Stress and deflection of helical spring

Notation: $F$ = axial load, $D$ = mean coil diameter, $d$ = wire diameter, $C = D/d$ (spring index), $n$ = number of active coils, $G$ = shear modulus.

Stress

A section of the wire is subjected to a direct shear force $F$ and a torque $T = FD/2$.

        F
        |
    ---(  )---   wire section: shear F + torque T = F D/2
       D/2

Torsional shear stress: $\tau_t = \dfrac{16T}{\pi d^3} = \dfrac{8FD}{\pi d^3}$

Direct shear stress: $\tau_d = \dfrac{F}{\pi d^2/4} = \dfrac{4F}{\pi d^2}$

Maximum stress (inner fibre) by simple addition:

$$\tau = \frac{8FD}{\pi d^3}\left(1 + \frac{d}{2D}\right) = K_s\,\frac{8FD}{\pi d^3}, \quad K_s = 1 + \frac{1}{2C}$$

Wahl correction factor

Because the wire is curved, the stress is higher at the inner side of the coil, and the direct shear is not uniform. A. M. Wahl included the curvature effect and direct shear in one factor:

$$K = \frac{4C-1}{4C-4} + \frac{0.615}{C}$$ $$\tau_{max} = K\,\frac{8FD}{\pi d^3} = K\,\frac{8FC}{\pi d^2}$$

For $C = 6$: $K = 1.2525$. For small $C$ (tight coils) $K$ is large; for static loads of ductile wire $K_s$ may be used, but for fatigue $K$ is used.

Deflection (strain energy method)

Strain energy due to torsion of the wire (length $l = \pi D n$, polar moment $J = \pi d^4/32$):

$$U = \frac{T^2 l}{2GJ} = \frac{(FD/2)^2(\pi D n)}{2G(\pi d^4/32)} = \frac{4F^2D^3 n}{G d^4}$$

Work done by load $= \tfrac{1}{2}F\delta$. Equating, $\tfrac12 F\delta = U$:

$$\delta = \frac{8FD^3 n}{G d^4} = \frac{8FC^3 n}{G d}$$

Stiffness

$$k = \frac{F}{\delta} = \frac{G d^4}{8D^3 n} = \frac{G d}{8C^3 n}$$

Stiffness increases with the fourth power of the wire diameter and falls with the cube of coil diameter.