---
title: "Modulus of toughness and resilience (250 kN)"
subject: "Civil Engineering Materials"
chapter: 1
key: "semester_3/civil_engineering_materials:numerical-toughness-resilience-300mm"
source: https://www.bctengineer.com/q/semester_3/civil_engineering_materials/numerical-toughness-resilience-300mm
site: BCT Engineering
---

# Modulus of toughness and resilience (250 kN)

Civil Engineering Materials · Chapter 1: Introduction to Civil Engineering Material

- Asked in: 2070 Chaitra (old course), 8 marks
- Video: [Difference between Modulus of Toughness and Modulus of Resilience](https://www.youtube.com/watch?v=ECJwQDHw5ao) — Civil Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/civil_engineering_materials/numerical-toughness-resilience-300mm
- Whole chapter: https://www.bctengineer.com/md/semester_3/civil_engineering_materials/chapter-1

## Question

A mild steel specimen of 10 mm diameter and 300 mm long, which resists the maximum tensile load of 250 kN at 2 mm diameter. If the material resists 120 kN yield load by elongating 8 mm, then what is the modulus of toughness (printed 'roughness') and resilience?

## Short answer

Original area = 78.54 mm²; yield stress = 1527.89 N/mm², ultimate stress = 3183.10 N/mm², strain = 8/300 = 0.02667. Modulus of resilience Ur = ½ × 1527.89 × 0.02667 = 20.37 N/mm² (MJ/m³); modulus of toughness Ut ≈ (1527.89+3183.10)/2 × 0.02667 = 62.81 N/mm² (MJ/m³).

## Answer

**Given:** $d = 10$ mm, $L = 300$ mm, $P_u = 250$ kN, $P_y = 120$ kN, elongation $\delta = 8$ mm.

**Assumptions:** the stresses are based on the original area $A_0$ (the 2 mm diameter at maximum load is the neck diameter and does not change $A_0$). The question gives only one elongation (8 mm), so this strain is used for both moduli. The modulus of resilience is the elastic energy per unit volume up to yield, and the modulus of toughness is taken as the approximate area under the curve (average of yield and ultimate stress times strain). The data give stresses far above those of real mild steel, but they are used as given.

$$
\begin{aligned}
A_0 &= \frac{\pi}{4}(10)^2 = 78.54\ \text{mm}^2\\
\sigma_y &= \frac{120\times10^3}{78.54} = 1527.89\ \text{N/mm}^2\\
\sigma_u &= \frac{250\times10^3}{78.54} = 3183.10\ \text{N/mm}^2\\
\varepsilon &= \frac{8}{300} = 0.02667
\end{aligned}
$$

**Modulus of resilience** (elastic limit taken at yield):

$$
U_r = \tfrac12\sigma_y\varepsilon = \tfrac12(1527.89)(0.02667) = 20.37\ \text{N/mm}^2
$$

**Modulus of toughness:**

$$
U_t \approx \frac{\sigma_y+\sigma_u}{2}\,\varepsilon = \frac{1527.89+3183.10}{2}(0.02667) = 62.81\ \text{N/mm}^2
$$

**Answer: Modulus of resilience = 20.37 N/mm² (MJ/m³); modulus of toughness ≈ 62.81 N/mm² (MJ/m³).**

## Check yourself

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

1. A0 for the 10 mm bar is
   A. 314.16 mm^2
   B. 3.14 mm^2
   C. 157.08 mm^2
   D. 78.54 mm^2
   Answer: D. pi/4 x 10^2 = 78.54. (Key point: Area)

2. Yield stress is
   A. 1527.89 N/mm^2
   B. 120 N/mm^2
   C. 3183.10 N/mm^2
   D. 763.9 N/mm^2
   Answer: A. 120000/78.54. (Key point: Yield stress)

3. Modulus of resilience is
   A. 62.81 N/mm^2
   B. 20.37 N/mm^2
   C. 40.74 N/mm^2
   D. 1527.89 N/mm^2
   Answer: B. 0.5 x 1527.89 x 0.02667. (Key point: Resilience)

4. Modulus of toughness (approx.) is
   A. 20.37 N/mm^2
   B. 125.6 N/mm^2
   C. 62.81 N/mm^2
   D. 41.9 N/mm^2
   Answer: C. Average stress 2355.5 x 0.02667. (Key point: Toughness)

## Close variants set in other exams

- [Modulus of toughness and resilience (300 kN)](https://www.bctengineer.com/question.md?key=semester_3%2Fcivil_engineering_materials%3Anumerical-toughness-resilience-400mm) — Civil Engineering Materials, chapter 1, asked 2×

## Related in Civil Engineering Materials

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- [Modulus of toughness of medium carbon steel](https://www.bctengineer.com/question.md?key=semester_3%2Fcivil_engineering_materials%3Anumerical-modulus-toughness-415) — Civil Engineering Materials, chapter 1

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