https://zhuanlan.zhihu.com/p/97700824
\sigma_{ij}=C_{ijkl}\varepsilon_{kl}
其中i,j,k,l∈[1,3]
写成矩阵乘法形式就是:(把俺累坏了...)
注意:弹性常数矩阵的4个下标我用逗号,分隔开了,前两个逗号表示应力\sigma的下标,后两个逗号表示应变\varepsilon的下标
\left[
\begin{array} {c}
\sigma_{11} \\
\sigma_{12} \\
\sigma_{13} \\
\sigma_{21} \\
\sigma_{22} \\
\sigma_{23} \\
\sigma_{31} \\
\sigma_{32} \\
\sigma_{33} \\
\end{array}
\right]
=
\left[
\begin{array}{ccccccccc}
C_{11,11} & C_{11,12} & C_{11,13} & C_{11,21} & C_{11,22} & C_{11,23} & C_{11,31} & C_{11,32} & C_{11,33} \\
C_{12,11} & C_{12,12} & C_{12,13} & C_{12,21} & C_{12,22} & C_{12,23} & C_{12,31} & C_{12,32} & C_{12,33} \\
C_{13,11} & C_{13,12} & C_{13,13} & C_{13,21} & C_{13,22} & C_{13,23} & C_{13,31} & C_{13,32} & C_{13,33} \\
C_{21,11} & C_{21,12} & C_{21,13} & C_{21,21} & C_{21,22} & C_{21,23} & C_{21,31} & C_{21,32} & C_{21,33} \\
C_{22,11} & C_{22,12} & C_{22,13} & C_{22,21} & C_{22,22} & C_{22,23} & C_{22,31} & C_{22,32} & C_{22,33} \\
C_{23,11} & C_{23,12} & C_{23,13} & C_{23,21} & C_{23,22} & C_{23,23} & C_{23,31} & C_{23,32} & C_{23,33} \\
C_{31,11} & C_{31,12} & C_{31,13} & C_{31,21} & C_{31,22} & C_{31,23} & C_{31,31} & C_{31,32} & C_{31,33} \\
C_{32,11} & C_{32,12} & C_{32,13} & C_{32,21} & C_{32,22} & C_{32,23} & C_{32,31} & C_{32,32} & C_{32,33} \\
C_{33,11} & C_{33,12} & C_{33,13} & C_{33,21} & C_{33,22} & C_{33,23} & C_{33,31} & C_{33,32} & C_{33,33} \\
\end{array}
\right]
\left[
\begin{array} {c}
\varepsilon_{11} \\
\varepsilon_{12} \\
\varepsilon_{13} \\
\varepsilon_{21} \\
\varepsilon_{22} \\
\varepsilon_{23} \\
\varepsilon_{31} \\
\varepsilon_{32} \\
\varepsilon_{33} \\
\end{array}
\right]
用Voigt标记
11→1 \\
12→2 \\
13→2 \\
21→4 \\
22→5 \\
23→6 \\
31→7 \\
32→8 \\
33→9 \\
\left[
\begin{array} {c}
\sigma_{1} \\
\sigma_{2} \\
\sigma_{3} \\
\sigma_{4} \\
\sigma_{5} \\
\sigma_{6} \\
\sigma_{7} \\
\sigma_{8} \\
\sigma_{9} \\
\end{array}
\right]
=
\left[
\begin{array}{ccccccccc}
C_{11} & C_{12} & C_{13} & C_{14} & C_{15} & C_{16} & C_{17} & C_{18} & C_{19} \\
C_{21} & C_{22} & C_{23} & C_{24} & C_{25} & C_{26} & C_{27} & C_{28} & C_{29} \\
C_{31} & C_{32} & C_{33} & C_{34} & C_{35} & C_{36} & C_{37} & C_{38} & C_{39} \\
C_{41} & C_{42} & C_{43} & C_{44} & C_{45} & C_{46} & C_{47} & C_{48} & C_{49} \\
C_{51} & C_{52} & C_{53} & C_{54} & C_{55} & C_{56} & C_{57} & C_{58} & C_{59} \\
C_{61} & C_{62} & C_{63} & C_{64} & C_{65} & C_{66} & C_{67} & C_{68} & C_{69} \\
C_{71} & C_{72} & C_{73} & C_{74} & C_{75} & C_{76} & C_{77} & C_{78} & C_{79} \\
C_{81} & C_{82} & C_{83} & C_{84} & C_{85} & C_{86} & C_{87} & C_{88} & C_{89} \\
C_{91} & C_{92} & C_{93} & C_{94} & C_{95} & C_{96} & C_{97} & C_{98} & C_{99} \\
\end{array}
\right]
\left[
\begin{array} {c}
\varepsilon_{1} \\
\varepsilon_{2} \\
\varepsilon_{3} \\
\varepsilon_{4} \\
\varepsilon_{5} \\
\varepsilon_{6} \\
\varepsilon_{7} \\
\varepsilon_{8} \\
\varepsilon_{9} \\
\end{array}
\right]
Cauthy应力,Cauthy应变都是对称滴!于是可以划去第7~9行和第7~9列!
现在想2点:
1)Cauchy应变是对称滴,并且“作用效果相同”。(举个例子:\varepsilon_4, \varepsilon_7对于\sigma_4的“贡献”同等,所以C_{44}=C_{47},同理,C_{45}=C_{48}, C_{46}=C_{49}...
2) 想想矩阵乘法,应力第4~6行本来是9+9个元素分别相乘再相加。现在变成6+6个元素分别相乘再相加,结合1)得到的C_{44}=C_{47}, C_{45}=C_{48}, C_{46}=C_{49}应变应该×2!!!(所谓“工程剪切应变”,本质上就是这么来的,即:Voigt记法从9×9变为6×6时得到的)
于是得到:
\left[
\begin{array} {c}
\sigma_{1} \\
\sigma_{2} \\
\sigma_{3} \\
\sigma_{4} \\
\sigma_{5} \\
\sigma_{6}
\end{array}
\right]
=
\left[
\begin{array}{cccccc}
C_{11} & C_{12} & C_{13} & C_{14} & C_{15} & C_{16} \\
C_{21} & C_{22} & C_{23} & C_{24} & C_{25} & C_{26} \\
C_{31} & C_{32} & C_{33} & C_{34} & C_{35} & C_{36} \\
C_{41} & C_{42} & C_{43} & C_{44} & C_{45} & C_{46} \\
C_{51} & C_{52} & C_{53} & C_{54} & C_{55} & C_{56} \\
C_{61} & C_{62} & C_{63} & C_{64} & C_{65} & C_{66} \\
\end{array}
\right]
\left[
\begin{array} {c}
\varepsilon_{1} \\
\varepsilon_{2} \\
\varepsilon_{3} \\
2\varepsilon_{4} \\
2\varepsilon_{5} \\
2\varepsilon_{6}
\end{array}
\right]
考虑对称性,以及1)xyz方向的正应变之间耦合(想想泊松效应~) 2)正应变和切应变不耦合(也就是:切应变不会导致任何正应力)3)不同切应变之间不耦合(也就是:\varepsilon_{xy}只会导致\sigma_{xy}而不会导致\sigma_{xz},\sigma_{yz}
\left[
\begin{array} {c}
\sigma_{1} \\
\sigma_{2} \\
\sigma_{3} \\
\sigma_{4} \\
\sigma_{5} \\
\sigma_{6}
\end{array}
\right]
=
\left[
\begin{array}{cccccc}
C_{11} & C_{12} & C_{13} & 0 & 0 & 0 \\
C_{21} & C_{22} & C_{23} & 0 & 0 & 0 \\
C_{31} & C_{32} & C_{33} & 0 & 0 & 0 \\
0 & 0 & 0 & C_{44} & 0 & 0 \\
0 & 0 & 0 & 0 & C_{55} & 0 \\
0 & 0 & 0 & 0 & 0 & C_{66} \\
\end{array}
\right]
\left[
\begin{array} {c}
\varepsilon_{1} \\
\varepsilon_{2} \\
\varepsilon_{3} \\
2\varepsilon_{4} \\
2\varepsilon_{5} \\
2\varepsilon_{6}
\end{array}
\right]
说句题外话:这里如果考虑FCC/BCC/HCP的弹性常数矩阵各自有多少个独立常数,就是晶体物理学的知识了,暂且按下不表挖个坑~
关于怎样计算各自独立的弹性常数,怎样在不同坐标系之间变换(晶体/地面坐标系),怎样画哪个很好看的各向异性小花花,又是一个坑~
回归主题:对于【弹性力学】常用的【各向同性】【小变形】【线弹性】,我们不妨先想想:《材料力学》学过的“广义胡克定律”
\varepsilon_{1}=\frac{1}{E}(\sigma_{1}-v(\sigma_2+\sigma_3))
(为什么要想”应变={什么}应力”,而不是”应力={什么}应变”呢?),这是因为,”应变={什么}应力”更直观啊~
根据泊松比定义:单轴方向拉伸,另外两个方向应变负值除以单轴拉伸方向应变值;
那么:\varepsilon_{1}=\frac{1}{E}(\sigma_{1}-v(\sigma_2+\sigma_3))就是很顺理成章的了:将泊松比v作“权重”,又因为变形方向是相反的,所以\varepsilon_{1}=\frac{1}{E}(\sigma_{1}-v(\sigma_2+\sigma_3))!
对于剪切应变,公式很简单:(前面提过,剪切应变之间是不耦合的!):
\varepsilon_{4}=\frac{1}{G}\sigma_{4}
根据《材料力学》知识(如果你没学过,就先记住):
杨氏模量,剪切模量和泊松比知2推3:
G=\frac{E}{2(1+v)}
所以:广义胡克定律可以写作:
\left[
\begin{array} {c}
\varepsilon_{1} \\
\varepsilon_{2} \\
\varepsilon_{3} \\
\varepsilon_{4} \\
\varepsilon_{5} \\
\varepsilon_{6}
\end{array}
\right]
=
\frac{1}{E}
\left[
\begin{array}{cccccc}
1 & -v & -v & 0 & 0 & 0 \\
-v & 1 & -v & 0 & 0 & 0 \\
-v & -v & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & {2(1+v)} & 0 & 0 \\
0 & 0 & 0 & 0 & {2(1+v)} & 0 \\
0 & 0 & 0 & 0 & 0 & {2(1+v)} \\
\end{array}
\right]
\left[
\begin{array} {c}
\sigma_{1} \\
\sigma_{2} \\
\sigma_{3} \\
\sigma_{4} \\
\sigma_{5} \\
\sigma_{6}
\end{array}
\right]
用MATLAB计算其逆矩阵
%%
fclose all; close all; clear; clc;
%%
syms v E;
B = 1/E *[ 1, -v, -v, 0, 0, 0;
-v, 1, -v, 0, 0, 0;
-v, -v, 1, 0, 0, 0;
0, 0, 0, 2*(1+v), 0, 0;
0, 0, 0, 0, 2*(1+v), 0;
0, 0, 0, 0, 0, 2*(1+v);];
C = inv(B);
pretty(C);
弹性力学】常用的【各向同性】【小变形】【线弹性】的弹性常数矩阵就得到啦!
\left[
\begin{array} {c}
\sigma_{1} \\
\sigma_{2} \\
\sigma_{3} \\
\sigma_{4} \\
\sigma_{5} \\
\sigma_{6}
\end{array}
\right]
=\frac{E(1-v)}{(1+v)(1-2v)}
\left[
\begin{array}{cccccc}
1 & \frac{v}{(1-v)} & \frac{v}{(1-v)} & 0 & 0 & 0 \\
\frac{v}{(1-v)} & 1 & \frac{v}{(1-v)} & 0 & 0 & 0 \\
\frac{v}{(1-v)} & \frac{v}{(1-v)} & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{1-2v}{2(1-v)} & 0 & 0 \\
0 & 0 & 0 & 0 & \frac{1-2v}{2(1-v)} & 0 \\
0 & 0 & 0 & 0 & 0 & \frac{1-2v}{2(1-v)} \\
\end{array}
\right]
\left[
\begin{array} {c}
\varepsilon_{1} \\
\varepsilon_{2} \\
\varepsilon_{3} \\
2\varepsilon_{4} \\
2\varepsilon_{5} \\
2\varepsilon_{6}
\end{array}
\right]
\left[
\begin{array} {c}
\sigma_{1} \\
\sigma_{2} \\
\sigma_{3} \\
\sigma_{4} \\
\sigma_{5} \\
\sigma_{6}
\end{array}
\right]
=
\left[
\begin{array}{cccccc}
\frac{E(1-v)}{(1+v)(1-2v)} & \frac{E(v)}{(1+v)(1-2v)} & \frac{E(v)}{(1+v)(1-2v)} & 0 & 0 & 0 \\
\frac{E(v)}{(1+v)(1-2v)} & \frac{E(1-v)}{(1+v)(1-2v)} & \frac{E(v)}{(1+v)(1-2v)} & 0 & 0 & 0 \\
\frac{E(v)}{(1+v)(1-2v)} & \frac{E(v)}{(1+v)(1-2v)} & \frac{E(1-v)}{(1+v)(1-2v)} & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{E}{2(1+v)} & 0 & 0 \\
0 & 0 & 0 & 0 & \frac{E}{2(1+v)} & 0 \\
0 & 0 & 0 & 0 & 0 & \frac{E}{2(1+v)} \\
\end{array}
\right]
\left[
\begin{array} {c}
\varepsilon_{1} \\
\varepsilon_{2} \\
\varepsilon_{3} \\
2\varepsilon_{4} \\
2\varepsilon_{5} \\
2\varepsilon_{6}
\end{array}
\right]