Balancing PPT.pptx

EME 305
Dynamics of
Machinery
Dr. M.R.S. Satyanarayana, Professor
GITAM Institute of Technology, GITAM (Deemed to be University)
Gandhi Nagar, Rushikonda, Visakhapatnam-530045, Andhra Pradesh, INDIA.

Module - 3
(Balancing)
Dr. M.R.S. Satyanarayana, Professor

Module 3 – Balancing
▰ Syllabus:
 Static balancing,
 dynamic balancing in single plane and two plane,
 automobile wheel balancing,
 balancing of fan and balancing of linkages,
 effect of balancing on shaking and pin forces.
3
Prof. M.R.S. Satyanarayana, Dept. of Mechanical Engineering, GITAM Institute of Technology, GITAM, Visakhapatnam

▰ Introduction:
 The high speed of engines and other machines is a common phenomenon now-a-
days.
 It is, therefore, very essential that all the rotating and reciprocating parts should be
completely balanced as far as possible.
 If these parts are not properly balanced, the dynamic forces are set up.
 These forces not only increase the loads on bearings and stresses in the various
members, but also produce unpleasant and even dangerous vibrations.
 In this chapter we shall discuss the balancing of unbalanced forces caused by
rotating masses, in order to minimise pressure on the main bearings when an engine
is running.
4

Syllabus – Balancing
▰ Balancing of Rotating Masses:
Whenever a certain mass is attached to a rotating shaft, it exerts some
centrifugal force, whose effect is to bend the shaft and to produce vibrations in
it.
In order to prevent the effect of centrifugal force, another mass is
attached to the opposite side of the shaft, at such a position so as to balance the
effect of the centrifugal force of the first mass.
This is done in such a way that the centrifugal force of both the masses
are made to be equal and opposite.
The process of providing the second mass in order to counteract the
effect of the centrifugal force of the first mass, is called balancing of rotating
masses.
5

The following cases are important from the subject point of
view:
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 different masses rotating in the same plane.
4. Balancing of different masses rotating in different planes.
6

▰ Balancing of a Single Rotating Mass By a Single Mass Rotating in the
Same Plane:
7
 Consider a disturbing mass m1 attached to a
shaft rotating at ω rad/s as shown in Figure.
 Let r1 be the radius of rotation of the mass m1
(i.e. distance between the axis of rotation of the
shaft and the centre of gravity of the mass m1).
 The centrifugal force exerted by the mass m1
on the shaft,
FC1 = m1 ω2r1 …….. (i)

Same Plane:
8
This centrifugal force acts radially outwards
and thus produces bending moment on the
shaft.
In order to counteract the effect of this force, a
balancing mass (m2) may be attached in the
same plane of rotation as that of disturbing
mass (m1) such that the centrifugal forces due
to the two masses are equal and opposite.

Same Plane:
9
Let r2 = Radius of rotation of the balancing mass
m2 (i.e. distance between the axis of rotation of the
shaft and the centre of gravity of mass m2 ).
Centrifugal force due to mass m2,
FC2 = m2 ω2r2 ….. (ii)
By equating equations (i) and (ii),
m1ω2r1 = m2ω2r2 or m1r1 = m2r2

Same Plane:
10
Notes : 1. The product m2.r2 may be split
up in any convenient way. But the radius
of rotation of the balancing mass (m2) is
generally made large in order to reduce
the balancing mass m2.
2. The centrifugal forces are proportional
to the product of the mass and radius of
rotation of respective masses, because ω2
is same for each mass.

▰ Balancing of a Single Rotating Mass By Two Masses Rotating in Different
Planes
If a disturbing mass is balanced by introducing a two balancing masses it
should be in the two different planes, parallel to the plane of rotation of the disturbing
mass, in such a way that they satisfy the following two conditions of equilibrium.
1. The net dynamic force acting on the shaft is equal to zero. This requires that the line
of action of three centrifugal forces must be the same. In other words, the centre of the
masses of the system must lie on the axis of rotation. This is the condition for static
balancing.
2. The net couple due to the dynamic forces acting on the shaft is equal to zero. In other
words, the algebraic sum of the moments about any point in the plane must be zero.
These conditions together give dynamic balancing.
11

Planes
Case 1: When the plane of the disturbing mass lies in between the planes of the two
balancing masses
12
Let l1 = Distance between the planes A and L,
l2 = Distance between the planes A and M,
l = Distance between the planes L and M.
The centrifugal force exerted by the mass m
in the plane A, FC = mω2r
Similarly, the centrifugal force exerted by the
mass m1 in the plane L, FC1 = m1ω2r1
and, the centrifugal force exerted by the mass
m2 in the plane M, FC2 = m2ω2r2

Planes
13
Since the net force acting on the shaft must be equal to zero, FC=FC1+FC2
or mω2r = m1ω2r1 = m2ω2r2 => m.r =m1r1 + m2r2 …… (i)
take moments about P then
FC1l=FCl2 or m1ω2r1 l= mω2r l2 or m1r1l = mrl2 or m1r1 =mrl2/l …… (ii)
take moments about Q then
FC2l=FCl1 or m2ω2r2 l= mω2r l1 or m1r1l = mrl2 or m2r2 =mrl1/l …… (iii)
The equation (i) represents the condition for static balance,
The equations (ii) or (iii) represents the condition for dynamic balance and must be
satisfy to achieve dynamic balance.

Planes –
Case 2: When the plane of the disturbing mass lies on one end of the planes of the
balancing masses
14
Since the net force acting on the shaft must be equal to zero,
FC+FC2 =FC1 or
mω2r + m2ω2r2 = m1ω2r1
=> m.r +m2r2 = m1r1 …… (iv) Static balance

▰ take moments about P then
FC1l=FCl2 or
m1ω2r1 l= mω2r l2 or
m1r1l = mrl2 or m1r1 =mrl2/l …… (v) Dynamic balance
▰ take moments about Q then
FC2l=FCl1 or m2ω2r2 l= mω2r l1 or
m1r1l = mrl2 or m2r2 =mrl1/l …… (vi) Dynamic balance
15

▰ Balancing of Several Masses Rotating in the Same Plane
16
• Consider any number of masses (say four) of
magnitude m1, m2, m3 and m4 at distances of r1, r2,
r3 and r4 from the axis of the rotating shaft.
• Let θ1, θ2, θ3 and θ4 be the angles of these masses
with the horizontal line OX, as shown in Figure.
• Let these masses rotate about an axis through O and
perpendicular to the plane of paper, with a constant
angular velocity of ω rad/s.
• The magnitude and position of the balancing mass
may be found out analytically or graphically

1. Analytical method
• Find out the centrifugal force (or the product of the mass and its radius of
rotation) exerted by each mass on the rotating shaft.
• Resolve the centrifugal forces horizontally and vertically and find their sums,
i.e. H and V .
• The Sum of horizontal components of the centrifugal forces,
∑H = m1 r1 cos θ1 + m2r2 cos θ2 + . .
• and sum of vertical components of the centrifugal forces,
∑V = m2 r2 sin θ2 + m2 r2 sin θ2 + . . . . . .
17

1. Analytical method
• Magnitude of the resultant centrifugal force, FC = ( 𝐻)2 + ( 𝑉)
2
• If is the angle, which the resultant force makes with the horizontal, then
tan = ∑V / ∑H
• The balancing force is then equal to the resultant force, but in opposite
direction.
• Now find out the magnitude of the balancing mass, such that FC = m.r
where m = Balancing mass, and r = Its radius of rotation.
18

19
2. Graphical method
• Draw the space diagram with the positions of
the several masses, as shown in Figure (a)
Space Diagram.
• Find out the centrifugal force exerted by each
mass on the rotating shaft.
• Now draw the vector diagram with the
obtained centrifugal forces, in magnitude and
direction to some suitable scale.

20
2. Graphical method
• As per polygon law of forces, the closing side ae
represents the resultant force in magnitude and
direction, as shown in Figure (b) The balancing force is,
then, equal to the resultant force, but in opposite
direction.
• Now find out the magnitude of the balancing mass (m)
at a given radius of rotation (r), such that
m ω2 r = Resultant centrifugal force
or m.r = Resultant of m1.r1, m2.r2, m3.r3 and m4.r4

▰ Example 1
Four masses m1, m2, m3 and m4 are 200 kg, 300 kg, 240 kg and 260 kg respectively. The
corresponding radii of rotation are 0.2 m, 0.15 m, 0.25 m and 0.3 m respectively and the angles
between successive masses are 45°, 75° and 135°. Find the position and magnitude of the balance
mass required, if its radius of rotation is 0.2 m.
Given : m1 = 200 kg ; m2 = 300 kg ; m3 = 240 kg ; m4 = 260 kg ; r1 = 0.2 m ;r2 = 0.15 m ; r3 = 0.25
m ; r4 = 0.3 m ; θ1 = 0° ; θ2 = 45° ; θ3 = 45° + 75° = 120° ; θ4 = 45° + 75°+ 135° = 255° ; r = 0.2 m
Let m = Balancing mass, and θ = The angle which the balancing mass makes with m1.
Since the magnitude of centrifugal forces are proportional to the product of each mass and its
radius, therefore m1 r1 = 200 x 0.2 = 40 kg-m; m2 r2 = 300 x 0.15 = 45 kg-m;
m2 r2 = 240 x 0.25 = 60 kg-m; m4 r4 = 260 x 0.3 = 78 kg-m;
The problem may, now, be solved either analytically or graphically. But we shall solve the problem
by both the methods one by one.
21

▰ Example 1 Solution
22
Analytical method; The space diagram is shown in Figure
Resolving m1.r1, m2.r2, m3.r3 and m4.r4 horizontally,
∑H = m1 r1 cos θ1 + m2 r2 cos θ2 + m3 r3 cos θ3 + m4 r4 cos θ4
= 40 cos0° + 45 cos45° + 60 cos120° + 78 cos255°
= 40 + 31.8 − 30 − 20.2 = 21.6 kg-m
Now resolving vertically,
∑V = m1 r1 sin θ1 + m2 r2 sin θ2 + m3 r3 sin θ3 + m4 r4 sin θ4
= 40 sin 0° + 45 sin 45° + 60 sin 120° + 78 sin 255°
= 0 + 31.8+ 52 − 75.3 = 8.5 kg-m
tan θ’ = ∑V /∑ H=8.5/ 21.6=0.3935 or θ’=21.48°+1800 = 201.480
Resultant R = ( 𝐻)2 + ( 𝑉)
2
= (21.6)2 + (8.5)2 = 23.2 Kg-m
We know that m.r = R = 23.2 or m = 23.2 / r = 23.2 / 0.2 = 116 kg

23
Graphical method;
Draw the vector diagram from the magnitude and direction of
centrifugal forces, to some suitable scale is shown in Figure.
The closing side of the polygon ae represents the resultant force.
By measurement, we find that ae = 23 kg-m.
The balancing force is equal to the resultant force, but opposite in
direction as shown in Figure.
Since the balancing force is proportional to m.r, therefore
m × 0.2 = vector ea = 23 kg-m or m = 23/0.2 = 115 kg Ans.
By measurement we also find that the angle of inclination of the
balancing mass (m) from the horizontal mass of 200 kg, θ = 201°

▰ Balancing of Several Masses Rotating in Different Planes
When several masses revolve in different planes, they may be transferred to
a reference plane (briefly written as R.P.), which may be defined as the
plane passing through a point on the axis of rotation and perpendicular to it.
The effect of transferring a revolving mass (in one plane) to a reference
plane is to cause a force of magnitude equal to the centrifugal force of the
revolving mass to act in the reference plane, together with a couple of
magnitude equal to the product of the force and the distance between the
plane of rotation and the reference plane.
24

In order to have a complete balance of the several revolving masses in
different planes, the following two conditions must be satisfied :
1. The forces in the reference plane must balance, i.e. the resultant force
must be zero.
2. The couples about the reference plane must balance, i.e. the resultant
couple must be zero.
25

26
Take balancing masses mL and mM in planes
L and M, one of the planes, say L as the
reference plane (R.P.).
Now consider four masses m1, m2, m3 and m4
revolving in planes 1, 2, 3 and 4 respectively
as shown in figure (a).
The distances of all the other planes to the
left of the reference plane may be regarded as
negative, and those to the right as positive.

27
The relative angular end view positions of these masses
are shown in the figure (b). Tabulate the data as shown
in Table.
Plane Mass
(m)
Radius
(r)
m.r Dist. from
Plane L
Couple
(m.r.l)
1
L(R.P)
2
3
M
4
m1
mL
m2
m3
mM
m4
r1
rL
r2
r3
rM
r4
m1.r1
mL.rL
m2.r2
m3.r3
mM.rM
M4.r4
- l1
0
l2
l3
lM
l4
- m1.r1.l1
0
m2.r2.l2
m3.r3.l3
mM.rM.lM
m4.r4.l4

28
 A couple may be represented by a vector drawn
perpendicular to the plane of the couple.
 The couple C1 introduced by transferring m1 to the
reference plane through O is proportional to
m1.r1.l1 and acts in a plane through Om1 and
perpendicular to the paper.
 The vector representing this couple is drawn in the
plane of the paper and perpendicular to Om1 as
shown by OC1 in Figure (c).
 Similarly, the vectors OC2, OC3 and OC4 are
drawn perpendicular to Om2, Om3 and Om4
respectively and in the plane of the paper.

29
The couple vectors as discussed above, are turned
counterclockwise through a right angle for
convenience of drawing as shown in Figure (d).
Now the vectors OC2, OC3 and OC4 are parallel
and in the same direction as Om2, Om3 and Om4,
while the vector OC1 is parallel to Om1 but in
opposite direction.
Hence the couple vectors are drawn radially
outwards for the masses on one side of the
reference plane and radially inward for the masses
on the other side of the reference plane.

30
Now draw the couple polygon as shown in Figure (e).
The vector d’o’ represents the balanced couple. Since the
balanced couple CM is proportional to mM.rM.lM,
therefore
CM=mM.rM.lM =vector d’o’
or mM = vector d’o’ / mM.rM.lM
From this expression, the value of the balancing mass
mM in the plane M may be obtained, and the angle of
inclination of this mass may be measured from Figure
(b).
(e) Couple Polygon

31
Now draw the force polygon as shown in Figure (f).
The vector eo (in the direction from e to o) represents
the balanced force.
Since the balanced force is proportional to mL.rL,
therefore,
mL.rL = vector eo
or mL = vector eo / rL
From this expression, the value of the
balancing mass mL in the plane L may be obtained and
the angle of inclination α of this mass with the
horizontal may be measured from Figure (b).
(f) Force Polygon

▰ Example 2
A shaft carries four masses A, B, C and D of magnitude 200 kg, 300 kg, 400 kg
and 200 kg respectively and revolving at radii 80 mm, 70 mm, 60 mm and 80 mm
in planes measured from A at 300 mm, 400 mm and 700 mm. The angles between
the cranks measured anticlockwise are A to B 45°, B to C 70° and C to D 120°.
The balancing masses are to be placed in planes X and Y. The distance between
the planes A and X is 100 mm, between X and Y is 400 mm and between Y and D
is 200 mm. If the balancing masses revolve at a radius of 100 mm, find their
magnitudes and angular positions.
Given : mA = 200 kg ; mB = 300 kg ; mC = 400 kg ; mD = 200 kg ; rA = 80 mm
= 0.08m ; rB = 70 mm = 0.07 m ; rC = 60 mm = 0.06 m ; rD = 80 mm = 0.08 m ;
rX = rY = 100 mm = 0.1 m
32

33
Graphical method;
Let mX = Balancing mass placed in plane X,
and mY = Balancing mass placed in plane Y.
The position of planes and angular position
of the masses (assuming the mass A as
horizontal) are shown in Figure (a) and (b)
respectively.
Assume the plane X as the reference plane
(R.P.). The distances of the planes to the
right of plane X are taken as + ve while the
distances of the planes to the left of plane X
are taken as – ve.

34
The data may be tabulated as shown in Table
Plane
(1)
Mass
(m)Kg
(2)
Radius
(r)
(3)
m.r
Kg-m
(4)
Dist. from
Plane X
(5)
Couple (m.r.l)
Kg-m2
(6)
A
X(R.P)
B
C
Y
D
200
mX
300
400
mY
200
0.08
0.1
0.07
0.06
0.1
0.08
16
0.1 mX
21
24
0.1mY
16
- 0.1
0
0.2
0.3
0.4
0.6
- 1.6
0
4.2
7.2
0.04 mM
9.6

35
Graphical method;
Let mX = Balancing mass placed in plane X,
and mY = Balancing mass placed in plane Y.
The position of planes and angular position
of the masses (assuming the mass A as
horizontal) are shown in Figure (a) and (b)
respectively.
Assume the plane X as the reference plane
(R.P.). The distances of the planes to the
right of plane X are taken as + ve while the
distances of the planes to the left of plane X
are taken as – ve.

36
• The procedure of graphical determination of the balancing
masses mX and mY and their angular positions is given
here.
• draw the couple polygon from the data given in Table
(column 6) as shown in Figure (c) to some suitable scale.
• The vector d’o’ represents the balanced couple.
• Since the balanced couple is proportional to 0.04 mY,
therefore by measurement.
0.04 mY = vector d’o’ = 7.3 kg-m2 or mY = 182.5 kg

37
The angular position of the mass mY is
obtained by drawing OmY in Figure (b),
parallel to vector d’o’ .
By measurement, the angular position
of mY is θY = 12° in the clockwise
direction from mass mA (i.e. 200 kg ).

38
• draw the force polygon from the data given in Table
(column 4) as shown in Figure (d). The vector eo
represents the balanced force. Since the balanced
force is proportional to 0.1 mX, therefore by
measurement,
0.1mX = vector eo = 35.5 kg-m or mX = 355 kg
• The angular position of the mass mX is obtained by
drawing OmX in Figure (b), parallel to vector eo.
• By measurement, the angular position of mX is θX =
145° in the clockwise direction from mass mA (i.e.
200 kg ). Ans.

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