---
title: "Stress-strain diagram of mild steel"
subject: "Strength of Materials"
chapter: 2
key: "semester_4/strength_of_materials_mechanical:stress-strain-diagram-mild-steel"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/stress-strain-diagram-mild-steel
site: BCT Engineering
---

# Stress-strain diagram of mild steel

Strength of Materials · Chapter 2: Stress and strain – axial load

- Asked in: year not given, 6 marks
- Video: [STRESS-STRAIN CURVE FOR MILD STEEL(Engineering Stress And True Stress)](https://www.youtube.com/watch?v=uO-p274-RyA) — Kudratnama
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/stress-strain-diagram-mild-steel
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-2

## Question

Draw the stress-strain diagram of mild steel in a tension test and explain the important points on it. How does the curve of a brittle material such as cast iron differ?

## Short answer

Mild steel shows proportional limit, elastic limit, upper/lower yield, strain hardening, ultimate stress, and fracture with necking (ductile, large elongation). Cast iron has no yield point, negligible elongation (brittle), sudden flat fracture, and is much stronger in compression than tension.

## Answer

The stress-strain diagram plots nominal stress $\sigma = P/A_0$ against strain $\varepsilon = \delta/L_0$ for a specimen pulled in a universal testing machine.

```text
 stress
   |              U
   |        Y1   _.-"-._
   |         /\_/        '.
   |        /Y2             \ F
   |       /
   |      / P,E
   |     /
   |    /
   +------------------------- strain
```

(P = proportional limit, E = elastic limit, Y1/Y2 = upper/lower yield point, U = ultimate point, F = fracture.)

#### Important points
1. **Proportional limit (P):** the stress is proportional to strain up to here, so Hooke's law $\sigma = E\varepsilon$ holds. The slope of the straight line is the modulus of elasticity $E$ (about 200 GPa for steel).
2. **Elastic limit (E):** the specimen returns to its original length on unloading. It lies very close to P.
3. **Yield point (Y1, Y2):** the strain increases suddenly with almost no increase in stress (plastic flow). There is an upper yield point and a lower yield point; design uses the yield stress $\sigma_y$ (about 250 MPa for mild steel).
4. **Strain hardening:** after yielding the material regains strength and the stress rises again to the ultimate point.
5. **Ultimate stress (U):** the maximum nominal stress. Beyond U a local reduction in area (necking) starts.
6. **Fracture (F):** the specimen breaks. The nominal stress falls because it is calculated on the original area, although the true stress keeps rising.

Mild steel is ductile: the elongation at fracture is 20 to 30 percent and it shows a cup and cone fracture.

#### Cast iron (brittle material)
| Point | Mild steel | Cast iron |
|---|---|---|
| Yield point | Clear | None |
| Elongation | Large (20 to 30 %) | Very small (below 1 %) |
| Necking | Present | Absent |
| Fracture | Cup and cone | Flat, sudden |
| Strength | Equal in tension and compression | Much stronger in compression |

For materials with no clear yield point (aluminium, high-strength steel) the **0.2 % offset yield stress** is used.

## Check yourself

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

1. What does the slope of the straight part of the stress-strain curve (up to the proportional limit) represent?
   A. The yield stress
   B. The modulus of elasticity E
   C. The ultimate stress
   D. Poisson's ratio
   Answer: B. Up to the proportional limit sigma = E epsilon holds, so the slope is E (about 200 GPa for steel). (Key point: Proportional limit and E)

2. What happens at the yield point of mild steel?
   A. Stress falls sharply and the bar breaks
   B. Strain increases suddenly with almost no increase in stress
   C. Necking begins at the ultimate stress
   D. The bar returns to its original length on unloading
   Answer: B. At yield the material flows plastically: strain jumps while stress stays nearly constant. (Key point: Yield point)

3. Why does the nominal stress fall after the ultimate point before fracture?
   A. The material becomes weaker in the elastic range
   B. Stress is computed on the original area although necking reduces the real area
   C. The load on the bar is increased
   D. Strain hardening reverses
   Answer: B. Nominal stress uses A0; with necking the actual area shrinks so the load falls, though true stress keeps rising. (Key point: Stress drop before fracture)

4. Which is true of cast iron compared with mild steel?
   A. It has a clear yield point and cup-and-cone fracture
   B. It shows large elongation and necking
   C. It has no clear yield point, tiny elongation and sudden flat fracture
   D. It is stronger in tension than in compression
   Answer: C. Cast iron is brittle: no yield point, no necking, elongation below 1 %, flat sudden fracture, and stronger in compression. (Key point: Cast iron vs mild steel)

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