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Lamé parameters

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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\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.

Notes

  1. ^ Lame's first parameter (for two-dimensional solids), FXSolver.com
  2. ^ Lame's first parameter (for three-dimensions), FXSolver.com

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


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