The Wahl correction factor

The simple torsional formula for shear stress in a helical spring, τ = 8·F·D/(π·d³), assumes the wire is a straight torsion bar. A spring wire is curved, and the coil curvature crowds the stress onto the inner surface of the coil — exactly where springs fail. The Wahl correction factor accounts for that curvature plus the direct (transverse) shear:

Kᵥᵥ = (4C − 1) / (4C − 4) + 0.615 / C, where C = D / d

Here C is the spring index — the ratio of mean coil diameter to wire diameter. The corrected peak shear stress is then:

τ = Kᵥᵥ · 8·F·D / (π·d³)

How big is the correction?

The factor is largest for tightly-wound springs (low index) and approaches 1 as the coil gets relatively large. Because Kᵥᵥ depends only on the spring index C = D/d, you can tabulate it once and read it off for any spring:

Spring index C	Wahl factor Kᵥᵥ	Stress added vs uncorrected
4	1.404	+40%
5	1.311	+31%
6	1.253	+25%
8	1.184	+18%
10	1.145	+14%
12	1.119	+12%

Skipping the factor on a C = 4 spring under-predicts the real peak stress by 40% — enough to turn a “safe” design into one that yields or fatigues. A C = 4 coil is wound so tight (mean diameter only four wire-diameters across) that the inner fibre is severely crowded; by C = 12 the coil is loose enough that the same crowding adds barely a tenth. That is why the correction matters most exactly where springs are most highly stressed, and why most designers keep the index in the C = 4–12 band the compression spring calculator flags.

Worked example — with vs without the correction

Take a real spring: wire diameter d = 2 mm, mean coil diameter D = 16 mm, carrying an applied force F = 75 N. The spring index is

C = D / d = 16 / 2 = 8

First the uncorrected shear stress, treating the wire as a straight torsion bar:

τ = 8·F·D / (π·d³) = 8 · 75 · 16 / (π · 2³) = 382 MPa

Now the Wahl factor at C = 8:

Kᵥᵥ = (4·8 − 1)/(4·8 − 4) + 0.615/8 = 31/28 + 0.0769 = 1.184

Multiply through for the corrected peak stress on the inner fibre:

τ = Kᵥᵥ · 8·F·D / (π·d³) = 1.184 · 382 = 452 MPa

The correction adds about 70 MPa — an 18% increase over the straight-bar estimate, and that 18% sits precisely on the surface where a compression spring cracks. Plug the same d, D and force into the compression spring calculator and it returns the identical Kᵥᵥ = 1.184 and τ = 452 MPa, then compares that figure against the wire’s size-dependent allowable. Wind the same wire tighter to D = 10 mm (C = 5) and the correction jumps to 31%, because the inner fibre is now far more crowded — geometry alone, not load, drives how much stress you would have missed.

Why the inner fibre

When a curved wire is twisted, the material on the inside of the curve sees a shorter moment arm and higher shear strain than the outside. The transverse shear from the applied load adds to it on the same inner fibre. The combined effect is a stress concentration on the coil’s inner diameter, which the Wahl factor captures in a single multiplier.

When the correction is negligible

Read down the table and the trend is clear: Kᵥᵥ falls steeply from C = 4 to about C = 8, then flattens. Past a spring index of roughly C = 12 it has settled near 1.1 and barely moves — going from C = 12 to C = 16 shaves the factor by only about one percentage point. For a loose, large-diameter coil the wire is nearly straight relative to its own thickness, so the curvature crowding the table describes all but disappears. In that regime the correction is genuinely small, and a first-pass estimate that ignores it is off by only a few percent rather than the 40% you would miss on a tight spring.

That said, the correction never costs anything to apply, and the index is purely a geometric ratio — it does not change with load, free length or material, the way the spring rate does. The practical rule is simple: always include it, and recognise that it only becomes large once the index drops below about 8.

Wahl vs Bergsträsser

A common alternative is the Bergsträsser factor K_B = (C + 0.5)/(C − 0.75), which is simpler and gives nearly identical results — at C = 8 it returns 1.172 against Wahl’s 1.184, about 1% apart, and the two stay within roughly 1.5% of each other across the whole practical range C = 4–12. Some texts instead separate the pure curvature factor from the direct-shear factor and multiply them. Any of these is acceptable engineering practice; the difference is well inside the scatter of real wire properties, so it rarely changes a design decision. This site uses the Wahl factor.

The compression spring calculator applies the Wahl factor automatically and compares the corrected stress against the material’s size-dependent allowable, so you see immediately whether a design has margin — and the worked example above matches its output exactly.

Frequently asked questions

What is the Wahl correction factor?

It is a multiplier that corrects the simple torsional shear stress in a helical spring for the extra stress on the inner coil surface caused by curvature and direct shear: Kᵥᵥ = (4C−1)/(4C−4) + 0.615/C, where C = D/d.

Do I always need the Wahl factor?

For any real spring, yes — the uncorrected formula under-predicts the peak stress, and the inner fibre is where springs actually fail. Only ignore it for a rough first estimate.

What is the difference between the Wahl and Bergsträsser factors?

Both correct the same effect. Bergsträsser’s K_B = (C+0.5)/(C−0.75) is simpler and within about 1% of Wahl’s for typical indices; Wahl’s separates curvature and direct-shear terms. Either is acceptable.

Wahl or Bergsträsser — which should I use?

Use whichever your reference standardises on; the difference is engineering noise. At spring index C = 8 the Wahl factor is 1.184 and Bergsträsser’s is 1.172 — about 1% apart, and the gap stays under ~1.5% across the whole practical range C = 4–12. This site applies the Wahl factor, so worked numbers here match the compression spring calculator exactly.