Helical compression spring

Introduction

A helical compression spring is a torsion bar wound into a coil: the wire twists as the spring compresses, and its rate, stress and strength follow from four numbers, wire diameter, coil diameter, active coils and the wire's shear modulus, plus the strength of the drawn wire, which rises as the wire gets thinner. Shigley's treatment, used in almost every machine-design course, adds the curvature correction, the end geometry that fixes solid height, a stability limit against buckling and the surge frequency. This calculator follows it, reproducing the textbook's own example, so that a valve spring, a relief-valve spring or a shock-absorbing spring on deck machinery can be checked in a minute.

It is for mechanical engineers and students; it covers static loading, with fatigue left to the designer.

Formulas

Geometry and rate:

D = OD − d,   C = D/d   (keep 4 ≤ C ≤ 12)
N_t = N_a + 0 / 1 / 2 / 2   (plain, plain-ground, squared, squared-ground)
L_s = d(N_t + 1), d N_t, d(N_t + 1), d N_t   (same order)
k = d⁴ G / (8 D³ N_a),   y = F/k,   F_s = k (L₀ − L_s)

Stress and strength:

τ = K 8 F D / (π d³),   K_B = (4C + 2)/(4C − 3)  or  K_W = (4C − 1)/(4C − 4) + 0.615/C
S_ut = A / d^m   (music wire 2211 MPa·mm^m, m 0.145; hard-drawn 1783/0.190; oil-tempered 1855/0.187; Cr-V 2005/0.168; Cr-Si 1974/0.108; stainless 302 1867/0.146)
S_sy = 0.45 S_ut (music wire, hard-drawn), 0.50 S_ut (hardened and tempered alloys), 0.35 S_ut (stainless)
n_s = S_sy / τ   (≥ 1.2 at solid height recommended)

Stability and dynamics:

stable if L₀ < 2.63 D / α   (α 0.5 fixed–fixed, 0.707, 1.0 hinged, 2.0 fixed–free)
f = ½ √(k/m),   m = ρ (π d²/4) π D N_a

Assumptions and limitations

• Static axial load, round wire, Shigley's formulas; the A/d^m fit is valid only inside each wire's diameter range.
• The static torsional yield fractions are for springs without set removal; preset springs may be allowed higher stress.
• No fatigue analysis: cyclic loading needs Zimmerli endurance data with a Goodman or Gerber criterion.
• No temperature, corrosion or relaxation effects; the small variation of G with wire size is ignored.

Worked example

Shigley Example 10-1: a music-wire spring, wire 0.037 in (0.94 mm), outside diameter 7/16 in (11.11 mm), 12.5 total coils squared and ground, free length 1.25 in (31.75 mm), G = 11.75 Mpsi (81 GPa).

1. D = 0.4005 in (10.17 mm), C = 10.8, K_B = 1.124; N_a = 10.5, L_s = 0.4625 in (11.75 mm).
2. Rate k = d⁴G/(8D³N_a) = 4.08 lbf/in (714 N/m). Deflection to solid 0.7875 in → F_s = 3.21 lbf (14.3 N).
3. S_ut = 201/0.037^0.145 = 324 kpsi (2 231 MPa); S_sy = 0.45 × 324 = 146 kpsi.
4. Stress at solid: 1.124 × 8 × 3.21 × 0.4005/(π × 0.037³) = 72.6 kpsi (500 MPa); n_s = 2.0: the spring can be closed solid without setting.
5. Buckling: 2.63 × 0.4005/0.5 = 2.11 in > 1.25 in: stable between flat plates. Surge frequency about 260 Hz.

At a working force of 2 lbf the deflection is 0.49 in and the safety factor 3.2.

FAQ

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