Torsion Spring Calculator

Spring rate, initial tension, force at any length and stress in the body and the hooks for extension springs; torque, angle and stress for torsion springs. Pick the wire and see the spring stretch or wind up to scale.

TorqueMath.com · Torsion Spring Calculator
Wire material
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coils
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Load table

Wire sizes for the same rate

Wire dCoilsBodyC% limit

How it is calculated

An extension spring is wound with the coils pressed together and a built-in initial tension Fi: it does not move until the pull exceeds Fi. The hooks, not the body, usually fail first, because the wire is bent sharply where the hook leaves the body.

Mean diameter D = De − d, index C = D / d Active coils Na = Nb + G / E Spring rate k = G·d⁴ / (8·D³·Na) Force F = Fi + k · (L − L0) L0 = (2C − 1 + Nb)·d (full loop hooks) Body stress τ = KB · 8FD / (π d³) KB = (4C + 2)/(4C − 3) Hook bending (A) σA = F·[ KA·16D/(π d³) + 4/(π d²) ], KA = (4C² − C − 1)/(4C(C − 1)) Hook torsion (B) τB = KB2 · 8FD/(π d³), KB2 = (4C2 − 1)/(4C2 − 4), C2 = 2r2/d = 4 Initial tension τi = 231/e^(0.105C) ± 6.9·(4 − (C − 3)/6.5) MPa (preferred range) Allowable (static) body 45 %, hook bending 75 %, hook torsion 40 % of Sut (35 / 55 / 30 % for stainless and bronze)

A torsion spring is loaded by a moment that bends the wire, so the stress is bending, not shear. As it winds up the coil diameter shrinks and the body gets longer, so it needs an arbor with clearance.

Active turns Na = Nb + (l1 + l2) / (3π D) Rate k = d⁴·E / (10.8 · D · Na) per turn (10.8 includes coil friction) Moment M = k · θ Bending stress σ = Ki · 32 M / (π d³), Ki = (4C² − C − 1)/(4C(C − 1)) Wound diameter D' = D · Nb / (Nb + θ/360°), inside D'i = D' − d Body length L = d · (Nb + 1 + θ/360°) Allowable (static) Sy = 78 % Sut (music, hard drawn), 87 % (Cr-V, Cr-Si), 61 % (stainless, bronze)

Wire strength Sut = A / d^m and the moduli G and E follow Shigley's Mechanical Engineering Design (tables 10-4, 10-5, 10-7, 10-10) and EN 13906-2/-3. Keep the index C between 4 and 12. Results are for static or slowly cycled loads; fatigue-rated springs need lower stresses and shot peening.

This calculator designs and checks a helical torsion spring – the kind on clothes pegs, hinges, garage doors and lever returns – that is loaded by a twisting moment on its legs. Enter wire and outer diameter, body turns, leg lengths, material, the working and preload angles and the arbor diameter. It returns the torque per degree and per turn, torque and leg force at each angle, bending stress against the wire's limit, the maximum safe angle, how far the coil shrinks as it winds up and whether it still clears the arbor. Switch to extension mode for pulling springs.

Worked example: music wire torsion spring, 1.6 mm wire, 20 turns, wound 90°

  1. Music wire (E = 196,500 MPa), d = 1.6 mm, outer diameter 14 mm, 20 body turns, two 25 mm legs, arbor Ø8 mm, wound 90° from free.
  2. Mean diameter D = 12.4 mm, C = 7.75. The legs bend too, adding (l1 + l2)/(3πD) = 50/(3π × 12.4) = 0.43 turns: Na = 20.43.
  3. Rate k = d⁴·E/(10.8·D·Na) = 1.6⁴ × 196,500 / (10.8 × 12.4 × 20.43) = 471 N·mm per turn = 1.31 N·mm per degree.
  4. At 90°: M = 1.31 × 90 = 118 N·mm (1.04 lbf·in), or 4.7 N at the end of a 25 mm leg. At the 20° preload: 26 N·mm.
  5. Ki = (4C² − C − 1)/(4C(C − 1)) = 1.106. Bending stress σ = Ki·32M/(πd³) = 1.106 × 32 × 118 / (π × 1.6³) = 324 MPa. The limit is 78 % of Sut = 0.78 × 2065 = 1611 MPa, so the spring is at 20 % and safe up to about 448°.
  6. Wound 90° the mean diameter shrinks to 12.4 × 20/20.25 = 12.25 mm; the inside is 10.65 mm, leaving 2.6 mm around the Ø8 arbor.

The spring gives 1.31 N·mm per degree and 118 N·mm at 90°, at only 20 % of its stress limit. It still clears the Ø8 mm arbor when wound up; the arbor should stay below about 9.6 mm (90 % of the wound inside diameter).

FAQ

How do you calculate torsion spring torque?

Torque is the rate times the angle: M = k·θ. The rate per turn is k = d⁴·E/(10.8·D·Na), where d is the wire diameter, D the mean coil diameter and Na the active turns (body turns plus the effect of the legs). Divide by 360 for torque per degree.

Which way should a torsion spring be loaded?

In the direction that winds it up and closes the coils. Coiling leaves residual stress that helps when the spring is wound further but adds to the stress if it is unwound, so the allowable stress is lower when loaded in the unwinding direction.

Why does a torsion spring need an arbor clearance?

As the spring winds up the coil diameter gets smaller: D' = D·N/(N + θ/360°). If the arbor is too close the coils grip it, friction rises and the spring may break. Keep the arbor about 10 % smaller than the wound inside diameter.

What is the difference between left hand and right hand torsion springs?

It is the direction the coils are wound. A right-hand spring winds up when the legs are turned counter-clockwise looking at the end; a left-hand spring the opposite way. Pick the hand so that the load closes the spring; the rate and stress are the same.

Why is 10.8 used instead of 64 in the torsion spring formula?

Bending theory gives d⁴E/(64·D·N) per radian, which is d⁴E/(10.2·D·N) per turn. Measured springs are slightly softer because of friction between coils and on the arbor, so Shigley and the spring industry use 10.8.

How many degrees can a torsion spring rotate?

Until the bending stress reaches the static limit, about 78 % of the tensile strength for music wire and 87 % for chrome alloy wire. The calculator shows this as the maximum safe angle. More turns spread the rotation over more wire and allow a larger angle.

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