Template: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.
	[math]\displaystyle{ K=\, }[/math]	[math]\displaystyle{ E=\, }[/math]	[math]\displaystyle{ \lambda=\, }[/math]	[math]\displaystyle{ G=\, }[/math]	[math]\displaystyle{ \nu=\, }[/math]	[math]\displaystyle{ M=\, }[/math]	Notes
[math]\displaystyle{ (K,\,E) }[/math]			[math]\displaystyle{ \tfrac{3K(3K-E)}{9K-E} }[/math]	[math]\displaystyle{ \tfrac{3KE}{9K-E} }[/math]	[math]\displaystyle{ \tfrac{3K-E}{6K} }[/math]	[math]\displaystyle{ \tfrac{3K(3K+E)}{9K-E} }[/math]
[math]\displaystyle{ (K,\,\lambda) }[/math]		[math]\displaystyle{ \tfrac{9K(K-\lambda)}{3K-\lambda} }[/math]		[math]\displaystyle{ \tfrac{3(K-\lambda)}{2} }[/math]	[math]\displaystyle{ \tfrac{\lambda}{3K-\lambda} }[/math]	[math]\displaystyle{ 3K-2\lambda\, }[/math]
[math]\displaystyle{ (K,\,G) }[/math]		[math]\displaystyle{ \tfrac{9KG}{3K+G} }[/math]	[math]\displaystyle{ K-\tfrac{2G}{3} }[/math]		[math]\displaystyle{ \tfrac{3K-2G}{2(3K+G)} }[/math]	[math]\displaystyle{ K+\tfrac{4G}{3} }[/math]
[math]\displaystyle{ (K,\,\nu) }[/math]		[math]\displaystyle{ 3K(1-2\nu)\, }[/math]	[math]\displaystyle{ \tfrac{3K\nu}{1+\nu} }[/math]	[math]\displaystyle{ \tfrac{3K(1-2\nu)}{2(1+\nu)} }[/math]		[math]\displaystyle{ \tfrac{3K(1-\nu)}{1+\nu} }[/math]
[math]\displaystyle{ (K,\,M) }[/math]		[math]\displaystyle{ \tfrac{9K(M-K)}{3K+M} }[/math]	[math]\displaystyle{ \tfrac{3K-M}{2} }[/math]	[math]\displaystyle{ \tfrac{3(M-K)}{4} }[/math]	[math]\displaystyle{ \tfrac{3K-M}{3K+M} }[/math]
[math]\displaystyle{ (E,\,\lambda) }[/math]	[math]\displaystyle{ \tfrac{E + 3\lambda + R}{6} }[/math]			[math]\displaystyle{ \tfrac{E-3\lambda+R}{4} }[/math]	[math]\displaystyle{ \tfrac{2\lambda}{E+\lambda+R} }[/math]	[math]\displaystyle{ \tfrac{E-\lambda+R}{2} }[/math]	[math]\displaystyle{ R=\sqrt{E^2+9\lambda^2 + 2E\lambda} }[/math]
[math]\displaystyle{ (E,\,G) }[/math]	[math]\displaystyle{ \tfrac{EG}{3(3G-E)} }[/math]		[math]\displaystyle{ \tfrac{G(E-2G)}{3G-E} }[/math]		[math]\displaystyle{ \tfrac{E}{2G}-1 }[/math]	[math]\displaystyle{ \tfrac{G(4G-E)}{3G-E} }[/math]
[math]\displaystyle{ (E,\,\nu) }[/math]	[math]\displaystyle{ \tfrac{E}{3(1-2\nu)} }[/math]		[math]\displaystyle{ \tfrac{E\nu}{(1+\nu)(1-2\nu)} }[/math]	[math]\displaystyle{ \tfrac{E}{2(1+\nu)} }[/math]		[math]\displaystyle{ \tfrac{E(1-\nu)}{(1+\nu)(1-2\nu)} }[/math]
[math]\displaystyle{ (E,\,M) }[/math]	[math]\displaystyle{ \tfrac{3M-E+S}{6} }[/math]		[math]\displaystyle{ \tfrac{M-E+S}{4} }[/math]	[math]\displaystyle{ \tfrac{3M+E-S}{8} }[/math]	[math]\displaystyle{ \tfrac{E-M+S}{4M} }[/math]		[math]\displaystyle{ S=\pm\sqrt{E^2+9M^2-10EM} }[/math] There are two valid solutions. The plus sign leads to [math]\displaystyle{ \nu\geq 0 }[/math]. The minus sign leads to [math]\displaystyle{ \nu\leq 0 }[/math].
[math]\displaystyle{ (\lambda,\,G) }[/math]	[math]\displaystyle{ \lambda+ \tfrac{2G}{3} }[/math]	[math]\displaystyle{ \tfrac{G(3\lambda + 2G)}{\lambda + G} }[/math]			[math]\displaystyle{ \tfrac{\lambda}{2(\lambda + G)} }[/math]	[math]\displaystyle{ \lambda+2G\, }[/math]
[math]\displaystyle{ (\lambda,\,\nu) }[/math]	[math]\displaystyle{ \tfrac{\lambda(1+\nu)}{3\nu} }[/math]	[math]\displaystyle{ \tfrac{\lambda(1+\nu)(1-2\nu)}{\nu} }[/math]		[math]\displaystyle{ \tfrac{\lambda(1-2\nu)}{2\nu} }[/math]		[math]\displaystyle{ \tfrac{\lambda(1-\nu)}{\nu} }[/math]	Cannot be used when [math]\displaystyle{ \nu=0 \Leftrightarrow \lambda=0 }[/math]
[math]\displaystyle{ (\lambda,\,M) }[/math]	[math]\displaystyle{ \tfrac{M + 2\lambda}{3} }[/math]	[math]\displaystyle{ \tfrac{(M-\lambda)(M+2\lambda)}{M+\lambda} }[/math]		[math]\displaystyle{ \tfrac{M-\lambda}{2} }[/math]	[math]\displaystyle{ \tfrac{\lambda}{M+\lambda} }[/math]
[math]\displaystyle{ (G,\,\nu) }[/math]	[math]\displaystyle{ \tfrac{2G(1+\nu)}{3(1-2\nu)} }[/math]	[math]\displaystyle{ 2G(1+\nu)\, }[/math]	[math]\displaystyle{ \tfrac{2 G \nu}{1-2\nu} }[/math]			[math]\displaystyle{ \tfrac{2G(1-\nu)}{1-2\nu} }[/math]
[math]\displaystyle{ (G,\,M) }[/math]	[math]\displaystyle{ M - \tfrac{4G}{3} }[/math]	[math]\displaystyle{ \tfrac{G(3M-4G)}{M-G} }[/math]	[math]\displaystyle{ M - 2G\, }[/math]		[math]\displaystyle{ \tfrac{M - 2G}{2M - 2G} }[/math]
[math]\displaystyle{ (\nu,\,M) }[/math]	[math]\displaystyle{ \tfrac{M(1+\nu)}{3(1-\nu)} }[/math]	[math]\displaystyle{ \tfrac{M(1+\nu)(1-2\nu)}{1-\nu} }[/math]	[math]\displaystyle{ \tfrac{M \nu}{1-\nu} }[/math]	[math]\displaystyle{ \tfrac{M(1-2\nu)}{2(1-\nu)} }[/math]

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