Relation between Young's modulus and bulk modulus

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 $\mu = 0.3$, $K = E/(3\times0.4) = 0.833E$.