How is the bulk modulus related to Young's modulus and Poisson's ratio?

The bulk modulus ($K$), Young's modulus ($E$), and Poisson's ratio ($\nu$) are related by the following equation: $K = \frac{E}{3(1-2\nu)}$

This relationship can also be rearranged to express Young's modulus in terms of bulk modulus and Poisson's ratio:

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

Where:

$K$ is the Bulk Modulus (a measure of a material's resistance to uniform compression).
$E$ is Young's Modulus (a measure of a material's stiffness or its resistance to elastic deformation under linear stress).
$\nu$ is Poisson's Ratio (a measure of the ratio of lateral strain to axial strain).

The relationship $K = \frac{E}{3(1-2\nu)}$ is one of the fundamental equations in linear elasticity, connecting three key elastic constants.

Here is an explanation of the relationship and how it's derived from the general principles of material behavior:

📐 Elastic Constants and Their Roles

Constant	Symbol	Description
Bulk Modulus	$K$	Measures a material's resistance to change in volume (volumetric strain) when subjected to uniform hydrostatic (all-around) pressure. A high $K$ means the material is incompressible.
Young's Modulus	$E$	Measures a material's stiffness or its resistance to linear elastic deformation (stretching or compressing) under axial stress.
Poisson's Ratio	$\nu$	Measures the ratio of lateral strain to axial strain. It quantifies the tendency of a material to contract in the transverse direction when stretched in the axial direction.

💡 The Physical Meaning of the Relationship

The formula $K = \frac{E}{3(1-2\nu)}$ effectively shows that a material's resistance to compression ($K$) is determined by a combination of its stiffness ($E$) and how much it deforms sideways ($\nu$).

Key Insights from the Formula:

Poisson's Ratio ($\nu$) is Critical: The term $(1-2\nu)$ is in the denominator, meaning the value of $\nu$ places a fundamental limit on $K$.
- Incompressible Material (e.g., Rubber): If a material is perfectly incompressible, its volume will not change under pressure. This corresponds to a Poisson's ratio of $\nu = 0.5$. If you substitute $\nu=0.5$ into the formula:
  
  $K = \frac{E}{3(1-2(0.5))} = \frac{E}{3(0)} \to \infty$
  
  This shows that an incompressible material must have an infinite bulk modulus ($K \to \infty$).
- Limits on $\nu$: Since the bulk modulus ($K$) must be positive (it takes positive pressure to cause negative volume change), the denominator must be positive: $1 - 2\nu > 0$, which means $\nu < 0.5$. Since materials also cannot expand axially when compressed, the lower limit is $\nu > -1$.

🛠️ Derivation

The relationship is derived by comparing the constitutive equations that relate stress ($\sigma$) and strain ($\epsilon$) using different sets of elastic constants.