MATLAB

の矩形要素を持つ有限要素法を使用したポアソンの方程式の近似ここでは三角要素を使用した例が見つかりました。私は三角形の要素を長方形のものに置き換えるためにメッシュを生成する方法を変更しましたが、それらを統合する方法はわかりません。ここでの私のバージョンは次のとおりです。MATLAB

%3.6 femcode.m 

% [p,t,b] = squaregrid(m,n) % create grid of N=mn nodes to be listed in p 
% generate mesh of T=2(m-1)(n-1) right triangles in unit square 
m=6; n=5; % includes boundary nodes, mesh spacing 1/(m-1) and 1/(n-1) 
[x,y]=ndgrid((0:m-1)/(m-1),(0:n-1)/(n-1)); % matlab forms x and y lists 
p=[x(:),y(:)]; % N by 2 matrix listing x,y coordinates of all N=mn nodes 
nelems=(m-1)*(n-1); 
t=zeros(nelems,3); 
for e=1:nelems 
    t(e) = e + floor(e/6); 
    t(e, 2) = e + floor(e/6) + 1; 
    t(e, 3) = e + floor(e/6) + 6; 
    t(e, 4) = e + floor(e/6) + 7; 
end 
% final t lists 4 node numbers of all rectangles in T by 4 matrix 
b=[1:m,m+1:m:m*n,2*m:m:m*n,m*n-m+2:m*n-1]; % bottom, left, right, top 
% b = numbers of all 2m+2n **boundary nodes** preparing for U(b)=0 

% [K,F] = assemble(p,t) % K and F for any mesh of rectangles: linear phi's 
N=size(p,1);T=nelems; % number of nodes, number of rectangles 
% p lists x,y coordinates of N nodes, t lists rectangles by 4 node numbers 
K=sparse(N,N); % zero matrix in sparse format: zeros(N) would be "dense" 
F=zeros(N,1); % load vector F to hold integrals of phi's times load f(x,y) 

for e=1:T % integration over one rectangular element at a time 
    nodes=t(e,:); % row of t = node numbers of the 3 corners of triangle e 
    Pe=[ones(4,1),p(nodes,:),p(nodes,1).*p(nodes,2)]; % 4 by 4 matrix with rows=[1 x y xy] 
    Area=abs(det(Pe)); % area of triangle e = half of parallelogram area 
    C=inv(Pe); % columns of C are coeffs in a+bx+cy to give phi=1,0,0 at nodes 
    % now compute 3 by 3 Ke and 3 by 1 Fe for element e 
    grad=C(2:3,:);Ke=Area*grad'*grad; % element matrix from slopes b,c in grad 
    Fe=Area/3; % integral of phi over triangle is volume of pyramid: f(x,y)=1 
    % multiply Fe by f at centroid for load f(x,y): one-point quadrature! 
    % centroid would be mean(p(nodes,:)) = average of 3 node coordinates 
    K(nodes,nodes)=K(nodes,nodes)+Ke; % add Ke to 9 entries of global K 
    F(nodes)=F(nodes)+Fe; % add Fe to 3 components of load vector F 
end % all T element matrices and vectors now assembled into K and F 

% [Kb,Fb] = dirichlet(K,F,b) % assembled K was singular! K*ones(N,1)=0 
% Implement Dirichlet boundary conditions U(b)=0 at nodes in list b 
K(b,:)=0; K(:,b)=0; F(b)=0; % put zeros in boundary rows/columns of K and F 
K(b,b)=speye(length(b),length(b)); % put I into boundary submatrix of K 
Kb=K; Fb=F; % Stiffness matrix Kb (sparse format) and load vector Fb 

% Solving for the vector U will produce U(b)=0 at boundary nodes 
U=Kb\Fb % The FEM approximation is U_1 phi_1 + ... + U_N phi_N 

% Plot the FEM approximation U(x,y) with values U_1 to U_N at the nodes 
% surf(p(:,1),p(:,2),0*p(:,1),U,'edgecolor','k','facecolor','interp'); 
% view(2),axis equal,colorbar

私は三角形要素の上に統合するために使用されるコードのセクションを編集し始めたが、私は続行するかどうかはわかりませんか、それはさえのための同様の方法で行われている場合長方形の要素。

UPDATE

だから私はDohyunが私の限られた理解した上で提案し、ここで私が今持っているものだものを組み込むために試してみた：

m=12; n=12; % includes boundary nodes, mesh spacing 1/(m-1) and 1/(n-1) 
[x,y]=ndgrid((0:m-1)/(m-1),(0:n-1)/(n-1)); % matlab forms x and y lists 
p=[x(:),y(:)]; % N by 2 matrix listing x,y coordinates of all N=mn nodes 
nelems=(m-1)*(n-1); 
t=zeros(nelems,4); 
a=0; 
for e=1:nelems 
    t(e) = e + a; 
    t(e, 2) = e + a + 1; 
    t(e, 3) = e + a + m + 1; 
    t(e, 4) = e + a + m; 
    a = floor(e/n); 
end 
% final t lists 4 node numbers of all rectangles in T by 4 matrix 
b=[1:m,m+1:m:m*n,2*m:m:m*n,m*n-m+2:m*n-1]; % bottom, left, right, top 
% b = numbers of all 2m+2n **boundary nodes** preparing for U(b)=0 
N=size(p,1);T=nelems; % number of nodes, number of rectangles 
% p lists x,y coordinates of N nodes, t lists rectangles by 4 node numbers 
K=sparse(N,N); % zero matrix in sparse format: zeros(N) would be "dense" 
F=zeros(N,1); % load vector F to hold integrals of phi's times load f(x,y) 

for e=1:T % integration over one rectangular element at a time 
    nodes=t(e,:); % row of t = node numbers of the 3 corners of triangle e 
    Pe=[ones(4,1),p(nodes,:),p(nodes,1).*p(nodes,2)]; % 4 by 4 matrix with rows=[1 x y xy] 
    Area=abs(det(Pe(1:3,1:3))); % area of triangle e = half of parallelogram area 
    C=inv(Pe); % columns of C are coeffs in a+bx+cy to give phi=1,0,0 at nodes 
    % now compute 3 by 3 Ke and 3 by 1 Fe for element e 
    % grad=C(2:3,:); 
    % constantKe=Area*grad'*grad; % element matrix from slopes b,c in grad 
    for i=1:4 
    for j=1:4 
     syms x y 
     Kn = int(int(... 
       C(i,2)*C(j,2)+ ... 
       (C(i,2)*C(j,4)+C(i,4)*C(j,2))*y + ... 
       C(i,4)*C(j,4)*y^2 + ... 
       C(i,3)*C(j,3) + ... 
       (C(i,4)*C(j,3)+C(i,3)*C(j,4))*x + ... 
       C(i,4)*C(j,4)*x^2 ... 
       , x, Pe(1, 2), Pe(2, 2)), y, Pe(1, 3), Pe(3, 3)); 
     K(nodes(i),nodes(j)) = K(nodes(i),nodes(j)) + Kn; 
    end 
    end 

    Fe=Area/3; % integral of phi over triangle is volume of pyramid: f(x,y)=1 
    % multiply Fe by f at centroid for load f(x,y): one-point quadrature! 
    % centroid would be mean(p(nodes,:)) = average of 3 node coordinates 
    % K(nodes,nodes)=K(nodes,nodes)+Ke; % add Ke to 9 entries of global K 
    F(nodes)=F(nodes)+Fe; % add Fe to 4 components of load vector F 
end % all T element matrices and vectors now assembled into K and F 

% [Kb,Fb] = dirichlet(K,F,b) % assembled K was singular! K*ones(N,1)=0 
% Implement Dirichlet boundary conditions U(b)=0 at nodes in list b 
K(b,:)=0; K(:,b)=0; F(b)=0; % put zeros in boundary rows/columns of K and F 
K(b,b)=speye(length(b),length(b)); % put I into boundary submatrix of K 
Kb=K; Fb=F; % Stiffness matrix Kb (sparse format) and load vector Fb 

% Solving for the vector U will produce U(b)=0 at boundary nodes 
U=Kb\Fb; % The FEM approximation is U_1 phi_1 + ... + U_N phi_N 
U2=reshape(U',m,n); 
% Plot the FEM approximation U(x,y) with values U_1 to U_N at the nodes 
surf(U2)

私が代わりの定積分を使用することができると思います結果は元のプログラムと一致しません。 P1三角形要素の場合について

Result of my program

Result of original program

出典

2017-01-07 luckysori