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Qwestrum Engineering360 · Mechanical Engineering · Dynamics of Machines

Balancing of Reciprocating Masses

The reciprocating mass produces a primary force F_p = m·r·ω²·cosθ and a secondary force F_s = (m·r·ω²/n)·cos2θ, with n = L/r. Partial primary balance is by a rotating mass; secondary forces are handled by multi-cylinder phasing, per SS Rattan.
Exam tip: use ω in rad/s for dynamics formulas; N in rpm needs ω = 2πN/60. Do not confuse flywheel C_s with governor speed control.

Key formulas & points

Skim these first — then read the full notes below.

  • Primary force balanced by rotating balance mass
  • Secondary force reduced by multi-cylinder phasing (60°, 90° cranks)
  • Hammer blow on railways from unbalanced reciprocating mass

Topic details

Introduction

Reciprocating balancing extends rotating balance to the piston-crank mechanism, where the connecting rod makes the acceleration non-sinusoidal. Expanding piston acceleration gives a primary term at crank speed and a secondary term at twice crank speed.

Scope in B.Tech and GATE syllabus

SS Rattan explains that a rotating balance mass can cancel the primary force only along the line of stroke; balancing it fully introduces an equal unbalanced force perpendicular to the stroke, so only a fraction (typically 2/3) is balanced in locomotives. The residual gives rise to hammer blow on railway track — a classic numerical.

Why this topic matters in practice

Multi-cylinder engines balance primary and secondary forces and couples by choosing crank angles (e.g. 90° V8, 120° inline-three). Determining which forces/couples cancel for a given firing arrangement is the higher-order exam question.

Key relations & formulas

Fprimary=m⋅r⋅ω2cosθF_{primary} = m\cdot r\cdot \omega^{2} cos \theta
(primary unbalanced force)
Fsecondary=(m⋅r⋅ω2n)cos2θF_{secondary} = (m\cdot r\cdot \frac{\omega^{2}}{n}) cos 2\theta
(secondary, n = L/r)
Fnet≈m⋅r⋅ω2cosθ+(m⋅r⋅ω2n)cos2θF_{net} \approx m\cdot r\cdot \omega^{2} cos \theta + (m\cdot r\cdot \frac{\omega^{2}}{n}) cos 2\theta
(partial balance)
mbal=m⋅r⋅cos2(α2)m_{bal} = m\cdot r\cdot cos^{2}(\frac{\alpha}{2})
(balance mass for angle α between cranks)

Notation and sign conventions

Relation 1 —
Fprimary=m⋅r⋅ω2cosθF_{primary} = m\cdot r\cdot \omega^{2} cos \theta
Fprimary=m⋅r⋅ω2cosθF_{primary} = m\cdot r\cdot \omega^{2} cos \theta
(primary unbalanced force)
Write this relation with symbols exactly as in SS Rattan — Theory of Machines before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
Fsecondary=F_{secondary} =
Fsecondary=(m⋅r⋅ω2n)cos2θF_{secondary} = (m\cdot r\cdot \frac{\omega^{2}}{n}) cos 2\theta
(secondary, n = L/r)
Write this relation with symbols exactly as in SS Rattan — Theory of Machines before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
Fnet≈m⋅r⋅ω2cosθ+F_{net} \approx m\cdot r\cdot \omega^{2} cos \theta +
Fnet≈m⋅r⋅ω2cosθ+(m⋅r⋅ω2n)cos2θF_{net} \approx m\cdot r\cdot \omega^{2} cos \theta + (m\cdot r\cdot \frac{\omega^{2}}{n}) cos 2\theta
(partial balance)
Write this relation with symbols exactly as in SS Rattan — Theory of Machines before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 4 —
mbal=m⋅r⋅cos2m_{bal} = m\cdot r\cdot cos^{2}
mbal=m⋅r⋅cos2(α2)m_{bal} = m\cdot r\cdot cos^{2}(\frac{\alpha}{2})
(balance mass for angle α between cranks)
Write this relation with symbols exactly as in SS Rattan — Theory of Machines before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Fundamentals and definitions

Piston acceleration is a = ω²r(cosθ + cos2θ/n), where n = L/r is the connecting-rod-to-crank ratio. Multiplying by the reciprocating mass gives the inertia force with a primary part m·r·ω²·cosθ and a secondary part (m·r·ω²/n)·cos2θ.

Governing relations in practice

The primary force varies at crank speed; a rotating mass on the crank can oppose its component along the stroke, but that same mass adds an unbalanced component across the stroke. Hence only partial balance (a fraction c of the reciprocating mass) is used, minimising the overall swaying/hammer effect.

Design and analysis considerations

The secondary force oscillates at twice crank speed and is smaller by the factor 1/n; it cannot be balanced by a single rotating mass and instead relies on cylinder arrangement. In a four-cylinder inline engine the primary forces cancel but the secondary forces add — the reason for balance shafts.

Advanced theory and extensions

Hammer blow is the vertical component of the balancing mass's centrifugal force on a locomotive; it alternately increases and decreases wheel-rail load, limiting the speed. Computing hammer blow and the variation of tractive effort are standard problems.

Assumptions and validity limits

State assumptions explicitly before using any relation for balancing of reciprocating masses — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.
Wrong assumptions invalidate the entire solution even when the formula is correct. In Dynamics of Machines viva and GATE descriptive questions, listing valid assumptions often earns separate marks.

Step-by-step problem approach

1. Read the question and list given data with SI units (common in Dynamics of Machines papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to balancing of reciprocating masses.
4. Use equation 1:
Fprimary=m⋅r⋅ω2cosθF_{primary} = m\cdot r\cdot \omega^{2} cos \theta
.
5. Use equation 2:
Fsecondary=F_{secondary} =
.
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.

Applications & exam relevance

Balancing of Reciprocating Masses appears in engines, flywheels, and high-speed shafts. In Indian mechanical curricula this topic is tested because it connects theory to balancing, vibration, and rotational dynamics.
GATE and semester exams often combine balancing of reciprocating masses with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use balancing of reciprocating masses?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

• Ignoring the secondary force term entirely (it matters at high speed)
• Fully balancing the primary force, which just shifts unbalance perpendicular to the stroke
• Using the wrong n = L/r ratio (mixing up rod length and crank radius)
• Forgetting hammer blow acts vertically and alternates in sign around a revolution

Quick revision checklist

Before attempting balancing of reciprocating masses problems, confirm you can:
1. Primary force balanced by rotating balance mass
2. Secondary force reduced by multi-cylinder phasing (60°, 90° cranks)
3. Hammer blow on railways from unbalanced reciprocating mass
Revise the solved examples in SS Rattan — Theory of Machines and one previous-year GATE or university paper for this unit.

Worked examples

Try the problem first — open the solution when you are ready to check.

Primary inertia force

Problem

A reciprocating mass m = 1.2 kg has crank radius r = 0.05 m and rotates at ω = 100 rad/s. Find the maximum primary inertia force.

Solution

F_p(max) = m·r·ω² (at θ = 0) = 1.2 × 0.05 × 100² = 1.2 × 0.05 × 10000 = 600 N.

Conceptual check — Balancing of Reciprocating Masses

Problem

In a Dynamics of Machines semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of balancing of reciprocating masses." What should a complete answer include?

Practice questions

Most-asked interview and GATE questions for this topic — expand any item for a model answer.

  1. 1
    What is Balancing of Reciprocating Masses, and why does it appear in B.Tech / GATE syllabi?

    Model answer

    The reciprocating mass produces a primary force F_p = m·r·ω²·cosθ and a secondary force F_s = (m·r·ω²/n)·cos2θ, with n = L/r. Partial primary balance is by a rotating mass; secondary forces are handled by multi-cylinder phasing, per SS Rattan.
  2. 2
    State the relation F_primary = m·r·ω² cos θ and name each symbol.

    Model answer

    The governing relation is Fprimary=m⋅r⋅ω2cosθF_{primary} = m\cdot r\cdot \omega^{2} cos \theta. Write every symbol with SI units before substituting numbers.
  3. 3
    State the relation F_secondary = and name each symbol.

    Model answer

    The governing relation is Fsecondary=F_{secondary} =. Write every symbol with SI units before substituting numbers.
  4. 4
    State the relation F_net ≈ m·r·ω² cos θ + and name each symbol.

    Model answer

    The governing relation is Fnet≈m⋅r⋅ω2cosθ+F_{net} \approx m\cdot r\cdot \omega^{2} cos \theta +. Write every symbol with SI units before substituting numbers.
  5. 5
    State the relation m_bal = m·r·cos² and name each symbol.

    Model answer

    The governing relation is mbal=m⋅r⋅cos2m_{bal} = m\cdot r\cdot cos^{2}. Write every symbol with SI units before substituting numbers.
  6. 6
    Explain: Primary force balanced by rotating balance mass

    Model answer

    Primary force balanced by rotating balance mass — state the assumption range and one exam trap linked to this point.
  7. 7
    Explain: Secondary force reduced by multi-cylinder phasing (60°, 90° cranks)

    Model answer

    Secondary force reduced by multi-cylinder phasing (60°, 90° cranks) — state the assumption range and one exam trap linked to this point.
  8. 8
    Explain: Hammer blow on railways from unbalanced reciprocating mass

    Model answer

    Hammer blow on railways from unbalanced reciprocating mass — state the assumption range and one exam trap linked to this point.
  9. 9
    How would you correct this error in a viva: Ignoring the secondary force term entirely (it matters at high speed)?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  10. 10
    How would you correct this error in a viva: Fully balancing the primary force, which just shifts unbalance perpendicular to the stroke?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  11. 11
    How would you correct this error in a viva: Using the wrong n = L/r ratio (mixing up rod length and crank radius)?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  12. 12
    How would you correct this error in a viva: Forgetting hammer blow acts vertically and alternates in sign around a revolution?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.

Exams & GATE

  • 1
    SS Rattan Ch. 16 — 4-cylinder inline engine secondary force cancellation.
  • 2
    Avoid: Ignoring the secondary force term entirely (it matters at high speed)
  • 3
    Avoid: Fully balancing the primary force, which just shifts unbalance perpendicular to the stroke
  • 4
    Avoid: Using the wrong n = L/r ratio (mixing up rod length and crank radius)

📖 Standard books (India)

  • Theory of Machines — SS Rattan

    Read: Syllabus unit

    Kinematics, cams, governors, and balancing

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