Balancing of rotating and reciprocating masses

Why it matters

An unbalanced rotor or crank mechanism produces a shaking force that rotates with the shaft and grows with the square of speed. It loads the bearings, shakes the foundation, causes noise and fatigue, and spoils surface finish on machine tools. Balancing grinding wheels, fans, crankshafts and multi-cylinder engines is therefore part of making any high-speed machine work.

Key ideas

Centrifugal force of an eccentric mass. A mass m at radius r on a shaft turning at ω exerts a force F = m·r·ω² on the shaft, radially outward and rotating with it. Because ω² multiplies every force equally, balancing works with the products m·r (kg·m) and, for different planes, m·r·l (kg·m²).

Static balance (single plane). Several masses in one transverse plane are balanced when the vector sum of their m·r is zero, i.e. the centre of mass lies on the axis. One balancing mass placed opposite the resultant does it. A statically balanced rotor stays at rest in any angular position on knife edges.

Dynamic balance (several planes). Masses in different planes can have zero resultant force yet produce a rotating couple. Complete balance needs both:

• Σ m·r = 0 (no resultant force), and
• Σ m·r·l = 0 (no resultant couple), l measured from a chosen reference plane. Any unbalanced rotor can be balanced by two masses in two chosen planes. Taking one balancing plane as the reference plane removes its unknown from the couple equation, so the couple polygon gives the other balancing mass first, and the force polygon then gives the first.

Reciprocating masses. In a slider-crank the piston and the reciprocating part of the rod (mass m) need an accelerating force along the line of stroke:

• primary force m·ω²·r·cos θ, at crank frequency,
• secondary force m·ω²·r·cos 2θ/n, at twice crank frequency, where n = l/r. The primary force behaves like the component along the line of stroke of a mass m at the crank pin. Adding a rotating counter-mass opposite the crank cancels a fraction c of the primary force along the stroke but introduces c·m·ω²·r·sin θ perpendicular to it, so a single-cylinder engine can never be fully balanced. Typically c is chosen between 1/2 and 3/4 to share the unbalance between the two directions.

Multi-cylinder engines. Primary forces are represented by masses m at the actual cranks (radius r, speed ω). Secondary forces are represented by imaginary secondary cranks at angle 2θ, of radius r/(4n), turning at 2ω, which gives the same m·(2ω)²·(r/4n)·cos 2θ = m·ω²·r·cos 2θ/n. An engine is balanced when the force and couple polygons close for both primary and secondary cranks. An in-line four-cylinder engine with cranks at 0°, 180°, 180°, 0° is balanced for primary forces and couples but not for secondary forces; an in-line six is fully balanced for both.

Locomotives. In a two-cylinder locomotive, part of the reciprocating mass is balanced by masses in the driving wheels. The vertical component of this balance mass is unbalanced and adds to and subtracts from the rail load once per revolution (hammer blow). Partial balancing also causes a variation of tractive effort and a swaying couple.

Formulas

F = m·r·ω² — centrifugal force (N); m mass (kg), r radius of its centre of mass (m), ω (rad/s).

Σ m·r = 0 (vector) — static balance; Σ m·r·l = 0 (vector) — couple balance (dynamic, with the force condition).

Balancing mass in one plane: m_b·r_b = |Σ m·r|, placed at 180° to the resultant.

F_primary = m·ω²·r·cos θ — N; maximum m·ω²·r at θ = 0° and 180°.

F_secondary = m·ω²·r·cos 2θ/n — N; maximum m·ω²·r/n at θ = 0°, 90°, 180°, 270°.

Partial primary balance with fraction c (balance mass B at radius b, B·b = c·m·r): unbalanced along the stroke (1 − c)·m·ω²·r·cos θ, perpendicular to the stroke c·m·ω²·r·sin θ.

Hammer blow: B·ω²·b (N), with B the part of the balance mass that balances reciprocating parts.

Worked examples

Example 1 (standard). Three masses rotate in one plane: 5 kg at 0.2 m (0°), 4 kg at 0.25 m (90°) and 6 kg at 0.15 m (210°). Find the balancing mass at 0.2 m radius and its angle.

1. m·r values: 1.0 kg·m at 0°, 1.0 kg·m at 90°, 0.9 kg·m at 210°.
2. Horizontal: ΣX = 1.0 + 0 + 0.9 cos 210° = 1.0 − 0.779 = 0.221 kg·m.
3. Vertical: ΣY = 0 + 1.0 + 0.9 sin 210° = 1.0 − 0.45 = 0.55 kg·m.
4. Resultant: √(0.221² + 0.55²) = 0.593 kg·m at tan⁻¹(0.55/0.221) = 68.1°.
5. Balancing mass: m_b = 0.593/0.2 = 2.96 kg, at 68.1° + 180° = 248.1°.
6. 2.96 kg at 0.2 m, at 248.1°

Example 2 (GATE level). A single-cylinder engine runs at 600 rev/min. Reciprocating mass 20 kg, crank radius 0.1 m, connecting rod 0.45 m. (a) Find the primary and secondary unbalanced forces at θ = 30°. (b) Two-thirds of the reciprocating mass is balanced by a rotating mass at crank radius. Find the balance mass and the unbalanced forces along and perpendicular to the line of stroke at θ = 30°.

1. ω = 2π × 600/60 = 62.83 rad/s; m·ω²·r = 20 × 3948 × 0.1 = 7896 N; n = 0.45/0.1 = 4.5.
2. Primary: 7896 × cos 30° = 6838 N.
3. Secondary: 7896 × cos 60°/4.5 = 877 N.
4. Balance mass: B·b = c·m·r → B = (2/3) × 20 × 0.1/0.1 = 13.3 kg at crank radius, opposite the crank.
5. Along the stroke (primary): (1 − 2/3) × 7896 × cos 30° = 2279 N.
6. Perpendicular: (2/3) × 7896 × sin 30° = 2632 N.
7. (a) 6.84 kN primary, 0.877 kN secondary; (b) B ≈ 13.3 kg, 2.28 kN along and 2.63 kN across the line of stroke

Common mistakes

• Balancing forces only and forgetting the couple when masses are in different planes.
• Adding m·r values as scalars instead of vectors.
• Using n = r/l instead of l/r in the secondary force.
• Thinking a counterweight can fully balance a single-cylinder engine; it only moves the primary unbalance into the perpendicular direction.
• Forgetting the sign when the reference plane lies between masses (l is negative on the other side).
• Using rev/min in m·r·ω².

For GATE PI

Expect single-plane balancing (one balance mass and its angle), two-plane balancing of a shaft with masses at given positions, primary and secondary forces of a slider-crank at a given crank angle, and the effect of partial balancing. Practise resolving m·r into components quickly, and the secondary-force formula.

Quick check

1. A 2 kg mass at 0.1 m runs at 100 rad/s. What force does it exert on the shaft?
2. Is a rotor with zero resultant force necessarily balanced?
3. At what crank angles is the secondary force of a slider-crank a maximum?
4. In partial balancing, what fraction of the primary force appears perpendicular to the stroke at θ = 90° when c = 0.5?

Answers: 1. 2000 N; 2. no, it may still have an unbalanced couple; 3. 0°, 90°, 180° and 270°; 4. half of m·ω²·r.