DOM - Unit 1

ME8594-DYNAMICS OF MACHINES
UNIT -1

UNIT I
FORCE ANALYSIS - 12
■ Dynamic force analysis – Inertia force and Inertia torque– D Alembert’s principle –Dynamic
Analysis in reciprocating engines – Gas forces – Inertia effect of connecting rod– Bearing loads –
Crank shaft torque – Turning moment diagrams –Fly Wheels – Flywheels of punching presses-
Dynamics of Cam- follower mechanism.
■ 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.

Newton's Law
■ 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 direction. This law is
sometimes referred to as the action-reaction law, with F called the "action" and −F the "reaction"

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
 https://www.youtube.com/watch?v=D-YzoxtFBHs

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.
 https://www.youtube.com/watch?v=Fri3H_YcskI

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
 https://www.youtube.com/watch?v=nukQw1wavl4&list=PLdLe0dTcWW-
t1rz5BDBzZVv-wHy2MabGK&index=1

D’Alembert’s Principal
 D’Alembert’s principles states that the inertia force and
torques, and the external forces and torques, acting on a
body together result in statically equilibrium.
 https://www.youtube.com/watch?v=jvBas8yUgNI

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
 https://www.youtube.com/watch?v=j9Ztjs1tsi8&list=PLdLe0dTcWW-
t1rz5BDBzZVv-wHy2MabGK&index=2

• 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

https://www.youtube.com/watch?v=mSP1C5bmPj8

SESSION OBJECTIVES
• Slider-crank mechanism
• Displacement, velocity and acceleration of a
slider in slider-crank mechanism.
• Forces on an engine

Slider-Crank mechanism
Inner-Dead center Outer-Dead center

PQ = l cos f CQ= lsinf
OQ = rcos q
CQ = rsinq

ENGINE FORCE ANALYSIS
■ Newton’s law of motion:
■ First law of motion:
 Everybody continues in its states of it rest or of uniform motion in a
straight line unless an external resultant forces acts on it.
 Thus, the Newton’s law helps us to define a force.

Second law of motion
 The rate of change of momentum of a body is directly proportional to
the force acting on it and takes place in the direction of force.
 F =d dt (mv) = mdv dt
Where,
F=m.a
F= Force acting on the particle m=Mass of the particle v= Velocity of
the particle a=Acceleration of the particle

Third law of motion
 To every action there is an equal and opposite reaction.
 It means that the force of action and reaction between two bodies are
equal in magnitude but opposite in direction
 https://www.youtube.com/watch?v=_sr3hBxu614

Inertia Force
 Inertia force = - External (or accelerating) force= -m.a
Inertia Torque:
 Torque, 𝑇∞ 𝑑 𝑑𝑡 (𝐼.𝜔)
 Since I is constant, there fore 𝑇 = 𝐼 𝑑𝜔 𝑑𝑡 = 𝐼.𝛼
Where,
■ I= Mass moment of inertia of the body
■ 𝜔=Angular velocity of the body
■ 𝛼= Angular acceleration of the body
■ T= External (or accelerating) torque
 https://www.youtube.com/watch?v=-Fz3LzbzKrQ

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 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.

Velocity and Acceleration of the Reciprocating
Parts in Engines
■ The velocity and acceleration of the reciprocating parts of the steam engine or
internal combustion engine (briefly called as I.C. engine) may be determined by
graphical method or analytical method.
■ The velocity and acceleration, by graphical method, may be determined by one
of the following constructions:
 1. Klien’s construction,
 2. Ritterhaus’s construction,
 3. Bennett’s construction.
■ We shall now discuss these constructions, in detail, in the following pages.

Approximate 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 Figure 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 timetseconds, during
which the crank has turned through an angle θ.

Approximate analytical method for velocity and
acceleration of the piston
 https://www.youtube.com/watch?v=RsuGFDcoB2Y&list
=PLdLe0dTcWW-t1rz5BDBzZVv-
wHy2MabGK&index=3

Approximate analytical method for
velocity and acceleration of the piston
■ Motion of a crank and connecting rod of a reciprocating steam
engine. 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.

Approximate analytical method for velocity and
acceleration of the piston
Velocity of the piston
From the figure

Angular velocity and acceleration of the
connecting rod
■ Consider the motion of a connecting rod and a crank as shown in Fig. 15.7.From the
geometry of the figure, we find that
CQ = l sinφ=r sinθ

Angular velocity and acceleration of the
connecting rod

DYNAMIC FORCE ANALYSIS
Velocity and Acceleration of Reciprocating Part
 https://www.youtube.com/watch?v=RsuGFDcoB2Y&list=
PLdLe0dTcWW-t1rz5BDBzZVv-wHy2MabGK&index=3

 Displacement of piston (x) = r[(1-cos θ)+sin2θ /2n]
 Velocity of Piston (Vp) =
 Acceleration of Piston (Ap) =
 Angular velocity of connecting Rod (ωc ) =
 Angular acceleration of connecting Rod (Ac) = - ω2 Sinθ/n

Notation
Representation
Forces Units
r Crank radius Mtr
l Length of connecting rod Mtr
Angle made in crank with IDC Degree
Inclination of C.R with line of stroke Degree
Ratio of
Degree
n=l/r crank radius to length of C.R -

Session objectives
• Motion analysis of slider Crank mechanism
• How to calculate the displacement, velocity and acceleration.
• Forces acting in a slider Crank mechanism
• Torque in an engine

Slider Crank mechanism- Analysis
• Displacement of piston,x =
• Velocity of piston, v =
• Acceleration of piston, a =

Motion Analysis Problem
 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 m,
l = 1.5 m
N = 180 rpm
q = 40 Deg.

To find
v
a
q = ? When a=0
Solution
r=0.3 m, l = 1.5 m, N = 180 rpm, q = 40 Deg.
ω= π * N / 60 = π * 180/60 = 18.85 rad/s
Velocity of piston:

Solution
R=0.3 m, l = 1.5 m, N = 180 rpm, q = 40 Deg.
 = π * N / 60 = π* 180/60 = 18.85 rad/s
Acceleration of piston:

Angular displacement:
Formulae

Problems on Approximate analytical method for
velocity and acceleration of the piston
 .In a slider crank mechanism, the length of the crank and
connecting rod are 150 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
To Find: Velocity and acceleration of the slider and Angular Velocity and
Angular Acceleration of the connecting rod
Solution:
Using Formula:
 Velocity of Piston (Vp) =
 Acceleration of Piston (Ap) =

Solution for Velocity and acceleration of the
slider
We know that ratio of the length of connecting rod and crank,
n= l/r= 0.6 / 0.15 = 4

Solution for Angular Velocity and Angular
Acceleration of the connecting rod

Forces on Reciprocating Parts
 Piston Effort FP = FL ± FI ± WR RF
 Net Load on Piston (single acting Cylinder) FL = p × A1
 Double acting Cylinder FL =p1 × A1 p2 × A2
 Inertia Force 𝐹I= 𝑚𝜔2 𝑟[ 𝑐𝑜𝑠𝜃 + 𝑐𝑜𝑠2 𝜃 /2 𝑛]
 Force acting along C.R 𝐹 𝑄 = 𝐹 𝑃 cos∅

 Normal reaction 𝐹 𝑁 = 𝐹 𝑄 𝑠𝑖𝑛∅ = 𝐹 𝑃 𝑡𝑎𝑛∅
 Crank pin effort 𝐹 𝑇 = 𝐹 𝑄sin ( 𝜃 + ∅)
 Thrust on crankshaft 𝐹 𝐵 = 𝐹 𝑄 cos( 𝜃 + ∅)
 Crank effort 𝑇 = 𝐹 𝑇 × 𝑟 = [ 𝐹𝑃 cos 𝜃sin( 𝜃 + ∅)] × 𝑟

Notation
Representation
Forces Units
𝑚𝑅 Mass of reciprocating parts Kg
𝑊𝑅 Weight of reciprocating part Kg
𝐹𝐼 Inertia force N
𝐹𝐿 Net load on piston N
𝐹𝑃 Piston effort N
𝐹𝑄 Force acting along C.R N
𝐹𝑁 Normal reaction N
𝐹𝑇 Crank pin effort N
𝐹𝐵 Thrust on crankshaft N
𝑇 Crank effort N-m
𝑝 Net pressure in cylinder N/m2

Forces in engine
1. Piston effort FP
2. Force along connecting
rod FQ
3. Thrust on cylinder
walls FN
4. Thrust on crankshaft
bearings FB
5. Crank-pin
Effort/Tangential force
FT

Forces in engine
1. Piston effort FP
FL = Gas forces/Load
= P1A1 – P2 (A1-A2)
FI = Inertia forces
= ma
Rf = Friction forces
WR = weight of piston

Forces in engine
2. Force along connecting rod
FQ

Forces in engine
3.Thrust on cylinder
walls FN

Forces in engine
4. Thrust on crankshaft
bearings

Forces in engine
5.Crank pin effort
6. Turning Moment on crank
shaft
T = FT * r

Engine force analysis problems
1. Displacement of piston,x =
2. Velocity of piston, v =
3. Acceleration of piston, a =
4. Rel. Between q & f =>

Dynamic force analysis of reciprocating engine -
Formulae
5. Piston effort,
– Gas forces
– Inertia forces
– Frictional forces, RF = m .N

Dynamic force analysis of reciprocating engine -
Formulae
6. Thrust on connecting Rod,
7. Thrust on cylinder walls,
8. Thrust on crank shaft bearings,
9. Crank – Pin effort
10. Torque , T = FT x r

Problems on Forces on Reciprocating Parts
 The crank-pin circle radius of a horizontal engine is 300 mm. The mass of
the reciprocating parts is 250 kg. When the crank has travelled 60° from
I.D.C., the difference between the driving and the back pressures is 0.35
N/mm2. The connecting rod length between centres is 1.2 m and the
cylinder bore is 0.5 m. If the engine runs at 250 r.p.m. and if the effect of
piston rod diameter is neglected, calculate : 1. pressure on slide bars, 2.
thrust in the connecting rod, 3. tangential force on the crank-pin, and 4.
turning moment on the crank shaft

Given: r = 300 mm = 0.3 m ; mR = 250 kg; θ = 60°; p1 – p2 = 0.35 N/mm2; l =1.2 m ;D=
0.5 m = 500 mm ; N= 250 r.p.m. orω= 2π× 250/60 = 26.2 rad/s
To find :1. pressure on slide bars, 2. thrust in the connecting rod, 3. tangential force on the
crank-pin, and 4. turning moment on the crank shaft
Using Formulas
• Net load on the piston =Fp=FL- (P1-P2)П/4*D2
• Ratio of length of Connecting rod and Crank n=l/r
• Inertia Force 𝐹𝐼 = 𝑚𝜔2𝑟[𝑐𝑜𝑠𝜃 + 𝑐𝑜𝑠2𝜃 /2𝑛]
• Piston Effort FP = FL ± FI
• Pressure on slide bars sinφ =sinθ /n and FN =FP tanφ
• Thrust in the connecting rod FQ=Fp/cos φ
• Tangential force on the crank-pin FT = FQ sin (θ+φ)
• Turning moment on the crank shaft T=FT *r

Solution
 First of all, let us find out the piston effort (FP).
 We know that net load on the piston,

Pressure on slide bars
Let φ = Angle of inclination of the connecting rod to the line of stroke.
We know that, sinφ =sinθ /n= sin 60/4=0.866 /4= 0.2165
∴ φ = 12.5°
We know that pressure on the slide bars,
FN =FPtanφ = 49.424 × tan 12.5°
= 10.96 kN Ans.

Thrust in the connecting rod and Tangential force on
the crank-pin
We know that thrust in the connecting rod,
FQ = FP/ cosφ
= 49.424/cos 12.5°
=50.62 kNAns
We know that tangential force on the crank pin,
FT =FQ sin (θ+φ)
= 50.62 sin (60+12.5 )
= 48.28 kNAns

.Turning moment on the crank shaft
■ We know that turning moment on the crank shaft,
T=FT *r
=48.28* 0.3
=14.484 kN-m Ans

 The crank-pin circle radius of a horizontal engine is 300 mm. The mass of
the reciprocating parts is 250 kg. When the crank has travelled 60° from
I.D.C., the difference between the driving and the back pressures is 0.35
N/mm2. The connecting rod length between centres is 1.2 m and the
cylinder bore is 0.5 m. If the engine runs at 250 r.p.m. and if the effect of
piston rod diameter is neglected, calculate :i. pressure on slide bars, ii.
Thrust in the connecting rod, iii. Tangential force on the crank-pin, and iv
turning moment on the crank shaft.
Solution

Given:
Crank radius r = 300 mm
m = 250 kg;
q = 60;
p1 – p2 = 0.35 N/mm^2 ;
l = 1.2m;
Cylinder Bore, D = 0.5m = 500mm
Speed in rpm, N = 250rpm
To find;
i. FN Thrust on side walls
ii. FQ Thrust on connecting rod
iii. FT Crank-pin effort/ Tang.
force
iv. T, Torque

1. n = l/r = 1.2/0.3 = 4
2. Gas forces,

3. Angle bet. Connecting rod and horizontal f
4. Pressure on slide bars/ Thrust on walls FN
5. Thrust in connecting rod FQ

6. Crank-Pin effort FT
7. Turning moment on crank shaft T

Vertical engine
Top-Dead center
Bottom-Dead center
L

Problem on vertical engine
A vertical petrol engine 100 mm diameter and 120 mm stroke has a connecting rod
250 mm long. The mass of the piston is 1.1 kg. The speed is 2000 r.p.m. On the
expansion stroke with a crank 20° from top dead centre, the gas pressure is 700
kN/m2. Determine:
1. Net force on the piston
2. Resultantt load on the gudgeon pin,
3. Thrust on the cylinder walls, and
4. Speed above which, other things remaining same, the gudgeon pin load would be
reversed in direction.

Given:
D = 100 mm = 0.1 m
L = 120 mm = 0.12 m or
r = L/2 = 0.06 m ;
l = 250 mm = 0.25 m ;
mR = 1.1 kg ;
N = 2000 r.p.m. Or
ω = 2 π × 2000/60 = 209.5 rad/s
q = 20°;
p = 700 kN/m2
To Find
1.Net force on the piston
2.Resultant load on the gudgeon pin,
3. Thrust on the cylinder walls, and
4. Speed above which, other things remaining same, the gudgeon pin load would be reversed in
direction

1. Net force on piston
– Gas Forces
– Inertia Forces
– Final piston effort

2. Resultantt load on the gudgeon pin,(connecting rod thrust)
3. Thrust on the cylinder walls,

4. Speed above which, the gudgeon pin load would be reversed in direction.
– Gudgeon pin load i.e. FQ will be negative if FP is negative, And FP will
be negative when FL and W are less than FI => FI > FL + W

EQUIVALENT DYNAMICAL SYSTEM
Session-04

• 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. Consider a rigid body, having its centre of gravity at G

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,

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 equation (vi).

 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 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.
Solution.
Given : h = 650 mm = 0.65 m ; l1 = 650–25 = 625 mm = 0.625 m ; m =
37.5 kg ; tp = 1.87 s
First of all, let us find the radius of gyration (kG) of the connecting rod
(considering it is a compound pendulum), about an axis passing
through its centre of gravity, G. We know that for a compound
pendulum, time period of oscillation (tp),

It is given that one of the masses is located at the small end centre. Let the
other mass is located at a distance l2 from the centre of gravity G, as shown
in Fig. 15.19. We know that, for a dynamicallyequivalentsystem,
l1.l2 = (kG)2

Session-05
CORRECTION COUPLE TO BE APPLIED TO MAKE TWO MASS
SYSTEM DYNAMICALLY EQUIVALENT

CORRECTION COUPLE TO BE APPLIED TO MAKE TWO MASS
SYSTEM DYNAMICALLY EQUIVALENT
Correction couple to be applied to make the two-mass system
dynamically equivalent

 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

Turning moment diagram
20.07.2020-25.07.2020

4 stroke engine – Turning moment diagram

Purpose
• Turning moment diagram is a graphical representation of turning moment or
torque (along Y-axis) versus crank angle (X-axis) for various positions of
crank.
• The area under the TMD gives the work don per cycle.
• The work done per cycle when divided by the crank angle per cycle gives
the mean torque Tm.
• The mean torque Tm multiplied by the angula velocity of the crank gives the
power consumed by the machine or developed by an engine.

Fly wheels
The Fly wheel stores the excessive
mechanical energy from the engine.
It stores the energy during power stroke
and releases it during the other strokes.
Two types
Disk type
Rim type

Types of flywheels
• Disc type
• I = mD2/8
• Rim type
• I = mD2/4

Important Terms
1. Maximum fluctuation of speed
■ It is the difference between the maximum & minimum speeds in a cycle.
N1 – N2
2. Coefficient of fluctuation of speed
■ It is the ratio of maximum fluctuation of speed to the mean speed
■ It is expressed as percentage of mean speed

Energy stored in flywheel
E = mk2w2Cs
Or Iw2Cs
Where,
m = mass (kg)
k = radius of gyration (m) [ k2 = D/4 for rim type k2 = D/8 for disc type]
w = angular velocity of flywheel (rad/s)
Cs = coefficient of fluctuation of speed (no unit)
E = excessive energy (J)
I = mass Moment of inertia (kg-m2)

Energy in fly wheels problems
Formulae
• Excess energy, stored in flywheel, E = mk2w2Cs or Iw2Cs
• Work done /cycle = area under the curve
• Work done /cycle = Tmean x (angle for one cycle)
• i.e. Net Area = Tmean x 4p (for 4 stroke engine)

 A single cylinder 4 stroke gas engine develops 18.4 KW at 300
rpm with work done by the gases during the expansion being 3
times the work done on the gases during compression.The
work done during the suction & exhaust strokes is negligible.
The total fluctuation of speed is 2% of the mean. The TMD
may be assumed to be triangular in shape. Find the mass
moment of inertia of the flywheel.

Given:
Power = 18.4kw
N = 300rpm
WE= 3xWC
Cs = 0.02
4 stroke engine
i.e q = 4p (2revs)
I = ?

■ w = 2pN/60 = 2p * 300/60 = 31.416 rad/s
■ Power = Tmean x w
 18400 = Tmean x 31.416
 Tmean = 585.7 N-m
■ Work done = Tmean x q = WE – WC
 585.7 x 4p = 3WC – WC
 WC = 585.7 x 4p /2
= 3680 N-m
 WE. = 3x 3678.61
= 11040 N-m

■ But,. WE= ½ x p x Tmax
 Tmax = 11040 x 2/ p
 Tmax = 7028.3 N-m
■ Similar triangles
x = 2.88 rad
■ Excess energy in flywheel,

• Mass Moment of inertia
• e = Iw2Cs
• I = 9276.7 / (31.416)2 x 0.02
• I = 470 kg-m2

Energy in multi cylinder engine

 The turning moment diagram for a petrol engine is drawn to the
following scales : Turning moment, 1 mm = 5 N-m ; crank angle, 1
mm = 1°. The turning moment diagram repeats itself at every half
revolution of the engine and the areas above and below the mean
turning moment line taken in order are 295, 685, 40, 340, 960, 270
mm2. The rotating parts are equivalent to a mass of 36 kg at a radius
of gyration of 150 mm. Determine the coefficient of fluctuation of
speed when the engine runs at 1800 r.p.m.

Given
m = 36kg,
k= 150 mm,
N = 1800 rpm
1 mm ‘T’ = 5Nm
1 mm ‘q’ = 1o
1 mm2 = 5 x p/180 = 0.0872 N-m

■ We need the maximum fluctuation of energy to find Cs
=> DE is found from the turning moment diagram

Cs can be found using the formula

 A shaft fitted with a flywheel rotates at 250 r.p.m. and drives a machine. The
torque of machine varies in a cyclic manner over a period of 3 revolutions. The
torque rises from 750 N-m to 3000 N-m uniformly during ½ revolution and
remains constant for the following revolution. It then falls uniformly to 750 N-m
during the next ½ revolution and remains constant for one revolution, the cycle
being repeated thereafter.
 Determine the power required to drive the machine and percentage fluctuation in
speed, if the driving torque applied to the shaft is constant and the mass of the
flywheel is 500 kg with radius of gyration of 600 mm.

Given
N = 250 rpm
i.e  = 2pN/60 = 26.18 rad/s
Rev. = 3 i.e. q = 6p
m = 500 kg
k = 600 mm = 0.6m

From the turning moment diagram,
Work done = Area under curve
. = OAEF+ABG +
BGHC + CHD
= 6p x 750 + ½ x p x 2250
+ 2p x 2250 + ½x px 2250
= 11250 p N-m
But, the work done can also be
taken as,
Work done = Tmean x 6p
Tmean = 11250p / 6p = 1875N-m
Power = Tmean x  = 1875 x 26.18
Power = 49.125 kW

Find area above the Tmean line.
Energy, DE= BLM + BMNC + CNP
 We need LM and NP
 Consider similar triangles,
LM = 0.5p
For finding NP
NP = 0.5p
Energy, DE= ½ x0.5px1125 + 2p x 1125
+ ½x 0.5p x 1125
DE= 8837 N-m

Flywheel problems
27.07.2020-01.08.2020

 The turning moment curve for an engine is represented by the
equation,T = (20 000 + 9500 sin 2θ – 5700 cos 2θ) N-m, where θ is
the angle moved by the crank from inner dead centre. If the resisting
torque is constant, find:
 1. Powerr developed by the engine ; 2. Moment of inertia of
flywheel in kg-m2, if the total fluctuation of speed is not exceed
1% of mean speed which is 180 r.p.m; and 3. Angular
acceleration of the flywheel when the crank has turned through
45° from inner dead centre.

Given :
T = (20 000 + 9500 sin 2θ – 5700 cos 2θ) N-m ;
N = 180 r.p.m. or
ω = 2π × 180/60 = 18.85 rad/s
TMD

• Work done per revolution
Tmean= Work / q = 40000p/2p
=20000 N-m
Power = Tmean x q = 20000 x 18.85
Power = 377 kW

• The turning moment diagram for one stroke (i.e. half revolution of the crankshaft)
is shown. Since at points B and D, the torque exerted on the crankshaft is equal to
the mean resisting torque on the flywheel

• Maximum fluctuation of energy
• But, Maximum fluctuation of energy can also be written as,
DE = I2Cs
11078 = I x (18.85)2 x 0.01
I = 3121 kg-m2

• Angular acceleration at 45 degree
 Excess torque at 45 degree
 Excess torque at 45 degree can also be written as
Tmean = I x a
a = 9500/3151 = 3.044 rad/s2

The turning moment diagram for a multicylinder engine has been
drawn to a scale 1 mm = 600 N-m vertically and 1 mm = 3°
horizontally. The intercepted areas between the output torque curve
and the mean resistance line, taken in order from one end, are as
follows :
+ 52, – 124, + 92, – 140, + 85, – 72 and + 107 mm2, when the
engine is running at a speed of 600 r.p.m. If the total fluctuation of
speed is not to exceed ± 1.5% of the mean, find the necessary mass
of the flywheel of radius 0.5 m

 A three cylinder single acting engine has its cranks set equally at 120°
and it runs at 600 r.p.m. The torque-crank angle diagram for each
cycle is a triangle for the power stroke with a maximum torque of 90
N-m at 60° from dead centre of corresponding crank. The torque on
the return stroke is sensibly zero. Determine : 1. power developed. 2.
coefficient of fluctuation of speed, if the mass of the flywheel is 12 kg
and has a radius of gyration of 80 mm, 3. coefficient of fluctuation of
energy, and 4. maximum angular acceleration of the flywheel.

Given
N = 600 r.p.m.
ω = 2 π × 600/60 = 62.84 rad /s;
Tmax = 90 N-m;
m = 12 kg;
k = 80 mm = 0.08 m
TMD

■ Work done/cycle (Area under curve)
3 x (Area of triangle)
3 x ½ x p x 90 = 424 N-m
■ Tmean = Work / q
= 424 / 2p
= 67.5 N-m
■ Tmean = (Tmax + Tmin)/2
Tmin = 2 x Tmean – Tmax
Tmin = 45 N-m
■ The turning moment diagram
can be represented as,

■ Finding the area of the multi-cylinder TMD

• Fluctuation in energy
• Maximum = E + 5.89
• Minimum = E-5.89
• Maximum fluctuation in energy DE = E + 5.89 – (E – 5.89)
= 11.78 N-m

■ Coefficient of fluctuation of energy
■ Maximum angular acceleration of flywheel
 Flywheel reaches maximum acceleration when maximum energy is available
 Maximum energy is available when the difference between maximum torque available and
the torque needed for operation is maximum.
 That is,. when the torque is at Tmax, energy to flywheel is Tmax - Tmean

Dimensions of the Flywheel Rim
Consider a rim of the flywheel as shown.
■ Let D = Mean diameter of rim in metres,
■ R = Mean radius of rim in metres,
■ A = Cross-sectional area of rim in m2,
■ ρ = Density of rim material in kg/m3,
■ N = Speed of the flywheel in r.p.m.,
■ ω = Angular velocity of the flywheel in rad/s,
■ v = Linear velocity at the mean radius in m/s
= ω .R = π D.N/60, and
■ σ = Tensile stress or hoop stress in N/m2 due to the centrifugal force.

Formula to remember
■ Centrifugal / Hoop stress. s = r x v2
■ linear Velocity ,. v = kw or Rw

Dimensions of the Flywheel Rim
Consider a rim of the flywheel as shown.
■ Let D = Mean diameter of rim in metres,
■ R = Mean radius of rim in metres,
■ A = Cross-sectional area of rim in m2,
■ ρ = Density of rim material in kg/m3,
■ N = Speed of the flywheel in r.p.m.,
■ ω = Angular velocity of the flywheel in rad/s,
■ v = Linear velocity at the mean radius in m/s
= ω .R = π D.N/60, and
■ σ = Tensile stress or hoop stress in N/m2
. due to the centrifugal force.

Formula to remember
■ Centrifugal / Hoop stress. s = r x v2
■ linear Velocity ,. v = kw or Rw = pDN/60
■ Coefficient of energy, CE = DE/Work done
■ Volume of Rim ,. VOLUME = pDA

The turning moment diagram for a multi-cylinder engine has been drawn to a scale
of 1 mm to 500 N-m torque and 1 mm to 6° of crank displacement. The intercepted
areas between output torque curve and mean resistance line taken in order from one
end, in sq. mm are – 30, + 410, – 280, + 320, – 330, + 250, – 360, + 280, – 260 sq.
mm, when the engine is running at 800 r.p.m. The engine has a stroke of 300 mm
and the fluctuation of speed is not to exceed ± 2% of the mean speed. Determine a
suitable diameter and cross-section of the flywheel rim for a limiting value of the
safe centrifugal stress of 7 Mpa. The material density may be assumed as 7200
kg/m3. The width of the rim is to be 5 times the thickness.

Given
N = 800 r.p.m. or
ω = 2π × 800 / 60 = 83.8 rad/s;
Stroke = 300 mm ;
σ = 7 MPa = 7 × 106 N/m2 ;
ρ = 7200 kg/m3
Cs = ± 2% = 4% = 0.04

■ Diameter of the flywheel rim
Let D = Diameter of the flywheel rim in metres, and
v = Peripheral velocity of the flywheel rim in m/s.
We know that centrifugal stress (σ),
7 × 106 = ρ.v2 = 7200 v2
v2 = 7 × 106/7200 = 972.2
∴ v = 31.2 m/s
But,
v = π D.N/60
∴ D = v × 60 / π N
= 31.2 × 60/π × 800
D = 0.745 m

■ Cross-section of the flywheel rim
Let t = Thickness of the flywheel rim
b = Width of the flywheel rim
= 5 t …(Given)
∴ Cross-sectional area of rim,
A = b.t = 5 t × t = 5 t 2
■ Turning moment diagram- Scale
– turning moment scale is 1 mm = 500 N-m and
– crank angle scale is 1 mm = 6° = π /30 rad,
Therefore
– 1 mm2 on the turning moment diagram
= 500 × π / 30 = 52.37 N-m

■ Energy Variation
■ Mass of flywheel

■ Dimensions of Flywheel
We know that mass of the flywheel rim (m),
m = Volume × density = π D.A.ρ
604 = π × 0.745 × 5t2 × 7200
= 84 268 t 2
∴ t2 = 604 / 84 268 = 0.007 17 m2
t = 0.085 m = 85 mm
and b = 5t = 5 × 85
b = 425 mm

REFERENCES
 1. V.Ramamurthi, “Mechanics of Machines”, Narosa Publishing House, 2002.
 2. Khurmi, R.S.,”Theory of Machines”, 14th Edition, S Chand Publications,
2005.
 Module 1 - Lecture 1 - Rigid Body Motion – YouTube
 Module 1 - Lecture 2 - Rigid Body Motion – YouTube
 Module 1 - Lecture 3 - Dynamic Force Analysis of Mechanisms -
YouTube

DOM - Unit 1

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