Springs Notes

Last reviewed 16 Sept 2026 · 6 min read

Springs and their uses

A spring is an elastic member that deflects under load and recovers when the load is removed. Springs are used to absorb shock (vehicle suspension, railway buffers), store energy (clocks, toys), apply force (valve springs, brakes, clutches) and measure force (spring balances).

Type	Main loading in the material	Typical use
Close-coiled helical (tension or compression)	Torsion of the wire	Valve springs, buffers, balances
Open-coiled helical	Torsion + bending	Special mechanisms
Torsion (spiral, helical torsion)	Bending	Clocks, clothes pegs, door hinges
Leaf (laminated, carriage)	Bending	Truck and railway wagon suspension
Disc (Belleville)	Bending	Short-stroke, high-load applications

Stiffness (spring constant, spring rate) $k$ = load per unit deflection: $k = W / δ$ (N/mm). Flexibility = $1/ k$ . Proof load = the maximum load a spring can carry without permanent set; the corresponding deflection is the proof deflection; energy stored up to it is the proof resilience.

Close-coiled helical spring under axial load

A spring is close-coiled when the helix angle is so small (under about 10°) that each coil is nearly in a plane perpendicular to the axis. An axial load $W$ then produces an almost pure twisting moment $T = W R$ in the wire, where $R$ is the mean coil radius.

Notation: wire diameter $d$ , mean coil radius $R$ (mean diameter $D = 2 R$ ), number of active coils $n$ , modulus of rigidity $G$ . Length of wire $l = 2 π R n$ .

Shear stress

From the torsion formula for the wire:

τ = \frac{16 T}{π d ^{3}} = \frac{16 W R}{π d ^{3}} = \frac{8 W D}{π d ^{3}}

Deflection

The twist of the wire $θ = \frac{T l}{G J}$ ; the axial deflection $δ = R θ$ :

δ = \frac{W R \cdot 2 π R n \cdot R}{G \cdot π d ^{4} /32} = \frac{64 W R ^{3} n}{G d ^{4}}

FormulaClose-coiled helical spring

τ = \frac{16 W R}{π d ^{3}} = \frac{8 W D}{π d ^{3}}, δ = \frac{64 W R ^{3} n}{G d ^{4}} = \frac{8 W D ^{3} n}{G d ^{4}}

k = \frac{W}{δ} = \frac{G d ^{4}}{64 R ^{3} n} = \frac{G d ^{4}}{8 D ^{3} n}

Energy stored: $U = \frac{1}{2} W δ = \frac{32 W ^{2} R ^{3} n}{G d ^{4}}$ .

Exam TipWhat changes stiffness

$k \propto d^{4} / (D^{3} n)$ . Doubling the wire diameter makes the spring 16 times stiffer; doubling the coil diameter makes it 8 times more flexible; doubling the number of coils halves the stiffness. Cutting a spring into two equal halves doubles the stiffness of each half.

Wahl's correction factor

The simple formula ignores direct shear $W / (π d^{2} /4)$ and the curvature of the wire (the inner side of the coil is more highly stressed, as in a curved beam). With spring index $C = D / d$ :

τ_{ma x} = K_{w} \frac{8 W D}{π d ^{3}}, K_{w} = \frac{4 C - 1}{4 C - 4} + \frac{0.615}{C}

For $C$ = 6, $K_{w} = 23/20 + 0.1025 = 1.2525$ . Spring index is usually kept between about 4 and 12.

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Springs