Shear stress in springs: why the wire works in torsion

When you compress or stretch a helical spring, instinct says the wire is being pulled like a rubber band. In practice something quite different happens: every cross-section of the wire is twisted about its own axis. A spring is essentially a torsion bar wound into a helix, and that is why the stress that governs the design is a shear stress, not a tensile stress.

Grasping this distinction is what separates a correct calculation from a guess. The shear stress decides whether the spring survives millions of cycles or breaks in the first week, whether it can be made from thinner and cheaper wire, or whether it needs a nobler steel. This guide walks through the physics, the formula with the Wahl correction, how torsion springs differ, the allowable stress, the size effect, and a full worked example.

Why the spring works in torsion

Picture a single coil cut by a plane passing through the axis of the wire. The axial load F, applied along the spring's axis, acts at a distance equal to the mean coil radius (D/2) from the centre of the wire. That lever arm creates a torque T = F · D/2 that twists the wire's cross-section. It is exactly the same mechanism as a drive shaft under torque.

In torsion, shear stress is maximum at the wire surface and zero at the centre. Because the coil is curved, the inner side (facing the spring axis) travels a shorter path than the outer side, so the shear distribution concentrates precisely on the inner fibre. That is where the spring fails. This is why every serious calculation targets the stress on the inner surface of the coil, not an average.

The shear stress formula with the Wahl correction

Starting from the pure torsion of a circular bar, the shear stress in the wire can be written from the torque T = F · D/2 and the polar moment of inertia. Rearranging the terms leads to the bare term, τ0, which already includes the dominant torsional shear:

That term, however, underestimates the real stress for two reasons: there is also a direct shear from the transverse force, and there is a stress concentration due to the curvature of the coil. The Wahl factor, Kw, corrects both effects at once and depends only on the spring index C = D/d. The corrected stress, which is the one you compare against the material, becomes:

Notice the d³ in the denominator. Stress grows with the cube of the inverse wire diameter: shaving 10% off the wire raises the stress by about 37%. That is why wire diameter is the most sensitive variable in the entire design. Springs with a low index C (very tight coils) have a high Kw, which penalises the stress; indices between 6 and 10 are usually the best compromise between manufacturability and stress level.

Torsion springs are different: bending, not shear

Here lies a common mistake. A torsion spring does not twist the wire, despite the name. In it the load is a moment applied about the axis of the helix, which winds or unwinds the coil. That load bends the wire rather than twisting it. The dominant stress is therefore a normal bending stress, not a shear stress.

The bending stress on the inner fibre of a torsion spring follows the form of bending of a circular-section beam, corrected by an inner-fibre concentration factor, Ki:

The practical consequence matters: because the material resists bending better than shear, torsion springs allow higher working stresses than compression springs made from the same wire. When reading an allowable-stress table, it is therefore mandatory to know whether it refers to shear (compression or extension) or to bending (torsion). Swapping one for the other yields designs that are either dangerous or needlessly heavy.

Allowable stress: a fraction of the tensile strength

You do not compare the working stress against an absolute number, but against a fraction of the material's ultimate tensile strength, Sut. Every spring steel has its own Sut, and the allowable stress is defined as a percentage of it. As an engineering reference, for steel-wire compression springs working in shear the ranges below apply:

• Static or light service: up to about 45% of Sut for the working stress.
• Fatigue or cyclic service: drop to somewhere between 30% and 40% of Sut, depending on the required cycle count.
• Torsion springs (bending): allow higher values, typically between 60% and 80% of Sut, because they work in normal stress.
• Presetting: yielding the spring beyond its elastic limit during manufacture creates favourable residual stresses and lets you raise the static allowable.
• Shock, corrosion or elevated temperature: apply extra margin and reduce the allowable fraction further.

The safety factor is simply the ratio between the adopted allowable stress and the calculated working stress. A value between 1.2 and 1.5 is common for well-known static service; uncertain loads or fatigue call for larger margins.

The size effect: thin wire is stronger

A point that surprises beginners: Sut is not a constant of the material, it depends on the wire diameter. Cold drawing, which pulls the wire through successive dies, refines the microstructure and work-hardens the steel. The thinner the wire, the more passes and the more work-hardening, so thin wire is far stronger per square millimetre than thick wire.

For music wire, Sut is around 2200 to 2400 MPa for diameters near 0.5 mm and falls to something like 1600 to 1800 MPa for wires several millimetres thick. This means the allowable stress is itself a function of diameter. Using the same Sut for every diameter is a mistake that over-designs thin springs and under-designs thick ones.

A full worked example

Let us size the spring from this guide's preset. Data: working force F = 150 N, outer diameter OD = 25 mm, wire diameter d = 2.5 mm. The mean diameter is D = OD − d = 22.5 mm, and the index is C = D/d = 22.5 / 2.5 = 9.

First the Wahl factor. With C = 9: Kw = (4·9 − 1)/(4·9 − 4) + 0.615/9 = 35/32 + 0.0683 = 1.094 + 0.068 ≈ 1.16.

Now the bare term. The denominator is π · d³ = π · 2.5³ = π · 15.625 = 49.09 mm³. The numerator is 8 · F · D = 8 · 150 · 22.5 = 27000 N·mm. So τ0 = 27000 / 49.09 ≈ 550 MPa.

Applying the correction: τ = Kw · τ0 = 1.16 · 550 ≈ 638 MPa. This is the real working stress on the inner fibre of the coil.

Now the verdict. For music wire at this diameter, take Sut ≈ 2000 MPa and static service at 45%: allowable stress ≈ 0.45 · 2000 = 900 MPa. Since 638 MPa is well below 900 MPa, the spring is safe, with a safety factor of 900/638 ≈ 1.4. There is room to raise the load or slightly reduce the wire.

Stress, spring rate and the d³ versus d⁴ distinction

The wire diameter governs two quantities at the same time, but with different exponents, and understanding this is the heart of spring design. The spring rate grows with the fourth power of the diameter, while the stress falls with the cube of the diameter:

In other words, thickening the wire makes the spring stiffer (more N per mm) and, at the same time, lowers the working stress. It looks like a double win, but there is a price: a stiffer spring also builds more force for the same deflection, which pushes the stress back up. The designer's game is to balance d, D and the coil count to reach the desired rate while keeping the stress within the allowable. Because the exponents 3 and 4 differ, there is almost always a combination that satisfies both requirements at once.

On molas.app.br, every time you adjust the wire diameter, the outer diameter or the coil count, the tool recomputes the working stress with the Wahl correction and shows the safety margin against the chosen material, in real time. You see immediately whether the spring is loose, at the limit or rejected, without opening a spreadsheet.