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

# True stress and true strain relations

Material Science · Chapter 3: Mechanical Properties and their tests

- Asked in: year not given, 6 marks
- Video: [Engineering stress and True Stress | Difference | Relation](https://www.youtube.com/watch?v=M43sgbaqH5A) — MatProc
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/material_science/true-stress-strain-relations
- Whole chapter: https://www.bctengineer.com/md/semester_3/material_science/chapter-3

## Question

Define true stress and true strain. Derive the relations between true and engineering stress and strain, stating the limits of validity. Why does the true stress-strain curve keep rising after the ultimate tensile strength?

## Short answer

True stress is load/instantaneous area; true strain is ln(Li/L0). Up to necking, ε_T = ln(1+e) and σ_T = σ(1+e), using volume constancy. Beyond necking, these relations fail because strain is non-uniform; true stress is computed from neck area. The true stress curve continues rising because the area reduction outpaces load drop due to strain hardening.

## Answer

**True stress** $\sigma_T$ is the load divided by the *instantaneous* cross-sectional area $A_i$; **true strain** $\varepsilon_T$ is the sum of incremental strains, each based on the instantaneous length.

$$
\sigma_T = \frac{P}{A_i}, \qquad \varepsilon_T = \int_{L_0}^{L_i}\frac{dL}{L} = \ln\frac{L_i}{L_0}
$$

#### Derivation
Engineering strain is $e = \dfrac{L_i - L_0}{L_0}$, so $\dfrac{L_i}{L_0} = 1 + e$. Therefore

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

Plastic deformation conserves volume: $A_0 L_0 = A_i L_i$, so $A_i = \dfrac{A_0}{1+e}$. Then

$$
\sigma_T = \frac{P}{A_i} = \frac{P}{A_0}(1+e) = \sigma\,(1 + e)
$$

#### Validity
Both equations hold only up to the start of necking (uniform deformation). After necking the strain is no longer uniform, so volume constancy cannot be applied over the whole gauge length. Beyond necking, $\sigma_T$ is found from the measured neck area, and $\varepsilon_T = \ln(A_0/A_i)$.

#### Flow curve
In the plastic region $\sigma_T = K\varepsilon_T^{\,n}$, where $K$ is the strength coefficient and $n$ the strain-hardening exponent (0.1-0.5 for metals). Necking begins at $\varepsilon_T = n$.

#### Why true stress keeps rising
After the UTS the engineering stress falls, because the load falls while the original area $A_0$ is still used in the calculation. But the neck area shrinks even faster than the load, so the load per actual area ($\sigma_T$) keeps increasing, due to strain hardening, until fracture.

| Basis | Engineering | True |
|---|---|---|
| Area used | Original | Instantaneous |
| Curve after UTS | Falls | Rises |
| Strain | $\Delta L/L_0$ | $\ln(L_i/L_0)$ |

## Check yourself

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

1. Which relation links true strain and engineering strain?
   A. eps_T = e
   B. eps_T = ln(1+e)
   C. eps_T = e^2
   D. eps_T = 1+e
   Answer: B. Integrating dL/L gives ln(Li/L0) = ln(1+e). (Key point: True strain relation)

2. Which holds up to necking?
   A. sigma_T = sigma/(1+e)
   B. sigma_T = sigma(1-e)
   C. sigma_T = sigma(1+e)
   D. sigma_T = sigma + e
   Answer: C. Volume constancy gives A_i = A_0/(1+e), so sigma_T = sigma(1+e). (Key point: True stress relation)

3. Why are these equations valid only up to necking?
   A. Elasticity ends
   B. Strain is no longer uniform after necking
   C. Volume increases
   D. Load rises
   Answer: B. After necking deformation is localised and volume constancy cannot be applied along the gauge. (Key point: Validity)

4. Why does true stress keep rising after UTS?
   A. Area is assumed constant
   B. Load rises
   C. Neck area shrinks faster than load falls, with strain hardening continuing
   D. Temperature rises
   Answer: C. Load per actual area keeps increasing because area falls faster than load. (Key point: Rising true stress)

## Related in Material Science

- [Tensile test and stress-strain diagram](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Atensile-test-engineering-stress-strain) — Material Science, chapter 3
- [Strain-hardening exponent and true stress](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Astrain-hardening-exponent-numerical) — Material Science, chapter 3
- [Tensile test data of a steel bar](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Atensile-test-numerical-steel-bar) — Material Science, chapter 3
- [Ductile and brittle stress-strain behaviour](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Aductile-vs-brittle-stress-strain) — Material Science, chapter 3

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