---
title: "Elastic constants and triaxial loading"
subject: "Strength of Materials"
chapter: 2
key: "semester_4/strength_of_materials_mechanical:elastic-constants-triaxial-strains"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/elastic-constants-triaxial-strains
site: BCT Engineering
---

# Elastic constants and triaxial loading

Strength of Materials · Chapter 2: Stress and strain – axial load

- Asked in: year not given, 4+4 marks
- Video: [Volumetric strain | Elastic constants | Simple Stresses & Strains | Strength of Materials | GATE](https://www.youtube.com/watch?v=e8Xa4Qk3dOE) — KN Prasad Academy (NITian)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/elastic-constants-triaxial-strains
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-2

## Question

(a) Define Poisson's ratio and bulk modulus, and derive the relation E = 3K(1 - 2v). (b) A steel block 100 mm x 80 mm x 50 mm (dimensions along x, y, z) carries stresses sigma_x = +100 MPa (tensile), sigma_y = -60 MPa (compressive) and sigma_z = +40 MPa (tensile) on its faces. Take E = 200 GPa and Poisson's ratio v = 0.3. Find the strains in the three directions, the change in volume, and the values of K and G.

## Short answer

Poisson’s ratio ν = -lateral/axial strain. Bulk modulus K = p/(-εv). E = 3K(1-2ν) from hydrostatic strain. For given stresses: εx = 5.3×10⁻⁴, εy = -5.1×10⁻⁴, εz = 1.4×10⁻⁴; ΔV = +64 mm³; K = 166.7 GPa; G = 76.9 GPa.

## Answer

#### (a) Definitions and derivation
- **Poisson's ratio** $\nu$ is the ratio of lateral strain to axial strain in a uniaxial stress state: $\nu = -\varepsilon_{lat}/\varepsilon_{axial}$ (0.25 to 0.33 for metals).
- **Bulk modulus** $K$ is the ratio of hydrostatic (uniform) pressure to the volumetric strain it produces: $K = p/(-\varepsilon_v)$.

Generalized Hooke's law for a cube under three stresses:

$$
\varepsilon_x = \frac{1}{E}\left[\sigma_x - \nu(\sigma_y+\sigma_z)\right]
$$

and similarly for $\varepsilon_y,\varepsilon_z$. Under hydrostatic pressure $p$, $\sigma_x=\sigma_y=\sigma_z=-p$:

$$
\varepsilon_x=\varepsilon_y=\varepsilon_z = \frac{-p(1-2\nu)}{E}
$$

$$
\varepsilon_v = \varepsilon_x+\varepsilon_y+\varepsilon_z = -\frac{3p(1-2\nu)}{E}
\;\Rightarrow\; K = \frac{p}{-\varepsilon_v} = \frac{E}{3(1-2\nu)}
$$

$$
E = 3K(1-2\nu)
$$

#### (b) Numerical
Strains:

$$
\begin{aligned}
\varepsilon_x &= \frac{100 - 0.3(-60+40)}{200000} = \frac{106}{200000} = 5.3\times10^{-4}\\
\varepsilon_y &= \frac{-60 - 0.3(100+40)}{200000} = \frac{-102}{200000} = -5.1\times10^{-4}\\
\varepsilon_z &= \frac{40 - 0.3(100-60)}{200000} = \frac{28}{200000} = 1.4\times10^{-4}
\end{aligned}
$$

Volumetric strain:

$$
\varepsilon_v = (5.3 - 5.1 + 1.4)\times10^{-4} = 1.6\times10^{-4}
$$

Check: $\varepsilon_v = \frac{1-2\nu}{E}(\sigma_x+\sigma_y+\sigma_z) = \frac{0.4\times80}{200000} = 1.6\times10^{-4}$.

$$
V = 100\times80\times50 = 4\times10^5\ \text{mm}^3, \qquad \Delta V = \varepsilon_v V = 64\ \text{mm}^3\ \text{(increase)}
$$

Elastic constants:

$$
K = \frac{E}{3(1-2\nu)} = \frac{200}{3\times0.4} = 166.7\ \text{GPa}, \qquad G = \frac{E}{2(1+\nu)} = \frac{200}{2.6} = 76.9\ \text{GPa}
$$

**Answer: $\varepsilon_x = 5.3\times10^{-4}$, $\varepsilon_y = -5.1\times10^{-4}$, $\varepsilon_z = 1.4\times10^{-4}$; $\Delta V = +64\ \text{mm}^3$; $K = 166.7$ GPa; $G = 76.9$ GPa.**

## Check yourself

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

1. How is Poisson's ratio defined?
   A. Negative ratio of lateral strain to axial strain in uniaxial stress
   B. Ratio of axial stress to lateral stress
   C. Ratio of shear stress to shear strain
   D. Ratio of pressure to volumetric strain
   Answer: A. nu = -epsilon_lat / epsilon_axial under uniaxial stress. (Key point: Poisson's ratio definition)

2. In the derivation, what is the strain in each direction under hydrostatic pressure p?
   A. -p/E in each direction
   B. -p(1-2v)/E in each direction
   C. -p(1+2v)/E in each direction
   D. -3p(1-2v)/E in each direction
   Answer: B. Each strain is [-p - v(-p - p)]/E = -p(1-2v)/E; the sum gives eps_v = -3p(1-2v)/E and hence K = E/[3(1-2v)]. (Key point: Derivation of E = 3K(1-2v))

3. How is epsilon_x found for the block?
   A. [sigma_x + v(sigma_y + sigma_z)]/E
   B. [sigma_x - v(sigma_y + sigma_z)]/E, using signed stresses
   C. sigma_x/E only
   D. [sigma_x - v sigma_y]/E ignoring sigma_z
   Answer: B. Generalized Hooke's law: (100 - 0.3(-60+40))/200000 = 5.3 x 10^-4. (Key point: Strain in x)

4. What are the volumetric strain and the value of G for the block?
   A. 1.6 x 10^-4 and G = 76.9 GPa
   B. 1.6 x 10^-4 and G = 166.7 GPa
   C. 5.3 x 10^-4 and G = 76.9 GPa
   D. 3.2 x 10^-4 and G = 166.7 GPa
   Answer: A. eps_v = (1-2v)(sum of stresses)/E = 1.6 x 10^-4, and G = E/[2(1+v)] = 76.9 GPa (K = 166.7 GPa). (Key point: Volumetric strain and K, G)

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