Balance of Machinery

Unbalanced rotating masses produce a centrifugal force $mr\omega^2$ that rotates with the shaft and loads the bearings. Balancing removes it.

Static vs dynamic balancing

Dynamic balance includes static balance, but static balance does not guarantee dynamic balance.

Conditions for complete balance (several masses in different planes)

1. The vector sum of all the centrifugal forces is zero:

$$\sum mr\omega^2 = 0 \;\Rightarrow\; \sum mr = 0$$

2. The vector sum of the couples about any reference plane is zero:

$$\sum mr\,l = 0$$

where $l$ is the axial distance of each plane from the reference plane. Both conditions are solved graphically (force and couple polygons) or by resolving into horizontal and vertical components.

A single rotating mass in one plane cannot be balanced by one mass in a different plane because a couple remains. It needs two masses in two planes, so that both force and couple are balanced.

Set-up

     L        A         M
     |        |         |
     |  a     |   c     |
   --+--------+---------+-----> shaft axis
     m_L      m         m_M
     r_L      r         r_M
        |---------b----------|

Let the distances be $a$ (L to A), $c$ (A to M) and $b = a + c$ (L to M). The balancing masses are $m_L$ at radius $r_L$ and $m_M$ at radius $r_M$. All three masses lie in one axial plane through the shaft (same line).

Case (i) A between L and M

Force balance:

$$m_L r_L + m_M r_M = m r \quad(\text{balancing masses on the side opposite to } m)$$

Couple balance about plane L:

$$m_M r_M\, b = m r\, a \;\Rightarrow\; m_M r_M = m r\,\frac{a}{b}$$

Substituting in the force equation,

$$m_L r_L = m r - m r\frac{a}{b} = m r\,\frac{c}{b}$$

Both balancing masses act in the same direction, opposite to $m$ (at $180^\circ$ to it).

Case (ii) A outside L and M

Say A lies beyond M (order L, M, A), with A at distance $a$ from L and M at $b$ ($a > b$). Couple balance about plane L:

$$m_M r_M\, b = m r\, a \;\Rightarrow\; m_M r_M = m r\,\frac{a}{b}\ (\text{larger than } mr)$$

Let the direction of $m$ be positive. The couple equation fixes $m_M$ opposite to $m$. For the forces, with signs ($+$ same direction as $m$):

$$m_L r_L - m_M r_M + m r = 0 \;\Rightarrow\; m_L r_L = m_M r_M - m r = m r\,\frac{a-b}{b}$$

The result is positive, so $m_L$ acts in the same direction as $m$, and $m_M$ acts opposite to $m$.

Summary: The balancing masses follow from $\sum F = 0$ and $\sum M = 0$. When A is between the planes, both balancing masses are opposite to $m$. When A is outside, the plane nearer to A carries a larger mass opposite to $m$, and the farther plane carries a mass in the same direction as $m$.

For masses in one plane, balance requires $\sum mr = 0$ (vector sum). Take mass A along the x-axis ($0^\circ$) and find the resultant of the $mr$ vectors by resolving.

$$\sum H = \sum mr\cos\theta = 1.539\ \text{kg·m}, \qquad \sum V = \sum mr\sin\theta = 0.646\ \text{kg·m}$$

Resultant

$$R = \sqrt{(\sum H)^2 + (\sum V)^2} = \sqrt{(1.539)^2 + (0.646)^2} = 1.670\ \text{kg·m}$$ $$\tan\alpha = \frac{\sum V}{\sum H} = \frac{0.646}{1.539} \Rightarrow \alpha = 22.8^\circ \text{ (measured from A)}$$

Balancing mass

The balancing mass $m_b$ must cancel this resultant, so it acts opposite to it ($180^\circ$ away):

$$m_b r_b = R \Rightarrow m_b = \frac{1.670}{0.2} = 8.35\ \text{kg}$$ $$\theta_b = 22.8^\circ + 180^\circ = 202.8^\circ \text{ from mass A, in the same sense as B, C, D}$$

Answer: Balancing mass $= 8.35$ kg at 200 mm radius, at $202.8^\circ$ from mass A.

Method: Take plane L as the reference plane. For complete balance, $\sum mr = 0$ (forces) and $\sum mr\,x = 0$ (couples). Use the angle of A as $0^\circ$. Unknowns: $m_L$, $m_M$ and their angles.

Couple about plane L

$$\sum mrx\cos\theta = -2.478, \qquad \sum mrx\sin\theta = -0.413\ \ (\text{kg·m}^2)$$

The couple of the balancing mass in M about L must cancel this. Plane M is at $x_M = 1.70$ m, so

$$m_M r_M x_M = \sqrt{-2.478^2 + -0.413^2} = 2.512\ \text{kg·m}^2$$ $$m_M r_M = \frac{2.512}{1.70} = 1.478\ \text{kg·m} \Rightarrow m_M = \frac{1.478}{0.15} = 9.85\ \text{kg}$$ $$\theta_M = \text{angle of } (-\sum mrx\cos\theta,\ -\sum mrx\sin\theta) = 9.5^\circ$$

Force balance

The sum of the four given masses gives $\sum H = -0.381$, $\sum V = 1.782$ (kg·m). Adding plane M, the remainder to be cancelled by L is

$$m_L r_L\cos\theta_L = -1.077, \qquad m_L r_L\sin\theta_L = -2.025$$ $$m_L r_L = \sqrt{-1.077^2 + -2.025^2} = 2.294\ \text{kg·m} \Rightarrow m_L = \frac{2.294}{0.15} = 15.29\ \text{kg}$$ $$\theta_L = 242.0^\circ$$

Answer: Plane L: $m_L = 15.29$ kg at $242.0^\circ$ from A. Plane M: $m_M = 9.85$ kg at $9.5^\circ$ from A (both at 150 mm radius, same sense as the given angles).