Unit2 lectbyvrk dynofmachinery

DYNAMICS OF MACHINERY
STATIC AND DYNAMIC BALANCING
What are the reasons for unbalanced forces?
Unbalanced forces are set up in high speed engines due to the following reasons:
(a) Rotating masses: Rotating masses if not balanced, produce centrifugal forces
which act as unbalanced forces and cause undesirable vibrations and noise.
(b) Reciprocating masses: reciprocating masses if not balanced, produce inertia
forces which act as unbalanced forces and cause vibrations and noise.
What is balancing?
Balancing is the process of designing or modifying machinery, so that the unbalance in
the machinery is reduced to an acceptable level and if possible completely
eliminated.
Why balancing is necessary?
If rotating and reciprocating masses are not balanced, they produce centrifugal /
inertia forces which will cause excessive vibration, noise, wear and tear of the
system. Hence balancing is very essential.

What are the types of balancing?
Balancing
Rotating masses Reciprocating masses
Single
Different Primary force Secondary force
Combined rotating and
reciprocating masses

Define static balancing.
A system is said to be in static balance if the net dynamic force acting on the system is
zero. Σ F = 0.
Define dynamic balancing.
A system is said to be in dynamic balance if the net dynamic forces as well as net
dynamic couples are equal to zero respectively.
What are the various types of balancing of rotating masses?
(a) Balancing of a single rotating mass by a single mass rotating in the same plane
(static balancing)
(b) Balancing of a single rotating mass by two masses rotating in different planes
(dynamic balancing)
(c) Balancing of several masses rotating in the same plane (static balancing)
(d) Balancing of several masses rotating in different planes (dynamic balancing)

What is rocking?
Sometimes it may not be possible to introduce a single balancing mass in the same
plane of rotation of the disturbing mass. Hence the balancing mass is provided in a
plane parallel to the plane of rotation of the disturbing mass. This balances the
forces, but produces an unbalanced couple. This unbalanced couple tends to rock
the shaft in its bearings. The shaft is subjected to bending. This phenomenon is
called rocking of the shaft.
Define reference plane.
Reference plane is a plane passing through a point on the axis of rotation and
perpendicular to it. It is used as a reference for analysis and balancing.
What are the conditions to be satisfied in reference plane for complete balance in the
case of balancing of several masses in different planes?
(a) Forces in the reference plane must balance i.e. resultant force must be zero.
(b) Couples about the reference plane must balance i.e. resultant couple must be
zero.

How will you balance several masses rotating in different planes?
When several masses rotate in different planes, they may be transferred to a plane
known as reference plane (R P) and hence the problem is reduced to that of
several masses rotating in the same plane.
This transfer will introduce the following effects in the reference plane:
(a) An unbalanced force equal and parallel to the centrifugal force produced by the
rotating mass (F)
(b) An unbalanced couple whose magnitude is equal to the product of the centrifugal
force and the distance between the two planes (F l)
Why are the projection of hands in opposite direction in watches and clocks?
Projections are made such that mass occupied in the projected length is same as mass
occupied in the needle from the pivot point. Hans are thus balanced.

BALANCING OF ROTATING MASSES
How would you balance several rotating masses in the same plane by analytical
method?
Σ H = Σ mi ri cos θi
Σ V = Σ mi ri sin θi
Resultant force magnitude , R = *(Σ H)2 +(Σ V)2 ]
Direction θ’ = Σ V / Σ H
Inclination of balancing mass with horizontal θ R = 180 + θ’
Let Value of radius of rotation of balancing mass be “ r “
Balancing mass be “ m “
The relation used to find “ m “ or “ r “ is “ m = R / r “

BALANCING OF ROTATING MASSES

BALANCING OF RECIPROCATING MASSES
What do you mean by balancing of reciprocating masses?
It is the dynamic balancing of inertia force and couple (shaking couple) due to
reciprocating masses to avoid vibrations of reciprocating engines.
What will happen if reciprocating masses are not balanced?
As the stroke of the reciprocating parts is proportional to radius of crank, the radius of
crank acts as an eccentricity of reciprocating masses. Due to this eccentricity
inertia force is produced causing vibrations of the engine. Even if there is no
unbalanced force due to inertia effect, there will be unbalanced couple called
shaking couple causing vibrations of the engine.

BALANCING OF RECIPROCATING MASSES
Why complete balancing of reciprocating mass is not possible?
To balance the rotating mass at crank pin, balancing mass is connected dynamically
opposite to the crank pin. But the vertical component of the force produced by the
balancing mass is unbalanced and produces hammer blow. In reciprocating
engines, unbalanced forces in the direction of line of stroke are more dangerous
than the forces perpendicular to the line of stroke. Therefore reciprocating masses
balanced only to an extent (fraction “ c “ of reciprocating masses) to avoid
hammer blow. Complete balancing of reciprocating masses are not possible.
Distinguish between the nature of unbalanced force due to reciprocating mass and
that due to rotating mass.
Inertia forces ( both primary and secondary) due to reciprocating masses are constant
in direction but vary in magnitudes.
Centrifugal forces due to rotating masses are constant in magnitude, but vary in
direction.

BALANCING A SINGLE CYLINDER ENGINE

PRIMARY AND SECONDARY UNBALANCED FORCES
I

BALANCING MULTI CYLINDER ENGINES
Why are the cranks of a twin cylinder locomotives placed at right angles to each
other?
The cranks of a twin cylinder locomotives placed at right angles to each other because
(a) More uniform turning moment is obtained.
(b) Engine can be started easily after stopping in any position.
What do you mean by outside cylinder locomotive?
In outside cylinder locomotive, two cylinders are placed outside the driving
wheels, one on each side of the driving wheels to obtain uniform turning moment.
Define tractive force.
Tractive force is the resultant unbalanced primary force along the line of stroke.
Variation of tractive force (effort) of an engine is caused by the resultant unbalanced
primary force due to two cylinders along the line of stroke of the locomotive
engine.

Define tractive force.
Tractive force is the resultant unbalanced primary force along the line of stroke.
Variation of tractive force (effort) of an engine is caused by the resultant unbalanced
primary force due to two cylinders along the line of stroke of the locomotive
engine.
Mention the expression for tractive force. At what angles of inclination of the crank
with the line of stroke, the tractive force reaches a maximum value?
Tractive force FT = (1-c) m ω2 r (cos θ – sin θ)
It is maximum at θ =1350 and θ = 315 0
FT max = ±√2 (1-c) m ω2 r

Define swaying couple.
The unbalanced primary forces along the line of stroke for two cylinders which are
separated by a distance and thus constitute a couple about the centre line of the
locomotive engine between the two cylinders. This couple tends to sway the
engine alternately in clockwise and anticlockwise direction. Hence this couple is
called swaying couple.
Mention the expression for swaying couple. At what angles of inclination of the crank
with the line of stroke, the swaying couple reaches a maximum value?
Swaying couple CS = (1-c) m ω2 r (a/2) (cos θ + sin θ)
It is maximum at θ =450 and θ = 225 0

Define hammer blow.
The unbalanced force perpendicular to the line of stroke produces variation in
pressure on the rails which results in hammering acting action on the rails. The
maximum value of this vertical unbalanced force is called hammer blow. This is
caused by the mass provided to balance the reciprocating mass.
Mention the expression for hammer blow. At what angles of inclination of the crank
with the line of stroke, the hammer blow reaches a maximum value?
Hammer Blow, FH = B ω2 b sin θ; where B is balancing mass, b is radius of rotation of
balancing mass.
It is maximum at θ = 900 and θ = 270 0
FH max = ± B ω2 b
Effect of hammer blow: Causes variation in pressure between the wheel and the rail.

Mention the expression for permissible value of angular speed ω of the wheel to avoid
lifting from the track.
Permissible value of angular speed ω of the wheel to avoid lifting of the wheel from
the track , ω = [P/(Bb)] 1/2
Where B is balancing mass, b is radius of rotation of the balancing mass.
What do you mean by multi cylinder engine?
When two or more cylinders are there in an engine, then it is called multi cylinder
engine. These engines are broadly classified according to the arrangement of
cylinders with respect to each other and crank shaft.
Mention the important types of multi cylinder engines.
Multi cylinder engine
Inline Radial V engine

What do you mean by radial engine? Why are they preferred?
Radial engine is one which has pistons arranged in a circle about the crank centre.
In radial engines, connecting rods (equal to number of cylinders) are connected to a
common crank. Unlike in inline engines, the plane of rotation of crank (common
crank) is same. Therefore there is no unbalanced primary or secondary couple.
What is V-12 engine? Why is it used in some luxury cars?
The engine has two banks of 6 cylinders each, the banks inclined to each other at an
angle. We know that in 6 cylinder inline engine, each bank of cylinders is self
balancing and hence the whole engine is balanced. The engine has two banks of 6
cylinders each, the banks inclined to each at an angle. Each bank with 6 cylinders 4
stroke inline engines is self balanced. Hence the whole engine is balanced and
hence v-12 engine is used in some luxury cars.

What are the various methods of force balancing of linkages?
(a) Method of static balancing
(b) Method of principle vectors
(c) Method of linearly independent vectors.
(d) Use of cam driven masses.
(e) Addition of an axially symmetric duplicate mechanism
Mention the types of balancing machine.
(a) Static balancing machines: These machines measure the static unbalance only.
(b) Dynamic balancing machines: These machines measure the dynamic unbalance
only.
(c) Universal balancing machines: These machines are capable of measuring both
static and dynamic unbalance.

How are the cylinders arranged in uncoupled three cylinder locomotives?
In an uncoupled three cylinder locomotive, there are three cylinders out of which two
are outside cylinders and one is inside cylinder. On each side of the two driving
wheels beyond the wheel, one outside cylinder lies. The inside cylinder lies in
between the centre of the two driving wheels to obtain uniformity in turning
moment. The angle between the cranks of each cylinder is 1200 .
FIG.

What do you mean by inline engine?
In line engine is one in which the piston axes form a single plane coincident with the
crank shaft and in which pistons are all on the same side of the crank shaft.
What are the conditions to be satisfied for the complete balance of multi cylinder in
line engines?
(a) Primary forces must balance i.e. primary force polygon must close.
(b) Primary couples must balance i.e. primary couple polygon must close.
(c) Secondary forces must balance i.e. secondary force polygon must close.
(d) Secondary couples must balance i.e. secondary couple polygon must close.

FIRING ORDER
What do you mean by firing order?
The sequence in which charge is ignited inside the the engine cylinders is called firing
order. There are different firing orders for multi cylinder engines
For two stroke engines, interval between the cranks = 360 / n
For four stroke engines, interval between the cranks = 720 / n
The cranks are assumed to rotate in clockwise direction and hence, they are
numbered in anticlockwise as per firing order.

FIRING ORDER
Draw primary and secondary crank positions of four cylinder two stroke in line engine
for the firing order 1-5-2-3-4.
1
5
2
3
4 2
5
4
31
Primary Crank positions Secondary Crank positions

STATIC BALANCING DYNAMIC BALANCING

PIVOTED CRADLE BALANCING MACHINES
PIVOTED CRADLE BALANCING MACHINE: This machine measures both static and
dynamic unbalance of rotating parts. It is similar to dynamic balancing machine
except that in this no preliminary static balancing of the machine part is required.
Hence it is also called Universal Balancing Machine.
FIG

PIVOTED CRADLE BALANCING MACHINES
PIVOTED CRADLE BALANCING MACHINE: This machine measures both static and
dynamic unbalance of rotating parts. It is similar to dynamic balancing machine
except that in this no preliminary static balancing of the machine part is required.
Hence it is also called Universal Balancing Machine.
It works on the principle that a given system of revolving masses can be balanced by
introducing two balancing masses in arbitrarily chosen planes of motion. First two
planes say L and M are selected. By using these planes, the balancing effects are to
be measured and necessary corrections are applied.
The rotating part is mounted on the cradle in such a manner that the axis of the pivots
P-P lies on one of the chosen planes say L. The purpose of this that the out of
balance effect in this plane can not cause oscillations of the cradle. The out of
balance effect in the other plane M produces a moment (Fm a sin θ) in the plane of
oscillation and rocks the cradle about the axis P-P. The maximum value of this
moment Fm a can be measured so that the amount of unbalance (wm r) in plane M
can be obtained. Thereafter the cradle is slided along the guide, so that the axis of
oscillation lies in plane M and the amount and angular position of the unbalance
(wl r) in plane L can be obtained.

Problem 1: Given: Four masses attached to a shaft in the same plane
m1 m2 m3 m4 Unit
Mass 200 300 240 260 kg
Radius 0.2 0.15 0.25 0.3 m
Angle between successive masses
m1-m2 m2-m3 m3-m4
θ 450 75 0 135 0
Balancing mass is provided at radius 0.2 m
Find the magnitude and angular position of the balancing mass

Problem 2
Four masses A,B,C,D are attached to a rotating shaft
A B C D Unit
Mass 20 10 8 kg
Radius 50 62.5 100 75 mm
Distance between planes of successive masses
A and B B and C C and D
600 600 600 mm
Shaft is in complete balance
Find (a) the mass of A; (b) Angular positions of A, B, C, and D

Problem 3
A B C D Unit
Mass 200 300 400 200 kg
Radius 80 70 60 80 mm
Distance
from Plane A 0 300 400 700 mm
Angle between cranks measured anticlockwise
θ 450 70 0 120 0
Balancing masses in planes X and Y
Distance between planes
A and X X and Y Y and D
100 400 200 mm
Balancing masses at radius 100 mm
Find (a) the balancing masses;
(b) Angular positions of the balancing masses.

Problem 4
Four masses A,B,C,D are to be balanced.
A B C D Unit
Mass 30 50 50 kg
Radius 180 240 120 150 mm
Angle between planes
B and C B and D C and D BD and CD are
θ 90 0 210 0 120 0 in same sense
Planes B and C are 300 mm apart
Find (a) the mass of rotor A; (b) angular position of A;
(c ) Position of plane A and D .

Problem 5
Four masses A,B,C,D are in order along the axis. Shaft and masses are in
complete balance.
A B C D Unit
Mass 30 50 40 kg
Radius 180 240 120 150 mm
Angular spacing
C and B D and B
θ 90 0 210 0
Planes B and C are 300 mm apart
Find (a) the mass of rotor A; (b) angular position of rotor A;
(c ) Position of plane A and D .

Problem 6
A B C D Unit
Mass 200 300 400 200 kg
Radius 80 70 60 80 mm
Distance
from Plane A 0 300 400 700 mm
Angle between cranks measured anticlockwise
θ 450 70 0 120 0
Balancing masses in planes X and Y
A and X X and Y Y and D
100 400 200 mm
Balancing masses at radius 100 mm
Find (a) the balancing masses;
(b) Angular positions of the balancing masses.

Problem 7
A shaft carries Four masses A,B,C,D in parallel planes.
Shaft is in complete balance
A B C D Unit
Mass 18 21.5 kg
Eccentricity 80 60 60 80 mm
Angle between masses
B and C B and A
θ 100 0 190 0 in same sense
A and B B and C
100 200 mm
Find (a) mass of A and D; (b) Distance between A and D;
(c ) Angular position of mass at D.

Problem 8
Four Cylinder Vertical Engine
Crank radius r = 225 mm
Plane of rotation of 1st crank from 3rd crank at 600 mm
Plane of rotation of 2nd crank from 3rd crank at 300 mm
Plane of rotation of 4th crank from 3rd crank at 300 mm
Reciprocating masses of 1st cylinder 100 kg.
Reciprocating masses of 2nd cylinder 120 kg.
Reciprocating masses of 4th cylinder 100 kg.
Find (a) Reciprocating mass for 3rd cylinder;
(b) Angular positions of 1st, 2nd,3rd, and 4th cylinders for complete
balance.

Problem
9
Four Cylinder Vertical Engine
Crank radius r = 150 mm
Plane of rotation of 1st crank from 3rd crank at 400 mm
Plane of rotation of 2nd crank from 3rd crank at 200 mm
Plane of rotation of 4th crank from 3rd crank at 200 mm
Reciprocating masses of 1st cylinder 50 kg.
Reciprocating masses of 2nd cylinder 60 kg.
Reciprocating masses of 4th cylinder 50 kg.
Find (a) Reciprocating mass for 3rd cylinder;
(b) Angular positions of 1st, 2nd,3rd, and 4th cylinders for complete
balance.

Problem10
Six Cylinder Vertical 4 stroke inline engine
Firing order 1-4-2-6-3-5
Stroke length L = 2 r = 0.1 m
Connecting rod length l = 0.2 m
Pitch distance between cylinder centre lines 100, 100, 150, 100, 100 mm
Reciprocating mass per cylinder mR = 1 kg.
Engine speed N = 3000 rpm
Find Unbalanced primary and secondary forces and couples. Consider
plane midway between cylinders 3 and 4 as reference.

Problem11
Three Cylinder inline IC engine
Cranks are set at angle = 120 0
Pitch of the cylinders = 1 m
Stroke length of the piston L = 0.6 m
Reciprocating mass for inside cylinder = 300 kg
Reciprocating mass for outside cylinder = 260 kg
Distance between plane of rotation of balancing mass from inside crank = 0.8 m
Percentage of reciprocating masses to be balanced, c = 0.4
Radius of balancing masses = 0.6 m
Find (a) magnitudes and (b) angular positions of balancing masses; (c ) Hammer
blow/wheel at axle speed = 6 rps.

Problem12
Uncoupled outside cylinder
Mass of rotating parts per cylinder = 360 kg.
Mass of reciprocating parts per cylinder m R= 300 kg.
Angle between cranks 90 0
Crank radius r = 0.3 m
Cylinder centres = 1.75 m
Radius of balancing masses = 0.75 m
Wheel centres = 1.45 m
All the rotating masses and 2/3 of reciprocating masses are balanced
in planes of driving wheels.
Find (a) magnitudes and (b) angular positions of balancing masses; (c )Limiting
speed in kph ( for the wheel to lift off the rails) if load on each driving wheel is 30
kN and diameter of the wheel is 1.8 m
(d) Swaying couple at the above speed.

Problem13
Two cylinder inside uncoupled locomotive
Mass of rotating parts per cylinder = 250 kg.
Distance between cylinders = 0.6 m
Radius of balancing masses = 1 m
Distance between driving wheels = 1.5 m
Diameter of driving wheels = 2 m
Speed of locomotive = 80 km/hr
Find (a) Hammer blow; (b ) Maximum variation in tractive effort; (c ) Maximum
swaying couple.

Problem14
Two cylinder inside uncoupled locomotive
Distance between cylinders = 0.65 m
Distance between driving wheels = 1.6 m
Diameter of driving wheels = 1.8 m
Maximum hammer blow = 45000 N at Speed of locomotive = 100 km/hr
Find (a) Fraction of reciprocating masses to be balanced c; (b) Variation in
tractive effort; (c ) Maximum swaying couple.

Problem15
Four cylinder inline engine
Speed N = 1800 rpm
Crank radius r = 0.06m
Connecting rod length l = 0.24 m
Spacing between cylinders = 150 mm
Mass of reciprocating parts per cylinder m R= 1.5 kg.
Angular positions of cranks 1-4-2-3 at 90 o intervals in end view.
Cylinders are numbered in sequence 1-2-3-4 from one end.
Find Unbalanced primary and secondary forces and couples with reference to
central plane of engine.

Unit 2: Problem 1- A 270 cm long shaft carries three pulleys, two at
its ends and the third at its mid point. The two end pulleys have
masses 48 kg and 20 kg and their centre of gravities are 1.5 cm and
1.25 cm respectively from the axis of the shaft, the middle pulley has
a mass of 56 kg and its CG is 1.5 cm from the shaft axis. The pulleys
are so keyed to the shaft that the assembly is in static balance. The
shaft rotates at 300 rpm in two bearings 180 cm apart with equal
over hangings on either side. Determine the relative angular position
of the pulleys.
Given Data: Length of shaft L = 2.7 m
Let A, C, and E be the pulleys at left end, middle and right end
respectively.
Let B and D be the bearings; Length between bearings = 1.8 m
Mass (kg) Radius (m) CentrifugalFo
rce /ω2
A 48 0.015 0.72
C 56 0.015 0.84
E 20 0.0125 0.25
To find : relative angular position of the pulleys
50

Plane Diagram Space Diagram
2.70
1.35
0.45
A C E
B D
C
Table
Mass
m (kg)
Radius
r (m)
Centrifugal
Force /ω2
=mr
Angle
θ
mr cos θ ‘ mr sin θ
a 48 0.015 0.72 θA 0.72cosθA 0.72sinθA
c 56 0.015 0.84 0 0.84 0
e 20 0.0125 0.25 θE 0.25cosθE 0.25sinθE
Sum = 0 0
Solving the two equations, 0.72cosθA+0.84+0.25cosθE = 0, and
0.72sinθA+0.25sinθE = 0 you may find it difficult!!!
Solving by familiar equation for a triangle,
51
C
E
A

EA
‘ a = 72
‘ e = 25
‘ c =84
2.70
1.35
0.45
A C E
B D
C
E
A
a c e
Magnitude 72 84 25
A C E Total
cos 0.595 -0.346 0.96
53.5 110.2 16.3 180
180+53.5 180-16.3
Angle deg 233.5 163.7
52
Angular position of mass A= 163.7 o
Angular position of mass E= 233.5 o

Unit 2: Problem 2- A, B, C, and D are four masses carried by a rotating shaft at radii
100 mm, 150 mm, 150 mm, and 200 mm respectively. The planes in which the masses
rotate are spaced at 500 mm apart and the magnitude of the masses B, C, and D are 9
kg, 5 kg and 4 kg respectively. Find the required mass A and the relative angular
settings of the four masses so that the shaft shall be in complete balance.
Given Data A B C D
Mass m (kg) ma 9 5 4
Radius r (m) 0.1 0.15 0.15 0.2
Distance between planes = 0.5 m
B
1.50
1.00
0.50
A(RP) CB D
53
D
C
B
A

D
C
‘ c = 0.75
‘ d = 1.20
‘ b =0.675
B
C
D
Table:
Plane Mass Radius Force/ω2
Distance from
RP Couple/ω2 Angle
m (kg) r (m)
mr (kg
m) l (m)
mrl (kg
m2) θ (deg)
A (RP) mA 0.1 0.1mA 0 0
B 9 0.15 1.35 0.5 0.675 0
C 5 0.15 0.75 1 0.75 θ C = 65.4
D 4 0.2 0.8 1.5 1.2 θD = 214.6
54

Plane Mass Radius Force/ω2 Angle
m (kg) r (m) mr (kg m) θ (deg) mr cosθ mr sinθ
A (RP) mA 0.1 0.1mA
B 9 0.15 1.35 0 1.35 0
C 5 0.15 0.75 65.4 0.311897 0.6821
D 4 0.2 0.8 214.6 -0.657824 -0.455
Sum = 1.004074 0.2268
Resultant= 0.1 mA = 1.029
Angle of resultant θ= 12.72
Angle of equilibrant θA =180+12.72 192.72deg
mA= 10.29kg
D
C
B
A
Results
Plane Mass Radius Angle
m (kg) r (m) θ (deg)
A (RP) 10.29 0.1 192.7
B 9 0.15 0
C 5 0.15 65.4
D 4 0.2 214.6
55

Unit 2: Problem 3- The following data refer to two cylinder uncoupled locomotive
Rotating mass per cylinder = 300 kg;
Reciprocating mass per cylinder = 330 kg;
Distance between wheels = 1.5 m;
Distance between cylinders (pitch) = 600 mm;
Diameter of treads of driving wheels = 1.8 m;
Crank radius = 325 mm;
Radius of centre of balance mass = 650 mm;
Locomotive speed = 60 km/hr;
Angle between cylinder cranks = 90 o ;
Dead load on each wheel = 40 kN;
Determine
(a) Balancing mass required in the planes of driving wheels if whole of the revolving
mass and 2/3 of the reciprocating mass are to be balanced.
(b) (b) Swaying couple;
(c) Variation in tractive force;
(d) Maximum and minimum pressure on the wheels; and
(e) Maximum speed of the locomotive without lifting the wheels from the rails.
56

Given data: mo = 300 kg; mr = 330 kg; mT = mo+ (2/3)mr = 520 kg;
Distance between wheels, lW = 1.5 m; Distance between cylinders, lc= a = 0.6 m;
D = 1.8 m, R = D/2 = 0.9 m; Velocity, v = ωR = 60x1000/3600= 16.667 m/s;
(1 km = 1000 m and 1 hr = 3600 sc); ω = 16.667/0.9 = 18.52 rad/s;
Plane Mass Radius Force/ω2
Distance
from RP Couple/ω2 Angle
m (kg) r (m) mr (kg m) l (m) mrl (kg m2) θ (deg)
A (RP) mA 0.65 0.65mA 0 0 -
B 520 0.325 169 0.45 76.05 0
C 520 0.325 169 1.05 177.45 90
D mD 0.65 0.65mD 1.5 0.975mD θD = 246.8
C
B
1.50
1.05
0.45
A(RP) CB D
Wheel1 Cyl.1 Cyl.2 Wheel2
A D
C
B
57

Plane Couple/ω2 Angle
mrl (kg m2) θ (deg) mrl cosθ mrl sinθ
A (RP) 0 -
B 76.05 0 76.05 0
C 177.45 90 -0.112192 177.45
D 0.975mD θD
SUM = 75.93781 177.45
θ = Tan -1 (177.45/75.94)= 66.8 deg
θD 180+66.8= 246.8 deg
R= 193.0156
mD =193.0156/0.975 = 198 kg
Table:
Plane Mass Radius Force/ω2 Angle mr cosθ mr sinθ
m (kg) r (m) mr (kg m) θ (deg)
A (RP) mA 0.65 0.65mA -
B 520 0.325 169 0 169 0
C 520 0.325 169 90 -0.106849 169
D 198 0.65 128.7 246.8 -50.49516 -118.4
SUM= 118.398 50.62
θ= 23.13919
θA = 203.1392deg
R= 128.765kgm
mA= R/0.65= 198kgm
58

59
Maximum and Minimum Pressure on wheels:

60
Unit 2: Problem 4- The firing order of a six cylinder vertical four stroke in line engine is
1-4-2-6-3-5. The piston stroke is 80 mm and the length of each connecting rod is 180
mm. the pitch distance between the cylinder centre lines are 80 mm, 80 mm, 120 mm,
8 mm, and 80 mm respectively. The reciprocating mass per cylinder is 1.2 kg and the
engine speed is 2400 rpm. Determine the out of balance primary and secondary forces
and couples of the engine taking a plane midway between the cylinder 3 and 4 as
reference point.
Given Data: n = 6; l = 0.18 m; m r= 1.2 kg; L = 0.08 m;
N = 2400 rpm; ω = 2πN/60 = 251.33 rad/s;
To find: UPF, UPC, USF, and USC
Interval (angle) between the cranks,θ = 4π/n = 4x180/6 = 120 o
2,5
1,6
4,3
654321
0.120.08 0.08 0.080.08
Space Diagram for Primary Forces
1,6
4,32,5
654321
0.120.08 0.08 0.080.08
Space Diagram for Secondary Forces

61
Table: (for Primary forces and couples)
Plane Mass Radius Force/ω2 Distance from RP Couple/ω2 Angle
1 1.2 0.04 0.048 -0.22 -0.01056 90
2 1.2 0.04 0.048 -0.14 -0.00672 330
3 1.2 0.04 0.048 -0.06 -0.00288 570
4 1.2 0.04 0.048 0.22 0.00288 210
5 1.2 0.04 0.048 0.14 0.00672 690
6 1.2 0.04 0.048 0.06 0.01056 450
Table: (for Secondaryforces and couples)
Plane Mass Radius Force/ω2 Distance from RP Couple/ω2 Angle
1 1.2 0.04 0.048 -0.22 -0.01056 90
2 1.2 0.04 0.048 -0.14 -0.00672 570
3 1.2 0.04 0.048 -0.06 -0.00288 1050
4 1.2 0.04 0.048 0.22 0.00288 330
5 1.2 0.04 0.048 0.14 0.00672 1290
6 1.2 0.04 0.048 0.06 0.01056 810
Do graphically and check answers: UPC=0; UPF=0; USC=0; USF=0.
See analytical answer also.

62
Table: (for Primary couples) Table: (for Primary forces )
PlaneCouple/ω2 Angle mrl cosθ mrl sinθ Plane Force/ω2 Angle mr cosθ mr sinθ
mrl (kg m2) θ (deg) (kg m2) (kg m2) mr (kg m) θ (deg) (kg m) (kg m)
1 -0.01056 90 0.00000 -0.01056 1 0.048 90 0.00000 0.04800
2 -0.00672 330 -0.00582 0.00336 2 0.048 330 0.04157 -0.02400
3 -0.00288 570 0.00249 0.00144 3 0.048 570 -0.04157 -0.02400
4 0.00288 210 -0.00249 -0.00144 4 0.048 210 -0.04157 -0.02400
5 0.00672 690 0.00582 -0.00336 5 0.048 690 0.04157 -0.02400
6 0.01056 450 0.00000 0.01056 6 0.048 450 0.00000 0.04800
SUM= 0.00000 0.00000 SUM= 0.00000 0.00000
Table: (for Secondary couples) Table: (for Secondary forces)
PlaneCouple/ω2 Angle mrl cosθ mrl sinθ Plane Force/ω2 Angle mr cosθ mr sinθ
mrl (kg m2)
θ
(deg) (kg m2) (kg m2) mr (kg m) θ (deg) (kg m) (kg m)
1 -0.01056 90 0.00000 -0.01056 1 0.048 90 0.00000 0.04800
2 -0.00672 570 0.00582 0.00336 2 0.048 570 -0.04157 -0.02400
3 -0.00288 1050 -0.00249 0.00144 3 0.048 1050 0.04157 -0.02400
4 0.00288 330 0.00249 -0.00144 4 0.048 330 0.04157 -0.02400
5 0.00672 1290 -0.00582 -0.00336 5 0.048 1290 -0.04157 -0.02400
6 0.01056 810 0.00000 0.01056 6 0.048 810 0.00000 0.04800
SUM= 0.00000 0.00000 SUM= 0.00000 0.00000
Analytical Answer
USC=0; USF=0
UPC=0; UPF=0;

Unit2 lectbyvrk dynofmachinery

Unit2 lectbyvrk dynofmachinery

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Unit2 lectbyvrk dynofmachinery