Coiled Spring

Calculate torsional stress, deflection, and spring rate for helical coil springs.

Inputs

W

Applied Load

D

Mean Coil Diameter

d

Wire Diameter

N_e

Active Coils

Compression: Ne = N − 2; Tension: Ne = N

G

Shear Modulus

GPa

Steel ≈ 80 GPa; Phosphor bronze ≈ 40 GPa

Formula Interpretation

Variables

W—applied load (N)

D—mean coil diameter (mm)

d—wire diameter (mm)

k—stress correction factor

c—spring index (c = D/d, typically 4–10)

Ne—number of active coils

G—shear modulus (MPa)

δ—deflection (mm)

① Torsional Stress

\tau = k\,\dfrac{8DW}{\pi d^3} \quad \text{(MPa)}

τ is the maximum shear stress in the wire cross-section; $W$ is the applied load; $D$ is the mean coil diameter; $d$ is the wire diameter; $k$ is the Wahl curvature correction factor.

② Wahl Correction Factor

k = \dfrac{4c-1}{4c-4}+\dfrac{0.615}{c}

$c$ is the spring index ( $c$ = $D$ / $d$ ), typically 4–10. The correction factor $k$ accounts for stress concentration due to inner-surface curvature of the helix.

③ Deflection

\delta = \dfrac{8N_e D^3 W}{G d^4} \quad \text{(mm)}

$N_e$ is the number of active coils; $G$ is the shear modulus (MPa). Deflection scales linearly with coil count and cubically with mean diameter.

④ Spring Rate

\kappa = \dfrac{W}{\delta} = \dfrac{G d^4}{8N_e D^3} \quad \text{(N/mm)}

Spring rate $\kappa$ (stiffness) is the force required to produce unit deflection: $\kappa$ = $W$ / $\delta$ .

⑤ Spring Index

c = \dfrac{D}{d}

$c$ = $D$ / $d$ indicates the relative slenderness of the spring. Larger $c$ gives a softer spring with less curvature stress, but may cause lateral buckling.

Knowledge Points

The spring rate (stiffness) $\kappa$ is the ratio of the applied load to the resulting deflection (N/mm). For compression springs with ground ends, the total coil count N = $N_e$ + 2; for tension springs $N_e$ = N. The shear stress is higher on the inner surface of the helix than the outer surface — the Wahl factor $k$ corrects for this stress concentration.

Worked Example

A helical compression spring has mean coil diameter $D = 50\,\text{mm}$ , wire diameter $d = 5\,\text{mm}$ , $N_e = 10$ active coils, and shear modulus $G = 80\,\text{GPa}$ . Find the deflection, Wahl factor, torsional stress, and spring rate when a load of $W = 100\,\text{N}$ is applied.

Step 1 — Spring index and Wahl factor (Formula ②)

c = \dfrac{D}{d} = \dfrac{50}{5} = 10 \quad k = \dfrac{4\times10-1}{4\times10-4}+\dfrac{0.615}{10} = 1.14

Step 2 — Deflection $\delta$ (Formula ③)

\delta = \dfrac{8\times10\times50^3\times100}{80{,}000\times5^4} = \dfrac{10^9}{5\times10^7} = 20 \text{ mm}

Step 3 — Torsional stress $\tau$ (Formula ①)

\tau = 1.14\times\dfrac{8\times50\times100}{\pi\times5^3} \approx 116 \text{ MPa}

Step 4 — Spring rate $\kappa$ (Formula ④)

\kappa = \dfrac{W}{\delta} = \dfrac{100}{20} = 5 \text{ N/mm}

Therefore: deflection $\delta = 20\,\text{mm}$ , torsional stress $\tau \approx 116\,\text{MPa}$ , spring rate $\kappa = 5\,\text{N/mm}$ .

Extended Knowledge

•The shear modulus G governs spring stiffness: carbon steel ≈ 80 GPa, stainless steel ≈ 73 GPa, phosphor bronze ≈ 40 GPa, beryllium copper ≈ 48 GPa.
•The spring index c is typically kept between 4 and 10: too small makes winding difficult and stress concentration severe; too large risks lateral buckling.
•For compression springs, the solid (fully compressed) height H₀ = d × N. Designs must ensure maximum working deflection remains less than H₀ to avoid coil clash.
•Fatigue life depends on maximum shear stress and stress amplitude. Under cyclic loading, verify fatigue safety factor using the corrected stress amplitude — a factor of at least 1.3 is typically required.
•Helical springs are used in automotive suspensions, industrial fixtures, valve springs, and precision instrument vibration dampers. Selection requires balancing load capacity, stroke, fatigue life, and environmental corrosion resistance.