Theory of machines_static and dynamic force analysis

Static and Dynamic Force
Analysis
Prepared by :-Prof K N Wakchaure
S.R.E.S.College of Engineering, Kopargaon.
Savitribai Phule PUNE UNIVERSITY
Theory of Machines I

Force
 In physics, a force is any interaction which tends to change the motion of an object.
 In other words, a force can cause an object with mass to change its velocity (which
includes to begin moving from a state of rest), i.e., to accelerate.
 Force can also be described by intuitive concepts such as a push or a pull.
 A force has both magnitude and direction, making it a vector quantity. It is measured in the
SI unit of newtons and represented by the symbol F.
 The original form of Newton's second law states that the net force acting upon an object is
equal to the rate at which its momentum changes with time.
 If the mass of the object is constant, this law implies that the acceleration of an object is
directly proportional to the net force acting on the object, is in the direction of the net
force, and is inversely proportional to the mass of the object.
 As a formula, this is expressed as:
 Related concepts to force include: thrust, which increases the velocity of an object; drag,
which decreases the velocity of an object; and torque which produces changes in rotational
speed of an object. In an extended body, each part usually applies forces on the adjacent
parts; the distribution of such forces through the body is the so-called mechanical stress.
 Pressure is a simple type of stress. Stress usually causes deformation of solid materials, or
flow in fluids.
 Aristotle famously described a force as anything that causes an object to undergo
"unnatural motion"
• Simple pendulum
• Theory and analysis of
Compound Pendulum
• Concept of equivalent
length of simple pendulum,
• Bifilar suspension,
• Trifilar suspension.
• Dynamics of reciprocating
engines: Two mass statically
and dynamically equivalent
system,
• correction couple,
• static and dynamic force
analysis of reciprocating
engine mechanism
• (analytical method only),
• Crank shaft torque,
• Introduction to T-θ
diagram.
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Types of Forces
 There are different types of forces that act in different ways on structures such as
bridges, chairs, buildings, in fact any structure.
 The main examples of forces are shown below. Study the diagram and text and then
draw a diagram/pictogram to represent each of these forces.
 A Static Load : A good example of this is a person seen on the left. He is holding a
stack of books on his back but he is not moving. The force downwards is STATIC.
 A Dynamic Load : A good example of a dynamic load is the person on the right. He is
carrying a weight of books but walking. The force is moving or DYNAMIC.
 Internal Resistance : The person in the diagram is sat on the mono-bicycle and the
air filled tyre is under great pressure. The air pressure inside it pushes back against
his/her weight.
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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Types of Forces
 Tension : The rope is in “tension” as the two people pull on it. This stretching puts
the rope in tension.
 Compression : The weight lifter finds that his body is compressed by the weights he
is holding above his head.
 Shear Force : A good example of shear force is seen with a simple scissors. The two
handles put force in different directions on the pin that holds the two parts together.
The force applied to the pin is called shear force.
 Torsion : The plastic ruler is twisted between both hands. The ruler is said to be in a
state of torsion.
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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Laws of Motion
 Newton's laws of motion are three physical laws that together laid the foundation
for classical mechanics. They describe the relationship between a body and the
forces acting upon it, and its motion in response to said forces. They have been
expressed in several different ways over nearly three centuries, and can be
summarised as follows.
 The three laws of motion were first compiled by Isaac Newton in his Philosophiæ
Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy),
first published in 1687
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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First law: When viewed in an inertial reference frame, an object either
remains at rest or continues to move at a constant velocity, unless
acted upon by an external force.
Second law: The vector sum of the forces F on an object is equal to the mass m
of that object multiplied by the acceleration vector a of the object:
F = ma.
Third law: When one body exerts a force on a second body, the second body
simultaneously exerts a force equal in magnitude and opposite in
direction on the first body.

Moment of Inertia
 Moment of inertia is the mass property of a rigid body that determines the torque
needed for a desired angular acceleration about an axis of rotation.
 Moment of inertia depends on the shape of the body and may be different around
different axes of rotation. A larger moment of inertia around a given axis requires
more torque to increase the rotation, or to stop the rotation, of a body about that
axis.
 Moment of inertia depends on the amount and distribution of its mass, and can be
found through the sum of moments of inertia of the masses making up the whole
object, under the same conditions.
 For example, if ma + mb = mc, then Ia + Ib = Ic.
 In classical mechanics, moment of inertia may also be called mass moment of
inertia, rotational inertia, polar moment of inertia, or the angular mass.
 When a body is rotating around an axis, a torque must be applied to change its
angular momentum. The amount of torque needed for any given change in angular
momentum is proportional to the size of that change.
 Moment of inertia may be expressed in terms of kilogram-square metres (kg·m2) in
SI units and pound-square feet (lbm·ft2) in imperial or US units.
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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Moment of Inertia
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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 Four objects racing down a plane while rolling without slipping. From back to front:
spherical shell (red), solid sphere (orange), cylindrical ring (green) and solid cylinder
(blue). The time for each object to reach the finishing line depends on their moment
of inertia.

Moment of Inertia
 Suppose a body of mass m is made to rotate about an axis z passing
through the body's center of mass.
 The body has a moment of inertia Icm with respect to this axis. The
parallel axis theorem states that if the body is made to rotate instead
about a new axis z′ which is parallel to the first axis and displaced from
it by a distance d, then the moment of inertia I with respect to the new
axis is related to Icm by
 Explicitly, d is the perpendicular distance between the axes z and z′.
 The parallel axis theorem can be applied with the stretch rule and
perpendicular axis theorem to find moments of inertia for a variety of
shapes.
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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Moment of Inertia
 The second moment of area, also known as moment of inertia of plane area, area
moment of inertia, or second area moment, is a geometrical property of an area which
reflects how its points are distributed with regard to an arbitrary axis.
 The second moment of area is typically denoted with either an for an axis that lies in the
plane or with a for an axis perpendicular to the plane. Its unit of dimension is length to
fourth power, L4.
 In the field of structural engineering, the second moment of area of the cross-section of
a beam is an important property used in the calculation of the beam's deflection and the
calculation of stress caused by a moment applied to the beam.
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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Simple Pendulum
 A simple pendulum, in its simplest form, consists of heavy bob suspended at the end
of a light inextensible and flexible string. The other end of the string is fixed at O, as
shown in Fig.
Let L = Length of the string,
m = Mass of the bob in kg,
W = Weight of the bob in newtons
= m.g, and
θ = Angle through which the string
is displaced.
• Simple pendulum
Compound Pendulum
system,
engine mechanism
(analytical method only),
diagram.
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Periodic time,
Frequency of oscillation,
11

Compound Pendulum
 When a rigid body is suspended vertically, and it oscillates with a small amplitude
under the action of the force of gravity, the body is known as compound pendulum,
as shown in Fig.
 Let m = Mass of the pendulum in kg,
 W = Weight of the pendulum in newtons = m.g ,
 k = Radius of gyration about an axis through the
centre of gravity G and perpendicular to the plane of motion,
 h = Distance of point of suspension O from the centre of gravity G
of the body.
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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Compound Pendulum
 If the pendulum is given a small angular displacement θ, then the couple tending to
restore the pendulum to the equilibrium position OA• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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Since θ is very small
mass moment of inertia about the axis of suspension O,
∴ Angular acceleration of the pendulum,
angular acceleration is directly proportional to angular displacement,
therefore the pendulum executes simple harmonic motion.
We know that the periodic time

Compound Pendulum
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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We know that, the periodic time

Compound Pendulum
 A small flywheel of mass 85 kg is suspended in a vertical plane as a compound
pendulum. The distance of centre of gravity from the knife edge support is 100 mm
and the flywheel makes 100 oscillations in 145 seconds. Find the moment of inertia
of the flywheel through the centre of gravity.
 Given : m = 85 kg ; h = 100 mm = 0.1 m
 Since the flywheel makes 100 oscillations in 145 seconds, therefore frequency of
oscillation,
 n = 100/145 = 0.69 Hz
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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kG= 0.2061 m
I =3.6 kg-m2

Compound Pendulum
 A connecting rod is suspended from a point 25 mm above the centre of small end,
and 650 mm above its centre of gravity, its mass being 37.5 kg. When permitted to
oscillate, the time period is found to be 1.87 seconds. Find the mass moment of
inertia when pendulum is located at the small end.
 Given : h = 650 mm = 0.65 m ; l1 = 650 – 25 = 625 mm= 0.625 m ; m = 37.5 kg ; tp =
1.87 s
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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kG= 0.377 m
I =5.32 kg-m2

A connecting rod is suspended from a point 25 mm
above the centre of small end, and 650 mm above its
centre of gravity, its mass being 37.5 kg. When
permitted to oscillate, the time period is found to be
1.87 seconds. Find the mass moment of inertia when
pendulum is located at the small end.
17

Compound Pendulum
 The connecting rod of an oil engine has a mass of 60 kg, the distance between the
bearing centres is 1 metre. When suspended vertically with a knife-edge through the
small end, it makes 100 oscillations in 190 seconds and with knife-edge through the
big end it makes 100 oscillations in 165seconds. Find the moment of inertia of the
rod in kg-m2 and the distance of C.G. from the small end centre.
 Solution. Given : m = 60 kg ; h1 + h2 = 1 m.
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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When the axis of oscillation coincides with the small end centre,
then frequency of oscillation,
n1 = 100/190 = 0.526 Hz
When the axis of oscillation coincides with the big end centre, the
frequency of oscillation,
n2 = 100/165 = 0.606 Hz
h1 = 0.767 m
kG=0.316m
I=60 × 0.1 = 6 kg-m2

The connecting rod of an oil engine has a mass of 60
kg, the distance between the bearing centres is 1
metre. When suspended vertically with a knife-edge
through the small end, it makes 100 oscillations in
190 seconds and with knife-edge through the big end
it makes 100 oscillations in 165seconds. Find the
moment of inertia of the rod in kg-m2 and the
distance of C.G. from the small end centre.
Solution. Given : m = 60 kg ; h1 + h2 = 1 m.
19

Equivalent length of simple pendulum
 Equations of periodic time of simple pendulum and compound pendulum are given
below• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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Periodic time of simple pedulum,
Periodic time of compound pendulum
By comparing above equation we see that the equivalent length of a simple pendulum,
which gives the same frequency as compound pendulum, is
equivalent length of simple pendulum (L) depends upon the distance between the point of
suspension and the centre of gravity (G), therefore L can be changed by changing the
position of point of suspension. This will, obviously, change the periodic time of a compound
pendulum. The periodic time will be minimum if L is minimum.

Bifilar Suspension
 The moment of inertia of a body may be determined experimentally by an apparatus
called bifilar suspension.
 The body whose moment of inertia is to be determined (say AB) is suspended bytwo
long parallel flexible strings as shown in Fig.
 When the body is twisted through a small angle θ about a vertical axis through the
centre of gravity G, it will vibrate with simple harmonic motion in a horizontal plane.
• Simple pendulum
Compound Pendulum
• Bifilar suspension
system,
engine mechanism
diagram.
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• Let m=Mass of the body,
• W=Weight of the body in newtons = m.g,
• kG=Radius of gyration about an axis through the centre of
gravity,
• I=Mass moment of inertia of the body about a vertical axis
through 2G.,
• l=Length of each string,
• x=Distance of A from G (i.e. AG),
• y=Distance of B from G (i.e. BG),θ=Small angular
displacement of the body from the equilibrium position in
the horizontal plane,
• φA and φB=Corresponding angular displacements of the
strings,
• andα=Angular acceleration towards the equilibrium
position.

22
Bifilar Suspension

Bifilar Suspension
 When the body is stationary, the tension in the strings are given by
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When the body is displaced from its equilibrium position in a
horizontal plane through a small angle θ, then the angular
displacements of the strings are given by
Component of tension TA and TB in the horizontal plane

Bifilar Suspension
 When the body is stationary, the tension in the strings are given by
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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When the body is displaced from its equilibrium
position in a horizontal plane through a small angle
θ, then the angular displacements of the strings are
given by
Component of tension TA and TB in the
horizontal plane

Bifilar Suspension
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These components of tensions TA and TB are equal and opposite in direction, which gives rise
to a couple. The couple or torque applied to each string to restore the body to its initial
equilibrium position, i.e. restoring torque
and accelerating (or disturbing) torque
for equilibrium condition Restoring torque = accelerating
torque Hence..

Bifilar Suspension
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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A small connecting rod of mass 1.5 kg is suspended in a
horizontal plane by two wires 1.25 m long. The wires are
attached to the rod at points 120 mm on either side of the
centre of gravity. If the rod makes 20 oscillations in 40
seconds, find the radius of gyration and the mass moment of
inertia of the rod about a vertical axis through the centre of
gravity.
Given : m = 1.5 kg ; l = 1.25 m ; x = y = 120 mm = 0.12 m
Since the rod makes 20 oscillations in 40 s, therefore
frequency of oscillation ,
n = 20/40 = 0.5 Hz
kG=0.107 m
I= 0.017 kg-m2

 A small connecting rod of mass 1.5 kg is suspended in a horizontal
plane by two wires 1.25 m long. The wires are attached to the rod
at points 120 mm on either side of the centre of gravity. If the rod
makes 20 oscillations in 40 seconds, find the radius of gyration and
the mass moment of inertia of the rod about a vertical axis through
the centre of gravity.
 Given : m = 1.5 kg ; l = 1.25 m ; x = y = 120 mm = 0.12 m
 Since the rod makes 20 oscillations in 40 s, therefore frequency of
oscillation ,
 n = 20/40 = 0.5 Hz
27

Trifilar Suspension
 Trifilar Suspension (Torsional Pendulum)It is also used to find the moment of inertia
of a body experimentally. The body (say a disc or flywheel) whose moment of inertia
is to be determined is suspended by three long flexible wires A, Band C, as shown in
Fig. When the body is twisted about its axis through a small angle θ and then
released, it will oscillate with simple harmonic motion.
 Let m=Mass of the body in kg,
 W=Weight of the body in newtons = m.g,
 kG=Radius of gyration about an axis through c.g.,
 I=Mass moment of inertia of the disc about an axis
through O and perpendicular to it = m.k2,
 l=Length of each wire,
 r=Distance of each wire from the axis of the disc,
 θ=Small angular displacement of the disc,
 φ=Corresponding angular displacement of the wires,
 andα=Angular acceleration towards the equilibrium position.
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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29

Trifilar Suspension
 For small angular deflection of disk ,
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Since the three wires are attached symmetrically with respect to the axis, therefore
the tension in each wire will be one-third of the weight of the body.
∴ Tension in each wire = m.g/3
Component of the tension in each wire perpendicular to r
∴ Torque applied to each wire to restore the body to its initial equilibrium
position i.e. restoring torque
Total restoring torque applied to three wires,
disturbing torque
From above two equations

Trifilar Suspension
 For small angular deflection of disk ,
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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Since the three wires are attached symmetrically with
respect to the axis, therefore the tension in each wire will
be one-third of the weight of the body.
∴ Tension in each wire = m.g/3
Component of the tension in each wire perpendicular to r
∴ Torque applied to each wire to restore the body to its
initial equilibrium position i.e. restoring torque
Total restoring torque applied to three wires,
disturbing torque
From above two equations

Trifilar Suspension
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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We know that periodic time,

Trifilar Suspension
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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In order to find the radius of gyration of a car, it is suspended with its axis vertical from
three parallel wires 2.5 metres long. The wires are attached to the rim at points
spaced120° apart and at equal distances 250 mm from the axis. It is found that the
wheel makes 50 torsional oscillations of small amplitude about its axis in170 seconds.
Find the radius of gyration of the wheel.
Given : l = 2.5 m; r = 250 mm = 0.25 m;
n = 50/170 = 5/17 Hz
kG = 0.268 m = 268 mm

Trifilar Suspension
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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A connecting rod of mass 5.5 kg is placed on a horizontal platform whose mass is 1.5
kg. It is suspended by three equal wires, each 1.25 m long, from a rigid support. The
wires are equally spaced round the circumference of a circle of 125 mm radius. When
the c.g. of the connecting rod coincides with the axis of the circle, the platform makes
10 angular oscillations in 30seconds. Determine the mass moment of inertia about an
axis through its c.g.
Given : m1 = 5.5 kg ; m2 = 1.5 kg ; l = 1.25 m ; r = 125 mm = 0.125 m
n = 10/30 = 1/3 Hz
kG = 0.168 m
I= 0.198 kg-m2

35
A connecting rod of mass 5.5 kg is placed on a horizontal platform
whose mass is 1.5 kg. It is suspended by three equal wires, each
1.25 m long, from a rigid support. The wires are equally spaced
round the circumference of a circle of 125 mm radius. When the
c.g. of the connecting rod coincides with the axis of the circle, the
platform makes 10 angular oscillations in 30seconds. Determine
the mass moment of inertia about an axis through its c.g.
Given : m1 = 5.5 kg ; m2 = 1.5 kg ; l = 1.25 m ; r = 125 mm = 0.125
m
n = 10/30 = 1/3 Hz

Numericals
 A small flywheel having mass 90 kg is suspended in a vertical plane as a compound pendulum. The
distance of centre of gravity from the knife edge support is 250 mm and the flywheel makes
50oscillations in 64 seconds. Find the moment of inertia of the flywheel about an axis through the
centre of gravity.
 [Ans. 3.6 kg-m2]
 The connecting rod of a petrol engine has a mass 12 kg. In order to find its moment of inertia it is
suspended from a horizontal edge, which passes through small end and coincides with the small end
centre. It is made to swing in a vertical plane, such that it makes 100 oscillations in 96 seconds. If the
point of suspension of the connecting rod is 170 mm from its c.g., find : 1. radius of gyration about an
axis through its c.g., 2. moment of inertia about an axis through its c.g., and 3. length of the equivalent
simple pendulum.
 [Ans. 101 mm ; 0.1224 kg-m2 ; 0.23 m]
 A connecting rod of mass 40 kg is suspended vertically as a compound pendulum. The distance
between the bearing centres is 800 mm. The time for 60 oscillations is found to be 92.5 seconds when
the axis of oscillation coincides with the small end centre and 88.4 seconds when it coincides with the
big end centre. Find the distance of the centre of gravity from the small end centre, and the moment of
inertia of the rod about an axis through the centre of gravity.[Ans. 0.442 m ; 2.6 kg-m2]11.The following
data were obtained from an experiment to find the moment of inertia of a pulley by bifilar suspension
:Mass of the pulley = 12 kg ; Length of strings = 3 m ; Distance of strings on either side of centre of
gravity = 150 mm ; Time for 20 oscillations about the vertical axis through c.g. = 46.8 seconds
Calculate the moment of inertia of the pulley about the axis of rotation.
 [Ans. 0.1226 kg-m2]
36

Inertia force
 The inertia force is an imaginary force, which when acts upon a rigid body, brings it
in an equilibrium position. It is numerically equal to the accelerating force in
magnitude, but opposite in direction.
 Mathematically,
 Inertia force = – Accelerating force = – m.a
where m = Mass of the body, and a=Linear acceleration of the centre
of gravity of the body.
 Similarly, the inertia torque is an imaginary torque, which when applied upon the
rigid body, brings it in equilibrium position.
 It is equal to the accelerating couple in magnitude but opposite in direction.
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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Analytical Method for Velocity and
Acceleration of the Piston
 Consider the motion of a crank and
connecting rod of a reciprocating steam
engine as shown in Fig.
 Let OC be the crank and PC the
connecting rod.
 Let the crank rotates with angular
velocity of ω rad/s and the crank turns
through an angle θ from the inner dead
centre (briefly written as I.D.C).
 Let x be the displacement of a
reciprocating body P from I.D.C. after
time t seconds, during which the crank
has turned through an angle θ
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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Let
 l=Length of connecting rod between the centres,
 r=Radius of crank or crank pin circle,
 φ=Inclination of connecting rod to the line of stroke PO,
 And n=Ratio of length of connecting rod to the radius of
crank = l/r.

• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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 Velocity of the piston
 From the geometry
From triangles CPQ and CQO,
CQ = l sin φ = r sin θ or l/r = sin θ/sin φ∴
n = sin θ/sin φ or sin φ = sin θ/n

• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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 From the geometry
From triangles CPQ and CQO,
CQ = l sin φ = r sin θ or l/r = sin θ/sin φ∴
n = sin θ/sin φ or sin φ = sin θ/n

• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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Expanding the above expression by binomial theorem, we get
Substituting the value of (1 – cos φ) in equation

• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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Differentiating equation (iv) with respect to θ,
∴ Velocity of P with respect to O or velocity of the piston P,
Substituting the value of dx/dθ from equation
Velocity of the piston

• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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 Acceleration of the piston
Since the acceleration is the rate of change of
velocity, therefore acceleration of the piston P,
Acceleration of the piston

• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
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Displacement of the piston

Acceleration of the connecting rod
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
46
 Angular displacement of connecting rod
Consider the motion of a connecting rod and
a crank as shown in figure. From the
geometry of the figure, we find that
CQ = l sin φ = r sin θ
Differentiating both sides with respect to time t,
Since the angular velocity of the connecting rod PC is same as the angular velocity of point
P with respect to C and is equal to dφ/dt, therefore angular velocity of the connecting rod

• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
47
 Angular velocity of connecting rod
Angular velocity of connecting rod

• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
48
 Angular acceleration of connecting rod
αPC = Angular acceleration of P with respect to PC
differentiating equation of angular velocity of connecting rod

• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
49
 Angular acceleration of connecting rod
Since is small as compared to , therefore it may be neglected.
unity is small as compared to n2, hence the
term unity may be neglected.

• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
50
Angular acceleration of connecting rod
Angular displacement of connecting rod

Formulae to be remember
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
51
Displacement of the piston
Angular acceleration of connecting rod
Angular displacement of connecting rod

Numericals
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
52
 If the crank and the connecting rod are 300 mm and 1 m long respectively and the
crank rotates at a constant speed of 200 r.p.m., determine:1. The crank angle at
which the maximum velocity occurs, and 2. Maximum velocity of the piston.
 Solution. Given : r = 300 mm = 0.3 m; l = 1 m; N = 200 r.p.m. or ω = 2 π × 200/60 =
20.95 rad/s
 n=l/r = 1/0.3 = 3.33
 For maximum velocity of the piston,
Ans:
θ = 75º
Vpmax = = 6.54 m/s

Numericals
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
53
 The crank and connecting rod of a steam engine are 0.3 m and 1.5 m in length.
The crank rotates at 180 r.p.m. clockwise. Determine the velocity and
acceleration of the piston when the crank is at 40 degrees from the inner dead
centre position. Also determine the position of the crank for zero acceleration of
the piston.
 Given : r = 0.3; l = 1.5 m ; N = 180 r.p.m. or ω = π × 180/60 = 18.85 rad/s; θ = 40°
 n = l/r = 1.5/0.3 = 5
Ans:
Velocity of the piston =4.19m/s
Acceleration of the piston =5.35m/s2
Position of the crank for zero acceleration of the piston, ap=0
θ1 = 79.27° or 280.73°

Numericals
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
54
 In a slider crank mechanism, the length of the crank and connecting rod are150
mm and 600 mm respectively. The crank position is 60° from inner dead centre.
The crank shaft speed is 450 r.p.m. (clockwise). Using analytical method,
determine: 1. Velocity and acceleration of the slider, and 2. Angular velocity and
angular acceleration of the connecting rod.
 Given: r = 150 mm = 0.15 m; l = 600 mm = 0.6 m; θ = 60°; N = 400 r.p.m or ω = π ×
450/60 = 47.13 rad/s
 n = l/r = 0.6/0.15 = 4
 Velocity of the slider =6.9m/s
 acceleration of the slider= 124.94 m/s2
 angular velocity of the connecting rod =5.9rad/s
 angular acceleration of the connecting rod =481 rad/s2

Numericals
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
55
 In a slider crank mechanism, the length of the crank and connecting rod are 100 mm and 400
mm respectively. The crank rotates uniformly at 600 r.p.m. clockwise. When the crank has
turned through45° from the inner dead centre, find, by analytical method : 1. Velocity and
acceleration of the slider,2. Angular velocity and angular acceleration of the connecting rod.
 [Ans. 5.2 m/s; 279 m/s2; 11 rad/s; 698 rad/s2]
 A petrol engine has a stroke of 120 mm and connecting rod is 3 times the crank length. The
crank rotates at 1500 r.p.m. in clockwise direction. Determine: 1. Velocity and acceleration of
the piston, and 2. Angular velocity and angular acceleration of the connecting rod, when the
piston had travelled one-fourth of its stroke from I.D.C.
 [Ans. 8.24 m/s, 1047 m/s2; 37 rad/s, 5816 rad/s2]
 The stroke of a steam engine is 600 mm and the length of connecting rod is 1.5 m. The crank
rotates at 180 r.p.m. Determine: 1. velocity and acceleration of the piston when crank has
travelled through an angle of 40° from inner dead centre, and 2. the position of the crank for
zero acceleration of the piston.
 [Ans. 4.2 m/s, 85.4 m/s2; 79.3° from I.D.C]

Dynamic force analysis of reciprocating
engine mechanism
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
56
 The various forces acting on the reciprocating
parts of a horizontal engine are shown in Fig.
The expressions for these forces, neglecting
the weight of the connecting rod, may be
derived as discussed below :
Piston effort: It is the net force acting on the
piston or crosshead pin, along the line of stroke.
It is denoted by FP in Fig.
Let mR = Mass of the reciprocating parts, e.g. piston, crosshead pin or gudgeon pin etc., in
kg, and
WR = Weight of the reciprocating parts in newtons = mR.g
We know that acceleration of the reciprocating parts,

engine mechanism
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
57
Accelerating force or inertia force of the reciprocating
parts,
On the other hand, the inertia force due to retardation of the reciprocating parts, helps the
force on the piston. Therefore, Piston effort,
In a horizontal engine, the reciprocating parts are
accelerated from rest, during the latter half of the stroke
(i.e. when the piston moves from inner dead centre to outer
dead centre).
It is, then, retarded during the latter half of the stroke (i.e. when the piston moves from outer
dead centre to inner dead centre).
The inertia force due to the acceleration of the reciprocating parts, opposes the force on
the piston due to of pressures in the cylinder on the two sides of the piston.
The –ve sign is used when the piston is accelerated, and +ve sign
is used when the piston is retarded.

engine mechanism
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
58
In a double acting reciprocating steam engine, net load on
the piston,
where
p1, A1=Pressure and cross-sectional area on the back end
side of the piston,
p2, A2=Pressure and cross-sectional area on the crank end
side of the piston,
a=Cross-sectional area of the piston rod.
If ‘p’ is the net pressure of steam or gas on the piston and D is diameter of the piston, then
Net load on the piston,
In case of a vertical engine, the weight of the reciprocating parts assists the piston effort during the
downward stroke (i.e. when the piston moves from top dead centre to bottom dead centre) and opposes
during the upward stroke of the piston (i.e. when the piston moves from bottom dead centre to top dead
centre).
∴ Piston effort,

engine mechanism
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
59
Force acting along the connecting rod: It is denoted by FQ in
Fig.. From the geometry of the figure, we find that
We know that

engine mechanism
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
60
Thrust on the sides of the cylinder walls or normal reaction on the guide
bars. It is denoted by FN in Fig.From the figure, we find that

engine mechanism
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
61
The force acting on the connecting rod FQ may be
resolved into two components,
• one perpendicular to the crank
• other along the crank.
The component of FQ perpendicular to the crank
is known as crank-pin effort and it is denoted by
FT in Fig.
The component of FQ along the crank produces a
thrust on the crank shaft bearings and it is
denoted by FB in Fig.
Resolving FQ perpendicular to the crank,
Crank-pin effort and thrust on crank shaft bearings:
and resolving FQ along the crank,

engine mechanism
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
62
Crank effort or turning moment or torque on the
crank shaft: The product of the crank-pin effort (FT) and
the crank pin radius (r) is known as crank effort or turning
moment or torque on the crank shaft

Equivalent Dynamical System
• Simple pendulum
Compound Pendulum
engines
• Two mass statically and
dynamically equivalent
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
63
In order to determine the motion of a rigid body, under the action of
external forces, it is usually convenient to replace the rigid body by two
masses placed at a fixed distance apart, in such a way that,
Consider a rigid body, having its centre of gravity at
G, as Let
m = Mass of the body,
kG = Radius of gyration about its centre of gravity G,
m1 and m2 = Two masses which form a
dynamical equivalent system,
l1 = Distance of mass m1 from G,
l2 = Distance of mass m2 from G,
L = Total distance between the masses m1 and m2.

• Simple pendulum
Compound Pendulum
engines
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
64
In order to determine the motion of a rigid body, under the action of
external forces, it is usually convenient to replace the rigid body by two
masses placed at a fixed distance apart, in such a way that,
1. the sum of their masses is equal to the total mass of the
body ;
2. the centre of gravity of the two masses coincides with
that of the body ;
3. the sum of mass moment of inertia of the masses about
their centre of gravity is equal to the mass moment of
inertia of the body.
When these three conditions are satisfied, then it is said to be an equivalent dynamical system.

• Simple pendulum
Compound Pendulum
engines
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
65
From equations (i) and (ii),
Substituting the value of m1 and m2 in equation (iii), we have
This equation gives the essential condition of placing the two masses, so that the system
becomes dynamical equivalent. The distance of one of the masses (i.e. either l1 or l2) is
arbitrary chosen and the other distance is obtained from above equation.

Numericals
• Simple pendulum
Compound Pendulum
engines
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
66
 The connecting rod of a gasoline engine is 300 mm long between its centres. It has a mass
of 15 kg and mass moment of inertia of 7000 kg-mm2. Its centre of gravity is at 200 mm
from its small end centre. Determine the dynamical equivalent two-mass system of the
connecting rod if one of the masses is located at the small end centre.
 l = 300 mm; m = 15 kg; I = 7000 kg-mm2;l1 = 200 mm
m1 = Mass placed at the small end centre, and
m2 = Mass placed at a distance l2 from the centre of gravity G.

Numericals
• Simple pendulum
Compound Pendulum
engines
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
1/25/2017
PROF. K N WAKCHAURE
67
 A connecting rod is suspended from a point 25 mm above the centre ofsmall end, and 650
mm above its centre of gravity, its mass being 37.5 kg. When permitted to oscil-late, the
time period is found to be 1.87 seconds. Find the dynamical equivalent system
constituted of two masses, one of which is located at the small end centre.
 h = 650 mm = 0.65 m ; l1 = 650 – 25 = 625 mm= 0.625 m ; m = 37.5 kg ; tp = 1.87 s

Correction Couple to be Applied to Make
Two Mass System Dynamically Equivalent
• Simple pendulum
Compound Pendulum
engines
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
1/25/2017
PROF. K N WAKCHAURE
68
 when two masses are placed arbitrarily, then the conditions (i) and (ii) of dynamic
equivalent will only be satisfied. But the condition (iii) is not possible to satisfy. This
means that the mass moment of inertia of these two masses placed arbitrarily, will differ
than that of mass moment of inertia of the rigid body.
Consider two masses, one at A and the other at D be placed arbitrarily, as shown in Fig..
Let
l3=Distance of mass placed at D from G,
I1=New mass moment of inertia of the two masses;
k1=New radius of gyration;α=Angular acceleration of the body;
I=Mass moment of inertia of a dynamically equivalent system;
k=Radius of gyration of a dynamically equivalent system.
We know that the torque required to accelerate the body,
T=I.α = m (k)2α
Similarly, the torque required to accelerate the two-mass system placed arbitrarily,
T1=I1.α = m (k1)2 α

Correction Couple to be Applied to Make
Two Mass System Dynamically Equivalent
• Simple pendulum
Compound Pendulum
engines
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
1/25/2017
PROF. K N WAKCHAURE
69
 when two masses are placed arbitrarily, then the conditions (i) and (ii) of dynamic
equivalent will only be satisfied. But the condition (iii) is not possible to satisfy. This
means that the mass moment of inertia of these two masses placed arbitrarily, will differ
than that of mass moment of inertia of the rigid body.
We know that the torque required to accelerate the body,
T=I.α = m (k)2 α...(i)
the torque required to accelerate the two-mass system placed arbitrarily,
T1=I1.α = m (k1)2 α...(ii)
∴ Difference between the torques required to accelerate the two-mass system and the torque
required to accelerate the rigid body,
T'=T1–T = m (k1)2 α – m (k)2 α = m [(k1)2 – (k)2] α...(iv)
The difference of the torques T' is known as correction couple.
This couple must be applied, when the masses are placed arbitrarily to make the system
dynamical equivalent. This, of course, will satisfy the condition (iii).

Correction Couple
• Simple pendulum
Compound Pendulum
engines
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
1/25/2017
PROF. K N WAKCHAURE
70  A connecting rod of an I.C. engine has a mass of 2 kg and the distance between the centre
of gudgeon pin and centre of crank pin is 250 mm. The C.G. falls at a point 100 mm from
the gudgeon pin along the line of centres. The radius of gyration about an axis through the
C.G. perpendicular to the plane of rotation is 110 mm. Find the equivalent dynamical
system if only one of the masses is located at gudgeon pin. If the connecting rod is replaced
by two masses, one at the gudgeon pin and the other at the crank pin and the angular
acceleration of the rod is 23 000 rad/s2 clockwise, determine the correction couple applied
to the system to reduce it to a dynamically equivalent system.
Equivalent dynamical system
It is given that one of the masses is located at the gudgeon pin.
Let the other mass be located at a distance l2 from the centre of gravity.
We know that for an equivalent dynamical system.
Since the connecting rod is replaced by two masses located at the two centres (i.e. one at the
gudgeon pin and the other at the crank pin),
therefore, l = 0.1 m, and l3 = l – l1 = 0.25 – 0.1 = 0.15 m
Let k1 = New radius of gyration. We know that (k1)2= l1.l3 = 0.1 × 0.15 = 0.015 m2

Crank Shaft Torque
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
1/25/2017
PROF. K N WAKCHAURE
71
Inertia Forces in a Reciprocating Engine, Considering the Weight of Connecting
Rod
mc=Mass of the connecting rod,
l=Length of the connecting rod,
l1=Length of the centre of gravity of the connecting rod
from P
r= radius of crank
S=2*r =stroke of piston
Ɵ=crank angle
ⱷ= angle made by connecting rod
ap= acceleration of piston
n=l/r =obliquity ratio
N= crank rotation in RPM
ω= angular velocity of crank rad/s
αpc= angular acceleration of crank rad/s2=

Crank Shaft Torque
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
1/25/2017
PROF. K N WAKCHAURE
72
When we consider mass of connecting rod to find torque acting on crank shaft,
then there are three torques acting on crank shaft
T1= Torque due to masses of reciprocating parts
T2=Torque due correction couple acting on connecting rod
T3= Torque due mass acting at crank pin

Crank Shaft Torque
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
1/25/2017
PROF. K N WAKCHAURE
73
The –ve sign is used when the piston is accelerated(0-180), and
+ve sign is used when the piston is retarded(180-360).
T1= Torque due to masses of reciprocating parts
mR=reciprocating masses =mp+m1
mp= mass of piston
m1= mass at piston pin/gudgeon pin
ap=acceleration of piston
If ‘p’ is the net pressure of steam or gas on the piston and D is diameter of the piston, then
Net load/gas force on the piston,
𝑚 𝑟 = 𝑚 𝑝 + 𝑚1

Crank Shaft Torque
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
1/25/2017
PROF. K N WAKCHAURE
74
T2=Torque due correction couple acting on connecting rod
𝑇′ = 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑢𝑝𝑙𝑒
𝑻 𝟐 =
𝑻′
∗ 𝑶𝑵
𝑷𝑵
𝑂𝑁 = 𝑟𝑐𝑜𝑠(Ɵ
𝑃𝑁 = 𝑙𝑐𝑜𝑠(ⱷ
𝑇′ = 𝑚 𝑐 𝑘1
2
−𝑘2 α
𝑚 𝑐 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑛𝑔 𝑟𝑜𝑑
𝑘1
2
= 𝑙1 ∗ 𝑙3

Crank Shaft Torque
• Simple pendulum
Compound Pendulum
system,
engine mechanism
diagram.
THEORY OF MACHINES-
I
UNIT-
II
1/25/2017
PROF. K N WAKCHAURE
75
T3= Torque due mass acting at crank pin
𝑚3 =
𝑚 𝑐 𝑙1
𝑙1 + 𝑙3
𝑻 𝟑 = 𝒎 𝟑 ∗ 𝒈 ∗ 𝑶𝑵
𝑂𝑁 = 𝑟𝑐𝑜𝑠(Ɵ
𝑻 𝒕𝒐𝒕𝒂𝒍 = 𝑻 𝟏 + 𝑻 𝟐 + 𝑻 𝟑

TURNING MOMENT DIAGRAM
76
Turning Moment Diagram for a Single Cylinder Double Acting Steam Engine
The turning moment diagram (also known as crank effort diagram) is the graphical representation of
the turning moment or crank-effort for various positions of the crank.

1/25/2017
PROF. K N WAKCHAURE
77
Turning moment diagram for a four stroke cycle internal combustion engine.

1/25/2017
PROF. K N WAKCHAURE
78
Turning moment diagram for a multi-cylinder engine.

Friction
 at every joint in a machine, force of friction arises due to the relative motion
between two parts and hence some energy is wasted in overcoming the
friction.
 The friction between the wheels and the road is essential for the car to move
forward.
79

Types of Friction
 1. Static friction: It is the friction, experienced by a body, when at rest.
 2. Dynamic friction. It is the friction, experienced by a body, when in motion. The dynamic
friction is also called kinetic friction and is less than the static friction.
 It is of the following three types :
 (a) Sliding friction. It is the friction, experienced by a body, when it slides over another
body.
 (b) Rolling friction. It is the friction, experienced between the surfaces which has balls or
rollers interposed between them.
 (c) Pivot friction. It is the friction, experienced by a body, due to the motion of rotation as
in case of foot step bearings.
80

Laws of Static Friction
81

Laws of Kinetic or Dynamic Friction
82

Coefficient of Friction
83

Limiting Angle of Friction
84

Angle of Repose
1/25/2017
PROF. K N WAKCHAURE
85

Friction in turning pairs
86

Friction axis
87

Friction axis
88

Friction axis
89

Thank you…
90

Theory of machines_static and dynamic force analysis

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Theory of machines_static and dynamic force analysis