Balancing is Important in Machineries. Classified in to Balancing of Rotating and reciprocating Masses.

1. Unit 2 Module 2
Balancing of Reciprocating Masses
B. VARUN
Assistant Professor(Sr.Gr)
Department of Mechanical Engineering

2. Shaking Force
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 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

3. Primary and Secondary Unbalanced Forces of Reciprocating
Masses
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Considering a reciprocating engine mechanism
Let
m = Mass of the reciprocating parts,
l = Length of the connecting rod PC,
r = Radius of the crank OC,
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.

4. Acceleration
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 𝑎𝑝 = 𝜔2
𝑟 𝑐𝑜𝑠𝜃 +
cos 2𝜃
𝑛
 Inertia Force FI = Mass * Acceleration
 𝐹𝐼 = 𝑚 𝜔2𝑟 𝑐𝑜𝑠𝜃 +
cos 2𝜃
𝑛
 Primary Unbalanced Force = 𝑚 𝜔2
𝑟 𝑐𝑜𝑠𝜃
 Secondary Unbalanced Force = 𝑚 𝜔2
𝑟
cos 2𝜃
𝑛

5. Partial Balancing of Locomotives
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 The locomotives, usually, have two cylinders with cranks placed
at right angles to each other in order to have uniformity in
turning moment diagram.

6. Effect of Partial Balancing of Reciprocating Parts of Two
Cylinder Locomotives
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 Due to the partial balancing of the reciprocating parts, there is
an unbalanced primary force along the line of stroke and also an
unbalanced primary force perpendicular to the line of stroke.
1. Variation in tractive force along the line of stroke ;
2. Swaying couple.

7. Variation of Tractive Force
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 The resultant unbalanced force due to the two cylinders, along
the line of stroke, is known as tractive force.
Tractive Force= ± 𝟐 𝟏 − 𝒄 𝒎 𝒓 𝝎𝟐

8. Swaying Couple
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 The unbalanced forces along the line of stroke for the two
cylinders constitute a couple about the centre line between the
cylinders. This couple has swaying effect about a vertical axis,
and tends to sway the engine alternately in clockwise and
anticlockwise directions. Hence the couple is known as swaying
couple.
Swaying Couple= ±
𝒂
𝟐
𝟏 − 𝒄 𝒎 𝒓 𝝎𝟐

9. Hammer Blow
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 The maximum magnitude of the unbalanced force along the
perpendicular to the line of stroke is known as hammer blow.
Hammer Blow = 𝑩 𝝎𝟐
𝒃

10. Problem no 1
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The following data refer to a two cylinder
uncoupled locomotive
Rotating mass per cylinder : 280 kg
Reciprocating mass per cylinder : 300 kg
Distance between wheels : 1400mm
Distance between cylinder centers : 600mm
Diameter of treads of driving wheels :
1800mm
Crank radius :300mm
Radius of centre of balanced mass = 620
mm
Locomotive speed : 50km/hr
Angle between cylinder cranks : 90
Dead load on each wheels : 3.5 tonne
Determine :
 Balancing Mass required in the planes
of driving wheels if whole of revolving
and 2/3 of the reciprocating mass to
be balanced
 Swaying Couple
 Variation in the tractive force
 Maximum and minimum pressure on
the rails

11. Problem No 2
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The following data apply to an outside cylinder
uncoupled locomotive :
Rotating mass per cylinder = 360 kg ;
Reciprocating parts per cylinder= 300 kg ;
Angle between cranks = 90° ;
Crank radius = 0.3 m ;
Cylinder centres = 1.75 m ;
Radius of balance masses = 0.75 m ;
Wheel centres = 1.45 m.
If whole of the rotating and two-thirds of
reciprocating parts are to be balanced
in planes of the driving wheels,
find :
1. Magnitude and angular positions of
balance masses,
2. Speed in kilometres per hour at which
the wheel will lift off the rails when the
load on each driving wheel is 30 kN and
the diameter of tread of driving wheels
is 1.8 m, and
3. Swaying couple at speed arrived at in
(2) above.

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