---
title: "Spring materials and yield strength estimation"
subject: "Machine Design II"
chapter: 6
key: "semester_7/machine_design_ii:spring-materials-yield-strength"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/spring-materials-yield-strength
site: BCT Engineering
---

# Spring materials and yield strength estimation

Machine Design II · Chapter 6: Spring design

- Asked in: year not given, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/spring-materials-yield-strength
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-6

## Question

List the common spring materials and the factors governing their selection. How are the tensile strength and the torsional yield strength of spring wire estimated? Estimate them for music wire of 2 mm diameter (A = 2211 MPa mm^m, m = 0.145).

## Short answer

Common materials: music wire, hard‑drawn wire, oil‑tempered, chrome‑vanadium, stainless steel. Selection factors: strength, fatigue, temperature, corrosion, cost. Tensile strength S_ut = A/d^m; torsional yield S_sy ≈ 0.45 S_ut (music wire). For 2 mm music wire: S_ut ≈ 2000 MPa, S_sy ≈ 900 MPa.

## Answer

#### Common spring materials
| Material | Features | Use |
|---|---|---|
| Hard-drawn wire | Cheapest, low strength | Low-stress static springs |
| Music wire (high-carbon) | Highest tensile strength, good fatigue | Small precision springs |
| Oil-tempered wire | General purpose, hardened and tempered | Valve, clutch, general springs |
| Chrome-vanadium | High fatigue, shock resistance, up to 220 °C | Valve springs |
| Chrome-silicon | Very high strength, high temperature | Heavy-duty valve springs |
| Stainless steel (302) | Corrosion resistant | Food, marine |
| Phosphor bronze, beryllium copper | Electrical conductivity, corrosion resistance | Electrical contacts |

**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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does the tensile strength of drawn spring wire vary with diameter, S_ut = A/d^m?
   A. Thicker wire is stronger
   B. Strength is independent of d
   C. Thinner wire is stronger because it is more worked
   D. Strength grows with d^m
   Answer: C. Thin wire is more cold-worked, so S_ut = A/d^m is larger for smaller d. (Key point: Tensile strength vs diameter)

2. What fraction of S_ut is taken as torsional yield strength S_sy for music wire (cold-drawn carbon steel)?
   A. 0.67
   B. 0.45
   C. 0.50
   D. 0.25
   Answer: B. Cold-drawn carbon steels use S_sy about 0.45 S_ut; hardened and tempered steels use 0.50. (Key point: Torsional yield estimate)

3. For music wire of 2 mm with A = 2211 and m = 0.145, S_ut and S_sy are about
   A. 1000 MPa and 450 MPa
   B. 2000 MPa and 900 MPa
   C. 2000 MPa and 1340 MPa
   D. 2211 MPa and 995 MPa
   Answer: B. S_ut = 2211/2^0.145 = 2000 MPa and S_sy = 0.45 x 2000 = 900 MPa. (Key point: Music wire example)

4. Why is the torsional yield strength, not tensile, used in spring design?
   A. Spring wire fails in torsion
   B. Spring wire never yields in tension
   C. Tension tests are impossible on wire
   D. Torsion gives a higher strength value
   Answer: A. The wire of a helical spring is mainly stressed in torsion, so shear strength is the relevant limit. (Key point: Why torsional strength)

## Related in Machine Design II

- [Helical spring under fluctuating load](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aspring-fatigue-soderberg-design) — Machine Design II, chapter 6
- [Design of a helical compression spring](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Acompression-spring-design-static) — Machine Design II, chapter 6
- [Surge and natural frequency of valve spring](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aspring-surge-natural-frequency) — Machine Design II, chapter 6
- [Stress and deflection of helical spring](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Ahelical-spring-stress-deflection-derive) — Machine Design II, chapter 6
- [Extension springs, ends and spring combinations](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aextension-spring-ends-series-parallel) — Machine Design II, chapter 6
- [Reliability from stress-strength interference](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Astress-strength-interference-reliability) — Machine Design II, chapter 4

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