Balancing of a Single Rotating Mass By Two Masses Rotating in Different Planes

          In the previous article we have discussed that by introducing a single balancing mass in the same plane of rotation as that of disturbing mass, the centrifugal forces are balanced. These 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 in its bearing. 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 equal to zero. In other words, the algebraic sum of the moments about any point in the plane must be zero.

          The above conditions (1) and (2) together gives Dynamic Balancing. The following two possibilities may arise while attaching the two balancing masses:

1. The plane of the disturbing mass may be in between the planes of the two balancing masses, and

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

          Consider a disturbing mass 'm' lying in a plane 'A' to be balanced by two rotating masses m₁ and m₂ lying in two different planes 'L' and 'M' as shown in Fig., Let r, r₁ and r₂ be the radii of rotation of the masses in planes A, L and M respectively.

           Let        l₁ = Distance between the planes A and L,

                         l₂ =  Distance between the planes A and M, and

                         l = Distance between the planes L and M.

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

                                   F_C = m . r . ω²

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

                                   F_{C 1} = m₁ . r₁ . ω²

     and, the centrifugal force exerted by the mass 'm2' in the plane 'M', 

                                   F_C2 = m₂ . r₂ . ω²

          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, therefore

                                                F_C = F_{C 1} + F_C2

                                             m . r = m₁ . r₁ + m₂ . r_{2      ..... (1)}

          Now in order to find the magnitude of balancing force in the plane 'L', take moments about 'P' which is the point of intersection of the plane 'M' and the axis of rotation. Therefore

                                               F_{C 1} X l = F_C X l₂

                                             m₁ . r₁ . l = m . r . l_{2             ...... (2)}

          Similarly, in order to find the balancing force in plane 'M', take moments about 'Q' which is the point of intersection of the plane 'L' and the axis of rotation. Therefore

                                               F_C2 X l = F_C X l₁

                                           m₂ . r₂ . l = m . r . l_{1              ........ (3)}

The equation (1) represents the condition for static balance, but in order to achieve dynamic balancing, equations (2) and (3) must also be satisfied.

2. When the plane of the disturbing mass lies on one end of the planes of the 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., As discussed above, the following conditions must be satisfied in order to balance the system, i.e., 

                                              F_C + F_C2 = F_{C 1} + F_C2

                                     m . r + m₂ . r₂ = m₁ . r₁ + m₂ . r_{2        ..... (4)}

          Now, to find the balancing force in the plane 'L', take moments about 'P' which is the point of intersection of the plane 'M' and the axis of rotation. Therefore

                                               F_{C 1} X l = F_C X l₂

                                             m₁ . r₁ . l = m . r . l_{2             ...... (5)}

          Similarly, to find the balancing force in the plane 'M', take moments about 'Q' which is the point of intersection of the plane 'L' and the axis of rotation. Therefore

                                               F_C2 X l = F_C X l₁

                                           m₂ . r₂ . l = m . r . l_{1              ........ (6)}

Balancing Of Several Rotating Masses In The Single plane

          Let us Consider four masses of magnitude m₁, m₂, m₃ and m₄ are at a distance of r₁, r₂, r₃ and r₄ from the axis of rotating shaft. Let θ₁, θ₂, θ₃ and θ₄ be the angles of these masses with the horizontal line OX, as shown in Fig., Consider these masses rotate with a constant velocity of 'ω' rad/sec, about an axis through 'O' and perpendicular to the plane of paper.

          The magnitude and position of the balancing mass may be found out by using two methods, they are analytical method and graphical method. These methods are discussed below:

1. Analytical Method:

1. First we have to find out centrifugal force exerted by each mass on the rotating shaft.

2. We have to resolve the centrifugal forces horizontally and vertically, and find their sums, i.e., ΣH and ΣV. We know that 

Sum of horizontal components of the centrifugal forces, 

                      ΣH = m₁ . r₁. Cos θ1 + m₂ . r₂. Cos θ2  ………

and sum of vertical components of the centrifugal forces, 

                      ΣV = m₁ . r₁. Sin θ1 + m₂ . r₂. Sin θ2  ………

3. Magnitude of the resultant centrifugal force, 

4. If is the angle, which the resultant force makes with the horizontal, then 

5. The balancing force is then equal to the resultant force, but in opposite direction. 

6. Magnitude of balancing mass, such that 

                         F_c = m . r

       Where     m = Balancing Mass, 

                        r   =  It's radius of rotation

2. Graphical Method:

1. Draw space diagram of several masses based on their positions as shown in above Fig., 

2. We have to find out centrifugal force exerted by each mass on the rotating shaft.

3. Now draw the vector diagram with the centrifugal forces we obtained.

4. Here 'ab' represents the centrifugal force exerted by the mass m₁ (i.e., m₁ . r₁ ) in magnitude and direction to some suitable scale. 

5. Similarly for other masses m₂, m₃ and m₄, draw 'bc', 'cd', and 'de'.

5. The balancing force is then equal to the resultant force, but in opposite direction. 

6. Magnitude of balancing mass (m) at a given radius of rotation (r), such that 

                         F_c = m . ω². r = Resultant centrifugal force

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

Balancing of Several Masses Rotating in Different Planes

When several masses revolve in different planes, they can be transferred to a 'Reference plane' (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 equals 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. In order to have a complete balance of the several revolving masses in different planes, the following conditions must be satisfied:

1. The resultant force must be zero (i.e All the forces in the reference plane must be balanced)

2. The resultant couple must be zero (i.e The couple about the reference plane must be balanced)

Let us now consider four masses revolving in different planes 1, 2, 3 and 4 respectively as shown in Fig., (a), and their relative angular positions are shown in Fig (b).,

          The magnitude of the balancing masses m_A and m_B in planes A and B may be obtained as discussed below.
1. Take one of the plane say A, as reference plane (R.P).
2. The distances of the other planes to the left of the reference plane is taken as negative, and those are present on the right as Positive.
3. Tabulate the planes data in the same order from left to right as shown in below table.

4. A couple may be represented by a vector drawn perpendicular to the plane of couple. Couple C₁ is obtained by transferring m₁ to the reference plane through O. The couple obtained is m₁ . r₁ . l₁ and it acts in a plane through and perpendicular to the paper. The vector representing this couple is drawn in the plane of the paper and perpendicular to Om₁ as OC₁ shown by in Fig., Similarly the remaining vectors for remaining masses is calculated and shown in Fig.,
5. The couple vectors as discussed above, are turned counter clockwise through a right angle for convenience of drawing without changing relative positions.
6. Now draw the couple polygon as shown in Fig., The couples about the reference plane must be balance. i.e., the resultant couple must be zero.

7. Now draw the force polygon as shown in Fig., The forces in the reference plane must balance. i.e., the resultant forces must be zero.

From the above expression, the value of balancing mass m_A in the plane 'A' may be obtained and the angle of inclination of this mass with the horizontal may be measured from Fig., (Angular positions of masses) 

BALANCING OF ROTATING MASSES

          We have already discussed that whenever a certain extra amount of mass is attached to a rotating shaft, it produces (or exerts) some centrifugal force. Due to this force the shaft will bend and produces 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 result of the centrifugal force of the primary mass. This can be done in such a way that the centrifugal force of both the masses are made to be equal and opposite. The method of providing the second mass so as to counteract the effect of the centrifugal force of the first mass, is termed as Balancing Of Rotating Masses.

The following cases are more important from the subject point of view:

1. Balancing of a single rotating mass by a single mass rotating within the same plane.

2. Balancing of a single rotating mass by two masses rotating in different planes.

3. Balancing of various masses rotating in the same plane.

4. Balancing of various masses rotating in different planes.

We will discuss these cases, in detail, in the following posts.

EXERCISES - Balancing Of Rotating masses

EXERCISES

1. Define Balancing? Write a short notes on Static Balancing? and Dynamic Balancing? State necessary conditions to achieve them?

2. Why is balancing of rotating parts (or masses) necessary for highspeed engines?

3. Explain how a single revolving mass is balanced by two masses revolving in different planes? 

4. Explain the method of balancing a single rotating mass by another single rotating mass in same plane?

5. Explain how the different masses rotating in different planes are balanced?

6. Explain the method of balancing of different masses revolving in same plane?

Balancing of Rotating Masses -- OBJECTIVE TYPE QUESTIONS

Balancing Of Rotating masses

Objective type questions

1. The balancing of Rotating and Reciprocating parts of the engine is necessary when it runs at

(a) Slow speed                                         (b) Medium speed 

(c) High speed                                         (d) No speed

2. Which of statements is true for static balancing of shaft,

(a) The net dynamic forces acting on the shaft is zero           
(b)  The net couple due to dynamic forces acting on the shaft is zero

(c) Both (a) and (b)                          (d) None of the above statements

3. For Dynamic balancing of shaft,

(a) The net dynamic forces acting on the shaft is zero           
(b)  The net couple due to dynamic forces acting on the shaft is zero

(c) Both (a) and (b)                          (d) None of the above statements

4. Which of the following statements is correct about the balancing of a mechanicalsystem?

(a) If it is under Static balance, then there will be dynamic balance too           
(b)  If it is under Dynamic balance, then there will be Static balance too

(c) Both Dynamic and Static balance have to be achieved separately
(d) None of the above mentioned statements

5. A distributing mass m₁ attached to a rotating shaft may be balanced by a single mass m₂ attached in the same plane of ratation as that of m₁ such that 

(a) m₁.r₂ = m₂.r₁           
(b)  m₁.r₁ = m₂.r₂

(c) m₁. m₂ = r₁.r₂

6. Which of the following statements are associated inorder to have a complete balancing of the several revolving masses in different planes 

(a) The resultant couple must be zero           
(b) The resultant force must be zero

(c) Both resultant force and couple must be zero
(d) None of the above

7. Which of the following statements are associated with complete dynamic balancing of rotating systems? 

(1) The resultant couple due to all inertia forces is zero           
(2) The support reactions due to forces are zero but not due to couples

(3) The system is automatically statically balanced
(4) Centre of masses of the system lies on the axis of rotation

(a) 1, 2 and 3 only           (b) 2, 3 and 4 only

(c) 1, 3 and 4 only           (d) 1, 2, 3 and 4 

Balancing of Reciprocating Masses -- Introduction

Introduction

          We have discussed in previous chapters, the various forces acting on the reciprocating parts of an engine. The result of all the forces performing on the body of the engine due to inertia forces only is known as unbalanced force or shaking force.  Thus if the resultant of all the forces due to inertia effects is zero, then there'll be no unbalanced force, however even then an unbalanced couple or shaking couple will be present. 

          Consider a horizantal reciprocating engine mechanism as shown in below Fig.,

Let   F_R  = Force required to accelerate the reciprocating parts,

         F_I   = Inertia force due to reciprocating parts,

         F_N  = Force on the sides of the cylinder walls or normal force acting on the cross-head guides, and

         F_B  = Force acting on the crankshaft bearing or main bearing.

          Since F_R and F_I are equal in magnitude however act in opposite direction, therefore they balance one another. The horizantal component of F_B (i.e F_BH) acting along the line of reciprocation is also equal and opposite to F_I. This force F_{BH =} F_U is an unbalanced force or shaking force and required to be properly balanced.

         The force on the sides of the cylinder walls (F_N) and the vertical component of  F_B(i.e F_BV) are equal and opposite and thus form a shaking couple of magnitude  F_N X x _or F_BV X x. 

          From the above statements we tend to see that the imapct (or) effect of the reciprocating parts is to produce a shaking force and a shaking couple. Since the shaking force and a shaking couple differ in magnitude and direction during the engine cycle, therefore they cause terribly objectionable vibrations.

          Thus the purpose (or  aim) of balancing the reciprocating masses is to eliminate the shaking force and a shaking couple. In most of the mechanisms, we can produce the shaking force and a shaking couple by adding appropriate balancing mass, But it is usually not practical to eliminate them completely. In otherwords, the reciprocating masses are only partially balanced. 

Note : The masses rotating witht he crankshaft are normally balanced and they do not transmit any unbalanced or shaking force on the body of the engine.