VTU-NPTEL-NMEICT Project Progress Reportnptel.vtu.ac.in/VTU-NMEICT/MV/Module 6.pdf · VTU-NPTEL-NMEICT Project Progress Report . ... Maxwell’s reciprocal theorem: Maxwell’s reciprocal

MODULE-VI --- MULTI DOF 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

VI

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

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CONTENTS

Sl. No. DISCRETION

1. Lecture Notes (Multi -DOF)

2. Quadrant -2

a. Animations.

b. Videos.

c. Illustrations.

3. Quadrant -3

a. Wikis.

b. Open Contents

4. Quadrant -4

a. Problems.

b Self Assigned Q & A.


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VI. Multi DOF

Lecture Notes

EQUATIONS OF MOTION:

The equations of motion for a two degree-of-freedom system are derived using the free body

diagram method or an energy method.

The free-body diagram method is the same as for SDOF systems, except that multiple free-

body diagrams or equations may be used. Newton’s law (∑F=ma) is applied to the free-body

diagram of a particle. The equations ∑F=ma and ∑M=Iα are applied to a free-body diagram

of a rigid body undergoing planar motion with rotation about a fixed axis through zero. For a

rigid body undergoing planar motion, D-Alembert’s principle can be applied as ∑Fext=∑Feff

and (∑MA)ext=(∑MA )eff where A is any point. The system of effective forces is a force

equal to applied at the mass center and a moment equal to Iα.

INFLUENCE COEFFICIENTS

It is the influence of unit displacement at one point on the forces at various points of a multi-

DOF system. (Or) it is the influence of unit Force at one point on the displacements at various

points of a multi-DOF system.

The equations of motion of a multi-degree freedom system can be written in terms of

influence coefficients. A set of influence coefficients can be associated with each of matrices

involved in the equations of motion.

[M]{��} + [K] {x} = [0]

For a simple linear spring the force necessary to cause unit elongation is referred as stiffness

of spring. For a multi-DOF system one can express the relationship between displacement at

a point and forces acting at various other points of the system by using influence coefficients

referred as stiffness influence coefficients

The equations of motion of a multi-degree freedom system can be written in terms of inverse

of stiffness matrix referred as flexibility influence coefficients.

Matrix of flexibility influence coefficients = [ K ]−1

The elements corresponds to inverse mass matrix are referred as flexibility mass/inertia

coefficients.

Matrix of flexibility mass/inertia coefficients = [ M ]−1


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The flexibility influence coefficients are popular as these coefficients give elements of

inverse of stiffness matrix. The flexibility mass/inertia coefficients give elements of inverse

of mass matrix

STIFFNESS INFLUENCE COEFFICIENTS:

For a multi-DOF system one can express the relationship between displacement at a point and

forces acting at various other points of the system by using influence coefficients referred as

stiffness influence coefficients.

{F} = [K]{x}

[K] = �𝐾11 𝐾12 𝐾13𝐾21 𝐾22 𝐾23𝐾31 𝐾32 𝐾33

�

Where, k11, ……..k33 are referred as stiffness influence coefficients

K11 - stiffness influence coefficient at point 1 due to a unit deflection at point 1

k21 - stiffness influence coefficient at point 2 due to a unit deflection at point 1


1. Flexibility influence coefficients.

It is the influence of unit Force at one point on the displacements at various points of a

multi-DOF system, the relationship between force and displacement is given by,

{F} = [K]{x}

{x}= [K]-1 {F}

{x} = [αij]{F}

[α𝑖𝑗] = �α11 α12 α13α21 α22 α23α31 α32 α33

�

Where, [αij] - Matrix of Flexibility influence coefficients given by

α 11, ……..a33 are referred as stiffness influence coefficients

α 11 -flexibility influence coefficient at point 1 due to a unit force at point 1

α 21 - flexibility influence coefficient at point 2 due to a unit force at point 1



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2. Maxwell’s reciprocal theorem:

Maxwell’s reciprocal theorem states the that deflection at any point in the system due to a

unit load acting at any other of the same system is equal to the deflection at the second point

due to a unit load at the first point.

According to this statement,

∴ α𝑖𝑗 = α𝑗𝑖 (Maxwell’s reciprocal theorem)

To prove this theorem considering a simply supported beam with concentrated loads. F1 and

F2 as shown in fig: below.

Fig: Maxwell’s reciprocal theorem

The work done is obtained when the system is deformed by the application of force.

Now suppose force F1 is applied at point 1 gradually from zero its maximum value. Then

deflection at point 1 is α11F1.

∴ 𝑤𝑜𝑟𝑘𝑑𝑜𝑛𝑒 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 1 = 12𝐹1(𝛼11𝐹1) =

12𝐹12𝛼11 −−− (1)

when the force F2 is gradually applied at point 2, the workdone will be,


12𝐹22𝛼22 −−− (2)

But when F2 is applied, there will be an additional deflection at point 1 due to F2, which is

equal to 𝛼12𝐹2. since already a force W1 is acting at point i, the workdone by force F1

corresponding to deflection 𝛼12𝐹2 at point 1 will be 𝐹1𝐹2𝛼12. Hence, total workdone in the

first mode.

(𝐹𝐷)1 = 12𝐹12𝛼11 +

12𝐹22𝛼22 + 𝐹1𝐹2𝛼12 −−− (3)

Similarly workdone in the second mode

(𝐹𝐷)2 = 12𝐹12𝛼22 +

12𝐹22𝛼11 + 𝐹1𝐹2𝛼21 −−− (3)

From the equation (1) & (2) we have

α12 = α21

In general we can write

∴ 𝛂𝒊𝒋 = 𝛂𝒋𝒊 Proof of the theorem.

F1 F2

2 1


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MODAL ANALYSIS

This is a technique by which the equations of motion (EOM), which are originally expressed

in physical coordinates, are transformed to modal coordinates using the eigenvalues and

eigenvectors gotten by solving the undamped frequency eigenproblem. The transformed

equations are called modal equations. In a mathematical context, modal coordinates are also

called generalized coordinates, or principal coordinates. For structural dynamics they can be

interpreted as response amplitudes of orthonormalized vibration modes.

The distinguishing feature of modal equations is that for an undamped system they uncouple.

Consequently, each modal equation may be solved independently of the others. Once

computed, modal responses may be transformed back to physical coordinates and superposed

to produce the physical response of the original system. The method is a particular case of

what is known in applied mathematics as orthogonal projection methods. Instead of tackling

the original equations directly, they are projected onto another space (the ‘modal space” in

the case of dynamics) in which equations decouple.

1. Modal analysis on undamped:

Two degree of freedom system it is possible to uncouple the equations of motion of an n-

DOF provided we know beforehand the normal modes or eigenvectors of the system. A

modal matrix [U] is referred to a square matrix where in each column represents an

eigenvector.

Thus for an n-degree of freedom system,

[U] =

⎣⎢⎢⎢⎢⎢⎢⎡

⎩⎪⎪⎨

⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭

⎪⎪⎬

⎪⎪⎫

1 ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

2

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑟

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑠

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑛⎦⎥⎥⎥⎥⎥⎥⎤

−−(1)

The transpose of the above matrix can be written as


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[U]′ =

⎣⎢⎢⎢⎢⎢⎡[𝑋1[𝑋1⋯

[𝑋1⋯

[𝑋1⋯

[𝑋1

𝑋2𝑋2⋯𝑋2⋯𝑋2⋯𝑋2

⋯⋯⋯⋯⋯⋯⋯⋯

𝑋𝑟𝑋𝑟⋯𝑋𝑟⋯𝑋𝑟⋯𝑋𝑟

⋯⋯⋯⋯⋯⋯⋯⋯

𝑋𝑠𝑋𝑠⋯𝑋𝑠⋯𝑋𝑠⋯𝑋𝑠

⋯⋯⋯⋯⋯⋯⋯⋯

𝑋𝑛]1𝑋𝑛]2⋯𝑋𝑛]𝑟⋯𝑋𝑛]𝑠⋯𝑋𝑛]𝑛⎦

⎥⎥⎥⎥⎥⎤

−−(2)

Let the differential equation of motion for an n-degree of freedom undamped system be,

[M]{��} + [K] {x} = [0] ---- (3)

To decouple these equations, let us the linear transformation

[X] = [U] {y} ---- (4)

Where {y} are the principal coordinates and can be found out by pre-multiplying the above

equation by [U]-1 to get

[U]-1 {x} = [U]-1[U] {y}

{y} = [U]-1 {x} ---- (5)

Substituting Equation (3) in (4),

[M] [U]{��} + [K] [U]{y} = [0] ---- (6)

Pre multiply the above equation by [U]’ to get

[U]’ [M] [U]{��}+ [U]’ [K] [U] {y}= [0] ---- (7)

The terms [U]’ [M] [U] & [U]’ [K] [U] in the above equation are each a diagonal matrix since

the off diagonal terms which involve rth row and Sth column express the orthogonality

relationship which are zero.

{X}r’ [M] {X}S = 0, r ≠ s

And {X}r’ [K] {X}S = 0, r ≠ s ---- (8)

Terms [U]’ [M] [U] & [U]’ [K] [U] in the above equation are each a diagonal which involve

rth row and rth column express the gives generalized relationship which are zero.

{X}r’ [M] {X}r = Mr

And {X}r’ [K] {X}r = Mr ---- (9)

As we know that equation of motion


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⎣⎢⎢⎢⎡𝑀10⋯0⋯0

0𝑀2⋯0⋯0

⋯⋯⋯⋯⋯⋯

00⋯𝑀3⋯0

⋯⋯⋯⋯⋯⋯

00⋯0⋯𝑀4

⎦⎥⎥⎥⎤

⎩⎪⎨

⎪⎧��1��2⋯��𝑟⋯��𝑛⎭⎪⎬

⎪⎫

+

⎣⎢⎢⎢⎡𝐾10⋯0⋯0

0𝐾2⋯0⋯0

⋯⋯⋯⋯⋯⋯

00⋯𝐾3⋯0

⋯⋯⋯⋯⋯⋯

00⋯0⋯𝐾4

⎦⎥⎥⎥⎤

⎩⎪⎨

⎪⎧𝑦1𝑦2⋯𝑦𝑟⋯𝑦𝑛⎭⎪⎬

⎪⎫

=

⎩⎪⎨

⎪⎧0

0⋯0⋯0⎭⎪⎬

⎪⎫

−−(10)

� \ 𝑀𝑟

\ � {��} + �

\ 𝐾𝑟

\ � {y} = {0} −−(11)

Further, it can easily be seen that

𝐾𝑟 = λ𝑟𝑀𝑟 −−(12)

To prove the above relationship we write the equation

𝐾 {𝑋}𝑟 = λ𝑟𝑀 {𝑋}𝑟 (13)

Pre multiply the above equation by transpose of r th mode to get

{𝑋}𝑟′ [𝐾] {𝑋}𝑟 = λ𝑟 {𝑋}′𝑟[𝑀]{𝑋}𝑟 (14)

The above equation from

𝐾𝑟 = λ𝑟𝑀𝑟 (15)

Substituting equation (12) in Equation (11)

� \ 𝑀𝑟

\ � {��} + �

\ λ𝑟𝑀𝑟 \

� {y} = {0}

� \ 𝑀𝑟

\ � {��} + �

\ 𝜔𝑟2𝑀𝑟 \

� {y} = {0} −−− (16)

Where λ𝑟 = 𝜔𝑟2= the eigenvalue for the rth mode

Thus the above equations as detailed below are n-uncoupled differential equations of motion for an n-degree of freedom system in terms of the principal coordinates y.

��𝑟 + 𝜔𝑟2𝑦𝑟 = 0 (𝑟 = 1,2,3 … … . .𝑛) −−− (17) The solution of the above equation is

𝑦𝑟 = 𝐴𝑟 𝑐𝑜𝑠𝜔𝑟𝑡 + 𝐵𝑟 𝑠𝑖𝑛𝜔𝑟𝑡 (𝑟 = 1,2,3 … … . .𝑛) −−− (18) The above eqution can also be written as

{𝑦} = {𝐴𝑐𝑜𝑠𝜔𝑡 + 𝐵𝑠𝑖𝑛𝜔𝑡} −−− (19) From the Eq. (4) and Eq. (18)

�

𝑋1𝑋2⋯𝑋𝑛

� = [U ]�

𝐴1 𝑐𝑜𝑠𝜔1𝑡 + 𝐵1 𝑠𝑖𝑛𝜔1𝑡 𝐴2 𝑐𝑜𝑠𝜔2𝑡 + 𝐵2 𝑠𝑖𝑛𝜔2𝑡

⋯ ⋯ ⋯ ⋯ ⋯ 𝐴𝑛 𝑐𝑜𝑠𝜔𝑛𝑡 + 𝐵𝑛 𝑠𝑖𝑛𝜔𝑛𝑡

� −−− (20)


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Which give the vibratory response of un-damped free vibrations. 𝐴𝑟 𝑎𝑛𝑑 𝐵𝑟 (r=

1,2,…..n) can be obtained from the initial conditions.

Equation (4) can also be seen to be expanded and put in the folloing form which is more

convenient at times.

⎩⎪⎨

⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬

⎪⎫

=

⎩⎪⎨


⎪⎫

1

𝑦1 +

⎩⎪⎨


⎪⎫

2

𝑦2 +

⎩⎪⎨


⎪⎫

𝑟

𝑦𝑟 +

⎩⎪⎨


⎪⎫

𝑛

𝑦𝑛 −−− (21)

Where y1, y2, …… yn are as in eq. (18). The above equation can be written in short as

{X} = {X}1𝑦1 + {X}2𝑦2 + {X}𝑟𝑦𝑟 + {X}𝑛𝑦𝑛 −−− (21)

2. Model analysis: damped

Two degree of freedom system it is possible to uncouple the equations of motion of an n-DOF provided we know beforehand the normal modes or eigenvectors of the system. A modal matrix [U] is referred to a square matrix where in each column represents an eigenvector. Thus for an n-degree of freedom system,

[U] =

⎣⎢⎢⎢⎢⎢⎢⎡

⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

1 ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

2

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑟

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑠

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑛⎦⎥⎥⎥⎥⎥⎥⎤

−−(1)


[U]′ =

⎣⎢⎢⎢⎢⎢⎡[𝑋1[𝑋1⋯

[𝑋1⋯

[𝑋1⋯

[𝑋1

𝑋2𝑋2⋯𝑋2⋯𝑋2⋯𝑋2

⋯⋯⋯⋯⋯⋯⋯⋯


⋯⋯⋯⋯⋯⋯⋯⋯


⋯⋯⋯⋯⋯⋯⋯⋯


⎥⎥⎥⎥⎥⎤

−−(2)

Let the differential equation of motion for an n-degree of freedom damped system be, [M]{��} +[C] {��} + [K] {x} = [0] ---- (3)

Where [C] is the damping matrix

[C] = �

𝑐11𝑐21⋯⋯𝑐𝑛1

𝑐12𝑐22⋯⋯𝑐𝑛2

⋯⋯⋯⋯⋯

⋯⋯⋯⋯⋯

⋯⋯⋯⋯⋯

𝑐1𝑛𝑐2𝑛⋯⋯𝑐𝑛𝑛

� −−(4)


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Apart from static and dynamic coupling amongst the generalized coordinates X1, X2, … Xn

there exists now damping coupling also. The equations can get uncoupled with regard to

damping matrix contains only the diagonal terms, i.e., if

𝑐𝑖𝑗 = 0 𝑖 ≠ 𝑗


[X] = [U] {y} ---- (5)



[M] [U]{��} +[c] [U]{��} + [K] [U]{y} = [0] ---- (6)


[U]’ [M] [U]{��}+[U]’[c] [U]{��} + [U]’ [K] [U] {y}= [0] ---- (7)




{X}r’ [M] {X}S = 0, r ≠ s

And {X}r’ [K] {X}S = 0, r ≠ s ---- (8)



{X}r’ [M] {X}r = Mr

And {X}r’ [K] {X}r = Mr ---- (9)


⎣⎢⎢⎢⎡𝑀10⋯0⋯0

0𝑀2⋯0⋯0

⋯⋯⋯⋯⋯⋯

00⋯𝑀3⋯0

⋯⋯⋯⋯⋯⋯

00⋯0⋯𝑀4

⎦⎥⎥⎥⎤

⎩⎪⎨

⎪⎧��1��2⋯��𝑟⋯��𝑛⎭⎪⎬

⎪⎫

+

⎣⎢⎢⎢⎡𝐶10⋯0⋯0

0𝐶2⋯0⋯0

⋯⋯⋯⋯⋯⋯

00⋯𝐶3⋯0

⋯⋯⋯⋯⋯⋯

00⋯0⋯𝐶4

⎦⎥⎥⎥⎤

⎩⎪⎨

⎪⎧��1��2⋯��𝑟⋯��𝑛⎭⎪⎬

⎪⎫

+

⎣⎢⎢⎢⎡𝐾10⋯0⋯0

0𝐾2⋯0⋯0

⋯⋯⋯⋯⋯⋯

00⋯𝐾3⋯0

⋯⋯⋯⋯⋯⋯

00⋯0⋯𝐾4

⎦⎥⎥⎥⎤

⎩⎪⎨


⎪⎫

=

⎩⎪⎨

⎪⎧0

0⋯0⋯0⎭⎪⎬

⎪⎫

−−(10)

� \ 𝑀𝑟

\ � {��} + �

\ 𝐶𝑟

\ � {y} + �

\ 𝐾𝑟

\ � {y} = {0} −−(11)


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𝐾𝑟 = λ𝑟𝑀𝑟 −−(12)


𝐾 {𝑋}𝑟 = λ𝑟𝑀 {𝑋}𝑟 (13)


The term [U]’[c] [U] usually does not reduce to a diagonal matrix unless we ase the concept

of proportional damping, i.e., [c] being proportional to [M] or to [K] or to a linear

combination of both.

Linear combination of both

[C]= α[M]+β[K] (14)

Where α and β are constants.

Then [U]’[c] [U] = α [U]’ [M] [U] +β [U]’ [K] [U]

=� \ α𝑀𝑟 \

� + � \

β λ𝑟𝑀𝑟 \

�

=� \

(α + β 𝜔𝑟2)𝑀𝑟 \

� (15)

Expressing (α + β 𝜔𝑟2) in terms of the modal damping ξ𝑟

(α + β 𝜔𝑟2) = 2ξ𝑟𝜔𝑟


� \ 𝑀𝑟

\ � {��} + �

\ � 2ξ𝑟𝜔𝑟�𝑀𝑟

\ � {��𝑟} + �


� {y} = {0} −−(16)


��𝑟 + �2ξ𝑟𝜔𝑟��𝑟 + 𝜔𝑟2𝑦𝑟 = 0 (𝑟 = 1,2,3 … … . . 𝑛) −−− (17) The solution of the above equation is

𝑦𝑟 = 𝑒−𝜁𝜔𝑛t� 𝐴𝑟𝑐𝑜𝑠��1 − 𝜁 2�𝜔𝑟t + 𝐵𝑟𝑠𝑖𝑛��1 − 𝜁 2�𝜔𝑟t �

(𝑟 = 1,2,3 … … . . 𝑛) −−− (18) From the Eq. (4) and Eq. (18)

�


� = [U ]�

𝑦1𝑦2⋯𝑦𝑛� −−− (20)


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Which give the vibratory response of damped free vibrations. 𝐴𝑟 𝑎𝑛𝑑 𝐵𝑟 (r= 1,2,…..n)

can be obtained from the initial conditions.



⎩⎪⎨


⎪⎫

=

⎩⎪⎨


⎪⎫

1

𝑦1 +

⎩⎪⎨


⎪⎫

2

𝑦2 +

⎩⎪⎨


⎪⎫

𝑟

𝑦𝑟 +

⎩⎪⎨


⎪⎫

𝑛

𝑦𝑛 −−− (21)


{X} = {X}1𝑦1 + {X}2𝑦2 + {X}𝑟𝑦𝑟 + {X}𝑛𝑦𝑛 −−− (21)


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QUADRANT-2

Animations(related to Multi DOF) • http://www.acs.psu.edu/drussell/Demos/multi-dof-springs/multi-dof-

springs.html • http://ae-www.usc.edu/Books/yang-

sssd/demos/Example%2017%20modes%20of%20vibration%5Cmodes.htm • http://www.lmsintl.com/modal-analysis • http://www.youtube.com/watch?v=-qRYaZP0938

Videos: (related to Multi DOF)

• http://www.youtube.com/watch?v=904QIyzM29o • http://www.youtube.com/watch?v=t3381jmX3u8 • http://freevideolectures.com/Course/3129/Structural-Dynamics/38 • http://www.learnerstv.com/video/Free-video-Lecture-24094-engineering.htm • http://freevideolectures.com/Course/2684/Mechanical-Vibrations/20 • http://www.youtube.com/watch?v=OxcCPTc_bXw • http://www.cosmolearning.com/video-lectures/modal-analysis-damped-11554/ • http://www.learningace.com/doc/2579161/c3aa48e3a29f5abf68b2ec8c286d869a/m

od-7-lec-4-modal-analysis-damped • http://mechanical-engineering.in/forum/videos/view-502-lecture-20-mod-7-lec-4-

modal-analysis-damped/


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http://www.acs.psu.edu/drussell/Demos/multi-dof-springs/multi-dof-springs.html

http://www.acs.psu.edu/drussell/Demos/multi-dof-springs/multi-dof-springs.html

http://ae-www.usc.edu/Books/yang-sssd/demos/Example%2017%20modes%20of%20vibration%5Cmodes.htm

http://ae-www.usc.edu/Books/yang-sssd/demos/Example%2017%20modes%20of%20vibration%5Cmodes.htm

http://www.youtube.com/watch?v=904QIyzM29o

http://www.youtube.com/watch?v=t3381jmX3u8

http://freevideolectures.com/Course/3129/Structural-Dynamics/38

http://www.learnerstv.com/video/Free-video-Lecture-24094-engineering.htm

http://freevideolectures.com/Course/2684/Mechanical-Vibrations/20

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

http://www.cosmolearning.com/video-lectures/modal-analysis-damped-11554/

http://www.learningace.com/doc/2579161/c3aa48e3a29f5abf68b2ec8c286d869a/mod-7-lec-4-modal-analysis-damped

http://www.learningace.com/doc/2579161/c3aa48e3a29f5abf68b2ec8c286d869a/mod-7-lec-4-modal-analysis-damped

http://mechanical-engineering.in/forum/videos/view-502-lecture-20-mod-7-lec-4-modal-analysis-damped/

http://mechanical-engineering.in/forum/videos/view-502-lecture-20-mod-7-lec-4-modal-analysis-damped/

http://www.lmsintl.com/modal-analysis

http://www.youtube.com/watch?v=-qRYaZP0938


ILLUSTRATIONS

1. Explain the different types of influence coefficient Methods.

Solution) 1. STIFFNESS INFLUENCE COEFFICIENTS:




{F} = [K]{x}

[K] = �𝐾11 𝐾12 𝐾13𝐾21 𝐾22 𝐾23𝐾31 𝐾32 𝐾33

�








{F} = [K]{x}

{x}= [K]-1 {F}

{x} = [αij]{F}

[α𝑖𝑗] = �α11 α12 α13α21 α22 α23α31 α32 α33

�







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2. Explain Maxwell’s reciprocal theorem

Solu) Maxwell’s reciprocal theorem states the that deflection at any point in the system due

to a unit load acting at any other of the same system is equal to the deflection at the second

point due to a unit load at the first point.










12𝐹12𝛼11 −−− (1)



12𝐹22𝛼22 −−− (2)




first mode.

(𝐹𝐷)1 = 12𝐹12𝛼11 +

12𝐹22𝛼22 + 𝐹1𝐹2𝛼12 −−− (3)


(𝐹𝐷)2 = 12𝐹12𝛼22 +

12𝐹22𝛼11 + 𝐹1𝐹2𝛼21 −−− (3)


α12 = α21


F1 F2

2 1


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3. Explain briefly modal analysis on Undamped Free vibration.

Solution)


[U] =

⎣⎢⎢⎢⎢⎢⎢⎡

⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

1 ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

2

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑟

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑠

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑛⎦⎥⎥⎥⎥⎥⎥⎤

−−(1)


[U]′ =

⎣⎢⎢⎢⎢⎢⎡[𝑋1[𝑋1⋯

[𝑋1⋯

[𝑋1⋯

[𝑋1

𝑋2𝑋2⋯𝑋2⋯𝑋2⋯𝑋2

⋯⋯⋯⋯⋯⋯⋯⋯


⋯⋯⋯⋯⋯⋯⋯⋯


⋯⋯⋯⋯⋯⋯⋯⋯


⎥⎥⎥⎥⎥⎤

−−(2)

Let the differential equation of motion for an n-degree of freedom undamped system be, [M]{��} + [K] {x} = [0] ---- (3)


[X] = [U] {y} ---- (4)



[U]-1 {x} = [U]-1[U] {y}

{y} = [U]-1 {x} ---- (5)


[M] [U]{��} + [K] [U]{y} = [0] ---- (6)


[U]’ [M] [U]{��}+ [U]’ [K] [U] {y}= [0] ---- (7)




{X}r’ [M] {X}S = 0, r ≠ s


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And {X}r’ [K] {X}S = 0, r ≠ s ---- (8)



{X}r’ [M] {X}r = Mr

And {X}r’ [K] {X}r = Mr ---- (9)


⎣⎢⎢⎢⎡𝑀10⋯0⋯0

0𝑀2⋯0⋯0

⋯⋯⋯⋯⋯⋯

00⋯𝑀3⋯0

⋯⋯⋯⋯⋯⋯

00⋯0⋯𝑀4

⎦⎥⎥⎥⎤

⎩⎪⎨

⎪⎧��1��2⋯��𝑟⋯��𝑛⎭⎪⎬

⎪⎫

+

⎣⎢⎢⎢⎡𝐾10⋯0⋯0

0𝐾2⋯0⋯0

⋯⋯⋯⋯⋯⋯

00⋯𝐾3⋯0

⋯⋯⋯⋯⋯⋯

00⋯0⋯𝐾4

⎦⎥⎥⎥⎤

⎩⎪⎨


⎪⎫

=

⎩⎪⎨

⎪⎧0

0⋯0⋯0⎭⎪⎬

⎪⎫

−−(10)

� \ 𝑀𝑟

\ � {��} + �

\ 𝐾𝑟

\ � {y} = {0} −−(11)


𝐾𝑟 = λ𝑟𝑀𝑟 −−(12)


𝐾 {𝑋}𝑟 = λ𝑟𝑀 {𝑋}𝑟 (13)


{𝑋}𝑟′ [𝐾] {𝑋}𝑟 = λ𝑟 {𝑋}′𝑟[𝑀]{𝑋}𝑟 (14)




� \ 𝑀𝑟

\ � {��} + �

\ λ𝑟𝑀𝑟 \

� {y} = {0}

� \ 𝑀𝑟

\ � {��} + �


� {y} = {0} −−− (16)



��𝑟 + 𝜔𝑟2𝑦𝑟 = 0 (𝑟 = 1,2,3 … … . .𝑛) −−− (17)


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The solution of the above equation is 𝑦𝑟 = 𝐴𝑟 𝑐𝑜𝑠𝜔𝑟𝑡 + 𝐵𝑟 𝑠𝑖𝑛𝜔𝑟𝑡 (𝑟 = 1,2,3 … … . .𝑛) −−− (18)

The above eqution can also be written as {𝑦} = {𝐴𝑐𝑜𝑠𝜔𝑡 + 𝐵𝑠𝑖𝑛𝜔𝑡} −−− (19)

From the Eq. (4) and Eq. (18)

�


� = [U ]�



� −−− (20)





⎩⎪⎨


⎪⎫

=

⎩⎪⎨


⎪⎫

1

𝑦1 +

⎩⎪⎨


⎪⎫

2

𝑦2 +

⎩⎪⎨


⎪⎫

𝑟

𝑦𝑟 +

⎩⎪⎨


⎪⎫

𝑛

𝑦𝑛 −−− (21)


{X} = {X}1𝑦1 + {X}2𝑦2 + {X}𝑟𝑦𝑟 + {X}𝑛𝑦𝑛 −−− (21)


Sol) Two degree of freedom system it is possible to uncouple the equations of motion of an n-DOF provided we know beforehand the normal modes or eigenvectors of the system. A modal matrix [U] is referred to a square matrix where in each column represents an eigenvector. Thus for an n-degree of freedom system,

[U] =

⎣⎢⎢⎢⎢⎢⎢⎡

⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

1 ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

2

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑟

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑠

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑛⎦⎥⎥⎥⎥⎥⎥⎤

−−(1)


[U]′ =

⎣⎢⎢⎢⎢⎢⎡[𝑋1[𝑋1⋯

[𝑋1⋯

[𝑋1⋯

[𝑋1

𝑋2𝑋2⋯𝑋2⋯𝑋2⋯𝑋2

⋯⋯⋯⋯⋯⋯⋯⋯


⋯⋯⋯⋯⋯⋯⋯⋯


⋯⋯⋯⋯⋯⋯⋯⋯


⎥⎥⎥⎥⎥⎤

−−(2)


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[C] = �

𝑐11𝑐21⋯⋯𝑐𝑛1

𝑐12𝑐22⋯⋯𝑐𝑛2

⋯⋯⋯⋯⋯

⋯⋯⋯⋯⋯

⋯⋯⋯⋯⋯


� −−(4)




𝑐𝑖𝑗 = 0 𝑖 ≠ 𝑗


[X] = [U] {y} ---- (5)



[M] [U]{��} +[c] [U]{��} + [K] [U]{y} = [0] ---- (6)


[U]’ [M] [U]{��}+[U]’[c] [U]{��} + [U]’ [K] [U] {y}= [0] ---- (7)




{X}r’ [M] {X}S = 0, r ≠ s

And {X}r’ [K] {X}S = 0, r ≠ s ---- (8)



{X}r’ [M] {X}r = Mr

And {X}r’ [K] {X}r = Mr ---- (9)



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⎣⎢⎢⎢⎡𝑀10⋯0⋯0

0𝑀2⋯0⋯0

⋯⋯⋯⋯⋯⋯

00⋯𝑀3⋯0

⋯⋯⋯⋯⋯⋯

00⋯0⋯𝑀4

⎦⎥⎥⎥⎤

⎩⎪⎨

⎪⎧��1��2⋯��𝑟⋯��𝑛⎭⎪⎬

⎪⎫

+

⎣⎢⎢⎢⎡𝐶10⋯0⋯0

0𝐶2⋯0⋯0

⋯⋯⋯⋯⋯⋯

00⋯𝐶3⋯0

⋯⋯⋯⋯⋯⋯

00⋯0⋯𝐶4

⎦⎥⎥⎥⎤

⎩⎪⎨

⎪⎧��1��2⋯��𝑟⋯��𝑛⎭⎪⎬

⎪⎫

+

⎣⎢⎢⎢⎡𝐾10⋯0⋯0

0𝐾2⋯0⋯0

⋯⋯⋯⋯⋯⋯

00⋯𝐾3⋯0

⋯⋯⋯⋯⋯⋯

00⋯0⋯𝐾4

⎦⎥⎥⎥⎤

⎩⎪⎨


⎪⎫

=

⎩⎪⎨

⎪⎧0

0⋯0⋯0⎭⎪⎬

⎪⎫

−−(10)

� \ 𝑀𝑟

\ � {��} + �

\ 𝐶𝑟

\ � {y} + �

\ 𝐾𝑟

\ � {y} = {0} −−(11)


𝐾𝑟 = λ𝑟𝑀𝑟 −−(12)


𝐾 {𝑋}𝑟 = λ𝑟𝑀 {𝑋}𝑟 (13)






[C]= α[M]+β[K] (14)



=� \ α𝑀𝑟 \

� + � \

β λ𝑟𝑀𝑟 \

�

=� \


� (15)




� \ 𝑀𝑟

\ � {��} + �


\ � {��𝑟} + �


� {y} = {0} −−(16)


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�


� = [U ]�

𝑦1𝑦2⋯𝑦𝑛� −−− (20)





⎩⎪⎨


⎪⎫

=

⎩⎪⎨


⎪⎫

1

𝑦1 +

⎩⎪⎨


⎪⎫

2

𝑦2 +

⎩⎪⎨


⎪⎫

𝑟

𝑦𝑟 +

⎩⎪⎨


⎪⎫

𝑛

𝑦𝑛 −−− (21)


{X} = {X}1𝑦1 + {X}2𝑦2 + {X}𝑟𝑦𝑟 + {X}𝑛𝑦𝑛 −−− (21)

5. Find the infuence coefficient for the system shown in figure (1).

Fig: (1)

Solution:

First Influence coefficients are:

𝛼11 =1𝐾

&𝛼12 = 𝛼13 =1𝐾

By Maxwell reciprocal theorem

𝛼21 = 𝛼12; 𝛼13 = 𝛼31

3m 2m m

2K K 3K


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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

6. Find the fundamental vibration for the system shown in figure (2), by using

Maxwell’s reciprocal theorem.

𝐼 = 4 × 10−7𝑚4; 𝐸 == 1.96 × 1011 𝑁/𝑚2


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

𝛂𝟑𝟑 =116𝑘

m2=50 kg m1=100 kg

180mm 180mm 300mm

Fig: (2)


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𝛼12 = 𝛼21 =𝑙2(3𝐿 − 𝑙)

6𝐸𝐼= 4.599 × 10−8 𝑚/𝑁

𝜶𝟏𝟏 = 𝟐.𝟒𝟕 × 𝟏𝟎−𝟖𝒎𝑵

; 𝜶𝟐𝟐 = 𝟏.𝟏𝟒𝟖 × 𝟏𝟎−𝟖𝒎𝑵

; 𝜶𝟏𝟐 = 𝜶𝟐𝟏 = 𝟒.𝟓𝟗𝟗 × 𝟏𝟎−𝟖𝒎𝑵

7. Find the fundamental vibration for the system shown in figure (3), by using Maxwell’s reciprocal theorem. E196 GPa; I=10 -6m 4; m1=40 kg; m2=20 kg.

Fig: (3) Solution: Where 𝛼11,𝛼12,𝛼21 & 𝛼22 are the influence coefficients are calculated by using following figs:

Fig: (a)

Fig: (b)

1 2

m1

m2

160mm 180mm 180mm

80mm

1

L1=160mm L2=260mm

L=420mm

2 m2

160mm 180mm


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Fig: (b)

𝜶𝟏𝟏 =𝑙12𝑙2

2

3𝐸𝐼𝑙=

0.162 × 0.262

3 × 196 × 109 × 10−6 × 0.42= 𝟕.𝟎𝟎𝟕𝟒𝟓 × 𝟏𝟎−𝟗

𝒎𝑵

𝜶𝟏𝟐 =𝑏𝑥(𝑙2 − 𝑏2 − 𝑥2)

3𝐸𝐼𝑙= 𝟔.𝟗𝟎𝟕𝟒𝟓 × 𝟏𝟎−𝟗

𝒎𝑵

= 𝛼21

𝜶𝟐𝟐 =𝑙12𝑙2

2

3𝐸𝐼𝑙=

0.242 × 0.182

3 × 196 × 109 × 10−6 × 0.42= 𝟕.𝟓𝟓𝟔𝟖𝟓 × 𝟏𝟎−𝟗

𝒎𝑵

𝛼11 = 7.00745 × 10−9𝑚𝑁

; 𝛼12 = 𝛼21 = 6.90745 × 10−9𝑚𝑁

; 𝛼22 = 7.55685 × 10−9𝑚𝑁


Maxwell’s reciprocal theorem..

Fig: (4)

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


L=420mm

2 m2

L1=240

L2=180mm

4m 2m

3m

5K

7K 5K


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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



Where 1𝐾𝑒

= 17𝑘

+ 15𝑘

= 1235𝑘

9. For the un-damped two DOF system Shown in figure (5) with the generalized coordinates X1, X2, determine

(a) The principal coordinates, and (b) The ensuing vibrations of the system for the initial conditions.

Fig: (5)

{X1}t=0 = 1, {�� R1}t=0 = 0

{X2}t=0 = 2, {�� R2}t=0 = 0

Take m1 = m2 = m and k1 = k2 = 0

Solution:

[𝑀] = �𝑚 00 𝑚� , [𝑘] = �2𝑘 −𝑘

−𝑘 𝑘 �

𝛂𝟑𝟑 =12

35𝑘

m1 m2

K2 K1

X2 X1


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[𝑑] = [𝑀]−1 [𝑘] = �

2𝑘𝑚

−𝑘𝑚

−𝑘𝑚

𝑘𝑚

�

The frequency equation W.K.T, in this case after expanding and re-arranging becomes

λ2 −3𝑘𝑚

λ +𝑘2

𝑚2 = 0

λ1 = 0.382 𝑘 𝑚� λ2 = 2.618 𝑘 𝑚�

𝜔1 = �λ1 = 0.618�𝑘 𝑚� 𝑟𝑎𝑑/𝑠𝑒𝑐

𝜔2 = �λ2 = 1.618�𝑘 𝑚� 𝑟𝑎𝑑/𝑠𝑒𝑐 Mode shapes are obtained from the matrix equations, we get

⎣⎢⎢⎢⎡

2𝑘𝑚 − λ𝑖

−𝑘𝑚

−𝑘𝑚

𝑘𝑚 − λ𝑖⎦

⎥⎥⎥⎤�𝑋1𝑋2

� = �00�

For different values of λ𝑖 and these are as follows

�𝑋1𝑋2�1

= � 11.618� , �𝑋1𝑋2

�2

= � 1−0.618�

[𝑈] = � 1 1

1.618 −0.618� The inverse of above matrix can be obtained as

[𝑈]−1 =1

2.236�0.618 11.618 −1�

From equation

�𝑦1𝑦2� =

12.236

�0.618 11.618 −1� �

𝑋1𝑋2�

Giving 𝑦1 = 0.2277𝑋1 + 0.447𝑋2𝑦2 = 0.724𝑋1 − 0.447𝑋2

� −−−−(1)

As the principal coordinates are. Initial condition is

�𝑋1𝑋2�𝑡=0

= �12� , ��1��2�𝑡=0

= �00�

Vibratory response of the system from Eq. is

�𝑋1𝑋2� = [𝑈] �𝐴1 𝑐𝑜𝑠𝜔1𝑡 + 𝐵1 𝑠𝑖𝑛𝜔1𝑡

𝐴2 𝑐𝑜𝑠𝜔2𝑡 + 𝐵2 𝑠𝑖𝑛𝜔2𝑡� −−− (2)

Form Eq. (1) and (2) we have


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�12� = [𝑈] �𝐴1𝐴2�

�00� = [𝑈] �𝐵1𝜔1𝐵2𝜔2�� (3)

Pre-multiplying both the sides of the above equation by [U]-1 we get

�𝐴1𝐴2� = [𝑈]−1 �12� =

12.236 �

0.618 11.618 −1� �

12�

�𝐴1𝐴2� = � 1.171

−0.171�

�𝐵1𝐵2� = �00�

� −−− (4)

Therefore

�𝑋1𝑋2� = � 1 1

1.618 −0.618�

⎩⎪⎨

⎪⎧ 1.171𝑐𝑜𝑠0.618�

𝐾 𝑚� 𝑡

−0.171𝑐𝑜𝑠1 .618�𝐾 𝑚� 𝑡

⎭⎪⎬

⎪⎫

−−−−− (5)

𝑋1 = 1.171𝑐𝑜𝑠0.618�𝐾 𝑚� 𝑡 − 0.171𝑐𝑜𝑠1 .618�

𝐾 𝑚� 𝑡

𝑋2 = 1.895𝑐𝑜𝑠0.618�𝐾 𝑚� 𝑡 − 0.106𝑐𝑜𝑠1 .618�

𝐾 𝑚� 𝑡


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QUADRANT-3

Wikis: (related to Multi DOF) • http://en.wikipedia.org/wiki/Degrees_of_freedom_(mechanics) • http://en.wikipedia.org/wiki/Vibration • http://en.wikipedia.org/wiki/Degrees_of_freedom • http://wiki.answers.com/Q/What_is_multi_degree_of_freedom_suspension_syste

m • http://en.wikipedia.org/wiki/Modal_analysis_using_FEM • http://en.wikipedia.org/wiki/Modal_analysis

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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http://en.wikipedia.org/wiki/Degrees_of_freedom_(mechanics)

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

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

http://wiki.answers.com/Q/What_is_multi_degree_of_freedom_suspension_system

http://wiki.answers.com/Q/What_is_multi_degree_of_freedom_suspension_system

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

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


QUADRANT-4

Problems


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

2. Find the fundamental vibration for the system shown in figure (2), by using Maxwell’s

reciprocal theorem.

𝛂𝟑𝟑 =116𝑘

3m 2m m

2K K 3K


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𝐼 = 4 × 10−7𝑚4; 𝐸 == 1.96 × 1011 𝑁/𝑚2


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

𝜶𝟏𝟏 = 𝟐.𝟒𝟕 × 𝟏𝟎−𝟖𝒎𝑵

; 𝜶𝟐𝟐 = 𝟏.𝟏𝟒𝟖 × 𝟏𝟎−𝟖𝒎𝑵

; 𝜶𝟏𝟐 = 𝜶𝟐𝟏 = 𝟒.𝟓𝟗𝟗 × 𝟏𝟎−𝟖𝒎𝑵


Fig: (3)

m2=50 kg m1=100 kg

180mm 180mm 300mm

Fig: (2)

1 2 m2

160mm 180mm 180mm

80mm


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Solution: Where 𝛼11,𝛼12,𝛼21 & 𝛼22 are the influence coefficients are calculated by using following figs:

Fig: (a)

Fig: (b)

Fig: (b)

𝜶𝟏𝟏 =𝑙12𝑙2

2

3𝐸𝐼𝑙=

0.162 × 0.262

3 × 196 × 109 × 10−6 × 0.42= 𝟕.𝟎𝟎𝟕𝟒𝟓 × 𝟏𝟎−𝟗

𝒎𝑵

𝜶𝟏𝟐 =𝑏𝑥(𝑙2 − 𝑏2 − 𝑥2)

3𝐸𝐼𝑙= 𝟔.𝟗𝟎𝟕𝟒𝟓 × 𝟏𝟎−𝟗

𝒎𝑵

= 𝛼21

𝜶𝟐𝟐 =𝑙12𝑙2

2

3𝐸𝐼𝑙=

0.242 × 0.182

3 × 196 × 109 × 10−6 × 0.42= 𝟕.𝟓𝟓𝟔𝟖𝟓 × 𝟏𝟎−𝟗

𝒎𝑵

𝛼11 = 7.00745 × 10−9𝑚𝑁

; 𝛼12 = 𝛼21 = 6.90745 × 10−9𝑚𝑁

; 𝛼22 = 7.55685 × 10−9𝑚𝑁

m1 1

L1=160mm L2=260mm

L=420mm

L=420mm

2 m2

160mm 180mm

2 m2

L1=240

L2=180mm


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4. Find the fundamental vibration for the system shown in figure (4), by using Maxwell’s

reciprocal theorem..

Fig: (4)

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



Where 1𝐾𝑒

= 17𝑘

+ 15𝑘

= 1235𝑘

4m 2m

3m

5K

7K 5K


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5. For the un-damped two DOF system Shown in figure (5) with the generalized coordinates

X1, X2, determine (c) The principal coordinates, and (d) The ensuing vibrations of the system for the initial conditions.

Fig: (5)

{X1}t=0 = 1, {�� R1}t=0 = 0

{X2}t=0 = 2, {�� R2}t=0 = 0

Take m1 = m2 = m and k1 = k2 = 0

Solution:

[𝑀] = �𝑚 00 𝑚� , [𝑘] = �2𝑘 −𝑘

−𝑘 𝑘 �

[𝑑] = [𝑀]−1 [𝑘] = �

2𝑘𝑚

−𝑘𝑚

−𝑘𝑚

𝑘𝑚

�


λ2 −3𝑘𝑚

λ +𝑘2

𝑚2 = 0

λ1 = 0.382 𝑘 𝑚� λ2 = 2.618 𝑘 𝑚�

𝜔1 = �λ1 = 0.618�𝑘 𝑚� 𝑟𝑎𝑑/𝑠𝑒𝑐


⎣⎢⎢⎢⎡


−𝑘𝑚

−𝑘𝑚


⎥⎥⎥⎤�𝑋1𝑋2

� = �00�


𝛂𝟑𝟑 =12

35𝑘

m1 m2

K2 K1

X2 X1


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�𝑋1𝑋2�1

= � 11.618� , �𝑋1𝑋2

�2

= � 1−0.618�

[𝑈] = � 1 1


[𝑈]−1 =1

2.236�0.618 11.618 −1�

From equation

�𝑦1𝑦2� =

12.236

�0.618 11.618 −1� �

𝑋1𝑋2�

Giving 𝑦1 = 0.2277𝑋1 + 0.447𝑋2𝑦2 = 0.724𝑋1 − 0.447𝑋2

� −−−−(1)


�𝑋1𝑋2�𝑡=0

= �12� , ��1��2�𝑡=0

= �00�





�12� = [𝑈] �𝐴1𝐴2�

�00� = [𝑈] �𝐵1𝜔1𝐵2𝜔2�� (3)


�𝐴1𝐴2� = [𝑈]−1 �12� =

12.236 �

0.618 11.618 −1� �

12�

�𝐴1𝐴2� = � 1.171

−0.171�

�𝐵1𝐵2� = �00�

� −−− (4)

Therefore

�𝑋1𝑋2� = � 1 1

1.618 −0.618�

⎩⎪⎨

⎪⎧ 1.171𝑐𝑜𝑠0.618�

𝐾 𝑚� 𝑡

−0.171𝑐𝑜𝑠1 .618�𝐾 𝑚� 𝑡

⎭⎪⎬

⎪⎫

−−−−− (5)


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𝑋1 = 1.171𝑐𝑜𝑠0.618�𝐾 𝑚� 𝑡 − 0.171𝑐𝑜𝑠1 .618�

𝐾 𝑚� 𝑡

𝑋2 = 1.895𝑐𝑜𝑠0.618�𝐾 𝑚� 𝑡 − 0.106𝑐𝑜𝑠1 .618�

𝐾 𝑚� 𝑡


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Frequently asked Questions.











(e) The principal coordinates, and (f) The ensuing vibrations of the system for the initial conditions.

1 2 m2

160mm 180mm 180mm

80mm

4m 2m

3m

5K

7K 5K


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{X1}t=0 = 1, {�� R1}t=0 = 0

{X2}t=0 = 2, {�� R2}t=0 = 0

Take m1 = m2 = m and k1 = k2 = 0

Self Answered Question & Answer


Solution) 1. STIFFNESS INFLUENCE COEFFICIENTS:




{F} = [K]{x}

[K] = �𝐾11 𝐾12 𝐾13𝐾21 𝐾22 𝐾23𝐾31 𝐾32 𝐾33

�








{F} = [K]{x}

{x}= [K]-1 {F}

{x} = [αij]{F}

[α𝑖𝑗] = �α11 α12 α13α21 α22 α23α31 α32 α33

�

m1 m2

K2 K1

X2 X1


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Solu) Maxwell’s reciprocal theorem states the that deflection at any point in the system due

to a unit load acting at any other of the same system is equal to the deflection at the second

point due to a unit load at the first point.










12𝐹12𝛼11 −−− (1)



12𝐹22𝛼22 −−− (2)




first mode.

(𝐹𝐷)1 = 12𝐹12𝛼11 +

12𝐹22𝛼22 + 𝐹1𝐹2𝛼12 −−− (3)

F1 F2

2 1


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(𝐹𝐷)2 = 12𝐹12𝛼22 +

12𝐹22𝛼11 + 𝐹1𝐹2𝛼21 −−− (3)


α12 = α21




Solution)


[U] =

⎣⎢⎢⎢⎢⎢⎢⎡

⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

1 ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

2

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑟

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑠

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑛⎦⎥⎥⎥⎥⎥⎥⎤

−−(1)


[U]′ =

⎣⎢⎢⎢⎢⎢⎡[𝑋1[𝑋1⋯

[𝑋1⋯

[𝑋1⋯

[𝑋1

𝑋2𝑋2⋯𝑋2⋯𝑋2⋯𝑋2

⋯⋯⋯⋯⋯⋯⋯⋯


⋯⋯⋯⋯⋯⋯⋯⋯


⋯⋯⋯⋯⋯⋯⋯⋯


⎥⎥⎥⎥⎥⎤

−−(2)

Let the differential equation of motion for an n-degree of freedom undamped system be, [M]{��} + [K] {x} = [0] ---- (3)


[X] = [U] {y} ---- (4)



[U]-1 {x} = [U]-1[U] {y}

{y} = [U]-1 {x} ---- (5)


[M] [U]{��} + [K] [U]{y} = [0] ---- (6)



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[U]’ [M] [U]{��}+ [U]’ [K] [U] {y}= [0] ---- (7)




{X}r’ [M] {X}S = 0, r ≠ s

And {X}r’ [K] {X}S = 0, r ≠ s ---- (8)



{X}r’ [M] {X}r = Mr

And {X}r’ [K] {X}r = Mr ---- (9)


⎣⎢⎢⎢⎡𝑀10⋯0⋯0

0𝑀2⋯0⋯0

⋯⋯⋯⋯⋯⋯

00⋯𝑀3⋯0

⋯⋯⋯⋯⋯⋯

00⋯0⋯𝑀4

⎦⎥⎥⎥⎤

⎩⎪⎨

⎪⎧��1��2⋯��𝑟⋯��𝑛⎭⎪⎬

⎪⎫

+

⎣⎢⎢⎢⎡𝐾10⋯0⋯0

0𝐾2⋯0⋯0

⋯⋯⋯⋯⋯⋯

00⋯𝐾3⋯0

⋯⋯⋯⋯⋯⋯

00⋯0⋯𝐾4

⎦⎥⎥⎥⎤

⎩⎪⎨


⎪⎫

=

⎩⎪⎨

⎪⎧0

0⋯0⋯0⎭⎪⎬

⎪⎫

−−(10)

� \ 𝑀𝑟

\ � {��} + �

\ 𝐾𝑟

\ � {y} = {0} −−(11)


𝐾𝑟 = λ𝑟𝑀𝑟 −−(12)


𝐾 {𝑋}𝑟 = λ𝑟𝑀 {𝑋}𝑟 (13)


{𝑋}𝑟′ [𝐾] {𝑋}𝑟 = λ𝑟 {𝑋}′𝑟[𝑀]{𝑋}𝑟 (14)





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� \ 𝑀𝑟

\ � {��} + �

\ λ𝑟𝑀𝑟 \

� {y} = {0}

� \ 𝑀𝑟

\ � {��} + �


� {y} = {0} −−− (16)



��𝑟 + 𝜔𝑟2𝑦𝑟 = 0 (𝑟 = 1,2,3 … … . .𝑛) −−− (17) The solution of the above equation is

𝑦𝑟 = 𝐴𝑟 𝑐𝑜𝑠𝜔𝑟𝑡 + 𝐵𝑟 𝑠𝑖𝑛𝜔𝑟𝑡 (𝑟 = 1,2,3 … … . .𝑛) −−− (18) The above eqution can also be written as

{𝑦} = {𝐴𝑐𝑜𝑠𝜔𝑡 + 𝐵𝑠𝑖𝑛𝜔𝑡} −−− (19) From the Eq. (4) and Eq. (18)

�


� = [U ]�



� −−− (20)





⎩⎪⎨


⎪⎫

=

⎩⎪⎨


⎪⎫

1

𝑦1 +

⎩⎪⎨


⎪⎫

2

𝑦2 +

⎩⎪⎨


⎪⎫

𝑟

𝑦𝑟 +

⎩⎪⎨


⎪⎫

𝑛

𝑦𝑛 −−− (21)


{X} = {X}1𝑦1 + {X}2𝑦2 + {X}𝑟𝑦𝑟 + {X}𝑛𝑦𝑛 −−− (21)


Sol) Two degree of freedom system it is possible to uncouple the equations of motion of an n-DOF provided we know beforehand the normal modes or eigenvectors of the system. A modal matrix [U] is referred to a square matrix where in each column represents an eigenvector. Thus for an n-degree of freedom system,


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[U] =

⎣⎢⎢⎢⎢⎢⎢⎡

⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

1 ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

2

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑟

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑠

⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨


⎪⎪⎬

⎪⎪⎫

𝑛⎦⎥⎥⎥⎥⎥⎥⎤

−−(1)


[U]′ =

⎣⎢⎢⎢⎢⎢⎡[𝑋1[𝑋1⋯

[𝑋1⋯

[𝑋1⋯

[𝑋1

𝑋2𝑋2⋯𝑋2⋯𝑋2⋯𝑋2

⋯⋯⋯⋯⋯⋯⋯⋯


⋯⋯⋯⋯⋯⋯⋯⋯


⋯⋯⋯⋯⋯⋯⋯⋯


⎥⎥⎥⎥⎥⎤

−−(2)



[C] = �

𝑐11𝑐21⋯⋯𝑐𝑛1

𝑐12𝑐22⋯⋯𝑐𝑛2

⋯⋯⋯⋯⋯

⋯⋯⋯⋯⋯

⋯⋯⋯⋯⋯


� −−(4)




𝑐𝑖𝑗 = 0 𝑖 ≠ 𝑗


[X] = [U] {y} ---- (5)



[M] [U]{��} +[c] [U]{��} + [K] [U]{y} = [0] ---- (6)


[U]’ [M] [U]{��}+[U]’[c] [U]{��} + [U]’ [K] [U] {y}= [0] ---- (7)




{X}r’ [M] {X}S = 0, r ≠ s

And {X}r’ [K] {X}S = 0, r ≠ s ---- (8)


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{X}r’ [M] {X}r = Mr

And {X}r’ [K] {X}r = Mr ---- (9)


⎣⎢⎢⎢⎡𝑀10⋯0⋯0

0𝑀2⋯0⋯0

⋯⋯⋯⋯⋯⋯

00⋯𝑀3⋯0

⋯⋯⋯⋯⋯⋯

00⋯0⋯𝑀4

⎦⎥⎥⎥⎤

⎩⎪⎨

⎪⎧��1��2⋯��𝑟⋯��𝑛⎭⎪⎬

⎪⎫

+

⎣⎢⎢⎢⎡𝐶10⋯0⋯0

0𝐶2⋯0⋯0

⋯⋯⋯⋯⋯⋯

00⋯𝐶3⋯0

⋯⋯⋯⋯⋯⋯

00⋯0⋯𝐶4

⎦⎥⎥⎥⎤

⎩⎪⎨

⎪⎧��1��2⋯��𝑟⋯��𝑛⎭⎪⎬

⎪⎫

+

⎣⎢⎢⎢⎡𝐾10⋯0⋯0

0𝐾2⋯0⋯0

⋯⋯⋯⋯⋯⋯

00⋯𝐾3⋯0

⋯⋯⋯⋯⋯⋯

00⋯0⋯𝐾4

⎦⎥⎥⎥⎤

⎩⎪⎨


⎪⎫

=

⎩⎪⎨

⎪⎧0

0⋯0⋯0⎭⎪⎬

⎪⎫

−−(10)

� \ 𝑀𝑟

\ � {��} + �

\ 𝐶𝑟

\ � {y} + �

\ 𝐾𝑟

\ � {y} = {0} −−(11)


𝐾𝑟 = λ𝑟𝑀𝑟 −−(12)


𝐾 {𝑋}𝑟 = λ𝑟𝑀 {𝑋}𝑟 (13)






[C]= α[M]+β[K] (14)



=� \ α𝑀𝑟 \

� + � \

β λ𝑟𝑀𝑟 \

�


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=� \


� (15)




� \ 𝑀𝑟

\ � {��} + �


\ � {��𝑟} + �


� {y} = {0} −−(16)





�


� = [U ]�

𝑦1𝑦2⋯𝑦𝑛� −−− (20)





⎩⎪⎨


⎪⎫

=

⎩⎪⎨


⎪⎫

1

𝑦1 +

⎩⎪⎨


⎪⎫

2

𝑦2 +

⎩⎪⎨


⎪⎫

𝑟

𝑦𝑟 +

⎩⎪⎨


⎪⎫

𝑛

𝑦𝑛 −−− (21)


{X} = {X}1𝑦1 + {X}2𝑦2 + {X}𝑟𝑦𝑟 + {X}𝑛𝑦𝑛 −−− (21)


3m 2m m

2K K 3K


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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



𝐼 = 4 × 10−7𝑚4; 𝐸 == 1.96 × 1011 𝑁/𝑚2

𝛂𝟑𝟑 =116𝑘

m2=50 kg m1=100 kg

180mm 180mm 300mm

Fig: (2)


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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

𝜶𝟏𝟏 = 𝟐.𝟒𝟕 × 𝟏𝟎−𝟖𝒎𝑵

; 𝜶𝟐𝟐 = 𝟏.𝟏𝟒𝟖 × 𝟏𝟎−𝟖𝒎𝑵

; 𝜶𝟏𝟐 = 𝜶𝟐𝟏 = 𝟒.𝟓𝟗𝟗 × 𝟏𝟎−𝟖𝒎𝑵


Fig: (3) Solution: Where 𝛼11,𝛼12,𝛼21 & 𝛼22 are the influence coefficients are calculated by using following figs:

Fig: (a)

1 2

m1

m2

160mm 180mm 180mm

80mm

1

L1=160mm L2=260mm

L=420mm

2 m2

160mm 180mm


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Fig: (b)

Fig: (b)

𝜶𝟏𝟏 =𝑙12𝑙2

2

3𝐸𝐼𝑙=

0.162 × 0.262

3 × 196 × 109 × 10−6 × 0.42= 𝟕.𝟎𝟎𝟕𝟒𝟓 × 𝟏𝟎−𝟗

𝒎𝑵

𝜶𝟏𝟐 =𝑏𝑥(𝑙2 − 𝑏2 − 𝑥2)

3𝐸𝐼𝑙= 𝟔.𝟗𝟎𝟕𝟒𝟓 × 𝟏𝟎−𝟗

𝒎𝑵

= 𝛼21

𝜶𝟐𝟐 =𝑙12𝑙2

2

3𝐸𝐼𝑙=

0.242 × 0.182

3 × 196 × 109 × 10−6 × 0.42= 𝟕.𝟓𝟓𝟔𝟖𝟓 × 𝟏𝟎−𝟗

𝒎𝑵

𝛼11 = 7.00745 × 10−9𝑚𝑁

; 𝛼12 = 𝛼21 = 6.90745 × 10−9𝑚𝑁

; 𝛼22 = 7.55685 × 10−9𝑚𝑁



Fig: (4)

Solution:

L=420mm

2 m2

L1=240

L2=180mm

4m 2m

3m

5K

7K 5K


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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



Where 1𝐾𝑒

= 17𝑘

+ 15𝑘

= 1235𝑘


(g) The principal coordinates, and (h) The ensuing vibrations of the system for the initial conditions.

𝛂𝟑𝟑 =12

35𝑘


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Fig: (5)

{X1}t=0 = 1, {�� R1}t=0 = 0

{X2}t=0 = 2, {�� R2}t=0 = 0

Take m1 = m2 = m and k1 = k2 = 0

Solution:

[𝑀] = �𝑚 00 𝑚� , [𝑘] = �2𝑘 −𝑘

−𝑘 𝑘 �

[𝑑] = [𝑀]−1 [𝑘] = �

2𝑘𝑚

−𝑘𝑚

−𝑘𝑚

𝑘𝑚

�


λ2 −3𝑘𝑚

λ +𝑘2

𝑚2 = 0

λ1 = 0.382 𝑘 𝑚� λ2 = 2.618 𝑘 𝑚�

𝜔1 = �λ1 = 0.618�𝑘 𝑚� 𝑟𝑎𝑑/𝑠𝑒𝑐


⎣⎢⎢⎢⎡


−𝑘𝑚

−𝑘𝑚


⎥⎥⎥⎤�𝑋1𝑋2

� = �00�


�𝑋1𝑋2�1

= � 11.618� , �𝑋1𝑋2

�2

= � 1−0.618�

[𝑈] = � 1 1


[𝑈]−1 =1

2.236�0.618 11.618 −1�

m1 m2

K2 K1

X2 X1


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From equation

�𝑦1𝑦2� =

12.236

�0.618 11.618 −1� �

𝑋1𝑋2�

Giving 𝑦1 = 0.2277𝑋1 + 0.447𝑋2𝑦2 = 0.724𝑋1 − 0.447𝑋2

� −−−−(1)


�𝑋1𝑋2�𝑡=0

= �12� , ��1��2�𝑡=0

= �00�





�12� = [𝑈] �𝐴1𝐴2�

�00� = [𝑈] �𝐵1𝜔1𝐵2𝜔2�� (3)


�𝐴1𝐴2� = [𝑈]−1 �12� =

12.236 �

0.618 11.618 −1� �

12�

�𝐴1𝐴2� = � 1.171

−0.171�

�𝐵1𝐵2� = �00�

� −−− (4)

Therefore

�𝑋1𝑋2� = � 1 1

1.618 −0.618�

⎩⎪⎨

⎪⎧ 1.171𝑐𝑜𝑠0.618�

𝐾 𝑚� 𝑡

−0.171𝑐𝑜𝑠1 .618�𝐾 𝑚� 𝑡

⎭⎪⎬

⎪⎫

−−−−− (5)

𝑋1 = 1.171𝑐𝑜𝑠0.618�𝐾 𝑚� 𝑡 − 0.171𝑐𝑜𝑠1 .618�

𝐾 𝑚� 𝑡

𝑋2 = 1.895𝑐𝑜𝑠0.618�𝐾 𝑚� 𝑡 − 0.106𝑐𝑜𝑠1 .618�

𝐾 𝑚� 𝑡


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VTU-NPTEL-NMEICT Project Progress Reportnptel.vtu.ac.in/VTU-NMEICT/MV/Module 6.pdf · VTU-NPTEL-NMEICT Project Progress Report . ... Maxwell’s reciprocal theorem: Maxwell’s reciprocal