mechanics of machine 2

2. Mehran University College of Engineering and
Technology Khairpur
Mechanics of machine 2
BALANCING
Abdul Ahad Noohani (MUCET KHAIRPUR)

3. BALANCING

4. 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,
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.

5. 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..

6. STATIC 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.

7. 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 bicycle or motorcycle tire and wheel,
a thin flywheel,
an airplane propeller,
an individual turbine blade-wheel (but not the
entire turbine)

8. 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.
Despite this fact, auto tires are sometimes statically
balanced.
More often they are dynamically balanced and will be
discussed under that topic.

10. 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

11. 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 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

12. 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.

13. Some examples of devices which require dynamic
balancing are:
• rollers
• crank-shafts
• camshafts
• axles
• clusters of multiple gears, motor rotors, turbines,
propeller shafts.

16. Balancing of Several Masses Rotating in the Same
Plane

17. 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 magnitude
of the balance mass required, if its radius of rotation is
0.2 m

18. solution

20. 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.

22. ∴ Resultant,

23. We know that
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,

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

25. 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).

26. Balancing of Several Masses Rotating
in Different Planes
reference plane: the plane passing through a
point on the axis of rotation and perpendicular to it.
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 inthe 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.

28. 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.

30. 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.

31. 1. First of all, draw the couple polygon from the
data given in Table
By measurement, the angular position of mY

32. Now draw the force polygon from the data given
By measurement, the angular position of mX

33. Example 21.4.
A, B, C and D are four masses carried by a
rotating shaft
at 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
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 the
shaft shall be in complete balance

36. UNBALANCE LEC # 02
The inertia force is an imaginary force, which when acts
upon a rigid body, brings it in an equilibrium position.
The resultant of all the forces acting on the body of the
engine due to inertia forces only is known as unbalanced
force or shaking force.
if the resultant of all the forces due to inertia effects is
zero, then there will be no unbalanced force, but even then
an unbalanced couple or shaking couple will be present

37. 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 force
and as shaking couple by adding appropriate balancing mass, but
it is usually not practical to eliminate them completely.

38. HORIZONTAL RECIPROCATING ENGINE MECHANISM

39. 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.
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.

40. 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.
but it is usually not practical to eliminate them completely.
In other words, the reciprocating masses are only partially
balanced.

41. LECTURE #03

42. PRIMARY AND SECONDARY UNBALANCED FORCES OF
RECIPROCATING MASSES
Consider a reciprocating engine
mechanism:
acceleration of the reciprocating parts is approximately given
by the expression,

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

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

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

46. 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.

48. Partial Balancing of Unbalanced Primary Force in a
Reciprocating Engine:
The primary unbalanced force may be considered as the component of the
centrifugal force

49. Let
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

50. Condition for balancing the primary force

51. vertical component centrifugal force :
but the centrifugal force has also a vertical component of
Magnitude:
This force remains unbalanced.
.

52. The maximum value of this force is
when : θ is 90° and 270°
which is same as the maximum value of the primary force

53. 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.

54. 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:

55. If the balancing mass is required to balance the
revolving masses as well as reciprocating masses,
then:

56. 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
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

59. Residual : What is left over