d) To verify and compare the modal experimental results with numerical simulation.

2.1 LITERATURE REVIEW

a) Impact Testing

With the ability to compute FRF measurements in an FFT analyzer, impact testing became
popular during the late 1970s as a fast, convenient, and relatively low cost way of finding the
mode shapes of a structure. To perform an impact test, all that is needed is an impact hammer
with a load cell attached to its head to measure the input force, a piezofilm sensor to measure
the response at a single fixed point, a two channel FFT analyzer to compute FRFs, and post
processing software for identifying and displaying the mode shapes in animation. Piezofilm is
a transducer with unique capabilities to produce voltage proportional to compressive or tensile
mechanical stress or strain. In a typical impact test, the piezofilm is attached to a single point
on the structure, and the hammer is used to impact it at points and directions as required to
define its mode shapes. FRFs are computed one at a time, between each impact point and the
fixed response point. Modal parameters are defined by curve fitting the resulting set of FRFs.
Figure 1 depicts the impact testing process.

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 Figure 1: Impact Testing

b) Piezoelectric Sensor

The ability of certain crystals to produce electricity by means of applying mechanical impact or
stress is called piezoelectricity. The piezoelectricity term is originally derived from Greek
word, piezein which means to squeeze or to press. It was first discovered by Pierre and Jacques
Curie in the latter by part of the 19th century. They found that minerals for example quartz and
tourmaline can convert mechanical energy to electrical. The sensor used to sense the signals
from the impact is called as piezoelectric sensor which can be used to sense movement,
vibration and shock measurement. Piezoelectric crystals for example quartz, when being
subjected to an external voltage, can change its own dimension by a small amount of
nanometres scales, besides also able to generate electrical charge with applied pressure. Such
phenomenon can be applied in numerous application such as production and detection of
sound, generation of high voltages, electronic frequency generation, and ultrafine focusing of
optical assemblies.

3
3.1 METHODOLOGY

Modes are a mathematical concept or construct, and are a convenient way of describing
resonant vibration. Modes can be determined both experimentally and analytically.
Analytically speaking, modes are solutions to differential equations of motion that describe the
linear, stationary vibration of a structure. Experimentally, all modal testing is done by
measuring ODSs, and then interpreting or post processing them in a specific manner to define
mode shapes.
In this study, the determination of elastic properties such as Poisson’s ration, young
modulus and shear modulus is based on the measurement of the two first resonant frequencies
that obtained from the experiment. Knowing the diameter, thickness and mass of the specimen,
with the knowledge of the vibration modes, natural frequencies the elastic constants of the
brass material can be determined.

a) Experimental Analysis

For experimental analysis, Impact Hammer software is used to record the vibration signals
from the impact force signal. Then, the signal from the piezofilm sensor will be analysed by
using MATLAB software to obtain the frequencies. In this case, MATLAB is used to utilise
measurements of dynamic input forces and output responses on the specimen.

b) Numerical Simulation Analysis

ANSYS software is used to simulate the numerical analysis. ANSYS is starts out with
knowledge about the structure geometry, boundary conditions and material characteristics. The
post processing is done to obtain the frequencies and the mode shapes of the involved material.

3.1.1 Equipment

1. Laptop with installed software (Impact Hammer and Matlab)

2. Data Acquisition
3. Impact hammer
4. Piezoelectric film sensor
5. Support foam

4
6. Circular plate specimen- brass (diameter: 350 mm)
7. Experimental location: Semi-anechoic room, acoustic lab

3.1.2 Experimental Set Up

Impact testing is performed to find mode shapes of the specimen. Modal analysis systems
composed of impact hammer, sensor such as transducers (piezofilm sensor), data acquisition
system and a host PC.

Figure 2: Experimental set up

3.2 PROCEDURES

1. The equipment are fixed in the appropriate place and checked to make sure it is
functioning properly and safe to be used. Experimental equipment being set up as in
Figure 2.
2. Four paces of foam supporters are used to support the specimen impact and are
positioned at the intersection of nodal diameters.
3. Piezoelectric film sensor been attached on the material outer edge of the specimen.
4. Prepared setting on Impact hammer software and click Run button.
5. Force is applied on the specimen with the impact hammer at the center of the material
(for flexural) and at 90 away from the center point of the material (for torsional).
a) Block Size = Min Sample Rate (Hz) = 3200

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 b) Channel Type (EJ) = Left with only Force (F) and Voltage (V)
6. Force applied and signals detected were automatically recorded by the software. UKM-
TDMS File Viewer is opened to view the file that auto-saved by software earlier.
7. From file selected on step #6, signal behaviour can be observed and amount of Force
(F) applied obtained from graph.
8. TDMS file were convert to TXT file. All the triggers was saved to TXT file. Then
STOP button is clicked.
9. Matlab software was run to analyse the signals. Using the Matlab coding, type in the
location and file name of TXT file converted earlier into Matlab coding (refer line #5)
and click RUN.
10. Three graphs obtained. Graph 1 - signal in the Time-domain, graph 2 - signal in
Frequency-domain, Graph 3 - I- kaz™ signal.
11. Step #8 is repeated to obtain graph for all triggers.

6
4.1 RESULT AND DISCUSSION

4.1.1 Experimental Result

In this experiment, the objective is to determine the material properties of Poisson’s ratio,
Young modulus and shear modulus of brass specimen from the identified natural frequency.
The material properties obtained from the experiment will be compared with the numerical
simulation data.

From the specimen, few triggers ranging from 1 to 6 were obtained. The triggers are
analysed using MATLAB coding to get the natural frequencies, f. A plot of time domain and
frequency domain were obtained. For example, the trigger 1 signal of brass specimen for
torsional and flexural modes of vibration signal is shown in Figure 3 and figure 4 below. The
frequency recorded is obtained from the highest peak in the frequency domain graph.

Figure 3: Time domain and frequency domain graph from trigger 1, torsional modes

7
 Figure 4: Time domain and frequency domain graph from trigger 1, flexural modes

In order to find the mechanical properties of the specimen, we apply three steps for the
calculation from the experimental data. The steps are as follows:

1) The Poisson’s ratio (𝜇) from the experimental values for the first and second natural
resonant frequencies is determined.
2) Two independent values for E (dynamic Young’s modulus) are calculated using the
Poisson’s ratio from step 1 and the first natural and second natural resonant
frequencies. E is determined as the average of the two independent calculations.
3) The value of G (dynamic Shear Modulus) is calculated using the Poisson’s ratio from
step 1 and the calculated value of E from step 2.

Poisson’s ratio can be determined by the frequency ratio of the torsional and flexural modes of
vibration signal (f1/f2) and the dimension ratio (t/R). It is known that the disk-shaped brass plate
has the value of thickness (t), radius (R) and mass (m) as follows:

𝑡 = 1.5 𝑚𝑚 𝑅 = 175 𝑚𝑚 𝑚 = 1195.74

Where,
f1 = first natural resonant frequency (Hz) of the plate (torsional resonant frequency).
f2 = second natural resonant frequency (Hz) of the plate (flexural resonant frequency)

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The value of 𝑓1 and 𝑓2 are obtained from the experiment. The data from the Impact hammer
software are translated to get the graph of time domain. By using the FFT, the frequency
domain graph is generated.

By referring to the ASTM E1876-15; Standard Test Method for Dynamic Young’s
Modulus, Shear Modulus and Poisson’s Ratio by Impulse Excitation of Vibration (for disk-
shaped specimens), the formulas for calculating 𝜇, 𝐸 𝑎𝑛𝑑 𝐺 are applied. The calculations are
summarized as follows for trigger 1. Calculation for trigger 2 to trigger 6 will be shown in
Appendix C.

Trigger 1

𝑓1 = 44.81 𝐻𝑧
𝑓2 = 77.81 𝐻𝑧
𝑓2 77.81 𝐻 𝑡 1.5 𝑚𝑚 = 8.5714 × 10−3 ≈ 0
𝑓1
= 44.81 𝐻𝑧 = 1.7364 𝑅 = 175 𝑚𝑚

Referring to the TABLE Poisson’s Ratio (μ) Values (as a function of f2/f1 and t/R), the value
of Poisson’s Ratio as interpolated in the graph.

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 Poisson's Ratio vs f2/f1
0.348
0.346 1.75, 0.346
0.344
0.342
y = 0.68x - 0.844
0.34
0.338
0.336
0.334
0.332
0.33
1.725, 0.329
0.328
1.72 1.725 1.73 1.735 1.74 1.745 1.75 1.755

For 𝑓2⁄𝑓1 = 1.736, µ = 0.336

a) To find the Young’s Modulus (𝑬)

𝐸1 + 𝐸2
𝐸= 2
2 2 2 2 2 2
𝐸1 = 37.6991𝑓1 2(1−𝜇 ) 𝐸2 = 37.6991𝑓2 2(1−𝜇 )
𝐾1 𝑡 3 𝐾2 𝑡 3

Where
𝐾1 = first natural geometric factor from Table A1.2a (using linear interpolation as necessary)
𝐾2 = second natural geometric factor from Table A1.3a (using linear interpolation as necessary)

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 K1 vs Poisson's Ratio
5.36
5.34 0.3, 5.34
5.32
5.3
y = -2.96x + 6.228
5.28
5.26
5.24
5.22
5.2
0.35, 5.192
5.18
0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36

For µ = 0.336, 𝐾1 = 5.233

K2 vs Poisson's Ratio
9.12
0.35, 9.111
9.1

9.08
y = 2.22x + 8.334
9.06

9.04

9.02

9 0.3, 9

8.98
0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36

For µ = 0.336, 𝐾2 = 9.080

37.6991(44.81)2(350)21195.74(1 − 0.3362)
𝐸1 = = 106.3466 𝐺𝑃𝑎
(5.233)2(1.5)3

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 37.6991(77.81)2(350)21195.74(1 − 0.3362)
𝐸2 =
= 106.5869 𝐺𝑃𝑎
(9.080)2(1.5)3
106.3466+106.5869
= 2 𝐺𝑃𝑎 = 106.4668 𝐺𝑃𝑎

b) To find the Shear Modulus (𝑮)

𝐸 106.4668 𝐺𝑃𝑎 = 39.8454 𝐺𝑃𝑎

𝐺=
[2(1 + 𝜇)] = [2(1 + 0.336)]

c) The value of Poisson’s Ratio, Young’s Modulus and Shear Modulus of the Brass
Plate

All the calculated results from the data of the six (6) triggers are considered to find the average
value of the mechanical properties of the material. The average value of the data is calculated
using the mean formula as given:

n
x  1  xj
n
j 1

The Poisson’s Ratio (𝝁)

𝜇1 = 0.336
𝜇2 = 0.347
𝜇3 = 0.300
𝜇4 = 0.319
𝜇5 = 0.322
𝜇6 = 0.319
1 6
𝜇= ∑ 0.336+0.347+0.300+0.319+0.322+0.319
𝜇𝑗 = = 0.327
6 𝑗=1 6

The Young’s Modulus (𝑬)

𝐸1 = 106.4668 𝐺𝑃𝑎
𝐸2 = 107.8157 𝐺𝑃𝑎
𝐸3 = 108.2520 𝐺𝑃𝑎
𝐸4 = 105.5979 𝐺𝑃𝑎
𝐸5 = 105.7453 𝐺𝑃𝑎
𝐸6 = 105.5979 𝐺𝑃𝑎
1 6
𝐸= ∑ 𝐸𝑗 =
106.4668+107.8157+108.2520+105.5979+105.7453+105.5979
= 106.5793 𝐺𝑃𝑎
6 𝑗=1 6

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The Shear Modulus (𝑮)

𝐺1 = 39.8454 𝐺𝑃𝑎
𝐺2 = 40.0207 𝐺𝑃𝑎
𝐺3 = 41.6354 𝐺𝑃𝑎
𝐺4 = 40.0295 𝐺𝑃𝑎
𝐺5 = 39.9944 𝐺𝑃𝑎
𝐺6 = 40.0295 𝐺𝑃𝑎
1 6
𝐺= ∑ 39.8454 +40.0207+41.6354+40.0295+39.9944+40.0295
𝐺𝑗 = = 40.2592 𝐺𝑃𝑎
6 𝑗=1 6

The overall result of natural frequencies, Poisson’s ratio, young modulus and shear modulus
for trigger 1 to 6 is presented in Table 1.

Table 1: Average value for f1 and f2 with material properties

Trigger f1 (Hz) f2 (Hz) μ E (GPa) G (GPa)

1 44.81 77.81 0.336 106.467 39.845

2 45.01 78.81 0.347 107.816 40.021
3 45.61 76.61 0.300 108.252 41.635
4 44.81 76.61 0.319 105.598 40.030
5 44.81 76.81 0.322 105.745 39.994
6 44.81 76.61 0.319 105.598 40.030
𝑿̅ 44.98 77.21 0.327 106.579 40.259

4.1.2 Numerical Simulation Result

The models developed in geometry need to be defined for the material properties and mesh
before the model can be analysed. First, the material properties were applied for brass. The
model was solved with a mesh consisting of 1922 elements and 13804 nodes.

Mesh Setting

Element size 10 mm

Nodes 13804
Elements 1933

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Material properties of material are given in table below:

Material Brass

Young modulus, E 102 GPa

Poisson ratio,  0.345

Shear modulus, G 37.918 GPa

Figure 5: Mesh of disk-shape plate

We can influence computational time of the analysis, when a range of frequencies or number
of mode shapes is specified. The type of solver and the solution method in program ANSYS is
selected automatically. The first nine mode shapes are shown in Figure 6 and the first nine
natural frequencies are shown in Table 1.

Table 1: Natural frequency of particular mode shapes

Mode Shape Frequency [Hz]
1.
2.
0.
3.
4.

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 5. 1.7521e-003
6. 2.7192e-003
7. 44.07
8. 44.072
9. 76.987

The nine modes of vibration are displacements and rotations of rigid bodies. Places with
maximum deflection, are shown in red and places with little or no deflection, are shown in
blue.

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

Figure 6: Mode shapes for modal frequency of (a) to (d) 0 Hz, (e) 0.0017 Hz, (f) 0.0027 Hz,
(g) 44.07, (h) 44.072 and (i) 76.987 Hz

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It is observed from the tabulated values that the deflection is dependent on the frequency. As
the frequency increases, the deflection is increased. The following bar chart indicates the
frequency at each calculated mode. It can be seen at the first six modes, there is no frequency
recorded until it reach mode shape 7.

Figure 7: Frequency at each calculated mode

4.1.3 Comparison between Experimental and Simulation Result

The percentage of difference between the result of experimental and ANSYS can be calculated
by below formula:-

Example calculation percentage of difference for frequency:

44.98 − 44.07
𝑓 − 𝑓𝑠𝑖𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛 | × 100% = 2.02%
| 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 | × 100% =
|
𝑓𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 44.98

Table 2 shows the comparison between experimental and numerical simulation (ANSYS) and
their percentage of difference.

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 Table 2: Experimental and numerical simulation frequency and stresses with percentage of
difference.

Numerical
Percentage of
Experimental Simulation
Difference
(ANSYS)

f1 44.98 Hz 44.07 Hz 2.02%

f2 77.21 Hz 76.99 Hz 0.28%

Young modulus, E 106.579 GPa 102 GPa 4.30%

Poisson ratio,  0.327 0.345 5.22%

Shear modulus, G 40.259 GPa 37.918 GPa 5.81%

From the table above shows the overall obtained result from both of the experimental and
numerical simulation (ANSYS). It can be shown that the percentage of differences between
experimental and numerical simulation (ANSYS) are small because the obtained value is less
than 10%. By comparing the detected modes (frequency), the obtained frequency and stresses
from the experimental and simulation is fairly close. Both of the value between experimental
and simulation are significant.

5.1 CONCLUSION

Each mode is defined by a natural (modal or resonant) frequency and a mode shape. If either
the material properties or the boundary conditions of a structure change, its modes will change.
For instance, if mass is added to a structure, it will vibrate differently. At or near the natural
frequency of a mode, the overall vibration shape (“operating deflection shape”) of a structure
will tend to be dominated by the mode shape of the resonance.

In conclusion, properties of brass material are achieved using piezofilm vibration

signals. From the experiment and simulation, a modal analysis tells us is at which frequency
the specimen will absorb all the energy applied to it by an excitation, and what the shape looks
like which corresponds to this frequency. It is observed from the simulation that the deflection
is dependent on the frequency. As the frequency increases, the deflection is increased.

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 By comparing the detected modes (frequency), the obtained stresses from the
experimental and simulation is fairly close. The obtained value for both approaches is slightly
different might due to an error occurred during the experiment that affects the data, i.e.
knocking force is not constant. Thus, affect the obtained vibration signal of the circular plate
brass. Therefore, a constant impact force is needed to reduce the error.

6.1 REFERENCES

ASTM E1876-15: Standard Test Method for Dynamic Young’s Modulus, Shear Modulus, and
Poisson’s Ratio by Impulse Excitation of Vibration
Bocko, J. et al., 2014. Modal Analysis of Circular Plates. Applied Mechanics and Materials,
611, pp.245–251. Available at: http://www.scientific.net/AMM.611.245.
He, J. & Fu, Z.-F., 2001. Modal Analysis. Modal Analysis, 117(10), p.291. Available at:
http://cedb.asce.org/cgi/WWWdisplay.cgi?9104604.
Hemmatnezhad, M., 2015. Experimental Vibration Investigation of Annular Plates. ,
01(January), pp.1–7.
Jhung, M.J., Choi, Y.H. & Ryu, Y.H., 2009. Free vibration analysis of circular plate with
eccentric hole submerged in fluid. Nuclear Engineering and Technology, 41(3), pp.355–
364.
Nuawi, M.Z., 2012. Mesokurtosis Zonal Nonparametric Signal Analysis for Dynamic
Characterisation of Metallic Material. , 24, pp.21–27.
Nuawi, M.Z. et al., 2013. Time domain analysis method of the impulse vibro-acoustic signal
for fatigue strength characterisation of metallic material. Procedia Engineering, 66,
pp.539–548. Available at: http://dx.doi.org/10.1016/j.proeng.2013.12.106.
Richardson, M.H., 1997. Is it a mode shape, or an operating deflection shape? Sound and
Vibration, (30th Anniversary Issue), pp.1–8.

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 APPENDIX A

Time Domain and Frequency Domain of Trigger 1 to Trigger 6

Torsional Mode

Trigger 1

19
Trigger 2

20
Trigger 3

21
Trigger 4

22
Trigger 5

23
Trigger 6

24
Flexural Mode

Trigger 1

25
Trigger 2

26
Trigger 3

27
Trigger 4

28
Trigger 5

29
Trigger 6

30
APPENDIX B

Table for Poisson’s ratio, K1 AND K2 values

31
APPENDIX C

Trigger 2

𝑓1 = 45.01 𝐻𝑧

𝑓2 = 78.81 𝐻𝑧

𝑓2 −3
= 175 𝑚𝑚 = 8.5714 × 10 ≈ 0
78.81 𝐻𝑧 𝑡 1.5 𝑚𝑚
𝑓1 = 45.01 𝐻𝑧 = 1.751 𝑅

Referring to the Table A1.1, (Poisson’s Ratio (μ) values (as a function of f2/f1 and t/r)), the
value of Poisson’s Ratio as interpolated in the graph.

Poisson's Ratio vs f2/f1

0.364
0.362 1.775, 0.362
0.36
0.358
y = 0.64x - 0.774
0.356
0.354
0.352
0.35
0.348
0.346 1.75, 0.346
0.344
1.745 1.75 1.755 1.76 1.765 1.77 1.775 1.78

For 𝑓2⁄𝑓1 = 1.751, µ = 0.347

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a) To find the Young’s Modulus (𝑬)

𝐸1 + 𝐸2
𝐸= 2
2 2 2 2 2 2
𝐸1 = 37.6991𝑓1 2(1−𝜇 ) 𝐸2 = 37.6991𝑓2 2(1−𝜇 )
𝐾1 𝑡 3 𝐾2 𝑡 3

Where
𝐾1 = first natural geometric factor from Table A1.2a (using linear interpolation as necessary)
𝐾2 = second natural geometric factor from Table A1.3a(using linear interpolation as necessary)

K1 vs Poisson's Ratio
5.36
5.34 0.3, 5.34

5.32
5.3
y = -2.96x + 6.228
5.28
5.26
5.24
5.22
5.2
0.35, 5.192
5.18
0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36

For µ = 0.347, 𝐾1 = 5.201

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 K2 vs Poisson's Ratio
9.12
0.35, 9.111
9.1

9.08
y = 2.22x + 8.334
9.06

9.04

9.02

9 0.3, 9

8.98
0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36

For µ = 0.347, 𝐾2 = 9.104

37.6991(45.01)2(350)21195.74(1 − 0.3472)
𝐸1 =
= 107.7842 𝐺𝑃𝑎
(5.201)2(1.5)3
37.6991(78.81)2(350)21195.74(1 − 0.3472)
𝐸2 =
= 107.8472 𝐺𝑃𝑎
(9.104)2(1.5)3
107.7842+107.8472
= 2 𝐺𝑃𝑎 = 107.8157 𝐺𝑃𝑎

b) To find the Shear Modulus (𝑮)

𝐸 107.8157 𝐺𝑃𝑎 = 40.0207 𝐺𝑃𝑎

𝐺=
[2(1 + 𝜇)] = [2(1 + 0.347)]

Trigger 3

𝑓1 = 45.61 𝐻𝑧

𝑓2 = 76.61 𝐻𝑧

𝑓2 −3
= 175 𝑚𝑚 = 8.5714 × 10 ≈ 0
76.61 𝐻𝑧 𝑡 1.5 𝑚𝑚
𝑓1
= 45.61 𝐻𝑧 = 1.680 𝑅

Referring to the Table A1.1, (Poisson’s Ratio (μ) values (as a function of f2/f1 and t/r)), the
value of Poisson’s Ratio as interpolated in the graph.

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 Poisson's Ratio vs f2/f1
0.314
0.312 1.7, 0.312
0.31
0.308
y = 0.6x - 0.708
0.306
0.304
0.302
0.3
0.298
1.675, 0.297
0.296
1.67 1.675 1.68 1.685 1.69 1.695 1.7 1.705

For 𝑓2⁄𝑓1 = 1.680, µ = 0.300

a) To find the Young’s Modulus (𝑬)

𝐸1 + 𝐸2
𝐸= 2
2 2 2 2 2 2
𝐸1 = 37.6991𝑓1 2(1−𝜇 ) 𝐸2 = 37.6991𝑓2 2(1−𝜇 )
𝐾1 𝑡 3 𝐾2 𝑡 3

Where
𝐾1 = first natural geometric factor from Table A1.2a (using linear interpolation as necessary)
𝐾2 = second natural geometric factor from Table A1.3a(using linear interpolation as necessary)

35
 For µ = 0.300, 𝐾1 = 5.340

For µ = 0.300, 𝐾2 = 9.000

37.6991(45.61)2(350)21195.74(1 − 0.3002)
𝐸1 =
= 108.6198 𝐺𝑃𝑎
(5.340)2(1.5)3
37.6991(76.61)2(350)21195.74(1 − 0.3002)
𝐸2 =
= 107.8841 𝐺𝑃𝑎
(9.000)2(1.5)3
108.6198+107.8841
= 2 𝐺𝑃𝑎 = 108.2520 𝐺𝑃𝑎

b) To find the Shear Modulus (𝑮)

𝐸 108.2520 𝐺𝑃𝑎 = 41.6354 𝐺𝑃𝑎

𝐺=
[2(1 + 𝜇)] = [2(1 + 0.300)]

Trigger 4

𝑓1 = 44.81 𝐻𝑧

𝑓2 = 76.61 𝐻𝑧

𝑓2 −3
= 175 𝑚𝑚 = 8.5714 × 10 ≈ 0
76.61 𝐻𝑧 𝑡 1.5 𝑚𝑚
𝑓1
= 44.81 𝐻𝑧
= 1.710 𝑅

Referring to the Table A1.1, (Poisson’s Ratio (μ) values (as a function of f2/f1 and t/r)), the
value of Poisson’s Ratio as interpolated in the graph.

36
 Poisson's Ratio vs f2/f1
0.33
1.725, 0.329
0.328
0.326
0.324 y = 0.68x - 0.844
0.322
0.32
0.318
0.316
0.314
0.312 1.7, 0.312
0.31
1.695 1.7 1.705 1.71 1.715 1.72 1.725 1.73

For 𝑓2⁄𝑓1 = 1.710, µ = 0.319

a) To find the Young’s Modulus (𝑬)

𝐸1 + 𝐸2
𝐸= 2
2 2 2 2 2 2
𝐸1 = 37.6991𝑓1 2(1−𝜇 ) 𝐸2 = 37.6991𝑓2 2(1−𝜇 )
𝐾1 𝑡 3 𝐾2 𝑡 3

Where
𝐾1 = first natural geometric factor from Table A1.2a (using linear interpolation as necessary)
𝐾2 = second natural geometric factor from Table A1.3a (using linear interpolation as necessary)

37
 K1 vs Poisson's Ratio
5.36
5.34 0.3, 5.34
5.32
5.3
y = -2.96x + 6.228
5.28
5.26
5.24
5.22
5.2
0.35, 5.192
5.18
0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36

For µ = 0.319, 𝐾1 = 5.284

38
 K2 vs Poisson's Ratio
9.12
0.35, 9.111
9.1

9.08
y = 2.22x + 8.334
9.06

9.04

9.02

9 0.3, 9

8.98
0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36

For µ = 0.319, 𝐾2 = 9.042

37.6991(44.81)2(350)21195.74(1 − 0.3192)
𝐸1 =
= 105.6930 𝐺𝑃𝑎
(5.284)2(1.5)3
37.6991(76.61)2(350)21195.74(1 − 0.3192)
𝐸2 =
= 105.5028 𝐺𝑃𝑎
(9.042)2(1.5)3
105.6930+105.5028
= 2 𝐺𝑃𝑎 = 105.5979 𝐺𝑃𝑎

b) To find the Shear Modulus (𝑮)

𝐸 105.5979 𝐺𝑃𝑎 = 40.0295 𝐺𝑃𝑎

𝐺=
[2(1 + 𝜇)] = [2(1 + 0.319)]

Trigger 5

𝑓1 = 44.81 𝐻𝑧

𝑓2 = 76.81 𝐻𝑧

𝑓2 −3
= 175 𝑚𝑚 = 8.5714 × 10 ≈ 0
76.81 𝐻𝑧 𝑡 1.5 𝑚𝑚
𝑓1 = 44.81 𝐻𝑧 = 1.714 𝑅

Referring to the Table A1.1, (Poisson’s Ratio (μ) values (as a function of f2/f1 and t/r)), the
value of Poisson’s Ratio as interpolated in the graph.

39
 Poisson's Ratio vs f2/f1
0.33
1.725, 0.329
0.328
0.326
0.324 y = 0.68x - 0.844
0.322
0.32
0.318
0.316
0.314
0.312 1.7, 0.312
0.31
1.695 1.7 1.705 1.71 1.715 1.72 1.725 1.73

For 𝑓2⁄𝑓1 = 1.714, µ = 0.322

a) To find the Young’s Modulus (𝑬)

𝐸1 + 𝐸2
𝐸= 2
2 2 2 2 2 2
𝐸1 = 37.6991𝑓1 2(1−𝜇 ) 𝐸2 = 37.6991𝑓2 2(1−𝜇 )
𝐾1 𝑡 3 𝐾2 𝑡 3

Where
𝐾1 = first natural geometric factor from Table A1.2a (using linear interpolation as necessary)
𝐾2 = second natural geometric factor from Table A1.3a (using linear interpolation as necessary)

40
 K1 vs Poisson's Ratio
5.36
5.34 0.3, 5.34
5.32
5.3
y = -2.96x + 6.228
5.28
5.26
5.24
5.22
5.2
0.35, 5.192
5.18
0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36

For µ = 0.322, 𝐾1 = 5.275

41
 K2 vs Poisson's Ratio
9.12
0.35, 9.111
9.1

9.08
y = 2.22x + 8.334
9.06

9.04

9.02

9 0.3, 9

8.98
0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36

For µ = 0.322, 𝐾2 = 9.049

37.6991(44.81)2(350)21195.74(1 − 0.3222)
𝐸1 =
= 105.8269 𝐺𝑃𝑎
(5.275)2(1.5)3
37.6991(76.81)2(350)21195.74(1 − 0.3222)
𝐸2 =
= 105.6636 𝐺𝑃𝑎
(9.049)2(1.5)3
105.8269+105.6636
= 2 𝐺𝑃𝑎 = 105.7453 𝐺𝑃𝑎

b) To find the Shear Modulus (𝑮)

𝐸 105.7453 𝐺𝑃𝑎 = 39.9944 𝐺𝑃𝑎

𝐺=
[2(1 + 𝜇)] = [2(1 + 0.322)]

Trigger 6

𝑓1 = 44.81 𝐻𝑧

𝑓2 = 76.61 𝐻𝑧

𝑓2 −3
= 175 𝑚𝑚 = 8.5714 × 10 ≈ 0
76.61 𝐻𝑧 𝑡 1.5 𝑚𝑚
𝑓1 = 44.81 𝐻𝑧 = 1.710 𝑅

Referring to the Table A1.1, (Poisson’s Ratio (μ) values (as a function of f2/f1 and t/r)), the
value of Poisson’s Ratio as interpolated in the graph.

42
 Poisson's Ratio vs f2/f1
0.33
1.725, 0.329
0.328
0.326
0.324 y = 0.68x - 0.844
0.322
0.32
0.318
0.316
0.314
0.312 1.7, 0.312
0.31
1.695 1.7 1.705 1.71 1.715 1.72 1.725 1.73

For 𝑓2⁄𝑓1 = 1.710, µ = 0.319

a) To find the Young’s Modulus (𝑬)

𝐸1 + 𝐸2
𝐸= 2
2 2 2 2 2 2
𝐸1 = 37.6991𝑓1 2(1−𝜇 ) 𝐸2 = 37.6991𝑓2 2(1−𝜇 )
𝐾1 𝑡 3 𝐾2 𝑡 3

Where
𝐾1 = first natural geometric factor from Table A1.2a (using linear interpolation as necessary)
𝐾2 = second natural geometric factor from Table A1.3a (using linear interpolation as necessary)

43
 K1 vs Poisson's Ratio
5.36
5.34 0.3, 5.34
5.32
5.3
y = -2.96x + 6.228
5.28
5.26
5.24
5.22
5.2
0.35, 5.192
5.18
0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36

For µ = 0.319, 𝐾1 = 5.284

44
 K2 vs Poisson's Ratio
9.12
0.35, 9.111
9.1

9.08
y = 2.22x + 8.334
9.06

9.04

9.02

9 0.3, 9

8.98
0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36

For µ = 0.319, 𝐾2 = 9.042

37.6991(44.81)2(350)21195.74(1 − 0.3192)
𝐸1 =
= 105.6930 𝐺𝑃𝑎
(5.284)2(1.5)3
37.6991(76.61)2(350)21195.74(1 − 0.3192)
𝐸2 =
= 105.5028 𝐺𝑃𝑎
(9.042)2(1.5)3
105.6930+105.5028
= 2 𝐺𝑃𝑎 = 105.5979 𝐺𝑃𝑎

b) To find the Shear Modulus (𝑮)

𝐸 105.5979 𝐺𝑃𝑎 = 40.0295 𝐺𝑃𝑎

𝐺=
[2(1 + 𝜇)] = [2(1 + 0.319)]

45