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Problems in Two Spacial Dimensions

Bilinear rectangular elements.

We need to discuss briefly an alternate method of finite element interpolation: continuous, bilinearinterpolation on rectangles. Suppose a rectangular grid of lines x = x

i, y = y

j, i ∈ N

n, j ∈ N

m(where Nk = {1, . . . , k}) is drawn in the plane forming a system of rectangles Rij = {(x, y) : xi ≤x ≤ xi+1, yj ≤ y ≤ yj+1}. Assume further that the domain D can be approximated by ∪I Rij ,where I ⊂ Nn × Nm. For example, D might be a rectangle D = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d}and the grid lines xi = a + b(i − 1)/n, yj = c + b( j − 1)/m, i ∈ Nn, j ∈ Nm. Although it wouldbe possible the subdivide each Rij into two triangles, it is natural to attempt to use the rectangulargrid itself. Let Re be a rectangle in the grid, and consider interpolating a function u defined onRe using the values of u = uj at the corners z

ej , j = 1, 2, 3, 4 of this rectangle. Let’s label the

corners of Re so that ze1 = (a

e, be) is the lower left, ze2 = (ce, be) the lower right, ze3 = (c

e, de) theupper right, and ze4 = (a

e, de) the upper left (so ae < ce, be < de). The interpolation is easilyaccomplished if we define bilinear function H ej (x, y) such that H

ei (z

ej ) = δ ij = 1, i = j, 0 otherwise.

If hex = (ce − ae), hey = (d

e − be) denote the side lengths of Re, then

H e =

H e1H e2

H e3H e4

=

(ce − x)(de − y)/(hex

hey

)(x− ae)(de − y)/(hexh

ey)

(x− ae)(y − be)/(hexhey)

(ce − x)(y − be)/(hexhey)

Using these functions we can write u =4

j=1 H ej uj = H

eT U, where U = [u1, . . . , u4]T . It is easy tosee that this representation is unique. (If there were two representations then the difference wouldbe a bilinear function f vanishing at all zej , and hence would be identically zero. Since f restrictedto the line through ze1, z

e2 is a linear function of x vanishing at x = a

e and f restricted to the linethrough ze2, z

e3 is a linear function of y vanishing at y = b

e, we must have f = α(x − ae)(y − be).Since f (ze3) = 0, αh

exh

ey = 0 ⇒ α = 0.)

The assembly process using rectangular elements is basically the same as that using triangles. Of course, the number of nodes per element is now 4, and the element stiffness and load matrices are

4 by 4 and 4 by 1 respectively.Let’s now consider an example from our textbook using bilinear elements.

function ex592a(nx,ny)

%----------------------------------------------------------------------------

% Modified textbook file EX5.9.2.m

% to solve the two-dimensional Laplace’s equation given as

% u,xx + u,yy =0, 0 < x < 5, 0 < y < 10

% u(x,0) = 0, u(x,10) = 100sin(pi*x/10),

% u(0,y) = 0, u,x(5,y) = 0

% using bilinear rectangular elements

%(see Fig. 5.9.2 for the finite element mesh)

%

% Variable descriptions

% k = element matrix

% f = element vector

% kk = system matrix

% ff = system vector

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% gcoord = coordinate values of each node

% nodes = nodal connectivity of each element

% index = a vector containing system dofs associated with each element

% bcdof = a vector containing dofs associated with boundary conditions

% bcval = a vector containing boundary condition values associated with

% the dofs in ’bcdof’

%----------------------------------------------------------------------------

%------------------------------------

% input data for control parameters

%------------------------------------

nx1=nx+1; ny1=ny+1;

nel=nx*ny; % number of elements

nnel=4; % number of nodes per element

ndof=1; % number of dofs per node

nnode=nx1*ny1; % total number of nodes in system

sdof=nnode*ndof; % total system dofsxl=0; xr=5;

yb=0; yt=10;

%---------------------------------------------

% input data for nodal coordinate values

% gcoord(i,j) where i->node no. and j->x or y

% input data for nodal connectivity for each element

% nodes(i,j) where i-> element no. and j-> connected nodes

%---------------------------------------------------------

[gcoord, nodes]=gridgen([0,0],[5,10],nx,ny,1)

gridplot(nx,ny,gcoord,nodes,1)

%-------------------------------------

% input data for boundary conditions

%-------------------------------------

bcdof=[1:nx1,... % base nodes

nx1+1:nx1:nx1*(ny1-2)+1,... % left side nodes

nx1*(ny1-1)+1:nx1*ny1]; % top nodes

sz=size(bcdof);

nbn=sz(2);

bcval=zeros(1,nbn); xx=linspace(xl,xr,nx1);

bcval(nbn-nx1+1:nbn)=100*sin(pi*(xx-xl)/10); % along y=yt, u(x,yt)=sin(pi*x/xr)

%-----------------------------------------% initialization of matrices and vectors

%-----------------------------------------

ff=zeros(sdof,1); % initialization of system force vector

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kk=zeros(sdof,sdof); % initialization of system matrix

index=zeros(nnel*ndof,1); % initialization of index vector

%-----------------------------------------------------------------

% computation of element matrices and vectors and their assembly

%-----------------------------------------------------------------

for iel=1:nel % loop for the total number of elements

for i=1:nnel

nd(i)=nodes(iel,i); % extract connected node for (iel)-th element

x(i)=gcoord(nd(i),1); % extract x value of the node

y(i)=gcoord(nd(i),2); % extract y value of the node

end

xleng = x(2)-x(1); % length of the element in x-axis

yleng = y(4)-y(1); % length of the element in y-axis

index=feeldof(nd,nnel,ndof);% extract system dofs associated with element

k=felp2dr4(xleng,yleng); % compute element matrix

kk=feasmbl1(kk,k,index); % assemble element matrices

end

%-----------------------------

% apply boundary conditions

%-----------------------------

[kk,ff]=feaplyc2(kk,ff,bcdof,bcval);

%----------------------------

% solve the matrix equation

%----------------------------

fsol=kk\ff;

%---------------------

% analytical solution

%---------------------

for i=1:nnode

x=gcoord(i,1); y=gcoord(i,2);

esol(i)=100*sinh(0.31415927*y)*sin(0.31415927*x)/sinh(3.1415927);end

%------------------------------------

% print both exact and fem solutions

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%------------------------------------

num=1:1:sdof;

store=[num’ fsol esol’]

%===============================================================================

function [k]=felp2dr4(xleng,yleng)

%-------------------------------------------------------------------

% Purpose:

% element matrix for two-dimensional Laplace’s equation

% using four-node bilinear rectangular element

%

% Synopsis:

% [k]=felp2dr4(xleng,yleng)

%% Variable Description:

% k - element stiffness matrix (size of 4x4)

% xleng - element size in the x-axis

% yleng - element size in the y-axis

%-------------------------------------------------------------------

% element matrix

k(1,1)=(xleng*xleng+yleng*yleng)/(3*xleng*yleng);

k(1,2)=(xleng*xleng-2*yleng*yleng)/(6*xleng*yleng);

k(1,3)=-0.5*k(1,1);

k(1,4)=(yleng*yleng-2*xleng*xleng)/(6*xleng*yleng);

k(2,1)=k(1,2); k(2,2)=k(1,1); k(2,3)=k(1,4); k(2,4)=k(1,3);k(3,1)=k(1,3); k(3,2)=k(2,3); k(3,3)=k(1,1); k(3,4)=k(1,2);

k(4,1)=k(1,4); k(4,2)=k(2,4); k(4,3)=k(3,4); k(4,4)=k(1,1);

%==================================================================

function [kk,ff]=feaplyc2(kk,ff,bcdof,bcval)

%----------------------------------------------------------

% Purpose:

% Apply constraints to matrix equation [kk]{x}={ff}

%

% Synopsis:% [kk,ff]=feaplybc(kk,ff,bcdof,bcval)

%

% Variable Description:

% kk - system matrix before applying constraints

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% ff - system vector before applying constraints

% bcdof - a vector containging constrained d.o.f

% bcval - a vector containing contained value

%

% For example, there are constraints at d.o.f=2 and 10

% and their constrained values are 0.0 and 2.5,

% respectively. Then, bcdof(1)=2 and bcdof(2)=10; and

% bcval(1)=1.0 and bcval(2)=2.5.

%-----------------------------------------------------------

n=length(bcdof);

sdof=size(kk);

for i=1:n

c=bcdof(i);

for j=1:sdof

kk(c,j)=0;

end

kk(c,c)=1;

ff(c)=bcval(i);

end

%=================================================================

function [index]=feeldof(nd,nnel,ndof)

%----------------------------------------------------------

% Purpose:

% Compute system dofs associated with each element

%

% Synopsis:

% [index]=feeldof(nd,nnel,ndof)%


% index - system dof vector associated with element "iel"

% iel - element number whose system dofs are to be determined

% nnel - number of nodes per element

% ndof - number of dofs per node

%-----------------------------------------------------------

edof = nnel*ndof;

k=0;

for i=1:nnel

start = (nd(i)-1)*ndof;

for j=1:ndofk=k+1;

index(k)=start+j;

end

end

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%=====================================================================================

function [k]=felp2dr4(xleng,yleng)

%-------------------------------------------------------------------

% Purpose:

% element matrix for two-dimensional Laplace’s equation

% using four-node bilinear rectangular element

%

% Synopsis:

% [k]=felp2dr4(xleng,yleng)

%


% k - element stiffness matrix (size of 4x4)

% xleng - element size in the x-axis

% yleng - element size in the y-axis

%-------------------------------------------------------------------

% element matrix

k(1,1)=(xleng*xleng+yleng*yleng)/(3*xleng*yleng);

k(1,2)=(xleng*xleng-2*yleng*yleng)/(6*xleng*yleng);

k(1,3)=-0.5*k(1,1);

k(1,4)=(yleng*yleng-2*xleng*xleng)/(6*xleng*yleng);

k(2,1)=k(1,2); k(2,2)=k(1,1); k(2,3)=k(1,4); k(2,4)=k(1,3);

k(3,1)=k(1,3); k(3,2)=k(2,3); k(3,3)=k(1,1); k(3,4)=k(1,2);

k(4,1)=k(1,4); k(4,2)=k(2,4); k(4,3)=k(3,4); k(4,4)=k(1,1);

%====================================================================================

function [kk]=feasmbl1(kk,k,index)%----------------------------------------------------------

% Purpose:

% Assembly of element matrices into the system matrix

%

% Synopsis:

% [kk]=feasmbl1(kk,k,index)

%


% kk - system matrix

% k - element matri

% index - d.o.f. vector associated with an element

%-----------------------------------------------------------

edof = length(index);

for i=1:edof

ii=index(i);

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for j=1:edof

jj=index(j);

kk(ii,jj)=kk(ii,jj)+k(i,j);

end

end

%=====================================================================================

function [grd, connec]=gridgen(low_left, up_right, nx, ny, ind)

%

% FUNCTION [GRD, CONNEC]=GRIDGEN(LOW_LEFT, UP_RIGHT, NX, NY, IND)

%

% Creates a uniform grid on the rectangle with lower left corner

% low_left=[a,b] and upper right corner up_right=[c,d] using nx

% intervals along the x-axis and ny intervals on the y-axis. If ind=1

% the coordinates of nx*ny rectangular elements are returned in grd, and

% the corresponding connectivity matrix is returned in connec. If ind=2

% each rectangle is divided into two triangles and connec gives the% connectivity matrix.

%

if ~((ind==1)|(ind==2))

error(’ ind must be 1 (rectangles) or 2 (triangles)’);

return;

end

ny1=ny+1; nx1=nx+1;

x=linspace(low_left(1),up_right(1),nx1);

y=linspace(low_left(2),up_right(2),ny1);

[xx,yy]=meshgrid(x,y); % arrays xx, yy are ny1 by nx1, xx has rows x, yy columns y

grd=zeros(ny1*nx1,2);

for row=1:ny1 %

for col=1:nx1node_num=(row-1)*nx1+col;

grd(node_num,:)=[xx(row,col),yy(row,col)];

end

end

conq=zeros(nx*ny,4);

for row=1:ny % nodes for rectangular elements

for col=1:nx

nel=nx*(row-1)+col;

n0=nx1*(row-1)+col; % node at lower left of element nel

conq(nel,1:4)=[n0,n0+1,n0+1+nx1,n0+nx1]; % counter clockwise numbering

end

end

if ind==1connec=conq;

else

for kk=1:nx*ny % divide each of the nx*ny rectangles into 2 triangles 4 3

cont(2*kk-1,1:3)=[conq(kk,4),conq(kk,1),conq(kk,3)]; % local nodes 4,1,3 ----

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cont(2*kk,1:3)=[conq(kk,2), conq(kk,3), conq(kk,1)]; % local nodes 2,3,1 ----

end % 1 2

connec=cont;

end

%=========================================================================

function gridplot(nx,ny,grd,connec,ind)

close all

nx1=nx+1; ny1=ny+1;

x=grd(1:nx1,1)’; y=grd(1:nx1:nx1*ny1-nx1+1,2)’;

hx=x(2)-x(1); hy=y(2)-y(1); % grid spacings

hold on

axis([-hx+x(1), x(nx1)+hx, -2*hy+y(1), y(ny1)+hy]);

for kk=1:ny1 plot(x,y(kk)*ones(1,nx1)); end % plot horizontal grid lines

for kk=1:nx1 plot(x(kk)*ones(1,ny1),y); end % plot vertical grid lines

ofx=0.1*hx; ofy=0.2*hy; % offsets

for kk=1:nx1*ny1str=sprintf(’%d’,kk);

text(grd(kk,1)+ofx,grd(kk,2)+ofy,str,’Color’,[1,0,0],’FontSize’,8);

end

if ind==2 % complete triangles

for kk=1:2:length(connec(:,1))

x1=[grd(connec(kk,2),1),grd(connec(kk,3),1)];

y1=[grd(connec(kk,2),2),grd(connec(kk,3),2)];

plot(x1,y1,’b-’);

end

end

if ind==1

for kk=1:length(connec(:,1))

xg=grd(connec(kk,1),1)+hx/2; yg=grd(connec(kk,1),2)+hy/2;str=sprintf(’%d’,kk);

text(xg,yg,str);

end

else


xg=grd(connec(kk,1),1)+hx/3; yg=grd(connec(kk,1),2)-hy/3;

str=sprintf(’%d’,kk);

text(xg,yg,str,’FontSize’,8);

end


xg=grd(connec(kk,3),1)+2*hx/3; yg=grd(connec(kk,3),2)+hy/3;

str=sprintf(’%d’,kk);

text(xg,yg,str,’FontSize’,8);end

end

text(x(1),y(1)-2*hy/3,’element numbers’,’FontSize’,8);

text(x(1),y(1)-4*hy/3,’node numbers’,’Color’,[1,0,0],’FontSize’,8);

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hold off

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