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.
	$K=$	$E=$	$\lambda =$	$G=$	$\nu =$	$M=$	Notes
$(K,E)$			$\frac{3K(3K-E)}{9K-E}$	$\frac{3KE}{9K-E}$	$\frac{3K-E}{6K}$	$\frac{3K(3K+E)}{9K-E}$
$(K,\lambda )$		$\frac{9K(K-\lambda )}{3K-\lambda }$		$\frac{3(K-\lambda )}{2}$	$\frac{\lambda }{3K-\lambda }$	$3K-2\lambda$
$(K,G)$		$\frac{9KG}{3K+G}$	$K-\frac{2G}{3}$		$\frac{3K-2G}{2(3K+G)}$	$K+\frac{4G}{3}$
$(K,\nu )$		$3K(1-2\nu )$	$\frac{3K\nu }{1+\nu }$	$\frac{3K(1-2\nu )}{2(1+\nu )}$		$\frac{3K(1-\nu )}{1+\nu }$
$(K,M)$		$\frac{9K(M-K)}{3K+M}$	$\frac{3K-M}{2}$	$\frac{3(M-K)}{4}$	$\frac{3K-M}{3K+M}$
$(E,\lambda )$	$\frac{E+3\lambda +R}{6}$			$\frac{E-3\lambda +R}{4}$	$\frac{2\lambda }{E+\lambda +R}$	$\frac{E-\lambda +R}{2}$	$R=\sqrt{E^2+9{\lambda }^2+2E\lambda }$
$(E,G)$	$\frac{EG}{3(3G-E)}$		$\frac{G(E-2G)}{3G-E}$		$\frac{E}{2G}-1$	$\frac{G(4G-E)}{3G-E}$
$(E,\nu )$	$\frac{E}{3(1-2\nu )}$		$\frac{E\nu }{(1+\nu )(1-2\nu )}$	$\frac{E}{2(1+\nu )}$		$\frac{E(1-\nu )}{(1+\nu )(1-2\nu )}$
$(E,M)$	$\frac{3M-E+S}{6}$		$\frac{M-E+S}{4}$	$\frac{3M+E-S}{8}$	$\frac{E-M+S}{4M}$		$S=\pm \sqrt{E^2+9M^2-10EM}$ There are two valid solutions. The plus sign leads to $\nu \ge 0$. The minus sign leads to $\nu \le 0$.
$(\lambda ,G)$	$\lambda +\frac{2G}{3}$	$\frac{G(3\lambda +2G)}{\lambda +G}$			$\frac{\lambda }{2(\lambda +G)}$	$\lambda +2G$
$(\lambda ,\nu )$	$\frac{\lambda (1+\nu )}{3\nu }$	$\frac{\lambda (1+\nu )(1-2\nu )}{\nu }$		$\frac{\lambda (1-2\nu )}{2\nu }$		$\frac{\lambda (1-\nu )}{\nu }$	Cannot be used when $\nu =0\iff \lambda =0$
$(\lambda ,M)$	$\frac{M+2\lambda }{3}$	$\frac{(M-\lambda )(M+2\lambda )}{M+\lambda }$		$\frac{M-\lambda }{2}$	$\frac{\lambda }{M+\lambda }$
$(G,\nu )$	$\frac{2G(1+\nu )}{3(1-2\nu )}$	$2G(1+\nu )$	$\frac{2G\nu }{1-2\nu }$			$\frac{2G(1-\nu )}{1-2\nu }$
$(G,M)$	$M-\frac{4G}{3}$	$\frac{G(3M-4G)}{M-G}$	$M-2G$		$\frac{M-2G}{2M-2G}$
$(\nu ,M)$	$\frac{M(1+\nu )}{3(1-\nu )}$	$\frac{M(1+\nu )(1-2\nu )}{1-\nu }$	$\frac{M\nu }{1-\nu }$	$\frac{M(1-2\nu )}{2(1-\nu )}$