Elastic Moduli

Property 1: For a given material, Young's modulus and shear modulus are related by:

$Y = 2G(1 + \mu)$

where $\mu$ is Poisson's ratio.

Derivation: Consider a cube of side $L$ subjected to a tensile force $F$ along one axis. The longitudinal strain is $\epsilon = \Delta L / L$. Due to Poisson effect, the lateral strain is $-\mu \epsilon$ (the sides contract). Now consider a shear deformation of the same cube. The shear strain $\theta$ is related to the tensile and compressive strains along the diagonals. For a pure shear, the principal axes are at $45^\circ$ to the faces. The tensile strain along one diagonal equals $\theta/2$, and the compressive strain along the other diagonal equals $-\theta/2$. Using Hooke's law for the diagonal direction (which experiences both axial stress and lateral contraction from the perpendicular direction), one obtains the relation. The full derivation involves tensor analysis, which is beyond Class 11 scope — the result is stated here for completeness.

Property 2: For a given material, Young's modulus and bulk modulus are related by:

$Y = 3B(1 - 2\mu)$

Derivation: Consider a cube under uniform hydrostatic pressure $\Delta p$. The stress on each face is $\Delta p$. The volumetric strain is $\Delta V/V = 3\epsilon$ (for small strains, where $\epsilon$ is the linear strain along each axis). But each face also experiences lateral contraction from the perpendicular stresses. Using the generalized Hooke's law for three dimensions:

$\epsilon_x = \frac{1}{Y}[\sigma_x - \mu(\sigma_y + \sigma_z)]$

For hydrostatic pressure, $\sigma_x = \sigma_y = \sigma_z = -\Delta p$. So:

$\epsilon_x = \frac{1}{Y}[-\Delta p - \mu(-2\Delta p)] = -\frac{\Delta p}{Y}(1 - 2\mu)$

The volumetric strain is $\Delta V/V = 3\epsilon_x = -\frac{3\Delta p}{Y}(1 - 2\mu)$.

But by definition, $B = -\Delta p / (\Delta V/V) = \frac{Y}{3(1 - 2\mu)}$.

Rearranging: $Y = 3B(1 - 2\mu)$.

Property 3: The three moduli are related by:

$\frac{9}{Y} = \frac{3}{G} + \frac{1}{B}$

Derivation: From Property 1: $Y = 2G(1 + \mu)$, so $\mu = \frac{Y}{2G} - 1$.

From Property 2: $Y = 3B(1 - 2\mu)$, so $\mu = \frac{1}{2}\left(1 - \frac{Y}{3B}\right)$.

Equating the two expressions for $\mu$:

$\frac{Y}{2G} - 1 = \frac{1}{2} - \frac{Y}{6B}$ …