Relation between E, G and Poisson's ratio

Result: $E = 2G(1 + \mu)$, where $G$ is the modulus of rigidity.

Setting up

Consider a square element ABCD of side $a$ subjected to pure shear: shear stress $\tau$ on its faces. The diagonals then carry equal tensile and compressive stresses of magnitude $\tau$ at $45^\circ$ (principal stresses $\sigma_1 = +\tau$ and $\sigma_2 = -\tau$).

   A────────────D       Shear τ on all faces
   │            │       shear strain φ = τ/G
   │    ╲       │       diagonal BD stretches
   B────────────C       diagonal AC shortens

Strain in the diagonal in terms of E and μ

The tensile principal stress along diagonal BD is $\tau$ and the compressive one along AC is $-\tau$. The strain of diagonal BD is

$$\varepsilon_{BD} = \frac{\tau}{E} - \mu\left(\frac{-\tau}{E}\right) = \frac{\tau}{E}(1 + \mu) \qquad (1)$$

Strain in the diagonal in terms of the shear strain

Keep face AB fixed. Under the shear strain $\phi$, face DC slides parallel to AB to the position D′C′, so every point on it moves by $DD' = CC' = a\phi$. Diagonal BD lengthens by the component of $DD'$ along BD, i.e.

$$\delta_{BD} = DD'\cos45^\circ = a\phi\cos45^\circ$$

and the original diagonal length is $BD = a\sqrt{2} = a/\cos45^\circ$. Therefore

$$\varepsilon_{BD} = \frac{a\phi\cos45^\circ}{a/\cos45^\circ} = \phi\cos^245^\circ = \frac{\phi}{2} = \frac{\tau}{2G} \qquad (2)$$

(using $\phi = \tau/G$).

Equating (1) and (2)

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

For steel with $\mu = 0.3$: $G = E/2.6 = 0.385E$.