Interactive Lecture
Static and Dynamic Balance of Rotating Machinery
by Dr Arnaldo Delli Carri
Static unbalance occurs when the centre of mass of a rotating system is displaced from the axis of rotation. Key characteristics:
Centrifugal Force: \( F = m \cdot r \cdot \omega^2 \)
Where:
Dynamic unbalance occurs when there are multiple unbalanced masses in different axial planes along the shaft. This causes:
Dynamic unbalance needs correction in at least two planes (two-plane balancing) to cancel both the net force and the net couple.
Explore single plane (static) unbalance. At 0 RPM, the shaft rotates to its stable position with the heavy side down.
Explore multi-plane (dynamic) unbalance with three adjustable masses. Balance is achieved using two correction planes near the bearings.
For complete dynamic balance, we must satisfy two conditions:
Force Balance: \( \sum MR = 0 \)
Moment Balance: \( \sum MR \cdot z = 0 \) (where \(z\) is each mass's axial position along the shaft)
This requires balance masses in two correction planes, typically near the bearings for maximum effectiveness.
Rotor balancing matters in any machine with a spinning shaft:
| Concept | Equation | Description |
|---|---|---|
| Static Balance | \(\sum m_i r_i = 0\) | Sum of mass-radius products equals zero |
| Dynamic Balance | \(\sum m_i r_i = 0\) in all planes | Balance in multiple planes |
| Centrifugal Force | \(F = m \cdot r \cdot \omega^2\) | Outward force on rotating mass |
| Balancing Mass | \(m_b r_b = \sqrt{(\sum m_i r_i \cos\theta_i)^2 + (\sum m_i r_i \sin\theta_i)^2}\) | Required balancing mass-radius product |
| Balancing Angle | \(\theta_b = 180° + \tan^{-1}\left(\frac{\sum m_i r_i \sin\theta_i}{\sum m_i r_i \cos\theta_i}\right)\) | 180° opposite the resultant unbalance (place the balance mass here) |
A rotating shaft has three unbalanced masses:
All masses are in the same plane. Calculate the magnitude and angle of the balance mass required at a radius of 75 mm.
Show Solution
Given: Mass 1: 50 g at 80 mm, 0°; Mass 2: 40 g at 60 mm, 90°; Mass 3: 30 g at 70 mm, 180°; all in one plane; balance radius \(r_b = 75\) mm.
Step 1: Calculate MR (mass times radius) for each mass
Mass 1: \(MR_1 = 50 \times 80 = 4000 \text{ g·mm}\) at 0°
Mass 2: \(MR_2 = 40 \times 60 = 2400 \text{ g·mm}\) at 90°
Mass 3: \(MR_3 = 30 \times 70 = 2100 \text{ g·mm}\) at 180°
Step 2: Resolve into x and y components
\(MR_{1x} = 4000 \cos(0°) = 4000 \text{ g·mm}\), \(MR_{1y} = 4000 \sin(0°) = 0 \text{ g·mm}\)
\(MR_{2x} = 2400 \cos(90°) = 0 \text{ g·mm}\), \(MR_{2y} = 2400 \sin(90°) = 2400 \text{ g·mm}\)
\(MR_{3x} = 2100 \cos(180°) = -2100 \text{ g·mm}\), \(MR_{3y} = 2100 \sin(180°) = 0 \text{ g·mm}\)
Step 3: Sum the components
\(\sum MR_x = 4000 + 0 - 2100 = 1900 \text{ g·mm}\)
\(\sum MR_y = 0 + 2400 + 0 = 2400 \text{ g·mm}\)
Step 4: Calculate resultant unbalance
\(MR_{resultant} = \sqrt{1900^2 + 2400^2} = \sqrt{3,610,000 + 5,760,000} = 3061 \text{ g·mm}\)
\(\theta = \tan^{-1}\left(\frac{2400}{1900}\right) = 51.6°\)
Step 5: Balance mass (opposite direction: 180° + 51.6° = 231.6°)
\(m_b = \frac{MR_{resultant}}{r_b} = \frac{3061}{75} = 40.8 \text{ g}\) at 231.6°
Answer: 40.8 g at 231.6° (at 75 mm radius).
A shaft 600 mm long has two unbalanced masses:
Calculate the balance masses required in correction planes at the bearings (0 mm and 600 mm from left) at a radius of 65 mm.
Show Solution
Given: shaft 600 mm long; Mass A 60 g at 70 mm, 0°, axial position \(z = 150\) mm; Mass B 50 g at 80 mm, 120°, axial position \(z = 450\) mm; correction planes at \(z = 0\) and \(z = 600\) mm, both at \(r_b = 65\) mm. (\(z\) is the axial distance from the left bearing; the plane labels L and R are the left and right correction planes.)
Step 1: Calculate MR values
\(MR_A = 60 \times 70 = 4200 \text{ g·mm}\) at 0°, axial position \(z = 150\) mm
\(MR_B = 50 \times 80 = 4000 \text{ g·mm}\) at 120°, axial position \(z = 450\) mm
Step 2: Resolve into components
Mass A: \(x = 4200\), \(y = 0\)
Mass B: \(x = 4000 \cos(120°) = -2000\), \(y = 4000 \sin(120°) = 3464\)
Step 3: Force balance equations
\(\sum MR_x = 0: (MR_L)_x + (MR_R)_x + 4200 - 2000 = 0\)
\(\sum MR_y = 0: (MR_L)_y + (MR_R)_y + 0 + 3464 = 0\)
Step 4: Moment balance about left bearing
\(\sum (MR \cdot z)_x = 0: (MR_R)_x \times 600 + 4200 \times 150 - 2000 \times 450 = 0\)
\((MR_R)_x = \frac{900,000 - 630,000}{600} = 450 \text{ g·mm}\)
\(\sum (MR \cdot z)_y = 0: (MR_R)_y \times 600 + 3464 \times 450 = 0\)
\((MR_R)_y = -\frac{3464 \times 450}{600} = -2598 \text{ g·mm}\)
Step 5: Calculate balance masses
The correction MR at each plane comes straight from the balance equations, so its own angle is where the balance mass goes (no extra 180° flip).
Right plane: \((MR_R)_x = 450\), \((MR_R)_y = -2598\), so \(MR_R = \sqrt{450^2 + 2598^2} = 2637 \text{ g·mm}\)
Balance mass at right: \(m_R = 2637/65 = 40.6 \text{ g}\) at \(\theta_R = \tan^{-1}\!\left(\frac{-2598}{450}\right) = 279.8°\) (fourth quadrant)
Left plane: From force balance, \((MR_L)_x = -2650\), \((MR_L)_y = -866\), so \(MR_L = \sqrt{2650^2 + 866^2} = 2788 \text{ g·mm}\)
Balance mass at left: \(m_L = 2788/65 = 42.9 \text{ g}\) at 198.1° (third quadrant)
Answer: Right plane 40.6 g at 279.8°; left plane 42.9 g at 198.1° (both at 65 mm radius).
A rotor has an unbalance of 80 g at a radius of 100 mm. The rotor operates at 1800 RPM. Calculate:
Show Solution
Given: \(m = 0.08 \text{ kg}\), \(r = 0.1 \text{ m}\), \(\omega = \frac{2\pi N}{60}\)
Part 1: At 1800 RPM
\(\omega = \frac{2\pi \times 1800}{60} = 188.5 \text{ rad/s}\)
\(F = m \cdot r \cdot \omega^2 = 0.08 \times 0.1 \times (188.5)^2 = 284.2 \text{ N}\)
Part 2: At 3600 RPM
\(\omega = \frac{2\pi \times 3600}{60} = 377.0 \text{ rad/s}\)
\(F = 0.08 \times 0.1 \times (377.0)^2 = 1137 \text{ N}\)
Conclusion: Doubling the speed quadruples the force because \(F \propto \omega^2\). The same unbalance that is a mild nuisance at 1800 rpm puts a 1137 N rotating load on the bearings at 3600 rpm — enough to cut bearing life sharply and start fatigue cracks in the shaft.
Answer: 284.2 N at 1800 rpm; 1137 N at 3600 rpm.