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ME 2302 Dynamics of Machinery 1 UNIT I - Force Analysis LECTURE-I (1) Introduction: If the acceleration of moving links in a mechanism is running with considerable amount of linear and/or angular accelerations, inertia forces are generated and these inertia forces also must be overcome by the driving motor as an addition to the forces exerted by the external load or work the mechanism does. (2) Newton’s Law: First Law Everybody will persist in its state of rest or of uniform motion (constant velocity) in a straight line unless it is compelled to change that state by forces impressed on it. This means that in the absence of a non-zero net force, the center of mass of a body either is at rest or moves at a constant ve loc ity. Second Law A body of mass m subject to a force F undergoes an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force and inversely proportional to the mass, i.e., F = ma. Alternatively, the total force applied on a body is equal to the time derivative of linear momentum of the body. Third Law The mutual forces of action and reaction between two bodies are equal, opposite and collinear. This means that whenever a first body exerts a force F on a second body, the second body exerts a force −F on the first body. F a nd −F are equal in magnitude and opposite in direction. This law is sometimes referred to as the action-reaction law, with F called the "action" and −F the "reaction" (3) Types of force Analysis: Equilibrium of members with two forces Equilibrium of members with three forces Equilibrium of membe rs with two forces and torque Equilibrium of members with two couples. Equilibrium of members with four forces. (4) Principle of Super Position: Sometimes the number of external forces and inertial forces acting on a mechanism are too much for graphical solution. In this case we apply the method of superposition. Using superposition the entire system is broken up into (n) problems, where n is the number of forces, by considering the external and inertial forces of each link individually. Response of a linear system to several forces acting simultaneously is equal to the sum of responses of the system to the forces individually. This approach is useful because it can be performed by graphically. (5) Free Body Diagram: A free body diagram is a pictorial representation often used by physicists and engineers to analyze the forces acting on a body of interest. A free body diagram shows all forces of all types acting on this body. Drawing such a diagram can aid in solving for the unknown forces or the equations of
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ME 2302 – Dynamics of Machinery1
UNIT – I - Force Analysis LECTURE-I
(1) Introduction: If the acceleration of moving links in a mechanism is running with considerable amount of
linear and/or angular accelerations, inertia forces are generated and these inertia forces also must be overcome
by the driving motor as an addition to the forces exerted by the external load or work the mechanism does.
(2) Newton’s Law: First Law Everybody will persist in its state of rest or of uniform motion (constant
velocity) in a straight line unless it is compelled to change that state by forces impressed on it. This means that
in the absence of a non-zero net force, the center of mass of a body either is at rest or moves at a constant
velocity. Second Law A body of mass m subject to a force F undergoes an acceleration a that has the same
direction as the force and a magnitude that is directly proportional to the force and inversely proportional to the
mass, i.e., F = ma. Alternatively, the total force applied on a body is equal to the time derivative of linear
momentum of the body. Third Law The mutual forces of action and reaction between two bodies are equal,
opposite and collinear. This means that whenever a first body exerts a force F on a second body, the second
body exerts a force −F on the first body. F and −F are equal in magnitude and opposite in direct ion. This law is
sometimes referred to as the action-reaction law, with F called the "action" and −F the "reaction"
(3) Types of force Analysis:
Equilibrium of members with two forces
Equilibrium of members with three forces
Equilibrium of members with two forces and torque
Equilibrium of members with two couples.
Equilibrium of members with four forces.
(4) Principle of Super Position: Sometimes the number of external forces and inertial forces acting on a
mechanism are too much for graphical solution. In this case we apply the method of superposition. Using
superposition the entire system is broken up into (n) problems, where n is the number of forces, by considering
the external and inertial forces of each link individually. Response of a linear system to several forces acting
simultaneously is equal to the sum of responses of the system to the forces individually. This approach is useful
because it can be performed by graphically.
(5) Free Body Diagram: A free body diagram is a pictor ial representation often used by physicists and
engineers to analyze the forces acting on a body of interest. A free body diagram shows all forces of all types
acting on this body. Drawing such a diagram can aid in solving for the unknown forces or the equations of
ME 2302 – Dynamics of Machinery
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motion of the body. Creating a free body diagram can make it easier to understand the forces, and torques or
moments, in relation to one another and suggest the proper concepts to apply in order to find the solution to a
problem. The diagrams are also used as a conceptual device to help identify the internal forces—for example,
shear forces and bending moments in beams—which are developed within structures.
(6) D’Alemberts Principle: D'Alembert's principle, also known as the Lagrange–d'Alembert principle , is a
statement of the fundamental classical laws of motion. It is named after its discoverer, the French physicist and
mathematician Jean le Rond d'Alembert. The principle states that the sum of the differences between the forces
acting on a system and the time derivatives of the momenta of the system itself along any virtual displacement
consistent with the constraints of the system is zero.
(7) Dynamic Analysis of Four bar Mechanism: A four-bar linkage or simply a 4-bar or four-bar is the
simplest movable linkage. It consists of four rigid bodies (called bars or links), each attached to two others by
single joints or pivots to form closed loop. Four-bars are simple mechanisms common in mechanical
engineering machine design and fall under the study of kinematics.
Dynamic Analysis of Reciprocating engines.
Inertia force and torque analysis by neglecting weight of connecting rod.
Velocity and acceleration of piston.
Angular velocity and Angular acceleration of connecting rod.
Force and Torque Analysis in reciprocating engine neglecting the weight of connecting rod.
Equivalent Dynamical System
Determination of two masses of equivalent dynamical system
(8) Turning Moment Diagram: The turning moment diagram is graphical representation of the turning moment or crank effort for various positions of crank.
LECTURE-2
ME 2302 – Dynamics of Machinery
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LECTURE-3
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Lecture-4
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LECTURE-5
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LECTURE-6
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LECTURE-7
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(1) Introduction:
Balancing is the process of eliminating or at least reducing the ground forces and/or moments. It is achieved by
changing the location of the mass centres of links. Balancing of rotating parts is a well known problem. A
rotating body with fixed rotation axis can be fully balanced i.e. all the inertia forces and moments. For
mechanism containing links rotating about axis which are not fixed, force balancing is possible, moment
balancing by itself may be possible, but both not possible. We generally try to do force balancing. A fully force
balance is possible, but any action in force balancing severe the moment balancing.
(2) Balancing of rotating masses: The process of providing the second mass in order to counteract the effect of
the centrifugal force of the first mass is called balancing of rotating masses.
(3) Static balancing: The net dynamic force acting on the shaft is equal to zero. This requires that the line of
action of three centrifugal forces must be the same. In other words, the centre of the masses of the system must
lie on the axis of the rotation. This is the condition for static balancing.
(4) Dynamic balancing: The net couple due to dynamic forces acting on the shaft is equal to zero. The
algebraic sum of the moments about any point in the plane must be zero.
(5) Various cases of balancing of rotating masses:
Balancing of a single rotating mass by single mass rotating in the same plane.
Balancing of a single rotating mass by two masses rotating in the different plane.
Balancing of a several masses rotating in single plane.
Balancing of a several masses rotating in different planes.
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(12) Balancing of single cylinder engine:
A single cylinder engine produces three main vibrations. In describing them we will assume that the
cylinder is vertical. Firstly, in an engine with no balancing counterweights, there would be an enormous
ME 2302 – Dynamics of Machinery
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vibration produced by the change in momentum of the piston, gudgeon pin, connecting rod and crankshaft once
every revolution. Nearly all single-cylinder crankshafts incorporate balancing weights to reduce this. While
these weights can balance the crankshaft completely, they cannot completely balance the motion of the piston,
for two reasons. The first reason is that the balancing weights have horizontal motion as well as vertical motion,
so balancing the purely vertical motion of the piston by a crankshaft weight adds a horizontal vibration. The
second reason is that, considering now the vertical motion only, the smaller piston end of the connecting rod
(little end) is closer to the larger crankshaft end (big end) of the connecting rod in mid-stroke than it is at the top
or bottom of the stroke, because of the connecting rod's angle. So during the 180° rotation from mid-stroke
through top-dead-center and back to mid-stroke the minor contribution to the piston's up/down movement from
the connecting rod's change of angle has the same direction as the major contribution to the piston's up/down
movement from the up/down movement of the crank pin. By contrast, during the 180° rotation from mid -stroke
through bottom-dead-center and back to mid-stroke the minor contribution to the piston's up/down movement
from the connecting rod's change of angle has the opposite direction of the major contribution to the piston's
up/down movement from the up/down movement of the crank pin. The piston therefore travels faster in the top
half of the cylinder than it does in the bottom half, while the motion of the crankshaft weights is sinusoidal. The
vertical motion of the piston is therefore not quite the same as that of the balancing weight, so they can't be
made to cancel out completely. Secondly, there is a vibration produced by the change in speed and therefore
kinetic energy of the piston. The crankshaft will tend to slow down as the piston speeds up and absorbs energy,
and to speed up again as the piston gives up energy in slowing down at the top and bottom of the stroke. This
vibration has twice the frequency of the first vibration, and absorbing it is one function of the flywheel. Thirdly,
there is a vibration produced by the fact that the engine is only producing power during the power stroke. In a
four-stroke engine this vibration will have half the frequency of the first vibration, as the cylinder fires once
every two revolutions. In a two-stroke engine, it will have the same frequency as the first vibration. This
vibration is also absorbed by the flywheel.
(13) Balancing of inertial forces in the multi-cylinder engine:
In multi-cylinder engines the mutual counteractions of the various components in the Crank shaft
assembly are one of the essential factors determining the selection of the Crank shafts configuration and with it
the design of the engine itself. The inertial forces are Balanced if the common centre of gravity for all moving
crankshaft-assembly components lies at the crankshaft's midpoint, i.e. if the crankshaft is symmetrical (as
viewed from the front). The crankshaft's symmetry level can be defined using geometrical representations of
1st- and 2nd-order forces (star diagrams). The 2nd order star diagram for the four-cylinder in-line engine is
asymmetrical, meaning that this order is characterized by substantial free inertial Forces. These forces can be
balanced using two countershafts rotating in opposite directions at double the rate of the crankshaft (Lanchester
system).
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(1) Introduction:
When a system is subjected to an initial disturbance and then left free to vibrate on its own, the resulting
vibrations are referred to as free vibrations .Free vibration occurs when a mechanical system is set off with an
initial input and then allowed to vibrate freely. Examples of this type of vibration are pulling a child back on a
swing and then letting go or hitting a tuning fork and letting it ring. The mechanical system will then vibrate at
one or more of its "natural frequencies" and damp down to zero.
(2) Basic elements of vibration system:
Mass or Inertia
External excitation
(3) Causes of vibration: Unbalance: This is basically in reference to the rotating bodies. The uneven
distribution of mass in a rotating body contributes to the unbalance. A good example of unbalance related
vibration would be the vibrating alert in our mobile phones. Here a small amount of unbalanced weight is
rotated by a motor causing the vibration which makes the mobile phone to vibrate. You would have
experienced the same sort of vibration occurring in your front loaded washing machines that tend to vibrate
during the spinning mode.
Misalignment: This is an other major cause of vibration particularly in machines that are driven by motors or
any other prime movers.
Bent Shaft: A rotating shaft that is bent also produces the the vibrating effect since it losses it rotation
Gears in the machine: The gears in the machine always tend to produce vibration, mainly due to their
meshing. Though this may be controlled to some extent, any problem in the gearbox tends to get enhanced
with ease.
Bearings: Last but not the least, here is a major contributor for vibration. In majority of the cases every initial
problem starts in the bearings and propagates to the rest of the members of the machine. A bearing devoid of
lubrication tends to wear out fast and fails quickly, but before this is noticed it damages the remaining
components in the machine and an initial look would seem as if something had gone wrong with the other
components leading to the bearing failure.
(4) Effects of vibration:
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(a)Bad Effects: The presence of vibration in any mechanical system produces unwanted noise, high stresses,
poor reliability, wear and premature failure of parts. Vibrations are a great source of human discomfort in the
form of physical and mental strains.
(b)Good Effects: A vibration does useful work in musical instruments, vibrating screens, shakers, relive pain in
physiotherapy.
(5) Methods of reduction of vibration:
-unbalance is its main cause, so balancing of parts is necessary.
-using shock absorbers.
(6) Types of vibratory motion:
Free Vibration
Forced Vibration
(7) Terms used vibratory motion: (a)Time period (or)period of vibration: It is the time taken by a vibrating
body to repeat the motion itself.time period is usually expressed in seconds. (b) Cycle: It is the motion
completed in one time period. (c) Periodic motion: A motion which repeats itself after equal interval of time.
(d)Amplitude (X) The maximum displacement of a vibrating body from the mean position.it is usually
expressed in millimeter. (e) Frequency (f) The number of cycles completed in one second is called frequency
(8) Degrees of freedom: The minimum number of independent coordinates required to specify the motion of a
system at any instant is known as D.O.F of the system.
(9) Single degree of freedom system:
The system shown in this figure is what is known as a Single Degree of Freedom system. We use the
term degree of freedom to refer to the number of coordinates that are required to specify completely the
configuration of the system. Here, if the position of the mass of the system is specified then accordingly the
position of the spring and damper are also identified. Thus we need just one coordinate (that of the mass) to
specify the system completely and hence it is known as a single degree of freedom system.
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(10) Two degree of freedom system:
A two degree of freedom system With reference to automobile applications, this is referred as quarter
car model. The bottom mass refers to mass of axle, wheel etc components which are below the suspension
spring and the top mass refers to the mass of the portion of the car and passenger. Since we need to specify both
the top and bottom mass positions to completely specify the system, this becomes a two degree of freedom
system.
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Forced Vibration:
When the body vibrates under the influence of external force, then the body is said to be under forced vibration.
Examples of forced vibration:
1. Ringing of electric bell.
2. Vibrat ion of various machines like air compressor, IC engines, Machine tools and mobile cranes.
Types of external Excitation:
1. Period ic forces,
3. Random Forces.
Period ic forces are further classified into harmonic and non-harmonic forces. Vibration because of impulsive forces is called as
transient. Earthquake and acoustic excitat ion are typical examples of random forces. In this chapter we would be analysing only
Frequency of Under Damped Forced Vibrations
Consider a system consisting of spring, mass and damper as shown in Fig. Let the system is acted upon by an external periodic
(i.e. simple harmonic) d isturbing force,
Fx=F cos.t
whereF = Static force, and
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When the system is constrained to move in vertical guides, it has only one degree of freedom. Let at sometime t, the
mass is displaced downwards through a distance x from its mean position.
Important formulas to be remembered:
When Damping is negligib le, then c = 0
At resonance = n. Therefore the angular speed at which the resonance occurs is
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Lecture-2
Magnification Factor or Dynamic Magnifier
It is the ratio of maximum displacement of the forced vibration (xmax ) to the deflection due to the static force F(xo).
We have proved in the previous article that the maximum d isplacement or the amplitude of forced vibrat ion,
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The magnification factor or dynamic magnifier gives the factor by which the static deflection produced by a force F (i.e.xo) must
be multiplied in order to obtain the maximum amplitude of the forced vibrat ion (i.e. x max) by the harmonic force Fcos.t
1. If there is no damping (i.e. if the vibration is undamped), then c = 0. In that case, magnificat ion factor,
2. At resonance, = n. Therefore magnification factor,
Problem-1
A single cylinder vertical petrol engine of total mass 300 kg is mounted upon a steel chassis frame and causes a vertical
static deflection of 2 mm. The reciprocating parts of the engine have a mass of 20 kg and move through a vertical stroke of 1 50
mm with simple harmonic motion. A dashpot is provided whose damping resistance is directly proportional to the velocity and
amounts to 1.5 kN per metre per second.
Considering that the steady state of vibration is reached; determine: 1. the amplitude of forced vibrations, when the driving shaft
of the engine rotates at 480 r.p.m, and 2. the speed of the driving shaft at which resonance will occur.
Given:
m = 300 kg; = 2 mm = 2 × 10–3 m ;m1 = 20 kg ; l = 150 mm= 0.15 m ; c = 1.5 kN/m/s = 1500 N/m/s ; N =
480 r.p.m. or 2π480 / 60 = 50.3 rad/s
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Lecture-3
Problem – 2
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?
Given:
m = 10 kg; s = 10 N/mm = 10 × 103N/m ; X5 = X1 / 10
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Lecture-4
Problem – 3
A single cylinder vert ical petrol engine of total mass 300 kg is mounted upon a steel chassis frame and causes a
vertical static deflection of 2 mm. under this load. Calcu late 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.
Given: m = 20 kg; c = 1000 N/m/s; F = 125 N ;f = 8 cycles/s
Frequency of free vibrations
We know that frequency of free v ibrations,
The crit ical damping to make the motion aperiodic is such that damped frequency is zero,
This means that the viscous damping force is 1023 N at a speed of 1 m/s. Therefore a v iscous damping force amounting to
approximately 1000 N at a speed of 1 m/s is just sufficient to make the motion aperiodic.
Amplitude of ultimate motion
and stiffness of the spring, s = m.g/ δ= 20 × 9.81 / 0.015 = 13.1 × 103 N/m
Amplitude of ult imate motion i.e. maximum amplitude of forced vibration
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Lecture-5
Problem-4
The time of free vibration of a mass hung from the end of a helical spring is 0.8 second. When the
mass is stationary, the upper end is made to move upwards with a displacement y metre such that y = 0.018 sin 2 t,
where t is the time in seconds measured from the beginning of the motion. Neglecting the mass of the spring and any
damping effects, determine the vertical distance through which the mass is moved in the first 0.3 second.
Given :tp= 0.8 s ; y = 0.018 sin 2 t
Let m = Mass hung to the spring in kg, and
s = Stiffness of the spring in N/m.
We know that time period of free v ibrations (tp),
If x metres is the upward displacement of mass m from its equilibrium position after time t seconds, the equation of motion is
given by
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Lecture-6
Vibration Isolation and Transmissibility
A litt le consideration will show that when an unbalanced machine is installed on the foundation, it produces vibration in
the foundation. In order to prevent these vibrations or to min imise the transmission of forces to the foundation, the machines are
mounted on springs and dampers or on some vibration isolating material, as shown in Fig. The arrangement is assumed to have
one degree of freedom, i.e. it can move up and down only.
It may be noted that when a periodic (i.e . simple harmonic) d isturbing force F cost is applied to a machine of mass m supported
by a spring of stiffness s, then the force is transmitted by means of the spring and the damper or dashpot to the fixed support or
foundation.
The ratio of the force t ransmitted (FT) to the force applied (F) is known as the isolation factor or transmissibility ratio of the
spring support.
We have discussed above that the force transmitted to the foundation consists of the following two forces:
1. Spring force or elastic force which is equal to s. xmax, and
2. Damping force which is equal to c..xmax.
Since these two forces are perpendicular to one another, as shown in Fig, therefore the force transmitted,
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When the damper is not provided, then c = 0, and
From above, we see that when /n1, is negative. This means that there is a phase difference of 180° between the
transmitted force and the disturbing force (F cos.t). The value of /nmust be greater than 2 if is to be less than 1 and it is
the numerical value of , independent of any phase difference between the forces that may exist which is important. It is
therefore more convenient to use equation (ii) in the following form, i.e.
Fig below is the graph for different values of damping factor c/cc to show the variation of transmissibility ratio ( ) against the
ratio /n.
1. When /n2, then all the curves pass through the point = 1 for all values of damping factor c/cc.
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2. When /n2, then > 1 for all values of damping factor c/cc. This means that the force transmitted to the foundation
through elastic support is greater than the force applied.
3. When /n2, then < 1 for all values of damping factor c/cc. This shows thatthe force transmitted through elastic support
is less than the applied force. Thus vibration isolationis possible only in the range of /n 2
We also see from the curves in Fig above that the damping is detrimental beyond /n 2 and advantageous only in
the region /n < 2. It is thus concluded that for the vibration isolation, dampers need not to be provided but in order to limit
resonance amplitude, stops may be provided.
Lecture-7
Problem-5
The mass of an electric motor is 120 kg and it runs at 1500 r.p.m. The armature mass is 35 kg and its C.G. lies
0.5 mm from the axis of rotation. The motor is mounted on five springs of neglig ible damping so that the force transmitted is
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one-eleventh of the impressed force. Assume that the mass of the motor is equally d istributed among the five springs. Determine :
1. stiffness of each spring; 2. dynamic force t ransmitted to the base at the operating speed; and 3. natural frequency of the
system.
Given m1 = 120 kg; m2 = 35 kg; r = 0.5 mm = 5 × 10–4 m; = 1 / 11; N = 1500 r.p.m. or = 2π× 1500 / 60 = 157.1
1. Stiffness of each spring
Let s = Combined stiffness of the spring in N-m, and n = Natural circular frequency of vibration of the machine in rad/s.
We know that transmissibility rat io ( ),
Lecture-8
Problem-6
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A machine has a mass of 100 kg and unbalanced reciprocating parts of mass 2 kg which move through a vertical stroke
of 80 mm with simple harmonic motion. The machine is mounted on four springs, symmetrically arranged with respect to centre
of mass, in such a way that the machine has one degree of freedom and can undergo vertical d isplacements only.
Neglecting damping, calculate the combined stiffness of the spring in order that the force transmitted to the foundation is 1 / 25
th of the applied force, when the speed of rotation of machine crank shaft is 1000 r.p.m.
When the machine is actually supported on the springs, it is found that the damping reduces the amplitude of successive free
vibrations by 25%. Fin: 1. the force transmitted to foundation at 1000 r.p.m., 2. the force transmitted to the foundation at
resonance, and 3. the amplitude of theforced vibration of the machine at resonance.
Given: m1 = 100 kg ;m2 = 2 kg ; l = 80 mm = 0.08 m ; = 1 / 25 ; N = 1000 r.p.m. or 2π1000 / 60 = 104.7 rad/s
Combined stiffness of springs
Let s = Combined stiffness of springs in N/m, and
n = Natural circu lar frequency of vibration of the machine in rad/s.
We know that transmissibility rat io ( ),
1. Force transmitted to the foundation at 1000 r.p.m.
Let FT = Force trans mitted, and
x1 = Init ial amplitude of vibrat ion.
Since the damping reduces the amplitude of successive free v ibrations by 25%, therefore final amplitude of vib ration,
x20.75 x1
We know that damping coefficient or damping fo rce per unit velocity,
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ca X 2m10.942100 = 188 N/m/s
and critical damping coefficient, cc 2m.n210020.5 = 4100 N/m/s
Actual value of transmissibility ratio,
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Lecture-9
Problem – 7
A single-cylinder engine of total mass 200 kg is to be mounted on an elastic support which permits vibratory
movement in vertical direct ion only. The mass of the piston is 3.5 kg and has a vertical reciprocating motion which may be
assumed simple harmonic with a stroke of 150 mm. It is desired that the maximum vibratory force transmitted through the elastic
support to the foundation shall be 600 N when the engine speed is 800 r.p.m. and less than this at all higher speeds.
1. Find the necessary stiffness of the elastic support, and the amplitude of vibration at 800 r.p.m., and
2. If the engine speed is reduced below 800 r.p.m. at what speed will the transmitted forceagain becomes 600 N?
Given :m1 = 200 kg ; m2 = 3.5 kg ; l = 150 mm = 0.15 mm or r = l/2 = 0.075 m ;FT = 600 N ; N = 800 r.p.m. or
2π800 / 60 = 83.8 rad/s
We know that the disturbing force at 800 r.p.m.,
F = Centrifugal force on the piston
m2. .r = 3.5 (83.8)
2 0.075 = 1843 N
xmax= Max.amplitude of vibrat ion in metres.
Since the max.vibratory force transmitted to the foundation is equal to the force on the elastic support neglecting damping),
therefore Max.v ibratory force trans mitted to the foundation,
FT = Force on the elastic support
= St iffness of elastic support × Max.amplitude of vibrat ion
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Lecture-10
Problem-8
A single cylinder vertical petrol engine of total mass 300 kg is mounted upon a steel chassis frame and causes a vertical
static deflection of 2 mm. The reciprocating parts of the engine have a mass of 20 kg and move through a vertical stroke of 150
mm with simple harmonic motion. A dashpot is provided whose damping resistance is directly proportional to the velocity and
amounts to 1.5 kN per metre per second.
Considering that the steady state of vibration is reached; determine: 1. the amplitude of forced vibrations, when the driving shaft
of the engine rotates at 480 r.p.m, and 2. the speed of the driving shaft at which resonance will occur.
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Given:
m = 300 kg; = 2 mm = 2 × 10–3 m ;m1 = 20 kg ; l = 150 mm= 0.15 m ; c = 1.5 kN/m/s = 1500 N/m/s ; N =
480 r.p.m. or 2π480 / 60 = 50.3 rad/s
Lecture-11
Problem-9
The mass of a single degree damped vibrating system is 7.5 kg and makes 24 free oscillations in 14 seconds when disturbed fro m
its equilibrium position. The amplitude of vibration reduces to 0.25 of its in itial 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 crit ical damping.
Given: m = 7.5 kg
Since 24 oscillations are made in 14 seconds, therefore frequency of free vibrations,
fn= 24/14 = 1.7 and n 2π.fn2π1.7 10.7 rad/s
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Lecture-12 Problem-10
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 oscillat ions.
Assuming that the damping force varies as the velocity, determin :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.
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Given: 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
Since the stiffness of each spring is 10 × 103 N/m and there are 3 springs, therefore total stiffness,
s3X10X10 3 30X10
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1) Governors.
2) Gyroscope.
Governors:
The function of a governor is to regulate the mean speed of an engine, when there are
variations in the load when the load on an engine increases, its speed decreases, therefore it becomes
necessary to increase the supply of working fluid. On the other hand, when the load on the engine
decreases, its speed increases and thus less working fluid is required. The governor automatically controls
the supply of working fluid to the engine with the varying load conditions and keeps the mean speed within
certain limits.
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Difference between a Flywheel and Governor:
The function of a flywheel in an engine is entirely different from that of a governor. It controls the speed variation caused by the fluctuations of the engine turning moment during each cycle of operation. It
does not control the speed variations caused by a varying load. The varying demand for power is met by the governor regulating the supply of working fluid.
Types of Governors
The governors may, broadly, be classified as 1. Centrifugal governors 2. Inertia governors.
Centrifugal governors:
The centrifugal governors are based on the balancing of centrifugal force on the rotating ballsby an
equal and opposite radial force, known as the controlling force.It consists of two balls ofequal mass, which are attached to the arms as shown in Fig. These balls are known as governorballs or fly balls.
The balls revolve with a spindle, whichis driven by the engine through bevel gears. The upperends
of the arms are pivoted to the spindle, so that theballs may rise up or fall down as they revolve about
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thevertical axis. The arms are connected by the links to asleeve, which is keyed to the spindle. This sleeve
revolveswith the spindle; but can slide up and down.The balls and the sleeve risewhen the spindle speedincreases, and falls when the speed decreases. In orderto limit the travel of the sleeve in upward and downwarddirections, two stops S, S are provided on thespindle. The sleeve isconnected by a bell crank
leverto a throttle valve. The supply of the working fluid decreaseswhen the sleeve rises and increases when it falls. When the load on the engine increases, the engine and the governor speed decreases. This results in
the decrease of centrifugal force on the balls. Hence the balls move inwards and the sleeve moves downwards. The downward movement of the sleeve operates a throttle valve at the other end of the bell crank lever to increase the supply of working fluid and thus the engine speed is increased. In this case, the
extra power output is provided to balance the increased load. When the load on the engine decreases, the engine and the governor speed increases, which results in the increase of centrifugal force on the balls.
Thus the balls move outwards and the sleeve rises upwards. This upward movement of the sleeve reduces the supply of the working fluid and hence the speed is decreased. In this case, the power output is reduced.
Lecture-2
Terms Used in Governors
1. Height of a governor. It is the vertical distance from the centre of the ball to a point where the axes of the arms (or arms produced) intersect on the spindle axis. It is usually denoted by h.
2. Equilibrium speed. It is the speed at which the governor balls, arms etc., are in complete equilibrium and the sleeve does not tend to move upwards or downwards. 3. Mean equilibrium speed. It is the speed at the mean position of the balls or the sleeve.
4. Maximum and minimum equilibrium speeds. The speeds at the maximum and minimum radius of rotation of the balls, without tending to move either way are known as maximum and minimum equilibrium
speeds respectively. Note : There can be many equilibrium speeds between the mean and the maximum and the mean and t he minimum equilibrium speeds.
5. Sleeve lift. It is the vertical distance which the sleeve travels due to change in equilibrium speed.
Watt Governor:
The simplest form of a centrifugal governor is a Watt governor, as shown in Fig. It isbasically a
conical pendulum with links attached to a sleeve of negligible mass. The arms of thegovernor may be connected to the spindle in the following three ways: 1. The pivot P may be on the spindle axis as shown in Fig. (a).
2. The pivot P may be offset from the spindle axis and the arms when produced intersect at O, as shown in Fig. (b).
3. The pivot P may be offset, but the arms cross the axis at O, as shown in Fig.(c).
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m = Mass of the ball in kg,
w = Weight of the ball in newtons = m.g, T = Tension in the arm in newtons,
= Angular velocity of the arm and ball about the spindle axis in rad/s,
r = Radius of the path of rotation of the ball i.e. horizontal distance from the centre of the ball to the spindle axis in metres,
FC = Centrifugal force acting on the ball in newtons = m..r, and
h = Height of the governor in metres.
Problem-1
Calculate the vertical height of a Watt governor when it rotates at 60 r.p.m. Also find the change in vertical height when its speed increases to 61 r.p.m.
Given: N1 = 60 r.p.m. ;N2 = 61 r.p.m. Initial height We know that initial height,
Change in vertical height We know that final height,
Change in vertical height = h1 – h2 = 0.248 – 0.24 = 0.008 m = 8 mm
Lecture-3
Porter Governor:
The Porter governor is a modification of a Watt’s governor, with central load attached to the sleeve as shown in Fig.(a). The load moves up and down the central spindle. This additional downward force
increases the speed of revolution required to enable the balls to rise to any predetermined level. Consider the forces acting on one-half of the governor as shown in Fig.(b).
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m = Mass of each ball in kg, w = Weight of each ball in newtons = m.g, M = Mass of the central load in kg,
W = Weight of the central load in newtons = M.g, r = Radius of rotation in metres, h = Height of governor in metres,
N = Speed of the balls in r.p.m.
FC = Centrifugal force acting on the ball in newtons = m.2.r,
T1 = Force in the arm in newtons, T2 = Force in the link in newtons,
α= Angle of inclination of the arm (or upper link) to the vertical, and
= Angle of inclination of the link (or lower link) to the vertical.
1. When the length of arms are equal to the length of links and the points P and D lie on the same
vertical line, then tanα= tanor q = tanα/ tan= 1
2. When the loaded sleeve moves up and down the spindle, the frictional force acts on it in a direction opposite to that of the motion of sleeve.
If F = Frictional force acting on the sleeve in newtons, then the equations (v) and (vi) may be written as
The + sign is used when the sleeve moves upwards or the governor speed increases and negative sign is
used when the sleeve moves downwards or the governor speed decreases. Problem – 2
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A Porter governor has equal arms each 250 mm long and pivoted on the axis of rotation.
Each ball has a mass of 5 kg and the mass of the central load on the sleeve is 25 kg. Theradius of rotation of the ball is 150 mm when the governor begins to lift and 200 mm when the governor is at maximum speed. Find the minimum and maximum speeds and range of speed of the governor. (AU 2006)
Gn :BP = BD = 250 mm = 0.25 m ; m = 5 kg ; M = 15 kg ; r1 = 150 mm = 0.15m; r 2 = 200 mm = 0.2 m
Minimum speed when r1 = BG = 0.15 m
Let N1 = Minimum speed. From Fig(a), we find that height of the governor,
hPG – 0.2 m
N1 = 133.8 r.p.m.
Maximum speed when r2 = BG = 0.2 m Let N2 = Maximum speed. From Fig(b), we find that height of the governor,
N2 = 154.5 r.p.m Range of speed
We know that range of speed= N2 – N1 = 154.4 – 133.8 = 20.7 r.p.m.
Lecture-4
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The Proell governor has the balls fixed at B and C to the extension of the links DF and EG,
as shown in Fig (a). The arms FP and GQ are pivoted at P and Q respectively. Consider the equilibrium of the forces on one-half of the governor as shown in Fig (b). The instantaneous centre (I) lies on the intersection of the line PF produced and the line from D drawn perpendicular to the spindle axis. The
perpendicular BM is drawn on ID.
Problem – 3 A governor of the Proell type has each arm 250 mm long. The pivots of the upper and lower arms are 25 mm from the axis. The central load acting on the sleeve has a mass of 25 kg and the each
rotating ball has a mass of 3.2 kg. When the governor sleeve is in mid-position, the extension link of the lower arm is vertical and the radius of the path of rotation of the masses is175 mm. The vertical height of
the governor is 200 mm.If the governor speed is 160 r.p.m. when in mid-position, find : 1. length of the extensionlink; and 2. tension in the upper arm.
Given: PF = DF = 250 mm = 0.25 m ;PQ = DH = KG = 25 mm = 0.025 m ; M = 25 kg ; m = 3.2 kg ; r = FG = 175 mm = 0.175 m ; h = QG = PK = 200 mm = 0.2 m ; N = 160 r.p.m.
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BF = BM – FM = 0.308 – 0.2 = 0.108 m = 108 mm
Tension in the upper arm
Problem – 4
A Proell governor has equal arms of length 300 mm. The upper and lower ends of the arms are pivoted on the axis of the governor. The extension arms of the lower links are each 80 mm long and parallel to the axis when the radii of rotation of the balls are 150 mm and 200 mm. The mass of each ball is
10 kg and the mass of the central load is 100 kg. Determine the range of speed of thegovernor.
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Lecture-5
Hartnell Governor
A Hartnell governor is a spring loaded governor as shown in Fig. 18.18. It consists of two bell crank levers pivoted at the points O,O to the frame. The frame is attached to the governor spindle and
therefore rotates with it. Each lever carries a ball at the end of the vertical arm OB and a roller at the end of the horizontal arm OR. A helical spring in compression provides equal downward forces on the two rollers
through a collar on the sleeve. The spring force may be adjusted by screwing a nut up or down on the sleeve.
Hartnell governor
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Minimum position:
Maximum Position:
Problem-5: In a spring loaded Hartnell type governor, the extreme radii of rotation of the balls are 80
mm and 120 mm. The ball arm and the sleeve arm of the bell crank lever are equal in length. The mass of each ball is 2 kg. If the speeds at the two extreme positions are 400 and 420 r.p.m., find : 1. the initial compression of the central spring, and 2. the spring constant. Given:
r1 = 80 mm = 0.08 m ;r2 = 120 mm = 0.12 m ; x = y ; m = 2 kg ; N1 = 400r.p.m. or = 2 π× 400/60 = 41.9 rad/s ; N2 =
420 r.p.m. or 2= 2 π× 420/60 = 44 rad/s
Initial compression of the central spring
For Minimum Position:
For Maximum Position
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Lecture-6
Problem-6:
In a spring controlled governorof the type, as shown in Fig, the mass of each ball is1.5 kg and the mass of the sleeve is 8 kg. The two arms ofthe bell crank lever are at right angles and their
lengthsare OB = 100 mm and OA = 40 mm. The distance ofthe fulcrum O of each bell crank lever from the axis ofrotation is 50 mm and minimum radius of rotation of thegovernor balls is also 50 mm. The correspondingequilibrium speed is 240 r.p.m. and the sleeve is requiredto lift 10 mm for an increase in
speed of 5 per cent. Findthe stiffness and initial compression of the spring. Given:
m = 1.5 kg; M = 8 kg; OB = x= 100 mm = 0.1 m; OA = y = 40 mm = 0.04 m; r = 50 mm
= 0.05 m; r1 = 50 mm = 0.05 m; N1 = 240 r.p.m. or1= 2π× 240/60 = 25.14 rad/s; h = 10 mm = 0.01 m; Increase in speed = 5%
The spring controlled governor of the type, as shown in Fig, has the pivots for the bell crank lever on the moving sleeve. The spring is compressed between the sleeve and the cap which is fixed to the end of the
governor shaft. The simplest way of analysing this type of governor is by taking moments about the instantaneous centre of all the forces which act on one of the bell crank levers. The minimum position of the governor is shown in Fig(a).
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The maximum position of the governor is shown in Fig.(b). From the geometry of thefigure,
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Hartung Governor
A spring controlled governor of the Hartung type is shown in Fig (a). In this type of
governor, the vertical arms of the bell crank levers are fitted with spring balls which compress against the
frame of the governor when the rollers at the horizontal arm press against the sleeve.
S = spring force, FC = Centrifugal force,
M = Mass on the sleeve, and x and y = Lengths of the vertical and horizontal arm of the bell crank lever resp.
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Lecture-7
Problem – 6 In a spring-controlled governor of the Hartung type, the length of the ball and sleeve arms
are 80 mm and 120 mm respectively. The total travel of the sleeve is 25 mm. In the mid position, each spring is compressed by 50 mm and the radius of rotation of the mass centres is 140 mm. Each ball has a
mass of 4 kg and the spring has a stiffness of 10 kN/m of compression. The equivalent mass of the governor gear at the sleeve is 16 kg. Neglecting the moment due to the revolving masses when the arms are inclined, determine the ratio of the range of speed to the mean speed of the governor. Find, also, the speed
in the mid-position.
Given: x = 80 mm = 0.08 mm; y = 120 mm = 0.12 m ;h = 25 mm = 0.025 m ; r = 140 mm = 0.14 m ; m = 4 kg ; s = 10 kN/m = 10 × 103N/m ; M = 16 kg ; Initial compression = 50 mm = 0.05 m
Mean speed of the governor
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Wilson-Hartnell Governor
A Wilson-Hartnell governor is a governor in which the balls are connected by a spring in
tension as shown in Fig. An auxiliary spring is attached to the sleeve mechanism through a lever by means
of which the equilibrium speed for a given radius may be adjusted. The main spring may be considered of
two equal parts each belonging to both the balls. The line diagram of a Wilson- Hartnell governor is shown
in Fig.
Sensitiveness of Governors Sensitiveness is defined as the ratio of the difference between the maximum and minimum equilibrium speeds to the mean equilibrium speed.
Stability of Governors
A governor is said to be stable when for every speed within the working range there is a definite
configuration i.e. there is only one radius of rotation of the governor balls at which the governor is in
equilibrium. For a stable governor, if the equilibrium speed increases, the radius of governor balls must
also increase.
Note: A governor is said to be unstable, if the radius of rotation decreases as the speed increases.
Isochronous Governors
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A governor is said to be isochronous when the equilibrium speed is constant (i.e. range of speed is
zero) for all radii of rotation of the balls within the working range, neglecting friction. The isochronism is
the stage of infinite sensitivity.
Haunting
A governor is said to be hunt if the speed of the engine fluctuates continuously above andbelow the mean speed. This is caused by a too sensitive governor which changes the fuel supplyby a large amount when a small change in the speed of rotation takes place.
Effort and Power of a Governor
The effort of a governor is the mean force exerted at the sleeve for a given percentage change of speed* (or lift of the sleeve). It may be noted that when the governor is running steadily, there is no force at the sleeve. But, when the speed changes, there is a resistance at the sleeve which opposes its motion. It is
assumed that this resistance which is equal to the effort, varies uniformly from a maximum value to zero while the governor moves into its new position of equilibrium.
The power of a governor is the work done at the sleeve for a given percentage change of speed. It is the product of the mean value of the effort and the distance through which the sleeve moves. Power = Mean effort × lift of sleeve
Lecture-8
Controlling force:
Governor running at a steady speed, the inward force acting on the rotating balls is known as
controlling force. It is equal and opposite to the centrifugal reaction. The Controlling force is given by
Controlling force, FC = m.2.r
The controlling force is provided by the weight of the sleeve and balls as in Porter governor and by the
spring and weight as in Hartnell governor (or spring controlled governor). When the graph between the
controlling force (F C) as ordinate and radius of rotation of the balls (r) as abscissa is drawn, then the graph
obtained is known as controlling force diagram. This diagram enables the stability and sensitiveness of the
governor to be examined and also shows clearly the effect of friction.
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Controlling Force Diagram for Porter Governor
where Φis the angle between the axis of radius of rotation and a line joining a given point (say A) onthe
curve to the origin O.
Controlling Force Diagram for Spring-controlled Governors
The controlling force diagram for the spring controlled governors is a straight line, as shownin Fig.
We know that controlling force,
F C = m.2.r or FC /r = m.2
For the governor to be stable, the controlling force (FC) must increase as the radius of rotation(r) increases,
i.e. FC / r must increase as r increases. Hence the controlling force line AB when produced must intersect
the controlling force axis below the origin, as shown in Fig. The relation between the controlling force (F C)
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and the radius of rotation (r) for the stability of spring controlled governors is given by the following
equation
Where a and b are constants.
The value of b in equation (i) may be made either zero or positive by increasing the initial tension of the
spring. If b is zero, the controlling force line CD passes through the origin and the governor becomes
isochronous because FC /r will remain constant for all radii of rotation. The relation between the controlling
force and the radius of rotation, for an isochronous governor is, therefore,
FC = a.r----------------------------------------------------------------------------- (ii)
If b is greater than zero or positive, then FC /r decreases as r increases, so that the equilibrium speed of the
governor decreases with an increase of the radius of rotation of balls, which is impracticable. Such a
governor is said to be unstable and the relation between the controlling force and the radius of rotation is,
therefore
FC = a.r+ b------------------------------------------------------------------------ (iii)
Coefficient of Insensitiveness
We have assumed the governor to be frictionless. In actual practice, there is always friction
in the joints and operating mechanism of the governor. Since the frictional force always acts in the opposite
direction to that of motion, therefore, when the speed of rotation decreases, the friction prevents the
downward movement of the sleeve and the radial inward movement of the balls. On the other hand, when
the speed of rotation increases, the friction prevents the upward movement of the sleeve and radial outward
movement of the balls.
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Lecture-9
Gyroscopic Couple:
Consider a disc spinning with an angular velocity rad/s about the axis of spin OX, in
anticlockwise direction when seen from the front, as shown in Fig(a). Since the plane in which the disc is
rotating is parallel to the plane YOZ, therefore it is ca lled plane of spinning. The planeXOZ is a horizontal
plane and the axis of spin rotates in a plane parallel to the horizontal plane aboutan axis OY. In other
words, the axis of spin is said to be rotating or processing about an axis OY. Inother words, the axis of spin
is said to be rotating or processing about an axis OY (which is perpendicular to both the axes OX and OZ)
at an angular velocity P rap/s. This horizontal plane XOZ is called plane of precession and OY is the axis
of precession.

Angular momentum of the disc= I.
Since the angular momentum is a vector quantity, therefore it may be represented by the vector ox , as
shown in Fig.(b). The axis of spin OX is also rotating anticlockwise when seen from the top about the axis
OY. Let the axis OX is turned in the plane XOZ through a small angle δradians to the position OX’, in time
δtseconds. Assuming the angular velocity to be constant, the angular
momentum will now be represented by vector OX’.
Change in angular momentum = I..δ
Rate of change of angular momentum
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The rate of change of angular momentum will result by the application of a couple to the disc, therefore the
couple applied to the disc causing precession
Problem – 7
A uniform disc of diameter 300 mm and of mass 5 kg is mounted on one end of an arm of
length 600 mm. The other end of the arm is free to rotate in a universal bearing. If the disc rotates about the arm with a speed of 300 r.p.m. clockwise, looking from the front, with what speed will it precess about the
vertical axis? Given: d = 300 mm or r = 150 mm = 0.15 m ;m = 5 kg ; l = 600 mm = 0.6 m ; N = 300 r.p.m. or π × 300/60 =
Lecture-10
Effect of the Gyroscopic Couple on an Aeroplane
The top and front views of an aeroplane are shown in Fig (a). Let engine or propeller rotates in the
clockwise direction when seen from the rear or tail end and the aeroplane takes a turn to the left.
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Aeroplane taking left turn
Notes: 1. when the aeroplane takes a right turn under similar conditions as discussed above, the effect of
the reactive gyroscopic couple will be to dip the nose and raise the tail of the aeroplane. 2. When the engine or propeller rotates in anticlockwise direction when viewed from the rear or tail end and the aeroplane takes a left turn, then the effect of reactive gyroscopic couple will be to dip the nose and
raise the tail of the aeroplane. 3. When the aeroplane takes a right turn under similar conditions as mentioned in note 2 above, theeffect of
reactive gyroscopic couple will be to raise the nose and dip the tail of the aeroplane. 4. When the engine or propeller rotates in clockwise direction when viewed from the front and theaeroplane takes a left turn, then the effect of reactive gyroscopic couple will be to raise the tail and dip
the noseof the aeroplane. 5. When the aeroplane takes a right turn under similar conditions as mentioned in note 4-above, theeffect of
reactive gyroscopic couple will be to raise the nose and dip the tail of the aeroplane.
Problem – 8 An aeroplane makes a complete half circle of 50 metres radius, towards left, when flying at
200 km per hr. The rotary engine and the propeller of the plane has a mass of 400 kg and a radius of gyration of 0.3 m. The engine rotates at 2400 r.p.m. clockwise when viewed from the rear. Find the gyroscopic couple on the aircraft and state its effect on it. Given :R = 50 m ; v = 200 km/hr = 55.6 m/s ; m = 400 kg ; k = 0.3 m ; N = 2400 r.p.m. or = 2π× 2400/60 = 251 rad/s
We know that mass moment of inertia of the engine and the propeller,
I = m.k 2= 400(0.3)2= 36 kg-m2
and angular velocity of precession,
P= v/R= 55.6/50 = 1.11 rad/s
We know that gyroscopic couple acting on the aircraft,
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C = I.P= 36 × 251.4 × 1.11 = 100 46 N-m = 10.046 kN-m
When the aeroplane turns towards left, the effect of thegyroscopic couple is to lift the nose upwards and tail downwards.
Terms Used in a Naval Ship
The top and front views of a naval ship are shown in Fig. The fore end of the ship iscalled bow and
the rear end is known as stern or aft. The left hand and right hand sides of the ship,when viewed from the stern are called port and star-board respectively. We shall now discuss theeffect of gyroscopic couple on
the naval ship in the following three cases: 1. Steering, 2. Pitching and 3. Rolling.
Lecture-11
Effect of Gyroscopic Couple on a Naval Ship during Steering
Steering is the turning of a complete ship in a curve towards left or right, while it moves forward. Consider the ship taking a left turn, and rotor rotates in the clockwise direction when viewed from the stern, as shown in Fig. The effect of gyroscopic couple on a naval ship during steering taking left or right turn
may be obtained in the similar way as for an aeroplane as discussed earlier.
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Effect of Gyroscopic Couple on a Naval Ship during Pitching
Effect of Gyroscopic Couple on a Naval Ship during Rolling
We know that, for the effect of gyroscopic couple to occur, the axis of precession should always be
perpendicular to the axis of spin. If, however, the axis of precession becomes parallel to the axis of spin, there will be no effect of the gyroscopic couple acting on the body of the ship. In case of rolling of a ship,
the axis of precession (i.e. longitudinal axis) is always parallel to the axis of spin for all positions. Hence, there is no effect of the gyroscopic couple acting on the body of a ship.
Problem – 9 The turbine rotor of a ship has a mass of 8 tonnes and a radius of gyration 0.6 m. It rotates at
1800 r.p.m. clockwise, when looking from the stern. Determine the gyroscopic couple, if the ship travels at 100 km/hr and steer to the left in a curve of 75 m radius. Given:
m = 8 t = 8000 kg; k = 0.6 m; N = 1800 r.p.m. or π × 1800/60 = 188.5 rad/s; v = 100 km/h = 27.8
m/s; R = 75 m
Mass moment of inertia of the rotor, I = m.k 2= 8000 (0.6)2= 2880 kg-m2
Angular velocity of precession,
P= v / R = 27.8 / 75 = 0.37 rad/s Gyroscopic couple,
C = I.P= 2880 × 188.5 × 0.37 = 200 866 N-m
When the rotor rotates in clockwise direction when looking from the stern and the ship steers to the left, the effect of the reactive gyroscopic couple is to raise the bow and lower the stern.
Problem – 10 The turbine rotor of a ship has a mass of 3500 kg. It has a radius of gyration of 0.45 m and a
speed of 3000 r.p.m. clockwise when looking from stern. Determine the gyroscopic couple and its effect upon the ship:
1. When the ship is steering to the left on a curve of 100 m radius at a speed of 36 km/h.
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2. When the ship is pitching in a simple harmonic motion, the bow falling with its maximum velocity. The
period of pitching is 40 seconds and the total angular displacement between the two extreme positions of pitching is 12 degrees. Given: m = 3500 kg; k = 0.45 m; N = 3000 r.p.m. or π× 3000/60 = 314.2 rad/s
Lecture-12
Stability of a Four Wheel Drive Moving in a Curved Path
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A little consideration will show that when the vehicle is running at high speeds, PI may be zero or even negative. This will cause the inner wheels to leave the ground thus tending to overturn the Automobile. In
order to have the contact between the inner wheels and the ground, the sum of P/2 and Q/2 must be less than W/4.
Problem – 10:
A four-wheeled trolley car of total mass 2000 kg running on rails of 1.6 mgauge, rounds a curve of 30 m radius at 54 km/h. The track is banked at 8°. The wheels have an external diameter of 0.7 m
and each pair with axle has a mass of 200 kg. The radius of gyration for each pair is 0.3 m. The height of centre of gravity of the car above the wheel base is 1 m. Determine, allowing for centrifugal force and gyroscopic couple actions, the pressure on each rail. Given: m = 2000 kg ;x = 1.6 m ; R = 30 m ; v = 54 km / h = 15 m / s ; = 8° ; dW= 0.7 m or rW= 0.35 m ; m1 = 200 kg ; k = 0.3 m ; h = 1 m
First of all, let us find the reactions RA and RB at the wheels A and B respectively. The various forces acting on the trolley car are shown in Fig.
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