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:
Other moduli describe the material's response (strain) to other kinds of stress: the shear modulus describes the response to shear, and Young's modulus describes the response to linear strain. For a fluid, only the bulk modulus is meaningful. For an anisotropic solid such as wood or paper, these three moduli do not contain enough information to describe its behaviour, and one must use the full generalized Hooke's law.
Strictly speaking, the bulk modulus is a thermodynamic quantity, and it is necessary to specify how the temperature varies in order to specify a bulk modulus: constant-temperature (), constant-enthalpy (adiabatic ), and other variations are possible. In practice, such distinctions are usually only relevant for gases.
|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)|
|Glass||3.5×1010 to 5.5×1010 Pa|
|Solid helium||5×107 Pa (approximate)|
For a gas, the adiabatic bulk modulus is approximately given by
For crystalline solids with a symmetry lower than cubic the bulk modulus is not the same in all directions and needs to be described with a tensor with more than one independent value. It is possible to study the tensor elements using powder diffraction under applied pressure.