How this calculation works
The familiar spring equation F = kx assumes the spring rate is already known. This page derives an ideal rate from wire diameter, coil diameter, active coils and shear modulus. It also estimates curved-wire shear stress at an entered load.
k = Gd⁴/(8D³Na); C = D/d; Kw = (4C−1)/(4C−4) + 0.615/C; τ = Kw 8FD/(πd³)A worked example
For 4 mm wire, 32 mm mean diameter, eight active coils and G = 80 GPa, the rate is 9.765625 N/mm. A 100 N load produces 10.24 mm ideal deflection. The spring index is eight, and the corrected stress is higher than the straight-wire torsion estimate.
Good to know
Free length, solid height, coil contact, end treatment, buckling and fatigue are not evaluated. A computed deflection does not demonstrate available travel. Material strength and allowable repeated stress must be assessed independently.
Reference: RoyMech — helical spring equations (opens in a new tab) ↗
What this model answers
The familiar spring equation F = kx assumes the spring rate is already known.
This page derives an ideal rate from wire diameter, coil diameter, active coils and shear modulus. It also estimates curved-wire shear stress at an entered load.
The Spring force energy and work calculator uses an already known spring rate to obtain force or energy; this page derives that ideal rate from active-coil geometry.
Keep the input conventions consistent
Use mean coil diameter, halfway between inside and outside coil diameters.
Count active coils rather than automatically using the total number of turns. The model accepts spring indices from three to twenty to bound its close-coiled approximation.
Read the result carefully
Wire diameter appears to the fourth power in stiffness, while mean diameter appears to the third power in the denominator.
Small dimensional changes can therefore dominate a load comparison. The Wahl factor increases nominal torsional stress to account approximately for curvature and direct shear.
Wire diameter and mean coil diameter have different roles in both rate and corrected stress. Use the mean diameter required by the formula, rather than substituting the outside coil diameter. Total manufactured coils can also differ from active coils; inactive end turns do not contribute equally to the modeled deflection.
Work through the numbers
For 4 mm wire, 32 mm mean diameter, eight active coils and G = 80 GPa, the rate is 9.765625 N/mm.
A 100 N load produces 10.24 mm ideal deflection. The spring index is eight, and the corrected stress is higher than the straight-wire torsion estimate.
Where the calculation stops
Free length, solid height, coil contact, end treatment, buckling and fatigue are not evaluated.
A computed deflection does not demonstrate available travel. Material strength and allowable repeated stress must be assessed independently.
A few questions, answered
Can total coils always replace active coils?
No. Inactive end turns affect overall geometry but do not contribute to the same deflection model.
What should I change in the comparison?
Double active coils at unchanged geometry; rate halves and load deflection doubles.