Balancing Of Rotating Masses

When a number of rotors are mounted on a shaft, It is very difficult to match the centre of gravity of masses with the axis of rotation. Due to an eccentricity, when shaft rotates at an angular velocity $\omega$, there is centrifugal force of magnitude equal to inertia force set up in the system.

If the resultant force is not equal to zero, then transverse vibrations will take place. Thus rotating masses must be completely balanced to avoid any kind of vibrations.

Types Of Balancing

1. Static Balancing :

• In case of static balancing, the resultant inertia force acting on system must be zero or force polygon must be closed.

• In case a number of masses are rotating in one plane, then static balancing is the main criteria for finding the resultant unbalanced force.

2. Dynamic Balancing :

• When a number of masses are rotating in different planes, then to find out the resultant force, all masses are transferred in one plane and force polygon is completed.

• But when the forces are transferred, in one plane, there are couples produced by each of them, which will act perpendicular to line of action of force.

• In this case, for complete balancing, along with resultant force, resultant couple acting on system must be zero.

• In other words, for dynamic balancing, force polygon as well as couple polygon must be closed.

Balancing Of Reciprocating Masses

1. Balancing of single cylinder I. C. Engine :

• 

• Let 'm' be the mass of reciprocating parts, 'l' be the length of connecting parts 'r' be the radius of crank.

  $n = \frac {l}{r}$

• Acceleration of reciprocating mass is given by $a = \omega^{2} r [cos \ \theta + \frac {cos 2 \ \theta } {n} ]$

• Then inertia force $F_{1} = ma$ $\quad \quad \quad \quad \quad \quad \quad \quad = m \omega^{2} r \ cos \ \theta + \frac { m \omega^{2} r \ cos \ 2 \ \theta } {n}$

  $\therefore F_{1} $ = $ Primary \ force + Secondary \ force.$

• In above case, primary unbalanced force is grater than secondary unbalanced force.

• In case of average speed I.C. Engines, effect of secondary forces is negligible. From above expression we can say primary forces are maximum, twice in one revolution (i.e. at $\theta = 0^{\circ}$ and $180^{\circ}$).

2. Balancing of primary force in reciprocating engines :

• For primary force balancing, we can assume mass 'm' is kept at the crank pin and revolving with speed $\omega$. Then, component of this force, in the line of centre of cylinder = Primary = $m \omega^{2} r \ cos \ \theta $

• Then to balance this force, we can assume mass (B) is kept of radial distance 'b' exactly opposite to the crank pin. Then this mass will have the centrifugal force, $Bb \omega^{2}$. Then, component of this force in horizontal direction will balance the primary force i.e.
  $Bb \omega^{2} \ cos \ \theta \ = \ m r \omega^{2} \ cos \ \theta $
  

$\therefore \ Bb \ = \ mr$ .....(1)

• But component of centrifugal force, due to mass B, perpendicular to the centre line of cylinder will remain unbalanced. Hence in case of single cylinder reciprocating engine, complete balancing of reciprocating mass is not possible.

• Then as compromise, let 'C' be the fraction of reciprocating masses balanced.
  From equation (1), we can write
  $Bb = Cmr$
  Then unbalanced force along he line of stroke
  $=m \omega^{2} r \ \cos \theta \ - \ B b \omega \ \cos \theta$
  $=m \omega^{2} r \ \cos \theta\ - \ C m r \omega^{2} \ \cos \ \theta$
  $=m \omega^{2} r \ \cos \ \theta \ [1 \ - \ C]$
  Then component of unbalanced force, perpendicular to the line of stroke
  $=\mathrm{Bb} \omega^{2} \ \sin \ \theta \ = \ \mathrm{C} \mathrm{mr} \omega^{2} \ \sin \ \theta$
  $\therefore$ Resultant unbalanced force at any instant
  $=\sqrt{ [ m \omega^{2} r \ \cos \ \theta \ (1 \ - \ C)^{2} ]+\left(\mathrm{Cmr} \omega^{2} \ \sin \ \theta\right)^{2}}$
  $=m \omega^{2} r \sqrt{(1-C)^{2} \cos ^{2} \theta+C^{2} \sin ^{2} \theta}$
  Direction of this resultant unbalanced force with horizontal,
  $\tan \alpha=\frac{\text { unbalanced force in vertical direction }}{\text { unbalanced force in horizontal direction }}$

3. Balancing of primary forces of multi cylinder in line engines :

• Multi-cylinder in line engine means the cylinder center lines are in the same plane and on the same side of center line of the crankshaft.

• For balancing of multi-cylinder in line engines, following two conditions must be satisfied for primary force balancing.
  (i) The algebraic sum of primary forces must zero, i.e. primary force polygon-must be closed.
  (ii) The algebraic sum of couple about any point due to primary force must be zero, i.e. primary couple polygon must be closed.

4. Balancing of secondary forces of multi cylinder in line engines :

(i) Secondary force polygon must be closed.
(ii) Secondary couple polygon must be closed.

• In case of two stroke engines,
  Angle between adjacent cranks (sequenced in the order of firing )
  $=\frac{2 \ \pi}{\text { No. of cylinders }}$
  Whereas,
  In force stroke engines,
  Angle between adjacent cranks (sequenced in the order of firing )
  $=\frac{4 \ \pi}{\text { No. of cylinders }}$