Clutch Design Calculator and Clutch Types Simulator
A clutch connects a driving shaft to a driven shaft and lets them be separated while running. A friction clutch does it by pressing surfaces together: while their speeds differ it slips, turning the speed difference into heat, and once they match it locks. This simulator designs the four friction clutches found in machine-design syllabuses and then runs a take-up so you can see the slip, the lock-up and the heat:
- the single-plate clutch of a manual car;
- the multi-plate clutch of motorcycles and automatic transmissions;
- the cone clutch, whose wedge multiplies the spring force;
- the centrifugal clutch of go-karts, mopeds and chainsaws, which engages by itself.
Torque capacity of a plate clutch: uniform pressure and uniform wear
On an annular face from inner radius ri to outer radius ro, pressed by an axial force F with friction coefficient μ, the torque is T = μ·F·n·Rf, where n is the number of friction faces and Rf the friction radius. Two pressure distributions are used:
| Theory | Pressure | Axial force | Friction radius Rf | Default car clutch |
|---|---|---|---|---|
| Uniform wear (design) | p·r = constant, peak at ri | 2π·pmax·ri(ro − ri) | (ro + ri)/2 | 330.8 N·m |
| Uniform pressure (new) | p = constant | p·π(ro2 − ri2) | ⅔(ro3 − ri3)/(ro2 − ri2) | 335.4 N·m |
The default car clutch has 228 × 150 mm organic facings (μ = 0.35) on both sides of the disc (n = 2) and a 5,000 N diaphragm spring. A run-in clutch has worn until the wear rate, proportional to pressure times sliding speed, is the same everywhere, so p·r is constant. That gives the lower torque and is the theory used for design. With the outer diameter and the allowable pressure fixed, the torque is greatest when d/D = 1/√3 ≈ 0.577.
How many plates does a multi-plate clutch need?
In a multi-plate clutch, plates splined to the driving basket alternate with lined plates splined to the driven hub. One actuating force passes through the whole pack, so every face carries the same normal force, and n = n1 + n2 − 1. The number of faces follows from the design torque:
n = ⌈ β·T / (μ·F·Rf) ⌉, F ≤ 2π·pmax·ri(ro − ri)
Worked example: 160 N·m at the clutch, reserve β = 1.3, plates 140 × 110 mm, paper lining in oil (μ = 0.12) and 3,000 N of spring force. One face carries 0.12 × 3,000 × 0.0625 = 22.5 N·m. The pack needs 208/22.5 = 9.24, so 10 faces: 5 lined discs and 6 steel plates. The peak pressure is 0.58 MPa, well inside the 2.76 MPa allowed. A dry sintered lining (μ = 0.30) would need only 4 faces. Wet clutches use many plates because oil lowers μ; in return the oil cools them and smooths the engagement.
Cone clutch: wedge action and self-locking
An axial force F on a cone of semi-angle α produces a normal force N = F/sin α, so T = μ·F·Rf/sin α. The default cone (mean diameter 180 mm, α = 20°, 1,000 N, μ = 0.35) carries 92.1 N·m, 2.92 times a flat face at the same radius. Pushing it home while it slides takes N(sin α + μ cos α) = 1,962 N. If tan α ≤ μ, the cone self-locks and must be pulled out with N(μ cos α − sin α). That is why α is kept a little above the friction angle tan−1μ. Gearbox synchronizers are small oil-lubricated cone clutches with α of about 6–7°.
Centrifugal clutch design factors
Each shoe of mass m, with its centre of mass at radius rg, is held off the drum by a spring preload Fs. It engages when mω2rg = Fs, at ωe = √(Fs/(m·rg)). Above that, T = z·μ·R·(mω2rg − Fs). The design factors, each a slider in the simulator, are:
- the shoe mass and its radius, which set the centrifugal force;
- the engagement speed, which sets the spring preload (a common rule puts it near three-quarters of the running speed);
- the drum radius, the number of shoes and the lining μ, which set the torque;
- the contact arc and the shoe width, which keep the lining pressure N/(θ·R·b) within its limit.
The default kart clutch has three 120 g shoes at 40 mm, engages at 2,800 rpm (Fs = 412.7 N) and carries 15.6 N·m at 3,600 rpm. Under a steady 10 N·m throttle the engine holds 3,336 rpm while the clutch slips and the kart catches up.
Slip energy, temperature rise and the take-up
With no engine or load torque, the heat of a take-up is E = I1I2(ω1 − ω2)2/[2(I1 + I2)], whatever the clutch torque. With the throttle held open the engine keeps supplying energy during the slip. The default car launch (1,400 kg, first gear, 1,500 rpm, 90 N·m of throttle, 1 s pedal release) locks in 0.65 s and makes 8.8 kJ of heat, a 2.2 °C bulk rise in 8 kg of cast iron. A 20 % hill raises that to 11.2 kJ. Dropping the clutch in 0.1 s stalls the engine, and a 2 s release roughly doubles the heat to 16.4 kJ.
Who uses this simulator?
Mechanical and automotive engineering students working clutch problems in machine design (uniform wear and uniform pressure, multi-plate, cone and centrifugal clutches); diploma and vocational learners studying the car clutch and its diaphragm spring; teachers who want to show a clutch slipping, locking and heating up on a projector; and designers who need a quick torque, pressure and plate-count check.
Frequently asked questions
How do you calculate the torque capacity of a plate clutch?
Multiply the friction coefficient, the axial clamp force, the number of friction faces and the friction radius: T = μ·F·n·Rf. By the uniform-wear theory, used for design, Rf = (ro + ri)/2. By the uniform-pressure theory, for a new clutch, Rf = (2/3)(ro³ − ri³)/(ro² − ri²), which is slightly larger. A car clutch disc lined on both sides has n = 2.
How is the number of plates in a multi-plate clutch determined?
Find the largest axial force the lining allows, F = 2π·pmax·ri·(ro − ri) under uniform wear, then the torque one face carries, μ·F·Rf. The number of faces is the design torque (engine torque times a reserve factor) divided by that, rounded up: n = ⌈β·T/(μ·F·Rf)⌉. Taking n even puts steel plates at both ends, giving n/2 lined discs and n/2 + 1 steel plates, since n = n₁ + n₂ − 1.
Why is uniform wear theory used for clutch design?
Wear is proportional to pressure times sliding speed, and sliding speed grows with radius, so a clutch wears in until p·r is constant. That theory gives a smaller torque than uniform pressure (about 1 to 2.5 % less for usual proportions), so designing with it is conservative for a clutch that has run in.
What is the semi-cone angle of a cone clutch and why does it matter?
The axial force F creates a normal force F/sin α on the cone, so the torque is μ·F·Rf/sin α: a smaller angle gives more torque from the same spring. But if tan α is less than or equal to μ the cone self-locks and must be pulled out with a force N(μ cos α − sin α). The angle is chosen a little above the friction angle tan⁻¹μ.
How does a centrifugal clutch work and at what speed does it engage?
Shoes on a hub turned by the engine are held off the drum by springs. The shoe engages when its centrifugal force m·ω²·r_g equals the spring preload Fs, so ωe = √(Fs/(m·r_g)). Above that each shoe presses with N = m·ω²·r_g − Fs and the torque is T = z·μ·R·(m·ω²·r_g − Fs), growing with the square of speed.
Explore Related Simulators
To see where the clutch sits in the driveline, open the Gearbox Simulator; automatic transmissions apply their multi-plate clutches to the sets in the Planetary Gear Simulator. The friction coefficient itself is explored in the Friction & Contact Forces simulator, the engine-side inertia in the Flywheel Dynamics simulator, and the friction that drives a belt in the Belt & Chain Drive simulator.
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