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

# Relation between Young's modulus and bulk modulus

Strength of Materials · Chapter 3: Simple Stress and Strain

- Asked in: 2078 Bhadra, 6 marks; 2071 Chaitra, 8 marks; 2069 Asar, 8 marks; 2066 Bhadra (old course), 6 marks (asked 4 times)
- Video: [Relation Between Bulk Modulus and Modulus of Elasticity/ Relation between E and K.](https://www.youtube.com/watch?v=Xy0Gk94nG-o) — GTU MIMP
- 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-k-relation
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3

## Question

Derive relationship between young's modulus and bulk modulus.

## Short answer

Under equal triaxial stress σ, volumetric strain ε_v = 3σ(1-2μ)/E. Since bulk modulus K = σ/ε_v, equating gives E = 3K(1-2μ), requiring μ < 0.5 for positive K.

## Answer

**Result:** $E = 3K(1 - 2\mu)$ where $E$ is Young's modulus, $K$ the bulk modulus and $\mu$ Poisson's ratio.

#### Definitions
- Bulk modulus: $K = \dfrac{\text{volumetric (hydrostatic) stress}}{\text{volumetric strain}} = \dfrac{p}{\varepsilon_v}$
- Volumetric strain $\varepsilon_v = \dfrac{\delta V}{V}$.

#### Derivation
Take a cube of side $L$ subjected to equal stress $\sigma$ on all six faces (uniform hydrostatic compression/tension), so $\sigma_x = \sigma_y = \sigma_z = \sigma$.

Strain in the x-direction due to all three stresses (Hooke's law with Poisson effect):

$$
\varepsilon_x = \frac{\sigma}{E} - \mu\frac{\sigma}{E} - \mu\frac{\sigma}{E} = \frac{\sigma}{E}(1 - 2\mu)
$$

By symmetry $\varepsilon_y = \varepsilon_z = \varepsilon_x$.

Volume $V = L^3$; for small strains the volumetric strain is the sum of the linear strains:

$$
\varepsilon_v = \varepsilon_x + \varepsilon_y + \varepsilon_z = \frac{3\sigma}{E}(1 - 2\mu)
$$

(Check: $V + \delta V = L^3(1+\varepsilon)^3 \approx L^3(1 + 3\varepsilon)$.)

From the definition of bulk modulus, with $p = \sigma$:

$$
K = \frac{\sigma}{\varepsilon_v} = \frac{\sigma}{\dfrac{3\sigma}{E}(1 - 2\mu)} = \frac{E}{3(1 - 2\mu)}
$$

$$
\boxed{E = 3K(1 - 2\mu)}
$$

**Notes:** since $K$ must be positive, $1 - 2\mu > 0$, so $\mu < 0.5$. For an incompressible material, $\mu = 0.5$ and $K \to \infty$. For example, for steel with $\mu = 0.3$, $K = E/(3\times0.4) = 0.833E$.

## Check yourself

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

1. What loading is used to derive the E-K relation?
   A. Uniaxial tension only
   B. Pure shear
   C. Equal stress σ on all six faces of a cube
   D. Torsion
   Answer: C. Bulk modulus is defined for hydrostatic stress, so σx = σy = σz = σ. (Key point: Setup)

2. Why is the x-strain σ(1 - 2μ)/E?
   A. Lateral stresses add to it
   B. Own stress gives σ/E and the two lateral stresses each reduce it by μσ/E
   C. Only the x-stress matters
   D. μ is multiplied by three
   Answer: B. Poisson contraction caused by the other two stresses subtracts 2μσ/E. (Key point: Strain)

3. How is εv obtained for small strains?
   A. εv = εx + εy + εz = 3σ(1 - 2μ)/E
   B. εv = εx^3
   C. εv = εx
   D. εv = 2εx
   Answer: A. (1+ε)^3 ≈ 1 + 3ε for small ε, so volumetric strain is the sum of the linear strains. (Key point: Volumetric strain)

4. What does E = 3K(1 - 2μ) imply for μ?
   A. μ > 0.5
   B. μ = 0 gives K = 0
   C. K is independent of μ
   D. μ < 0.5 for positive K; μ = 0.5 means incompressible (K infinite)
   Answer: D. K = E/[3(1-2μ)] is positive only if μ < 0.5 and goes to infinity at 0.5. (Key point: Result and limit)

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