---
title: "Tensile test engineering and true values"
subject: "Manufacturing and Production Processes"
chapter: 2
key: "semester_4/manufacturing_and_production_processes:tensile-test-true-stress-strain-numerical"
source: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/tensile-test-true-stress-strain-numerical
site: BCT Engineering
---

# Tensile test engineering and true values

Manufacturing and Production Processes · Chapter 2: Manufacturing Properties of Materials

- Asked in: year not given, 8 marks
- Video: [Understanding True Stress and True Strain](https://www.youtube.com/watch?v=AkX6JqlWRqc) — The Efficient Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/tensile-test-true-stress-strain-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/manufacturing_and_production_processes/chapter-2

## Question

A tensile test is carried out on a round steel specimen of initial diameter 12 mm and gauge length 50 mm. The following readings are recorded: load at yield 28.5 kN, maximum load 42.4 kN, load at fracture 33.0 kN. After fracture the gauge length is 64 mm and the diameter at the neck is 8.4 mm. Calculate (a) yield strength, (b) ultimate tensile strength, (c) percentage elongation, (d) percentage reduction in area, (e) true fracture stress and true fracture strain.

## Short answer

Initial area 113.10 mm2, fracture area 55.42 mm2. Yield strength = 28,500/113.10 = 252.0 MPa; UTS = 42,400/113.10 = 374.9 MPa; elongation = 28%; reduction in area = 51%; true fracture stress = 33,000/55.42 = 595.5 MPa; true fracture strain = ln(113.10/55.42) = 0.713.

## Answer

#### Given data

$d_0 = 12\ \text{mm},\ L_0 = 50\ \text{mm},\ L_f = 64\ \text{mm},\ d_f = 8.4\ \text{mm}$

Initial area:

$$
A_0 = \frac{\pi}{4}(12)^2 = 113.10\ \text{mm}^2
$$

Area at fracture:

$$
A_f = \frac{\pi}{4}(8.4)^2 = 55.42\ \text{mm}^2
$$

#### (a) Yield strength

$$
\sigma_y = \frac{P_y}{A_0} = \frac{28\,500}{113.10} = 252.0\ \text{MPa}
$$

#### (b) Ultimate tensile strength

$$
\sigma_{UTS} = \frac{P_{max}}{A_0} = \frac{42\,400}{113.10} = 374.9\ \text{MPa}
$$

#### (c) Percentage elongation

$$
\%EL = \frac{L_f - L_0}{L_0}\times 100 = \frac{64-50}{50}\times 100 = 28\%
$$

#### (d) Percentage reduction in area

$$
\%RA = \frac{A_0 - A_f}{A_0}\times 100 = \frac{113.10 - 55.42}{113.10}\times 100 = 51\%
$$

#### (e) True fracture stress and strain

True stress uses the instantaneous (smallest) area:

$$
\sigma_{f} = \frac{P_f}{A_f} = \frac{33\,000}{55.42} = 595.5\ \text{MPa}
$$

True strain at fracture (volume constant, so $L/L_0 = A_0/A$):

$$
\varepsilon_f = \ln\frac{A_0}{A_f} = \ln\frac{113.10}{55.42} = 0.713
$$

| Quantity | Value |
|---|---|
| Yield strength | 252.0 MPa |
| UTS | 374.9 MPa |
| Elongation | 28 % |
| Reduction in area | 51 % |
| True fracture stress | 595.5 MPa |
| True fracture strain | 0.713 |

The true fracture stress (595 MPa) is much higher than the engineering stress at fracture ($33\,000/113.1 = 291.8$ MPa) because of the large reduction of area at the neck.

**Answer:** $\sigma_y = 252$ MPa, UTS $= 374.9$ MPa, elongation $= 28\%$, reduction in area $= 51\%$, true fracture stress $= 595.5$ MPa, true fracture strain $= 0.713$.

## Check yourself

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

1. Which area is used to compute yield strength and UTS?
   A. Final neck area A_f
   B. Initial area A_0
   C. Average of A_0 and A_f
   D. Instantaneous area at yield
   Answer: B. Engineering strengths divide load by the original area A_0 = 113.10 mm^2. (Key point: Engineering stress area)

2. Yield load 28.5 kN on A_0 = 113.10 mm^2 gives a yield strength of about
   A. 252 MPa
   B. 375 MPa
   C. 505 MPa
   D. 126 MPa
   Answer: A. 28 500 / 113.10 = 252 MPa. (Key point: Yield strength value)

3. Why is the true fracture stress (595.5 MPa) much higher than the engineering fracture stress (291.8 MPa)?
   A. The fracture load is larger than the maximum load
   B. True stress uses the much smaller neck area at fracture
   C. True stress uses the gauge length
   D. The material hardens elastically
   Answer: B. True stress = P_f/A_f with the smallest (neck) area, so it is larger than P_f/A_0. (Key point: True fracture stress)

4. How is true fracture strain found from the data, and what is its value?
   A. ln(L_f/L_0) = 0.247
   B. ln(A_0/A_f) = 0.713
   C. (L_f - L_0)/L_0 = 0.28
   D. (A_0 - A_f)/A_0 = 0.51
   Answer: B. With constant volume, L/L_0 = A_0/A, so epsilon_f = ln(113.10/55.42) = 0.713. (Key point: True strain at fracture)

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