Theory of Machines : Balancing | Notes of Mechanical Engineering

1. BALANCING OF ROTATING MASSES

In any rotating system, having one or more rotating masses, if the centre of mass of the system does not lie on the axis of rotation the system is unbalanced.

When an unbalanced mass rotating about an axis it experiences a centrifugal force in a radially outward direction. This force is called as disturbing force of the system.

The magnitude of disturbing force is F_c = mrω²

where,       F_c = centrifugal force or disturbing force, N

m = mass of the rotating body, kg.

r = distance of centre of mass (C.G) from the axis of rotation, m

ω = angular speed of rotation, rad/s.

1. BALANCING OF SINGLE MASS ROTATING IN SINGLE PLANE

Many times, one or several masses are rotating in a single plane. The examples of such cases are: steam turbine rotors, impellers of centrifugal pumps, impellers of hydraulic turbines, etc.

                                                                                       Fig.1: Single Unbalanced Rotating Mass

During the rotation of shaft, a dynamic force (centrifugal force) equal to mrω² acts in a radially outward direction as shown in the above figure.

This unbalanced force results in increase in load on the bearings, increased bending moment on the shaft and vibrations of the system.

This dynamic force can be balanced by either of the following two methods.

• Balancing by single mass rotating in the same plane (Internal balancing)
• Balancing by two masses rotating in two different planes (External balancing)

3.1 Balancing by single mass rotating in the same plane (Internal balancing)

To balance the single rotating mass a counter mass or balancing mass ‘m_b’ is placed in the plane of rotation of the disturbing mass at a radius ‘r_b’ and exactly opposite to it, such that the centrifugal force due to the two masses are equal and opposite.

Mathematically,

mrω² = m_br_bω²        or     mr = m_br_b

This balancing of a disturbing mas by a single balancing mass in the same plane is known as Internal balancing.

3.2 Balancing by two masses rotating in different planes (External balancing):

If the balancing mass cannot be placed in the plane of rotation of the disturbing mass, then it is not possible to balance the disturbing mass by a single balancing mass.

If a single balancing mass is placed in a plane parallel to the plane of rotation of disturbing mass, the dynamic force (centrifugal force) can be balanced. However, this arrangement will introduce an unbalanced couple.

To achieve the complete balancing of the system, at least two balancing masses are required to be placed in two planes parallel to the plane of disturbing mass.

To achieve the complete balancing, two balancing masses ‘m_b1’ and ‘m_b2’ 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:

• The resultant dynamic force acting on the shaft must be equal to zero. For this, the line of action of three dynamic forces must be the same. This is the condition for static balancing.
• The resultant couple due to dynamic forces acting on the shaft must be equal to zero. In other words, the algebraic sum of the moments due to dynamic forces about any point in the plane must be zero. This is the condition for dynamic balancing.

1. BALANCING OF SEVERAL MASSES ROTATING IN SAME PLANE

If the combined mass centre of the system lies on the rotational axis then it is called as in static balance.

Consider masses attached in the same plane. Due to rotation, unbalanced force is mrω² in each mass

Fig. 2: Balancing of several masses rotating in same plane

Total unbalance force

  F = m₁r₁ω² + m₂r₂ω² + m₃r₃ω²

Here all forces are in same plane. Place another mass m_c at r_c with θ_c angle with respect to m₁, such that resultant force becomes zero.

⇒ F + m_cr_cω² = 0  

⇒ ∑mrω² + m_cr_cω² = 0

⇒ ∑mr + m_cr_c = 0

The above equation can be solved either mathematically or graphically.

Mathematical Solution

Divide each force into its x and y components,

                                                                     i.e.    ∑ mrcosθ + m_cr_ccos θ_c = 0     ⇒ m_cr_c cosθ_c = –∑mrcosθ       

and   ∑ mrsinθ + m_cr_csin θ_c = 0      ⇒ m_cr_csin θ_c = –∑mrsinθ      

1. BALANCING OF SEVERAL MASSES ROTATING IN DIFFERENT PLANE

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. Choose a Reference plane at O so that the distances of the planes 1, 2 and 3 from O are l₁, l₂ and l₃ respectively.

Fig. 3: (a) Balancing of several masses rotating in same plane,

(b) System with counter mass/balancing mass,

(c) Moment Polygon, (d) Force Polygon

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

1. BALANCING OF RECIPROCATING MASS

Almost all IC engines use reciprocating engines (slider-crank mechanism), it would produce reciprocating unbalance force.

Since acceleration in reciprocating engine is,

Inertia force produced due to this acceleration,

Here F_P = mrω²cosθ is primary unbalanced force acting along line of a stroke of the cylinder. It is due to SHM of parts. Maximum value of the primary force mrω^2.

Here

F_S is secondary unbalanced force also acting along line of a stroke of the cylinder. It is due to obliquity of arrangement.

6.1 Partial balancing: 

To minimize the effect of the unbalanced force, a compromise is, usually, made i.e. 2/3 of the reciprocating mass is balanced (or a value between one-half and three quarters). If c is fraction of the reciprocating mass that balanced, then

Primary force balanced by the mass = cmrω²cosθ

Primary force unbalanced by the mass = (1 – c)cmrω²cosθ

Unbalanced vertical force component = cmrω²sinθ

Resultant unbalanced force at any instant R

7. EFFECT OF PARTIAL BALANCING OF LOCOMOTIVES

In partial balancing of reciprocating engine there are two unbalanced forces acting on engine.

• Unbalanced force along the line of stroke F_H.
• Unbalanced force perpendicular to the line of stoke F_v.

The effect of F_H is to produce the variation of Tractive Force along the line of stroke and couple of such force is known as Swaying Couple.

The effect F_V, is to produce the variation of pressure on the rails which cause hammering action on the rails known as hammer blow.

7.1 Variation of tractive force (F_T):

A variation in the attractive force (effort) of an engine is caused by the unbalanced portion of the primary force which acts along the line of stroke of a locomotive engine.

The tractive force is maximum or minimum when

 θ = 135° or 315°

7.2 Swaying Couple (C_s):

Unbalanced primary forces along the lines of stroke are separated by a distance apart and thus, constitute a couple. This tends to make the leading wheels sway from side to side.

Fig.4: Swaying Couple

The swaying couple tends to sway the engine alternately in clockwise and anticlockwise direction about vertical axis.

Swaying couple = moments of forces about the engine centre line

The swaying couple is maximum or minimum when

θ = 45° or 225°

8. Hammer Blow:

The unbalanced force perpendicular to line of stroke produces the variation of pressure on the rails which causes hammering action on the rails which is called as hammer blow.

We know that unbalanced force along the perpendicular to the line of stroke due to balancing mass m_b at radius r_b is m_bω²r_b sin θ. This force is maximum when sinθ is unity i.e., when θ = 90° or 270°, Therefore F_VU(max) = m_bω²r_b.

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