DYNAMICS OF MACHINES UNIT-1 BY Mr.P.RAMACHANDRAN/AP/MECH/KIT/CBE

DYNAMICS OF MACHINES Prepared by:
Notes Prof. B. SHANMUGASUNDARAM, M.E, (Ph.D)
Associate Professor/Mech. Engg., KIT, Coimbatore.
KIT - KALAIGNARKARUNANIDHI INSTITUTE OF TECHNOLOGY
COIMBATORE – 641 402
DEPARTMENT OF MECHANICAL ENGINEERING
B.E – III YEAR - V SEMESTER
DYNAMICS OF MACHINES
UNIT I
FORCE ANALYSIS
Force
Force is a pull or push, which acts on a body changes or tends to change, the
state of rest or of uniform motion of the body. A force is completely characterized by
its point of application, its magnitude and direction.
Applied Force
The energy or effort provided to a machine to perform work is called applied
force.
An applied force is a force which is applied to an object by a person or
another object. If a person is pushing a desk across the room, then there is applied
force acting upon the object. The applied force is the force exerted on the desk by
the person. The applied forces are classified as active and reactive forces.
Constraint Forces
Constraints and those conditions that we put on the movement of a system so
that its motion gets restricted. Constraint reduces the degrees of freedom of a
system.
Constraint forces are the forces that are applied on a system to enforce a
constraint. A body is not always free to move in all directions. The restriction to the
free motion of body in any direction is called constraint.
Types of Constrained Motions
Following are the three types of constrained motions:
1. Completely Constrained Motion:
When the motion between a pair is limited to a definite direction irrespective of
the direction of force applied, then the motion is said to be a completely constrained
motion.
For example, the piston and cylinder (in a steam engine) form a pair and the
motion of the piston is limited to a definite direction (i.e., it will only reciprocate)

relative to the cylinder irrespective of the direction of motion of the crank, as shown
in Fig. 1.1. The motion of a square bar in a square hole, as shown in Fig. 1.2, and
the motion of a shaft with collars at each end in a circular hole, as shown in Fig. 1.3,
are also examples of completely constrained motion.
Fig. 1.1 Reciprocating Steam Engine
Fig. 1.2 Square Bar in a Square Hole Fig. 1.3 Shaft with Collars in a Circular Hole
2. Incompletely Constrained Motion:
When the motion between a pair can take place in more than one direction,
then the motion is called an incompletely constrained motion. The change in the
direction of impressed force may alter the direction of relative motion between the
pair.
A circular bar or shaft in a circular hole, as shown in Fig. 1.4, is an example of
an incompletely constrained motion as it may either rotate or slide in a hole. These
both motions have no relationship with the other.
Fig. 1.4 Shaft in a Circular Hole

3. Successfully Constrained Motion:
When the motion between the elements, forming a pair, is such that the
constrained motion is not completed by itself, but by some other means, then the
motion is said to be successfully constrained motion.
Consider a shaft in a foot-step bearing as shown in Fig. 1.5. The shaft may
rotate in a bearing or it may move upwards. This is a case of incompletely
constrained motion. But if the load is placed on the shaft to prevent axial upward
movement of the shaft, then the motion of the pair is said to be successfully
constrained motion. The motion of an I.C. engine valve (these are kept on their seat
by a spring) and the piston reciprocating inside an engine cylinder are also the
examples of successfully constrained motion.
Fig. 1.5 Shaft in a Foot Step Bearing
Static Force Analysis
If components of a machine accelerate, inertia forces are produced due to
their masses. However, if the magnitudes of these forces are small compared to the
externally applied loads, they can be neglected while analysis the mechanism. Such
an analysis is known as static force analysis.
For example, in lifting cranes, the bucket load and the static weight loads may
be quite high relative to any dynamic loads due to accelerating masses, and thus
static force analysis is justified.
Dynamic Force Analysis
When the inertia effect due to the mass of the components is also considered,
it is called dynamic force analysis.
Dynamic forces are associated with accelerating masses. As all machines
have some accelerating parts, dynamic forces are always present when machines
operate. In situations where dynamic forces are dominant or comparable with
magnitudes of external forces and operating speeds are high, dynamic analysis has
to be carried out.

For example, in case pf rotors which rotate at speeds more that 80000 rpm,
even the slightest eccentricity at the centre of mass from the axis of rotation
produces very high dynamic forces. This may lead to vibrations, wear, noise or even
machine failure.
Conditions for a body to be in static equilibrium and dynamic equilibrium
Necessary and sufficient conditions for static and dynamic equilibrium are:
1). Vector sum of all forces acting on a body is zero.
2). vector sum of the moments of all forces acting about any arbitrary point or axis is
zero.
First condition is sufficient condition for static equilibrium together with second
condition is necessary for dynamic equilibrium.
Equilibrium of Two Force Member
The member under the action of two force will be in equilibrium if,
1). The two forces are of same magnitude,
2). The forces act along the same line, and
3). The forces are in opposite direction.
Equilibrium of Three Force Member
A body or member will be in equilibrium under the action of three forces if,
1). The resultant of the forces is zero, and
2). The line of action of the forces intersect at a point.
Space Diagram
A space diagram is a graphical description of the system. It generally shows
the shape and size of the system, the weight, the externally applied loads, the
connection and the supports of the system.
Free Body Diagram
A free body diagram is a sketch of the isolated or free body which shows all
the pertinent weight forces, the externally applied loads, and the reaction from its
supports and connections acting upon it by the removed elements. The various
advantages of free body diagrams are:
1). Free body diagram assist in seeing and understanding all aspects of problem.
2). They help in planning the approach to the problem.
3). They make mathematical relations easier to the problem.

Inertia
The property of matter offering resistance to any change of its state of rest or
of uniform motion in a straight line is known as inertia.
Inertia Force and Inertia Torque
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.
D-Alembert’s Principle
Consider a rigid body acted upon by a system of forces. The system may be
reduced to a single resultant force acting on the body whose magnitude is given by
the product of the mass of the body and the linear acceleration of the centre of mass
of the body.
According to Newton’s second law of motion,
F = m.a … (i)
Where, F = Resultant force acting on the body,
m = Mass of the body, and
a = Linear acceleration of the centre of mass of the body.
The above equation may also be written as:
F – m.a = 0 ... (ii)
A little consideration will show, that if the quantity – m.a be treated as a force,
equal, opposite and with the same line of action as the resultant force F, and include
this force with the system of forces of which F is the resultant, then the complete
system of forces will be in equilibrium. This principle is known as D-Alembert’s
principle. The equal and opposite force – m.a is known as reversed effective force or
the inertia force (briefly written as FI). The equation (ii) may be written as,
F + FI = 0 ... (iii)

Thus, D-Alembert’s principle states that the resultant force acting on a body
together with the reversed effective force (or inertia force), are in equilibrium.
This principle is used to reduce a dynamic problem into an equivalent static
problem.
Principle of Superposition
The principle of superposition states that for linear systems the individual
responses to several disturbances or driving functions can be superposed on each
other to obtain the total response of the system.
Forces on the Reciprocating Parts of an Engine, Neglecting the Weight of the
Connecting Rod
The various forces acting on the reciprocating parts of a horizontal engine are
shown in Fig. 1.6. The expressions for these forces, neglecting the weight of the
connecting rod, may be derived as discussed below:
1. 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. 1.6.
Fig. 1.6 Forces on the Reciprocating Parts of an Engine
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,

Accelerating force or inertia force of the reciprocating parts,
It may be noted that 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 the difference of pressures in the cylinder on the two
sides of the piston. On the other hand, the inertia force due to retardation of the
reciprocating parts helps the force on the piston. Therefore,
Piston effort, PF = Net load on the piston Inertia force
= FL FI ... (Neglecting frictional resistance)
= FL FI - RF ... (Considering frictional resistance)
where RF = Frictional resistance.
The –ve sign is used when the piston is accelerated, and +ve sign is used
when the piston is retarded.
In a double acting reciprocating steam engine, net load on the piston,
FL = p1A1 – p2A2 = p1A1 – p2(A1 – a)
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.
Note:
1. 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,
FL = Pressure × Area
=
2. 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).

2. Force acting along the Connecting Rod:
It is denoted by FQ in Fig. 1.6. From the geometry of the figure, we find that,
3. Thrust on the Sides of the Cylinder Walls or Normal Reaction on the Guide
Bars:
It is denoted by FN in Fig. 1.6. From the figure, we find that,
4. Crank-Pin Effort and Thrust on Crank Shaft Bearings:
The force acting on the connecting rod FQ may be resolved into two
components, one perpendicular to the crank and the 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. 1.6. The component of FQ along the crank produces a thrust on
the crank shaft bearings and it is denoted by FB in Fig. 1.6.
Resolving FQ perpendicular to the crank,
and resolving FQ along the crank,

5. Crank Effort or Turning Moment or Torque on the Crank Shaft:
The product of the crankpin effort (FT) and the crank pin radius (r) is known as
crank effort or turning moment or torque on the crank shaft. Mathematically,

Problems
1. Find the inertia force for the following data of an I.C. engine.
Bore = 175 mm, stroke = 200 mm, engine speed = 500 r.p.m, length of
connecting rod = 400 mm, crank angle = 60° from T.D.C and mass of
reciprocating parts = 180 kg.

2. 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.

3. A vertical double acting steam engine has a cylinder 300 mm diameter and
450 mm stroke and runs at 200 r.p.m. The reciprocating part has a mass of 225
kg and the piston rod is 50 mm diameter. The connecting rod is 1.2 m long.
When the crank has turned through 125° from the top dead centre, the steam
pressure above the piston is 30 kN/m2
and below the piston is 1.5 kN/m2
.
Calculate the effective turning moment on the crank shaft.

4. The crank and connecting rod of a petrol engine, running at 1800 r.p.m. are
50 mm and 200 mm respectively. The diameter of the piston is 80 mm and the
mass of the reciprocating parts is 1 kg. At a point during the power stroke, the
pressure on the piston is 0.7 N/mm2
, when it has moved 10 mm from the inner
dead centre. Determine: 1). Net load on the gudgeon pin, 2). Thrust in the
connecting rod, 3). Reaction between the piston and cylinder, and 4). The
engine speed at which the above values become zero.

5. During a trial on steam engine, it is found that the acceleration of the piston
is 36 m/s2
when the crank has moved 30° from the inner dead centre position.
The net effective steam pressure on the piston is 0.5 N/mm2
and the frictional
resistance is equivalent to a force of 600 N. The diameter of the piston is 300
mm and the mass of the reciprocating parts is 180 kg. If the length of the crank
is 300 mm and the ratio of the connecting rod length to the crank length is 4.5,
find: 1). Reaction on the guide bars, 2). Thrust on the crank shaft bearings, and
3). Turning moment on the crank shaft.

6. 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). 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.

7. A horizontal steam engine running at 120 r.p.m. has a bore of 250 mm and a
stroke of 400 mm. The connecting rod is 0.6 m and mass of the reciprocating
parts is 60 kg. When the crank has turned through an angle of 45° from the
inner dead centre, the steam pressure on the cover end side is 550 kN/m2
and
that on the crank end side is 70 kN/m2
. Considering the diameter of the piston
rod equal to 50 mm, determine: 1). turning moment on the crank shaft, 2).
thrust on the bearings, and 3). acceleration of the flywheel, if the power of the
engine is 20 kW, mass of the flywheel 60 kg and radius of gyration 0.6 m.

Equivalent Dynamical System
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; and
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.
Correction Couple (or) Error in Torque
The error in torque (T) is given by,
T1
= m l1 (l - L) A
This couple must be applied, when the masses are placed arbitrarily to make
the system dynamically equivalent.
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, and 3. Bennett’s construction.
Klien’s Construction
Let OC be the crank and PC the connecting rod of a reciprocating steam
engine, as shown in Fig. 1.7 (a). Let the crank makes an angle with the line of
stroke PO and rotates with uniform angular velocity rad/s in a clockwise direction.
The Klien’s velocity and acceleration diagrams are drawn as discussed below:
Fig. 1.7 Klien’s Construction

Klien’s Velocity Diagram:
First of all, draw OM perpendicular to OP; such that it intersects the line PC
produced at M. The triangle OCM is known as Klien’s velocity diagram. In this
triangle OCM,
OM may be regarded as a line perpendicular to PO,
CM may be regarded as a line parallel to PC, and ... (since it is the same line)
CO may be regarded as a line parallel to CO.
We have already discussed that the velocity diagram for given configuration is
a triangle ocp as shown in Fig. 1.7 (b). If this triangle is revolved through 90°, it will
be a triangle oc1p1, in which oc1 represents vCO (i.e. velocity of C with respect to O or
velocity of crank pin C) and is parallel to OC,
op1 represents vPO (i.e. velocity of P with respect to O or velocity of cross-
head or piston P) and is perpendicular to OP, and
c1p1 represents vPC (i.e. velocity of P with respect to C) and is parallel to CP.
A little consideration will show, that the triangles oc1p1 and OCM are similar.
Therefore,
Thus, we see that by drawing the Klien’s velocity diagram, the velocities of
various points may be obtained without drawing a separate velocity diagram.
Klien’s Acceleration Diagram:
The Klien’s acceleration diagram is drawn as discussed below:
1. First of all, draw a circle with C as centre and CM as radius.
2. Draw another circle with PC as diameter. Let this circle intersect the previous
circle at K and L.
3. Join KL and produce it to intersect PO at N. Let KL intersect PC at Q. This forms
the quadrilateral CQNO, which is known as Klien’s acceleration diagram.
We have already discussed that the acceleration diagram for the given
configuration is as shown in Fig. 1.7 (c). We know that,

(i) o'c' represents ar
CO (i.e. radial component of the acceleration of crank pin C with
respect to O ) and is parallel to CO;
(ii) c'x represents ar
PC (i.e. radial component of the acceleration of crosshead or
piston P with respect to crank pin C) and is parallel to CP or CQ;
(iii) xp' represents at
PC (i.e. tangential component of the acceleration of P with respect
to C ) and is parallel to QN (because QN is perpendicular to CQ); and
(iv) o'p' represents aPO (i.e. acceleration of P with respect to O or the acceleration of
piston P) and is parallel to PO or NO.
A little consideration will show that the quadrilateral o'c'x p' [Fig. 1.7 (c)] is
similar to quadrilateral CQNO [Fig. 1.7 (a)]. Therefore,
Note:
1. The acceleration of piston P with respect to crank pin C (i.e. aPC) may be obtained
from:
2. To find the velocity of any point D on the connecting rod PC, divide CM at D1 in the
same ratio as D divides CP. In other words,
3. To find the acceleration of any point D on the connecting rod PC, draw a line from
a point D parallel to PO which intersects CN at D2.

4. If the crank position is such that the point N lies on the right of O instead of to the
left as shown in Fig. 1.7 (a), then the acceleration of the piston is negative. In other
words, the piston is under going retardation.
5. The acceleration of the piston P is zero and its velocity is maximum, when N
coincides with O. There is no simple graphical method of finding the corresponding
crank position, but it can be shown that for N and O to coincide, the angle between
the crank and the connecting rod must be slightly less than 90°. For most practical
purposes, it is assumed that the acceleration of piston P is zero, when the crank OC
and connecting rod PC are at right angles to each other.
Inertia Forces in a Reciprocating Engine, Considering the Weight of
Connecting Rod
In a reciprocating engine, let OC be the crank and PC, the connecting rod
whose centre of gravity lies at G. The inertia forces in a reciprocating engine may be
obtained graphically as discussed below:
1. First of all, draw the acceleration diagram OCQN by Klien’s construction. We know
that the acceleration of the piston P with respect to O,
acting in the direction from N to O. Therefore, the inertia force FI of the reciprocating
parts will act in the opposite direction as shown in Fig. 1.8.
Fig. 1.8 Inertia Forces is Reciprocating Engine, Considering the Weight of
Connecting Rod

2. Replace the connecting rod by dynamically equivalent system of two masses as
discussed. Let one of the masses be arbitrarily placed at P. To obtain the position of
the other mass, draw GZ perpendicular to CP such that GZ = k, the radius of
gyration of the connecting rod. Join PZ and from Z draw perpendicular to DZ which
intersects CP at D. Now, D is the position of the second mass.
Note:
The position of the second mass may also be obtained from the equation,
GP × GD = k2
3. Locate the points G and D on NC which is the acceleration image of the
connecting rod. This is done by drawing parallel lines from G and D to the line of
stroke PO. Let these parallel lines intersect NC at g and d respectively. Join gO and
dO. Therefore, acceleration of G with respect to O, in the direction from g to O,
and acceleration of D with respect to O, in the direction from d to O,
4. From D, draw DE parallel to dO which intersects the line of stroke PO at E. Since
the accelerating forces on the masses at P and D intersect at E, therefore their
resultant must also pass through E. But their resultant is equal to the accelerating
force on the rod, so that the line of action of the accelerating force on the rod, is
given by a line drawn through E and parallel to gO, in the direction from g to O. The
inertia force of the connecting rod FC therefore acts through E and in the opposite
direction as shown in Fig. 1.8. The inertia force of the connecting rod is given by,
where mC = Mass of the connecting rod.
A little consideration will show that the forces acting on the connecting rod
are:
(a) Inertia force of the reciprocating parts (FI) acting along the line of stroke PO,
(b) The side thrust between the crosshead and the guide bars (FN) acting at P and
right angles to line of stroke PO,
(c) The weight of the connecting rod (WC = mC.g),
(d) Inertia force of the connecting rod (FC),
(e) The radial force (FR) acting through O and parallel to the crank OC,
(f) The force (FT) acting perpendicular to the crank OC.
Now, produce the lines of action of FR and FN to intersect at a point I, known
as instantaneous centre. From I draw IX and IY, perpendicular to the lines of action
of FC and WC. Taking moment about I, we have,

The value of FT may be obtained from this equation and from the force
polygon as shown in Fig. 1.8, the forces FN and FR may be calculated. We know that,
torque exerted on the crankshaft to overcome the inertia of the moving parts = FT ×
OC
Note:
When the mass of the reciprocating parts is neglected, then FI is zero.

Problems
8. The crank and connecting rod lengths of an engine are 125 mm and 500 mm
respectively. The mass of the connecting rod is 60 kg and its centre of gravity
is 275 mm from the crosshead pin centre, the radius of gyration about centre
of gravity being 150 mm. If the engine speed is 600 r.p.m. for a crank position
of 45° from the inner dead centre, determine, using Klien’s or any other
construction 1). the acceleration of the piston; 2). the magnitude, position and
direction of inertia force due to the mass of the connecting rod.

9. The following data refer to a steam engine:
Diameter of piston = 240 mm; stroke = 600 mm; length of connecting rod = 1.5
m; mass of reciprocating parts = 300 kg; mass of connecting rod = 250 kg;
speed = 125 r.p.m; centre of gravity of connecting rod from crank pin = 500
mm; radius of gyration of the connecting rod about an axis through the centre
of gravity = 650 mm. Determine the magnitude and direction of the torque
exerted on the crankshaft when the crank has turned through 30° from inner
dead centre.

10. The connecting rod of an internal combustion engine is 225 mm long and
has a mass 1.6 kg. The mass of the piston and gudgeon pin is 2.4 kg and the
stroke is 150 mm. The cylinder bore is 112.5 mm. The centre of gravity of the
connection rod is 150 mm from the small end. Its radius of gyration about the
centre of gravity for oscillations in the plane of swing of the connecting rod is
87.5 mm. Determine the magnitude and direction of the resultant force on the
crank pin when the crank is at 40° and the piston is moving away from inner
dead centre under an effective gas presure of 1.8 MN/m2
. The engine speed is
1200 r.p.m.

Turning Moment Diagram
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. It is plotted on Cartesian co-ordinates, in which the turning moment is
taken as the ordinate and crank angle as abscissa.
Turning Moment Diagram for a Single Cylinder Double Acting Steam Engine
A turning moment diagram for a single cylinder double acting steam engine is
shown in Fig. 1.9. The vertical ordinate represents the turning moment and the
horizontal ordinate represents the crank angle. We have discussed that the turning
moment on the crankshaft,
where FP = Piston effort,
r = Radius of crank,
n = Ratio of the connecting rod length and radius of crank, and
= Angle turned by the crank from inner dead centre.
Fig. 1.9 Turning Moment Diagram for a Single Cylinder, Double Acting Steam
Engine
From the above expression, we see that the turning moment (T) is zero, when
the crank angle ( ) is zero. It is maximum when the crank angle is 90° and it is
again zero when crank angle is 180°.

This is shown by the curve abc in Fig. 1.9 and it represents the turning
moment diagram for outstroke. The curve cde is the turning moment diagram for
instroke and is somewhat similar to the curve abc.
Since the work done is the product of the turning moment and the angle
turned, therefore the area of the turning moment diagram represents the work done
per revolution. In actual practice, the engine is assumed to work against the mean
resisting torque, as shown by a horizontal line AF. The height of the ordinate aA
represents the mean height of the turning moment diagram. Since it is assumed that
the work done by the turning moment per revolution is equal to the work done
against the mean resisting torque, therefore the area of the rectangle aAFe is
proportional to the work done against the mean resisting torque.
Fig. 1.10 For Flywheel, have a look at your Tailor’s Manual Sewing Machine

Note:
1. When the turning moment is positive (i.e. when the engine torque is more than the
mean resisting torque) as shown between points B and C (or D and E) in Fig. 1.8,
the crankshaft accelerates and the work is done by the steam.
2. When the turning moment is negative (i.e. when the engine torque is less than the
mean resisting torque) as shown between points C and D in Fig. 1.8, the crankshaft
retards and the work is done on the steam.
3. If T = Torque on the crankshaft at any instant, and
Tmean = Mean resisting torque.
Then accelerating torque on the rotating parts of the engine = T – Tmean
4. If (T –Tmean) is positive, the flywheel accelerates and if (T – Tmean) is negative, then
the flywheel retards.
Turning Moment Diagram for a Four Stroke Cycle Internal Combustion Engine
A turning moment diagram for a four stroke cycle internal combustion engine
is shown in Fig. 1.11. We know that in a four stroke cycle internal combustion
engine, there is one working stroke after the crank has turned through two
revolutions, i.e. 720° (or 4 radians).
Fig. 1.11 Turning Moment Diagram for a Four Stroke Cycle Internal
Combustion Engine

Since the pressure inside the engine cylinder is less than the atmospheric
pressure during the suction stroke, therefore a negative loop is formed as shown in
Fig. 1.11. During the compression stroke, the work is done on the gases; therefore a
higher negative loop is obtained. During the expansion or working stroke, the fuel
burns and the gases expand; therefore a large positive loop is obtained. In this
stroke, the work is done by the gases. During exhaust stroke, the work is done on
the gases; therefore a negative loop is formed. It may be noted that the effect of the
inertia forces on the piston is taken into account in Fig. 1.11.
Turning Moment Diagram for a Multi-cylinder Engine
A separate turning moment diagram for a compound steam engine having
three cylinders and the resultant turning moment diagram is shown in Fig. 1.12. The
resultant turning moment diagram is the sum of the turning moment diagrams for the
three cylinders. It may be noted that the first cylinder is the high pressure cylinder,
second cylinder is the intermediate cylinder and the third cylinder is the low pressure
cylinder. The cranks, in case of three cylinders, are usually placed at 120° to each
other.
Fig. 1.12 Turning Moment Diagram for a Multi-Cylinder Engine
Fluctuation of Energy
The fluctuation of energy may be determined by the turning moment diagram
for one complete cycle of operation. Consider the turning moment diagram for a
single cylinder double acting steam engine as shown in Fig. 1.9. We see that the
mean resisting torque line AF cuts the turning moment diagram at points B, C, D and
E. When the crank moves from a to p, the work done by the engine is equal to the
area aBp, whereas the energy required is represented by the area aABp. In other
words, the engine has done less work (equal to the area aAB) than the requirement.

This amount of energy is taken from the flywheel and hence the speed of the
flywheel decreases. Now the crank moves from p to q, the work done by the engine
is equal to the area pBbCq, whereas the requirement of energy is represented by the
area pBCq. Therefore, the engine has done more work than the requirement. This
excess work (equal to the area BbC) is stored in the flywheel and hence the speed of
the flywheel increases while the crank moves from p to q.
Similarly, when the crank moves from q to r, more work is taken from the
engine than is developed. This loss of work is represented by the area CcD. To
supply this loss, the flywheel gives up some of its energy and thus the speed
decreases while the crank moves from q to r. As the crank moves from r to s, excess
energy is again developed given by the area DdE and the speed again increases. As
the piston moves from s to e, again there is a loss of work and the speed decreases.
The variations of energy above and below the mean resisting torque line are called
fluctuations of energy. The areas BbC, CcD, DdE, etc. represent fluctuations of
energy.
A little consideration will show that the engine has a maximum speed either at
q or at s. This is due to the fact that the flywheel absorbs energy while the crank
moves from p to q and from r to s. On the other hand, the engine has a minimum
speed either at p or at r. The reason is that the flywheel gives out some of its energy
when the crank moves from a to p and q to r. The difference between the maximum
and the minimum energies is known as maximum fluctuation of energy.
Determination of Maximum Fluctuation of Energy
A turning moment diagram for a multi-cylinder engine is shown by a wavy
curve in Fig.1.13. The horizontal line AG represents the mean torque line. Let a1, a3,
a5 be the areas above the mean torque line and a2, a4 and a6 be the areas below the
mean torque line. These areas represent some quantity of energy which is either
added or subtracted from the energy of the moving parts of the engine.
Let the energy in the flywheel at A = E, then from Fig. 1.13, we have,
Energy at B = E + a1
Energy at C = E + a1– a2
Energy at D = E + a1 – a2 + a3
Energy at E = E + a1 – a2 + a3 – a4
Energy at F = E + a1 – a2 + a3 – a4 + a5
Energy at G = E + a1 – a2 + a3 – a4 + a5 – a6
= Energy at A (i.e. cyclerepeats after G)
Let us now suppose that the greatest of these energies is at B and least at E.
Therefore,
Maximum energy in flywheel = E + a1
Minimum energy in the flywheel = E + a1 – a2 + a3 – a4
Maximum fluctuation of energy, E = Maximum energy – Minimum energy
= (E + a1) – (E + a1 – a2 + a3 – a4)
= a2 – a3 + a4

Fig. 1.13 Determination of Maximum Fluctuation of Energy
Fig. 1.14 Flywheel Stores Energy when the Supply is in Excess and Releases
Energy when Energy is in Deficit
Coefficient of Fluctuation of Energy
It may be defined as the ratio of the maximum fluctuation of energy to the
work done per cycle. Mathematically, coefficient of fluctuation of energy,

The work done per cycle (in N-m or joules) may be obtained by using the
following two relations:
The following table shows the values of coefficient of fluctuation of energy for
steam engines and internal combustion engines.
Table 1.1 Coefficient of Fluctuation of Energy (CE) for Steam and Internal
Combustion Engines

Flywheel
A flywheel used in machines serves as a reservoir, which stores energy
during the period when the supply of energy is more than the requirement, and
releases it during the period when the requirement of energy is more than the
supply.
In case of steam engines, internal combustion engines, reciprocating
compressors and pumps, the energy is developed during one stroke and the engine
is to run for the whole cycle on the energy produced during this one stroke. For
example, in internal combustion engines, the energy is developed only during
expansion or power stroke which is much more than the engine load and no energy
is being developed during suction, compression and exhaust strokes in case of four
stroke engines and during compression in case of two stroke engines. The excess
energy developed during power stroke is absorbed by the flywheel and releases it to
the crankshaft during other strokes in which no energy is developed, thus rotating
the crankshaft at a uniform speed. A little consideration will show that when the
flywheel absorbs energy, its speed increases and when it releases energy, the
speed decreases. Hence a flywheel does not maintain a constant speed; it simply
reduces the fluctuation of speed. In other words, a flywheel controls the speed
variations caused by the fluctuation of the engine turning moment during each cycle
of operation.
In machines where the operation is intermittent like punching machines,
shearing machines, riveting machines, crushers, etc., the flywheel stores energy
from the power source during the greater portion of the operating cycle and gives it
up during a small period of the cycle. Thus, the energy from the power source to the
machines is supplied practically at a constant rate throughout the operation.
Functions of Flywheel and Governor
The function of a governor in an engine is entirely different from that of a
flywheel. It regulates the mean speed of an engine when there are variations in the
load, e.g., when the load on the engine increases, it becomes necessary to increase
the supply of working fluid. On the other hand, when the load decreases, less
working fluid is required. The governor automatically controls the supply of working
fluid to the engine with the varying load condition and keeps the mean speed of the
engine within certain limits.
As discussed above, the flywheel does not maintain a constant speed; it
simply reduces the fluctuation of speed. It does not control the speed variations
caused by the varying load.

Difference between Flywheel and Governor:
S. No Flywheel Governor
1 The function of flywheel is to reduce the
fluctuations of speed during a cycle
above and below the mean value for
constant load from prime mover.
Its function is to control the mean
speed over a period for output
load variations.
2 It works continuously from cycle to cycle. It works intermittently, i.e., only
when there is change in the load.
3 It has no influence on mean speed of the
prime mover.
It has no influence over cyclic
speed fluctuations.
Coefficient of Fluctuation of Speed
The difference between the maximum and minimum speeds during a cycle is
called the maximum fluctuation of speed. The ratio of the maximum fluctuation of
speed to the mean speed is called the coefficient of fluctuation of speed.
The coefficient of fluctuation of speed is a limiting factor in the design of
flywheel. It varies depending upon the nature of service to which the flywheel is
employed.
Note:
The reciprocal of the coefficient of fluctuation of speed is known as coefficient
of steadiness and is denoted by m.

Energy Stored in a Flywheel
A flywheel is shown in Fig. 1.15. When a flywheel absorbs energy, its speed
increases and when it gives up energy, its speed decreases.
Fig. 1.15 Flywheel

Problems
11. The mass of flywheel of an engine is 6.5 tonnes and the radius of gyration
is 1.8 metres. It is found from the turning moment diagram that the fluctuation
of energy is 56 kN-m. If the mean speed of the engine is 120 r.p.m, find the
maximum and minimum speeds.

12. The flywheel of a steam engine has a radius of gyration of 1 m and mass
2500 kg. The starting torque of the steam engine is 1500 N-m and may be
assumed constant. Determine: 1). the angular acceleration of the flywheel, and
2). the kinetic energy of the flywheel after 10 seconds from the start.

13. A horizontal cross compound steam engine develops 300 kW at 90 r.p.m.
The coefficient of fluctuation of energy as found from the turning moment
diagram is to be 0.1 and the fluctuation of speed is to be kept within ± 0.5% of
the mean speed. Find the weight of the flywheel required, if the radius of
gyration is 2 metres.

14. 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.

15. 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.

16. 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 1/2 revolution and
remains constant for the following revolution. It then falls uniformly to 750 N-m
during the next 1/2 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.

17. During forward stroke of the piston of the double acting steam engine, the
turning moment has the maximum value of 2000 N-m when the crank makes an
angle of 80° with the inner dead centre. During the backward stroke, the
maximum turning moment is 1500 N-m when the crank makes an angle of 80°
with the outer dead centre. The turning moment diagram for the engine may be
assumed for simplicity to be represented by two triangles. If the crank makes
100 r.p.m. and the radius of gyration of the flywheel is 1.75 m, find the
coefficient of fluctuation of energy and the mass of the flywheel to keep the
speed within ± 0.75% of the mean speed. Also determine the crank angle at
which the speed has its minimum and maximum values.

18. 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.

19. A single cylinder, single acting, four stroke gas engine develops 20 kW at
300 r.p.m. The work done by the gases during the expansion stroke is three
times the work done on the gases during the compression stroke, the work
done during the suction and exhaust strokes being negligible. If the total
fluctuation of speed is not to exceed ± 2 per cent of the mean speed and the
turning moment diagram during compression and expansion is assumed to be
triangular in shape, find the moment of inertia of the flywheel.

20. The turning moment diagram for a four stroke gas engine may be assumed
for simplicity to be represented by four triangles, the areas of which from the
line of zero pressure are as follows: Suction stroke = 0.45 × 10–3
m2
;
Compression stroke = 1.7 × 10–3
m2
; Expansion stroke = 6.8 × 10–3
m2
; Exhaust
stroke = 0.65 × 10–3
m2
. Each m2
of area represents 3 MN-m of energy.
Assuming the resisting torque to be uniform, find the mass of the rim of a
flywheel required to keep the speed between 202 and 198 r.p.m. The mean
radius of the rim is 1.2 m.

21. The turning moment curve for an engine is represented by the equation, T
= (20 000 + 9500 sin 2 – 5700 cos 2 ) N-m, where is the angle moved by
the crank from inner dead centre. If the resisting torque is constant, find: 1).
Power 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.

22. A certain machine requires a torque of (5000 + 500 sin ) N-m to drive it,
where is the angle of rotation of shaft measured from certain datum. The
machine is directly coupled to an engine which produces a torque of (5000 +
600 sin 2 ) N-m. The flywheel and the other rotating parts attached to the
engine has a mass of 500 kg at a radius of gyration of 0.4 m. If the mean speed
is 150 r.p.m, find: 1). the fluctuation of energy, 2). the total percentage
fluctuation of speed, and 3). the maximum and minimum angular acceleration
of the flywheel and the corresponding shaft position.

23. The equation of the turning moment curve of a three crank engine is (5000
+ 1500 sin 3 ) N-m, where is the crank angle in radians. The moment of
inertia of the flywheel is 1000 kg-m2
and the mean speed is 300 r.p.m.
Calculate: 1). power of the engine, and 2). the maximum fluctuation of the
speed of the flywheel in percentage when (i) the resisting torque is constant,
and (ii) the resisting torque is (5000 + 600 sin ) N-m.

Dimensions of the Flywheel Rim
Consider a rim of the flywheel as shown in Fig. 1.16.
Fig. 1.16 Rim of a Flywheel
Let D = Mean diameter of rim in metres,
R = Mean radius of rim in metres,
A = Cross-sectional area of rim in m2
,
= Density of rim material in kg/m3
,
N = Speed of the flywheel in r.p.m.,
= Angular velocity of the flywheel in rad/s,
v = Linear velocity at the mean radius in m/s
= .R = DN/60, and
= Tensile stress or hoop stress in N/m2
due to the centrifugal force.
Consider a small element of the rim as shown shaded in Fig. 1.6. Let it subtends
an angle at the centre of the flywheel.
Volume of the small element = A × R
Mass of the small element, dm = Density × volume = AR
and centrifugal force on the element, acting radially outwards,
Vertical component of dF
Total vertical upward force tending to burst the rim across the diameter XY.

This vertical upward force will produce tensile stress or hoop stress (also
called centrifugal stress or circumferential stress), and it is resisted by 2P, such that
Equating equations (i) and (ii),
We know that mass of the rim,
From equations (iii) and (iv), we may find the value of the mean radius and
cross-sectional area of the rim.
Note:
If the cross-section of the rim is a rectangular, then
A = b × t
where b = Width of the rim, and
t = Thickness of the rim.

Problems
24. 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.

25. A single cylinder double acting steam engine develops 150 kW at a mean
speed of 80 r.p.m. The coefficient of fluctuation of energy is 0.1 and the
fluctuation of speed is ± 2% of mean speed. If the mean diameter of the
flywheel rim is 2 metre and the hub and spokes provide 5% of the rotational
inertia of the flywheel, find the mass and cross-sectional area of the flywheel
rim. Assume the density of the flywheel material (which is cast iron) as 7200
kg/m3
.

26. A multi-cylinder engine is to run at a speed of 600 r.p.m. On drawing the
turning moment diagram to a scale of 1 mm = 250 N-m and 1 mm = 3°, the
areas above and below the mean torque line in mm2
are : + 160, – 172, + 168, –
191, + 197, – 162.
The speed is to be kept within ± 1% of the mean speed of the engine.
Calculate the necessary moment of inertia of the flywheel. Determine the
suitable dimensions of a rectangular flywheel rim if the breadth is twice its
thickness. The density of the cast iron is 7250 kg/m3
and its hoop stress is 6
MPa. Assume that the rim contributes 92% of the flywheel effect.

27. The turning moment diagram of a four stroke engine may be assumed for
the sake of simplicity to be represented by four triangles in each stroke. The
areas of these triangles are as follows:
Suction stroke = 5×10–5
m2
; Compression stroke = 21×10–5
m2
;
Expansion stroke = 85×10–5
m2
; Exhaust stroke = 8×10–5
m2
.
All the areas excepting expression stroke are negative. Each m2
of area
represents 14 MN-m of work.
Assuming the resisting torque to be constant, determine the moment of
inertia of the flywheel to keep the speed between 98 r.p.m. and 102 r.p.m. Also
find the size of a rim-type flywheel based on the minimum material criterion,
given that density of flywheel material is 8150 kg/m3
; the allowable tensile
stress of the flywheel material is 7.5 MPa. The rim cross-section is rectangular,
one side being four times the length of the other.

28. An otto cycle engine develops 50 kW at 150 r.p.m. with 75 explosions per
minute. The change of speed from the commencement to the end of power
stroke must not exceed 0.5% of mean on either side. Find the mean diameter of
the flywheel and a suitable rim cross section having width four times the depth
so that the hoop stress does not exceed 4 MPa. Assume that the flywheel
stores 16/15 times the energy stored by the rim and the work done during
power stroke is 1.40 times the work done during the cycle. Density of rim
material is 7200 kg/m3
.

Flywheel in Punching Press
The function of a flywheel in an engine is to reduce the fluctuations of speed,
when the load on the crankshaft is constant and the input torque varies during the
cycle. The flywheel can also be used to perform the same function when the torque
is constant and the load varies during the cycle. Such an application is found in
punching press or in a riveting machine.
A punching press is shown diagrammatically in Fig. 1.17. The crank is driven
by a motor which supplies constant torque and the punch is at the position of the
slider in a slider-crank mechanism. From Fig. 1.17, we see that the load acts only
during the rotation of the crank from , when the actual punching
takes place and the load is zero for the rest of the cycle. Unless a flywheel is used,
the speed of the crankshaft will increase too much during the rotation of crankshaft
will increase too much during the rotation of crank from
and again from , because
there is no load while input energy continues to be supplied. On the other hand, the
drop in speed of the crankshaft is very large during the rotation of crank from
due to much more load than the energy supplied. Thus the
flywheel has to absorb excess energy available at one stage and has to make up the
deficient energy at the other stage to keep to fluctuations of speed within permissible
limits. This is done by choosing the suitable moment of inertia of the flywheel.
Fig. 1.17 Operation of Flywheel in a Punching Press

Let E1 be the energy required for punching a hole. This energy is determined
by the size of the hole punched, the thickness of the material and the physical
properties of the material.

Problems
29. A punching press is driven by a constant torque electric motor. The press
is provided with a flywheel that rotates at maximum speed of 225 r.p.m. The
radius of gyration of the flywheel is 0.5 m. The press punches 720 holes per
hour; each punching operation takes 2 second and requires 15 kN-m of
energy. Find the power of the motor and the minimum mass of the flywheel if
speed of the same is not to fall below 200 r. p. m.

30. A machine punching 38 mm holes in 32 mm thick plate requires 7 N-m of
energy per sq. mm of sheared area, and punches one hole in every 10
seconds. Calculate the power of the motor required. The mean speed of the
flywheel is 25 metres per second. The punch has a stroke of 100 mm.
Find the mass of the flywheel required, if the total fluctuation of speed is
not to exceed 3% of the mean speed. Assume that the motor supplies energy
to the machine at uniform rate.

31. A riveting machine is driven by a constant torque 3 kW motor. The moving
parts including the flywheel are equivalent to 150 kg at 0.6 m radius. One
riveting operation takes 1 second and absorbs 10 000 N-m of energy. The
speed of the flywheel is 300 r.p.m. before riveting. Find the speed immediately
after riveting. How many rivets can be closed per minute?

32. A punching press is required to punch 40 mm diameter holes in a plate of
15 mm thickness at the rate of 30 holes per minute. It requires 6 N-m of energy
per mm2
of sheared area. If the punching takes 1/10 of a second and the r.p.m.
of the flywheel varies from 160 to 140, determine the mass of the flywheel
having radius of gyration of 1 metre.

33. A punching machine makes 25 working strokes per minute and is capable
of punching 25 mm diameter holes in 18 mm thick steel plates having ultimate
shear strength 300 MPa. The punching operation takes place during 1/10th of a
revolution of the crankshaft.
Estimate the power needed for the driving motor, assuming a
mechanical efficiency of 95 percent. Determine suitable dimensions for the rim
cross-section of the flywheel, having width equal to twice thickness. The
flywheel is to revolve at 9 times the speed of the crankshaft. The permissible
coefficient of fluctuation of speed is 0.1.
The flywheel is to be made of cast iron having a working stress (tensile)
of 6 MPa and density of 7250 kg/m3
. The diameter of the flywheel must not
exceed 1.4 m owing to space restrictions. The hub and the spokes may be
assumed to provide 5% of the rotational inertia of the wheel.

DYNAMICS OF MACHINES UNIT-1 BY Mr.P.RAMACHANDRAN/AP/MECH/KIT/CBE

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DYNAMICS OF MACHINES UNIT-1 BY Mr.P.RAMACHANDRAN/AP/MECH/KIT/CBE