Balancing of rotating and reciprocating masses

Why it matters

An unbalanced rotor or crank train shakes its bearings and frame once (or twice) every revolution, with a force that grows with the square of speed. That is the cause of most machine vibration complaints: worn bearings, fatigue cracks, noise and poor surface finish on machine tools. Balancing rotors on a balancing machine and choosing crank arrangements in engines removes or reduces these shaking forces at the design stage.

Key ideas

Rotating unbalance. A mass m at radius r turning at ω needs a centripetal force m·ω²·r from the shaft; the shaft and bearings feel an equal outward force that rotates with the shaft. Because every mass on a shaft turns at the same ω, balancing works with the products m·r (for forces) and m·r·l (for couples, with l the axial distance from a reference plane), treated as vectors at the masses' angular positions.

Static and dynamic balance.

• Static balance: the centre of mass lies on the axis, Σ m·r = 0 (vector sum). The rotor stays at rest in any angular position on knife edges. One balancing mass in any plane can achieve this.
• Dynamic balance: in addition, the centrifugal forces produce no net couple, Σ m·r·l = 0. A long rotor (motor armature, crankshaft, turbine) needs this, and in general needs balancing masses in two planes.
• A statically balanced rotor can still be dynamically unbalanced: two equal masses 180° apart in different planes form a rotating couple.

Several masses in one plane. Add the m·r vectors (graphically by a force polygon or analytically by components). The balancing mass supplies an equal and opposite m·r.

Several masses in different planes (reference-plane method). Choose one balancing plane as the reference plane. Taking moments about it eliminates the balancing mass in that plane: the couple polygon (m·r·l) gives the balancing mass in the other plane. Then the force polygon (m·r), now including that mass, gives the balancing mass in the reference plane. Masses on opposite sides of the reference plane take opposite signs of l.

Reciprocating unbalance (single-cylinder). The inertia force of the reciprocating mass m acts 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 (n = l/r). The primary force is the component along the stroke of a force m·ω²·r rotating with the crank, so it behaves like a mass m at the crank pin, seen only along the stroke. A rotating counterweight cannot cancel a force that acts along one line only: balancing it fully along the stroke creates an equal unbalance at right angles. So a fraction c (usually 1/2 to 2/3) of the reciprocating mass is balanced, as a compromise, together with all of the rotating mass.

Multi-cylinder in-line engines. Treat each cylinder's primary force as a crank-pin mass at its crank angle, and each secondary force as an imaginary crank at angle 2θ rotating at 2ω with radius r/(4n). Primary balance requires the primary force and couple polygons to close; secondary balance needs the same with angles doubled. A four-cylinder in-line engine with cranks at 0°, 180°, 180°, 0° has primary forces and couples balanced but an unbalanced secondary force 4·m·ω²·r·cos 2θ / n, which is why some four-cylinder engines carry twin balancer shafts running at 2ω. An in-line six is fully balanced for primary and secondary forces and couples.

Locomotives (partial balancing of two-cylinder engines). Balancing part of the reciprocating mass with wheel counterweights causes three effects: variation of tractive effort (residual unbalance along the line of stroke), swaying couple about a vertical axis (the two cylinders' residual forces are a distance a apart), and hammer blow (the vertical component of the counterweight's centrifugal force, alternately pressing on and lifting off the rail). Hammer blow sets the speed at which a wheel would lift.

Formulas

F = m·ω²·r (centrifugal force of a rotating mass, N)

Single plane: Σ m·r·cos θ + m_b·r_b·cos θ_b = 0 and Σ m·r·sin θ + m_b·r_b·sin θ_b = 0 m_b·r_b = √((Σ m·r·cos θ)² + (Σ m·r·sin θ)²), at θ_b = tan⁻¹(Σ m·r·sin θ / Σ m·r·cos θ) + 180°

Two planes: Σ m·r·l = 0 (couples about the reference plane) and Σ m·r = 0 (forces), both as vectors.

Reciprocating (single cylinder, n = l/r):

• primary F_p = m·ω²·r·cos θ ; secondary F_s = m·ω²·r·cos 2θ / n
• partial balance fraction c: residual along stroke (1 − c)·m·ω²·r·cos θ; across the stroke c·m·ω²·r·sin θ
• balance mass at radius b: B·b = c·m·r (plus the rotating mass share m_rot·r)

Locomotive (two cylinders, cranks at 90°, cylinder centre distance a):

• variation of tractive effort = ±√2·(1 − c)·m·ω²·r
• swaying couple = ±(a/√2)·(1 − c)·m·ω²·r
• hammer blow = B·ω²·b

Worked examples

Example 1 (standard): three masses in one plane. Given: 10 kg at 100 mm and 0°; 8 kg at 150 mm and 90°; 6 kg at 120 mm and 210°. Find the balancing mass at 200 mm.

1. m·r values: 1.0, 1.2, 0.72 kg·m.
2. Horizontal: 1.0 cos 0° + 1.2 cos 90° + 0.72 cos 210° = 1.0 + 0 − 0.6235 = 0.3765 kg·m
3. Vertical: 0 + 1.2 − 0.72 × 0.5 = 0.84 kg·m
4. Resultant = √(0.3765² + 0.84²) = 0.9205 kg·m at tan⁻¹(0.84 / 0.3765) = 65.9°.
5. Balancing: m_b = 0.9205 / 0.2 = 4.60 kg at 65.9° + 180° = 245.9°. Answer: 4.60 kg at 200 mm radius, at 245.9°.

Example 2 (GATE level): two-plane balancing. Given: on a shaft, 5 kg at 200 mm radius, 0°, in a plane 0.3 m from plane L; 4 kg at 150 mm, 120°, 0.7 m from L. Balancing masses at 200 mm radius in plane L (0 m) and plane R (1.0 m).

1. Couples about L (m·r·l): 5 × 0.2 × 0.3 = 0.30 at 0°; 4 × 0.15 × 0.7 = 0.42 at 120° → components (−0.21, 0.364).
2. Sum: (0.30 − 0.21, 0 + 0.364) = (0.09, 0.364) kg·m². Plane R must give m_R × 0.2 × 1.0 = −(0.09, 0.364) → magnitude 0.375 kg·m at 256.1°, so m_R = 1.87 kg.
3. Forces (m·r): 1.0 at 0° → (1.0, 0); 0.6 at 120° → (−0.30, 0.520); plane R mass → (−0.09, −0.364). Sum: (0.61, 0.156).
4. Plane L: m_L × 0.2 = 0.630 kg·m at 194.3° → m_L = 3.15 kg. Answer: m_R ≈ 1.87 kg at 256°, m_L ≈ 3.15 kg at 194° (both at 200 mm).

Example 3 (reciprocating, partial balance). Reciprocating mass 20 kg, r = 150 mm, n = 4, 300 rpm, c = 2/3, θ = 30°.

1. ω = 31.42 rad/s ; m·ω²·r = 20 × 987.0 × 0.15 = 2961 N
2. Primary 2961 × cos 30° = 2564 N; secondary 2961 × cos 60° / 4 = 370 N (unbalanced).
3. With c = 2/3: along stroke (1/3) × 2564 = 855 N; across (2/3) × 2961 × sin 30° = 987 N; resultant 1306 N.
4. Balance mass at crank radius: B = (2/3) × 20 = 13.3 kg (for the reciprocating part). Answer: primary 2.56 kN and secondary 0.37 kN unbalanced; with 2/3 balanced the primary residual is 1.31 kN at θ = 30°.

Common mistakes

• Adding m·r values as scalars instead of vectors.
• Forgetting that masses on the other side of the reference plane have negative l.
• Calling a rotor balanced because it is statically balanced.
• Thinking a counterweight can fully balance a reciprocating mass; it only moves the unbalance across the stroke.
• Using ω in rpm.
• Forgetting the 2θ crank angles and 2ω speed for secondary forces.

For GATE ME

Expect: balancing mass for several masses in one plane; static versus dynamic balance; primary and secondary unbalanced forces; which multi-cylinder arrangements leave secondary forces or couples unbalanced; hammer blow and tractive-effort variation. Practise component tables for single- and two-plane problems.

Quick check

1. Two 2 kg masses at equal radius, 180° apart, in planes 0.5 m apart. Static and dynamic balance?
2. Frequency of the secondary force for an engine at 3000 rpm?
3. Why is only part of the reciprocating mass balanced in a single-cylinder engine?
4. What unbalance remains in a 0°-180°-180°-0° four-cylinder in-line engine?
5. A 3 kg mass at 0.1 m turns at 100 rad/s. Its centrifugal force?

Answers: 1. Statically balanced, dynamically unbalanced (rotating couple). 2. 100 Hz (twice the 50 Hz crank frequency). 3. Full balancing along the stroke would create an equal unbalance at right angles. 4. Secondary force. 5. 3 × 10⁴ × 0.1 = 3000 N.