Template:Elastic moduli

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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

The stiffness matrix (9 by 9, or 6 by 6 in Voigt notation) in Hooke's law (in 3D) can be parametrized by only two components for homogeneous and isotropic materials. One may choose whichever pair one prefers among the elastic moduli given below. Some of the possible conversions are listed in the table.

References[edit]

  • G. Mavko, T. Mukerji, J. Dvorkin. The Rock Physics Handbook. Cambridge University Press 2003 (paperback). ISBN 0-521-54344-4