me471_hwk05ans

ME 471 Problem Set 5Solutions

5.8) The weak form of the given initial boundary value problem is

∫ 3

0

(

v∂u

∂t+ 0.1

∂v

∂x

∂u

∂x

)

dx =

∫ 3

010v dx+ 0.1v(3, t)

∂u

∂x(3, t)

where u(0, t) = 100, ∂u/∂x(3, t) = 100 exp(−t), and u(x, 0) = 50. An m-file to solve this problem isgiven below

function prob5_8(xvec, nstep, dt)

% 1D transient heat conduction problem:

% INPUTS:

% xvec = grid points in [0,3], xvec=[x(1)=0,x(2),...,x(n+1)=3]

% nstep = Number of time steps

% dt = time increment

%

sdof=length(xvec); % number of nodes

n=sdof-1; % number of elements

nodes=[[1:n]’ [2:n+1]’]; % configuration matrix

mm=zeros(sdof,sdof); kk=zeros(sdof,sdof); ff=zeros(sdof,1);

for iel=1:n

gn=nodes(iel,:);

h=xvec(gn(2))-xvec(gn(1));

ke=elstif(h);

me=elmass(h);

fe=elload(h);

for ii=1:2 % assemble mass and stiffness

gn1=gn(ii);

ff(gn1)=ff(gn1)+fe(ii);

for jj=1:2

gn2=gn(jj);

mm(gn1,gn2)=mm(gn1,gn2)+me(ii,jj);

kk(gn1,gn2)=kk(gn1,gn2)+ke(ii,jj);

end

end

end

uold=init_val(xvec’); % set inital conditions

effrc=zeros(sdof,1); fnew=zeros(sdof,1);

t=0;

close all

hold on

plot(xvec, uold); % plot initial condition

for k=1:nstep % use implicit (backward difference) method

fnew=ff; fnew(sdof)=ff(sdof)+10*exp(-t-dt); % load vector at t+dt

effrc=mm*uold+dt*fnew; % effective forcing

aa=mm+dt*kk;

aa(1,:)=0; aa(1,1)=1; effrc(1)=100; % impose bc at x=0 and time t

fsol=aa\effrc;

uold=fsol;

if(rem(k,50)==0) plot(xvec,uold); end

t=t+dt;

end

esol=exac(xvec);

plot(xvec, uold,’r’, xvec, esol, ’o’);

hold off

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

% end of main program

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

function me=elmass(h)

% element mass matrix

me=(h/6)*[2 1; 1 2];

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

function ke=elstif(h)

% element stiffness matrix

ke=0.1*(1/h)*[1 -1; -1 1];

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

function fe=elload(h)

% element load vector

fe=5*h*[1;1];

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

function h=init_val(x)

h=50*ones(size(x));

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

function u=exac(x)

% steady state solution

u=100+300*x-50*x.^2;

5.13) This and Problem 5.15 require the solution of the initial boundary value problem

∂u

∂t−∆u = 0, 0 < x, y < 1

with u given on the entire boundary of the unit square. Using forward time differencing leads to them-file (based on Ex5.11.1 of the texbook)

function prob5_13()

nx=4; ny=2;

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 dofs

deltt=0.01; % time step size for transient analysis

stime=0.0; % initial time

ftime=1; % termination time

ntime=fix((ftime-stime)/deltt); % number of time increment

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

% input data for nodal coordinate values

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

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

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

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

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

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

% input data for boundary conditions

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

bcdof=[1:nx1, nx1+1:nx1:ny*nx1+1, ny*nx1+2:nnode-1, 2*nx1:nx1:nnode];

lendof=length(bcdof);

bcval=zeros(1,lendof);

bcval(lendof-ny+1:lendof)=200; % last ny nodes at 200

bcval(lendof-ny-nx+2:lendof-ny)=100; %next to last nx-1 nodes at 100

[(1:lendof)’,bcdof’, bcval’]

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

% initialization of matrices and vectors

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

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

fn=zeros(sdof,1); % initialization of effective system vector

fsol=zeros(sdof,1); % solution vector

sol=zeros(2,ntime+1); % vector containing time history solution

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

mm=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

nd(1)=nodes(iel,1); % 1st connected node for (iel)-th element

nd(2)=nodes(iel,2); % 2nd connected node for (iel)-th element

nd(3)=nodes(iel,3); % 3rd connected node for (iel)-th element

nd(4)=nodes(iel,3); % 4th connected node for (iel)-th element

x1=gcoord(nd(1),1); y1=gcoord(nd(1),2);% coord values of 1st node

x2=gcoord(nd(2),1); y2=gcoord(nd(2),2);% coord values of 2nd node

x3=gcoord(nd(3),1); y3=gcoord(nd(3),2);% coord values of 3rd node

x4=gcoord(nd(4),1); y4=gcoord(nd(4),2);% coord values of 4th node

xleng=x2-x1; % element size in x-axis

yleng=y4-y1; % element size in y-axis

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

k=felp2dr4(xleng,yleng); % compute element matrix (time-independent term)

m=felpt2r4(xleng,yleng); % compute element matrix (time-dependent term)

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

mm=feasmbl1(mm,m,index); % assemble element matrices

end

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

% loop for time integration

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

for in=1:sdof

fsol(in)=0.0; % initial condition

end

sol(1,1)=fsol(8); % store time history solution for node no. 8

sol(2,1)=fsol(9); % store time history solution for node no. 9

for it=1:ntime % start loop for time integration

fn=deltt*ff+(mm-deltt*kk)*fsol; % compute effective column vector

[mm,fn]=feaplyc2(mm,fn,bcdof,bcval); % apply boundary condition

fsol=mm\fn; % solve the matrix equation

sol(1,it+1)=fsol(8); % store time history solution for node no. 8

sol(2,it+1)=fsol(9); % store time history solution for node no. 9

end

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

% analytical solution at node 8

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

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

% plot the solution at nodes 8 and 9

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

time=0:deltt:ntime*deltt;

plot(time,sol(1,:),’*’);

xlabel(’Time’)

ylabel(’Solution at nodes’)

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

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 [m]=felpt2r4(xleng,yleng)

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

% Purpose:

% element matrix of transient term for two-dimensional Laplace’s

% equation using four-node bilinear rectangular element

%

% Synopsis:

% [m]=felpt2r4(xleng,yleng)

%


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



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

% element matrix

m=(xleng*yleng/36)*[4 2 1 2;

2 4 2 1;

1 2 4 2;

2 1 2 4];

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);

for j=1:edof

jj=index(j);

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

end

end

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)

%


% kk - system matrix before applying constraints

% 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:ndof

k=k+1;

index(k)=start+j;

end

end

5.15) Solving the same problem using Crank-Nicolson differencing leads to the m-file (based onEx5.11.5 of the texbook)

function prob5_15()

nx=4; ny=4;

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 dofs

deltt=0.01; % time step size for transient analysis

stime=0.0; % initial time

ftime=1; % termination time

ntime=fix((ftime-stime)/deltt); % number of time increment

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

% input data for nodal coordinate values

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

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

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

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

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

input(’press any key to continue’);

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

% input data for boundary conditions

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

bcdof=[1:nx1, nx1+1:nx1:ny*nx1+1, ny*nx1+2:nnode-1, 2*nx1:nx1:nnode];

lendof=length(bcdof);

bcval=zeros(1,lendof);

bcval(lendof-ny+1:lendof)=200; % last ny nodes at 200

bcval(lendof-ny-nx+2:lendof-ny)=100; %next to last nx-1 nodes at 100

bcval(lendof)=150; bcval(nx1)=100; bcval(lendof-ny-nx+1)=50; % average values at corners

[(1:lendof)’,bcdof’, bcval’]

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

% initialization of matrices and vectors

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

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

fn=zeros(sdof,1); % initialization of effective system vector

fsol=zeros(sdof,1); % solution vector

sol=zeros(2,ntime+1); % vector containing time history solution

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

mm=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

nd(1)=nodes(iel,1); % 1st connected node for (iel)-th element

nd(2)=nodes(iel,2); % 2nd connected node for (iel)-th element

nd(3)=nodes(iel,3); % 3rd connected node for (iel)-th element

nd(4)=nodes(iel,3); % 4th connected node for (iel)-th element

x1=gcoord(nd(1),1); y1=gcoord(nd(1),2);% coord values of 1st node

x2=gcoord(nd(2),1); y2=gcoord(nd(2),2);% coord values of 2nd node

x3=gcoord(nd(3),1); y3=gcoord(nd(3),2);% coord values of 3rd node

x4=gcoord(nd(4),1); y4=gcoord(nd(4),2);% coord values of 4th node

xleng=x2-x1; % element size in x-axis

yleng=y4-y1; % element size in y-axis

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

k=felp2dr4(xleng,yleng); % compute element matrix (time-independent term)

m=felpt2r4(xleng,yleng); % compute element matrix (time-dependent term)

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

mm=feasmbl1(mm,m,index); % assemble element matrices

end

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

% loop for time integration

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

for in=1:sdof

fsol(in)=0.0; % initial condition

end

sol(1,1)=fsol((nnode+1)/2); % store time history solution for middle node (assumes nnode is odd)

kn=2*mm+deltt*kk; % compute effective system matrix

for it=1:ntime % start loop for time integration

fn=deltt*ff+(2*mm-deltt*kk)*fsol; % compute effective column vector

[kn,fn]=feaplyc2(kn,fn,bcdof,bcval); % apply boundary condition

fsol=kn\fn; % solve the matrix equation

sol(1,it+1)=fsol((nnode+1)/2); %

end

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

% analytical solution at node 8

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

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

% plot the solution at nodes 8 and 9

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

time=0:deltt:ntime*deltt;

plot(time,sol(1,:));

xlabel(’Time’)

ylabel(’Solution at nodes’)

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

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)



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

% 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 [m]=felpt2r4(xleng,yleng)

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

% Purpose:

% element matrix of transient term for two-dimensional Laplace’s

% equation using four-node bilinear rectangular element

%

% Synopsis:

% [m]=felpt2r4(xleng,yleng)

%


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



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

% element matrix

m=(xleng*yleng/36)*[4 2 1 2;

2 4 2 1;

1 2 4 2;

2 1 2 4];

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);

for j=1:edof

jj=index(j);

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

end

end

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)

%


% kk - system matrix before applying constraints

% 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:ndof

k=k+1;

index(k)=start+j;

end

end

me471_hwk05ans

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