Công thức chuyển đổi
Vật liệu đàn hồi tuyến tính đẳng hướng đồng nhất có tính chất đàn hồi được xác định một cách độc nhất bởi bất cứ hai mô đun nào trong số này; do đó, nếu cho hai loại, bất cứ mô đun đàn hồi nào đều có thể được tính theo các công thức sau.
K
=
{\displaystyle K=\,}
E
=
{\displaystyle E=\,}
λ
=
{\displaystyle \lambda =\,}
G
=
{\displaystyle G=\,}
ν
=
{\displaystyle \nu =\,}
M
=
{\displaystyle M=\,}
Chú thích
(
K
,
E
)
{\displaystyle (K,\,E)}
K
{\displaystyle K}
E
{\displaystyle E}
3
K
(
3
K
−
E
)
9
K
−
E
{\displaystyle {\tfrac {3K(3K-E)}{9K-E}}}
3
K
E
9
K
−
E
{\displaystyle {\tfrac {3KE}{9K-E}}}
3
K
−
E
6
K
{\displaystyle {\tfrac {3K-E}{6K}}}
3
K
(
3
K
+
E
)
9
K
−
E
{\displaystyle {\tfrac {3K(3K+E)}{9K-E}}}
(
K
,
λ
)
{\displaystyle (K,\,\lambda )}
K
{\displaystyle K}
9
K
(
K
−
λ
)
3
K
−
λ
{\displaystyle {\tfrac {9K(K-\lambda )}{3K-\lambda }}}
λ
{\displaystyle \lambda }
3
(
K
−
λ
)
2
{\displaystyle {\tfrac {3(K-\lambda )}{2}}}
λ
3
K
−
λ
{\displaystyle {\tfrac {\lambda }{3K-\lambda }}}
3
K
−
2
λ
{\displaystyle 3K-2\lambda \,}
(
K
,
G
)
{\displaystyle (K,\,G)}
K
{\displaystyle K}
9
K
G
3
K
+
G
{\displaystyle {\tfrac {9KG}{3K+G}}}
K
−
2
G
3
{\displaystyle K-{\tfrac {2G}{3}}}
G
{\displaystyle G}
3
K
−
2
G
2
(
3
K
+
G
)
{\displaystyle {\tfrac {3K-2G}{2(3K+G)}}}
K
+
4
G
3
{\displaystyle K+{\tfrac {4G}{3}}}
(
K
,
ν
)
{\displaystyle (K,\,\nu )}
K
{\displaystyle K}
3
K
(
1
−
2
ν
)
{\displaystyle 3K(1-2\nu )\,}
3
K
ν
1
+
ν
{\displaystyle {\tfrac {3K\nu }{1+\nu }}}
3
K
(
1
−
2
ν
)
2
(
1
+
ν
)
{\displaystyle {\tfrac {3K(1-2\nu )}{2(1+\nu )}}}
ν
{\displaystyle \nu }
3
K
(
1
−
ν
)
1
+
ν
{\displaystyle {\tfrac {3K(1-\nu )}{1+\nu }}}
(
K
,
M
)
{\displaystyle (K,\,M)}
K
{\displaystyle K}
9
K
(
M
−
K
)
3
K
+
M
{\displaystyle {\tfrac {9K(M-K)}{3K+M}}}
3
K
−
M
2
{\displaystyle {\tfrac {3K-M}{2}}}
3
(
M
−
K
)
4
{\displaystyle {\tfrac {3(M-K)}{4}}}
3
K
−
M
3
K
+
M
{\displaystyle {\tfrac {3K-M}{3K+M}}}
M
{\displaystyle M}
(
E
,
λ
)
{\displaystyle (E,\,\lambda )}
E
+
3
λ
+
R
6
{\displaystyle {\tfrac {E+3\lambda +R}{6}}}
E
{\displaystyle E}
λ
{\displaystyle \lambda }
E
−
3
λ
+
R
4
{\displaystyle {\tfrac {E-3\lambda +R}{4}}}
2
λ
E
+
λ
+
R
{\displaystyle {\tfrac {2\lambda }{E+\lambda +R}}}
E
−
λ
+
R
2
{\displaystyle {\tfrac {E-\lambda +R}{2}}}
R
=
E
2
+
9
λ
2
+
2
E
λ
{\displaystyle R={\sqrt {E^{2}+9\lambda ^{2}+2E\lambda }}}
(
E
,
G
)
{\displaystyle (E,\,G)}
E
G
3
(
3
G
−
E
)
{\displaystyle {\tfrac {EG}{3(3G-E)}}}
E
{\displaystyle E}
G
(
E
−
2
G
)
3
G
−
E
{\displaystyle {\tfrac {G(E-2G)}{3G-E}}}
G
{\displaystyle G}
E
2
G
−
1
{\displaystyle {\tfrac {E}{2G}}-1}
G
(
4
G
−
E
)
3
G
−
E
{\displaystyle {\tfrac {G(4G-E)}{3G-E}}}
(
E
,
ν
)
{\displaystyle (E,\,\nu )}
E
3
(
1
−
2
ν
)
{\displaystyle {\tfrac {E}{3(1-2\nu )}}}
E
{\displaystyle E}
E
ν
(
1
+
ν
)
(
1
−
2
ν
)
{\displaystyle {\tfrac {E\nu }{(1+\nu )(1-2\nu )}}}
E
2
(
1
+
ν
)
{\displaystyle {\tfrac {E}{2(1+\nu )}}}
ν
{\displaystyle \nu }
E
(
1
−
ν
)
(
1
+
ν
)
(
1
−
2
ν
)
{\displaystyle {\tfrac {E(1-\nu )}{(1+\nu )(1-2\nu )}}}
(
E
,
M
)
{\displaystyle (E,\,M)}
3
M
−
E
+
S
6
{\displaystyle {\tfrac {3M-E+S}{6}}}
E
{\displaystyle E}
M
−
E
+
S
4
{\displaystyle {\tfrac {M-E+S}{4}}}
3
M
+
E
−
S
8
{\displaystyle {\tfrac {3M+E-S}{8}}}
E
−
M
+
S
4
M
{\displaystyle {\tfrac {E-M+S}{4M}}}
M
{\displaystyle M}
S
=
±
E
2
+
9
M
2
−
10
E
M
{\displaystyle S=\pm {\sqrt {E^{2}+9M^{2}-10EM}}}
Có hai nghiệm hợp lệ.
Dấu cộng dẫn đến
ν
≥
0
{\displaystyle \nu \geq 0}
.
Dấu trừ dẫn đến
ν
≤
0
{\displaystyle \nu \leq 0}
.
(
λ
,
G
)
{\displaystyle (\lambda ,\,G)}
λ
+
2
G
3
{\displaystyle \lambda +{\tfrac {2G}{3}}}
G
(
3
λ
+
2
G
)
λ
+
G
{\displaystyle {\tfrac {G(3\lambda +2G)}{\lambda +G}}}
λ
{\displaystyle \lambda }
G
{\displaystyle G}
λ
2
(
λ
+
G
)
{\displaystyle {\tfrac {\lambda }{2(\lambda +G)}}}
λ
+
2
G
{\displaystyle \lambda +2G\,}
(
λ
,
ν
)
{\displaystyle (\lambda ,\,\nu )}
λ
(
1
+
ν
)
3
ν
{\displaystyle {\tfrac {\lambda (1+\nu )}{3\nu }}}
λ
(
1
+
ν
)
(
1
−
2
ν
)
ν
{\displaystyle {\tfrac {\lambda (1+\nu )(1-2\nu )}{\nu }}}
λ
{\displaystyle \lambda }
λ
(
1
−
2
ν
)
2
ν
{\displaystyle {\tfrac {\lambda (1-2\nu )}{2\nu }}}
ν
{\displaystyle \nu }
λ
(
1
−
ν
)
ν
{\displaystyle {\tfrac {\lambda (1-\nu )}{\nu }}}
Không thể sử dụng khi
ν
=
0
⇔
λ
=
0
{\displaystyle \nu =0\Leftrightarrow \lambda =0}
(
λ
,
M
)
{\displaystyle (\lambda ,\,M)}
M
+
2
λ
3
{\displaystyle {\tfrac {M+2\lambda }{3}}}
(
M
−
λ
)
(
M
+
2
λ
)
M
+
λ
{\displaystyle {\tfrac {(M-\lambda )(M+2\lambda )}{M+\lambda }}}
λ
{\displaystyle \lambda }
M
−
λ
2
{\displaystyle {\tfrac {M-\lambda }{2}}}
λ
M
+
λ
{\displaystyle {\tfrac {\lambda }{M+\lambda }}}
M
{\displaystyle M}
(
G
,
ν
)
{\displaystyle (G,\,\nu )}
2
G
(
1
+
ν
)
3
(
1
−
2
ν
)
{\displaystyle {\tfrac {2G(1+\nu )}{3(1-2\nu )}}}
2
G
(
1
+
ν
)
{\displaystyle 2G(1+\nu )\,}
2
G
ν
1
−
2
ν
{\displaystyle {\tfrac {2G\nu }{1-2\nu }}}
G
{\displaystyle G}
ν
{\displaystyle \nu }
2
G
(
1
−
ν
)
1
−
2
ν
{\displaystyle {\tfrac {2G(1-\nu )}{1-2\nu }}}
(
G
,
M
)
{\displaystyle (G,\,M)}
M
−
4
G
3
{\displaystyle M-{\tfrac {4G}{3}}}
G
(
3
M
−
4
G
)
M
−
G
{\displaystyle {\tfrac {G(3M-4G)}{M-G}}}
M
−
2
G
{\displaystyle M-2G\,}
G
{\displaystyle G}
M
−
2
G
2
M
−
2
G
{\displaystyle {\tfrac {M-2G}{2M-2G}}}
M
{\displaystyle M}
(
ν
,
M
)
{\displaystyle (\nu ,\,M)}
M
(
1
+
ν
)
3
(
1
−
ν
)
{\displaystyle {\tfrac {M(1+\nu )}{3(1-\nu )}}}
M
(
1
+
ν
)
(
1
−
2
ν
)
1
−
ν
{\displaystyle {\tfrac {M(1+\nu )(1-2\nu )}{1-\nu }}}
M
ν
1
−
ν
{\displaystyle {\tfrac {M\nu }{1-\nu }}}
M
(
1
−
2
ν
)
2
(
1
−
ν
)
{\displaystyle {\tfrac {M(1-2\nu )}{2(1-\nu )}}}
ν
{\displaystyle \nu }
M
{\displaystyle M}