---
title: "Modulus of toughness of medium carbon steel"
subject: "Civil Engineering Materials"
chapter: 1
key: "semester_3/civil_engineering_materials:numerical-modulus-toughness-415"
source: https://www.bctengineer.com/q/semester_3/civil_engineering_materials/numerical-modulus-toughness-415
site: BCT Engineering
---

# Modulus of toughness of medium carbon steel

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

- Asked in: 2057 Chaitra (old course), 4 marks
- 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-modulus-toughness-415
- Whole chapter: https://www.bctengineer.com/md/semester_3/civil_engineering_materials/chapter-1

## Question

The yield stress of a medium carbon steel is found to be 415 N/mm$^2$. Its ultimate stress is 20% more than the yield stress. If the steel fractures at 2% of strain, find the modulus of toughness of the steel.

## Short answer

Yield stress 415 N/mm², ultimate stress 498 N/mm² (20% higher), fracture strain 0.02. Modulus of toughness Ut ≈ average stress × fracture strain = (415+498)/2 × 0.02 = 9.13 N/mm² (9.13 MJ/m³).

## Answer

**Given:** $\sigma_y = 415\ \text{N/mm}^2$, $\sigma_u = 1.20\times415 = 498\ \text{N/mm}^2$, fracture strain $\varepsilon_f = 2\% = 0.02$.

The modulus of toughness is the area under the stress-strain curve up to fracture. Taking it as the average of the yield and ultimate stress times the fracture strain:

$$
U_t \approx \frac{\sigma_y+\sigma_u}{2}\,\varepsilon_f = \frac{415+498}{2}\times0.02 = 456.5\times0.02 = 9.13\ \text{N/mm}^2
$$

**Answer: Modulus of toughness ≈ 9.13 N·mm/mm³ = 9.13 MJ/m³.**

## Check yourself

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

1. Ultimate stress is
   A. 435 N/mm^2
   B. 498 N/mm^2
   C. 83 N/mm^2
   D. 518 N/mm^2
   Answer: B. 1.2 x 415. (Key point: Ultimate)

2. Toughness is taken as
   A. yield stress x 0.02 only
   B. ultimate squared over 2E
   C. average of yield and ultimate stress x fracture strain
   D. sum of stresses x strain
   Answer: C. Approximate area under curve. (Key point: Method)

3. Modulus of toughness is
   A. 4.56 N/mm^2
   B. 18.3 N/mm^2
   C. 456.5 N/mm^2
   D. 9.13 N/mm^2
   Answer: D. 456.5 x 0.02. (Key point: Result)

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