Helical Spring Stress⚠ unverified
Mechanical / Springs · Shear stress in a helical spring (with Wahl/curvature factor)
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| F | F | N | 500.0 | Axial force |
| d | d | m | 0.005 | Wire diameter |
| D | D | m | 0.04 | Mean coil diameter |
| K | K | — | 1.2 | Stress correction factor |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| tau | τ | Pa | Shear stress |
The science & history
Understanding the Parameters
-
Axial force $F$ — the working load. Stress is proportional to $F$; the maximum operating load sets the peak stress that must stay below the allowable.
-
Wire diameter $d$ — enters as $1/d^3$: thicker wire lowers stress rapidly. This competes with the rate requirement (thicker wire also stiffens the spring), so wire size is a compromise between rate and stress.
-
Mean coil diameter $D$ — the moment arm; larger coils raise the torque $FD/2$ and hence the stress linearly.
-
Wahl factor $K$ — the correction that turns the idealised torsion stress into the real peak. It bundles two effects: the direct transverse shear superimposed on the torsion, and the curvature of the wire that crowds stress onto the inner fibre of the coil. It depends only on the spring index $C = D/d$ (see Spring Index): $K = \tfrac{4C-1}{4C-4} + \tfrac{0.615}{C}$. A default of 1.2 corresponds to $C \approx 8$; low-index springs need $K$ up to ~1.4 or more.
Registry note: $K$ is supplied as an input rather than computed from $C$; enter the Wahl value for your index. (For a static yield check some texts use a smaller shear-only factor $K_s = 1 + 0.5/C$; for fatigue use the full Wahl $K$.) See Known Issues.
Derivation (Approaching a Proof)
The wire of a helical spring is loaded primarily in torsion by the torque $T = FD/2$. The peak torsional shear stress in a round bar is $\tau_t = T r/J = T(d/2)/(\pi d^4/32) = 16T/(\pi d^3)$. Substituting $T = FD/2$:
$$\tau_t = \frac{16(FD/2)}{\pi d^3} = \frac{8 F D}{\pi d^3}.$$
Two corrections turn this ideal torsion stress into the real maximum:
-
Direct shear. The axial force $F$ also produces a uniform transverse shear $\approx F/A$ across the wire, which adds to the torsional stress on the inner coil fibre. This contributes a factor $\approx (1 + 0.5/C)$.
-
Curvature. Because the wire is curved into a coil, its inner fibre is shorter, concentrating shear there beyond what straight-bar torsion predicts.
Wahl (1929) combined both into a single factor depending only on the spring index:
$$K = \frac{4C - 1}{4C - 4} + \frac{0.615}{C}.$$
Multiplying gives the design stress $\tau = K\,\dfrac{8FD}{\pi d^3}$. As $C \to \infty$, $K \to 1$ and the formula reduces to pure torsion; at small $C$ the inner-fibre concentration dominates.
Dimensional check. $[\tau] = \dfrac{\text{N}\cdot\text{m}}{\text{m}^3} = \dfrac{\text{N}}{\text{m}^2} = \text{Pa}$ (K dimensionless). ✓
History and Development
The torsion basis of spring stress is classical; A. M. Wahl derived the curvature-plus-shear correction factor in 1929 and consolidated spring theory in Mechanical Springs (1944). The Wahl factor remains standard in Shigley and every spring-manufacturer design guide, used with the Goodman/Sines fatigue criteria (Goodman Line, Fatigue Endurance Limit) to set safe stress ranges for cyclically loaded springs.
Related Concepts: Spring Index, Wahl Correction Factor, Helical Spring Rate, Shear Stress, Goodman Line, Fatigue Endurance Limit, Spring Set Removal
Notes: $K$ from the spring index (Wahl for fatigue, $K_s = 1+0.5/C$ for static). Compare $\tau$ to the allowable shear (often $\sim0.5$–$0.65\,S_{ut}$; see Spring Set Removal). Shot-peening raises fatigue strength.