Dynamic force analysis of slider-crank mechanism

Why it matters

In engines, compressors, presses and crank-driven actuators, the forces in the connecting rod, crank pin, bearings and cylinder wall are set as much by inertia as by gas or load pressure. At high speed the inertia force of the piston can exceed the gas force over part of the cycle. Dynamic force analysis gives the turning moment the crank delivers at each angle, which feeds directly into flywheel design, bearing selection and balancing.

Key ideas

D'Alembert's principle. A body accelerating at a can be treated as if in equilibrium by adding an inertia force −m·a at its centre of mass (and an inertia couple −I_G·α). Dynamic analysis then becomes a static force analysis at each crank position.

Forces in a horizontal engine (in-line slider-crank). θ = crank angle from inner dead centre (the dead centre where the piston is farthest from the crank), φ = angle of the connecting rod (obliquity), n = l/r.

• Gas (load) force on the piston: F_g = p·A (net of back pressure for double-acting cylinders).
• Inertia force of reciprocating parts: F_i = m_R·a_P, opposing acceleration. In the first half of the outstroke the piston accelerates, so F_i opposes the gas force; in the second half it decelerates and F_i adds to it.
• Piston effort (net force along the line of stroke): F_P = F_g − F_i (minus friction if given; for a vertical engine add the weight m_R·g on the down-stroke).
• Force along the connecting rod: F_Q = F_P / cos φ.
• Side thrust on the cylinder wall: F_N = F_P·tan φ; a longer rod (larger n) reduces it.
• At the crank pin, F_Q splits into a tangential crank effort F_T = F_Q·sin(θ + φ) and a radial force F_R = F_Q·cos(θ + φ) on the main bearing.
• Turning moment on the crankshaft: T = F_T·r.

Reciprocating mass. m_R = piston + gudgeon pin + rings + the part of the connecting rod mass assigned to the small end.

Connecting rod as a dynamically equivalent system. A rigid body of mass m and radius of gyration k_G can be replaced by two point masses m₁ and m₂ on a line through G, at distances l₁ and l₂ from G on opposite sides, if:

• m₁ + m₂ = m (same mass);
• m₁·l₁ = m₂·l₂ (same centre of mass);
• l₁·l₂ = k_G² (same moment of inertia). Fixing m₁ at the small end decides l₂; the second mass then lands at a point other than the crank pin. For quick work the rod is split statically (masses inversely proportional to distances from G) into the small and big ends; this keeps mass and centre of mass but not inertia, so a small correction couple is needed for exact work.

Signs and the cycle. Over a revolution F_P, T and F_N change magnitude and sign. Plotting T against θ gives the turning moment diagram, the starting point for flywheel design. Friction is usually neglected in standard problems unless it is given.

Formulas

F_g = p·A (p in Pa, A = πD²/4 in m², F in N) a_P ≈ ω²·r·(cos θ + cos 2θ / n) (m/s²) F_i = m_R·a_P (N) F_P = F_g ∓ F_i (minus while the piston accelerates, plus while it decelerates; ± m_R·g for vertical engines) sin φ = sin θ / n F_Q = F_P / cos φ ; F_N = F_P·tan φ F_T = F_Q·sin(θ + φ) ; F_R = F_Q·cos(θ + φ) T = F_P·r·(sin θ + sin 2θ / (2√(n² − sin² θ))) (N·m; same as F_T·r) Equivalent system: m₁ + m₂ = m, m₁·l₁ = m₂·l₂, l₁·l₂ = k_G²

Worked examples

Example 1 (standard): forces at one crank angle. Given: horizontal engine, bore 100 mm, net gas pressure 1.2 MPa, reciprocating mass 1.5 kg, crank r = 75 mm, rod l = 300 mm, 1800 rpm, θ = 30° from inner dead centre (outstroke).

1. A = π(0.1)²/4 = 7.854 × 10⁻³ m² → F_g = 1.2 × 10⁶ × 7.854 × 10⁻³ = 9425 N
2. ω = 2π × 1800 / 60 = 188.5 rad/s, n = 4.
3. a_P = 188.5² × 0.075 × (cos 30° + cos 60° / 4) = 2665 × 0.991 = 2641 m/s²
4. F_i = 1.5 × 2641 = 3961 N (opposes the gas force; piston is accelerating).
5. F_P = 9425 − 3961 = 5463 N
6. sin φ = sin 30° / 4 = 0.125 → φ = 7.18°.
7. F_Q = 5463 / cos 7.18° = 5507 N ; F_N = 5463 × tan 7.18° = 688 N
8. F_T = 5507 × sin 37.18° = 3328 N ; T = 3328 × 0.075 = 249.6 N·m Answer: piston effort ≈ 5.46 kN, rod thrust ≈ 5.51 kN, side thrust ≈ 688 N, turning moment ≈ 250 N·m.

Example 2 (GATE level): equivalent masses of a connecting rod. Given: rod mass 2 kg, centre distance 250 mm, G is 75 mm from the big-end centre (so 175 mm from the small end), k_G = 100 mm. Replace the rod by two masses, one at the small-end centre.

1. l₁ = 175 mm (small end to G). l₂ = k_G² / l₁ = 100² / 175 = 57.1 mm on the other side of G.
2. m₁ = m·l₂ / (l₁ + l₂) = 2 × 57.1 / 232.1 = 0.492 kg at the small end.
3. m₂ = 2 − 0.492 = 1.508 kg at 57.1 mm from G towards the big end, i.e. 17.9 mm short of the crank pin.
4. Static split for comparison: small end 2 × 75 / 250 = 0.6 kg, big end 1.4 kg. This puts 0.6 kg at the small end against 0.49 kg in the exact system; the difference is why a correction couple is needed. Answer: 0.492 kg at the small end and 1.508 kg at 57.1 mm from G (dynamically equivalent); static split 0.6 kg / 1.4 kg.

Common mistakes

• Adding the inertia force to the gas force during the first half of the outstroke; it opposes it while the piston accelerates.
• Using F_P·r or F_Q·r as the turning moment; the perpendicular component is F_Q·sin(θ + φ).
• Forgetting to convert pressure from MPa or bar to Pa, or diameter from mm to m.
• Using the rpm value in ω²r.
• Ignoring the piston weight in a vertical engine.
• Assuming a two-mass split at the two ends is dynamically equivalent; it matches only mass and centre of mass.

For GATE ME

Expect: piston effort, thrust in the connecting rod, side thrust and crank effort or turning moment at a given crank angle; inertia force of reciprocating parts; dynamically equivalent two-mass systems. Practise the full chain F_g → F_i → F_P → F_Q → F_T → T at one angle, with the sign of F_i set by whether the piston is speeding up or slowing down.

Quick check

1. Piston effort 4 kN, φ = 10°. Force in the connecting rod?
2. At inner dead centre (θ = 0), what is the turning moment?
3. Which way does the reciprocating inertia force act in the second half of the outstroke?
4. Name the three conditions for dynamical equivalence.
5. How does a longer connecting rod change the side thrust?

Answers: 1. 4 / cos 10° = 4.06 kN. 2. Zero (sin θ = 0 and sin 2θ = 0). 3. Along the motion, adding to the gas force (the piston is decelerating). 4. Same total mass, same centre of mass, same moment of inertia about G. 5. It reduces it (smaller φ).