Mechanical Design & Machine Elements
Wahl Helical Spring Shear Stress (K_W) and Spring Rate Formula
Calculate spring index C = D / d, Wahl curvature-shear factor K_W = (4C - 1)/(4C - 4) + 0.615/C, inner-coil shear stress tau_max, and axial spring rate k = d^4*G / (8*D^3*N_a).
Account for both direct transverse shear and inner-coil curvature via Wahl factor K_W
Compressing a helical coil spring of wire diameter d and mean coil diameter D (spring index C = D / d) subjects the wire to pure torsion T = F * D / 2 PLUS direct transverse shear F AND curved-wire inner-fiber concentration! Wahl’s exact correction factor K_W = (4*C - 1) / (4*C - 4) + 0.615 / C amplifies nominal torsional stress 8*F*D / (PI()*d^3) at the inner coil surface, while axial spring rate obeys k = (d^4 * G) / (8 * D^3 * N_a).
Audit a music-wire spring (d = 4.0 mm in B2, mean diameter D = 24.0 mm in B3 -> index C = 6.0000; shear modulus G = 79,300 MPa in B4; active coils N_a = 8 in B5; axial force F = 300 N in B6): K_W = (23 / 20) + (0.615 / 6) = 1.1500 + 0.1025 = 1.2525! Peak inner-coil shear stress reaches 1.2525 * 286.4789 = 358.8148 MPa, and axial stiffness is k = (256 * 79,300) / (8 * 13,824 * 8) = 22.9456 N/mm.
| Helical Spring Metric (d=4, D=24, Na=8, F=300 N) | Formula | Verified Output |
|---|---|---|
| Spring Index (C) | D / d (Target 4 <= C <= 12) | 6.0000 |
| Wahl Correction Factor (K_W) | (4*C - 1)/(4*C - 4) + 0.615/C | 1.2525 (+25.25% Inner Peak) |
| Max Inner Shear Stress (tau_max) | K_W * 8 * F * D / (PI() * d^3) | 358.8148 MPa |
| Axial Spring Rate (k) | d^4 * G / (8 * D^3 * N_a) | 22.9456 N/mm |
Wahl Corrected Inner-Coil Shear Stress tau_max (MPa) Formula
Enforce D > d (spring index C > 1; physically D > d is required so wire coils do not overlap through the center axis and 4*C - 4 > 0!).
=LET(d_wr, B2, d_mn, B3, f_ax, B6, IF(OR(d_wr <= 0, d_mn <= d_wr, f_ax < 0), "Invalid Spring Geometry (Require D > d)", LET(c_ix, d_mn / d_wr, k_wh, (4 * c_ix - 1) / (4 * c_ix - 4) + 0.615 / c_ix, ROUND(k_wh * (8 * f_ax * d_mn) / (PI() * d_wr^3), 4))))Helical Spring Stiffness Rate k = (d^4 * G) / (8 * D^3 * N_a) Formula
Notice the 4th-power sensitivity to wire diameter d and inverse-cube sensitivity to coil diameter D: doubling wire diameter increases spring rate 16-fold!
=LET(d_wr, B2, d_mn, B3, g_sh, B4, n_act, B5, IF(OR(d_wr <= 0, d_mn <= d_wr, g_sh <= 0, n_act <= 0), "Invalid Spring Geometry (Require D > d)", ROUND((d_wr^4 * g_sh) / (8 * d_mn^3 * n_act), 4)))Try the example
Use the original values or change them to test your understanding.