---
title: "Relation between elastic moduli and Poisson's ratio"
subject: "Strength of Materials"
chapter: 3
key: "semester_3/strength_of_materials:modulus-e-g-k-relation"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/modulus-e-g-k-relation
site: BCT Engineering
---

# Relation between elastic moduli and Poisson's ratio

Strength of Materials · Chapter 3: Simple Stress and Strain

- Asked in: 2080 Baisakh, 6 marks; 2078 Kartik, 6 marks; 2072 Chaitra, 8 marks (asked 3 times)
- Video: [Elastic Constants and their Relationship | Lec-5](https://www.youtube.com/watch?v=sZi0W3pr8Jg) — Unacademy GATE & ESE - Civil & ES
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/modulus-e-g-k-relation
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3

## Question

Derive the relations between the elastic constants (Young's modulus E, modulus of rigidity G, bulk modulus K) and Poisson's ratio.

## Short answer

E = 2G(1+μ) from shear diagonal strain; E = 3K(1-2μ) from volumetric strain under hydrostatic stress. Eliminating μ gives E = 9KG/(3K+G) and μ = (3K-2G)/[2(3K+G)].

## Answer

The three elastic constants $E$, $G$, $K$ are connected with Poisson's ratio $\mu$ by:

$$
E = 2G(1 + \mu),\qquad E = 3K(1 - 2\mu),\qquad E = \frac{9KG}{3K + G},\qquad \mu = \frac{3K - 2G}{2(3K + G)}
$$

#### (a) Relation between E, G and μ
Take a square element under pure shear $\tau$. The diagonals carry principal stresses $+\tau$ (tension) and $-\tau$ (compression).

Strain of the tension diagonal from Hooke's law:
$$
\varepsilon = \frac{\tau}{E} + \mu\frac{\tau}{E} = \frac{\tau}{E}(1 + \mu)
$$

Geometry: the shear strain $\phi = \tau/G$ stretches the diagonal by $\varepsilon = \phi/2 = \dfrac{\tau}{2G}$.

Equating:
$$
\frac{\tau}{2G} = \frac{\tau}{E}(1 + \mu) \;\Rightarrow\; E = 2G(1 + \mu)\qquad (1)
$$

#### (b) Relation between E, K and μ
A cube under equal stress $\sigma$ on all faces has
$$
\varepsilon_x = \frac{\sigma}{E}(1 - 2\mu),\qquad \varepsilon_v = 3\varepsilon_x = \frac{3\sigma}{E}(1 - 2\mu)
$$
and $K = \sigma/\varepsilon_v$, so
$$
K = \frac{E}{3(1 - 2\mu)} \;\Rightarrow\; E = 3K(1 - 2\mu)\qquad (2)
$$

#### (c) Eliminating μ to relate E, G and K
From (1): $\mu = \dfrac{E}{2G} - 1$. Substitute in (2):
$$
E = 3K\left[1 - 2\left(\frac{E}{2G} - 1\right)\right] = 3K\left(3 - \frac{E}{G}\right)
$$
$$
E\left(1 + \frac{3K}{G}\right) = 9K \;\Rightarrow\; E = \frac{9KG}{3K + G}
$$

Also, equating (1) and (2): $2G(1 + \mu) = 3K(1 - 2\mu)$, so
$$
\mu = \frac{3K - 2G}{2(3K + G)}
$$

**Example:** for $E = 200$ GPa and $\mu = 0.3$: $G = \dfrac{200}{2.6} = 76.9$ GPa and $K = \dfrac{200}{3(0.4)} = 166.7$ GPa.

## Check yourself

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

1. Which set of relations is derived?
   A. E = 2G(1-μ), E = 3K(1+2μ), E = KG
   B. E = G + K
   C. E = 2G(1+μ), E = 3K(1-2μ), E = 9KG/(3K+G)
   D. E = 3G(1-2μ)
   Answer: C. Shear and hydrostatic analyses give the first two, and eliminating μ gives the third. (Key point: Relations)

2. In part (a) what gives the diagonal strain in terms of G?
   A. τ/G exactly
   B. Shear strain τ/G stretches the diagonal by half of it, τ/2G
   C. 2τ/G
   D. τ/3G
   Answer: B. Geometry gives diagonal strain φ/2. (Key point: Shear part)

3. What is the volumetric strain for equal stress σ on all faces?
   A. 3σ(1 - 2μ)/E
   B. σ(1 - 2μ)/E
   C. 3σ/E
   D. σ(1 + 2μ)/E
   Answer: A. εv = 3εx and εx = σ(1-2μ)/E. (Key point: Bulk part)

4. How is E = 9KG/(3K+G) obtained?
   A. Add the two equations
   B. Multiply E = 2G(1+μ) by K
   C. Set μ = 0.5
   D. Substitute μ = E/2G - 1 from (1) into (2) and solve for E
   Answer: D. Eliminating μ between the two relations gives E in terms of K and G only. (Key point: Elimination)

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