# 泊松比

$G = \frac{E} {2(1+\nu)}$

$(\lambda,\,G)$ $(E,\,G)$ $(K,\,\lambda)$ $(K,\,G)$ $(\lambda,\,\nu)$ $(G,\,\nu)$ $(E,\,\nu)$ $(K,\, \nu)$ $(K,\,E)$ $(M,\,G)$
$K=\,$ $\lambda+ \tfrac{2G}{3}$ $\tfrac{EG}{3(3G-E)}$ $\tfrac{\lambda(1+\nu)}{3\nu}$ $\tfrac{2G(1+\nu)}{3(1-2\nu)}$ $\tfrac{E}{3(1-2\nu)}$ $M - \tfrac{4G}{3}$
$E=\,$ $\tfrac{G(3\lambda + 2G)}{\lambda + G}$ $\tfrac{9K(K-\lambda)}{3K-\lambda}$ $\tfrac{9KG}{3K+G}$ $\tfrac{\lambda(1+\nu)(1-2\nu)}{\nu}$ $2G(1+\nu)\,$ $3K(1-2\nu)\,$ $\tfrac{G(3M-4G)}{M-G}$
$\lambda=\,$ $\tfrac{G(E-2G)}{3G-E}$ $K-\tfrac{2G}{3}$ $\tfrac{2 G \nu}{1-2\nu}$ $\tfrac{E\nu}{(1+\nu)(1-2\nu)}$ $\tfrac{3K\nu}{1+\nu}$ $\tfrac{3K(3K-E)}{9K-E}$ $M - 2G\,$
$G=\,$ $\tfrac{3(K-\lambda)}{2}$ $\tfrac{\lambda(1-2\nu)}{2\nu}$ $\tfrac{E}{2(1+\nu)}$ $\tfrac{3K(1-2\nu)}{2(1+\nu)}$ $\tfrac{3KE}{9K-E}$
$\nu=\,$ $\tfrac{\lambda}{2(\lambda + G)}$ $\tfrac{E}{2G}-1$ $\tfrac{\lambda}{3K-\lambda}$ $\tfrac{3K-2G}{2(3K+G)}$ $\tfrac{3K-E}{6K}$ $\tfrac{M - 2G}{2M - 2G}$
$M=\,$ $\lambda+2G\,$ $\tfrac{G(4G-E)}{3G-E}$ $3K-2\lambda\,$ $K+\tfrac{4G}{3}$ $\tfrac{\lambda(1-\nu)}{\nu}$ $\tfrac{2G(1-\nu)}{1-2\nu}$ $\tfrac{E(1-\nu)}{(1+\nu)(1-2\nu)}$ $\tfrac{3K(1-\nu)}{1+\nu}$ $\tfrac{3K(3K+E)}{9K-E}$