Upload
vuonghanh
View
226
Download
1
Embed Size (px)
Citation preview
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
VTU-NPTEL-N
MEICT Proj
ect
Page 1 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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.
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 2 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 3 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
k31 - stiffness influence coefficient at point 3 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
α 31 - flexibility influence coefficient at point 3 due to a unit force at point 1
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 4 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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,
∴ 𝑤𝑜𝑟𝑘𝑑𝑜𝑛𝑒 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 2 = 12𝐹2(𝛼22𝐹2) =
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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 5 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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 ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
2
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑟
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑠
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑛⎦⎥⎥⎥⎥⎥⎥⎤
−−(1)
The transpose of the above matrix can be written as
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 6 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
[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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 7 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
⎣⎢⎢⎢⎡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)
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 8 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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⎭⎪⎬
⎪⎫
=
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
1
𝑦1 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
2
𝑦2 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
𝑟
𝑦𝑟 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
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] =
⎣⎢⎢⎢⎢⎢⎢⎡
⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
1 ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
2
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑟
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑠
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑛⎦⎥⎥⎥⎥⎥⎥⎤
−−(1)
The transpose of the above matrix can be written as
[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 damped system be, [M]{��} +[C] {��} + [K] {x} = [0] ---- (3)
Where [C] is the damping matrix
[C] = �
𝑐11𝑐21⋯⋯𝑐𝑛1
𝑐12𝑐22⋯⋯𝑐𝑛2
⋯⋯⋯⋯⋯
⋯⋯⋯⋯⋯
⋯⋯⋯⋯⋯
𝑐1𝑛𝑐2𝑛⋯⋯𝑐𝑛𝑛
� −−(4)
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 9 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
To decouple these equations, let us the linear transformation
[X] = [U] {y} ---- (5)
Where {y} are the principal coordinates and can be found out by pre-multiplying the above
equation by [U]-1 to get
[M] [U]{��} +[c] [U]{��} + [K] [U]{y} = [0] ---- (6)
Pre multiply the above equation by [U]’ to get
[U]’ [M] [U]{��}+[U]’[c] [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
⎣⎢⎢⎢⎡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
⎦⎥⎥⎥⎤
⎩⎪⎨
⎪⎧𝑦1𝑦2⋯𝑦𝑟⋯𝑦𝑛⎭⎪⎬
⎪⎫
=
⎩⎪⎨
⎪⎧0
0⋯0⋯0⎭⎪⎬
⎪⎫
−−(10)
� \ 𝑀𝑟
\ � {��} + �
\ 𝐶𝑟
\ � {y} + �
\ 𝐾𝑟
\ � {y} = {0} −−(11)
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 10 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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)
Where λ𝑟 = 𝜔𝑟2= the eigenvalue for the rth mode
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
Substituting equation (15) in Equation (11)
� \ 𝑀𝑟
\ � {��} + �
\ � 2ξ𝑟𝜔𝑟�𝑀𝑟
\ � {��𝑟} + �
\ 𝜔𝑟2𝑀𝑟 \
� {y} = {0} −−(16)
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 + �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)
�
𝑋1𝑋2⋯𝑋𝑛
� = [U ]�
𝑦1𝑦2⋯𝑦𝑛� −−− (20)
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 11 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
Which give the vibratory response of 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⎭⎪⎬
⎪⎫
=
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
1
𝑦1 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
2
𝑦2 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
𝑟
𝑦𝑟 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
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)
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 12 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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/
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 13 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
ILLUSTRATIONS
1. Explain the different types of influence coefficient Methods.
Solution) 1. 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
k31 - stiffness influence coefficient at point 3 due to a unit deflection at point 1
2. 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
α 31 - flexibility influence coefficient at point 3 due to a unit force at point 1
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 14 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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.
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,
∴ 𝑤𝑜𝑟𝑘𝑑𝑜𝑛𝑒 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 2 = 12𝐹2(𝛼22𝐹2) =
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
F1 F2
2 1
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 15 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
∴ 𝛂𝒊𝒋 = 𝛂𝒋𝒊 Proof of the theorem.
3. Explain briefly modal analysis on Undamped Free vibration.
Solution)
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 ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
2
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑟
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑠
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑛⎦⎥⎥⎥⎥⎥⎥⎤
−−(1)
The transpose of the above matrix can be written as
[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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 16 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
⎣⎢⎢⎢⎡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)
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 17 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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)
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⎭⎪⎬
⎪⎫
=
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
1
𝑦1 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
2
𝑦2 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
𝑟
𝑦𝑟 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
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)
4. Explain briefly modal analysis on Undamped Free vibration.
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] =
⎣⎢⎢⎢⎢⎢⎢⎡
⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
1 ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
2
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑟
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑠
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑛⎦⎥⎥⎥⎥⎥⎥⎤
−−(1)
The transpose of the above matrix can be written as
[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)
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 18 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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)
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
To decouple these equations, let us the linear transformation
[X] = [U] {y} ---- (5)
Where {y} are the principal coordinates and can be found out by pre-multiplying the above
equation by [U]-1 to get
[M] [U]{��} +[c] [U]{��} + [K] [U]{y} = [0] ---- (6)
Pre multiply the above equation by [U]’ to get
[U]’ [M] [U]{��}+[U]’[c] [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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 19 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
⎣⎢⎢⎢⎡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
⎦⎥⎥⎥⎤
⎩⎪⎨
⎪⎧𝑦1𝑦2⋯𝑦𝑟⋯𝑦𝑛⎭⎪⎬
⎪⎫
=
⎩⎪⎨
⎪⎧0
0⋯0⋯0⎭⎪⎬
⎪⎫
−−(10)
� \ 𝑀𝑟
\ � {��} + �
\ 𝐶𝑟
\ � {y} + �
\ 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)
Where λ𝑟 = 𝜔𝑟2= the eigenvalue for the rth mode
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
Substituting equation (15) in Equation (11)
� \ 𝑀𝑟
\ � {��} + �
\ � 2ξ𝑟𝜔𝑟�𝑀𝑟
\ � {��𝑟} + �
\ 𝜔𝑟2𝑀𝑟 \
� {y} = {0} −−(16)
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 20 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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 + �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)
�
𝑋1𝑋2⋯𝑋𝑛
� = [U ]�
𝑦1𝑦2⋯𝑦𝑛� −−− (20)
Which give the vibratory response of 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⎭⎪⎬
⎪⎫
=
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
1
𝑦1 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
2
𝑦2 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
𝑟
𝑦𝑟 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
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)
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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 21 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
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
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
𝛂𝟑𝟑 =116𝑘
m2=50 kg m1=100 kg
180mm 180mm 300mm
Fig: (2)
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 22 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 23 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
8. Find the fundamental vibration for the system shown in figure (4), by using
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
By Maxwell reciprocal theorem
L=420mm
2 m2
L1=240
L2=180mm
4m 2m
3m
5K
7K 5K
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 24 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
Third Influence coefficients are:
𝛼33 =𝑓𝐾𝑒
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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 25 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
[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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 26 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
�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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 27 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 28 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
QUADRANT-4
Problems
1. 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
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
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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 29 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
𝐼 = 4 × 10−7𝑚4; 𝐸 == 1.96 × 1011 𝑁/𝑚2
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
𝜶𝟏𝟏 = 𝟐.𝟒𝟕 × 𝟏𝟎−𝟖𝒎𝑵
; 𝜶𝟐𝟐 = 𝟏.𝟏𝟒𝟖 × 𝟏𝟎−𝟖𝒎𝑵
; 𝜶𝟏𝟐 = 𝜶𝟐𝟏 = 𝟒.𝟓𝟗𝟗 × 𝟏𝟎−𝟖𝒎𝑵
3. 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)
m2=50 kg m1=100 kg
180mm 180mm 300mm
Fig: (2)
1 2 m2
160mm 180mm 180mm
80mm
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 30 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 31 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
4. Find the fundamental vibration for the system shown in figure (4), by using 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
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
Third Influence coefficients are:
𝛼33 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 17𝑘
+ 15𝑘
= 1235𝑘
4m 2m
3m
5K
7K 5K
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 32 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
�
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
𝛂𝟑𝟑 =12
35𝑘
m1 m2
K2 K1
X2 X1
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 33 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
�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
�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)
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 34 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 35 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
Frequently asked Questions.
1. Explain the different types of influence coefficient Methods.
2. Explain Maxwell’s reciprocal theorem
3. Explain briefly modal analysis on Undamped Free vibration.
4. Explain briefly modal analysis on Undamped Free vibration.
5. Find the fundamental vibration for the system shown in figure (2), by using
Maxwell’s reciprocal theorem.
6. 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.
7. Find the fundamental vibration for the system shown in figure (4), by using
Maxwell’s reciprocal theorem..
8. For the un-damped two DOF system Shown in figure (5) with the generalized coordinates X1, X2, determine
(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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 36 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
{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
1. Explain the different types of influence coefficient Methods.
Solution) 1. 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
k31 - stiffness influence coefficient at point 3 due to a unit deflection at point 1
2. 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
�
m1 m2
K2 K1
X2 X1
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 37 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
α 31 - flexibility influence coefficient at point 3 due to a unit force at point 1
3. 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.
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,
∴ 𝑤𝑜𝑟𝑘𝑑𝑜𝑛𝑒 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 2 = 12𝐹2(𝛼22𝐹2) =
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)
F1 F2
2 1
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 38 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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.
4. Explain briefly modal analysis on Undamped Free vibration.
Solution)
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 ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
2
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑟
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑠
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑛⎦⎥⎥⎥⎥⎥⎥⎤
−−(1)
The transpose of the above matrix can be written as
[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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 39 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
[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
⎣⎢⎢⎢⎡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)
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 40 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
� \ 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)
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⎭⎪⎬
⎪⎫
=
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
1
𝑦1 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
2
𝑦2 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
𝑟
𝑦𝑟 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
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)
5. Explain briefly modal analysis on Undamped Free vibration.
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,
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 41 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
[U] =
⎣⎢⎢⎢⎢⎢⎢⎡
⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
1 ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
2
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑟
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑠
⋮⋮⋮⋮⋮⋮⋮⋮ ⎩⎪⎪⎨
⎪⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑠⋮𝑋 𝑛⎭
⎪⎪⎬
⎪⎪⎫
𝑛⎦⎥⎥⎥⎥⎥⎥⎤
−−(1)
The transpose of the above matrix can be written as
[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 damped system be, [M]{��} +[C] {��} + [K] {x} = [0] ---- (3)
Where [C] is the damping matrix
[C] = �
𝑐11𝑐21⋯⋯𝑐𝑛1
𝑐12𝑐22⋯⋯𝑐𝑛2
⋯⋯⋯⋯⋯
⋯⋯⋯⋯⋯
⋯⋯⋯⋯⋯
𝑐1𝑛𝑐2𝑛⋯⋯𝑐𝑛𝑛
� −−(4)
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
To decouple these equations, let us the linear transformation
[X] = [U] {y} ---- (5)
Where {y} are the principal coordinates and can be found out by pre-multiplying the above
equation by [U]-1 to get
[M] [U]{��} +[c] [U]{��} + [K] [U]{y} = [0] ---- (6)
Pre multiply the above equation by [U]’ to get
[U]’ [M] [U]{��}+[U]’[c] [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)
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 42 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
⎣⎢⎢⎢⎡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
⎦⎥⎥⎥⎤
⎩⎪⎨
⎪⎧𝑦1𝑦2⋯𝑦𝑟⋯𝑦𝑛⎭⎪⎬
⎪⎫
=
⎩⎪⎨
⎪⎧0
0⋯0⋯0⎭⎪⎬
⎪⎫
−−(10)
� \ 𝑀𝑟
\ � {��} + �
\ 𝐶𝑟
\ � {y} + �
\ 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)
Where λ𝑟 = 𝜔𝑟2= the eigenvalue for the rth mode
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 \
�
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 43 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
=� \
(α + β 𝜔𝑟2)𝑀𝑟 \
� (15)
Expressing (α + β 𝜔𝑟2) in terms of the modal damping ξ𝑟
(α + β 𝜔𝑟2) = 2ξ𝑟𝜔𝑟
Substituting equation (15) in Equation (11)
� \ 𝑀𝑟
\ � {��} + �
\ � 2ξ𝑟𝜔𝑟�𝑀𝑟
\ � {��𝑟} + �
\ 𝜔𝑟2𝑀𝑟 \
� {y} = {0} −−(16)
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 + �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)
�
𝑋1𝑋2⋯𝑋𝑛
� = [U ]�
𝑦1𝑦2⋯𝑦𝑛� −−− (20)
Which give the vibratory response of 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⎭⎪⎬
⎪⎫
=
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
1
𝑦1 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
2
𝑦2 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
𝑟
𝑦𝑟 +
⎩⎪⎨
⎪⎧𝑋1𝑋2⋮𝑋𝑟⋮𝑋𝑛⎭⎪⎬
⎪⎫
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)
6. Find the infuence coefficient for the system shown in figure (1).
3m 2m m
2K K 3K
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 44 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
7. 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
𝛂𝟑𝟑 =116𝑘
m2=50 kg m1=100 kg
180mm 180mm 300mm
Fig: (2)
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 45 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
𝜶𝟏𝟏 = 𝟐.𝟒𝟕 × 𝟏𝟎−𝟖𝒎𝑵
; 𝜶𝟐𝟐 = 𝟏.𝟏𝟒𝟖 × 𝟏𝟎−𝟖𝒎𝑵
; 𝜶𝟏𝟐 = 𝜶𝟐𝟏 = 𝟒.𝟓𝟗𝟗 × 𝟏𝟎−𝟖𝒎𝑵
8. 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)
1 2
m1
m2
160mm 180mm 180mm
80mm
1
L1=160mm L2=260mm
L=420mm
2 m2
160mm 180mm
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 46 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
9. Find the fundamental vibration for the system shown in figure (4), by using
Maxwell’s reciprocal theorem..
Fig: (4)
Solution:
L=420mm
2 m2
L1=240
L2=180mm
4m 2m
3m
5K
7K 5K
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 47 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
Third Influence coefficients are:
𝛼33 =𝑓𝐾𝑒
Where 1𝐾𝑒
= 17𝑘
+ 15𝑘
= 1235𝑘
10. For the un-damped two DOF system Shown in figure (5) with the generalized coordinates X1, X2, determine
(g) The principal coordinates, and (h) The ensuing vibrations of the system for the initial conditions.
𝛂𝟑𝟑 =12
35𝑘
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 48 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
�
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�
m1 m2
K2 K1
X2 X1
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 49 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
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
�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
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 50 of 51
MODULE-VI --- MULTI DOF VIBRATION ENGINEERING 2014
Dr.MOHAMED HANEEF,PRINCIPAL ,GHOUSIA COLLEGE OF ENGINERING,RAMANAGARAM
VTU-NPTEL-N
MEICT Proj
ect
Page 51 of 51