# UNIT 6 INTRODUCTION TO BALANCING Introduction to

and b 2 are shown in Figure 6.4. 160 Theory of Machines The equations are written again by canceling out 2 g Z from both sides. W r W r W r 1 1 2 2. ...

UNIT 6 INTRODUCTION TO BALANCING

Introduction to Balancing

Structure 6.1

Introduction Objectives

6.2

Force on Shaft and Bearing due to Single Revolving Mass

6.3

Balancing of a Single Revolving Mass

6.4

Procedure for Balancing

6.5

External Balancing of Single Revolving Mass

6.6

Static and Dynamic Balancing

6.7

Several Masses Revolving in Same Transverse Plane

6.8

Balancing of Several Masses in Different Transverse Planes

6.9

Summary

6.1 INTRODUCTION In the system of rotating masses, the rotating masses have eccentricity due to limited accuracy in manufacturing, fitting tolerances, etc. A mass attached to a rotating shaft will rotate with the shaft and if the centre of gravity of the rotating mass does not lie on the axis of the shaft then the mass will be effectively rotating about an axis at certain radius equal to the eccentricity. Since the mass has to remain at that radius, the shaft will be pulled in the direction of the mass by a force equal to the centrifugal force due to inertia of the rotating mass. The rotating centrifugal force provides harmonic excitation to system which thereby causes forced vibration of the machines. We will discuss how such a force can be balanced to remove the effect of unbalance. The unbalance is expressed as product of mass and eccentricity.

Objectives After studying this unit, you should be able to understand 

what is unbalanced force, and the effect of this,

how is unbalanced force due to single rotating mass balanced, and

how is unbalanced force due to several rotating masses in the same plane determined?

6.2 FORCE ON SHAFT AND BEARING DUE TO SINGLE REVOLVING MASS Figure 6.1 shows a revolving mass attached to a horizontal shaft, which is supported by two bearings. The mass M is at a radius r from the axis of the shaft. The mass is attached to the shaft at a distance a from bearing on the left and at distance b from right hand bearing so that the span of the shaft between the bearings is a + b. The eccentricity is due to the tolerances assigned, the limited accuracy of the manufacturing machines and nonhomogenity of the material. The shaft is rotating with an angular velocity  rad/s. A dynamic force F will pull the shaft towards the connected mass M. The magnitude of F is given by F  M 2 r

. . . (6.1) 155

Theory of Machines

The force F will be a bending force on the shaft and will cause bending moment. Additional bending stress will be induced and reactions at bearings A and B will occur. The reactions RA and RB can be calculated by considering the equilibrium. They are RA 

b M 2 r ab

. . . (6.2)

RB 

a M 2 r ab

. . . (6.3) M 

M

r r A

B

F a

b

Figure 6.1 : Revolving Mass attached to a Horizontal Shaft

These reactions on bearings will rotate with the mass, hence will cause fatigue damage. The wear of bearing all over the circumference will also increase. The shaft will be subjected to bending moment whose maximum value will occur at the section where the mass M is connected to the shaft. The bending moment under the revolving mass will be M 

ab M 2 r ab

. . . (6.4)

when the reactions at the supports become more than the tolerable limits of bearings, the balancing is done. Example 6.1 A shaft of circular cross-section of diameter 50 mm is supported in two bearings at a distance of 1 m. A mass of 20 kg is attached to the shaft such that its centre of gravity is 5 mm from the axis. The mass is placed at a distance of 400 mm from left hand bearing. To avoid unequal wearing of bearings, the designer places the mass in the centre of the span. Calculate reactions at bearings, maximum bending moments and bending stresses if the shaft rotates at 750 rpm. Solution The force caused on shaft due to rotation = F 2 N 2  750   78.54 rad/s 60 60 W 2 F  r g



Use M  20 kg,   78.54 rad/s, r  5  10 3 m F  20  (78.5)2  5  10 3  616.85 N

Case I Mass at a = 400 mm from left hand (LH) bearing b = 1000 – 400 = 600 mm a + b = 1000 mm A – left, B – right hand bearing (Figure 6.1) 

RA 

156

b F ab

600  616.85 1000

. . . (i)

or

RA  370.1 N

. . . (ii)

and

RB  F  RA  616.85  370.1

. . . (iii)

or

RB  246.74 N

Introduction to Balancing

 Maximum BM,  RA . a  370.1  400 BM  148.04  103 Nmm

or

. . . (iv)

Bending stress is given by 32 BM

b 

d3

where d = 50 mm = shaft diemeter 32  148.04

b 

  (50)

3

 103

b  12.06 N/mm2

or

. . . (v)

Case II Mass at the centre of span, i.e. a = b = 500 mm 

RA  RB 

F 616.85   308.425 N 2 2

 Maximum BM,

F  a  308.425  500  154.213  103 Nmm 2

32 BM

b  

d3 32  154.213   (50)3

 103

b  17.35 N/mm2

or

. . . (vi)

. . . (vii)

6.3 BALANCING OF A SINGLE REVOLVING MASS Balancing is a process of the redistribution of the mass in the system such that the reactions at the bearings are within the tolerable limits of the bearings. There are two methods of achieving this. System Method The effects of an off-axis or eccentric mass connected to a rotating shaft, as brought out above, has to be nullified. One simple way by which this is achieved is by attaching another mass M1 at a radius r1, exactly opposite to M as shown in Figure 6.2. The shaft is rotating at an angular speed of . M

M

 r

A

r

B r1

r1 M1 M1

Figure 6.2 : Balancing of a Single Revolving Mass : System Method

157

Theory of Machines

The mass M1 and its radius are so chosen that it is equal to the centrifugal force due to M, i.e. F

M M 2  r  1 2 r1 g g

which means that M r  M1 r1

. . . (6.5)

If Eq. (6.5) is satisfied then the resultant force on the shaft and hence on bearing will be zero. Thus additional reaction or overload on the bearings is zero and BM = 0, hence no additional stress in the shaft will be induced. The system is now called internally balanced. Internal balance is achieved by adding a balancing mass exactly opposite to revolving mass which causes unbalance. Thus the disturbing and balancing masses (M and M1, respectively) are in the same plane for internal balance and they satisfy the condition given by Eq. (6.5). This method is used for balancing auto wheels, etc. Second Method In this method, instead of meutralising centrifugal force, the eccentricity or radius ‘r’ is reduced. By doing this, we intend to reduce the magnitude of the centrifugal force. This method is used for thicker discs like flywheel where it is possible to take out mass by shallow drilling. The side of the disc which consists of centre of gravity is called heavy side as shown in Figure 6.3(a). The opposite side to the heavy side is called light side. Since heavy side consists of more mass, the mass M1 as given by the Eq. (6.5) can be taken out by drilling a shallow hole of diameter ‘d’ and depth ‘b’ as given by the following relation  2 d b   M1 4

where ‘b’ is less than thickness of the disc and  is density of material. Heavy Side

+

+

r CG

CG

r1 +

Light Side

Shallow Hole

Figure 6.3 : Balancing of a Single Revolving Mass : Second Method

6.4 PROCEDURE FOR BALANCING A thin disc like flywheel or a car wheel may be mounted on an axle or a shaft. It can be rotated by hand. The side which comes down can be marked. It is rotated again. If the same side comes down, this is heavy side and opposite to this is light side. If there is no force measuring device and rotating device, some mass can be mounted whenever it is possible on light side. The care should be taken that the mounting distance is as large as possible. If marked heavy side again comes down, more mass can be mounted on the light side. This process is repeated till any side comes down. Now it is farely balanced.

158

If there is a machine like wheel balancing machine, it indicates the magnitude of mass and the location where balancing mass should be mounted. These methods are trial and error methods and are time consuming methods. This cannot be used in industries where time available per piece is less. The industries have balancing machines which have

rotating device and transducers to provided magnitude of balancing mass and its location. In practice, we never aim the perfect balancing. The machine or the component is balanced till reactions are within a tolerable limit of the bearings.

Introduction to Balancing

SAQ 1 (a)

What do you mean by unbalance and why it is due to?

(b)

What do you mean by balancing?

(c)

Why all the rotating systems are not balanced?

Example 6.2 In Example 6.1 find what weight of the balancing mass will achieve complete balance if the balancing mass has its centre of gravity at a distance of 7.5 mm from the axis of rotation. Will this be true for both positions of disturbing mass in Example 6.1. Solution Use (Eq. (18.5) with M  25 kg, r  5 mm, r1  7.5 mm 

25  5  M1  7.5

M1 

25  5  1  16.67 kg 7.5

. . . (i)

Since the Eq. (6.5) is independent of distance along the shaft the position of disturbing mass will not affect the magnitude of balancing mass. So balancing mass is same as at Eq. (i) for M at a = 400 mm or M in the centre of the span.

6.5 EXTERNAL BALANCING OF SINGLE REVOLVING MASS The single revolving mass W connected to the shaft at radius r causes the unbalance force and reactions at the support. However, if two masses W1 at radius r1 and W2 at radius r2 are attached to the shaft, respectively in the same axial plane then also balancing of force due to rotation of W can be achieved. The condition of balance in case as shown in Figure 6.3 will be W W W 2  r  1 2 r1  2 2 r2 g g g

The bending moments due to the forces due to rotating masses will be balanced if W W W 2  r a  1 2 r1 a1  2 2 r2 a2 g g g W W W 2  r b  1 2 r1 b1  2 2 r2 b2 g g g

a, b, a1, b1, a2 and b2 are shown in Figure 6.4.

159

Theory of Machines

The equations are written again by canceling out

2 from both sides. g

W r  W1 r1  W2 r2

. . . (6.6)

W r a  W1 r1 a1  W2 r2 a2

. . . (6.7)

W r b  W1 r1 b1  W2 r2 b2

. . . (6.8)

Thus the disturbing force or unbalanced force on the shaft is removed. There is no excess reaction at any of bearings A and B. This is known as external balancing. l

A

M

m

L W

W a

b

r

A a1

b1 a2

b2

r1

r2

W1 W2

W1

W2

Figure 6.4 : External Balancing of Single Revolving Mass

The external balancing with two rotating masses (Figure 6.4) is resorted to when it is not possible to introduce the balancing mass exactly opposite to disturbing mass in the same radial plane. It may be worthwhile to note that a single mass placed in the same axial plane but in a different radial plane may satisfy the condition that W r  W1 r1 (W2  0) but W and W1 will together cause a couple to act upon the shaft. This moment of the couple will tend to rock the shaft in the bearings. The balancing masses in the same axial plane but in two different radial planes can satisfy the conditions of zero force transverse to beam and zero moment. The Eqs. (6.6), (6.7) and (6.8) are such conditions. If we define three radial planes for three masses W, W1 and W as A, L and M, respectively and call distance between A and L as l and that between A and M as m then from Figure 6.3 it is seen that a = a1 + l and b = b2 + m. Then replacing a by (a1 + l) and b by (b2 + m) in Eqs. (6.7) and (6.8), respectively following are obtained. W r l  W2 r2 (l  m)

or

W2 r2  W r

l lm

. . . (6.9)

and

W1 r1  W r

m lm

. . . (6.10)

Note that same results may be obtained if we take moments about sections L and M of the shaft. Also note that Eq. (6.6) implies that reactions at supports are zero. Eqs. (6.9) and (6.10) are more convenient to use along with Eq. (6.6) for solving a problem on external balancing. Again note that (l + m) is the distance between two radial planes in which balancing masses are placed. We understand that the Eqs. (6.9) and (6.10) are applicable to a situation as shown in Figure 6.3 but if both L and M are on one side of plane A then also these equations are true but one of W1 and W2 will be on the same side of the shaft as W . (l + m) can be denoted by d, so that

and 160

W2 r2  W r

l d

. . . (6.11)

W1 r1  W r

m d

. . . (6.12)

Example 6.3

Introduction to Balancing

A mass of 100 kg is fixed to a rotating shaft so that distance of its mass centre from the axis of rotation is 228 mm. Find balancing masses in following two conditions : (a)

Two masses – one on left of disturbing mass at a distance of 100 mm and radius of 400 mm, and other on right at a distance of 200 mm and radius of 150 mm.

(b)

Two masses placed on right of the disturbing mass respectively at distances of 100 and 200 mm and radii of 400 and 200 mm

The masses are placed in the same axial plane. Solution For Case (a) see Figure 6.5. A W r = 228 L 100

M

200

r2 = 150

r1 = 400 W2

W1

Figure 6.5 : Figure for Example 6.3

r = 228 mm, l = 100 mm, m = 200 mm, d = l + m = 100 + 200 = 300 mm, W r1 = 400 mm, r2 = 150 mm,  100 kg, W1 = ?, W2 = ? g From Eq. (6.6) W r  W1 r1  W2 r2 100  228  W1  400  W2  150

(g cancels out)

. . . (i)

From Eq. (6.11) W2  150  100  228 

W2  50.67 kg

100 300

. . . (ii)

From Eq. (6.12) W1 

or

100  228 200  400 300

W1  38 kg

. . . (iii)

Check with Eq. (i) 22800 = 15200 + 7600

161

Theory of Machines

For Case (b) see Figure 6.6 W2

W

200 100

100

W1

Figure 6.6 : Figure for Example 6.3

From Eq. (6.11) W2 

or

100  228 100  150 100

(d  100)

W2  152 kg

. . . (iii)

From Eq. (6.12) W1 

or

100  228 200  400 100

W1  114 kg

. . . (iv)

From Eq. (18.6) 100  228  114  400  W2  150

or

22800  45600  W2 150 W2   152 kg

. . . (v)

The negative sign indicates W2 is on the other side of W1.

6.6 STATIC AND DYNAMIC BALANCING If the centre of gravity of all rotating masses is made to coincide with the axis of rotation, a state is achieved when bearings will carry no additional reaction. However, the masses may still cause some net bending moment on the shaft. Such bending moment will keep changing its plane and thus cause shaft to vibrate. Connecting the masses in such a way that bending moment is made to vanish will result in situation when shaft will not vibrate. The balancing when only centers of gravity of attached mass system lies on axis of rotation is known as static balancing. The balancing with centers of attached mass system made to coincide with axis of rotation and no net bending moment acting on shaft is called dynamic balancing. In dynamic balancing forces and moments both are to be balanced.

6.7 SEVERAL MASSES REVOLVING IN SAME TRANSVERSE PLANE

162

A number of masses weighing W1, W2, etc. may be connected to the shaft such that their respective centres of gravity are at distances of r1, r2, etc. Each of these masses may be placed at its own angular position in the transverse or radial planes as depicted in Figure 6.7(a). Each will exert centrifugal force which will be proportional to the product

of mass and radius (i.e. W r). If we wish to ascertain if the net effect will be an unbalanced force, then we draw a polygon of forces. If the force polygon does not close then the resultant unbalance is equal to the closing side of the polygon. In this case the closing side is 5, O. Thus a force equivalent to 5, O in the direction 5 to O will close the polygon. Hence balancing mass may be connected parallel to line joining 5 and O and the length of this side will be the product of mass and radius. W3

Introduction to Balancing

W2 3

W 4 r4 4 r2 r3

W 3 r3

W1 5

r5

r4

W4

W 5 r5

r1

2 a

W5

0

W 2 r2 W 1 r1

(a)

1

(b)

Figure 6.7 : Several Masses Revolving in same Transverse Plane

Example 6.4 Four masses W1, W2, W3 and W4 at radii of 225 mm, 175 mm, 250 mm and 300 mm are connected at angles of zero, 45o, 75o and 120o from horizontal line as shown. If the shaft rotates at 500 rpm, find what unbalanced force acts upon the shaft and at what angle from mass W1. If a mass to balance the system can be placed at a radius of 200 mm, find the weight of the mass. Take W1 = 1000 N, W2 = 1500 N, W3 = 1200 N and W4 = 800 N Solution W1 r1  2.25  105 W2 r2  2.625  105

W3 r3  3  105

W4 r4  2.4  105

Figure 6.7 shows the orientation of disturbing forces with magnitudes (proportional to W r). Force polygon is shown in Figure 6.8(b). From the polygon the unbalanced W r = 7.75  105 is at an angle of 207.5o.

W 4 r4

3

2.4

Wr

120o

W 3 r3

2.625 75o 120 75 45 2.25

W 2 r2

o

207.5

45o

W 1 r1

(a)

(b) Figure 6.8 : Figure for Example 6.4

163

Theory of Machines

The unbalanced force 

W 2  r g 2

7.75  105  2 500  3    10 9.81 60  

 2.166  105 N

If radius at which balancing mass is placed is 200 mm. Then

W r  7.75  105 W 

7.75  105 200

W  3875 N

Alternative method is to resolve W r along horizontal and vertical direction and find their resultant. W r ( H )  (2.25  2.625 cos 45  3 cos 75  2.4 cos 120) 105

 (2.25  1.856  0.7765  1.2) 105

= 3.6765  105 W r (V )  (2.625 sin 45  3 sin 75  2.4 sin 120) 105  (1.856  2.9  2.0785) 105

= 6.83  105 

W r  105 [W r ( H )]2  [W r (V )]2  105 [3.6765]2  [6.83]2

= 7.76  105   tan 1

W r (V ) 6.83  tan 1  tan 1 1.8583  61.7o W r (H ) 3.675

Note that W r is the resultant unbalance force which will act upward. The balancing mass will be placed opposite to it, i.e. downward as shown by broken line in Figure 18.7(a). 207.5o is the measured angle. The calculated value of the angle is 360 – 90 – 61.7 = 208.3o. Compare the Values From polygon construction W r  7.75,   207.5o

From calculation W r  7.76,   208.3o .

6.8 BALANCING OF SEVERAL MASSES IN DIFFERENT TRANSVERSE PLANES

164

To begin with we will name the planes in which masses revolve as A, B, C and D. The masses rotating in these planes are respectively Wa, Wb, Wc and Wd. The radii at which centers of gravity lie from axis of rotation in these planes are respectively ra, rb, rc and rd.

The angular separation between masses starting from A and B are ,  and . The balancing masses will be placed in two planes L and M which are between A and B and between C and D, respectively. The Figure 6.9 depicts the system. Distance between planes L and any of planes A, B, C and D is denoted by l with appropriate suffix. The same distances for plane M is denoted by m. Wm

B

A

C

D

L

Wc

M lb

la

Wb

lc

ld

 

Introduction to Balancing

 Wa

md ma

WL

mc

md

Wd

Figure 6.9 : Balancing of Several Masses in Different Transverse Plane

The method apparently is same as used in Section 6.4 in which balance masses in two planes L and M were found. So we need to repeat the procedure for mass in plane A and balancing mass in planes L and M. Thus we proceed in steps of planes A, B, C and D. A table of the kind shown below as Table 6.1 will be helpful. Rows will be dedicated to planes in which masses revolve which could be known or unknown. 2 is a g constant (Eq. (6.1)). Two planes L and M are chosen which are respectively at distances of l and m from the plane of revolving mass and balancing masses are placed in planes L and M as given by Eqs. (6.11) and (6.12). So the columns of the table will describe plane (A, B, C, etc.) weight W; radius r; force W r; distances l and m; balancing forces in L and M.

For a mass of weight W, revolving at radius r, the force is proportional to W r as

Table 6.1 : Calculation of Balancing Masses in Two Planes Plane

Weight W

Force 2 

Distance From

Balancing Force 

g

g

Wr A B

Wa Wb

ra rb

Wa ra Wb rb

2

Plane L l la lb

Plane M m ma mb

Plane L

Plane M

Wa ra ma

Wa ra la

d Wb rb mb d

d Wb rb lb d

The sign of the forces in last two columns will be decided by observation. Yet as a rule if a force is in the same direction as the disturbing force, then it will be positive and if in the opposite direction it will be negative. After the balancing forces have been calculated in planes L and M which will be parallel to forces in planes A and B, etc. they are combined to give a single resultant. Their inclination to force in plane A can be determined. The above procedure will be followed in solving the example. Example 6.5 In Figure 6.9 four masses Wa = 1000 N, Wb = 1500 N, Wc = 1200 N and Wd = 1300 N revolve respectively at radii of ra = 225 mm, rb = 175 mm, rc = 250 mm and rd = 300 mm in planes A, B, C and D. Two planes L and M are

165

Theory of Machines

selected to place balancing masses Wl and Wm at a radius of 600 mm. The masses Wb, Wc and Wd are respectively at angles of 45o, 75o and 135o from Wa and distances between planes are : la = 300 mm, lb = 375 mm, lc = 750 mm, ld = 1500 mm, ma = 1800 mm, mb = 875 mm, mc = 500 mm, md = 250 mm. Find the balancing masses and orientation of their radii from radius of mass Wa. Solution See Figure 6.9. Proceed as per Table 6.1. The distance between planes L and M, d = ld – md = 1500 – 875 = 825 mm. Plane

Weight W N

Distance of Plane From

Force 2 

Balancing Force 2  g

g Wr

Plane, L l (mm)

Plane, M m (mm)

Plane L Wrm/d Nmm  6.48  105

Plane M Wrl/d Nmm 1.08  105

225

2.25  105

300

1800

B

1.5  10

3

175

2.625  10

375

875

 3.675  105

 1.575  105

C

1.2  103

250

3.00  105

750

500

 2.4  105

 3.6  105

D

1.3  10

300

3.90  10

250

1.56  10

 9.36  105

A

103

3

5

5

1500

5

The last two columns show balancing forces for those in planes A and B, etc. Hence, these forces will act in opposite direction to Wa, Wb, etc. four forces in planes L and M will be equal to one force by a single rotating mass. Thus, two balancing masses – one each in planes L and M will be obtained. Balancing forces in L plane are shown in Figure 6.10, along with disturbing forces Wa, Wb, Wc and Wd. To find resultant of all balancing forces we go for their components along horizontal and vertical directions. The relevant angles are shown in figure. Wc Wd Wb

1.56  105 6.48  10

5

135 o

45 45o 29o

75o

45 75

Wa

151o

3.675  105 12.364  10

5

2.4  105

Figure 6.10 : Figure for Example 6.5

Wr ( H )  6.48  105  3.675  105 cos 45  2.4  105 cos 75  1.56  105 cos 45  (6.48  2.6  0.621  1.1) 105

 10.8  105 Wr (V )  3.675  105 sin 4.5  2.4  105 sin 75  1.56  105 sin 45

 (2.6  2.32  1.1) 105

166

 6.02  105

Introduction to Balancing

W r (Resultant)  [W r ( H )]2  [W r (V )]2

 116.64  36.24  105  12.364  105 Nmm r  600 mm, W 

With

12.364  105  2060.7 N 600

The angle with Wa  tan 1

W r (V ) 6.02  tan 1  tan 1 0.5574  29o W r (H ) 10.8

Balancing forces in M plane are shown in Figure 6.11 along with disturbing forces Wa, Wb, Wc and Wd. The resultant is found as above. Wr ( H )  (1.08  1.57  cos 135  3.6 cos 105  9.36 cos 45) 105

 (1.08  1.11  0.93  6.62) 105  5.66  105 Wr (V )  (1.57 sin 135  3.6 sin 105  9.36 sin 45) 105  (1.11  3.48  6.62) 105

 11.2  105

Wr ( R)  5.662  11.22  105  12.55  105

At r = 600 mm, the balancing mass in M plane will be W 

12.55  105  2091 N 600

This will be placed at angle  with direction of Wa   tan 1

11.2  tan 1 1.98  63.2o 5.66 Wc Wd Wb

1.08  105

Wa

45o o

1.57  10

5

105 135o

9.36  105 3.6  10

5

12.55  105

Figure 6.11 : Figure for Example 6.5

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Theory of Machines

Thus we see that for balancing the revolving masses in planes A, B, C and D we have to connect two masses of 2060.7 N and 2091 N, respectively which are between A and B and between C and D.

SAQ 2 (a)

What do you understand by balancing of revolving masses?

(b)

If not balanced what effects are induced on shaft bearing system due to unbalanced rotating masses.

(c)

How do you achieve balance of rotating masses which lie in parallel transverse planes of a shaft?

(d)

Five masses A, B, C, D and E revolve in the same plane at equal radii. A, B and C are respectively 10, 5 and 8 kg in mass. The angular direction from A are 60o, 135o, 210o and 270o. Find the masses D and E for complete balance.

(e)

A shaft carries three pulleys A, B and C at distance apart of 600 mm and 1200 mm. The pulleys are out of balance to the extent of 25, 20 and 30 N at a radius of 25 mm. The angular position of out of balance masses in pulleys B and C with respect to that in pulley A are 90o and 210o respectively. It is required that the pulleys be completely balanced by providing balancing masses revolving about axis of the shaft at radius of 125 mm. The two masses are to be placed in two transverse planes midway between the pulleys.

6.9 SUMMARY The masses that are connected to shaft and whose centers of gravity do not lie on axis of the rotation, revolve about the axis at constant radius. Moving in circular path they are subjected to centrifugal force which may cause bending stress in the shaft and rotating reactions in the bearing. To nullify their effects, revolving masses can be provided in the plane of disturbing mass or in some other parallel plane. If balancing masses are placed in the plane of unbalance for single mass, only one balancing plane is required otherwise two balancing planes shall be required. The balancing process requires that both the bending forces and moments on the shaft be made to vanish.

6.10 ANSWERS TO SAQs SAQ 2 (d)

Since the radii are equal, the forces are proportional to masses. The Figure 6.11 shows the forces and their orientation. Since the system is balance the force polygon must close whereby we can find the unknown forces and corresponding masses. We plot only masses hence sides of polygon will directly give the masses. In the force (proportional force) polygon known values are written inside the polygon and measured values are written outside. Mass D = 8 kg

168

Mass E = 6 kg

8

Introduction to Balancing

5

C

B 2.625 60o o

135

10 A

210o 270o D

E

Orientation of Forces c

8

8

d

b

5

6

10 e

a

Proportional Force Polygon Figure 6.11

SAQ 5 A

B

L

M

C Wb

25 mm

25 mm

25 mm

20 N

25 N

30 N

la

90o Wa

1200 mm

600 lb

300

210o

Wc lc

300 mb

1500 300

ma

mc 300

900 D = 1500 – 300 = 1200 mm

Plane

W (N)

r (mm)

Wr (Nmm)

Distance From

Balancing Force

L (l)

M (m)

L Wr m/d

M Wr l/d

A

25

25

625

300

900

 468.75

156.25

B

20

25

500

300

300

 125.00

 125.00

C

30

25

750

1500

300

+ 187.50

 375.00

169

Theory of Machines

L Plane Wr ( H )  468.75  187.5 cos 30

 631.13 Wr (V )  125  187.5 sin 30

 218.75 468.75 

30o

W 187.5 125.00

Wr ( R)  631.132  218.752

 668 Nmm

r  125 mm

W 

668  5.34 N 125

  tan 1

218.75  tan 1 0.347  19.14o 631.13

M Plane Wr ( H )  156.25  375 cos 30  481 Wr (V )  375 sin 30  156.25  31.25 Wr ( R)  4812  31.252  482 375.00

30o

125

r  125 mm

W 

482  3.86 N 125

tan  

31.25  0.065 481

  3.7o

170

W 156.25