---
title: "Strain-hardening exponent and true stress"
subject: "Material Science"
chapter: 3
key: "semester_3/material_science:strain-hardening-exponent-numerical"
source: https://www.bctengineer.com/q/semester_3/material_science/strain-hardening-exponent-numerical
site: BCT Engineering
---

# Strain-hardening exponent and true stress

Material Science · Chapter 3: Mechanical Properties and their tests

- Asked in: year not given, 6 marks
- Video: [Example: Calculating True Strain and Strain Hardening Exponent](https://www.youtube.com/watch?v=c4DiZpz7-9Q) — Dr Kidd Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/material_science/strain-hardening-exponent-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_3/material_science/chapter-3

## Question

(a) A metal shows true stress 450 MPa at true strain 0.05 and true stress 520 MPa at true strain 0.12 in the uniform plastic region. Assuming $\sigma_T = K\varepsilon_T^{\,n}$, find n and K. (b) Find the true strain at the start of necking, the true stress at necking and the engineering UTS. (c) An engineering stress of 400 MPa is measured at an engineering strain of 0.10 in the uniform region; find the true stress and true strain.

## Short answer

Using ln(520/450)/ln(0.12/0.05) gives n = 0.165; K = 450/(0.05^0.165) = 738 MPa. Necking starts at ε_T = n = 0.165, true stress = 738×(0.165^0.165) = 548 MPa, engineering UTS = 548/1.180 = 465 MPa. For e=0.10, σ_T = 400×1.10 = 440 MPa, ε_T = ln(1.10) = 0.0953.

## Answer

#### (a) n and K
Taking logs of the power law $\sigma_T = K\varepsilon_T^{\,n}$ for the two points:

$$
n = \frac{\ln(\sigma_2/\sigma_1)}{\ln(\varepsilon_2/\varepsilon_1)} = \frac{\ln(520/450)}{\ln(0.12/0.05)} = \frac{0.1449}{0.8755} = 0.165
$$

$$
K = \frac{\sigma_1}{\varepsilon_1^{\,n}} = \frac{450}{(0.05)^{0.165}} = 738\ \text{MPa}
$$

#### (b) Necking
Necking begins when the strain-hardening rate equals the stress, $d\sigma_T/d\varepsilon_T = \sigma_T$, which gives $\varepsilon_T = n$.

$$
\varepsilon_{T,neck} = n = 0.165
$$

$$
\sigma_{T,neck} = K n^{n} = 738 \times (0.165)^{0.165} = 548\ \text{MPa}
$$

Engineering UTS: $\sigma_T = \sigma(1+e)$ with $1 + e = e^{\varepsilon_T} = e^{0.165} = 1.180$, so

$$
\sigma_{UTS} = \frac{548}{1.180} = 465\ \text{MPa}
$$

#### (c) Engineering to true
$$
\sigma_T = \sigma(1+e) = 400 \times 1.10 = 440\ \text{MPa}
$$

$$
\varepsilon_T = \ln(1+e) = \ln 1.10 = 0.0953
$$

This is valid because $e = 0.10$ is less than the engineering strain at necking ($1.180 - 1 = 0.18$).

**Answer: n = 0.165, K = 738 MPa; necking at ε_T = 0.165, σ_T = 548 MPa, engineering UTS = 465 MPa; (c) σ_T = 440 MPa, ε_T = 0.0953.**

## Check yourself

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

1. How is n found from two points?
   A. Ratio of the stresses
   B. Difference of strains
   C. n = ln(sigma2/sigma1)/ln(eps2/eps1)
   D. n = sigma/eps
   Answer: C. Taking logs of the power law gives the slope n = 0.165. (Key point: Finding n)

2. At what true strain does necking begin?
   A. Equal to K
   B. Equal to n
   C. Equal to 1
   D. Equal to 2n
   Answer: B. Necking starts when dsigma/deps = sigma, giving eps_T = n. (Key point: Necking condition)

3. How is engineering UTS found from true stress at necking?
   A. Multiply by (1+e)
   B. Add e
   C. Divide by (1+e) = e^n, giving about 465 MPa
   D. Divide by n
   Answer: C. sigma_UTS = 548/1.180 = 465 MPa. (Key point: Engineering UTS)

4. True stress and strain for e = 0.10, sigma = 400 MPa?
   A. 440 MPa and 0.0953
   B. 400 MPa and 0.10
   C. 360 MPa and 0.0953
   D. 440 MPa and 0.10
   Answer: A. sigma_T = 400 x 1.1 = 440; eps_T = ln 1.1 = 0.0953. (Key point: Part c)

## Related in Material Science

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