# Lamé parameters

(Redirected from Lame elastic constants)

In continuum mechanics, the Lamé parameters (also called the Lamé coefficients or Lamé constants) 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; 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, and the notation μ is paired with the use of λ.

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

${\displaystyle {\boldsymbol {\sigma }}=2\mu {\boldsymbol {\varepsilon }}+\lambda \;\mathrm {tr} ({\boldsymbol {\varepsilon }})I}$

where σ is the stress, ε the strain tensor, ${\displaystyle \scriptstyle I}$ the identity matrix and ${\displaystyle \scriptstyle \mathrm {tr} (\cdot )}$ the trace function.

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 ${\displaystyle K=\lambda +(2/3)\mu }$.

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.

• 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 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.
${\displaystyle K=\,}$ ${\displaystyle E=\,}$ ${\displaystyle \lambda =\,}$ ${\displaystyle G=\,}$ ${\displaystyle \nu =\,}$ ${\displaystyle M=\,}$ Notes
${\displaystyle (K,\,E)}$ ${\displaystyle K}$ ${\displaystyle E}$ ${\displaystyle {\tfrac {3K(3K-E)}{9K-E}}}$ ${\displaystyle {\tfrac {3KE}{9K-E}}}$ ${\displaystyle {\tfrac {3K-E}{6K}}}$ ${\displaystyle {\tfrac {3K(3K+E)}{9K-E}}}$
${\displaystyle (K,\,\lambda )}$ ${\displaystyle K}$ ${\displaystyle {\tfrac {9K(K-\lambda )}{3K-\lambda }}}$ ${\displaystyle \lambda }$ ${\displaystyle {\tfrac {3(K-\lambda )}{2}}}$ ${\displaystyle {\tfrac {\lambda }{3K-\lambda }}}$ ${\displaystyle 3K-2\lambda \,}$
${\displaystyle (K,\,G)}$ ${\displaystyle K}$ ${\displaystyle {\tfrac {9KG}{3K+G}}}$ ${\displaystyle K-{\tfrac {2G}{3}}}$ ${\displaystyle G}$ ${\displaystyle {\tfrac {3K-2G}{2(3K+G)}}}$ ${\displaystyle K+{\tfrac {4G}{3}}}$
${\displaystyle (K,\,\nu )}$ ${\displaystyle K}$ ${\displaystyle 3K(1-2\nu )\,}$ ${\displaystyle {\tfrac {3K\nu }{1+\nu }}}$ ${\displaystyle {\tfrac {3K(1-2\nu )}{2(1+\nu )}}}$ ${\displaystyle \nu }$ ${\displaystyle {\tfrac {3K(1-\nu )}{1+\nu }}}$
${\displaystyle (K,\,M)}$ ${\displaystyle K}$ ${\displaystyle {\tfrac {9K(M-K)}{3K+M}}}$ ${\displaystyle {\tfrac {3K-M}{2}}}$ ${\displaystyle {\tfrac {3(M-K)}{4}}}$ ${\displaystyle {\tfrac {3K-M}{3K+M}}}$ ${\displaystyle M}$
${\displaystyle (E,\,\lambda )}$ ${\displaystyle {\tfrac {E+3\lambda +R}{6}}}$ ${\displaystyle E}$ ${\displaystyle \lambda }$ ${\displaystyle {\tfrac {E-3\lambda +R}{4}}}$ ${\displaystyle {\tfrac {2\lambda }{E+\lambda +R}}}$ ${\displaystyle {\tfrac {E-\lambda +R}{2}}}$ ${\displaystyle R={\sqrt {E^{2}+9\lambda ^{2}+2E\lambda }}}$
${\displaystyle (E,\,G)}$ ${\displaystyle {\tfrac {EG}{3(3G-E)}}}$ ${\displaystyle E}$ ${\displaystyle {\tfrac {G(E-2G)}{3G-E}}}$ ${\displaystyle G}$ ${\displaystyle {\tfrac {E}{2G}}-1}$ ${\displaystyle {\tfrac {G(4G-E)}{3G-E}}}$
${\displaystyle (E,\,\nu )}$ ${\displaystyle {\tfrac {E}{3(1-2\nu )}}}$ ${\displaystyle E}$ ${\displaystyle {\tfrac {E\nu }{(1+\nu )(1-2\nu )}}}$ ${\displaystyle {\tfrac {E}{2(1+\nu )}}}$ ${\displaystyle \nu }$ ${\displaystyle {\tfrac {E(1-\nu )}{(1+\nu )(1-2\nu )}}}$
${\displaystyle (E,\,M)}$ ${\displaystyle {\tfrac {3M-E+S}{6}}}$ ${\displaystyle E}$ ${\displaystyle {\tfrac {M-E+S}{4}}}$ ${\displaystyle {\tfrac {3M+E-S}{8}}}$ ${\displaystyle {\tfrac {E-M+S}{4M}}}$ ${\displaystyle M}$

${\displaystyle S=\pm {\sqrt {E^{2}+9M^{2}-10EM}}}$

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

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