4.1 Introduction to Balancing | unit 4 balancing

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Unit 4

Balancing

4.1 Introduction to Balancing

Often an unbalance of forces is produced in rotary or reciprocating machinery due to the inertia forces associated with the moving masses.

Balancing is the process of designing or modifying machinery so that the unbalance is reduced to an acceptable level and if possible is eliminated entirely.

A particle or mass moving in a circular path experiences a centripetal acceleration and a force is required to produce it.

An equal and opposite force acting radially outwards acts on the axis of rotation and is known as centrifugal force.

This is a disturbing force on the axis of rotation, the magnitude of which is constant but the direction changes with the rotation of the mass.

In a revolving rotor, the centrifugal force remains balanced as long as the center of the mass of the rotor lies on the axis of the shaft.

When the center of mass does not lie on the axis or there is an eccentricity, an unbalanced force is produced.

Need of Balancing

The high speed of engines and other machines is a common phenomenon now-a-days.

It is, therefore, very essential that all the rotating and reciprocating parts should be completely balanced as far as possible.

If these parts are not properly balanced, the dynamic forces are set up.

These forces not only increase the loads on bearings and stresses in the various members, but also produce unpleasant and even dangerous vibrations.

4.2 Static and Dynamic Balancing

Static Balancing

A system is said to be in static balance if the net dynamic force acting on the system is zero.

Σ F = 0

The net dynamic force acting on a shaft is zero. This requires that the line of action of three centrifugal forces must be same. In other words, the center of line of masses of the system must lie on the axis of the rotation

If the rotor is thin enough (longitudinally) as shown in Fig. the unbalance force can be assumed to be confined to one plane (the plane of the disc).

Such a case is known as “static” unbalance.

Such a system when mounted on a knife-edge as shown in Fig. will always come to rest in one position only – where the center of gravity comes vertically below the knife-edge point.

Dynamic Balancing

A system is said to be in dynamic balance if the net dynamic forces as well as net dynamic couples are equal to zero respectively.

Σ F = 0

Σ T = 0

The net couple due to dynamic forces acting on a shaft is equal to zero. In other words, the algebraic sum of moments about any point on the plane must be zero. i.e. the axis of shaft must coincide with the axis of rotation.

Consider the rotor shown in Fig. It is easily observed that mass distribution cannot be approximately confined to just one plane.

So, unbalance masses and hence unbalance forces are in general present all along the length of the rotor. Such a case is known as “dynamic unbalance”.

4.3 Balancing of Rotating Masses

Whenever a certain mass is attached to a rotating shaft, it exerts some centrifugal force, whose effect is to bend the shaft and to produce 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 effect of the centrifugal force of the first mass.

This is done in such a way that the centrifugal force of both the masses are made to be equal and opposite.

The process of providing the second mass in order to counteract the effect of the centrifugal force of the first mass, is called balancing of rotating masses.

4.4 Balancing of a single rotating mass by a single mass rotating in the same plane.

Consider a disturbing mass m1 attached to a shaft rotating at ω rad/s as shown in Fig.

Let r1 be the radius of rotation of the mass m1 (i.e. distance between the axis of rotation of the shaft and the center of gravity of the mass m1).

We know that the centrifugal force exerted by the mass m1 on the shaft,

This centrifugal force acts radially outwards and thus produces bending moment on the shaft.

In order to counteract the effect of this force, a balancing mass (m2) may be attached in the same plane of rotation as that of disturbing mass (m1) such that the centrifugal forces due to the two masses are equal and opposite.

Let r2 be the radius of rotation of the mass m2 (i.e. distance between the axis of rotation of the shaft and the center of gravity of the mass m2).

Therefore, centrifugal force exerted by the mass m2 on the shaft,

Equating both equations

In the previous arrangement for balancing gives rise to a couple which tends to rock the shaft in its bearings.

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.

The net couple due to dynamic forces acting on a shaft is equal to zero. In other words, the algebraic sum of moments about any point on the plane must be zero.

i.e. The system must be dynamic balance

4.5 Balancing of Several Masses Rotating in the Same Plane

Consider, any number of masses (say 4) of magnitude m1, m2, m3 and m4 at distances of r1, r2, r3, r4 from the axis of rotating shaft.

Let , , and be angles of these masses with horizontal line OX as shown in fig.

Let these masses rotate about an axis through O and perpendicular to page of paper, with a constant angular velocity rad/sec.

The magnitude and position of the balancing mass may be found out analytically or graphically as discussed below:

Analytical Method

The magnitude and direction of the balancing mass may be obtained, analytically, as discussed below:

First of all, find out the centrifugal force exerted by each mass on the rotating shaft

Resolve the centrifugal forces horizontally and vertically and find their sums, i.e. Σ H and Σ V.

We know that,

Summation of horizontal component of centrifugal forces,

Summation of vertical component of centrifugal forces,

The magnitude of resultant centrifugal force,

If,

is the angle which the resultant forces make with horizontal then,

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

Now, find the magnitude of the balancing mass such that,

Where,

= Balancing mass

= Radius of rotation

2. Graphical Method

The magnitude and position of the balancing mass may also be obtained graphically as discussed below:

First of all, draw the space diagram with the position of several masses, as shown in fig (a)

Find out the centrifugal force (or product of mass and radius of rotation) exerted by each mass on the rotating shaft.

Now, draw the vector diagram with the obtained centrifugal force, such that ab represents a centrifugal force exerted by the mass m1 (or m1 r1) in magnitude and direction with suitable scale. Similarly draw bc, cd and de to represent centrifugal forces of other masses m2, m3, and m4

Now, as per polygon law of forces, the closing side ae represents the resultant force in magnitude and direction, as shown in fig.

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

Now, find the magnitude of the balancing mass (m) at a given radius of rotation (r), such that,

Resultant Centrifugal force

Resultant of , , and

4.6 Balancing of Several Masses Rotating in Different Planes

When several masses revolve in different planes, they may be transferred to a reference plane (briefly written as 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 equal 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.

(a) Planer position of masses

In order to have a complete balance of the several revolving masses in different planes, the following two conditions must be satisfied :

1. The forces in the reference plane must balance, i.e. the resultant force must be zero.

2. The couples about the reference plane must balance, i.e. the resultant couple must be zero.

Let us now consider four masses m1, m2, m3 and m4 revolving in planes 1, 2, 3 and 4 respectively as shown in Fig. (a).

The relative angular positions of these masses are shown in the end view Fig. (b)

The magnitude of the balancing masses mL and mM in planes L and M may be obtained as discussed below:

Take one of the planes, say L, as the reference plane (R.P). The distance of all the other planes to the left of the R.P may be regarded as negative, and those to the right as positive.

Tabulate the data as shown in the table. The planes are tabulated in the same order in which they occur, reading from left to right

A couple may be represented by a vector drawn perpendicular to the plane of the couple. The couple C1 introduced by transferring m1 to th reference plane through O is proportional to

and acts in the plane through O

and perpendicular to the paper. The Vector representing this couple is drawn in the plane of the paper and perpendicular to O

as shown by OC1 in Fig (c). Similarly, the vectors, OC2, OC3 and OC4 are drawn perpendicular to O

, O

and O

respectively and in the plane of paper.

The couple vectors as discussed above, are turned counter clockwise through a right angle for convenience of drawing as shown in fig (d). We see that their relative positions remain unaffected. Now the vectors OC2, OC3 and OC4 are parallel and in the same direction as O

, O

and O

, while vector OC1 is parallel to O

but in opposite direction. Hence the couple vectors are drawn radially outwards for the masses on one side of the reference plane and radially inward for the masses on the other side of the reference plane.

Now draw the couple polygon as shown in fig (e). The vector d’o’ represents the balanced couple. Since the balanced couple CM is proportional to mM, rM, lM. Therefore

From this expression, the value of the balancing mass mM in the plane M may be obtained, and angle of inclination of this may be measured from fig (b)

Now, draw the force polygon as show in the fig (f). The vector eo (in the direction from e to o) represents the balanced force. Since the balanced force is proportional to mL.rL, therefore,

From this expression, the value of the balancing mass mL in the plane L may be obtained.

4.7 Balancing of reciprocating masses

Consider a horizontal reciprocating engine mechanism as shown in Fig.

FR = Force required to accelerate the reciprocating parts.

FI = Inertia force due to reciprocating parts,

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

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

Since FR and FI, are equal in magnitude but opposite in direction, therefore they balance each other.

The horizontal component of FB (i.e. FBH) acting along the line of reciprocation is also equal and opposite to FI.

This force FBH = FU is an unbalanced force or shaking force and required to be properly balanced.

The force on the sides of the cylinder walls (FN) and the vertical component of FB (i.e. FBV) are equal and opposite and thus form a shaking couple of magnitude FN x X or FBV x X.

From above, we see that the effect of the reciprocating parts is to produce a shaking force and a shaking couple. Since the shaking force and a shaking couple vary in magnitude and direction during the engine cycle, therefore they cause very objectionable vibrations.

Thus, the purpose of balancing the reciprocating masses is to eliminate the shaking force and a shaking couple. In most of the mechanisms, we can reduce the shaking force and a shaking couple by adding appropriate balancing mass, but it is usually not practical to eliminate them completely. In other words, the reciprocating masses are only partially balanced.

4.8 Primary and Secondary Unbalanced Forces of Reciprocating Masses

Consider a reciprocating engine mechanism as shown in Fig.

Let,

m = Mass of the reciprocating parts,

L = Length of the connecting rod PC,

r = Radius of the crank 0C,

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

ω = Angular speed of the crank,

n = Ratio of length of the connecting rod to the crank radius = L/r.

the acceleration of the reciprocating parts is approximately given by the expression

Therefore, inertia force due to reciprocating parts,

The horizontal component of the force exerted on the crank shaft bearing (i.e. FBH) is equal and opposite to inertia force (FI). This force is an unbalanced one and is denoted by FU.

Therefore, unbalanced force is,

Characteristics of primary and secondary unbalanced forces

The primary unbalanced force is maximum when θ = 0o or 180o. Thus, the primary force is maximum twice in one revolution of the crank. The maximum primary unbalanced force is given by,

The secondary unbalanced force is maximum when θ = 0o, 90o, 180o and 360o. Thus, the secondary force is maximum four times in one revolution of the crank. The maximum secondary unbalanced force is given by,

From above we see that secondary unbalanced force is 1/n times the maximum primary unbalanced force.

In case of moderate speeds, the secondary unbalanced force is so small that it may be neglected as compared to primary unbalanced force.

The unbalanced force due to reciprocating masses varies in magnitude but constant in direction while due to the revolving masses, the unbalanced force is constant in magnitude but varies in direction.

4.9 Balancing of Single Cylinder 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 Fig.

The primary force acts from O to P along the line of stroke. Hence, 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 a radius b, placed diametrically opposite to the crank pin C

A little consideration will show, that the primary 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) of magnitude ()

This force remains unbalanced. The maximum value of this force is equal to Bω2b when θ is 90° and 270°, which is same as the maximum value of the primary force mω2r

From the above discussion, we see that in the first case, the primary unbalanced force acts along the line of stroke whereas in the second case, the unbalanced force acts along the perpendicular to the line of stroke.

The maximum value of the 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 unbalanced force from the line of stroke to the perpendicular of line of stroke.

As a compromise, let a fraction ‘c’ of the reciprocating masses is balanced, such that: cmr = B.b

Therefore, Unbalanced force along the line of stroke

and unbalanced force along the perpendicular to the line of stroke

4.10 Balancing of Multi-Cylinder In-line Engine

Balancing of Primary Forces

The multi-cylinder engines with the cylinder center lines in the same plane and on the same side of the center line of the crankshaft, are known as In-line engines.

The following two conditions must be satisfied in order to give the primary balance of the reciprocating parts of a multi-cylinder engine:

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

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

The primary unbalanced force due to the reciprocating masses is equal to the component parallel to the line of stroke, of the centrifugal force produced by the equal mass placed at the crankpin and revolving with it.

Therefore, in order to give the primary balance of the reciprocating parts of a multi-cylinder engine, it is convenient to imagine the reciprocating masses to be transferred to their respective crankpins and to treat the problem as one of revolving masses.

Notes :

For a two-cylinder engine with cranks at 180°, condition (1) may be satisfied, but this will result in an unbalanced couple. Thus the above method of primary balancing cannot be applied in this case.

For a three-cylinder engine with cranks at 120° and if the reciprocating masses per cylinder are same, then condition (1) will be satisfied because the forces may be represented by the sides of an equilateral triangle. However, by taking a reference plane through one of the cylinder center lines, two couples with non-parallel axes will remain and these cannot vanish vectorially. Hence the above method of balancing fails in this case also.

For a four cylinder engine, similar reasoning will show that complete primary balance is possible.

For a multi-cylinder engine, the primary forces may be completely balanced by suitably arranging the crank angles, provided that the number of cranks are not less than four'

The closing side of the primary force polygon gives the maximum unbalanced primary force and the closing side of the primary couple polygon gives the maximum un-balanced primary couple

Balancing of Secondary Forces

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

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. 2 ω) as shown in Fig.

Thus, in multi-cylinder in-line engines, each imaginary secondary crank with a mass attached to the crankpin 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 secondary force polygon gives the maximum unbalanced secondary force and the closing side of the secondary couple polygon gives the maximum unbalanced secondary couple.

4.11 Balancing of Radial Engines (Direct and Reverse Cranks Method )

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 cranks (in radial or V-engines) is same, therefore there is no unbalanced primary or secondary couple. Consider a reciprocating engine mechanism as shown in Fig.

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 the primary forces

We have already discussed that primary force is

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 suppose 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.

We know that the centrifugal force acting on the primary direct and reverse crank

Hence, for primary effects the mass m of the reciprocating parts at P may be replaced by two masses at C and 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 line of stroke, are each equal to , but opposite in direction. Hence these components are balanced.

Considering the 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 2ω rad/s in the clockwise direction, while the crank OD' is the secondary reverse crank and rotates at 2ω rad/s in the anti CW direction as shown in Fig.

Maximum primary force acting on the frame of the compressor

The primary direct and reverse crank positions as shown in Fig (a) and (b), are obtained as discussed below:

Since θ=0° for cylinder 1, therefore both the primary direct and reverse cranks will coincide with the common crank.

Since θ = ±120° for cylinder 2, therefore the primary direct crank is 120° clockwise and the primary reverse crank is 120° anti-clockwise from the line of stroke of cylinder 2.

Since θ= ± 240° for cylinder 3, therefore the primary direct crank is 240° clockwise and the primary reverse crank is 240° anti-clockwise from the line of stroke of cylinder 3.

Maximum secondary force acting on the frame of the compressor

The secondary direct and reverse crank positions as shown in Fig. (a) and (b) are obtained as discussed below:

Since θ =0° and 2θ =0° for cylinder 1, therefore both the secondary direct and reverse cranks will coincide with the common crank.

Since θ = ±120° and 2θ = ± 240° for cylinder 2, therefore the secondary direct crank is 240° clockwise and the secondary reverse crank is 240° anticlockwise from the line of stroke of cylinder 2.

Since θ = ± 240° and 2θ = ± 480°, therefore the secondary direct crank is 480° or 120° clockwise and the secondary reverse crank is 480° or 120° anti-clockwise from the line of stroke of cylinder 3.

From Fig.(a), we see that the secondary direct cranks form a balanced system.

Therefore, there is no unbalanced secondary force due to the direct cranks. From Fig. (b) we see that the resultant secondary force is equivalent to the centrifugal force of a mass 3 m/2 attached at a crank radius of r/4n and rotating at a speed of 2ω rad/s in the opposite direction to the crank.2

4.12 Balancing of V-engines

Consider a symmetrical two cylinder V-engine as shown in Fig