---
title: "Tensile test numerical, 40 mm specimen"
subject: "Civil Engineering Materials"
chapter: 1
key: "semester_3/civil_engineering_materials:numerical-tensile-test-40mm"
source: https://www.bctengineer.com/q/semester_3/civil_engineering_materials/numerical-tensile-test-40mm
site: BCT Engineering
---

# Tensile test numerical, 40 mm specimen

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

- Asked in: 2065 Shrawan (old course), 12 marks
- Video: [Mechanics of materials Lecture | Stress Strain Diagram Problem | Tensile test Problem](https://www.youtube.com/watch?v=S0mL6KCRsds) — Michael Thomas Rex F
- 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-tensile-test-40mm
- Whole chapter: https://www.bctengineer.com/md/semester_3/civil_engineering_materials/chapter-1

## Question

The following observations were made during a tensile test on a mild steel specimen 40 mm diameter and 200 mm long. Given: 2 mm elongation with 40 kN load (within limit of proportionality); yield load = 160 kN; maximum load = 240 kN; length of specimen at fracture = 250 mm. Determine the modulus of toughness and resilience.

## Short answer

Area = 1256.64 mm²; elastic loading gives E = 31.83/0.01 = 3183.1 N/mm². Resilience Ur = σy²/(2E) = 127.32²/(2×3183.1) = 2.55 N/mm² (MJ/m³). Toughness Ut ≈ (127.32+190.99)/2 × 0.25 = 39.79 N/mm² (MJ/m³) using fracture strain (250‑200)/200=0.25.

## Answer

**Given:** $d = 40$ mm, $L_0 = 200$ mm; in the elastic range $P = 40$ kN gives $\delta = 2$ mm; $P_y = 160$ kN; $P_u = 240$ kN; $L_f = 250$ mm.

#### Step 1: area and stresses
$$
\begin{aligned}
A_0 &= \frac{\pi}{4}(40)^2 = 1256.64\ \text{mm}^2\\
\sigma_y &= \frac{160\times10^3}{1256.64} = 127.32\ \text{N/mm}^2\\
\sigma_u &= \frac{240\times10^3}{1256.64} = 190.99\ \text{N/mm}^2
\end{aligned}
$$

#### Step 2: modulus of elasticity (from the elastic reading)
$$
\sigma = \frac{40\times10^3}{1256.64} = 31.83\ \text{N/mm}^2,\quad \varepsilon = \frac{2}{200} = 0.01,\quad E = \frac{31.83}{0.01} = 3183.1\ \text{N/mm}^2
$$

#### Step 3: modulus of resilience (up to yield)
$$
U_r = \frac{\sigma_y^2}{2E} = \frac{(127.32)^2}{2(3183.1)} = 2.546\ \text{N/mm}^2
$$

#### Step 4: modulus of toughness
Strain at fracture: $\varepsilon_f = \dfrac{250-200}{200} = 0.25$. Taking the average of yield and ultimate stress over the whole strain:

$$
U_t \approx \frac{\sigma_y+\sigma_u}{2}\,\varepsilon_f = \frac{127.32+190.99}{2}(0.25) = 39.79\ \text{N/mm}^2
$$

**Answer: Modulus of resilience = 2.55 N/mm² (MJ/m³); modulus of toughness ≈ 39.79 N/mm² (MJ/m³).**

## Check yourself

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

1. A0 for 40 mm diameter is
   A. 5026.5 mm^2
   B. 628.3 mm^2
   C. 1256.64 mm^2
   D. 125.66 mm^2
   Answer: C. pi/4 x 40^2. (Key point: Area)

2. E from 40 kN at 2 mm over 200 mm is
   A. 31.83 N/mm^2
   B. 200000 N/mm^2
   C. 318.3 N/mm^2
   D. 3183.1 N/mm^2
   Answer: D. 31.83/0.01. (Key point: E)

3. Modulus of resilience is
   A. 2.546 N/mm^2
   B. 39.79 N/mm^2
   C. 25.46 N/mm^2
   D. 127.32 N/mm^2
   Answer: A. 127.32^2/(2 x 3183.1). (Key point: Resilience)

4. Fracture strain used is
   A. 0.01, giving 1.59
   B. 0.25, giving toughness about 39.79 N/mm^2
   C. 0.5, giving 79.6
   D. 0.25, giving 2.55
   Answer: B. (250-200)/200 = 0.25; average stress 159.16 x 0.25. (Key point: Toughness)

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