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Illustration of uniform compression

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

As an example, suppose an iron cannon ball with bulk modulus 160 GPa is to be reduced in volume by 0.5%. This requires a pressure increase of 0.005×160 GPa = 0.8 GPa (116,000 psi).

Contents

Definition

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 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.

Thermodynamic relation

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 (KT), constant-entropy (adiabatic KS), and other variations are possible. In practice, such distinctions are usually only relevant for gases.

For a gas, the adiabatic bulk modulus KS is approximately given by

 K_S=\gamma\, P

and the isothermal bulk modulus KT is approximately given by

 K_T=P\,

where

γ is the adiabatic index, sometimes called κ.
P is the pressure.


In a fluid, the bulk modulus K and the density ρ determine the speed of sound c (pressure waves), according to the formula

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

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

Anisotropy

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.

Selected values

Approximate bulk modulus (K) for common materials
Material Bulk modulus in GPa Bulk modulus in psi
Glass (see also diagram below table) 35 to 55 5.8×106
Steel 160 23×106
Diamond[1] 442 64×106
Influences of selected glass component additions on the bulk modulus of a specific base glass.[2]
Approximate bulk modulus (K) for other 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)
Solid helium 5×107 Pa (approximate)

References

  1. ^ Phys. Rev. B 32, 7988 - 7991 (1985), Calculation of bulk moduli of diamond and zinc-blende solids
  2. ^ Bulk modulus calculation of glasses
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,\,G) (E,\,G) (K,\,\lambda) (K,\,G) (\lambda,\,\nu) (G,\,\nu) (E,\,\nu) (K,\, \nu) (K,\,E) (M,\,G)
K=\, \lambda+ \frac{2G}{3} \frac{EG}{3(3G-E)} \lambda\frac{1+\nu}{3\nu} \frac{2G(1+\nu)}{3(1-2\nu)} \frac{E}{3(1-2\nu)} M - \frac{4G}{3}
E=\, G\frac{3\lambda + 2G}{\lambda + G} 9K\frac{K-\lambda}{3K-\lambda} \frac{9KG}{3K+G} \frac{\lambda(1+\nu)(1-2\nu)}{\nu} 2G(1+\nu)\, 3K(1-2\nu)\, G\frac{3M-4G}{M-G}
\lambda=\, G\frac{E-2G}{3G-E} K-\frac{2G}{3} \frac{2 G \nu}{1-2\nu} \frac{E\nu}{(1+\nu)(1-2\nu)} \frac{3K\nu}{1+\nu} \frac{3K(3K-E)}{9K-E} M - 2G\,
G=\, 3\frac{K-\lambda}{2} \lambda\frac{1-2\nu}{2\nu} \frac{E}{2(1+\nu)} 3K\frac{1-2\nu}{2(1+\nu)} \frac{3KE}{9K-E}
\nu=\, \frac{\lambda}{2(\lambda + G)} \frac{E}{2G}-1 \frac{\lambda}{3K-\lambda} \frac{3K-2G}{2(3K+G)} \frac{3K-E}{6K} \frac{M - 2G}{2M - 2G}
M=\, \lambda+2G\, G\frac{4G-E}{3G-E} 3K-2\lambda\, K+\frac{4G}{3} \lambda \frac{1-\nu}{\nu} G\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}
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The bulk modulus (K) of a substance measures the substance's resistance to uniform compression. It is defined as the pressure increase needed to cause a given relative decrease in volume. Its base unit is the pascal.

As an example, suppose an iron cannon ball with bulk modulus 160 GPa is to be reduced in volume by 0.5%. This requires a pressure increase of 0.005×160 GPa = 0.8 GPa (116,000 psi).

Contents

Definition

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 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.

Thermodynamic relation

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 (isothermal K_T), constant-entropy (adiabatic K_S), and other variations are possible. In practice, such distinctions are usually only relevant for gases.

For a gas, the adiabatic bulk modulus K_S is approximately given by

K_S=\gamma\, P

and the isothermal bulk modulus K_T is approximately given by

K_T=P\,

where

γ is the adiabatic index, sometimes called κ.
P is the pressure.

In a fluid, the bulk modulus K and the density ρ determine the speed of sound c (pressure waves), according to the Newton-Laplace formula

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

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

Measurement

It is possible to measure the bulk modulus using powder diffraction under applied pressure.

Selected values

Approximate bulk modulus (K) for common materials
Material Bulk modulus in GPa Bulk modulus in psi
Glass (see also diagram below table) 35 to 55 5.8×106
Steel 160 23×106
Diamond[1] 442 64×106
Approximate bulk modulus (K) for other 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)
Solid helium 5×107 Pa (approximate)

References

  1. ^ Phys. Rev. B 32, 7988 - 7991 (1985), Calculation of bulk moduli of diamond and zinc-blende solids
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,\,G) (E,\,G) (K,\,\lambda) (K,\,G) (\lambda,\,\nu) (G,\,\nu) (E,\,\nu) (K,\, \nu) (K,\,E) (M,\,G)
K=\, \lambda+ \tfrac{2G}{3} \tfrac{EG}{3(3G-E)} \tfrac{\lambda(1+\nu)}{3\nu} \tfrac{2G(1+\nu)}{3(1-2\nu)} \tfrac{E}{3(1-2\nu)} M - \tfrac{4G}{3}
E=\, \tfrac{G(3\lambda + 2G)}{\lambda + G} \tfrac{9K(K-\lambda)}{3K-\lambda} \tfrac{9KG}{3K+G} \tfrac{\lambda(1+\nu)(1-2\nu)}{\nu} 2G(1+\nu)\, 3K(1-2\nu)\, \tfrac{G(3M-4G)}{M-G}
\lambda=\, \tfrac{G(E-2G)}{3G-E} K-\tfrac{2G}{3} \tfrac{2 G \nu}{1-2\nu} \tfrac{E\nu}{(1+\nu)(1-2\nu)} \tfrac{3K\nu}{1+\nu} \tfrac{3K(3K-E)}{9K-E} M - 2G\,
G=\, \tfrac{3(K-\lambda)}{2} \tfrac{\lambda(1-2\nu)}{2\nu} \tfrac{E}{2(1+\nu)} \tfrac{3K(1-2\nu)}{2(1+\nu)} \tfrac{3KE}{9K-E}
\nu=\, \tfrac{\lambda}{2(\lambda + G)} \tfrac{E}{2G}-1 \tfrac{\lambda}{3K-\lambda} \tfrac{3K-2G}{2(3K+G)} \tfrac{3K-E}{6K} \tfrac{M - 2G}{2M - 2G}
M=\, \lambda+2G\, \tfrac{G(4G-E)}{3G-E} 3K-2\lambda\, K+\tfrac{4G}{3} \tfrac{\lambda(1-\nu)}{\nu} \tfrac{2G(1-\nu)}{1-2\nu} \tfrac{E(1-\nu)}{(1+\nu)(1-2\nu)} \tfrac{3K(1-\nu)}{1+\nu} \tfrac{3K(3K+E)}{9K-E}

Simple English

The bulk modulus (letter K) is a measure of the resistance of a substance to compression on all sides. Gases are easily compressed, while solids and liquid are only compressed with difficulty. It is measured by a measuring instrument when another object is placed under pressure.


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