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

Dr.MOHAMED HANEEF ,PRINCIPAL,GHOUSIA COLLEGE OF ENGINEERING ,RAMANAGARAM

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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 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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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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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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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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http://en.wikipedia.org/wiki/Vibration

http://www.brown.edu/Departments/Engineering/Courses/En4/notes_old/Freevibes/freevibes.html

http://www.brown.edu/Departments/Engineering/Courses/En4/notes_old/Freevibes/freevibes.html

http://iitg.vlab.co.in/?sub=62&brch=175&sim=1077&cnt=2

http://www.acs.psu.edu/drussell/Demos/absorber/DynamicAbsorber.html

http://scholar.cu.edu.eg/?q=anis/files/mpe_-_2013-02-10_-_introduction.pdf

http://www.youtube.com/watch?v=ydflDCPUgYo

http://nptel.iitg.ernet.in/Mech_Engg/IIT%20Guwahati/Vibration%20Engineering(Video).htm

http://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/

http://www.freestudy.co.uk/dynamics/holzer.pdf

http://www.cdeep.iitb.ac.in/nptel/Mechanical/Dynamics%20of%20Machines/Course_home12.5.html




http://www.youtube.com/watch?v=pTRb6LVWiUM


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


𝜶𝟑𝟐 = 𝜶𝟐𝟑 =𝟑𝟐𝐤

Third Influence coefficients are:


Where 1𝐾𝑒

= 1𝑘

+ 12𝑘

+ 13𝑘

= 116𝑘

𝛂𝟑𝟑 =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.

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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𝑦1 = 𝑚1𝑔 𝛼11 + 𝑚2 𝑔 𝛼12

𝑦2 = 𝑚1𝑔 𝛼12 + 𝑚2 𝑔 𝛼22


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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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


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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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




4m 2m

3m

5K

7K 5K


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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


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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𝛂𝟐𝟐 =𝐟𝐾𝑒

=𝟏

3𝑘4

=4

3𝑘


𝜶𝟑𝟐 = 𝜶𝟐𝟑 =𝟒𝟑𝐤



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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�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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method.

Fig: (9)

Solution:



𝛼11 =1

7𝐾




𝛼21 = 𝛼12; 𝛼13 = 𝛼31

𝛼11 = 𝛼12 = 𝛼13 = 𝛼21 = 𝛼31 =1

7𝐾

4m 2m

3m

5K

7K 5K


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By applying unit force on 2nd mass we get 1𝐾𝑒

=1

7𝑘+

15𝑘

= 12

35𝑘


=𝟏

35𝑘12

=12

35𝑘

Since mass three has not connected to mass 2, where

𝜶𝟑𝟐 = 𝜶𝟐𝟑 = 𝜶𝟏𝟏 =𝟏𝟕𝐤



Where 1𝐾𝑒

= 17𝑘

+ 15𝑘

= 1235𝑘


� −−(1)



−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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�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

�


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�


1 =𝑚 ω2

7𝐾11.37


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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

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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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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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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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



Ans. From the graph

𝝎𝒏𝟏 , = 𝟎.𝟎𝟎 𝒓𝒂𝒅/𝒔𝒆𝒄


𝝎𝒏𝟑 = 𝟏.𝟕𝟎 𝒓𝒂𝒅/𝒔𝒆𝒄


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9. Find the fundamental vibration for the system shown in figure (12), by using Dunkerley’s

Method.

Fig: (12)

Solution:


𝛼11 =1𝐾

& 𝛼12 = 𝛼13 =1𝐾

𝛼11 =1𝐾



Where 1𝐾𝑒

= 1𝑘

+ 12𝑘

= 32𝑘

𝛂𝟐𝟐 =𝟑𝟐𝐤



Where 1𝐾𝑒

= 1𝑘

+ 12𝑘

+ 13𝑘

= 116𝑘

By using Dunkerley’s method

1/ ωn2 = α11m1 + α22m2 + α33m3 + …………..


𝝎𝒏 = �1

α11m1 + α22m2 + α33m3

𝛂𝟑𝟑 =116𝑘

3m 2m m

2K K 3K


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𝜔𝑛 = �1

1𝐾 (3m) + 𝟑𝟐𝐤 (2m) + 116𝑘 (m)

𝛚𝐧 = 𝟎.𝟑𝟓𝟕�𝐊𝐦



Method.

𝐼 = 4 × 10−7 𝑚4; 𝐸 == 1.96 × 1011 𝑁/𝑚2


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


1/ ωn2 = α11m1 + α22m2


𝝎𝒏 = �1

α11m1 + α22m2

m2=50 kg m1=100 kg

180mm 180mm 300mm

Fig: (13)


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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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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



http://en.wikipedia.org/wiki/Dunkerley's_method



www.freestudy.co.uk/dynamics/holzer.pdf%E2%80%8E


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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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


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


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


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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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


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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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


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


𝐾 15.4mω2

𝐾 28.66mω2

𝐾



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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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


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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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


× 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


× 10−3ω2 19.55× 10−3ω2


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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× 10−3ω2 20.32× 10−3ω2



given by

�1

3.474.44

� = 4.58 × 10−3ω2 + 15.91 × 10−3ω2 + 15.91 × 10−3ω2

𝛚 = 𝟏𝟒.𝟕𝟕𝟔 𝐫𝐚𝐝/𝐬𝐞𝐜


method.

Fig: (8)

Solution:



𝛼11 =1

3𝐾




𝛼21 = 𝛼12; 𝛼13 = 𝛼31

𝛼11 = 𝛼12 = 𝛼13 = 𝛼21 = 𝛼31 =1

3𝐾

K

3K

K

3m

2m

4m


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1𝐾𝑒

=1

3𝑘+

1𝑘

= 4

3𝑘


=𝟏

3𝑘4

=4

3𝑘


𝜶𝟑𝟐 = 𝜶𝟐𝟑 =𝟒𝟑𝐤



Where 1𝐾𝑒

= 13𝑘

+ 1𝑘

+ 1𝑘

= 73𝑘


� −−(1)



−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


⎥⎥⎥⎥⎤

−−(1′)


𝛂𝟑𝟑 =7

3𝑘


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�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

�


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�


1 =𝑚 ω2

3𝐾14.232

𝝎 = 𝟎.𝟒𝟓𝟕�𝑲 𝒎


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method.

Fig: (9)

Solution:



𝛼11 =1

7𝐾




𝛼21 = 𝛼12; 𝛼13 = 𝛼31

𝛼11 = 𝛼12 = 𝛼13 = 𝛼21 = 𝛼31 =1

7𝐾


1𝐾𝑒

=1

7𝑘+

15𝑘

= 12

35𝑘


=𝟏

35𝑘12

=12

35𝑘


𝜶𝟑𝟐 = 𝜶𝟐𝟑 = 𝜶𝟏𝟏 =𝟏𝟕𝐤

4m 2m

3m

5K

7K 5K


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Where 1𝐾𝑒

= 17𝑘

+ 15𝑘

= 1235𝑘


� −−(1)



−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′)


�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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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�


1 =𝑚 ω2

7𝐾11.37

𝝎 = 𝟎.𝟖𝟎𝟔�𝑲 𝒎


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Method.

Fig: (12)

Solution:


𝛼11 =1𝐾

& 𝛼12 = 𝛼13 =1𝐾

𝛼11 =1𝐾



Where 1𝐾𝑒

= 1𝑘

+ 12𝑘

= 32𝑘

𝛂𝟐𝟐 =𝟑𝟐𝐤



Where 1𝐾𝑒

= 1𝑘

+ 12𝑘

+ 13𝑘

= 116𝑘


1/ ωn2 = α11m1 + α22m2 + α33m3 + …………..


𝛂𝟑𝟑 =116𝑘

3m 2m m

2K K 3K


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𝝎𝒏 = �1

α11m1 + α22m2 + α33m3

𝜔𝑛 = �1

1𝐾 (3m) + 𝟑𝟐𝐤 (2m) + 116𝑘 (m)

𝛚𝐧 = 𝟎.𝟑𝟓𝟕�𝐊𝐦



Method.

𝐼 = 4 × 10−7 𝑚4; 𝐸 == 1.96 × 1011 𝑁/𝑚2


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


1/ ωn2 = α11m1 + α22m2


m2=50 kg m1=100 kg

180mm 180mm 300mm

Fig: (13)


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𝝎𝒏 = �1

α11m1 + α22m2

𝜔𝑛 = �1

(2.47 × 10−8)(100) + (1.148 × 10−8)(50)



Method.

𝐸 = 2.1 × 1011 𝑁/𝑚2


𝐼 =𝜋𝑑4

64=𝜋(0.05)4

64= 3.066 × 10−7 𝑚4


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


1/ ωn2 = α11m1 + α22m2


m2=20 kg m1=10 kg

0.2mm 0.25mm

Fig: (14)

0.05m


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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


method.

Fig: (1)

Solution:


𝛼11 =1𝐾

& 𝛼12 = 𝛼13 =1𝐾


𝛼21 = 𝛼12; 𝛼13 = 𝛼31

𝛼11 = 𝛼12 = 𝛼13 = 𝛼21 = 𝛼31 =1𝐾



Where 1𝐾𝑒

= 1𝑘

+ 12𝑘

= 32𝑘

𝛂𝟐𝟐 =𝟑𝟐𝐤


𝜶𝟑𝟐 = 𝜶𝟐𝟑 =𝟑𝟐𝐤



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


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



method.

𝛂𝟑𝟑 =116𝑘


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𝐼 = 4 × 10−7 𝑚4; 𝐸 == 1.96 × 1011 𝑁/𝑚2


𝑦1 = 𝑚1𝑔 𝛼11 + 𝑚2 𝑔 𝛼12

𝑦2 = 𝑚1𝑔 𝛼12 + 𝑚2 𝑔 𝛼22


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


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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method.

𝐸 == 2.1 × 1011 𝑁/𝑚2


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


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


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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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


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



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given by

�1

1.782.214

� = 4.994ω2 + 8.98ω2 + 11.129ω2

ω = 0.448 rad/sec


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



4m 2m

3m

5K

7K 5K


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given by

�1

1.571.32

� = 1.6ω2 + 2.5ω2 + 2.147ω2

ω = 786 rad/sec


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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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


𝐾 13.45mω2

𝐾 24.70mω2

𝐾


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


𝐾 15.4mω2

𝐾 28.66mω2

𝐾



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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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:


𝐼2𝑜𝑟 𝐽2 = 250𝑘𝑔𝑚𝑚2; 𝐾𝑡2 == 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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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


× 10−3ω2 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


× 10−3ω2 15.39× 10−3ω2


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


× 10−3ω2 19.55× 10−3ω2


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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× 10−3ω2 20.32× 10−3ω2



given by

�1

3.474.44

� = 4.58 × 10−3ω2 + 15.91 × 10−3ω2 + 15.91 × 10−3ω2

𝛚 = 𝟏𝟒.𝟕𝟕𝟔 𝐫𝐚𝐝/𝐬𝐞𝐜


method.

Fig: (8)

Solution:



𝛼11 =1

3𝐾




𝛼21 = 𝛼12; 𝛼13 = 𝛼31

𝛼11 = 𝛼12 = 𝛼13 = 𝛼21 = 𝛼31 =1

3𝐾

K

3K

K

3m

2m

4m


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1𝐾𝑒

=1

3𝑘+

1𝑘

= 4

3𝑘


=𝟏

3𝑘4

=4

3𝑘


𝜶𝟑𝟐 = 𝜶𝟐𝟑 =𝟒𝟑𝐤



Where 1𝐾𝑒

= 13𝑘

+ 1𝑘

+ 1𝑘

= 73𝑘


� −−(1)



−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


⎥⎥⎥⎥⎤

−−(1′)


𝛂𝟑𝟑 =7

3𝑘


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�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

�


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�


1 =𝑚 ω2

3𝐾14.232

𝝎 = 𝟎.𝟒𝟓𝟕�𝑲 𝒎


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method.

Fig: (9)

Solution:



𝛼11 =1

7𝐾




𝛼21 = 𝛼12; 𝛼13 = 𝛼31

𝛼11 = 𝛼12 = 𝛼13 = 𝛼21 = 𝛼31 =1

7𝐾


1𝐾𝑒

=1

7𝑘+

15𝑘

= 12

35𝑘


=𝟏

35𝑘12

=12

35𝑘


𝜶𝟑𝟐 = 𝜶𝟐𝟑 = 𝜶𝟏𝟏 =𝟏𝟕𝐤

4m 2m

3m

5K

7K 5K


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Where 1𝐾𝑒

= 17𝑘

+ 15𝑘

= 1235𝑘


� −−(1)



−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′)


�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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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�


1 =𝑚 ω2

7𝐾11.37

𝝎 = 𝟎.𝟖𝟎𝟔�𝑲 𝒎


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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

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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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




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Kt1 Kt2 Kt3

Ans. From the graph

𝝎𝒏𝟏 , = 𝟎.𝟒𝟔 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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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



Ans. From the graph

𝝎𝒏𝟏 , = 𝟎.𝟎𝟎 𝒓𝒂𝒅/𝒔𝒆𝒄


𝝎𝒏𝟑 = 𝟏.𝟕𝟎 𝒓𝒂𝒅/𝒔𝒆𝒄


Method.

2K K 3K


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Fig: (12)

Solution:


𝛼11 =1𝐾

& 𝛼12 = 𝛼13 =1𝐾

𝛼11 =1𝐾



Where 1𝐾𝑒

= 1𝑘

+ 12𝑘

= 32𝑘

𝛂𝟐𝟐 =𝟑𝟐𝐤



Where 1𝐾𝑒

= 1𝑘

+ 12𝑘

+ 13𝑘

= 116𝑘


1/ ωn2 = α11m1 + α22m2 + α33m3 + …………..


𝝎𝒏 = �1

α11m1 + α22m2 + α33m3

𝜔𝑛 = �1

1𝐾 (3m) + 𝟑𝟐𝐤 (2m) + 116𝑘 (m)

𝛂𝟑𝟑 =116𝑘

3m 2m m


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𝛚𝐧 = 𝟎.𝟑𝟓𝟕�𝐊𝐦



Method.

𝐼 = 4 × 10−7 𝑚4; 𝐸 == 1.96 × 1011 𝑁/𝑚2


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


1/ ωn2 = α11m1 + α22m2


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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Method.

𝐸 = 2.1 × 1011 𝑁/𝑚2


𝐼 =𝜋𝑑4

64=𝜋(0.05)4

64= 3.066 × 10−7 𝑚4


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


1/ ωn2 = α11m1 + α22m2


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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𝛚𝐧 = 𝟐𝟕𝟏.𝟏𝟔𝟑𝟎 𝐫𝐚𝐝/𝐬𝐞𝐜

𝒇𝐧 =𝛚𝐧

𝟐𝝅= 𝟒𝟑.𝟏𝟓𝟔𝟗 𝐇𝐳


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module-vii –calculation of natural frequency vibration engineering

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Transcript of module-vii –calculation of natural frequency vibration engineering