Fundamental-Finite-Element-Analysis-and-Applications

PROBLEMS 583

LL-------:---,M;-zas~s ~ m

Figure 8.25.

aluminum with E = 70 GPa and p = 2710 kg/rrr'. The attached mass is 112 = 500 kg.

The nodal coordinates, in meters, are as follows:

x yz

1 O. 4. O.
2 -3. 2. 5.
3 -3. O. O.
4 3. O. O.
5 O. O. 5.

8.9 Determine free-vibration frequencies and mode shapes for the uniform beam sim-
ply supported at ends and spring supported in the middle as shown in Figure 8.26.
Assume the following numerical data:

L = 4m; "'p = 271Okglm3 ; E = 70 GPa; =112 350kg

1= 350.9 X 10-6 m"; A = 8.39 X 10-3 m2 ; =k 4x 105N/m

Assume the spring is massless.

I~

Figure 8.26. Beam supportedby a spring

584 TRANSIENTPROBLEMS

8.10 Determine free-vibration frequencies and mode shapes for the plane frame shown
in Figure 8.27. Note the diagonal brace can carry axial forces only. The dimensions
in meters are shown in the figure. Assume the following numerical data:

p =7850kglm3; E = 210 GPa; m = 350kg

Frame members: 1= 350.9 X 10- 6 m"; A = 8.39 X 10-3 m2 ;
Diagonal brace: A = 1.252 X 10-3 m2

y
Mass, ill

4

o

- -o - - - - - - - - -x
5
3

Figure 8.27.

Transient Vibrations

8.11 The left half of the beam shown in Figure 8.28 is supporting a rotating machine
operating at 1500 rpm. Due' to unbalance, the machine exerts a uniformly distributed
load of 800 lb/in that varies harmonically. Using two beam elements, determine the
time history of midspan displacement. Use the following numerical data:

L =150in; =E 30 x 106 psi;

t ! !q sinwIt
ttt~ ~ ~ t

I'" L/2 '~r~-- L/2 ~I

Figure 8.28.

PROBLEMS 585

8,12 The plane frame shown in Figure 8.29 is subjected to pressure q from a blast. A
simple idealization of the blast loading is that of a triangular pulse as shown in the
figure. Use the following numerical data:

q =200lb/in; td =0.02 s; L = 180 in; =E '30 x 106lb/in2

A = 100in2; 1= 1000in4 ; m = 0.03lb· s2/in

T q LLoad time history

L/-fi.

1

L -I

Figure 8.29. Plane frame with distributed load

CHAPTER NINE

p-FORMULATION

In conventional finite element formulation, each element is based on a specific set of inter-

polation functions. After choosing an element type, the only way to obtain a better solution

is to refine the model. This formulation is called the h-formulation, where h indicates the

generic size of an element. For one-dimensional problems h is the length of an element. For

two- and three-dimensional problems it can be thought ofas the diameter of the inscribing

circle or sphere. In general, the finite element solution converges to the exact solution as

h -? O. /

An alternative formulation, called the p-formulation, is presented in this chapter. Inthis

formulation, the elements are based on interpolation functions that involve very high order

terms. The initial finite element model is fairly coarse and is based primarily on geometric

considerations. Refined solutions are obtained by increasing the order of the interpolation

functions used in the formulation. Efficient interpolation functions have been developed so

thathigher order solutions can be obtained in a hierarchical manner from the lower order

solutions.

9.1 p-FORMULATION FOR SECOND-ORDER 1D BVP

In this section p-finite element equations are derived for the one-dimensional boundary
value problem of the following form:

=ddx ( lc(xd)-£;lr(X;-)) + p(x)£l(x) + q(x) 0;

where k(x), p(x), and q(x) are given functions of x and £l(x) is the solution variable. The
586

p-FORMULATION FOR SECOND-ORDER 10 BVP 587

Master element

-t------------ X

Figure 9.1. Two node p-element for second-order BVP

boundary conditions of the following form may be specified at one or more points:

EBC: u =specified

NBC: -du =au +[3

dx

where a and [3 are some specified constants.
. The basic element is a two-node element shown in Figure 9.1. The mapping to the
. master element is based on linear Lagrangian interpolation functions for the two-node line
element:

-dx =-x-I +x-2 =L- and

ds 2 2 2

Allowing for the possibility of natural boundary conditions at the element ends, over an
element we must satisfy

Note that in the natural boundary condition at XI the negative of the derivative is specified.
As explained in Chapter 3, when applying this element to physical problems, the sign
makes sense with the usual sign convention adopted for those problems. The weak form
can be written using standard steps of writing the weighted residual, integration by parts,
and incorporating the natural boundary conditions:

9.1.1 Assumed Solution Using Legendre Polynomials
Each element has two nodal degrees of freedom (u l and u2) ·and any number of additional
degrees of freedom identified as 81' 82, .... The first two interpolation functions are the
same as those used in the mapping and directly involve nodal unknowns. The remaining
interpolation functions (called the p-modesi are not associated with nodal values. We can

588 p-FORMULATION
¢o

---I----s
-1 1

\J' _I,

--~

Figure 9.2. Plots of first nine Legendre polynomials

choose any set of functions for defining p-modes as long as they are easy to differentiate
and integrate. Among several possibilities, p-modes written in terms of Legendre polyno-
mials have been used quite successfully. The Legendre polynomials can be generated by
the following recursive equation:

=(i + l)¢i+1(s) (2i + l)s¢i(s) - i¢i_l (s); =i 1,2, ...

¢o(s) = 1; ¢I(S)=S

,I

where ¢i(S) is the ith Legendre polynomial. The first nine Legendre polynomials are given
below explicitly. Figure 9.2 shows plots of these functions over the interval (-1, 1). Note
that the even Legendre polynomials have a value of 1 at the ends and the odd polynomials
have values of -1 and 1 at the ends. The polynomial values fluctuate evenly over the entire
interval. The Legendre polynomials also have the following orthogonality property that
can easily be verified by direct computation:

{2i~ 1=r1¢i(S)¢j(S)dS for i =j

J-l .0 for i:1' j

¢o =1

¢I =S

¢z = 3sz 1

2" -"2

¢3 = 5s3 - 3s

2" 2"

p-FORMULATION FOR SECOND-ORDER 10 BVP 589

35s4 l5s2 3

¢4= -8--T+ 8

63s5 35s 3 l5s

¢5= -8--4+8

23ls6 3l5s4 105s2 5

¢6 = 16 - 16 +16 - 16

. 429s7 693s 5 3l5s3 35s

¢7= 16-16+16-1"6

6435s 8 300s6 3465s4 3l5s2 35
¢8= 128-32+~-32+128

An important consideration in choosing appropriate functions for the p-modes is the re-
quirement that the solution be continuous over the entire domain. In the standard finite
element formulation the solution at a node common between two elements is made con-
tinuous by using the same nodal degree of freedom at the node. Since the p-modes are not
associated with a node, they must be chosen carefully. The simplest approach is to define
p-modes that are always zero at the element ends. Thus they do not influence the solu-
tion outside of the element and are unique to each element. Therefore the standard finite
element assembly procedure can be used to get the global finite element equations. The
numerical examples presented in the following section will further clarify this point.

From the plots of the Legendre polynomials, it is easy to see that we can get zero values

=at s ± I by subtracting two even or two odd polynomials from each other. Thus suitable

p-modes can be defined by the following equation:

i = 1,2, ...

Note the definition includes a factor that is chosen for convenience. The particular choice

makes the integral for the orthogonality property come out to 1. By direct computation, it

can be verified that the two p-modes satisfy the following orthogonality property:

r {1l dPi(s) dPj(s) ds = for i = j
J-I ds ds 0
for i *" j

Because of this orthogonality property, the element matrices can be constructed in a very
efficient manner. The first six p~modes are written explicity as follows and are also plotted
in Figure 9.3. It can easily be seen that all the p-modes are zero at the ends. Being simple
polynomials, they are easy to differentiate and integrate as well.

tf>i-l tf>i+! IfPj

2S 3s2 1 -1 -(s2 - 1)
22 22

5s3 3s 1~- -s(s2 - 1)
22
22

590 p-FORMULATION

Figure 9.3. First six p-modes

tPi- 1 tPi+1 ffPi-81-(5s4 - 6s2 + 1)
3s2 1
35s4 15s2 3 2
3
22 -8--4+8 3s(7s4 - 10s2 + 3)
5s3 3s
63ss 35s3 15s 8...[2
4
22 -8--4+8 l f i-16- ( 2 I s6 - 35s4 + 15s2 - 1)
2
35s4 15s2 3 -231-s6- -31-5s4+1-05-s2 - -5

5 -8--4+8 16 16 16 16

6 63s5 35s3 15s 429s7 693ss 319 35s -1I6 f-f2s(33s6 - 63s4 + 35s2 - 5)

-8--4+8 1:6-1:6+1:6-"16

It is interesting to contrast these p-modes with the high-order Lagrange interpolation func-
tions. For example, the middle six, interpolation functions for a seventh-order Lagrange
interpolation are as shown in Figure 9.4. All these functions have zero values at the ends
and thus could have been used as p-modes. However, the function values fluctuate widely
over the interval, and thus the functions are not as nicely behaved as the p-modes based on
Legendre polynomials.

~ L3 L4

-~, ~' -~,-1 -1 .

Ls L6 L7

-~, -~-~1 tlr -~,
-1

Figure 9.4. Second through seventh interpolation functions for a seventh-order Lagrange interpola-
tion

p-FORMULATION FOR SECOND-ORDER 10 BVP 591

9.;1.2 Element Equations

The assumed solution consists of two linear Lagrange interpolation functions and any num-
ber of p-modes:

lUI]... )~UIz T

:iN d

where up Uz are nodal degrees of freedom and 01'0z" .. are unknown parameters asso-
ciated with the p-modes. With the mapping the derivatives of the assumed solution with
respect to actual element coordinate x can be written as follows:

du _ du ds 2 du _ 2 (dNI dNz dP I

=dx - ds dx L ds - L ds ds ds

The weak form over the element is

l x, (q(x)w;(x) + p(x)u(x)w;(x) - k(x)u' (x)w;(x)) dx
Xl
=-k(xl )(- f31 - (Xlu(XI))wi(XI) + k(xz)(f3z + (Xzu(xz))w;(xz) 0

Note the prime indicates differentiation with respect to x. Using the mapping x = xes), all

terms inside the integral are expressed in terms .of s, and the weak form is

Jr-Il ( q(s)w;(s) + - - 2dU2dw.)L
k(s)L ds L ds' '"ids
p(s)u(s)wi(s)

=- k(xl)(-f31 - (XI u(xl))wi(xl) + k(xz)(f3z + (Xzu(xz))w;(xz) 0

The weighting functions Wi"are NI' Nz, PI' .... We get one equation with each weighting

function. Writing all equations together in a matrix form, we have

Noting that u(x l) = uI, u(x z) = uz, NI(-l) = 1, NI(l) = O;Nz(- l) = 0, Nz(l) = 1, and
P;(±l)
= 0, we have

592 p-FORMULATION

Substituting the assumed solution, we have

1: ~d'-k(X,)(-P,(qN+ pNN'd - kBB'd) -a,u,)[g]

+k(~)$, +a,~)m=m

Rearranging the terms by keepirig the terms involving unknown parameters on the left-hand
side, we have

t t kexI)U'1 £II] [keXI)!3I]
kBBTd!:.. ds _ pNNTd~ ds _ keX2~U'2U2 = (I qN!:.. ds +
k eX
LI 2 LI - ( ··· LI 2
O)!32
...

Define ,I

II L

kk = kBBT ?ds

-I -

II L
k = - pNNT -ds
2
P -I

Ir = IL
qN-ds
q -I 2

keXI)U'IUI] [-U'lkeXI)
0
keX2)U'2U2
0
=- 0
[ ··· ...

p-FORMULATION FOR SECOND-ORDER 10 BVP 593

Thus the element equations are

As explained in detail in Chapter 3, during assembly, the natural boundary condition terms
are incorporated directly into the global equations as follows:

NBC atdof i: Add to global k Add to global r

ki], i) = leU, i) - a;le(x) rei) = rei) + le(,-":;)f3;

Thus at the element level the equations are simply

,9.1.3 Numerical Examples

Example 9.1 Using only one element model, shown in Figure 9.5, obtain the finite ele-
ment solution of the following boundary value problem. Consider 0, I, and 2 p-modes.

=d + 0; O<x<1

dx
(ru") (x - l)u

u(O) = 1; u'(l) = 6

For this problem

k(x) = r;p(x) = x-I; q(x) = 0
EBCatx=O:u= 1
NBC at x = 1: a = 0, f3 = 6

Element nodes: 0,1; L= 1

Change of variables: xes) = s + 1

... 2

lees) = !(s -i- 1)2; s+1 . q(s) = 0

pes) = -2- - 1;

With no p-modes we get the standard linear element equations as follows:

Interpolation functions, NT = ( 1 ~ s s ~ 1 )

UI Uz
:ct ':~~..; ..~__... ~.~ -
~-
XI= 0
xz=·l

-1-------------,- x

Figure 9.5.

594 p-FORMULATION

BT = dNTIdx = (VL)dNTIds = (-1 1)

L\ (!~kk = Ws + I)2)BBT V2ds =

I 1)'L\ t t(skp = (1 -• + l))NNTLds = ( J
. 12 12

The complete element equations are as follows:

(~ (0)-~)(UI) = 2... u2 0

12

The natural boundary conditions are

dof a f3

u2 0 6

dof k(x) -k(x)a k(x)[3

u2 1 0 6

The global equations after incorporating the NBC are

(~

/

The essential boundary condition is

dof Value

Incorporating the EBC, the final system of equations is as follows:

(f2)(u2) =(¥)

Solution of global equations: {ul -? 1, u2 -? IS}

The solution over elements is as follows:

Element 1

Nodes: {xI -? 0,x2 -? I}

Interpolation functions: NT = {t(2 - 2x), x}

dofvalues: aT ={I, IS}
Solution: u(x) =NTa = 14x + 1

p-FORMULATION FOR SECOND-ORDER 1D BVP 595

With one p-mode the element equations are as follows:

~ l~?Interpolation functions, NT =( I s s+l
2 -2 -2(s-. - 1))

BT = -dN T = ( 2 I dNT =(-1 1 {6s)
L)-
dx ds
u:k, =LWlH 1)')BB'Ld, =
-3"1 --.(I6

1 1

3" -.(6

1 4

-.(6 -.(6 5"

11

4 12

11

12· 12

1

TO

Note that the first 2 x 2 submatrix in these two matrices are exactly the same as those for

the previous case of no p-modes. This is a direct result of orthogonality of the·p-modes. In
an actual implementation of p-formulation, we can take advantage of this fact and compute
only the new terms that are added to the equations when considering higher p-modes. The
complete element equations are as follows:

7 (: H~)-41 13

12 -10-.(6

-41 5 z.y'f 8~)

13 12 -5-

-10-.(6 z.y'f 0

-5- 9

TO

The superscript over 8 indicates the element number while the subscript indicates the pa-
rameter number.

With only a single element the global equations before boundary conditions are the
same as the equations for the first element:

[0)=0.583333

-0.25
-0.25 -00..533260579293)[ I] 0
0.416667 Ll
LiZ
AI)[ -0.530723 0.326599 0.9
0

The natural boundary conditions are

dof k(x) -k(x)a k(x){3

LiZ 1. 0 6.

596 p-FORMULATION
The global equations after incorporating NBC are

0.583333 -0.25 -00..533260579293)( £II] ( 06.)

-0.25 8i0.416667 Uz ==
[ -0.530723
0.326599 0.9 l) 0

The essential boundary condition is

dof Value

Incorporating the EBC, the final system of equations is as follows:

8i0.416667 uz1) ) (6.25 )
0.530723
( 0.326599
0.326599)( ==
0.9

8iSolution of global equations: {£II -7 1., Uz -7 20.3168, l -7 -6.7830l}
)

The solution over elements is as follows:

Element 1
Nodes: {XI -7 0, Xz -7 1}

Y61 dC2x-1l-1}
Interpolation functions: NT (' { 2'C2 - 2x), X, z z

dof values: dT == {I., 20.3168, -6.78301}
Solution: uCx) == NTd == -16.6149x2 + 35.9317x + 1.

With two p-modes the element equations are as follows:

~ ~ ~Interpolation functions: NT == ( 1 S, S; 1 .j[Cs2 _ 1) .j[SCs2 _ 1))

B d:: d:(~)T == ==T Y6s ..jIC3sZ - 1))

== ( -1

1 -31 11

3 -{6 -3{fii

-31 11 1
3' {6 3{fii

- ( 6I 1 4 2

{6 '5 -{l5

1I 2 16
-3{fii 3{fii -{l5
2T

p-FORMULATION FOR SECOND-ORDER 1D BVP 597

I I _-..II I
4 12
10 6{10

II ~ I I 1 0
12 12
(1- !(s + l))NNTLds = - 5Y6 I
kp = -I - 14-{i5
-1-..0II I
I
10
- 5Y6

I 0 I I
6{10 - 14-{i5
42

The complete element equations are as follows:

7 -4I 13 I

12 -1OY6 -6{10

-4I 5 -z-5..I-I [~H~]1
12
9 3{10
13 z...{f
10 9-{f
-1OY6 -5- 14 8~) 0

I 1 9-{f II

-6{10 3{10 14 14

Once again, note that the lower p-mode equations are contained in the higher ones.
The global equations before the boundary conditions are

0.583333 -0.25 -0.530723 (0]I] 0
-0.25 0.416667 0.326599
( -0.530723 0.326599 0.9 -00..0150524700946][ LLliZ
-0.0527046 0.105409 0.497955
0.497955 8\1) = 0

0.785714 8~l) 0

The natural boundary conditions are'

LiZ 1. -k(x)a: k(x){3

o 6.

The global equations after incorporating the NBC are

0.583333 -:0.25 -0.530723 -00..0150257400946][. LLIZlI] (06.]
-0.25 0.416667 0.326599
[ -0.530723 0.326599 0.9 0.497955 8\1) = 0
-0.0527046 0.105409 0.497955
0.785714 8~l) 0

The essential boundary condition is
dof Value

598 p-FORMULATION

Incorporating the EBC, the final system of equations is as follows:

=0.4 16667

0.326599
0.326599 00..419075945059) [ 2 ] (60..25530723 )
0.9
oUil)
( 0.105409 0.497955 0.785714 oil) 0.0527046

Solution of global equations:

{ul -? 1., u2 -? 20.9794, (\) -? -8.47513, (I) -? 2.62375 }
0\ 02

Solution over elements is as follows:

Element 1

Nodes: {XI -? 0, x2 -? 1}

Interpolation functions:

-l)}NT =
-I(2 _ 2x) X 2(2x - 1)2 - 2 ~(2x - 1)3 - 2(2x

22 - 2
{ 2 " {6' -{16

=dof values: dT (1., 20.9794, -8.47513, 2.62375)
= =Solution: u(x) NTd 16.594h.3 - 45.6508x2 + 49.0361x + 1.

These solutions, together with the one obtained by using three p-modes, are shown in

Figure 9.6. The solutions are clearly converging. The plot of the first derivative shows that

there is still room for further improvement. The natural boundary condition is not satisfied

yet either but the solution is getting closer. The first derivative values at X = 1 are as

follows: i

Linear solution Computed u' (l )
With 1 p-mode
With 2 p-modes 14.
With 3 p-modes 2.70186
7.51665
5.17712

u du/dx -3p

20 60
50 --2p

15 40 - l p

30 -Lin.
10 20 .~

x 10+I----------="="~"'-'''='''--'-'''-"" x

0.2 0.4 0.6 0.8

Figure 9.6.

p-FORMULATION FOR SECOND-ORDER 1D BVP 599

Figure 9.7. Heat transfer through a circular fin

Example 9.2 AIm x 1 m plate is at a temperature of 100°F. It is to be cooled by using

3-cm-long and 0.25-cm-diameter aluminum fins (k = 237 W/m- °C) with a center-to-center

distance of 0.6 em. A typical fin is as shown in Figure 9.7. The ambient air temperature
is 30°C and average convection coefficient is 35 W/m2 • °C. Determine the temperature

distribution through the fins. What is the rate of heat transfer from the entire finned surface

of the plate?

"Assuming that the surrounding temperature Too is uniform, the problem can be treated

asone dimensional with the governing differential equation as follows:

-ddx (kx1" d1dTx-) - hP(T - T ) =0; O<x<L
00

at x=O and - dT =M(T - T) at x =L
kJid.x-
00

where kx = thermal conductivity, h = convection coefficient, P = 2rrr = surface area per

unit length over which convection takes place, r =radius of the fin, A =rrr2 = area of cross
section, and Too = ambient air temperature. Once the temperature distribution is mown,

the total heat loss from a fin can be computed using the following equation:

LLHeat loss = hP(T(x) - Too) dx

Comparing with the general form, we have

p =-hP;

EBC atz =0: NBC atx = L: Q!,=-Mk_A.' f3 = MToo

kA

The problem is solved using two elements with two p-modes each. The complete solution
details areas follows:

Derivation of element equations: 1~- -(s?- '- 1)
Element nodes: 0, L; L = L
Change of variables: xes) = !L(s + 1) 22

k(s) = kA; pes) = -hP; q(s) = hPToo

Interpolation functions, NT = ( 1 ; s s+1
2

600 p-FORMULATION

=(~)dNT {f(3t-1))
=(T "i1. 1 -Y6s
0
BT=dN -

dx L ds LL

kA -IkA: 0

I:

=1\ -IkA: kA 0 0

I:

kk !UcA)BBTLds = 2kA

0 0 T 0

0 0 0 2kA

T

hPL /lPL /lPL /lPL
"3 6 - 2-f6 6-{16

/lPL /lPL /lPL /lPL
6
=1\k p ~(hP)NNTLds = "3 - 2-f6 -6ViQ
/lPL 0
- 2-f6 /lPL /lPL

- 2-f6 "5

.l1f.b... /lPL 0 /lPL
6-{16 - 6-{16
21

IIT _

-Y6r q -
-I 1_ {h~PTc'oL ~hPT'coL - hPTcoL }
'i(hPTco)NLds -
,0

The complete element equations are asfollows:

!;d + /lPL 6/lPL- Ik:A hPL /lPL - 2 -/lPT~L
-hP-Ty~L-
L3 !;d + !lPL - 2-f6 [~+6-{16
/lPL hPT~L
6!lPL- Ik:A L3 hPL
-6-{16 - -f6
- 2/l-Pf6L !lPL - 2-f6 0 8~)
2kA + hPL 0
- 2-f6
L5
/lPL /lPL 2kA + /lPL
0 L 21
6-{16 -6-{16
=p(x) -0.549779; =q(x) 16.4934
=k(x) 0.00465348;

Two-element solution:

Nodal locations: {O, 0.015, 0.03}

Element 1:

Element nodes: (Xl -? 0, x2 -? 0.015}

0.312981 -0.308858 -0.00168335 I

-0.308858 0.312981 -0.00168335 -00.,000000443344663388]( TTz ] [ 00..11223377 ]

[ -0.00168335 -0.00168335 0.622114 o 8il ) = -0.101001
0.000434638 -0.000434638
o 0.620857 8~1) 0

Element 2:
Element nodes: {x2 -? 0.015, x3 -? 0.03}

0.312981 -0.308858 -0.00168335 -00.0,00000443344663388][ TT32] [ 00..11223377 ]
-0.308858 0.312981
[ -0.00168335 -0.00168335 =o 8i2) -0.101001
-0.00168335
0.000434638 -0.000434638 0.622114 O. 620857 8~2) 0 .

o

p-FORMULATION FOR SECOND-ORDER 10 BVP 601

Global equations before boundary conditions:

0.312981 -0.308858 o -0.00168335 o 0.000434638 o TI 0.1237
0.625962 0.2474
-0.308858 -0.308858 -0.00168335 -0.00168335 -0.000134638 0.000434638 T2 0.1237
-0.308858 -0.101001
o -0.00168335 0.312981 o -0.00168335 o -0o.000434638 0T\1,1 -0.101001
-0.00168335 o o 0\'1
-0.00168335 -0.000434638 o 0.622114 o o o o
oill o
o 0.000434638 -0.00168335 o 0.622114 0.620857 0.620857 oi'l
o
0.000434638 o o o o
o
o -0.000434638

Natural boundary conditions:

dof a (3

T3 -0.147679 4.43038

dof k(x) -k(x)a k(x){3

T3 0.00465348 0.000687223 0.0206167

Global equations after incorporating NBC:

0.312981 -0.308858 o -0.00168335 o 0.000434638 o TI 0.1237
0.625962 0.2474
-0.308858 -0.308858 -0.00168335 -0.00168335 -0.000434638 0.000434638 T2 0.144317
-0.308858 -0.101001
o -0.00168335 0.313668 o -0.00168335 o -0o.000434638 T, -0.101001
-0.00168335 o 0\11
-0.00168335 -0.000434638 o 0.622114 o o o
o 0\'1 o
o 0.000434638 -0.00168335 o 0.622114 0.620857
o o oill
0.000434638 o o o o 0.620857
o 0~1
o -0.000434638

Essential boundary condition:

dof Value
T1 100

Incorporating the EBC, the final system of equations is as follows:

0.625962 -0.308858 -0.00168335 -0.00168335 -0.000434638 0.000434638 T2 31.1332
-0.308858 0.313668 0 -0.00168335 0 -0.000434638 T3 0.144317
-0.00168335 0 0.622114 0 0\1) 0.0673339
-0.00168335 0 0 0 0
-0.000434638 -0.00168335 0 0.622114 0.620857 0 0\2) -0.101001
0 0 0 0 0 -0.0434638
0.000434638 0 0.620857 O~l)
-0.000434638 0~2) O.

Solution of global equations:

(I) . (?)
{T 100, T2 97.1818, T3 96.1534,01 -7 0.371193,01- -7 0.360785,
-7 -7 -7
I

02(I) -7 -0.00197291, 02(2) -7 -0.00071994)

602 p-FORMULATION

Solution over element 1:

Nodes: {xI -t 0, x2 -t 0.015}
Interpolation functions:

!(l -NT ={ 66.6667(21:- 0.015)), !(66.6667(2x - 0.015) + 1),

6666.67(2x - 0.015)2 - ~ 740741.(2x - 0.015)3 - 166.667(2x - 0.015)}

~' ~

dofvalues: dT = {100, 97.1818, 0.371193, -0.00197291}
Solution: T(x) = NTd = -3697.13x3 + 4124.23x2 - 248.912x + 100.

Solution over element 2:

Nodes: {xI -t 0.015, x2 -t 0.03}
Interpolation functions:

!(l -NT ={ 66.6667(2x - 0.045)), !(66.6667(2x - 0.045) + 1),

6666.67(2x - 0.045)2 - 740741.(2x - 0.045)3 - l66.667(2x - 0.045)}
~

dofvalues: dT = {97.l818, 96.1534, 0.360785, -0.00071994}

Solution: T(x) = NTd = -1349.13x3 + 4018.8x2 - 247.281x + 99.9913

Solution summary:

Range "Solution

1 0 5; x 5; 0.015 -3697.13x3 + 4124.23x2 - 248.912x + 100.

2 0.015 5; x 5; 0.03 -1349.13x3 + 4018.8x2 - 247.28lx + 99.9913

A plot of the solution is shown in Figure 9.8.

{-3697.13x3 +4124.232 - 248.912x+ 100., -1349.13x3 +4018.82 - 247.281x+ 99.9913}

T
100

99

98

0.005 0.01 0.015 .
Figure 9.8. Temperaturedistributionin the fin

p-FORMULATION FOR SECOND-ORDER 10 BVP 603

The heat loss over the length of a fin is determined as follows:

iL
hP(T(x) - Too) dx

Jo(0.015

= 0.549779(-3697.13;2 + 4l24.23~ - 248.9l2x + 100. - 30) dx

10.03

+ 0.549779(-1349.13;2 + 40l8.8~ - 247.281x + 99.9913 - 30) dx

0.015

= 1.11297

The following analytical solution for heat loss is available for the problem:

= _~ 8inh[mL] + b/(mk) Cosh[mL]
m yhF/(kA);"V hPkA (Tb - Too) Cosh[m]L + b/(mk) 81. nh[mL]

1.07569

The finite element solution compares fairly well with the analytical solution. To compute
the heat loss from the entire plate surface, we need to determine the total number of fins
and multiply the heat loss per fin by this number:

1mx 1m

Number of fins = (0.006m)(0.006m) =27,778 fins

Heat loss through all fins = 27778 x 1.113 = 30,917 W

As mentioned earlier, because of the orthogonality properties of p-modes, additional p-
modes simply add new rows and columns to the element k matrix. The existing entries
in the matrix do not change. By ordering the global equations as we have done in this
example, the global K matrix also retains this property. Thus adding new p-modes simply
requires computation of the new element rows and columns and their assembly into the
existing global equations. This is clearly demonstrated in the following global equations'
obtained with two elements and zero, one, and two p-modes.

With no p-modes we get the following global equations:

0.312981 -0.308858 0 ][Tl ] [0.1237]
o-0.308858 0.625962 -0.308858 Tz = 0.2474
-0.308858 0.312981
[ T3 ' ... 0.1237

With one p-mode for each element two new rows and columns are ridded to the global
equations. Note the first 3 x 3 system of equations is exactly the same as that with no

p-modes:

0.312981 -0.308858 0 -0.00168335 .0 t,
-0.308858 0.625962 -0.308858 -0.00168335 -0.00168335
-0.00168335 Tz [ 00.12243774
0 -0.308858 0.312981 0 0
-0.00168335 -0.00168335 0 0.622114 T3 = 0.1237
-0.00168335 -0.00168335 0 0.622114
0 8l1) -0.101001

8~) -0.101001

604 p-FORMULATION - - Linear be
T - - 1 P mode TI
- - 2p modes
100 - 3p modes
- - 4 p modes
99
98
97

x
0.005 0.01 0.015 0.02 0.025 0.03

Figure 9.9. Comparison of solutions using one element with zero to four p-modes

With two p-modes for each element the first 5 x 5 system of equations is exactly the same 9.
TI
as that with one p-mode:
CI
0.312981 -0.308858 0 -0.00168335 0 0.000434638 0 TI 0.1237 ell
-0.308858 0.625962 -0.308858 -0.00168335 -0.00168335 -0.000434638 0.000434638 T2 0.2474
-0.00168335 -0.000434638 T, 0.1237 WI
0 -0.308858 0.312981 0 0 0 0\11 -0.101001
-0.00168335 -0.00168335 0 0.622114 0 0 0 -0.101001 wI
·0 -0.00168335 -0.00168335 0 0.622114 0 0 0\21 0 an
-0.000434638 0 0 0 0.620857 0.620857 0 in
0.000434638 -0.000434638 0 0 0 0~1) si:
0 0.000434638 6~21

Thus, taking advantage of this hierarchical nature of element and global equations, it is
possible to create very efficient computer implementations based on the p-formulation.

The solution convergence can be demonstrated by evaluating solutions with increasing
number of p-modes. Figure 9.9 shows solutions obtained from using only one element
with zero through four p-modes. Except for the linear solution, all others are practically
indistinguishable from each other.

9.2 p-FORMULATION FOR SECOND-ORDER 20 BVP

In this section we consider p-formulation for the general two-dimensional boundary value
problem first introduced in Chapter 5. The differential equation defined over an arbitrary
two-dimensional area is as follows:

where kxCx, y), k/x, y), p(x,y), andq(x, y) are known functions defined over the area. The
solution variable is u(x, y). The area of the solution domain is denoted by A and its boundary
by C. The boundary is defined in terms of coordinate c and an outer unit normal to the

p-FORMULATION FOR SECOND·ORDER 20 BVP 605

Thery n, x and y components of the normal vector are denoted by nx and ny:

= ~/12x
n (n n= x + =2
) ,'Inl 1
/1 Y
y

undary conditions are of the following types:

Essential boundary condition specified on portion of the boundary indicated by Ce:

u(e) specified on C,

Natural boundary condition specified on portion of the boundary indicated by C :
Il

au

kx7o3x /1x
=au

k 7o3n
== + au a(e)u(e) + {3(e) on C
Il
ky7o3y /1y

p-Mode Assumed Solution

ution domain is discretized into arbitrary four-sided quadrilaterals. As discussed in
r 6, a quadrilateral element is mapped to a 2 x 2 square element called the master
t. The mapping between the actual element coordinates and the master element is
as follows:

N, (s, t), N2(s, t), ... are serendipity interpolation functions for the master element
y j' ••• are the coordinates of the key points of the actual quadrilateral element. The
ration functions for mapping a quadrilateral with straight sides and one with all
irved, are as follows:

rpolation functions for mapping when all four sides are straight lines: four key points
ure 9.10):

t(-l+s)(-l+t) I
2
-t(l+s)(-l+t)
N= 1

t (1 +s)(l +t)

-t (-1 +s)(l +t)

I- 2 .~I

Figure 9.10.

606 p-FORMULATION

Interpolation functions for mapping when all four sides are quadratic curves: eight key
points (Figure 9.11):

-t (-1 +s)(-1 + t)(1 + s + t) I s

1. (-1 +s2)(-1 + t) 2

2 1

t (-1 +t)(1-S2+ t+ s t) I·

--} (1 +s)(-1 +r2)

tN=
(1 +s)(1 +t)(-1 +s + t)

--} (-1 +S2)(1 +t)

t (-1 + s) (1 + s - t) (1 + t)

-}(-I+s)(-I+t2)

Figure 9.11.

The assumed solution for the p-formulation is developed in terms of s, t for the 2 X 2
master element. Each element has four corner nodal degrees of freedom up u2' u3' and u4
regardless of the number of key points used for mapping. Thus, even if an element has all

sides that are curved and is defined with eight key points for mapping, it still has only four
corner degrees of freedom. Linear Lagrange interpolation functions are used to interpolate

between the corner nodal unknowns. Higher order solutions are written by including a
number of p-modes identified with parameters 01'02' .... Thus the assumed solution is
written as follows:

/

u(s,t)=Nlul+· .. +N4U4+PI81+P202+···=(NI ... N4 PI ... ) U4

°1

=NI ~(1 - s)(1 - t); N2 = ~(s + 1)(1 - t); N3 = ~(s + 1)(t + 1);
N4 = ~(1 - s)(t + 1)

The p-modes must be chosen in such a way that during assembly it is possible to maintain
continuity of the assumed solution over the entire solution domain. With the standard finite
element formulation the continuity of the solution was maintained by simply matching ap-
propriate nodal degrees of freedom at a common element interface during assembly. Since
now the solution depends on the p-modes as well, we must account for nonzero p-modes
across common element interfaces during assembly. To make this process computationally
tractable, the p-modes are chosen carefully so that they are either zero on all four element
sides or nonzero on one side of the element only. Thus the p-modes are classified as side
modes and internalmodes. The side modes are zero along three sides of the element and
the internal modes are zero along all four sides of the element. The internal modes are in-
dependent for each element and do not require any special consideration during assembly.

p-FORMULATION FOR SECOND-ORDER 20 BVP 607

The 'side-mode parameters along the common interface between the two elements must be
matched during assembly to maintain continuity of the assumed solution. The process will
become clear when we consider numerical examples.

The one-dimensional p-modes based on Legendre polynomials can easily be extended
to define both the side and the internal modes. The one-dimensional p-modes in terms of s
are as follows:

1 i = 1,2, ...
Pi(s) == .y2(2i + 1) (!/Ji+l (s) - !/Ji-l(s));

where !/Ji are the Legendre polynomials. The functions Pi(s) are 0 at both s = -1 and s = 1.
A suitable side mode for side 1-2 (t = -1; see Figure 9.10) of the master element must

have a value of 1 on side 1-2 and zero on all other sides. This can be accomplished by

multiplying Fj(s) by (i - t)l2. Thus suitable p-modes for side 1-2 of the element can be

written as follows: .

p,~1-2) _ (1 - t)12 . i = 1,2, ...
I - .y2(2i + 1) (!/Ji+l (s) - !/Ji-l(s)),

Following a similar reasoning, suitable p-modes for the other three sides of the element
can be written as follows:

p,(2-3) _ (l + s)12 t _ rl): i =1,2, '"
I - .y2(2i + 1) (!/Ji+J () !/Ji-J ( )),

p,~3-4) _ (1 + t)12 s _ S' i =1,2, ...
I - .y2(2i + 1) (!/Ji+l () !/Ji-l ( )),

p,(4-1) _ (1 - s)12 . i = 1,2, ...

I - .y2(2i + 1) (!/Ji+l (t) - !/Ji-l(t)),

The internal modes can be defined by simply taking the product of one-dimensional p-
modes in the s and the t directions:

i = 1,2, ... ; j = 1,2, ...

= =Denoting the order of the assumed solution by n, with n 1 for linear, n 2 for quadratic,

etc., an assumed solution of any order can be written systematically as follows:

Linear solution: n = 1. No p-modes:

=(NT ~(l - s)(l - t) ~(s + 1)(1 - t) ~(s + l)(t + 1) ~(l - s)(t + 1))

Quadratic solution: n = 2. Four nodal modes + 4 side modes: p t - l)

NT = (Nl N2 N3 N4 p/-2 Pt-3 p[.-4
=Cubic solution: n 3. Four nodal modes + 8 side modes:

._--------------------------

608 p·FORMULATION

Higher order solutions: n :2: 4. Four nodal modes +4(n -1) side modes + (n - 2)(n - 3)/2

internal modes:

NT =(N N2 N3 N4 pl-2 pf-3 pr- 4 p~-I
I

Pi-2 p]:-3 pi-4 Pi-I ... pll p12 ... )

The total number of interpolation functions increases rapidly with n, For example, with

=n 5 we have 4 nodal modes, 16 side modes, and 3 internal modes for a total of 27 interpo-

lation functions for each element. However, the computational effort does not increase that
rapidly because of orthogonality properties of the Legendre polynomials used in defining
these modes.

Example 9.3 Write explicit expressions for a fourth-order interpolation for a 2 x 2 master
element.

=Here n 4 and we have 4 nodal modes, 12 side modes, and 1 internal mode for a

total of 17 interpolation functions. Using the above equations the expressions for these
interpolation functions are as follows:

Nodal modes ~(1 -.s)(1 - t) ~(s + 1)(1 - t) ~(s + 1)(t + 1) 1)~(1 - s)(t +
Side modes: 1st set
Side modes: 2nd set _(3~2 ~) (1 - t) (s + 1)(¥ - ~) (3t-~)(t+1) (1-s)(¥-~)
Side modes: 3rd set
Internalmode 2{6 2{6 2{6 2{6

(5r - ¥) (1 - t) (s + 1)(?f - ¥) (5r-¥)(t+1) (1- s)(?f - ¥)

2{l6 2{l6 2{l6 n2{l6
n(s + 1)(3r ./~2 +
(3st - 2r + ~) (1 - t) (35$4 _21$2 + 1) (1 ~ s) (3f - 2~2 +
8 48
2{l4 2...[14
2...[14 2...[14
~ e s2 _ ~)et2 _~)

6 2 222

9.2.2 Finite Element Equations

The finite element equations consist of exactly the same terms as those derived in Chap-
ter 5. The only difference is that now we are using different interpolation functions. Thus
the element equations are as follows:

where

=d (u 1 u2 Lt3 u4 01 02 ... )T

IIkk = BeBT dA; IIkp = - pNNT dA;

A A

p-FORMULATION FOR SECOND-ORDER 2D BVP 609

=N vector of interpolation functions

N; = vector of interpolation functions along a given element side

BT = '!a!:!x.1 aaNx, ... ) and

[ '!a!:!y.1 ayaN2 ' •••

The quantities kcr and rfJ are a result of applied natural boundary conditions and will affect
only those elements that share the boundary with the NBC. For the interior elements these

terms will be zero. Theinterpolation functions are written over the master element in terms
of s, t. However, the element equations need x and y derivatives of these interpolation func-
tions. As shown in Chapter 6 using the mapping xes, t) and yes, t) and employing the chain
rule of differentiation, the derivatives of the ith interpolation can be written as follows:

J = aasx =7afix) ( JlI detJ = laxay _ axaYI
( qazs qazt as at at as
-J?]

aaNx; ) _ J- T [ qa!!s.J.) _ 1 ( 7ayfi
[ q!!.J. - q!!.J. - detJ _ill

~ at &

The matrix of derivatives of the interpolation functions B can be written as follows:

BT = [ a~1 a~, a:;, )_
aN"
'!!:!.1 aaN;, ay ,
ay

BT __1_ 122 '!!a:!s.1 - 121 '!!a:!t.1 1 aN, 1 '!!a!:tJ. 122.aaNs" - 121 aaNt" )
- detJ [ -J12'!!a:!s.1 -l-"J11'!!a:!t.1
22fu - 21

- JaN, + Jla1Na,t - J aaNs" + J aaNt"
12fu 12 II

In order to carry out integration, the given functions kx' ky' p, and q must be converted into
thes, t form by using the mapping functions. Frequently it is assumed that these quantities

are constant over an element, in which case they can simply be taken out of the integral
sign. All integrands are expressed in terms of s, t and, therefore, the above element matrices

can be established by evaluating integrals of the following form:

r rkk = l

BeB T detl dsdt
)-1 )-1 ,

r tl

kp =- J-I J-1pNNT detl dsdt

r rl

r q = J-I J-I qNdetldsdt

610 p-FORMULATION

Using an m x n Gauss quadrature formula the element matrices are then evaluated numeri-
cally as follows:

tkk = J(-II J-I BCBTdetl dsdt
L: L:III 1J

= wjwjB(sj> tj)C(sj> t)BT(Sj. t) detl(sj. t)

j=1 j=l

k =- (I (I pNNT detl dsdt
p J:.I J-I

L: L:III n

= - WjWjP(Sj. tj)N(sj' tj)NT (Sj' t) detl(sj' tj)

j=1 j=1

r, =J(-Il J(-I1qNdetldsdt

L: L:m IJ

= WjW jq(Sj'tj)N(sj. t) detl(sj' tj)

j=1 j=1

where (Sj. t) are the Gauss points and wj and wj are the corresponding weights.

Elements involving a boundary where a nonzero natural boundary condition is specified
require computation of the ka matrix and the rf3 vector:

where c is a coordinate along the boundary and the integration is along the boundary of
the element. In general, the element boundary is an arbitrary curve in the x, y coordinate
system. After mapping each boundary curve, there is either a horizontal or a vertical line
representing a side of the master square element. The boundary coordinate c for each side
is mapped to a coordinate -1 s a s 1, as shown in Figure 9.12. The mapping for each side
x(a) and y(a) can be obtained from the element mapping as follows:

For side 1: a = s; t = -1
For side 2: S = 1; a = t
For side 3: a = -s; t = 1

For side 4: S = -1; a = -t

The boundary integrals along the first side of the element can now be written as follows:

rf3 =J(-II f3NJe da

where Ie is the Jacobian of the side:

p-FORMULATION FOR SECOND-ORDER 2D BVP 611

Master element y Side 3
t

a ~..JL-----x

1---2-----1

Figure 9.12. Boundary coordinates on the master and the actual element

Of course, the known quantities a and /3, if not constant, must be expressed in terms of

a using the mapping for the side. Now these integrals can easily be evaluated using one-

dimensional Gauss quadrature formulas:

r II II

rf3 = /3NJc da = w;/3Ca)NcCa)JcCa)

Ll i=l

where ai are the Gauss points and Wi are the corresponding weights.

Notation for the p-Mode Parameters The p-mode parameters are denoted by 0 with

subscripts and superscripts. For the side modes the subscripts indicate the mode number.

The superscripts consist of two numbers indicating the node numbers of the line segment

on which that side mode is nonzero. Thus O~-51 indicates the second side mode with

nonzero value on line 3-5 of the element. For the internal modes the subscript indicates

the mode number while the superscript indicates the element number. Thus o~ indicates

the second internal mode for the third element.

Example 9.4 Eight-Node Quadrilateral Element Consider solution of the following
equation over a quarter-circle annular domain using only one element as shown in Figure
9.13:

azu azu

2o~r +2oy-z +3!i+4 =·0

au =5u + 6 on side 3-4-5
2
an

u = 100 on side 1-2-3

612 p-FORMULATION

I y Actual element
85
2
6
1 6

1--2------l 4

2 3
7

0

x

02468

Figure 9.13. Eight-node masterelement and actualelement

Comparing the given equation to the general form, we have lex == ley == 2, P == 3, and q 4.
On side 3-4-5 we have a natural boundary condition with a == 5 andf3 == 6. On side 1-2-3
we have an essential boundary condition with u == 100.

The nodal coordinates are as follows:

12 34 56 78
x 2 5. 8 4...[2 0 O. 0 ...[2

y 0 O. o 4...[2 8 5. 2 ...[2

Use second-order interpolation (n == 2) to evaluate element equations:
Interpolation functions for mapping:

{i(1 - s)(1- t) - i(l - s2)(1 - t) - i(l - s)(1 - t2), 1(1 - s2)(1 - t),

i(s + 1)(1 - t) - i(1 - s2)(1 - t) - i(s + 1)(1 - t2).1(s + 1)(1 - p),
i(s + 1)(t + 1) - i(1 - S2)(t + 1) - i(s + 1)(1 - t2), 1(1 - s2)(t + 1),
i(1 - s)(t + 1) - i(1 - S2)(t + 1) - i(1 - s)(1 - t2), 1(1 - s)(1 - t2)}

Interpolation functions for assumed solution:

p-FORMULATION FOR SECOND-ORDER 2D BVP 613

aaNsT :={t-1 1-t t+1 ~(-t-1) '
4' 4' 4 '4

21 ~[32s(1 - t), 3t2/2 - 3/2 , 21 ~[32s(t + 1),
2...[6

aNT { Ts - 1' 1 s+1 1- s
at:= 4(-s-1), -4-' -4-'

fie~ ~3s2/2 - 3/2
2...[6
, 2~ 2s + 1) 3s2/2 - 3/2 , 2 ~f2i(l _ )}
t, 2...[6 st

The element equations are as follows:
z Element coordinates:

({2,O) {5.,0.} {8,O} (4-yz,4-YZ} {O,8} {O.,5.} {O,2} (-YZ, -YZ})

3st2 3st2 5t2 5t2 3st 3s
xes, t) := - - + - - - + - - - - 2.5t + - + 3.53553
-YZ 2 -YZ 2 2 -YZ

Yes, t) := -3s-t2 + -3st2 - -5t2 + -5t2 + -3st + 2.5t + -3s + 3.53553
-YZ 2 -YZ 2 2 -YZ

25J- 3-t2f+i32t2 _2It-+f.1i.. -3-YZts + 3ts - 12': - 5-YZt + 5t - .

1:= ( _3-t2f+i32t2 +2It-+f.1i.. -3.f2ts + 3ts + 1': - 5-YZt + 5t + 2 5

2.

detl:= 1.86396st2 + 3.106~t2 + 9s + 10.6066
-YZ

Given element data: kx := 2; ky :=2; p :=3; q ix: 4

Use 2 x 2 Gauss quadrature for integration.

Gauss point « {s ~ -0.57735, t ~ -0.57735}; Weight « 1.

NT := {O.622008, 0.166667, 0.0446582, 0.166667, -0.321975,
-0.086273, -0.086273, -0.321975}

aNT
7); := (-0.394338 0.394338 0.105662 -0.105662

-0.557678 -0.204124 -0.149429 0.204124).

aNT
a t := ( -0.394338 -0.105662 0.105662 0.394338.:

0.204124 "':0.149429 -0.204124 -0.557678)

614 p-FORMULATION

r T =(0.317444 -0.137753) ; det] =7.60918
0.11203 0.36538

= (BT -0.0708591 0.139736 0.0189866 -0.0878632 -0.20515 -0.0442137 -0.0193168 0.14162 )

-0.188261 0.00557075 0.0504442 0.132246 0.0121062 -0.0774664 -0.0913233 -0.180896

0.615781 -0.166645 -0.164998 -0.284138 0.186541 0.269621 0.282473 0.365553
0.297627 0.0446525 -0.175634 -0.435236 -0.10059 -0.0488202 0.285825
-0.166645 0.0446525 0.0442111 0.0761344 -0.0499836 -0.0722447 -0.0756885 -0.0979496
0.0761344 0.383637 -0.0967861 -0.157965 -0.553429
-0.164998 -0.175634 0.298678 -0.475472
-0.435236 -0.0499836 0.298678 0.642721 0.123766 0.0434829 0.11797
=kk -0.284138 -0.10059 -0.0722447 -0.0967861 0.123766 0.121076 0.12066 0.209776
0.186541 -0.0488202 -0.0756885 . -0.157965 0.0434829 0.12066 0.132599 0.803218
-0.0979496 -0.553429 -0.475472 0.11797 0.209776
0.269621 0.285825

0.282473

0.365553

-8.83185 -2.36649 -0.634098 -2.36649 4.5717 1.22498 1.22498 4.5717
-0.634098 -0.169906 -0.634098 1.22498 0.328234 0.328234 1.22498
-2.36649 -0.169906 -0.0455262 -0.169906 0.328234 0.0879499 0.0879499 0.328234
-0.634098 -0.169906 -0.634098 1.22498 0.328234 0.328234 1.22498
-0.634098 -2.36649 -0.634098 -0.634098 -2.36649
1.22498 0.328234 1.22498 -0.634098 -0.169906 -0.169906 -0.634098
=k p -2.36649 0.328234 0.0879499 0.328234 -0.634098 -0.169906 -0.169906 -0.634098
4.5717 0.328234 0.0879499 0.328234 -2.36649 -0.634098 -0.634098 -2.36649
1.22498 0.328234 1.22498
1.22498

1.22498

4.5717

r~ = (18.9319 5.07279 1.35925 5.07279 -9.79987 -2.62587 -2.62587 -9.79987)

Similarly, evaluating at the remaining Gauss points and adding contributions,

1.61408 -0.592297 -0.72701 -0.294777 -0.85304 0.523564 0.206715 -0.553646
-0.592297 1.29396 0.0253502 -0.72701 0.149446 -0.523564 0.49688 0.553646
-0.72701 0.0253502 1.29396 -0.592297 0.49688 -0.523564 0.149446 0.553646
-0.294777 0.206715 -0.85304
-0.85304 -0.72701 -0.592297 1.61408 2.04632 0.523564 0.592297 -0.553646
0.149446 0.49688 0.206715 -0.25158 -0.25158 0.25158
0.523564 0.5235q<l- -0.25158 2.04632 0.134714
0.206715 -0.523564 -0.523564 -0.85304- 0.592297 1.21948 0.25158 0.25158
-0.553646 0.49688 0.149446 -0.553646 0.25158 -0.25158 1.95799
0.553646 0.553646
0.134714

-10.866 -7.76142 -3.88071 -5.433 7.60461 4.75288 3.80231 6.65403
-20.1797 -10.0899 -3.88071 11.4069 12.3575 5.70346 4.75288
-7.76142 -10.0899 -20.1797 -7.76142 5.70346 12.3575 11.4069 4.75288
-10.866 3.80231 4.75288 7.60461 6.65403
-3.88071 -3.88071 -7.76142 -7.76142 -6.98528 -3.88071 -4.65685
11.4069 5.70346 3.80231 -6.98528 -10.0899 -6.98528 -3.88071
=kp -5.433 12.3575 12.3575 4.75288 -3.88071 -6.98528 -7.76142 -4.65685
7.60461 5.70346 11.4069 7.60461 -4.65685 -3.88071 -4.65685 -5.433
4.75288 4.75288 6.65403
4.75288

3.80231

6.65403

r~ =(37.2548 55.8823 55.8823 37.2548 -38.0231 -45.6277 -38.0231 -30.4184)

The natural boundary conditions are as follows:

Specified NBC values for side 2: ex =5; f3 =6

Interpolation functions for mapping:

{o o}0 1 - a + ~(a2 _ 1) 1 _ a2 a + 1 + ~(a2 - 1) 0 0
"2 2 ' '2 2 ',,

p-FORMULATION FOR SECOND-ORDER 2D BVP 615

Interpolation functions for solution:

oI - a a+I 0 0 3a2/2-Y-6 3/2 "0 O}
{,
2' 2'"

x(a) =-4V2a2 + 4a2 - 4a + 4V2; yea) =-4V2a2 + 4a2 + 4a + 4V2

4fJc = (-8V2a + 8a '-- + (-8V2a + 8a + 4t

Value in mapped coordinate: a(a) =5; [3(a) = 6
Gauss point = -0.57735; Weight = 1.
N[ = {O., 0.788675, 0.211325, 0., 0., -0.408248, 0., O.}

Jc = 6.2706; a = 5.;[3 = 6.

O. O. O. O. O. O. o. O. O.
-29.6728
O. 19.5018 5.2255 O. O. -10.0949 O. O. -7.9508

O. 5.2255 1.40017 O. O. - 2.70492 O. O. O.
k = O. O.
O. O. o. O. O. O. o. 15.3598
O.
Q' O. O. O. O. O. O. O. O. O.

O. -10.0949 -2.70492 O. O. 5.2255 O. O. O.
-7.9508
O. O. o. o. O. O. O. O. -29.6728
O. O. O. O. o. O. O. O.
O.
Gauss point =0.57735; Weight = 1. O.
=N[ {O., 0.211325, 0.788675, 0., 0., -0.408248, 0., O.} 15.3598
O.
Jc = 6.2706; a = 5.;[3 = 6. O.

O. O. O. O. O. O. O. O.

O. 1.40017 5.2255 O. O. - 2.70492 O. O.

O. 5.2255 19.5018 O. O. -10.0949 O. O.

ka =~:~: -~: ~:~: O. O. O.
O. O. O.

o. - 2.70492 -10.0949 O. O. 5.2255 O. O.

O. O. O. O. O. O. O. O.

O. O. O. O. O. O. O. O.

Adding contributions from all Gauss points,

O. O. O. O. O. O. O. O. O.

O. 20.902 10.451 O. O. -12.7998 O. O. -37.6236

O. 10.451 20.902 O. O. -12.7998 O. O. -37.6236

O. O. O. o. O. O. O. O. O.
O. O. o. o. o. O. ; '/3 = O.
O. O.
O. -12.7998 -12.7998 o. O. 10.451 O. O.
30.7195

O. O. O. O. O. O. O. O. O.

O. O. O. O. O. O. O. O. O.

616 p-FORMULATION
Thus the complete element equations are as follows:

-9.25191 -8.35372 -4.60772 -5.72777 6.75157 5.27645 4.00902 6.10039 ul 37.2548
-8.35372 2.01626 0.386502 -4.60772 11.5564 -0.965882 6.20034 5.30653 u3 18.2586
-4.60772 0.386502 2.01626 -8.35372 6.20034 -0.965882 11.5564 5.30653 18.2586
-5.72777 -9.25191 4.00902 6.75157 6.10039 Us 37.2548
-4.60772 -8.35372 -5.71511 5.27645 -3.28842 -4.40527 -38.0231
6.75157 11.5564 6.20034 4.00902 -7.23686 -7.23686 -7.23686 -3.746 u7 -14.9081
5.27645 -0.965882 5.27645 -3.28842 -5.71511 -4.40527 -38.0231
4.00902 6.20034 -0.965882 6.75157 -4.40527 1.58063 -4.40527 -3.47501 8\1,31 -30.4184
6.10039 5.30653 11.5564 6.10039 -7.23686
5.30653 -3.746 8\3.51

8\5,71

8\1,71

Since there are no other elements, these are the global equations. The essential boundary
conditions are as follows:

On element 1, side 1, specified value = 100
{UI = 100, u3 = 100,0\1,31 = O}

(See the following section for details on setting parameter values corresponding to a
givenEBC.)

Global equations after the EBC:

[ -82.3051367~2 -8.35372 5.30653 -0.965882 11.5564 Us
-9.25191 6.10039 5.27645 6.75157
5.30653 -3.47501 -4.40527 [ *W3Bl]u7
-0.965882 6.10039 -3.746 -3.746 -7.23686
11.5564 5.27645 -4.40527 1.58063 -5.71511 0\1,71 =
6.75157 1070.8
-7.23686 -1171.11

0\3.5 1 -445.964

0\5,71 -1058.96

Solving the final system of global equations, we get

{Us = -392.717, u7 = -943.4,01[1,71 = [3,51 = -558.828,01(5,7) = }
-1263.19,01 -41.996

The element solution is as follows:
The dof values for the element:

{ul 100, u3 = 100, Us = -392.717, u7 = -943.4,01(1,3) = 0,
0\3,5) = -558.828,0\5,7) = _41.996,0\1,71 = -1263,19}

aT = (100 100 -392.717 -943.4 0 -558.828 ,-41.996 -1263.19)

Element solution at {s -7 -0.57735, t -7 -0.57735}
Location: {3.02856, 1.14181}

NT = (0.622008 0.166667 0.0446582 0,166667

-0.321975 -0.086273 -0.086273 -0,321975)

p-FORMULATION FOR SECOND-ORDER 20 BVP 617

:' a~T =(-0.394338 0.394338 0.105662 -0.105662

-0.557678 -0.204124 -0.149429 0.204124)

faNitT = (-0.394338 -0.105662 0.105662 0.394338

0.204124 -0.149429 -0.204124 -0.557678)

J- T =(0.317444 -0.137753)

0.11203' 0.36538

a::BJ = =(-0.0708591 0.139736 0.0189866 -0.0878632

-0.20515 -0.0442137 -0.0193168 0.14162)

B; a:= T =(~0.188261 0.00557075 0.0504442 0.132246

y 0.0121062 -0.0774664 -0.0913233 -0.180896)

u =NTd =362.646
au
=B~. d = -71.052
-ax

:; = BJd = 112.791
Solution at any other location over the element can be computed in a similar manner.

9.2.3 Assembly of Element Equations

As seen from the previous section, in the p-formulation there are three types of degrees of
freedom: the nodal degrees of freedom, side-mode parameters, and internal-mode parame-
ters. The internal modes do not influence solution outside of the element because they are
zero along the element sides. However, side modes are nonzero on one of the sides of an
element. Therefore, to maintain solution continuity, during assembly we must assign the
same global parameter to the common sides between elements. This will add contributions
of elements to the appropriate-parameters in a manner similar to that for the nodal degrees
of freedom. The following example numerically illustrates the procedure.

Example 9.5 Consider solution of the Laplace equation over a rectangular domain using
two elements as shown in Figure 9.14.

0< x < 3; O<y<l

=Write element equations with order of interpolation ri 2. Assemble these equations to

form the global system of equations. '

For this problem kx = ky = 1, P = q = O. There are no natural boundary conditions. The
element equations can be obtained easily using the procedure discussed in the previous

section. With the second-order interpolation, there will be four side modes for each ele-

ment, no internal modes, and the element equations will be 8 x 8. The complete model has

618 p-FORMULATION

54

o
------------ x
o3
Figure 9.14. Two-element model

six nodes and seven unique element sides. Thus the global system will be 13 X 13. Using
the convention for numbering the nodal and the p-mode parameters as described earlier,
the global degrees offreedom are as follows:

dof = { £II' £1 2, £13, £14, u5' (l,2) (l,6) (2,3) (2,5) (3,4) (4,5) 15,6}}

£16, 81 .61 .61 .81 .61 .81 .61

The subscript 1 on the 6 terms indicates that all these parameters are associated with the
first set of side modes. As mentioned earlier, the superscripts indicate node numbers of
lines over which the mode has a nonzero value.

Global equations at the start of the element assembly process are

o 0 000 0 0 0 0 0 0 0 0

0000000 0 0 0 0 0 0

o0 0 0 0 0 0 0 0 0 0 0 0

o 0 0 0 0 000 0 0 0 0 0

o 0 0 0 0 0 0 0 000 0 0

o 0 0 0 O. 0 ~. 0 0 0 0 0 0
o 0 0 0 0 0 0 0 0 0 000
o 0 0 0 0 0 000 0 0 0 0

o0 0 0 0 0 0 0 0 0 0 0 0

o 0 0 0 0 0 0 0 0 0 000
o0 0 0 0 0 0 0 0 0 0 0 0
o0 0 0 0 0 0 0 0 0 0 0 0
o0 0 0 0 0 0 0 0 0 0 0 0

The equations for element 1 are as follows:

0.666667 -0.166667 -0.333333 -0.166667 -0.204124 0.204124 0.204124 -0.204124 o
-0.166667 0.666667 -0.166667 -0.333333 o
-0.333333 -0.166667 -0.204124 -0.204124 0.204124 0.204124 o
-0.166667 -0.166667 0.666667 o
-0.204124 -0.333333 -0.166667 0.666667 0.204124 -0.204124 -0.204124 0.204124 o
-0.204124 0.204124 o
0.204124 -0.204124 0.204124 0.204124 0.204124 0.204124 -0.204124 -0.204124 o
0.204124 -0.204124 -0.204124 o
-0.204124 0.204124 -0.204124 -0.204124 0.833333 o 0.166667 o
0.204124
0.204124 o 0.833333 o 0.166667

0.166667 o 0.833333 o

o 0.166667 o 0.833333



620 p-FORMULATION

Global locations to which this element contributes: (2, 3,4, 5, 9, 11, 12, lO)
Assembly locations for

k~ [2,2] [2,3] [2,4] [2,5] [2,9] [2, 11] [2, 12] [2,10]
[3,2] [3,3] [3,4] [3,5] [3,9] [3, 11] [3, 12] [3, 10]
[4,2] [4,3] [4,4] [4,5] [4,9] [4, 11] [4, 12] [4,10]
[5,2] [5,3] [5,4] [5,5] [5,9] [5, 11] [5, 12] [5, 10]
[9,2] [9,3] [9,4] [9,5] [9,9] [9, 11] [9, 12] [9, 10]
[11,2] [11,3] [II, 4] [11,5] [II, 9] [11, 11] [11, 12] [11, 10]
[12,2] [12, 3] [12,4] [12,5] [12,9] [12, 11] [12, 12] [12, 10]
[10,2] [10,3] [10,4] [10,5] [10,9] [10,11] [10, 12] [10, 10]

and for

r~ 2
3
4
5
9
11
12
10

Global equations after assembling this element:

"I

0.666667 -0.166667 0 0 -0.333333 -0.166667 -0.204124 -0.204124 0 0.204124 0 0 0.204124 "2
-0.166667 1.5 0.166667 -0.416667 -0.75 -0.333333 -O.2QJU24 0.204)24 -0.408248 -0.306186 0.102062 0.408248 0.204124 "3
0.166667 0.833333 -0.583333 -0.416667 -0.408248 -0.102062 0.408248 0 "4
0 -0.583333 0 0 0,1 0.102062 -0.102062 -0.408248 0 "5
0 -0.416667 -0.416667 0,833333 0.166667 0 0 0 0.'108248 0.102062 -0.'108248
-0.333333- -0.75 0 0.166667 1,5 -0.166667 0.304124 0.204124 0408248 -0306186 n 102062 0 -0.204124 F ~.1"16.210
-0.166667 0 0.666667 0.2041'2.4 -0.2041'2.4 0 0.204124 -0.204124
-0,204124 -0.333333 0 0 -0.166667 0.204124 0.833333 0 0 0 0 )~3611 = 0
-0.204124 -0.204124 -0.408248 0 0.204124 -0.204124 0 0 0 0.166667 0 0 0.166667
0.102062 0 0.204124 0 0.B33333 0.666667 0 0 -0.166567 0
0 0.204124 -0.102062 0.408248 0.408248 0 0 0 0 0 0
0.204124 -0.408248 0.408248 0.102062 0.204124 0.166667 0 ~2S 0.583333 0 0 .1>.51 ~
0 -0.306186 0 -0.102062 -0.306186 0 -0.166667 0.583333 . 1.41667 0.666667 0
-0.408248 0,102062 0 0 0 0 0
0 0.102062 0 0 0.166667 0 0 0 ./3.41 0
0.204124 -0.408248 -0.20412'1 0 0 0
0.408248 -0.204124 0 .10 ~
0.204124
0.833333 4•51

./5.61
I

Since there are no other elements left, these are the global equations for the system.

9.2.4 Incorporating Essential Boundary Conditions

For the second-order boundary value problem being considered in this chapter, the essential

=boundary condition is u specified along portion of the boundary. In the conventional

h-formulation, the specified value is incorporated into the global equations by assigning
appropriate values to the nodal degrees of freedom. If the specified value is constant along
a line, the nodes at the ends of the line are simply assigned this value. The interpolation
functions ensure that the solution along the entire line is equal to the specified value. If
the specified value is not constant but is a function of x, y, then the essential boundary

p-FORMULATION FOR SECOND-ORDER 20 BVP 621

condition is only approximated. The quality of the approximation depends on the order of
interpolation and can be improved by placing more nodes on the line.

In the p-formulation the situation is a little more complicated. In addition to the nodal
degrees of freedom, we must assign values to the side-mode parameters that are nonzero
for the side with the essential boundary condition. If the specified value is'constant, then the
node at the ends of the line are assigned the specified value and all side-mode parameters
for that line are set to zero. This will ensure that the solution is equal to the specified value

along the entire line. The was the situation in Example 9.4 and hence we assigned U = 100

to the end nodes and set the associated side-mode parameter to O. However, if the specified
value is a function, then we must find suitable values for the side-mode parameters as well.

=As an example consider a situation where u u(e) is specified along line 1-2 of an ele-

ment, where e is a coordinate along the line. Over the mapped element each line is repre-
sented in terms of a coordinate -1 ::; a ::; 1, as illustrated in Figure 9.12. The specified value
in terms of the mapped coordinate is u(a). The assumed solution along this line is written

as u =N~d, where N; is a vector of interpolation functions and d = lUI' Uz, o)-Z, otZ, ... )

is'a vector of unknown parameters for this line. The values of nodal parameters are simply
the specified function values at the ends. Thus

=ul = u(-I) and Uz u(l)

Our next task is to determine the parameter values o[-Z, o1-z, ... such that the solution

along the line is as close to u(a) as possible. To determine the values of these parameters,
we define the total squared error between the interpolated solution and the specified value
as follows:

11 ?

e = (u(a) -N~drJeda

-I

where Je is the Jacobian for the side introduced as a result of mapping. Writing N~
(N!, Nz' PI' Pz' ... ), the error can be written as follows:

where superscripts 'on 0 terms are dropped for convenience. The necessary condition for
the error to be a minimum is that the partial derivative of e with respect to the unknown
parameters 0; be zero. Thus we get the following equations for determining these parame-
ters:

i = 1,2, ...

Writing all equations together in a matrix form and rearranging terms, we get the following
system of equations:

622 p-FORMULATION.

1:1 ][~I] _PIP3Jeda ·.. (I:t(u(a) -Nlul -N2U2)PtJedaj

I I P2..P.3Jeda....". :2 - I I (£I(a) - Nt £II - N2u2)P2Je da

...

In compact form these equations can be written as follows:

t tLl ppTi, da 6 =LI LIPi, da :=:} AD =b

where

p T = (PI P2 ••. ), vector of p-modes that are nonzero for the side

=DT (8 1 82 ... ), vector of parameters corresponding to modes in vector P
=Ti u(a) - N, £II - N2u2
A = L(lI ppT i, da

b = J(-1l (u(a) -Nlu l - N2L0.)PJc d a = (1 IlPJcda

J-l

The integrals in matrix A and vector b are evaluated numerically using the one-dimensional
Gaussian quadrature. Solution of these equations give appropriate side-mode parameter
values to enforce the essential boundary condition along a given side of an element. The
procedure is further illustrated through the following numerical example.

Example 9.6 Consider the two-element model shown in Figure 9.14. The following es-
sential boundary condition is specified along side 3-4 of the domain:

= =£I y sin(7Ql/6) along x 3; O<y<l

Determine the appropriate parameter values to incorporate this boundary condition if a

third-order interpolation is to be used for developing element equations.

With the third-order interpolation, there will be two sets of side modes for a total of

eight side modes for each element. The unknown parameters along side 3-4 are as follows:

{u3' £14, ,,3-4 ,.3-4
ul ,02 }

y (y;)=On element 2, side 2, specified value sin

Mappm. g funncottiions a1ong thee sSlide: {-I 2- -a' -a2+ -1}

Mapped coordinates: x(a) =3; yea) = ~ + ~; Je = ~

n)Specified value in mapped coordinates: u(a) =(~ + ~) sin (~ (~ + ~)

p-FORMULATION FOR SECOND-ORDER 2D BVP 623

The-specified values at the two end nodes are as follows:

°Left end £lC-1) =

Right end £lCI) = ~.

Thus

=Values at ends (£l3, u4 ) {O, ~}

{I-aa+1}Noodaal imterpoIan'on fu'ncti•ons: -2-' -2-

-a -u=~CI) + (~ + ~) sin (~ (~ + ~) 7T)

4 2 2 6 2 2

2 3 Sa' Sa }

{ 3a
2.;ToT, ,
p-Modes for the side: pT = 2..{g 2,

Gauss point = -0.774597; Weight = 0.555556
i, =0.5; ti = -0.0497041; pT = {-0.244949, 0.244949)

A = ( 0.0166667 -0.0166667). b = ( 0.00338194)
-0.0166667 0.0166667' -0.00338194

Gauss point = 0.; Weight = 0.888889

s, =0.5; Ii = -0.12059; pT = (-0.612372, 0.)

A = (0.166667 0.); b = (0.0328206)

0. O. 0.

Gauss point = 0.774597; Weight-= 0.555556

t, =0.5; u = -0.0460909; pT =(-0.244949, -0.244949)

A = (0.0166667 0.0166667). b = (0.00313609)

0.0166667 0.0166667' 0.003.13609.

Adding contributions from all Gauss points:

0.2 O. )(013,41) (0.0393386 )
( O. 0.0333333 0~,41 = -0.000245849

Solution gives {013,41, 0~,41} = {0.196693, -0.00737546}
{U3' U4, 01[3,41,02[3,4)} = {0, 21,0.196693, -0.00737546}·

Known values from EBC:

{ U3 = 0, £l4 = 2I ,01{3,41 = 0.196693,02{3,41 = -0.00737546 }

624 p-FORMULATION

9.2.5 Applications

Example 9.7 Heat Flow in an L-Shaped Body Consider two-dimensional heat flow

over an L-shaped body with thermal conductivity k = 45 W/m· 'C shown in Figure 9.15.
The bottom is maintained at To = 110'C. Convection heat loss takes place on the top where
=the ambient air temperature is 20'C and the convection heat transfer coefficient is h

55 W/m2 • 'C. The right side is insulated. The left side is subjected to heat flux at a uniform

= =rate of qL 8000W/m2• Heat is generated in the body at a rate of Q 5 x 106W/m3•

Determine the temperature distribution in the body.
The simplest finite element model using quadrilateral elements is the two-element

model shown in Figure 9.16.

With n =2 the finite element solution is as follows.

Equations for element 1 are as follows:

Natural boundary conditions

Specified NBC values for side 3: 0: = -55; f3 = 1100

25.0962 19.9038 -6.05769 -38.9423 -29.6765 -5.65267 4.2395 5.65267 937.5
19.9038 70.0962 -38.9423 -51.0577 -43.8082 5.65267 32.5028 31.0897 937.5
-6.05769 -38.9423 -27.9265 -16.2514 733.5
-38.9423 -51.0577 43.5846 0.590385 14.1317 -20.4909 -8.1422 -20.4909 733.5
-29.6765 -43.8082 0.590385 88.5846 59.353 20.4909 -10.3846 -34.6154 -765.466
-5.65267 14.1317 59.353 57.6923 4.61538 38.9423 -688.919
4.2395 5.65267 20.4909 4.61538 96.0577 5.76923 24.2308 -598.9
5.65267 32.5028 -20.4909 -8.1422 -10.3846 5.76923 35.4942 96.0577 -688.919
31.0897 -27.9265 -20.4909 -34.6154 38.9423 24.2308
-16.2514

y (m)
0.03

0.015 go

o

--~-----~~----- x (m)

o 0.03 0.06

Figure 9.15. L-shaped body

6Y 5

x
Figure 9.16. Two element model

p-FORMULATION FOR SECOND-ORDER 20 BVP 625

Equations for element 2 are as follows:

Natural boundary conditions

Specified NBC values for side 2: a =-55; f3 = 1100
Specified NBC values for side 3: a = -55; f3 = 1100
Specified NBC values for side 4: a =0; f3 =8000

22.5 0 -11.25 -11.25 -9.18559 1.05988 9.18559 -7.41913 T1 817.5
0 44.725 -22.6375 -22.5 -18.3712 -1.95135 18.3712 14.8383 T4 741.75
-11.25 -22.6375 -0.275 -6.89743 -18.0344 3.53292 T5 725.25
-11.25 -22.5 32.925 9.18559 -8.84878 -10.952
-9.18559 -18.3712 -0.275 33.2 18.3712 8.12571 0 -8.65385 oPTA6l 801.
1.05988 -1.95135 18.3712 33.75 6.05769 16.875 -688.919
9.18559 18:3712 9.18559 8.12571 6.05769 54.2375 1.44231 0\4.51 -605.636
-7.41913 14.8383 -6.89743 -8.84878 0 1.44231 33.475 1.15385 -675.447
-18.0344 -10.952 -8.65385 16.875 1.153?5 46.875 0\5.61 -667.486

3.53292 0\1.61

The global equations after assembling all elements are

{ TI> TZ' 01II.ZI} = (110, 110, OJ

Known values from EBC:
{T1 = 110, Tz = 110, OJlI,Z} = 0}

Global equations after EBC:

43.5846 0.590385 0 0 -16.2514 0 -20-4909 -27.9265 0 0 T3 5683.5
0.590385 133.31 -22.6375 -22.5 -38.8621 14.8383 20.4909 -8.1422 -1.95135 18.3712 T4 11375.3
-22.6375 -0.275 3.53292 0 -6.89743 -18.0344 Ts 1962.75
0 -22.5 32.925 9.18559 -10.952 0 0 -8.84878 T6 2038.5
0 -38.8621 -0.275 33.2 18.3712 -8.65385 38.9423 0 8.12571 0\1,41 -4409.08
-16.2514 14.8383 9.18559 . '18.3712 129.808 46.875 0 . 24.2308 6.05769 0 148.618
0 20.4909 3.53292 -10.952 -8.65385 96.0577 0 16.875 1.15385 op.61 -688.919
-20.4909 -8.1422 38.9423 0 5.76923 5.76923 -4640.56
-27.9265 -1.95135 0 0 24.2308 0 0 35.4942 0 0 0\'-31 -722.223
18.3712 0 0 6.05769 16.875 0 0 0 0 -1685.86
0 -6.89743 8.12571 1.15385 0 54.2375 1.44231 0\3,41
0 -18.0344 -8.84878 0 1.44231 33.475
0\4.51

0\5.61

626 p-FORMULATION

Solving the final system of global equations, we get

{TJ = 129.195, T4 = 138.48, T5 = 162.431, T6 = (IAI = -12.0338,
160.634,81

8\1.61 = ..:..15.6763,8\2.31= -4.97674, 8\3AI = 11.6987,8\4,5) = -5.63838,8\5.6 1= 4.39332}

The solution for element 1 is as follows:

dof values for the element

{TI = 110, T2 = 110, T3 = 129.195, T4 = 138.48,81lI,21 = 0,
8\2,31= -4.97674, 8\3AI = 11.6987,8\IAI = -12.0338}

dT = (110 110 129.195 138.48 0 -4.97674 11.6987 -12.0338)

xyT er/e« erro,

0.0375 0.0075 123.545 -199.205 1310.77

The solution for element 2 is as follows:

dof values for the element

{TI = 110, T4 = 138.48, T5 = 162.431, T6 = 160.634,
8\IAI = -12.0338,8\4.51= _5.63838,8\5.61=4.39332, 8\1.61= -15.6763}

=dT (110 138.48 162.431 160.634 -12.0338 -5.63838 4.39332 -15.6763)

xy T aT/ax aT/Oy

0.D15 0.01875 151.752 72.87239 1210.35

Nodal Solution Summary:

dof x y Value

TI 0 0 110

T2 0.06 0 110

T3 0.06 0.D15 129.195

T4 0.03 0.015 138.48

T5 0.03 0.03 162.431

T6 0 0.D3 160.634

Element Solution Summary:

xy T aT/ax aT/ay

1 0.0375 0.0075 123.545 -199.205 1310.77
2 0.015 0.01875 151.752 -2.87239 1210.35

=With n 3 the following finite element solution is obtained:

p-FORMULATION FOR SECOND-ORDER 20 BVP 627

Nodal Solution Summary:

dof x y Value

T1 0 0 110
0 110
T2 0.06 0.015 125.645
0.Ql5 138.336
T3 0.06 0.03 161.689
T 0.03 159.31
. 0.03
4

Ts 0.03

T6 0

Element Solution Summary: T er/e» oT/oy

xy 124.347 -220.205 1502.78
1 0:0375 0.0075 151.263 -52.3684 1258.08
2 0.015 0.01875

With n = 4 the following finite element solution is obtained:

Nodal Solution Summary:

dof x y Value

T1 0 0 110

T2 0.06 0 110

T3 0.06 0.015 126.672

T4 0.03 0.015 139.179

Ts 0.03 0.03 161.022
0.03 157.838
T6 0

Element Solution Summary:

x T er/e« oT/oy

1 0.0375 0.0075 124.811 -260.161 1481.78
2 0.015 ·0:01875 151.195 -23.981 1318.8

Example 9.8 Seepage through Soil The problem of determining the amount of water
that seeps through dams or from underneath sheet piles can be formulated in terms of the
following equation:

~ (Ie orjJ) + ~ (Ie orjJ) - 0

ox .r ox oy y oy -

where rjJ(x, y) is the hydraulic head (or hydraulic potential) arid lex and ley are coefficients
of permeability in the x and y directions. Typical units for rjJ are meters and those for lex
and ley are rn/day. The fluid velocity components in the x and y directions are related to the

hydraulic head as follows:

orjJ orjJ
= - lxeo-x and = - lyeo-y
]I ]I

x y

628 p-FORMULATION

8¢j8n=O

8rp/8n=O

8rp/8n=O

Impermeable

Figure 9.17. Seepage through soil underneath a sheet pile wall

A typical situation is illustrated in Figure 9.17. The analytical model consists of the soil

on the downstream side. On the impermeable sides the no-flow condition is expressed in

=terms of the normal derivative of ¢ being zero. On the top ¢ hd , the hydraulic head on

the downstream side. On the left side the boundary condition on the soil below the pile is

¢ =h", the hydraulic head on the upstream side.

As a numerical example, consider the following numerical values:

a =b = 10m; h" = 10m; kx =Iey = 1 mls

The solution domain is divided into four elements, as shown in Figure 9.18. A solution
using a second-order (n = 2) p-formulation is as follows:

Equations for element 1:

0.666667 -0.166667 -0.333333 -0.166667 -0.204124 0.204124 0.204124 -0.204124 tP 1 0
-0.166667 0.666667 -0.166667 -0.333333 -0.2p4124 -0.204124 0.204124 0.204124 tP4 0
-0.333333 -0.166667 -0.204124 -0.204124 0.204124 tP5 0
-0.166667 -0.166667 0.666667 0.204124 -0.204124 tP2 0
-0.204124 -0.333333 -0.166667 0.666667 0.204124 0.204124 0.166667 -0.204124 0
-0.204124 0.204124 0.833333 0 0 0 oPAl 0
0.204124 -0.204124 0.204124 0.204124 0 0.833333 0.833333 0.166667 0
0.204124 -0.204124 -0.204124 0.166667 0 0 0 0\4.51 0
-0.204124 0.204124 -0.204124 -0.204124 0 0.166667 0.833333
0.204124 0\l51
0.204124
0\1.21

3Y 6 9

x
Figure 9.18. Four-element model for seepage under a sheet pile wall

p-FORMULATION FOR SECOND-ORDER 20 BVP 629

Equations for element 2:

0.666667 -0.166667 -0.333333 -0.166667 -0.204124 0.204124 0.204124 -0.204124 ¢2 0
-0.166667 0.666667 -0.166667 -0.333333 -0.204124 -0.204124 0.204124 0.204124 ¢5 0
-0.333333 -0.166667 -0.204124 -0.204124 0.204124 ¢6 0
-0.166667 -0.166667 0.666667 0.204124 -0.204124 ¢3 0
-0.204124 -0.333333 -0.166667 0.666667 0.204124 0.204124 0.166667 -0.204124 8\2.51 0
-0.204124 0.204124 0.833333 0 0 0 0
0.204124 -0.204124 0.204124 0.204124 0 0.833333 0.833333 0.166667 8\5.61 0
0.204124 -0.204124 -0.204124 0.166667 0 0 0 0
-0.204124 0.204124. -0.204124 -0.204124 0 0.166667 0.833333 8\3.61
0.204124
0.204124 8\2.31

Equations for element 3:

0.666667 -0.166667 -0.333333 -0.166667 -0.204124 0.204124 0.204124 -0.204124 ¢4 0
-0.166667 0.666667 -0.166667 -0.333333 -0.204124 -0.204124 0.204124' 0.204124 ¢7 0
-0.333333 -0.166667 -0.204124 -0.204124 0.204124 ¢8 0
-0.166667 -0.166667 0.666667 0.204124 -0.204124 ¢5 0
-0.204124 -0.333333 -0.166667 0.666667 0.204124 0.204124 0.166667 -0.204124 8\4.71 0
-0.204124 0.204124 0.833333 0 0 0 0
0.204124 -0.204124 0.204124 0.204124 0 0.833333 0.833333 0.166667 8\7.81 0
0.204124 -0.204124 -0.204124 0.166667 0 0 0 0
-0.204124 0.204124 -0.204124 -0.204124 0 0.166667 0.833333 8\5.8)
0.20412;+
0.204124 8\4.51

Equations for element 4:

0.666667 -0.166667 -0.333333 -0.166667 -0.204124 0.204124 0.204124 -0.204124 ¢5 0
-0.166667 0.666667 -0.166667 -0.333333 -0.204124 -0.204124 0.204124 0.204124 ¢8 0
-0.333333 -0.166667 ":0.204124 -0.204124 0.204124 ¢9 0
-0.166667 -0.166667 0.666667 0.204124 -0.204124 ¢6 0
-0.204124 -0.333333 -0.166667 0.666667 0.204124 0.204124 0.166667 -0.204124 8\5.8) 0
-0.204124 0.204124 0.833333 0 0 0 0
0.204124 -0.204124 0.204124 0.204124 0 0.833333 0.833333 0.166667 8\8.91 0
0.204124 -0.204124 -0.204124 0.166667 0 0 0 0
-0.204124 0.204124 -0.204124 -0.204124 0 0.166667 0.833333 8\6.9)
0.204124
0.204124 8\5.61

Essential boundary conditions:

On element 1, side 4, specified value = 10:
{rP2' rPP 6111.21} = {1O, 1O,OJ

On element 2, side 3, specified value = l:

{rP6' rP3' 6113.61} = {1,!, OJ
On element 4, side 3, specified value = 1:

{rP9' rP6' 6116.91} = [l, 1, OJ

630 p-FORMULATION

Known values from EBC:

{ °}¢1 = 10, ¢2 = 10, ¢3 = 1, ¢6 = 1, ¢9 = 1, 81{~ = 0,81~ =0,81~ =

Global equations after EBC:

"'I

1.33333 -0.333333 -0.166667 -0.333333 -0,204124 0 0,204124 -0.'108248 -0.204124 0 0.204124 0.204124 0 "5 5.
-0.333333 2.66667 -0.333333 -0.333333 0,204124 -0.'108248 -0.408248 -0.408248 -0.'108248 0.204124 0,204124 1.66667
-0.166667 -0.166667 0.204124 0.20412'1 -0.204124 0 "s'h
-0.333333 -0.333333 0,666667 0 0 0 0.204124 -0.204124 0 0.204124 -0.204124 -0.204124 0
-0.204124 -0.333333 -0.166667 1.33333 0 0 0 0,204124 0.204124 -0..108248 0 ,IIAI OS
0,833333 0 0.166667 O,2o.l124 0 0
0 0.204124 0 0 0 0.8333)] 0 0 0 0 0 0 0 ,1~31 0
0.204124 0.204124 0 0 0,166667 1.66667 0 0 0.166667 0 0.166667 0 204124
-0.408248 -0.408248 0 0 0 0 0 0 0 0 0 i2.SI 1.63299
-0.204124 -OA08248 O.2QJ1124 0.204124 0 0 0 1.66667 0 0 0 0 0 -4.08248
0 0.204124 0 0 0 0 0.833333 0 0,166667 0 ~t4.51
0.204124 0.20412" -0.204124 0.204124 0 0.166667 0 0 0 0 0.S33333 0
0.204124 -0.408248 0 -0,4082,18 0 0 0 0 0.166667 1.66667 0 0 0 ,/4,71 -204114
0 -0,408248 0.204124 -0.204124 0 0 0 0.166667 0 0 1.66667 0-1, 66667 -0.'108248
-0.204124 0 0 0 0 0 ~15.61
0.204124 -0.204124 0.166667 0 0 0
0.204124 0 /5.SI 0
0
0.833333 ./7,SI

.!S,91

Solving·the final system of global equations, we get

{¢4 =6.12921, ¢s =4.66596, ¢7 =5.08248, ¢s = 3.96195,
8\IAI = 0,0857292,8\2,31= 1.44833,8\2,SI = 1.36347,8\4,SI = -0.887807, 8\4.7l = 0.54473,
8\S,61= -0,708827, 8\S,SI = 0.440829, 8\7.S} = -0,251281, 8\S,91 -0.0306799}

Solution for element 1:
dof values for the element

{¢1 = 10, ¢4 =6.12921, ¢s =4.66596, ¢2 = 10,8\IAJ =0,0857292,
8\4.S} = -0.887807, 8\2,SJ = 1.36347,8\1.21= O}

dT = (10 6.12921 4.66596 10 0,0857292 -0.887807 1.36347 0)

xy¢ J¢/Jx J¢/Jy

2.5 2.5 7.5269 -0.811749 -0.302817

Solution for element 2:

dof values for the element

{¢2 = 10, ¢s = 4.66596, ¢6 = 1, ¢3 = (2.S) = {S,6} = -0,708827,
1,81 1.36347,81

8\3.61=0,8\2,31= 1.44833}

°=dT (10 4.66596 1 1 1.36347 -0,708827 1.44833)

xy¢ J¢/Jx J¢/ay

1 2.5 7.5 3.52259 -0,269207 -1.09961

Solution for element 3:
dof values for the element

p-FORMULATION FOR SECOND-ORDER 2D BVP 631

s- {¢4 = 6.12921,¢7 = 5.08248,¢s = 3.96195,¢s = 4.66596, 01M = 0.54473,
or·S
} = -0.251281, o\s.S} = 0.440829,0\4.S} = -0.887807}

aT = (6.12921 5.08248 3.96195 4.66596

0.54473 -0.251281 0.440829 -0.887807)

xy a¢/ax a¢/ay

7.5 2.5 5.00691 -0.253032 -0.245653

Solution for element 4:

dof values for the element

. IS.SI {S.9}
{¢s = 4.66596,¢s = 3.96195,¢9 = 1,¢6 = 1,01 = 0.440829,01 = -0.0306799,

0\6.9} =0, 0\S.6) =-0.708827}

aT = (4.66596 3.96195 0.440829 -0.0306799 0 -0.708827)

xy a¢/ax a¢/ay

7.5 7.5 2.74843 -0.153456 -0.608801

Computing a¢/ax and a¢/ay at several points within each element, the velocity field shown
in Figure 9.19 is obtained.

The following tables summarize solutions obtained at the element centers by increasing
the order of interpolation from 1 to 6. Complete calculations of these solutions can be

=seen on the book web site. The n 1 solution corresponds to the conventional rectangular

f

j j
I
8 /,
I

,I .' ./

i
6 .- , , .'

4 ~~ /~ , ./

.,"."

2

,

0
24 68

Figure 9.19. Velocity field for seepage under a sheet pile wall

632 p-FORMULATION .

element with bilinear solution and no p-modes.

xy ¢ 8¢/8x 8¢/8y

1 2.5 2.5 7.83036 -0.867857 -0.173571
2 2.5 7.5 4.19821 -0.520714 -1.27929
3 7.5 2.5 5.03393 -0.250714 -0.289286
4 7.5 7.5 2.65536 -0.0964286 -0.662143

Solution summary with 12 = 1:

xy ¢ 8¢/8x 8¢/8y

1 2.5 7.83036 -0.867857 -0.173571
2 2.5 7.5 4.19821 -0.520714 -1.27929
3 7.5 2.5 5.03393 -0.250714 -0.289286
4 7.5 7.5 2.65536 -0.0964286 -0.662143

Solution summary with 12 =2:

xy ¢ 8¢/8x 8¢/8y

1 2.5 2.5 7.5269 -0.811749 -0.302817
2 2.5 7.5 3.52259 -0.269207 -1.09961
3 7.5 2.5 5.00691 -0.253032 -0.245653
4 7.5 7.5 2.74843 -0.153456 -0.608801

Solution summary with 12 = 3:

xy ¢ 8¢/8x 8¢/8y

1 2.5 2.5 7.51643 -0.804721 -0.279215
2 2.5 7.5 3.54361 -0.240051 -1.06869
3 7.5 2.5 5.00409 -0.243475 -0.245966
4 7.5 7.5 2.73426 -0.140519 -0.642457

Solution summary with n =4:

xy ¢ 8¢/8x 8¢/8y

1 2.5 2.5 7.52003 -0.825071 -0.312494
2 2.5 7.5 3.73481 -0.189595 -0.996245
3 7.5 2.5 4.94684 -0.226552 -0.242614
4 7.5 7.5 2.63491 -0.0857831 -0.641989

Solution summary with n =5:

xy¢ 8¢/8x 8¢/8y

1 2.5 2.5 7.48739 -0.863053 -0.276887
2 2.5 7.5 3.68376 -0.316262 -1.10231
3 7.5 2.5 4.9193 -0.228476 -0.247952
4 7.5 7.5 2.62783 -0.0930979 -0.617761