Bulk modulus

Bulk modulus values for some example substances
Water	2.2×10⁹ Pa (value increases at higher pressures)
Air	1.42×10⁵ Pa (adiabatic bulk modulus)
Air	1.01×10⁵ Pa (constant temperature bulk modulus)
Steel	1.6×10¹¹ Pa
Glass	3.5×10¹⁰ to 5.5×10¹⁰ Pa
Solid helium	5×10⁷ Pa (approximate)

The bulk modulus (K) of a substance essentially measures the substance's resistance to uniform compression. It is defined as the pressure increase needed to effect a given relative decrease in volume.

As an example, suppose an iron cannon ball with bulk modulus 160 GPa (gigapascal) is to be reduced in volume by 0.5%. This requires a pressure increase of 0.005×160 GPa = 0.8 GPa. If the cannon ball is subjected to a pressure increase of only 100 MPa, it will decrease in volume by a factor of 100 MPa/160 GPa = 0.000625, or 0.0625%.

The bulk modulus K can be formally defined by the equation:

$K=-V\frac{\partial p}{\partial V}$

where p is pressure, V is volume, and ∂p/∂V denotes the partial derivative of pressure with respect to volume. The inverse of the bulk modulus gives a substance's compressibility.

Other moduli describe the material's response (strain) to other kinds of Hooke's law.

Strictly speaking, the bulk modulus is a gases.

For a gas, the adiabatic bulk modulus $K S$ is approximately given by

$K_S=\kappa\, p$

where

κ is the adiabatic index, sometimes called γ.

p is the pressure.

In a pressure waves), according to the formula

$c=\sqrt{\frac{K}{\rho}}.$

Solids can also sustain transverse waves, for these one additional shear modulus, is needed to determine wave speeds.

References

Bulk Elastic Properties on hyperphysics at Georgia State University

^ Bulk modulus calculation of glasses

v • d • e Elastic moduli for homogeneous isotropic materials

Bulk modulus ( $K$ ) | P-wave modulus ( $M$ )

Conversion formulas
Homogeneous isotropic linear elastic materials have their elastic properties uniquely determined by any two moduli among these, thus given any two, any other of the elastic moduli can be calculated according to these formulas.
	$(\lambda,\,\mu)$	$(E,\,\mu)$	$(K,\,\lambda)$	$(K,\,\mu)$	$(\lambda,\,\nu)$	$(\mu,\,\nu)$	$(E,\,\nu)$	$(K,\, \nu)$	$(K,\,E)$
$K=\,$	$\lambda+ \frac{2\mu}{3}$	$\frac{E\mu}{3(3\mu-E)}$			$\lambda\frac{1+\nu}{3\nu}$	$\frac{2\mu(1+\nu)}{3(1-2\nu)}$	$\frac{E}{3(1-2\nu)}$
$E=\,$	$\mu\frac{3\lambda + 2\mu}{\lambda + \mu}$		$9K\frac{K-\lambda}{3K-\lambda}$	$\frac{9K\mu}{3K+\mu}$	$\frac{\lambda(1+\nu)(1-2\nu)}{\nu}$	$2\mu(1+\nu)\,$		$3K(1-2\nu)\,$
$\lambda=\,$		$\mu\frac{E-2\mu}{3\mu-E}$		$K-\frac{2\mu}{3}$		$\frac{2 \mu \nu}{1-2\nu}$	$\frac{E\nu}{(1+\nu)(1-2\nu)}$	$\frac{3K\nu}{1+\nu}$	$\frac{3K(3K-E)}{9K-E}$
$\mu=\,$			$3\frac{K-\lambda}{2}$		$\lambda\frac{1-2\nu}{2\nu}$		$\frac{E}{2+2\nu}$	$3K\frac{1-2\nu}{2+2\nu}$	$\frac{3KE}{9K-E}$
$\nu=\,$	$\frac{\lambda}{2(\lambda + \mu)}$	$\frac{E}{2\mu}-1$	$\frac{\lambda}{3K-\lambda}$	$\frac{3K-2\mu}{2(3K+\mu)}$					$\frac{3K-E}{6K}$
$M=\,$	$\lambda+2\mu\,$	$\mu\frac{4\mu-E}{3\mu-E}$	$3K-2\lambda\,$	$K+\frac{4\mu}{3}$	$\lambda \frac{1-\nu}{\nu}$	$\mu\frac{2-2\nu}{1-2\nu}$	$E\frac{1-\nu}{(1+\nu)(1-2\nu)}$	$3K\frac{1-\nu}{1+\nu}$	$3K\frac{3K+E}{9K-E}$

Continuum mechanics

This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia article "Bulk_modulus". A list of authors is available in Wikipedia.