UNIT-IV FORCE ANALYSIS STATIC AND DYNAMICS.pdf

UNIT – 4 - FORCE ANALYSIS
1.1 INTRODUCTION
The subject Dynamics of Machines may be defined as that branch of
Engineering-science, which deals with the study of relative motion between the
various parts of a machine, and forces which act on them. The knowledge of this
subject is very essential for an engineer in designing the various parts of a machine.
A machine is a device which receives energy in some available form and
utilises it to do some particular type of work.
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.
1.2 NEWTON’ S LAW :
First Law
Everybody will persist in its state of rest or of uniform motion (constant
velocity) in a straight line unless it is compelled to change that state by forces
impressed on it. This means that in the absence of a non-zero net force, the center
of mass of a body either is at rest or moves at a constant velocity.
Second Law
A body of mass m subject to a force F undergoes an acceleration a that
has the same direction as the force and a magnitude that is directly
proportional to the force and inversely proportional to the mass, i.e., F = ma.
Alternatively, the total force applied on a body is equal to the time derivative
of linear momentum of the body.
Third Law
The mutual forces of action and reaction between two bodies are equal,
opposite and collinear. This means that whenever a first body exerts a force F on
a second body, the second body exerts a force −F on the first body. F and −F are
equal in magnitude and opposite in direction. This law is sometimes referred to as
the action-reaction law, with F called the "action" and −F the "reaction"
1.3 TYPES OF FORCE ANALYSIS:
Equilibrium of members with two forces
Equilibrium of members with three forces
of members with two couples.
Equilibrium of members with four forces.

1.3.1 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.
1.3.2 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.
1.4 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. Fourbars are simple
mechanisms common in mechanical engineering machine design and fall
under the study of kinematics.
Dynamic Analysis of Reciprocating engines.
Inertia force and torque analysis by neglecting weight of connecting rod.
Velocity and acceleration of piston.
Angular velocity and Angular acceleration of connecting rod.
Force and Torque Analysis in reciprocating engine neglecting the
weight of connecting rod.
Equivalent Dynamical System
Determination of two masses of equivalent dynamical system
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 equilib-rium position. It is equal to the
accelerating couple in magni-tude but opposite in direction.
1.4.1 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
F = Resultant force acting on the body, m
Mass of the body, and
Linear acceleration of the centre of mass of the
a body.
The equation (i) may also be written as:
F – m.a = 0
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 DAlembert’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
+ 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.
1.4.2 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.
We shall now discuss these constructions, in detail, in the following pages.
1.5 KLIEN’S CONSTRUCTION
Let OC be the crank and PC the connecting rod of a reciprocating steam
engine, as shown in Fig. 15.2 (a). Let the crank makes an angle θ with the line of
stroke PO and rotates with uniform angular velocity ω rad/s in a clockwise
direction. The Klien’s velocity and acceleration diagrams are drawn as discussed
below:

(a)
Klien’s acceleration diagram. (b) Velocity diagram. (c) Acceleration
diagram.
Fig. 15.2. Klien’s construction.
1.5.1 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 ...(Q 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. 15.2 (b). If this triangle is revolved through 90°, it
will be a triangle oc1 p1, in which oc1 represents vCO (i.e. velocity of C with
respect to O or velocity of crank pin C) and is paralel 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.
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.
1.5.2 Klien’s acceleration diagram
The Klien’s acceleration dia-gram is drawn as discussed below:
First of all, draw a circle with C as centre and CM as radius.
Draw another circle with PC as diameter. Let this circle
intersect the previous circle at K and L.
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. 15. 2 (c). We know that

r
(i) o'c' represents a ( i.e. radial component of the acceleration of crank pin C with respect
CO
to O ) and is parallel to CO; (ii) c'x represents ar
crosshead or piston P
(i.e. radial component of the acceleration of
PC
with respect to crank pin C) and is parallel to CP or CQ;
t
xp' represents a (i.e. tangential component of the acceleration of P with respect
to C )
PC and is parallel to QN (because QN
is perpendicular to CQ); and
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. 15.2 (c)] is similar to
quadrilateral CQNO [Fig. 15.2 (a)]. Therefore,
o
′
c
′
c
′
x xp
′
o
′
p
′
ω2(a constant)
OC CQ QN NO

1.6 SOLVED PROBLEMS
1. The crank and connecting rod of a reciprocating engine are 200 mm and 700 mm
respectively. The crank is rotating in clockwise direction at 120 rad/s. Find with the he lp of
Klein’s construction: 1. Velocity and acceleration of the piston, 2. Velocity and acceleration of
the mid point of the connecting rod, and 3. Angular velocity and angular acceleration of the
connecting rod, at the instant when the crank is at 30° to I.D.C. (inner dead centre).
Solution. Given: OC = 200 mm = 0.2 m ; PC = 700 mm = 0.7 m ; ω = 120 rad/s
Fig. 15.5
The Klein’s velocity diagram OCM and Klein’s acceleration diagram CQNO as shown in Fig. 15.5 is
drawn to some suitable scale, in the similar way as discussed in Art. 15.5. By measurement, we find that
OM = 127 mm = 0.127 m ; CM = 173 mm = 0.173 m ; QN = 93 mm = 0.093 m ; NO = 200 mm
0.2 m
Velocity and acceleration of the piston
We know that the velocity of the piston P,
vP = ω × OM = 120 × 0.127 = 15.24 m/s Ans. and
acceleration of the piston P,
aP = ω 2
× NO = (120)2
× 0.2 = 2880 m/s 2
Ans.
Velocity and acceleration of the mid-point of the connecting rod
In order to find the velocity of the mid-point D of the connecting rod, divide CM at D1 in the same
ratio as D divides CP. Since D is the mid-point of CP, therefore D1 is the mid-point of
CM, i.e. CD1 = D1M. Join OD1. By measurement,
OD1 = 140 mm = 0.14 m
Velocity of D, vD = ω × OD1 = 120 × 0.14 = 16.8 m/s Ans.
In order to find the acceleration of the mid-point of the connecting rod, draw a line DD2 parallel to
the line of stroke PO which intersects CN at D2. By measurement,
OD2 = 193 mm = 0.193 m
Acceleration of D,
aD = ω 2
× OD2 = (120)2
× 0.193 = 2779.2 m/s 2
Ans.
Angular velocity and angular acceleration of the connecting rod
We know that the velocity of the connecting rod PC (i.e. velocity of P with respect
to C), vPC = ω × CM = 120 × 0.173 = 20.76 m/s

1.7 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
Fig. 15.7. 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 .
Fig. 15.7. 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,
= 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.
Velocity of the piston:

Since the acceleration is the rate of change of velocity, therefore acceleration of the piston P,
1.8 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 sin φ = r sin θ

The negative sign shows that the sense of the acceleration of the connecting rod is such
that it tends to reduce the angle φ..
2.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.
Solution. Given : r = 150 mm = 0.15 m ; l = 600 mm = 0.6 m ; θ = 60°; N = 400 r.p.m or
= π × 450/60 = 47.13 rad/s
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

1.9 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. 15.8. 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. 15.8.
Fig. 15.8. 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.
Force acting along the connecting rod. It is denoted by FQ in Fig. 15.8. From the
geom-etry of the figure, we find that
FP
FQ cosφ

Thrust on the sides of the cylinder walls or normal reaction on the guide bars. It
is denoted by FN in Fig. 15.8. From the figure, we find that
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. 15.8. The component of FQ along the crank produces a
thrust on the crank shaft bearings and it is denoted by FB in Fig. 15.8. Resolving FQ
perpendicular to the crank,
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. Mathematically,

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
length between
centre the cylinder bore is 0.5 m. If the engine runs at 250 r.p.m. and if the effect of
piston ro neglected, calculate : 1. bars, 2. thrust in the connecting rod, 3. tangential force
on the cr the crank shaft.
The connecting rod pressure on slide pin, and 4. Turning moment of the shaft
Solution. 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. or ω = 2 π × 250/60 = 26.2 rad/s
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
sin φ sin θ
that, sin 60 0.866 0.2165 n
4 4
∴
φ = 12.5°
We know that pressure on the slide bars,
FN = FP tan φ = 49.424 × tan 12.5° = 10.96 kN Ans.
Thrust in the connecting rod
We know that thrust in the connecting rod,
F
F P 49.424 50.62 kN Ans.
Q
cosφ cos 12.5
Tangential force on the crank-pin
We know that tangential force on the crank pin,
FT FQ sin ( θ φ ) 50.62 sin (60 12. 5 ) 48.28 kN Ans.
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.
4. The crank
200 mm respectively. The diameter of the piston is 80 mm and the mass of the recipr and
is 1 kg. At a point during the power stroke, the pressure on the piston is 0.7 N/mm h connecting
10 mm from the inner dead centre. Determine : 1. rod of a
Thrust in the connecting rod, 3. Reaction between the piston and cylinder, and speed petrol engine,
at which the above values become zero. running at
1800
r.p.m.are 50 mm and
-ing
2
, when it Net load on the
gudgeon pin, 2.
4. The engine
Solution. Given : N = 1800 r.p.m. or ω = 2π × 1800/60 = 188.52 rad/s ; r = 50 mm = 0.05
m; l = 200 mm ; D = 80 mm ; mR = 1 kg ; p = 0.7 N/mm2
; x = 10 mm
Net load on the gudgeon pin We know that load on the piston,

When the piston has moved 10 mm from the inner dead centre, i.e. when P1P = 10 mm,
the crank rotates from OC1 to OC through an angle θ as shown in Fig. 15.10. By
measurement, we find that *θ = 33°.
We know that ratio of lengths of connecting rod and crank,
= l/r = 200 /50 = 4
and inertia force on the reciprocating parts,
Thrust in the connecting rod
We know that thrust in the connecting rod,
F
F P 1849
Reaction between the piston and cylinder
We know that reaction between the piston and cylinder,
FN FP tan φ 1849 tan 7.82 254 N Ans.
Engine speed at which the above values will become zero
A little consideration will show that the above values will become zero, if the inertia force
on the reciprocating parts (FI) is equal to the load on the piston (FL). Let ω 1 be the speed in
rad/s, at which FI = FL .

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:
Net force on the piston, 2. Resultant load on the gudgeon pin,
Thrust on the cylinder walls, and 4. Speed above which, other things
re-maining same, the gudgeon pin load would be reversed in direction.
Solution. 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
= 2 π × 2000/60 = 209.5 rad/s ; θ = 20°; p = 700 kN/m2
1. Net force on the piston
The configuration diagram of a vertical engine is shown in Fig.
15.11. We know that force due to gas pressure,
We know that for a vertical engine, net force on the piston,
FP FL− FI WR FL − FI mR .g
5500 − 3254 1.1 9.81 2256.8 N Ans.
Resultant load on the gudgeon pin
Let φ = Angle of inclination of the connecting rod to the line of stroke. We know
that,
sin φ = sin θ / n = sin 20°/4.17 = 0.082
φ = 4.7°
We know that resultant load on the gudgeon pin,

Thrust on the cylinder walls
We know that thrust on the cylinder walls,
FN FP tan φ 2256.8 tan 4. 7 185.5 N Ans.
Speed, above which, the gudgeon pin load would be reversed in direction
Let N1 = Required speed, in r.p.m.
The gudgeon pin load i.e. FQ will be reversed in direction, if FQ becomes negative. This is
only possible when FP is negative. Therefore, for FP to be negative, FI must be greater than (FL +
W R),
6. 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 inne r dead centre, the steam pressure on the
cover end side is 550 kN/m 2 and that on the crank end side is 70 kN/m 2 . Considering the
diameter of the piston rod equal to 50 mm, determine:
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.
Solution. Given : N = 120 r.p.m. or ω = 2π × 120/60 = 12.57 rad/s ; D = 250 mm = 0.25 m
;
L = 400 mm = 0.4 m or r = L/2 = 0.2 m ; l = 0.6 m ; mR = 60 kg ; θ = 45° ; d = 50 mm = 0.05 m ;
p1 = 550 kN/m2
= 550 × 10 3
N/m2
; p2 = 70 kN/m2
= 70 × 10 3
N/m2
Turning moment on the crankshaft
First of all, let us find the net load on the piston (FP).
We know that area of the piston on the cover end side,

Thrust on the bearings
We know that thrust on the bearings,
3. Acceleration of the flywheel
Given: P = 20 kW = 20 × 10 3 W; m = 60 kg ; k = 0.6 m Let α
= Acceleration of the flywheel in rad/s2.
We know that mass moment of inertia of the flywheel,
I = m.k2 = 60 × (0.6) 2 = 21.6 kg-m2
∴ Accelerating torque, TA = I.α = 21.6 α N-m
...(i)

1.10 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,
the sum of their masses is equal to the total mass of the body ;
the centre of gravity of the two masses coincides with that of the body ; and
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, as shown in Fig. 15.14.
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).

7 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
adynamicallyequivalentsystem,
l1.l2 = (kG)2
1.11 CORRECTION COUPLE TO BE APPLIED TO MAKE TWO MASS SYSTEM
DYNAMICALLY EQUIVALENT

In Art. 15.11, we have discussed the conditions for equivalent dynamical system of two
bodies. A little consideration will show that when two masses are placed arbitrarily*, then
the conditions (i) and (ii) as given in Art. 15.11 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.
Fig. 15.21. Correction couple to be
applied to make the two-mass system dynamically equivalent.
Consider two masses, one at A and the other atD be placed arbitrarily, as shown in Fig.
15.21. 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;
= Angular acceleration of the body;
I = Mass moment of inertia of a dynamically equivalent system; kG
Radius of gyration of a dynamically equivalent system. We
know that the torque required to accelerate the body,
T = I. = m (kG)2 ...(i)
Similarly, the torque required to accelerate the two-mass system placed arbitrarily,
T1 = I1. = m (k1)2 ...(ii)
Difference between the torques required to accelerate the two-mass system and the
required to accelerate the rigid body,
torque
T' = T1–T = m (k1)2 – m (kG)2 = m [(k1)2 – (kG)2] ...(iii)
The difference of the torques T' is known as correction couple. This couple must be
applied, when the masses are placed arbitrarilyto make the system dynamical equivalent.
This, of course, will satisfy the condition (iii)
8.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/s
2
clockwise,
determine the correction couple applied to the system to reduce it to a dynamically
equivalent system.
Solution.Given :m = 2kg ;l =250 mm =0.25 m ;l1 =100 mm= 0.1m ; kG = 110mm= 0.11m ;
= 23 000 rad/s2
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.
Correction couple
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
Correction couple,
1.12 INERTIA FORCES IN A RECIPROCATING ENGINE, CONSIDERING THE WEIGHT
OFCONNECTING 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
graphicallyas discussed below:
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,

aPO = aP =
2
× NO,
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. 15.22.
Fig. 15.22. Inertia forces is reciprocating engine, considering the weight of connecting rod.
Replace the connecting rod by dynamically equivalent system of two masses as
discussed in Art. 15.12. 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
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,
= a = 2 × aG O G gO
and acceleration of D with respect to O, in the direction from d to O,
2
×
= a = aD O D dO
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 accelerang 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.
15.22. The inertia force of the connecting rod is given by
FC = mC ×
2
× gO ...(i)
where mC = Mass of the connecting rod.
A little consideration will show that the forces acting on the connecting rod are :
Inertia force of the reciprocating parts (FI ) acting along the line of stroke PO,
The side thrust between the crosshead and the guide bars (FN) acting at P and
right angles to line of stroke PO,
Theweightoftheconnectingrod
(WC = mC.g),
Inertia force of the connecting rod (FC),
The radial force (FR) acting through O and parallel to the crank OC,
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 I X and I Y , perpendicular to the lines of action of FC and
WC. Taking moments about I, we have
FT × IC = FI × IP + FC × I X + WC × I Y ...(ii) The value of FT may be
obtained from this equation and from the force polygon as
shown in Fig. 15.22, 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
1.12.1 Analytical Method for Inertia Torque
The effect of the inertia of the connecting rod on the crankshaft torque may be obtained as
discussed in the following steps:
Fig. 15.23. Analytical method for inertia torque.

The mass of the connecting rod (mC) is divided into two masses. One of the mass
is placed at the crosshead pin P and the other at the crankpin C as shown in Fig. 15.23, so that
the centre of gravity of these two masses coincides with the centre of gravity of the rod G.
Since the inertia force due to the mass at C acts radially outwards along the crank
OC, therefore the mass at C has no effect on the crankshaft torque.
The inertia force of the mass at P may be obtained as
follows: Let 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.
In deriving the equation (ii) of the torque exerted on the crankshaft, it is assumed
that one of the two masses is placed at C and the other at P. This assumption does not satisfy
the condition for kinetically equivalent system of a rigid bar. Hence to compensate for it, a
correcting torque is neces- sary whose value is given by

The correcting torque T' may be applied to the system by two equal and opposite
forces FYacting through P and C. Therefore,

The total torque exerted on the crankshaft due to the inertia of the moving parts is the algebraic
sum of T I , T C and T W.
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.

Solution. Given : r = OC = 125 mm ; l = PC = 500 mm; mC = 60 kg ; PG = 275 mm ;
mC = 60 kg ; PG = 275 mm ; kG = 150 mm ; N = 600 r.p.m. or = 2 × 600/60 = 62.84 rad/s
= 45°
Acceleration of the piston
Let aP = Acceleration of the piston.
First of all, draw the configuration diagram OCP, as shown in Fig. 15.24, to some suitable
scale, such that
OC = r = 125 mm ; PC = l = 500 mm ; and = 45°.
Now, draw the Klien’s acceleration diagram OCQN, as shown in Fig. 15.24, in the same
manner as already discussed. By measurement,
NO = 90 mm = 0.09 m
Acceleration of the piston, aP = 2 × NO = (62.84)2 × 0.09 = 355.4
m/s Ans.
The magnitude, position and direction of inertia force due to the mass of the connecting rod
The magnitude, postition and direction of the inertia force may be obtained as follows:
Replace the connecting rod by dynamical equivalent system of two
masses, assuming that one of the masses is placed at P and the other mass at D.
The position of the point D is obtained as discussed in Art. 15.12.
Locate the points G and D on NC which is the acceleration
image of the connecting rod. Let these points are g and d on NC. Join gO and dO.
By measurement,
gO = 103 mm = 0.103 m

Acceleration of G, aG =
2
× gO, acting in the direction from g to O.
From point D, draw DE parallel to dO. Now E is the point
through which the inertia force of the connecting rod passes. The magnitude of
the inertia force of the connecting rod is given by
FC = mC ×
2
× gO = 60 × (62.84)
2
× 0.103 = 24 400 N = 24.4
kN Ans. (iv) From point E, draw a line parallel to gO, which shows the position of the
inertia force of
the connecting rod and acts in the opposite direction of gO.
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.
Solution. Given : D = 240 mm = 0.24 m ; L = 600 mm or r = L/2 = 300 mm = 0.3 m ; l
= 1.5 m ; mR = 300 kg ; mC = 250 kg ; N = 125 r.p.m. or = 2 ×
125/60 = 13.1 rad/s ; GC = 500 mm = 0.5 m ; kG = 650 mm = 0.65 m ; = 30°
The inertia torque on the crankshaft may be determined by graphical method or analytical
method as discussed below:
Graphical method
First of all, draw the configuration diagram OCP, as shown in Fig. 15.25, to some suitable
scale, such that
OC = r = 300 mm ; PC = l = 1.5 m ; and angle POC = = 30°.

Fig. 15.25
Now draw the Klien’s acceleration diagram OCQN, as shown in Fig. 15.25, and
complete the figure in the similar manner as discussed in Art. 15.14.
By measurement; NO = 0.28 m ; gO = 0.28 m ; IP = 1.03 m ; I X = 0.38 m ; I Y
= 0.98 m, and IC = 1.7 m.
We know that inertia force of reciprocating parts,
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/m
2
. The engine speed is 1200
r.p.m.
Solution. Given : l = PC = 225 mm = 0.225 m; mC = 1.6 kg; m R = 2.4 kg; L =
150 mm or r = L/2 = 75 mm = 0.075 m ; D = 112.5 mm = 0.1125 m ; PG = 150 mm ; kG =
87.5 mm = 0.0875 m ; = 40° ; p = 1.8 MN/m2 = 1.8 × 106 N/m2 ; N = 1200 r.p.m. or = 2
× 1200/60 = 125.7 rad/s
First of all, draw the configuration diagram OCP, as shown in Fig. 15.27 to some suitable
scale, such that OC = r = 75 mm ; PC = l = 225 mm ; and = 40°.
Now, draw the Klien’s acceleration diagramOCQN. Complete the diagram in the same
manner as discussed earlier. By measurement,
NO = 0.0625 m ; gO = 0.0685 m ; IC = 0.29 m ; IP = 0.24 m ; IY = 0.148 m ; and IX =
0.08 m
We know that force due to gas pressure,

Let us now find the values of FN and FR in magnitude and direction. Draw the force
polygon as shown in Fig. 15.25.
By measurement, FN = 3550 N; and FR = 7550 N
The magnitude and direction of the resultant force on the crank pin is given by FQ ,
which is the resultant of FR and FT.
By measurement, FQ = 13 750 N Ans.

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