---
title: "Modulus of toughness and resilience (300 kN)"
subject: "Civil Engineering Materials"
chapter: 1
key: "semester_3/civil_engineering_materials:numerical-toughness-resilience-400mm"
source: https://www.bctengineer.com/q/semester_3/civil_engineering_materials/numerical-toughness-resilience-400mm
site: BCT Engineering
---

# Modulus of toughness and resilience (300 kN)

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

- Asked in: 2066 Shrawan (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-400mm
- Whole chapter: https://www.bctengineer.com/md/semester_3/civil_engineering_materials/chapter-1

## Question

A mild steel specimen of 10 mm diameter and 400 mm long, which resists the maximum tensile load of 300 kN at 2 mm diameter. If the material resists 150 kN yield load by elongating 10 mm, then what is the modulus of toughness (printed 'roughness') and resilience?

## Short answer

Original area = 78.54 mm²; yield stress = 1909.86 N/mm², ultimate stress = 3819.72 N/mm², strain = 10/400 = 0.025. Modulus of resilience = ½ × 1909.86 × 0.025 = 23.87 N/mm² (MJ/m³); modulus of toughness ≈ (1909.86+3819.72)/2 × 0.025 = 71.62 N/mm² (MJ/m³).

## Answer

**Given:** $d = 10$ mm, $L = 400$ mm, $P_u = 300$ kN, $P_y = 150$ kN, elongation $\delta = 10$ mm.

**Assumptions:** 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 (10 mm), so this strain is used for both moduli. The elastic energy is taken 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).

$$
\begin{aligned}
A_0 &= \frac{\pi}{4}(10)^2 = 78.54\ \text{mm}^2\\
\sigma_y &= \frac{150\times10^3}{78.54} = 1909.86\ \text{N/mm}^2\\
\sigma_u &= \frac{300\times10^3}{78.54} = 3819.72\ \text{N/mm}^2\\
\varepsilon &= \frac{10}{400} = 0.025
\end{aligned}
$$

**Modulus of resilience:**

$$
U_r = \tfrac12\sigma_y\varepsilon = \tfrac12(1909.86)(0.025) = 23.87\ \text{N/mm}^2
$$

**Modulus of toughness:**

$$
U_t \approx \frac{\sigma_y+\sigma_u}{2}\,\varepsilon = \frac{1909.86+3819.72}{2}(0.025) = 71.62\ \text{N/mm}^2
$$

**Answer: Modulus of resilience = 23.87 N/mm² (MJ/m³); modulus of toughness ≈ 71.62 N/mm² (MJ/m³).**

## Check yourself

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

1. Strain used is
   A. 10/300
   B. 400/10
   C. 0.25
   D. 10/400 = 0.025
   Answer: D. Elongation over length. (Key point: Strain)

2. Modulus of resilience is
   A. 23.87 N/mm^2
   B. 71.62 N/mm^2
   C. 47.7 N/mm^2
   D. 1909.86 N/mm^2
   Answer: A. 0.5 x 1909.86 x 0.025. (Key point: Resilience)

3. Modulus of toughness (approx.) is
   A. 23.87 N/mm^2
   B. 71.62 N/mm^2
   C. 143.2 N/mm^2
   D. 95.5 N/mm^2
   Answer: B. (1909.86+3819.72)/2 x 0.025. (Key point: Toughness)

## Close variants set in other exams

- [Modulus of toughness and resilience (250 kN)](https://www.bctengineer.com/question.md?key=semester_3%2Fcivil_engineering_materials%3Anumerical-toughness-resilience-300mm) — Civil Engineering Materials, chapter 1, asked 2×

## Related in Civil Engineering Materials

- [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
- [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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