Mehran University College of Engineering and Technology Khairpur

Mechanics of machine 2Mechanics of machine 2

BALANCINGBALANCING

Abdul Ahad Noohani (MUCET KHAIRPUR)Abdul Ahad Noohani (MUCET KHAIRPUR)

BALANCINGBALANCINGBALANCINGBALANCING

Balancing

� 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.

� To eliminate all shaking forces and shaking moments

� Any link or member that is in pure rotation can, theoretically, � Any link or member that is in pure rotation can, theoretically, be perfectly balanced to eliminate all shaking forces and shaking moments.

� A rotating member can be balanced either statically or dynamically.

� Rotating parts can, and generally should, be designed to be inherently balanced by their geometry.

� However, the vagaries(unpredictable change or action) of production tolerances guarantee that there will still be some small unbalance in each part. Thus a balancing procedure will have to be applied to each part after manufacture

� The amount and location of any imbalance can be measured quite accurately and compensated for by adding or removing material in the correct locations..

STATIC BALANCESTATIC BALANCESTATIC BALANCESTATIC BALANCE� The requirement for static balance is simply that the sum of all

forces on the moving system (including d‘ Alembert inertial forces) must be zero.

• Despite its name, static balance does apply to things in motion.

• The unbalanced forces of concern are due to the accelerations of masses in the system.

• An other name for static balance is single-plane balance, which means that the masses which are generating the inertia forces are in, or nearly in, the same plane.

• It is essentially a two-dimensional problem.

� Some examples of common devices which meet this criterion, and thus can successfully be statically balanced, are:

� a single gear or pulley on a shaft,a single gear or pulley on a shaft,

� a bicycle or motorcycle tire and wheel,

� a thin flywheel,

� an airplane propeller,

� an individual turbine blade-wheel (but not the entire turbine)

� The common denominator among these devices is that they are all short in the axial direction compared to the radial direction, and thus can be considered to exist in a single plane.

� An automobile tire and wheel is only marginally suited to static balancing as it is reasonably thick in the axial direction compared to its diameter.direction compared to its diameter.

� Despite this fact, auto tires are sometimes statically balanced.

� More often they are dynamically balanced and will be discussed under that topic.

� Note that the only forces acting on this system are the inertia forces.

� For balancing, it does not matter what external forces may be acting on the system.

� External forces cannot be balanced by making any changes to the system's internal geometry.

� For balancing, it also does not matter how fast the system is rotating

Dynamic Balance

� Any rotating object or assembly which is relatively long in the axial direction compared to the radial direction requires dynamic balancing for complete balance.

� Dynamic balance is sometimes called two plane balanceDynamic balance is sometimes called two plane balance

� It require that two criteria to be met

� The sum of forces must be zero

� And the sum of moments must also be zero

To correct dynamic imbalance requires either adding or removing the right amount of mass at the proper angular locations in two correction planes separated by some distance along the shaft.

This will create the necessary counter forces to statically balance the system and also provide a counter couple to cancel the unbalanced moment.

Some examples of devices which require dynamic balancing are:

• rollers

• crank-shafts

• camshafts• camshafts

• axles

• clusters of multiple gears, motor rotors, turbines, propeller shafts.

Balancing of Several Masses Rotating in the Same Balancing of Several Masses Rotating in the Same Balancing of Several Masses Rotating in the Same Balancing of Several Masses Rotating in the Same

PlanePlanePlanePlane

Example 21.1. Four masses m1, m2, m3 and m4 are 200 kg, 300 kg,

240 kg and 260 kg respectively.

The corresponding radii of rotation are 0.2 m, 0.15 m,

0.25 m and 0.3 m respectively

and the angles between successive masses are 45°, 75°

and 135°.

Find the position and magnitudeof the balance mass required, if its radius of rotation is

0.2 m

solution

Since the magnitude of centrifugal forces are proportional to the product of each mass and its radius, Therefore

The problem may, now, be solved either analytically or graphically. But we shall solve the problem by both the methods one by one.

∴∴∴∴ Resultant,

We know that

Since ′θ is the angle of the resultant R Since ′θ is the angle of the resultant R from the horizontal mass of 200 kg, therefore the angle of the balancing mass from the horizontal mass of 200 kg,

1. First of all, draw the space diagram showing the positions of all the given masses as shown in Fig (a)

2. Since the centrifugal force of each mass is proportional to the product of the mass and radius, therefore

3. Now draw the vector diagram with the above values, to some suitable scale as, shown in Fig. (b).

Balancing of Several Masses Rotating Balancing of Several Masses Rotating Balancing of Several Masses Rotating Balancing of Several Masses Rotating

in Different Planesin Different Planesin Different Planesin Different Planes

reference plane: the plane passing through apoint on the axis of rotation and perpendicular to it.

In order to have a complete balance of the severalIn order to have a complete balance of the severalrevolving masses in different planes, the following twoconditions 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.

Example 21.2

A shaft carries four masses A, B, C and D of magnitude 200 kg, 300 kg,400 kg and 200 kg respectively

and revolving at radii 80 mm, 70 mm, 60 mm and 80 mm in planes measured from A at 300 mm, 400 mm and 700 mm.

The angles between the cranks measured anticlockwise are A to B 45°, B to C 70°and C to D 120°.

The balancing masses are to be placed in planes X and Y. The distance between the planes A and X is 100 mm, between X and Y is 400mm and between Y and D is 200 mm.

If the balancing masses revolve at a radius of 100 mm, find their magnitudes and angular positions.

The distances of the planes to the right of plane X are taken as + ve while the distances of the planes to the left of plane X are taken as – ve.

1. First of all, draw the couple polygon from the data given in Table

By measurement, the angular position of mY

Now draw the force polygon from the data given

By measurement, the angular position of mX

Example 21.4.A, B, C and D are four masses carried by a

rotating shaftat radii 100,125, 200 and 150 mm respectively. The planes in which the masses revolve are spaced 600 mm apart and the mass of B, C and spaced 600 mm apart and the mass of B, C and D are 10 kg, 5 kg, and 4 kg respectively.Find the required mass A and the relative angular settings of the four masses so that theshaft shall be in complete balance

UNBALANCE

� The inertia force is an imaginary force, which when actsupon a rigid body, brings it in an equilibrium position.

LEC # 02

� The resultant of all the forces acting on the body of theengine due to inertia forces only is known as unbalancedforce or shaking force.

� if the resultant of all the forces due to inertia effects iszero, then there will be no unbalanced force, but even thenan unbalanced couple or shaking couple will be present

� the inertia torque is an imaginary torque, which when applied upon the rigid body, brings it in equilibrium position.

�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 forceand as shaking couple by adding appropriate balancing mass, butit is usually not practical to eliminate them completely.

HORIZONTAL RECIPROCATING ENGINE MECHANISM

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

� 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 or FBV × x.

� Effect of the reciprocating parts is to produce a shaking forceand a shaking couple.

� Since the shaking force and a shaking couple vary inmagnitude and direction during the engine cycle, thereforethey cause very objectionable vibrations.

� but it is usually not practical to eliminate them completely.In other words, the reciprocating masses are only partially balanced.

LECTURE #03

PRIMARY AND SECONDARY UNBALANCED FORCES OF RECIPROCATING MASSES

Consider a reciprocating engine

mechanism:

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

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

�The primary unbalanced force is maximum, when θ = 0° or 180°.

�Thus, the primary force is maximum twice in onerevolution of the crank.

�The maximum primary unbalanced force is given by

� The secondary unbalanced force is maximum, whenθ = 0°, 90°,180° and 360°.

� Thus, the secondary force is maximum four timesin one revolution of the crank.in one revolution of the crank.

� The maximum secondary unbalanced force is givenby

� From above we see that maximum 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.

Partial Balancing of Unbalanced Primary Force in a

Reciprocating Engine:

The primary unbalanced force may be considered as the component of the

centrifugal force

Let The mass B at a radius b balances this unbalance force when acted in The mass B at a radius b balances this unbalance force when acted in

opposite direction, placed diametrically opposite to the crank pin C.

We know that centrifugal force due to mass B,

horizontal component of this force,

The primary force is balanced, if

Condition for balancing the primary force

� but the centrifugal force has also a vertical component ofMagnitude:

This force remains unbalanced.

vertical component centrifugal force :

This force remains unbalanced..

� The maximum value of this force is

when : θ is 90° and 270°

� which is same as the maximum value of the primary force

�From the above discussion,

� In the first case, the primary unbalanced force acts

along the line of stroke.

� 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:

� above method of balancing changes 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:

Unbalanced force along the line of stroke:

and unbalanced force along the perpendicular to the line of

stroke

Resultant unbalanced force at any instant:

If the balancing mass is required to balance the

revolving masses as well as reciprocating masses,

then:

Example 22.1 A single cylinder reciprocating engine has

� speed 240 rpm

� stroke 300 mm

� mass of reciprocating parts 50 kg

� mass of revolving parts at 150 mm radius 37 kg.

If two-third of the reciprocating parts and all the revolving If two-third of the reciprocating parts and all the revolving

parts are to be balanced

find :

1. The balance mass required at a radius of 400 mm.

2. The residual unbalanced force when the crank has

rotated 60° from top dead centre

Residual : What is left over