Vibration Analysis Unit

UNIT OUTCOME
• At the end of this unit, the students should be able to compute the frequency of free
vibrations.
SYLLABUS
Basic features of vibratory systems – Degrees of freedom – single degree of freedom – Free
vibration– Equations of motion – Natural frequency – Types of Damping – Damped vibration–
Torsional vibration of shaft – Critical speeds of shafts – Torsional vibration – Two and three
rotor torsional systems.

BASIC FEATURES OF VIBRATORY
SYSTEM
i. Vibrations: When elastic bodies such as a spring, a beam and shaft are displaced from
the equilibrium position by the application of external forces, and then released, they
execute a vibratory motion.

SYSTEM
i. Period of vibration or time period: It is the time
interval after which the motion is repeated
itself. The period of vibration is usually
expressed in seconds.
ii. Cycle: It is the motion completed during one
time period.
iii. Frequency: It is the number of cycles described
in one second. In S.I. units, the frequency is
expressed in hertz (briefly written as Hz)
which is equal to one cycle per second.

SYSTEM
i. Causes of Vibrations:
• Unbalanced forces: Produced within the
machine due to wear and tear.
• External excitations: Can be periodic or random
ii. Resonance:
• When the frequency of the external or applied
force is equal to the natural frequency resonance
occurs.
Vertical Shaking Accident and Cause Investigation of 39-story Office Building

SYSTEM
Components of vibratory system:
i. Spring/Restoring element:
• Its denoted by k or s;
• SI unit – N/m
ii. Dashpot/Damping component
• Its denoted by c;
• SI unit – N/m/s
iii. Mass/Inertia component
• Its denoted by m;
• SI unit – kg

DEGREES OF FREEDOM
• The minimum number of independent
coordinates required to determine
completely the position of all parts of a
system at any instant of time defines the
degree of freedom of the system.
1 DOF 2 DOF

TYPES OF VIBRATIONS
1. Free or Natural Vibrations: When no external force acts on the body, after
giving it an initial displacement, then the body is said to be under free or natural
vibrations. The frequency of the free vibrations is called free or natural
frequency.
2. Forced vibrations When the body vibrates under the influence of external force,
the the body is said to be under forced vibrations.The vibrations have the same
frequency as the applied force
3. Damped vibrations: When there is a reduction in amplitude over every cycle of
vibration, due to frictional resistance, the motion is said to be damped vibration.

TYPES OF VIBRATIONS
1. Longitudinal Vibrations:
Parallel to axis of shaft
2. Transverse Vibrations:
Approx. Perpendicular to axis
of shaft
3. Torsional Vibrations:
Moves in circles about axis of
shaft
Longitudinal Transverse Torsional

SPRING - STIFFNESS
d = Static deflection of spring
in meters.

NATURAL FREQUENCY OF FREE
VIBRATIONS
Equilibrium method – Longitud. Vibrations
• Restoring force
 W – (sd + sx)
 - sx
• Accelerating force
m(d2x/dt 2)
• Equating both
m(d2x/dt 2) + sx = 0
d2x/dt 2 + (s/m) x = 0
• SHM equation
d2x/dt 2 + w2 x = 0
• Angular Velocity:

W = mg =
sd
SF = 0

NATURAL FREQUENCY OF FREE
VIBRATIONSEquilibrium method
• Time period
 tp = 2p/w
• Natural frequency
 A
 B
• Deflection
 s= W/A = Ee = E x (d/l)
 d = Wl/ AE

Rayleigh’s method
In this method, the maximum kinetic energy at the mean position is equal to the
maximum potential energy (or strain energy) at the extreme position. Assuming the
motion executed by the vibration to be simple harmonic, then

NATURAL FREQUENCY OF
TRANSVERSE VIBRATIONS
Transverse vibrations
<= Same as Longitudinal
Vibrations

FORMULA
1. Natural frequency of longitudinal and transverse vibrations:
Where,
fn = Natural frequency. (Hz)
tp = Time period (s)
s = Stiffness (N/m)
m = Mass. (kg)
g = acceleration due to gravity (m/s2)
d = Static deflection. (m)

A cantilever shaft 50 mm diameter and 300 mm long has a disc
of mass 100 kg at its free end. The Young's modulus for the
shaft material is 200 GN/m 2 . Determine the frequency of
longitudinal and transverse vibrations of the shaft

FORMULA
2. Static deflection in beams,
Where,
fn = Natural frequency. (Hz)
tp = Time period (s)
s = Stiffness (N/m)
m = Mass. (kg)
g = acceleration due to gravity (m/s2)
d = Static deflection. (m)

FORMULA
2. Static deflection in beams,

1. A shaft of length 0.75 m, supported freely at the ends, is carrying a body of
mass 90 kg at 0.25 m from one end. Find the natural frequency of transverse
vibration. Assume E = 200 GN/m2 and shaft diameter = 50mm.
Given
l = 0.75 m ;
m = 90 kg ;
a = AC = 0.25 m ;
E = 200 GN/m2 = 200 × 109 N/m2
d = 50 mm = 0.05 m

• Moment of inertia of shaft
• Static deflection

A flywheel is mounted on a vertical shaft as shown in Fig. 23.8.
The both ends of the shaft are fixed and its diameter is 50 mm.
The flywheel has a mass of 500 kg. Find the natural frequencies
of longitudinal and transverse vibrations. Take E = 200 GN/m2
.

EFFECT OF MULTIPLE LOADS/ENERGY
(OR RAYLEIGH’S) METHOD
• To find the natural frequency of a
beam on which multiple loads are
acting, we find the deflection caused
by each load separately and find the
total effect.
Multiple loads on SSB

EFFECT OF SELF
WEIGHT/DUNKERLEY’S METHOD
• The weight of a beam is taken as an UDL acting on top of existing loads.

DEFLECTION UNDER SELF
WEIGHT/UDL

A shaft 50 mm diameter and 3 metres long is simply supported at the ends and carries three
loads of 1000 N, 1500 N and 750 N at 1 m, 2 m and 2.5 m from the left support. The
Young's modulus for shaft material is 200 GN/m2. Find the frequency of transverse
vibration.
Given
d = 50 mm = 0.05 m ;
l = 3 m,
W1 = 1000 N ;
W2 = 1500 N ;
W3 = 750 N;
E = 200 GN/m2 = 200 × 109 N/m2

• Static deflections due to point loads,.
1000 N load
1500 N load

• Static deflections due to point loads
750 N load
• Frequency of transverse vibrations

2. Calculate the natural frequency of a shaft 20 mm diameter
and 0.6 m long carrying a mass of 1 kg a its mid-point. The
density of the shaft material is 40 Mg/m3 and Young’s
modulus is 200 GN/m2. Assume the shaft to be freely
supported.
Given
d = 20 mm = 0.02 m ;
l = 0.6 m ;
m1 = 1 kg ;
ρ = 40 Mg/m3
= 40 × 106 g/m3
= 40 × 103 kg/m3
E = 200 GN/m2
= 200 × 109 N/m2

• Mass per unit length
• Static deflection due to point load

• Static deflection due to self weight-UDL
• Natural frequency of shaft

ASSIGNMENT
Calculate the natural frequency of a shaft 30 mm diameter and 0.6 m long carrying a mass of 1
kg a its mid-point and 2kg at 0.4m from the left end. The density of the shaft material is 30
Mg/m3 and Young’s modulus is 200 GN/m2. Assume the shaft to be freely supported.

CRITICAL/WHIRLING
SPEED OF SHAFT
• The centre of gravity of the pulley or gear mounted on a shaft is at certain distance
from the axis of rotation.
• Due to this, the shaft is subjected to centrifugal force which will bend the shaft
which will further increase the distance of centre of gravity from the axis of rotation.
• The bending of shaft not only depends upon the distance between C.G of the shaft
and gears, but also depends upon the speed at which the shaft rotates.
• The speed at which the shaft runs so that the additional deflection of the shaft
from the axis of rotation becomes infinite, is known as critical or whirling
speed.

1. A SHAFT 1.5 M LONG, SUPPORTED IN FLEXIBLE BEARINGS AT THE
ENDS CARRIES TWO WHEELS EACH OF 50 KG MASS. ONE WHEEL IS
SITUATED AT THE CENTRE OF THE SHAFT AND THE OTHER AT A
DISTANCE OF 375 MM FROM THE CENTRE TOWARDS LEFT. THE SHAFT
IS HOLLOW OF EXTERNAL DIAMETER 75 MM AND INTERNAL
DIAMETER 40 MM. THE DENSITY OF THE SHAFT MATERIAL IS 7700
KG/M3 AND ITS MODULUS OF ELASTICITY IS 200 GN/M2. FIND THE
LOWEST WHIRLING SPEED OF THE SHAFT, TAKING INTO ACCOUNT THE
MASS OF THE SHAFT.
Given:
l = 1.5 m ;
m1 = m2 = 50 kg ;
d1 = 75 mm = 0.075 m ;
d2 = 40 mm = 0.04 m ;
ρ = 7700 kg/m3
E = 200 GN/m2
= 200 × 109 N/m2

• Mass per unit length

• Deflection of beam at C
• Deflection of beam at D

• Static deflection due to weight of shaft
• Critical speed of shaft
Nc = 32.4 x 60 = 1944 rpm

2. . A vertical shaft of 5 mm diameter is 200 mm long and is supported in long bearings at
its ends. A disc of mass 50 kg is attached to the centre of the shaft. Neglecting any increase
in stiffness due to the attachment of the disc to the shaft, find the critical speed of rotation
Given
d = 5 mm = 0.005 m ;
l = 200 mm = 0.2 m ;
m = 50 kg ;
E = 200 GN/m2 = 200 × 109 N/m2

• Static deflection of shaft (fixed shaft-Long bearings)
d = Wa3b3/3EIl3
= 50 x 9.81 x (0.1)3 x(0.1)3 / 3 x200 × 109x 30.7 x 10-12
= 3.33 x 10-3 m
• Frequency of the shaft
fn
• Critical speed of shaft
Nc = 8.64 x 60 = 518.4 rpm ~ 520 rpm

A vertical steel shaft 15 mm diameter is held in long bearings 1
metre apart and carries at its middle a disc of mass 15 kg. The
eccentricity of the centre of gravity of the disc from the centre of
the rotor is 0.30 mm. The modulus of elasticity for the shaft
material is 200 GN/m2 and the permissible stress is 70 MN/m2.
Determine : 1. The critical speed of the shaft and 2. The range of
speed over which it is unsafe to run the shaft. Neglect the mass of
the shaft

DAMPING
• The vibrations are resisted by friction
force.
• This frictional resistance is called as
damping force.
• The damping element of a body is
represented by the damping
coefficient ‘c’.

DAMPING TYPES
• Based on the medium of force resistance.
• Coulomb’s damping (Dry friction)
• Viscous damping (Fluid damping)
• Based on the exciting force
• Free damped vibrations
• Forced damped vibrations

FREQUENCY OF FREE DAMPED
VIBRATIONS
• A viscous damping system

FORCES ON SYSTEM
• Damping force
• Spring force
• Inertia force
= sx

VIBRATIONS
• Equation of motion
• Solving the differential equation,
Put, x = ekt

• Solving the differential equation,
VIBRATIONS
• The roots are,

TYPES OF DAMPING
1. Over damping (Real roots)
• Real roots occur when,
• In over damping the mass moves slowly to
the equilibrium position.

TYPES OF DAMPING
2. Under damping (imaginary roots)
• Imaginary roots or complex conjugate
occur when,
• Underdamped systems are the most
applicable in real life.
• The system oscillates (at reduced
frequency compared to the undamped
case) with the amplitude gradually
decreasing to zero.
Frequency of Underdamped
system

TYPES OF DAMPING
3. Critical damping (equal roots)
• Critical damping occurs when roots are
equal
• Critically damped systems are find in
automobile suspensions.
• The system returns to equilibrium as
quickly as possible without oscillating.
=

FREQUENCY (F) VS. CIRCULAR
FREQUENCY (W)
• Frequency
 It is the number of cycles (repeated
movements) per second.
 Unit is s-1 or Hz
The expression for natural frequency is
• Circular Frequency
 It is the number of radians ()
per second.
 Unit is rad/s-1
The expression for circular
natural frequency is
1 cycle = 2p rad

CRITICAL DAMPING COEFFICIENT
• Critical Damping coefficient ( cc)
• The damping coefficient at which critical damping
occurs is called as critical damping coefficient

DAMPING RATIO
• The ratio of the actual damping coefficient (c) to the critical damping coefficient (cc)
is known as damping factor or damping ratio.
• The damping factor is the measure of the relative amount of damping in the existing
system with that necessary for the critical damped system
Damping factor, z

LOGARITHMIC DECREMENT
• It is defined as the natural logarithm of the
amplitude reduction factor.
• The amplitude reduction factor is the ratio of any
two successive amplitudes on the same side of the
mean position.

The following data are given for a vibratory system with viscous damping:
Mass = 2.5 kg ; spring constant = 3 N/mm and the amplitude decreases to 0.25
of the initial value after five consecutive cycles. Determine the damping
coefficient of the damper in the system
Given :
M = 2.5 kg ;
s = 3 N/mm = 3000 N/m ;
x6 = 0.25 x1
• Natural circular frequency

• Amplitude reduction factor
We know that,

• Logarithmic decrement
c = 7.65 Nm/s

FORMULAE
• Damped Vibration frequency
• Undamped Vibration frequency
• Critical damping coefficient
• Logrithmic decrement
• Damping ratio
z

1.
2. The measurements on a mechanical vibrating system show that it has a mass of
8 kg and that the springs can be combined to give an equivalent spring of
stiffness 5.4 N/mm. If the vibrating system have a dashpot attached which
exerts a force of 40 N when the mass has a velocity of 1 m/s, find :
1. critical damping coefficient,
2. damping factor,
3. logarithmic decrement, and
4. ratio of two consecutive amplitudes.
Given :
m = 8 kg ;
s = 5.4 N/mm = 5400 N/m
Since the force exerted by dashpot is 40 N, and the mass has a velocity of 1 m/s ,
therefore
Damping coefficient (actual),
c = 40 N/m/s

• Critical damping coefficient
• Damping factor

• Ratio of two consecutive amplitude (Amplitude reduction factor)

A coil of spring stiffness 4 N/mm supports vertically a mass of 20 kg at
the free end. The motion is resisted by the oil dashpot. It is found that
the amplitude at the beginning of the fourth cycle is 0.8 times the
amplitude of the previous vibration. Determine the damping force per
unit velocity. Also find the ratio of the frequency of damped and
undamped vibrations
3.
Given :
s = 4 N/mm = 4000 N/m ;
m = 20 kg
x4 = 0.8 x3
x3/x4 = 1/0.8 = 1.25
To find :
i. Damping force per unit velocity (c)
ii. Ratio of frequencies (wd/wn)

c = 20 N/m/s

• Ratio of frequencies
= 0.999

ASSIGNMENT
The mass of a single degree damped vibrating system is 7.5 kg and
makes 24 free oscillations in 14 seconds when disturbed from its
equilibrium position. The amplitude of vibration reduces to 0.25 of its
initial value after five oscillations. Determine :
1. stiffness of the spring,
2. logarithmic decrement, and
3. damping factor, i.e. the ratio of the system damping to critical
damping.

A machine of mass 75 kg is mounted on springs and is fitted
with a dashpot to damp out vibrations. There are three springs
each of stiffness 10 N/mm and it is found that the amplitude of
vibration diminishes from 38.4 mm to 6.4 mm in two complete
oscillations. Assuming that the damping force varies as the
velocity, determine : 1. the resistance of the dashpot at unit
velocity ; 2. the ratio of the frequency of the damped vibration
to the frequency of the undamped vibration ; and 3. the
periodic time of the damped vibration.

A mass suspended from a helical spring vibrates in a viscous
fluid medium whose resistance varies directly with the speed.
It is observed that the frequency of damped vibration is 90
per minute and that the amplitude decreases to 20 % of its
initial value in one complete vibration. Find the frequency of
the free undamped vibration of the system.

LINEAR VIBRATIONS VS TORSIONAL
VIBRATIONS
• Frequency of linear vibrations
• Where,
s = spring Stiffness (N/m)
(Force to deflect the spring by 1
m)
m = mass (kg)
(Resistance to linear motion)
• Frequency of torsional
vibrations
• Where,
q = Torsional Stiffness (N-m)
(Torque to rotate the spring by 1
rad)
I = Mass moment of inertia (kg-m2)
(Resistance to rotary motion)

TORSIONAL STIFFNESS
• Stiffness
Force per unit length
s = W/d
Where,
W = Load (N)
d = Static deflection (m)
• Torsional Stiffness
Torque per unit angle(rad)
q =
Where,
T = Torque (N-m)
q = angular displacement (rad)
C or G = Rigidity modulus (N/m2)
J = Polar moment of inertia (m4)

A flywheel is mounted on a vertical shaft as shown in figure. The both ends of a shaft are
fixed and its diameter is 50 mm. The flywheel has a mass of 500 kg and its radius of
gyration is 0.5 m. Find the natural frequency of torsional vibrations, if the modulus of
rigidity for the shaft material is 80 GN/m2.
Given
d = 50 mm = 0.05 m ;
m = 500 kg ;
k = 0.5m;
C = 80 GN/m2
= 80 × 109 N/m2
• Polar moment of inertia

• Torsional stiffness of length l1
• Torsional stiffness of length l2
Total torsional stiffness: q = q1 + q2
= 56 x 103 + 84 x 103
= 140 x 103 N-m

• Mass Moment of inertia of shaft
• Natural frequency of torsional vibrations
f n = 5.35 Hz

TWO ROTOR SYSTEM
• In a two Rotor system, a shaft held in bearings
carries a rotor at each end, it can vibrate
torsionally such that the two rotors move in the
opposite directions.
• Thus some length of the shaft is twisted in one
direction while the rest is twisted in the other.
• The section which does not undergo any twist
is called the node.

TWO ROTOR SYSTEM
• In a two rotor system,the frequency of both the parts of the shaft are same. And so,

THREE ROTOR SYSTEM
• Consider , the two rotors A and B are
fixed to the ends of the shaft. and the
rotor C is in between. Let the rotors A
!nd B rotate in the same direction and
C in the opposite direction and the
node points lie at D and E as shown in

THREE ROTOR SYSTEM
• When A and C are rotating in same
direction and B is rotating in opposite
direction, node is formed in between C
and B

THREE ROTOR SYSTEM
• When B and C are rotating in same
direction and A is rotating in opposite
direction, node is formed in between C
and A

TORSIONALLY EQUIVALENT SHAFT
• To find natural frequency of rotors
attached to shaft with varying
diameter, it is converted into an
equivalent shaft of a nominal
diameter.
• A torsionally equivalent shaft is one
which has the same torsional
stiffness as that of the stepped shaft We know that
T/q = GJ/l => q = Tl/GJ

TORSIONALLY EQUIVALENT SHAFT• Total twist in shaft
=>

TORSIONALLY EQUIVALENT SHAFT
• If the diameter of the first section of the stepped shaft ‘d1’ is taken as the diameter
of the ‘torsionally equivalent shaft’ , d, then
Put d = d1

1. A steel shaft 1.5 m long is 95 mm in diameter for the first 0.6 m of its length, 60 mm in
diameter for the next 0.5 m of the length and 50 mm in diameter for the remaining 0.4m of
its length. The shaft carries two flywheels at. two ends, the first having a mass of 900 kg
and 0.85 m radius of gyration located at the 95 mm diameter end and the second having a
mass of 700 kg and 0.55 m radius of gyration Located at the other end. Determine the
location of the node and the natural frequency of free torsional vibration of the system. The
modulus af rigidity of shaft material may be taken as 80 GN/m2.
Given:
L = 1.5 m;
d1 = 95 mm= 0.095 m;
l1 = 0.6 m;
d2 = 60 mm= 0.06 m.
l2 = 0.5 m ;
d3 = 50 mm= 0.05 m ;
13 = 0.4 m ;
mA = 900 kg;
kA = 0.85 m;
mb = 700 kg;
kB= 0.5 m
C = 80 GN/m2
= 80 x 109 N/m2

• Length of torsionally equivalent shaft of diameter d1
• Location of node
Mass moment of inertia of A
Mass Moment of inertia of B

• Position of node on original shaft

• Natural frequency of torsional vibrations
Polar Moment of inertia
Natural frequency
fn = 3.35 Hz

Stepped shaft to Torsional shaft
0.85m
T.S S.S
la/(0.095)4 = 0.6/(0.095)4 +0.25/(0.06)4
la = (7,366.43 + 19,290.1235) x (0.095)4
la = 2.17 m
Torsional shaft to Stepped shaft
2.2m
T.S S.S
2.2/(0.095)4 = 0.6/(0.095)4 +l2/(0.06)4
27,010.2286 = 7,366.43 +l2/(0.06)4
19,643.8 x (0.06)4 = l2
l2= 0.255 m
Total distance from A in stepped shaft
0.6+0.255 = 0.855 m

2. A steel shaft ABCD 1.5 m long has flywheel at its ends A and D. The mass of the
flywheel A is 600 kg and has a radius of gyration of 0.6 m. The mass of the flywheel
D is 800 kg and has a radius of gyration of 0.9 m. The connecting shaft has a
diameter of 50 mm for the Portion AB which is 0.4 m long ; and has a diameter of
60 mm for the portion BC which is 0.5 m long ; and has a diameter of d mm for the
portion CD which is 0.6 m long. Detennine :
1. the diameter 'd' of the portion CD so that the node of the torsional vibration of
the system will be at the centre of the length BC; and
2. The natural frequency of the torsional vibrations. The modulus of rigidity for
the shaft material is 80 GN/m2
Given :
L = 1.5 m ;
mA = 600 kg ;
kA = 0.6 m ;
mD= 800 kg ;
kD = 0.9 m ;
d1 = 50 mm = 0.05 m ;
L1 = 0.4 m ;
d2 = 60 mm = 0.06 m ;
l2= 0.5 m ;
d3 = d ;
13 = 0.6 m ;
C = 80 GN/m2 = 80 x109 N/m2

• Length of equivalent shaft
• Mass moment of inertia of Flywheel A
• Mass moment of inertia of Flywheel A

But, we know that length of torsionally
equivalent shaft is,
l = lA + lD = 0.52 + 0.173 = 0.693m

• The diameter of shaft CD
0.693
d = 0.0917 m = 91.7 mm
fnA
Polar moment of inertia
Natural frequency

0.25 m
30 mm 40 mm
50 mm
0.4 m 0.6 m

2 ROTOR
1 1
3 Rotor
1 1
IclcIala

1. A single cylinder oil engine drives directly a centrifugal pump. The rotating mass
of the engine, flywheel and the pump with the shaft is equivalent to a three rotor
system as shown in Fig. The mass moment, of inertia of the rotors A, B and C are
0.15, 0.3 and 0.09 kg-m2. Find the natural frequency of the torsional vibration. The
modulus of rigidity for the shaft material is 84 kN/m2
Given :
IA = 0.15 kg-m2 ;
IB = 0.3 kg-m2 ;
IC = 0.09 kg-m2 ;
d = 70 mm =0.07 m;
l1= 1.5 m;
l2 = l m;
C= 84 kN/mm2 = 84 x 109 N/m2

• Polar moment of inertia
• Natural frequency of single node system

• Natural frequency of two node system

FREE TORSIONAL VIBRATIONS
OF A GEARED SYSTEM

Vibration Analysis Unit

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Vibration Analysis Unit