# ഫലകം:Elastic moduli

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 ${\displaystyle \nu =0\Leftrightarrow \lambda =0}$
${\displaystyle (\lambda ,\,M)}$ ${\displaystyle {\tfrac {M+2\lambda }{3}}}$ ${\displaystyle {\tfrac {(M-\lambda )(M+2\lambda )}{M+\lambda }}}$ ${\displaystyle {\tfrac {M-\lambda }{2}}}$ ${\displaystyle {\tfrac {\lambda }{M+\lambda }}}$
${\displaystyle (G,\,\nu )}$ ${\displaystyle {\tfrac {2G(1+\nu )}{3(1-2\nu )}}}$ ${\displaystyle 2G(1+\nu )\,}$ ${\displaystyle {\tfrac {2G\nu }{1-2\nu }}}$ ${\displaystyle {\tfrac {2G(1-\nu )}{1-2\nu }}}$
${\displaystyle (G,\,M)}$ ${\displaystyle M-{\tfrac {4G}{3}}}$ ${\displaystyle {\tfrac {G(3M-4G)}{M-G}}}$ ${\displaystyle M-2G\,}$ ${\displaystyle {\tfrac {M-2G}{2M-2G}}}$
${\displaystyle (\nu ,\,M)}$ ${\displaystyle {\tfrac {M(1+\nu )}{3(1-\nu )}}}$ ${\displaystyle {\tfrac {M(1+\nu )(1-2\nu )}{1-\nu }}}$ ${\displaystyle {\tfrac {M\nu }{1-\nu }}}$ ${\displaystyle {\tfrac {M(1-2\nu )}{2(1-\nu )}}}$
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