Free · Physics models
Helical Torsion Spring Geometry and Stress
Relate coil geometry, applied torque, angular deflection and bending stress for an ideal round-wire spring.
- Formula & worked example
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Calculator inputs
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How to use this calculator
- Enter the known values in the units shown. Results update as you type.
- Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
- Use the worked example to check the method. Reset restores the starting fields.
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Formula and method
A 10 mm mean coil, 1 mm wire, 11 mm body and two 30 mm legs under 0.01 N·m torque yield about 0.3404 rad deflection at 200 GPa.
C = D/d; θ = 64 T D N/(E d⁴); kθ = T/θ.
Worked example
Enter these known values and leave the other values blank.
- Spring diameter (D)
- 10 mm
- Wire diameter (d)
- 1 mm
- Length (L)
- 11 mm
- Length, first end (l1)
- 30 mm
- Length, second end (l2)
- 30 mm
- Applied force (F)
- 1 N
- Arm (r)
- 10 mm
- Young modulus (E)
- 200 GPa
- Spring index (C)
- 10
- Number of active turns (Na)
- 10.637
- Numbers of full turns (Nb)
- 10
- Ends contribution (Ne)
- 0.6366
- Torque (M)
- 0.01 N·m
- Stress correction factor, inner (Ki)
- 1.0806
- Stress correction factor, outer (Ko)
- 0.9295
- Stress (σ)
- 0.11006 GPa
- Angular displacement (θ)
- 0.3404 rad
- Spring rate (k)
- 0.02938 N·m/rad
Assumptions and limitations
- The wire bends under torsional spring loading; use Young’s modulus E, not the shear modulus of a straight torsion bar.
- Mean coil diameter must exceed wire diameter and body length must exceed wire diameter. Positive torque and force represent load magnitudes.
- This ideal elastic estimate omits contact, end restraint, fatigue, permanent set and manufacturing tolerances.
- Enter known quantities and clear the unknown to use available inverse branches. Locks retain supplied values. Consistency messages cover fully visible equations; identities with hidden dependencies may require a manual consistency check. Preset values are illustrative approximations, and unit selectors preserve the underlying quantity.
- Stress-factor inverses select the physical spring-index branch C > 1. The inner factor exceeds one and the outer factor lies strictly between one half and one.
Common questions
How do I solve for a different quantity?
Clear that quantity and supply enough independent known values. Some outputs are one-way or tied to a selected preset, so they cannot determine every input.
References
Explicit physical domains for retained model. For physical spring index C > 1, the outer stress-factor inverse must select the minus-square-root numerator. The captured plus branch returns negative C. Substitution at C=10 gives Ko=409/440; the corrected inverse returns C=10.