Relation between elastic moduli and Poisson's ratio

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.