TOM UNIT-V Vibrations.pptx

UNIT – V
VIBRATIONS
Dr. S.Baskaran
Associate Professor
Department of Mechanical Engineering
Madanapalle Institute of Technology & Science
Madanapalle

2
UNIT V: VIBRATIONS
Introduction, degree of freedom, types of vibrations, free natural vibrations, Newton
method and energy method for single degree of freedom. Damped vibrations- under
damped, critically damped; and over damped systems, forced vibrations with
and without damping in single degree of freedom; Vibration isolation and
transmissibility, Torsional vibrations - two and three rotor systems.

3
Course Objectives
 To study basic concepts and terminology used in vibration analysis
 To study the basic concepts of degrees of freedom and elementary parts of
vibrating systems.
 To study the various classifications of vibrations
 To mathematically model real-world mechanical vibration problems
 To determine the effect of vibration on the performance and safety of systems,
and to control its effects.

4
Course Outcomes
Upon completion of this course the student will be able to:
 Understand the importance of study of vibration
 Give various classifications of vibration
 Know the concepts of vibration and single degree of freedom systems.
 Know the concepts of noises and the ways to control it
 Apply Newton’s equation of motion and energy methods to model basic
vibrating mechanical systems

Why Study Vibration ??
Tacoma Narrows bridge during wind-induced vibration. The bridge
opened on July 1, 1940, and collapsed on November 7, 1940.
Earthquake induced vibration
Turbine blade failure
5

Objectionable vibrations in a machine cause the loosening of parts, its
malfunctioning, or its eventual failure
The vibrational motions of engines, electric motors, other mechanical devices
which are usually the results of imbalance in rotating parts, uneven friction,
meshing gear, etc. are typically unwanted
Jet Engines
6
Washing Machine
Turbines
Wind Mill

Desirable Vibrations
Musical Instruments
Time keeping
Surface Finishing
Shakers, Sievers, Mixers
Conveyors, Hoppers, Pneumatic drills
Vibrators
7

9
Effect of Vibrations
Vibration in automobiles or in a train can cause discomfort to passengers
The structure or machine component subjected to vibration can fail due to
material
Fatigue resulting from the cyclic variation of the induced stress
Vibration causes more rapid wear of machine parts such as bearings and gears and
also creates excessive noise
In turbines, vibrations cause mechanical failures
The structures designed to support heavy centrifugal machines, like motors and
turbines, or reciprocating machines, like steam and gas engines and reciprocating
pumps, are also subjected to vibration
In machines, vibration causes fasteners such as nuts to become loose
In metal cutting processes, vibration can cause chatter, which leads to a poor
surface finish

Vibrations occurs due to…
Existed unbalanced forces
Foundation is not Good
Loose fittings in fasteners (reciprocating parts/ pumps)
Stopping and starting of aero plane/ train
Uneven surfaces (Ex. Speed breakers)
Existence of the other forces (wind)
Sudden forces, earth quake
Why to study Vibrations…..
Minimize the un balanced forces
Provide comfort / smooth running
Minimize the fatigue failures
Minimize the unwanted or dangerous vibrations 10

11
Importance of the study of Vibration
Vibrations can lead to excessive deflections and failure on the machines and
structures.
 To reduce vibration through proper design of machines and their mountings.
To improve the efficiency of certain machining, casting, forging & welding
processes.
To stimulate earthquakes for geological research and conduct studies in design
of nuclear reactors.

12
Mechanical Vibration
Any motion that repeats itself after an interval of time is called vibration or
oscillation. Vibration is a periodic or oscillatory motion of an object or a set of
objects.
The theory of vibration deals with the study of oscillatory motions of bodies and
the forces associated with them.
When elastic bodies 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.
Example: Swinging of a pendulum
Motion of a plucked string

Degree of Freedom (DOF)
Minimum number of independent coordinates required to determine completely the
positions of all parts of a system at any instant of time.
Examples of single degree-of-freedom systems:
14

Examples of second degree-of-freedom systems:
15

Examples of three degree-of-freedom systems:
16

Infinite number of degrees of freedom system are termed continuous or
distributed systems
Finite number of degrees of freedom are termed discrete or lumped parameter
systems
More accurate results obtained by increasing number of degrees of freedom
17
Examples of Infinite number of degrees-of-freedom system:

ELEMENTS OFA VIBRATORY SYSTEM
The elements that constitute a vibratory system are
(1) Mass
(2) Spring
(3) Damper
(4) Excitation
The first three elements describe the physical system.
For example, it can be said that a given system consists of a mass, a spring, and a damper arranged as
shown in the figure.
Energy may be stored in the mass and the spring and dissipated in the damper in the form of heat. Energy
enters the system through the application of an excitation. As shown in Figure, an excitation force is
applied to the mass m of the system.
Involves transfer of potential energy to kinetic energy and vice versa.
18

Important terms used in Vibratory motion
1.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.
2. Cycle. It is the motion completed during one time period.
3.Frequency. It is the number of cycles described in one second.
In S.I. units, the frequency is expressed in hertz (i.e. Hz)
which is equal to one cycle per second.
4.Amplitude. It is the maximum displacement of a body
from its mean position.
5. Resonance. It occurs when the frequency of the external force
coincides with one of the natural frequencies of the system. 19

Simple Harmonic Motion (SHM)
20

Simple Harmonic Motion – Spring Mass System
21

Types of vibratory motion
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.
A system is left to vibrate on its own after an initial disturbance and no external
force acts on the system.
E.g. simple pendulum
22

Simple Pendulum – Periodic Motion (SHM)
24

25
Forced vibrations
When the body vibrates under the influence of external force, then the body is
said to be under forced vibrations (or) A system that is subjected to a
repeating external force.
The external force applied to the body is a periodic disturbing force created by
unbalance.
Examples of these types of vibration include a
Washing machine shaking due to an imbalance
Transportation vibration caused by an engine or uneven road
Vibration of a building during an earthquake
Oscillation arises from diesel engines

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
resistances to the motion.
When any energy is lost or dissipated in
friction or other resistance during oscillations
26

Undamped vibrations
When there is no reduction in amplitude over every cycle of vibration, the motion
is said to be undamped vibration.
27

28
Types of Free Vibrations
1. Longitudinal vibrations
2. Transverse vibrations
3. Torsional vibrations

Longitudinal vibrations
When the particles of the shaft or disc moves parallel to the axis of
the shaft, as shown in Figure, then the vibrations are known as
longitudinal vibrations.
In this case, the shaft is elongated and shortened alternately and
thus the tensile and compressive stresses are induced alternately in
the shaft.
29

Transverse vibrations
When the particles of the shaft or disc move approximately
perpendicular to the axis of the shaft, as shown in Figure, then
the vibrations are known as transverse vibrations. In this case,
the shaft is straight and bent alternately and bending stresses are
induced in the shaft.
Examples:
30

Torsional vibrations
When the particles of the shaft or disc move in a circle
about the axis of the shaft, as shown in Figure, then the
vibrations are known as torsional vibrations.
In this case, the shaft is twisted and untwisted alternately
and the torsional shear stresses are induced in the shaft.
31

Natural Frequency of Free Longitudinal Vibrations
Equilibrium Method
Consider a constraint (i.e. spring) of negligible mass in an
unstrained position, as shown in Fig.
Let s = Stiffness of the constraint. It is the force required
to produce unit displacement in the direction of vibration.
It is usually expressed in N/m.
m = Mass of the body suspended from the constraint in kg
W = Weight of the body in newtons = mg
δ = Static deflection of the spring in metres due to weight
W newtons
x = Displacement given to the body by the external force,
in metres.
32

In the equilibrium position, as shown in Figure,
the gravitational pull W = mg
which is balanced by a force of spring,
such that W = sδ
By applying an external force, assume the body
is displaced vertically by a distance ‘x’, from the
equilibrium position as shown in Figure.
On the release of external force, the unbalanced
forces and acceleration imparted to the body are
related by Newton Second Law of motion.
33

therefore after time t,
Equating equations (i) and (ii), the equation of motion of the body of mass m after time t is
or
Spring force
Newton Second Law of motion
35

We know that the fundamental equation of simple harmonic motion is
36

Natural Frequency of Free Longitudinal Vibrations
Energy method
 A system is said to be conservative if no energy is lost due to
friction or energy-dissipating nonelastic members.
 Since the energy of a vibrating system is partly potential and
partly kinetic, the sum of these two energies remains
constant.
 In extreme positions: Shaft possesses maximum potential
energy and no kinetic energy.
 In the mean position: Shaft possesses maximum kinetic
energy and no potential energy. 38

We know that the kinetic energy is due to the motion of the body
and the potential energy is with respect to a certain datum
position which is equal to the amount of work required to move
the body from the datum position.
In the free vibrations, no energy is transferred to the system or from
the system. Therefore the summation of kinetic energy and potential
energy must be a constant quantity which is same at all the times.
i.e. K.E + P.E = Constant
39

∵ P.E. = Mean force × Displacement
40

Natural Frequency of Free Transverse Vibrations
Consider a shaft of negligible mass, whose
one end is fixed and the other end carries a
body of weight W, as shown in Figure.
Let s = Stiffness of shaft,
δ = Static deflection due to weight of
the body,
x = Displacement of body from mean
position after time t.
m = Mass of body = W/g
Restoring force = – s.x
Accelerating force =
42

Problem 1:
Acantilever 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/m2. Determine the
frequency of longitudinal and transverse vibrations of the shaft.
Solution:
Given data: d = 50 mm = 0.05 m ; l = 300 mm = 0.03 m ;
m = 100 kg ; E = 200 GN/m2 = 200 ×109 N/m2
Wanted data: Frequency of longitudinal and transverse vibrations
Formula : Frequency of longitudinal and Transverse vibration
43

We know that static deflection of the shaft due to longitudinal vibration,
Cross-sectional area of the shaft,
Moment of inertia of the shaft,
Frequency of longitudinal vibration,
44

We know that static deflection of the shaft due to Transverse vibration,
Frequency of Transverse vibration,
45

Natural Frequency of the various types of beams
and under various load conditions
46
Type of beam Diagram Deflection (δ) Natural Frequency
Spring or shaft suspended vertically whose
one end is fixed and the other end is free
which carrying a heavy disc (Longitudinal
vibration)
Mass of spring or shaft neglected
Mass of spring or shaft
considered
Cantilever beam with a point load W at the
free end (Transverse vibration)
(at the free end)

47
Cantilever beam with a uniformly distributed
load of w per unit length
f = 0.56 (EI/wl4)0.5
Simply supported beam with an eccentric
point load W
Simply supported beam with a central point
load W
Simply supported beam with a uniformly
distributed load of w per unit length
(at the centre)

48
Fixed beam with an eccentric point load W
Fixed beam with a central point load W
Fixed beam with a uniformly distributed load
of w per unit length
Shaft carrying a number of point loads and a
uniformly distributed load
(Dunkerley’s empirical formula)

Problem 2:
Ashaft of length 0.75 m, simply 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 = 50 mm.
Solution:
Given data: l = 0.75 m ; m = 90 kg ; a = AC = 0.25 m ; d = 50 mm = 0.05 m;
E = 200 GN/m2 = 200 ×109 N/m2
Wanted data: Natural frequency of transverse vibration
Formula : Frequency of Transverse vibration
49

Static deflection at the load point (i.e. at point C),
We know that moment of inertia of the shaft,
Frequency of Transverse vibration
50

Problem 3:
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.
Solution:
Given data: d = 50 mm = 0.05 m, l = 3 m,
W1 = 1000 N, W2 = 1500 N, W3 = 750 N, E = 200 GN/m2
Wanted data: Natural frequency of transverse vibration
Formula : Frequency of Transverse vibration
= 200 ×109 N/m2
51

The static deflection due to a point load W,
We know that moment of inertia of the shaft,
Static deflection due to a load of 1000 N,
Frequency of Transverse vibration
52

Critical or Whirling speed of a shaft
 Critical or whirling speed is defined as the
speed at which a rotating shaft will tend to
vibrate violently in the transverse direction if
the shaft rotates in horizontal direction.
 In other words, the whirling or critical speed
is the speed at which resonance occurs.
53

m = Mass of the rotor,
e = Initial distance of centre of gravity of the rotor from the centre line of the bearing or shaft axis,
when the shaft is stationary,
y =Additional deflection of centre of gravity of the rotor when the shaft starts rotating at ω rad/s,
s = Stiffness of the shaft i.e. the load required per unit deflection of the shaft.
Since the shaft is rotating at ω rad/s, therefore centrifugal force acting radially outwards through
G causing the shaft to deflect is given by
The shaft behaves like a spring. Therefore the force resisting the deflection, y = s y
For the equilibrium position,
57

We know that circular frequency,
,
when ωn = ωc , the value of y becomes infinite. Therefore ωc
whirling speed.
Critical or whirling speed,
is the critical or
If Nc is the critical or whirling speed in r.p.s., then
Hence the critical or whirling speed is the same as the natural frequency of
transverse vibration but its unit will be revolutions per second.
58

59
Problem 4:
Calculate the whirling speed of a shaft 20 mm diameter and 0.6 m long carrying a
mass of 1 kg at 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.
Solution:
Given data: 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
Wanted data: Whirling speed of a shaft
Formula : Whirling speed of a shaft
(Whirling speed of a shaft in r.p.s. is equal to the
frequency of transverse vibration in Hz)
= 200 ×109 N/m2

Static deflection due to 1 kg of mass at the centre,
Static deflection due to mass of the shaft,
Mass of the shaft per metre length,
Frequency of transverse vibration,
Whirling speed of a shaft in r.p.s. is equal to the frequency of transverse vibration in Hz ,
60

61
Problem 5:
;
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.
Solution:
Given data: 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
Formula : Whirling speed of a shaft
Wanted data: Whirling speed of a shaft taking into account the mass of the shaft
(Whirling speed of a shaft in r.p.s. is equal to
the frequency of transverse vibration in Hz)

The static deflection due to a load W
Static deflection due to a mass of 50 kg at C,
Static deflection due to a mass of 50 kg at D,
Mass of the shaft per metre length,
62

Static deflection due to uniformly distributed load or mass of the shaft,
Frequency of transverse vibration,
The whirling speed of shaft (Nc) in r.p.s. is equal to the frequency of transverse vibration in Hz,
therefore Nc = 32.4 r.p.s. = 32.4 × 60 = 1944 r.p.m.
63

The Diminishing of vibrations with time is called
Damping. Damping is the decrease in amplitude with time
due to the resistance of the medium to the vibration.
Damping occurs progressively as energy is taken out of
the system by another force such as friction.
Examples:
 Internal forces of a spring
 Viscous force in a fluid
 Shock absorber in a car
 Amass oscillates under water
 Oscillation of a metal plate in the magnetic field
64
Damped Vibrations (Viscous Damping)
Damper or dash-pot
At resonance condition
Damped condition
Undamped condition

Let m = Mass suspended from the spring,
s = Stiffness of the spring,
x = Displacement of the mass from the mean position at time t,
δ = Static deflection of the spring = m.g/s, and
c = Damping coefficient or the damping force per unit velocity.
The equation of motion
Frequency of Free Damped Vibrations (Viscous Damping)
65

Frequency of Free Damped Vibrations (Viscous Damping)
This is a differential equation of the second order. Assuming a solution of the form x = ekt
where k is a constant to be determined. Now the above differential equation reduces to
66

(i) When the roots are real (overdamping i.e No Oscillation)
The roots k1 and k2 are real but negative. This is a case of overdamping or large damping and the
mass moves slowly to the equilibrium position. This motion is known as aperiodic. When the
roots are real, the most general solution of the differential equation is
67

(ii) When the roots are complex conjugate (underdamping i.e System oscillates with
amplitude decreasing exponentially overtime)
the radical (i.e. the term under the square root) becomes negative. The two roots k1 and k2 are then
known as complex conjugate. This is a most practical case of damping and it is known as
underdamping or small damping. The two roots are
68

(iii) When the roots are equal (critical damping i.e No Oscillation)
the radical becomes zero and the two roots k1 and k2 are equal. This is a case of critical damping. In other
words, the critical damping is said to occur when frequency of damped vibration (fd) is zero (i.e. motion is
aperiodic). This type of damping is also avoided because the mass moves back rapidly to its equilibrium
position, in the shortest possible time. For critical damping, equation may be written as
The critical damping coefficient is the amount of damping required for a system to be critically damped. 69

Comparison of damped vibrations
70

Damping factor or damping ratio
The ratio of the actual damping coefficient (c) to the critical
damping coefficient (cc) is known as damping factor or damping ratio.
Mathematically,
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 logarithmic decrement represents the rate at which the amplitude of a free damped vibration
decreases.
71

Problem 6:
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.
Solution:
Given data: m = 2.5 kg ; s = 3 N/mm = 3000 N/m ; x6 = 0.25 x1
Wanted data: Damping coefficient of the damper (c)
Formula :
, ,
72

Problem 7:
An instrument vibrates with a frequency of 1 Hz when there is no damping. When
the damping is provided, the frequency of damped vibrations was observed to be 0.9
Hz. Find 1. the damping factor, and 2. logarithmic decrement.
Solution:
Given data: fn = 1 Hz ; fd = 0.9 Hz
Wanted data: Damping factor, Logarithmic decrement
Formula :
,
75

77
Problem 8:
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.
Solution:
Given data: s = 4 N/mm = 4000 N/m ; m = 20 kg
Wanted data: Damping force, Ratio of the frequency of damped
and undamped vibrations
Formula :
,

80
Problem 9:
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 dash-pot 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.
Solution:
Given data:m = 75 kg; s = 10 N/mm = 10 ×103 N/m; x1 = 38.4 mm = 0.0384 m; x3 = 6.4 mm =
0.0064 m
Wanted data: Resistance of the dash-pot, Ratio of the frequency of damped and undamped
vibrations, Periodic time of the damped vibration
Formula :
, ,

Forced vibrations without damping in single degree of freedom
Forced vibrations
83

Forced vibrations with damping in single degree of freedom
84

Magnification Factor or Dynamic Magnifier (D)
It is the ratio of maximum displacement of the forced vibration (xmax) to the deflection due to the
static force F(xo).
86

Problem 10:
A mass of 10 kg is suspended from one end of a helical spring, the other end being
fixed. The stiffness of the spring is 10 N/mm. The viscous damping causes the
amplitude to decrease to one-tenth of the initial value in four complete oscillations.
If a periodic force of 150 cos 50 t N is applied at the mass in the vertical direction,
find the amplitude of the forced vibrations. What is its value of resonance?
Solution:
Given data: m = 10 kg ; s = 10 N/mm = 10 × 103 N/m ; x5= x1/10
Wanted data: Amplitude of the forced vibrations, Resonance
Formula :
, 87

90
Problem 11:
A body of mass 20 kg is suspended from a spring which deflects 15 mm under this load.
Calculate the frequency of free vibrations and verify that a viscous damping force amounting to
approximately 1000 N at a speed of 1 m/s is just-sufficient to make the motion aperiodic.
If when damped to this extent, the body is subjected to a disturbing force with a maximum value
of 125 N making 8 cycles/s, find the amplitude of the ultimate motion.
Solution:
Given data: m = 20 kg ; δ = 15 mm = 0.015 m ; c = 1000 N/m/s ; F = 125 N ; f = 8 cycles/s
Wanted data: Frequency of free vibrations, V
erify a viscous damping force, Amplitude of the
ultimate motion
Formula :
, ,

Vibration Isolation and Transmissibility
The ratio of the force transmitted (FT) to the force applied (F) is known as the isolation factor or
transmissibility ratio of the spring support.
93

Torsional vibrations
When the particles of the shaft or disc move in a circle
about the axis of the shaft, as shown in Figure, then the
vibrations are known as torsional vibrations.
In this case, the shaft is twisted and untwisted alternately
and the torsional shear stresses are induced in the shaft.
95

Natural Frequency of Free Torsional Vibrations (Single Rotor System)
Consider a shaft of negligible mass whose one end is fixed and the other end
carrying a disc as shown in Figure.
Let θ = Angular displacement of the shaft from mean position after time t in
radians,
m = Mass of disc in kg,
I = Mass moment of inertia of disc in kg-m2
k = Radius of gyration in metres,
q = Torsional stiffness of the shaft in N-m.
= m.k2,
96

97
If Inertia of the shaft is considered

Free Torsional Vibrations (Two Rotor Systems)
99

Free Torsional Vibrations (Three Rotor Systems)
101

Torsionally Equivalent Shaft
103
d1, d2, d3 – diameters for the
lengths l1, l2, l3 respectively
θ1, θ2, θ3 – Angle of twist for
the lengths l1, l2, l3
respectively
θ – Total angle of twist
J1, J2, J3 – Polar moment of
the shafts of
d1, d2, d3
inertia for
diameters
respectively
Assume d = d1

Problem 12:
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 0.4 m 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 of rigidity of shaft material may
be taken as 80 GN/m2.
Solution:
Assume d1 = 95 mm, find the equivalent length of shaft
104

Problem 13:
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. Determine :
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.
Solution:
Assume d1 = 50 mm, find the equivalent length of shaft
107

Problem 14:
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 Figure. 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/mm2.
Solution:
111

TOM UNIT-V Vibrations.pptx

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to TOM UNIT-V Vibrations.pptx

Similar to TOM UNIT-V Vibrations.pptx (20)

Recently uploaded

Recently uploaded (20)

TOM UNIT-V Vibrations.pptx