---
title: "Tensile test data of a steel bar"
subject: "Material Science"
chapter: 3
key: "semester_3/material_science:tensile-test-numerical-steel-bar"
source: https://www.bctengineer.com/q/semester_3/material_science/tensile-test-numerical-steel-bar
site: BCT Engineering
---

# Tensile test data of a steel bar

Material Science · Chapter 3: Mechanical Properties and their tests

- Asked in: year not given, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/material_science/tensile-test-numerical-steel-bar
- Whole chapter: https://www.bctengineer.com/md/semester_3/material_science/chapter-3

## Question

A tensile test on a steel bar of diameter 12.5 mm and gauge length 50 mm gave the following data: load of 18 kN produced an elongation of 0.0367 mm (within the elastic range); yield load 36 kN; maximum load 52 kN; fracture load 41 kN. After fracture the gauge length was 62.5 mm and the minimum diameter at the neck was 8.5 mm. Determine (a) Young's modulus, (b) yield strength, (c) ultimate tensile strength, (d) engineering and true fracture stress, (e) percentage elongation and percentage reduction in area, (f) modulus of resilience.

## Short answer

Using A0 = π(12.5)²/4 = 122.72 mm², E = (18000/122.72)/(0.0367/50) = 200 GPa; yield strength = 36000/122.72 = 293.4 MPa; UTS = 52000/122.72 = 423.7 MPa; engineering fracture stress = 41000/122.72 = 334.1 MPa; true fracture stress = 41000/56.75 = 722.5 MPa; % elongation = (62.5-50)/50×100 = 25%; % RA = (122.72-56.75)/122.72×100 = 53.8%; resilience = (293.4)²/(2×200e3) = 0.215 MJ/m³.

## Answer

#### Given
$d_0 = 12.5$ mm, $L_0 = 50$ mm, $d_f = 8.5$ mm, $L_f = 62.5$ mm.

Original area:

$$
A_0 = \frac{\pi}{4}(12.5)^2 = 122.72\ \text{mm}^2
$$

Final (neck) area:

$$
A_f = \frac{\pi}{4}(8.5)^2 = 56.75\ \text{mm}^2
$$

#### (a) Young's modulus
$$
\sigma = \frac{18000}{122.72} = 146.7\ \text{MPa}, \qquad e = \frac{0.0367}{50} = 7.34\times10^{-4}
$$

$$
E = \frac{\sigma}{e} = \frac{146.7}{7.34\times10^{-4}} = 1.998\times10^{5}\ \text{MPa} \approx 200\ \text{GPa}
$$

#### (b) Yield strength
$$
\sigma_y = \frac{36000}{122.72} = 293.4\ \text{MPa}
$$

#### (c) Ultimate tensile strength
$$
\sigma_{u} = \frac{52000}{122.72} = 423.7\ \text{MPa}
$$

#### (d) Fracture stress
$$
\sigma_f = \frac{41000}{122.72} = 334.1\ \text{MPa (engineering)}
$$

$$
\sigma_{Tf} = \frac{41000}{56.75} = 722.5\ \text{MPa (true)}
$$

#### (e) Ductility
$$
\%\,EL = \frac{62.5 - 50}{50}\times100 = 25\%
$$

$$
\%\,RA = \frac{122.72 - 56.75}{122.72}\times 100 = 53.8\%
$$

#### (f) Modulus of resilience
$$
U_r = \frac{\sigma_y^2}{2E} = \frac{(293.4)^2}{2 \times 1.998\times10^{5}} = 0.215\ \text{MJ/m}^3
$$

(Since 1 MPa = 1 MN/m$^2$ = 1 MJ/m$^3$, the result is in MJ/m$^3$.)

| Quantity | Value |
|---|---|
| E | 200 GPa |
| Yield strength | 293.4 MPa |
| UTS | 423.7 MPa |
| Fracture stress (eng. / true) | 334.1 / 722.5 MPa |
| % elongation / % RA | 25% / 53.8% |
| Modulus of resilience | 0.215 MJ/m³ |

**Answer: E = 200 GPa, σy = 293.4 MPa, σu = 423.7 MPa, σf = 334.1 MPa (true 722.5 MPa), elongation 25%, reduction in area 53.8%, Ur = 0.215 MJ/m³.**

## Check yourself

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

1. What is A_0 for d = 12.5 mm?
   A. 122.7 mm^2
   B. 156.3 mm^2
   C. 61.4 mm^2
   D. 490.9 mm^2
   Answer: A. A_0 = pi/4 x 12.5^2 = 122.72 mm^2. (Key point: Original area)

2. How is E found from the elastic data?
   A. Yield stress / elongation
   B. Stress (18000/A0) / strain (0.0367/50) = about 200 GPa
   C. UTS / strain
   D. Load / elongation
   Answer: B. E = 146.7 MPa / 7.34e-4 = about 2e5 MPa. (Key point: Young's modulus)

3. Why is true fracture stress larger than engineering?
   A. It uses final neck area 56.75 mm^2 (722.5 MPa)
   B. It uses a higher load
   C. It uses gauge length
   D. It uses the UTS
   Answer: A. 41000/56.75 = 722.5 MPa vs 334.1 MPa on original area. (Key point: True fracture stress)

4. What are %EL and %RA?
   A. 53.8% and 25%
   B. 25% and 25%
   C. 25% and 53.8%
   D. 12.5% and 53.8%
   Answer: C. %EL = 12.5/50 = 25%; %RA = (122.72-56.75)/122.72 = 53.8%. (Key point: Ductility)

## 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
- [Tensile properties and effect of temperature](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Aproperties-from-tensile-test-and-temperature) — Material Science, chapter 3

## Related in other subjects

- [Tensile test numerical, 40 mm specimen](https://www.bctengineer.com/question.md?key=semester_3%2Fcivil_engineering_materials%3Anumerical-tensile-test-40mm) — Civil Engineering Materials, chapter 1
- [Tensile test engineering and true values](https://www.bctengineer.com/question.md?key=semester_4%2Fmanufacturing_and_production_processes%3Atensile-test-true-stress-strain-numerical) — Manufacturing and Production Processes, chapter 2
- [Total elongation of stepped-load steel bar](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asteel-bar-total-elongation-abcd) — Strength of Materials, chapter 3
- [Elongation of a stepped steel bar](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Astepped-bar-axial-elongation) — Strength of Materials, chapter 2