03 Balancing of Machinery Ppt(1).pdf .

Department of Mechanical & Manufacturing Engineering, MIT, Manipal 1 of 44
Chapter 3
Balancing of Machinery

• Introduction
 When man invented the wheel, he very quickly learnt that if it wasn’t
completely round and if it didn’t rotate evenly about it’s central axis,
then he had a problem!
 What the problem he had?
 The wheel would vibrate causing damage to itself and it’s support
mechanism and in severe cases, is unusable.
 A method had to be found to minimize the problem. The mass had to
be evenly distributed about the rotating centerline so that the resultant
vibration was at a minimum.

• Unbalance
 The condition which exists in a rotor when vibratory force or
motion is imparted to its bearings as a result of centrifugal
forces is called unbalance or the uneven distribution of mass
about a rotor’s rotating centerline.

• Balancing
 Balancing is the technique of correcting or eliminating
unwanted inertia forces or moments in rotating or reciprocating
masses and is achieved by changing the location of the mass
centers.
 The objectives of balancing an engine are to ensure:
1. That the center of gravity of the system remains stationery during a
complete revolution of the crank shaft and
2. That the couples involved in acceleration of the different moving
parts balance each other.

• Types of balancing
 Static Balancing:
i) Static balancing is a balance of forces due to action of gravity.
ii) A body is said to be in static balance when its centre of gravity is in the
axis of rotation.
 Dynamic balancing:
i) Dynamic balance is a balance due to the action of inertia forces.
ii) A body is said to be in dynamic balance when the resultant moments
or couples, which involved in the acceleration of different moving parts is
equal to zero.
iii) The conditions of dynamic balance are met, the conditions of static
balance are also met.

• BALANCING OF ROTATING MASSES
 When a mass moves along a circular path, it experiences a
centripetal acceleration and a force is required to produce it.
 An equal and opposite force called centrifugal force acts
radially outwards and is a disturbing force on the axis of
rotation. The magnitude of this remains constant but the
direction changes with the rotation of the mass.
 In a revolving rotor, the centrifugal force remains balanced as
long as the centre of the mass of rotor lies on the axis of
rotation of the shaft.

 When this does not happen, there is an eccentricity and an
unbalance force is produced. This type of unbalance is
common in steam turbine rotors, engine crankshafts, rotors of
compressors, centrifugal pumps etc.

 Balancing involves redistributing the mass which may be
carried out by addition or removal of mass from various
machine members. Balancing of rotating masses can be of
1. Balancing of a single rotating mass by a single mass rotating
in the same plane.
2. Balancing of a single rotating mass by two masses rotating in
different planes.
3. Balancing of several masses rotating in the same plane
4. Balancing of several masses rotating in different planes

• Balancing of a Single Rotating Mass by a Single Mass
Rotating in the Same Plane

Consider a disturbing mass m1 which is attached to a
shaft rotating at ω rad/s.
r = radius of rotation of the mass m
B= M r/ b is the balancing mass which has to be
attached diagonally opposite to M
Where b is radius of rotation of balancing mass B

• Balancing of a Single Rotating Mass by Two Masses
Rotating in Different Planes.
 There are two possibilities while attaching two balancing
masses:
– 1. The plane of the disturbing mass may be in between the planes of
the two balancing masses.
– 2. The plane of the disturbing mass may be on the left or right side of
two planes containing the balancing masses.

• Balancing of Several Masses Rotating in the Same Plane

• Graphical Approach:
Step 1: Draw the space diagram with the positions of the several masses, as
shown.
Step 2: Find out the centrifugal forces or product of the mass and radius of
rotation exerted by each mass.
Step 3: Now draw the vector diagram with the obtained centrifugal forces or
product of the masses and radii of rotation. To draw vector diagram take a
suitable scale. Let ab, bc, cd, de represents the forces Fc1, Fc2, Fc3 and Fc4
on the vector diagram. Draw ‘ab’ parallel to force Fc1 of the space diagram,
at ‘b’ draw a line parallel to force Fc2. Similarly draw lines cd, de parallel to
Fc3 and Fc4 respectively.

Step 4: As per polygon law of forces, the closing side ‘ae’
represents the resultant force in magnitude and direction as
shown in vector diagram.
Step 5: The balancing force is then , equal and opposite to the
resultant force.
Step 6: Determine the magnitude of the balancing mass ( m ) at a
given radius of rotation ( r ).

• Balancing of Several Masses Rotating in Different Planes

 In order to have a complete balance of the several revolving
masses in different planes,
1. the forces in the reference plane must balance, i.e., the resultant force
must be zero and
2. the couples about the reference plane must balance i.e., the resultant
couple must be zero.
 Example:
Consider four masses m1, m2, m3 and m4 attached to the
rotor at radii r1, r2, r3 and r4 respectively. The masses m1, m2, m3
and m4 rotate in planes 1, 2, 3 and 4 respectively.

 Choose a reference plane at ‘O’ so that the distance of the
planes 1, 2, 3 and 4 from ‘O’ are L1, L2 , L3 and L4 respectively.
The reference plane chosen is plane ‘L’. Choose another plane
‘M’ between plane 3 and 4 as shown.
 Plane ‘M’ is at a distance of Lm from the reference plane ‘L’.
The distances of all the other planes to the left of ‘L’ may be
taken as negative( -ve) and to the right may be taken as
positive (+ve).

 The magnitude of the balancing masses mL and mM in planes L
and M may be obtained by following the steps given below.
1. Tabulate the given data as shown after drawing the sketches
of position of planes of masses and angular position of
masses. The planes are tabulated in the same order in which
they occur from left to right.

2. Construct the couple polygon first. (The couple polygon can
be drawn by taking a convenient scale) Add the known
vectors and considering each vector parallel to the radial line
of the mass draw the couple diagram. Then the closing vector
will be ‘mM rM LM’.

3. Now draw the force polygon (The force polygon can be drawn
by taking a convenient scale) by adding the known vectors
along with ‘mM rM’. The closing vector will be ‘mL rL’. This
represents the balanced force.

Exercises
1. A shaft carries 5 masses at positions A, B, C, D & E.
Magnitude of the masses are 20 kg, 25 kg, 30 kg and 24 kg and
20 kg respectively. The masses are at radii of 250 mm, 160 mm,
100 mm, 200 mm ad 300 mm respectively. The angular positions
measured in the anticlockwise direction from A are respectively,
OA - 0°, 45°, 90°, 120° and 240°. Determine the unbalanced
force acting on the spindle at a speed of 240 rpm and calculate
magnitude and angular position of balancing mass to be attached
at a radius of 180 mm.

2. A shaft carries 5 masses at positions A, B, C, D & E. Magnitude of the
masses are 20 kg, 25 kg, 30 kg and 24 kg and 20 kg respectively. The
masses are at radii of 250 mm, 160 mm, 100 mm, 200 mm ad 300 mm
respectively. The angular positions measured in the anticlockwise
direction from A are respectively, OA - 0°, 45°, 90°, 120° and 240°.
Determine the unbalanced force acting on the spindle at a speed of 240
rpm and calculate magnitude and angular position of balancing mass to
be attached at a radius of 180 mm.

3. A rotating shaft carries four unbalanced masses 18 Kg, 14 Kg, 16 Kg,
12 Kg at radii 5 cm, 6 cm, 7cm and 6cm respectively. The 2nd, 3rd and
4th masses revolve in planes 8cms, 16cms and 28cms respectively
measured from the plane of the first mass and are angularly located at
60°, 135° and 270° respectively measured clockwise from the first mass
looking from this mass end of the shaft. The shaft is dynamically
balanced by two masses, both located at 5cm radii and revolving in
planes midway between those of 1st and 2nd masses and midway
between those of 3rd and 4th masses. Determine the magnitudes of the
masses and their respective angular positions.

4. A shaft carries four masses A, B, C and D wh.ch are
placed in parallel planes perpendicular to the longitudinal axis. The
unbalanced masses at planes B and C are 3.6 kg and 2.5 kg
respectively and both are assumed to be concentrated at a radius of 25
mm while the masses in planes A and D are both at radius of 40 mm.
The angle between the planes B and C is 100° and that between B and
A is 190°, both angles being measured in counterclockwise direction
from the plane B. The planes containing A and B are 250 mm apart and
those containing B and C are 500 mm. If the shaft is to be completely
balanced, determine
(i) Masses at the planes A and D.
(ii) The distance between the planes C and D.
(iii) The angular position of mass D.

5. A shaft is attached with four masses A, B, C and D of
magnitudes W N, 300N, 500 N and 400 N at radii of 18, 24,12,15
cms. The planes containing masses B and C are 30 cm apart.
The angle between B and C is 90° and C is vertical. B and C
make 210°and 120° respectively with D when measured in the
same sense. For perfect balance of the shaft find:-
a.) Weight and angular position of mass A.
b.) Positions of planes A and D.

• Balancing of Reciprocating Masses
 Any lack of balance of a engine and other machine is
responsible for dynamic forces which increases the bearing
loads and stresses in the members of the engine or machine.
 These dynamic forces may even set up the unpleasant and
even dangerous vibrations in the members of the machine.
Hence the reciprocating parts should also be as completely
balanced as possible.

Fp =
𝑅
𝑔
𝜔2𝑟 cos 𝜃 +
𝑅
𝑔
𝜔2𝑟
cos 2𝜃
𝑛

• Note:
1) Un-balanced force along the line of stroke is 1 − 𝑐
𝑅
𝑔
𝜔2
𝑟 cos 𝜃.
2) Un-balanced force perpendicular to the line of stroke is 𝑐
𝑅
𝑔
𝜔2
𝑟 sin 𝜃
3) If the balancing weight is required to balance the revolving parts of
W N as well as the forces of the reciprocating weight then B b = Wr +
cRr or B b = (W + cR) r

• COMPLETE BALANCING OF RECIPROCATING PARTS
• Conditions to be fulfilled:
1. Primary forces must balance i.e., primary force polygon is
enclosed.
2. Primary couples must balance i.e., primary couple polygon is
enclosed.
3. Secondary forces must balance i.e., secondary force polygon is
enclosed.
4. Secondary couples must balance i.e., secondary couple polygon
is enclosed.
• Usually, it is not possible to satisfy all the above conditions
fully for multi-cylinder engine. Mostly some unbalanced
force or couple would exist in the reciprocating engines.

1. A single cylinder reciprocation engine has a stroke of 300 mm and Is
running at 250 rpm. The mass of reciprocating parts is 5 kg and mass of
revolving parts at 150 mm radius Is 3.7 kg. If two third of the reciprocating
parts and all the revolving parts are to be balanced, find: a) the balancing
mass required at a radius of 400 mm. b) the residual unbalanced force when
the crank has rotated 60° from top dead centre.
2. A single cylinder horizontal diesel engine has a stroke of 400 mm. length
of the connecting rod is 900 mm. the revolving parts are equivalent to 600 N
at the crank radius. Weight of the piston is 420 N. Weight of the connecting
rod is 330 N. Centre of gravity of connecting rod is 300 mm from the crank
pin. Revolving balancing masses are attached at a radius of 250 mm so that
the whole of revolving parts and half of the reciprocating parts mutt be
balanced. Find the magnitude of balancing masses and un-balanced forces
on the engine when the speed is 1200 rpm.

• Balancing of Multi-cylinder Inline Engines
• Procedure: -
1. Draw end view and elevation of primary crank.
2. Select reference plane.
3. Prepare the tabular column.
4. Draw the primary couple polygon.
5. Draw primary force polygon.
• If there is any un-balance secondary force or couple, all the five
steps given above must be repeated for the secondary
parameters.

• Note:
 When the primary crank rotates through an angle θ from its top
dead centre position (TDC), the corresponding secondary
crank will rotate through 2θ from TDC position.
 The component of closing vector along the line of stroke gives
the values of un-balanced forces.

1. A four crank engine has two outer cranks set out 120° to each other
and their reciprocating masses are each equivalent to a weight of 4000
N. Radius of each crank is 0.3 m. Ratio of the length of connecting rod
to crank radius is four. Speed of the engine is 300 rpm. Distances
between planes of rotation of adjacent cranks are 0.45 m, 0.75 m and
0.45 m. If the engine should be in complete primary balance find
a)The reciprocating mass and angular positions of inner cranks.
b)Calculate maximum un-balanced secondary forces and couples by
taking engine central plane as reference plane.

2. The successive cranks of a five-cylinder in-line engine are 144°
apart. The spacing between cylinder center lines is 400 mm. The
lengths of the crank and the connecting rod are 100 mm and 450 mm
respectively. The reciprocating mass for each cylinder is 20 kg. The
engine speed is 630 rpm. Determine the maximum values of the
primary and secondary forces and couples and the position of the
central crank at which these occur.
3. In a four cylinder in-line IC engine, the mass of reciprocating parts of
cylinder number 1 and 4 are 100 kg and are 60° apart. The crank radius
is 150 mm, length of connecting rod is 450 mm and engine speed is
1200 rpm. Find, the mass of cylinders 2 and 3 when the engine is in
complete primary balance. Also find the unbalance secondary forces
and couple. The cylinders are placed 600 mm apart.

• Balancing of reciprocating masses in a V-Engine
𝐹𝑝𝑣 =
2𝑅
𝑔
𝜔2𝑟 𝑐𝑜𝑠2𝛼 𝑐𝑜𝑠𝜃
𝐹𝑝ℎ =
2𝑅
𝑔
𝜔2𝑟 𝑠𝑖𝑛2𝛼 𝑠𝑖𝑛 𝜃
𝐹𝑠𝑣 =
2𝑅
𝑔
𝜔2
𝑟
𝑛
cos 2 𝛼 𝑐𝑜𝑠2𝜃 𝑐𝑜𝑠𝛼
𝐹𝑠ℎ =
2𝑅
𝑔
𝜔2
𝑟
𝑛
sin 2 𝛼 𝑠𝑖𝑛2𝜃 𝑠𝑖𝑛𝛼

• A twin cylinder V-engine has centre lines of the cylinders at 90° to each
other and connecting rods are connected to a common crank. Weight
of the Reciprocating mass of each cylinder is 15 N. Radius of the crank
is 0.08 m and length of the connecting rod is 0.4 m. Show that the
engine may be balanced for primary forces, by means of a revolving
balancing mass. If the engine is running at 2000 rpm, what is the
maximum value of resultant secondary force?

• Direct and Reverse Crank Method of Balancing
 This method can be used for finding the primary and secondary
forces in a multi cylinder V-engines and radial engines.

• NOTE
i. Primary force due to each direct (or reverse) crank =
𝑅 2
𝑔
𝜔2
𝑟 cosine of angle with the vertical.
ii. Secondary force due to each direct (or reverse) crank =
𝑅
2𝑔
𝜔2 𝑟
𝑛
× cosine of angle with the vertical.
iii. Maximum primary (or secondary) force = (Fdirect + Freverse)
iv. Minimum primary (or secondary) force = (Fdirect – Freverse)

1. In a three cylinder radial engine, all three connecting rods operate on
a single crank. The cylinder center lines are at 120°. Mass of
reciprocating parts of each cylinder is 2 Kg. Crank radius is 80 mms.
Length of connecting rod is 280 mm. Speed of the engine is 2000 rpm.
Determine the following with regard to inertia of reciprocating parts.
a) Balance weight to be attached at 90 mm radius to obtain primary
balance.
b) Magnitude of secondary un-balanced force.

2. The piston of a 60° V-engine have stroke of 0.12 m. The two
connecting rods operate on a common crank pin and each is 0.24 m. If
the mass of reciprocating parts is 1.2 Kg per cylinder and the crank shaft
speed is 3600 rpm, determine the maximum value of primary and
secondary forces.
3. In a five cylinder radial engine the cylinders are equally spaced. Mass
of reciprocating parts per cylinder is 1 Kg. Stroke length is 0.1 m and
length of connecting rod is 0.15 m. When the engine rotates at 3000 rpm
find the maximum un-balanced Primary and secondary forces.

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