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Assembly Process

We will now begin to look at the fundamentals of finite element programming. Since the ideas usedin 1D problems (ODEs) are applied to problems in 2D and 3D, we can continue to use our modelproblem as a basis of discussion.

Review of Model Problem. Suppose that a, b , f are given functions and consider the ordinarydifferential equation: Find u ∈ V p = {w ∈ C 2([0, 1]) : w(0) = p} such that

−(au) + bu = f, 0 < x < 1, = d/dx, (au + cu)|x=1 = q

where p, q, c are given constants. A boundary condition like the one at x = 1 with c = 0 is calleda natural boundary condition for this ODE (for example, if u is the temperature in a wall and athe wall conductivity au(1) is the heat flux into the wall). I will present here an approach to theassembly process which starts with the global weak form representation of the solution. Anotherapproach is possible starting from a weak form developed for an individual element.

Connectivity Matrix. We assume that the basic interval [0,1] is divided into subintervals or elementsusing the grid (or mesh or node) points x1 = 0 < x2 < . . . , xne+1 = 1. In our text, these nodesare generally stored in an array called gcoord. We denote the left and right end points (nodes) of

a typical element I e = [xe1, x

e2]. The correspondence between element end points and grid points is

simply xe1 = xe, xe2 = xe+1. This can get slightly more complex for elements with interior points so

the correspondence is often stored in a ne × k connectivity matrix, say C , where n is the number of elements, k is the number of points per element, and C (e, i) is the global node number the ith nodeof element e. For the current problem (using linear elements)

Connectivity Matrix, 1D linear elements: C =

1 22 3...

ne ne + 1

Our text stores the connectivity matrix in an array generally called nodes. In two and three dimen-sions the geometry is more complex. For example, if our domain is the rectangle [0, 1] × [0, 1], we

may create a grid using lines parallel to the axes: x1 = 0 < x2 < . . . , xn+1 = 1, y1 = 0 < y2 <. . . , ym+1 = 1 . The elements in this case can be either ne = n ·m rectangles or ne = 2n ·m triangles.After numbering these elements in a suitable way, the connectivity matrix will have ne rows andeither 4 or 3 columns.

Review of Weak Formulation. We start from the weak form of the equation. To find the weakform, we proceed in the usual fashion. We define the residual r = −(au) + bu − f , and note thatif u ∈ V p is a solution of the differential equation, and w ∈ V 0 = {w ∈ C

2([a, b]) : w(0) = 0}

10

rwdx = 0.

Noting that (au)w = (auw) − auw, we find

A(u, w) =

1

0

(auw + buw) dx =

1

0

fwdx + (q − cu)w|x=1 = (f, w) + (q − cu(1))w(1).

where the right hand boundary condition has been used.

Numerical Solution. The above expression holds for all test functions w and is equvalent to theoriginal equation for smooth functions. We generate a numerical approximation as follows: Let V 0,n

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be a linear subspace of V 0 of finite dimension n we seek an approximate (or trial) solution u of theform û+v with û ∈ V p a fixed function satisfying the boundary condition at x = 0 and v is an elementof V 0,n. At the same time, we take the test functions to lie in V 0,n. (This is the Galerkin method. In1D, the dimension n of the approximating subspace will be ne in cases where the solution functionis specified at one point of the interval. If two natural boundary conditions are given, n = ne + 1.)For the present, we restrict considerations to the case of continuous piecewise linear functions. This

means that in each element I e = [xe1, x

e2] (where for linear elements x

ej = xe+j−1, j = 1, 2), the

restriction of the trial function to I e is given by u(x)|I e = ue1H

e1 (x) + u

e2H

e2 (x), with

H e1 (x) = (xe2 − x)/he,

H e2(x) = (x − xe1)/he,

where he = xe2−x

e1 is the length of the element, and the H

es are called element shape functions. From

the global point of view, the trial function u can be expressed in the form u = ne+1

j=1 φjuj, where

φj are the tent functions, φj = H j1 on element j , φj = H

j−12 on element j − 1, and φj = 0 otherwise

(u1 = p is given). Also from this point of view, the requirement that the weak form hold for all testfunctions vanishing at x = 0 is equivalent the requirement that it hold with w = φi, i = 2, . . . , n .To make the book keeping somewhat simpler, we will temporalily ignore the fact that w = φ1 is notallowable (and that also that u1 = p is already known). Thus, we’ll allow all possible φi for w. Inthis case, we can write the ne + 1 linear equations obtained by substituting w = φi into the weak

form as follows 10

aΦ(Φ)T + bΦΦT dx

U =

10

f Φ dx + au(1)Φ(1),

where Φ = [φ1, φ2, . . . , φne+1]T , U = [u1, u2, . . . , une+1]

T , and of course, Φ(1) = [0, 0, . . . , 0, 1]T ,and au(1) = −cune+1 + q . This system of equations can be written in the form KU = F whereK is called the (global) stiffness or system matrix and F is called the (global) load vector.Remember that the system K U = F cannot be inverted to give U = K −1F since the first equation(corresponding to w = φ1) must first be eliminated.

Element Stiffness and Load Matricies. Using the fact that 10

g dx = ne

e=1

I e

g dx for anyfunction g , we write the weak form as

nee=1

I e

aΦ

(Φ

)T

+ bΦΦT

dx

U =

nee=1

I e

f Φ dx + au

(1)Φ(1)

When Φ is restricted to element I e (we write Φ|I e

for this matrix) only two of the tent functions arenon-zero: φe = H e1 and φe+1 = H

e2 , that is

Φ|I e =

0...0

H e1H e2

0...

...

← component e← component e + 1

...

We see that on a single element it is possible to focus on a 2 by 2 submatrix of the global n × nmatricies.

The contribution to the stiffness matrix from element I e is the 2 by 2 submatrix with components

K ers =

xe2

xe1

(a(H er )(H es )

+ bH erH es ) dx, e = 1, . . . , ne, r,s = 1, 2,

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where, remaining consistant with our definition of the global stiffenss matrix K , we have ignoredthe fact that in element 1 the test function w must vanish at x = x1 = 0 so that w = H

11 is not

admissable. The advantage of doing this is clear – all element stiffnesses are the same. At the endof the computation, we will account for our inclusion of this additional term. The contribution tothe load vector from element e is

F e

r = xe2xe1

f H e

r dx, e = 1, . . . , ne, r = 1, 2,

where we have not included the additional contribution au(1)δe,neδr, 2 so as to have the same formfor all the element load components. This omission will have to be corrected at the conclusion of computation prior to solving for the unknown trial function components uj . In matrix form

K e =

I e

aH e(H T e )

+ bH eH T e dx

, F e =

I e

fH e dx, e = 1, . . . , ne

where H e = [H e1 , H

e2 ]

T .

The element stiffness and load matricies can be computed independently of one another. For exam-ple, if the coefficients a, b and f are all constant, then

K e

=

a

he 1 −1−1 1

+

bhe

6 2 1

1 2

, F e

= f

he

2 1

1

Let us recapitulate the two simplifying modifications we have introduced. First, we have ignored therestriction that the test function is required to vanish at x = x1 = 0 in element 1. The motivation forthis was to gain uniformity in the definition of the element stiffness matricies. Second, in element newe have omitted the additional contribution to the load vector arising from the boundary condition –again to obtain uniformity in the definition of the load vector components. We can summarize thesetwo modifications by saying that we have ignored all boundary conditions. After constructing theglobal matricies K, F , we will introduce the necessary corrections to insure the boundary conditionsare properly imposed.

Assembly Process and Matrix Inflation. To compute the global stiffness matrix, we first have toexpand or inflate the 2 by 2 element matricies back size (ne + 1) × (ne + 1) by padding with zeros.

Then we construct the overall stiffness or system matrix simply by adding together the inflatedmatricies – recall that the global matrices are just the sums of the (expanded) element local stiffnessmatrices. A similar procedure is used with the 2 by 1 load matrices which are inflated to ( ne + 1)×1vectors and summed to obtain the global load vector.

The process of inflating the element matricies and adding to the global or system matricies is muchsimpler then it sounds. We simply loop through all the elements, determining for each element whatglobal nodes correspond to its local nodes. Determination of how the local stiffness and load vectorsof individual elements contribute to the global matrices can easily be made using the connectivitymatrix. We see, for example, that local node 1 of element 2 corresponds to global node 2 and localnode 2 corresponds to global node 3. This shows that F 21 should be added to F 2 and F

22 should be

added to F 3. Similarily, K 211 should be added to K 22, K

212 should be added to K 23, K

222 to K 33,

and K 221 to K 32. Proceding in a similar fashion we find the global or system equations are (we take

n = ne + 1 in the equations below).

K 111u1 + K 112u2 = F

1

K 121u1 + (K 122 + K

211)u2 + K

212u3 = F

12 + F

21

K 221u2 + (K 222 + K

311)u3 + K

312u4 = F

22 + F

31

... =...

K n21un−1 + K n22un = F

n2

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or in matrix form

K 111 K 112 0 0 · · · 0

K 121 (K 122 + K

211) K

212 0 · · · 0

0 K 221 (K 222 + K

311) K

312 · · · 0

... · · ·...

0 0 0 0 · · · K n22

u1u2u3...

un

=

F 11F 12 + F

21

F 22 + F 31

...

F n2

The the loop for assembly of the stiffness and load vectors is programmed in the m-file belowborrowed from our text. Before examining this file, lets look at the computation of the matrixindex which is used in the assembly loop. Again this is provided in a file from our text. I haveprovided some additional comments intended to clarify the program.

function [index]=feeldof1(iel,nnel,ndof)

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

% Purpose:

% Compute system dofs associated with each element in one-

% dimensional problem

%

% Synopsis:% [index]=feeldof1(iel,nnel,ndof)

%

% Variable Description:

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

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

% nnel - number of nodes per element (nnel=2 for linear elements)

% ndof - number of dofs per node (ndof=1 for scalar problems)

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

edof = nnel*ndof; % =2 for linear elements, scalar problems

start = (iel-1)*(nnel-1)*ndof; % =iel-1 for linear elements, scalar problems

for i=1:edof

index(i)=start+i;

end

% for linear elements, scalar problems: index(1)=iel, index(2)=iel+1

Now we’re ready for the assembly loop. Again I’ve added some extra comments to the program fromthe text.

function [kk,ff]=feasmbl2(kk,ff,k,f,index)

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

% Purpose:% Assembly of element matrices into the system matrix &

% Assembly of element vectors into the system vector

%

% Synopsis:

% [kk,ff]=feasmbl2(kk,ff,k,f,index)

%

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% kk - system matrix

% ff - system vector

% k - element matrix

% f - element vector

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

%-----------------------------------------------------------edof = length(index); % =2 for linear, scalar problems

for i=1:edof

ii=index(i); % index=[iel,iel+1] for element iel

ff(ii)=ff(ii)+f(i); %[ff(iel);ff(iel+1)]=[f(1);f(2)]

for j=1:edof

jj=index(j);

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

end

end

For the equation we’re considering, the index matrix for element I e would be [e, e+1], so the looplimits would be 1:2.

Imposition of Boundary Conditions. The global stiffness and load matrices must now be modifiedso as to account for the boundary conditions which we have neglected up to this point. We will treatthe left and right hand conditions separately.

Left Boundary. First, we note that the test function w = φ1 should not have been used inK , and this means that the first row of K should be deleted. Second, note that u1 = p isgiven so that all the terms ki1u1 = ki1 p, i = 2, . . . , ne should be moved from the left side of the equation K U = F and added the the ith components of the load vector on the right side.These two modifications would change K to an ne × ne matrix, but would entail modificationof all load components. The same end can be achieved in a much simpler way as follows

1. Zero out row 1 of K , then set k11 = 1.

2. Redefine F 1 = p.

3. Now K remains (ne + 1) × (ne + 1), and the first equation becomes u1 = p, i.e., thisequation enforces the boundary condition at x = 0.

Right Boundary. Now, it is the final component of the load vector which must be modified.We have omitted the terms q − cu(1) = q − cun

efrom F n

e. The remedy is straight forward,

and falls into two categories.

c = 0 In this case we simply use the replacement F ne→ F n

e+ q .

c = 0 Now we must not only use the modification above F ne → F ne + q , but also must modifyknene as follows knene → knene + c

We are now ready to view some of the Text book files

function ex351a(n)

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

% EX3.5.1 -- Modified version of Text file

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% to solve the ordinary differential equation given as

% a u’’ + b u’ + c u = 1, 0 < x < 1

% u(0) = 0 and u(1) = 0

% using n linear elements

%

% Variable descriptions

% k = element matrix% f = element vector

% kk = system matrix

% ff = system vector

% 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

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

nel=n; % number of elements

nnel=2; % number of nodes per element

ndof=1; % number of dofs per node

nnode=nel+1; % total number of nodes in system

sdof=nnode*ndof; % total system dofs

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

% input data for nodal coordinate values

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

gcoord=linspace(0,1,nel+1);

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

% input data for nodal connectivity for each element

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

nodes=[[1:nel]’ [2:nel+1]’];

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

% input data for coefficients of the ODE

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

acoef=1; % coefficient ’a’ of the diff eqn

bcoef=-3; % coefficient ’b’ of the diff eqn

ccoef=2; % coefficient ’c’ of the diff eqn

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

% input data for boundary conditions

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

bcdof(1)=1; % first node is constrained

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bcval(1)=0; % whose described value is 0

bcdof(2)=nel+1; % last node is constrained

bcval(2)=0; % whose described value is 0

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

% initialization of matrices and vectors

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

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

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

nl=nodes(iel,1); nr=nodes(iel,2); % extract nodes for (iel)-th element

xl=gcoord(nl); xr=gcoord(nr);% extract nodal coord values for the elementeleng=xr-xl; % element length

index=feeldof1(iel,nnel,ndof);% extract system dofs associated with element

k=feode2l(acoef,bcoef,ccoef,eleng); % compute element matrix

f=fef1l(xl,xr); % compute element vector

[kk,ff]=feasmbl2(kk,ff,k,f,index); % assemble element matrices and vectors

end

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

% Global stiffness and load matrices have been assembled

% now apply boundary conditions

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

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

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

% solve the matrix equation

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

fsol=kk\ff;

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

% analytical solution

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

c1=0.5/exp(1);

c2=-0.5*(1+1/exp(1));

for i=1:nnode

x=gcoord(i);

esol(i)=c1*exp(2*x)+c2*exp(x)+1/2;

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end

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

% print both exact and fem solutions

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

fprintf(’%10s %10s %10s %10s\n’,’x’, ’fem soln’,’exac soln’,’error’);fprintf(’%10.2f %10.6f %10.6f %10.6f\n’,[gcoord; fsol’; esol; abs(fsol’-esol)]);

plot(gcoord,fsol’,’k’, gcoord, esol, ’ko’);

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

function [index]=feeldof1(iel,nnel,ndof)

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

% Purpose:

% Compute system dofs associated with each element in one-

% dimensional problem

%% Synopsis:

% [index]=feeldof1(iel,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;

start = (iel-1)*(nnel-1)*ndof;

for i=1:edof

index(i)=start+i;

end

function [k]=feode2l(acoef,bcoef,ccoef,eleng)

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

% Purpose:

% element matrix for (a u’’ + b u’ + c u)

% using linear element

%

% Synopsis:

% [k]=feode2l(acoef,bcoef,ccoef,eleng)

%


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% k - element matrix (size of 2x2)

% acoef - coefficient of the second order derivative term

% bcoef - coefficient of the first order derivative term

% ccoef - coefficient of the zero-th order derivative term

% eleng - element length

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

% element matrix

a1=-(acoef/eleng); a2=bcoef/2; a3=ccoef*eleng/6;

% k=a1[1 -1; -1 1]+a2[-1 1;-1 1]+a3[2 1;1 2]

k=[ a1-a2+2*a3 -a1+a2+a3;...

-a1-a2+a3 a1+a2+2*a3];

function [f]=fef1l(xl,xr)

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

% Purpose:

% element vector for f(x)=1% using linear element

%

% Synopsis:

% [f]=fef1l(xl,xr)

%


% f - element vector (size of 2x1)

% xl - coordinate value of the left node

% xr - coordinate value of the right node

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

% element vector

eleng=xr-xl; % element length

f=[ eleng/2; eleng/2];

function [kk,ff]=feasmbl2(kk,ff,k,f,index)

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

% Purpose:

% Assembly of element matrices into the system matrix &

% Assembly of element vectors into the system vector

%

% Synopsis:

% [kk,ff]=feasmbl2(kk,ff,k,f,index)

%


% kk - system matrix

% ff - system vector

% k - element matrix

% f - element vector

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

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

edof = length(index);

for i=1:edof

ii=index(i);ff(ii)=ff(ii)+f(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 containing constrained d.o.f

% bcval - a vector containing contained value

%

% For example, for homogeneous BC at x=0, 1,

% bcdof=[1, ne+1], bcval=[0,0]

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

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

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