Đọc thêm Mô_đun_khối

Mô đun đàn hồi đối với vật liệu đẳng hướng đồng nhất
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}