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ArticlePublished 11 Jul 2026Updated 12 Jul 20264 min readBy Kevin Jogin
KEVOS® Knowledge Library · Engineering → Mechanical Engineering

Engineering / Mechanical Engineering

Springs

A helical spring is a torsion bar wound into a package: the wire twists, the coil deflects, and four numbers — wire, diameter, coils, material — set everything. This page runs one spring from geometry to stress to stored energy.

  • Reading time · 4 min
  • 7 sections
  • One spring worked end to end
  • Wahl factor included
mean coil diameter DFFwire drunning example: d = 4, D = 32, C = 8, 8 active coils
Doc №KL-ENG-MECH-018
SectionEngineering → Mechanical Engineering
Sheet1 of 1
DrawnKEVOS®
Date2026-07-11

§1Geometry and the spring index

Wire diameter d, mean coil diameter D, active coils n — plus one ratio that flavours everything else.

spring index C = Dd  — aim for 4 to 12; the sweet spot near 8

Below C ≈ 4 the coil is tight to wind and its inner fibre badly overworked; above 12 it is floppy, tangle-prone and hard to guide. Active coils are those free to twist — closed and ground end coils sit out (§4). The running example throughout this page: d = 4 mm, D = 32 mm (C = 8), n = 8 active coils, spring steel with G = 79.3 GPa.

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§2Rate and deflection

The wire is a torsion bar of length π D n; wind-up at the wire becomes travel at the coil. The rate falls out in one line.

k = G d⁴8 D³ n  δ = Fk = 8 F D³ nG d⁴
Example 1 — the running spring’s rate

k = 79 300 × 4⁴ / (8 × 32³ × 8) = 9.68 N/mm; under 180 N it closes δ = 180/9.68 = 18.6 mm. The exponents are the design levers: d⁴ over D³n — thicken the wire and the spring stiffens violently; add coils or diameter and it softens.

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§3Shear stress and the Wahl factor

The basic stress is the torsion formula in disguise; the Wahl factor pays for the coil’s curvature, which crowds the shear onto the inner fibre.

τ = K_w 8 F Dπ d³  K_w = 4C − 14C − 4 + 0.615C
Example 2 — stress under the 180 N load

C = 8 gives K_w = 31/28 + 0.0769 = 1.184. τ = 1.184 × 8 × 180 × 32 / (π × 4³) = 271 MPa — comfortable for static duty in spring steel; a fatigue application would want this held well lower and the surface shot-peened. Note K_w grows as C shrinks: at C = 4 the penalty is 40 %, another vote against tight coils.

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§4Ends, solid height and buckling

End treatment sets how the spring stands and how many coils actually work; two housekeeping checks keep it honest in service.

Ends. Closed-and-ground is the standard for compression springs — square seating, and total coils ≈ active + 2. Plain ends save grinding but seat crookedly. Solid height is total coils × d: the running spring’s 10 × 4 = 40 mm; design so working travel never coils it solid, or the load path becomes metal-to-metal and the rate becomes infinite. Buckling: a slender compression spring bows like the Columns page’s strut — keep free length under about 4 × D unguided, or run it over a rod or in a bore (and accept the friction) beyond that.

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§5Stored energy

U = ½ k δ² = ½ F δ

The running spring at 180 N holds U = ½ × 9.68 × 18.6² = 1.67 J — the triangle under its force–deflection line, the same area picture as the spring-work integral on the Algebra and Equations page. Energy scales with δ², so a soft, long-travel spring out-stores a stiff short one at equal peak force — the design fork between a striker spring and a hold-down.

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§6Extension, torsion and leaf springs

Three cousins, one paragraph each — the helical mathematics carries into the first two nearly unchanged.

Extension springs obey the same k and τ formulas but are wound with initial tension: no deflection occurs until the preload is overcome, and the hooks — bent, stress-raised, unpeenable — are where they fail; generous hook radii matter more than wire grade. Torsion springs load the wire in bending, not shear: rate is a moment per degree, and they should always be worked in the wind-up direction so service load tightens the coil onto its arbor. Leaf springs are stacked cantilever beams — the Beams page in laminated form — trading the helical’s compactness for load capacity and built-in friction damping.

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§7Quick reference

The working core of the page on one card rack.

Rate

k = Gd⁴/8D³n

Stress

τ = K_w·8FD/πd³

K_w = (4C−1)/(4C−4) + 0.615/C

Index

C = D/d, aim 4–12

Housekeeping

solid ht = coils × d

free length ≤ 4D unguided

Energy

U = ½kδ²

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