vtusolutionvtusolution.in/uploads/9/9/9/3/99939970/mechanical_vibrations[10me... · resonance is...

vtuso

lution

.in

06ME 64 - Mechanical Vibrations, Dr.T.V.Givindaraju Introduction 1

Visvesvaraya Technological University

E-notes

Dr. .V. Govindaraju

Principal, Shirdi Sai Engineering College, Bangalore

06ME 64 - MECHANICAL VIBRATIONS UNIT - 1

Introduction: When an elastic body such as, a spring, a beam and a shaft are displaced from the

equilibrium position by the application of external forces, and then released, they execute a vibratory

motion, due to the elastic or strain energy present in the body. When the body reaches the

equilibrium position, the whole of the elastic or stain energy is converted into kinetic energy due to

which the body continues to move in the opposite direction. The entire KE is again converted into

strain energy due to which the body again returns to the equilibrium position. Hence the vibratory

motion is repeated indefinitely.

Oscillatory motion is any pattern of motion where the system under observation moves back and

forth across some equilibrium position, but dose not necessarily have any particular repeating

pattern.

Periodic motion is a specific form of oscillatory motion where the motion pattern repeats itself with a

uniform time interval. This uniform time interval is referred to as the period and has units of seconds

per cycle. The reciprocal of the period is referred to as the frequency and has units of cycles per

second. This unit of combination has been given a special unit symbol and is referred to as Hertz

(Hz)

Harmonic motion is a specific form of periodic motion where the motion pattern can be describe by

either a sine or cosine. This motion is also sometimes referred to as simple harmonic motion.

Because the sine or cosine technically used angles in radians, the frequency term expressed in the

units radians per seconds (rad/sec). This is sometimes referred to as the circular frequency. The

relationship between the frequency in Hz (cps) and the frequency in rad/sec is simply the

relationship. 2π rad/sec.

Natural frequency is the frequency at which an undamped system will tend to oscillate due to initial

conditions in the absence of any external excitation. Because there is no damping, the system will

oscillate indefinitely.

Damped natural frequency is frequency that a damped system will tend to oscillate due to initial

conditions in the absence of any external excitation. Because there is damping in the system, the

system response will eventually decay to rest.

Resonance is the condition of having an external excitation at the natural frequency of the system. In

general, this is undesirable, potentially producing extremely large system response.

Degrees of freedom: The numbers of degrees of freedom

that a body possesses are those necessary to completely

define its position and orientation in space. This is useful in

several fields of study such as robotics and vibrations.

Consider a spherical object that can only be positioned

somewhere on the x axis.

UNIT - 1

VIBRATIONS

Mechanical Vibrations[10ME72]

Dept. Of ME, ACE1

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

06ME 64 - Mechanical Vibrations, Dr.T.V.Givindaraju Introduction 2

This needs only one dimension, ‘x’ to define the position to the centre of gravity so it has one degree

of freedom. If the object was a cylinder, we also need an angle ‘θ’ to define the orientation so it has

two degrees of freedom.

Now consider a sphere that can be positioned in

Cartesian coordinates anywhere on the z plane. This

needs two coordinates ‘x’ and ‘y’ to define the

position of the centre of gravity so it has two

degrees of freedom. A cylinder, however, needs the

angle ‘θ’ also to define its orientation in that plane

so it has three degrees of freedom.

In order to completely specify the position and

orientation of a cylinder in Cartesian space, we

would need three coordinates x, y and z and three

angles relative to each angle. This makes six

degrees of freedom. A rigid body in space has

(x,y,z,θx θy θz).

In the study of free vibrations, we will be

constrained to one degree of freedom.

Types of Vibrations: Free or natural vibrations: A free vibration is one that occurs naturally with no energy being added

to the vibrating system. The vibration is started by some input of energy but the vibrations die away

with time as the energy is dissipated. In each case, when the body is moved away from the rest

position, there is a natural force that tries to return it to its rest position. Free or natural vibrations

occur in an elastic system when only the internal restoring forces of the system act upon a body.

Since these forces are proportional to the displacement of the body from the equilibrium position, the

acceleration of the body is also proportional to the displacement and is always directed towards the

equilibrium position, so that the body moves with SHM.

Figure 1. Examples of vibrations with single degree of freedom.

Note that the mass on the spring could be made to swing like a pendulum as well as bouncing up and

down and this would be a vibration with two degrees of freedom. The number of degrees of freedom

of the system is the number of different modes of vibration which the system may posses.

Mechanical Vibrations [10ME72]

Dept. Of ME, ACE 2

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

06ME 64 - Mechanical Vibrations, Dr.T.V.Givindaraju Introduction

3

The motion that all these examples perform is called SIMPLE HARMONIC MOTION (S.H.M.).

This motion is characterized by the fact that when the displacement is plotted against time, the

resulting graph is basically sinusoidal. Displacement can be linear (e.g. the distance moved by the

mass on the spring) or angular (e.g. the angle moved by the simple pendulum). Although we are

studying natural vibrations, it will help us understand S.H.M. if we study a forced vibration

produced by a mechanism such as the Scotch Yoke.

Simple Harmonic Motion

The wheel revolves at ω radians/sec and the pin forces

the yoke to move up and down. The pin slides in the slot

and Point P on the yoke oscillates up and down as it is

constrained to move only in the vertical direction by the

hole through which it slides. The motion of point P is

simple harmonic motion. Point P moves up and down so

at any moment it has a displacement x, velocity v and an

acceleration a.

The pin is located at radius R from the centre of the

wheel. The vertical displacement of the pin from the

horizontal centre line at any time is x. This is also the

displacement of point P. The yoke reaches a maximum

displacement equal to R when the pin is at the top and –R

when the pin is at the bottom.

This is the amplitude of the oscillation. If the wheel rotates at ω radian/sec then after time t seconds

the angle rotated is θ = ωt radians. From the right angle triangle we find x = R Sin(ωt) and the graph

of x - θ is shown on figure 3a.

Velocity is the rate of change of distance with time. The plot is also shown on figure 3a.

v = dx/dt = ωR Cos(ωt).

The maximum velocity or amplitude is ωR and this occurs as the pin passes through the horizontal

position and is plus on the way up and minus on the way down. This makes sense since the

tangential velocity of a point moving in a circle is v = ωR and at the horizontal point they are the

same thing.

Acceleration is the rate of change of velocity with time. The plot is also shown on figure 3a.

a = dv/dt = -ω2R Sin(-ω2

R)

The amplitude is ω2R and this is positive at the bottom and minus at the top (when the yoke is about

to change direction)

Since R Sin(ωR) = x; then substituting x we find a = -ω2 x

This is the usual definition of S.H.M. The equation tells us that any body that performs sinusoidal

motion must have an acceleration that is directly proportional to the displacement and is always

directed to the point of zero displacement. The constant of proportionality is ω2. Any vibrating body

that has a motion that can be described in this way must vibrate with S.H.M. and have the same

equations for displacement, velocity and acceleration.

Figure 2


Dept. Of ME, ACE3

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


4

FIGURE 3a FIGURE 3b

Angular Frequency, Frequency and Periodic time

ω is the angular velocity of the wheel but in any vibration such as the mass on the spring, it is called

the angular frequency as no physical wheel exists.

The frequency of the wheel in revolutions/second is equivalent to the frequency of the vibration. If

the wheel rotates at 2 rev/s the time of one revolution is ½ seconds. If the wheel rotates at 5 rev/s the

time of one revolution is 1/5 second. If it rotates at f rev/s the time of one revolution is

1/f. This

formula is important and gives the periodic time.

Periodic Time T = time needed to perform one cycle.

f is the frequency or number of cycles per second.

It follows that: T = 1/f and f =

1/T

Each cycle of an oscillation is equivalent to one rotation of the wheel and 1 revolution is an angle of

2π radians.

When θ = 2π and t = T.

It follows that since θ = ωt; then 2π = ωT

Rearrange and θ = 2π

/T. Substituting T = 1/f, then ω =2πf


4Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


5

Equations of S.H.M.

Consider the three equations derived earlier.

Displacement x = R Sin(ωt).

Velocity v = dx/dt = ωR Cos(ωt) and Acceleration a = dv/dt = -ω2R Sin(ωt)

The plots of x, v and a against angle θ are shown on figure 3a. In the analysis so far made, we

measured angle θ from the horizontal position and arbitrarily decided that the time was zero at this

point.

Suppose we start the timing after the angle has reached a value of φ from this point. In these cases, φ is called the phase angle. The resulting equations for displacement, velocity and acceleration are then

as follows.

Displacement x = R Sin(ωt + φ).

Velocity v = dx/dt = ωR Cos(ωt + φ).

Acceleration a = dv/dt = -ω2R Sin((ωt + φ).

The plots of x, v and a are the same but the vertical axis is displaced by φ as shown on figure 3b. A

point to note on figure 3a and 3b is that the velocity graph is shifted ¼ cycle (90o) to the left and the

acceleration graph is shifted a further ¼ cycle making it ½ cycle out of phase with x.

Forced vibrations: When the body vibrates under the influence of external force, then the body is

said to be under forced vibrations. The external force, applied to the body is a periodic disturbing

force created by unbalance. The vibrations have the same frequency as the applied force.

(Note: When the frequency of external force is same as that of the natural vibrations, resonance

takes place)

Damped vibrations: When there is a reduction in amplitude over every cycle of vibration, the motion

is said to be damped vibration. This is due to the fact that a certain amount of energy possessed by

the vibrating system is always dissipated in overcoming frictional resistance to the motion.

Types of free vibrations:

Linear / Longitudinal vibrations: When the disc is displaced vertically downwards by an external

force and released as shown in the figure 4, all the particles of the rod and disc move parallel to the

axis of shaft. The rod is elongated and shortened alternately and thus the tensile and compressive

stresses are induced alternately in the rod. The vibration occurs is know as Linear/Longitudinal

vibrations.

Transverse vibrations: When the rod is displaced in the transverse direction by an external force and

released as shown in the figure 5, all the particles of rod and disc move approximately perpendicular

to the axis of the rod. The shaft is straight and bends alternately inducing bending stresses in the rod.

The vibration occurs is know as transverse vibrations.

Torsional vibrations: When the rod is twisted about its axis by an external force and released as

shown in the figure 6, all the particles of the rod and disc move in a circle about the axis of the rod.

The rod is subjected to twist and torsional shear stress is induced. The vibration occurs is known as

torsional vibrations.


5Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


6

A

O

B

O

BA

DISC

ROD

DISC

ROD

B

O

A

ROD

DISC

FIGURE 4

FIGURE 5

FIGURE 6

Oscillation of a floating body:

You may have observed that some bodies floating

in water bob up and down. This is another example

of simple harmonic motion and the restoring force

in this case is buoyancy.

Consider a floating body of mass M kg. Initially it is

at rest and all the forces acting on it add up to zero.

Suppose a force F is applied to the top to push it

down a distance x. The applied force F must

overcome this buoyancy force and also overcome

the inertia of the body.

Buoyancy force:

The pressure on the bottom increases by ∆p = ρ g x.

The buoyancy force pushing it up on the bottom is Fb and this increases by ∆p A.

Substitute for ∆p and Fb = ρ g x A

Inertia force:

The inertia force acting on the body is Fi = M a

Balance of forces:

The applied force must be F = Fi + Fb -this must be zero if the vibration is free.

0 = Ma + ρ g x A

xM

Aga

ρ−=

This shows that the acceleration is directly proportional to displacement and is always directed

towards the rest position so the motion must be simple harmonic and the constant of proportionality

must be the angular frequency squared.


Dept. Of ME, ACE 6

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


7

M

Agf

M

Ag

M

Ag

n

ρ

ππ

ω

ρω

ρω

2

1

2

2

==

=

=

Example: A cylindrical rod is 80 mm diameter and has a mass of 5 kg. It floats vertically in water of

density 1036 kg/m3. Calculate the frequency at which it bobs up and down. (Ans. 0.508 Hz)

Principal of super position: The principal of super position means that, when TWO or more

waves meet, the wave can be added or subtracted.

Two waveforms combine in a manner, which simply adds their

respective Amplitudes linearly at every point in time. Thus, a

complex SPECTRUM can be built by mixing together different

Waves of various amplitudes.

The principle of superposition may be applied to waves whenever

two (or more) waves traveling through the same medium at the

same time. The waves pass through each other without being

disturbed. The net displacement of the medium at any point in

space or time, is simply the sum of the individual wave

dispacements.

General equation of physical systems is:

)(tFkxxcxm =++ &&& - This equation is for a

linear system, the inertia, damping and spring force are linear

function xandxx &&& , respectively. This is not true case of non-

linear systems.

)()()( tFxfxxm =++ &&& φ - Damping and spring

force are not linear functions of xandx&

Mathematically for linear systems, if 1x is a solution of;

)(1 tFkxxcxm =++ &&&

and 2x is a solution of;

)(2 tFkxxcxm =++ &&&

then )( 21 xx + is a solution of;

)()( 21 tFtFkxxcxm +=++ &&&

Law of superposition does not hold good for non-linear systems.

If more than one wave is traveling through the medium: The resulting net wave is given by the

Superposition Principle given by the sum of the individual waveforms”


7Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


8

Beats: When two harmonic motions occur with the same amplitude ‘A’ at different frequency is

added together a phenomenon called "beating" occurs.

The resulting motion is:

( ) ( ) ( )[ ]tftfAyyy 2121 2cos2cos ππ +=+=

with trigonometric manipulation, the above

equation can be written as:

tff

tff

Ay2

2cos2

2cos2 2121 +×

−= ππ

The resultant waveform can be thought of as a wave with frequency fave = (f1 + f2)/2 which is

constrained by an envelope with a frequency of fb = |f1 - f2|. The envelope frequency is called the beat

frequency. The reason for the name is apparent if you listen to the phenomenon using sound waves.

(Beats are often used to tune instruments. The desired frequency is compared to the frequency of the instrument. If a beat

frequency is heard the instrument is "out of tune". The higher the beat frequency the more "out of tune" the instrument

is.)

Fourier series: decomposes any periodic function or periodic signal into the sum of a

(possibly infinite) set of simple oscillating functions, namely sines and cosines (or complex

exponentials).

Fourier series were introduced by Joseph Fourier (1768–1830) for the purpose of solving the

heat equation in a metal plate.

The Fourier series has many such applications in electrical engineering, vibration analysis,

acoustics, optics, signal processing, image processing, quantum mechanics, thin-walled shell

theory,etc.

( )tfAy 11 2cos π=

( )tfAy 22 2cos π=


8Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


9

Sin & Cos functions

J. Fourier, developed a periodic function in terms of series of Sines and Cosines.

The vibration results obtained experimentally can be analysed analytically.

If x(t) is a periodic function with period T, the Fourier Series can be written as:

............3sin2sinsin

............3cos2coscos2

)(

321

321

++++

++++=

tbtbtb

tatataa

tx o

ωωω

ωωω


9Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


10

Refer PPT – for more Problems

( )( )

smm

Cos

tCosAvVelocity

/999

963.15020

=

××=

+== φωω

( )

( )2

2

2

/1712

963.15020

smm

Sin

tSinAaonAccelerati

−=

××−=

+−== φωω


10Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

06ME 64 - Mechanical Vibrations, Dr.T.V.Givindaraju Undamped Free Vibrations

1

δx

m

W

W=kδ

Unstrained postion

kx

m x

m O

x

k(δ+x)

mg

m

k= Stiffness spring

FIGURE 7

Visvesvaraya Technological University

E-notes

Dr. .V. Govindaraju

Principal, Shirdi Sai Engineering College, Bangalore

06ME 64 - MECHANICAL VIBRATIONS UNIT - 2

Undamped Free Vibrationis:

NATURAL FREQUENCY OF FREE LONGITUDINAL VIBRATION

Equilibrium Method: Consider a body of mass ‘m’ suspended from a spring of negligible mass as

shown in the figure 4.

Let m = Mass of the body

W = Weight of the body = mg

K = Stiffness of the spring

δ = Static deflection of the spring due to ‘W’

By applying an external force, assume the body

is displaced vertically by a distance ‘x’, from the

equilibrium position. On the release of external

force, the unbalanced forces and acceleration

imparted to the body are related by Newton

Second Law of motion.

∴ The restoring force = F = - k × x (-ve sign indicates, the restoring force ‘k.x’ is

opposite to the direction of the displacement ‘x’)

By Newton’s Law; F = m × a

2

2

dt

xdmxkF =−=∴

∴ The differential equation of motion, if a body of mass ‘m’ is acted upon by a restoring force ‘k’

per unit displacement from the equilibrium position is;

SHMrepresentsequationThisxm

k

dt

xd−=+ 0

2

2

SHMform

kx

dt

xd−==+ 22

2

2

0 ωω

The natural period of vibration is Seck

mT π

ω

π2

2==

The natural frequency of vibration is sec/2

1cycles

m

kf n

π=

From the figure 7; when the spring is strained by an amount of ‘δ’ due to the weight W = mg

δ k = mg

UNIT 2

UNDAMPED FREE VIBRATIONS


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


2

Static Equilibrium

postion

spring

x

m

k= Stiffness

FIGURE 8

Hence δ

g

m

k=

cpsorHzg

f nδπ2

1=∴

Energy method: The equation of motion of a conservative system may be established from energy

considerations. If a conservative system set in motion, the mechanical energy in the system is

partially kinetic and partially potential. The KE is due to the velocity of mass and the PE is due to

the stain energy of the spring by virtue of its deformation.

Since the system is conservative; and no energy is transmitted to the system and from the system in

the free vibrations, the sum of PE and KE is constant. Both velocity of the mass and deformation of

spring are cyclic. Thus, therefore be constant interchange of energy between the mass and the

spring. (KE is maximum, when PE is minimum and PE is maximum, when KE is minimum - so system goes through cyclic

motion)

KE + PE = Constant

[ ] )1(0 −=+PEKEdt

d

We have

2

2

2

1

2

1

==

dt

dxmvmKE -(2)

Potential energy due to the displacement is equal to

the strain energy in the spring, minus the PE change

in the elevation of the mass.

( )

( ) )3(2

1 2

0

0

−=−+=

−=∴

∫

∫

kxdxmgkxmg

dxmgdxforecespringTotalPE

x

x

Equation (1) becomes

=

+= 2

2

1

2

0kxx

kxPE

0

02

1

2

2

2

2

=

+

=

+

dt

dxkx

dt

xdm

kxdt

dx

dt

d

002

2

==

+

dt

dxORkx

dt

xdmEither

[ ]

SHMrepresentsEquationxm

k

dt

xd

kxxmkxdt

xdm

timeofvaluesallforzerobecandt

dxvelocityBut

−=+⇒

=+=+∴

0

00

.

2

2

2

2

&&


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


3

sec/2

1

sec2

cyclesm

kfvibrationoffrequencyNatural

andk

mTperiodTime

nπ

π

==

==∴

(The natural frequency is inherent in the system. It is the function of the system parameters 'k' and 'm' and it is

independent of the amplitude of oscillation or the manner in which the system is set into motion.)

Rayleigh’s Method: The concept is an extension of energy method. We know, there is a constant

interchange of energy between the PE of the spring and KE of the mass, when the system executes

cyclic motion. At the static equilibrium position, the KE is maximum and PE is zero; similarly when

the mass reached maximum displacement (amplitude of oscillation), the PE is maximum and KE is

zero (velocity is zero). But due to conservation of energy total energy remains constant.

Assuming the motion executed by the vibration to be simple harmonic, then;

x = A Sinωt

x = displacement of the body from the mean position after time 't' sec and

A = Maximum displacement from the mean position

tSinAx ω=&

At mean position, t = 0; Velocity is maximum

( ) ( )

=⇒

=

=

=

=

=

=∴

==

=∴

==

=∴

•

δ

g

m

k

gmkδ

m

k

π f

m

k ω

k m ω

PEKE We know

kA P.E Maximum

Axxk P.E Maximum

Aωm E Maximum K.

Axdt

dx v

n

/

Q

2

1

2

1

2

1

2

1

21

2

maxmax

2

max

2

max

22

max

max

max ω


Dept. Of ME, ACE 3

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


4

T V G

1. Determine the natural frequency of the spring-mass system, taking mass of the spring into

account.

k= Spring

stiffness

m

x

y

l dy

We know that PE + KE = Constant

( )

222

3

2

22

l

0

2

2

22

l

0

22

32

1

32

1x m

2

1

32

1x m

2

1

2

1x m

2

1

dy 2

1x m

2

1 EK

xm

mm

x

l

l

x

dyyl

x

xl

y

ss &&&

&&

&&

&&

+=+=

+=

+=

+=∴

∫

∫

ρ

ρ

ρ

( )

sec/

3

][

3

2

1

0

3

03

0;''

032

1

2

1 22

radiansm

m

kf

OR

mlcpsm

m

kf

xm

m

kx

equationalDifferentixxm

mxxk

KEPEdt

dttorespectwithatingDifferenti

xm

mxk

s

n

s

s

n

s

s

s

+

=∴

=

+

=∴

=

+

+

−=

++

=+

=

++

ρπ

&&

&&&&

&

Let l = Length of the spring under

equilibrium condition

ρ = Mass/unit length of the spring

ms = Mass of the spring = ρ × l

Consider an elemental length of 'dy' of the

spring at a distance 'y' from support.

∴ Mass of the element = ρ dy

At any instant, the mass 'm' is displaced by a

distance 'x' from equilibrium position.

springmass

2

(KE) and (KE) of sum theis

instant, at this system theof EK

k x 2

1 E P =∴


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


5

T V G

2. Determine the natural frequency of the system shown in figure by Energy and Newton's

method.

kx/2

F1

θ°

= x

/2

x/2

x

Io m1

m

Disc

Newton's Method: Use x, as co-ordinate

sec38

2

38

2

2

1

04

1

8

38

1

1

1

rad/mm

kfor

cpsmm

kf

kxmm

x

n

n

+=

+=∴

=+

+

π

&&

When mass 'm' moves down a distance 'x'

from its equilibrium position, the center

of the disc if mass m1 moves down by x

and rotates though and angle θ.

( )

013

0

22

1

16

3

2

1

44

1

8

1

2

1

2

1

42

1

2

1

)()(

2

2

1

2

21

2

2

22

12

12

22

12

=++⇒

=+

=

+=

++=

++=

+=

=⇒

=∴

xxkxxmxxm

PEKEdt

d

xkPE

xmxm

r

xrmxmxm

Ix

mxm

KEKEKE

r

x

rx

o

rotTr

&&&&&&&

&&

&&&

&&

&

&&

θ

θ

θ

62

52

422

2)3()2()1(

3

222

;

1;

12

1

11

1

11

−−−=

−−−=

−−+−=

−−=

−−+=

−−=

Fxmr

xI

rFrxmr

xI

kxFxm

xm

r

xbyreplaceandandinngSubstituti

rFFrI

kxFF

xmmdiscfor

Fxmmmassor

o

o

o

&&&&

&&&&

&&&&

&&&&

&&

&&

&&

θ

θ


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


6


method. Assume the cylinder rolls on the surface without slipping.

a) Energy method:

When mass 'm' rotates through an angle θ, the center of the roller move

distance 'x'

k

θ°

Io

mr

xx

Roller/Disc

kx

Fr

Newton’s method:

sec/3

2

3

2

2

1

02

3

2

1

2

.equationtorqueusing

radm

kforcps

m

kf

xkxm

xmxkxm

xmF

rFI

Fkxxm

Fam

nn

r

ro

r

==∴

=+

−−=

−=

−=

+−=

=∑

π

θ

&&

&&&&

&&

&&

&&

Hzorcpsmm

kf

radmm

kf

kxmm

mx

rmIkx

xmr

xI

xm

andsequationAdding

n

n

oo

83

2

2

1

sec/83

2

042

}2

1{

22

22

)6()4(

1

1

11

2121

+=

+=

=+

++

=−−=+

π

&&

&&&&&&

( )

m

kffrequencyNatural

xkxm

PEKEdt

d

xkPE

xmKE

r

xrmxm

Ixm

KEKEKE

n

o

rotTr

3

2

2

1

02

3

0

2

1

4

3

2

1

2

1

2

1

2

1

2

1

)()(

2

2

2

222

22

π

θ

==

=+⇒

=+

=

=∴

+=

+=

+=

&&

&

&&

&&

MechMechanical Vibrations[10Me72]

Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


7

T V G


method.

Energy Method: Use θ or x as coordinate

( )

orHzcpsrmI

krORcps

mm

krad

mm

kffrequencyNatural

xmmxk

xxmmxxk

PEKEdt

d

xkPE

xmxmKE

r

xrmxm

Ixm

KEKEKE

o

n

o

rotTr

2

2

11

1

1

2

21

2

2

22

12

22

2

1

2

2

2

1sec/

2

2

02

1

024

1

2

1

0

2

1

4

1

2

1

2

1

2

1

2

1

2

1

2

1

)()(

++=

+==

=

++

=

++⇒

=+

=

+=∴

+=

+=

+=

ππ

θ

&

&&&&&

&&

&&

&&

Newton’s Method: Consider motion of the disc with ‘θ’ as coordinate.

For the mass ‘m’ : .)1(−−= rFxm &&

For the disc : )2(. 2 −−−= θθ rkrFI ro&&

Substitute (1) in (2):

( )

=

+=

+=

+=∴

=++

−−=

−−=

21

1

2

2

2

2

22

22

2

2

1

2

2

2

1

2

1

sec/

0

rmIcpsmm

k

HzorcpsrmI

rk

radrmI

rkf

rkrmI

rkrmI

rkxrmI

o

o

o

n

o

o

o

π

π

θθ

θθθ

θθ

&&

&&&&

&&&&

k

θ°

x

Io m1

m

Disc

x=rθ

F

krθ


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


8

5. Determine the natural frequency of the system shown in figure 5 by Energy and Newton's

method. Assume the cylinder rolls on the surface without slipping.

Energy Method:

( )

( )

( )

( )

( ) ( )Hzorcps

rm

arkrad

rm

arkffrequencyNatural

arkrm

arkrm

PEKEdt

d

arkxkPE

rmKE

rmrm

Ixm

KEKEKE

n

o

rotTr

2

2

2

2

22

22

222

22

2222

22

3

4

2

1sec/

3

4

043

022

3

0

2

12

4

3

2

1

2

1

2

1

2

1

2

1

)()(

+=

+==

=++

=++⇒

=+

+=

=

=∴

+=

+=

+=

π

θθ

θθθθ

θ

θ

θθ

θ

&&

&&&&

&

&&

&&

Newton’s Method: Considering combined translation and rotational motion as shown in Figure

‘a’.

Hence it must satisfy:

( )

( )

( )

( )

( )

( ) ( )Hzorcps

rm

arkrad

rm

arkffrequencyNatural

arkrm

aararrkrm

aarkarkrrm

r

aarkrFrmand

arkFrm

aaxkrF

aboutTorqueIand

axkFxm

directionxinForcexm

n

o

2

2

2

2

22

2

2

2

3

4

2

1sec/

3

4

0)(43

0])([43

4)(43

addThen 2.by (2) and 2by (1)equation Multiply

)2(22

1

)1(2

2

''

)(2

+=

+==

=++

=++++

+−+−=

−+−=

−+−−=

+−=

=

+−−=

=

∑

∑

π

θ

θθ

θθθ

θθ

θθ

θ

θθ

θ

&&

&&

&&

&&

&&

&&

&&

&&

Refer PPT – For more problems

kθ°

xIo

m

a

r

k

+x

x

c

c'

(x+aθ)

o

o'

F

2k(x+aθ)

a(Appx)

FIGURE 5

Figure 'a'


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


9

C

θ°

θ°

l

l

A

B

Wsinθ

WWcosθ

T

Oscillation of a simple pendulum

O

Oscillation of a simple pendulum: Figure shows the arrangement of simple pendulum, which consists of a light, inelastic

(inextensible), flexible string of length 'l' with heavy bob of weight W (m×g)

suspended at the lower end and the upper end is fixed at 'O'. The bob oscillates freely

in a vertical plane.

The pendulum is in equilibrium, when the bob is at 'A'. If the bob is brought at B or C

and released, it will start oscillating between B and C with 'A' as mean position.

Let θ be a very small angle (′ 4o), the bob will have SHM.

Consider the bob at 'B', the forces acting on the bob are:

i) weight of the bob = W = mg ⇒ acting downwards vertically.

ii) tension 'T' in the string

The two components of the weight 'W'

i) along the string = W cosθ

ii) normal to the string = W sinθ

The component W sinθ acting towards 'A'

will be unbalanced and will give rise to an

acceleration 'a' in the direction of 'A'.

Acceleration of the body with SHM is given by; [a = -ω2 × Distance form

centre]

Numerically, a = ω2 × Arc AB -(2)

From (1) and (2)

Second's pendulum: is defined as that pendulum which has one beat per second. Thus

the time period for second's pendulum will be equal to two seconds.

( )1

sin small; very is Since

sinsin

sinMassForce on Accelerati

−=

=

=∴

=

==

===∴

L

ABArcg

Radius

ABArc lengthg

ga

gm

mg

mWa

θ

θθθ

θθ

θ

2

T

22

==

==

=

n oscillatiolf of the Beat is haBeat

g

LTnoscillatioofperiodTime

L

g

πω

π

ω


Dept. Of ME, ACE 9

Vtusolution.in

Vtusolution.in

vtuso

lution

.in


10

θ°

θ°

G

W Sinθ

W = m g

W Cosθ

��

O

Equilibrium position

A

Point of suspension

l

Compound pendulum:

When a rigid body is suspended

vertically, and it oscillates with a small

amplitude under the effect of force of

gravity, the body is known as

compound pendulum.

Let W = weight of the pendulum

= mass × g = m × g N

kG = radius of gyration

l = distance from point of suspension

to 'G' (CG) of the body.

The components of the force W, when the

pendulum is given a small angular

displacement 'θ' are:

1. W cosθ - along the axis of the body.

2. W sinθ - along normal to the axis of the body.

The component W sinθ acting towards

equilibrium position (couple tending to

restore) of the pendulum:

C = W sinθ × l = m g l sinθ

Since θ is very small sinθ = θ

∴ C = m g l θ

Mass moment of inertia about the axis of suspension 'O':

I = IG + m l2 (parallel axis theorem)

= m (kG2+ l

2)

( )

sec222 period that timeknow We

pendulum theofon acceleratiAngular

22

22

22

lg

lk

onAccelerati

ntDisplacemeT

onAccelerati

ntDisplaceme

lg

lk

lkm

mgl

I

C

G

G

G

+====

=+

=

+=

==∴

πα

θππ

α

θ

α

θ

θα

α

ll

k

l

l

lk

lgn

T

G

G

+=+

=

+===

222

G

22

kL

pendulum simple of

length equivalent thependulum, simpleith equation w above theComparing

2

11 n oscillatio ofFrequency

π


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

1

E-notes Dr. Ajit Prasad S L

Professor in Mechanical Engineering

PES College of Engineering

Mandya

Damped Free Vibrations

Single Degree of Freedom Systems

Introduction:

Damping – dissipation of energy.

For a system to vibrate, it requires energy. During vibration of the system, there will be

continuous transformation of energy. Energy will be transformed from potential/strain to

kinetic and vice versa.

In case of undamped vibrations, there will not be any dissipation of energy and the

system vibrates at constant amplitude. Ie, once excited, the system vibrates at constant

amplitude for infinite period of time. But this is a purely hypothetical case. But in an

actual vibrating system, energy gets dissipated from the system in different forms and

hence the amplitude of vibration gradually dies down. Fig.1 shows typical response

curves of undamped and damped free vibrations.

Types o damping:

(i) Viscous damping

In this type of damping, the damping resistance is proportional to the relative velocity

between the vibrating system and the surroundings. For this type of damping, the

differential equation of the system becomes linear and hence the analysis becomes easier.

A schematic representation of viscous damper is shown in Fig.2.

Here, F α x& or xcF &= , where, F is damping resistance, x& is relative velocity and c is the

damping coefficient.

UNIT - 3UNIT - 3

DAMPED FREE VIBRATIONS


1Dept. Of ME , ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

2

(ii) Dry friction or Coulomb damping

In this type of damping, the damping resistance is independent of rubbing velocity and is

practically constant.

(iii) Structural damping

This type of damping is due to the internal friction within the structure of the material,

when it is deformed.

Spring-mass-damper system:

Fig.3 shows the schematic of a simple spring-mass-damper system, where, m is the mass

of the system, k is the stiffness of the system and c is the damping coefficient.

If x is the displacement of the system, from Newton’s second law of motion, it can be

written

kxxcxm −−= &&&

Ie 0=++ kxxcxm &&& (1)

This is a linear differential equation of the second order and its solution can be written as st

ex = (2)

Differentiating (2), stsex

dt

dx== &

stesx

dt

xd 2

2

2

== &&

Substituting in (1), 02 =++ stststkecseems

( ) 02 =++ stekcsms

Or 02 =++ kcsms (3)

Equation (3) is called the characteristic equation of the system, which is quadratic in s.

The two values of s are given by

m

k

m

c

m

cs −

±−=

2

2,122

(4)


2Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

3

The general solution for (1) may be written as

tstseCeCx 21

21 += (5)

Where, C1 and C2 are arbitrary constants, which can be determined from the initial

conditions.

In equation (4), the values of s1 = s2, when m

k

m

c=

2

2

Or, nm

k

m

cω==

2 (6)

Or nmc ω2= , which is the property of the system and is called critical

damping coefficient and is represented by cc.

Ie, critical damping coefficient = nc mc ω2=

The ratio of actual damping coefficient c and critical damping coefficient cc is called

damping factor or damping ratio and is represented by ζ.

Ie, cc

c=ζ (7)

In equation (4), m

c

2 can be written as n

c

c m

c

c

c

m

cωζ .

22=×=

Therefore, ( ) [ ] nnnns ωζζωωζωζ 1.. 222

2,1 −±−=−±−= (8)

The system can be analyzed for three conditions.

(i) ζ > 1, ie, c > cc, which is called over damped system.

(ii) ζ = 1. ie, c = cc, which is called critically damped system.

(iii) ζ < 1, ie, c < cc, which is called under damped system.

Depending upon the value of ζ, value of s in equation (8), will be real and unequal, real

and equal and complex conjugate respectively.

(i) Analysis of over-damped system (ζ > 1).

In this case, values of s are real and are given by

[ ]ns ωζζ 12

1 −+−= and [ ]ns ωζζ 12

2 −−−=

Then, the solution of the differential equation becomes

tt nn

eCeCxωζζωζζ

−−−

−+−

+=1

2

1

1

22

(9)

This is the final solution for an over damped system and the constants C1 and C2 are

obtained by applying initial conditions. Typical response curve of an over damped system

is shown in fig.4. The amplitude decreases exponentially with time and becomes zero at t

= ∞.


3Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

4

(ii) Analysis of critically damped system (ζ = 1).

In this case, based on equation (8), s1 = s2 = -ωn

The solution of the differential equation becomes

tststeCeCx 21

21 +=

Ie, tt nn teCeCx

ωω −−+= 21

Or, ( ) tnetCCxω−

+= 21 (10)

This is the final solution for the critically damped system and the constants C1 and C2 are

obtained by applying initial conditions. Typical response curve of the critically damped

system is shown in fig.5. In this case, the amplitude decreases at much faster rate

compared to over damped system.


4Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

5

(iii) Analysis of under damped system (ζ < 1).

In this case, the roots are complex conjugates and are given by

[ ]njs ωζζ 2

1 1−+−=

[ ]njs ωζζ 2

2 1−−−=

The solution of the differential equation becomes

tjtj nn

eCeCxωζζωζζ

−−−

−+−

+=22 1

2

1

1

This equation can be rewritten as

+=

−−

−

−tjtj

t nnn eCeCex

ωζωζζω

22 1

2

1

1 (11)

Using the relationships

θθθ sincos iei +=

θθθ sincos iei −=−

Equation (11) can be written as

{ } { }[ ]tjtCtjtCex nnnn

tn ωζωζωζωζζω 22

2

22

1 1sin1cos1sin1cos −−−+−+−=−

Or ( ){ } ( ){ }[ ]tCCjtCCex nn

tn ωζωζζω 2

21

2

21 1sin1cos −−+−+=−

(12)

In equation (12), constants (C1+C2) and j(C1-C2) are real quantities and hence, the

equation can also be written as

{ } { }[ ]tBtAex nn

tn ωζωζζω 22 1sin1cos −+−=−

Or, ( ){ }[ ]1

2

1 1sin φωζζω+−=

−teAx n

tn (13)

The above equations represent oscillatory motion and the frequency of this motion is

represented by nd ωζω 21−= (14)

Where, ωd is the damped natural frequency of the system. Constants A1 and Φ1 are

determined by applying initial conditions. The typical response curve of an under damped

system is shown in Fig.6.


5Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

6

Applying initial conditions,

x = Xo at t = 0; and 0=x& at t = 0, and finding constants A1 and Φ1,

equation (13) becomes

−+−

−= −−

ζ

ζωζ

ζ

ζω2

12

2

1tan1sin

1te

Xx n

to n (15)

The term to ne

X ζω

ζ

−

− 21 represents the amplitude of vibration, which is observed to decay

exponentially with time.


6Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

7

LOGARITHMIC DECREMENT

Referring to Fig.7, points A & B represent two successive peak points on the response

curve of an under damped system. XA and XB represent the amplitude corresponding to

points A & B and tA & tB represents the corresponding time.

We know that the natural frequency of damped vibration = nd ωζω 21−= rad/sec.

Therefore, π

ω

2

d

df = cycles/sec

Hence, time period of oscillation =

ndd

ABf

ttωζ

π

ω

π21

221

−===− sec (16)

From equation (15), amplitude of vibration

XA = Anto e

X ζω

ζ

−

− 21

XB = Bnto eX ζω

ζ

−

− 21

Or, ( ) ( )ABnBAn tttt

B

A eeX

X −−−==

ζωζω

Using eqn. (16), 21

2

ζ

πζ

−= e

X

X

B

A

Or, 21

2log

ζ

πζ

−=

B

Ae

X

X


7Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

8

This is called logarithmic decrement. It is defined as the logarithmic value of the ratio of

two successive amplitudes of an under damped oscillation. It is normally denoted by δ.

Therefore, δ = 21

2log

ζ

πζ

−=

B

Ae

X

X (17)

This indicates that the ratio of any two successive amplitudes of an under damped system

is constant and is a function of damping ratio of the system.

For small values of ζ, πζδ 2≈

If X0 represents the amplitude at a particular peak and Xn represents the amplitude after

‘n’ cycles, then, logarithmic decrement = δ = =1

0logX

Xe =

2

1logX

Xe ……

n

n

eX

X 1log −=

Adding all the terms, nδ = n

n

eX

X

X

X

X

X 1

2

1

1

0 ......log −×

Or, n

eX

X

n

0log1

=δ (18)


8Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

9

Solved problems

1) The mass of a spring-mass-dashpot system is given an initial velocity 5ωn, where

ωn is the undamped natural frequency of the system. Find the equation of motion

for the system, when (i) ζ = 2.0, (ii) ζ = 1.0, (i) ζ = 0.2.

Solution:

Case (i) For ζ = 2.0 – Over damped system

For over damped system, the response equation is given by

tt nn

eCeCxωζζωζζ

−−−

−+−

+=1

2

1

1

22

Substituting ζ = 2.0, [ ] [ ] tt nn eCeCx

ωω 73.3

2

27.0

1

−−+= (a)

Differentiating, t

n

t

nnn eCeCx

ωω ωω 73.3

2

27.0

1 73.327.0−−

−−=& (b)

Substituting the initial conditions

x = 0 at t = 0; and nx ω5=& at t = 0 in (a) & (b),

0 = C1 + C2 (c)

5ωn = -0.27 ωn C1 – 3.73 ωn C2 (d)

Solving (c) & (d), C1 = 1.44 and C2 = -1.44.

Therefore, the response equation becomes

[ ] [ ]( )tt nn eex

ωω 73.327.044.1

−−−= (e)

Case (ii) For ζ = 1.0 – Critically damped system

For critically damped system, the response equation is given by

( ) tnetCCxω−

+= 21 (f)

Differentiating, ( ) tt

nnn eCetCCx

ωωω −−++−= 221

& (g)


x = 0 at t = 0; and nx ω5=& at t = 0 in (f) & (g),

C1 = 0 and C2 = 5ωn

Substituting in (f), the response equation becomes

( ) t

nnetx

ωω −= 5 (h)

Case (iii) For ζ = 0.2 – under damped system

For under damped system, the response equation is given by

{ }[ ]1

2


−teAx n

tn

Substituting ζ = 0.2, ( ){ }[ ]1

2.0

1 98.0sin φωω+=

−teAx n

tn (p)


9Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

10

Differentiating,

( ){ }[ ] ( )1

2.0

11

2.0

1 98.0cos98.098.0sin2.0 φωωφωω ωω+++−=

−−teAteAx n

t

nn

t

nnn& (q)


x = 0 at t = 0; and nx ω5=& at t = 0 in (p) & (q),

A1sinΦ1 = 0 and A1 cosΦ1 = 5.1

Solving, A1 = 5.1 and Φ1 = 0

Substituting in (p), the response equation becomes

( ){ }[ ]tex n

tn ωω98.0sin1.5

2.0−= (r)

2) A mass of 20kg is supported on two isolators as shown in fig.Q.2. Determine the

undamped and damped natural frequencies of the system, neglecting the mass of the

isolators.

Solution:

Equivalent stiffness and equivalent damping coefficient are calculated as

30000

13

3000

1

10000

1111

21

=+=+=kkkeq

300

4

100

1

300

1111

21

=+=+=CCCeq

Undamped natural frequency = sec/74.1020

1330000

radm

keq

n ===ω

cpsf n 71.12

74.10==

π

Damped natural frequency = nd ωζω 21−=

1745.0

2013

300002

4300

2=

××

==mk

C

eq

eqζ


10Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

11

sec/57.1074.101745.01 2radd =×−=∴ω

Or, cpsf d 68.12

57.10=

×=

π

3) A gun barrel of mass 500kg has a recoil spring of stiffness 3,00,000 N/m. If the

barrel recoils 1.2 meters on firing, determine,

(a) initial velocity of the barrel

(b) critical damping coefficient of the dashpot which is engaged at the end of the

recoil stroke

(c) time required for the barrel to return to a position 50mm from the initial

position.

Solution:

(a) Strain energy stored in the spring at the end of recoil:

mNkxP −=××== 2160002.13000002

1

2

1 22

Kinetic energy lost by the gun barrel:

222 2505002

1

2

1vvmvT =××== , where v = initial velocity of the barrel

Equating kinetic energy lost to strain energy gained, ie T = P,

216000250 2 =v

v = 29.39m/s

(b) Critical damping coefficient = mNkmCc sec/2449550030000022 −=×==

(c) Time for recoiling of the gun (undamped motion):

Undamped natural frequency = srm

kn /49.24

500

300000===ω

Time period = sec259.029.24

22===

π

ω

πτ

n

Time of recoil = sec065.04

259.0

4==

τ

Time taken during return stroke:

Response equation for critically damped system = ( ) tnetCCxω−

+= 21

Differentiating, ( ) t

n

t nn etCCeCxωω ω −−

+−= 212&

Applying initial conditions, x = 1.2, at t = 0 and 0=x& at t = 0,

C1 = 1.2, & C2 = 29.39

Therefore, the response equation = ( ) tetx 49.2439.292.1 −+=

When x = 0.05m, by trial and error, t = 0.20 sec

Therefore, total time taken = time for recoil + time for return = 0.065 + 0.20 = 0.265 sec

The displacement – time plot is shown in the following figure.


11Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

12

4) A 25 kg mass is resting on a spring of 4900 N/m and dashpot of 147 N-se/m in

parallel. If a velocity of 0.10 m/sec is applied to the mass at the rest position, what

will be its displacement from the equilibrium position at the end of first second?

Solution:

The above figure shows the arrangement of the system.

Critical damping coefficient = nc mc ω2=

Where srm

kn /14

25

4900===ω

Therefore, mNcc sec/70014252 −=××=

Since C< Cc, the system is under damped and 21.0700

147===

cc

cζ


12Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

13

Hence, the response equation is ( ){ }[ ]1

2


−teAx n

tn

Substituting ζ and ωn, ( ){ }[ ]1

21421.0

1 1421.01sin φ+−= ×− teAx t

( ){ }[ ]1

94.2

1 7.13sin φ+= − teAx t

Differentiating, ( ){ }[ ] ( )1

94.2

11

94.2

1 7.13cos7.137.13sin94.2 φφ +++−= −− teAteAx tt&

Applying the initial conditions, x = 0, at t = 0 and smx /10.0=& at t = 0

Φ1 = 0

( ){ }[ ] ( )1111 cos7.13sin94.210.0 φφ AA +−=

Since, Φ1 = 0, 0.10 = 13.7 A1; A1 = 0.0073

Displacement at the end of 1 second = ( ){ }[ ] mex 494.2 105.37.13sin0073.0 −− ×==

5) A rail road bumper is designed as a spring in parallel with a viscous damper.

What is the bumper’s damping coefficient such that the system has a damping ratio

of 1.25, when the bumper is engaged by a rail car of 20000 kg mass. The stiffness of

the spring is 2E5 N/m. If the rail car engages the bumper, while traveling at a speed

of 20m/s, what is the maximum deflection of the bumper?

m

k

c

Solution: Data = m = 20000 kg; k = 200000 N/m; 25.1=ζ

Critical damping coefficient =

mNkmcc sec/1024.12000002000022 5 −×=××=××=

Damping coefficient mNCC C sec/1058.11024.125.1 55 −×=××=×= ζ


kn /16.3

20000

200000===ω

Since 25.1=ζ , the system is over damped.

For over damped system, the response equation is given by

tt nn

eCeCxωζζωζζ

−−−

−+−

+=1

2

1

1

22

Substituting ζ = 1.25, [ ] [ ] tt nn eCeCx

ωω 0.2

2

5.0

1

−−+= (a)


13Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

14

Differentiating, t

n

t

nnn eCeCx

ωω ωω 0.2

2

5.0

1 0.25.0−−

−−=& (b)


x = 0 at t = 0; and smx /20=& at t = 0 in (a) & (b),

0 = C1 + C2 (c)

20 = -0.5 ωn C1 – 2.0 ωn C2 (d)

Solving (c) & (d), C1 = 4.21and C2 = -4.21

Therefore, the response equation becomes

[ ] [ ]( )meex tt ×−×− −= 32.658.121.4 (e)

The time at which, maximum deflection occurs is obtained by equating velocity equation

to zero.

Ie, 00.25.00.2

2

5.0

1 =−−=−− t

n

t

nnn eCeCx

ωω ωω&

Ie, 061.2665.6 32.658.1 =+− −− ttee

Solving the above equation, t = 0.292 secs.

Therefore, maximum deflection at t = 0.292secs,

Substituting in (e), [ ] [ ]( )meex 292.032.6292.058.121.4 ×−×− −= , = 1.99m.

6) A disc of a torsional pendulum has a moment of inertia of 6E-2 kg-m2 and is

immersed in a viscous fluid. The shaft attached to it is 0.4m long and 0.1m in

diameter. When the pendulum is oscillating, the observed amplitudes on the same

side of the mean position for successive cycles are 90, 6

0 and 4

0. Determine (i)

logarithmic decrement (ii) damping torque per unit velocity and (iii) the periodic

time of vibration. Assume G = 4.4E10 N/m2, for the shaft material.

Shaft dia. = d = 0.1m

Shaft length = l = 0.4m

Moment of inertia of disc = J = 0.06 kg-m2.

Modulus of rigidity = G = 4.4E10 N/m2

Solution: The above figure shows the arrangement of the system.

(i) Logarithmic decrement = 405.04

6log

6

9log === eeδ

(ii) The damping torque per unit velocity = damping coefficient of the system ‘C’.


14Dept. Of ME, ACE

Mechanical Vibrations [10ME72]Mechanical Vibrations [10ME72]

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

15

We know that logarithmic decrement = 21

2

ζ

πζδ

−= , rearranging which, we get

Damping factor 0645.0405.04

405.0

4 2222=

+=

+=

πδπ

δζ

Also, CC

C=ζ , where, critical damping coefficient = JkC tC 2=

Torsional stiffness = radmNd

l

G

l

GIk

p

t /1008.132

1.0

4.0

104.4

32

64104

−×=×

××

=×==ππ

Critical damping coefficient = radmNJkC tC /50906.01008.122 6 −=××==

Damping coefficient of the system = radmNCC C /8.320645.0509 −=×=×= ζ

(iii) Periodic time of vibration = 21

221

ζω

π

ω

πτ

−===

nddf

Where, undamped natural frequency = sec/6.424206.0

1008.1 6

radJ

kt

n =×

==ω

Therefore, sec00148.00645.016.4242

2

2=

−×=

πτ

7) A mass of 1 kg is to be supported on a spring having a stiffness of 9800 N/m. The

damping coefficient is 5.9 N-sec/m. Determine the natural frequency of the system.

Find also the logarithmic decrement and the amplitude after three cycles if the

initial displacement is 0.003m.

Solution:


kn /99

1

9800===ω

Damped natural frequency = nd ωζω 21−=

Critical damping coefficient = mNmc nc sec/19899122 −=××=××= ω

Damping factor = 03.0198

9.5===

cc

cζ

Hence damped natural frequency = sec/99.9899013.011 22radnd =×−=−= ωζω


15Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

16

Logarithmic decrement = 188.003.01

03.02

1

2

22=

−

××=

−=

π

ζ

πζδ

Also, n

eX

X

n

0log1

=δ ; if x0 = 0.003,

then, after 3 cycles, 3

0 003.0log

3

1188.0,log

1

Xie

X

X

ne

n

e ×==δ

ie, me

X3

188.033 1071.1003.0 −

××==

8) The damped vibration record of a spring-mass-dashpot system shows the

following data.

Amplitude on second cycle = 0.012m; Amplitude on third cycle = 0.0105m;

Spring constant k = 7840 N/m; Mass m = 2kg. Determine the damping constant,

assuming it to be viscous.

Solution:

Here, 133.00105.0

012.0loglog

3

2 === eeX

Xδ

Also, 21

2

ζ

πζδ

−= , rearranging, 021.0

133.04

133.0

4 2222=

+=

+=

πδπ

δζ

Critical damping coefficient = mNkmcc sec/4.2507840222 −=××=××=

Damping coefficient mNCC C sec/26.54.250021.0 −=×=×= ζ

9) A mass of 2kg is supported on an isolator having a spring scale of 2940 N/m and

viscous damping. If the amplitude of free vibration of the mass falls to one half its

original value in 1.5 seconds, determine the damping coefficient of the isolator.

Solution:


kn /34.38

2

2940===ω

Mechanical Vibrations [10ME72]Mechanical Vibrations[10ME72]

16Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

17

Critical damping coefficient = mNmc nc sec/4.15334.38222 −=××=××= ω

Response equation of under damped system = ( ){ }[ ]1

2


−teAx n

tn

Here, amplitude of vibration =tneA

ζω−

1

If amplitude = X0 at t = 0, then, at t = 1.5 sec, amplitude = 2

0X

Ie, 0

0

1 XeA n =×−ζω

or 01 XA =

Also, 2

05.1

1

XeA n =

×−ζω or

2

05.134.38

0

XeX =× ××−ζ or

2

15.134.38 =××−ζe

Ie, 25.134.38 =××ζe , taking log, 012.069.05.134.38 =∴=×× ζζ

Damping coefficient mNCC C sec/84.14.153012.0 −=×=×= ζ


17Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Forced Vibrations Dr. M.A.Kamoji

KLE CET Belgaum

Introduction: In free un-damped vibrations a system once disturbed from its initial

position executes vibrations because of its elastic properties. There is no

damping in these systems and hence no dissipation of energy and hence it

executes vibrations which do not die down. These systems give natural

frequency of the system.

In free damped vibrations a system once disturbed from its position will

execute vibrations which will ultimately die down due to presence of

damping. That is there is dissipation of energy through damping. Here one

can find the damped natural frequency of the system.

In forced vibration there is an external force acts on the system. This

external force which acts on the system executes the vibration of the system.

The external force may be harmonic and periodic, non-harmonic and

periodic or non periodic. In this chapter only external harmonic forces acting

on the system are considered. Analysis of non harmonic forcing functions is

just an extension of harmonic forcing functions.

Examples of forced vibrations are air compressors, I.C. engines, turbines,

machine tools etc,.

Analysis of forced vibrations can be divided into following categories as per

the syllabus.

1. Forced vibration with constant harmonic excitation

2. Forced vibration with rotating and reciprocating unbalance

3. Forced vibration due to excitation of the support

A: Absolute amplitude

UNIT - 4

FORCESUNIT - 4

FORCED VIBRATIONS


1Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

B: Relative amplitude

4. Force and motion transmissibility

For the above first a differential equation of motion is written. Assume a

suitable solution to the differential equation. On obtaining the suitable

response to the differential equation the next step is to non-dimensional the

response. Then the frequency response and phase angle plots are drawn.

1. Forced vibration with constant harmonic excitation

From the figure it is evident that spring force and damping force oppose the

motion of the mass. An external excitation force of constant magnitude acts

on the mass with a frequency ω. Using Newton’s second law of motion an

equation can be written in the following manner.

Equation 1 is a linear non homogeneous second order differential equation.

The solution to eq. 1 consists of complimentary function part and particular

1tSinωFkxxcxm o −−−−−−=++ &&&


2Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

integral. The complimentary function part of eq, 1 is obtained by setting the

equation to zero. This derivation for complementary function part was done

in damped free vibration chapter.

The complementary function solution is given by the following equation.

Equation 3 has two constants which will have to be determined from the

initial conditions. But initial conditions cannot be applied to part of the

solution of eq. 1 as given by eq. 3. The complete response must be

determined before applying the initial conditions. For complete response the

particular integral of eq. 1 must be determined. This particular solution will

be determined by vector method as this will give more insight into the

analysis.

Assume the particular solution to be

Differentiating the above assumed solution and substituting it in eq. 1

[ ] 3φtωξ1SineAx 2n2tζω

2cn −−−+−=

−

2xxx pc −−−−−+=

( ) 4φωtXSinxp −−−−−=

( )πφωtXSinωx

2

πφωtωXSinx

2

p

p

+−=

+−=

&&

&

( )

( ) 50πφωtXSinmω

2

πφωtxcωφωtkXSintSinωF

2

o

−−−−−=+−−

+−−−−


3Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Fig. Vector representation of forces on system subjected to forced vibration

Following points are observed from the vector diagram

1. The displacement lags behind the impressed force by an angle Φ.

2. Spring force is always opposite in direction to displacement.

3. The damping force always lags the displacement by 90°. Damping

force is always opposite in direction to velocity.

4. Inertia force is in phase with the displacement.

The relative positions of vectors and heir magnitudes do not change with

time.

From the vector diagram one can obtain the steady state amplitude and

phase angle as follows


4Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

The above equations are made non-dimensional by dividing the numerator

and denominator by K.

where, is zero frequency deflection

Therefore the complete solution is given by

( ) ( )[ ] 6cωmωkFX222

0 −−−+−=

( )[ ] 7mωkcωtanφ 2-1 −−−−=

( )[ ] ( )[ ]

( )( )

92

nωω1

nωω2ξ

tan 1-φ

8

ωω2ξωω1

XX

2

n

22

n

st

−−−−

=

−−−

+−

=

kFX ost =

pc xxx +=

[ ]( )

( )[ ] ( )[ ]10

ωω2ξωω1

φωtSinX

φtωξ1SineAx

2

n

22

n

st

2n

2tζω

2n

−−−

+−

−+

+−= −


5Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

The two constants A2 and φ2 have to be determined from the initial

conditions.

The first part of the complete solution that is the complementary function

decays with time and vanishes completely. This part is called transient

vibrations. The second part of the complete solution that is the particular

integral is seen to be sinusoidal vibration with constant amplitude and is

called as steady state vibrations. Transient vibrations take place at damped

natural frequency of the system, where as the steady state vibrations take

place at frequency of excitation. After transients die out the complete

solution consists of only steady state vibrations.

In case of forced vibrations without damping equation 10 changes to

Φ2 is either 0° or 180° depending on whether ω<ωn or ω>ωn

Steady state Vibrations: The transients die out within a short period of

time leaving only the steady state vibrations. Thus it is important to know

the steady state behavior of the system,

Thus Magnification Factor (M.F.) is defined as the ratio of steady state

amplitude to the zero frequency deflection.

[ ]( )

( )11

ωω1

ωtSinXφtωSinAx

2

n

st2n2 −−−

−++=

( )[ ] ( )[ ]12

ωω2ξωω1

1

X

XM.F.

2

n

22

nst

−−−

+−

==

( )( )

132

nωω1

nωω2ξ

tan 1-φ −−−−

=


6Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Equations 12 and 13 give the magnification factor and phase angle. The

steady state amplitude always lags behind the impressed force by an angle

Φ. The above equations are used to draw frequency response and phase

angle plots.

Fig. Frequency response and phase angle plots for system subjected forced

vibrations.

Frequency response plot: The curves start from unity at frequency ratio of

zero and tend to zero as frequency ratio tends to infinity. The magnification

factor increases with the increase in frequency ratio up to 1 and then

decreases as frequency ratio is further increased. Near resonance the

amplitudes are very high and decrease with the increase in the damping

ratio. The peak of magnification factor lies slightly to the left of the

resonance line. This tilt to the left increases with the increase in the damping

ratio. Also the sharpness of the peak of the curve decreases with the increase

in the damping.


7Dept. OF ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Phase angle plot: At very low frequency the phase angle is zero. At

resonance the phase angle is 90°. At very high frequencies the phase angle

tends to 180°. For low values of damping there is a steep change in the phase

angle near resonance. This decreases with the increase in the damping. The

sharper the change in the phase angle the sharper is the peak in the

frequency response plot.

The amplitude at resonance is given by equation 14

The frequency at which maximum amplitude occurs is obtained by

differentiating the magnification factor equation with respect to frequency

ratio and equating it to zero.

Also no maxima will occur for or

2. Rotating and Reciprocating Unbalance

Machines like electric motors, pumps, fans, clothes dryers, compressors

have rotating elements with unbalanced mass. This generates centrifugal

type harmonic excitation on the machine.

The final unbalance is measured in terms of an equivalent mass mo rotating

with its c.g. at a distance e from the axis of rotation. The centrifugal force is

proportional to the square of frequency of rotation. It varies with the speed

142ξXX

FX cω

str

or

−−−=

=

152ξ1ω

ω2

n

p−−−−=

⟩

2

1ξ 707.0⟩ξ


Dept. Of ME, ACE 8

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

of rotation and is different from the harmonic excitation in which the

maximum force is independent of the frequency.

Let mo = Unbalanced mass

e = eccentricity of the unbalanced mass

M= Total mass of machine including

unbalanced mo

mo makes an angle ωt with ref. axis.

Moeω2 is the centrifugal force that acts

radially outwards.

Equation of motion is

The solution of following equation 2 is given by

( ) ( )

1t sinωeωmkxxcxM

dt

dxckxesinωsx

dt

dm

dt

xdmM

2o

2

2

o2

2

o

−−−=++

−−=++−

&&&

2t sinωFkxxcxm o −−−=++ &&&

[ ]( )

( )[ ] ( )[ ]2n

22

n

st

2n

2tζω

2

ωω2ξωω1

φωtSinX

φtωξ1SineAx n

+−

−+

+−= −


9Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Compare eq. 1 with eq.2 the only change is Fo is replaced by mo eω2

The transient part of the solution remains the same. The only change is in the

steady state part of the solution.

Therefore the steady state solution of eq.1 can be written as

The above equation reduces to dimensionless form as

The phase angle equation and its plot remains the same as shown below

( )

222

2o

k

cω

k

Mω1

keωmwhere...X

φωtXsinx

+

−

=

−=

2

n

22

n

2

n

2o

ω

ω2ξ

ω

ω1

ω

ω

M

eωm

X

+

−

=

−

= −

2

n

n1

ω

ω1

ω

ω2ξ

tanφ


10Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Frequency response and phase angle plots (Unbalance)

At low speeds is small, hence all response curves start

from zero.

At resonance ω / ωn = 1, therefore

And amplitude is limited due to damping present in the system. Under these

conditions the motion of main mass (M-mo) lags that of the mass mo by 90°.

When ω / ωn is very large the ratio X/(moe/M) tends to unity and the main

mass (M-mo) has an amplitude of X= moe/M . This motion is 180° out of

phase with the exciting force. That is when unbalanced mass moves up, the

main mass moves down and vice versa.

Problem 1

A counter rotating eccentric weight exciter is used to produce the forced

oscillation of a spring-supported mass as shown in Fig. By varying the speed

2222ξξξξ1111

MMMMeeeemmmm

XXXXoooo =


11Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

of rotation, a resonant amplitude of 0.60 cm was recorded. When the speed

of rotation was increase considerably beyond the resonant frequency, the

amplitude appeared to approach a fixed value of 0.08 cm. Determine the

damping factor of the system.

Dividing one by the other

(Answer)

Problem 2

A system of beam supports a mass of 1200 kg. The motor has an unbalanced

mass of 1 kg located at 6 cm radius. It is known that the resonance occurs at

2210 rpm. What amplitude of vibration can be expected at the motors

operating speed of 1440 rpm if the damping factor is assumed to be less than

0.1

Solution:

Given: M = 1200 kg, mo= 1 kg, eccentricity = e = 0.06m, Resonance at 2210

rpm, Operating speed = 1440 rpm, ξ = 0.1, X = ?.

0.6cm0.6cm0.6cm0.6cm2222ξξξξMMMMeeeemmmm

XXXXoooo

==

0.08cm0.08cm0.08cm0.08cmMMMMeeeemmmmXXXX oooo ==

0.06670.06670.06670.0667ξξξξ =

ssss / / / /radradradrad 231.43231.43231.43231.4360606060

NNNN 2222ππππωωωωnnnn == s150.79rad/60

N 2πω

op==

0.652rω

ω

n

==

2222nnnn22222222nnnn nnnn2222ooooωωωωωωωω2222ξξξξωωωω

ωωωω1111

ωωωωωωωω

MMMMeeeeωωωωmmmmXXXX

+

−

=

2


12Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Substituting the appropriate values in the above eq.

X = 0.036 mm (Answer)

However, if ξ is made zero, the amplitude X = 0.037 mm (Answer)

This means if the damping is less than 0.1, the amplitude of vibration will be

between 0.036 mm and 0.037 mm. (Answer)

Problem 3

An eccentric mass exciter is used to determine the vibratory characteristics

of a structure of mass 200 kg. At a speed of 1000 rpm a stroboscope showed

the eccentric mass to be at the bottom position at the instant the structure

was moving downward through its static equilibrium position and the

corresponding amplitude was 20 mm. If the unbalance of the eccentric is

0.05 kg-m, determine, (a) un damped natural frequency of the system (b) the

damping factor of the structure (c) the angular position of the eccentric at

1300 rpm at the instant when the structure is moving downward through its

equilibrium position.

Solution:

Given: M = 200 kg, Amplitude at 1000 rpm = 20 mm, moe = 0.05 kg-m

At 1000 rpm the eccentric mass is at the bottom when the structure was

moving downward – This means a there is phase lag of 90° (i.e., at

resonance). At resonance ω = ωn.

(Answer) (Answer)

0.00625ξ

2ξ

1

M

em

X

s104.72rad/60

N 2πωω

o

n

=

=

===

oooo 2222nnnnnnnn1111176.89176.89176.89176.89φφφφ

ωωωωωωωω1111ωωωωωωωω2222ξξξξ

tantantantan

=

−

= −φφφφ


13Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Problem 4

A 40 kg machine is supported by four springs each of stiffness 250 N/m.

The rotor is unbalanced such that the unbalance effect is equivalent to a

mass of 5 kg located at 50mm from the axis of rotation. Find the amplitude

of vibration when the rotor rotates at 1000 rpm and 60 rpm. Assume

damping coefficient to be 0.15

Solution:

Given: M = 40 kg, mo= 5 kg, e = .05 m, ξ = 0.15, N = 1000 rpm and 60 rpm.

When N = 1000 rpm

X at 1000 rpm = 6.26 mm (Solution)

When N = 60 rpm

Using the same eq. X at 60 rpm = 14.29 mm (Solution)

Problem 5

A vertical single stage air compressor having a mass of 500 kg is mounted

on springs having a stiffness of 1.96×105 N/m and a damping coefficient of

0.2. The rotating parts are completely balanced and the equivalent

reciprocating parts have a mass of 20 kg. The stroke is 0.2 m. Determine the

5rad/sM

kω

s104.67rad/60

N 2πω

n ==

==

20.934ω

ω

n

=


ωωωω1111

ωωωωωωωω


+

−

=

2

1.256ω

ω

6.28rad/s60

N 2πω

n

=

==

Mechanical Vibrations [10Me72]

14Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

dynamic amplitude of vertical motion and the phase difference between the

motion and excitation force if the compressor is operated at 200 rpm.

Solution

Given: M = 500 kg, k = 1.96×105 N/m, ξ = 0.2, mo = 20 kg, stroke = 0.2 m, N

= 200 rpm, X = ?.

Stroke = 0.2 m, i.e. eccentricity e = stroke/2 = 0.1 m

Using the equations X = 10.2 mm and φ = 105.9° (Solution)

(a)

(Solution)

(b)

X = 0.15 cm and φ = 169° (Solution)

(c)

(d)

0.175715.5

125

c

cξ

s/m715.54N2064002kM2c

s/m125N4500c

c

c

===

−=×==

−==


ωωωω1111

ωωωωωωωω


+

−

=

2

−

= − 2222nnnnnnnn1111ωωωωωωωω1111ωωωωωωωω2222ξξξξ

tantantantanφφφφ

0.357cmM 2ξ

emX

170.74rpmπ2

ω60N

17.88rad/s20

6400

M

kω

or

nr

n

==

=×

×=

===

( ) ( ) 12.4NkXX cωF

9.6N100.156400kX

7.85N100.1560

400π2125X cω

22

2

2

=+=

=××=

=××××

×=

−

−


15Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Conclusions on rotating and reciprocating unbalance • Unbalance in machines cannot be made zero. Even small

unbalanced mass can produce high centrifugal force. This

depends on the speed of operation.

• Steady state amplitude is determined for a machine subjected

unbalanced force excitation.

• For reciprocating machines, the eccentricity can be taken as

half the crank radius.

• Frequency response plot starts from zero at frequency ratio

zero and tends to end at unity at very high frequency ratios.

2. Response of a damped system under the harmonic

motion of the base In many cases the excitation of the system is through the support or the base

instead of being applied through the mass. In such cases the support will be

considered to be excited by a regular sinusoidal motion.

Example of such base excitation is an automobile suspension system excited

by a road surface, the suspension system can be modeled by a linear spring

in parallel with a viscous damper. such model is depicted in Figure 1.

There are two cases: (a) Absolute Amplitude of mass m

(b) Relative amplitude of mass m

(a) Absolute Amplitude of mass m

m

Y=Ysinωt

K C

X


16Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

It is assumed that the base moves harmonically, that is

where Y denotes the amplitude of the base motion and ω represents the

frequency of the base excitation

Substituting Eq.2 in Eq. 1

The above equation can be expressed as

where Y[k2 + (c ω )

2]

1/2 is the amplitude of excitation force.

Examination of equation 3 reveals that it is identical to an Equation

developed during derivation for M.F. The solution is:

( )φφφφωtωtωtωtXsinXsinXsinXsinxxxx −=

=++ tFkxxcxm o ωsin&&&


17Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Therefore the steady state amplitude and phase angle to eq. 3 is

The above equations can be written in dimensionless form

as follows

( )[ ] ( )[ ]2222nnnn22222222nnnn ooooωωωωωωωω2222ξξξξωωωωωωωω1111

kkkkFFFFXXXX+−

=

( )φφφφ----ααααωtωtωtωtXsinXsinXsinXsinxxxx +=

2222nnnn22222222nnnn2222nnnnωωωωωωωω2222ξξξξωωωω

ωωωω1111

ωωωωωωωω2222ξξξξ1111

YYYYXXXX

+

−

+

=

−

= − 2222nnnnnnnn1111ωωωωωωωω1111ωωωωωωωω2222ξξξξ

tantantantanφφφφ


18Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

The motion of the mass m lags that of the support by an

angle (φ – α) as shown by equation 6.

Equation 5 which gives the ratio of (X/Y) is also known as motion

transmissibility or displacement transmissibility.

Fig. gives the frequency response curve for motion transmissibility.

−

−

=− −− nnnn11112222nnnnnnnn1111ωωωωωωωω2222ξξξξtantantantan


tantantantanααααφφφφ


19Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

1. For low frequency ratios the system moves as a rigid body and X/Y

�1.

2. At resonance the amplitudes are large

3. For very high frequency ratios the body is almost stationary (X/Y � 0)

It will seen later that the same response curve is also used for Force

Transmissibility.

(b) Relative Amplitude of mass m

Here amplitude of mass m relative to the base motion is considered. The

equations are basically made use in the

Seismic instruments. If z represents the relative motion of the mass w.r.t.

support,

Substituting the value of x in eq.

The above equation is similar to the equation developed for rotating and

reciprocating unbalances. Thus the relative steady state amplitude can

written as

zyx

yxz

+=

−=

tY ωsin2mωkzzczm =++ &&&

2222nnnn22222222nnnn nnnnωωωωωωωω2222ξξξξωωωω

ωωωω1111

ωωωωωωωω

YYYYZZZZ

+

−

=

2 ( )( )

=

− 21

2

n

n

ωω

ωωξφ 1-tan


20Dept. OF ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Thus eq. 7 and eq.8 are similar to the one developed during the study

of rotating an reciprocating unbalances. Frequency and phase response plots

will also remain same.

Problem 1

The support of a spring mass system is vibrating with an amplitude of 5 mm

and a frequency of 1150 cpm. If the mass is 0.9 kg and the stiffness of

springs is 1960 N/m, Determine the amplitude of vibration of mass. What

amplitude will result if a damping factor of 0.2 is included in the system.

Solution:

Given: Y = 5 mm, f = 1150cpm, m =0.9 kg, k = 1960 N/m, X = ?

ξ = 0.2, then X = ?

When ξ = 0, X = 0.886 mm (Solution)

When ξ = 0.2, X = 1.25 mm (Solution)

46.67rad/s0.9

1960

m

kωn ===

s120.43rad/1150/60π2fπ2ω =××=××=

2.58rω

ω

n

==


ωωωω1111


YYYYXXXX

+

−

+

=


21Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Observe even when damping has increased the amplitude has not decreased

but it has increased.

Problem 2

The springs of an automobile trailer are compressed 0.1 m under its own

weight. Find the critical speed when the trailer is travelling over a road with

a profile approximated by a sine wave of amplitude 0.08 m and a

wavelength of 14 m. What will be the amplitude of vibration at 60 km/hr.

Solution:

Given: Static deflection = dst = 0.1 m, Y = 0.08 m, γ = 14 m, Critical Speed

= ?, X60 = ?.

Critical speed can be found by finding natural frequency.

Corresponding V = 22.06 m/s = 79.4 km/hr

Amplitude X at 60 km/hr

V60 = 16.67 m/s

When damping is zero X at 60km/hr = 0.186 m (Solution)

1.576cps2π

ωf

9.9rad/s.1

9.81

d

g

m

k

nn

st

n

==

====ω

22.06m/s1.57614fwavelengthV n =×=×=

.756rω

ω

n

==

7.48rad/sf 2πω

1.19cps14

16.67wavelengthvelocityf

==

===


ωωωω1111


YYYYXXXX

+

−

+

=


22Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Problem 3

A heavy machine of 3000 N, is supported on a resilient foundation. The

static deflection of the foundation due to the weight of the machine is found

to be 7.5 cm. It is observed that the machine vibrates with an amplitude of 1

cm when the base of the machine is subjected to harmonic oscillations at the

undamped natural frequency of the system with an amplitude of 0.25 cm.

Find (a) the damping constant of the foundation (b) the dynamic force

amplitude on the base (c) the amplitude of the displacement of the machine

relative to the base.

Solution

Given: mg = 3000 N, Static deflection = dst = 7.5 cm, X = 1 cm,

Y = 0.25 cm, ω = ωn, ξ = ?, Fbase = ?, Z = ?

(a)

Solving for ξ = 0.1291

(c)

Note: Z = 0.00968 m, X = 0.001m, Y = 0.0025 m, Z is not equal to

X-Y due phase difference between x, y, z.

Using the above eq. when ω = ωn, the

( )( )2

2

n

2ξ

2ξ14

0.0025

0.010

Y

X

ωω

+===

=

s/m903.05Nkm2ξξcc c −=×==

2

n

22

n

2

n

ω

ω2ξ

ω

ω1

ω

ω

Y

Z

+

−

=


23Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

relative amplitude is Z = 0.00968 m (Solution)

(b)

(Solution)

Problem 4

The time of free vibration of a mass hung from the end of a helical spring is

0.8 s. When the mass is stationary, the upper end is made to move upwards

with displacement y mm given by y = 18 sin 2πt, where t is time in seconds

measured from the beginning of the motion. Neglecting the mass of spring

and damping effect, determine the vertical distance through which the mass

is moved in the first 0.3 seconds.

Solution:

Given: Time period of free vibration = 0.8 s., y = 18 sin 2πt,

ξ = 0, x at the end of first 0.3 s. = ?

where, Y = 18 mm, and ω = 2π rad/s.

The complete solution consists of Complementary function and Particular

integral part.

where,

( ) ( )

388.5NF

3.65rad/sωω

kZZ cωF

d

n

22

d

=

==

+=

t kYsinωkxxm

kykxxm

=+

=+

&&

&&

( )φαωtXsinx

tBsinωtAcosωx

xxx

p

nnc

pc

−+=

+=

+=

2

nω

ω1

YX

−

=

( )

( ) 1ω

ωif,180αφ

1ω

ωif0,αφ

n

o

n

⟩

=−

⟨=−

0αφand

0.8ω

ω

2πω

0.8

2π

τ

2πω

n

n

=−

=

=

==


24Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

The complete solution is given by

Substituting the initial conditions in the above eq. constants A and B can be

obtained

0αφand

0.8ω

ω

2πω

0.8

2π

τ

2πω

n

n

=−

=

=

==

t sinω

ω

ω1

YxHence,

2

n

p

−

=

t sinω

ω

ω1

YtBsinωtAcosωx

2

n

nn

−

++=

−

−=

=

2

n

n

ω

ω1

ω

ωY

Band

0Agives

0tat0,x

0tat0;x

==

==

&


25Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Thus the complete solution after substituting the values of A and B

when t = 0.3 s, the value of x from the above eq.

is x = 19.2 mm (Solution)

Conclusions on Response of a damped system under the

harmonic motion of the base

• Review of forced vibration ( constant excitation force and rotating

and reciprocating unbalance ).

• Steady state amplitude and phase angle is determined when the base

is excited sinusoidally. Derivations were made for both absolute and

relative amplitudes of the mass.

4. Vibration Isolation and Force Transmissibility

• Vibrations developed in machines should be isolated from the

foundation so that adjoining structure is not set into vibrations. (Force

isolation)

• Delicate instruments must be isolated from their supports which may

be subjected to vibrations. (Motion Isolation)

• To achieve the above objectives it is necessary to choose proper

isolation materials which may be cork, rubber, metallic springs or

other suitable materials.

−

−

= tsinωω

ωt sinω

ω

ω1

Yx n

n2

n


26Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

• Thus in this study, derivations are made for force isolation and

motion isolation which give insight into response of the system and

help in choosing proper isolation materials.

Transmissibility is defined as the ratio of the force transmitted to the

foundation to that impressed upon the system.

The force transmitted to the base is the sum of the spring force and damper

force. Hence, the amplitude of the transmitted force is:

1sin −−−=++ tFkxxcxm o ω&&&

( )φω −= tXSinxp


27Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Substituting the value of X from Eq. 2 in Eq. 3 yields


28Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Hence, the force transmission ratio or transmissibility, TR is given by

Eq. 6 gives phase angle relationship between Impressed force and

transmitted force.

−

−





29Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

• Curves start from unity value of transmissibility pass through

unit value of transmissibility at (ω / ωn ) = and after that they

Tend to zero as (ω / ωn ) → .


30Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

• Response plot can be divided into three regions depending on

the control of spring, damper and mass.

• When (ω / ωn ) is large, it is mass controlled region. Damping

in this region deteriorates the performance of machine.

• When (ω / ωn ) is very small, it is spring controlled region.

• When (ω / ωn ) ranges from 0.6 to , it is damping controlled

region.

• For effective isolation (ω / ωn ) should be large. It means it will

have spring with low stiffness (hence large static deflections).

Motion Transmissibility

Motion transmissibility is the ratio of steady state amplitude of mass m (X)

to the steady amplitude (Y) of the supporting base.

The equations are same as that of force transmissibility. Thus the frequency

response and phase angle plots are also the same.

Problem 1

A 75 kg machine is mounted on springs of stiffness k=11.76×106 N/m with a

damping factor of 0.2. A 2 kg piston within the machine has a reciprocating

motion with a stroke of 0.08 m and a speed of 3000 cpm. Assuming the

motion of the piston to be harmonic, determine the amplitude of vibration of

machine and the vibratory force transmitted to the foundation.

Solution:

Given: M = 75 kg, ξ = 0.2, mo = 2 kg, stroke = 0.08 m, N = 3000 cpm, X = ?,

FT = ?.


ωωωω1111


YYYYXXXX

+

−

+

=

−

−





31Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Using the above eq. X = 1.25 mm (Solution)

Using the above eq. FT = 2078 N (Solution)

rad/srad/srad/srad/s12512512512575757575

1010101011.7611.7611.7611.76mmmmkkkkωωωω

5555nnnn =×

==

rad/srad/srad/srad/s314.16314.16314.16314.166060606030003000300030002222ππππωωωω =

×=

mmmm0.040.040.040.042222

0.080.080.080.08eeee2.512.512.512.51ωωωωωωωωnnnn ===

2222nnnn22222222nnnn nnnnooooωωωωωωωω2222ξξξξωωωω

ωωωω1111

ωωωωωωωω

MMMMeeeemmmm

XXXX

+

−

=

2

NNNN7900790079007900eeeeωωωωmmmmFFFF 2222oooooooo ==


32Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Solution by Complex Algebra Here steady state solution is obtained for a system subjected to constant

excitation force using complex algebra

Let the harmonic forcing function be represented by in complex form as

Now the equation of motion becomes

Actual excitation is given by real part of F(t), the response will also

be given by real part of x(t), where x(t) is a complex quantity

satisfying the differential equation 1.

Assuming a particular solution as

Substituting eq. 3 in eq. 2

Multiplying the numerator and denominator of eq.4 by

And separating real and imaginary parts

1sin −−−=++ tFkxxcxm o ω&&&

( ) ttttiiiiooooeeeeFFFFttttFFFF ω=

2−−−=++ ttttiiiiooooeeeeFFFFkxkxkxkxxxxxccccxxxxmmmm ω&&&

( ) 3−−−= tiωXeXeXeXettttxxxx pppp( )

4444icicicicωωωωmmmmωωωωkkkk

FFFFXXXX 2222oooo −−−+−

=

( )[ ]icicicicωωωωmmmmωωωωkkkk 2222 −−


33Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Multiplying and dividing the above expression by

Thus the steady state solution becomes

The real part of the above equation 8 gives the steady state

response

Dividing the numerator and denominator of this eq. by k

( ) ( )5555

ωωωωccccmmmmωωωωkkkkccccωωωωiiii

ωωωωccccmmmmωωωωkkkkmmmmωωωωkkkkFFFFXXXX 22222222222222222222222222222222 2222oooo −−−

+−−

+−

−=

( ) 2222222222222222 ωωωωccccmmmmωωωωkkkk +−

( ) ( ) ( )

+−

−

+−

−

+−

= 22222222222222222222222222222222 22222222222222222222oooo ωωωωccccmmmmωωωωkkkkccccωωωωiiii

ωωωωccccmmmmωωωωkkkkmmmmωωωωkkkk

ωωωωccccmmmmωωωωkkkkFFFFXXXX

( )6666eeee

ωωωωccccmmmmωωωωkkkkFFFFXXXX iiiiφφφφ2222222222222222oooo −−−

+−

= −

7777mmmmωωωωkkkkccccωωωωtantantantanφφφφwhere,where,where,where, 22221111 −−−

−

= −

( )( ) 8888eeee

ωωωωccccmmmmωωωωkkkkFFFFxxxx φφφφ----ttttiiii2222222222222222oooopppp −−−

+−

= ω

( )

( )9999

ωωωωccccmmmmωωωωkkkkφφφφωtωtωtωtcoscoscoscosFFFFxxxx 2222222222222222oooopppp −−−

+−

−=


Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Problem 1

A torsional system consists of a disc of mass moment of inertia J = 10 kg-

m2, a torsional damper of damping coefficient c = 300 N-m-s/rad, and steel

shaft of diameter 4 cm and length 1 m (fixed at one end and attached to the

disc at the other end). A steady angular oscillation of amplitude 2o is

observed when a harmonic torque of magnitude 1000 N-m is applied to the

disc. (a) Find the frequency of the applied torque, (b) find the maximum

torque transmitted to the support. Assume modulus of rigidity for the steel

rod to be 0.83×1011 N/m2.

Solution:

Given: J = 10 kg-m2, c = 300 N-m-s/rad, l = 1 m, d = 4 cm,

Θ = 2o, To = 1000 N-m, G = 0.83×1011 N/m

2, ω = ?, Tmax=?

Stiffness of the shaft is given by

( )[ ] ( )[ ]2222nnnnωωωωωωωω2222ξξξξ22222222nnnnωωωωωωωω1111 kkkkooooFFFFXXXX+−

=

radradradradmmmmNNNN20860208602086020860

3232323211110.040.040.040.04ππππ101010100.830.830.830.83

LLLLGIGIGIGI

θθθθTTTTkkkk

444411111111pppptttt −=×

×××===

rad/srad/srad/srad/s45.6745.6745.6745.6710101010

20860208602086020860JJJJkkkkωωωω ttttnnnn ===

( )[ ] ( )[ ]2222nnnnωωωωωωωω2222ξξξξ22222222nnnnωωωωωωωω1111 kkkkooooFFFFXXXX+−

=

( )[ ] ( )[ ]2222nnnnωωωωωωωω2222ξξξξ22222222nnnnωωωωωωωω1111 ttttkkkkooooTTTTθθθθ+−

=

0.035rad0.035rad0.035rad0.035rad180180180180

ππππ2222radiansradiansradiansradiansininininθθθθ =×

=

1.3691.3691.3691.3690.0350.0350.0350.03520860208602086020860

1000100010001000θθθθkkkk

TTTTttttoooo =×

=

34Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

( ) 2222nnnn222222222222ωωωωωωωωrrrrwhere,where,where,where,1.8761.8761.8761.876rrrr4444ξξξξrrrr1111 ==+−

0.3280.3280.3280.32845.6745.6745.6745.67101010102222300300300300

2J2J2J2Jωωωωccccξξξξ nnnn =

××==

00000.8760.8760.8760.8761.569r1.569r1.569r1.569rrrrr 2222 =−−

1.4161.4161.4161.416ωωωωωωωωnnnn = rad/srad/srad/srad/s64.6864.6864.6864.68ωωωω =

2222nnnn22222222nnnn2222nnnnooootrtrtrtrωωωωωωωω2222ξξξξωωωω

ωωωω1111


TTTTTTTT

+

−

+

= mmmm1000N1000N1000N1000NTTTTtrtrtrtr −=


35Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Vibration Measuring Instruments

Need for vibration measuring instruments

1.To detect shifts in the natural frequencies – could indicate a possible

failure or need for maintenance

2.To select operational speeds to avoid resonance

3.Theoretically estimated values may be different from the actual

values due to assumptions.

4.To design active vibration isolation systems

5.To validate the approximate model

6.To identify the system in terms of mass, stiffness and damper.

When a transducer is used in conjunction with another device to measure

vibrations it is called vibration pickup.

Commonly used vibration pickups are seismic instruments.

If the seismic instrument gives displacement of the vibrating body – It is

known as VIBROMETER.

If the seismic instrument gives velocity of the vibrating body – It is known

as VELOMETER.

If the seismic instrument gives acceleration of the vibrating body – It is

known as ACCELEROMETER.

SEISMIC INSTRUMENT


UNIT - 5

VIBRATION MEASURING INSTRUMENTS

1Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Vibrating body is assumed to have a Harmonic Motion given by

Eq. of motion for the mass m can be written as

Defining relative displacement as

Substituting this value of x in equation 2

Steady state solution of eq. 4 is given by

1−−−= ttttYsinYsinYsinYsinyyyy ω

( ) ( ) 22220000yyyyxxxxkkkkyyyyxxxxccccxxxxmmmm −−−=−+−+ &&&&

zzzzyyyyxxxxor,or,or,or,3333yyyyxxxxzzzz

+=

−−−=

4444------------ttttYsinYsinYsinYsinmmmmωωωωkzkzkzkzzzzzcccczzzzmmmmyyyymmmmkzkzkzkzzzzzcccczzzzmmmm 2222 ω=++

−=++

&&&

&&&&&

( )φφφφωtωtωtωtZsinZsinZsinZsinzzzz −=

[ ] ( )

−=

−

=

=

+−

=

+

−

=

−−

2

1

2

n

n1

n222

2

2

n

22

n

2

n

r1

r2ξtan

ω

ω1

ω

ω2ξ

tanφ

ω

ωrwhere

r2ξr1

r

ω

ω2ξ

ω

ω1

ω

ω

Y

Z


2Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

The frequency response plot is shown in the fig. The type of instrument is

determined by the useful range of frequency

The phase angle plot shown below indicates the phase lag of the seismic

mass with respect to vibrating base of machine

0 1 2 3 4

0

1

2

3

4

Accele

rom

ete

r

Vib

rom

ete

r

an

d

Ve

lom

ete

r

ζ = 0 .0

ζ = 0 .1

ζ = 0 .2

ζ = 0 .3

ζ = 0 .4

ζ = 0 .5

ζ = 0 .7 0 7

ζ = 1

Z/Y

ω / ωn ( r )

0 1 2 3 4 5

0

2 0

4 0

6 0

8 0

1 0 0

1 2 0

1 4 0

1 6 0

1 8 0

ξ = 1 .0

ξ = 0 .7 0 7ξ = 0 .5

ξ = 0 .2

ξ = 0 .1

ξ = 0

Pha

se

an

gle

, φ

ω /ωr ( r )


33Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Vibrometers Used for measurement of displacement of vibrating body.

It means when Z / Y ~ 1, the observed reading on the scale directly gives the

displacement of the vibrating body. For this to happen r = ω / ωn ≥ 3.

Thus when r = ω / ωn ≥ 3, Z / Y ~ 1, but there is a phase lag. Z lags behind Y

by an angle φ or by time lag of t = φ / ω. This time lag is not important if

the input consists of single harmonic component.

Thus for vibrometers the range of frequency lies on the right hand side of

frequency response plot. It can also be seen from the plot a better

approximation can be obtained if ξ is less than 0.707.

Also for fixed value of ω, and for the value of ωn

must be small. It means mass must be very large and stiffness must be small.

This results in bulky instrument which may not be desirable in many

applications.

Accelerometer

Accelerometer measures the acceleration of a vibrating body.

They are widely used for measuring acceleration of vibrating bodies and

earthquakes. Integration of acceleration record provides displacement and

velocity.

( )φφφφωtωtωtωtZsinZsinZsinZsinzzzz −=

[ ] ( ) Y

Z1

r2ξr1

r222

2

==

+−ttttYsinYsinYsinYsinyyyy ω=

[ ] ( )

[ ] ( ) 2222nnnn22222222 222222222222 222222222222 2222ωωωω

YYYYωωωωZZZZ Y,Y,Y,Y,rrrrZZZZthenthenthenthen

1111rrrr2222ξξξξrrrr1111

1111rrrr2222ξξξξrrrr1111

rrrrYYYYZZZZ

==

=

+−

+−

=


4Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

The expression is equal to the acceleration amplitude of the

Vibrating body. The amplitude recorded by the instrument is proportional to

the acceleration of vibrating body since ωn is constant for the instrument.

For an accelerometer the ratio << 1 and the range of operation lies

on the extreme left hand side of the frequency

response curve.

Since << 1, the natural frequency must be high. That is mass

must be as low as possible. Thus the instrument is small in size and is

preferred in vibration measurement.

Frequency ratio, r should be between 0 and 0.6

Damping ratio, ξ , should be less than 0.707

Problem 1 The static deflection of the vibrometer mass is 20 mm. The

instrument when attached to a machine vibrating with a frequency of 125

cpm records a relative amplitude of 0.03 cm. Determine (a) the amplitude of

vibration (b) the maximum velocity of vibration (c) the maximum

acceleration.

Solution:

Given: dst = 20 mm, f = 125 cpm, Z = 0.03 cm, Y = ?,

Vmax = ?, a = ?, ξ = 0.0

YYYYωωωω 2222

0.590.590.590.59ωωωωωωωωrrrr

rad/srad/srad/srad/s13.0913.0913.0913.0960606060NNNN2222ππππωωωωrad/srad/srad/srad/s22.1522.1522.1522.150.020.020.020.02

9.819.819.819.81ddddggggωωωω nnnn ststststnnnn==

=====

[ ] ( )0000ξξξξasasasasrrrr1111

rrrrrrrr2222ξξξξrrrr1111

rrrrYYYYZZZZ 22222222222222222222 2222

=−

=

+−

=

2222222222222222

cm/scm/scm/scm/s95.9595.9595.9595.95YYYYωωωωononononAcceleratiAcceleratiAcceleratiAcceleratiMaxMaxMaxMaxcm/scm/scm/scm/s0.7330.7330.7330.733YYYYωωωωVelocityVelocityVelocityVelocityMaxMaxMaxMax

0.56cm0.56cm0.56cm0.56cm0.5340.5340.5340.5340.030.030.030.03YYYY

0.5340.5340.5340.5340.590.590.590.5911110.590.590.590.59

YYYYZZZZ

==

==

==

=−

=


5Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Critical speed of shaft

• There are many engineering applications in which shafts carry disks (

turbines, compressors, electric motors, pumps etc.,)

• These shafts vibrate violently in transverse directions at certain speed

of operation known as critical speed of shaft.

• Among the various causes that create critical speeds, the mass

unbalance is the most important.

• The unbalance cannot be made zero. There is always some unbalance

left in rotors or disks.

• Whirling is defined as the rotation of the plane containing the bent

shaft about the bearing axis.

• The whirling of the shaft can take place in the same or opposite

direction as that of the rotation of the shaft.

• The whirling speed may or may not be equal to the rotation speed.

Critical speed of a light shaft having a single disc – without

damping


6Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Consider a light shaft carrying a single disc at the centre in deflected

position. ‘S’ is the geometric centre through which centre line of shaft

passes. ‘G’ is centre of gravity of disc. ‘O’ intersection of bearing centre

line with the disc ‘e’ is distance between c.g. ‘G’ and the geometric centre

‘S’. ‘d’ is displacement of the geometric centre ‘S’ from the un deflected

position ‘O’. ‘k’ is the stiffness of the shaft in the lateral direction.

The forces acting on the disc are

- Centrifugal force ‘mω2(d+e)’ acts radially outwards at ‘G’

- Restoring force ‘kd’ acts radially inwards at ‘S’

- For equilibrium the two forces must be equal and act along the

same line

• Deflection ‘d’ tends to infinity when ω = ωn

• ‘d’ is positive below the critical speed, the disc rotates with

heavy side outside when ω < ωn

• ‘d’ is negative above the critical speed, the disc rotates with

light side outside when ω > ωn

• When ω >> ωn, ‘d → - e’ , which means point ‘G’ approaches

point ‘O’ and the disc rotates about its centre of gravity.

( )

( )2

r1

er

ω

ω1

eω

ω

mωk

emωd

1kredmω

2

2

2

n

2

n

2

2

2

−−−−

=

−

=−

=

−−−=+

Mechanical Vibrations 10ME72[

7Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Problem 1

A rotor having a mass of 5 kg is mounted midway on a 1 cm shaft supported

at the ends by two bearings. The bearing span is 40 cm. Because of certain

manufacturing inaccuracies, the centre of gravity of the disc is 0.02 mm

away from the geometric centre of the rotor. If the system rotates at 3000

rpm find the amplitude of steady state vibrations and the dynamic force

transmitted to the bearings. Assume the rotor to be simply supported. Take E

= 1.96×1011

N/m2.

Solution:

Given: m = 5 kg, d = 1 cm, l = 40 cm, e = 0.02 mm,

N = 3000 rpm, E = 1.96×1011

N/m2 , d = ?, Simply supported

In the above eq. e and ω are known, ωn has to be found out in order to find

d. To find ωn, stiffness has to be determined.

For a simply supported shaft the deflection at the mid point is given by the

following equation.

−

=2

n

2

n

ω

ω1

ω

ω

e

d

N/m720000.464

0.01π101.9648

l

48EI

δ

mgk,

48EI

mglδ

3

411

3

3

=×

××××====


8Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

- Sign implies displacement is out of phase with centrifugal force

Critical speed of light shaft having a single disc – with damping

O’ is intersection of bearing centre line with the disc

’S’ is geometric centre of the disk

2.168ω

ωrad/s,120

m

kω rad/s,314.16

60

Nπ2ω

n

n ====××

=

( )0.023mm0.021.1558d1.1558,

2.1681

2.168

ω

ω1

ω

ω

e

d2

2

2

n

2

n−=×−=−=

−=

−

=

N0.828bearingeachonload

N1.656kdbearingonDynamic

=

==


9Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

‘G’ is centre of gravity of disc

The forces acting on the disc are

1. Centrifugal force at G along OG produced

2. Restoring force kd at S along SO

3. Damping force cωd at S in a direction opposite to velocity at S

4. The points O,S and G no longer lie on straight line

Let,

OG = a,SG = e,OS = d,∟GOS = α ∟GSA = φ

From the geometry

Eliminating a and α from equations 3 and 4 with the help of equations 1 and

2

2ecosφdacosα

1esinφasinα

−−−+=

−−−=

40asinαmωdcω

0,Y

30amωkd

0,X

2

2

−−−=+−

=

−−−=+−

=

∑

∑

( )

( ) 60esinφmωdcω

50ecosφdmωkd2

2

−−−=+−

−−−=++−


10Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Squaring and adding the equations 7 and 8

From the curves or the equation it can be seen that (Discussions of speeds

above and below critical speed)

(a) Φ ~ 0 when ω << ωn (Heavy side out)

(b) 0 < φ < 90° when ω < ωn (Heavy side out)

8emω

dcωsinφeq.6From

7emω

dmωkdcosφeq.5From

2

2

2

−−−=

−−−−

=

( ) ( )222

2

cωmωk

mω

e

d

+−

=

( ) ( )9

r2ξr1

r

ω

ω2ξ

ω

ω1

ω

ω

e

d222

2

2

n

2

n

2

n−−−

+−

=

+

−

=

10r1

r2ξtan

ω

ω1

ω

ω2ξ

tanφ2

1

2

n

n1 −−−−

=

−

= −−


11Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

(c) Φ = 90° when ω = ωn

(d) 90° < Φ < 180° when ω > ωn (Light side out)

Mechanical Vibratiosn [10ME72]

12Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

(e) Φ ~ 180° when ω >> ωn, (Light side out, disc rotates about its

centre of gravity)

Critical speed of shaft may be placed above or below the operating speed.

1. If the unit is to operate at high speeds, that do not vary widely, the critical

speed may be below the operating speed, and the shaft is then said to be

flexible.

2 In bringing the shaft up to the operating speed, the critical speed must be

passed through. If this is done rapidly, resonance conditions do not have

chance to build up.

3. If the operating speed is low or if speeds must vary through wide ranges,

the critical speed is placed over the operating speed and the shaft is said to

be rigid or stiff.

4. Generally the running speed must at least 20% away from the critical

speed.

Problem 1

A disk of mass 4 kg is mounted midway between the bearings

Which may be assumed to be simply supported. The bearing span is 48 cm.

The steel shaft is 9 mm in diameter. The c.g. of the disc is displaced 3 mm

from the geometric centre. The equivalent viscous damping at the centre of


13Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

the disc- shaft may be taken as 49 N-s/m. If the shaft rotates at 760 rpm, find

the maximum stress in the shaft and compare it with the dead load stress in

the shaft when the shaft is horizontal. Also find the power required to drive

the shaft at this speed. Take E = 1.96 × 1011

N/m2.

Solution:

Given: m = 4kg, l = 48 cm, e = 3 mm, c = 49 N-s/m, N = 760rpm

E = 1.96 × 1011

N/m2, smax = ?. Dia = 9 mm, E = 1.96 × 10

11 N/m

2.

d = 0.017 m (solution)

The dynamic load on the bearings is equal to centrifugal force of the disc

which is equal to the vector sum of spring force and damping force.

The total maximum load on the shaft under dynamic conditions is the sum of

above load and the dead load.

The load under static conditions is

The maximum stress, due to load acting at the centre of a simply supported

shaft is

0.074km2

cξ

79.5rad/s60

2ππω

82.8rad/sm

kω

27400N/ml

48EIk

n

3

==

==

==

==

2

n

2

n

2

n

ω

ω2ξ

ω

ω1

ω

ω

e

d

+

−

=

( ) ( ) ( ) 470NcωkddcωkdF2222

dy =+=+=

( ) 509.2N9.814470Fmax =×+=

39.2N9.814Fs =×=


14Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

The total maximum stress under dynamic conditions

Maximum stress under dead load

--------------------------------------------------------------------------------------------

F10168diaπ24

64diaFl

2I

diaMs 4

4×=

×××

××=

×

×=

28

max

4

max N/m108.55F10168s ×=××=

274

max N/m106.5939.210168s ×=××=

( )

90W60

NT2πPower

m1.125NddcωTtorqueDamping

==

−===


15Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

1

Unit - 6

Vibrations of Two Degree of Freedom Systems

Dr. T. Jagadish. Professor for Post Graduation, Department of Mechanical Engineering,

Bangalore Institute of Technology, Bangalore

Introduction

A two degree of freedom system is one that requires two coordinates to completely describe

its equation of motion. These coordinates are called generalized coordinates when they are

independent of each other. Thus system with two degrees of freedom will have two equation of

motion and hence has two frequencies.

A two degree freedom system differs from a single degree of freedom system in that it has

two natural frequencies and for each of these natural frequencies there correspond a natural state of

vibration with a displacement configuration known as NORMAL MODE. Mathematical terms

related to these quantities are known as Eigen values and Eigen vectors. These are established from

the two simultaneous equation of motion of the system and posses certain dynamic properties

associated.

A system having two degrees of freedom are important in as far as they introduce to the

coupling phenomenon where the motion of any of the two independent coordinates depends also on

the motion of the other coordinate through the coupling spring and damper. The free vibration of two

degrees of freedom system at any point is a combination of two harmonics of these two natural

frequencies.

Under certain condition, during free vibrations any point in a system may execute harmonic

vibration at any of the two natural frequencies and the amplitude are related in a specific manner and

the configuration is known as NORMALMODE or PRINCIPAL MODE of vibration. Thus system

with two degrees of freedom has two normal modes of vibration corresponding two natural

frequencies.

Free vibrations of two degrees of freedom system:

Consider an un-damped system with two degrees of freedom as shown in Figure 6.1a, where

the masses are constrained to move in the direction of the spring axis and executing free vibrations.

The displacements are measured from the un-stretched positions of the springs. Let x1 and x2 be the

displacement of the masses m1 and m2 respectively at any given instant of time measured from the


UNIT - 6

SYSTEMS WITH TWO DEGREES OF FREEDOMS

Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

2

equilibrium position with x2 > x1. Then the spring forces acting on the masses are as shown in free

body diagram in Figure 6.1b

(a) (b)

Figure 6.1

Based on Newton’s second law of motion ∑ƒ =

For mass m1

for mass (2)

The solution for x1 and x2 are obtained by considering that they can have harmonic vibration under

steady state condition. Then considering the case when the mass m1 execute harmonic vibration at

frequency ω1 and the mass m2 execute harmonic vibration at frequency ω2 then we have

x1 = X1 sin ω1t, and x2 = X2 sin ω2t ----------- (3)

Where X1 and X2 are the amplitudes of vibrations of the two masses under steady state conditions.

Substituting equation (3) into equation (1) we have

- m1ω12X1 sin ω1t + (k1 + k2) X1 sin ω1t = k2 X2 sin ω2t

Therefore X1 = k2 sinω2t

X2 (k1 + k2 ) – mω12 sinω1t

Since X1 and X2 are the amplitude of two harmonic motions, their ratio must be constant and

independent of time. Therefore sinω2t / sinω1t = C a constant.

.. m x

.. m1 x1 + k1 x1 - k2x2 + k2x1 = 0

.. m1 x1 + (k1 + k2) x1 = k2x2 ------------ (1)

= ..

m1 x1 - k1 x1 + k2(x2 –x1)

.. m2 x2 + k3 x2 + k2x2 - k2x1 = 0

.. m2 x2 + (k2 + k3) x2 = k2x1 ------------ (2)

= ..

m2 x2 - k3 x2 - k2(x2 –x1)

k1x1

k2(x2 - x1)

k3x2

..

m1x1

..

m2x2

m1

m2 m

k2

k1

x1

x2

k3


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

3

Consider if C > 1. Then at time t = π/2ω1 , sinω1t will be sinω1 x π/2ω1 = sin π/2 = 1

Therefore sinω2t / sinω1t > 1 or sinω2t > 1 which is impossible. Hence C > 1 is not possible.

Similarly it can be shown that C < 1 is also not possible. Thus the only possibility is that C = 1

Hence sinω2t / sinω1t = 1 which is only possible if ω2 = ω1 = ω. Hence the two harmonic motion

have to be of the same frequency. Thus the solution of equation (1) and (2) can be

x1 = X1 sin ωt, and x2 = X2 sin ωt ----------- (4)

------------------ (5)

Substitute equation (4) and (5) into the equation (1) and (2)

- m1ω2X1 sin ωt + (k1 + k2) X1 sin ωt = k2 X2 sin ωt ----------------- (6)

- m2 ω2X2 sin ωt + (k2 + k3) X2 sin ωt = k2 X1 sin ωt. ----------------- (7)

Canceling the common term sin ωt on both the sides and re arranging the terms we have from

equation (6)

X1/X2 = k2 / (k1 + k2 – m1ω2) ---------------- (8)

X1/X2 = [(k2 + k3) – m2ω2] / k2 ----------- (9)

Thus equating equation (8) and (9) we have

X1/X2 = k2 / (k1 + k2 – m1ω2) = [(k2 + k3) – m2ω

2] / k2 -------------- (10)

Cross multiplying in equation (10) we have

(k1 + k2 – m1ω2) (k2 + k3 – m2ω

2) = k2

2 on simplification we get

m1m2 ω4 – [m1 (k2 + k3) + m2 (k1 + k2)] ω

2 + [k1k2 + k2k3 + k3k1] = 0 ---------- (11)

The above equation (11) is quadratic in ω2 and gives two values of ω

2 and therefore the two positive

values of ω correspond to the two natural frequencies ωn1 and ωn2 of the system. The above equation

is called frequency equation since the roots of the above equation give the natural frequencies of the

system.

Now considering m1 = m2 = m and k1 = k3 = k

Then the frequency equation (11) becomes

m2ω

4 – 2m (k + k2) ω

2 + (k

2 + 2kk2) = 0 ------------------ (12)

Let: ω2 = λ ∴λ

2 = ω

4, ∴ m

2 λ

2 – 2 m (k + k2) λ + (k

2 + 2kk2) = 0 ------------------ (13)

The roots of the above equation (13) are as follows: Let a = m2, b = -2 m (k + k2); c = (k

2 + 2kk2)

∴ λ1,2 = [- b ± √(b2 – 4ac)] / 2a

λ1,2 = [- (-2m) (k + k2) ± √[-2m (k+k2)]2 – 4 (m

2) (k

2 + 2kk2)]/2m

2

= [+ 2m (k +k2)] / 2m2 ± [√4m

2[(k

2 + k2

2 + 2 kk2) – (k

2 + 2kk2)]/4m

4 = (k+ k2) /m ± √(k2

2/m

2)

= (k +k2) /m ± k2/m

.. x1= - ω

2X1 sin ωt

.. x2= - ω

2X2 sin ωt


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

4

Thus λ1 = (k + k2) /m – k2 / m = k/m. Then ωn12 = K/m ∴∴∴∴ωωωωn1 = √√√√(k/m) ----------------- (14)

and ∴λ2 = (k + 2k2) /m Thus ωn22 = (k + 2k2) /m. Then ∴∴∴∴ωωωωn2 = √√√√[(k + 2k2) /m] --- (15)

ωn1 is called the first or fundamental frequency or 1st mode frequency, ωn2 is called the second or 2

nd

mode frequency. Thus the number of natural frequencies of a system is equal to the number of

degrees of freedom of system.

Modes Shapes: From equation (10) we have X1/X2 = k2/(k+k2) -mω2 = (k2 + k) - mω

2/k2 ---(16)

Substitute ωn1 = √(k/m) in any one of the above equation (16).

(X1/X2)ωn1 = k2 / (k+ k2 – m(k/m)) or ((k2 + k) – m(k/m))/k2 = k2/k2 = 1

(X1/X2)ωn1 = 1 ----------------- (17)

Similarly substituting ωn2 = √[(k + 2k2) /m] in any one of the above equation (16).

(X1/X2)ωn2 = k2 / (k + k2 – m(k+ 2k2)/m) or ((k2 + k) – m(k+ 2k2)/m))/k2 = - k2/k2 = -1

(X1/X2)ωn2 = -1 ----------------- (18)

The displacements X1 and X2 corresponding to the two natural frequency of the system can be

plotted as shown in Figure 6.2, which describe the mode in which the masses vibrate. When the

system vibrates in principal mode the masses oscillate in such a manner that they reach maximum

displacements simultaneously and pass through their equilibrium points simultaneously or all

moving parts of the system oscillate in phase with one frequency. Since the ratio X1/X2 is important

rather than the amplitudes themselves, it is customary to assign a unit value of amplitude to either X1

or X2. When this is done, the principal mode is referred as normal mode of the system.

(a) (b) (c)

Figure – 6.2

ωn1 = √(k/m) ωn2 = √[(k + 2k2) /m]

1st Mode 2

nd Mode

m1

m2

k3

k2

k1

x1

x2

x1 = 1

x2 = 1

Node

x1 = 1

x2 = -1


Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

5

It can be observed from the figure – 6.2b when the system vibrates at the first frequency, the

amplitude of two masses remain same. The motion of both the masses are in phase i.e., both the

masses move up or down together, the length of the middle spring remains constant, this spring

(coupling spring) is neither stretched nor compressed. It moves rigid bodily with both the masses and

hence totally ineffective as shown in Figure 6.3a. Even if the coupling spring is removed the two

masses will vibrate as two single degree of freedom systems with ωn = √(K/m).

When the system vibrates at the second frequency the displacement of the two masses have

the same magnitude but with opposite signs. Thus the motions of m1 and m2 are 1800 out of phase,

the midpoint of the middle spring remains stationary for all the time. Such a point which experiences

no vibratory motion is called a node, as shown in Figure 6.3b which is as if the middle of the

coupling spring is fixed

When the two masses are given equal initial displacements in the same direction and

released, they will vibrate at first frequency. When they are given equal initial displacements in

opposite direction and released they will vibrate at the second frequency as shown in Figures 6.3a

and 6.3b

(a) (b)

Figure – 6.3

ωn1 = √(k/m) ωn2 = √[(k + 2k2) /m]

1st Mode 2

nd Mode

If unequal displacements are given to the masses in any direction, the motion will be superposition

of two harmonic motions corresponding to the two natural frequencies.

m1

m2

k3

k1

x1

x2

m1

m2

k3

k1

x1

x2

= - T1

θ


Dept Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

6

Problems

1. Obtain the frequency equation for the system shown in Figure – 6.4. Also determine the natural

frequencies and mode shapes when k1 = 2k, k2 = k, m1 = m and m2 = 2m.

(a) (b)

Figure – 6.4.

Solution

Consider two degrees of freedom system shown in Figure 6.4a, where the masses are constrained to

move in the direction of the spring axis and executing free vibrations. The displacements are

measured from the un-stretched positions of the springs. Let x1 and x2 be the displacement of the

masses m1 and m2 respectively at any given instant of time measured from the equilibrium position

with x2 > x1. Then the spring forces acting on the masses are as shown in free body diagram in

Figure 6.4b


For mass m1

for mass (2)

The solution for x1 and x2 are obtained by considering that they can have harmonic vibration under

steady state condition. Then considering the case when the masses execute harmonic vibration at

frequency ω. Thus if x1 = X1 sin ωt, and x2 = X2 sin ωt ----------- (3)

k1x1

k2(x2 - x1)

..

m1x1

..

m2x2

m1

m2

k2

k1

x1

x2

.. m x

.. m1 x1 + k1 x1 - k2x2 + k2x1 = 0

.. m1 x1 + (k1 + k2) x1 = k2x2 ------------ (1)

= ..

m1 x1 - k1 x1 + k2(x2 –x1)

.. m2 x2 + k2x2 - k2x1 = 0

.. m2 x2 + k2x2 = k2x1 ------------ (2)

= ..

m2 x2 - k2(x2 –x1)


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

7

Then we have x1= - ω2X1 sin ωt, x2 = - ω

2 X2 sin ωt ------------------ (4)

Substitute equation (3) and (4) into the equation (1) and (2) we get

- m1ω2X1 sin ωt + (k1 + k2) X1 sin ωt = k2 X2 sin ωt ----------------- (5)

- m2 ω2X2 sin ωt + k2X2 sin ωt = k2 X1 sin ωt. ----------------- (6)

From equation (5) we have X1/X2 = k2/[(k1 + k2) – m1ω2] -------------- (7)

From equation (6) we have X1/X2 = [ k2 – m2ω2] / k2 -------------- (8)

Equating (7) and (8)

k2 / (k1 + k2 – m1ω2) = [k2 – m2ω

2] /k2

k22 = (k1 + k2 – m1ω

2) (k2 – m2ω

2)

k22 = (k1 + k2) k2 – m1ω

2 k2 – m2ω

2 (k1 + k2) + m1 m2ω

4

m1 m2 ω4 - ω

2 [m1 k2 + m2 (k1 + k2)] + k1 k2 = 0 --------------- (9)

letting ω2 = λ m1 m2 λ

2 – λ [m1k2 + m2 (k1 + k2)] + k1k2 = 0 ---------------------- (10)

Equation (10) is the frequency equation of the system which is quadratic in λ and hence the solution

is

λ = [[m1k2 + m2 (k1 + k2)] ± √ [[{m1k2 + m2(k1+k2)}2]- 4 m1 m2k1k2]] / 2m1m2

To determine the natural frequencies Given k1 = 2 k, k2 = k and m1 = m, m2 = 2m

λ = [mk + 2m (2k +k) ± √[mk + 6mk)2 – 4m 2mk

2k]] / 2m . 2m

= [7mk ± √[(7mk)2 – 4 (4m

2k

2)]] / 4m

2

= [7mk ± √(49m2k

2 – 16m

2k

2] / 4m

2

λ = [7mk ± 5.744 mk] /4m2 Thus λ1 = [7mk - 5.744 mk]/4m

2 and λ2 = [7mk + 5.744 mk] /4m

2

λ1 = ωn12 = [7 mk – 5.744 mk] /4m

2 = 1.255 mk /4m

2 = 0.3138 k/m Thus ωωωωn1 = 0.56 √√√√(k/m)

λ2 = ωn22 = [7mk + 5.744 mk] /4m

2 = 3.186 k/m. Thus ωωωωn2 = 1.784 √√√√(k/m)

Substituting the values of frequencies into the amplitude ratio equation as given by equation (7) and

(8) one can determine the mode shapes:

FOR THE FIRST MODE:

Substituting ωn12 = 0.3138 K/m into either of the equation (7) or (8) we get first mode shape:

I.e. X1/X2 = k2/[(k1 + k2) – m1ω2] -------------- (7) X1/X2 = [ k2 – m2ω

2] / k2 -------------- (8)

X1/X2 = k / [(2k + k – m ω2] = k/ [3k –m. ωn1

2]

X1/X2 = k/ [3k –2m. 0.3138k/m] = 1/(3 – 0.3138) =1/2.6862 = 0.3724

Thus we have X1/X2 = 0.3724. Then If X1 = 1, X2 = 2.6852


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

FOR THE SECOND MODE:

Substituting ωn22 = 3.186 K/m into either of the equation (7) or (8) we get first mode shape:

I.e. X1/X2 = k2/[(k1 + k2) – m1ω2]

X1/X2 = k / [(2k + k – m ω2] = k/ [3k

X1/X2 = k/ [3k –2m. 3.186k/m] = 1/(3

X1/X2 = (k – 2m.3.186 k/m)/k = [1

Thus we have X1/X2 = -5.37. Then If

MODE SHAPE FOR FIRST MODE

Derive the frequency equation for a double pendulum shown in figure.6.

frequency and mode shapes of the double pendulum when m

Consider two masses m1 and m

figure 6.6. Assume the system vibrates in vertical plane with small amplitude under which it only

has the oscillation.

Let θ1 and θ2 be the angle at any given instant of time with the vertical and x

horizontal displacement of the masses m

m1

m2

k2

k1

x1

x2

= 3.186 K/m into either of the equation (7) or (8) we get first mode shape:

] -------------- (7) X1/X2 = [ k2 – m2ω2] / k2 --------------

] = k/ [3k –m. ωn12]

] = 1/(3 – 3.186) = 1/ (-0.186) = - 5.37 or

)/k = [1-2(3.186)] = - 5.37

5.37. Then If X1 = 1, X2 = - 0.186

SHAPE FOR FIRST MODE SECOND MODE

ωn12 = 0.3138 K/m ωn2

2 = 3.186 K/m

Figure – 6.5.

Derive the frequency equation for a double pendulum shown in figure.6.6. Determine the natural

frequency and mode shapes of the double pendulum when m1 = m2 = m l1 = l2 = l

Figure 6.6

and m2 suspended by string of length l1 and l

. Assume the system vibrates in vertical plane with small amplitude under which it only

be the angle at any given instant of time with the vertical and x

horizontal displacement of the masses m1 and m2 from the initial vertical position respectively.

x1 = 1

x2 = 2.6852

x1 = 1

x2 = -0.186

Node

8

= 3.186 K/m into either of the equation (7) or (8) we get first mode shape:

-------------- (8)

SECOND MODE

= 3.186 K/m

. Determine the natural

= l

and l2 as shown in the

. Assume the system vibrates in vertical plane with small amplitude under which it only

be the angle at any given instant of time with the vertical and x1 and x2 be the

om the initial vertical position respectively.


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

For small angular displacement we have sin

Figure.6.

Figure 6.7hows the free body diagram for the two masses. For equilibrium under static condition the

summation of the vertical forces should be equal to zero. Thus we have

At mass m1 T1 cosθ1 = mg + T

At mass m2 T2 cosθ2 = mg -------

For smaller values of θ we have

T2 = m2g ----- (4) and T1 = m

When the system is in motion, t

derived by applying Newton Second Law of motion

Then we have for mass m1

For mass m2

Substituting the expression for T

equation (1) into the above equation (6) and (7) we have


= .. m2 x2 + T2 sin θ2 = 0 ----- (7)

- T2 sin θ2 ..

m2 x2

.. m1x1

+ [(m1+ m2) g](x1/l1) = m

+ [{(m1+ m2)/l1} + m2/l2]gx ..

m1x1

.. m1 x1 +T1 sin θ1 = T2 sin θ2

.. m1 x1 = -T1 sin θ1 + T2 sin θ2

For small angular displacement we have sinθ1 = x1 / l1 and sinθ2 = (x2 – x1) / l2

Figure.6.7 Free body diagram

hows the free body diagram for the two masses. For equilibrium under static condition the

summation of the vertical forces should be equal to zero. Thus we have

= mg + T2 cosθ2 ----- (2)

------- (3)

cosθ = 1. Then the above equations can be written as

= m1g + m2g T1 = (m1 + m2)g ----- (5)

the differential equation of motion in the horizontal direction can be

derived by applying Newton Second Law of motion.

Substituting the expression for T2 and T1 from equation (4) and (5) and for sinq

equation (1) into the above equation (6) and (7) we have

(7)

) = m2 g[(x2 – x1)/l2

gx1 = (m2g/l2) x2 --------- (8)

----- (6)

9

--- (1)

hows the free body diagram for the two masses. For equilibrium under static condition the

= 1. Then the above equations can be written as

ion of motion in the horizontal direction can be

from equation (4) and (5) and for sinq1 and sinq2 from

Mechanical Vibrations[10ME72]Mechanical Vibrations [10ME72]

Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

10


Equations (8) and (9) represent the governing differential equation of motion. Thus assuming the

solution for the principal mode as

Substitute in (10) into equation (8) and (9) and cancelling the common term sinωt we have

[-m1ω2+{(m1+m2)/l1+m2/l2}g]A=(m2g /l2)B --(11)

[-m2ω2+(m2g/l2)]B = (m2g/l2)A ---------- (12)

From equation (11) we have

A/B=(m2g/l2)/[{(m1+m2)/l1+m2/l2}g]-m1ω2 -- (13)

From equation (12) we have

A/B = [(m2g / l2) -m2ω2] / (m2g / l2) ------ (14)

Equating equation (13) and (14) we have

A/B = (m2g/l2)/[{(m1+m2)/l1 + m2/l2}g-m1ω2] = [(m2g/l2) -m2ω

2]/(m2g/l2)

[{(m1+ m2)/l1+m2/l2}g-m1ω2][(m2g/l2) -m2ω

2] = (m2g / l2)

2 ------- (15)

Equation (15) is a the quadratic equation in w2 which is known as the frequency equation.

Solving for ω2 we get the natural frequency of the system.

Particular Case:

When m1 = m2 = m and l1 = l2 = l

Then equation (13) will be written as A/B = (mg/l)/[(3mg/l)-mω2] = (g/l)/[(3g/l) - ω

2]

A/B= 1/[3 – (ω2l/g)] ------ (16)

and equation (14) will be written as A/B = [1 – (ω2l/g)] ------ (17)

Equating equation (16) and (17) we get A/B = 1/[3- (ω2l/g)] = [1- (ω

2l/g)]

[3- (ω2l/g)] * [1- (ω

2l/g)] = 1 or (3g- ω

2l)*(g- ω

2l) = g

2

3g2 – 3glω

2 – glω

2 +l

2ω

4 = g

2 or l

2ω

4 – 4glω

2 + 2g

2 = 0 or

ω4 – (4g/l) ω

2 + (2g

2/l

2) = 0 ------- (18)

letting λ = ω2 in equation (18) we get λ

2 – (4g/l)λ + (2g

2/l

2) = 0 ----- (19)

Which is a quadratic equation in l and the solution for the equation (19) is

.. m2x2

+ m2 g(x2 – x1)/l2

+ (m2g/l2) x1 = (m2g/l2) x2 ------ (9) .. m2x2

.. x1 = - ω

2Asinωt and

.. x2

= - ω2Bsinωt -------- (10)


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

λ1,2 = (2g/l) + √[(4g2/l

2) – (2g

2/l

2

λ1 = (g/l)(2 -√2) = 0.5858(g/l) -----

Since λ=ω2 then the natural frequency

and ωn2 = √ l2 = 1.8478√(g/l) thus

Substituting ωn1 and ωn2 from equation (23) and (24) into either of the equation (16) or (17) we

the mode shape

FOR THE FIRST MODE:

Mode shapes for the first natural frequency

I mode from equation (16) A/B= 1/[3

(A/B)1=1/[3-ωn12l/g] =1/[(3-{(g/l)(2

Thus when A =1 B = 2.4142

Also from equation (17) A/B = [1

For ωn1 = 0.7654 √(g/l) or ωn1

(A/B)1= (1- ωn12

l/g) = [1- {(g/l)(2

or (A/B)1 = (1–2+√2 ) = √2 -

Thus when A =1 B = 2.4142

Modes shape is shown in figure-6.8


Mode shapes for second natural frequency

II mode from equation (16) is given by

(A/B)2=1/[3-ωn12l/g] =1/[(3-{(g/l)(2+

(A/B)2 = 1/(- 0.4142) = - 2.4142 or Thus when A =1 B =

Also from equation (17) A/B = [1

ωn2 = 1.8478√(g/l) or ωn22 = (g/

(A/B)2= (1- ωn12

l/g) = [1- {(g/l)(2+

(A/B)2 = (1–2-√2 ) = -(1+√2) =

Thus when A =1 B = -0.4142

Modes shape is shown in figure-6.

2) or λ1,2 = (g/l)(2 ±√2) --------------------

----- (21) and λ2 = (g/l)(2 +√2) = 3.4142(g/l) ------

then the natural frequency ωn1 = √ l1 = 0.7654√(g/l) thus ωωωωn1 = 0.7654

(g/l) thus ωωωωn2 = 1.8478√(g/l) ---------- (24)

from equation (23) and (24) into either of the equation (16) or (17) we

Mode shapes for the first natural frequency ωn1 = 0.7654√(g/l) or ωn12 = (g/l)(2

I mode from equation (16) A/B= 1/[3 – (ω2l/g)]

{(g/l)(2- √2)*l/g}] = 1/(3-2+√2) =1/(1+√2) = 1/ 2.4142 = 0.4142

A/B = [1 – (ω2l/g)]

n12 = (g/l)(2 - √2)

{(g/l)(2- √2)}l/g]

-1 = 0.4142

6.8

Figure- 6.8

Mode shapes for second natural frequency ωn2 = 1.8478√(g/l) or ωn22 = (g/l)(2+

is given by A/B=1/[3–(ω2l/g)]

{(g/l)(2+√2)*l/g}] = 1/(3-2-√2) =1/(1-√2)

2.4142 or Thus when A =1 B = -0.4142

A/B = [1–(ω2l/g)]

= (g/l)(2+√2)

{(g/l)(2+√2)}l/g]

= - 2.4142

6.9

Figure- 6.9

11

-------------------- (20)

------ (22)

= 0.7654√(g/l) --- (23)

from equation (23) and (24) into either of the equation (16) or (17) we get

= (g/l)(2 - √2)

= 1/ 2.4142 = 0.4142

= (g/l)(2+√2)


Dept . Of ME. ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

12

Determine the natural frequencies of the coupled pendulum shown in the figure – 6.10. Assume that

the light spring of stiffness ‘k’ is un-stretched and the pendulums are vertical in the equilibrium

position.

Figure – 6.10.

Solution:

Considering counter clockwise angular displacement to be positive and taking the moments about

the pivotal point of suspension by D.Alembert’s principle we have

----------------- (1)

----------------- (2)

Equation (1) and (2) can also be written as

----------------- (3)

----------------- (4)

Equation (3) and (4) are the second order differential equation and the solution for θ1 and θ2 are

obtained by considering that they can have harmonic vibration under steady state condition. Then

considering the case when the masses execute harmonic vibration at frequency ω

Thus if θ1 = Asin ωt, and θ2 = Bsin ωt ----------- (5)

Substitute equation (5) into the equation (3) and (4) and canceling the common terms we get

(- ml2ω

2 + mgl + ka

2)A = ka

2B ----------------- (6)

(- ml2ω

2 + mgl + ka

2)B = ka

2A ----------------- (7)

From equation (6) we have A/B = ka2/ [mgl + ka

2 – ml

2ω

2] -------------- (8)

From equation (7) we have A/B = [mgl + ka2 – ml

2ω

2] / ka

2 -------------- (9)

Equating (8) and (9)

A/B = ka2/ [mgl + ka

2 – ml

2ω

2] = [mgl + ka

2 – ml

2ω

2] / ka

2

[mgl + ka2 – ml

2ω

2]

2 = [ka

2]2 ------------------- (10) or

mgl + ka2 – ml

2ω

2 = + ka

2 ω

2 = ( mgl + ka

2 + ka

2) / ml

2 -------- (11)

θ θ θ θ2

m

m m

θθθθ1

k

mg

l

a

mg

ka(θθθθ1 - θθθθ2)

m

ka(θθθθ1 - θθθθ2)

= - mglθ2 + ka(θ1 – θ2) ..

ml2

θ2

+ (mgl + ka)θ1 = kaθ2 ..

ml2

θ1

+ (mgl + ka)θ2 = kaθ1 ..

ml2

θ2

= - mglθ1 – ka(θ1 – θ2) ..

ml2

θ1


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

ω1,2 = √[( mgl + ka2 + ka

2) / ml

2]

ω1 = √[( mgl + ka2 - ka

2) / ml

2]

ω2 = √[( mgl + ka2 + ka

2) / ml

2] =

Substituting the values of frequencies


FOR THE FIRST MODE:

Substituting ωn12 = g/l into either of the equation (8) or (9) we get first mode shape:


2 – ml

2ω

2] = ka

A/B = 1


Substituting ωn22 = [(g/l) + (2ka

shape:


2 – ml

2ω

2] = ka

= ka2/ [mgl + ka

2 – mlg - 2ka

2] = (ka

Mode shapes at these two natural frequencies are as shown in figure

MODE SHAPES AT TWO DIFFERENT FREQUENCIES

FIRST MODE

ωωωωn12 = g/l A/B = 1

Figure-6.10 Mode Shapes at first frequency

] -------- (12)

= √(g/l) -------- (13)

] = √[(g/l) + (2ka2/ml

2)] -------- (14)

Substituting the values of frequencies into the amplitude ratio equation as given by equation (8) and


= g/l into either of the equation (8) or (9) we get first mode shape:

] = ka2/ [mgl + ka

2 – ml

2g/l] = ka

2/ [mgl + ka

2 –

= [(g/l) + (2ka2/ml

2)] into either of the equation (8) or (9) we get second mode

] = ka2/ [mgl + ka

2 – ml

2[(g/l) + (2ka

2/ml

2)]]

] = (ka2/ -ka

2) = -1 Thus A/B = -1

Mode shapes at these two natural frequencies are as shown in figure- 6.10

MODE SHAPES AT TWO DIFFERENT FREQUENCIES

SECOND MODE

= g/l A/B = 1 ωωωωn22 = [(g/l) + (2ka

2/ml

at first frequency Figure-6.11 Mode Shapes at second frequency

13

into the amplitude ratio equation as given by equation (8) and

= g/l into either of the equation (8) or (9) we get first mode shape:

mlg] = ka2/ ka

2

)] into either of the equation (8) or (9) we get second mode

SECOND MODE

/ml2)] A/B = -1

at second frequency


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

14

m m2

l2 l1

Derive the equation of motion of the system shown in figure 6.12. Assume that the initial tension ‘T’

in the string is too large and remains constants for small amplitudes. Determine the natural

frequencies, the ratio of amplitudes and locate the nodes for each mode of vibrations when m1 = m2

= m and l1= l2 = l3 = l.

Figure 6.12.

At any given instant of time let y1 and y2 be the displacement of the two masses m1 and m2

respectively. The configuration is as shown in the figure 6.13.

(a) (b)

Figure 6.13. Figure 6.14

The forces acting on the two masses are shown in the free body diagram in figure 6.14(a) and (b)

From figure 6.13 we have sinθ1 = (y1/l1) sinθ2 = [(y1 – y2)/l2] and sinθ3 = (y2/l3)

For small angle we have sinθ1 = θ1 = (y1/l1), sinθ2 = θ2 = [(y1 – y2)/l2] and sinθ3 = θ3 = (y2/l3)

and cosθ1 = cosθ2 = cosθ3 = 1.0 Thus the equation of motion for lateral movement of the masses

For the mass m1

or

For the mass m2

or

sin

θθθθ

θθθθ1

θθθθ2

θθθθ3

θθθθ3

3y1

2y2

2

m1

θθθθ1

θθθθ2

T Tcosθθθθ

T

Tcosθθθθ

T(sinθθθθ1+ T

T

Tcosθθθθ Tcosθθθθ

Tsinθθθθ

.. y2

θθθθ2

θθθθ3

m2

y1

..

m1y1 + [(T/l1) + (T/l2)]y1 = (T/l2)y2 ---- (1) ..

.. m1y1 = - (Tsinθ1 + Tsinθ2) = - T (θ1 + θ2)

m1y1 = - T [(y1 /l1) + (y1- y2) /l2] ..

.. m2y2 = (Tsinθ2 – Tsinθ3)

m2y2 = T[(y1- y2) /l2 - (y2 /l3)] ..

y2 + [(T/l2) +(T/l3)]y2 = (T/l2)y1 ----- (2)


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

15

Assuming harmonic motion as y1 =Asinωt and y2 = Bsinωt ------- (3) and substituting this

into equation (1) and (2) we have [-m1ω2 + (T/l1) + (T/l2)] A = (T/l2) B ----- (4)

[-m2ω2 + (T/l2) + (T/l3)] B = (T/l2) A ----- (5)

Thus from equation (4) we have A/B = (T/l2) / [(T/l1) + (T/l2) - m1ω2] ----- (6)

and from equation (5) we have A/B = [(T/l2) + (T/l3) - m2ω2] / (T/l2) ----- (7)

Equating equation (6) and (7) we have A/B=(T/l2)/[(T/l1)+(T/l2)- m1ω2]/[(T/l2)+(T/l3)-m2ω

2]/(T/l2)

Thus we have [(T/l1)+(T/l2)–m1ω2][(T/l2)+(T/l3)-m2ω

2] = (T

2/l2

2) -------------- (8)

Equation (8) is the equation on motion which is also known as frequency equation. Solving this

equation gives the natural frequencies of the system.

Particular Case: When m1 = m2 = m and l1 = l2 = l3 = l then equation (6) can be written as

A/B = (T/l)/[(T/l)+(T/l)-mω2] = (T/l)/[(2T/l)-mω

2] ---------- (9)

and equation (7) can be written as A/B = [(T/l)+(T/l) -mω2]/(T/l) = [(2T/l) - mω

2]/(T/l) ---- (10)

Equating equation (9) and (10) we have [(2T/l - mω2]

2 = (T/l)

2 -------- (11)

Thus 2T/l - mω2 = + (T/l) ------- (12) Therefore we have ω

2 = [(2T+T)]/ml ------ (13)

ωn1 = √[(2T-T]/ml] = √(T/ml) -------- (14) and ωn2 = √[(2T+T]/ml] = √(3T/ml) --------- (15)

Substituting equation (14) and (15) into either of the equation (9) or (10) we have the ratio of

amplitudes for the two natural frequencies. For the first natural frequency ωn1 = √(T/ml) then from

equation (9) we have (A/B)ωn1 = (T/l)/[(2T/l)-mω2] = (T/l)/[(2T/l) – m(T/ml)] = (T/l)/(T/l) = +1

or from equation (10) we have (A/B)ωn1 = [(2T/l) - mω2]/(T/l) = [(2T/l) – m(T/ml)]/(T/l)

Thus (A/B)ωn1 = (T/l)/(T/l) = +1

For the second natural frequency ωn2 = √(3T/ml) then from equation (9) we have

(A/B)ωn2 = (T/l)/[(2T/l)-mω2] = (T/l)/[(2T/l) – m(3T/ml)] = (T/l)/(-T/l) = -1

Thus (A/B)ωn2 = (T/l)/(-T/l) = -1 Then the mode shape will be as shown in figure 6.15(a) and (b)

Figure 6.15(a) Figure 6.15(b)

First Mode ωn1=√(T/ml), (A/B)ωn1 = +1 Second Mode ωn2=√(3T/ml), (A/B)ωn1 = -1

+1 +1 m1 m2 m1

+1

m2 -1


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

16

θθθθ1 J1

θθθθ2 J2

Kt1

Kt2

Torsional Vibratory systems

Derive the equation of motion of a torsional system shown in figure 6.16. Let J1 and J2 be the mass

moment of inertia of the two rotors which are coupled by shafts having torsional stiffness of Kt1 and

Kt2.

Figure 6.16 Two Degree of Freedom Figure 6.17 Free Body Diagram

torsional system

If θ1 and θ2 are the angular displacement of the two rotors at any given instant of time, then the shaft

with the torsional stiffness Kt1 exerts a torque of Kt1θ1 and the shaft with the torsional stiffness Kt2

exerts a torque of Kt2(θ2- θ1) as shown in the free body diagram figure 6.17

Then by Newton second law of motion we have for the mass m1

or

for the mass m2

or

Equation (1) and (2) are the governing Equations of motion of the system.

Equivalent Shaft for a Torsional system

In many engineering applications we find shaft of different diameters as shown in Figure 6.18 are in

use.

Figure-6.18 Stepped shaft

Disc-1

Disc-2

Kt1θθθθ1 Jθθθθ1 ..

Jθθθθ2

..

θθθθ2

θθθθ1

Kt2(θθθθ2 - θθθθ1)

.. J1θ1 = - K1θ1 + K2(θ2 – θ1)

.. J1θ1+( K1+ K2)θ1 = K2θ2 ------ (1)

.. J2θ2 = - K2(θ2 – θ1)

.. J2θ2 + K2θ2 = K2θ1 -------- (2)

d1 d2 d3 d4 J1θθθθ J2 J3 J4

1m2 2 3 4

L1 L2 L3 L4

Ja JB


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

17

For vibration analysis it is required to have an equivalent system. In this section we will study how

to obtain the torsionally equivalent shaft. Let θ be the total angle of twist in the shaft by application

of torque T, and θ1, θ2, θ3 and θ4 be twists in section 1, 2, 3 and 4 respectively. Then we have

θ = θ1 + θ2 + θ3 + θ4

From torsion theory we have,

T = Gθ Where J = pd4/32 Polar moment of inertia of shaft.

J L

Thus θ = θ1 + θ2 + θ3 + θ4 will be

θ = TL1 + TL2 + TL3 + TL4

J1G1 J2G2 J3G3 J4G4

If material of shaft is same, then the above equation can be written as

θ = 32T [ L1 + L2 + L3 + L4 ]

πG [ d14 d2

4 d3

4 d4

4]

If de and Le are equivalent diameter and lengths of the shaft, then:

Le = [ L1 + L2 + L3 + L4 ]

de [ d14 d2

4 d3

4 d4

4]

Le= L1[ de] + L2[ de ] + L3[ de ] + L4[ de ]

[d14] [d2

4] [d3

4] [d4

4]

Equivalent shaft of the system shown in Figure- 6.19

Figure – 6.19 Equivalent shaft of the system shown in figure – 6.18

Definite and Semi-Definite Systems

Definite Systems

A system, which is fixed from one end or both the ends is referred as definite system. A definite

system has nonzero lower natural frequency. A system, which is free from both the ends, is referred

as semi-definite system. For semi-definite systems, the first natural frequency is zero.

Various definite linear and a torsional systems are shown in figure-6.19

de

Le

JA JB


Dept. Of ME, ACEDept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

18

Figure-6.19 Various definite systems

Semi Definite or Degenerate System Systems

Systems for which one of the natural frequencies is equal to zero are called semi definite systems.

Various definite linear and a torsional systems are shown in figure-6.20

Figure-6.20 Semi-Definite systems

Problem to solve

Derive the equation of motion of a torsional system shown in figure 6.21.

Figure-6.21 Two Rotor System

m1

m2

x1

k2

x2

k1

K1

J1 θθθθ1

J2

K2

θθθθ2

m1 m2 K K K

x1 x2

θθθθ2 K

J2 J1

θθθθ1

m m K

θθθθ2

K

J2 J1

x2 l2x1


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

19

Generator

Gears

Shaft-1

J1

J2

Jg1

Jg2

Kt1

Kt2

Turbine

Shaft-2

Vibration of Geared Systems

Consider a Turbo-generator geared system is shown in the figure 6.22.

Figure-6.22: Turbo-Generator Geared System.

The analysis of this system is complex due to the presence of gears. Let ‘i’ be the speed ratio of the

system given by

i = Speed of Turbine

Speed of Generator

First step in the analysis of this system is to convert the original geared system into an equivalent

rotor system. Which is done with respect to either of the shafts.

When the Inertia of Gears is Neglected

The basis for this conversion is to consider the energies i.e. the kinetic and potential energy for the

equivalent system should be same as that of the original system. Thus if θ1 and θ2 are the angular

displacement of the rotors of moment of inertia J1 and J2 respectively then neglecting the inertia of

the gears the Kinetic and Potential energy of the original system are given by

Since θ2 = iθ1 Then the above equations can be written as

Thus the above equation shows that the original system can be converted into equivalent system with

respect to the first shaft as shown in figure- 6.23

Figure-6.23 Turbo-generator geared system neglecting the inertia of gears

U = 1/2 kt1θ12+1/2 kt2θ2

2

T =1/2 J1θ12

+ 1/2 J2θ22

. .

T =1/2 J1θ12

+ 1/2 J2(iθ2)2 = 1/2 J1θ1

2 + 1/2 (i

2J2)θ1

2

. . . .

U = 1/2 kt1θ12+1/2 kt2(iθ2)

2 = 1/2 kt1θ1

2+1/2 (i

2 kt2)θ1

2

Generator

J1

kt1

Turbine

i2kt2

J2


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

20

l1

m, J G

K2 K1

Which is obtained by multiplying the inertia and stiffness of the second shaft by i2 and keeping this

part of the system in series with the first part. Thus the stiffness of this equivalent two rotor system is

Thus the frequency of the system is given by

When the Inertia of Gears is Considered

If the inertia of the gears is not negligible then the equivalent system with respect to the first shaft

can be obtained in the same manner and finally we have the three rotor system as shown in figure-

6.24

Figure-6.24 Considering the inertia of gears

CO-ORDINATE COUPLING AND PRINCIPAL COORDINATES.

Consider a two degree of freedom system as shown in the figure- 6.25. The vibration is restricted in

plane of paper.

Figure-6.25. Two degree of freedom system

If m is the mass of the system, J is the Mass Moment of Inertia the system and G is the centre of

gravity. k1 and k2 are the stiffness of the springs which are at a distance ‘l1’ and ‘l2’ from the line

passing through the centre of gravity of the mass.

Then the system has two generalized co-ordinates, x is in Cartesian and θ is in Polar co-ordinate

systems when it is vibrating.

kte = i2 kt1kt2 / (kt1 + i

2 kt2)

ωn = kte(J1 + i2 J2) / i

2 J1J2 rad/sec

J1 J2

Jg1+i2Jg2

Generator

kt1 i2kt2

Turbine


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

21

At any given instant of time for a small disturbance the system occupy the position as shown in

figure-6.26(a).

Figure-6.26 (a) system under vibration (b) displacements at the springs

If ‘x’ is the displacement at the center of gravity of the system. Then the amount of displacements

that take place at the left spring is (x–l1θ)and at the right spring is (x+ l2θ) which is as shown in

figur-6.26(b).

At any given instant of time when the body is displaced through a rectilinear displacement ‘x’ and an

angular displacement ‘θ’ from its equilibrium position. The left spring with the stiffness k1 and the

right spring with the stiffness k2 are compressed through (x–l1θ) and (x+l2θ) from their equilibrium

position, The forces acting on the system is as shown in the free body diagram in figure-6.26. The

differential equation of motion of the system in ‘x’ and ‘θ’ direction are written by considering the

forces and moments in their respective direction.

Thus we have the equation of motion

Rearranging the above two equation we have

Since J = mr2 The above two equation can also be written as

Letting [(k1 + k2)/m = a, (k1l1 - k2l2)/ m = b and (k1l12 + k2l2

2)/ mr

2 = c

Thus substituting these into equation (5) and (6) we have

G

K1 K2

m,J

l1 l2

Static equilibrium

(x – l1θθθθ) (x + l2θθθθ)

θθθθ

x

G

l1 l2

k1(x-l1θθθθ) k2(x + l2θθθθ)

- k (x+l θθθθ)

..

Jθ =+k1(x-l1θ)l1 - k2(x+l2θ)l2 ----- (2)

..

mx + (k1 + k2)x = ( k1l1 - k2l2)θ ----- (3) Jθ + (k1l12 + k2l2

2) θ = (k1l1- k2l2) x ----- (4)

..

θ + [(k1l12

+ k2l22)/ mr

2] θ = [(k1l1- k2l2)/ mr

2] x ----- (6)

..

..

mx = - k1(x - l1θ) - k2(x + l2θ) ----- (1)

..

x + [(k1 + k2)/m]x = [( k1l1 - k2l2)/ m]θ ----- (5)


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

22

The above two differential equation (7) and (8) are coupled with respect to the coordinates in which

‘b’ is called the coupling coefficient or coordinate coupling.

Since if b=0 the two coordinate coupling equations (7) and (8) are independent of each other. The

two equations are then decoupled and each equation may be solved independently of the other. Such

a coordinate are called PRINCIPAL COORDINATE OR NORMAL COORDINATES.

Therefore the two i.e. rectilinear and angular motions can exists independently of each other with

their natural frequency as √a and √c.

Thus for the case of decoupled system when b=0 then (k1l1 – k2l2)/m = 0 or k1l1 – k2l2 = 0 or k1l1 =

k2l2. Then the natural in rectilinear and angular modes are ωnl = √a and ωna = √c

ωnl = √a = √(k1 + k2)/m and ωna = √c = √(k1l12

+ k2l22)/ mr

2

In general for a two degree of freedom under damped free vibration the equation of motion can be

written in the matrix form as

m11 m12 x1 c11 c12 x1 k11 k12 x1 0

+ + =

m21 m22 x2 c21 c22 x2 k21 k22 x2 0

Which reveal the type of coupling present in the system as Dynamic or Mass Coupling exist if the

mass matrix is non diagonal matrix. Where as stiffness or static Coupling exist if the stiffness matrix

is non diagonal. Where as damping Coupling exist if the damping matrix is non diagonal.

The system is dynamically decoupled when the mass matrix exists is a diagonal matrix.

m11 0 x1 c11 c12 x1 k11 k12 x1 0

+ + =

0 m22 x2 c21 c22 x2 k21 k22 x2 0

The system is damped decoupled when the damping matrix exists is a diagonal matrix.

m11 m12 x1 c11 0 x1 k11 k12 x1 0

+ + =

m21 m22 x2 0 c22 x2 k21 k22 x2 0

..

x + ax = bθ ----- (7)

θ + c θ = (b/r2) x ----- (8)

..

..

..

.

.

..

.

.

..

..

.

.


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

23

The system is statically decoupled when the stiffness matrix exists is a diagonal matrix.

m11 m12 x1 c11 c12 x1 k11 0 x1 0

+ + =

m21 m22 x2 c21 c22 x2 0 k22 x2 0

Dynamic or Mass Coupling:

If there is some point ‘C’ in the system along which a force is applied to the system produces pure

translation along the line of action of force as shown in figure-6.27


Then the equation of motion is


The above equation can be written in matrix form as

When k2l4 – k1l3 = 0 or k2l4 = k1l3 then the system is statically decoupled but dynamically coupled in

which the equation of motion will be which was

Mxc + meθ = -k1(xc-l3θ) - k2(xc+l4θ) ..

..

Jθ + mexc + (k2l4 - k1l3)xc + (k1l32 + k2 l4

2) θ = 0

..

..

Mxc + meθ + (k1+ k2)xc + (k2l4- k2l3)θ = 0 ..

..

Jθ + mexc= -k1(xc-l3θ) - k2(xc+l4θ) ..

..

.. xc

θ

.. (k1+ k2) (k2l4- k1l3)

(k2l4- k1l3) (k1l32+ k2l4

2)

+

xc

θ

0

0

=

M me

me J

(xc-l3θθθθ)(xc+ l4θθθθ)

θθθθ

xc

C

k1(x-l3θθθθ) k2(x+l4θθθθ)

G

l3 l4

Static equilibrium line

G

k1 k2

M,J

l1 l2

C

l3 l4

e e

..

..

.

.


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

24

Static/Stiffness and Dynamic/Mass Coupling: If there is a point ‘C’ in the system along which a

displacement produces pure translation along the line of action of spring force as shown in

figure-6.28


Then the equation of motion is

and


The above equation can be written in matrix form as

In which both the mass matrix and stiffness matrix are non-diagonal matrix hence the system is both

statically and dynamically coupled.

.. xc

θ

.. (k1+ k2) 0

0 (k1l32+ k2l4

2)

+

xc

θ

0

0

=

M me

me J

xc

(xc+ lθθθθ) θθθθ

C

k1xc k2(xc + lθθθθ)

G

l1 l2


G

k1 k2

M,J

l1 l2

C

l

l

Jθ + ml1xc + k2lxc + k1l2

θ = 0 ..

..

Mxc + ml1θ + (k1+ k2)xc + k2l = 0 ..

..

Mxc + ml1θ = - k1xc - k2(xc+lθ) ..

..

Jθ + m l1xc= -k2(xc+ lθ)l ..

..

+

.. xc

θ

.. (k1+ k2) k2l

k2l k2l2

xc

θ

0

0

=

M ml1

ml1 J


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

25

θ x K1

L1 L2

K2

Problem

Determine the normal mode of vibration of an automobile shown in figure-6.29 simulated by a

simplified two degree of freedom system with the following numerical values m = 1460 kg,

L1 = 1.35m, L2 = 2.65 m, K1 = 4.2x105N/m, K2 = 4.55x10

5 N/m and J=mr

2 where r= 1.22 m

Figure-6.29

Automobile can be modeled as shown in figure -6.30


Let at any given instant of time the translatory displacement be ‘x’ and an angular displacement be

‘θ’ from its equilibrium position of the automobile. Then the left spring with the stiffness k1 and the

right spring with the stiffness k2 are compressed through (x–l1θ) and (x+l2θ) from their equilibrium

position, The forces acting on the system are as shown in the free body diagram in figure-6.30(b).

The differential equation of motion of the automobile in ‘x’ and ‘θ’ direction are written by

considering the forces and moments in their respective direction.

Thus we have the equation of motion

G

K1 K2

m,J

l1 l2


(x – l1θθθθ) (x + l2θθθθ)

θθθθ

x

G

l1 l2

k1(x-l1θθθθ) k2(x + l2θθθθ)

- k (x+l θθθθ)

..

Jθ =+k1(x-l1θ)l1 - k2(x+l2θ)l2 ----- (2)

..

mx = - k1(x - l1θ) - k2(x + l2θ) ----- (1)


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

26


Equation (3) and (4) are the second order differential equation and the solution for x and θ are

obtained by considering that they can have harmonic vibration under steady state condition. Then

considering the case when the system execute harmonic vibration at frequency ω

Thus if x = Asinωt, and θ = Bsin ωt ---- (5) Substitute equation (5) into the equation (3) and (4) and

canceling the common term sinωt we get

[- mω2 + (k1 + k2)] A = (k1l1 – k2l2) B ------- (6)

[- Jω2 + (k1l1

2 + k2l2

2)] B = (k1l1 – k2l2) A ---- (7)

From equation (6) we have A/B = (k1l1-k2l2)/ [(k1 + k2) – mω2] ---------- (8)

From equation (7) we have A/B = [(k1l12 + k2l2

2) - Jω

2] / (k1l1 – k2l2) ---- (9)

Equating (8) and (9) A/B = (k1l1-k2l2)/ [(k1 + k2) – mω2] = [(k1l1

2 + k2l2

2) - Jω2] / (k1l1 – k2l2)

[(k1+k2)–mω2][(k1l1

2+k2l2

2)-Jω

2]=(k1l1-k2l2)

2 Further Simplification will give

mJω4–[J(k1+k2) + m(k1l1

2+k2l2

2)]ω

2 + k1k2(l1+l2)

2 = 0 --------- (10)

Substituting the value of m, J, k1, k2, l1, l2 into the above equation (10) we have

3.173x106ω

4– 4.831x10

9ω

2+1.72x10

12=0 --- (11) or ω

4– 1.523x103ω

2+5.429x10

5=0 ---- (12)

Letting ω2 =λ we have λ

2– 1.523x10

3λ+5.429x10

5 = 0 ---- (13)

Equation (13) is quadratic equation in λ. Thus solving equation (13) we get two roots which are

λ1= 569.59, λ2= 953.13 Since ω2

=λ we have ω = √λ. Thus ω1=23.86 rad/sec and ω2=30.87 rad/sec

Thus fn1 = 3.797 Hz and fn2 = 4.911 Hz

Un-damped Dynamic Vibration Absorber

Consider a two degree of freedom system with a forcing function F1 = Fosinωt as shown in

figure-6.31(a).

(a) (b)

Figure- 6.31(a) Two degree of freedom system with forcing function F1 on mass 1

..

mx + (k1 + k2)x = ( k1l1 - k2l2)θ ----- (3)

Jθ + (k1l12 + k2l2

2) θ = (k1l1- k2l2) x ----- (4)

..

m1

m2

k2

k1

x1

x2

F1

m1

m2

k2 (x2 –

x )

k1x1

F1 = Fosinωωωωt


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

27

Let x1 and x2 be the displacement of the masses m1 and m2 respectively at any given instant of time

measured from the equilibrium position with x2 > x1. Then the spring forces acting on the masses are

as shown in free body diagram in Figure 6.31(b)


For mass m1 we have

for mass (2)

The solution for x1 and x2 are obtained by considering that the masses execute harmonic vibration at

frequency w. Thus if x1 = X1 sin ωt, and x2 = X2 sin ωt ----- (3)

Then we have and --------- (4)

Substituting equation (3) and (4) into the equation (1) and (2) we get

-m1ω2X1 sinωt + (k1+ k2)X1 sinωt = k2X2 sin ωt + Fosinωt ----------------- (5)

- m2ω2X2 sinωt + k2X2 sinωt = k2 X1 sin ωt -------------- (6)

Canceling the common term sinwt on both the sides of equation 95) and (6) we have

[(k1+ k2) - m1ω2]X1 - k2X2 = Fo ----------- (7)

k2 X1 - [k2- m2ω2]X2 = 0----- (8)

Solving for X1 and X2 by cramer’s rule

-------- (9) -------- (10)

where ∆ is the determinant of characteristic equations.

-------- (11)

Solving the above determinant we get ------- (12)

..

x1 = - ω2X1sinωt

..

x2 = - ω2X2sinωt

.. m x

.. m2 x2 + k2x2 - k2x1 = 0

.. m2 x2 + k2x2 = k2x1 ------------ (2)

= ..

m2 x2 - k2(x2 –x1)

= ..

m1 x1 - k1 x1 + k2(x2 –x1) + Fosinωt

.. m1 x1 + k1 x1 - k2x2 + k2x1 = Fosinωt ------------ (1)

{ }{ } 2

2

2

22

2

121KωmKωm)K(K∆ −−−+=

∆

ωmK

K

0

F

X

2

22

20

1

−

−

=∆

0

F

K

ωm)K(K

X

0

2

2

121

2

−

−+

=

0ωmK

K

K

ωm)K(K∆

2

22

2

2

2

121 ≠−

−

−

−+=


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

28

If the two vibratory masses are considered separately as shown in Figure- 6.32, the mass 1 is a main

system and mass 2 is an secondary system. This system can be used as Dynamic vibration absorber

or Tuned damper by using the amplitude Equations (9) and (10).

Figure- 6.32

If the system has to be used as a Dynamic vibration absorber, then the amplitude of vibration of

mass m1 should be equal to zero, i.e X1=0.

-------- (13) Then we have

Fo(k2 – m2ωωωω2) = 0 since Fo cannot be equal to zero we have k2 – m2ωωωω

2 = 0

ωωωω2 = k2 /m2 or ωωωω

= √ k2 /m2 rad/sec -------- (14)

The above Eqn. is the natural frequency of secondary or absorber system.

0∆

ωmK

K

0

F

X

2

22

20

1=

−

−

=0

ωmK

K

0

F

2

22

20=

−

−

m1

m2

k2

k1

x1

x2

F1 Main System

Secondary System

Dept. Of ME, ACE

Mechanical vibraMechanical Vibrations [10ME72]Vtusolution.in

Vtusolution.in

vtuso

lution

.inMechanical Vibrations (ME 65)

CHAPTER-8

MULTI DEGREE OF FREEDOM SYSTEMS

Topics covered:

Influence co-efficents

Approximate methods

(i) Dunkerley’s method

(ii) Rayleigh’s method

Influence co-efficients

Numerical methods

(i) Matrix iteration method

(ii) Stodola’s method

(iii) Holzar’s method

1. Influence co-efficents

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 co-efficients. A set of influence co-efficents can be associated with each of

matrices involved in the equations of motion.

[ ]{ } [ ]{ } [ ]0xKxM =+&&

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 co-efficents referred as stiffness influence coefficents

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

inverse of stiffness matrix referred as flexibility influence co-efficients.

Matrix of flexibility influence co-efficients = [ ] 1−K

The elements corresponds to inverse mass matrix are referred as flexibility

mass/inertia co-efficients.

Matrix of flexibility mass/inertia co-efficients =[ ] 1−M

The flexibility influence co-efficients are popular as these coefficents give elements

of inverse of stiffness matrix. The flexibility mass/inertia co-efficients give elements

of inverse of mass matrix

UNIT - 7

NUMERICAL METHODS FOR MULTI DEGREE FREEDOM OF SYSTEMS


1Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.inVTU e-learning Course ME65 Mechanical Vibrations

Prof. S. K. Kudari, Principal APS College of Engineering, Bengaluru-28.

2

Stiffness influence co-efficents.

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 co-

efficents referred as stiffness influence coefficents.

{ } [ ]{ }xKF =

[ ]

=

333231

322221

131211

kkk

kkk

kkk

K

wher, 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

Example-1.

Obtain the stiffness coefficients of the system shown in Fig.1.

I-step:

Apply 1 unit deflection at point 1 as shown in Fig.1(a) and write the force equilibrium

equations. We get,

2111KKk +=

221Kk −=

0k31

=

m1

K1

m2

K2

x1=1 Unit

x2=0

m3

K3

x3=0

k11

k21

k31

x2=1 Unit

m1

K1

m2

K2

x1=0

m3

K3

x3=0

K12

k22

k32

m1

K1

m2

K2

x1=0

x2=0

m3

K3

x3=1 Unit

k13

k23

k33

(a) (b) (c)

Fig.1 Stiffness influence coefficients of the system


Dept. Of ME, ACE 2

Vtusolution.in

Vtusolution.in

vtuso

lution



3

II-step:

Apply 1 unit deflection at point 2 as shown in Fig.1(b) and write the force equilibrium

equations. We get,

212-Kk =

3222KKk +=

331-Kk =

III-step:

Apply 1 unit deflection at point 3 as shown in Fig.1(c) and write the force equilibrium

equations. We get,

0k13

=

323-Kk =

333Kk =

[ ]

=

333231

322221

131211

kkk

kkk

kkk

K

[ ]( )

( )

+

+

=

33

3322

221

KK-0

K-KKK-

0K-KK

K

From stiffness coefficients K matrix can be obtained without writing Eqns. of motion.

Flexibility influence co-efficents.

{ } [ ]{ }xKF =

{ } [ ] { }FKx1−

=

{ } [ ]{ }Fαx =

where, [ ]α - Matrix of Flexibility influence co-efficents given by

[ ]

=

333231

322221

131211

ααα

ααα

ααα

α

wher, α11, ……..α33 are referred as stiffness influence coefficients

α11-flexibility influence coefficient at point 1 due to a unit force at point 1

α21- flexibility influence coefficient at point 2 due to a unit force at point 1

α31- flexibility influence coefficient at point 3 due to a unit force at point 1

Example-2.

Obtain the flexibility coefficients of the system shown in Fig.2.


3Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



4

I-step:

Apply 1 unit Force at point 1 as shown in Fig.2(a) and write the force equilibrium

equations. We get,

1

312111K

1ααα ===

II-step:

Apply 1 unit Force at point 2 as shown in Fig.2(b) and write the force equilibrium

equations. We get,

21

3222K

1

K

1αα +==

III-step:

Apply 1 unit Force at point 3 as shown in Fig.2(c) and write the force equilibrium

equations. We get,

21

23K

1

K

1α +=

321

33K

1

K

1

K

1α ++=

Therefore,

1

1331122111K

1ααααα ======

21

233222K

1

K

1ααα +===

321

33K

1

K

1

K

1α ++=

(b) (c)

Fig.2 Flexibility influence coefficients of the system

x2=α22

m1

K1

m2

K2

x1=α12

m3

K3

x3=α32

F1=0

F2=1

F3=0

m1

K1

m2

K2

x1=α13

x2=α23

m3

K3

x3=α33

F1=0

F2=1

F3=0

(a)

m1

K1

m2

K2

m3

K3

x3=α31

F1=0

F2=1

F3=0

x1=α11

x2=α21


4Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



5

For simplification, let us consider : KKKK321

===

K

1

K

1ααααα

1

1331122111=======

K

2

K

1

K

1ααα

233222=+===

K

3

K

1

K

1

K

1α

33=++=

[ ]

=

333231

322221

131211

ααα

ααα

ααα

α

[ ]

=

321

221

111

K

1α

[ ] [ ] 1Kα

−=

In Vibration analysis if there is need of [ ] 1K

−one can use flexibility co-efficent matrix.

Example-3

Obtain of the Flexibility influence co-efficents of the pendulum system shown in the

Fig.3.

I-step:

Apply 1 unit Force at point 1 as shown in Fig.4 and write the force equilibrium

equations. We get,

m

m

l

l

m

l

Fig.3 Pendulum system

m

m

l

l

m

l

F=1

T θ

α11

Fig.4 Flexibility influence

co-efficents

Mechanical Vibrations 10ME72]

5Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



6

lθ sin T =

3mgm)mg(mθ cos T =++=

3mg

1θ tan =

θ sinθ tan small, is θ =

l

αθ sin

11=

θ sin lα11

=

3mg

lα

11=

Similarly apply 1 unit force at point 2 and next at point 3 to obtain,

5mg

lα

22=

the influence coefficients are:

5mg

lααααα

1331122111======

6mg

11lααα

233222===

6mg

11lα

33=

Approximate methods

In many engineering problems it is required to quickly estimate the first

(fundamental) natural frequency. Approximate methods like Dunkerley’s method,

Rayleigh’s method are used in such cases.

(i) Dunkerley’s method

Dunkerley’s formula can be determined by frequency equation,

[ ] [ ] [ ]0KMω2 =+−

[ ] [ ] [ ]0MωK2 =+−

[ ] [ ] [ ] [ ]0MKIω

1 1

2=+−

−

[ ] [ ][ ] [ ]0MαIω

1

2=+−

For n DOF systems,


6Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



7

[ ]0

m.00

.

0.m0

0.0m

α.αα

.

α.αα

α.αα

1.00

.

0.10

0.01

ω

1

n

2

1

nnn2n1

2n2221

1n1211

2=

+

−

[ ]0

mαω

1.mαmα

.

mα.mαω

10

mα.mαmαω

1

nnn22n21n1

n2n2222

n1n2121112

=

+−

+−

+−

...

Solve the determinant

( )

( ) [ ]0...mα...mmααmmαα

ω

1mα...mαmα

ω

1

nnn313311212211

1n

2nnn222111

n

2

=++++

+++−

−

(1)

It is the polynomial equation of nth degree in (1/ω2). Let the roots of above Eqn. are:

2

n

2

2

2

1ω

1...... ,

ω

1 ,

ω

1

0...ω

1

ω

1......

ω

1

ω

1

ω

1

ω

1

ω

1...... ,

ω

1

ω

1 ,

ω

1

ω

1

1n

22

n

2

2

2

1

n

2

2

n

22

2

22

1

2

=−

+++−

=

−

−

−

− (2)

Comparing Eqn.(1) and Eqn. (2), we get,

=

+++

2

n

2

2

2

1ω

1......

ω

1

ω

1( )

nnn222111mα...mαmα +++

In mechanical systems higher natural frequencies are much larger than the

fundamental (first) natural frequencies. Approximately, the first natural frequency is:

≅

2

1ω

1( )


The above formula is referred as Dunkerley’s formula, which can be used to estimate

first natural frequency of a system approximately.

The natural frequency of the system considering only mass m1 is:

1

1

111

1nm

K

mα

1ω ==

The Dunkerley’s formula can be written as:

2

nn

2

2n

2

1n

2

1ω

1......

ω

1

ω

1

ω

1+++≅ (3)


7Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



8

where, ..... ,ω ,ω2n1n

are natural frequency of single degree of freedom system

considering each mass separately.

The above formula given by Eqn. (3) can be used for any mechanical/structural

system to obtain first natural frequency

Examples: 1

Obtain the approximate fundamental natural frequency of the system shown in Fig.5

using Dunkerley’s method.

Dunkerley’s formula is:

≅

2

1ω

1( )

nnn222111mα...mαmα +++ OR

2

nn

2

2n

2

1n

2

1ω

1......

ω

1

ω

1

ω

1+++≅

Any one of the above formula can be used to find fundamental natural frequency

approximately.

Find influence flexibility coefficients.

K

1ααααα

1331122111=====

K

2ααα

233222===

K

3α

33=

Substitute all influence coefficients in the Dunkerley’s formula.

≅

2

1ω

1( )


m

K

m

K

x1

x2

m

K

x3

Fig.5 Linear vibratory system


8Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



9

≅

2

1ω

1

K

6m

K

3m

K

2m

K

m=

++

K/m0.40ω1=

rad/s

Examples: 2

Find the lowest natural frequency of the system shown in Figure by Dunkerley’s

method. Take m1=100 kg, m2=50 kg

VTU Exam July/Aug 2006 for 20 Marks

Obtain the influence co-efficents:

EI

1.944x10α

-3

11=

EI

9x10α

-3

22=

≅

2

nω

1( )

222111mαmα +

rad/s 1.245ωn

=

(ii) Rayleigh’s method

It is an approximate method of finding fundamental natural frequency of a system

using energy principle. This principle is largely used for structural applications.

Principle of Rayleigh’s method

Consider a rotor system as shown in Fig.7. Let, m1, m2 and m3 are masses of rotors on

shaft supported by two bearings at A and B and y1, y2 and y3 are static deflection of

shaft at points 1, 2 and 3.

1 2

180 120

m1 m2

Fig.6 A cantilever rotor system.


9Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



10

For the given system maximum potential energy and kinetic energies are:

∑=

=n

1i

iimaxgym

2

1V (4)

∑=

=n

1i

2

iimaxym

2

1T &

where, mi- masses of the system, yi –displacements at mass points.

Considering the system vibrates with SHM,

i

2

iyωy =&

From above equations

∑=

=n

1i

2

ii

2

maxym

2

ωT (5)

According to Rayleigh’s method,

maxmaxTV = (6)

substitute Eqn. (4) and (5) in (6)

∑

∑

=

==n

1i

2

ii

n

1i

ii

2

ym

gym

ω (7)

The deflections at point 1, 2 and 3 can be found by.

gmαgmαgmαy3132121111

++=


++=


++=

Eqn.(7) is the Rayleigh’s formula, which is used to estimate frequency of transverse

vibrations of a vibratory systems.

1 2 3

m1 m2 m3

A B

y1 y2

y3

Fig.7 A rotor system.


10Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



11

Examples: 1

Estimate the approximate fundamental natural frequency of the system shown in Fig.8

using Rayleigh’s method. Take: m=1kg and K=1000 N/m.

Obtain influence coefficients,

2K

1ααααα

1331122111=====

2K

3ααα

233222===

2K

5α

33=

Deflection at point 1 is:


++=

( )2000

5g

2K

5mg122

2K

mgy

1==++=



++=

( )2000

11g

2K

11mg362

2K

mgy

2==++=



++=

( )2000

13g

2K

13mg562

2K

mgy

3==++=

Rayleigh’s formula is:

2m

2K

2m

K

x1

x2

m

K

x3



11Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



12

∑

∑

=

==n

1i

2

ii

n

1i

ii

2

ym

gym

ω

2

222

2

2

g2000

132

2000

112

2000

52

g2000

132x

2000

112x

2000

52x

ω

+

+

++

=

rad/s 12.41ω =

Examples: 2

Find the lowest natural frequency of transverse vibrations of the system shown in

Fig.9 by Rayleigh’s method.

E=196 GPa, I=10-6 m4, m1=40 kg, m2=20 kg

VTU Exam July/Aug 2005 for 20 Marks

Step-1:

Find deflections at point of loading from strength of materials principle.

For a simply supported beam shown in Fig.10, the deflection of beam at distance x

from left is given by:

( ) b)(l xfor bxl6EIl

Wbxy

222 −≤−−=

For the given problem deflection at loads can be obtained by superposition of

deflections due to each load acting separately.

1 2

m1 m2

B

160 80 180

A

Fig.9 A rotor system.

b x

l

W

Fig.10 A simply supported beam


12Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



13

Deflections due to 20 kg mass

( ) ( ) EI

0.2650.180.160.42

6EI0.42

x0.18x0.169.81x20y

222'

1=−−=

( ) ( ) EI

0.290.180.240.42

6EIx0.42

x0.18x0.249.81x20y

222'

2=−−=

Deflections due to 40 kg mass

( ) ( ) EI

0.5380.160.260.42

6EIx0.42

x0.16x0.269.81x40y

222''

1=−−=

( ) ( ) EI

0.530.160.180.42

6EIx0.42

x0.16x0.189.81x40y

222''

2=−−=

The deflection at point 1 is:

EI

0.803yyy

''

1

'

11=+=

The deflection at point 2 is:

EI

0.82yyy

''

2

'

22=+=

∑

∑

=

==n

1i

2

ii

n

1i

ii

2

ym

gym

ω

( )( ) ( )22

2

20x0.8240x0.803

20x0.8240x0.8039.81ω

+

+=

rad/s 1541.9ωn

=


13Dept. Of ME,ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



14

Numerical methods

(i) Matrix iteration method

Using this method one can obtain natural frequencies and modal vectors of a vibratory

system having multi-degree freedom.

It is required to have ω1< ω2<……….< ωn

Eqns. of motion of a vibratory system (having n DOF) in matrix form can be written

as:

[ ]{ } [ ]{ } [ ]0xKxM =+&&

where,

{ } { } ( )φωtsinAx += (8)

substitute Eqn.(8) in (9)

[ ]{ } [ ]{ } [ ]0AKAMω2 =+− (9)

For principal modes of oscillations, for rth

mode,

[ ]{ } [ ]{ } [ ]0AKAMωrr

2

r=+−

[ ] [ ]{ } { }r2

r

rA

ωAMK

11=

−

[ ]{ } { }r2

r

rA

ωAD

1= (10)

where, [ ]D is referred as Dynamic matrix.

Eqn.(10) converges to first natural frequency and first modal vector.

The Equation,

[ ] [ ]{ } { }r

2

rrAωAKM =

−1

[ ]{ } { }r

2

rr1AωAD = (11)

where, [ ]1

D is referred as inverse dynamic matrix.

Eqn.(11) converges to last natural frequency and last modal vector.

In above Eqns (10) and (11) by assuming trial modal vector and iterating till the Eqn

is satisfied, one can estimate natural frequency of a system.

Examples: 1

Find first natural frequency and modal vector of the system shown in the Fig.10 using

matrix iteration method. Use flexibility influence co-efficients.

Find influence coefficients.

2K

1ααααα

1331122111=====

2K

3ααα

233222===


14Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



15

2K

5α

33=

[ ]

=

333231

322221

131211

ααα

ααα

ααα

α

[ ] [ ]

==−

531

331

111

2K

1Kα

1

First natural frequency and modal vector

[ ] [ ]{ } { }r2

r

rA

ωAMK

11=

−

[ ]{ } { }r2

r

rA

ωAD

1=

Obtain Dynamic matrix [ ] [ ] [ ]MKD1−

=

[ ]

=

=

562

362

122

2K

m

100

020

002

531

331

111

2K

mD

Use basic Eqn to obtain first frequency

[ ]{ } { }12

r

1A

ωAD

1=

Assume trial vector and substitute in the above Eqn.

Assumed vector is: { }

=

1

1

1

u1

First Iteration

[ ]{ } =1

uD

1

1

1

562

362

122

2K

m

=

2.6

2.2

1

2K

5m

As the new vector is not matching with the assumed one, iterate again using the new

vector as assumed vector in next iteration.


15Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



16

Second Iteration

[ ]{ } =2

uD

2.6

2.2

1

562

362

122

2K

m

=

3.13

2.55

1

K

4.5m

Third Iteration

[ ]{ } =3

uD

3.133

2.555

1

562

362

122

2K

m

=

3.22

2.61

1

K

5.12m

Fourth Iteration

[ ]{ } =4

uD

3.22

2.61

1

562

362

122

2K

m

=

3.23

2.61

1

K

5.22m

As the vectors are matching stop iterating. The new vector is the modal vector.

To obtain the natural frequency,

[ ]

3.22

2.61

1

D

=

3.23

2.61

1

K

5.22m

Compare above Eqn with with basic Eqn.

[ ]{ } { }12

1

1A

ωAD

1=

K

5.22m

ω

1

2

1

=

m

K

5.22

1ω2

1=

m

K0.437ω

1= Rad/s

Modal vector is:

{ }

=

3.23

2.61

1

A1


16Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



17

Method of obtaining natural frequencies in between first and last one

(Sweeping Technique)

For understanding it is required to to clearly understand Orthogonality principle of

modal vectors.

Orthogonality principle of modal vectors

Consider two vectors shown in Fig.11. Vectors { } { }b and a are orthogonal to each

other if and only if

{ } { } 0baT

=

{ } 0b

baa

2

1

21=

{ } 0b

b

10

01aa

2

1

21=

{ } [ ]{ } 0bIaT

= (12)

where, [ ]I is Identity matrix.

From Eqn.(12), Vectors { } { }b and a are orthogonal to each other with respect to

identity matrix.

Application of orthogonality principle in vibration analysis

Eqns. of motion of a vibratory system (having n DOF) in matrix form can be written

as:

[ ]{ } [ ]{ } [ ]0xKxM =+&&

{ } { } ( )φωtsinAx +=

[ ]{ } [ ]{ } [ ]0AKAMω2 =+− 11

x1

x2

{ }

=2

1

b

bb

{ }

=2

1

a

aa

Fig.11 Vector representation graphically


17Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



18

[ ]{ } [ ]{ }11 AKAMω2 =

If system has two frequencies ω1 and ω2

[ ]{ } [ ]{ }11 AKAMω2

1= (13)

[ ]{ } [ ]{ }22 AKAMω2

2= (14)

Multiply Eqn.(13) by { }T

2A and Eqn.(14) by { }T

1A

{ } [ ]{ } { } [ ]{ }1

T

21

T

2

2

1AKAAMAω = (15)

{ } [ ]{ } { } [ ]{ }2

T

12

T

1

2

1AKAAMAω = (16)

Eqn.(15)-(16)

{ } [ ]{ } 0AMA2

T

1=

Above equation is a condition for mass orthogonality.

{ } [ ]{ } 0AKA2

T

1=

Above equation is a condition for stiffness orthogonality.

By knowing the first modal vector one can easily obtain the second modal vector

based on either mass or stiffness orthogonality. This principle is used in the matrix

iteration method to obtain the second modal vector and second natural frequency.

This technique is referred as Sweeping technique

Sweeping technique

After obtaining { }11

ω and A to obtain { }22

ω and A choose a trial vector { }1

V

orthogonal to { }1

A ,which gives constraint Eqn.:

{ } [ ]{ } 0AMV1

T

1=

{ } 0

A

A

A

m00

0m0

00m

VVV

3

2

1

3

2

1

321=

( ) ( ) ( ){ } 0AmVAmVAmV333222111

=++

( ) ( ) ( ){ } 0VAmVAmVAm333222111

=++

321βVαVV +=

where α and β are constants

−=

11

22

Am

Amα

−=

11

33

Am

Amβ

Therefore the trial vector is:


18Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



19

( )

+

=

3

2

32

3

2

1

V

V

βVαV

V

V

V

=

3

2

1

V

V

V

100

010

βα0

[ ]{ }1VS=

where [ ]S= is referred as Sweeping matrix and { }1V is the trial vector.

New dynamics matrix is:

[ ] [ ][ ]SDDs

+

[ ]{ } { }22

2

1sA

ωVD

1=

The above Eqn. Converges to second natural frequency and second modal vector.

This method of obtaining frequency and modal vectors between first and the last one

is referred as sweeping technique.

Examples: 2

For the Example problem 1, Find second natural frequency and modal vector of the

system shown in the Fig.10 using matrix iteration method and Sweeping technique.

Use flexibility influence co-efficients.

For this example already the first frequency and modal vectors are obtained by matrix

iteration method in Example 1. In this stage only how to obtain second frequency is

demonstrated.

First Modal vector obtained in Example 1 is:

{ }

=

=

3.23

2.61

1

A

A

A

A

3

2

1

1

[ ]

=

100

020

002

M is the mass matrix

Find sweeping matrix

[ ]

=

100

010

βα0

S

2.612(1)

2(2.61)

Am

Amα

11

22 −=

−=

−=


19Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



20

1.6152(1)

1(3.23)

Am

Amβ

11

33 −=

−=

−=

Sweeping matrix is:

[ ]

=

100

010

1.615-2.61-0

S

New Dynamics matrix is:

[ ] [ ][ ]SDDs

+

[ ]

−

−−

=

=

1.890.390

0.110.390

1.111.610

K

m

100

010

1.615-2.61-0

531

362

122

2K

mD

s

First Iteration

[ ]{ } { }22

2

1sA

ωVD

1=

=

=

−

−−

8.14

1

9.71-

K

0.28m

2.28

0.28

2.27-

K

m

1

1

1

1.890.390

0.110.390

1.111.610

K

m

Second Iteration

=

=

−

−−

31.54

1-

21.28-

K

0.5m

15.77

0.50-

10.64-

K

m

8.14

1

9.71-

1.890.390

0.110.390

1.111.610

K

m

Third Iteration

=

=

−

−−

15.38

1-

8.67-

K

3.85m

59.52

3.85-

33.39-

K

m

31.54

1-

21.28-

1.890.390

0.110.390

1.111.610

K

m

Fourth Iteration

=

=

−

−−

13.78

1-

8.98-

K

2.08m

28.67

2.08-

18.68-

K

m

15.38

1-

8.67-

1.890.390

0.110.390

1.111.610

K

m

Fifth Iteration

=

=

−

−−

13.5

1

7.2-

K

1.90m

25.65

1.90-

13.68-

K

m

13.78

1

8.98-

1.890.390

0.110.390

1.111.610

K

m

Sixth Iteration


20Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



21

=

=

−

−−

13.43

1-

7.08-

K

1.87m

25.12

1.87-

13.24-

K

m

13.5

1-

7.2-

1.890.390

0.110.390

1.111.610

K

m

K

1087m

ω

1

2

2

=

m

K

1.87

1ω2

1=

m

K0.73ω

1=

Modal vector

{ }

=

1.89

0.14-

1-

A2

Similar manner the next frequency and modal vectors can be obtained.


21Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



22

(ii) Stodola’s method

It is a numerical method, which is used to find the fundamental natural frequency and

modal vector of a vibratory system having multi-degree freedom. The method is

based on finding inertia forces and deflections at various points of interest using

flexibility influence coefficents.

Principle / steps

1. Assume a modal vector of system. For example for 3 dof systems:

=

1

1

1

x

x

x

3

2

1

2. Find out inertia forces of system at each mass point,

1

2

11xωmF = for Mass 1

2

2


3

2


3. Find new deflection vector using flexibility influence coefficients, using the

formula,

++

++

++

=

′

′

′

333322311

233222211

133122111

3

2

1

αFαFαF

αFαFαF

αFαFαF

x

x

x

4. If assumed modal vector is equal to modal vector obtained in step 3, then solution

is converged. Natural frequency can be obtained from above equation, i.e

If

′

′

′

≅

3

2

1

3

2

x

x

x

x

x

x1

Stop iterating.

Find natural frequency by first equation,

1331221111αFαFαFx ++==′ 1

5. If assumed modal vector is not equal to modal vector obtained in step 3, then

consider obtained deflection vector as new vector and iterate till convergence.

Example-1

Find the fundamental natural frequency and modal vector of a vibratory system shown

in Fig.10 using Stodola’s method.

First iteration


22Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



23

1. Assume a modal vector of system { }1

u =

=

1

1

1

x

x

x

3

2

1

2. Find out inertia forces of system at each mass point

2

1

2

112mωxωmF ==

2

2

2

222mωxωmF ==

2

3

2

33mωxωmF ==

3. Find new deflection vector using flexibility influence coefficients

Obtain flexibility influence coefficients of the system:

2K

1ααααα

1331122111=====

2K

3ααα

233222===

2K

5α

33=

1331221111αFαFαFx ++=′

Substitute for F’s and α,s

2K

5mω

2K

mω

K

mω

K

mωx

2222

1=++=′

2332222112αFαFαFx ++=′


2K

11mω

2K

3mω

2K

6mω

K

mωx

2222

2=++=′

3333223113αFαFαFx ++=′


2K

13mω

2K

5mω

2K

6mω

K

mωx

2222

3=++=′

4. New deflection vector is:

=

′

′

′

13

11

5

2K

mω

x

x

x2

3

2

1

=

′

′

′

2.6

2.2

1

2K

5mω

x

x

x2

3

2

1

={ }2

u

The new deflection vector{ } { }12

uu ≠ . Iterate again using new deflection vector { }2

u

Second iteration

22Dept. OF ME, ACE

MechanicalMechanical Vibrations [10ME72]

23

Vtusolution.in

Vtusolution.in

vtuso

lution



24

1. Initial vector of system { }2

u =

=

′

′

′

2.6

2.2

1

x

x

x

3

2

1


2

1

2

112mωxωmF =′=′

2

2

2

22mω4xωmF 4.=′=′

2

3

2

332.6mωxωmF =′=′

3. New deflection vector,

1331221111αFαFαFx ′+′+′=′′


2K

9mω

2K

2.6mω

2K

4.4mω

K

mωx

2222

1=++=′′

2332222112αFαFαFx ′+′+′=′′


2K

23mω

2K

7.8mω

2K

13.2mω

K

mωx

2222

2=++=′′

3333223113αFαFαFx ′+′+′=′′


2K

28.2mω

2K

13mω

2K

13.2mω

K

mωx

2222

3=++=′′


=

′′

′′

′′

28.2

23

9

2K

mω

x

x

x2

3

2

1

=

′′

′′

′′

3.13

2.55

1

2K

9mω

x

x

x2

3

2

1

={ }3

u


uu ≠ . Iterate again using new deflection vector{ }3

u

Third iteration


u =

=

′′

′′

′′

3.13

2.55

1

x

x

x

3

2

1


2

1

2

112mωxωmF =′′=′′

2

2

2

225.1mωxωmF =′′=′′


24Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



25

3. new deflection vector,

1331221111αFαFαFx ′′+′′+′′=′′′


2K

10.23mω

2K

3.13mω

2K

5.1mω

K

mωx

2222

1=++=′′′

2332222112αFαFαFx ′′+′′+′′=′′′


2K

26.69mω

2K

9.39mω

2K

15.3mω

K

mωx

2222

2=++=′′′

3333223113αFαFαFx ′′+′′+′′=′′′


2K

28.2mω

2K

16.5mω

2K

15.3mω

K

mωx

2222

3=++=′′′


=

′′′

′′′

′′′

33.8

26.69

10.23

2K

mω

x

x

x2

3

2

1

=

′′′

′′′

′′′

3.30

2.60

1

2K

10.23mω

x

x

x2

3

2

1

={ }4

u

The new deflection vector { } { }34

uu ≅ stop Iterating

Fundamental natural frequency can be obtained by.

1=2K

10.23mω2

m

K0.44ω = rad/s

Modal vector is:

{ }

=

3.30

2.60

1

A1

2

3

2

333.13mωxωmF =′′=′′


25Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



26

Example-2

For the system shown in Fig.12 find the lowest natural frequency by Stodola’s method

(carryout two iterations)

July/Aug 2005 VTU for 10 marks

Obtain flexibility influence coefficients,

3K

1ααααα

1331122111=====

3K

4ααα

233222===

3K

7α

33=

First iteration

1. Assume a modal vector of system { }1

u =

=

1

1

1

x

x

x

3

2

1


2

1

2

114mωxωmF ==

2

2

2

222mωxωmF ==

2

3

2

33mωxωmF ==

3. New deflection vector using flexibility influence coefficients,

1331221111αFαFαFx ++=′

3K

7mω

3K

mω

3K

2mω

3K

4mωx

2222

1=++=′

4m

3K

2m

K

x1

x2

m

K

x3



26Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



27

2332222112αFαFαFx ++=′

3K

16mω

3K

4mω

3K

8mω

3K

4mωx

2222

2=++=′

3333223113αFαFαFx ++=′

3K

19mω

3K

7mω

3K

8mω

3K

4mωx

2222

3=++=′


=

′

′

′

19

16

7

3K

mω

x

x

x2

3

2

1

=

′

′

′

2.71

2.28

1

3K

7mω

x

x

x2

3

2

1

={ }2

u


uu ≠ . Iterate again using new deflection vector { }2

u

Second iteration


u =

=

′

′

′

2.71

2.28

1

x

x

x

3

2

1


2

1

2

114mωxωmF =′=′

2

2

2

224.56mωxωmF =′=′

2

3

2

332.71mωxωmF =′=′

3. New deflection vector

1331221111αFαFαFx ′+′+′=′′

3K

11.27mω

3K

2.71mω

3K

4.56mω

3K

4mωx

2222

1=++=′′

2332222112αFαFαFx ′+′+′=′′

3K

33.08mω

3K

10.84mω

3K

18.24mω

3K

4mωx

2222

2=++=′′

3333223113αFαFαFx ′+′+′=′′

3K

41.21mω

3K

18.97mω

3K

18.24mω

3K

4mωx

2222

3=++=′′



27Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



28

=

′′

′′

′′

41.21

33.08

11.27

3K

mω

x

x

x2

3

2

1

=

′′

′′

′′

3.65

2.93

1

K

3.75mω

x

x

x2

3

2

1

={ }3

u

Stop Iterating as it is asked to carry only two iterations. The Fundamental natural

frequency can be calculated by,

12K

3.75mω2

=

m

K0.52ω =

Modal vector,

{ }

=

3.65

2.93

1

A1

Disadvantage of Stodola’s method

Main drawback of Stodola’s method is that the method can be used to find only

fundamental natural frequency and modal vector of vibratory systems. This method is

not popular because of this reason.


28Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



29

(iii) Holzar’s method

It is an iterative method, used to find the natural frequencies and modal vector of a

vibratory system having multi-degree freedom.

Principle

Consider a multi dof semi-definite torsional semi-definite system as shown in Fig.13.

The Eqns. of motions of the system are:

0θ(θKθJ21111

=−+ )&&

0θ(θKθ(θKθJ32212122

=−+−+ ))&&

0θ(θKθ(θKθJ43323233

=−+−+ ))&&

0θ(θKθJ34344

=−+ )&&

The Motion is harmonic,

( )ωtsinφθii

= (17)

where i=1,2,3,4

Substitute above Eqn.(17) in Eqns. of motion, we get,

)φ(φKφJω21111

2 −= (18)

)φ(φK)φ(φKφJω32212122

2 −+−=

)φ(φK)φ(φKφJω43323233

2 −+−=

)φ(φKφJω34344

2 −+ (19)

Add above Eqns. (18) to (19), we get

0φJω4

1i

ii

2 =∑=

For n dof system the above Eqn changes to,

0φJωn

1i

ii

2 =∑=

(20)

The above equation indicates that sum of inertia torques (torsional systems) or inertia

forces (linear systems) is equal to zero for semi-definite systems.

θ1 K1 θ2 θ3 K2 K3 θ4

J4 J3

J2 J1

Fig.13 A torsional semi-definite system


29Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



30

In Eqn. (20) ω and φi both are unknowns. Using this Eqn. one can obtain natural

frequencies and modal vectors by assuming a trial frequency ω and amplitude φ1 so

that the above Eqn is satisfied.

Steps involved

1. Assume magnitude of a trial frequency ω

2. Assume amplitude of first disc/mass (for simplicity assume φ1=1

3. Calculate the amplitude of second disc/mass φ2 from first Eqn. of motion

0)φ(φKφJω21111

2 =−=

1

11

2

12K

φJωφφ −=

4. Similarly calculate the amplitude of third disc/mass φ3 from second Eqn. of motion.

0)φ(φK)φ(φKφJω32212122

2 =−+−=

0)φ(φK)φK

φJω(φKφJω

3221

1

11

2

1122

2 =−+−−=

0)φ(φKφJωφJω32211

2

22

2 =−+−=

22

2

11

2

322φJωφJω)φ(φK +=−

2

22

2

11

2

23K

φJωφJω-φφ

+= (21)

The Eqn (21) can be written as:

2

2

1i

2

ii

23K

ωφJ

-φφ∑

==

5. Similarly calculate the amplitude of nth disc/mass φn from (n-1)th Eqn. of motion

is:

n

n

1i

2

ii

1-nnK

ωφJ

-φφ∑

==

6. Substitute all computed φi values in basic constraint Eqn.

0φJωn

1i

ii

2 =∑=

7. If the above Eqn. is satisfied, then assumed ω is the natural frequency, if the Eqn is

not satisfied, then assume another magnitude of ω and follow the same steps.

For ease of computations, Prepare the following table, this facilitates the calculations.

Table-1. Holzar’s Table

MMechanical Vibrations [10ME72]

30Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



31

1 2 3 4 5 6 7 8

ω

S No J φ Jω

2φ

K

Example-1

For the system shown in the Fig.16, obtain natural frequencies using Holzar’s method.

Make a table as given by Table-1, for iterations, follow the steps discussed earlier.

Assume ω from lower value to a higher value in proper steps.

Table-2. Holzar’s Table for Example-1

1 2 3 4 5 6 7 8

ω

S No J φ Jω

2φ

K

I-iteration

0.25

1 1 1 0.0625 0.0625 1 0.0625

2 1 0.9375 0.0585 0.121 1 0.121

3 1 0.816 0.051 0.172

II-iteration

0.50

1 1 1 0.25 0.25 1 0.25

2 1 0.75 0.19 0.44 1 0.44

3 1 0.31 0.07 0.51

III-iteration

0.75

1 1 1 0.56 0.56 1 0.56

2 1 0.44 0.24 0.80 1 0.80

3 1 -0.36 -0.20 0.60

IV-iteration

1.00 1 1 1 1 1 1 1

2 1 0 0 1 1 1

∑ φ2Jω ∑ φ2

JωK

1

θ1 K1 θ2 θ3 K2

J3 J2 J1

Fig.14 A torsional semi-definite system

∑ φ2Jω ∑ φ2

JωK

1


31Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



32

3 1 -1 -1 0

V-iteration

1.25

1 1 1 1.56 1.56 1 1.56

2 1 -0.56 -0.87 0.69 1 0.69

3 1 -1.25 -1.95 -1.26

VI-iteration

1.50

1 1 1 2.25 2.25 1 2.25

2 1 -1.25 -2.82 -0.57 1 -0.57

3 1 -0.68 -1.53 -2.10

VII-iteration

1.75

1 1 1 3.06 3.06 1 3.06

2 1 -2.06 -6.30 -3.24 1 -3.24

3 1 1.18 3.60 0.36

Table.3 Iteration summary table

ω

0 0

0.25 0.17

0.5 0.51

0.75 0.6

1 0

1.25 -1.26

1.5 -2.1

1.75 0.36

∑ φ2Jω


32Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



33

The values in above table are plotted in Fig.15.

From the above Graph, the values of natural frequencies are:

rad/s 1.71ω

rad/s 1ω

rad/s 0ω

3

2

1

=

=

=

Definite systems

The procedure discussed earlier is valid for semi-definite systems. If a system is

definite the basic equation Eqn. (20) is not valid. It is well-known that for definite

systems, deflection at fixed point is always ZERO. This principle is used to obtain the

natural frequencies of the system by iterative process. The Example-2 demonstrates

the method.

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00

-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

∑ φ2Jω

Frequency, ω

Fig.15. Holzar’s plot of Table-3


33Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



34

Example-2

For the system shown in the figure estimate natural frequencies using Holzar’s

method.

July/Aug 2005 VTU for 20 marks

Make a table as given by Table-1, for iterations, follow the steps discussed earlier.

Assume ω from lower value to a higher value in proper steps.

Table-4. Holzar’s Table for Example-2

1 2 3 4 5 6 7 8

ω

S No J φ Jω

2φ

K

I-iteration

0.25

1 3 1 0.1875 0.1875 1 0.1875

2 2 0.8125 0.1015 0.289 2 0.1445

3 1 0.6679 0.0417 0.330 3 0.110

4 0. 557

II-iteration

0.50

1 3 1 0.75 0.75 1 0.75

2 2 0.25 0.125 0.875 2 0.437

3 1 -0.187 -0.046 0.828 3 0.27

4 -0.463

III-iteration

0.75

1 3 1 1.687 1.687 1 1.687

2 2 -0.687 -0.772 0.914 2 0.457

3 1 -1.144 -0.643 0.270 3 0.090

4 -1.234

IV-iteration

1.00 1 3 1 3 3 1 3

2 2 -2 -4 -1 2 -0.5

J 2J

3K 2K K

3J

Fig.16 A torsional system

∑ φ2Jω ∑ φ2

JωK

1


34Dept. Of ME,ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



35

3 1 -1.5 -1.5 -2.5 3 -0.833

4 -0.667

V-iteration

1.25

1 3 1 4.687 4.687 1 4.687

2 2 -3.687 -11.521 -6.825 2 -3.412

3 1 -0.274 -0.154 -6.979 3 -2.326

4 2.172

VI-iteration

1.50

1 3 1 6.75 6.75 1 6.75

2 2 -5.75 -25.875 -19.125 2 -9.562

3 1 3.31 8.572 -10.552 3 -3.517

4 7.327

1 2 3 4 5 6 7 8

ω

S No J φ Jω

2φ

K

VII-iteration

1.75

1 3 1 9.18 9.18 1 9.18

2 2 -8.18 -50.06 -40.88 2 -20.44

3 1 12.260 37.515 -3.364 3 -1.121

4 13.38

VIII-iteration

2.0

1 3 1 12 12 1 12

2 2 -11 -88 -76 2 -38

3 1 -27 108 32 3 10.66

4 16.33

IX-iteration

2.5

1 3 1 18.75 18.75 1 18.75

2 2 -17.75 -221.87 -203.12 2 -101.56

3 1 83.81 523.82 320.70 3 106.90

4 -23.09

∑ φ2Jω ∑ φ2

JωK

1


35Dept.Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution



36

Table.5 Iteration summary table

ω φ4

0 0

0.25 0.557

0.5 -0.463

0.75 -1.234

1 -0.667

1.25 2.172

1.5 7.372

1.75 13.38

2 16.33

2.5 -23.09

The values in above table are plotted in Fig.17.

From the above Graph, the values of natural frequencies are:

rad/s 2.30ω

rad/s 1.15ω

rad/s 0.35ω

3

2

1

=

=

=

0.0 0.5 1.0 1.5 2.0 2.5

-20

-10

0

10

20

Dis

pla

ce

me

nt,

φ4

Frequency,ω

1ω

2ω

3ω

Fig.17. Holzar’s plot of Table-5


36Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

1

Unit - 7

Vibration of Continuous System

Dr. T. Jagadish. Professor for Post Graduation, Department of Mechanical Engineering, Bangalore

Institute of Technology, Bangalore

Continuous systems are tore which have continuously distributed mass and Elasticity. These continuous

systems are assumed to be homogeneous and isotropic obeying hook law within the elastic limit. Since to

specify the position of every point in the continuous systems an infinite number of coordinates is required

therefore a continuous systems is considered to have infinite number of degrees of freedom. Thus there will

be infinite natural frequencies

Free vibration of continuous system is sum of the principal or normal modes. For the normal mode

vibration, every particle of the body performs simple harmonic motion at the frequency corresponding to

the particular root of the frequency equation, each particles passing simultaneously through its respective

equilibrium position. If the elastic curve resulting due to vibration motion starts coinciding exactly with

one of the normal modes, only that normal mode will be produced. If the elastic curve of the system under

which the vibration is started, is identical to any one of the principal mode shapes, then the system will

vibrate only in that principal mode.

Vibration of String.

A string stretched between two support is shown in figure-1 is an infinite degree of freedom system.

Figure-1

dx X

Y

y

x dx

T

θθθθ

θθθθ + (δ + (δ + (δ + (δθθθθ/δ/δ/δ/δx) dx

T

UNIT - 7

MULTIDEGREE VIBRATIONS SYSTEMSMULTI DEGREE FREEDOM OFSYSTEMS


Dept. Of ME, ACE

UNIT - 8

MODAL ANALYSIS AND CONDITION MONITORING

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

2

Let the tension T in the string be large so that for small displacement or amplitude the tension in the

string remains constant throughout the string. Consider an element of length dx at a distance ‘x’ from the

left end. At any instant of time the element of the string be displaced through a distance ‘y’ from the

equilibrium position. Then the tension at both the ends of this element is ‘T’. If θ the angle at the left end

of the element makes with the horizontal x- axis then the angle at the right end of the element is

θ + (δθ/δx)dx

The components of these two tensions at the ends of the element along x-axis balance each others.

The components along Y – axis � �� sin � Since θ is small

� � �� 1�

If ‘ρ' is mass per unit length, then the mass of the element is ‘ρdx’ and the differential equation of motion

along Y-axis according to Newton’s second law of motion is,

Mass x Acceleration = Sum of all the forces.

�� Therefore � ��

��

But we know slope � � �� then we have

� �� or ��

��

�� or

��

�� ⁄ ��

��

��

�� --- (2)

Where � !� �⁄ ------ (3) is the velocity of wave propagation along the string. Solution of equation (2)

can be obtained by assuming Y through separation of variables to be a product of two functions in the

form. " ��, $� � % �� & �$� � � � ��4�

()"(�) � & (

)"(�) *��

()"($) � % (

)"($) � � � �5�

Substituting equation (5) in to equation (2) we have


Dept. Of ME, ACE 2

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

3

& �� ,��

��-�� ./

�,

��

-�� -��

0�,��

�-

��-�� ---- (6)

Left hand side of equation (6) is a function of x alone and the right hand side is a function of t alone. These

two can only be equal if each one of these expression is a constant. This constant may be positive, zero or

negative. If it is a positive constant or zero then there is no vibratory motion. Hence this constant has to be

negative constant and equal to -�2

Then 0�,��

�-

��-�� ) � � � � � �7�

)%()"(�) � 2

) � 0 *�� 1& ()&($) � 2) � 0

()"(�) � �4� �) % � 0 *�� (

)&($) � 2) � 0 � � � �� 8�

With the general solution Y � A sin 89� � : ;.< 89� � � � ��9�

G � D sin $ � @;.< $ � � � ��10� Substituting equation (9) and (10) into equation (4) we have

" ��, $� � �A sin 40� � : ;.< 40� �D sin $ � @;.< $� � � � �11�

Since 2 can have infinite values, being an infinite degree of freedom system the general solution can be

written as

" ��, $� � ∑ BAisin 8C9 � � :� ;.< 8C9 � D�Disin �$ � @�;.< �$� � ��11�∞EF�

The arbitrary constant A, B, D and E have to be determined from the boundary condition and the initial

conditions.

For the string of length ‘L’ stretched between two fixed points. The boundary conditions are

y(0,t) = y(L, t) = 0

The condition y (0, t) = 0 will require B = 0 so the solution will be


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

4

" ��, $� � �A sin 40� �D sin $ � @;.< $� � � � �12�

The condition y (L, t) = 0 them fields the equation 0 � BA sin 89H D�D sin $ � @;.< $�

Since I<��2$ � @;.<J$ K 0, then we have BA sin 89H D � 0

or B89H D � �L for n = 1, 2,3 ………. ----- (13)

Since M � ;/O ------------ (14) and

� 2LM � 2L;/P ---------- (15). Where λ is the wave length of oscillation.

Substituting equation (15) into equation (13) we have

B�QR/P9 DH � �L or 2H/O � � -------- (16)

Each ‘n’ represents normal mode vibration with natural frequency determined from the equation

2HM/; � � therefore M � �;/2H = S)T√�/V for n = 1,2,3…….. ------ (17)al with the

distribution % � B sin �LH � D ----- (18)

Thus the general solution is given by

" ��, $� � W�I� sin �t � @� cos �t�∞

��1 sin S\T � � � � �19�

With � S\�T � � � �20�

With the initial condition of Y(x,0) and Y (x, 0) the constants Dn and En can be evaluated.

Longitudinal vibration of Rod/Bar


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

The bar/rod considered for the analysis is assumed to be thin and a uniform cross sectional area throughout

its length. Due to axial force the displacement ‘u’ along the bar will be a function of both position ‘x’ and

time ‘t’. The bar/rod has infinite number of natural frequencies and modes of vibration

Consider an element of length dx at a distance ‘x’ from the left end of the bar as shown in the

figure -2.

Figure

At any instant of time during vibration let ‘P’ be the axial force at the left end of the element. Then the

axial force at the right end of the element is P+(

displacement at x + dx will be u+(∂u/

by an amount of (∂u/∂x)dx and hence the unit strain is

But from Hook’s law the ratio of unit stress to the unit strain is equal to the Modulus of Elasticity ‘E

Hence we have ∂u/∂x = P/AE

In which ‘A’ is the cross sectional area of the bar/rod.

have AE ∂2u/∂x2

= ∂P/∂x ------ (2)

Considering the Dynamic Equilibrium of the element from Newton’s second law of motions me have.

(Mass) X (Acceleration) of the Element = (Unbalanced Resultant External force)

(∂P/∂x) dx = Ar dx ∂2u/∂t2

= ------ (3)

Where ρ is the density of rod/bar mass per unit volume, substituting equation

EAdx (∂2u/∂x2

) = rAdx ∂2u/∂t2

∂2u/∂x2

= (r/E) ∂2u/∂t2



bar/rod has infinite number of natural frequencies and modes of vibration


igure -2 Displacement in rod/bar member.


axial force at the right end of the element is P+(∂p/∂x)dx and If ‘u’ is the displacement at x, then the

u/∂x)dx. Then the element dx in the new position has changed in length

x)dx and hence the unit strain is ∂u/∂x.

But from Hook’s law the ratio of unit stress to the unit strain is equal to the Modulus of Elasticity ‘E

x = P/AE ------- (1)

‘A’ is the cross sectional area of the bar/rod. Differentiating the equation (1) w

(2) But (∂P/∂x) dx is the unbalanced force


(Mass) X (Acceleration) of the Element = (Unbalanced Resultant External force)

(3)

ass per unit volume, substituting equation (2) into (3)

5



bar/rod has infinite number of natural frequencies and modes of vibration



x)dx and If ‘u’ is the displacement at x, then the

x)dx. Then the element dx in the new position has changed in length

But from Hook’s law the ratio of unit stress to the unit strain is equal to the Modulus of Elasticity ‘E’,

Differentiating the equation (1) with respect to x we


(2) into (3) for ∂P/∂x we have


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

6

∂2u/∂x2

= [1/(E/r)] ∂2u/∂t2

--------------- (4)

∂2u/∂x2

= (1/c2) ∂2

u/∂t2 --------------- (5)

Where c = √(E/r) ------ (6) is the velocity of propagation of the displacement or stress wave in the rod/bar.

Solution of equation (5) can be obtained by assuming u through separation of variables to be a product of

two functions in the form u(x, t) = U(x) G(t) ------ (7)

∂2u/∂x2

=G(∂2U/∂x2

) and ∂2u/∂t2

=U(∂2G/∂t2

) --(8)

Substituting equation (8) in to equation (5) we have G(∂2U/∂x2

) = (U/c2)(∂2

G/∂t2)

Then (c2/U)(∂2

U/∂x2) = (1/G)(∂2

G/∂t2) ----- (9)

Left hand side of equation (9) is a function of x alone and the right hand side is a function of t alone. These

two can only be equal if each one of these expression is a constant.

This constant may be positive, zero or negative. If it is a positive constant or zero then there is no vibratory

motion. Hence this constant have to be negative constant and equal to - w2 . Then we have

(c2/U)(∂2

U/∂x2)=(1/G)(∂2

G/∂t2)= - ω

2 ------- (10)

(c2/U)(∂2

U/∂x2)= - w

2 and (1/G)(∂2

G/∂t2)= - ω

2

∂2U/∂x2

+(ω/c)2U = 0 and ∂2

G/∂t2+ω

2G = 0 - (11)

With the general solution U = Asin(ω/c)x + Bcos(ω/c)x ----- (12) G = Dsinωt + Ecosωt ------ (13)

Substituting equation (12) and (13) into equation (7) we have

u(x, t) = [Asin(ω/c)x + Bcos(ω/c)x] * [Dsinωt + Ecosωt] --------- (14)

Since ω can have infinite values, as the system being an infinite degree of freedom the general solution can

be written as

u(x, t) =∑i∞

=1[Aisin(ωi/c)x + Bicos(ωi/c)x] [Disinωit + Eicosωit ] -------- (15)


conditions.


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

Example:

Determine the natural frequencies and mode shape of a bar/rod when both the ends are free

Solution: When the bar/rod with both the ends being free then the stress and strain at the ends are zero

The boundary conditions are ∂u/∂x = 0 at x= 0 and x= L

u(x, t) = [Asin(ω/c)x + Bcos(ω/c)x] [Dsin

∂u/∂x = [(Aω/c)cos(ω/c)x + (Bω/c)sin(

(∂u/∂x)x=0=(Aω/c)[Dsinwt + Ecosωt] = 0

(∂u/∂x) x=L = (ω/c)[Acos(ωL/c) + Bsin(

Since the equation (4) and (5) must be true for any time ‘t’

From equation (4) A must be equal to zero.

equation (5) is satisfied only when sin(

Since c = √(E/ρ) then we have ωnL

since the natural frequency is fn = ωn

fn = (n/2L) √(E/ρ) ------------- (7)

Each ‘n’ represents normal mode vibration with natural frequency ‘f

equation (7).

Torsional vibration

Consider an element of length dx at a distance x from one end of the shaft as shown in the figure

Determine the natural frequencies and mode shape of a bar/rod when both the ends are free

When the bar/rod with both the ends being free then the stress and strain at the ends are zero

∂x = 0 at x= 0 and x= L ------- (1)

/c)x] [Dsinωt + Ecosωt] ------------ (2)

/c)sin(ω/c)x] [Dsinωt + Ecosωt] ------------ (3)

t] = 0 --- (4)

L/c) + Bsin(ωL/c)] [Dsinωt + Ecosωt] = 0 ------ (5)

Since the equation (4) and (5) must be true for any time ‘t’

From equation (4) A must be equal to zero. Since B must be finite value in order to have vibration then

sin(ωnL/c) = 0 or ωnL/c =sin-1

(0) = nπ

L√(ρ/E) = nπ Thus ωn = (nπ/L) √(E/ρ) ------

n/2π = [(nπ/L) √(E/ρ)]/ 2π

Each ‘n’ represents normal mode vibration with natural frequency ‘fn’ determined from the

sional vibration of circular shaft.

Consider an element of length dx at a distance x from one end of the shaft as shown in the figure

7

Determine the natural frequencies and mode shape of a bar/rod when both the ends are free.

When the bar/rod with both the ends being free then the stress and strain at the ends are zero

Since B must be finite value in order to have vibration then

(6)

’ determined from the

Consider an element of length dx at a distance x from one end of the shaft as shown in the figure -3


Dept. Of ME, ACE 7

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

8

Figure -3 Torque acting on the element of the circular shaft

At any given instant of time during vibration let T be the torque at left end of the element. Then the torque

at the right end of the element is given by T+(∂T/∂x)dx.

If θ is the angular twist of the shaft at the distance x, then θ+(∂θ/∂x)dx is the angular twist of the

shaft at the distance x+dx. Therefore the angular twist in the element of length dx is (∂θ/∂x)dx.

If ‘J’ is the polar moment of inertia of the shaft and ‘G’ the modulus of rigidity then the angular twist in the

element of length dx is given by torsion formula

dθ = (T/GJ) dx ------- (1) or GJ(dθ/dx) = T ------- (2)

Differentiating with respect to x we have GJ(d2θ/dx

2) = dT/dx ------- (3)

In which GJ is the torsional stiffness of the shaft. Since the torque on the two face of the element being T

and T+(∂T/∂x)dx the net torque on the element will be (∂T/∂x)dx --- (4)

Substituting equation (3) for ∂T/∂x in to equation (4) we have (∂T/∂x)dx = GJ(∂2θ/∂x2

)dx ------- (5)

Considering the dynamic equilibrium of the element one can obtain the equation of motion by equating the

product of mass moment of inertia ‘Jρdx’ and the acceleration ‘∂2θ/∂t2

’ to the net torque acting on the

element we have.

(Jρdx) ∂2θ/∂t2

= GJ(∂2θ/∂x2

)dx

∂2θ/∂t2

= (G/ρ)(∂2θ/∂x2

) or

∂2θ/∂x2

= (ρ/G) ∂2θ/∂t2

∂2θ/∂x2

= [1/(G/ρ)] ∂2θ/∂t2

∂2θ/∂x2

= (1/c2) ∂2

θ/∂t2 --------- (6)

Where c = √(G/ρ) ------ (7) is the velocity of wave propagation in which r is the density of the shaft in

mass per unit volume.


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

9

Solution of equation (6) can be obtained by assuming q through separation of variables to be a product of

two functions in the form. θ(x, t) = Θ(x) G(t) ------ (7)

∂2θ/∂x2

=G(∂2Θ/∂x2) and ∂2

θ/∂t2=Θ(∂2

G/∂t2)---(8)

Substituting equation (8) in to equation (6) we have G(∂2Θ/∂x2) = (Θ/c

2)(∂2

G/∂t2)

Then (c2/Θ)(∂2Θ/∂x2

) = (1/G)(∂2G/∂t2

) ----- (9)

Left hand side of equation (9) is a function of ‘Θ’ alone and the right hand side is a function of ‘t’ alone.

These two can only be equal if each one of these expression is a constant. This constant may be positive,

zero or negative. If it is a positive constant or zero then there is no vibratory motion. Hence this constant

has to be negative constant and equal to - ω2

(c2/Θ)(∂2Θ/∂x2

)=(1/G)(∂2G/∂t2

) = -ω2

-- (10)

(c2/Θ)(∂2Θ/∂x2

)= - ω2

and (1/G)(∂2G/∂t2

)= - ω2

∂2Θ/∂x2+(ω/c)

2Θ = 0 and ∂2G/∂t2

+ω2G = 0--(11)

With the general solution Θ = Asin(ω/c)x + Bcos(ω/c)x ----- (12) G = Dsinωt + Ecosωt ------ (13)

Substituting equation (12) and (13) into equation (7) we have

θ(x, t) = [Asin(ω/c)x + Bcos(ω/c)x] [Dsinωt + Ecosωt] --------- (14)

Since ω can have infinite values, as the system being an infinite degree of freedom the general solution can

be written as

θ(x, t) =∑i∞

=1[Aisin(ωi/c)x + Bicos(ωi/c)x] [Disinωit + Eicosωit ] -------- (15)


conditions.

Lateral Vibration of beams

To derive the differential equation of motion for lateral vibration of beams one has to consider the forces

and moments acting on the beam. Let V and M are the shear forces and bending moments respectively

acting on the beam with p(x) represents the intensity of lateral loading per unit length of the beam.


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

10

Consider a beam of length ‘L’ and moment of inerter ‘I’ subjected to distributed lateral load of p(x) N per

unit length. At any section x-x at a distance x from one end consider an element of length dx as shown in

the figure-4. The force and the moments acting on the element of length dx are as shown in the figure-5

Figure-4 Beam Figure-5 Force and Moments

For static equilibrium the summation of the forces and moment should be equal to zero. Thus summing the

vertical force in the Y – direction and equating to zero we have

V + p(x) dx – [V +(∂V/∂x) dx] = 0

[p(x) - (∂V/∂x)] dx = 0 since dx ≠ 0

p(x) - (∂V/∂x) = 0 or ∂V/∂x = p(x) ---- (1)

Which state that the rate of change of shear force along the length of the beam is equal to the load per unit

length.

Summation of the moment about any point should be equal to zero. Thus summing the moments about the

right bottom corner of the element considering the clockwise moment as positive we have

M + p(x) dx2/2 + V dx – [M +(∂M/∂x) dx] = 0 or p(x) dx

2/2 + V dx – (∂M/∂x) dx = 0

Since dx is small neglecting the higher order terms we have

[V – (∂M/∂x)] dx = 0 Since dx ≠ 0 we have

V – (∂M/∂x) = 0 ∂M/∂x = V ----- (2)

Which state that the rate of change of bending moment along the length of the beam is equal to the shear

force.

From equation (1) and (2) we have ∂2M/∂x2

= ∂V/∂x = p(x) ----- (3)


Dept.. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

11

But we known that the bending moment is related to curvature by flexure formula and for the coordinates

indicated we have EI(d2y/dx

2) = M ----- (4)

Substituting equation (4) into equation (3) we have ∂2[EI(d

2y/dx

2) ]/∂x2

= p(x)

EI (d4y/dx

4) = p(x) ----- (5)

For the dynamic equilibrium when the beam is having transverse vibrations about its static equilibrium

position under its own weight and due to the distributed load per unit length must be equal to the inertia

load due its mass and acceleration. By assuming harmonic motion the inertia force ρωy2

is in the same

direction as that of the distributed load p(x).

Thus we have p(x) = ρω2y ------ (6)

In which ρ is the mass density per unit length of the beam. Thus substituting equation (6) into equation (5)

for p(x) we have EI(d4y/dx

4) = ρω

2y or EI(d

4y/dx

4) - ρω

2y = 0 ------ (7)

If the flexural rigidity EI is constant then equation (7) can be written as d4y/dx

4 - (ρω

2/EI)y = 0 ---- (8)

Letting β4 = ρω

2/EI ------ (9) d

4y/dx

4 – β

4y = 0 ------ (10)

The above equation (10) is a fourth order differential equation for the lateral vibration of a uniform cross

section beam. The solution of the above equation (10) can be obtained by assuming the displacement ‘y’ of

the form y = eax

---- (11)

Which will satisfy the fourth order differential equation (10) when a = ± β and a = ± iβ ---- (12)

Since e±βx

= coshβx ± sinhβx ---- (13) and e±ibx

= cosβx ± sinβx ---- (14)

Then the solution of the above equation (10) will be in the form

Y=Acoshβx+Bsinhβx+Ccosβx+Dsinβx -- (15) Which is readily established

The natural frequencies of lateral vibration of beams are found from equation (9) given by

βn4 = ρωn

2/EI Thus ωn

2 = βn

4 EI/ρ or ωn

= βn

2√EI/ρ

ωn = (βnL)

2√(EI/ρL4) ------ (16) Where the number βn depends on the boundary conditions of the beam.

The following table lists the numerical values of (βnL)2

for typical end boundary conditions of the beam.

Mech Vibrations [10Me72]Mechanical Vibrations [10ME72]

Dept. Of ME , ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

12

Beam Condition First Mode Second Mode Third Mode

Simply supported 9.87 39.5 88.9

Cantilever 3.52 22.0 61.7

Free-Free 22.4 61.7 121.0

Clemped -Clemped 22.4 61.7 121.0

Champed hinged 15.4 50.0 104.0

Hinged -Fre 0 15.4 50.0

Example-1: Determine the first three natural frequencies of a Rectangular Cantilever Beam for the

following data Length L = 1 m Breath B = 0.03 m Depth D = 0.04 m Young's Modulus

E = 2x1011

N/m2 Density r = 7850.0 Kg/m

3

Solution: For a cantilever beam the boundary conditions are at x= 0 displacement and slope are zero

i.e. y = 0 and dy/dx = 0 ------ (1) and at x= L the shear force ‘V’ and the bending moment ‘M’ is zero

i.e. d2y/dx

2 = 0 and d

3y/dx

3 = 0 -------- (2)

Substituting these boundary conditions into the general solution

Y=Acoshβx + Bsinhβx + Ccosβx + Dsinβx --- (3)

we obtain (y)x=0 = A + C = 0 then A = - C – (4)

(dy/dx)x=0= β(Asinhβx + Bcoshβx – Csinβx + Dcosβx)x=0 = 0

β(B + D) = 0 since β ≠ 0 then B = -D --- (5)

(d2y/dx

2)x=L= β

2(AcoshβL + BsinhβL – CcosβL – DsinβL) = 0

Since β2 ≠ 0 and A = -C and B = -D and we have A(coshβL + cosβL) + B(sinhβL + sinβL) = 0 --- (6)

A/B =-(sinhβL + sinβL)/(coshβL + cosβL) ----- (7)

(d3y/dx

3)x=L= β

3(AsinhβL + BcoshβL + CsinβL - DcosβL) = 0

Again Since β2 ≠ 0 and A= -C and B= -D and we have A(sinhβL - sinβL) + B(coshβL + cosβL) = 0 ---- (8)

A/B =-(coshβL + cosβL)/(sinhβL - sinβL) ----- (9)


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

13

From equation (7) and (9) we have A(coshβL+cosβL) + B(sinhβL+sinβL) = 0 --- (10)

(sinhβL + sinβL) (coshβL + cosβL)

A/B = ---------------------- = ------------------------

(coshβL + cosβL) (sinhβL - sinβL)

Which reduces to coshβLcosβL + 1 = 0 ---------- (11)

Equation (11) is satisfied by a number of values of βL, corresponding to each normal mode of vibration,

which for the first three modes these values are tabulated in the table as

(β1l)2 = 3.52, (β2l)

2 = 22.00, (β3l)

2 = 61.70

Area A = 0.0012 m2 Moment of Inertia I = 1.600E-07m

4

Density per unit length (ρ*A) = ρL = 7850x 0.0012 = 9.4200 Kg/m

Circular Frequency

ωn1 = (β1l)2√(EI)/(ρl

4) = 2.052E+02 rad/sec

ωn2 = (β2l)2√(EI)/(ρl

4) = 1.282E+03 rad/sec

ωn3= (β3l)2√(EI)/(ρl

4) = 3.596E+03rad/sec

Natural Frequencies

fn1 = ωn1/2π = 32.639 Hz

fn2 = ωn2/2π = 203.994 Hz

fn3 = ωn3/2π = 572.111 Hz

Example-2: Determine the first three natural frequencies of a Rectangular Cantilever Beam for the

following data Length L = 1 m Breath B = 0.03 m Depth D = 0.04 m Young's Modulus E = 2.00E+11 N/m2

Density r = 7850.0 Kg/m3

Solution: For a cantilever beam the boundary conditions are at x= 0 displacement and slope are zero


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

14

i.e. y = 0 and dy/dx = 0 ------ (1)

and at x= L the shear force ‘V’ and the bending moment ‘M’ is zero

i.e. d2y/dx

2 = 0 and d

3y/dx

3 = 0 -------- (2)

Substituting these boundary conditions into the general solution

Y=Acoshβx + Bsinhβx + Ccosβx + Dsinβx --- (3)

we obtain (y)x=0 = A + C = 0 then A = - C – (4)

(dy/dx)x=0= β(Asinhbx + Bcoshbx – Csinbx + Dcosbx)x=0 = 0

β(B + D) = 0 since β ≠ 0 then B = -D --- (5)

(d2y/dx

2)x=L= β

2(AcoshβL + BsinhβL – CcosβL – DsinβL) = 0

Since β2 ≠ 0 and A = -C and B = -D and we have

A(coshβL + cosβL) + B(sinhβL + sinβL) = 0 ---- (6)

A/B =-(sinhβL + sinβL)/(coshβL + cosβL) ----- (7)

(d3y/dx

3)x=L= β

3(AsinhβL + BcoshβL + CsinβL - DcosβL) = 0

Again Since β2 ≠ 0 and A = -C and B = -D and we have

A(sinhβL - sinβL) + B(coshβL + cosβL) = 0 ----- (8)

A/B =-(coshβL + cosβL)/(sinhβL - sinβL) ----- (9)

From equation (7) and (9) we have

A(coshβL+cosβL)+B(sinhβL+sinβL)= 0 --- (10)

(sinhβL + sinβL) (coshβL + cosβL)

A/B = ---------------------- = ------------------------

(coshβL + cosβL) (sinhβL - sinβL)

Which reduces to coshβLcosβL + 1 = 0 ---------- (11)


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtuso

lution

.in

15

Equation (11) is satisfied by a number of values of βL, corresponding to each normal mode of vibration,

which for the first three modes these values are tabulated in the table as

(β1l)2 = 3.52, (β2l)

2 = 22.00, (β3l)

2 = 61.70

Area A = 0.0012 m2 Moment of Inertia I = 1.600E-07m

4

Density per unit length (ρ*A) = ρL = 7850x 0.0012 = 9.4200 Kg/m

Circular Frequency

ωn1 = (β1l)2√(EI)/(ρl

4) = 2.052E+02 rad/sec

ωn2 = (β2l)2√(EI)/(ρl

4) = 1.282E+03 rad/sec

ωn3= (β3l)2√(EI)/(ρl

4) = 3.596E+03rad/sec

Natural Frequencies

fn1 = ωn1/2π = 32.639 Hz

fn2 = ωn2/2π = 203.994 Hz

fn3 = ωn3/2π = 572.111 Hz


Dept. Of ME, ACE

Vtusolution.in

Vtusolution.in

vtusolutionvtusolution.in/uploads/9/9/9/3/99939970/mechanical_vibrations[10me... · resonance is...

Documents