Eshelby's tensor for three dimensional mesh
2 ビュー (過去 30 日間)
古いコメントを表示
Dear all,
Seasons' greetings.
I am trying to find out the Eshelby's tensor for superellipsoid system. I am following the protocol of finding flux integral from this link https://www3.nd.edu/~nancy/Math20550/Matlab/Assignments/SurfaceIntegrals/surfaceintegrals.html#27
The code returns me the symbolic coefficients, rather than any numeric value.
Please help.
%%%%%%%%%%%%%%% designing superellipsoid %%%%%%%%%%%%%%%%%
%declare constants%
a1=25; a2=25; a3=25; epsilon1=1; epsilon2=1; epsilon3=1;
n=100;
%declare variables%
etamax=pi/2; etamin=-pi/2;
wmax=pi; wmin=-pi;
deta=(etamax-etamin)/n;
dw=(wmax-wmin)/n;
[i,j] = meshgrid(1:n+1,1:n+1);
eta = etamin + (i-1) * deta; w = wmin + (j-1) * dw;
%%%%%%%%%%%%%%% crating symbolic variables %%%%%%%%%%%%%%%%%
syms eta w
ellip=[(a1.*sign(cos(eta)).*abs(cos(eta)).^epsilon1.*sign(cos(w)).*abs(cos(w)).^epsilon1),(a2.*sign(cos(eta)).*abs(cos(eta)).^epsilon2.*sign(sin(w)).*abs(sin(w)).^epsilon2),(a3.*sign(sin(eta)).*abs(sin(eta)).^epsilon3)];
F=[(a1.*sin(eta).*cos(w)) (a2.*sin(eta).*sin(w)) (a3.*cos(eta))];
nds=simplify(cross(diff(ellip,eta),diff(ellip,w)));
Fpar=subs(F,[(a1.*sin(eta).*cos(w)) (a2.*sin(eta).*sin(w)) (a3.*cos(eta))],ellip);
Fj=@(Fpar,nds)Fpar.*transpose(nds);
flux=(symint2(Fj,eta,-pi,pi,w,-(pi/2),(pi/2)));
0 件のコメント
回答 (0 件)
参考
カテゴリ
Help Center および File Exchange で Calculus についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!