مدول حجمی

از نسبت این دو عبارت ضریب کشسانی حجمی حاصل می‌شود که به صورت زیر تعریف می‌شود:

K=-V{\frac {\partial P}{\partial V}}

از آنجایی که عبارت $dv/v$ بدون بعد است می‌توان واحد $K$ را بر حسب فشار بیان کرد.

منابع

"Bulk Elastic Properties". hyperphysics. Georgia State University.

Cohen, Marvin (1985). "Calculation of bulk moduli of diamond and zinc-blende solids". Phys. Rev. B 32: 7988–7991.

Bibcode 1985PhRvB..32.7988C. doi:10.1103/PhysRevB.32.7988.

Fluegel, Alexander. "Bulk modulus calculation of glasses". glassproperties.com

رابطه‌های تبدیل مدول‌ها به یکدیگر
خواص کشسانی مواد کشسان خطی همگن و همسانگرد را می‌توان با داشتن دو مدول دلخواه به طور کامل و منحصر به فردی تعیین کرد. بنابراین با در دست داشتن دو مدول و با استفاده از فرمول‌های زیر می‌توان سایر مدول‌ها را محاسبه کرد.
	$K=\,$	$E=\,$	$\lambda =\,$	$G=\,$	$\nu =\,$	$M=\,$	توضیحات
$(K,\,E)$	$K$	$E$	${\tfrac {3K(3K-E)}{9K-E}}$	${\tfrac {3KE}{9K-E}}$	${\tfrac {3K-E}{6K}}$	${\tfrac {3K(3K+E)}{9K-E}}$
$(K,\,\lambda )$	$K$	${\tfrac {9K(K-\lambda )}{3K-\lambda }}$	$\lambda$	${\tfrac {3(K-\lambda )}{2}}$	${\tfrac {\lambda }{3K-\lambda }}$	$3K-2\lambda \,$
$(K,\,G)$	$K$	${\tfrac {9KG}{3K+G}}$	$K-{\tfrac {2G}{3}}$	$G$	${\tfrac {3K-2G}{2(3K+G)}}$	$K+{\tfrac {4G}{3}}$
$(K,\,\nu )$	$K$	$3K(1-2\nu )\,$	${\tfrac {3K\nu }{1+\nu }}$	${\tfrac {3K(1-2\nu )}{2(1+\nu )}}$	$\nu$	${\tfrac {3K(1-\nu )}{1+\nu }}$
$(K,\,M)$	$K$	${\tfrac {9K(M-K)}{3K+M}}$	${\tfrac {3K-M}{2}}$	${\tfrac {3(M-K)}{4}}$	${\tfrac {3K-M}{3K+M}}$	$M$
$(E,\,\lambda )$	${\tfrac {E+3\lambda +R}{6}}$	$E$	$\lambda$	${\tfrac {E-3\lambda +R}{4}}$	${\tfrac {2\lambda }{E+\lambda +R}}$	${\tfrac {E-\lambda +R}{2}}$	$R={\sqrt {E^{2}+9\lambda ^{2}+2E\lambda }}$
$(E,\,G)$	${\tfrac {EG}{3(3G-E)}}$	$E$	${\tfrac {G(E-2G)}{3G-E}}$	$G$	${\tfrac {E}{2G}}-1$	${\tfrac {G(4G-E)}{3G-E}}$
$(E,\,\nu )$	${\tfrac {E}{3(1-2\nu )}}$	$E$	${\tfrac {E\nu }{(1+\nu )(1-2\nu )}}$	${\tfrac {E}{2(1+\nu )}}$	$\nu$	${\tfrac {E(1-\nu )}{(1+\nu )(1-2\nu )}}$
$(E,\,M)$	${\tfrac {3M-E+S}{6}}$	$E$	${\tfrac {M-E+S}{4}}$	${\tfrac {3M+E-S}{8}}$	${\tfrac {E-M+S}{4M}}$	$M$	$S=\pm {\sqrt {E^{2}+9M^{2}-10EM}}$ 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 $\nu =0\Leftrightarrow \lambda =0$
$(\lambda ,\,M)$	${\tfrac {M+2\lambda }{3}}$	${\tfrac {(M-\lambda )(M+2\lambda )}{M+\lambda }}$	$\lambda$	${\tfrac {M-\lambda }{2}}$	${\tfrac {\lambda }{M+\lambda }}$	$M$
$(G,\,\nu )$	${\tfrac {2G(1+\nu )}{3(1-2\nu )}}$	$2G(1+\nu )\,$	${\tfrac {2G\nu }{1-2\nu }}$	$G$	$\nu$	${\tfrac {2G(1-\nu )}{1-2\nu }}$
$(G,\,M)$	$M-{\tfrac {4G}{3}}$	${\tfrac {G(3M-4G)}{M-G}}$	$M-2G\,$	$G$	${\tfrac {M-2G}{2M-2G}}$	$M$
$(\nu ,\,M)$	${\tfrac {M(1+\nu )}{3(1-\nu )}}$	${\tfrac {M(1+\nu )(1-2\nu )}{1-\nu }}$	${\tfrac {M\nu }{1-\nu }}$	${\tfrac {M(1-2\nu )}{2(1-\nu )}}$	$\nu$	$M$

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مدول حجمی

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