This derivation expresses Young's Modulus ($E$) in terms of the Bulk Modulus ($K$) and Shear Modulus ($G$). The process involves combining two fundamental sets of constitutive equations: the relationship between stress ($\sigma_{ij}$) and strain ($\epsilon_{ij}$) using $K$ and $G$, and the same relationship using $E$ and Poisson's ratio ($\nu$).
We begin with the general equations for an isotropic elastic material:
Stress in terms of Bulk Modulus ($K$) and Shear Modulus ($G$):
$$ \sigma_{ij} = K \epsilon_{kk} \delta_{ij} + 2 G \kappa_{ij} \quad \text{(Eq. 1)} $$
where:
Strain in terms of Young's Modulus ($E$) and Poisson's Ratio ($\nu$):
$$ \epsilon_{ij} = \frac{1}{E} \left[ (1 + \nu) \sigma_{ij} - \nu \sigma_{kk} \delta_{ij} \right] \quad \text{(Eq. 2)} $$
where $\sigma_{kk} = \sigma_{11} + \sigma_{22} + \sigma_{33}$ is the hydrostatic stress (trace of the stress tensor).
Take the trace of Equation 2 (set $i=j=k$ and sum):
$$ \epsilon_{kk} = \frac{1}{E} \left[ (1 + \nu) \sigma_{kk} - \nu \sigma_{kk} \delta_{kk} \right] $$
Since the sum of the Kronecker delta is $\delta_{kk} = \delta_{11} + \delta_{22} + \delta_{33} = 3$:
$$ \epsilon_{kk} = \frac{1}{E} \left[ (1 + \nu) \sigma_{kk} - 3 \nu \sigma_{kk} \right] $$
$$ \epsilon_{kk} = \frac{1}{E} \left[ (1 + \nu - 3 \nu) \sigma_{kk} \right] $$
$$ \epsilon_{kk} = \frac{1}{E} (1 - 2 \nu) \sigma_{kk} $$
The Bulk Modulus ($K$) is defined by $K = \frac{\sigma_{kk}}{3 \epsilon_{kk}}$. Rearranging the trace equation gives:
$$ \sigma_{kk} = \frac{E}{1 - 2 \nu} \epsilon_{kk} $$
Substituting this into the definition of $K$:
$$ K = \frac{1}{3 \epsilon_{kk}} \left( \frac{E}{1 - 2 \nu} \epsilon_{kk} \right) $$
$$ \mathbf{K = \frac{E}{3 (1 - 2 \nu)}} \quad \text{(Eq. A)} $$