Theory of Machines: Unit V: Balancing and Vibration

Balancing of several masses rotating in the same plane (balancing of several coplanar rotating masses)

Balancing and Vibration - Theory of Machines

If several masses are rigidly attached to a shaft at different radii in one plane perpendicular to the shaft and the shaft is made to rotate, each mass will set up out-of-balance force on the shaft.

BALANCING OF SEVERAL MASSES ROTATING IN THE SAME PLANE (BALANCING OF SEVERAL COPLANAR ROTATING MASSES)

If several masses are rigidly attached to a shaft at different radii in one plane perpendicular to the shaft and the shaft is made to rotate, each mass will set up out-of-balance force on the shaft. In this case, complete balance can be obtained by placing only one balancing mass in the same plane.

Illustration: Consider any number of masses (say, three masses) of magnitude m1, m2 and m3 at distances of r1, r2 and r3 from the axis of the rotating shaft. Their relative positions are indicated by angles θ1, θ2 and θ3 as shown in Fig.12.3(a).


The magnitude and position of the balancing mass can be obtained analytically or graphically as discussed below.

1. Analytical Method

Step 1: First of all, find out the centrifugal force (or mass moments i.e., the product of the mass and its radius of rotation) exerted by each mass on the rotating shaft.


• Since all the masses are connected to the shaft, all will have the same angular velocity o, we need not to calculate the actual magnitude of centrifugal force, therefore FC = mr.

Step 2: Resolve the centrifugal forces horizontally and vertically. Then find their sums Σ FH and Σ ΕV


Step 3: Find the magnitude of the resultant centrifugal force (FCR).


Step 4: The balancing force is equal to the resultant force, but in opposite direction. Now find out the magnitude of the balancing mass, such that


where mB = Balancing mass in kg, and

rB = Its radius of rotation in metres.

Step 5: The angle (θR) made by the resultant force FCR with the horizontal is given by


We know that the balancing mass is placed in the opposite direction to the resultant force direction. Therefore, the angle (θB) made by the balancing mass with the horizontal is given by


2. Graphical Method

Step 1: First, draw the space diagram with the position of the several masses, as shown in Fig.12.3(a).

Step 2: Find out the centrifugal force (ie., the product of the mass and its radius of rotation) exerted by each mass on the rotating shaft.


Step 3: Now draw the force polygon with the obtained centrifugal forces by polygon law of forces, such that vector oa represents the centrifugal force exerted by the mass m1 (or m1 r1) in magnitude and direction to some suitable scale.

Similarly, draw vectors ab and bc to represent centrifugal forces of other masses m2 and m3 (or m2 r2 and m3 r3).

Step 4: According to polygon law of forces, the closing side vector oc (from o to c and not from c to o) represents the resultant force in magnitude and direction, as shown in Fig.12.3(b). The balancing force is equal to the resultant force, but in opposite direction. Now find out the magnitude of the balancing mass må at a given radius of rotation rB, such that

mB rB Resultant centrifugal force = Vector oc.

• It may be noted that space diagrams need not to be drawn to the scale. Since the diagram is only to show the angle of inclination of several masses, therefore the angle of inclinations must be marked accurately.

Step 5: To obtain the position of balancing mass in space diagram, draw a line parallel to the resultant force but in opposite direction. Now measure the angle of inclination of this line in the same sense from the horizontal line OX, which gives the angle of inclination of the balancing mass (θB).

Note 

If the force polygon closes in loop, then the system is called the balanced system.

Example 12.1 

A rigid-rotor has all its unbalance in one plane and can be considered to consist of three masses m1 = 5 kg, m2 = 3 kg at an angle 165° counter clockwise from mɲ, and m3 = 8 kg at angle 85° clockwise from my. The radii r1 = 20 cm, r2 = 8 cm, r3 = 14 cm. Determine the balancing mass required at a radius of 10 cm. Specify the location of this mass with respect to m1.

Given data: 

m1 = 5 kg; m2 = 3 kg; m3 = 8 kg; r1 = 20 cm = 0.2 m; r2 = 8 cm = 0.08 m; r3 = 14 cm = 0.14 m; rB = 10 cm = 0.1 m.

Solution: 

This problem can be solved by both analytical and graphical methods, as below.

Method 1: Analytical Method

Step 1: To find the centrifugal forces

We know that the magnitude of centrifugal force is proportional to the product of mass and its radius of rotation, therefore


Step 2: To resolve the centrifugal forces horizontally and vertically.

Resolving the centrifugal force horizontally, we get


Here the angles θ1, θ2 and θ3 should be measured in the same direction i.e., either in clockwise direction or in counter clockwise direction. Assuming the mass m1 lies horizontally, we get θ1 = 0°. Given that m2 is at an angle 165° counter clockwise from m1, therefore θ2 = 165°. But m3 is at an angle of 85° measured clockwise from m1, therefore θ3 = 360° - 85° = 275°.


Resolving the centrifugal forces vertically, we get


Step 3: To find the magnitude of the resultant centrifugal force

Resultant centrifugal force is given by


Step 4: To find the magnitude of balancing mass

We know that the balancing mass mB rB must be equal to resultant centrifugal force by magnitude.


Step 5: To find angle made by the balancing mass

We know that the angle made by resultant centrifugal force with the horizontal line,


Method 2: Graphical Method

Step 1: The space/configuration diagram with the position of the given three masses can be drawn as shown in Fig. 12.4(a).

Step 2: The centrifugal force exerted by each mass are found as


Step 3: Now the force polygon i.e., vector diagram can be drawn to some suitable scale (say, 1 cm = 0.2 kg.m), as shown in Fig. 12.4(b).

[First draw the vector oa that represents m1 r1 = 1 kg.m parallel to OA. Then from a draw vector ab that represents m2 r2 = 0.24 kg.m parallel to OB; from b draw vector bc that represents m3 r3 = 1.12 kg.m parallel to OC.]

Step 4: We know that the closing side of the force polygon (vector co) represents the resultant force by magnitude and direction.

By measurement from force polygon, we get

FCR = vector co = 1.4 kg.m


Step 5: Angle of inclination of balancing mass is determined by drawing a line parallel to vector co in the space diagram, as shown in Fig.12.4(a) (in dotted lines). Now 0B can be measured from space diagram as θB = 130° from m1 measured in counter clockwise direction Ans.

Theory of Machines: Unit V: Balancing and Vibration : Tag: : Balancing and Vibration - Theory of Machines - Balancing of several masses rotating in the same plane (balancing of several coplanar rotating masses)


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ME3491 4th semester Mechanical Dept | 2021 Regulation | 4th Semester Mechanical Dept 2021 Regulation