Lamé parameters

In linear elasticity, the Lamé parameters or Lamé constants are the two parameters

• λ, also called Lamé's first parameter[1] [2]
• μ, the shear modulus or Lamé's second parameter (also referred to as G)

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

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

where σ is the stress, ε the strain tensor, $\scriptstyle I$ the identity matrix and $\scriptstyle\mathrm{tr}(\cdot)$ 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 $K = \lambda + (2/3)\mu$.

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.

References

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

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

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

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