Helical Spring Deflection⚠ unverified
Mechanical / Springs · Compute the axial deflection of a helical spring
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| F | F | N | 1.0 | Applied axial force |
| d | d | m | 1.0 | Wire diameter |
| D | D | m | 1.0 | Mean coil diameter |
| N | N | — | 1.0 | Number of active coils (dimensionless) |
| G | G | Pa | 79000000000.0 | Shear modulus of the wire material, in pascals (Pa). Default is 79e9 |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | δ | m | Axial deflection, in metres (m) |
The science & history
Understanding the Parameters
-
Axial force $F$ — deflection is directly proportional to force: the defining property of a linear spring. Plotting $F$ against $\delta$ gives a straight line whose slope is the spring rate.
-
Wire diameter $d$ — appears as $1/d^4$, so a thicker wire deflects far less for the same load — the same fourth-power leverage that governs the rate, seen from the compliance side.
-
Mean coil diameter $D$ — enters as $D^3$: larger coils deflect much more, because the longer moment arm and longer wire per turn both add compliance.
-
Active coils $N$ — deflection grows linearly with the number of working coils; each coil contributes its own twist, and they add up in series.
-
Shear modulus $G$ — a stiffer (higher-$G$) wire material deflects less. Again $G$, not $E$, because the wire works in torsion.
Derivation (Approaching a Proof)
The deflection follows directly from treating the coil wire as a torsion bar (full derivation in Helical Spring Rate). The axial force $F$ applies a torque $T = FD/2$ to the wire; by Castigliano's theorem the axial deflection equals the derivative of the stored torsional strain energy with respect to $F$:
$$\delta = \frac{\partial U}{\partial F}, \qquad U = \int \frac{T^2}{2GJ}\,\mathrm{d}\ell,$$
with wire length $\ell = \pi D N$ and polar moment $J = \pi d^4/32$. Evaluating,
$$\delta = \frac{T(D/2)\ell}{GJ} = \frac{F (D/2)^2 (\pi D N)}{G(\pi d^4/32)} = \frac{8 F D^3 N}{G d^4}.$$
Equivalently, dividing $F$ by the spring rate $k = Gd^4/(8D^3N)$ gives the same result — the deflection and rate formulas are two views of one linear spring. The direct-shear and curvature contributions to deflection are small (a few percent for typical $C$) and are omitted here.
Dimensional check. $[\delta] = \dfrac{\text{N}\cdot\text{m}^3}{(\text{N/m}^2)\,\text{m}^4} = \dfrac{\text{N}\cdot\text{m}^3 \cdot \text{m}^2}{\text{N}\cdot\text{m}^4} = \text{m}$. ✓
History and Development
Like the rate formula, the deflection expression comes from the classical torsion-bar analysis of helical springs formalised by A. M. Wahl (1944) and standard in Shigley. It is used to set free length, working loads, and to confirm the spring does not reach its solid height (fully compressed) before the design load — a key failure mode in valve trains and mechanisms.
Related Concepts: Helical Spring Rate, Helical Spring Stress, Spring Index, Spring Energy, Spring Natural Frequency
Notes: Linear-spring result ($\delta = F/k$). Ensure the deflection plus the mounted (preload) deflection stays below the available stroke to solid height. Uses $G$ (torsion), active coils only.