Lesson 3.4: Balancing of Rotating & Reciprocating Masses

Balancing is crucial for smooth operation of machines. GATE and PSU exams often test concepts of static and dynamic balancing, reciprocating engine vibration, and mass forces.

🔹 1. Introduction

• Definition: Balancing is the process of eliminating or reducing shaking forces and moments in rotating or reciprocating machinery.

• Importance:

  • Reduces vibration

  • Prevents bearing & shaft failure

  • Improves machine life and efficiency

• Types of Imbalance:

  1. Static Imbalance: Center of mass not on axis → produces vertical shaking

  2. Dynamic Imbalance: Unequal mass distribution along shaft → causes rotation vibration

🔹 2. Balancing of Rotating Masses

• Single Rotating Mass:

  • Mass $m$ at radius $r$ → centrifugal force F=mrω2F = m r \omega^2F=mrω2

  • Static balancing: Place counterweights so center of mass aligns with axis

• Two or More Rotating Masses:

  • Vector sum of centrifugal forces = 0 → balanced

  • Graphical Method: Mass polygon or vector diagram

  • Analytical Method: Solve equations for counterweights

• Applications: Rotors, fans, turbines, grinding machines

🔹 3. Balancing of Reciprocating Masses

• Reciprocating parts: Pistons, connecting rods → produce primary and secondary forces

• Primary Force (F1):

F1=mrω2F_1 = m r \omega^2F1​=mrω2

• Secondary Force (F2): Due to non-uniform acceleration of connecting rod

F2=mrω2rlcos⁡2θF_2 = m r \omega^2 \frac{r}{l} \cos 2\thetaF2​=mrω2lr​cos2θ

Where $l$ = length of connecting rod

• Balancing Methods:

  1. Partial Balancing: Some secondary forces left unbalanced (common in engines)

  2. Complete Balancing: All primary & secondary forces balanced → used in high-speed engines

• Hammer Blow & Fluctuating Force: Result of incomplete balancing → design consideration

🔹 4. Reciprocating Engine Example

• Single-cylinder engine: m = 2 kg, r = 0.1 m, l = 0.3 m, ω = 100 rad/s

• Primary force: F1=2∗0.1∗1002=2000F_1 = 2*0.1*100^2 = 2000F1​=2∗0.1∗1002=2000 N

• Secondary force: F2=2∗0.1∗1002∗0.1/0.3∗cos⁡2θ=666.7cos⁡2θF_2 = 2*0.1*100^2 * 0.1/0.3 * \cos 2\theta = 666.7 \cos 2\thetaF2​=2∗0.1∗1002∗0.1/0.3∗cos2θ=666.7cos2θ N

Observation: Primary force much higher → counterweights used on crankshaft

🔹 5. Solved Examples (PYQ Style)

Example 1 (GATE ME 2017):
Single rotating mass, m = 5 kg, r = 0.2 m, ω = 50 rad/s. Find balancing mass at opposite side.

• Centrifugal force F=mrω2=5∗0.2∗502=2500NF = m r ω^2 = 5*0.2*50^2 = 2500 NF=mrω2=5∗0.2∗502=2500N

• Place counterweight of 5 kg opposite to eliminate shaking

Example 2 (PSU Exam):
Two-cylinder reciprocating engine, cylinders 180° apart, m = 1.5 kg, r = 0.08 m. Find primary shaking force.

• F=mrω2(1−cosθ)F = m r ω^2 (1 – cos θ)F=mrω2(1−cosθ) (use standard formula)

🔹 6. Practice Exercises

1. Find counterweight required to balance a single rotating mass.

2. Compute primary and secondary forces for a reciprocating engine (given m, r, l, ω).

3. Explain why multi-cylinder engines have reduced vibration.

4. Draw vector diagram for two rotating masses at 90° apart.

5. Discuss consequences of incomplete balancing in high-speed machinery.

🔹 7. Summary

• Purpose: Reduce vibration and shaking forces

• Rotating Masses: Centrifugal forces, static & dynamic balancing

• Reciprocating Masses: Primary & secondary forces, partial & complete balancing

• Applications: Engines, turbines, compressors, reciprocating machines