The formula for Poisson's ratio ($\nu$) in terms of the Bulk Modulus ($K$) and the Shear Modulus ($G$) is: $\nu = \frac{3K - 2G}{6K + 2G}$. This relationship is a core result in Linear Elasticity for isotropic materials and is derived by eliminating Young's Modulus ($E$) from the two fundamental equations that link the pairs of elastic moduli:

1. Bulk Modulus: $K = \frac{E}{3(1 - 2\nu)}$
2. Shear Modulus: $G = \frac{E}{2(1 + \nu)}$

The steps to combine and solve these equations to isolate $\nu$ .

1. Starting Relationships

The derivation begins with the two established relationships:

1. Bulk Modulus ($K$) relation (from hydrostatic stress):

   $$ K = \frac{E}{3(1 - 2\nu)} $$

2. Shear Modulus ($G$) relation (from pure shear stress):

   $$ G = \frac{E}{2(1 + \nu)} $$

2. Express $E$ in terms of $G$ and $\nu$

From the Shear Modulus relationship, isolate $E$:

$$ G = \frac{E}{2(1 + \nu)} $$

$$ \implies \mathbf{E = 2G(1 + \nu)} $$

(Equation 3)

3. Express $E$ in terms of $K$ and $\nu$

From the Bulk Modulus relationship (Equation 1), isolate $E$:

$$ K = \frac{E}{3(1 - 2\nu)} $$

$$ \implies \mathbf{E = 3K(1 - 2\nu)} $$

(Equation 4)