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Springs and clutches: storing energy and controlling drive

How springs store energy and how clutches connect and disconnect a drive, with the main types of each and worked calculations for spring rate, spring stress and clutch torque.

On this page
  1. Springs
  2. Clutches

Springs and clutches look unrelated, but they work together in many machines. A spring stores energy by deflecting and gives it back, keeps parts pressed together, and absorbs shock. A clutch connects and disconnects a drive while it is running, and in most clutches it is a spring that provides the clamping force.

Compression springSingle-plate clutchFLMean coil diameter, Dwire, dF = k·xEngineFlywheelFriction discPressure plateSpringsCoverOutput togearboxReleasebearing
Left: a helical compression spring shortens in proportion to the force, F = kx. Right: in a friction clutch, springs clamp the friction disc between the flywheel and the pressure plate, so the engine drives the output shaft; pushing the release bearing frees the disc.

Springs

Types of spring

TypeHow it is loadedTypical uses
Helical compressionPushed shorterValves, suspensions, pens, press tools, clutches
Helical extensionPulled longer; hooks or loops at the endsDoors, trampolines, return springs on levers
TorsionTwisted about the coil axisClothes pegs, hinges, mouse traps, lids
LeafBent like a beamTruck and trailer suspension
Disc (Belleville)Conical washers pressed flat; very stiff, compactBolted joints, clutches, valves, preloading bearings
Wave springWavy flat wire; short and compactBearing preload, compact assemblies
Gas springCompressed gas in a cylinder; nearly constant forceCar tailgates, office chairs, machine guards

Spring rate

Within its working range, a spring’s deflection is proportional to the force (Hooke’s law):

F=k x,k=G d48 D3 nF = k\,x, \qquad k = \frac{G\,d^4}{8\,D^3\,n}
SymbolMeaningUnit
FFForceN
kkSpring rate (stiffness)N/m
xxDeflectionm
GGShear modulus of the wire (about 80 GPa for spring steel)Pa
ddWire diameterm
DDMean coil diameterm
nnNumber of active coilsnone

The rate depends on the wire diameter to the fourth power, so a small change in wire size changes the spring a lot. Springs in series (one after the other) have a combined rate 1/k=1/k1+1/k21/k = 1/k_1 + 1/k_2; springs in parallel (side by side) add: k=k1+k2k = k_1 + k_2.

Spring stress

The wire of a helical spring is mainly twisted, so the stress is shear:

τ=Kw 8 F Dπ d3,Kw=4C−14C−4+0.615C,C=Dd\tau = K_w\,\frac{8\,F\,D}{\pi\,d^3}, \qquad K_w = \frac{4C - 1}{4C - 4} + \frac{0.615}{C}, \qquad C = \frac{D}{d}

KwK_w is the Wahl factor, which adds the extra stress on the inside of the coil caused by its curvature. The spring index CC is usually kept between about 4 and 12: below 4 the spring is hard to coil, above 12 it tangles and buckles.

Worked example: a compression spring

A spring is coiled from 3 mm music wire (G = 80 GPa) with a mean diameter of 24 mm and 8 active coils. Find its rate, the force at 15 mm of deflection and the shear stress at that force.

Step 1: spring rate.

k=80×109×0.00348×0.0243×8=7320 N/m=7.32 N/mmk = \frac{80 \times 10^9 \times 0.003^4}{8 \times 0.024^3 \times 8} = 7320\ \text{N/m} = 7.32\ \text{N/mm}

Step 2: force.

F=7.32×15=110 NF = 7.32 \times 15 = 110\ \text{N}

Step 3: stress. C = 24 / 3 = 8, so Kw=31/28+0.615/8=1.18K_w = 31/28 + 0.615/8 = 1.18.

τ=1.18×8×110×0.024π×0.0033=295 MPa\tau = 1.18 \times \frac{8 \times 110 \times 0.024}{\pi \times 0.003^3} = 295\ \text{MPa}
Answerk = 7.3 N/mm; F = 110 N; τ ≈ 295 MPa, well below the roughly 850 MPa allowable for 3 mm music wire

Spring materials

Music wire and other cold-drawn carbon spring steels are the cheapest and strongest for small springs. Oil-tempered and chrome–silicon wires are used for larger springs and higher temperatures, such as engine valve springs. Stainless steel (302, 17-7PH) is used where corrosion matters, and phosphor bronze or beryllium copper where the spring also conducts electricity.

Clutches

A clutch lets a driving shaft (such as an engine or motor) connect to and disconnect from a driven shaft while it is turning.

TypeHow it worksTypical uses
Single-plate friction (dry)Springs clamp a friction disc between the flywheel and a pressure plateCars, motorcycles (some), small machinery
Multi-plate (wet)Many thin plates in oil share the torqueMotorcycles, automatic gearboxes, tractors
Cone clutchFriction on a conical surface; high torque for its sizeSynchronisers in manual gearboxes
CentrifugalShoes fly out against a drum as speed risesChainsaws, go-karts, scooters
Dog clutchInterlocking teeth; no slip, engaged at low relative speedGearbox selectors, machine tools
One-way (freewheel, sprag)Drives one way, overruns the otherBicycle hubs, starter motors, conveyors
ElectromagneticA coil pulls an armature against a friction faceCar air-conditioning compressors, printers
Slip (torque limiter)Slips above a set torqueOverload protection on conveyors and drives

How a single-plate clutch works

When engaged, springs (usually a single diaphragm spring) push the pressure plate towards the flywheel, clamping the friction disc between them. Friction on both faces of the disc carries the engine torque to the disc, whose hub is splined to the gearbox input shaft. Pressing the pedal pushes the release bearing against the spring fingers, the pressure plate moves back, and the disc spins freely.

Clutch torque capacity

Using the uniform-wear assumption, which suits a clutch that has bedded in:

T=μ F Rm N,Rm=Ro+Ri2T = \mu\,F\,R_m\,N, \qquad R_m = \frac{R_o + R_i}{2}
SymbolMeaningUnit
TTTorque the clutch can carry before slippingN·m
μ\muCoefficient of friction of the lining (about 0.25–0.4 dry)none
FFClamping (spring) forceN
RoR_o, RiR_iOuter and inner radius of the friction liningm
NNNumber of friction surfaces (2 for a single plate)none

Worked example: torque capacity of a clutch

A single-plate clutch has linings with outer and inner radii of 100 and 60 mm, μ = 0.3, and a spring force of 4000 N.

Rm=0.100+0.0602=0.080 m,T=0.3×4000×0.080×2=192 N⋅mR_m = \frac{0.100 + 0.060}{2} = 0.080\ \text{m}, \qquad T = 0.3 \times 4000 \times 0.080 \times 2 = 192\ \text{N·m}
AnswerThe clutch slips above about 192 N·m

The clutch is usually sized to carry 1.2 to 1.5 times the engine’s maximum torque, so it doesn’t slip in normal use but still slips before something more expensive breaks.

Key points

  • Spring force is proportional to deflection: F = kx, with k depending on d⁴ / D³.
  • Helical spring wire is in shear; the Wahl factor corrects for coil curvature.
  • A friction clutch’s torque is μ × clamp force × mean radius × number of surfaces.
  • Springs provide the clamping force in most clutches.

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