Design of single-plate and multi-plate clutches

Why it matters

The clutch connects a running engine to a stationary gearbox smoothly and must then carry full engine torque without slipping. A clutch that is too small slips and burns; one that is too large is heavy, has high inertia and makes gear changes harder. Sizing the friction faces, spring force and number of plates is a standard automotive design task.

Key ideas

Single-plate dry clutch (most cars): one friction disc, lined on both sides, is clamped between the flywheel and the pressure plate by a diaphragm spring (or coil springs). It therefore has two friction surfaces. Pressing the pedal moves the release bearing and lifts the pressure plate off the disc.

Multi-plate clutch (motorcycles, automatic transmissions, heavy vehicles): alternate plates are splined to the driving and driven members. With n₁ plates on one member and n₂ on the other, the number of pairs of contact surfaces is n = n₁ + n₂ − 1. More surfaces give more torque within a small diameter. Wet (oil-immersed) clutches have a lower μ (about 0.06–0.1) but run cooler and wear less than dry ones (μ about 0.3–0.4 for organic linings).

Two design assumptions for an annular face (outer radius R₁, inner radius R₂):

• Uniform pressure: valid for a new clutch, whose faces are flat and well bedded. The friction radius is R_f = (2/3)·(R₁³ − R₂³)/(R₁² − R₂²).
• Uniform wear: after initial running, wear is proportional to pressure × rubbing velocity, so p·r = constant. Pressure is highest at the inner radius. The friction radius is the simple mean, R_f = (R₁ + R₂)/2. This gives a slightly lower torque for the same axial force, so it is the safer and usual design assumption; the maximum pressure (at R₂) must stay within the lining's allowable value.

Design steps: find engine torque, multiply by a service factor (typically 1.2–1.5 to allow for wear and overload; take it from the data book), choose a lining and its μ and allowable pressure, choose R₂/R₁ (often 0.6–0.8), then find the radii, the axial (spring) force and, for multi-plate clutches, the number of pairs.

Heat: each engagement dissipates the energy difference between the engine and the slower driven side as heat. Repeated slipping (hill starts, towing) overheats a dry clutch, so area, plate mass and ventilation are also checked.

Cone clutch: for a cone of semi-angle α, the axial force needed is reduced because the normal force is W/sin α: T = μ·W·R_m / sin α. Used in synchronisers in gearboxes.

Formulas

Torque: T = n·μ·W·R_f Uniform pressure: W = π·p·(R₁² − R₂²), R_f = (2/3)·(R₁³ − R₂³)/(R₁² − R₂²) Uniform wear: W = 2π·p_max·R₂·(R₁ − R₂), R_f = (R₁ + R₂)/2 Pairs of surfaces: n = n₁ + n₂ − 1 (n = 2 for a single-plate clutch) Cone clutch: T = μ·W·R_m / sin α Torque from power: T = 60·P / (2π·N)

Symbols: T = torque capacity (N·m or N·mm); n = number of pairs of friction surfaces; μ = coefficient of friction (–); W = axial spring force (N); R_f = friction (effective) radius (mm); R₁, R₂ = outer and inner radii of the lining (mm); p = uniform pressure (MPa); p_max = maximum pressure, at R₂ (MPa); n₁, n₂ = plates on driving and driven members; R_m = mean cone radius (mm); α = cone semi-angle; P = power (W); N = speed (rpm).

Worked examples

Example 1 (standard). A car's single-plate clutch has lining radii 150 mm and 100 mm, μ = 0.3 and a total spring force of 4 kN. Find the torque capacity by uniform wear and uniform pressure, and the maximum lining pressure.

1. Two friction surfaces: n = 2.
2. Uniform wear: R_f = (150 + 100)/2 = 125 mm; T = 2 × 0.3 × 4000 × 125 = 300 000 N·mm = 300 N·m.
3. p_max = W / (2π·R₂·(R₁ − R₂)) = 4000 / (2π × 100 × 50) = 0.127 MPa, at the inner radius.
4. Uniform pressure: R_f = (2/3) × (150³ − 100³)/(150² − 100²) = (2/3) × 190 = 126.7 mm; T = 2 × 0.3 × 4000 × 126.7 = 304 N·m.
5. Design on 300 N·m (uniform wear). With a service factor of 1.3, the clutch suits an engine torque up to about 230 N·m.

Example 2 (GATE level). A wet multi-plate motorcycle clutch must transmit 25 kW at 3000 rpm. Friction faces have R₁ = 60 mm and R₂ = 40 mm, μ = 0.1, maximum permissible pressure 0.4 MPa. Assume uniform wear. Find the number of plates.

1. T = 60 × 25 000 / (2π × 3000) = 79.58 N·m.
2. Maximum axial force: W = 2π × 0.4 × 40 × (60 − 40) = 2011 N.
3. Torque per pair of surfaces: μ·W·R_f = 0.1 × 2011 × 50 = 10 053 N·mm = 10.05 N·m.
4. Pairs needed: 79.58 / 10.05 = 7.92, so n = 8 pairs.
5. Plates: n₁ + n₂ − 1 = 8, so n₁ + n₂ = 9 plates (for example 5 on the drum and 4 on the hub).

Common mistakes

• Forgetting that a single-plate clutch has two friction surfaces.
• Counting plates instead of pairs of contact surfaces in a multi-plate clutch.
• Using the mean radius with uniform pressure, or the 2/3 formula with uniform wear.
• Placing the maximum pressure at the outer radius; under uniform wear it is at the inner radius.
• Writing the axial force as 2π·R·p; force is pressure times an area, π·p·(R₁² − R₂²) for uniform pressure.
• Mixing N·m and N·mm.

For GATE ME

Expect: torque capacity of single- and multi-plate clutches under uniform pressure and uniform wear; ratio of the two; axial force for a given torque; number of plates; maximum pressure; and cone-clutch torque. Practise the derivations of R_f for both assumptions, since conceptual questions often ask which assumption applies to a new or worn clutch.

Quick check

1. A multi-plate clutch has 4 driving and 3 driven plates. How many pairs of surfaces?
2. R₁ = 100 mm, R₂ = 60 mm. Friction radius under uniform wear?
3. Which assumption gives the larger torque for the same force?
4. Where is the pressure highest under uniform wear?
5. μ = 0.3, W = 2 kN, R_f = 0.1 m, single plate. Torque?

Answers: 1. 6. 2. 80 mm. 3. Uniform pressure. 4. At the inner radius. 5. 2 × 0.3 × 2000 × 0.1 = 120 N·m.