# Lamé parameters

In continuum mechanics, Lamé parameters (also called the Lamé coefficients, Lamé constants or Lamé moduli) are two material-dependent quantities denoted by λ and μ that arise in strain-stress relationships.[1] In general, λ and μ are individually referred to as Lamé's first parameter and Lamé's second parameter, respectively. Other names are sometimes employed for one or both parameters, depending on context. For example, the parameter μ is referred to in fluid dynamics as the dynamic viscosity of a fluid(not the same units); whereas in the context of elasticity, μ is called the shear modulus,[2]: p.333  and is sometimes denoted by G instead of μ. Typically the notation G is seen paired with the use of Young's modulus E, and the notation μ is paired with the use of λ.

In homogeneous and isotropic materials, these define Hooke's law in 3D,

${\displaystyle {\boldsymbol {\sigma }}=2\mu {\boldsymbol {\varepsilon }}+\lambda \;\operatorname {tr} ({\boldsymbol {\varepsilon }})I,}$
where σ is the stress, ε the strain tensor, I the identity matrix and tr the trace function. Hooke's law may be written in terms of tensor components using index notation as
${\displaystyle \sigma _{ij}=2\mu E_{ij}+\lambda \delta _{ij}E_{kk},}$
where σij is the stress tensor, Eij the strain tensor, and δij the Kronecker delta.

The two parameters together constitute a parameterization of the elastic moduli for homogeneous isotropic media, popular in mathematical literature, and are thus related to the other elastic moduli; for instance, the bulk modulus can be expressed as K = λ + 2/3μ. Relations for other moduli are found in the (λ, G) row of the conversions table at the end of this article.

Although the shear modulus, μ, must be positive, the Lamé's first parameter, λ, can be negative, in principle; however, for most materials it is also positive.

The parameters are named after Gabriel Lamé. They have the same dimension as stress and are usually given in the pressure unit [Pa].

• K. Feng, Z.-C. Shi, Mathematical Theory of Elastic Structures, Springer New York, ISBN 0-387-51326-4, (1981)
• G. Mavko, T. Mukerji, J. Dvorkin, The Rock Physics Handbook, Cambridge University Press (paperback), ISBN 0-521-54344-4, (2003)
• W.S. Slaughter, The Linearized Theory of Elasticity, Birkhäuser, ISBN 0-8176-4117-3, (2002)

## References

1. ^ "Lamé Constants". Weisstein, Eric. Eric Weisstein's World of Science, A Wolfram Web Resource. Retrieved 2015-02-22.
2. ^ Jean Salencon (2001), "Handbook of Continuum Mechanics: General Concepts, Thermoelasticity". Springer Science & Business Media ISBN 3-540-41443-6
Conversion formulae
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.
${\displaystyle K=\,}$  ${\displaystyle E=\,}$  ${\displaystyle \lambda =\,}$  ${\displaystyle G=\,}$  ${\displaystyle \nu =\,}$  ${\displaystyle M=\,}$  Notes
${\displaystyle (K,\,E)}$  ${\displaystyle {\tfrac {3K(3K-E)}{9K-E}}}$  ${\displaystyle {\tfrac {3KE}{9K-E}}}$  ${\displaystyle {\tfrac {3K-E}{6K}}}$  ${\displaystyle {\tfrac {3K(3K+E)}{9K-E}}}$
${\displaystyle (K,\,\lambda )}$  ${\displaystyle {\tfrac {9K(K-\lambda )}{3K-\lambda }}}$  ${\displaystyle {\tfrac {3(K-\lambda )}{2}}}$  ${\displaystyle {\tfrac {\lambda }{3K-\lambda }}}$  ${\displaystyle 3K-2\lambda \,}$
${\displaystyle (K,\,G)}$  ${\displaystyle {\tfrac {9KG}{3K+G}}}$  ${\displaystyle K-{\tfrac {2G}{3}}}$  ${\displaystyle {\tfrac {3K-2G}{2(3K+G)}}}$  ${\displaystyle K+{\tfrac {4G}{3}}}$
${\displaystyle (K,\,\nu )}$  ${\displaystyle 3K(1-2\nu )\,}$  ${\displaystyle {\tfrac {3K\nu }{1+\nu }}}$  ${\displaystyle {\tfrac {3K(1-2\nu )}{2(1+\nu )}}}$  ${\displaystyle {\tfrac {3K(1-\nu )}{1+\nu }}}$
${\displaystyle (K,\,M)}$  ${\displaystyle {\tfrac {9K(M-K)}{3K+M}}}$  ${\displaystyle {\tfrac {3K-M}{2}}}$  ${\displaystyle {\tfrac {3(M-K)}{4}}}$  ${\displaystyle {\tfrac {3K-M}{3K+M}}}$
${\displaystyle (E,\,\lambda )}$  ${\displaystyle {\tfrac {E+3\lambda +R}{6}}}$  ${\displaystyle {\tfrac {E-3\lambda +R}{4}}}$  ${\displaystyle {\tfrac {2\lambda }{E+\lambda +R}}}$  ${\displaystyle {\tfrac {E-\lambda +R}{2}}}$  ${\displaystyle R={\sqrt {E^{2}+9\lambda ^{2}+2E\lambda }}}$
${\displaystyle (E,\,G)}$  ${\displaystyle {\tfrac {EG}{3(3G-E)}}}$  ${\displaystyle {\tfrac {G(E-2G)}{3G-E}}}$  ${\displaystyle {\tfrac {E}{2G}}-1}$  ${\displaystyle {\tfrac {G(4G-E)}{3G-E}}}$
${\displaystyle (E,\,\nu )}$  ${\displaystyle {\tfrac {E}{3(1-2\nu )}}}$  ${\displaystyle {\tfrac {E\nu }{(1+\nu )(1-2\nu )}}}$  ${\displaystyle {\tfrac {E}{2(1+\nu )}}}$  ${\displaystyle {\tfrac {E(1-\nu )}{(1+\nu )(1-2\nu )}}}$
${\displaystyle (E,\,M)}$  ${\displaystyle {\tfrac {3M-E+S}{6}}}$  ${\displaystyle {\tfrac {M-E+S}{4}}}$  ${\displaystyle {\tfrac {3M+E-S}{8}}}$  ${\displaystyle {\tfrac {E-M+S}{4M}}}$  ${\displaystyle S=\pm {\sqrt {E^{2}+9M^{2}-10EM}}}$

There are two valid solutions.
The plus sign leads to ${\displaystyle \nu \geq 0}$ .

The minus sign leads to ${\displaystyle \nu \leq 0}$ .

${\displaystyle (\lambda ,\,G)}$  ${\displaystyle \lambda +{\tfrac {2G}{3}}}$  ${\displaystyle {\tfrac {G(3\lambda +2G)}{\lambda +G}}}$  ${\displaystyle {\tfrac {\lambda }{2(\lambda +G)}}}$  ${\displaystyle \lambda +2G\,}$
${\displaystyle (\lambda ,\,\nu )}$  ${\displaystyle {\tfrac {\lambda (1+\nu )}{3\nu }}}$  ${\displaystyle {\tfrac {\lambda (1+\nu )(1-2\nu )}{\nu }}}$  ${\displaystyle {\tfrac {\lambda (1-2\nu )}{2\nu }}}$  ${\displaystyle {\tfrac {\lambda (1-\nu )}{\nu }}}$  Cannot be used when ${\displaystyle \nu =0\Leftrightarrow \lambda =0}$
${\displaystyle (\lambda ,\,M)}$  ${\displaystyle {\tfrac {M+2\lambda }{3}}}$  ${\displaystyle {\tfrac {(M-\lambda )(M+2\lambda )}{M+\lambda }}}$  ${\displaystyle {\tfrac {M-\lambda }{2}}}$  ${\displaystyle {\tfrac {\lambda }{M+\lambda }}}$
${\displaystyle (G,\,\nu )}$  ${\displaystyle {\tfrac {2G(1+\nu )}{3(1-2\nu )}}}$  ${\displaystyle 2G(1+\nu )\,}$  ${\displaystyle {\tfrac {2G\nu }{1-2\nu }}}$  ${\displaystyle {\tfrac {2G(1-\nu )}{1-2\nu }}}$
${\displaystyle (G,\,M)}$  ${\displaystyle M-{\tfrac {4G}{3}}}$  ${\displaystyle {\tfrac {G(3M-4G)}{M-G}}}$  ${\displaystyle M-2G\,}$  ${\displaystyle {\tfrac {M-2G}{2M-2G}}}$
${\displaystyle (\nu ,\,M)}$  ${\displaystyle {\tfrac {M(1+\nu )}{3(1-\nu )}}}$  ${\displaystyle {\tfrac {M(1+\nu )(1-2\nu )}{1-\nu }}}$  ${\displaystyle {\tfrac {M\nu }{1-\nu }}}$  ${\displaystyle {\tfrac {M(1-2\nu )}{2(1-\nu )}}}$