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 

Primary and secondary Unbalanced Forces of Reciprocating masses

          Consider a reciprocating engine mechanism as shown in Below Fig., 

                    Let m = Mass of reciprocating parts,

                           l = Length of connecting rod PC, 

                           r = Radious of the crank OC, 

                          Θ = Angle of inclination of the crank with the line of stroke PO,

                          ω = Angular speed of the crank,

                      η = Ratio of length of the connecting rod to the crank radious = l/r

          We have already discussed in previous articles that the acceleration of the reciprocating parts is approximatey given by expression, 

          Inertia force due to reciprocating parts or force required to accelerate the reciprocating parts,

We have discussed in the previous article that the horizantal component of the force exerted on the crank shaft bearing (i.e F_BH ) is equal and opposite to inertia force (F1). This force is an unbalanced one and is denoted by F_U  .

Unbalanced force 

          The expression (m.ω^2 .r cos Θ) is known as Primary unbalnced force and (m.ω^2 .r [cos 2Θ]/n ) is called secondary unbalanced forces.

Partial Balancing of Unbalanced Primary Force in a reciprocating Engine

          The primary unbalanced force may be considered as the component of the centrifugal force produced by a rotating mass 'm' placed at the crank radius 'r' as shown in below Fig.,

          The primary force act from O to P along the line of stroke. Hence, the balancing of primary force is considered as equivalent to the balancing of mass 'm' rotating at the crank radius 'r'. This is balanced by having a mass B at radius b, placed diametrically opposite to the crank pin C. 

We know that centrifugal force due to mass B, 

                              = B. w^2. b.
and horizantal component of this force acting in opposite direction of primary force
                              = B. w^2. b cos θ.
The primary force is balnced if 
                    B. w^2. b cos θ = m w^2. r cos θ 
                                                 or
                     B.b = m. r
          A little consideration will show, that the primare force is completely balanced if B.b = m. r , but the centrifugal force produced due to the revolving mass B, has also a vertical  component perpendicular to the line of stroke having magnitude B. w^2. b sin θ. This force remains unbalnced. The maximum of this force occurs at value of θ = 90° or 270°. 
          From the above discussion we know that there are two unbalced force, one along the line of stroke where as the second unbalnced force acts along the perpendicular to the line of stroke. The maximum valve of force remains same  in both the cases.It is thus obvious, that the effect of the above method of balancing is to change the direction of the maximum unbalnced force from the line of stroke to the perpendicular of line of stroke. A compromise let a fraction 'c' of the reciprocationg masses is balnced such that

Partial Balancing of Locomotives

          The locomotives, usually, have two cylinders with cranks located at right angles (90°) to each other in order to have uniformly turning moment diagram and also the engine can be started easily after stopping in any position. Balance masses are placed on the wheels in both types. 

        In coupled locomotive, wheels are coupled by connecting their crank pins with coupling rods. As the coupling rod revolves with the crank pin, its proportionate mass can be considered as a revolving mass which can be completely balanced. 

Locomotive engines depending upon the location of cylinders they, are classified in two types they are:

           1. Inside cylinder engine and 2. Outside cylinder engine. 

          In the Inside cylinder locomotives, the two cylinders are placed in between the planes of two driving wheels as shown in Fig., Whereas in the Outside cylinder locomotives, the two cylinders are placed outside the driving wheels one on each side of the driving wheel, as shown in Fig., 

          They are further classified as Coupled and Uncoupled. If two or more pairs of wheels are coupled together to increase the adhesive force between the wheels and the track, it is called as coupled locomotive. otherwise, it is called as uncoupled locomotive.
          A single or uncoupled locomotive is one, in which the effort is transmitted to one pair of the wheels only; Whereas in coupled locomotives, the driving wheels are connected to the leading and trailing wheel by an outside connecting rod.
          Thus, whereas in uncoupled locomotive, there are four planes for consideration, two of the cylinders and two of the driving wheels, In coupled locomotives there are six planes, two of cylinders, two of coupling rods and two of wheels. The planes which contain the coupling rod masses lie outside the planes that contain the balance (counter) masses. Also, in case of coupled locomotives, the mass required to balance the reciprocating parts is distributed among all the wheels which are coupled. Thus, results in a reduced Hammer Blow.

Effect of Partial Balancing of Reciprocating Parts of Two Cylinder Locomotives

          We have discussed in the previous article that the reciprocating parts are only partially balanced. Due to this partial balancing of the reciprocating parts, there is an unbalanced primary force along the line of stroke and also an unbalanced primary force perpendicular to the line of stroke. 

          Locomotive engines operate at low speed and the ratio of length of connecting rod to radius of crank is generally large enough to neglect the effect of secondary force. 

The effect of an unbalanced primary force along the line of stroke is to produce:

1. Variation in tractive force along the line of stroke,  and 

2. Swaying couple

          The effect of an unbalanced primary force perpendicular to the line of stroke is to produce variation in pressure on the rails, which results in hammering action on the rails. The maximum magnitude of the unbalanced force along the perpendicular to the line of stroke is known as Hammer blow. 

Swaying couple

2. Swaying couple:

        The unbalanced portions of forces acting along the line of stroke are distance 'l' apart. These forces constitutes a couple about YY axis, which tends to make the leading wheels sway from side to side. This couple is known as swaying couple.

Note: In order to reduce the magnitude of the swaying couple, revolving balancing masses are introduced. But, as discussed in the previous article, the revolving balancing masses cause unbalanced forces to act at right angles to the line of stroke. These forces vary the downward pressure of the wheels on the rails and cause oscillation of the locomotive in a vertical plane about a horizontal axis. Since a swaying couple is more harmful than an oscillating couple, therefore a value of 'C' from 2/3 to 3/4, in two-cylinder locomotive with two pairs of coupled wheels, is usually used. But in large four cylinder locomotives with three or more pairs of coupled wheels, the value of 'C' is taken as 2/5.

Balancing of Secondary Forces of Multi-cylinder In-line Engines

          When the connecting rod is not long (i.e. when the obliquity of the connecting rod is considered), then the secondary distributing force due to the reciprocating mass arises.

We have discussed in previous articles, that the secondary force,

          As in case of primary forces, the secondary forces may be considered to be equivalent to the component, parallel to the line of stroke, of the centrifugal force produced by an equal mass placed at the imaginary crank of length r/4n and revolving at twice the speed of the actual crank (i.e. 2ω )  as shown in fig.,

          Thus in multi-cylinder in-line engines, each imaginary secondary crank with a mass attached to the crank pin is inclined to the line of stroke at twice the angle of the actual crank. The values of the secondary forces and couples may be obtained by considering the revolving mass. This is done in the similar way as discussed for primary forces. The following two conditions must be satisfied in order to give a complete secondary balance of an engine:

1. The algebraic sum of the secondary forces must be equal to zero. In other words, the secondary force polygon must close, and

2.   The algebraic sum of the couples about any point in the plane of the secondary forces must be equal to zero. In other words, the secondary couple polygon must close.

Note: The closing side of the polygon gives the maximum unbalanced secondary force and the closing side of the secondary couple polygon gives the maximum unbalanced secondary couple.

Balancing of Radial Engines (Direct and Reverse Cranks Method )

          The direct and reverse crank balancing method (also known as Contra-rotating mass balancing) aren't separate methods but rather one way of modelling the inertial effects of a reciprocating mass. The method of direct and reverse cranks is used in balancing of radial or V-Engines, in which the connecting rods are connected to a common crank. Since the plane of rotation of the various crank (in radial or V-Engines) is same, therefore there is no unbalanced primary or secondary couple. This is convenient for machines that have unaligned stroke centerlines (such as V-Engines and Rotary engines) since it eliminates the need to use "complicated Trigonometry".

          Consider a reciprocating engine mechanism as shown in above Fig (a)., Let the crank OC (known as the direct crank) rotates uniformly at 'ω' radians per second in a clockwise direction. Let at any instant the crank makes an angle 'θ'  with the line of stroke OP. The indirect or reverse crank OC' is the image of the direct crank OC, when seen through the mirror placed at the line of stroke. A little consideration will show that when the direct crank revolves in a clockwise direction, the reverse crank will revolve in the anticlockwise direction. We shall now discuss the primary and secondary forces due to the mass (m) of the reciprocating parts at P.

Considering Primary Forces: 
          We have already discussed that primary force is m.ω ².r cosθ . This force is equal to the component of the centrifugal force along the line of stroke, produced by a mass (m) placed at the crank pin C. Now let us support that the mass (m) of the reciprocating parts is divided into two parts, each equal to m/2 . 

          It is assumed that m/2 is fixed at the Direct Crank (termed as primary direct crank) pin C and m/2 at the Reverse crank (termed as primary reverse crank) pin C' , as shown in Fig (b).,

Hence, for primary effects of the mass m of the reciprocating parts at P may be replaced by two masses at C each of magnitude m/2.

Note:   The component of the centrifugal forces of the direct and reverse cranks, in a direction perpendicular to the stroke, are each equal to (m/2).ω ².r sinθ, but opposite in direction. are balanced.

Considering Secondary Forces:
          We know that the secondary force

         In the similar way as discussed above, it will be seen that for the secondary effects, the mass (m) of the reciprocating parts may be replaced by two masses (each m/2 placed at D and D' such that OD=OD'= r/4n. The crank OD is the secondary direct crank and rotates at 2 rad/s in the clockwise direction, while the crank OD' is the secondary reverse crank and rotates at 2 rad/s in the anticlockwise direction as shown in Fig (c).

Balancing of V-engines

Consider a symmetrical two cylinder V-Engine as shown in Fig., The common crank OC is driven by two connecting rods PC and QC. The lines of stroke OP and OQ are inclined to the vertical OY, at an angle α as shown in Fig.,

Let                          m= Mass of reciprocating parts per cylinder,

                                l = Length of connecting rod, 

                                r = Radius of crank,

                                n= Ratio of length of connecting rod to crank radius = l/r

                                θ = Inclination of crank to the vertical at any instant,

                                ω  = Angular velocity of crank.

We know that inertia force due to reciprocating parts of the cylinder 1, along the line of stroke, 

and the inertia force due to reciprocating parts of the cylinder 2, along the line of stroke,

The balancing of V-Engines is only considered for primary and secondary forces as discussed below:

Considering primary forces:

Considering secondary forces:

EXERCISES - Balancing Of Reciprocating masses

EXERCISES

1. Define Balancing? Write a short note on primary and secondary balancing?

2. Explain why only a part of the unbalanced force due to reciprocating masses is balanced by revolving mass.

3. Derive the following expressions, for an uncoupled two-cylinder locomotive engine:
(a) variation is tractive force  (b) Swaying couple; and (c) Hammer blow

4. Explain the method of balancing a number of masses rotating in one plane by another mass rotating in the same plane
5. What are in-line engines? How are they balanced? It is possible to balance them completely? 

6. Explain the 'direct and reverse crank' method for determining unbalanced forces in radial engines.

7. Discuss the balancing of V-Engines.

8. Explain the method of balancing a single rotating mass by another mass in same plane.
9. Five masses A,B,C,D and E Rotate in the same plane at equal radii. The masses A, B and C are 10 kg, 5 kg, and 8 kg respectively. The angular position of masses B, C, D and E measured in the same direction from A are 60°,  135°, 210° and 270° respectively. Find the masses D and E for complete balance.


For ANSWERS CLICK HERE

OBJECTIVE QUESTIONS - Balancing Of Reciprocating masses

Balancing Of Reciprocating masses

Objective type questions

1. The primary unbalanced forces is maximum when the angle of inclination of the crank with the line of stroke is

(a) 0°           (b)  90°

(c) 180°        (d) 360° 

2. The partial balancing means

(a) Perpendicular to axis 

(b) Parallel to its axis 

(c) In a circle about the axis

3. When a body is subjected to transverse vibrations, the stress induced in a body will be

(a) Shear stress          (b) Tensile stress          (c) Compressive stress

4. In a locomotive, the ratio of the connecting rod length to the crank radius is kept very large in order to

(a) Minimise the effect of primary forces     (b) Minimise the effect of secondary forces

(c) Have perfect balancing                                 (d) Start the locomotive quickly 

5. The swaying couple is maximum or minimum when the angle of inclination of the crank to the line of stroke ( θ ) is equal to

(a) 45° and 135°               (b) 90° and 135°  

(c) 135° and 225°             (d) 45° and 225° 

6. The tractive force is maximum or minimum when the angle of inclination of the crank to the line of stroke ( θ ) is equal to

(a) 90° and 225°               (b) 135° and 180°  

(c) 180° and 225°             (d) 135° and 315° 

7. The swaying couple is due to the 

(a) Primary unbalanced forces              (b) Secondary unbalanced forces

(c) Two cylinders of locomotive            (d) Partial balancing 

8. In a locomotive, the maximum magnitude of the unbalanced force along the perpendicular to the line of stroke, is known as 

(a) Tractive force                                      (b) Swaying couple

(c) Hammer blow                                     (d) None of these

9. The effect of hammer blow in a locomotive can be reduced by 

(a) Decreasing the speed                (b) Using two or three pairs of wheels coupled together

(c) Balancing whole of the reciprocating parts              (d)  Both (a) and (b)

10. Multi-cylinder engines are desirable because 

(a) Only balancing problems are reduced      (b) Only flywheel size is reduced

(c) Both (a) and (b)                                            (d) None of these

11. When the primary direct crank of a reciprocating engine makes an angle with the line of stroke, then the secondary direct crank will make an angle of . . . . . . . with the line of stroke. 

(a)  θ/2                   (b)  θ                    (c)  2θ                  (d)  3θ

12. Secondary forces in reciprocating mass on engine frame are 

(a) Of same frequency as of primary forces
(b) Twice the frequency as of primary forces

(c) Four times the frequency as of primary forces

(d) None of these

13. The Secondary unbalanced force produced by the reciprocating parts of a certain cylinder of a given engine with crank radius 'r' and connecting rod length 'l' can be considered as equal to primary unbalanced force produced by the same weight having 

(a) An equivalent crank radius r²/4l and rotating at twice the speed of the engine
(b) r²/4l as equivalent crank radius and rotating at engine speed

(c) Equivalent crank length of r²/4l and rotating at engine speed

(d) None of these

14. Which of the statement is correct?

(a) In any engine, 100% of the reciprocating masses can be balanced dynamically
(b) In the case of balancing of multicylinder engine, the value of secondary force is higher than the value of the primary force

(c) In case of balancing of multimass rotating systems, dynamic balancing can be directly

(d) None of these