Advanced Spring
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▸Formula · assumptions · example(DIN EN 13906-1 · IS 7906 · ASTM A228 (Music Wire))
Formula
\tau = K_w \frac{8 F D}{\pi d^3}, \quad K_w = \frac{4C-1}{4C-4} + \frac{0.615}{C}, \quad k = \frac{G d^4}{8 D^3 n_a}
Wahl curvature stress concentration factor Kw, torsional shear stress τ, and linear spring rate k for helical compression springs.
Assumptions
- · Spring index C = D/d maintained within standard manufacturing bounds (4 ≤ C ≤ 12)
- · Shear modulus G remains constant within operating temperature limit (G ≈ 79.3 GPa for spring steel)
- · Clash clearance: minimum 10% deflection margin maintained above solid height under peak load
Worked example
Worked example — Helical Spring (d = 3 mm, D = 24 mm, na = 8 coils, F = 250 N)
Calculates spring index C = 8.0, Wahl factor Kw = 1.184, shear stress τ = 558 MPa, spring rate k = 15.1 N/mm, and active deflection δ = 16.5 mm.