How to use this calculator
- Enter the wire and coil geometry. Enter the wire diameter d and the mean coil diameter D. The spring index C = D/d should be roughly 4–14.
- Enter the active coils. Enter the number of active body coils Nₐ that wind up under load.
- Enter the applied moment. Enter the moment M applied about the coil axis to wind the spring.
- Set the material modulus. Enter Young’s modulus E — about 207 GPa for music wire, 193 GPa for stainless.
- Read the results. Read the deflection angle, the Wahl-corrected inner-fibre bending stress, the spring index and the rate per turn.
How it works
A helical torsion spring is loaded by a moment about the coil axis, and that
moment puts the wire in bending — not torsion, despite the name.
Because the wire bends, the stiffness depends on the material’s Young’s
modulus E, not the shear modulus G that governs a compression spring. The
rate per turn is
R = E·d⁴ / (10.8·D·Nₐ)
where d is the wire diameter, D the mean coil diameter
and Nₐ the active coils. The wind-up under a moment M is
simply the moment divided by the rate, expressed in turns:
revolutions = M / R, and the angle is revolutions × 360°.
The bending stress is corrected for coil curvature with the Wahl bending
factor Kb = (4C² − C − 1) / (4C(C − 1)), where C = D/d is the
spring index. This is the bending counterpart of the shear Wahl factor
Kᵥᵥ used on compression springs. The peak stress, at the inner fibre
of the wire, is
σ = Kb · 32·M / (π·d³)
As the spring winds up, the body coils tighten and the inside diameter shrinks, so allow clearance over any arbor the spring rides on. Formulas follow standard spring-design practice (e.g. Shigley, Mechanical Engineering Design, chapter 10).
Worked example
Verified against the live calculator
Music wire, d = 2 mm, mean coil D = 20 mm (so the spring
index C = 10), Nₐ = 6 active coils, E = 207 GPa,
with an applied moment M = 1 N·m. The rate per turn is
R = 207000 × 2⁴ / (10.8 × 20 × 6) ≈ 2555.6 N·mm/turn, i.e. about
2.556 N·m/turn. The wind-up is
1000 / 2555.6 ≈ 0.391 turns, giving a deflection angle of
≈ 140.9°. The Wahl bending factor is
Kb = 389/360 ≈ 1.0806, so the inner-fibre bending stress is
σ = 1.0806 × 32 × 1000 / (π × 2³) ≈ 1376 MPa. Those are exactly the
numbers the calculator shows when you load this page.
Frequently asked questions
How do I calculate the deflection of a torsion spring?
Find the rate per turn R = E·d⁴/(10.8·D·Nₐ), where E is Young’s modulus, d the wire diameter, D the mean coil diameter and Nₐ the active coils. The wind-up is the applied moment divided by the rate, in turns: revolutions = M / R, and the angle is revolutions × 360°. Enter your values above and the calculator solves it.
Why does a torsion spring use Young’s modulus E and not the shear modulus G?
Despite the name, a helical torsion spring loads its wire in bending, not torsion — the coil winds tighter as the arms rotate. Bending stiffness depends on Young’s modulus E, so the rate uses E (a compression spring, which twists its wire, uses the shear modulus G instead).
What is the Wahl bending correction factor Kb?
It corrects the bending stress for coil curvature, which concentrates stress on the inner fibre of the wire: Kb = (4C² − C − 1)/(4C(C − 1)), where C = D/d is the spring index. It is the bending counterpart of the shear Wahl factor Kᵥᵥ used for compression springs. Skipping it under-predicts the peak stress.
Where is the maximum stress in a torsion spring?
At the inner fibre of the wire — the surface facing the coil axis. The curvature correction Kb is greater than 1 precisely because that inner surface sees a higher bending stress than a straight beam would: σ = Kb · 32·M / (π·d³).
Does the coil diameter change under load?
Yes. As the spring winds up, the body coils tighten, so the spring’s inside diameter shrinks. Allow clearance over the arbor or shaft the spring works on so the coils do not bind as the inside diameter reduces at full deflection.
Does this work in metric and imperial?
Yes — toggle SI/Imperial in the header. The moment is entered in N·m or lbf·ft, lengths in mm or inches, and stress is shown in MPa or ksi. The deflection angle is in degrees either way.
Method & assumptions
- The wire is loaded in bending, so the rate uses Young’s modulus E and the bending Wahl factor Kb — different from a compression spring, whose wire twists (shear modulus G, shear Wahl factor Kᵥᵥ).
- The maximum stress is at the inner fibre of the coil; the body coils tighten and the inside diameter shrinks as the spring winds up, so allow arbor clearance.
- Static loading only — fatigue, set and stress relaxation are not modelled.
- End/arm leverage, friction against the arbor and pre-wind are not included; for a safety-critical design, verify against the governing standard and your supplier’s certified data, and have a professional engineer review it.