Dynamic force analysis of slider-crank mechanism

Why it matters

The connecting rod, gudgeon pin, crank pin and main bearings of an engine are sized from the forces in the slider-crank mechanism, and the torque the crankshaft delivers at each crank angle comes from the same analysis. At high engine speed the inertia of the piston becomes as large as the gas force, so a static analysis that ignores it can be wrong by a factor of two.

Key ideas

D'Alembert's principle. Add an inertia force −m·a (opposite to the acceleration) to each moving mass, and an inertia couple −I·α to each rotating link, then treat the mechanism as if it were in static equilibrium. This turns a dynamics problem into a statics one.

Forces on the piston (horizontal engine, crank speed ω constant).

• Gas force F_L: the net force from the gas pressure on the piston (pressure × area; for double-acting engines subtract the force on the other side, whose area is reduced by the piston-rod area).
• Inertia force of the reciprocating parts F_I = m_R·a_P, where a_P = ω²r(cos θ + cos 2θ / n). It opposes the acceleration: near inner dead centre (IDC) the piston accelerates towards the crank, so the inertia force acts towards the cylinder head and reduces the effort; near outer dead centre it adds to it.
• Piston effort (net force along the line of stroke): F_P = F_L − F_I (also ± m_R·g for a vertical engine, and minus friction if it is given).

Resolving the piston effort. Let φ be the obliquity of the connecting rod (angle with the line of stroke), sin φ = sin θ / n.

• Force along the connecting rod: F_Q = F_P / cos φ.
• Side thrust on the cylinder wall: F_N = F_P·tan φ. This is what wears the cylinder on the thrust side.
• At the crank pin, F_Q has a tangential component (crank effort) F_T = F_Q·sin(θ + φ) and a radial component F_R = F_Q·cos(θ + φ) that loads the main bearings.
• Turning moment on the crankshaft: T = F_T·r. It is zero at both dead centres (θ = 0° and 180°).

Equivalent dynamical system for the connecting rod. The rod has mass m and moment of inertia m·k_G² about its centre of mass G. It can be replaced by two point masses on its axis if: the masses add up to m, their centre of mass is at G, and their moment of inertia about G equals m·k_G² (so l₁·l₂ = k_G², where l₁ and l₂ are their distances from G). In practice the rod is usually split into masses at the gudgeon pin and the crank pin, which satisfies the first two conditions but not the third; a correction couple accounts for the difference. With masses at the ends:

• reciprocating part (added to the piston mass): m·(distance of G from crank pin) / l;
• rotating part (added to the crank pin, balanced by counterweights): m·(distance of G from gudgeon pin) / l.

Speed effect. F_I grows with ω². At low speed the crank torque follows the gas force; at high speed the inertia force reshapes the torque curve and can make the piston effort near IDC negative, which puts the connecting rod in tension. This always happens near top dead centre between the exhaust and intake strokes, where the gas force is small.

Formulas

F_L = p·π·D² / 4

• F_L: gas force on the piston (N); p: net gas pressure (Pa); D: bore (m).

F_I = m_R·ω²·r·(cos θ + cos 2θ / n)

• F_I: inertia force of reciprocating parts (N); m_R: reciprocating mass (kg); ω: crank speed (rad/s); r: crank radius (m); θ: crank angle from IDC; n = l/r. Approximate (first two harmonics), good for n of about 3.5 or more.

F_P = F_L − F_I (horizontal engine)

• F_P: piston effort (N), positive from the cylinder head towards the crank. For a vertical engine with the cylinder above the crank, add m_R·g (weight acts towards the crank).

sin φ = sin θ / n, F_Q = F_P / cos φ, F_N = F_P·tan φ

• φ: obliquity of the connecting rod; F_Q: thrust in the rod (N); F_N: side thrust on the cylinder wall (N).

F_T = F_Q·sin(θ + φ), F_R = F_Q·cos(θ + φ)

• F_T: crank effort, tangential to the crank circle (N); F_R: radial force on the crankshaft bearings (N).

T = F_T·r = F_P·r·(sin θ + sin 2θ / (2·√(n² − sin²θ)))

• T: turning moment on the crankshaft (N·m). Both forms are exact for the given F_P.

m₁ + m₂ = m, m₁·l₁ = m₂·l₂, l₁·l₂ = k_G²

• Equivalent dynamical two-mass system of a rod: m₁, m₂ (kg) at distances l₁, l₂ (m) from G; k_G: radius of gyration about G (m).

Worked examples

Example 1 (standard: forces and torque at one crank angle). A horizontal engine has bore 100 mm, stroke 120 mm, connecting rod 240 mm, reciprocating mass 1.5 kg and runs at 2400 rpm. At θ = 30° from IDC on the expansion stroke the net gas pressure is 2.0 MPa. Find the piston effort, the thrust in the rod, the side thrust, the crank effort and the turning moment.

1. r = 0.06 m, n = 240 / 60 = 4, ω = 2π × 2400 / 60 = 251.33 rad/s, ω² = 63 165 s⁻².
2. Gas force: F_L = 2.0 × 10⁶ × π × 0.1² / 4 = 15 708 N.
3. Inertia force: F_I = 1.5 × 63 165 × 0.06 × (cos 30° + cos 60° / 4) = 5684.9 × (0.8660 + 0.125) = 5684.9 × 0.9910 = 5634 N.
4. Piston effort: F_P = 15 708 − 5634 = 10 074 N.
5. Obliquity: sin φ = 0.5 / 4 = 0.125, φ = 7.18°, cos φ = 0.9922, tan φ = 0.1260.
6. Thrust in rod: F_Q = 10 074 / 0.9922 = 10 154 N; side thrust: F_N = 10 074 × 0.1260 = 1269 N.
7. Crank effort: F_T = 10 154 × sin(37.18°) = 10 154 × 0.6043 = 6136 N.
8. Turning moment: T = 6136 × 0.06 = 368 N·m. Check: 10 074 × 0.06 × (0.5 + 0.8660 / (2 × 3.969)) = 604.4 × 0.6091 = 368 N·m.

Example 2 (GATE level: speed at which inertia cancels the gas force). For the same engine (m_R = 1.5 kg, r = 0.06 m, n = 4), the net gas force at IDC is 10 kN. At what speed is the piston effort at IDC zero?

1. At θ = 0: F_I = m_R·ω²·r·(1 + 1/n) = 1.5 × 0.06 × 1.25 × ω² = 0.1125·ω².
2. Set F_I = F_L: 0.1125·ω² = 10 000, so ω² = 88 889 and ω = 298.1 rad/s.
3. N = 60ω / 2π = 2847 rpm. Above this speed the connecting rod is in tension at IDC even with 10 kN of gas force.

Example 3 (rod split). A 2.0 kg connecting rod 240 mm long has its centre of mass 80 mm from the crank-pin centre. Mass treated as reciprocating = 2.0 × 80 / 240 = 0.667 kg; mass treated as rotating = 2.0 × 160 / 240 = 1.333 kg.

Common mistakes

• Adding the inertia force to the gas force at every angle. It opposes the acceleration: it subtracts near IDC and adds near ODC.
• Using the gas pressure without the area, or with the diameter instead of the area, or gauge and absolute pressures mixed.
• Taking the crank effort as F_P·sin θ, which ignores the obliquity of the rod.
• Forgetting that the rod's reciprocating share is set by the distance of G from the crank pin (not from the gudgeon pin).
• Using rpm in place of rad/s in ω².
• Leaving out the weight of reciprocating parts in a vertical engine when the problem gives it.

For GATE ME

Questions ask for the inertia force at a crank angle, piston effort, thrust in the connecting rod, side thrust, crank effort or turning moment, and sometimes the equivalent two-mass system of a connecting rod. Many are multi-step with linked answers, so keep intermediate results to four significant figures. Practise example 1 at θ = 0°, 90° and 150° to see how the forces change.

Quick check

1. Which way does the inertia force of the reciprocating parts act just after IDC?
2. Turning moment at a dead centre?
3. Obliquity when θ = 90° and n = 5?
4. If engine speed doubles, how does the inertia force change?
5. Side thrust if F_P = 8 kN and φ = 10°?

Answers: 1. Towards the cylinder head (opposing the piston's acceleration towards the crank). 2. Zero. 3. sin φ = 1/5, φ = 11.5°. 4. It becomes four times larger. 5. 8000 × tan 10° = 1411 N.