module-vii –calculation of natural frequency vibration engineering
Transcript of module-vii –calculation of natural frequency vibration engineering
MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
VTU-NPTEL-NMEICT
Project Progress Report
The Project on Development of Remaining Three Quadrants to NPTEL Phase-I under grant in aid NMEICT, MHRD, New Delhi
DEPARTMENT OF MECHANICAL ENGINEERING,
GHOUSIA COLLEGE OF ENGINEERING,
RAMANARAM -562159
Subject Matter Expert Details
SME Name : Dr.MOHAMED HANEEF
PRINCIPAL, VTU SENATE MEMBER
Course Name:
Vibration engineering
Type of the Course
web
Module
VII
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
CONTENTS
Sl. No. DISCRETION
1. Lecture Notes (Calculation of Natural Frequency).
2. Quadrant -2
a. Animations.
b. Videos.
c. Illustrations.
3. Quadrant -3
a. Wikis.
b. Open Contents
4. Quadrant -4
a. Problems.
b. Self Answered Question & Answer.
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
Module-VII
CALCULATION OF NATURAL FREQUENCY 1. Rayleigh’s Method: This is the energy method to find the frequency. This method is used to find the natural frequency of the
system when transverse point loads are acting on the beam or shaft. Good estimate of fundamental
frequency can be made by assuming the suitable deflection curve for the fundamental mode. The
maximum kinetic energy is equated to maximum potential energy of the system to determine the natural
frequency.
𝜔𝑛 = �𝑔∑𝑚𝑖𝑦𝑖∑𝑚𝑖𝑦𝑖2
Where: 𝑦1 = 𝑚1𝑔 𝛼11 + 𝑚2 𝑔 𝛼12
𝑦2 = 𝑚1𝑔 𝛼12 + 𝑚2 𝑔 𝛼22
2. Stodola Method: The stodola method may be set up in the following tabular form as follows, assuming an arbitrary set of
values for the fundamental principal mode, the inertia force acting on each mass is equal to the product of
the assumed deflection and the square of the natural frequency as shown in row 2. The spring force in row
3 is equal to the total inertia force acting on each spring. Row 4 is obtained by dividing row 3, term by
term, by their respective spring constants. The calculated deflections in row 5 are found by adding the
deflections due to the springs, with the mass near the fixed end having the least deflection and so on. The
calculated deflections are then compared with the assumed deflections. This process is continued until the
calculated deflections are proportional to the assumed deflections. When this is true the assumed
deflections will represent the configuration of the fundamental principal mode of vibration of the system.
k1 m1 k2 m2 K3 M3 1.Assumed deflection
2.Inertia force
3.Spring force
4.Spring deflection
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
5.Calculated deflection
3. Matrix Iteration Method: This is an iterative procedure that leads to the principal modes of vibration of a system and its natural
frequencies.
Displacements of the masses are estimated, from which the matrix equation of the system is written. The
influence coefficients of the system are substituted into the matrix equation which is then expanded.
Normalization of the displacement and expansion of the matrix is repeated. This process is continued until
the first mode repeat itself to any desired degree of accuracy.
For the next higher modes and natural frequencies, the orthogonality principle is used to obtain a new
matrix equation that is free from any lower modes the iterative procedure is repeated.
4. Holzer Method & Dunkerley’s Method:
i) Holzer Method
Begin the Holzer tabulation with the column of position, indicating the masses of the system.
The second column is for the values of the different masses of the system; this information is
given. The third column is the product of mass and frequency squared. Displacement comes
next, and is obtained from the preceding row minus the total displacement at the end of the same
row. Column five is just the product of columns three and four. The total inertia force is inserted
in column six. It is equal to the sum of the total inertia force in the preceding row plus the inertia
force on the same row. The rest are plainly evident.
An initial displacement, usually equal to unity for convenience, is assumed. If the assumed
frequency happens to be one of the natural frequencies of the systems, the final total inertia force
on the system should be zero. This is because the system is having free vibration. If the final total
inertia force is not equal to zero, the amount indicates the discrepancy of the assumed frequency.
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
Positio
n m1 m1ω2 xi m1xiω2 �𝑚𝑖𝑥𝑖𝜔2
𝑖
1
𝑘𝑖𝑗 �𝑚𝑖𝑥𝑖𝜔2
𝑘𝑖𝑗
𝑖
1
Assumed frequency, ω =
ii) Dunkerley’s Method
The Dunkerley’s equation for multi dot system is given by
1/ ωn2 =1/ ω1
2 + 1/ ω22 + ω3
3 + …………..
Where ωn = fundamental natural frequency of the system.
ω1, ω2, ω3 etc. are the natural frequency of the system with each mass acting separately at its
point of application, in the absence of other masses, using influence coefficients Prof. Dunkerley
has suggested the following empirical equation.
1/ ωn2 = α11m1 + α22m2 + α33m3 + …………..
Dunkerley’s equation is used to determine the fundamental natural frequency of the system
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
QUADRANT-2 Animations (Animation links related ,Calculation of natural frequencies)
1. http://en.wikipedia.org/wiki/Vibration 2. http://www.brown.edu/Departments/Engineering/Courses/En4/notes_old/Freevibes/freevibes.htm
l 3. http://iitg.vlab.co.in/?sub=62&brch=175&sim=1077&cnt=2 4. http://www.acs.psu.edu/drussell/Demos/absorber/DynamicAbsorber.html 5. http://scholar.cu.edu.eg/?q=anis/files/mpe_-_2013-02-10_-_introduction.pdf
Videos (video links related , Calculation of natural frequencies)
• http://www.youtube.com/watch?v=ydflDCPUgYo • http://nptel.iitg.ernet.in/Mech_Engg/IIT%20Guwahati/Vibration%20Engineering(Video).htm • www.youtube.com/watch?v=FwN-webXqnE • http://freevideolectures.com/Course/2684/Mechanical-Vibrations/30 • http://www.cosmolearning.com/video-lectures/matrix-iteration-method-11564/ • www.freestudy.co.uk/dynamics/holzer.pdf • http://www.cdeep.iitb.ac.in/nptel/Mechanical/Dynamics%20of%20Machines/Course_home1
2.5.html • http://www.cdeep.iitb.ac.in/nptel/Mechanical/Dynamics%20of%20Machines/Course_home1
2.4.html • http://www.youtube.com/watch?v=pTRb6LVWiUM
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
ILLUSTRATIONS 1. Find the fundamental vibration for the system shown in figure (1), by using Rayleigh’s
method.
Fig: (1)
Solution:
First Influence coefficients are:
𝛼11 =1𝐾
& 𝛼12 = 𝛼13 =1𝐾
By Maxwell reciprocal theorem
𝛼21 = 𝛼12; 𝛼13 = 𝛼31
𝛼11 = 𝛼12 = 𝛼13 = 𝛼21 = 𝛼31 =1𝐾
Second Influence coefficients are:
𝛼22 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 1𝑘
+ 12𝑘
= 32𝑘
𝛂𝟐𝟐 =𝟑𝟐𝐤
By Maxwell reciprocal theorem
𝜶𝟑𝟐 = 𝜶𝟐𝟑 =𝟑𝟐𝐤
Third Influence coefficients are:
𝛼33 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 1𝑘
+ 12𝑘
+ 13𝑘
= 116𝑘
𝛂𝟑𝟑 =116𝑘
3m 2m m
2K K 3K
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
By using Rayleigh’s method 𝑦1 = 𝑚1𝑔 𝛼11 + 𝑚2 𝑔 𝛼12 + 𝑚3 𝑔 𝛼13𝑦2 = 𝑚1𝑔 𝛼21 + 𝑚2 𝑔 𝛼22 + 𝑚3 𝑔 𝛼23 𝑦3 = 𝑚1𝑔 𝛼31 + 𝑚2 𝑔 𝛼32 + 𝑚3 𝑔 𝛼33
� −−(1)
𝐲𝟏 = 3𝑚(9.81)1𝑘 + 2𝑚(9.81)
1𝑘 + 𝑚(9.81)
1𝑘 = 𝟓𝟖.𝟖𝟔
𝐦𝐤
𝐲𝟐 = 3𝑚(9.81)1𝑘 + 2𝑚(9.81)
32k + 𝑚(9.81)
32k = 𝟕𝟑.𝟓𝟕𝟓
𝐦𝐤
𝐲𝟐 = 3𝑚(9.81)1𝑘 + 2𝑚(9.81)
32k + 𝑚(9.81)
116𝑘 = 𝟕𝟔.𝟖𝟒
𝐦𝐤
Natural frequency is
𝝎𝒏 = �𝒈�𝑚1 𝑦1 + 𝑚2 𝑦2 + 𝑚3 𝑦3�𝑚1 𝑦1
2 + 𝑚2 𝑦22 + 𝑚3 𝑦3
2
𝜔𝑛 = �9.81 ��3m × 58.86 m
k � + �2m × 73.575 mk � + �𝑚 × 76.84 m
k ��
3𝑚�58.86 mk �
2 + 2𝑚 �73.575 m
k �2
+ 𝑚 �76.84 mk �
2
𝛚𝐧 = 𝟎.𝟑𝟖𝟔�𝐊𝐦
𝐫𝐚𝐝/𝐬𝐞𝐜
2. Find the fundamental vibration for the system shown in figure (2), by using Rayleigh’s
method.
𝐼 = 4 × 10−7 𝑚4; 𝐸 == 1.96 × 1011 𝑁/𝑚2
Static deflection at two points is given by
m2=50 kg m1=100 kg
180mm 180mm 300mm
Fig: (2)
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝑦1 = 𝑚1𝑔 𝛼11 + 𝑚2 𝑔 𝛼12
𝑦2 = 𝑚1𝑔 𝛼12 + 𝑚2 𝑔 𝛼22
First Influence coefficients are:
𝛼11 =𝐹𝑙3
3𝐸𝐼; 𝛼22 =
𝐹𝐿3
3𝐸𝐼 & 𝛼12 = 𝛼21 =
𝑙2(3𝐿 − 𝑙)6𝐸𝐼
𝛼11 =𝐹𝑙3
3𝐸𝐼=
𝑙3
3𝐸𝐼= 2.47 × 10−8 𝑚/𝑁
𝛼22 =𝐹𝐿3
3𝐸𝐼=
𝐿3
3𝐸𝐼= 1.148 × 10−8 𝑚/𝑁
𝛼12 = 𝛼21 =𝑙2(3𝐿 − 𝑙)
6𝐸𝐼= 4.599 × 10−8 𝑚/𝑁
By using Rayleigh’s method
𝑦1 = 100(9.81) �2.47 × 10−8� + 50(9.81) �4.599 × 10−8� = 4.85 × 10−5
𝑦2 = 100(9.81) �4.599 × 10−8� + 50(9.81) �1.148 × 10−8� = 1.044 × 10−4
𝝎𝒏 = �𝒈�𝑚1 𝑦1 + 𝑚2 𝑦2�𝑚1 𝑦1
2 + 𝑚2 𝑦22
𝜔𝑛 = �9.81[(100 × 4.85 × 10−5) + (50 × 1.044 × 10−4)]
100(4.85 × 10−5)2 + 50 (1.044 × 10−4)2
𝛚𝐧 = 𝟑𝟓𝟓.𝟒𝟕 𝐫𝐚𝐝/𝐬𝐞𝐜
3. Using Stodola method to find the fundamental mode of vibration and its natural frequency of the spring mass system shown in the figure given K1 = K2=K3=1 N/M & m1=m2= m3=1kg.
Trials K1=1K m1=1m K2=1K m2=1m K3=1K m3=1m 1. Assumed deflection 1 1 1 2. Inertia force 1 × ω2 1 × ω2 1 × ω2 3. Spring force 3ω2 2ω2 ω2 4. Spring deflection 3ω2 2ω2 ω2 5. Calculated deflection 3ω2 2ω2 ω2
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
Calculated deflection𝟑𝛚𝟐 1 1.66 2
1. Assumed deflection 1 1.66 2 2. Inertia force 1 × ω2 1.66
× ω2 2 × ω2
3. Spring force 4.66ω2 3.66ω2 2ω2 4. Spring deflection 4.66ω2 3.66ω2 2ω2 5. Calculated deflection 4.66ω2 8.32ω2 10.32ω2
Calculated deflection4.66𝛚𝟐 1 1.78 2.214
1. Assumed deflection 1 1.78 2.214 2. Inertia force 1 × ω2 1.78
× ω2 2.214 × ω2
3. Spring force 4.99ω2 3.99ω2 2.21ω2 4. Spring deflection 4.99ω2 3.99ω2 2.21ω2 5. Calculated deflection 4.994ω2 8.98ω2 11.129ω2
Calculated deflection4.66𝛚𝟐 1 1.799 2.24
The values are closed to previous values; hence the fundamental principle mode or frequency is
given by
�1
1.782.214
� = 4.994ω2 + 8.98ω2 + 11.129ω2
ω = 0.448 rad/sec
4. Using Stodola method to find the fundamental mode of vibration and its natural frequency for the branch system shown in fig:
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
Fig: (5)
Solution:
Trials K1=7K m1=4m K2=5K m2=3m K3=5K m3=2m 1. Assumed deflection 1 1 1 2. Inertia force 4 × ω2 3 × ω2 2 × ω2 3. Spring force 9ω2 3ω2 2ω2 4. Spring deflection 1.228ω2 0.6ω2 0.4ω2 5. Calculated deflection 1.22ω2 1.88ω2 1.6ω2
Calculated deflection1.22ω2 1 1.48 1.312
1. Assumed deflection 1 1.48 1.312 2. Inertia force 4 × ω2 4.3 × ω2 2.6 × ω2 3. Spring force 11.01ω2 4.389ω2 2.62ω2 4. Spring deflection 1.573ω2 0.8778ω2 0.52ω2 5. Calculated deflection 1.573ω2 2.45ω2 2.093ω2
Calculated deflection1.573ω2 1 1.557 1.33
1. Assumed deflection 1 1.557 1.33 2. Inertia force 4 × ω2 4.6 × ω2 2.66 × ω2 3. Spring force 11.31ω2 4.6ω2 2.66ω2 4. Spring deflection 1.615ω2 0.93ω2 0.523ω2 5. Calculated deflection 1.615ω2 2.54ω2 2.147ω2
Calculated deflection1.615ω2 1 1.57 1.32
4m 2m
3m
5K
7K 5K
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
The values are closed to previous values; hence the fundamental principle mode or frequency is
given by
�1
1.571.32
� = 1.6ω2 + 2.5ω2 + 2.147ω2
ω = 786 rad/sec
5. Determine the natural frequency of the system shown in fig: by using matrix iteration
method.
Fig: (8)
Solution:
We solve by influence coefficient method, the influence coefficient for the given system can by
written are as follows by applying unit force at the 1st mass.
𝛼11 =1
3𝐾
Then, 2nd & 3rd mass will simply move by the same amount due to the action of unit force at
first mass 𝛼12 = 𝛼13 = 13𝐾
By Maxwell reciprocal theorem
𝛼21 = 𝛼12; 𝛼13 = 𝛼31
𝛼11 = 𝛼12 = 𝛼13 = 𝛼21 = 𝛼31 =1
3𝐾
By applying unit force on 2nd mass we get
1𝐾𝑒
=1
3𝑘+
1𝑘
= 4
3𝑘
K
3K
K
3m
2m
4m
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝛂𝟐𝟐 =𝐟𝐾𝑒
=𝟏
3𝑘4
=4
3𝑘
By Maxwell reciprocal theorem
𝜶𝟑𝟐 = 𝜶𝟐𝟑 =𝟒𝟑𝐤
Third Influence coefficients are:
𝛼33 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 13𝑘
+ 1𝑘
+ 1𝑘
= 73𝑘
From the influence coefficient theory the equation of motion can be written as follows −𝑥1 = 𝛼11𝑚1�̈� 1 + 𝛼12𝑚2 �̈� 2 + 𝛼13𝑚3 �̈� 3−𝑥2 = 𝛼21𝑚1�̈� 1 + 𝛼22𝑚2 �̈� 2 + 𝛼23 𝑚3�̈� 3−𝑥3 = 𝛼31 𝑚1�̈� 1 + 𝛼32𝑚2 �̈� 2 + 𝛼33𝑚3 �̈� 3
� −−(1)
Substituting the values of present given problems & replacing �̈� 𝑖by −ω2𝑥 𝑖 the above equation can
be written as �̈� 𝑖= −ω2𝑥 𝑖
−𝑥1 = 1
3K 4𝑚(−ω2𝑥 1) +
13𝐾
2𝑚(−ω2𝑥 2) + 1
3𝐾𝑚(−ω2𝑥 3)
−𝑥2 = 1
3𝐾4𝑚(−ω2𝑥 1) +
43𝐾
2𝑚 (−ω2𝑥 2) + 4
3𝐾𝑚(−ω2𝑥 3)
−𝑥3 = 1
3𝐾 4𝑚(−ω2𝑥 1) +
43𝐾
2𝑚 (−ω2𝑥 2) + 7
3𝐾𝑚 (−ω2𝑥 3)⎦
⎥⎥⎥⎥⎤
−−(1′)
𝑥1 = 1
3𝐾 4𝑚ω2𝑥 1 +
13𝐾
2𝑚 ω2𝑥 2 + 1
3𝐾𝑚ω2𝑥 3
𝑥2 = 1
3𝐾4𝑚ω2𝑥 1 +
43𝐾
2𝑚 ω2𝑥 2 + 4
3𝐾𝑚ω2𝑥 3
𝑥3 = 1
3𝐾 4𝑚ω2𝑥 1 +
43𝐾
2𝑚 ω2𝑥 2 + 7
3𝐾𝑚 ω2𝑥 3⎦
⎥⎥⎥⎥⎤
−−(1′)
The above equation can be represented in matrix notations as follows
𝛂𝟑𝟑 =7
3𝑘
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
�𝑥1𝑥2𝑥3
� = 𝑚 ω2
⎣⎢⎢⎢⎢⎡
43𝐾
23𝐾
13𝐾
43𝐾
83𝐾
43𝐾
43𝐾
83𝐾
73𝐾⎦⎥⎥⎥⎥⎤
�𝑥1𝑥2𝑥3
�
�𝑥1𝑥2𝑥3
� =𝑚 ω2
3𝐾�4 2 14 8 44 8 7
� �𝑥1𝑥2𝑥3
�
To start the matrix iteration method we assume the following values,
1st Iteration:
𝑥1 = 1; 𝑥2 = 2; 𝑥3 = 4;
�124� =
𝑚 ω2
3𝐾�4 2 14 8 44 8 7
� �124� =
𝑚 ω2
3𝐾�123648� = 12 �
𝑚 ω2
3𝐾� �
134� = �
4𝑚 ω2
𝐾� �
134�
2nd Iteration:
𝑥1 = 1; 𝑥2 = 3; 𝑥3 = 4;
�134� =
𝑚 ω2
3𝐾�4 2 14 8 44 8 7
� �134� =
𝑚 ω2
3𝐾�144456� = 14 �
𝑚 ω2
3𝐾� �
13.143
4�
3rd Iteration:
𝑥1 = 1; 𝑥2 = 3.143; 𝑥3 = 4;
�1
3.1434
� =𝑚 ω2
3𝐾�4 2 14 8 44 8 7
� �1
3.1434
� =𝑚 ω2
3𝐾�14.28645.14457.144
� = 14.286 �𝑚 ω2
3𝐾� �
13.16004.040
�
4th Iteration:
𝑥1 = 1; 𝑥2 = 3.16; 𝑥3 = 4;
�1
3.164.040
� =𝑚 ω2
3𝐾�4 2 14 8 44 8 7
� �1
3.164.040
� =𝑚 ω2
3𝐾�14.32057.2857.28
� = 14.32 �𝑚 ω2
3𝐾� �
13.162
4�
Since the values are obtained to initial values the first mode of the fundamental frequency is
1 =𝑚 ω2
3𝐾14.232
𝝎 = 𝟎.𝟒𝟓𝟕�𝑲 𝒎
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
6. Determine the natural frequency of the system shown in fig: by using matrix iteration
method.
Fig: (9)
Solution:
We solve by influence coefficient method, the influence coefficient for the given system can by
written are as follows by applying unit force at the 1st mass.
𝛼11 =1
7𝐾
Then, 2nd & 3rd mass will simply move by the same amount due to the action of unit force at
first mass 𝛼12 = 𝛼13 = 17𝐾
By Maxwell reciprocal theorem
𝛼21 = 𝛼12; 𝛼13 = 𝛼31
𝛼11 = 𝛼12 = 𝛼13 = 𝛼21 = 𝛼31 =1
7𝐾
4m 2m
3m
5K
7K 5K
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
By applying unit force on 2nd mass we get 1𝐾𝑒
=1
7𝑘+
15𝑘
= 12
35𝑘
𝛂𝟐𝟐 =𝐟𝐾𝑒
=𝟏
35𝑘12
=12
35𝑘
Since mass three has not connected to mass 2, where
𝜶𝟑𝟐 = 𝜶𝟐𝟑 = 𝜶𝟏𝟏 =𝟏𝟕𝐤
Third Influence coefficients are:
𝛼33 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 17𝑘
+ 15𝑘
= 1235𝑘
From the influence coefficient theory the equation of motion can be written as follows −𝑥1 = 𝛼11𝑚1�̈� 1 + 𝛼12𝑚2 �̈� 2 + 𝛼13𝑚3 �̈� 3−𝑥2 = 𝛼21𝑚1�̈� 1 + 𝛼22𝑚2 �̈� 2 + 𝛼23 𝑚3�̈� 3−𝑥3 = 𝛼31 𝑚1�̈� 1 + 𝛼32𝑚2 �̈� 2 + 𝛼33𝑚3 �̈� 3
� −−(1)
Substituting the values of present given problems & replacing �̈� 𝑖by −ω2𝑥 𝑖 the above equation can
be written as �̈� 𝑖= −ω2𝑥 𝑖
−𝑥1 = 1
3K 4𝑚(−ω2𝑥 1) +
13𝐾
2𝑚(−ω2𝑥 2) + 1
3𝐾𝑚(−ω2𝑥 3)
−𝑥2 = 1
3𝐾4𝑚(−ω2𝑥 1) +
43𝐾
2𝑚 (−ω2𝑥 2) + 4
3𝐾𝑚(−ω2𝑥 3)
−𝑥3 = 1
3𝐾 4𝑚(−ω2𝑥 1) +
43𝐾
2𝑚 (−ω2𝑥 2) + 7
3𝐾𝑚 (−ω2𝑥 3)⎦
⎥⎥⎥⎥⎤
−−(1′)
𝑥1 = 1
7𝐾 4𝑚ω2𝑥 1 +
17𝐾
3𝑚 ω2𝑥 2 + 1
7𝐾2𝑚ω2𝑥 3
𝑥2 = 1
7𝐾4𝑚ω2𝑥 1 +
1235𝐾
3𝑚 ω2𝑥 2 + 1
7𝐾2𝑚ω2𝑥 3
𝑥3 = 1
7𝐾 4𝑚ω2𝑥 1 +
17𝐾
3𝑚 ω2𝑥 2 + 12
35𝐾2𝑚 ω2𝑥 3⎦
⎥⎥⎥⎥⎤
−−(1′)
𝛂𝟑𝟑 =12
35𝑘
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
The above equation can be represented in matrix notations as follows
�𝑥1𝑥2𝑥3
� = 𝑚 ω2
⎣⎢⎢⎢⎢⎡
47𝐾
37𝐾
27𝐾
47𝐾
3635𝐾
27𝐾
47𝐾
37𝐾
2435𝐾⎦
⎥⎥⎥⎥⎤
�𝑥1𝑥2𝑥3
�
�𝑥1𝑥2𝑥3
� =𝑚 ω2
7𝐾�4 3 24 7.2 24 3 4.8
� �𝑥1𝑥2𝑥3
�
To start the matrix iteration method we assume the following values,
1st Iteration:
𝑥1 = 1; 𝑥2 = 2; 𝑥3 = 3;
�123� =
𝑚 ω2
7𝐾�4 3 24 7.2 24 3 4.8
� �123� =
𝑚 ω2
7𝐾�
1624.424.4
� = 16 �𝑚 ω2
3𝐾� �
11.5251.525
�
2nd Iteration:
𝑥1 = 1; 𝑥2 = 1.525; 𝑥3 = 1.525;
�1
1.5251.525
� =𝑚 ω2
7𝐾�4 3 24 7.2 24 3 4.8
� �1
1.5251.525
� =𝑚 ω2
7𝐾�11.62518.0315.89
� = 11.625 �𝑚 ω2
3𝐾� �
11.551.36
�
3rd Iteration:
𝑥1 = 1; 𝑥2 = 1.55; 𝑥3 = 1.36;
�1
1.551.36
� =𝑚 ω2
7𝐾�4 3 24 7.2 24 3 4.8
� �1
1.551.36
� =𝑚 ω2
7𝐾�11.3717.8813.88
� = 11.37 �𝑚 ω2
3𝐾� �
11.5721.33
�
4th Iteration:
𝑥1 = 1; 𝑥2 = 1.572; 𝑥3 = 1.33;
�1
1.5721.33
� =𝑚 ω2
7𝐾�4 2 14 8 44 8 7
� �1
1.5721.33
� =𝑚 ω2
7𝐾�
11.3725.138
17.09344� = 11.37 �
𝑚 ω2
3𝐾� �
11.58
1.327�
Since the values are obtained to initial values the first mode of the fundamental frequency is
1 =𝑚 ω2
7𝐾11.37
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝝎 = 𝟎.𝟖𝟎𝟔�𝑲 𝒎
7. By using Holzer method, find the natural frequencies of the system in fig 10: assume
K=1N/m; m=1 kg.
Fig: (10)
Solution:
Assumed
frequency Position mi miω2 xi m1xiω2 �𝑚𝑖𝑥𝑖𝜔2
𝑖
1
𝑘𝑖𝑗 �𝑚𝑖𝑥𝑖𝜔2
𝑘𝑖𝑗
𝑖
1
Trial-1
𝜔 = 0.1
𝜔2 = 0.01
1 1 0.01 1.0 0.01 0.01 1 0.01
2 2 0.02 0.99 0.0198 0.0298 1 0.0298
3 4 0.04 0.96 0.0384 0.0682 3 0.0227
4 ∞ 0.93
Trial-2
𝜔 = 0.2
𝜔2 = 0.04
1 1 0.04 1 0.04 0.04 1 0.04
2 2 0.08 0.96 0.077 0.117 1 0.117
3 4 0.16 0.84 0.1324 0.249 3 0.083
4 ∞ 0.76
Trial-3
𝜔 = 0.4
𝜔2 = 0.16
1 1 0.16 1 0.16 0.160 1 0.160
2 2 0.32 0.84 0.269 0.429 1 0.429
3 4 0.64 0.411 0.264 0.693 3 0.231
K
3K
K
m
2m
4m
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
4 ∞ 0.180
Trial-4
𝜔 = 0.5
𝜔2 = 0.25
1 1 0.25 1.0 0.25 0.25 1 0.25
2 2 0.5 0.75 0.375 0.625 1 0.625
3 4 1 0.125 0.125 0.750 3 0.250
4 ∞ -0.125
Trial-5
𝜔 = 0.60
𝜔2 = 0.36
1 1 0.36 1.0 0.36 0.36 1 0.36
2 2 0.72 0.64 0.46 0.82 1 0.820
3 4 1.44 -0.18 -0.259 0.561 3 0.187
4 ∞ -0.367
Trial-6
𝜔 = 0.80
𝜔2 = 0.64
1 1 0.64 0.1 0.64 0.640 1 0.64
2 2 1.28 0.36 0.461 1.101 1 0.101
3 4 2.56 -0.741 -1.90 -0.80 3 -0.267
4 ∞ -0.474
Trial-7
𝜔 = 1.0
𝜔2 = 1.0
1 1 1 1 1 1 1 1
2 2 2 0 0 1 1 1
3 4 4 -1 -4 -3 3 -1
4 ∞ 0
Trial-8
𝜔 = 1.2
𝜔2 = 1.44
1 1 1.44 1 1.44 1.44 1 1.44
2 2 2.88 -0.44 -1.27 0.17 1 0.17
3 4 5.76 -0.61 -3.51 -3.34 3 -1.11
4 ∞ 0.50
Trial-9
𝜔 = 1.4
𝜔2 = 1.96
1 1 1.96 1.0 1.96 1.96 1 1.96
2 2 3.92 0.96 -3.76 -1.80 1 -1.80
3 4 7.84 0.84 6.58 4.78 3 1.59
4 ∞ -0.75
Plot the graph with the assumed frequencies against the displacement in the rectangle box to get
𝜔𝑛1 , 𝜔𝑛2 & 𝜔𝑛3
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
Kt1 Kt2 Kt3
Ans. From the graph
𝝎𝒏𝟏 , = 𝟎.𝟒𝟔 𝒓𝒂𝒅/𝒔𝒆𝒄
𝝎𝒏𝟐 = 𝟏.𝟎𝟎 𝒓𝒂𝒅/𝒔𝒆𝒄
𝝎𝒏𝟑 = 𝟏.𝟑𝟓 𝒓𝒂𝒅/𝒔𝒆𝒄
8. By using Holzer method, find the natural frequencies of the system in fig 10: assume Kt1=
Kt1= Kt2= Kt3=1N/m; J1= J2= J3= kg.
Solution:
Assumed
frequency Position Ji Jiω2 θi Jiθiω2
�𝐽𝑖θ𝑖𝜔2
𝑖
1
𝑘𝑡𝑖𝑗 �𝐽𝑖θ𝑖𝜔2
𝑘𝑡𝑖𝑗
𝑖
1
Trial-1
𝜔 = 0.5
𝜔2 = 0.25
1 1 0.25 1 0.25 0.25 1 0.25
2 1 0.25 -0.75 0.1875 0.4375 1 0.4375
3 1 0.25 0.3125 0.0781 0.515
Trial-2
𝜔 = 0.75
1 1 0.562 1 0.562 0.562 1 0.5625
2 1 0.562 0.438 0.2461 0.801 1 0.8086
Fig: (11)
J1
or
J2
or
J3
or
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝜔2 = 0.5625 3 1 0.562 -0.3705 0.208 0.6305
Trial-3
𝜔 = 1
𝜔2 = 1
1 1 1 1 1 1 1 1
2 1 1 0 0 1 1 1
3 1 1 -1 -1 0
Trial-4
𝜔 = 1.5
𝜔2 = 2.25
1 1 2.25 1 2.25 2.25 1 2.25
2 1 2.25 -1.25 -2.8125 -0.5625 1 -0.5625
3 1 2.25 -0.6875 -1.5469 -2.1094
Trial-5
𝜔 = 2
𝜔2 = 4
1 1 4 1 4 4 1 4
2 1 4 -3 -12 -8 1 -8
3 1 4 5 20 12
Plot the graph with the assumed frequencies against the displacement in the rectangle box to get
𝜔𝑛1 , 𝜔𝑛2 & 𝜔𝑛3
Ans. From the graph
𝝎𝒏𝟏 , = 𝟎.𝟎𝟎 𝒓𝒂𝒅/𝒔𝒆𝒄
𝝎𝒏𝟐 = 𝟏.𝟎𝟎 𝒓𝒂𝒅/𝒔𝒆𝒄
𝝎𝒏𝟑 = 𝟏.𝟕𝟎 𝒓𝒂𝒅/𝒔𝒆𝒄
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
9. Find the fundamental vibration for the system shown in figure (12), by using Dunkerley’s
Method.
Fig: (12)
Solution:
First Influence coefficients are:
𝛼11 =1𝐾
& 𝛼12 = 𝛼13 =1𝐾
𝛼11 =1𝐾
Second Influence coefficients are:
𝛼22 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 1𝑘
+ 12𝑘
= 32𝑘
𝛂𝟐𝟐 =𝟑𝟐𝐤
Third Influence coefficients are:
𝛼33 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 1𝑘
+ 12𝑘
+ 13𝑘
= 116𝑘
By using Dunkerley’s method
1/ ωn2 = α11m1 + α22m2 + α33m3 + …………..
Natural frequency is
𝝎𝒏 = �1
α11m1 + α22m2 + α33m3
𝛂𝟑𝟑 =116𝑘
3m 2m m
2K K 3K
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝜔𝑛 = �1
1𝐾 (3m) + 𝟑𝟐𝐤 (2m) + 116𝑘 (m)
𝛚𝐧 = 𝟎.𝟑𝟓𝟕�𝐊𝐦
𝐫𝐚𝐝/𝐬𝐞𝐜
10. Find the fundamental vibration for the system shown in figure (13), by using Dunkerley’s
Method.
𝐼 = 4 × 10−7 𝑚4; 𝐸 == 1.96 × 1011 𝑁/𝑚2
First Influence coefficients are:
𝛼11 =𝐹𝑙3
3𝐸𝐼; 𝛼22 =
𝐹𝐿3
3𝐸𝐼
𝛼11 =𝐹𝑙3
3𝐸𝐼=
𝑙3
3𝐸𝐼= 2.47 × 10−8 𝑚/𝑁
𝛼22 =𝐹𝐿3
3𝐸𝐼=
𝐿3
3𝐸𝐼= 1.148 × 10−8 𝑚/𝑁
By using Dunkerley’s method
1/ ωn2 = α11m1 + α22m2
Natural frequency is
𝝎𝒏 = �1
α11m1 + α22m2
m2=50 kg m1=100 kg
180mm 180mm 300mm
Fig: (13)
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝜔𝑛 = �1
(2.47 × 10−8)(100) + (1.148 × 10−8)(50)
𝛚𝐧 = 𝟑𝟒𝟗.𝟎𝟎𝟔 𝐫𝐚𝐝/𝐬𝐞𝐜
QUADRANT-3
Wikis: (This includes wikis related to Calculation of Natural Frequencies)
• http://en.wikipedia.org/wiki/Rayleigh%E2%80%93Ritz_method: • http://en.cyclopaedia.net/wiki/Rayleigh-Ritz • http://elearning.vtu.ac.in/14/enotes/Mech%20vib/8-SKKudari.pdf • http://elearning.vtu.ac.in/P6/enotes/ME65/Unit8-SKK.pdf • http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors • www.freestudy.co.uk/dynamics/holzer.pdf • http://www.cdeep.iitb.ac.in/nptel/Mechanical/Dynamics%20of%20Machines/Course_ho
me12.5.html
• http://en.wikipedia.org/wiki/Dunkerley's_method
• http://www.cdeep.iitb.ac.in/nptel/Mechanical/Dynamics%20of%20Machines/Course_home12.4.html
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
Open Contents: • Mechanical Vibrations, S. S. Rao, Pearson Education Inc, 4th edition, 2003. • Mechanical Vibrations, V. P. Singh, Dhanpat Rai & Company, 3rd edition, 2006. • Mechanical Vibrations, G. K.Grover, Nem Chand and Bros, 6th edition, 1996 • Theory of vibration with applications ,W.T.Thomson,M.D.Dahleh and C
Padmanabhan,Pearson Education inc,5th Edition ,2008 • Theory and practice of Mechanical Vibration : J.S.Rao&K,Gupta,New Age International
Publications ,New Delhi,2001
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
QUADRANT-4
Problems 1. Find the fundamental vibration for the system shown in figure (2), by using Rayleigh’s
method.
𝐸 == 2.1 × 1011 𝑁/𝑚2
Static deflection at two points is given by
𝑦1 = 𝑚1𝑔 𝛼11 + 𝑚2 𝑔 𝛼12
𝑦2 = 𝑚1𝑔 𝛼12 + 𝑚2 𝑔 𝛼22
𝐼 =𝜋𝑑4
64=𝜋(0.05)4
64= 3.066 × 10−7 𝑚4
First Influence coefficients are:
𝛼11 =𝐹𝑙3
3𝐸𝐼; 𝛼22 =
𝐹𝐿3
3𝐸𝐼 & 𝛼12 = 𝛼21 =
𝑙2(3𝐿 − 𝑙)6𝐸𝐼
𝛼11 =𝐹𝑙3
3𝐸𝐼=
𝑙3
3𝐸𝐼= 4.1 × 10−7 𝑚/𝑁
𝛼22 =𝐹𝐿3
3𝐸𝐼=
𝐿3
3𝐸𝐼= 4.75 × 10−7 𝑚/𝑁
𝛼12 = 𝛼21 =𝑙2(3𝐿 − 𝑙)
6𝐸𝐼= 1.19 × 10−7 𝑚/𝑁
By using Rayleigh’s method
𝑦1 = 10(9.81) �4.1 × 10−7� + 20(9.81) �1.19 × 10−7� = 2.737 × 10−5m
𝑦2 = 10(9.81) �1.19 × 10−7� + 20(9.81) �4.75 × 10−7� = 10.486 × 10−5m
m2=20 kg m1=10 kg
0.2mm 0.25mm
Fig: (3)
0.05m
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝝎𝒏 = �𝒈�𝑚1 𝑦1 + 𝑚2 𝑦2�𝑚1 𝑦1
2 + 𝑚2 𝑦22
𝜔𝑛 = �9.81[(10 × 2.737 × 10−5) + (20 × 10.486 × 10−5)]
10(2.737 × 10−5)2 + 20 (10.486 × 10−5)2
𝛚𝐧 = 𝟑𝟏𝟗.𝟖 𝐫𝐚𝐝/𝐬𝐞𝐜
𝒇𝐧 =𝛚𝐧
𝟐𝝅= 𝟓𝟎.𝟗𝟐 𝐇𝐳
2. Using Stodola method to find the fundamental mode of vibration and its natural frequency for the branch system shown in fig:
Fig: (6)
Solution:
Trials K1=3K m1=m K2=2K m2=2m K3=K m3=3m 1. Assumed deflection 1 1 1 2. Inertia force 𝑚ω2 2𝑚ω2 3𝑚ω2 3. Spring force 6𝑚ω2 5𝑚ω2 3𝑚ω2 4. Spring deflection 6mω2
3𝐾
=2mω2
𝐾
5mω2
2𝐾
=2.5mω2
𝐾
3mω2
𝐾
5. Calculated deflection 2mω2
𝐾 4.5mω2
𝐾 7.5mω2
𝐾
m
K
3K
2K
3m
2m
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
Calculated deflection2mω2
𝐾
1 2.25 3.75
1. Assumed deflection 1 2.25 3.75 2. Inertia force mω2 4.5mω2 11.25mω2 3. Spring force 16.75mω2 15.75mω2 11.25mω2 4. Spring deflection 5.5mω2
𝐾 7.87mω2
𝐾 11.25mω2
𝐾
5. Calculated deflection 5.5mω2
𝐾 13.45mω2
𝐾 24.70mω2
𝐾
Calculated deflection4.66𝛚𝟐 1 2.41 4.42
1. Assumed deflection 1 2.41 4.42 2. Inertia force mω2 4.82mω2 13.26mω2 3. Spring force 19.08𝑚ω2 18.08mω2 13.26mω2 4. Spring deflection 6.36mω2
𝐾 9.04mω2
𝐾 13.26mω2
𝐾
5. Calculated deflection 6.36mω2
𝐾 15.4mω2
𝐾 28.66mω2
𝐾
Calculated deflection4.66𝛚𝟐 1 2.42 4.5
The values are closed to previous values; hence the fundamental principle mode or frequency is
given by
�1
2.424.5
� = 6.36mω2
𝐾+ 15.4mω2
𝐾+ 28.66mω2
𝐾
ω = 0.395 �K𝑚
rad/sec
Model shape:
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
0.1
Kt1 Kt2 Kt3
3. Using Stodola method to find the fundamental mode of vibration and its natural frequency for the torsional branch system shown in fig:
𝐼1𝑜𝑟 𝐽1 = 300 𝑘𝑔𝑚𝑚2; 𝐾𝑡1 == 3.75 × 105 𝑁.𝑚/𝑟𝑎𝑑
𝐼2𝑜𝑟 𝐽2 = 250𝑘𝑔𝑚𝑚2; 𝐾𝑡2 == 1.25 × 105 𝑁.𝑚/𝑟𝑎𝑑
𝐼3𝑜𝑟 𝐽3 = 125 𝑘𝑔𝑚𝑚2; 𝐾𝑡3 == 1.25 × 105 𝑁.𝑚/𝑟𝑎𝑑
Solution:
Where J or I = Mass moment of inertia of rotor assume that the system is vibrating at one of its
principal modes with natural frequency 𝜔 and that the motion is periodic.
m
K
3K
2K
3m
2m
2.42
4.5
Fig: (7)
J1
or
J2
or
J3
or
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
Trials 𝐾𝑡1 == 3.75× 105
𝐽1 = 300 𝐾𝑡2 == 1.25× 105
𝐽2 = 250 𝐾𝑡3 == 1.25× 105
𝐽3 = 125
1. Assumed
deflection(𝜃𝑖) 1 1 1
2. Inertia torque (𝑇𝑖) 300ω2 250ω2 125ω2 3. Spring force 675ω2 375ω2 125ω2 4. Spring deflection 675ω2
3.75 × 105
= 1.8× 10−3ω2
3 × 10−3ω2 1 × 10−3ω2
5. Calculated deflection 1.8× 10−3ω2 4.8
× 10−3ω2 5.8× 10−3ω2
Calculated deflection= 1.8 × 10−3ω2 1 2.67 3.2
1. Assumed deflection 1 2.67 3.2 2. Inertia force 300ω2 667.5ω2 400ω2 3. Spring force 1367.5ω2 1067.5ω2 400ω2 4. Spring deflection 3.65 × 10−3ω2 8.54
× 10−3ω2 3.2× 10−3ω2
5. Calculated deflection 3.65× 10−3ω2 12.19
× 10−3ω2 15.39× 10−3ω2
Calculated deflection3.65 × 10−3ω2 1 3.34 4.22
1. Assumed deflection 1 3.34 4.22 2. Inertia force 300ω2 835ω2 527.5ω2 3. Spring force 1662.5ω2 1362.5ω2 527.5ω2 4. Spring deflection 4.43 × 10−3ω2 10.9
× 10−3ω2 4.22× 10−3ω2
5. Calculated deflection 4.43× 10−3ω2 15.33
× 10−3ω2 19.55× 10−3ω2
Calculated deflection4.43 × 10−3ω2 1 3.46 4.41
1. Assumed deflection 1 3.46 4.41 2. Inertia force 300ω2 865ω2 551.25ω2 3. Spring force 1716.25𝑚ω2 1416.25ω2 551.25ω2 4. Spring deflection 4.58 × 10−3ω2 11.33
× 10−3ω2 4.41× 10−3ω2
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
5. Calculated deflection 4.58× 10−3ω2 15.91
× 10−3ω2 20.32× 10−3ω2
Calculated deflection4.43 × 10−3ω2 1 3.47 4.44
The values are closed to previous values; hence the fundamental principle mode or frequency is
given by
�1
3.474.44
� = 4.58 × 10−3ω2 + 15.91 × 10−3ω2 + 15.91 × 10−3ω2
𝛚 = 𝟏𝟒.𝟕𝟕𝟔 𝐫𝐚𝐝/𝐬𝐞𝐜
4. Determine the natural frequency of the system shown in fig: by using matrix iteration
method.
Fig: (8)
Solution:
We solve by influence coefficient method, the influence coefficient for the given system can by
written are as follows by applying unit force at the 1st mass.
𝛼11 =1
3𝐾
Then, 2nd & 3rd mass will simply move by the same amount due to the action of unit force at
first mass 𝛼12 = 𝛼13 = 13𝐾
By Maxwell reciprocal theorem
𝛼21 = 𝛼12; 𝛼13 = 𝛼31
𝛼11 = 𝛼12 = 𝛼13 = 𝛼21 = 𝛼31 =1
3𝐾
K
3K
K
3m
2m
4m
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
By applying unit force on 2nd mass we get
1𝐾𝑒
=1
3𝑘+
1𝑘
= 4
3𝑘
𝛂𝟐𝟐 =𝐟𝐾𝑒
=𝟏
3𝑘4
=4
3𝑘
By Maxwell reciprocal theorem
𝜶𝟑𝟐 = 𝜶𝟐𝟑 =𝟒𝟑𝐤
Third Influence coefficients are:
𝛼33 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 13𝑘
+ 1𝑘
+ 1𝑘
= 73𝑘
From the influence coefficient theory the equation of motion can be written as follows −𝑥1 = 𝛼11𝑚1�̈� 1 + 𝛼12𝑚2 �̈� 2 + 𝛼13𝑚3 �̈� 3−𝑥2 = 𝛼21𝑚1�̈� 1 + 𝛼22𝑚2 �̈� 2 + 𝛼23 𝑚3�̈� 3−𝑥3 = 𝛼31 𝑚1�̈� 1 + 𝛼32𝑚2 �̈� 2 + 𝛼33𝑚3 �̈� 3
� −−(1)
Substituting the values of present given problems & replacing �̈� 𝑖by −ω2𝑥 𝑖 the above equation can
be written as �̈� 𝑖= −ω2𝑥 𝑖
−𝑥1 = 1
3K 4𝑚(−ω2𝑥 1) +
13𝐾
2𝑚(−ω2𝑥 2) + 1
3𝐾𝑚(−ω2𝑥 3)
−𝑥2 = 1
3𝐾4𝑚(−ω2𝑥 1) +
43𝐾
2𝑚 (−ω2𝑥 2) + 4
3𝐾𝑚(−ω2𝑥 3)
−𝑥3 = 1
3𝐾 4𝑚(−ω2𝑥 1) +
43𝐾
2𝑚 (−ω2𝑥 2) + 7
3𝐾𝑚 (−ω2𝑥 3)⎦
⎥⎥⎥⎥⎤
−−(1′)
𝑥1 = 1
3𝐾 4𝑚ω2𝑥 1 +
13𝐾
2𝑚 ω2𝑥 2 + 1
3𝐾𝑚ω2𝑥 3
𝑥2 = 1
3𝐾4𝑚ω2𝑥 1 +
43𝐾
2𝑚 ω2𝑥 2 + 4
3𝐾𝑚ω2𝑥 3
𝑥3 = 1
3𝐾 4𝑚ω2𝑥 1 +
43𝐾
2𝑚 ω2𝑥 2 + 7
3𝐾𝑚 ω2𝑥 3⎦
⎥⎥⎥⎥⎤
−−(1′)
The above equation can be represented in matrix notations as follows
𝛂𝟑𝟑 =7
3𝑘
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
�𝑥1𝑥2𝑥3
� = 𝑚 ω2
⎣⎢⎢⎢⎢⎡
43𝐾
23𝐾
13𝐾
43𝐾
83𝐾
43𝐾
43𝐾
83𝐾
73𝐾⎦⎥⎥⎥⎥⎤
�𝑥1𝑥2𝑥3
�
�𝑥1𝑥2𝑥3
� =𝑚 ω2
3𝐾�4 2 14 8 44 8 7
� �𝑥1𝑥2𝑥3
�
To start the matrix iteration method we assume the following values,
1st Iteration:
𝑥1 = 1; 𝑥2 = 2; 𝑥3 = 4;
�124� =
𝑚 ω2
3𝐾�4 2 14 8 44 8 7
� �124� =
𝑚 ω2
3𝐾�123648� = 12 �
𝑚 ω2
3𝐾� �
134� = �
4𝑚 ω2
𝐾� �
134�
2nd Iteration:
𝑥1 = 1; 𝑥2 = 3; 𝑥3 = 4;
�134� =
𝑚 ω2
3𝐾�4 2 14 8 44 8 7
� �134� =
𝑚 ω2
3𝐾�144456� = 14 �
𝑚 ω2
3𝐾� �
13.143
4�
3rd Iteration:
𝑥1 = 1; 𝑥2 = 3.143; 𝑥3 = 4;
�1
3.1434
� =𝑚 ω2
3𝐾�4 2 14 8 44 8 7
� �1
3.1434
� =𝑚 ω2
3𝐾�14.28645.14457.144
� = 14.286 �𝑚 ω2
3𝐾� �
13.16004.040
�
4th Iteration:
𝑥1 = 1; 𝑥2 = 3.16; 𝑥3 = 4;
�1
3.164.040
� =𝑚 ω2
3𝐾�4 2 14 8 44 8 7
� �1
3.164.040
� =𝑚 ω2
3𝐾�14.32057.2857.28
� = 14.32 �𝑚 ω2
3𝐾� �
13.162
4�
Since the values are obtained to initial values the first mode of the fundamental frequency is
1 =𝑚 ω2
3𝐾14.232
𝝎 = 𝟎.𝟒𝟓𝟕�𝑲 𝒎
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
5. Determine the natural frequency of the system shown in fig: by using matrix iteration
method.
Fig: (9)
Solution:
We solve by influence coefficient method, the influence coefficient for the given system can by
written are as follows by applying unit force at the 1st mass.
𝛼11 =1
7𝐾
Then, 2nd & 3rd mass will simply move by the same amount due to the action of unit force at
first mass 𝛼12 = 𝛼13 = 17𝐾
By Maxwell reciprocal theorem
𝛼21 = 𝛼12; 𝛼13 = 𝛼31
𝛼11 = 𝛼12 = 𝛼13 = 𝛼21 = 𝛼31 =1
7𝐾
By applying unit force on 2nd mass we get
1𝐾𝑒
=1
7𝑘+
15𝑘
= 12
35𝑘
𝛂𝟐𝟐 =𝐟𝐾𝑒
=𝟏
35𝑘12
=12
35𝑘
Since mass three has not connected to mass 2, where
𝜶𝟑𝟐 = 𝜶𝟐𝟑 = 𝜶𝟏𝟏 =𝟏𝟕𝐤
4m 2m
3m
5K
7K 5K
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
Third Influence coefficients are:
𝛼33 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 17𝑘
+ 15𝑘
= 1235𝑘
From the influence coefficient theory the equation of motion can be written as follows −𝑥1 = 𝛼11𝑚1�̈� 1 + 𝛼12𝑚2 �̈� 2 + 𝛼13𝑚3 �̈� 3−𝑥2 = 𝛼21𝑚1�̈� 1 + 𝛼22𝑚2 �̈� 2 + 𝛼23 𝑚3�̈� 3−𝑥3 = 𝛼31 𝑚1�̈� 1 + 𝛼32𝑚2 �̈� 2 + 𝛼33𝑚3 �̈� 3
� −−(1)
Substituting the values of present given problems & replacing �̈� 𝑖by −ω2𝑥 𝑖 the above equation can
be written as �̈� 𝑖= −ω2𝑥 𝑖
−𝑥1 = 1
3K 4𝑚(−ω2𝑥 1) +
13𝐾
2𝑚(−ω2𝑥 2) + 1
3𝐾𝑚(−ω2𝑥 3)
−𝑥2 = 1
3𝐾4𝑚(−ω2𝑥 1) +
43𝐾
2𝑚 (−ω2𝑥 2) + 4
3𝐾𝑚(−ω2𝑥 3)
−𝑥3 = 1
3𝐾 4𝑚(−ω2𝑥 1) +
43𝐾
2𝑚 (−ω2𝑥 2) + 7
3𝐾𝑚 (−ω2𝑥 3)⎦
⎥⎥⎥⎥⎤
−−(1′)
𝑥1 = 1
7𝐾 4𝑚ω2𝑥 1 +
17𝐾
3𝑚 ω2𝑥 2 + 1
7𝐾2𝑚ω2𝑥 3
𝑥2 = 1
7𝐾4𝑚ω2𝑥 1 +
1235𝐾
3𝑚 ω2𝑥 2 + 1
7𝐾2𝑚ω2𝑥 3
𝑥3 = 1
7𝐾 4𝑚ω2𝑥 1 +
17𝐾
3𝑚 ω2𝑥 2 + 12
35𝐾2𝑚 ω2𝑥 3⎦
⎥⎥⎥⎥⎤
−−(1′)
The above equation can be represented in matrix notations as follows
�𝑥1𝑥2𝑥3
� = 𝑚 ω2
⎣⎢⎢⎢⎢⎡
47𝐾
37𝐾
27𝐾
47𝐾
3635𝐾
27𝐾
47𝐾
37𝐾
2435𝐾⎦
⎥⎥⎥⎥⎤
�𝑥1𝑥2𝑥3
�
�𝑥1𝑥2𝑥3
� =𝑚 ω2
7𝐾�4 3 24 7.2 24 3 4.8
� �𝑥1𝑥2𝑥3
�
𝛂𝟑𝟑 =12
35𝑘
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
To start the matrix iteration method we assume the following values,
1st Iteration:
𝑥1 = 1; 𝑥2 = 2; 𝑥3 = 3;
�123� =
𝑚 ω2
7𝐾�4 3 24 7.2 24 3 4.8
� �123� =
𝑚 ω2
7𝐾�
1624.424.4
� = 16 �𝑚 ω2
3𝐾� �
11.5251.525
�
2nd Iteration:
𝑥1 = 1; 𝑥2 = 1.525; 𝑥3 = 1.525;
�1
1.5251.525
� =𝑚 ω2
7𝐾�4 3 24 7.2 24 3 4.8
� �1
1.5251.525
� =𝑚 ω2
7𝐾�11.62518.0315.89
� = 11.625 �𝑚 ω2
3𝐾� �
11.551.36
�
3rd Iteration:
𝑥1 = 1; 𝑥2 = 1.55; 𝑥3 = 1.36;
�1
1.551.36
� =𝑚 ω2
7𝐾�4 3 24 7.2 24 3 4.8
� �1
1.551.36
� =𝑚 ω2
7𝐾�11.3717.8813.88
� = 11.37 �𝑚 ω2
3𝐾� �
11.5721.33
�
4th Iteration:
𝑥1 = 1; 𝑥2 = 1.572; 𝑥3 = 1.33;
�1
1.5721.33
� =𝑚 ω2
7𝐾�4 2 14 8 44 8 7
� �1
1.5721.33
� =𝑚 ω2
7𝐾�
11.3725.138
17.09344� = 11.37 �
𝑚 ω2
3𝐾� �
11.58
1.327�
Since the values are obtained to initial values the first mode of the fundamental frequency is
1 =𝑚 ω2
7𝐾11.37
𝝎 = 𝟎.𝟖𝟎𝟔�𝑲 𝒎
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
6. Find the fundamental vibration for the system shown in figure (12), by using Dunkerley’s
Method.
Fig: (12)
Solution:
First Influence coefficients are:
𝛼11 =1𝐾
& 𝛼12 = 𝛼13 =1𝐾
𝛼11 =1𝐾
Second Influence coefficients are:
𝛼22 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 1𝑘
+ 12𝑘
= 32𝑘
𝛂𝟐𝟐 =𝟑𝟐𝐤
Third Influence coefficients are:
𝛼33 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 1𝑘
+ 12𝑘
+ 13𝑘
= 116𝑘
By using Dunkerley’s method
1/ ωn2 = α11m1 + α22m2 + α33m3 + …………..
Natural frequency is
𝛂𝟑𝟑 =116𝑘
3m 2m m
2K K 3K
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝝎𝒏 = �1
α11m1 + α22m2 + α33m3
𝜔𝑛 = �1
1𝐾 (3m) + 𝟑𝟐𝐤 (2m) + 116𝑘 (m)
𝛚𝐧 = 𝟎.𝟑𝟓𝟕�𝐊𝐦
𝐫𝐚𝐝/𝐬𝐞𝐜
7. Find the fundamental vibration for the system shown in figure (13), by using Dunkerley’s
Method.
𝐼 = 4 × 10−7 𝑚4; 𝐸 == 1.96 × 1011 𝑁/𝑚2
First Influence coefficients are:
𝛼11 =𝐹𝑙3
3𝐸𝐼; 𝛼22 =
𝐹𝐿3
3𝐸𝐼
𝛼11 =𝐹𝑙3
3𝐸𝐼=
𝑙3
3𝐸𝐼= 2.47 × 10−8 𝑚/𝑁
𝛼22 =𝐹𝐿3
3𝐸𝐼=
𝐿3
3𝐸𝐼= 1.148 × 10−8 𝑚/𝑁
By using Dunkerley’s method
1/ ωn2 = α11m1 + α22m2
Natural frequency is
m2=50 kg m1=100 kg
180mm 180mm 300mm
Fig: (13)
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝝎𝒏 = �1
α11m1 + α22m2
𝜔𝑛 = �1
(2.47 × 10−8)(100) + (1.148 × 10−8)(50)
𝛚𝐧 = 𝟑𝟒𝟗.𝟎𝟎𝟔 𝐫𝐚𝐝/𝐬𝐞𝐜
8. Find the fundamental vibration for the system shown in figure (14), by using Dunkerley’s
Method.
𝐸 = 2.1 × 1011 𝑁/𝑚2
Static deflection at two points is given by
𝐼 =𝜋𝑑4
64=𝜋(0.05)4
64= 3.066 × 10−7 𝑚4
First Influence coefficients are:
𝛼11 =𝐹𝑙3
3𝐸𝐼; 𝛼22 =
𝐹𝐿3
3𝐸𝐼 & 𝛼12 = 𝛼21 =
𝑙2(3𝐿 − 𝑙)6𝐸𝐼
𝛼11 =𝐹𝑙3
3𝐸𝐼=
𝑙3
3𝐸𝐼= 4.1 × 10−7 𝑚/𝑁
𝛼22 =𝐹𝐿3
3𝐸𝐼=
𝐿3
3𝐸𝐼= 4.75 × 10−7 𝑚/𝑁
By using Dunkerley’s method
1/ ωn2 = α11m1 + α22m2
Natural frequency is
m2=20 kg m1=10 kg
0.2mm 0.25mm
Fig: (14)
0.05m
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝝎𝒏 = �1
α11m1 + α22m2
𝜔𝑛 = �1
(4.1 × 10−7)(10) + (4.75 × 10−7)(20)
𝛚𝐧 = 𝟐𝟕𝟏.𝟏𝟔𝟑𝟎 𝐫𝐚𝐝/𝐬𝐞𝐜
𝒇𝐧 =𝛚𝐧
𝟐𝝅= 𝟒𝟑.𝟏𝟓𝟔𝟗 𝐇𝐳
Frequently asked Questions.
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Self Answered Question & Answer
1. Find the fundamental vibration for the system shown in figure (1), by using Rayleigh’s
method.
Fig: (1)
Solution:
First Influence coefficients are:
𝛼11 =1𝐾
& 𝛼12 = 𝛼13 =1𝐾
By Maxwell reciprocal theorem
𝛼21 = 𝛼12; 𝛼13 = 𝛼31
𝛼11 = 𝛼12 = 𝛼13 = 𝛼21 = 𝛼31 =1𝐾
Second Influence coefficients are:
𝛼22 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 1𝑘
+ 12𝑘
= 32𝑘
𝛂𝟐𝟐 =𝟑𝟐𝐤
By Maxwell reciprocal theorem
𝜶𝟑𝟐 = 𝜶𝟐𝟑 =𝟑𝟐𝐤
Third Influence coefficients are:
𝛼33 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 1𝑘
+ 12𝑘
+ 13𝑘
= 116𝑘
3m 2m m
2K K 3K
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By using Rayleigh’s method 𝑦1 = 𝑚1𝑔 𝛼11 + 𝑚2 𝑔 𝛼12 + 𝑚3 𝑔 𝛼13𝑦2 = 𝑚1𝑔 𝛼21 + 𝑚2 𝑔 𝛼22 + 𝑚3 𝑔 𝛼23 𝑦3 = 𝑚1𝑔 𝛼31 + 𝑚2 𝑔 𝛼32 + 𝑚3 𝑔 𝛼33
� −−(1)
𝐲𝟏 = 3𝑚(9.81)1𝑘 + 2𝑚(9.81)
1𝑘 + 𝑚(9.81)
1𝑘 = 𝟓𝟖.𝟖𝟔
𝐦𝐤
𝐲𝟐 = 3𝑚(9.81)1𝑘 + 2𝑚(9.81)
32k + 𝑚(9.81)
32k = 𝟕𝟑.𝟓𝟕𝟓
𝐦𝐤
𝐲𝟐 = 3𝑚(9.81)1𝑘 + 2𝑚(9.81)
32k + 𝑚(9.81)
116𝑘 = 𝟕𝟔.𝟖𝟒
𝐦𝐤
Natural frequency is
𝝎𝒏 = �𝒈�𝑚1 𝑦1 + 𝑚2 𝑦2 + 𝑚3 𝑦3�𝑚1 𝑦1
2 + 𝑚2 𝑦22 + 𝑚3 𝑦3
2
𝜔𝑛 = �9.81 ��3m × 58.86 m
k � + �2m × 73.575 mk � + �𝑚 × 76.84 m
k ��
3𝑚�58.86 mk �
2 + 2𝑚 �73.575 m
k �2
+ 𝑚 �76.84 mk �
2
𝛚𝐧 = 𝟎.𝟑𝟖𝟔�𝐊𝐦
𝐫𝐚𝐝/𝐬𝐞𝐜
2. Find the fundamental vibration for the system shown in figure (2), by using Rayleigh’s
method.
𝛂𝟑𝟑 =116𝑘
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝐼 = 4 × 10−7 𝑚4; 𝐸 == 1.96 × 1011 𝑁/𝑚2
Static deflection at two points is given by
𝑦1 = 𝑚1𝑔 𝛼11 + 𝑚2 𝑔 𝛼12
𝑦2 = 𝑚1𝑔 𝛼12 + 𝑚2 𝑔 𝛼22
First Influence coefficients are:
𝛼11 =𝐹𝑙3
3𝐸𝐼; 𝛼22 =
𝐹𝐿3
3𝐸𝐼 & 𝛼12 = 𝛼21 =
𝑙2(3𝐿 − 𝑙)6𝐸𝐼
𝛼11 =𝐹𝑙3
3𝐸𝐼=
𝑙3
3𝐸𝐼= 2.47 × 10−8 𝑚/𝑁
𝛼22 =𝐹𝐿3
3𝐸𝐼=
𝐿3
3𝐸𝐼= 1.148 × 10−8 𝑚/𝑁
𝛼12 = 𝛼21 =𝑙2(3𝐿 − 𝑙)
6𝐸𝐼= 4.599 × 10−8 𝑚/𝑁
By using Rayleigh’s method
𝑦1 = 100(9.81) �2.47 × 10−8� + 50(9.81) �4.599 × 10−8� = 4.85 × 10−5
𝑦2 = 100(9.81) �4.599 × 10−8� + 50(9.81) �1.148 × 10−8� = 1.044 × 10−4
𝝎𝒏 = �𝒈�𝑚1 𝑦1 + 𝑚2 𝑦2�𝑚1 𝑦1
2 + 𝑚2 𝑦22
𝜔𝑛 = �9.81[(100 × 4.85 × 10−5) + (50 × 1.044 × 10−4)]
100(4.85 × 10−5)2 + 50 (1.044 × 10−4)2
𝛚𝐧 = 𝟑𝟓𝟓.𝟒𝟕 𝐫𝐚𝐝/𝐬𝐞𝐜
m2=50 kg m1=100 kg
180mm 180mm 300mm
Fig: (2)
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
3. Find the fundamental vibration for the system shown in figure (2), by using Rayleigh’s
method.
𝐸 == 2.1 × 1011 𝑁/𝑚2
Static deflection at two points is given by
𝑦1 = 𝑚1𝑔 𝛼11 + 𝑚2 𝑔 𝛼12
𝑦2 = 𝑚1𝑔 𝛼12 + 𝑚2 𝑔 𝛼22
𝐼 =𝜋𝑑4
64=𝜋(0.05)4
64= 3.066 × 10−7 𝑚4
First Influence coefficients are:
𝛼11 =𝐹𝑙3
3𝐸𝐼; 𝛼22 =
𝐹𝐿3
3𝐸𝐼 & 𝛼12 = 𝛼21 =
𝑙2(3𝐿 − 𝑙)6𝐸𝐼
𝛼11 =𝐹𝑙3
3𝐸𝐼=
𝑙3
3𝐸𝐼= 4.1 × 10−7 𝑚/𝑁
𝛼22 =𝐹𝐿3
3𝐸𝐼=
𝐿3
3𝐸𝐼= 4.75 × 10−7 𝑚/𝑁
𝛼12 = 𝛼21 =𝑙2(3𝐿 − 𝑙)
6𝐸𝐼= 1.19 × 10−7 𝑚/𝑁
By using Rayleigh’s method
𝑦1 = 10(9.81) �4.1 × 10−7� + 20(9.81) �1.19 × 10−7� = 2.737 × 10−5m
𝑦2 = 10(9.81) �1.19 × 10−7� + 20(9.81) �4.75 × 10−7� = 10.486 × 10−5m
𝝎𝒏 = �𝒈�𝑚1 𝑦1 + 𝑚2 𝑦2�𝑚1 𝑦1
2 + 𝑚2 𝑦22
𝜔𝑛 = �9.81[(10 × 2.737 × 10−5) + (20 × 10.486 × 10−5)]
10(2.737 × 10−5)2 + 20 (10.486 × 10−5)2
m2=20 kg m1=10 kg
0.2mm 0.25mm
Fig: (3)
0.05m
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝛚𝐧 = 𝟑𝟏𝟗.𝟖 𝐫𝐚𝐝/𝐬𝐞𝐜
𝒇𝐧 =𝛚𝐧
𝟐𝝅= 𝟓𝟎.𝟗𝟐 𝐇𝐳
4. Using Stodola method to find the fundamental mode of vibration and its natural frequency of the spring mass system shown in the figure given K1 = K2=K3=1 N/M & m1=m2= m3=1kg.
Trials K1=1K m1=1m K2=1K m2=1m K3=1K m3=1m 6. Assumed deflection 1 1 1 7. Inertia force 1 × ω2 1 × ω2 1 × ω2 8. Spring force 3ω2 2ω2 ω2 9. Spring deflection 3ω2 2ω2 ω2 10. Calculated deflection 3ω2 2ω2 ω2
Calculated deflection𝟑𝛚𝟐 1 1.66 2
6. Assumed deflection 1 1.66 2 7. Inertia force 1 × ω2 1.66
× ω2 2 × ω2
8. Spring force 4.66ω2 3.66ω2 2ω2 9. Spring deflection 4.66ω2 3.66ω2 2ω2 10. Calculated deflection 4.66ω2 8.32ω2 10.32ω2
Calculated deflection4.66𝛚𝟐 1 1.78 2.214
6. Assumed deflection 1 1.78 2.214 7. Inertia force 1 × ω2 1.78
× ω2 2.214 × ω2
8. Spring force 4.99ω2 3.99ω2 2.21ω2 9. Spring deflection 4.99ω2 3.99ω2 2.21ω2 10. Calculated deflection 4.994ω2 8.98ω2 11.129ω2
Calculated deflection4.66𝛚𝟐 1 1.799 2.24
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
The values are closed to previous values; hence the fundamental principle mode or frequency is
given by
�1
1.782.214
� = 4.994ω2 + 8.98ω2 + 11.129ω2
ω = 0.448 rad/sec
5. Using Stodola method to find the fundamental mode of vibration and its natural frequency for the branch system shown in fig:
Fig: (5)
Solution:
Trials K1=7K m1=4m K2=5K m2=3m K3=5K m3=2m 6. Assumed deflection 1 1 1 7. Inertia force 4 × ω2 3 × ω2 2 × ω2 8. Spring force 9ω2 3ω2 2ω2 9. Spring deflection 1.228ω2 0.6ω2 0.4ω2 10. Calculated deflection 1.22ω2 1.88ω2 1.6ω2
Calculated deflection1.22ω2 1 1.48 1.312
6. Assumed deflection 1 1.48 1.312 7. Inertia force 4 × ω2 4.3 × ω2 2.6 × ω2 8. Spring force 11.01ω2 4.389ω2 2.62ω2 9. Spring deflection 1.573ω2 0.8778ω2 0.52ω2 10. Calculated deflection 1.573ω2 2.45ω2 2.093ω2
4m 2m
3m
5K
7K 5K
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
Calculated deflection1.573ω2 1 1.557 1.33
6. Assumed deflection 1 1.557 1.33 7. Inertia force 4 × ω2 4.6 × ω2 2.66 × ω2 8. Spring force 11.31ω2 4.6ω2 2.66ω2 9. Spring deflection 1.615ω2 0.93ω2 0.523ω2 10. Calculated deflection 1.615ω2 2.54ω2 2.147ω2
Calculated deflection1.615ω2 1 1.57 1.32
The values are closed to previous values; hence the fundamental principle mode or frequency is
given by
�1
1.571.32
� = 1.6ω2 + 2.5ω2 + 2.147ω2
ω = 786 rad/sec
6. Using Stodola method to find the fundamental mode of vibration and its natural frequency for the branch system shown in fig:
Fig: (6)
Solution:
Trials K1=3K m1=m K2=2K m2=2m K3=K m3=3m 6. Assumed deflection 1 1 1
m
K
3K
2K
3m
2m
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
7. Inertia force 𝑚ω2 2𝑚ω2 3𝑚ω2 8. Spring force 6𝑚ω2 5𝑚ω2 3𝑚ω2 9. Spring deflection 6mω2
3𝐾
=2mω2
𝐾
5mω2
2𝐾
=2.5mω2
𝐾
3mω2
𝐾
10. Calculated deflection 2mω2
𝐾 4.5mω2
𝐾 7.5mω2
𝐾
Calculated deflection2mω2
𝐾
1 2.25 3.75
6. Assumed deflection 1 2.25 3.75 7. Inertia force mω2 4.5mω2 11.25mω2 8. Spring force 16.75mω2 15.75mω2 11.25mω2 9. Spring deflection 5.5mω2
𝐾 7.87mω2
𝐾 11.25mω2
𝐾
10. Calculated deflection 5.5mω2
𝐾 13.45mω2
𝐾 24.70mω2
𝐾
Calculated deflection4.66𝛚𝟐 1 2.41 4.42
6. Assumed deflection 1 2.41 4.42 7. Inertia force mω2 4.82mω2 13.26mω2 8. Spring force 19.08𝑚ω2 18.08mω2 13.26mω2 9. Spring deflection 6.36mω2
𝐾 9.04mω2
𝐾 13.26mω2
𝐾
10. Calculated deflection 6.36mω2
𝐾 15.4mω2
𝐾 28.66mω2
𝐾
Calculated deflection4.66𝛚𝟐 1 2.42 4.5
The values are closed to previous values; hence the fundamental principle mode or frequency is
given by
�1
2.424.5
� = 6.36mω2
𝐾+ 15.4mω2
𝐾+ 28.66mω2
𝐾
ω = 0.395 �K𝑚
rad/sec
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
0.1
Kt1 Kt2 Kt3
Model shape:
7. Using Stodola method to find the fundamental mode of vibration and its natural frequency for the torsional branch system shown in fig:
𝐼1𝑜𝑟 𝐽1 = 300 𝑘𝑔𝑚𝑚2; 𝐾𝑡1 == 3.75 × 105 𝑁.𝑚/𝑟𝑎𝑑
𝐼2𝑜𝑟 𝐽2 = 250𝑘𝑔𝑚𝑚2; 𝐾𝑡2 == 1.25 × 105 𝑁.𝑚/𝑟𝑎𝑑
𝐼3𝑜𝑟 𝐽3 = 125 𝑘𝑔𝑚𝑚2; 𝐾𝑡3 == 1.25 × 105 𝑁.𝑚/𝑟𝑎𝑑
Solution:
Where J or I = Mass moment of inertia of rotor assume that the system is vibrating at one of its
principal modes with natural frequency 𝜔 and that the motion is periodic.
m
K
3K
2K
3m
2m
2.42
4.5
Fig: (7)
J1
or
J2
or
J3
or
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
Trials 𝐾𝑡1 == 3.75× 105
𝐽1 = 300 𝐾𝑡2 == 1.25× 105
𝐽2 = 250 𝐾𝑡3 == 1.25× 105
𝐽3 = 125
6. Assumed
deflection(𝜃𝑖) 1 1 1
7. Inertia torque (𝑇𝑖) 300ω2 250ω2 125ω2 8. Spring force 675ω2 375ω2 125ω2 9. Spring deflection 675ω2
3.75 × 105
= 1.8× 10−3ω2
3 × 10−3ω2 1 × 10−3ω2
10. Calculated deflection 1.8× 10−3ω2 4.8
× 10−3ω2 5.8× 10−3ω2
Calculated deflection= 1.8 × 10−3ω2 1 2.67 3.2
6. Assumed deflection 1 2.67 3.2 7. Inertia force 300ω2 667.5ω2 400ω2 8. Spring force 1367.5ω2 1067.5ω2 400ω2 9. Spring deflection 3.65 × 10−3ω2 8.54
× 10−3ω2 3.2× 10−3ω2
10. Calculated deflection 3.65× 10−3ω2 12.19
× 10−3ω2 15.39× 10−3ω2
Calculated deflection3.65 × 10−3ω2 1 3.34 4.22
6. Assumed deflection 1 3.34 4.22 7. Inertia force 300ω2 835ω2 527.5ω2 8. Spring force 1662.5ω2 1362.5ω2 527.5ω2 9. Spring deflection 4.43 × 10−3ω2 10.9
× 10−3ω2 4.22× 10−3ω2
10. Calculated deflection 4.43× 10−3ω2 15.33
× 10−3ω2 19.55× 10−3ω2
Calculated deflection4.43 × 10−3ω2 1 3.46 4.41
6. Assumed deflection 1 3.46 4.41 7. Inertia force 300ω2 865ω2 551.25ω2 8. Spring force 1716.25𝑚ω2 1416.25ω2 551.25ω2 9. Spring deflection 4.58 × 10−3ω2 11.33
× 10−3ω2 4.41× 10−3ω2
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
10. Calculated deflection 4.58× 10−3ω2 15.91
× 10−3ω2 20.32× 10−3ω2
Calculated deflection4.43 × 10−3ω2 1 3.47 4.44
The values are closed to previous values; hence the fundamental principle mode or frequency is
given by
�1
3.474.44
� = 4.58 × 10−3ω2 + 15.91 × 10−3ω2 + 15.91 × 10−3ω2
𝛚 = 𝟏𝟒.𝟕𝟕𝟔 𝐫𝐚𝐝/𝐬𝐞𝐜
8. Determine the natural frequency of the system shown in fig: by using matrix iteration
method.
Fig: (8)
Solution:
We solve by influence coefficient method, the influence coefficient for the given system can by
written are as follows by applying unit force at the 1st mass.
𝛼11 =1
3𝐾
Then, 2nd & 3rd mass will simply move by the same amount due to the action of unit force at
first mass 𝛼12 = 𝛼13 = 13𝐾
By Maxwell reciprocal theorem
𝛼21 = 𝛼12; 𝛼13 = 𝛼31
𝛼11 = 𝛼12 = 𝛼13 = 𝛼21 = 𝛼31 =1
3𝐾
K
3K
K
3m
2m
4m
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
By applying unit force on 2nd mass we get
1𝐾𝑒
=1
3𝑘+
1𝑘
= 4
3𝑘
𝛂𝟐𝟐 =𝐟𝐾𝑒
=𝟏
3𝑘4
=4
3𝑘
By Maxwell reciprocal theorem
𝜶𝟑𝟐 = 𝜶𝟐𝟑 =𝟒𝟑𝐤
Third Influence coefficients are:
𝛼33 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 13𝑘
+ 1𝑘
+ 1𝑘
= 73𝑘
From the influence coefficient theory the equation of motion can be written as follows −𝑥1 = 𝛼11𝑚1�̈� 1 + 𝛼12𝑚2 �̈� 2 + 𝛼13𝑚3 �̈� 3−𝑥2 = 𝛼21𝑚1�̈� 1 + 𝛼22𝑚2 �̈� 2 + 𝛼23 𝑚3�̈� 3−𝑥3 = 𝛼31 𝑚1�̈� 1 + 𝛼32𝑚2 �̈� 2 + 𝛼33𝑚3 �̈� 3
� −−(1)
Substituting the values of present given problems & replacing �̈� 𝑖by −ω2𝑥 𝑖 the above equation can
be written as �̈� 𝑖= −ω2𝑥 𝑖
−𝑥1 = 1
3K 4𝑚(−ω2𝑥 1) +
13𝐾
2𝑚(−ω2𝑥 2) + 1
3𝐾𝑚(−ω2𝑥 3)
−𝑥2 = 1
3𝐾4𝑚(−ω2𝑥 1) +
43𝐾
2𝑚 (−ω2𝑥 2) + 4
3𝐾𝑚(−ω2𝑥 3)
−𝑥3 = 1
3𝐾 4𝑚(−ω2𝑥 1) +
43𝐾
2𝑚 (−ω2𝑥 2) + 7
3𝐾𝑚 (−ω2𝑥 3)⎦
⎥⎥⎥⎥⎤
−−(1′)
𝑥1 = 1
3𝐾 4𝑚ω2𝑥 1 +
13𝐾
2𝑚 ω2𝑥 2 + 1
3𝐾𝑚ω2𝑥 3
𝑥2 = 1
3𝐾4𝑚ω2𝑥 1 +
43𝐾
2𝑚 ω2𝑥 2 + 4
3𝐾𝑚ω2𝑥 3
𝑥3 = 1
3𝐾 4𝑚ω2𝑥 1 +
43𝐾
2𝑚 ω2𝑥 2 + 7
3𝐾𝑚 ω2𝑥 3⎦
⎥⎥⎥⎥⎤
−−(1′)
The above equation can be represented in matrix notations as follows
𝛂𝟑𝟑 =7
3𝑘
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
�𝑥1𝑥2𝑥3
� = 𝑚 ω2
⎣⎢⎢⎢⎢⎡
43𝐾
23𝐾
13𝐾
43𝐾
83𝐾
43𝐾
43𝐾
83𝐾
73𝐾⎦⎥⎥⎥⎥⎤
�𝑥1𝑥2𝑥3
�
�𝑥1𝑥2𝑥3
� =𝑚 ω2
3𝐾�4 2 14 8 44 8 7
� �𝑥1𝑥2𝑥3
�
To start the matrix iteration method we assume the following values,
1st Iteration:
𝑥1 = 1; 𝑥2 = 2; 𝑥3 = 4;
�124� =
𝑚 ω2
3𝐾�4 2 14 8 44 8 7
� �124� =
𝑚 ω2
3𝐾�123648� = 12 �
𝑚 ω2
3𝐾� �
134� = �
4𝑚 ω2
𝐾� �
134�
2nd Iteration:
𝑥1 = 1; 𝑥2 = 3; 𝑥3 = 4;
�134� =
𝑚 ω2
3𝐾�4 2 14 8 44 8 7
� �134� =
𝑚 ω2
3𝐾�144456� = 14 �
𝑚 ω2
3𝐾� �
13.143
4�
3rd Iteration:
𝑥1 = 1; 𝑥2 = 3.143; 𝑥3 = 4;
�1
3.1434
� =𝑚 ω2
3𝐾�4 2 14 8 44 8 7
� �1
3.1434
� =𝑚 ω2
3𝐾�14.28645.14457.144
� = 14.286 �𝑚 ω2
3𝐾� �
13.16004.040
�
4th Iteration:
𝑥1 = 1; 𝑥2 = 3.16; 𝑥3 = 4;
�1
3.164.040
� =𝑚 ω2
3𝐾�4 2 14 8 44 8 7
� �1
3.164.040
� =𝑚 ω2
3𝐾�14.32057.2857.28
� = 14.32 �𝑚 ω2
3𝐾� �
13.162
4�
Since the values are obtained to initial values the first mode of the fundamental frequency is
1 =𝑚 ω2
3𝐾14.232
𝝎 = 𝟎.𝟒𝟓𝟕�𝑲 𝒎
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
9. Determine the natural frequency of the system shown in fig: by using matrix iteration
method.
Fig: (9)
Solution:
We solve by influence coefficient method, the influence coefficient for the given system can by
written are as follows by applying unit force at the 1st mass.
𝛼11 =1
7𝐾
Then, 2nd & 3rd mass will simply move by the same amount due to the action of unit force at
first mass 𝛼12 = 𝛼13 = 17𝐾
By Maxwell reciprocal theorem
𝛼21 = 𝛼12; 𝛼13 = 𝛼31
𝛼11 = 𝛼12 = 𝛼13 = 𝛼21 = 𝛼31 =1
7𝐾
By applying unit force on 2nd mass we get
1𝐾𝑒
=1
7𝑘+
15𝑘
= 12
35𝑘
𝛂𝟐𝟐 =𝐟𝐾𝑒
=𝟏
35𝑘12
=12
35𝑘
Since mass three has not connected to mass 2, where
𝜶𝟑𝟐 = 𝜶𝟐𝟑 = 𝜶𝟏𝟏 =𝟏𝟕𝐤
4m 2m
3m
5K
7K 5K
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
Third Influence coefficients are:
𝛼33 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 17𝑘
+ 15𝑘
= 1235𝑘
From the influence coefficient theory the equation of motion can be written as follows −𝑥1 = 𝛼11𝑚1�̈� 1 + 𝛼12𝑚2 �̈� 2 + 𝛼13𝑚3 �̈� 3−𝑥2 = 𝛼21𝑚1�̈� 1 + 𝛼22𝑚2 �̈� 2 + 𝛼23 𝑚3�̈� 3−𝑥3 = 𝛼31 𝑚1�̈� 1 + 𝛼32𝑚2 �̈� 2 + 𝛼33𝑚3 �̈� 3
� −−(1)
Substituting the values of present given problems & replacing �̈� 𝑖by −ω2𝑥 𝑖 the above equation can
be written as �̈� 𝑖= −ω2𝑥 𝑖
−𝑥1 = 1
3K 4𝑚(−ω2𝑥 1) +
13𝐾
2𝑚(−ω2𝑥 2) + 1
3𝐾𝑚(−ω2𝑥 3)
−𝑥2 = 1
3𝐾4𝑚(−ω2𝑥 1) +
43𝐾
2𝑚 (−ω2𝑥 2) + 4
3𝐾𝑚(−ω2𝑥 3)
−𝑥3 = 1
3𝐾 4𝑚(−ω2𝑥 1) +
43𝐾
2𝑚 (−ω2𝑥 2) + 7
3𝐾𝑚 (−ω2𝑥 3)⎦
⎥⎥⎥⎥⎤
−−(1′)
𝑥1 = 1
7𝐾 4𝑚ω2𝑥 1 +
17𝐾
3𝑚 ω2𝑥 2 + 1
7𝐾2𝑚ω2𝑥 3
𝑥2 = 1
7𝐾4𝑚ω2𝑥 1 +
1235𝐾
3𝑚 ω2𝑥 2 + 1
7𝐾2𝑚ω2𝑥 3
𝑥3 = 1
7𝐾 4𝑚ω2𝑥 1 +
17𝐾
3𝑚 ω2𝑥 2 + 12
35𝐾2𝑚 ω2𝑥 3⎦
⎥⎥⎥⎥⎤
−−(1′)
The above equation can be represented in matrix notations as follows
�𝑥1𝑥2𝑥3
� = 𝑚 ω2
⎣⎢⎢⎢⎢⎡
47𝐾
37𝐾
27𝐾
47𝐾
3635𝐾
27𝐾
47𝐾
37𝐾
2435𝐾⎦
⎥⎥⎥⎥⎤
�𝑥1𝑥2𝑥3
�
�𝑥1𝑥2𝑥3
� =𝑚 ω2
7𝐾�4 3 24 7.2 24 3 4.8
� �𝑥1𝑥2𝑥3
�
𝛂𝟑𝟑 =12
35𝑘
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
To start the matrix iteration method we assume the following values,
1st Iteration:
𝑥1 = 1; 𝑥2 = 2; 𝑥3 = 3;
�123� =
𝑚 ω2
7𝐾�4 3 24 7.2 24 3 4.8
� �123� =
𝑚 ω2
7𝐾�
1624.424.4
� = 16 �𝑚 ω2
3𝐾� �
11.5251.525
�
2nd Iteration:
𝑥1 = 1; 𝑥2 = 1.525; 𝑥3 = 1.525;
�1
1.5251.525
� =𝑚 ω2
7𝐾�4 3 24 7.2 24 3 4.8
� �1
1.5251.525
� =𝑚 ω2
7𝐾�11.62518.0315.89
� = 11.625 �𝑚 ω2
3𝐾� �
11.551.36
�
3rd Iteration:
𝑥1 = 1; 𝑥2 = 1.55; 𝑥3 = 1.36;
�1
1.551.36
� =𝑚 ω2
7𝐾�4 3 24 7.2 24 3 4.8
� �1
1.551.36
� =𝑚 ω2
7𝐾�11.3717.8813.88
� = 11.37 �𝑚 ω2
3𝐾� �
11.5721.33
�
4th Iteration:
𝑥1 = 1; 𝑥2 = 1.572; 𝑥3 = 1.33;
�1
1.5721.33
� =𝑚 ω2
7𝐾�4 2 14 8 44 8 7
� �1
1.5721.33
� =𝑚 ω2
7𝐾�
11.3725.138
17.09344� = 11.37 �
𝑚 ω2
3𝐾� �
11.58
1.327�
Since the values are obtained to initial values the first mode of the fundamental frequency is
1 =𝑚 ω2
7𝐾11.37
𝝎 = 𝟎.𝟖𝟎𝟔�𝑲 𝒎
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
10. By using Holzer method, find the natural frequencies of the system in fig 10: assume
K=1N/m; m=1 kg.
Fig: (10)
Solution:
Assumed
frequency Position mi miω2 xi m1xiω2 �𝑚𝑖𝑥𝑖𝜔2
𝑖
1
𝑘𝑖𝑗 �𝑚𝑖𝑥𝑖𝜔2
𝑘𝑖𝑗
𝑖
1
Trial-1
𝜔 = 0.1
𝜔2 = 0.01
1 1 0.01 1.0 0.01 0.01 1 0.01
2 2 0.02 0.99 0.0198 0.0298 1 0.0298
3 4 0.04 0.96 0.0384 0.0682 3 0.0227
4 ∞ 0.93
Trial-2
𝜔 = 0.2
𝜔2 = 0.04
1 1 0.04 1 0.04 0.04 1 0.04
2 2 0.08 0.96 0.077 0.117 1 0.117
3 4 0.16 0.84 0.1324 0.249 3 0.083
4 ∞ 0.76
Trial-3
𝜔 = 0.4
𝜔2 = 0.16
1 1 0.16 1 0.16 0.160 1 0.160
2 2 0.32 0.84 0.269 0.429 1 0.429
3 4 0.64 0.411 0.264 0.693 3 0.231
4 ∞ 0.180
Trial-4 1 1 0.25 1.0 0.25 0.25 1 0.25
K
3K
K
m
2m
4m
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝜔 = 0.5
𝜔2 = 0.25
2 2 0.5 0.75 0.375 0.625 1 0.625
3 4 1 0.125 0.125 0.750 3 0.250
4 ∞ -0.125
Trial-5
𝜔 = 0.60
𝜔2 = 0.36
1 1 0.36 1.0 0.36 0.36 1 0.36
2 2 0.72 0.64 0.46 0.82 1 0.820
3 4 1.44 -0.18 -0.259 0.561 3 0.187
4 ∞ -0.367
Trial-6
𝜔 = 0.80
𝜔2 = 0.64
1 1 0.64 0.1 0.64 0.640 1 0.64
2 2 1.28 0.36 0.461 1.101 1 0.101
3 4 2.56 -0.741 -1.90 -0.80 3 -0.267
4 ∞ -0.474
Trial-7
𝜔 = 1.0
𝜔2 = 1.0
1 1 1 1 1 1 1 1
2 2 2 0 0 1 1 1
3 4 4 -1 -4 -3 3 -1
4 ∞ 0
Trial-8
𝜔 = 1.2
𝜔2 = 1.44
1 1 1.44 1 1.44 1.44 1 1.44
2 2 2.88 -0.44 -1.27 0.17 1 0.17
3 4 5.76 -0.61 -3.51 -3.34 3 -1.11
4 ∞ 0.50
Trial-9
𝜔 = 1.4
𝜔2 = 1.96
1 1 1.96 1.0 1.96 1.96 1 1.96
2 2 3.92 0.96 -3.76 -1.80 1 -1.80
3 4 7.84 0.84 6.58 4.78 3 1.59
4 ∞ -0.75
Plot the graph with the assumed frequencies against the displacement in the rectangle box to get
𝜔𝑛1 , 𝜔𝑛2 & 𝜔𝑛3
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
Kt1 Kt2 Kt3
Ans. From the graph
𝝎𝒏𝟏 , = 𝟎.𝟒𝟔 𝒓𝒂𝒅/𝒔𝒆𝒄
𝝎𝒏𝟐 = 𝟏.𝟎𝟎 𝒓𝒂𝒅/𝒔𝒆𝒄
𝝎𝒏𝟑 = 𝟏.𝟑𝟓 𝒓𝒂𝒅/𝒔𝒆𝒄
11. By using Holzer method, find the natural frequencies of the system in fig 10: assume Kt1=
Kt1= Kt2= Kt3=1N/m; J1= J2= J3= kg.
Solution:
Assumed
frequency Position Ji Jiω2 θi Jiθiω2
�𝐽𝑖θ𝑖𝜔2
𝑖
1
𝑘𝑡𝑖𝑗 �𝐽𝑖θ𝑖𝜔2
𝑘𝑡𝑖𝑗
𝑖
1
Trial-1
𝜔 = 0.5
𝜔2 = 0.25
1 1 0.25 1 0.25 0.25 1 0.25
2 1 0.25 -0.75 0.1875 0.4375 1 0.4375
3 1 0.25 0.3125 0.0781 0.515
Trial-2
𝜔 = 0.75
1 1 0.562 1 0.562 0.562 1 0.5625
2 1 0.562 0.438 0.2461 0.801 1 0.8086
Fig: (11)
J1
or
J2
or
J3
or
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝜔2 = 0.5625 3 1 0.562 -0.3705 0.208 0.6305
Trial-3
𝜔 = 1
𝜔2 = 1
1 1 1 1 1 1 1 1
2 1 1 0 0 1 1 1
3 1 1 -1 -1 0
Trial-4
𝜔 = 1.5
𝜔2 = 2.25
1 1 2.25 1 2.25 2.25 1 2.25
2 1 2.25 -1.25 -2.8125 -0.5625 1 -0.5625
3 1 2.25 -0.6875 -1.5469 -2.1094
Trial-5
𝜔 = 2
𝜔2 = 4
1 1 4 1 4 4 1 4
2 1 4 -3 -12 -8 1 -8
3 1 4 5 20 12
Plot the graph with the assumed frequencies against the displacement in the rectangle box to get
𝜔𝑛1 , 𝜔𝑛2 & 𝜔𝑛3
Ans. From the graph
𝝎𝒏𝟏 , = 𝟎.𝟎𝟎 𝒓𝒂𝒅/𝒔𝒆𝒄
𝝎𝒏𝟐 = 𝟏.𝟎𝟎 𝒓𝒂𝒅/𝒔𝒆𝒄
𝝎𝒏𝟑 = 𝟏.𝟕𝟎 𝒓𝒂𝒅/𝒔𝒆𝒄
12. Find the fundamental vibration for the system shown in figure (12), by using Dunkerley’s
Method.
2K K 3K
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
Fig: (12)
Solution:
First Influence coefficients are:
𝛼11 =1𝐾
& 𝛼12 = 𝛼13 =1𝐾
𝛼11 =1𝐾
Second Influence coefficients are:
𝛼22 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 1𝑘
+ 12𝑘
= 32𝑘
𝛂𝟐𝟐 =𝟑𝟐𝐤
Third Influence coefficients are:
𝛼33 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 1𝑘
+ 12𝑘
+ 13𝑘
= 116𝑘
By using Dunkerley’s method
1/ ωn2 = α11m1 + α22m2 + α33m3 + …………..
Natural frequency is
𝝎𝒏 = �1
α11m1 + α22m2 + α33m3
𝜔𝑛 = �1
1𝐾 (3m) + 𝟑𝟐𝐤 (2m) + 116𝑘 (m)
𝛂𝟑𝟑 =116𝑘
3m 2m m
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝛚𝐧 = 𝟎.𝟑𝟓𝟕�𝐊𝐦
𝐫𝐚𝐝/𝐬𝐞𝐜
13. Find the fundamental vibration for the system shown in figure (13), by using Dunkerley’s
Method.
𝐼 = 4 × 10−7 𝑚4; 𝐸 == 1.96 × 1011 𝑁/𝑚2
First Influence coefficients are:
𝛼11 =𝐹𝑙3
3𝐸𝐼; 𝛼22 =
𝐹𝐿3
3𝐸𝐼
𝛼11 =𝐹𝑙3
3𝐸𝐼=
𝑙3
3𝐸𝐼= 2.47 × 10−8 𝑚/𝑁
𝛼22 =𝐹𝐿3
3𝐸𝐼=
𝐿3
3𝐸𝐼= 1.148 × 10−8 𝑚/𝑁
By using Dunkerley’s method
1/ ωn2 = α11m1 + α22m2
Natural frequency is
𝝎𝒏 = �1
α11m1 + α22m2
𝜔𝑛 = �1
(2.47 × 10−8)(100) + (1.148 × 10−8)(50)
m2=50 kg m1=100 kg
180mm 180mm 300mm
Fig: (13)
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝛚𝐧 = 𝟑𝟒𝟗.𝟎𝟎𝟔 𝐫𝐚𝐝/𝐬𝐞𝐜
14. Find the fundamental vibration for the system shown in figure (14), by using Dunkerley’s
Method.
𝐸 = 2.1 × 1011 𝑁/𝑚2
Static deflection at two points is given by
𝐼 =𝜋𝑑4
64=𝜋(0.05)4
64= 3.066 × 10−7 𝑚4
First Influence coefficients are:
𝛼11 =𝐹𝑙3
3𝐸𝐼; 𝛼22 =
𝐹𝐿3
3𝐸𝐼 & 𝛼12 = 𝛼21 =
𝑙2(3𝐿 − 𝑙)6𝐸𝐼
𝛼11 =𝐹𝑙3
3𝐸𝐼=
𝑙3
3𝐸𝐼= 4.1 × 10−7 𝑚/𝑁
𝛼22 =𝐹𝐿3
3𝐸𝐼=
𝐿3
3𝐸𝐼= 4.75 × 10−7 𝑚/𝑁
By using Dunkerley’s method
1/ ωn2 = α11m1 + α22m2
Natural frequency is
𝝎𝒏 = �1
α11m1 + α22m2
𝜔𝑛 = �1
(4.1 × 10−7)(10) + (4.75 × 10−7)(20)
m2=20 kg m1=10 kg
0.2mm 0.25mm
Fig: (14)
0.05m
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MODULE-VII –CALCULATION OF NATURAL FREQUENCY VIBRATION ENGINEERING 2014
𝛚𝐧 = 𝟐𝟕𝟏.𝟏𝟔𝟑𝟎 𝐫𝐚𝐝/𝐬𝐞𝐜
𝒇𝐧 =𝛚𝐧
𝟐𝝅= 𝟒𝟑.𝟏𝟓𝟔𝟗 𝐇𝐳
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