Fundamental-Finite-Element-Analysis-and-Applications

ASSEMBLY OF ELEMENT EQUATIONS 33

The pressure will be accounted for in elements 2 and 4. Choosing node 4 as the first
node for element 2, the pressure 'term will be applied to side 1 of the element. The normal
pressure is in the direction opposite to the outer normal to the surface and therefore it must
be assigned a negative sign. For element 4, node 6 is chosen as the first node and thus the
. pressure on this element is also on side 1. The complete computations of element equations
and their assembly are as follows. All numerical quantities are in pounds and inches.

The equations for element 1 are as follows:

h = 0.25; E = 10000; v = 0.2 2083.33 01
104 16.7 10416.7
0
Plane stress constitutive matrix, C = [ 208~.33 o 4166.67

Nodal coordinates:

Element Node Global Node Number x Y

1 1 O. O.
2 3 2. O.
3 4 2. 1.5

xI = 0.;x2 = 2.;x3 = 2.
YI = 0.;Y2 = 0';Y3 = 1.5
Using these values, we get

bl=-1.5; b2 = 1.5; b3 = O.
cj =0.; c2 = -2.; <s = 2.

Element area, A = 1.5 o· 1

- 0.5 0 0.5 0 O. 0.666667
BT = 0 o. 0 0 O.
-0.666667
[ 0.·· -0.5 -0.666667 0.5 0.666667

Thus the element stiffness matrix is:

976.563 0 -976.563 260.417 0 -260.417
390.625 520.833 -390.625 -520.833 0
0 520.833 1671.01 -781.25 -694.444
-390.625 2126.74 260.417
lc = hABCB T :::; -976.563 -520.833 -781.25 520.833 -1736.11
260.417 -694.444 520.833 694.444
0 -1736.11 0
0 260.417 0 1736.11

-260.417

Complete element equations:

976.563 0 -976.563 260.417 0 -260.417 llj O.
0 390.625 520.833 -390.625 -520.833 0 O.
520.833 1671.01 -781.25 -694.444 vI O.
-976.563 -390.625 2126.74 260.417 ll3 O.
260.417 -520.833 -781.25 520-,833 -1736.11 v3 O.
0 -694.444 520.833 694.444 ll4 O.
0 -1736.11 0 v4
-260.417 260.417 0 1736.11

34 FINITE ELEMENTMETHOD:THE BIG PICTURE

The element contributes to (1,2,5,6,7,8) degrees of freedom.
1
2
5

Locations for element contributions to a global vector: 6

and to a global matrix: [1, 1] [1,2] [1,5] 7 [1, 8]
[2, 1] [2,2] [2, 5] 8 [2, 8]
[5, 1] [5,2] [5,5] [5,8]
[6, 1] [6,2] [6,5] [1, 6] [1,7] [6,8]
[2, 6] . [2,7]
[7, 1] [7,2] [7,5] [5,6] [5,7] [7,8]
[8, 1] [8,2] [8,5] [6,6] [6,7] [8,8]

[7,6] [7,7]
[8,6] [8,7]

Adding element equations into appropriate locations, we have

976.563 0 0 0 -976.563 260.417 0 -260.417 0 0 0 0 ul 0
0 0
0 390.625 0 0 520.833 -390.625 -520.833 0 0000 VI 0
0 0
000 0 00 0 0000 U2 0
-976.563 0
260.417 000 0 00 0 0000 v2 0
u3 0
o. 520.833 0 0 1671.01 -781.25 -694.444 260.417 0 0 0 0 v3 0
0
-260.417 -390.625 0 0 -781.25 2126.74 520.833 -1736.11 0 0 0 0 U4 0
0 0
0 -520.833 0 0 -694.444 520.833 694.444 0 0 0.' 0 0 V4
0
0 0 0 0 260.417 -1736.11 0 1736.11 0 0 0 0 u5
v5
000 0 00 0 0000
U6
000 0 00 0 0000
v6
000 0 00 0 0000

000 0 00 0 0000

The equations for element 2 are as fdllows: 2083.33 41J67 ]
10416.7
h = 0.25; E = 10000; v = 0.2
0
[10416.7

Plane stress constitutive matrix, C = 2083.33
.0

Nodal coordinates:

Element Node Global Node Number x Y

"1 4 2. 1.5
2 2 O. 2.
3 1 O. O.

Xl = 2,;x2 = 0';X3 = O.
Yl = 1.5; Y2 = 2.; Y3 = O.
Using these values, we get

b, =2.; b2 = - 1 . 5 ; b3 = -0.5
(;1 =0.; c2 = 2.; c3 = -2.

ASSEMBLY OF ELEMENT EQUATIONS 35

=Element area, A 2.

0.5 0 -0.3750 -0.125 -0o.5 J

B1' = 0 O. 0 0.5 ·0
[ O. 0.5 0.5 -0.375 -0.5 -0.125

Thus the element stiffne-ss--m~-a-tr-ix~-i-s ---~------~-----------------------~-------------------------------------------- -!

1302.08 0 -976.563 260.417 -325.521 -260.417
520.833 520.833 -390.625 -520.833 -130.208
0 520.833 1253.26 -585.938 -276.693
-390.625 1595.05 65.1042
k =hABCB1' = -976.563 -520.833 -585.938 325.521 -1204.43
260.417 -130.208 -276.693 325.521 602.214
-1204.43 195.313 195.313
-325.521 65.1042 1334.64

-260.417

Load vector due to distributed load on side 1 (nodes (4, 2)):
Specified load components: qn = -20; qt = 0
End nodal coordinates: ( {2., 1.5} (O., 2.}), giving side length L = 2.06155

Components of unit normal to the side: nx = 0.242536; ny = 0.970143

Using these values, we get

r~=(-1.25 -5. -1.25 -5. 0 0)

Complete element equations:

1302.08 0 -976.563 260.417 -325.521 -260.417 u4 -1.25
0 520.833 520.833 -390.625 -520.833 -130.208 -5.
520.833 1253:26 -585.938 -276.693 v4 -1.25
-976.563 -390.625 1595.05 65.1042 -5.
260.417 -520.833 -585_.938 325.521 -1204.43 Uz
-130.208 -276.693 325.521 602.214 Vz O.
-325.521 -1204.43 195.313 195.313 O.
-260.417 65.1042 1334.64 - ul

VI

The element contributes to {7,8, 3, 4, 1, 2} degrees of freedom.

7
8
3
Locations for element contributions to a global vector: 4

1
2

and to a global matrix: [7,7] [7, 8] [7,3] [7,4] [7, 1] [7,2]
[8,7] [8,8] [8,3] -[8,4] [8, 1] [8,2]

[3,7] [3,8] [3,3] [3,4]' [3, 1] [3,2]
[4,7] [4,8] [4,3] [4,4]. [4, 1] [4,2]
[1,7] [1,8] [1, 3] [1, 4] - [1, 1] [1,2]
[2,7] [2, 8] [2,3] [2,4] [2, 1] [2,2]

36 FINITE ELEMENTMETHOD:THE BIG PICTURE

Adding element equations into appropriate locations, we have

1578.78 195.313 -276.693 325.521 -976.563 260.417 -325.521 -781.25 0 0 0 0 1/, 0
0
195.313 1725.26 65.1042 -1204.43 520.833 -390.625 -781.25 -130.208 0 0 0 0 v, -1.25
-5.
-276.693 65.1042 1253.26 -585.938 0 0 -976.563 520.833 0 0 0 0 1/2
0
325.521 -1204.43 -585.938 1595.05 0 0 260.417 -390.625 0 0 0 0 v2 0
-1.25
-976.563 520.833 0 0 1671.01 -781.25 -694.444 260.417 0 0 0 0 1/3 -5.

260.417 -390.625 0 0 -781.25 2126.74 520.833 -1736.11 0 0 0 0 v3 0

-325.521 -781.25 -976.563 260.417 -694.444 520.833 1996.53 0 0000 1/4 0
v4
-781.25 -130.208 520.833 -390.625 260.417 -1736.11 0 2256.94 0 0 0 0 1/5 0
0
000 00 00 0 0000 "s

000 00 00 0 0000 1/6

000 00 00 0 0000 v6

000 00 00 0 0000

The remaining elements can be processed in exactly the same manner. After assembling
all elements, the global equations for the model are as follows:

1578.78 195.313 -276.693 325.521 -976.563 260.411 -325,521 -781.25 0 0 0 0 "I 0
19S.313 1725.26 65.1042 -1204.43 520.833 -390.625 -781.25 -130.20B 0 0 0 VI 0
-276.693 0 -976.563 0 0 0 .0
325.521 65.1042 1253,26 -585.938 0 0 520.833 0 0 0 0 1/2 -1.25
-976.563 -1204.43 -585.938 1595.05 312.5. 0 260.417 -390.625 -651.042 260.417 -325.521 0 '2 -5.
260.417 -520.833 -lI7I.8S 520.833 -260.417 -781.25 "3 0
-325.521 520.833 0 0 -520.833 4166.67 520.833 0 0 -651.042 -781.25
-781.25 -390.625 0 0 -1111.88 510.833 520.833 _3385.42 0 0 260.417 -130.208 ='3 0
-181.25 -976.563 260.417 -3385.42 3125. -520.833 169271 -781.25 -1041.67
0 -130.208 520.833 -390.625 520.833 520.833 -520.833 -781.25 2864.58 5~O.833 "4 -~5
0 0 0 -651.042 -260.417 4166.67 -1041.67 520.833 520.833 -260.417 v4 -10.
0 0 0 0 -781.25 0 0 260.417 -2604.17 2018.23 "5 0
0 0 0 260.417 -130.208 0 0 260.417 v5 0
0 0 0 0 -325.52J -651JM2 0 -2604.17 "6 -1.25
0 -781.25 520.833 260,417 v6 -5.
-260.417 0
2994.79

• MathematicafMATLAB Implementation 1.7 on the Book Web Site:
Plane stress example assembly

1.3 BOUNDARY CONDITIONS AND NODAL SOLUTION
;'

A variety of boundary conditions were specified for the problems considered in the previ-
ous two sections. For the truss problem, some nodes were located at the supports, which
means that the displacements at these nodes must be zero. For the heat flow problem con-
ditions to be satisfied along the boundaries involved specified temperature, convection,
insulation, or heat flux. For the stress analysis problem, on some surfaces loading was.
applied while other surfaces were attached to fixed supports.

In the finite element analysis, some of the boundary conditions are incorporated directly
into the element equations. These are known as the natural boundary' conditions. Examples
of natural boundary conditions encountered in the previous sections are convection, insu-
lation, and heat flux boundary conditions for the heat flow problem and applied surface
loading for the structural problems. A review of element equations and assembly proce-
dures discussed in the previous sections should make it clear that these types of boundary
conditions have been taken into account while formulating and assembling the element
equations ..

The boundary conditions that are not incorporated directly into the element equations
are known as the essential boundary conditions. For heat flow problems these involve
specified temperatures along some boundaries. For stress analysis problems nodal sup-

BOUNDARY CONDITIONS AND NODAL SOLUTION 37

purts represent essential boundary conditions. Once again, a review of element equations

and assembly procedures discussed in the previous sections should clearly indicate that

these types of boundary conditions have not been taken into account while formulating

and assembling the element equations. .

Precise reasoning for splitting the boundary conditions into two categories-natural and

essential-will become clear after studying the mathematical foundations of the finite el-

ement method in the following chapter. At this stage our main concern is to see how to

actually incorporate boundary conditions into the finite element equations. Obviously we

must take care of all applicable boundary conditions when solving a given problem. We

start with the element equations and their assembly. The boundary conditions that can be

incorporated through the element equations are handled first. The remaining boundary con-

ditions are the essential boundary conditions. that are taken into account using procedures

discussed in this section.

In the assembly process it is assumed that all nodal degrees of freedom (dot) are un-

known, However, due to essential boundary conditions, some of these degrees of freedom

may have a zero or some other specified value. Therefore, introduction of essential bound-

ary conditions into the global equations involves substituting the known values into the

vector of nodal variables and malting appropriate changes to the equations.'

There are three methods for accomplishing this. The first method involves rearrange-

ment of equations but has an advantage that the final system of equations is smaller than

the global system. The second method keeps the original order of variables but does re-

quire several modifications to the equations. The third method is an approximate method

but requires the fewest changes to the equations. On some computers rearranging a large

system of equations is very inefficient and hence the approximate method may be more

economical. For hand calculations obviously the first method is preferred. Most examples

presented in this text use this first approach. The details of these methods follow.

1.3.1 Essential Boundary Conditions by Rearranging Equations

As an illustration, assume that a finite element model from a heat flow problem results in
the following global systerrrof equations: .

1000 10 20 30 40 50 100
10 2000 21 31 41 51
20 21 3000 32 42 52 110
30 31 32 4000 43 53 120
40 41 42 43 5000 54 130
50 51 52 53 54 6000 140
150

Consider the situation when some of the nodes are located on sides over which the tem-

= = =peratures are known. As an illustration, say T1 5, T4 -7, and Ts O. Thus we only

have three nodal unknown temperatures remaining, namely 0.,7;, and T6• These three un-

knowns can be obtained by solving the corresponding equations, 2, 3, and 6. Introducing

the known values in the nodal degrees of freedom vector and retaining only the equations

Since there are only three unknowns, these equations can be rearranged by moving the
known values to the right-hand side. The first column is multiplied by 5, the fourth column
by -7, and the fifth column by 0, and hence we get

)[T202010 2 + 5 [1200) 7 [3312) + 0 [4412)

[ 51 T3 )
320100 5512 - = [111200)
52 6000
T6 50 53 54 150

Moving all constant vectors to the right-hand side, we have

)[T[ 2 =[111200) + [--15000) + [221274)
220010 320100 5521 )
51 52 6000 T3

T6 150 . -250 371

Adding all terms on the right-hand-side, the final 3 X 3 system of equations is as follows:

)[T[ 2 = [ 227474)
220010 320100 5521 )
51 152 6000 T3

T6 271

The solution of this system of equations gives

(T2 -? 0.136559, T3 -? 0.0796266, T6 -? 0.0433158)

Note that in these computations, when the specified value is zero, nothing gets added to the
right-hand side. Thus imposing a zero value for a nodal degree of freedom simply requires
removing both the corresponding row and the column from the system of equations. In this
case it is not necessary to even assemble the corresponding rows and columns and therefore
such degrees of freedom can be eliminated even before the assembly process begins. This
is an important observation to save computational effort, especially for structural problems
where zero nodal displacements are common due to supports.

You may be curious as to why we retained equations 2, 3, and 6 and removed the other
three. We have three unknowns left but why can we not use any three equationsto solve for
the three remaining unknowns? For a precise answer to this question one needs to study
the mathematical basis of the finite element method presented in the following chapter.
However, a simple reasoning based on physical grounds is as follows. When a temperature

BOUNDARY CONDITIONS AND NODAL SOLUTION 39

is specified at a node, there must be a corresponding unknown heat source at that node
to maintain that temperature. Similarly, for a structural problem, when a displacement is

specified at a node, there actually is an unknown reaction corresponding to this degree of

freedom. Thus, when we insert known values into the nodal variables vector, for equa-
tions to remain consistent, unknown heat source or reaction terms must be inserted into

the right-hand-side vector for every specified nodal degree of freedom. Denoting theses

unknowns as R1, R4, and Rs for the illustrative example, a consistent system of equations

after introducing given essential boundary conditions is as follows:

1000 10 20 30 40 50 5 R1 + 100
10 2000 21 31 41 51 110
20 21 3000 32 42 52 t;
30 31 32 4000 43 53 120
40 41 42 43 5000 54 T3
50 51 52 53 54 6000 sR,4 + 130
-7 + 140

0 150

T6

This system represents six equations in six unknowns, and if we want, we can solve these
equations directly for the six unknowns. However, since we do not need the newly intro-
duced unknowns, it is advantageous to reduce the system of equations to three for com-
puting the three unknown temperatures. However, the only way to solve for the remaining
nodal unknowns is to use equations 2, 3, and 6. Any other combination of three equations
will not yield the desired result. The first three equations, as an example, will have four

unknowns Rp Tz, T3, and T6 and thus are not suitable for the required solution .

• MathematicafMATIAB Implementation 1.8 on the Book Web Site:
Imposing essential boundary conditions

1.3.2 Essential Boundary Conditions by Modifying Equations

The method ofrearrangement of equations to handle essential boundary conditions is at-
tractive because the final system of equations is smaller than the global system. However,
on some computers rearranging a large system of equations is very inefficient, and thus it
is desirable to develop a method of incorporating essential boundary conditions that does
not require rearrangement of equations.

A simple alternative is to modify each equation that corresponds to a specified nodal
value as follows:

(i) Insert 1 at the diagonal location and zero at all off-diagonal locations in the row
corresponding to the specified degree of freedom.

(ii) Insert the known nodal value to the corresponding location in the right-hand side.

Using this approach for the illustrative example used in the previous section, we get the
following system of global equations:

40 FINITE ELEMENT METHOD: THE BIG PICTURE

10 o 00 o t; 5
t; 110
10 2000 21 31 41 51
T3 :::: 120
20 21 3000 32 42 52 T4 -7
Ts 0
o0 o 10 o T6 150

o0 001 o
52 53 54
50 51 6000

Solving this system of equations for all six nodal values gives the following solution that
is exactly the same as before:

{Tl :::: 5., Tz :::: 0.136559, T3 :::: 0.0796266, T4 :::: -7., Ts :::: 0, T6 :::: 0.0433l58}

One drawback of the procedure so far is that the resulting coefficient matrix is not sym-
metric anymore. For large problems this is a serious drawback. It is possible to rectify the
situation by making the corresponding column entries 0 and subtracting the products of
terms in these columns with the known values from the right-hand side in a manner similar
to that done in the previous case:

1 0 000 0 T1 5 0 0 05
0 2000 21 0 0 51 t; 110 10 31 41 277
0 21 3000 0 0 52
0 0 0 10 0 T3 :::: 120 -5x 20 ': (-7) x 32 -Ox 42 :::: 244
0 0 001 0 T4 -7 0 0 0 -7
0 51 52 0 0 6000
Ts 0 0 0 00

T6 150 50 53 54 271

{Tl :::: 5., Tz :::: 0.136559, T3 :::: 0.0796266, T4 :::: -7., Ts :::: 0, T6 :::: 0.0433158}

For imposing zero values of the nodal variables the process works very well because no

modifications to the right-hand side ~re necessary. However, for nonzero values the proce-

dure loses some of its simplicity. The following procedure, though approximate, is much

simpler to implement.

1.3.3 Approximate Treatment of Essent.ial Boundary Conditions

The previous two methods may. be inefficient to implement on some computers because
they require several modifications to the system of equations. A simple but approximate
alternative is to modify each equation corresponding to a specified nodal value as follows:

(i) Insert a large number at the diagonal corresponding to the known degree of free-
dom.

(ii) Add a value equal to this large number times the known nodal value to the right-
hand side.

Using this approach for the illustrative example used in the previous section and with 1010
as an arbitrarily chosen large number, we get the following system of global equations:

BOUNDARY CONDITIONS AND NODAL SOLUTION 41

1010 10 20 30 40 50 T! 5 X 1010
10 2000 21
20 21 3000 31 41 51 Tz no
30 31 32
40 41 42 32 42 52 13 120
50 51 52 1010 43 53 -7xlO lO
T4
43 1010 54 Ts '0
T6
53 54 6000 150

Solving this system of equations for all six nodal values gives the following solution that
is essentially the same as before:

(T! = 5., Tz = 0.136559, T3 = 0.0796266, T4 = -7., Ts = 8.97177x10-9, T6 = 0.0433158)

To see the reason why this procedure works, consider the first equation of the modified
system:

,'

10000000000T! + lOTz + 20T3 + 30T4 + 40Ts + 50T6 = 50000000000

The two large terms dominate the equation and contribution of remaining terms is negligi-
ble. Thus the equation is effectively

10000000000T! '" 50000000000

giving

=Similar
Ts = O.
reasoning shows that the fourth equation gives T4 -7 and the fifth equation
The other equations give the remaining unknowns. The method gives reasonable

answers as long as the large number added to the diagonal is at least an order of magnitude

greater than the largest element in the coefficient matrix. Using the number 106 instead of

1010, we get the following solution that shows little more error in the specified values but

may still be considered reasonable:

(T! =5.0002, Tz = 0.136559, T3 =0.0796257,
14 =-7.56016, 15 =0.0000897165, T6 = 0.0433147)

1.3.4 Computation of Reactions to Verify Overall Equilibrium

For structural problems, reactions at the supports provide a quick and simple check on the
validity of the analysis. For these problems the finite element equations represent equilib-
rium equations. Thus for overall equilibrium the sum of applied loads must be equal and
opposite to the sum of all reactions at the supports. Common blunders, such as employing
wrong or inconsistent units or careless computational errors, can easily be discovered if
reactions are available.

The concept of equilibrium is also applicable to heat flowproblems where the total heat
flowing into the body must be equal to that going out of the body.

After solving for the nodal unknowns, the reactions can be computed by using the equa-
tions that were deleted when the essential boundary conditions were being imposed. Thus

42 FINITE ELEMENTMETHOD:THE BIG PICTURE

for the illustrative example the reactions can be computed from the first, fourth, and fifth
equations. Retaining these rows and inserting all nodal values, the reactions are obtained
from the following computation:

5

0.136559
(1000 50] (R' 100]30

40
10 20 30 40 0.0796266 = R4 ++ 130
31 32 4000 43 53 -7
41 42 43 5000 54 0 Rs + 140

0.0433158

Rearranging and performing computations, we get the following reactions:

5

1~ ;~J( 0.136559
10 20 30
RR4I ) = [.103000 31 32 4000 0.0796266 - 110300) = (4-2679957.01.29 )
Rs 40 41 42 43 140
5000 54 -7 [ -229.718

o

0.0433158

The following examples illustrate the use of reactions in validating solution consistency.

• MathematicalMATLAB Implementation 1.9 on the Book Web Site:
Imposing essential boundary conditions

Example 1.7 Five-Bar Truss The following global equations for five-bar plane truss
were developed in Example 1.4. Incorporate essential boundary conditions into the system
and obtain the final reduced system/of equations in terms of the remaining unknowns.
Solve for nodal values. Compute reactions and verify overall equilibrium.

32600.2 76067.2 -32600.2 -76067.2 0 0 0 0 III 0
76067.2 297490. -76067.2 -177490. 0 -120000 0 0 VI 0
-32600.2 -76067.2 243089. -32998.3 -177490. -76067.2 112 0
-76067.2 -177490. 119136. 119136. 32998.3 32998.3 -76067.2 -32600.2 v2 -150000.
-32998.3 243089. 152998. -32998.3 -120000 0 113 0
0 0 -32998.3 -32998.3 0 0 v3 0
0 -120000 32998.3 32998.3 ..:..120000 152998. 297490. 76067.2 114 0
0 -177490. -32998.3 0 76067.2 32600.2 v4 0
0 0 -76067.2 -76067.2 0
0 -32600.2 0

Nodes 1 and 4 of the truss have pin supports. Thus the essential boundary conditions are
as follows:

Node dof Value

1 ul 0
0
VI

4 u4 0
v4 0

BOUNDARY CONDITIONS AND NODAL SOLUTION 43

After deleting equations (1, 2, 7, 8), we have

[ -32600.2 -76067.2 243089. 119136. -32998.3 32998.3 -177490. 0
-76067.2 -177490. 119136. 243089. 32998.3 -32998.3 -76067.2
-32998.3 .32998.3 152998. -32998.3 0
0 0 32998.3 -32998.3 152998. -120000
-120000 -32998.3 0 =[ -15~~]-76067.2] ll2
0
-32600.2 v2
0 ll3
0 v3
0

0

Since all entries in columns (1,2,7, 8) are multiplied by zero nodal values, they can be /'

"

removed. The final global system of equations is thus as follows: II {j ~.'
'{J 'y:P;'i' -

243089. 119136. -32998.3 32998.3] [U2] = [ 0 ]
119136. 243089. 32998.3 -32998.3 v2 -150000.
[ -32998.3 32998.3 152998.
32998.3 -32998.3 -32998.3 u3 0
-32998.3
152998. v3 0

Solving the final system of global equations, we get the following nodal values:

(U2 = 0.538954, v2 = -0.953061, u3 = 0.264704, v3 = -0.264704)

Complete Table of Nodal Values

_U v

10 0
2 0.538954 -0.953061
3 0.264704 -0.264704
40 0

Computation of Reactions

Equation numbers of dof with specified values: (1,2,7, 8)
Extracting equations (1,2,7, 8) from the global system, we have

[3726600607..22 76067.2 -32600.2 -76067.2 0 0 0 llj
0 297490. -76067.2 -177490. 0 -120000 0
0 -177490. -76067.2 -120000 297490. VI
0 -76067.2 0 0 76067.2
0 -32600.2 0 7606t] +0]ll2
[R'V2 _ R2 +0.
ll3 - R3 +0.

32600.2 v3 R4 +0.

ll4

v4

44 FINITEELEMENTMETHOD:THE BIG PICTURE

Substituting the nodal values and rearranging,

RR21] _ [3726600607..22 76067.2 -32600.2 -76067.2 o oo o
-76067.2 -177490. o -120000 o o
297490. -177490. -76067.2 o 297490.
[ R3 - 0 -76067.2 -120000 7606~'2] 0.538954
o -32600.2 -0.953061
R4 0 o o o 76067.2 0.264704
32600.2 -0.264704

o
o

Carrying out computations, the reactions are as follows:

Label dof Reaction

RI ul 54926.7
R2 vI 159927.
R3 u4 -54926.7
R4 v 4 -9926.67

Sum of reactions:

dof: u 0
dof: v 150000.

There is no applied load in the x direction. Since the sum of reactions in the horizontal
direction is zero, the equilibrium is satisfied in this direction. There is an applied load of
150000 in the -y direction which balances with the sum of reactions in the y direction,
indicating that equilibrium is satisfied in this direction as well. Hence the solution satisfies
the overall equilibrium.

/

Example 1.8 Heat Flow through a Square Duct The following global equations for
heat flow through a square duct are developed in Example 1.5. Incorporate essential bound-
ary conditions into the system and obtain the final reduced system of equations in terms
of the remaining unknowns. Solve for nodal values. Compute reactions and verify overall
energy balance:

-07] T'] 8~]1.4 0
0 -0.7

0 4.56667 1.58333 0 -2.1 T2
0 1.58333 3.51667 0.35 -1.4 T3 = 81.

-0.7 0 0.35 3.15 -2.8 T4 0

-0.7 -2.1 -1.4 -2.8 7. Ts 0

Nodes 1 and 4 are on the inside face that is maintained at a temperature of 300·C. Thus the
essential boundary conditions are as follows:

Node dof Value

1 300
4 300

BOUNDARY CONDITIONS AND NODALSOLUTION 45

= =Incorporating these boundary conditions means setting TI T4 300 and removing equa-

tions 1 and 4 from the global system:

300

( 00 4.56667 1.58333 0 ~1)-2.1)Tz ['1
-0.7 1.58333 3.51667 0.35
-2.1 -1.4 -2.8 -1.4 T3 =
7 300

Ts

Multiplying the first column by 300 and the fourth column by 300, the equations can be
written as follows:

-2.1)(T [0) (0) [81)41..5568636373 31..5518636373 -1.4z)

( -2.1 -1.4 7 T3
+ 300 0 + 300 0.35 = 81

Ts -0.7 -2.8 0

Moving constant vectors to the right-hand side, we get the final system of equations as
follows:

-2.1)(T4.56667 1.58333 z) = [ 81. )
1.58333 3.51667 -1.4 -24.
( -2.1 T3

-1.4 7. Ts 1050.

Solving the final system of global equations, we get the following nodal values:

(Tz = 93.5466, T3 = 23.8437, t; = 182.833}

Complete Table of Nodal Values

T

I 300
2 93.5466
3 23.8437
4 300
5 182.833

Computation of Reactions

Equation numbers of dof with specified values: (1, 4)
Extracting these equations from the global system (the one prior to adjustment for es-
sential boundary conditions), we have

1.4 0 0 -0.7
( -0.7 0 0.35 3.15

46 FINITE ELEMENTMETHOD:THE BIG PICTURE
Substituting the nodal values and rearranging,

)=((R1 -10..47 o0 -0.7 -07) 933.504066]
R2 o 0.35 3.15 -2'8 23.8437

. 300 .

182.833

Table of reactions:

Label dof Reaction

s, T1 82.0171

R2 T4 231.414

Sum of reactions: 313.431

The sum of reactions represents the total heat energy put into the system. This must be
balanced by the heat loss due to convection. The convection heat loss takes place only on
side 2-3 of the model. The average temperature of this side is

Tavg = !(T2 + T3) = 58.695

Assuming this temperature to be constant for the entire O.3-m length of the side,

Convection heat loss from side 2-3 = hL(Tavg - T,,,) = 313.43

This heat loss is exactly the same as the sum of reactions, indicating that the solution
satisfies overall energy balance.

Example 1.9 Stress Analysis of a Bracket The following global equations for stress
analysis of a bracket are developed in Example 1.6. Incorporate essential boundary con-
ditions into the system and obtain tllli" final reduced system of equations in terms of the
remaining unknowns. Solve for nodal values. Compute reactions and verify overall equi-
librium:

1578.78 195.313 -276.693 325.52t -976.563 260.417 -325.521 -781.25 0 0 0 0 "I 0
195.313 1725.26 -1204.43 520.833 -390.625 -781.25 0 0 0 0 VI 0
-276.693 65.1042 -585.938 0 -976.563 -130.208 0 0 0 0 "2 -1.25
-325.521 65.IO'l2 1253.26 0 0 520.833 0 0 0 0 V2 -5.
-976.563 -1204.43 -585.938 1595.05 3125. 0 260.417 -651.042 260.417 -325.521 0
26OA17 0 -520.833 -1171.88 -390.625 520.833 -260.417 -781.25 -781.25 "3 0
-325,521 520.833 0 0 -520.833 4166.67 520.833 0 -651.042 -130.208 '3 -25
-390.625 0 -1171.88 520.833 520.833 0 0 260.417 -10.
-781.25 -781.25 -976.563 260.417 -3385.42 3125. -3385.42 1692.71 0 -1041.67 520.833 1/4 0
0 -130.20B 520.833 -390.625 520.833 -520.833 -520.833 -781.25 -781.25 520.833 0
0 0 -651.042 520.833 4166.67 -1041.67 2864.58 2018.23 v"-260.411 -1.25
0 0 0 0 -260.417 0 0 260.417 520.833 0 260.417 "5 -5.
0 0 0 0 260.417 -781.25 0 0 -2604.17 -2604.17 '5
0 0 0 -325.521 -130.208 -651.042 260.417
0 0 -781.25 520.833 -260.417 0 "6
2994.79 "6

Nodes 1 and 2 are on the fixed end of the bracket. Thus the essential boundary conditions
are as follows:

Node dof Value

1 uj 0
vj 0

2 u2 0

v2 0

BOUNDARY CONDITIONS AND NODALSOLUTION 47

Incorporating these boundary conditions means removing equations 1,2,3, and 4 from the
global system. Furthermore, since the specified values are 0, the corresponding columns
carralso be removed. Thus, after incorporating the essential boundary conditions, the final
system of equations is as follows:

3125. -520.833 -1171.88 520.833 -651.042 260.417 -325.521 -781.25 u3 0
-520.833 4166.67 520.833 -3385.42 520.833 -260.417 -781.25 -130.208 v3 0
-1171.88 3125. -520.833 0 -651.042 u4 -2.5
520.833 0 0 520.833 v4 -10.
520.833 -3385.42 . -520.833 4166.67 1692.71 0 260.417 -260.417 0
-651.042 0 0 -781.25 -1041.67 Us 0
520.833 0 0 -781.25 2864.58 260.417 -1.25
260.417 -260.417 -1041.67 520.833 520.833 -2604.17 )/5 -5.
-325.521 -781.25 -651.042 260.417 -2604.17 2018.23 u6
-781.25 -130.208 520.833 -260.417 260.417 0 )/6
0 2994.79

Solving the final system of global equations, we get the following nodal values:

={U3 -0.0103553, v3 = -0.0255297, u4 =0.00472765, v4 =-0.0247357,
=u5 -0.0131394, "s = -0.0554931, u6 =0.0000838902, "e = -0.0555664}

Complete Table of Nodal Values

uv

10 0

20 0

3 -0.0103553 -0.0255297

4 0..00472765 -0.0247357

5 -0.0131394 -0.0554931

6 0.0000838902 -0.0555664

Computation of Reactions - .

Equation numbers of dof with specified values: {I, 2, 3, 4}
Extracting equations {I, 2, 3, 4} from the global system, we have

"I

VI

U2

V2

[""'" ['", ]195.313 1725.26
195.313 -276.693 325.521 -976.563 260.417 -325.521 -781.25 0 0 0 "3
!] ".325.521 -1204.43
65.1042 -1204.43 520.833 -390.625 -781.25 -130.208 0 0 0 V3 _ Rz + O.
- R3 -1.25
-276.693 65.1042 1253.26 -585.938 0 0 -976.563 520.833 0 0 0
v. R. -5.
-585.938 1595.05 0 0 260.417 -390.625 0 0 0

"5

Vs

"6
v6

48 FINITEELEMENTMETHOD: THE BIG PICTURE

Substituting the nodal values and rearranging,

o
o
o
o

RRzI ) _ (1517985.7.3813 195.313 -276.693 325.521 -976.563 260.417 -325.521 -781.25 oo 00 00 O~o) --00..00215053259573 [ 0O· )
( R3 - -276.693 1725.26 65.1042 -1204.43 520.833 -390.625 -781.25 -130.208
-585.938 -976.563 o 0 0 0.00472765 - -_1.
R4 325.521 65.1042 1253.26 o o 520.833 o 0 0
-1204.43 -585.938 1595.05 o o 260.417 -390.625 -0.0247357 52.5

-0.0131394

-0.0554931

0.0000838902

-0.0555664

Table of reactions:

Label dof Reaction

Rj uj 21.25
Rz vj 4.10648
R3 Uz -16.25
R4 Vz 15.8935

Sum of reactions:

dof: U 5.
dof: v 20.

The sum of reactions must be balanced by the applied loading. The loading is applied
normal to the top surface. To determine the total load and its components in the horizontal
and vertical directions, we need its length and normal vector. The end nodal coordinates of
this line are

Yz = 2

The length and components of the unit normal to the line are computed as follows:

L = ~(xz _xj)z + (yz - Yjf= 4.12311

nx = (yz - Yj)/L = 0.242536; ny = -(xz - xj)IL = 0.970143

The total applied load and its components in the coordinate directions can now be deter-
mined as follows:

Total load =qhL =-20.6155
Horizontal component =qhLnx =-5.

Vertical component = qh:Lny = -20.

These are equal and opposite to the sum of reactions. Hence the solution satisfies the overall
equilibrium.

ELEMENTSOLUTIONS AND MODELVALIDITY 49

1.4 ELEMENT SOLUTIONS AND MODEL VALIDITY

Once the solution for all nodal degrees of freedom is available, we can go back to the
element equations and develop a complete solution over each element. The process simply
requires substituting computed nodal values into the assumed element solution defined
during derivation of element equations. Any secondary quantities, such as derivatives or
integrals involving the solution, can be obtained by performing necessary computations
on the element solutions. The examples presented in the following sections clarify the
procedure for different elements introduced earlier in this chapter.

1.4.1 Plane Truss Element

As will be seen in Chapters 2 and 4, the fundamental assumption made in deriving the plane
truss element equations is that the axial displacement varies linearly over the element. This
means that, in terms of nodal degrees of freedom, the displacement along the element axis
is written as follows:

O:5,S:5,L

where, as shown in Figure 1.21, s is a local coordinate that runs along the element length

with s = 0 at the first node of the element to s = L at the second node and L = length

of the element. The terms dl and dz are the axial displacements at the element ends. The
relationship between the local axial displacements and the global nodal degrees offreedom

is as follows.

["']~(~~) = (~
Axial displacements: Ins 0 mo') VI -= Td
Transformation matrix: 0 Is

Ins~JT -- C s00
0 Is

where Is and Ins are the direction cosines of the element axis as defined earlier. The axial
strain is simply the first derivative of the axial displacement, giving constant strain over the
element as follows:

Using Hooke's law, the axial stress is obtained by multiplying strain by the modulus of
elasticity:

(F = EE

50 FINITE ELEMENTMETHOD:THE BIG PICTURE Local dof

y
Global dof

x

Figure 1.21. Global and local degrees of freedom for a plane truss element

Since axial stress is equal to axial force divided by the area, the axial force in the element
is

F =erA

The sign convention used in these equations assumes that the tension is positive and the
compression is negative.

Example 1.10 Five-Bar Truss The following nodal values for a five-bar plane truss

were obtained in Example 1.7. Determine axial strains, axial stresses, and axial forces in

different elements of the truss. ,/

uv

10 0

2 0.538954 -0.953061

3 0.264704 -().264704

40 0

The displacements are in inches, loads in pounds, and stresses in pounds per inch squared.
The solution for element 1 is as follows:

Element nodal coordinates: first node (node #1): (0,0); Second node (node #2):
(1500.,3500.)

= = = =Xl 0; YI 0; x2 1500.; Y2 3500.
= =L ~ (x2 - x l )2 + (Y2 - YI)2 3807.89

= = = =Directio. n cosines: Is x2 L- Xl 0.393919; Ins Y2 -L YI 0.919145

,.... ..r::

ELEMENT SOLUTIONS AND MODEL VALIDITY 51

Global to local transformation matrix,

T _ (0.393919 0.919145 0 0)

- 0 0 0.393919 0.919145

gl~al [~] 0.53~954Element nodal displacements in
coordinates, d = =[ ]

v2 -0.953061

£.Ele~en~ 10Calc~~~i~.~!eS, dL=':.~~ -0.6~·3697nodal displacements in ) I
e ifd~ ~d?Axial displacements at element ends,
=::Q:!!63§27J':"d. '0
cL,._.j I) (1-
r;(C.'-J Cd \LEAx=ia2l s0tra0in0, E0'=0(;d2A- =d,~)/L = q~\.l i)
-0.000174'295

I

Axial stress, ap.:= J{@'!/:"-34.8591; Axial force = A(J" =-139436..
v~ .'

For any other elementthe calculations follow exactly the same pattern.

The solution summary is as follows:

Stress Axial force

1 -34.8591 -139436.
2 -629994 -25199.8
3 -10.5881 -31764.4
4 -10.5881 -31764.4
,5 22.4608 44921.7

~ MathematicafMATLAB Implementation 1.10 on the Book Web Site:
Plane truss element results

~ MathematicafMATLAB Implementation 1.11 on the Book Web Site:
Complete solution ofa plane truss

1.4.2 Triangular Element for Two-Dimensional Heat Flow

The finite element equations for a triangular element for two-dimensional steady-state heat
flow are based on the assumption of linear temperature distribution over the element. In
terms of nodal temperatures, the temperature distribution over a typical element is written
as follows:

52 FINITE ELEMENTMETHOD:THE BIG PICTURE
where

CI = X3 - X2; C2 = X t - X3; C3 =X2 -XI;

11 =X2Y3 - X3Y2; =12 x3Yt - x 1Y3; =13 XIY2 - X2YI;

The quantities Ni , i = 1, 2, 3, are linear triangle interpolation functions. The vector d is

the vector of nodal quantities that are all known at this stage. Knowing the coordinates at'
the three element nodes, the interpolation functions can easily be written for each element.
Multiplying these interpolation functions by the nodal temperature computed from the so-
lution of global equations gives the temperature distribution T(x, y) over each element. This
expression can be differentiated or integrated in a usual manner to obtain other quantities
of interest.

Note that for each element a different expression for the solution T(x, y) is obtained.
The finite element assembly process matches the degrees of freedom at the common nodes
between the elements. Therefore, T must be continuous across common element bound-
aries. However, since T is a linear function of X and y, its derivatives are constant. Thus
over each element the derivative values could be different, and in general two different
derivative values will be computed at a common interface between the two elements. The
magnitude of this discontinuity can be used as a guide to determine if the finite element
discretization is suitable or needs further refinement.

Example 1.11 neat Flow through a Square Duct The following nodal temperature
values for heat flow through a square duct were obtained in Example 1.8. Determine the
temperature distribution over each element and its x and y derivatives. By comparing the
magnitudes of these derivatives across common element interfaces, comment on suitability
of the finite element mesh.

Node Temperature

1 300
2 93.5466
3 23.8437
4 300
5 182.833

The temperature is in degrees Celsius and heat flow in watts per meter.
The solution for element 1 is as follows:

Element nodes: First node (node #1): {O., O.}; Second node (node #2): {O.2, O.}; Third
node (node #5): {O.I, O.I}

XI =0.;x2 =0.2;x3 =0.1
YI =0.;Y2 =0';Y3 =0.1

ELEMENT SOLUTIONS AND MODEL VALIDITY 53

Using these values, we get

bl =-0.1; bz =0.1; b3 =O.
cj =-0.1;
Cz = -0.1; u,c3 =0.2
II =0.02; = O.
I z =0.;

Element area, A =.0.01

Substituting these into the formulas for triangle interpolation functions, we get

Interpolation functions, NT = (-5.x - 5.y + 1., 5.x - 5.y, 1O.y)

From a global solution the temperatures at the element nodes are

(from nodes (I, 2, 5)), dT = {300, 93.5466, 182.833}

Thus the temperature distribution over the element, T(x, y) = NTd = -1032.27x-

139.406y + 300.

Differentiating with respect to x and y, aT/ax =-1032.27 and oT/oy =-139.406

For any other element the calculations follow exactly the same pattern.
The solution summary is as follows:

T(x,y) aT/ox oT/oy

1 -1032.27x - 139.406y + 300. -1032.27 -139.406

2 -1125.2x - 232:343y + 318.587 -1125.2 -232.343

3 -1171.67x - 209.109y + 320.911 -1171.67 -209.109

4 300. -1171.67x -1171.67 o

As seen from the model shown in Figure 1.18, elements 1 and 2 have a common side
between them. However, the derivative values for the two elements show significant dif-
ference. Elements 3 and 4 also share a side. The x derivatives for the two elements are the
same; however, their y derivatives are vastly different. TIns large difference in the deriva-
tive values shows thatthe finite element mesh used for the analysis is very coarse. A much
finer mesh is needed for any meaningful results. As mentioned earlier, this coarse mesh is
used here to illustrate all necessary computations in detail.

~ MathematicafMATLAB Implementation 1.12 on the Book Web Site:
Heat flow element results .

• MathematicalMATLAB Implementation 1.13 on the Book Web Site:
Complete solution of a heat flow problem

54 FINITEELEMENTMETHOD:THE BIG PICTURE

1.4.3 Triangular Element for Two-Dimensional Stress Analysis

The derivation of the plane stress triangular element equations is based on the assumption
of linear displacements over the element. In terms of nodal quantities, the displacements
over an element can be written as follows.

=U(x,y) N, uj + N2u2 + N3u3
=V(X,y) Njv I +N2v2 +N3v3

Uj

vI

=( ~J( U(X, y) )
V(X,y)
n, 0 N2 0 N3 U2' =.NTa

0 Nj 0 N2 0 V2

U3

V3

=where Nj , i 1, 2, 3, are the triangle interpolation functions (same as those used for the

triangular element for heat flow). Knowing the coordinates at the three element nodes,

the interpolation functions can easily be calculated for each element. Multiplying these

interpolation functions by the nodal displacements computed from the solution of global

equations gives the horizontal and vertical displacement distributions u(x, y) and vex, y)

over each element. These expression can be differentiated or integrated in a usual manner

to obtain other quantities of interest. '.

As discussed later in Chapter 7, the three components of in-plane element strains are

computed directly by differentiating displacement functions as follows:

o b2 0 b3

cj 0 c2 0
bj c2 b2 c3

The stresses can be computed as follows:

a =C€

where C is the appropriate constitutive matrix. For homogeneous, isotropic elastic materi-
als under plane stress conditions

0]E v1 v 0
1
C=
1- [ o 0 l;:v

where E = Young's modulus and v = Poisson's ratio.

For plane stress' problems, the three out-of-plane strain and stress components are as
follows:

ELEMENTSOLUTIONS AND MODEL VALIDITY 55

v(~ + oy) ='YyZ 0; 'Yix = 0

EZ=-·E; TZ.T =0

~=O;

The complete vectors of strains and stresses are ordered as follows:

= oy(J" (~

From a design point of view, one is often interested in principal stresses or effective (von
Mises) stress. The principal stresses can be computed by determining eigenvalues of the
following 3 x 3 matrix of stress components:

Principal stresses = eigenvalues[S]

The effective (von Mises) stress can be computed from the component stresses using the
following equation:

a;, = ~~(a:T - oy)2 + (oy - ~)2 + (~- ~)2 + 6(-s, + T;z + T~T)

Similar to the heat flow problem, for each element different expressions for displacements
u(x,y) and vex, y) are obtained. The finite element assembly process matches the degrees of
freedom at the common nodes between the elements. Therefore, the displacements must be
continuous across common element boundaries. However, since displacements are linear
functions of x and y, their derivatives, and hence strains and stresses, are constant. Thus
over each element the stress values could be different, and in general two different stress
values will be computed at a common interface between the two elements. The magnitude
of this discontinuity can be used as a guide to determine if the finite element discretization
is suitable or needs furtherrefinement,

Example 1.12 Stress Analysis of a Bracket The following nodal displacement values
for the bracket problem were obtained in Example 1.9. Determine displacement distribu-
tion over each element. Using these, compute element strains, stresses, principal stresses,
and effective stress. By comparing the magnitudes of stresses across common element in-
terfaces, comment on the suitability of the finite element mesh.

uv

10 0

20 0

3 -0.0103553 -0.0255297

4 0.00472765 -0.0247357

5 -0.0131394 -0.0554931

6 0.0000838902 -0.0555664

56 FINITEELEMENTMETHOD:THE BIG PICTURE

Solution for Element 1

h =0.25; E = 10000; v =0.2

10416.7 2083.33 0 1
Plane stress constitutive matrix, C = 2083 10416.7
[ 0.33 0

o 4166.67

Element nodes: first node (node 1): {O., O.}; second node (node 3): {2.,O.}; third node
(node 4): {2., 1.5}

XI =0.; X2 =2.;
YI =0.;
Y2 =0.;
Using these values, we get

bl =-1.5; b2 = 1.5; b3 =O.
c2 =-2.; c3 =2.
cI =0.;
II 3.; 12 =0.; 13 =0.

Element area: A = 1.5

~.- 0.5 ~.5 -~.666667~· ~0. 6666671"\

BT = 0
[ O. -0.5 -0.666667 0.5 0.666667 )

Substituting these into the formulas for triangle interpolation functions, we get

/

Interpolation functions, {I. - 0.5x, 0.5x - 0:666667y, 0.666667y}

NT =( 1. - 0.5x O' 0.5x - 0.666667y 0 0.666667y 0)
o 1. - 0.5x 0 0.5x 0.666667y 0.666667y
o

From a global solution the displacements at the element nodes are

{displacements at nodes (I, 3, 4)):
dT ={O,0, -0.0103553, -0.0255297, 0.00472765, -0.0247357}

The displacement distribution over the element is

Y))U(X, NTd = ( 0.0100553y - 0.00517764X)
( vex, y) 0.000529362y - 0.0127648x

In-plane strain components, € =BTd =(-0.00517764 0.000529362 -0.00270956)
In-plane stress components, a =C€ =(-52.8309 -5.27256 -11.2898)

ELEMENT SOLUTIONS AND MODEL VALIDITY 57

Computing out-of-plane strain and stress components using appropriate formulas, the
complete strain and stress vectors are as follows:

€T = (-0.00517764 0.000529362 0.00116207 -0.00270956 0 0)
aT = (-52.8309 -5.27256 0 -11.2898 0 0)

Substituting these stress components into appropriate formulas,

Principal stresses = (0 -2.72856 -55.3749)

Effective stress (von Mises) = 54.0623

Proceeding in exactly the same manner, we can get solutions for all elements. A summary
of the solution at element centers follows:

Coordinate Displacement Stresses Principal Stresses Effective Stress

1 1.33333 -0.00187588 -52.8309 0 54.0623
0.5 -0.0167551 -5.27256 -2.72856
0 -55.3749
-11.2898
0 0

2 0.666667 0.00157588 24.6232 67.2393 92.0659
1.16667 -0.00824522 4.92464 0
0
-51.5326 -37.6915
0
0

3 3.33333 . -0.0078036 -14.6533 0 18.3167
0.333333 -0.0455297 -3.66334 0
0 -18.3167
-7.32667
0
0

4 2.66667 -0.00184791 3.10223 26.3357 38.0742
0.833333 -0.0352772 5.91407 0
0
-21.7822 ·-17.3194
0
0

As seen from the model shown in Figure 1.26, elements 1 and 4 have a common side
between them. However, the effective stress values for the two elements show significant
difference. Elements 3 and 4 also share a side. The effective stresses for these two elements
are quite different as well. This large difference in the stress values shows that the finite

58 FINITE ELEMENTMETHOD: THE BIG PICTURE

element mesh used for the analysis is very coarse. A much finer mesh is needed for any
meaningful results. As mentioned earlier, this coarse mesh is used here to illustrate all
necessary computations in detail.

• MathematicafMATLAB Implementation 1.14 on the Book Web Site:
Plane stress element results

• MathematicafMATLAB Implementation 1.15 on the Book Web Site:
Complete solution ofa plane stress problem

1.5 SOLUTION OF LINEAR EQUATIONS

As seen from the previous section, the global system of equations after boundary conditions
consists of n equations in n unknowns:

Kd=R

Most scientific calculators have functions for solving linear system of equations. These
are useful for solving small textbook problems. Computer algebra systems (Mathematica,
MATLAB, MAPLE, etc.) have sophisticated built-in functions for solving large systems
of linear equations. For example, using Mathematica, a system of equations can be solved
by first defining matrix K and right-hand-side vector R and then issuing the following
command:

LinearSolve [K, RJ

For completeness two basic methods fbr solution of a large-scale linear system of equations
are briefly discussed in this section. Detailed treatment of these methods is beyond the
scope of this text.

1.5.1 Solution Using Choleski Decomposition

An efficient method to solve the system of equations is to first decompose the matrix K
into a product of lower and upper triangular matrices. For symmetric matrices the so-
called Choleski decomposition is very efficient. In this decomposition a symmetric matrix
K is decomposed into a product of a lower triangular matrix L and its transpose LT in the
following form:

K=LLT 0 III 121 131 ]'.,In2
0 122 132
[,n 0 0 0]o . In3
121 122 0 LT = 0 0 1~3
L = 1~1 1~2 1~3 1~, ' l~n
000
.:.; In3

SOLUTION OF LINEAR EQUATIONS 59

Such a decomposition is always possible for nonsingular symmetric matrices. Assuming
this decomposition is known, wecan solve the system of equations in two steps as follows:

Kd =R =? LLT d = R

=If we define LTd y, then

, 1: l 01 0 0

LLT d = R =? Ly =R =? 1~1 1z~z2 133

[
1111 ll1z 1113

Since L is a lower triangular matrix, the first row gives the solution for the first variable
as YI = r/l l l · Knowing this, we can solve for Yz = (rz - lzIYI)llzz from the second equa-
tion. Thus, proceeding forward from the first equation, we can solve for all intermediate '
variables Yi' This is known asforward elimination.

_ ri - L"Jik--1IIikYk. i = 1,2, ... , Ii
Yi - 1.. '

II

[I_n~ 1.,]["'] nLTd =y =?
Now we can solve for the actual unknowns di as follows:

lz1 131 ... ll1z dz Yz
lzz ... 1113 d3 = Y3
0 132
1~3 ...

: '.
1 ~1I d0 0 0 ...
' ;11

ll

=In this system the last equation has only one unknown, giving dll y/l,l1l • The second

to the last equation is used- next to get dll _ l , and thus, working backward from the last
equation, we can solve for all unknowns. This is known as backward substitution.

=i n, n - 1, ... , 1

The procedure is especially efficient if solutions for several different right-hand sides are
needed. Since in this case the major computational task of decomposing the matrix into
LLT form is not necessary for every right-hand side, one simply needs to perfortn a forward
elimination and a backward substitution for each different right-hand side.

ChoJeski Decomposition The Choleski decomposition algorithm can be developed
by considering direct multiplication of terms in the Land L T matrices and equating them
to the corresponding terms in the K matrix:

60 FINITEELEMENTMETHOD:THE BIG PICTURE

k 11 k21 k31 k"l III 0 0 I"~]['!' 121 131 ... 1I,",2, ]
1'21 ~2 k32 k"2 121 122 0 122 132 1"2
k~2 k33 0 1~3 ...
=J( LLT ::=} k~1 k"3 = I~I 1~2 1~3
...
:
k"l ",,2 k"3 0 0 ... I,:"
k"" I'd 1"2 1"3

By direct multiplication we can see that the entries in the first row of the L matrix can be
established as follows:

Row 1: {k::kll =Iii::=} III =

~.I _ _ 1k21
111/21 ::=} 121 -
-

-II

t :ki! i = 2, ... , n

lei! = II IIi! ::=} Ii! =

_ 11

Once entries in row 1 are known, we can proceed to row 2 and establish the entries there
by direct multiplication as follows:

Row 2: le22 =I~I + 1~2 ::=} 122 = ~le22 -/~1

= =k32 121'31 + '22/32 ::=} 132 k~-3-,?,---"'-2''1-'/'3'1'-
22

J, - .J2I' i! + '22/1"2 ::=} '1"2 _ k"2 -'2I'i! . 3, ... , n
'i2 - ', '
-
22

Proceeding in a similar manner, we canwrite the following general formulas for entries in

the ith row of the L matrix: I -

Notice that the procedure requires taking the square root of terms corning from the diagonal
of the J( matrix and then division by this term to establish other terms in a given row. This
is called the pivot element. Clearly these operations will fail if there is a negative or zero
pivot. However, as long as the matrix J( is nonsingular, it is always possible to rearrange
the system of equations to avoid a negative or zero pivot.

The computational steps can be written in the form of the following algorithm:

Choleski Decomposition Algorithm

For i = 1, n

For j = 1, i-I

SOLUTION OF LINEAR EQUATIONS 61

end

Example 1.13 Find the solution of the following system of equations using Choleski
decomposition:

[; l~ ~ -l~][~~]_ [~]
6 2 11 -5 ad3, - 7
3 -10 -5 21 8

We first generate the LLT decomposition of the coefficient matrix using the above algo-
rithm:

30 o
][3 1 2 1] [9 6 3]K =LLT = 1 -{IT 3
20
o o 0 -{IT 0 --{IT = 3 12 2 -10
o 0 0 ..;7 -..;7 6 2 11 -5
..;7
[ 1 --{IT -{7 { i o 0 0 {i
3 -10 -5 21

Using the L matrix, the solution for intermediate variables (forward elimination) is as fol-

lows: .

30 o

1 -{IT o
2 0 ..;7
[ 1 --{IT -..;7

Row 1: 3Yl =5 ===:> Y1 = ~
Row 2:
Row 3: 3mlYI + -{ITh = 6 ===:> h =
Row 4:
2Yl + 0Yz + ..;7Y3 =7 ===:> Y3 = 3~
= =lYI + (--{IT)y:i + (-..;7)Y3 + {iY4 8 ===:> Y4 3~

Using the L T matrix, the solution for actual variables (backward substitution) is as follows:

5

3"

13

3 {IT

11

3-17

43

3-{i

62 FINITE ELEMENTMETHOD: THE BIG PICTURE

Row 4: --y(;L;2. d4 --.-:3!-11.2 -----.-' d4 -- 161
Row 3:
Row 2: --y'7I d3 + (--v'7I )d4 =.3l.0L ====> d3 = 342Z3
Row 1:
':rt{fidz + Od3 + (-{fi)d4 = 3~ ====> dZ =

3d l + 1dZ + 2d3 + Id4 = ~ ====> d, = -7i~

1.5.2 Conjugate Gradient Method

The well-known conjugate gradient method for solution of unconstrained optimization
problems can be used to develop an iterative method for solution of a linear system of
equations. The method has been used effectively for solution of nonlinear finite element
problems that require repeated solutions of large systems of equations. Consider a sym-
metric system of n X n linear equations expressed in matrix form as follows:

[' ' kJZ Kd=R
kZJ kF
":,1 "liZ k'"r] ["]10.11 dz _ rz

.. cL I~I

Finding the solution of this system of equations is equivalent to minimizing the following
quadratic function:

minf(d) = 4dTKd - dTR

,/
This can easily be seen by using the necessary conditions for the minimum, namely, the
gradient of the function must be zero:

where

Since K is a symmetric matrix, the transpose of the first term in each row is same as the
second term. Thus the gradient can be expressed as follows:

SOLUTION OF LINEAR EQUATIONS 63

Noting that

'eeI:f] =[10: 10: ·.·-...·00:] . . .=identity matnx

reT 0 0 ... 1
II

we see that the gradient of the quadratic function is

g(d) = Kd-R

A condition of zero gradient implies Kd - R = 0, which clearly is the given linear system of
equations. Thus solution of a linear system of equations is equivalent to finding a minimum

of the quadratic function f.

Basic Conjugate Gradient Method In the conjugate gradient method, the minimum

of f is located by starting from an assumed solution d(O) (say with all variables equal to

zero) and performing iterations as follows:

k =0, 1,...

where a(k) is known as the step length and h(k) is the search direction. At the kth iteration
the search direction h(k) is determined as follows:

The scalar multiplier fJ determines the portion of the previous direction to be added to

determine the new direction. According to the Fletcher-Reeves formula,

After establishing the search direction, the minimization problem reduces to finding a(k) in
order to

For the quadratic function f we have

64 FINITE ELEMENTMETHOD:THE BIG PICTURE
Expanding,

!Cd(k) + (P)h(k)) = ~Cd(k){Kd(k) + ~a(k)(d(k){Kh(k)
+ ~a(k)(h(k){Kd(k) + ~(a(k))2Ch(k){Kh(k) - (d(k){R - a(kJch(k){R

For the minimum the derivative of this expression with respect to a(k) must be equal to
zero, giving the following equation for the step length:

Noting the symmetry of K, the first two terms can be combined and moved to the right-hand
side to give

giving

Since f3 is usually small, h(k) "" _g(k), and thus for computational efficiency the step length

expression is slightly modified as follows: "

Basic Conjugate Gradient Algorithm The computational steps can be organized
into the following algorithm.

= =Choose a starting point d(O). Compute g(O) Kd(O) - R. Set h(O) _g(O). Compute
=reO) (g(O){g(O). Choose a convergence tolerance parameter E. Set k O.

1. If -,J-k) .s E, stop; d(k) is the desired solution. Otherwise, continue.
2." Compute the step length:

3. Next point:

SOLUTION OF LINEAR EQUATIONS 65

4. Update the quantities for the next iteration:

= + +g(k+l) K(d(k) o:(k)/z(k») _ R == g(k) o:(k)Z(k)

,.(k+l) = (g(k+l){g(k+l)

fP) = ,.(k+l)h·(k)

=-s"'"/z(k+1) + (3(k)/z(k)

5. Set k = k: + 1 and go to step 1.

Example 1.14 Find the solution of the following system of equations using the basic
conjugate gradient method:

Starting point, d(O) = (0,0,0, O)

= =g(O) Kd(O) -R (-5., -6., -7., -8.)
= =/z(0) _g(O) (5.,6.,7., 8.)
= = =,.(0) g(O)Tg(O) 174.; (3(0) 0

Iteration 1:

z(O) = Klz(O) :::; {129., 21., 79.,88.)

= =Step length, 0:(0) ,.(O)/h(O)Tz(O) 0.0857988
=Next point, d(1) d(O) + 0:(0)/z(0) == (0.428994,0.514793,0.600592,0.686391)
= =g(l) +g(O) o:(O)z(O) (6:06805,-4.19822, -0.221893, -0.449704)

,.(1) = g(I)Tg(l) = 54.6978

= =(3(1) ,.(1)/,.(0) 0.314355
= =/z(l) +-g(l) (3(0)/z(0) (':4.49627,6.08435,2.42238, 2.96454)

Iteration 2:

= =z(l) KIz(l) (1.21451,34.7228, -2.98549, -24.1888)

Step length, o:m = ,.(I)/Iz(l)Tz(l) = 0.431153

Next point, d(2) = d(1) + o:(I)/z(l) = (-1.50959, 3.13808, 1.64501, 1.96456)

= =g(2) +g(l) o:(I)Z(I) (6.59169, 10.7726, -1.50909, -10.8788)
= =,.(2) gC2)TgC2) 280.125
= =(3(2) r(2)/,.(I) 5.12132
= =1z(2) +_g(2) (3(l}Jz(l) (-29,6185, 20.3873,13.9149,26.0612)

Iteration 3:

Z(2) =KIz(2) = (-43.7321,-76.9897,-114.179, 184.981)

= =Step length, 0:(2) ,.(2)//z(2)Tz(2) 0.0947098

66 FINITE ELEMENTMETHOD:THE BIG PICTURE

Next point, d(3) == d(2) + OP)h(2) == (-4.31475,5.06896,2.96288, 4.43281)
g(3) == g(2) + (1'(2)Z(2) == (2.44982, 3.48091, -12.323, 6.64076)

r(3) == g(3)Tg<3) == 214.073
/3(3) == r(3)/r(2) == 0.764207

h(3) == _g(3) + /3(2)h(2) == {-25.0845, 12.0992,22.9568, 13.2754]

Iteration 4:

z(3) == [(h(3) == (-11.8961, -16.903, 59.8392, -32.2469]
Step length, (1'(3) == r(3)/h(3)TZ(3) == 0.205934
Next point, d(4) == d(3) +(1'(3)h(3) == {-9.48052, 7.56061, 7.69048, 7. 16667}

g(4) == g(3) + (1'(3)Z(3) == (7.10543 X 10-15, -9.76996 X 10-15,0.,2.57572 x 10- 14 )

r(4) == g(4)Tg(4) == 8.09371 X 10- 28
/3(4) == r(4)/r(3) == 3.78081 X 10- 30
+h(4) == _g(4) /3(3)h(3)

== {~7.10543 x 10- 15,9.76996 X 10- 15,8.67954 X 10- 29, -2.57572 x 1O-14}

Solution converged after four iterations:

g == {7.10543 X 10- 15, -9.76996 X 10-15,0.,2.57572 x 10-141

r == gTg == 8.09371 X 10-28
Solution, d == {-9.48052, 7.56061, 7.69048, 7.16667}

Preconditioned Conjugate Gradient When the coefficient matrix K is ill-

conditioned, the convergence of the conjugate gradient method could be very slow. In-

troducing a suitably chosen preconditioning matrix P improves convergence. To see this,

we multiply both sides of the system of equations by the inverse of the preconditioning

matrix to get ;'

Clearly, if P == K, we have the solution of the system of equations. Thus, intuitively, we
can expect that, ifP is a matrix that is reasonably close to K, then the method will converge
faster. The matrix P is known as the preconditioning matrix. Several different techniques
have been proposed for establishing this matrix, Two popular choices will be discussed
later in the section. First we develop the computational procedure for the solution of equa-
tions when the preconditioning matrix is included.

In general, the product P"!K does not produce a symmetric matrix, even if both matrices
are individually symmetric. In order to get a symmetric system of equations, we introduce
the following decomposition of the P matrix,

and a new vector of intermediate variables y,

SOLUTldN OF LINEAR EQUATIONS 67

Introducing these, the original system of equations is modified as follows:

Thus we have a system of equations as follows:

= =where K E-1!W-T and k E-1R.

The matrix K is a symmetric matrix, At this point we can use the standard conjugate

gradient method to find a solution for intermediate variables y:

= R;g(k) Ky(k) _ =ii(k) _g(k) + j3(k)h(k-l)

=j3(k) rn(k){ -(k) = fn(k){ -(k)

l.5 g . a(k) 15 g
(h(k){iih(k)
[g(k-J){g(k-l) ,

=y(k+l) y(k) + a(k)h(k)

Multiplying vector y(k+l) by E-T gives the solution of the original problem:

A serious drawback of the procedure so far is that it requires a decomposition of the pre-

conditioning matrix p-l = E-TE-l, which is a very time consuming task for a general

matrix. Fortunately, it is possible to rewrite the above expressions so that only p-l appears

in the equations and thus there is no need to explicitly carry out the decomposition.

Substituting for K and k into the g(k) expression, we have

Using this, we can write ii(k) as follows:

ii(k) = -=..::g(k) + j3(k)h(k-l) = _E-1g(k) + j3(k)h(k-l)

Multiplying both sides by E-T ,

= h.(k) =+E-Tii(k) == h(k) _E-TE-1g(k) j3(k)E-Th(k-l) => +_p-lg(k) j3(k)h(k-l)

Similarly, other expressions in the algorithm can be rewritten as follows:

, [g(k) { g(k) = [g(k) { E-TE-1g(k) . [g(k) { p-lg(k)
j3(kl = = --=::...........:-:;;--=---

[g(k-J){glk-l) [g(k-J){E-TE-1g(k-l) [g(k-J){p-lg(k-J)

a(k) = (g(k){g(k) r:[g(k){p-l g(k) '[g(kl]T I g(k)

= = .:::::--=...,,,,--..:::--

(h(k){E-lKE-Ti'L0') (E-Th(kl {!W-Tii(k) (h (k){ Kh(k)

=y(k+1) y(k) + a(k)h(k)

68 FINITEELEMENTMETHOD:THE BIG PICTURE

Multiplying the expression for the intermediate variables by E-T , we have

Preconditioned Conjugate Gradient (PCG) Algorithm The computational steps
can be organized into the following algorithm:

Initialization

=Choose a starting point d(D). Set g(D) Kd(D) - R.

=Solve for W(D): Pw(D) g(D).

Set h(D) = -W(D).

=Compute r(D) (g(D){W(D).

Choose a convergence tolerance parameter E. Set k = O.

1. If ,J.k) ~ E, stop; d(k) is the desired solution. Otherwise, continue.
2. Compute the step length:

3. Next point:

= +d(k+l) d(k) a(k)h(k)

I'

4. Update the quantities for the next iteration:

= +g(k+l) g(k) a(k)z(k)
=Solve for W(k+l): PW(k+l) g(k+l)

r(k+l) = (g(k+l){W(k+l)

=j3(K) ,J.k+I)/,J.k)

= +h(k+l) _W(k+l) j3(k)h(k)

5. Set k = k + 1 and go to step 1.

Jacobi Preconditioning One of the simplest choices of the preconditioning matrices
is the so-called Jacobi preconditioning. Here the P matrix is a diagonal matrix with entries
. equal to the diagonal elements of the J( matrix. Thus p-I is established as follows:

-I _ 1. P:IJ-·1 =0' i =1= j; i, j = 1,... , n
Pii - k ..' '

II

SOLUTION OF LINEAR EQUATIONS 69

Example 1.15 Find the solution of the following system of equations using the peG
method with Jacobi preconditioning:

93
3 12
[ 6· 2
3 -10

~ Tl~. 2~.9. 0 0 0 ]

Preconditioning matrix, P =. [

Starting point, d(O! =. (0,0,0, O)

g(O) =. Kd(O) -R =. (-5, -6, -7, -8)

=Solving Pw(O) =. g(O), we have w(O) {-0.555556, -0.5, -0.636364, -0.380952}.
= =h(O) _w(O) (0.555556,0.5,0.636364, 0.380952)

r(O) =. g(O)T w(O) = 1.10873; /3(0)= 0

Iteration 1:

z(O) = Kh(O) =. {11.461, 5.12987, 9.42857, 1.48485}

=Step length, a(O) r(O)/h(O)Tz(O) =. 0.85689

Next point, d(l) =. d(O) + a(O)h(O) = (0.47605,0.428445,0.545294, 0.326434}

= =g(l) +g(O) a(O)z(O) (4.82085, -1.60426,1.07925, -6.72765)
=Solving PW(I) =. g(l), we have w(l) {0.53565, -0.133689, 0.0981136, -0.320364}
=r(l) =. g(l)Tw(1) 5.05795

/3(l) =. r(l)/r(O) =. 0.380871
-s'"h(l) =.
+/3(OWO) =. {-0.324055, 0.324124, 0.144259, 0.465458}

Iteration 2:

= =z(1) Kh(l) (0.317806, :::1.44873, -2.03652, 4.83991}

Step length, a(1) =. r(l)Ih(l)Tz(l) = 3.64817

=Next point, d(2) d(J) + a(l)h(l) =. (-0.706158, 1.61091, 1.07158, 2.02451)

g(2) = g(1) + a(J)z(l) = {5.98026, -6.8895, -6.35033, 1O-.9292}

Solving PW(2) =. g<2), we have W(2) =. (0.664474, -0.574125, -0.577303, 0.520438}

r(2) =. g(2)T liP) =. 17.2832

= =/3(2) r(2)lr(l) 3.41704
=h(2) _g(2) +/3(l)h(J) =. (-1.77178, 1.68167, 1.07024, 1.0i005}

Iteration 3:
Z(2) =. Kh(2) = (-1.26943,6.30468, -0.84494, -5.01222)
Step length, a(2) =. r(2)lh(2)TZ(2)·= 2.62505
Next point, d(3) =. d(2) + a(2)h(2) = (-5.35718, 6.02538, 3.881O~, 4.83344)

70 FINITE ELEMENTMETHOD: THE BIG PICTURE

= =g(3) +g(2) aP)z(2) (2.64795,9.66064, -8.56835, -2.22814)
Solving Pw(3) =g(3), we have W(3) = (0.294216,0.805053, -0.778941, -0.106102)
= =r(3) g(3)T w(3) 15.467
= =/3(3) r(3)lr(2) 0.894918
= =h(3) +_g<3) /3(2)h(2) (-1. 87982, 0.699903, 1.73672, 1.06371 j

Iteration 4:

= =Z(3) Kh(3) (-1.20719, -4.40425, 3.90628,1.0158)
= =Step length, aP) r(3) Ih(3)Tz(3) 2.19348
= =Next point, d(4) d(3) +,aP)h(3) (-9.48052,7.56061,7.69048, 7.16667j
= +g(4) g(3) aP)z(3)

=(-7.99361 X 10- 15, 1.42109 X 10- 14,8.88178 X 10- 15, -3.73035 x 1O-14j
= =Solving PW(4) g(4), we have W(4) (-8.88178 X 10- 16,1.18424 X 10- 15,8.07435 X

10- 16, -1.77636 x 10- 15)

= =r(4) g(4)T w(4) 9.73646 X 10-29
= =/3(4) r(4)jr<3) 6.29497 x 10-30
= +h(4) _g(4) /3(3lh<3)

=(8.88178 X 10- 16, -1.18424 X 10- 15, -8.07435 X 10- 16,1.77636 x 1O-15j

The solution converged after four iterations:

=g (-7.99361 X 10- 15, 1.42109 X 10- 14,8.88178 X 10- 15, -3.73035 x 1O-14j
= =r gTg 9.73646 X 10-29

Solution, d = (-9.48052,7.56061,7.69048, 7. 16667j

I

Incomplete Choleski Preconditioning Currently the most popular preconditioning
method is that based on partial or incomplete Choleski factorization of the K matrix. In this
method a lower triangular matrix L is constructed by considering only elements in a limited .
band around the main diagonal of matrix K. Denoting this band by m, the incomplete
Choleski decomposition algorithm is as follows:

For i = 1, n
For j = 1, i-I
If (i - j :$ m) then

=Iij (leij - I:kj=-IIIikI jk )/Ijj

=oOtherwise Iij

end

Iii = ~leii - I:i-:,11 Itk

end

With this lower triangular L matrix, the vector h in the PCG algorithm is established using
forward elimination and backward substitution as follows:

SOLUTION OF LINEAR EQUATIONS 71

= =Ph. g :::::::> LLT/z g
=Forward elimination for intermediate variablesy: Ly g

=Back substitution for h: LTh y

Example 1.16 Find the solution of the following system of equations using the PCG
method with incomplete Choleski decomposition:

[ 1;~ ~ -1~][~~] =[~]
6 2 11 -5 x3 7
3 -10 -5 21 x4 8

The solution using a band m = 2 (two terms in each row around the diagonal) for generating
the preconditioning matrix is as follows:

Lower triangular factor of preconditioning matrix,

3. 0 o
3.31662
L _ 1. O. o o]

- [ 2. -3.01511 2.64575 0,
2.8875
o -1.88982

=Starting point, d(O) {O, 0, 0, O}
= =g(O) Kd(O) -R (-5, -6, -7, -8}
= =Solving LLTw(O) g(O), we have w(O) {1.28557, -1.98131, -1.77103, -1.74611}
= =/z(0) _w(O) {-1.28557, 1.98131, 1.77103, 1.74611}

r(O) =g(O)TW(O) = 11.7637;(3(0) =.0

Iteration 1:

z(O) = K/Z(O) = {10.2383, 6., 7., 4.1433}

Step length, a(O) = r(O)1/z(O)Tz(O) =, 1.73367 .

= =Next point, d(1) +d(O) a(O)h(O) {-2.22874, 3.43493, 3.07037, 3.02717}
= =g(l) +g(O) a(O)z(O) {12.7498, 4.402,5.13567, -0.816894}
= =Solving LLTW(1) g(l), we have W(l) {2.02045, -0.322185, -0.744609, -0.369609}

r(l) = g(1)T w(l)= 20.82

= =(3(1) r(1)lr(0) 0.654181
=/z(l) +_g(l) (3(O)h(O) (-2.86144,1.61832,1.90318, 1.51188}

Iteration 2:

z(l) = K/z(l) = {-4.9433, -0.476917, -0.556404, -2.53399}

= =Step length, a(l) r(l)lh(l)Tz(l) 2.45428
= =Next point, d(2) d(1) + a(l)/z(l) (-9.25152, 7.40674, 7.7~131, 6.73774}
= =g(2) +g(l) a(l)z(1) {0.617597, 3.23151, 3.7701, -7.036}
Solving LLTW(2) =g(2), we havew(2) = {-0.166593, 0.0693565, 0.318143, -0.226273}

72 FINITE ELEMENTMETHOD: THE BIG PICTURE

=r(2) = g(2)TW(2) 2.91273
= =/3(2) r(2)/r(1) 0.1399
=h(2) _g(2) +/3(1)h(1) = (-0.233724, 0.157047, -0.0518876, 0.437785)

Iteration 3:

Z(2) =Kh(2) = (-0.630348, -3.29823, -3.84794, 7.18128)
= =Step length, aP) r(2)/h(2)Tz(2) 0.979771
Next point, d(3) =d(2) + aP)!l'2) = (-9.48052, 7.56061, 7.69048, 7.16667)
= +g(3) g(2) a(2)z(2)

=(1.77636 X 10- 15, -1.77636 X 10- 15,8.88178 X 10- 16,4.44089 X 10-151
=Solving LLTw(3) = g(3), we have ",(3) (1.03298 X 10- 16, 1.10676 X 10- 17, 1.35579 X

10- 16,2.49022 X 10- 16 )
r(3) = g(3)TW(3) = 1.39013 X 10-30

= =/3(3) r(3)/r(2) 4.77262 x 10-31

+h(3) = _g(3) /3(2)h(2)

= (-1.03298 X 10-16, -1.10676 X 10- 17, -1.35579 X 10- 16, -2.49022 x 10- 16)

The solution converged after three iterations:

=g (1.77636 X 10- 15, -1.77636 X 10- 15,8.88178 X 10- 16,4.44089 x 10- 15 )
= =r gTg 1.39013' X 10-30

Solution,d = {-9.48052, 7.56061, 7.69048, 7.166671

1.6 MULTIPOiNTCONSTRAINTS

/!

In some finite element modeling situations it becomes necessary to introduce constraints
between several different degrees of freedoms. Such constraints are known as multipoint
constraints and in general are expressed as follows:

cn d1 + c12d2 + + c1ndn = ql
C21d1 + c 22d2 +
+ c 2ndn = q2

where cij ' i, j = 1,2, ... , and qi' i = 1, 2, ... , are specified constants and d.; i = 1, 2, ... ,
are the nodal degrees of freedom. In matrix form the constraints equations can be expressed
as follows:

Cd=q

where with m constraints C is an m X n matrix and q is an m x 1 vector.
A common situation is modeling an inclined roller support. As is clear from Figure

1.22, in this case the displacement component normal to the incline is zero but neither

MULTIPOINT CONSTRAINTS 73

y
2

2

------------ x

Figure 1.22. Inclined roller support

its horizontal or vertical displacement is zero, and hence this boundary condition cannot
be incorporated using techniques discussed earlier. With the support malcing an angle a
with the horizontal, the component of ul normal to the support is ul sin(a) and that of VI is
vI cos(a). Thus we must solve the global system of equations with the following multipoint
constraint between the degrees of freedom at this node:

UI sin(a) + vI cos(a) = 0

Another common use of multipoint constraints is in creating valid meshes when tran-
sitioning from a coarse to a fine mesh. As discussed earlier, the mesh consisting of three
four-node quadrilateral elements in Figure 1.23 is invalid because node 4, which forms a
corner of elements 2 and 3, is not attached to one of the four corners of element 1. A valid
mesh can be obtained if the displacements at node 4 are related to the displacements at
nodes 3 and 5. For the case when the node 4 is located at the midway point on line 3-5
the mesh compatibility can be achieved by defining two multipoint constraints defining
displacements at node 4 as averages of the displacements at nodes 3 and 5. Thus we solve
the system of equations with..the two following multipoint constraints:

= =u4 -U3-+2U-s ===? u3 - 2u4 + Us 0

v4 = -v3-+2vs- ===? v3 - 2v4 + vs-= 0

y2 5 8

---------x

Figure 1.23. Multipoint constraints to create valid mesh in transition region

74 FINITEELEMENTMETHOD:THE BIG PICTURE

Figure 1.24. Rigid element in a finite element mesh

As a final example that requires use of multipoint constraints, we consider a situation
in which some portions of a finite element model are much stiffer as compared to the rest.
To avoid numerical difficulties in solving the resulting system of equations, it generally is
more efficient to treat the stiff regions as completely rigid. Figure 1.24 shows a simple ex-
ample in which the dark shaded triangle is assumed to be rigid. The nodes at the comers of
this triangle must displace in such a manner that the original shape is maintained. A planar
rigid body has only three degrees offreedom (two translations and a planar rotation). Thus
.the six degrees of freedom of the element must be related by three constraint equations.

For a triangle the three constraint equations can be written by recognizing the fact that
the lengths of the sides of the triangle must remain unchanged between the initial configu-
ration and the deformed configuration. In the initial configuration coordinates of the nodes
of the rigid triangle are (x" YI)' (x2, Y2)' and (x3, Y3 ). In the deformed configuration these

coordinates will be (XI + UI' Yl + VI)' (x2 + £12, Y2 + v2), and (x3 + u 3' Y3 + v3)' Equating

lengths of sides obtained from these coordinates gives us the following three equations:

/

f(-£1 1+£12 XI +X2)2 + (-VI +V2 -YI + y2 = (X2 _xl)2 + (Y2 - YI)2.
f f(-£12 + U3 - X2 + x 3)2 + (-V2 + V3 - Y2 + y3 = (X3 - x 2)2 + (Y3 - y2

(U I - U3 + XI - x 3)2 + (VI - V3 + YI - Y3)2 = (XI X3)2 + (YI - Y3)2

Expanding these equations and for small displacements neglecting the displacement

squared terms (uI, U I U2'" .), we get the following equations:

2u 1x I - 2u2x I - 2U IX 2 + 2U 2X2 + 2v IYI - 2v2YI - 2V IY2 + 2V2Y2 = 0 ,

=+ + +2U 2X2 - 2U 3X2 - 2U 2X3 2U 3X3 2V2Y2 - 2V 3Y2 - 2V2Y3 2V 3Y3 0

yi -2ulxI - 2u3x I - 2U IX3 + 2U 3X3 + 2vI 2v3YI - 2V IY3 + 2V 3Y3 = 0

-Choosing £II' VI' and u2 as the three independent degrees of freedom (usually called the
master degrees of freedom), we can solve these equations for the remaining three degrees

of freedom (called the slave degrees of freedom). The resulting equations are written in the

MULTIPOINT CONSTRAINTS 75

uz=l, Ul=VI=a
3

Figure 1.25. Rigid-body modes for a plane triangle
form of the following multipoint constraints:

or

z+ X -Xl
---Uz -
YI - Yz
azXl -Xl+ VI Vz =

---U

YI - Yz

=aY3 -Yz
---U l + Yl -Y3 - U3

YI -Yz ---Uz

Yl - Yz

a+ +X -X

_I_ _3 U
v X -X V =
_3_ _1 U? -
YI - Yz: I Yl - Yz - 3

The first column of this transformation matrix represents a rigid-body mode that sets ul = 1

=and vI Uz = a.Similarly, the second column represents a mode with VI = 1, ul = Uz = a

and the third column Uz = 1, ul = vI = a. These three rigid-body modes are shown in

Figure 1.25.
An arbitrary rigid zone in a finite element model can be handled by breaking the shape

into rigid triangles and then using the constraints for each of the triangles. The idea can

also be extended to three-dimensional tetrahedrals each consisting of four triangular faces.

1.6.1 Solution Using Lagrange Multipliers

In the presence of multipoint constraints, the solution of global equations obviously be-
comes complicated. The minimization form introduced in the previous section allows us
to use standard optimization techniques for constrained problems. Thus we set up the fol-
lowing optimization problem for solution of nodal degrees of freedom:

Findd to

Minimize f = !dT Kd - dTR

Subject to Cd - q = 0

76 FINITEELEMENT METHOD:THE BIG PICTURE
44

3
2

o5

-3 p

o (m)
5

Figure 1.26. Five-bar truss with inclined support

A standard optimization technique for solution of this problem is to introduce m Lagrange
multipliers Ai' one for each equality constraint, and define an equivalent unconstrained
problem as follows (for details see Bhatti [95]):

Find d and A to

Minimize L = !dT[(d -dTR + AT(Cd - q)

where L is known as the Lagrangian. The necessary conditions for a minimum of the La-
grangian are that its derivatives with respect to d, and Ai be zero. Carrying out differentia-
tions, the resulting equations are as follows:

-aaLd=O~[. (d-R+CTA=O
aL /
aA-=O~Cd-q=O

Writing the two sets of equations together, the complete system of linear equations can be
expressed as follows:

This system of linear equations can be solved using any of the methods discussed previ-
ously in this chapter. For structural problems a Lagrange multiplier can be interpreted as
the force necessary to apply the specified constraint.

Example 1.17 Truss with an Inclined Support Consider the five-bar pin-jointed
structure shown in Figure 1.26. All members have the same cross-sectional area and are of

the same material, E =70 GPa, and A = 10-3 m2. The load P = 20 leN. The dimensions in

meters are shown in the figure.
For numerical calculations, the N . mm units are convenient. The displacements will be

in mm and the stresses in MPa. The complete computations are as follows:

MULTIPOINT CONSTRAINTS 77

specified Nodal Loads

Node dof Value

3 u3 20000.

Equations for Element 1

E =70000;A = 1000

Element Node Global Node Number x Y
5000. -3000.
1 1
2 3

Xl = 0; YI = 0; Xz = 5000.; Yz = -3000.

L =~ (xz - xI)z + (yz - YI)Z =5830.95

Direction cosines:

=1 Xz -Lxl = 0.857493,' m = Yz L-YI =-05. 14496
S S

Substituting into the truss element equations, we get

8827.13 -5296.28 -8827.13 5296.28](U I ] (0.]
-5296.28 311'J.77 5296.28 -3177.77 vI _ O.

( -8827.13 5296.28 8827.13 -5296.28 u3 - O.
. 5296.28 -3177.77 -5296.28 3177.77 v3 O.

The element contributes tQ{I, 2, 5, 61 global degrees offreedom.

Adding element equations into appropriate locations, we have

8827.13 -5296.28 0 0 -8827.13 5296.28 0 0 ul 0

-5296.28 3177.77 0 0 5296.28 -3177:77 0 0 vI 0

0 0 00 0 0 0 0 Uz 0

0 0 00 00 0 0 Vz = 0
-8827.13 5296.28 0 0 8827.13 -5296.28 0 0 20000.
u3

5296.28 -3177.77 0 0 -5296.28 3177.1'7 0 0 v3 0

0 0 00 0 0 0 0 u4 0

0 0 00 0 0 0 0 v4 0

Processing the other four elements in exactly the same manner, we get the following equa-
tions:

78 FINITE ELEMENT METHOD:THE BIG PICTURE

17654.3 0 0 0 -8827.13 5296.28 -8827.13 -5296.28 !II 0
0 29688.9 0 -23333.3 5296.28 -3177.77 -5296.28 -3177.77 0
0 14000. 0 -14000. VI 0
0 0 .0 0 0 0 0 0
-23333.3 0 23333.3 . 8827.13 0 0 0 !lz 20000.
-8827.13 0 -5296.28 0 0 Vz 0
5296.28 5296.28 -14000. 0 -5296.28 14844.4 0 -11666.7 0
-3177.77 0 0 0 0 22827.1 5296.28 ! l3 0
-8827.13 -5296.28 0 0 -11666.7 5296.28 14844.4
-5296.28 -3177.77 0 v3

!l4

v4

The essential boundary conditions are

Node dof Value

2 112 0
v2 0

Then we remove (3,4) rows and columns. After adjusting for essential boundary condi-
tions, we have

17654.3 0 -8827.13 5296.28 -8827.13 -5296.28 ul 0
0 29688.9 5296.28 -3177.77 -5296.28 -3177.77 0
5296.28 8827.13 -5296.28 vI 20000.
-8827.13 -3177.77 14844.4 0 0 0
5296.28 -5296.28 -5296.28 0 -11666.7 u3 0
-3177.77 0 0 22827.1 0
-8827.13 0 -11666.7 5296.28 5296.28 v3
-5296.28 14844.4
u4

v4

=The multipoint constraint due to inclined support at node 1 is ul sin(7f/6)+ VI cos(7f/6) O.

The augmented global equations with the Lagrange multiplier are as follows:

17654.3 0 -8827.13 5296.28 -8827.13 -5296.28 1/2 ul 0
.,f3 vI 0
0 29688.9 5296.28 -3177.17 -5296.28 -3177.77- T
-8827.13 5296.28 8827.13 -5296.28 0 u3 20000.
-5296.28 14844.4 00 0
5296.28 -3177.77 0 v3 = 0
-8827.13 -5296.28 0 0 0 -11666.7 0
-5296.28 -3177.77 0 -11666.7 u4 0
22827.1 5296.28 0 v4 0
1/2 .,f3 0 0 it 0
5296.28· 14844.4
2
00

Solving the final system of global equations, we get

{U I = 5.14286, vI = -2.96923, u3 = 16.8629, v3 = 12.788,
u4 = -1.42857, v4 = 11.7594, it = 80000.}

Solution for Element 1
Nodal coordinates

Element Node Global Node Number x y

1 1 0 o
2 3 5000.
-3000.

MULTIPOINT CONSTRAINTS 79

XI == 0; Yj == 0; X2 = 5000.; Y2 zi: -3000.

= =L ~ (x2 - XI)2 + (Y2 - YI)2 5830.95

Direction cosines:

TIs = X -x = 0.857493; = =Ins Y2 L-Yj -05. 14496

Global-to-local transformation matrix:

T = (0o.857493 -0.514496 0 0)
0 0.857493 -0.514496

Element nodal displacements in global coordinates:

Ut]d = vI =(5-2.1.946298263 ]
(u3 16.8629

v3 12.788
Element nodal displacements in local coordinates:

d =Td == (5.93762)

I 7,88048

E =70000; A = 1000
Axial strain, E = (d2 - dj)IL = 0.000333197
Axial stress, o: =EE = 23.3238

Axial force, erA = 23323.8

Ina similar manner, we can compute the solutions over the remaining elements:

. Stress Axial Force

1 23.3238 23323.8

- .2 23.3238· 23323.8

3 69.282 69282.

4 -20. -20000.

5 -12. -12000.

1.6.2 Solution Using Penalty Function

The Lagrange multiplier method of imposing constraints has two drawbacks. First, it
requires adding new rows and columns to the global system of equations that for large
systems may be inefficient. Second, the resulting system has zeros on the diagonal corre-
sponding to the constraint equations. Some simple equation. solvers that assume nonzero
diagonal terms may not work for this system. Another standard technique of imposing
constraints, the so-called penalty function approach, does not have these two drawbacks.

In this method a large penalty number f1 is chosen and the equivalent unconstrained
problem is defined as follows (for details see Bhatti [95]):

80 FINITE ELEMENTMETHOD: THE BIG PICTURE

Findd to

=Minimize ¢ ~dTKd - dTR + ~fJ.(Cd - ql(Cd - q)

With fJ. being large, the minimization process forces the constraints to be satisfied. The
necessary conditions for the minimum result in the following system of equations:

Rearranging terms, the system of linear equations can be expressed as follows:

This system of linear equations can be solved using any of the methods discussed previ-
ously in this section.

The performance of the method depends on the value chosen for the penalty parameter

u. Large values, say of the order of fJ. = 1010, give accurate solutions; however, the resulting

system of equations may be ill-conditioned. ITfJ. values are small as compared to other
terms in the global equations, the solution will not satisfy the constraints very accurately.
A general rule of thumb is to set fJ. equal to 105 times the largest number in the global K
matrix.

Example 1.18 Truss Supporting a Rigid Plate A plane truss is designed to support a
rigid triangular plate as shown in Figure 1.27. All members have the same cross-sectional
area A = 1 in2 and are of the same material, E = 29,000 ksi. The load P = 20 kips. The
dimensions in ft are shown in the figure. Note there is no connection between the diagonal
members where they cross each other.

The model consists of five nodes and thus the global system of equations before bound-
/'

ary conditions will be 10 x 10. The equations for the six truss elements are written as in
the previous examples and assembled in the usual manner to give the following system of
equations:

2
20

r'"~J

o (ft)
50
o 25

Figure 1.27. Truss supporting a rigid plate

I MULTIPOINT CONSTRAINTS 81

142.693 36.8215 0 0 -96.6667 0 -46.0268 -36.8215 0 0 "1 0
0
36.8215 150.29 0 '-120.833 0 0 -36.8215 -29.4572 0 0 vI 0
0
0 0 142.693 -36.8215 -46.0268 36.8215 -96.6667 O. 0 0 "2 0
0
0 -120.833 -36.8215 150.29 36.8215 -29.4572 O. O. 0 0 v2 0
-96.6667 0 -46.0268 -36.8215 O. : -20
0 36.8215 142.693 150.29 O. o. 0 0 "3 0
0 36.8215 0 0 v3 -20
-29.4572 -36.8215 -120.833

-46.0268 -36.8215 -96.6667 O. O. O. 142.693 36.8215 0 0 "4
-36.8215 -29.4572 O. O.
O. -120.833 36.8215 150.29 0 0 v4

0 0 0 0 0 0 0 0 0 0 "5
0 0 0 0 0 0 0 0 0 0 v5

The essential' boundary conditions at node 1 (u l = vI = 0) are incorporated by removing
the corresponding rows and columns in the usual way. Node 3 also has zero vertical dis-

placement. However! since this node is connected to the rigid plate as well, the boundary

condition v3 = 0 will be imposed later as part of the multipoint constraints. Removing the

first two rows and columns, the global system of equations is as follows: .

Kd=R ~

142.693 -36.8215 -46.0268 36.8215 -96.6667 O. 00 u2 O.
-36.8215 150.29 36.8215 -29.4572 O. O. 00 O.
-46.0268 36.8215 142.693 -36.8215 O. O. 00 "2 O.
-29.4572 150.29 O. -120.833 00 O.
36.8215 -36.8215 36.8215 00 u3 O.
-96.6667 O. O. O. 142.693 150.29 00 v3 -20.
O. O. -120.833 36.8215 0 00 u4 O.
O. 0 0 0 00 v4 -20.
0 0 0 0 0
0 0 0 Us

Vs

The rigid plate is connected between nodes 3, 5, and 4. Treating u3 ' "s- and Us as indepen-

dent degrees offreedom, the multipoint constraints are as follows:

Expanding and rearranging, we have

To this list we must also add the roller support constraint that "3 = O. Thus the complete set
of constraint equations, expanded to include all degrees of freedom present in the global
equations, are

82 FINITEELEMENTMETHOD:THE BIG PICTURE

Uz

o -~ ~ Vz
~] t1 =mCd.=q~ [0~o0 -1
o0 -1 0 0
1 0 0
o 0 1

-1
o0 1 00 0

Us

Vs

To use the penalty function approach, we choose the penalty parameter j1 equal to lOs times
the largest number in the global K matrix:

j1 = 150.29 X lOs = 1.5029 X 107

Incorporating the constraints into the global equations with this value of u, the final system

of equations is as follows:

( K + p C T C j d = R + p CT q =

0.00142693 -0.000368215 -0.000460268 0.000368215 -0.000966667 O. 0 0 "2 O.
0.0015029 0.000368215 -0.000294572 O. O. 0 0 "2
-0.000368215 0.000368215 -187.863 O. O. 234.829 187.863 1/3 O.
-234.827 -450.87 O. 150.289 187.863 150.29
-0.000460268 -0.000294572 -187.863 0.000368215 150.29 0 "3 o.
O. O. -150.289 -150.289 0 0 "4
liP 0.000368215 O. 150.289 0.000368215 0 -385.119 -187.863 "4 O.
-0.000966667 o. O. 187.863 0 -187.863 -150.29 1/5 O.
234.829 150.29 150.29 -20.
O. 0 187.863 0 "5
0 O.
0 -20.

0

Solving this system of linear equations, we get

(Uz = 0.172845, Vz = 0.076446, u3 =-0.139174, v3 = 3.99227 X 10-6,
U4 = 0.292292, v4 6.0391~·X 10-6, Us = 0.29229, Vs = -0.539324)

Substituting these values into the constraint equations, we can see that the constraints are
reasonably satisfied:

6

Cd = 11..3636034756 xx'11O0--6 ] ", [00]

[ 2.0469 X 10-6 0
3.99227 X 1O~6 0

Knowing the nodal values, the element solutions can be computed in the usual manner:

Strain Stress Axial Force

1 0.000318525 9.23722 9.23722

2 -0.000463913 -13.4535 -13.4535

3 0.000594098 17.2289 17.2289

4 0.000398156 11.5465 11.5465

5 -0.000509888 -14.7868 -14.7868

6 8.52877 x 10-9 0.000247334 0.000247334