MECHANICS OF MACHINES PART 3

BALANCING

1.1 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.

When an unbalanced system is rotating, periodic linear and/or torsional

forces are generated which are perpendicular to the axis of rotation. The
periodic nature of these forces is commonly experienced as vibration. These
off-axis vibration forces may exceed the design limits of individual
machine elements, reducing the service life of these parts. For instance, a
bearing may be subjected to perpendicular torsion forces that would not
occur in a nominally balanced system, or the instantaneous linear forces may
exceed the limits of the bearing. Such excessive forces will cause failure in
bearings in short time periods. Shafts with unbalanced masses can be bent
by the forces and experience fatigue failure.

Under conditions where rotating speed is very high even though the mass is
low, as in gas turbines or jet engines, or under conditions where rotating
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speed is low but the mass is high, as in ship propellers, balance of the
rotating system should be highly considered, because it may generate large
vibrations and cause failure of the whole system.

Balancing of rotating bodies is important to avoid vibration. In heavy

industrial machines such as gas turbines and electric generators, vibration
can cause catastrophic failure, as well as noise and discomfort. In the case of
a narrow wheel, balancing simply involves moving the center of gravity to
the center of rotation. For a system to be in complete balance, both force and
couple polygons should be close in order to prevent the effect of centrifugal
force. It is important to design the machine parts wisely so that the
unbalance is reduced up to the minimum possible level or eliminated
completely.

In this topic, we shall discuss the balancing of unbalanced forces caused by

rotating masses, in order to minimize pressure on the main bearings when an
engine is running.

1.2 Balancing of Rotating Masses

Often an unbalance of force is produced in rotary or reciprocating

machinery due to the inertia forces associated with the moving masses.
Whenever a 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
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centrifugal forces 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.
Balancing is, therefore, the process of designing or modifying machinery so
that the unbalance is reduced to an acceptable level and if possible is
eliminated entirely.

Fig. 1.1: Rotor with Unbalanced Mass

 A particle or mass moving in a circular path experiences a centripetal

acceleration and a force is required to produce it. An equal and
opposite force acting radially outwards acts on the axis of rotation and
is known as centrifugal force [Fig. 1.1(a)]. This is a disturbing force
on the axis of rotation, the magnitude of which is 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 the rotor lies on the axis of the shaft. When
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the centre of mass does not lie on the axis or there is an eccentricity,
an unbalanced force is produced.

 [Fig. 1.1(b)]. This type of unbalance is very common. For example, in

steam turbine rotors, engine crankshafts, rotary compressors and
centrifugal pumps.

 Most of the serious problems encountered in high-speed machinery

are the direct result of unbalanced forces. These forces exerted on the
frame by the moving machine members are time varying, impart
vibratory motion to the frame and produce noise. Also, there are
human discomfort and detrimental effects on the machine
performance and the structural integrity of the machine foundation.

 The most common approach to balancing is by redistributing the mass

which may be accomplished by addition or removal of mass from
various machine members.

 There are two basic types of unbalance, namely:

(i) rotating unbalance; and,

(ii) reciprocating unbalance – which may occur separately or in

combination.

Static Balancing:

A system of rotating masses is said to be in static balance if the combined

mass centre of the system lies on the axis of rotation.
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Static balance occurs when the center of gravity of an object is on the axis of
rotation. The object can therefore remain stationary, with the axis horizontal,
without the application of any braking force. It has no tendency to rotate due
to the force of gravity. This is seen in bike wheels where the reflective plate
is placed opposite the valve to distribute the center of mass to the center of
the wheel. Other examples are grindstones, discs or car wheels. Verifying
static balance requires the freedom for the object to rotate with as little
friction as possible.

This may be provided with sharp, hardened knife edges, adjusted to be both
horizontal and parallel. Alternatively, a pair of free-running ball bearing
races is substituted for each knife edge, which relaxed the horizontal and
parallel requirement. The object is either axially symmetrical like a wheel or
must be provided with an axle. It is slowly spun, and when it comes to rest,
it will stop at a random position if statically balanced. If not, an adhesive or
clip on weight is securely attached to achieve balance.

Dynamic Balance

A rotating system of mass is in dynamic balance when the rotation does not
produce any resultant centrifugal force or couple. The system rotates without
requiring the application of any external force or couple, other than that
required to support its weight. If a system is initially unbalanced, to avoid
the stress upon the bearings caused by the centrifugal couple, counter
balancing weights must be added.
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This is seen when a bicycle wheel gets a buckled rim. The wheel will not
rotate to a preferred position but because some rim mass is offset there is a
wobbling couple leading to a dynamic vibration. If the spokes on this wheel
cannot be adjusted to center the rim, an alternative method can be used to
provide dynamic balance.

1.4 Types of Balancing:

There are main two types of balancing conditions

(i) Balancing of rotating masses

(ii) Balancing of reciprocating masses

(i) 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 forces 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.

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.
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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.

We shall now discuss these cases, in detail, as follows.

1.5 Balancing of a single rotating mass by a single mass rotating in the

same plane

A disturbing mass m1 attached to a shaft rotating at ω rad/s as shown in

Fig.1.2. 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)

We know that the centrifugal force exerted by the mass m1 on the

shaft is given by:

FC1 = m1. ω 2
. r1 Eqn. (1)

This centrifugal force acts radically outwards and thus produces bending
moments on the shaft. In order to counteract the effect of this force a
balancing mass (m2 such that the centrifugal force due to the two masses are
equal and opposite.
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Fig. 1.2: Balancing of a Single Rotating Mass by a Single Mass Rotating

in the Same Plane.

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)

Therefore the centrifugal force due to the mass m2 is given by:

2
FC2 = m2. ω . r2 ………………Eqn. (2)

Equating equations (1) and (2), we get:

m1.ω2.r2 = m2.ω2.r2

or, m1.r1 = m2.r2 …………………….Eqn. (3)

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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
2
radius of rotation of respective masses, because ω is the same for
each mass.

1.6 Balancing of a Single Rotating Mass by Two Masses Rotating in

Different Planes

By introducing a single balancing mass in the same plane of rotation as that

of disturbing mass, the centrifugal force is balanced. In other words, the two
forces are equal in magnitude and opposite in direction. But this type of
arrangement for balancing gives rise to a couple which tends to rock the
shaft it its bearings. Therefore, in order to put the system in complete
balance, two balancing masses are placed in 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.
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The conditions (1) and (2) together give dynamic balancing. The following
two possibilities may arise while attaching the two balancing masses;

(i) The plane of the disturbing mass may be in between the planes of the
two balancing masses; and,
(ii) The plane of the disturbing mass may lie on the left or right of the two
planes containing the balancing masses.

We shall now discuss both the above cases one by one.

1.7 When the plane of the disturbing mass lies in between the planes of
the two balancing masses.

Consider a disturbing mass m lying in plane A to be balanced by two

rotating masses m1 and m2 lying in two different planes L and M as shown in
Fig. 1.3.

Let,

r, r1 and r2 be the radii of rotation of the masses in planes A, L and M

respectively;

l1 = distance between the planes A and L,

l2 = distance between the planes A and M,

L = distance between the planes L and M.

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Fig. 1.3: Balancing of a Single Rotating Mass by Two Rotating Masses

in Different Planes when the Plane of Single Rotating Mass Lies in
Between the Planes of Two Balancing Masses

We know the centrifugal force exerted by the mass m in the plane A, is

given by:

FC= m.ω2.r…………………..……. Eqn. (4)

Similarly, the centrifugal force exerted by the mass m1 in the plane L, is

given by:

FC1 = m1. ω2. r1 ………………..…. Eqn. (5)

And, the centrifugal force exerted by the mass m2 in the plane M, is given
by:

FC2 = m2.ω2. r2…………………….. Eqn. (6)

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Since the net force acting on the shaft must be equal to zero, therefore the
centrifugal force on the disturbing mass must be equal to the sum of the
centrifugal forces on the balancing masses. Hence,

FC = FC1 + FC2

Or, m.ω 2.r = m1.ω 2.r1 + m2.ω 2.r2 ……………Eqn (7)

Therefore,

m.r = m1.r1 + m2.r2 ………………….…Eqn (8)

In order to find the magnitude of balancing force in the plane L (or the
dynamic force at the bearing Q of the rotating shaft), we take moments about
P which is the point of intersection of the plane M and the axis of rotation.

Therefore,

FC1*l = FC*l2

Or, m1.ω2.r1*l = m.ω2.r*l2 ……………Eqn (9)

Therefore,

m1.r1.l = m.r.l2

or, m1.r1 = m.r*l2/l …………… Eqn (10)

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Similarly, in order to find the balancing force in plane M (or the dynamic
force at the bearing P of a shaft), we take moments about plane Q, which is
the point of intersection of the plane L and the axis of rotation.

Therefore,

FC2*l = FC*l1

Or, m2.ω 2.r2*l = m.ω2.r*l1

Therefore,
m2.r2.l = m.r.l1
or, m2.r2 = m.r*l1/l …………… (11)

It may be noted that equation (7) represents the condition for static balance,
but in order to achieve dynamic balance, equations (10) and (11) must also
be satisfied.
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1.8 When the plane of the disturbing mass lies on one end of the planes
of the balancing masses

Fig. 1.4. Balancing of a single rotating mass by two rotating masses in

different planes, when the plane of single rotating mass lies at one end of
the planes of balancing masses.

In this case, the mass m lies in the plane A and the balancing masses lie in
the planes L and M as shown in fig. 1.4. As already discussed above, the
following conditions must be satisfied in order to balance the system, i.e.

FC + FC2 = FC1

Or, m.ω2.r+m2.ω2.r2 = m1.ω2.r1

Therefore,

m.r + m2.r2 = m1.r1 ………………(12)

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Now, to determine the balancing force in the plane L (or the dynamic force
at the bearing Q of a shaft), we take moments about P which is the point of
intersection of the plane M and the axis of rotation.

Therefore,

FC1*l = FC*l2

Or, m1.ω2.r1*l = m.ω2.r*l2

m1.r1.l = m.r.l2

Or, m1.r1 = m.r*l2/l……………… (13)

Eqn. (13) is the same as equation (6)

Similarly, to determine the balancing force in the plane M (or the dynamic
force at the bearing P of a shaft), we take moments about Q, which is the
point of intersection of the plane L and the axis of rotation.

Therefore,

FC2*l = FC*l1

Or, m2.ω2.r2*l = m.ω2.r*l2

Or, m2.r2.l = m.r.l1

or m2.r2 = m.r*l1/l………… (14)

This equation is the same as equation (11)

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1.9 Balancing of Several Masses Rotating in the Same Plane

Consider any number of masses, say four, of magnitude m1, m2, m3 and m4 at
distances r1, r2, r3 and r4 from the axis of the rotating shaft. Let θ 1, θ 2,
θ 3, and θ 4, be the angles of these masses with the horizontal line OX, as
shown in Fig. 1.5 (a). 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.

(a) Space diagram. (b) Vector diagram.

Fig. 1.5: Balancing of several masses rotating in the same plane.

This problem can be solved either analytically or graphically. Since

graphical method is simpler, we shall it.
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The magnitude and position of the balancing mass may be found out
graphically as discussed below:

 First of all, draw the space diagram with the positions of the several
masses, as shown in Fig. 1.5 (a).

 Find out the centrifugal force (or product of the mass and radius of
rotation) exerted by each mass on the rotating shaft.

 Then draw the vector diagram with the obtained centrifugal forces (or
the product of the masses and their radii of rotation), such that length
ab represents the centrifugal force exerted by the mass m1 (or m1.r1)
in magnitude and direction to some suitable scale. Similarly, draw
lengths bc, cd and de to represent centrifugal forces of other masses
m2, m3 and m4 (or m2.r2, m3.r3 and m4.r4).
Now, as per polygon law of forces, the closing side ae represents the
resultant force in magnitude and direction, as shown in Fig. 1.5 (b).

 The balancing force is, then, equal to 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.r.ω2 = Resultant centrifugal force

or m.r = Resultant of m1.r1, m2.r2, m3.r3 and m4.r4

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 (In general for graphical solution, vectors m 1.r1, m2.r2, m3.r3, m4.r4,
etc., are added. If they close in a loop, the system is balanced.
Otherwise, the closing vector will be giving mc.rc. Its direction
identifies the angular position of the countermass relative to the other
mass.)

Example 1
Four masses m1, m2, m3 and m4 are 200kg, 300kg, 240kg and 260kg
respectively. The corresponding radii of rotation are 0.2m, 0.15m, 0.25m
and 0.3m respectively and the angles between successive masses are 450, 750
and 1350. Determine the position and magnitude of the balance mass
required, if its radius of rotation is 0.2m.

Solution

Given: m1 = 200kg; m2 = 300kg; m3 = 240kg; m4 = 260kg;

r1 = 0.2m; r2 = 0.15m; r3 = 0.25m; r4=0.3m;

θ1 = 00; θ = 450; θ = 450+750 = 1200; θ = 450+750+1350 =2550; r = 0.2m

2 3 4

Let m = Balancing mass and

θ = The angle which the balancing mass makes with
the reference plane.

Since the magnitude of centrifugal forces are proportional to the product of

each mass and its radius, therefore
m1.r1 = 200*0.2 = 40kg-m
m2.r2 =300*0.15 = 45kg-m
m3.r3 = 240*0.25 = 60kg-m
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m4.r4 = 260*0.3 = 78kg-m

The problem may, now, be solved either analytically or graphically. But we
shall solve the problem graphically.
Solution

Fig. 1.6: Angular position of masses

Solution

Using graphical method, the magnitude and the position of the balancing
mass may also be found as shown below:

1. First of all, the space diagram is drawn showing the positions of all
the given masses as shown in Fig. 1.7 (a).
2. Since the centrifugal force of each mass is proportional to the product
of the mass and radius, therefore

m1.r1 = 200*0.2 = 40 kg-m

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m2.r2 = 300*0.15 = 45 kg-m

m3.r3= 240*0.25 = 60 kg-m

m4.r4 = 260*0.3 = 78 kg-m

3. Now draw the vector diagram with the above values, to some suitable
scale, as shown in fig. 17 (b). The closing side of the polygon ae
represents the resultant force. By measurement, we find that ae =
23kg-m.

Fig. 1.7

4. The balancing force is equal to the resultant force. Since the balancing
force is proportional to m.r, therefore,

m× 0.2 = vector ea= 23 kg-m

or, mC = 23/0.2

mC = 115 kg.
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By measurement we also find that the angle of inclination of the balancing

mass (m) from the horizontal or positive X-axis, is given by,

θC = 201°.

2.0 Dynamic Balancing

When several masses rotate in different planes, the centrifugal forces, in

addition to being out of balance, also form couples.

Note:

A couple, in mechanics, is a pair of equal parallel forces that are opposite

in direction. A couple consists of two parallel forces that are equal in
magnitude, opposite in sense and do not share a line of action. It does not
produce any translation, only rotation. The turning effect, or moment, of a
couple is measured by the product of the magnitude of either force and the
perpendicular distance between the action lines of the forces. The resultant
force of a couple is zero. BUT, the resultant of a couple is not zero; it is a
pure moment.

These masses revolve in different planes, and 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. As
shown in Fig. 18, the effect of transferring a revolving mass (in one plane) to
a reference plane is to create a force equal to:

l.m1.r1.ω2
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This force has a magnitude equal to the centrifugal force m1.r1.ω2 of the
revolving mass, m1, 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. In order to have a complete
balance of the several revolving masses in different planes, the following
two conditions must be satisfied:

(a) All the forces in the reference plane must balance, i.e. the resultant
force must be zero.
(b) The couples about the reference plane must balance, i.e. the resultant
couple must be zero.

Therefore, a system of rotating masses is in dynamic balance when the

resultant centrifugal force as well as resultant couple does not exist in the
system.

In the analysis that follows, the mr and mrl products (rather that mrω2 and
mrlω2 values), usually referred to as force and couple, respectively are used
to draw force and couple polygons. These quantities are used since it is more
convenient to do so.

Fig. 1.8 shows two masses m1, and m2 masses revolving diametrically
opposite each other in different planes such that,

m1r1 = m2r2.

The centrifugal forces are balanced, but an unbalanced couple of magnitude,

m1r1l (= m2r2l) is introduced.

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The couple acts in a plane that contains the axis of rotation and the two
masses. Thus, the couple is of constant magnitude but variable direction.

Fig. 1.8

Let us now consider four masses m1, m2, m3 and m4 revolving in planes 1, 2,
3 and 4 respectively as shown in fig. 1.9(a). The relative angular positions of
these masses are shown in the end view Fig. 1.9 (b).
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Fig. 1.9:

The magnitude of the balancing masses mL and mM in planes L and M may

be obtained as discussed below:

(i) Take one of the planes, say L as the reference plane (R.P.). 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.
(ii)Tabulate the data as shown in table 1. The planes are tabulated in the
same order in which they occur, reading from left to right.

Table 1:

Plane Mass (m) Radius (r) Cent.force Distance Couple÷ ω2

÷ ω2 (m.r) from plane (m.r.l)
L (l)
(1) (2) (3) (4) (5) (6)

1 m1 r1 m1.r1 -l1 -m1.r1.l1

L(R.P.) mL rL mL.rL 0 0
2 m2 r2 m2.r2 l2 m2.r2.l2
3 m3 r3 m3.r3 l3 m3.r3.l3
M mM rM mM.rM lM mM.rM.lM
4 m4 r4 m4.r4 l4 m4.r4.l4
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(iii) 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 fig. 1.9 (c). Similarly, the
vectors OC2, OC3 and OC4 are drawn perpendicular to Om2, Om3 and
Om4 respectively and in the plane of the paper.

Fig. 1.9: Balancing of several masses rotating in different planes

(iv) The couple vectors as discussed above, are turned clockwise

through a right angle for convenience of drawing as shown in fig. 1.9
(d). We see that their relative positions remain unaffected. Now the
vectors OC2, OC3 and OC4 are 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.
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(v) Now draw the couple polygon as shown in fig. 1.9. (e).The vector ḋ ȯ
represents the balanced couple. Since the balanced couple CM is
proportional to mM.rM.lM, therefore,

C M =m M .r M .l M =vector ḋ ȯ

vector ḋ Ȯ
Or, mM =
r M . lM

(vi) 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 fig. 1.9 (b).
(vii) Now draw the force polygon as shown in fig. 1.9 (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,
vector eo
m L .r L =vector eo∨m L =
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 fig. 1.9 (b).

2.1 Balancing of Several Masses Rotating in the different

Planes

Let there be a rotor revolving with a uniform angular velocity ω. m1, m2 and
m3 are the masses attached to the rotor at radii r1, r2 and r3 respectively. The
masses m1, m2 and m3 rotate in planes 1, 2 and 3 respectively.
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First, a reference plane at O is chosen so that the distances of the planes 1, 2

and 3 from O are l1, l2 and l3 respectively.

Second, transference of each unbalanced force to the reference plane

introduces the like number of forces and couples.

Hence,

(i) the unbalanced forces in the reference planes are m1r1ω2, m2r2ω2 and
m3r3ω2 acting radially outwards, and,
(ii) the unbalanced couples in the reference plane are m1r1ω2l1, m2r2ω2l2
and m3r3ω2l3 which may be represented by vectors parallel to the
respective force vectors, i.e., parallel to the respective radii of m1, m2
and m3.

For complete balancing of the rotor, the resultant force and resultant couple
both should be zero, i.e.,

m1r1ω2 + m2r2ω2 + m3r3ω2 = 0…….…………………. (15) and,

m1r1ω2l1 + m2r2ω2l2 + m3r3ω2l3 = 0…....……………… (16)

If the Eqs (15) and (16) are not satisfied, then there are unbalanced forces
and couples. A mass placed in the reference plane may satisfy the force
equation but the couple equation is satisfied only by two equal forces in
different transverse planes.
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Example 3:

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.

Solution:

mA = 200 kg rA = 80 mm θA = 0° lA = -100 mm

mB = 300 kg rB= 70 mm θB = 45° lB = 200 mm

mC = 400 kg rC = 60 mm θC = 45° +70° = 115° lC = 300 mm

mD = 200 kg rD = 80 mm θD = 115° + 120° = 235° lD = 600 mm

rX = rY = 100 mm lY = 400 mm

mX = Balancing mass placed in plane X, and

Let
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 Fig. 2.0 (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.
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(a) Position of planes (b) Angular position of masses

Fig. 2.0

mArAlA = 200 x 0.08 x (-0.1) = -1.6 kg.m2 mArA = 200 x 0.08 = 16 kg.m

mBrBlB = 300 x 0.07 x 0.2 = 4.2 kg.m2 mBrB = 300 x 0.07 = 21 kg.m
mCrClC = 400 x 0.06 x 0.3 = 7.2 kg.m2 mCrC = 400 x 0.06 = 24 kg.m
mDrDlD = 200 x 0.08 x 0.6 = 9.6 kg.m2 mDrD = 200 x 0.08 = 16 kg.m

Using graphical method, the balancing masses mX and mY and their angular
positions may be determined graphically as discussed below:
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Table 2

Mass (m) Radius Cent.force ÷ ω2 Distance from Couple ÷ ω2

Plane Angle
kg (r)m (mr) kg-m Ref. Plane (l) m (mrl) kg-m2
A 0° 200 0.08 16 – 0.1 –1.6
X (R.P.) θX mX 0.1 0.1 mX 0 0
B 45° 300 0.07 21 0.2 4.2
C 115° 400 0.06 24 0.3 7.2
Y θY mY 0.1 0.1 mY 0.4 0.04 mY
D 235° 200 0.08 16 0.6 9.6

(i) First of all, the couple polygon is drawn from the data given in Table 2
(column 7) as shown in Fig. 2.1 (a) 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,

0.04mY = vector d′o′ = 73 kg-m2

Or, mY = 182.5 kg

The angular position of the mass mY is obtained by drawing OmY in Fig. 2.0
(b), parallel to vector d′o′.

By measurement, the angular position of mY is given by,

θY = 12° in the clockwise direction from mass mA (i.e. 200 kg).

Therefore,

θY = 360o - 12° = 348o

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(a) Couple Polygon (b) Force Polygon

Fig. 2.1

(iii) Now draw the force polygon from the data given in Table 2 (column
5) as shown in Fig. 2.1 (b). 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 Fig. 2.0
(b), parallel to vector eo.
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By measurement, the angular position of mX is θX = 145° in the clockwise

direction from mass mA (i.e. 200 kg), so that:

θX = 360°– 145° = 215°.