10.5 Design of Helical, Leaf & Torsion Springs

Why Springs Are a Torsion Problem

A close-coiled helical spring looks as if it should be analysed in bending, but it is not. When an axial load $W$ is applied to a coil of mean diameter $D$, the load acts at a radius $D/2$ from the wire axis, applying a torque to the wire:

$T = \frac{WD}{2}$

The wire therefore carries torsional shear, and the design follows directly from the torsion formula for a circular section.

Shear Stress

Applying $\tau = 16T/\pi d^3$ with the torque above gives the uncorrected shear stress:

$\tau = \frac{16T}{\pi d^3} = \frac{8WD}{\pi d^3}$

This expression neglects two real effects:

1. Direct shear. The load $W$ also produces a direct transverse shear over the wire cross-section.
2. Curvature. The inner fibre of the coil has a shorter length than the outer, concentrating stress at the inside of the coil.

Both are captured by the Wahl correction factor:

$K_w = \frac{4C - 1}{4C - 4} + \frac{0.615}{C}$

where $C = D/d$ is the spring index. The corrected maximum shear stress, occurring at the inner surface of the coil, is

$\boxed{\tau_{\max} = K_w\,\frac{8WD}{\pi d^3}}$

In the Wahl factor, the first term accounts for curvature and the second for direct shear.

Spring index

$C = \frac{D}{d}$

Because $K_w$ falls as $C$ rises, a low spring index means a large stress correction — a spring index of 4 gives $K_w \approx 1.40$, while an index of 10 gives only $1.14$.

Deflection and Stiffness

Applying the angle-of-twist relation over the developed wire length $\pi D n$, where $n$ is the number of active coils:

$\boxed{\delta = \frac{8WD^3 n}{G d^4}}$

So the spring rate or stiffness is

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

The sensitivity to wire diameter

Stiffness varies with the fourth power of wire diameter and inversely with the cube of coil diameter. A 10% increase in wire diameter raises stiffness by about 46%; a 10% increase in coil diameter reduces it by about 25%. When a spring is too soft, changing the wire gauge is far more effective than any other adjustment.

Strain energy

$U = \tfrac{1}{2}W\delta = \tfrac{1}{2}k\delta^2$

This is the energy a spring stores and returns, and it is what makes springs useful as shock absorbers and energy accumulators.

Combinations of Springs

Series springs are softer than either individual spring; parallel springs are stiffer than either. Two identical springs of stiffness $k$ give $k/2$ in series and $2k$ in parallel.

A useful check on which case applies: if the springs are joined end to end so the same force passes through both, they are in series. If both are compressed between the same two plates, they are in parallel — including the common concentric arrangement of a small spring nested inside a larger one.

Ends, Coils and Solid Length

Squared and ground ends are standard for compression springs because they seat squarely and apply load along the axis without inducing bending. Solid length is the compressed length when all coils touch, and the free length must exceed the solid length by the maximum working deflection plus a clash allowance of about 15%.

Buckling and Surge

Buckling

A compression spring behaves like a column and buckles when it is too slender. The governing ratio is the free length to mean coil diameter:

Surge

When a spring is compressed rapidly, a wave of compression travels along the coils at the material's velocity of sound and reflects back. If the excitation frequency approaches the spring's natural frequency, surge occurs and stresses rise dramatically. The fundamental natural frequency of a spring with one end fixed is

$f_n = \frac{1}{2}\sqrt{\frac{k\,g}{W_s}}$

where $W_s$ is the weight of the active coils. Design practice is to keep the natural frequency at least 15 to 20 times the operating frequency, which is why valve springs in high-speed engines use variable pitch or nested springs to break up the resonance.

Leaf Springs

A leaf spring is a bending element, unlike the helical spring. It comprises a stack of plates of decreasing length, approximating a beam of uniform strength — the classic diamond-shaped plate cut into strips and stacked.

For a semi-elliptical leaf spring of span $2L$, width $b$, thickness $t$ and $n$ leaves, carrying a central load $2W$:

$\sigma_b = \frac{6WL}{n b t^2}, \qquad \delta = \frac{6WL^3}{n E b t^3}$

Leaves are classified as full length (the master leaf plus extra full-length leaves, carrying the eye) and graduated. Because the two groups deflect equally but have different stiffnesses, the full-length leaves carry a higher stress than the graduated ones:

$\sigma_F = \frac{18WL}{bt^2(3n_F + 2n_G)}, \qquad \sigma_G = \frac{12WL}{bt^2(3n_F + 2n_G)}$

so $\sigma_F = 1.5,\sigma_G$. Nipping — pre-bending the graduated leaves to a smaller radius so that an initial gap exists — equalises the stresses and is standard practice in vehicle suspension design.

Torsion Springs

A helical torsion spring is loaded by a moment about its coil axis. Here the wire carries bending, not torsion — the opposite of the compression spring. For a spring of $n$ coils:

$\sigma_b = K\frac{32M}{\pi d^3}, \qquad \theta = \frac{64 M D n}{E d^4}$

where $\theta$ is the angular deflection in radians and $K$ a curvature correction.

Materials

Springs are almost always fatigue-loaded, so surface finish matters enormously. Shot peening induces residual compressive stress at the surface and can raise fatigue life several-fold, and it is routine on vehicle suspension and valve springs.