---
title: "Steel block under triaxial loads: dimension and volume change"
subject: "Strength of Materials"
chapter: 3
key: "semester_3/strength_of_materials:steel-block-triaxial-loads-change-dimensions"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/steel-block-triaxial-loads-change-dimensions
site: BCT Engineering
---

# Steel block under triaxial loads: dimension and volume change

Strength of Materials · Chapter 3: Simple Stress and Strain

- Asked in: 2074 Asoj, 8 marks
- Video: [Problem on Volumetric Strain of rectangular bar (BRB)](https://www.youtube.com/watch?v=LW8QrMxbTDw) — Sampurna Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/steel-block-triaxial-loads-change-dimensions
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3

## Question

A block of steel $300\ \text{mm}\times150\ \text{mm}\times100\ \text{mm}$ is subjected to axial loads as shown in figure below. Find the change in the dimensions of the bar and change in volume for the material of the block. Take $E_s = 200\ \text{GN/m}^2$ and poisson's ratio ($\sigma$) = 0.30. [Figure: block with dimensions 300 mm (along Z), 150 mm (along Y) and 100 mm (along X); forces: 500 kN along Z (pulling at the top and bottom), 700 kN along X (tensile at both ends), 900 kN along Y (compressive).]

## Short answer

Compute stresses σ_x=15.56, σ_y=-30.00, σ_z=33.33 MPa. Use generalized Hooke’s law with E=200 GPa, μ=0.30 to find strains, then dimension changes: +0.0073 mm in x, -0.0335 mm in y, +0.0565 mm in z. Volume change = +170.0 mm³.

## Answer

Take x = 100 mm, y = 150 mm, z = 300 mm. Tension is positive. $E = 200\times10^3$ N/mm², $\mu = 0.30$.

#### Stresses (load ÷ area of the loaded face)
$$
\begin{aligned}
\sigma_z &= \frac{500\times10^3}{100\times150} = 33.333\ \text{N/mm}^2\ (\text{tension})\\
\sigma_x &= \frac{700\times10^3}{150\times300} = 15.556\ \text{N/mm}^2\ (\text{tension})\\
\sigma_y &= -\frac{900\times10^3}{100\times300} = -30.000\ \text{N/mm}^2\ (\text{compression})
\end{aligned}
$$

#### Strains (generalised Hooke's law)
$$
\begin{aligned}
\varepsilon_x &= \frac{1}{E}\left[\sigma_x - \mu(\sigma_y+\sigma_z)\right] = \frac{1}{200\times10^3}\left[15.556 - 0.3(-30.000+33.333)\right] = 7.2778\times10^{-5}\\
\varepsilon_y &= \frac{1}{E}\left[\sigma_y - \mu(\sigma_x+\sigma_z)\right] = -2.2333\times10^{-4}\\
\varepsilon_z &= \frac{1}{E}\left[\sigma_z - \mu(\sigma_x+\sigma_y)\right] = 1.8833\times10^{-4}
\end{aligned}
$$

#### Change in dimensions
$$
\begin{aligned}
\delta_x &= \varepsilon_x\,(100) = +0.00728\ \text{mm}\\
\delta_y &= \varepsilon_y\,(150) = -0.03350\ \text{mm}\\
\delta_z &= \varepsilon_z\,(300) = +0.05650\ \text{mm}
\end{aligned}
$$

#### Change in volume
$$
\varepsilon_v = \varepsilon_x+\varepsilon_y+\varepsilon_z = 3.7778\times10^{-5},\qquad
\delta V = \varepsilon_v V = 3.7778\times10^{-5}\times(100\times150\times300) = +170.00\ \text{mm}^3
$$

Check with $\varepsilon_v = \dfrac{(1-2\mu)}{E}(\sigma_x+\sigma_y+\sigma_z) = \dfrac{0.4}{200\times10^3}(18.889) = 3.7778\times10^{-5}$.

**Answer: $\delta_x = +0.0073$ mm, $\delta_y = -0.0335$ mm, $\delta_z = +0.0565$ mm; $\delta V = +170.0$ mm³ (an increase).**

## Check yourself

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

1. How is each stress found?
   A. Load divided by the area of the face it acts on
   B. Load divided by block volume
   C. Load divided by the area of the largest face for all
   D. Load times area
   Answer: A. For example sigma_z = 500e3/(100x150) = 33.33 N/mm^2. (Key point: Stress calc)

2. Which expression gives strain in x?
   A. epsilon_x = sigma_x/E only
   B. epsilon_x = [sigma_x - mu(sigma_y + sigma_z)]/E
   C. epsilon_x = [sigma_x + mu(sigma_y + sigma_z)]/E
   D. epsilon_x = mu sigma_x/E
   Answer: B. Stresses in the other two directions produce lateral strain of opposite sign. (Key point: Hooke's law)

3. How is the compressive load in Y treated?
   A. As a negative stress, -30 N/mm^2
   B. As positive
   C. Ignored
   D. As zero strain
   Answer: A. Tension is positive, so compression is negative. (Key point: Sign)

4. What is the change in volume?
   A. +170 mm^3 (an increase), from sum of strains x V
   B. -170 mm^3
   C. 0
   D. +0.17 mm^3
   Answer: A. epsilon_v = 3.78e-5 times V = 4.5e6 mm^3 = 170 mm^3. (Key point: Volume)

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