inertia force and inertia torque in reciprocating engine and inertia force of reciprocating masses pdf free download
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514 Theory of Machines
15
Features
Inertia
1. Introduction.
2. Resultant Effect of a System of
Forces Acting on a Rigid Body.
Forces in
3. D-Alembert’s Principle.
4. Velocity and Acceleration of
the Reciprocating Parts in
Reciprocating
Engines.
5. Klien’s Construction.
6. Ritterhaus’s Construction.
Parts
7. Bennett’s Construction.
8. Approximate Analytical
Method for Velocity and
Acceleration of the Piston.
15.1. Introduction
9. Angular Velocity and
The inertia force is an imaginary force, which when
Acceleration of the Connecting
acts upon a rigid body, brings it in an equilibrium position. It
Rod.
is numerically equal to the accelerating force in magnitude,
10. Forces on the Reciprocating
but opposite in direction. Mathematically,
Parts of an Engine Neglecting
Weight of the Connecting Rod.
Inertia force = – Accelerating force = – m.a
11. Equivalent Dynamical System.
where m = Mass of the body, and
12. Determination of Equivalent
a = Linear acceleration of the centre
Dynamical System of Two
of gravity of the body.
Masses by Graphical Method.
13. Correction Couple to be
Similarly, the inertia torque is an imaginary torque,
Applied to Make the Two Mass
which when applied upon the rigid body, brings it in equilib-
Systems Dynamically
rium position. It is equal to the accelerating couple in magni-
Equivalent.
tude but opposite in direction.
14.Inertia Forces in a
Reciprocating Engine
15.2. Resultant Effect of a System of Forces
Considering the Weight of
Acting on a Rigid Body
Connecting Rod.
Consider a rigid body acted upon by a system of
15. Analytical Method for Inertia
forces. These forces may be reduced to a single resultant force
Torque.
514
Chapter 15 : Inertia Forces in Reciprocating Parts 515
F whose line of action is at a distance h from the centre of
gravity G. Now let us assume two equal and opposite forces
(of magnitude F ) acting through G, and parallel to the
resultant force, without influencing the effect of the
resultant force F, as shown in Fig. 15.1.
A little consideration will show that the body is
now subjected to a couple (equal to F × h) and a force,
equal and parallel to the resultant force F passing through
G. The force F through G causes linear acceleration of the
Fig. 15.1. Resultant effect of a system of
c.g. and the moment of the couple (F × h) causes angular
forces acting on a rigid body.
acceleration of the body about an axis passing through G
and perpendicular to the point in which the couple acts.
Let α = Angular acceleration of the rigid body due to couple,
h = Perpendicular distance between the force and centre of gravity of the
body,
m = Mass of the body,
k = Least radius of gyration about an axis through G, and
I = Moment of inertia of the body about an axis passing through its centre
of gravity and perpendicular to the point in which the couple acts
2
= m.k
We know that
Force, F = Mass × Acceleration = m.a ...(i)
2 2
and F.h = m.k .α = I.α ...(ii)
...(Im = .k )
From equations (i) and (ii), we can
find the values of a and α, if the values of F,
m, k, and h are known.
15.3. 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.
The above picture shows the reciprocating parts
According to Newton’s second law of
of a 19th century oil engine.
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 equation (i) 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
516 Theory of Machines
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 F ). The equation (ii) may be written as
I
F + F = 0 ...(iii)
I
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.
15.4. 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.
15.5. 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.
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,
...( It is the same line.)
CM may be regarded as a line parallel to PC, and
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
Chapter 15 : Inertia Forces in Reciprocating Parts 517
as shown in Fig. 15.2 (b). If this triangle is revolved through 90°, it will be a triangle oc p , in which
1 1
oc represents v (i.e. velocity of C with respect to O or velocity of crank pin C) and is paralel to OC,
1 CO
op represents v (i.e. velocity of P with respect to O or velocity of cross-head or piston P)
1 PO
and is perpendicular to OP, and
c p represents v (i.e. velocity of P with respect to C) and is parallel to CP.
1 1 PC
A little consideration will show, that the triangles oc p and OCM are similar. Therefore,
1 1
oc op c p
11 11
===ω (a constant)
OC OM CM
vv v
CO PO PC
== =ω
or
OC OM CM
∴ v = ω × OC ; v = ω × OM, and v = ω × CM
CO PO PC
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 dia-
gram 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 inter-
sect 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. 15. 2 (c). We know that
r
(i) o'c' represents (i.e. radial component of the acceleration of crank pin C with respect
a
CO
to O ) and is parallel to CO;
r
(ii) c'x represents (i.e. radial component of the acceleration of crosshead or piston P
a
PC
with respect to crank pin C) and is parallel to CP or CQ;
t
(iii) xp' represents (i.e. tangential component of the acceleration of P with respect to C )
a
PC
and is parallel to QN (because QN is perpendicular to CQ); and
(iv) o'p' represents a (i.e. acceleration of P with respect to O or the acceleration of piston
PO
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,
oc′′ c′x xp′ op ′ ′
2
== = =ω (a constant)
OC CQ QN NO
518 Theory of Machines
rr t
a aaa
CO PC PC PO 2
=== =ω
or
OC CQ QN NO
rr22
∴
aO =ω ×C;a = ω ×CQ
CO PC
t22
=ω × ;and = ω ×
aQN a NO
PC PO
Thus we see that by drawing the Klien’s acceleration diagram, the acceleration of various
points may be obtained without drawing the separate acceleration diagram.
Notes: 1. The acceleration of piston P with respect to crank pin C (i.e. a ) may be obtained from:
PC
′′
cp a
22
PC
=ω or =ω
CN CN
2
∴ a = ω × CN
PC
2. To find the velocity of any point D on the connecting rod PC, divide CM at D in the same ratio as D
1
divides CP. In other words,
CD CD
1
=
CM CP
∴ Velocity of D, v = ω × OD
D 1
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 D .
2
2
∴ Acceleration of D, a = ω × OD
D 2
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. 15.2 (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.
15.6. Ritterhaus’s Construction
Let OC be the crank and PC the connecting rod of a rciprocating steam engine, as shown in
Fig. 15.3. 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 Ritterhaus’s velocity and acceleration diagrams are
drawn as discussed below:
Fig. 15.3. Ritterhaus’s construction.
Ritterhaus’s velocity diagram
Draw OM perpendicular to the line of stroke PO, such that it intersects the line PC produced
at M. The triangle OCM is known as Ritterhaus’s velocity diagram. It is similar to Klien’s velocity
diagram.
Chapter 15 : Inertia Forces in Reciprocating Parts 519
∴ Velocity of C with respect to O or the velocity of crank pin C,
v = v = ω × OC
CO C
Velocity of P with respect to O or the velocity of crosshead or piston P,
v = v = ω × OM
PO P
and velocity of P with respect to C, v = ω × CM
PC
Ritterhaus’s acceleration diagram
The Ritterhaus’s acceleration diagram is drawn as discussed below:
1. From point M, draw MK parallel to the line of stroke PO, to interect OC produced at K.
2. Draw KQ parallel to MO. From Q draw QN perpendicular to PC.
3. The quadrilateral CQNO is known as Ritterhaus’s acceleration diagram. This is similar
to Klien’s acceleration diagram.
∴ Radial component of the acceleration of C with respect to O or the acceleration of crank
pin C,
r 2
aa==ω×OC
CO C
Radial component of the acceleration of the crosshead or piston P with respect to crank
pin C,
r 2
=ω ×
aCQ
PC
Tangential component of the acceleration of P with respect to C,
t 2
aQ =ω ×N
PC
and acceleration of P with respect to O or the acceleration of piston P,
2
aa==ω×NO
PO P
Notes : 1. The acceleration of piston P with respect to crank pin C is given by
2
a = ω × CN
PC
2. To find the velocity of any point D on the connecting rod PC, divide CM at D in the same ratio as D
1
divides CP. In other words,
CD CD
1
=
CM CP
∴ Velocity of D v = ω × OD
D 1
3. To find the acceleration of any point D on the connecting rod PC, draw DD parallel to the line of
2
stroke PO, which intersects CN at D . The acceleration of D is given by
2
2
a = ω × OD
D 2
15.7. Bennett’s Construction
Let OC be the crank and PC the connecting rod of reciprocating steam engine, as shown in
Fig. 15.4. Let the crank makes an angle θ with the line of stroke PO and rotates with uniform angular
velocity ω rad/s in the clockwise direction. The Bennett’s velocity and acceleration diagrams are
drawn as discussed below:
Bennett’s velocity diagram
When the crank OC is at right angle to the line of stroke, it occupies the postition OC and the
1
crosshead P moves to the position P , as shown in Fig. 15.4. Now, produce PC to intersect OC at M.
1 1
The triangle OCM is known as Bennett’s velocity diagram. It is similar to Klien’s velocity diagram.
520 Theory of Machines
Fig. 15.4. Bennett’s construction.
∴ Velocity of C with respect to O or the velocity of crank pin C,
v = v = ω × OC
CO C
Velocity of P with respect to O or the velocity of crosshead or piston P,
v = v = ω × OM
PO P
and velocity of P with respect to C, v = ω × CM
PC
Bennett’s acceleration diagram
The Bennett’s acceleration diagram is drawn as discussed below:
1. From O, draw OL perpendicular to P C (i.e. position of connecting rod PC when crank is
1 1 1
at right angle). Mark the position of point L on the connecting rod PC such that CL = C L .
1 1
2. From L, draw LK perpendicular to PC and from point K draw KQ perpendicular to the line
of stroke PO. From point C, draw CN perpendicular to the line of stroke PO. Join NQ. A little
consideration will show that NQ is perpendicular to PC.
3. The quadrilateral CQNO is known as Bennett’s acceleration diagram. It is similar to
Klien’s acceleration diagram.
∴ Radial component of the acceleration of C with respect to O or the acceleration of the
crank pin C,
r 2
aa==ω×OC
CO C
Radial component of the acceleration of the crosshead or piston P with respect to crank
pin C,
r 2
aC =ω ×Q
PC
Tangential component of the acceleration of P with respect to C,
t 2
=ω ×
aQN
PC
and acceleration of P with respect to O or the acceleration of piston P,
2
aa==ω×NO
PO P
Notes : 1. The acceleration of piston P with respect to crank pin C is given by
2
aC =ω ×N
PC
2. The velocity and acceleration of any point D on the connecting rod PC may be obtained in the similar
way, as discussed in the previous articles, i.e.
Velocity of D, v = ω × OD
D 1
2
and Acceleration of D, a = ω × OD
D 2
Chapter 15 : Inertia Forces in Reciprocating Parts 521
Example 15.1. 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 help 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
1. Velocity and acceleration of the piston
We know that the velocity of the piston P,
v = ω × OM = 120 × 0.127 = 15.24 m/s Ans.
P
and acceleration of the piston P,
2 2 2
a = ω × NO = (120) × 0.2 = 2880 m/s Ans.
P
2. 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 D in the
1
same ratio as D divides CP. Since D is the mid-point of CP, therefore D is the mid-point of CM, i.e.
1
CD = D M. Join OD . By measurement,
1 1 1
OD = 140 mm = 0.14 m
1
∴ Velocity of D, v = ω × OD = 120 × 0.14 = 16.8 m/s Ans.
D 1
In order to find the acceleration of the mid-point of the connecting rod, draw a line DD
2
parallel to the line of stroke PO which intersects CN at D . By measurement,
2
OD = 193 mm = 0.193 m
2
∴ Acceleration of D,
2 2 2
a = ω × OD = (120) × 0.193 = 2779.2 m/s Ans.
D 2
3. 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),
v = ω × CM = 120 × 0.173 = 20.76 m/s
PC
522 Theory of Machines
∴ Angular acceleration of the connecting rod PC,
v 20.76
PC
ω= = = 29.66 rad/s Ans.
PC
PC 0.7
We know that the tangential component of the acceleration of P with respect to C,
t22 2
aQ =ω ×N = (120) × 0.093= 1339.2 m/s
PC
∴ Angular acceleration of the connecting rod PC,
t
a 1339.2
2
PC
α= = = 1913.14 rad/s
Ans.
PC
0.7
PC
Example 15.2. 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 Ritterhaus’s construction, determine 1. Velocity and accelera-
tion of the slider, 2. Velocity and acceleration of point D on the connecting rod which is 150 mm from
crank pin C, and 3. angular velocity and angular acceleration of the connecting rod.
Solution. Given : OC = 150 mm = 0.15m ; PC = 600 mm = 0.6 m ; CD = 150 mm = 0.15 m ;
N = 450 r.p.m. or ω = 2π × 450/60 = 47.13 rad/s
The Ritterhaus’s velocity diagram OCM and acceleration diagram CQNO, as shown in
Fig. 15.6, is drawn to some suitable scale in the similar way as discussed in Art. 15.6. By measure-
ment, we find that
OM = 145 mm = 0.145 m ; CM = 78 mm = 0.078 m ; QN = 130 mm = 0.13 m ; and
NO = 56 mm = 0.056 m
Fig. 15.6
1. Velocity and acceleration of the slider
We know that the velocity of the slider P,
vO =ω×M = 47.13× 0.145 = 6.834 m/s Ans.
P
and acceleration of the slider P,
22 2
Ans.
aN =ω ×O= (47.13) × 0.056= 124.4 m/s
P
2. Velocity and acceleration of point D on the connecting rod
In order to find the velocity of point D on the connecting rod, divide CM at D in the same
1
ratio as D divides CP. In other words,
CD
CD CD 150
1
== or CD× CM=× 78= 19.5 mm
1
CM CP CP 600
Join OD . By measurement, OD = 145 mm = 0.145 m
1 1
∴ Velocity of point D,
vO =ω×D = 47.13× 0.145 = 6.834 m/s Ans.
D1
Chapter 15 : Inertia Forces in Reciprocating Parts 523
In order to find the acceleration of point D on the connecting rod, draw DD parallel to the
2
line of stroke PO. Join OD . By measurement, we find that OD = 120 mm = 0.12 m.
2 2
∴ Acceleration of point D,
22 2
Ans.
aO =ω ×D = (47.13) × 0.12 = 266.55 m/s
D2
3. Angular velocity and angular acceleration of the connecting rod
We know that the velocity of the connecting rod PC (or the velocity of point P with respect
to C ),
vC =ω×M = 47.13× 0.078 = 3.676 m/s
PC
∴ Angular velocity of the connecting rod,
v 3.676
PC
ω= = = 6.127 rad/s Ans.
PC
PC 0.6
We know that the tangential component of the acceleration of P with respect to C,
t22 2
aQ =ω ×N = (47.13) × 0.13 = 288.76 m/s
PC
∴ Angular acceleration of the connecting rod PC,
t
a 288.76
PC 2
α= = = 481.27 rad/s Ans.
PC
PC 0.6
15.8. 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,
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.
Velocity of the piston
From the geometry of Fig. 15.7,
′′ ′′ ′
xP== P OP−OP=() PC+CO−(PQ+QO)
PQ=φ l cos ,
...
=+() lr−(lcosφ+rcosθ)
and QO=θ r cos
524 Theory of Machines
l
=−rl (1 cosθ)+ (1− cosφ)=r
(1−θ cos )+ (1− cosφ)
r
=−rn (1 cosθ)+ (1− cosφ) ...(i)
From triangles CPQ and CQO,
CQ = l sin φ = r sin θ or l/r = sin θ/sin φ
∴ n = sin θ/sin φ or sin φ = sin θ/n ...(ii)
1
1
2
2
sin θ
2
2
We know that, cos φ= =
() 1s−φ in 1 −
2
n
Expanding the above expression by binomial theorem, we get
2
1sin θ
cos φ= 1− × + .....
...(Neglecting higher terms)
2
2
n
2
sin θ
or 1c−φ os = ...(iii)
2
2n
Substituting the value of (1 – cos φ) in equation (i), we have
2
2
sin θ sin θ
xr== (1−θ cos )+ n× r (1−θ cos )+
..(iv)
2
2n
2 n
Differentiating equation (iv) with respect to θ,
dx 1 sin 2θ
==rr sinθ+ × 2 sin θ. cos θ sinθ+
... (v)
2n
dn θ 2
(2 sinθθ .cos=sin2θ)
∴ Velocity of P with respect to O or velocity of the piston P,
dx dx dθ dx
vv= = =× =×ω
PO P
dt dθθ dt d
...( Ratio of change of angular velocity=θdd /t=ω)
Substituting the value of dx/dθ from equation (v), we have
sin 2θ
vv==ω.sr inθ+
PO P ...(vi)
2n
Note: We know that by Klien’s construction,
vO =ω×M
P
Comparing this equation with equation (vi), we find that
sin 2θ
OM = r sinθ+
2n
Acceleration of the piston
Since the acceleration is the rate of change of velocity, therefore acceleration of the piston P,
dv dv dθ dv
PP P
a== × = ×ω
P
dt dθθ dt d
Chapter 15 : Inertia Forces in Reciprocating Parts 525
Differentiating equation (vi) with respect to θ,
dv cos 2θ× 2
cos 2θ
P
=ω.. rr cos θ+ =ω
cosθ+
dθ 2 n
n
dv
P
Substituting the value of in the above equation, we have
dθ
cos 2θθ cos 2
2
ar =ω.. × ω=ωr
cos θ+ cos θ +
P ...(vii)
nn
Notes: 1. When crank is at the inner dead centre (I.D.C.), then θ = 0°.
cos 0° 1
22
ar =ω.. =ωr
cos 0°+ 1+
∴ P
nn
2. When the crank is at the outer dead centre (O.D.C.), then θ = 180°.
cos 2×° 180 1
22
ar =ω.. cos 180°+ =ωr −1+
∴ P
nn
As the direction of motion is reversed at the outer dead centre therefore changing the sign of the above
expression,
1
2
ar =ω . 1 −
P
n
Above picture shows a diesel engine. Steam engine, petrol engine and diesel engine, all
have reciprocating parts such as piston, piston rod, etc.
15.9. 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
526 Theory of Machines
l
r sin θ
... n =
∴ sin φ= × sinθ =
r
ln
Differentiating both sides with respect to time t,
dd φθ cos θ cosθ
dθ
cos φ× = × = × ω ... =ω
dt
dt n dt n
Since the angular velocity of the connecting rod PC is same as the angular velocity of point P
with respect to C and is equal to dφ/dt, therefore angular velocity of the connecting rod
dφθ cos ω ω cosθ
ω= = × = ×
PC
dt n cosφφ n cos
1
1
2
2
sin θ
sin θ
2
...
2 sinφ=
We know that, cos φ= =
() 1sin 1 −
−φ
2 n
n
ωθ cos ω cosθ
ω= × = ×
∴
PC
1
1
nn
221/2
2
(s n−θin)
2
sin θ
n
1 −
2
n
ωθ cos
=
...(i)
221/2
(s n−θin)
Angular acceleration of the connecting rod PC,
d() ω
PC
α = Angular acceleration of P with respect to C =
PC
dt
We know that
dd ()ωω() dθd()ω
PC PC PC
=×=×ω
...(ii)
dt dθθ dt d
...(ddθ= /t ω)
Now differentiating equation (i), we get
cos
d() ω d ωθ
PC
=
221/2
(s n−θin )
dd θθ
2 2 1/2 2 2 –1/ 2
1
(nn −θ sin )(−sinθ)−(cosθ)× ( −sinθ× ) −2sinθcosθ
2
=ω
22
n−θ sin
221/2 2 2–1/2 2
(snn −θin)(−sinθ)+(s−inθ) sinθcosθ
=ω
22
n−θ sin
221/2 2 2–1/2 2
(sin) ( sin) cos
nn −θ − − θ θ
=−ω sin θ
22
n−θ sin
22 2
(s n−θin)−cosθ
sin 221/2
=−ω θ
...Dividing and multiplying by (n−θ sin )
223/2
(s n−θin)
Chapter 15 : Inertia Forces in Reciprocating Parts 527
2
−ω sin θ −ω sin θ (n − 1)
22 2
==
n−θ (sin + cosθ)
223/2 2 23/2
(snn −θin ) ( −θ sin )
22
...( sin θ+ cos θ= 1)
22
d() ω −ω sinθ (n − 1)
PC
∴α = ×ω=
PC
...From equation (ii) ...(iii)
223/2
dθ
(s n−θin)
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 2
Notes: 1. Since sin θ is small as compared to n , therefore it may be neglected. Thus, equations (i) and (iii) are
reduced to
22
ωθ cos −ω sinθ (n− 1)
ω= ,andα =
PC PC
3
n
n
2
2. Also in equation (iii), unity is small as compared to n , hence the term unity may be neglected.
2
−ω sin θ
∴ α=
PC
n
Example 15.3. If the crank and the connecting rod are 300 mm and 1 m long respectively
and the crank rotates at a constant speed of 200 r.p.m., determine:1. The crank angle at which the
maximum velocity occurs, and 2. Maximum velocity of the piston.
Solution. Given : r = 300 mm = 0.3 m ; l = 1 m ; N = 200 r.p.m. or ω = 2 π × 200/60 = 20.95 rad/s
1. Crank angle at which the maximum velocity occurs
Let θ = Crank angle from the inner dead centre at which the maximum
velocity occurs.
We know that ratio of length of connecting rod to crank radius,
n = l/r = 1/0.3 = 3.33
and velocity of the piston,
sin 2θ
sin θ+
vr =ω.
P ...(i)
2 n
For maximum velocity of the piston,
dv
2cos 2θ
P
=ω 0. ie. .r cosθ+ =0
dθ 2n
2 2
or n cos θ + 2 cos θ – 1 = 0 ...( cos 2θ= 2 cos θ− 1)
2
2 cos θ + 3.33 cos θ – 1 = 0
2
−± 3.33 (3.33)+ 4× 2× 1
cos θ= = 0.26
∴ ...(Taking + ve sign)
22 ×
or θ = 75º Ans.
2. Maximum velocity of the piston
Substituting the value of θ = 75° in equation (i), maximum velocity of the piston,
sin150° 0.5
vr =ω. sin 75°+ = 20.95× 0.3 0.966 + m/s
P(max)
23 n .33
= 6.54 m/s Ans.
528 Theory of Machines
Example 15.4. 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.
Solution. Given : r = 0.3; l = 1.5 m ; N = 180 r.p.m. or ω = π × 180/60 = 18.85 rad/s; θ = 40°
Velocity of the piston
We know that ratio of lengths of the connecting rod and crank,
n = l/r = 1.5/0.3 = 5
∴ Velocity of the piston,
sin 80°
sin 2θ
sin 40°+
vr =ω. sinθ+ = 18.85× 0.3 m/s
P
25 ×
2n
=4.19 m/s Ans.
Acceleration of the piston
We know that acceleration of piston,
cos 2θ° cos80
22 2
ar =ω . = (18.85) × 0.3 m/s
cos θ+ cos 40°+
P
n 5
2
= 85.35 m/s
Ans.
Position of the crank for zero acceleration of the piston
Let θ = Position of the crank from the inner dead centre for zero acceleration
1
of the piston.
We know that acceleration of piston,
cos 2θ
2
1
ar =ω . cosθ+
P
1
n
2
ω .r
a =
or ...( 0)
0(=θ ncos+cos2θ)
P
11
n
∴ n cos θ + cos 2θ = 0
1 1
2 2
5 cos θ + 2 cos θ – 1 = 0 or 2 cos θ + 5 cos θ – 1 = 0
1 1 1 1
2
−±55 +4×1×2
cos 0.1862
θ= =
∴ ...(Taking + ve sign)
1
22 ×
or θ = 79.27° or 280.73° Ans.
1
Example 15.5. 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
ω = π × 450/60 = 47.13 rad/s
1. 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
Chapter 15 : Inertia Forces in Reciprocating Parts 529
∴ Velocity of the slider,
sin 120°
sin 2θ
sin 60°+
vr =ω. sinθ+ = 47.13× 0.15 m/s
P
24 ×
2n
=6.9 m/s Ans.
and acceleration of the slider,
cos 2θ° cos 120
22 2
=ω . = (47.13) × 0.15 m/s
ar cos θ+ cos 60°+
P
n 4
2
124.94 m/s Ans.
=
2. Angular velocity and angular acceleration of the connecting rod
We know that angular velocity of the connecting rod,
ωθ cos 47.13× cos 60°
Ans.
ω= = = 5.9 rad/s
PC
n 4
and angular acceleration of the connecting rod,
22
ωθ sin (47.13)× sin 60°
2
α= = = 481 rad/s
Ans.
PC
n 4
15.10. 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 :
1. Piston effort. It is the net force acting on the piston or crosshead pin, along the line of
stroke. It is denoted by F in Fig. 15.8.
P
Fig. 15.8. Forces on the reciprocating parts of an engine.
Let m = Mass of the reciprocating parts, e.g. piston, crosshead pin or
R
gudgeon pin etc., in kg, and
W = Weight of the reciprocating parts in newtons = m .g
R R
We know that acceleration of the reciprocating parts,
cos 2θ
2
aa==ω .r
cos θ+
RP
n
530 Theory of Machines
∴ Accelerating force or inertia force of the reciprocating parts,
cos 2θ
2
Fm==.. a mω.r
cos θ+
IRR R
n
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
Connecting rod of a petrol engine.
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, F = Net load on the piston Inertia force
P
=FF ...(Neglecting frictional resistance)
LI
=− FF R ...(Considering frictional resistance)
LI F
where R = Frictional resistance.
F
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,
F = p A – p A = p A – p (A – a)
L 1 1 2 2 1 1 2 1
where p , A = Pressure and cross-sectional area on the back end side of the
1 1
piston,
p , A = Pressure and cross-sectional area on the crank end side of the
2 2
piston,
a = Cross-sectional area of the piston rod.
Notes : 1. If ‘p’ is the net pressure of steam or gas on the piston and D is diameter of the piston, then
π
2
=×pD ×
Net load on the piston, F = Pressure × Area
L
4
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).
∴ Piston effort, FF=± F W−R
PL I R F
2. Force acting along the connecting rod. It is denoted by F in Fig. 15.8. From the geom-
Q
etry of the figure, we find that
F
P
F =
Q
cos φ
The acceleration of the reciprocating parts by Klien’s construction is,
2
a = ω × NO
P
2
∴ F = m . ω × NO
I R.
Chapter 15 : Inertia Forces in Reciprocating Parts 531
2
sin θ
cos φ= 1−
We know that
2
n
F
P
∴ F =
Q
2
sin θ
1 −
2
n
3. Thrust on the sides of the cylinder walls or normal reaction on the guide bars. It is
denoted by F in Fig. 15.8. From the figure, we find that
N
F
F
P P
F =
FF=φ sin= × sinφ=F tanφ ...
NQ P Q
φ
cos
cos φ
4. Crank-pin effort and thrust on crank shaft bearings. The force acting on the connecting
rod F may be resolved into two components, one perpendicular to the crank and the other along the
Q
crank. The component of F perpendicular to the crank is known as crank-pin effort and it is denoted
Q
by F in Fig. 15.8. The component of F along the crank produces a thrust on the crank shaft bearings
T Q
and it is denoted by F in Fig. 15.8.
B
Resolving F perpendicular to the crank,
Q
F
P
FF=θ sin (+φ)= × sin (θ+φ)
TQ
cos φ
and resolving F along the crank,
Q
F
P
FF=θ cos (+φ)= × cos (θ+φ)
BQ
cos
φ
5. Crank effort or turning moment or torque on the crank shaft. The product of the crank-
pin effort (F ) and the crank pin radius (r) is known as crank effort or turning moment or torque on
T
the crank shaft. Mathematically,
F sin (θ+φ)
P
Crank effort, TF=×r= ×r
T
cos φ
(sin cos cos sin )
F θφ+ θ φ
P
=× r
cos φ
sin
φ
sin θ+ cos θ×
=× Fr
P
cos φ
=θFr (sin+ cosθ tanφ)× ...(i)
P
We know that l sin φ = r sin θ
r sin θ l
...
sin φ= sin θ= n =
r
ln
2
sin θ 1
222
cos 1 sin 1 sin
and φ= − φ = − = n − θ
2
n
n
sinφθ sin n sinθ
∴ tan φ= = × =
22 2 2
cos φ n
nn −θ sin −θ sin
532 Theory of Machines
Substituting the value of tan φ in equation (i), we have crank effort,
cosθθ sin
sin θ+
TF=×r
P
22
n−θ sin
sin 2θ
sin θ+
=× Fr
P ...(ii)
22
2sin
n−θ
θθ= θ
...( 2 cos sin sin 2 )
2 2 2
Note: Since sin θ is very small as compared to n therefore neglecting sin θ, we have,
sin 2θ
Crank effort, TF=×r =F×OM
sinθ+
PP
2n
We have seen in Art. 15.8, that
sin 2θ
OM = r sin θ+
2n
Therefore, it is convenient to find OM instead of solving the large expression.
Example 15.6. 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.
Solution. Given : D =175 mm ; L = 200 mm = 0.2 m or r = L / 2 = 0.1 m ; N = 500 r.p.m. or
ω = 2π × 500/60 =52.4 rad/s ; l = 400 mm = 0.4 m ; m = 180 kg
R
The inertia force may be calculated by graphical method or analytical method as discussed
below:
1. Graphical method
First of all, draw the Klien’s acceleration diagram OCQN to some suitable scale as shown in
Fig. 15.9. By measurement,
ON = 38 mm = 0.038 m
∴ Acceleration of the reciprocating parts,
2
a = ω × ON
R
2
= (52.4) × 0.038 = 104.34 m/s
We know that inertia force,
F = m × a = 180 × 104.34 N
I R R
= 18 780 N = 18.78 kN Ans.
2. Analytical method
We know that ratio of lengths of connecting rod and crank,
n = l / r = 0.4 / 0.1 = 4
cos 2θ
2
∴ Inertia force,
Fm=ω..r cos θ+
IR
Fig. 15.9
n
cos 120°
2
=× 180 (52.4)× 0.1 cos 60°+ = 18 530 N
4
= 18.53 kN Ans.
Superfluous data.
Chapter 15 : Inertia Forces in Reciprocating Parts 533
Example 15.7. 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
2
the driving and the back pressures is 0.35 N/mm . 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.
2
Solution. Given: r = 300 mm = 0.3 m ; m = 250 kg; θ = 60°; p – p = 0.35 N/mm ;
R 1 2
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 (F ).
P
We know that net load on the piston,
ππ
22
Fp=− ( p ) ×D= 0.35× (500)= 68730 N
L1 2
44
...( Force = Pressure × Area)
Ratio of length of connecting rod and crank,
nl==/1 r .2/0.3=4
and accelerating or inertia force on reciprocating parts,
cos 2θ
2
Fm=ω.crosθ+
IR
n
cos 120°
2
== 250 (26.2) 0.3 19306 N
cos 60°+
4
∴ Piston effort, F = F – F = 68 730 – 19 306 = 49 424 N = 49.424 kN
P L I
1. Pressure on slide bars
Let φ = Angle of inclination of the connecting rod to the line of stroke.
sinθ° sin 60 0.866
We know that, sin φ= = = = 0.2165
n 44
∴ φ = 12.5°
We know that pressure on the slide bars,
F = F tan φ = 49.424 × tan 12.5° = 10.96 kN Ans.
N P
2. Thrust in the connecting rod
We know that thrust in the connecting rod,
F 49.424
P
F== = 50.62 kN Ans.
Q
cosφ° cos 12.5
3. Tangential force on the crank-pin
We know that tangential force on the crank pin,
sin ( ) 50.62 sin (60 12.5 ) 48.28 kN
FF=θ+φ= °+ °= Ans.
TQ
4. Turning moment on the crank shaft
We know that turning moment on the crank shaft,
TF=×r= 48.28× 0.3= 14.484 kN-m Ans.
T
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