Spring materials and yield strength estimation

Common spring materials

Selection factors: strength and fatigue limit, resistance to stress relaxation and sag, operating temperature, corrosion, electrical or magnetic needs, wire availability and cost, and size (thin wire is stronger).

Estimation of strength

The tensile strength of drawn wire depends on diameter (thin wire is more heavily worked and stronger):

$$S_{ut} = \frac{A}{d^{\,m}}$$

where $A$ and $m$ are constants for the material and $d$ is in mm. Springs fail in torsion, so the torsional yield strength is taken as a fraction of $S_{ut}$:

• cold-drawn carbon steels (music wire, hard-drawn): $S_{sy} \approx 0.45\,S_{ut}$
• hardened and tempered steels (oil-tempered, alloy wires): $S_{sy} \approx 0.50\,S_{ut}$
• the ultimate shear strength is $S_{su} \approx 0.67\,S_{ut}$; it is used in fatigue (Goodman) calculations, while static design uses $S_{sy}$.

The allowable design shear stress is $S_{sy}$ divided by the factor of safety (often 1.2 to 1.5).

Example: music wire, $d = 2$ mm

$$S_{ut} = \frac{2211}{2^{0.145}} = \frac{2211}{1.1057} = 2000\ \text{MPa}$$ $$S_{sy} = 0.45\times2000 = 900\ \text{MPa}$$

Answer: $S_{ut} \approx 2000$ MPa and $S_{sy} \approx 900$ MPa.