Fundamental-Finite-Element-Analysis-and-Applications

FIN"ITE ELEMENT COMPUTATIONS INVOLVING MAPPEDELEMENTS 433

Computationof element matricesat (s ~ -0.57735, t ~ 0.57735) with weight =1.,

J = (1.78868 0.211325) . detJ =1.

0.394338 0.605662 '

NT = (0.166667 0.0446582 0.166667 0.622008)

a:: = (-0.105662 0.105662 0.394338 -0.394338)

a~T = (-0.394338 -0.105662 0.105662 0.394338)

B T "= ( 0.0915064 0.105662 0.197169 -0.394338)

-0.683013 -0.211325 0.105662 0.788675

c=(~ ~)

0.483253 0.163675 -0.0360844 -0.610844]
0.0669873 0.0193376 -0.25
0.163675 0.0193376
lek = ( -0.0360844 -0.25 0.0889156 -0.0721688

-0.610844 -0.0721688 0.933013

Computationof element matrices at (s ~ 0.57735, t ~ -0.57735) with weight = 1.,

J =(1.21132 0.788675). detl = 1.

0.105662. 0.894338 '

NT =(0.166667" 0.622008 0.166667 0.0446582)

&aNT = (-0)94338 0.394338 0.105662 -0.105662)

a~T = (-0.105662 -0.394338 0.394338 0.105662)

BT = (-0.341506 0.394338 0.0528312 -0.105662)

0.183013 -0.788675 0.394338 " 0.211325

nc=(~

0.266747 -0.413675 0.0360844 0.110844 ]
0.933013 -0.269338 -0.25
k = -0.413675
k ( 0.0360844 -0.269338 0.161084 0.0721688
0.110844 -0.25 0.0721688 0.0669873

434 MAPPEDELEMENTS

=Computation of element matrices at (s -? 0.57735,t -? 0.57735} with weight 1.,

] = (1.78868 0.788675). det] = 1.28868
0.394338 0.894338 '

NT = (0.0446582 0.166667 0.622008 0.166667)

&()NT = (-0.105662 0.105662 0.394338 -0.394338)

a:T
= (-0.105662 -0.394338 0.394338 0.105662)

BT =(-0.0409965 0.193998 0.153001 -0.306002)

-0.0819931 -0.612005 0.306002 0.387995

c _(2 0)
- 01

0.0129954 0.0441676 -0.0484994 -0.00866359]
0.0441676 0.579672 -0.164836 -0.459003
-0.164836 0.181002 0.0323329
k" =[ -0.0484994 -0.459003 0.0323329 0.435334

-0.00866359

Summing contributions from all points, we get

1.09091 -0.227273 -0.136364 -0.727273 ]
-1.18182
k = -0.227273 {81818 -0.409091
-0.409091 0.454545 0.0909091
k [ -0.136364 -1.18182 0.0909091 1.81818
-0.727273

Example 6.18 Eight-Node Quadrilateral Element Evaluate matrix kp for the eight-
node element shown in Figure 6.34. Assume p = 5 over the element.

sY Actual element

I

2

1

1-----2--1 x

Figure 6.34. Eight-node master element and actual element

FINITE ELEMENT COMPUTATIONS INVOLVING MAPPED ELEMENTS 435

xy

1 2. O.

2 4. O.

3 8. O.

4 5.65685 5.65685

5 o. 8.

6 O. 4.

7 O. 2.

8 1.41421 1.41421

Mapping to the master element,

xes, t) = -0.5ti +.0.5i - 0.62132t2s - 1.5ts + 2.12132s - 1.03553t2 - 2.t + 3.03553

yes, t) = 0.5ts2 + 0.5s2 - 0.62132t2s + 1.5ts + 2. 12132s - 1.03553t2 + 2.t + 3.03553
] _ (-0.62132t2 - l.st - 1.5t + l.s + 2.12132 -0.5s2 - 1.24264ts - 1.5s - 2.07107t - 2.)
- -0.62132t2 + l.st + 1.5t + l.s + 2.12132 0.5s2 - 1.24264ts + 1.5s - 2.07107t + 2.

det] = 1.s3 + 1.86396t2s2 + 5.12132s2 + 6.0061t2s + 1O.364s+ 3.72792t2 + 8.48528

The matrix kp is to be evaluated as follows:

Using 3 X 3 Gauss quadrature, we need to evaluate the integrand at the nine Gauss points
and then sum the contributions from each Gauss point to get the integral:

Point =Weight wi X
s

1 -0.774597 -0.774597 0.308642
0.493827
2 -0.774597 0 0.308642
0.493827
3 -0.774597 0.774597 0.790123
'0.493827
4 0 -0.774597 0.308642
0.493827
50 0 0.308642

60 0.774597

7 0.774597 -0.774597

8 0.774597 0

9 0.774597 0.774597

Computation of the kp matrix at point 1 is as follows:

= =Point {s -7 -0.774597, t -7 -0.774597}; Weight 0.308642

I = ( 1.53583 -0.279447) det] = 3.18182
0.412036 1.99676;

436 MAPPEDELEMENTS

NT == (0.432379 0.354919 -0.1 0.0450807 -0.032379 0.0450807 -0.1 0.354919 )

0.917974 0.753521 -0.212308 0.0957097 -0.0687431 0.0957097 -0.212308 0.753521
0.618529 -0.174273 0.0785635 -0.0564279 0.0785635 -0.174273 0.618529
0.753521 -0.174273 -0.0221356 -0.0221356 -0.174273
0.0785635 0.0491022 0.00997888 0.0158988 0.00997888 0.0491022 0.0785635
-0.212308 -0.0221356 -0.00716729 -0.00716729 -0.00716729 -0.0221356 -0.0564279
~0.0564279 0.00997888 0.00997888 0.0785635
=kp 0.0957097 0.0158988 -0.0221356 0.00514787 -0.0221356 0.0158988 -0.174273
-0.0687431 0.0785635 -0.0221356 0.0785635 -0.00716729 0.0785635 -0.0221356 0.618529
-0.174273
0.0957097 0.0491022 0.0158988 0.0491022
0.618529 -0.174273 -0.0564279 -0.174273
-0.212308

0.753521

Computation of the k p matrix at point 2 is as follows:

Point == (s -7 -0.774597, t -7 OJ; Weight == 0.493827

] == ( 1.34672 -11..11338811); det] =3.06543
1.34672

NT =(-0.1 0.2 -0.1 0.112702 -0.1 0.2 -0.1 0.887298)

0.0756895 -0.151379 0.0756895 -0.0853034 0.0756895 -0.151379 0.0756895 -0.671592
-0.151379 0.302758 -0.151379 0.170607 -0.151379 0.302758 -0.151379 1.34318

0.0756895 -0.151379 0.0756895 -0.0853034 0.0756895 -0.151379 0.0756895 -0.671592
-0.0853034 0.170607 -0.0853034 0.0961383 -0.0853034 0.170607 -0.0853034 0.756895

kp == 0.0756895 -0.151379 0.0756895 -0.0853034 0.0756895 -0.151379 0.0756895 -0.671592
0.302758 -0.151379 0.170607 -0.151379 0.302758 -0.151379 1.34318
-0.151379
0.0756895 -0.151379 0.0756895 -0.0853034 0.0756895 -0.151379 0.0756895 -0.671592
1.34318 -0.671592 0.756895 -0.671592 1.34318 -0.671592 5.95902
-0.671592

After computing and summing contributions from all nine points, the lep matrix for the
element is as follows:

4.60149 -8.71619 1.40948 -11.5208 ;' 3.01679 -8.69358 3.0281 -6.13668
37.5578 -2.2417 28.3616 -8.69358 17.3872 -8.69358 17.5481
-8.71619 -2.2417 11.076 -5.06893 -0.209149 -8.69358 -6.11406
28.3616 -5.06893 59.6034 -5.06893 28.3616 3.01679 17.6349
1.40948 -8.69358 -0.209149 -5.06893 11.076 -2.2417 -11.5208 -6.11406
17.3872 -8.69358 28.3616 -2.2417 37.5578 17.5481
lep = -11.5208 -8.69358 -8.71619 1.40948 -6.13668
3.01679 17.5481 3.01679 -11.5208 1.40948 17.5481 -8.71619 18.4849
-6.11406 17.6349 -6.11406
-8.69358 4.60149
-6.13668
3.0281

-6.13668

6.4.4 Evaluation of Boundary Integrals

Elements involving a boundary where a nonzero natural boundary condition is specified
require computation of the lea matrix and the 1'p 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

FINITEELEMENTCOMPUTATIONS INVOLVING MAPPEDELEMENTS 437

Master element dc dy
t

L

la dx

a x

1---2-----1

Figure 6.35. Boundary coordinates on the master and the actual element

system. Once mapped, each boundary curve is either a horizontal or a vertical line repre-
senting a side of the master square element. The boundary coordinate e for .each side is
mapped to a coordinate -1 -s a ::; 1 for each side, as shown in Figure 6.35. The mapping
for each side x(a) and yea) 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

To write interpolation functions N; along a side, we substitute the appropriate s, t values
in the interpolation functions N(s, t). The only remaining task is to relate the differential
length de along an element side to the differential line segment da along the side of the
master element. This can be done easily by noting that

de =~dXZ +dl

Dividing both sides by darse. have

( ddaX)2 + (ddya)2 =} de =Jc da

where Jc is called the Jacobian of a side,

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

11rf3 = [3NJcda
-1

438 MAPPEDELEMENTS

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:

2:lea "
=- L(l1 aNeN:Ie cia =- j=1 wja(ajWcCajW:(aj)Je(a)

2:J-liJ = "
( 1 =
cia wj{3(a jWe(a)Ie(a)
{3NJe

1 j=1

where aj are the one-dimensional Gauss points and w j are the corresponding weights.

Example 6.19 Four-Node Quadrilateral Element Evaluate the matrix lea for the four-
node quadrilateral element shown in Figure 6.36. Assume a is equal to 1.35 over side 3-4
of the element. The interpolation functions are

NT = H(s - l)(t - 1), -!(s + l)(t - 1), !(s + l)(t + 1), -!(s - 1)(t + l)}

Mapping to the master element,

xes. t) = -t2s + 3s + t + 3

-2 -2 -2

yes, t) = ts + -4s + -34t + -43
-4

For the NBC on side 3, a = -s and t = 1. Thus

N: = (0 9. l; a a ; 1)

=x(a) 2 - 2a; yea) = 23" - 2a"

cix _ -2' dy = -21"; I = {l7
cia - , cia
e2

Master element Y Actual element 3
(4,2)
t
(0, OJ
I

2

1

1----2----1

Figure 6.36. Four-node master element and actual element

FINITEELEMENTCOMPUTATIONS INVOLVING MAPPEDELEMENTS 439

Using two Gauss points, the matrix k", for the side is established as follows:

Gauss point = -0.57735; Weight = 1.; Jc = Yl7/2

N~ = (0 0 0.788675 0.211325)

~ [~ 0 0
00

k. 0 -1.73111 j463849]

0 -0.463849 -0.124288

~Gauss point = 0.57735; Weight = 1.; t, =

N~ = (0 0 0.211325 0.788675)

~ [~ 0 0
00

k. 0 -0.124288 j,463849]

0 -0.463849 -1.73111

Summing contributions from all Gauss points,

~ [~ 0 0 j,927699]
00
k. 0 -1.8554
0 -0.927699 -1.8554

Example 6.20 Eight-Node Quadrilateral Element Compute vector rfJ for the eight-

node element shown in Figure 6.37. Assume f3 is equal to 23.4 over side 7-8-1 of the

element.

I sY Actual element
s
2

1

1----2----1 x
Figure 6.37. Eight-node master element and actual element

440 MAPPEDELEMENTS

xY

1 2. O.

2 4. O.

3 8. O.

4 5.65685 5.65685

5 O. 8.

6 O. 4.

7 O. 2.

8 1.41421 1.41421

Interpolation functions and their derivatives are

NT == { - i(s - l)(t - l)(s + t + 1), 1(S2 - l)(t - 1), i(t - 1)(-s2 + ts + t + 1),
-1(s + 1)(t2 - 1), i(s + l)(t + l)(s + t - 1), -1(s2 - l)(t + 1),
*(s - l)(s - t + l)(t + 1), ~(s - 1)(t2 - I)}

Mapping to the master element,

xes, t) == -0.5ts2 + 0.5s 2 - 0.62132t2s - 1.5ts + 2.12132s - 1.03553t2 - 2.t + 3.03553
Yes, t) == 0.5ts2 + 0.5s 2 - 0.62132t2s + 1.5ts + 2.12132s - 1.03553t 2 + 2.t + 3.03553

We now compute rf3 resulting from the NBC for side 4 with fJ == 23.4. For this side a == -t

and s == -1. Thus

I'

N T == (a-+21- + 21:(a2 - 1) 0 0 0 o0 -12--a + -1(a2 - 1) 1 - a2 )
c 2

x(a) == -0.414214a2 + La + 1.41421; yea) == -0.414214a2 - La + 1.41421

dx
da == 1. - 0.828427a

dy
da == -0. 828427a -1.

Jc == ~(-0. 828427a - 1.)2 + (1. - 0.828427a)2

Using three Gauss points, the vector rf3 for the side is established as follows:
Gauss point == -0.774597; weight == 0.555556; Jc == 1.68034:

N~ = (-0.0872983 0 0 0 0 0 0.687298 0.4)
r~=(-1.90698 0 0 0 0 0 15.0137 8.73778)

COMPLETE MATHEMATICA AND MATLAB SOLUTIONS OF 2D BVP INVOLVING MAPPED ELEMENTS 441

Gauss point = 0; weight = 0.888889; Ie = 1.41421:
N~ = (0 0 0 0 0 0 0 1)

rJ =(0 0 0 0 0 0 0 29.4156) .

Gauss point = 0.774597; weight = 0.555556; Ie = 1.68034:
N~ .= (0.687298 0 0 0 0 0 -0.0872983 0.4 )

rJ = (15.0137 0 0 0 0 0 -1.90698 8.73778)

Summing contributions from all Gauss points,

'rJ =(13.1067 0 0 0 0 0 13.1067 46.8912)

6.5 COMPLETE MATHEMATICA AND MATLAB SOLUTIONS OF 20 BVP
INVOLVING MAPPED ELEMENTS

As seen from the previous sections, for any arbitrary shaped element in the x, y coordinate
system, the element equations are developed by first creating a mapping as follows.

Xes, t) =Nl xl + N2x2 + + N"xll
yes, t) =NIYl + N2Yz + + NIlYIl

where N, are the interpolation functions for the appropriate master element and xj and Yj
are the nodal coordinates. The Jacobian matrix of the mapping is as follows:

f= ax
as
( aays

The actual finite element equations are

where using mapping the three terms involving area integrals are evaluated as follows:

JJ i: i:kk = BCBT dA = T detl ds dt

BCB

A 1:J22 aaNs ,'_J21 aaNt"

L: L:m 1l

= wjwjB(si' t)C(s" tj)BT (si' tj) detl(sj' t)

j=1 j=1

BT _ _1_ J22 '!!a!..Js.. - J21 '!!a!..Jt.. Jq2!2!.aJs.. - J a N2

217ft

- detl [ - I '!!a!..Js.. + I '!!!..J.. - I 12q!a!.Js.. +I aN, - I12 ~as + I aaNt"
12 11 at
llat 11

442 MAPPEDELEMENTS

0)c =(kox ky
ff f: f:lcp = -
pNNT dA = - pNNT det] ds dt

A

=- I I111 11
wiwjP(Si> tj)N(Si' tj)NT (Si' t) det](sj' t)
j=l j=l

where Sj, tj are the Gauss points and wi> wj are the corresponding weights.
The two natural boundary condition terms involve line integrals:

where Ie is the Jacobian of the side,

Ie = (ddaX)2+ (dy)2

da

,/

where ai are the Gauss points and wi are the corresponding weights.
Calculations involving mapped elements are clearly very tedious to perform by hand.

However, it is not too difficult to create MATLAB and Mathematica functions to develop

the element equations numerically. The following implementations are for four-node and

eight-node quadrilateral elements. They can be used as templates to create functions for

other mapped elements.

MATLAB Implementation: Four-Node Quadrilateral Element for 2D BVP The

analysis of two-dimensional boundary value problems using mapped elements can be

performed conveniently by writing three MATLAB functions, one for defining the ele-

ment lck + lcp and rq vectors, one for evaluating natural boundary terms, and the third for

computing the element solution. The BVPQuad4Element employs 2 x 2 integration. The

BVPQuad4NBCTerm employs two-point integration. The functions can be modified easily

to use any other integration formula. The BVPQuad4Results function computes results at

the point of the element that corresponds to the origin of the master element (s = t = 0).

The function returns the location of this point (in terms of x, y coordinates) and the solution

and its x and y derivatives at this point. The function can easily be modified to compute

results at any other point.

COMPLETE MATHEMATICA AND MATLAB SOLUTIONS OF 2D BVP INVOLVING MAPPEDELEMENTS 443

MatlabFiles\Chap6\BVPQuad4Element.m

function [ke , rq] = BVPQuad4Element (kx , ky, p, q, coord)"

% [ke, rq] BVPQuad4Element(kx, ky, p, q, coord)
%Generates for a 4 node quadrilateral element for 2d BVP
%kx, ky, p, q = parameters defining the BVP
%coord coordinates at the element ends

%Use 2x2 integr~tion. Gauss point locations and weights

pt=1/sqrt(3);
gpLocs = [-pt,-pt; -pt,pt; pt,-pt; pt,pt];
gpWts = [1,1,1,1];
kk=zeros(4); kp=zeros(4); rq=zeros(4,1);
for i=1:length(gpWts)

s = gpLocs(i, 1); t = gpLocs(i, 2); w = gpWts(i);
n = [(1/4)*(1 - s)*(1 - t), (1/4)*(s + 1)*(1 -t),

(1/4)*(s + 1)*(t + 1), (1/4)*(1 - s)*(t + 1)];
dns=[(-1 + t)/4, (1 - t)/4, (1 + t)/4, (-1 - t)/4];
dnt=[(-1 + s)/4, (-1 - s)/4; (1 + s)/4, (1 - s)/4];
x = n*coord(:,1); y = n*coord(:,2);
dxs = dns*coord(:,1); dxt = dnt*coord(:,1);
dys = dns*coord(:,2); dyt = dnt*coord(:,2);
J = [dxs, dxt; dys, dyt]; detJ = det(J);
bx = (J(2, 2)*dns - J(2, 1)*dnt)/detJ;
by = (-J(1, 2)*dns + J(1, 1)*dnt)/detJ;
b = [bx; by];
c = [kx, 0; 0, ky];
kk = kk + detJ*w* b'*c*b;
kp = kp - detJ*w*p * n'*n;
rq = rq + detJ*w*q * n';
end
ke=kk+kp;

MatlabFiles\Chap6\BVPQuad4NBCTerm.m

function [ka, rb] = BVPQuad4NBCTerm(side, alpha, beta, coord)

% [ka, rb] = BVPQuad4NBCTerm(side, alpha, beta, coord)
%Generates kalpha and rbeta when NBC is specified along a side
%side = side over which the NBC is specified
%alpha and beta = coefficients specifying the NBC
%coord = coordinates at the element ends

%Use 2 point integration. Gauss point locations and weights

pt=-1/sqrt(3);
gpLocs = [-pt, pt];
gpWts = [1,1];
ka=zeros(4); rb=zeros(4,1);

444 MAPPEDELEMENTS

for i=1:length(gpWts)
a = gpLocs(i); w = gpWts(i);
switch (side)
case 1
n = [(1 - a)/2, (1 + a)/2, 0, 0];
dna = [-1/2, 1/2, 0, 0];
case 2
n = [0, (1 - a)/2, (1 + a)/2, 0];
dna = [0, -1/2, 1/2, 0];
case 3
n = [0, 0, (1 - a)/2, (1 + a)/2];
dna = [0, 0, -1/2, 1/2];
case 4
n = [(1 + a)/2, 0, 0, (1 - a)/2];
dna = [1/2, 0, 0, -1/2];
end
dxa = dna*coord(:,1); dya = dna*coord(:,2);
Jc=sqrt(dxa-2 + dya-2);

ka = ka - alpha*Jc*w*n'*n;
rb = rb + beta*Jc*w*n';
end

MatlabFiles\Chap6\BVPQuad4Results.m

function results = BVPQuad4Results(coord, dn)

%results = BVPQuad4Result~(coord, dn)
%Computes element solution for a quadrilateral element for 2D BVP
%coord = nodal coordinates
%dn = nodal solution
%The solution is computed at the element center
%The output variables are loc, u and its x and y derivatives

s = 0; t = 0;

n = [(1/4)*(1 - s)*(1 - t), (1/4)*(s + 1)*(1 -t),
(1/4)*(s + 1)*(t + 1), (1/4)*(1 - s)*(t + 1)];

dns=[(-1 + t)/4, (1 - t)/4, (1 + t)/4, (-1 - t)/4];
dnt=[(-1 + s)/4, (-1 - s)/4, (1 + s)/4, (1 - s)/4];
x = n*coord(:,1); y = n*coord(:,2);
dxs = dns*coord(:,1); dxt = dnt*coord(:,1);
dys = dns*coord(:,2); dyt = dnt*coord(:,2);
J = [dxs, dxt; dys, dyt]; detJ = det(J);
bx = (J(2, 2)*dns - J(2, 1)*dnt)/detJ;
by = (-J(1, 2)*dns + J(1, 1)*dnt)/detJ;
b = [bx; by];
results=[x, y, n*dn, bx*dn, by*dn];

COMPLETE MATHEMATICA AND MATLAB SOLUTIONS OF 20 BVP INVOLVING MAPPED ELEMENTS 445

y (em) To Insulated

2.~ t x (em)
6
2

1.5 qL

1

0.5

o

o2 4

Figure 6.38. .Lshaped body

Example 6.21 Heat Flow in an L-Shaped Body Using Quad4 Elements Con-

sider two-dimensional heat flow over an L-shaped body with thermal conductivity k =

45 W/m' °C shown in Figure 6.38. 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 convec-

tion 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 = 8000 W/m2. Heat is generated in the
body at a rate of Q = 5 x 106 W1m3 • Determine the temperature distribution in the body.

To demonstrate the use of the functions presented in this section, we obtain a finite
element solution of the problem using only two quadrilateral elements, as shown in Fig-
ure 6.39. Obviously we do not expect very good results. The mesh is chosen to simplify
calculations. The steps are exactly those used in other MATLAB implementations.

MatlabFiles\Chap6\LHeatQua'!4Ex.m

%Heat flow through an L-shaped body using quad4 elements

h = 55; tf= 20; htf = h*tf;
kx = 45; ky = 45; Q = 5*10-6; ql = 8000; to = 110;

Y
12

x

Figure 6.39. Two four-node quadrilateral element model

446 MAPPEDELEMENTS

nodes = 1.5/100*[0,2; 2, 2; 2, 1; 4, 1; 4, 0; 0, 0];
lmm = [6, 3, 2, 1; 6, 5, 4, 3];
debe = [5:6]; ebcVals=tO*ones(length(debc),1);
dof=length(nodes); elems=size(lmm,1);
K=zeros (dof ) ; R = zeros (dof , 1) ;
% Generate equations for each element and assemble them.
for i=1:elems

1m = lmm(i,:);
[k, r] = BVPQuad4Element(kx, ky, 0, Q, nodes(lm,:»;
K(lm, 1m) = K(lm, 1m) + k;
R(lm) = R(lm) + r;
end

%Compute and assemble NBC contributions

1m = lmm(1,:);
[k, r] = BVPQuad4NBCTerm(4, 0, ql, nodes(lm,:));
K(lm, 1m) = K(lm, 1m) + k;
R(lm) = R(lm) + r;
[k, r] = BVPQuad4NBCTerm(2, -h, htf, nodes(lm,:));
K(lm, 1m) = K(lm, 1m) + k;
. R(lm) = R(lm) + r;

[k, r] = BVPQuad4NBCTerm(3, -h, htf, nodes(lm,:));
K(lm, 1m) = K(lm, 1m) + k;
R(lm) = R(lm) + r;

1m = lmm(2,: ).;

[k, r] = BVPQuad4NBCTerm(3, -h, htf, nodes(lm,:));

K(lm, 1m) = K(lm, 1m) + k; ;'

R(lm) = R(lm) + r; .

%Nodal solution ...
d = NodalSoln(K, R, debe, ebcVals)

results=[];

for i=1:elems

results = [results; BVPQuad4Results(nodes(lmm(i,:),:),
d(lmm(i,:)))];

end

format short g
results

» LHeatQuad4Ex

d '=

153.3936
142.9067
132.8533
124.5394

COMPLETE MATHEMATICA AND MATLAB SOLUTIONS OF 20 BVP INVOLVING MAPPED ELEMENTS 447

. 110.0000 0.01875 134.79 -90.82 1187.7
110.0000 0.0075 119.35 -92.377 1338.8

results =

0.015
0.0375

Mathematica tmptementetion: Eight-Node Quadrilateral Element for 2D BVP

The functions for eight-node quadrilateral elements are almost identical to those for the
four node quadrilaterals. The only changes are the dimensions of the matrices and that we
use three points for integration and eight node serendipity interpolation functions.

Needs["NumericalMath'Oauesdanlluadrtrture '"J:

BVPQuad8Element[kx_, ky., p_, q., {{xL, yL}, {x2_, y2_}, {x3..., ys.], {x4..., y4.:},

{x5..., y5_}, [xd., y6_}, {x7_, y7 .}, {x8..., y8_}}] :=

Module[{sloe, swt s, tloe, twts, gpLoes, gpWts, n, s, t, dns, dnt, xst, yst,

J, npt, Jpt, dnsPt, dntPt, detJ, bx, by, bt, b, c, kk, kp, rq},

{sloe, swts} = Transpose[GaussianQuadratureWeights[3, -1,1]];

={tloe, twts} Transpose[GaussianQuadratureWeights[3, -1,1]];

gpLoes = Flatten[Outer[List, sloe, tloe], 1];

=gpWts Flatten[Outer[Times, swts, tms], 1];

n=

{(114)* (1 - s) * (1 - t) - (1/4) * (1 - s-2) * (1 - t)
-(1/4) * (1 - s) * (1 - t-2), (112) * (1 - s-2) * (1 - t), (114) * (s + 1) * (1 - t)

-(1/4) * (1 - s-2) * (1 - t) - (114) * (s + 1) * (1 - t-2),

(1/2) * (s + 1) * (1 - t-2), (1/4) * (s + 1) * (t + 1) - (114) * (1- s-2) * (t + 1)
-(1/4) * (s+ 1) * (1- t-2), (1/2) * (1- s-2) * (t + 1), (114) * (1- s) * (t + 1)
-(114) * (1 - s-2) * (t + 1) - (114)* (1- s) * (1 - t-2),

(112)~' (1 - s) * (1 - t-2)};

[dns, dnt} = [D]n, s], D[n, t]};
xst =n.{x1, x2, x3, x4,.x5, x6, x7, x8};
yst =n.{y1, y2, y3, y4, y5, y6, y7, y8};
J = {{D[xst, s], D[xst, t]}, {D[yst, a], D[yst, t]}};
.kk = kp = Table[O, {8}, {8}];

rq = Table[O, {8}]; ..

Do[pt = [s -7 gpLoes[[i, 1]], t ~ gpLoes[[i, 2]]};
w = gpWts[[i]];
[npt, Jpt, dnsPt, dntPt} = in, J, dns, dnt}l.pt;
=detJ det[Jpt];
bx = (Jpt[[2, 2]]dnsPt - Jpt[[2, 1J]dntPt)/detJ;
=by (-Jpt[[1, 2]]dnsPt + Jpt[[1, 1]]dntPt)/deq;
bt = [bx, by}; b = Transpose[bt];
e = {{kx, OJ, to, ky));

kk+ = detJwb.c.bt;

kp+ = -detJ wp Outer[Times, npt, npt];

448 MAPPEDELEMENTS

rq+ == detJwqnpt, Ii, 1, Length[gpWts]J
];

[kk + kp, rq}

BVPQuad8NBCTerm[side_,0;_,/3-, ((xL, yL), (x2_, y2_), Ix3-, y3_}, [x-l., y4_),
[xb., y5_}, [xfi., y6_}, [x7 _, y7 .], (x8_, y8_llJ :==

Module[(k, r, pts, wts, n, s, t, a, xa, ya, Jc, Jcpt),
k == Table[O, (8), [8}];r == Table[O, (8j];
(pts, wtsj == Transpose[GaussianQuadratureWeights[2, -1, 1]];
n==

(1/4) * (1 - s) * (1 - t) - (1/4) * (1 - s-2) * (1 - t)
-(V4) * (1 - s) * (1 - t-2), (V2) * (1 - s-2) * (1 - t),

(V4) * (s + 1) * (1 - t) - (1/4) * (1 - s-2) * (1 - t)
-(V4) * (s + 1) * (1- t-2), (V2) * (s + 1) * (1- t-2),

(1/4) * (s + 1) * (t + 1) - (1/4) * (1- s-2) * (t + 1)
-(1/4) * (s + 1) * (1- t-2), (1/2) * (1 - s-2) * (t + 1),

(1/4) * (1 - s) * (t + 1) - (1/4) * (1 - s-2) * (t + 1)
-(1/4) * (1 - s) * (1 - t -2), (V2) * (1 - s) * (1 - t -2)};

n == Switch[side, 1, nJ.(s --) a, t --) -1},
2, ts/, (s --) 1, t --) a},
3, nJ.(s --) -a, t --) 1),
4, nJ.(s --) -1, t --) -a}

];
xa == n.(x1, x2, x3, x4, x5, x6, x7, x8);
ya == n.(y1, y2, y3, y4, y5, y6, y7, y8);
Jc == Sqrt[D[xa, ar2 + D[ya, aJ-2J;
Do[(npt, Jcpt) == [n, Jel/.l--) pts[[i]];

k+ == -0; * Jept * wts[[iJJ Outer[Times, npt, npt];
r+ == f3 * Jcpt * wts[(iJ]npt,

(i, 1, Length[pts])
];
(k,r}
];

BVPQuad8Results[({xL, yt.], (x2_, y2_), [x3_,y3_}, (x4_, y4_),

Ix5-, y5_), (x6..., y6_), (x7 _, y7 _I, [xfs., y8_)},d.] :==

Module[(pt, n, s, t, dns, dnt, bx, by, xst, yst, J, npt, Jpt, dnsPt, dntPt, detJ, loe},
pt == [s --) 0, t --) OJ;

n==

((1/4) * (1 - s) * (1 - t) - (1/4) * (1 - s-2) * (1 - t)
-(V4) * (1 - s) * (1 - t-2), (V2) * (1 - s-2) * (1 - t),
(1/4) * (s + 1) * (1 - t) - (V4) * (1 - s-2) * (1 - t)
-(1/4) * (s + 1) * (1- t-2), (V2) * (s + 1) * (1- t-2),
(1/4) * (s + 1) * (t + 1) - (1/4) * (1 - s-2) * (t + 1)
-(V4) * (s + 1) * (1 - t-2), (V2) * (1 - s-2) * (t + 1),

COMPLETE MATHEMATICA AND MATLAB SOLUTIONS OF 2D BVP INVOLVING MAPPEDELEMENTS 449

(1/4) * (1- s) * (t + 1) - (1/4) * (1- s-2) * (t + 1)
-(V4) * (1 - s) * (1 .: t-2), (V2) * (1 - s) * (1 - t-2)};
[dns, dnt) = [D]n, s], D[n,t]);
xst =n.{x1, x2, x3, x4, x5, x6, x7, x8};

=yst n.{y1, y2, y3, y4, y5, y6, y7, y8);

J = ({D[xst, s], D[xst, t]), {D[yst, s], D[yst, t]}};
[Loc, npt, Jpt, dnsPt, dntPt) =

detJ = det[Jpt];
bx = (Jpt[[2, 2]] dnsPt - Jpt[[2, 1]] dntPt)/detJ;
by = (-Jpt[[1, 2]] dnsPt + Jpt[[1, 1]] dntPt)/detJ;
floc, npt.d, bx.d, by.d)

Example 6.22 Heat Flow in an L·Shaped Body Using QuadS Elements To demon-
cstrate the use of the functions presented in this section, we obtain a finite element solution
of the problem using only two quadrilateral elements, as shown in Figure 6.40.

Clear[k, r];

h = 55; tf = 20;htf = h * tf;
kx = ky = 45; Q= 5 * 10-6; ql = 8000; to = 110;
nodes = 1.5/100{{O, 2}, (1, 2), (2, 2), {2, 1.5}, (2, 1), {3, 1},{4, 1},

(4, 0.5), (4, 0), {2, OJ, to, O),{O, 1},{1,0.5}};

=Im[1] (11, 13,5,4,3,2, 1, 12);

Im[2] = (11, 10, 9, 8, 7, 6, 5, 13);
Do[{k[i], r[i]) = BVPQuad8Element[kx, ky, 0, Q, nodes[[lm[i]]]], (i, 1,2)];
{k[1], r[1]}+ = BVPQuad8NBCTerm[4, 0, ql, nodes[[lm[1]]]];

(k[1], r[1])+ =BVPQuad8NBCTerln[2, -h, htf, nodes[[lm[1]]]];

=(k[1], r[1])+ BVPQuad8NBCTerm[3, -:h, htf, nodes[[lm[1]]]];

(k[2], r[2])+ = BVPQuad8NBCTerm[3, -h, htf, nodes[[lm[2]]]];
R = Table[O, (13)];

x

Figure 6.40. Two eight-node quadrilateral element model

450 MAPPEDELEMENTS

K == Table[O, (13j, {13}];
Do[K[[lm[iJ, Im[iJ]J+ == k[iJ; R[[lm[iJ]J+ == r[i], (i, 1, 2)];

Essential boundary conditions are as follows:

debc == Range[9, 11J;
ebcVals == Table[tO, {3lJ;

df == Complement[Range[13J, debc]

{l, 2, 3, 4, 5, 6, 7,8,12, l3}

Kf == K[[df, dfJJ

57.3302 -33.3087 23.1028 -36.1508 30.4563 0 0 0 -46.6746 -16.5079
-33.3087 95.8127 -20.4516 3.65079 -28.0159 0 0 0 3.1746 -5.07937
0 0 0 -19.3651
23.1028 -20.4516 58.5885 -68.0829 33.5653 0 0 0 -28.4127 16.3492
-36.1508 3.65079 -68.0829 156.24 -82.6067 -85.3722 51.9123 -90.6349 42.0635 -156.349
257.914 133.749 -27.277 14.6032 65.3968
30.4563 -28.0159 33.5653 -82.6067 -85.3722 -27.277 70.6397 -103.968 -33.6508 -56.0317
0 0 0 0 14.6032 -103.968 258.889 0 101.111
0 0 0 0 51.9123 0 0 0 0 -23.1746
0 0 0 0 -90.6349 65.3968 -56.0317 101.111 0 353.968
-33.6508
-46.6746 3.1746 -28.4127 42.0635 -156.349 125.968
-16.5079 -5.07937 -19.3651 16.3492 -23.1746

Rf == R([df]J - K[[df, debcJJ.ebcVals

{-2627.56, 2762.08, -2665.76,4411.4, -11968.1, 12022., -7457.,20925.,5732.3, 30884.9}

The solution for nodal unknowns can now be obtained by using the LinearSolve function
as follows:

dfVals == LinearSolve[Kf, RfJ ,/
(156.445,150.755,149.201,144.228,133.843, 123.998, 121.754, 119.151, 144.677, 129.134

The complete vector of nodal values, in the original order established by the node number-
ing, is as obtained by combining these values with those specified as essential boundary
conditions as follows:

d == Table(O, (13JJ;
d((debcJ] == ebcVals;
d[(dfJ] == dfVals;

d

[156.445,150.755,149.201,144.228,133.843,123.998, 121.754, 119.151,
110, 110, 110, 144.677, 129.134)

Using the BVPQuad8Results function, the solution and its x and y derivatives for each
element can now be computed as follows:

BVPQuad8Results[nodes[[lm[1JJJ, d([lm[1J]JJ
{(0.015,0.01875),147.025, -255.209, 960.97}

TRIANGULAR ELEMENTS BY COLLAPSING QUADRILATERALS 451

BVPQuad8Results[nodes[[lm[2]]]' d[[lm[2]]]]
((0.0375, 0.0075}, 122.242, -221.837, 1155.06}

6.6 TRIANGULAR ELEMENTS BY COLLAPSING QUADRILATERALS

By collapsing one side of a quadrilateral, it is possible to map triangular areas into 2 x 2
square areas. As seen in Figure 6.41, this is accomplished by considering all nodes on side
4 of the master area to be physically located at the same point. Thus for a straight-sided
triangle nodes 1 and 4 are coincident. The mapping in terms of its three physical nodes is

as follows:

Ws -xes,t) = l)(t - 1) + ~(l - s)(t + 1))xJ + ~(-s - l)(t - 1)x2 + ~(s + l)(t + 1)x3

=yes, t) (~(s - 1)(t -1) + ~(1 - s)(t + 1))YJ + ~(-s -1)(t -1)Y2 + ~(s + 1)(t + 1)Y3

For a triangle with all three sides curved nodes 1,7, and 8 are coincident. The mapping in
terms of its six physical nodes is as follows:

Master element y Triangular element
t
4,1
I
x
2

1

2

Master element Y Triangular element
t6
5
I
7,8,1 x
2

1

2

Figure 6.41. Mapping triangular elements into quadrilaterals

452 MAPPEDELEMENTS

= !(1-xes, t) Ws -l)(s - t + l)(t + 1) + s)(t -l)(s + t + 1) + !(s - 1)(t2 -1))x\

+ 4(s2 - 1)(t -1)x2 + !(t - 1)(-s2 + ts + t + 1)x3 + 4(-s - 1)(t2 -1)x4

4(1 -+ !(s + l)(t + l)(s + t - l)xs + s2)(t + 1)x6

= !(1-yes, t) (!(s -l)(s - t + l)(t + 1) + s)(t - l)(s + t + 1) + 4(s - 1)(t2 -1))y\

4(+ 4(S2 - l)(t - 1)Y2 + !(t - 1)(-s2 + ts + t + 1)Y3 + -s - 1)(t2 - 1)Y4

4(1 -+ !(s + l)(t + l)(s + t - 1)ys + s2)(t + 1)Y6

One key advantage of this procedure is that we do not need to develop separate interpo-
lation functions and computer programs for triangular elements. Any computer implemen-
tation of quadrilateral elements can be used to handle triangular elements as well simply
by repeating the nodes that define the last side of the quadrilateral.

6.7 INFINITE ELEMENTS

Many practical problems involve computational domains that are not well defined and
are essentially infinite in one or more directions. One such situation was encountered in
Chapter 5 when modeling fluid flow around an object. There are no definite boundaries
of the flow region. It was suggested to simply extend the boundaries far enough from the
object so that assumption of uniform flow is valid. Another common situation that involves
infinite boundaries is analysis of structures supported on ground. In most such cases part
of the ground must be included in the finite element model. However, usually there is no
rigid boundary that defines the extent of ground that should be included in the model. In all
these situations the results will not be reliable if the computational domain is too small. On
the other hand, making the computational domain very large wastes computational effort.
The so-called infinite elements, presented in this section, are designed to be used as the last
layer of elements in the direction where the boundary is not well defined.

6.7.1 One-Dimensional BVP

The key idea in developing an infinite element is to create an appropriate mapping such that
an infinite element is mapped to a standard master element. Consider mapping of an infinite
line to a standard three-node master line as shown in Figure 6.42. We need a mapping so

=that the point s = -.1 on the masterline is mapped to Xl's = 0 is mapped to x2' and s = 1 is

mapped to x 00. An appropriate mapping can be developed by starting with a relationship
in the following form:

where ao' a\, a2, and f(s) are to be determined to give the desired mapping. The coefficienn
ao and a l can be established by the requirement of mapping x\ and x2 to s = -1 and s = 0

INFINITE ELEMENTS 453

Three node master line

Infinite line

Node No e 2 Node 3 ' 1 - -8_ - - - -
-1 s
Xl X2 X3-700

--------X

Figure 6.42. Master line for mapping an actual infinite line

respectively, as follows.

Ats =-1: Xl =ao - at + azf(-l)
Ats =0:
Xz = ao+ azf(O)

Solving the two equations for ao and aI' we get

Substituting these, the mapping takes the following form:

=Xes) Xz - azf(O) + [-xl + Xz + azf(-1) - azf(O)]s + azf(s)

Rearranging terms,

Xes) =Xz + S(Xz - Xi) + az[s(f(-1) - f(O)) - f(O) + f(s)]

= = =The next task is to choose f(s) and az so that X 00 when s 1. Clearly, f(s) 1/(1 - s)

will do the job, giving

Tdihsetacnoceeffbiceitewnet eanz tchaen tbweocnhoodseens, air.be.i,traarzil=y. A common choice is to set az equal to twice the
2(xz - Xl)' Using this value of az, the mapping

takes the following simple form:

. 2s (s + 1) ..
xes) = -s--Xl l - -s--l-xz

Note the element has three nodes. However, only the first two, which are at finite locations,
are used in the mapping.

The assumed solution over the mapped element is written using the standard quadratic
interpolation functions:

u(s) =( -sZ - -s

2 2

454 MAPPEDELEMENTS

The derivative of this solution with respect to x is calculated using the chain rule as usual
for the mapped elements:

-du = du dx du =-I d-u = -1 ( s - 1
-dx -
ds -dx- :ds: = } J ds J2

where

The element equations for a one-dimensional element for the ID BVP considered in Chap-
ter 3 can be developed in a usual manner:

Because of singularity in J, the integrals cannot be evaluated in the usual sense. However,
since the intention of these elements is to approximate the behavior of a region that is far
away from the actual region of interest, this fact is ignored. In numerical implementations
integrations are typically carried outusing a two-point Gauss quadrature formula. Thus the
element equations are

= =(kk +kp)d rq ::=} kd r

Mathematica Functions for Solution of One-Dimensional Boundary Value
Problem Involving Infinite Domain The following functions are developed for
performing computations for both the standard and the infinite elements for the one-
dimensional boundary value problem. The functions are a little more general than those
presented in Chapter 3. The same function can handle a linear, quadratic, or infinite ele-
ment. The coefficients k, p, and q can be any arbitrary functions of x. The integrations are
performed using the two-point Gauss quadrature.

BVP1DElement[k_, p~ q_,nodes., type_ : Linear] :=
Module[{n, nm, s, XS, dxs, ks, ps, qs, kk, kp, ke, re, gpw},
Switch[type,

Linear, n =nm ={(1 - s)/2, (1 + s)/2J,
Quadratic, n =nm ={-s/2 + s-2I2, 1- s-2, s/2 + s-2I2},
=Infinite, n {-s/2 + s-2I2, 1- s-2, sl2 + s-2/2};

nm = {(2 * s)/(-1 + s), -(1 + s)/(-1 + s)}

INFINITE ELEMENTS 455

];

gpw = ((s- > -1/Sqrt[3], s- > 1/Sqrt[3]), (1, 1))1/N;

xs = run.nodes; J = D[xs, s];

[ks, ps, qs) = [k, p, q)l.x -7 xs;

b = D[n, s]/J;

kk = ks Otiter[Times, b, b]J;

kk = Apply[Plus, MapThread[#2 * kkI.#1&, gpw]];

kp = -ps Outer[Times, n, n]J; v

kp = Apply[Plus, MapThread[#2 * kp/.#1&, gpw]];

ke = kk+kp;

re = qsnJ;

re = Apply[Plus, MapThread[#2 * re/.#1&, gpvj];

[ke, re)] .

Clear[x]; BVP1DElementSolution[nodes_, nodalSoL, t ype, : Linear] :=

Module[(n. run, e, xs, u, x1, x2},

xt = no:ies[[1]];

Swit ch]type,
Linear, n = run = ((1 - s)/2, (1 + s)/2);

x2 = nodes[[2]],
Quadratic, n = run = {-s/2 + s-2I2, 1 - s-2, s/2 + s-2I2);

x2 = nodes[[3]],
Infinite, n = (-s/2 + s-2I2, 1- s-2, s/2 + s-2(2);

run = {(2 * s)/(-1 + s), -(1 + s)/(-1 + s));
=x2 00

];

xs = nm.nodes;

n = n/.Solve[xs == x, s][[1]];
u =Simplify[n.nodalSol];

(x1 < x < x2, ul]

Example 6.23 ID JBVPwith ][nfinite Domain Consider solution of the following one-
dimensional boundary value problem:

d2u 2 l<x<oo
u(oo) = 0
dJ? = i3;

u(l) = 1;

can easily be verified that the following is the exact solution for this problem:
I

=uexact ~

problem is of the type considered in Chapter 3 with

k = 1; p=o; q=-XJ2

456 MAPPED ELEMENTS

23 4
Us
Ul U2 U3 U4
•
Q •••
5
234
x==6
x==l x==2 x==3 x==4

Figure 6.43. Three linear and one infinite element model

We consider the finite element solution of this problem using three standard linear elements

and one infinite element. The finite element model is as shown in Figure 6.43. The model

has six degrees offreedom. Only five are shown on the figure. The degree of freedom £16 is
atx == 00.

k == 1;p == 0; q == -21x-3;
nodes == (1,2,3,4, 6);

Do[lm[i] == Ii, i + 1};
(ke[i], re[iJ} == BVP1DElement[k, p, q, nodes[[lm[iJJJ, Linear], [L, 1, 3}J;

lm[4] == (4, 6,6);
(ke(4], re(4]) == BVPiDElement(k, p, q, nodes[[(4, 6)]], Infinite];
K == Table(O, [6}, (6)]; R == Table[O, (6l];
Do(K((lm[i], lm(i]]]+ == keji.];
R((lm(i]J)+ == re(i], [L, 1,4)];

The essential boundary conditions are as follows:

debe == (1, 6); ,I
ebcVals == (1, OJ;

df == Complement[Range[6], debe]

(2,3,4,5)

Kf == K[(df, df]J

2. -1. 0 0]
-1. 2. -1. 0

[ o -1. 1.72222 -0.777778
0 -0.777778 0.888889
o

Rf == R[(df]] - K[(df, debc]].ebcVals

[0.660606, -0.0836104, -0.0329453, -0.0427856}

The solution for nodal unknowns can now be obtained by using the LinearSolve function
as follows:

dfVals == LinearSolve[Kf, Rf]//N
(0.49616,0.331713,0.250877, 0.171384)

INFINITE ELEMENTS 457

The complete vector of nodal values, in the original order established by the node number-
ing, is as obtained by combining these values with those specified as essential boundary
conditions as follows:

=d Table[O, {6}];

d[[debc]] = ebcVals;

=d[[df]] dfVals;

d

{I, 0.49616, 0.331713, 0.250877, 0.171384, OJ

The complete solution for each element can be computed as follows:

=res[1] BVPiDEleinentSolution[nodes[[lm[1]]], d[[lm[1]]], Linear]

(I < x < 2, 1.50384 - 0.50384x)

=res[2] BVP1DElementSolution[nodes[[lm[2]]], d[[lm[2]]], Linear]

(2 < x < 3,0.825052 - 0.164446x)

res[3] = BVPiDElementSolution[nodes[[lm[3]]], d[[lm[3]]], LinearJ

(3 < x < 4, 0.574221 - 0.080836x)

res[4] = BVP1DElementSolution[nodes[[{4, 5}]],d[[lin[4]]], Infinite]

{4 0.869317X-2.47376}
< x < 00, (x-2.)2 .

In Figure 6.44 the finite element solution is compared with the exact solution. This plot is
generated as follows:

u Exact

0.8 - - PEA
0.6
0.4 x
0.2
4 6 8 10
o2 Figure 6.44.

458 MAPPEDELEMENTS

=eso1 App1y[Which, Flatten[Tab1e[res[iJ, (i, 1,4}])J;

P1ot[(1/x, Eva1uate[eso1]}, [x, 1, 10), P1otSty1e -? (Hue[0.5], Hue[0.9]},
AxesLabe1 -? ("x", "u"], AxesOrigin -? (1, OJ,
Flot.Legend -? {"Exact", "FEN'}, LegendShadow -? None,

LegendPosition -? (0, Oil;

6.7.2 Two-Dimensional BVP

The ideas presented for the one-dimensional problem can be easily extended to two-
dimensional problems. The simplest infinite element for a two-dimensional boundary
value problem is a six-node element shown in Figure 6.45. In the infinite direction each
side has three nodes, with nodes 5 and 6 located at infinity. The master element is a
usual 2 x 2 square element used with other mapped elements. Note that, since the nodes
at infinity are not used explicitly in the mapping functions, the node-numbering scheme
is different than our usual convention of moving counterclockwise around the element.
From the figure it is clear that in the t direction we can use the standard linear Lagrange
interpolation function:

Mapping functions in the t direction: 1- t 1+t
-2-;
2

In the s direction we use the mapping functions developed for the one-dimensional infinite
element:

Mapping functions in the s direction: 2s -(s-+-1)
s -1;
s-1

Taking the product of these functions, we have the following mapping of the infinite ele-
ment to the 2 x 2 master element(

_ s(1- t) set + 1) _ (s + 1)(1 - t) _ (s + l)(t + 1)
x(s, t) - s - 1 xl + S - 1 x2
2(S - 1) x3 "2(S - 1) x4

and

_ s(1 - t) s(t + 1) _ (s + 1)(1 - t) _ (s + l)(t + 1)
Y(s, t) - s _ 1 Yl + S _ 1 Yz
2(s _ 1) Y3 2(s _ 1) Y4

The assumed solution over the master element is written using the product of quadratic
Lagrange interpolation functions in the s direction and the linear function in the t direction"

u == NI £!I + '" + N6£!6 == NTd

!(s - 1)s(1 - t)

!(s - l)s(t + 1)
-~(s - 1)(s + 1)(1 - t)
N= -~(s - 1)(s + l)(t + 1)

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

PROBLEMS 459

Master element y Infinite element
2
t4
5
T -------x

2

1

Figure 6.45. Mapped infinite element

Figure 6.46. Finite element mesh terminated with infinite elements on the left and the bottom

Using these interpolation functions, the element equations can be derived in the usual man-
ner. As mentioned already, the integrals in the element equations are singular. However,
since only crude approximations are desired, in actual implementations integrations are
typically carried out using a 2 X 2 Gauss quadrature. A typical finite element mesh in-
volving infinite elements is as shown in Figure 6.46. All other elements are the standard
quadrilateral elements. Onlythe last layer of elements in the x and y directions are the
infinite elements.

PROBLEMS

Integration Using Change of Variables
6.1 Show that the following integral over a four-sided region bounded by the curves
shown in Figure 6.47, evaluates to the value given:

If(x-y)dA;ry =-22-30
Axy

460 MAPPEDELEMENTS

Use the following change of variables:

x(s • t) = ts 2 - S2 - 3ts + 7-8s + -23
-8 -8
-8

y(s • t) = st2 - t2 - 3st + -78t + -2s + 2
-8 -8 -8

Verify that with this change of variables, in terms of (s, r), the bounding curves are

x
3 y=2+"2

=7 -2x
y =7x- -

2

Figure 6.47.

6.2 Show that the following integral over a four-sided region bounded by the curves
shown in Figure 6.48 evaluates to the value given:

If(xy)dA.'Y = 11521 /
A.ry

c2 : y = 7 - 2x;

Use the following change of variables:

x(s, t) = "32s - 2t: + 1; y(s,t) = st + 23"t +2
-"2

{I, I}

2

{I, -I}

Figure 6.48.

PROBLEMS 461

Verify that with this change of variables, in terms of (s, t), the bounding curves are

and therefore the given quadrilateral is mapped to the 2 x 2 square as shown.

6.3 A function f = x2y is to be integrated over a quadrilateral shown in Figure 6.49.

The sides of the quadrilateral are defined by the following equations:

The following change of variables is proposed for evaluation of the integral:

xes, t) =23s - 2t + I; Yes, t) = 't4s - 4s: + "94t + 47:

detJ= -38s +8-t +41-3

(a) Verify that with this change of variables, in terms of (s, t), the bounding curves
are

and therefore the given quadrilateral is mapped to the 2 x 2 square as shown.
(b) Show that the integral is

If_ ? dAxy = 39101 = 30.3667

(x-y)

A,,,,

y=4 {I, II
4

o '" 3 2
23
-1 (I, -1)
-1 0

Figure 6.49.

Numericial Integration

Use numerical integration to evaluate the integral in Problem 6.2. Use I x I and 2 x 2
Gaussian integration. Show all calculations.

Use numerical integration to evaluate the integral in Problem 6.3. Use I x I and 2 x 2
Gaussian integration. Show all calculations.

462 MAPPEDELEMENTS

Mapping Quadrilaterals

6.6 Develop an appropriate mapping to map the quadrilateral shown in Figure 6.50 to
the 2 x 2 square master area. Compute the determinant of the Jacobian matrix and
use it to demonstrate that the mapping is good. The coordinates of the key points
are as follows:

xy

1 8. O.

2 5.65685 5.65685

3 O. 8.

40 3

52 0

Master area 4 y 3 Quadrilateral
t 8
6 x
I 44 8
2
2 0

1 5
024 6
2

Figure 6.50.

,I

6.7 Develop an appropriate mapping to map the quadrilateral shown in Figure 6.51 to
the 2 x 2 square master area. Compute the determinant of the Jacobian matrix and
use it to demonstrate that the mapping is good.

6

2.5
2
5

1.5

3 0.5 1 1.5 2
Figure 6.51.

PROBLEMS 463

6.8 A quadrilateral element with one curved side and the other three sides straight is

mapped to a 2 x 2 square master area using five-node serendipity interpolation func-
tions. The mapping is computed as follows:

xes, t) :::: O.828427ts2 - O.828427s2 + 1.5ts - 2.5s - 2.32843t + 3.32843
yes, t) :::: O.828427ts2 - O.828427s2 - 1.25ts + 2.75s - 2.07843t + 3.57843

Assume that the solution over the element is also written using the same five-node
serendipity interpolation functions. Compute x and y derivatives of the interpolation
functions at x :::: y :::: 4.

6.9 Use 2 x 2 Gauss quadrature to integrate f :::: x2y over the quadrilateral shown in

Figure 6.52.:

Master area YQuadrilateral
t 4

T 3

2 0 x
0 23
1

1--2----1

Figure 6.52.

6.10 A four-node quadrilateral element is shown in Figure 6.53.

Y(0, 3j 2 Actual element Master element

·4 4t 3

_ _ _ _ _ _ _(} OJ T

Figure 6.53. 2

1

1-----2-----1

(a) Develop an appropriate mapping to map the actual element into the 2 x 2

master element shown in the figure.

(b) Verify that the mapping is good.

(c) Compute aNlay, where N3(s, t) is the third interpolation function for the mas-

ter element. Give the value of the derivative at node 3 of the element.

464 MAPPEDELEMENTS

fit(d) Compute 64N2N3 dA, where N2(s, t) and N3(s. t) are the second and the

third interpolation functions for the master element and A is the area of the
actual element, Use numerical integration with a 1 x 1 Gauss quadrature for-
mula. No credit will be given if any other integration method is used, even if
the integral is correct.

fc(e) Compute 4N3c de, where c is the line 3-4 of the element and N3c is the third

interpolation function for the master element expressed in a coordinate that
runs along this line. Use numerical integration with a two-point Gauss quadra-
ture formula. No credit will be given if any other integration method is used,
even if the integral is correct.

6.11 A 2D boundary value problem is being solved by using eight-node serendipity el-
ements. An element near a boundary is as shown in the Figure 6.54. The nodal
coordinates are as follows:

xy

1 6. O.

2 4.24264 4.24264

3 O. 6.

4 O. 3.

50 0

6 3 -2
-4
2
73

8 4.5 -2.

=(a) Determine the vector rf3 as a result of the natural boundary condition aT/an

100 on side 5-6-7 of the element, For simplicity use only a one-point integra-
tion. ,I

For the following parts use the following mapping:

=x 2.87 + 2.25s - 0.62s2 + 1.37t + 0.75st - 0.62s2t

y = 1.12 - 2.5s - 0.62s2 + 3.12t - 0.5st - 0.62s2t

o 123 4 5

Figure 6.54.

PROBLEMS 465

Also assume that after complete assembly and considerations of boundary
conditions the following nodal solution is obtained:

d = (0 0 0 0 1 0 0 0)

(b) Determine the solution T at the centroid of the element.
(c) Determine the x derivative of the solution aT/ax at node 5 of the element.

Finite Element Computations over Mapped Elements

6.12 Evaluate matrix kk for the four-node quadrilateral element shown in Figure 6.55.

= =Assume lex 0.5 and ley 1.3 over the element. Use 2 X 2 Gaussian quadrature.

Master element y4Actual element

t (-1,4}
...•........... .,..312,3}
T
V<· /·:'·{··:·j)~3, II
2s
x+---l,.~----
1 IO,O}

1----2--~

Figure 6.55.

6.13 Compute matrix ka: for the eight-node element shown in Figure 6.56. Assume a is
equal to 2 over side 1-2-3 of the element.

12 345 678

x 1 I 0 3 5 9 2 3

2 2 2 4 2

.y 5 3 0 2 5 3 II

2 2 2 4'

5

.r------x

0.5 I 1.5
Figure 6.56.

466 MAPPEDELEMENTS

6.14 The square duct shown in Figure 6.57 carries hot gases at a temperature of 300°C.
The duct is insulated by a layer of square fiberglass that has a thermal conductivity
of 0.04 W/m . DC. The outside surface of the fiberglass is subjected to convection

with It = 15 W/m2 . °C and Too = 30°C. Take advantage of symmetry and model

only one-eighth of the section, shown in dark shade. Assume that the temperature
on the inside surface is same as that of the hot gases. Using only one four-node
quadrilateral finite element, determine the nodal temperatures.

30°C 3

Figure 6.57.

6.15 The square duct shown in Figure 6.58 carries hot gases at a temperature of 300°C.

The duct is insulated by a layer of circular fiberglass that has a thermal conductivity

of 0.04 W/m . DC. The outside surface of the fiberglass is subjected to convection

= =with It 15 W/m2 • °C and Too 20°C. Determine heat loss through the insulation.

Take advantage of symmetry and model only one-eighth of the section, shown in

dark shade. Use only one finite element.

/ 4 56

12 3 78

x 1. 0.92388 0.707107 0.603553 0.5 0.5 0.5 0.75
y O. 0.382683 0.707107 0.603553 0.5 0.25 O. O.

3

Figure 6.58.

CHAPTER SEVEN .;

&&

ANALYSIS OF ELASTIC SOLIDS

The problem of determining stresses and strains in elastic solids subjected to loading and
temperature changes is considered in this chapter. In addition to externally applied load-
ing, the solid may be subjected to body forces that are distributed over its volume. Inertia
forces and self-weight are the two common examples of the body forces. The fundamental
concepts from elasticity are reviewed in Section 7.1. Using these concepts, the governing
differential equations in terms of stresses and displacements are derived in Section 7.2.
These equations are much more complicated than the one considered in Chapters 5 and 6.
However, since the equations are still of the second order, all the fundamental concepts
introduced in the earlier chapters are still applicable. Section 7.3 presents the general form
of finite element equations for analysis of elastic solids. Specific elements for analysis of
plane stress and plane strain.problems are presented in Section 7.4.

7.1 FUNDAMENTAL CONCEPTS IN ELASTICITY

7.1.1 Stresses
Internal forces are produced in a body when it is subjected to external loading. These
internal forces are characterized in terms of stresses. The stress at a point P is defined
by passing a plane section through that point and considering force per unit area on the
differential element. Denoting the area of this differential element by M and the force on
it by DJi', the stress vector at point P for the plane whose unit normal is 1l is written as
follows:

467

468 ANALYSIS OF ELASTIC SOLIDS

The normal and shear stresses are written by considering components of the force in the
normal (I1Fn ) and tangential (11Ft ) directions of the plane as follows:

Normal stress: a: (P) == lim I1FII
Shear stress: II AA->O M

TII(P) . 1M1Ft
==hm

Note that the definition of stress is intimately tied to the plane section that is characterized
in terms of its unit normal n. Since an infinite number of planes can be passed through
a given point P, the stress state at the point must clearly state the plane section being
considered. The usual practice is to define stresses in terms of planes whose normals are
along the coordinate directions. For example, in a cartesian coordinate system, the stresses
on a plane whose normal is along the x axis (the y-z plane) are as follows.

Normal stress: a:(P) == . I-M1Fx
lim
x
AA->O

Shear stress: T (P) == lim 11Ft
M
x AA->O

The shear stress is further expressed in terms of the two in-plane force components in the
y and z directions:

Shear stress components: I1F
T ~,(P) == lim A / ;

-J AA->O ut1

Similarly, we can define the stresses on the planes whose normals are along the y axis (the
x-z plane) and along the z-axis (the x-y plane). The complete stress state at point P can then
be described in terms of nine stres~tcomponents:

Stress vector on plane with normal along x axis: T
Stress vector on plane with normal along y axis:
Stress vector on plane with normal along z axis: t.t== (a:t T.t)' Txz )

ty == (Tyx OJ T
T yz )

tz == (Tzx TZy ~)T

These stress components are shown in Figure 7.1 on a differential block. On the three hid-
den faces of the block the outer normals are in the opposite directions; thus the directions
of the stresses will be opposite to the ones shown in the figure. Writing the stress vectors
as rows in a 3 x 3 matrix, the so-called stress tensor is defined as follows:

Using the moment equilibrium condition, it can be shown that the following shear stress
components are equal:

FUNDAMENTAL CONCEPTS IN ELASTICITY 469

x

Figure 7.1. Notation for stress components in rectangular coordinates

Thus the stress tensor S is a symmetric matrix and there are six unique stress components.
In the finite element formulation, it is convenient to arrange the six stress components in a
vector form as follows:

Stresses on an Inclined Plane It is possible to relate stresses on any inclined plane

to those on the planes normal to the coordinate axes. To derive this relationship, consider
an infinitesimal wedge (tetrahedron), shown in Figure 7.2. The stress vectors on the three
vertical planes are -tx' -ty' and -tz• The stress vector on the inclined plane with unit normal

n = (nx ny lIlz is denoted by tIl. If the area of the inclined plane is dA, then from

geometry the areas of the wedge perpendicular to the coordinate axes are, respectively,
Ilx dA, Ily dA, and Ilz dA. The equilibrium of forces acting on the tetrahedron gives

or

z

Figure 7.2. Stresses on an inclined plane

470 ANALYSIS OF ELASTIC SOLIDS

Thus the stress vector on any inclined plane can be obtained by multiplying the stress
tensor S with the normal vector for the plane:

This is an important result because it shows that the six component stresses 0;,•... , Tzx are
enough to uniquely determine the stress state with respect to any plane passing through
that point. The components of the stress vector til in the direction of the normal to the plane
is by definition the normal stress on the plane, and hence

(Til = t~n == nTt ll

Finally, using the vector sum, the shear stress on the plane can be written as follows:

Example 7.1 Stress components at a point in a solid are given as follows:

0;, = 9; OJ. = -3; ~ =3; =Txy -2; Txz = 0; =Tyz 2N/m2

Compute the stress vector, normal stress, and shear stress on a plane that passes through
(1,0,0), (0,2,0), and (0. 0, 1).

The first task is to compute the unit normal for the plane that passes through the given
points. This can be done by defining two vectors, one from point 1 to 2 and the other from
point 1 to 3. The normal vector is then the cross product of the two vectors. Finally the unit
normal vector n is obtained by dividing the normal vector by its length:

Vector from point 1 to 2: v12 = t"o, 2, OJ - (1, 0, OJ = (-1,2, OJ ~1}=(2,1,2j
Vector from point 1 to 3: o.Vl3 = (0,0, I] - (1, OJ = (-1,0, Ij

Normal vector: VII=VI2XV13={I~ ~l,-I=~ ~I,I=~

Unit normal vector: Length = Y22 + 12 + 22 = 3

n -- {23' 31' 32}

From the given stress components, the stress tensor is

[ 9 -2 0)S = -2 -3 2 Nlm2
023

The stress vector on the plane with unit normal n is

til =Sn = {¥,-1, J)

FUNDAMENTAL CONCEPTS IN ELASTICITY 471

The normal and shear stresses on the plane can now be computed to get

Normal stress on the plane: 2{26
Shear stress on the plane:
=Til - 3 -

Applied Surface Forces The relationship between the usual stress components on
planes perpendicular to the coordinate axes and those acting on an inclined plane was
derived by considering an arbitrarily placed tetrahedron. If the tetrahedron is taken near the
boundary of a solid such that the inclined plane is along the boundary, then the components
of the stress vector til on the inclined plane must be equal to the applied forces on the
boundary. Denoting. the components of the applied forces on the boundary by qx' qy' and
qz' the applied surface forces are related to stresses as follows:

qx = CT.tnx + T.tYny + Txznz

qy = Tyxnt. + O:vny + TYZnZ

qz =T;:.tnx + TZyny + o;;nz

Principal Directions and Principal Stresses Principal directions are the unit nor-
mals for the planes over which there are no shear stresses and thus the normal stresses
are maximum. Since the off-diagonal terms in the stress tensor S are the shear stresses, it
follows that the principal planes are those that make the S matrix diagonal. Thus principal
stresses and directions can be determined by solving the following eigenvalue problem:

(8 - (T[)ll p = 0

where (T is a principal stress, I is a 3 x 3 identity matrix, and np is a normal vector for the

=principal plane. The eigenvalues are determined by setting det(S - (T1) O. For each eigen-

value the corresponding eigenvector is determined by solving the above equation for IIp '
The principal stresses ar~ .ordered according to their algebraic values:

Maximum: (T1 ; Intermediate: 02; Minimum: (T3

It can be shown that the maximum shear stress on any plane is given as follows:

= OJ(Tj -

Tmax - - 2 -

where (TI is the maximum principal stress and OJ is the minimum principal stress. This T max

acts on a plane inclined at 45" to the directions of both (Tj and (T3'

Example 7.2 Stress components at a point in a solid are given as follows:

=CT.t = 10; OJ == -7; 0;; == 5; T.t)' -3;

Compute principal stresses and their directions.

472 ANALYSIS OF ELASTICSOLIDS
From the given stress components, the stress tensor is

S== 10 -3 02)N/m2
-3 -7
[o 25

Mathematica's Eigensystem command is used to determine the eigenvalues and corre-
sponding eigenvectors of a matrix:

Eigensystem[((10., -3. OJ, (-3, -7, 2). (O. 2, 5)}]

10.5353 -7.81721 5.2819 )

( (-0.982474, 0.175309. 0.0633422) {-0.164108. -0.974648, 0.152084} (-0.0883982, -0.139024. -0.986336)

Ordering the eigenvalues from highest to lowest values, we have the following principal
stresses and unit normals for the principal planes:

Principal Stresses' Normal Vectors for Principal Planes

10.5353 -0.982474
0.175309
0.0633422

2 5.2819 -0.0883982
-0.139024
-0.986336

3 -7.81721 -0.164108
-0.974648
I'
0.152084

7.1.2 Stress Failure Criteria

The ultimate and yield strengths of engineering materials are determined from the tests

conducted on specimens loaded in one direction. In order to use this information for gen-

eral three-dimensional solids, we must convert the three-dimensional stress state into an

equivalent value that is representative of failure stress for that body. The three most com-

monly used stress failure criteria are as follows. .

Maximum Shear Stress Failure-Tresca Criterion According to this criterion, the
material failure occurs when the maximum shear stress at any plane in the material reaches
the value of shear stress as determined by a uniaxial tension test. The maximum shear
stress on any plane is

== if! - if3
2
T max

where if! is the maximum principal stress and 03 is the minimum principal stress. In a
uniaxial test if! is the applied axial stress and 03 == 0; therefore the maximum shear stress

FUNDAMENTAL CONCEPTS IN ELASTICITY 473

is'0"/2, where 0"1is the failure stress of the material in a uniaxial test. Thus, according to
the maximum shear stress failure criterion, the stress failure occurs when

Using this criteria, the factor of safety can be defined as follows:

Factor of safety = 0"12 == 0"
_1_
_ _1 _
T max OJ - OJ

Example 7.3 If a material has yield strength 0"1 = 50 MPa, what is the factor of safety
using the maximum shear stress failure criterion if the state of stress at a point in a solid is
given as follows: .

0:, =5; OJ, =-18; ~ =0; =Txy 15; T xz =0; =T yZ OMPa
Component stresses = (5., -18., 0.,15.,0., O.}

= =Principal stresses (12.4011,0., -25.4011}; T max 18.9011

Factor of safety = 0"12 = 1.32268
_1_

T max

Von Misf!s Failure Criterion The von Mises criterion is the most commonly used
stress failure criterion for metals. It assumes that the stress failure occurs when the octahe-
dral shear stress value is equal to the octahedral shear stress at yielding in a uniaxial tensile
test.

An octahedral plane is a plane -that makes equal angles with the principal stress direc-
tions. The shear stress on the octahedral plane is known as the octahedral shear stress.
Using stress transformation for an inclined plane, it can be shown that on an octahedral
plane, the stresses are as follows:

Octahedral normal stress: = OJ+0"2+0j
Octahedral shear stress:
O"oct 3

Toct = ~~ (OJ -02)2 + (0"2 - OJ)2 + (OJ - O"j)2

where O"j > 0"2 > OJ are the principal stresses. The octahedral shear stress can also be
expressed in terms of normal and shear stress components as follows:

Using this equation, the octahedral shear stress at yielding in a uniaxial tensile test (0;, =

0"1' OJ, = ... =Tzr =0) is

474 ANALYSIS OF ELASTICSOLIDS

Thus, according to the von Mises criterion, the stress failure occurs when

or

The quantity on the left-hand side is usually referred to as the effectivestressor von Mises
stress and is denoted by O"e:

Using this criterion, the factor of safety can be defined as follows:

Factor of safety = O"f

O"e

=Example7.4 If a material has yield strength O"f 50 MFa, what is the factor of safety

using the von Mises failure criterion if the state of stress at a point in a solid is given as
follows:

0:,=5; 0;.=-18; ~=O; Txy=15; Txz=O;

Component stresses = (5., -18./ 0.,15.,0., O.)

cr" = 33.3766; I 0"

Factor of safety = J = 1.49805
cr"

Mohr-Coulomb Failure Criterion In brittle materials the failure strengths in tension
and compression tests are usually different. According to the Mohr-Coulomb criterion, the
failure of brittle materials is predicted when

0"1_0"3=1

Oif a;,f

where 0"1 is the maximum principal stress, 0"3is the minimum principal stress, Oif is failure
stress in the uniaxial tension test, and O"ef is failure stress in uniaxial compression. Note
that proper sign of 0"1 and 03 must be used in this equation while Oif and O"e! are always
positive.

Using this criterion, the factor of safety can be defined as follows:

Factor of safety = I 1 I

0"1 Oif - 0"3 a;,f

FUNDAMENTAL CONCEPTS IN ELASTICITY 475

=Example 7.5 If a material has failure strength in uniaxial tension Oif 15MPa and that in
=uniaxial compression a;,f lOOMPa, what is the factor of safety using the Mohr-Coulomb

failure criterion if the state of stress at a point in a solid is given as follows:

<r,,=5; ay=-18; 0;;=0; Txy=15; ,Txz=O; Tyz=OMPa

Principal stresses = {12.4011,0., -25.401 I}; Factor of safety =0.925285

7.1.3 Strains
When subjected to loading, a point originally at (x, y, z) displaces to (x + u, Y + v, Z + w).
The changes (u, v, w) are the displacement components in the coordinate directions and are

collected together in a displacement vector it = (u v w/. The displacement of a point

does not indicate the actual deformation of the solid at that point since the displacement
could simply be a rigid-body motion of the solid. To correctly characterize deformations
in solids, we consider an infinitesimal cube in the original configuration and determine
its change in shape and size in the displaced position. The change in size is characterized
by changes in the lengths of the sides of the cube. The change in shape is determined by
considering the change in the angle between two intersecting lines from the original 90°.

To characterize changes in length, consider a differential line segment AB parallel to
the x axis in the original configuration. After displacement the line moves to a position
indicated by A'B', as shown in Figure 7.3. The displacement of point A is u and that of
the neighboring point B is u + duo Assume point A is at (x, y, z). Since the line originally

is assumed to be parallel to the x axis, point B is at (x + dx, y, z). Due to displacement, the
coordinates of point A' are (x + u, Y + v, Z + w) and that of point B are (x + dx + u + du,
y + v + dv, z + w + dw). From these coordinates the original and the deformed lengths can

be computed as follows:

LAB =dx
LA'B' = ~ dv2 + dw2 + (-du - dx)2

Since line AB is parallel to the x axis, the increments in displacements are functions only

z

y

A' _ _ _
A . B'

B

x

Figure 7.3. Deformation of a differential line segment originally parallel to the x axis

476 ANALYSIS OF ELASTIC SOLIDS

of x, and therefore

du = 8£1 dx; 8v dw = 8w dx
8x dv = 8x dx;
8x

Substituting these expressions, the deformed length LA' B' can be written as follows:

( -8v )2d2 + -8w2 d2 + ( 8-£-1d x - d x)2
8x 8x 8x

or

LA'B' = dx ( 88Vx )2 + (88Wx )2 + (1 + 88£x1)2

The normal strain in the x direction is then defined as the change in length of line AB
divided by its original length:

So far this expression is exact. To simplify further, we assume small displacements and that
the displacement derivatives are much smaller as compared to 1. Under this assumption

«(1. (8V)2 + (8W)2 + 8£1)2
8x ;' 8x 8x

and thus neglecting the first two terms under the radical sign, we get

8£1
8xE 1':;1-

X

Proceeding in a similar manner, with lines parallel to y and z axes, the normal strains in the
y and z directions can be written as follows:

8w
8zE 1':;1-

Z

The normal strains quantify changes in the lengths of a differential cube. The changes in its
angles, which are originally 90°, are quantified in terms of shear strains. Consider two lines
AB and AC, originally lying in the x-y plane, meeting at 90° at point A. After displacement
the lines move to A'B' and A' C', as shown in Figure 7.4. The point A at (x, y, z) is displaced

to point A' at (x + £I, Y+ v, z). Since we are interested in determining change in angles alone,

and we have already made the assumption that the displacements are small, the change in
lengths is ignored. Thus the point B' moves up by increment in displacement dv relative to

Figure 7.4. Change in the angle between two differential line segments in the x-y plane

pointA' and point C' moves to the right by duo The total change in the original right angle
is thus

du dv

Angle change = dy + dx

Along AB, dy = dz = 0, and therefore

av
dv= -dx

ax

= =Similarly, along AC, dx dz 0, and therefore

du = au dy
ay

Using these expressions and defining the angle change as the shear strain 'Y."" we have

au av

='Yxy ay + ax

Proceeding in a similar fashion by taking lines in the y-z and z-x planes, the following shear

strains can be defined:

av aw au aw
yzx=a-z +a-x
'YyZ = az + ay;

The complete vector of strain-displacement relationships is written as follows:

Ex aaxll

Ey al'
E_ BY
E= c
all'
'Y.ry
t:
0'Z
£ay!!+illa:x.
'Y:.<
ill:. + aalyl'

B:

£e!!: + aalxl'

478 ANALYSIS OF ELASTIC SOLIDS

7.1.4 Constitutive Equations

The stresses and strains are related through constitutive equations. The simplest form of
the constitutive equations is the linear elasticity, which is a generalization of the familiar
Hooke's law. Assuming stresses are proportional to strains, the most general form of linear
elastic constitutive equations is as follows:

eTx CII C12 Cl3 Cl4 Cl5 Cl6 Ex
eTy C21 C22 C23 C24 C25 C26 Ey

eTz C31 Cn C33 C34 C35 C36 Ez

l.t)' C41 C42 C43 C44 C45 C46 'Y.t)'
lyz C51 C52 C53 C54 C55 C56 I'YZ
lzx C61 C62 C63 C64 C65 C66 I'z.<

or

(J" = CE

The 36 coefficients Cij are called elastic constants and must be established somehow for
each material. Several assumptions are introduced to reduce the number of elastic con-
stants in the constitutive equations. It is almost universally assumed in linear elasticity that
the constitutive matrix C is symmetric. With this assumption the number of constants is
reduced to 21. This number can be reduced further if the material properties are symmetric
about some known planes. For example, if the material properties are symmetric about the
xy plane, then it can be shown that the C matrix has the following form with 13 independent
coefficients:

0;, Cl1 Cl2 C\3 C14 . 0 0 Ex
Ey
OJ C12 C;22 C23 C24 0 0 Ez
0 0
=~ C13 C23 C33 C34 0 0 I'xy

lxy C14 C24 C34 C44 I'YZ

lyZ 0 0 0 0 C55 C56 I'zx

Tzt. 0 0 0 0 C56 C66

If the material properties are symmetric about all three coordinate planes, then the number
of constants reduces to 9 and the material is called orthotropic. Wood is commonly treated
as an orthotropic material. The constitutive equations for an orthotropic material are usually
expressed as follows:

0:< Cl1 C12 C13 0 0 0 Ex

OJ Cl2 C22 C23 0 0 0 Ey
=~ C13 C23 C33 0 0 0
0 0 Ez
l.t)' 0 0 0 C44
I'xy
lyZ 0 0 0 0 C55 0
I'YZ
lzx 0 0 0 0 0 C66
I'z.<

Finally, if the material properties are symmetric about every plane, then the material is
called isotropic and only two independent constants are needed to completely define the

FUNDAMENTAL CONCEPTS IN ELASTICITY 479

constitutive equations. The constants are identified as E, the Young's modulus of elastic-
ity, and v, the Poisson's ratio. The constitutive equations for an isotropic material are as

follows:

0:" I-v v v 0 0 0 Ex
0;,
v I-v v 0 0 0 Ey
0; v 0 0 0
E v I-v Ez
T.ty (1 + y)(1 - 2v) 0 -12-2-v 0 0 rty
0 0

TyZ 0 0 0 0 1-2v 0 YyZ
~

Tzx 0 0 0 0 0 -12-2-v Yzx

where G is known as the shear modulus and is related to E and v as follows:

G E -
= -+ v)
2(1

For an orthotropic material,

1 _ V.Q' -~ 0 0 0
Ey E,
E;

_3:£ 1 _3£ 0 0 0

Ex By E,

-~ -~ 1 0 0 0
0 0
D= E, Ey E;
0
0 0 1
Grt

0 0 0 0 10
GJ•

0 0 0 0 0I
Gr.;

where Ex is Young's modulus in the x direction, Vxy is Poisson's ratio relating Ex to cr/Ey,
etc. The D matrix is symmetric and therefore

=Vxy vyx vxz = vz.< =v yz vzy
Ey s, Ez Ex
Ez Ey

480 ANALYSIS OF ELASTICSOLIDS

The constitutive equations give a relationship between stress and strain at a point. IT the
constitutive equations do not change from one point to another point in the material,
then the material is known as homogeneous; otherwise it is inhomogeneous. In practice,
for most linear elasticity problems it is assumed that the material is homogeneous and
isotropic. Thus the material behavior is usually described in terms of E and y values that
are constant for the entire solid.

7.1.5 Temperature Effects and Initial Strains

In general, it is assumed that prior to loading there are no stresses and strains in the body.
However, there are situations when there are initial strains that must be taken into consid-
eration. The most common situations causing initial strains in solids are the temperature
and moisture changes.

A temperature rise of t.T results in a uniform strain that depends on the coefficient of
thermal expansion a of the material. The temperature change does not cause shear strains.
Thus the initial strain vector due to temperature change is as follows:

at.T

at.T

EO = at.T
0

o
o

The stresses are caused by the difference in the total strains and the initial strains:

/

where EO is the vector of initial strains. In the inverse form

E =Du + EO

7.2 GOVERNING DIFFERENTIAL EQUATIONS

Consider an arbitrary solid supported in a stable manner under the influence of externally
applied forces on its surface such as pressure and concentrated loads. In addition, the solid
may be subjected to applied forces distributed over the entire volume of the body. These
forces are known as body forces. Typical examples of body forces are self-weight of the ob-
ject and centrifugal forces developed in rotating objects. The governing differential equa-
tions can be developed easily by considering equilibrium of forces acting on an isolated
part of the object. This approach gives three equilibrium equations in terms of stresses,
These equations are derived in the first section. No constitutive equations are used in de-
riving these equations and therefore they are applicable to all materials.

The most common finite element formulations are based on displacements as primary
unknowns. To use this formulation, the stress equilibrium equations must be expressed

GOVERNING DIFFERENTIAL EQUATIONS 481

in terms of displacements using the strain-displacement and constitutive equations.
The displacement equilibrium "equations presented in this section assume linear strain-
displacement relations and linear elastic stress-strain behavior. Thus the finite element
formulation based on these equations, presented in the following sections, is applicable
only to small-displacement problems that are restricted to linear elastic behavior.

7.2.1 Stress Equilibrium Equations

If a body is in equilibrium, then any arbitrary part of the body must also be in equilibrium.

The surface S of an arbitrary part isolated from a solid is subjected to stress vector tn =

= =(tlU tny tll Z {. The normal to the surface is 1l (l1x l1y I1z{. The body forces b

(bx by b,{ aredistributed over the entire volume V of the part. The equilibrium of the
isolated part requires that

II IIItnxdS+ bxdV =0

sv

II IIItnydS + bydV =0

sv

II IIItnydS + bydV= 0

sv

Since tlU = o;,l1x + Txyl1y + T.<z;l1z the first equation can be written as follows:

We can use the divergence theorem to convert surface integrals to volume integrals. The
divergence theorem is a special case of the Green-Gauss (integration-by-parts) theorem:

If g =1, then ag/ax =0, and we have

SimilarIy,

and

482 ANALYSIS OF ELASTIC SOLIDS

Using the divergence theorem, the surface integrals can be written in terms of volume
integrals, giving

Since this holds for any arbitrary region, the integrand must be zero:

Proceeding in the same manner with the other two equations, we get the following equa-
tions in terms of stresses:

7.2.2 Governing Differential Equations in Terms of Displacements

The governing differential equations can be expressed in terms of displacements by using
the constitutive equations and the strain-displacement relationships. For a linear elastic
isotropic material we have

CT,; I-v v v 0 0 0 Ex
0
OJ, vI' I-v v 0 0 0 ~
0
~ E v v I-v 0 0 Ez
== (l + v)(l - 2v) 0 0 0 -12-2-v 0 Yxy
Txy

T yZ 0 0 0 0 -12-2-v 0 YyZ

Tzx 0 0 0 0 0 -12-2-v Yzx

Substituting strain-displacement equations Ex == 8u/8x, ... into the constitutive equations
and carrying out matrix multiplication, we get

a: == (l E - 2v) ((1 - v)-88ux + v88-yv +v88-Wz )

x + v)(l

.a:.== E 2v) (v88-ux + (1 - v)88-yv + v88-Wz )
y (1 + v)(1 -

E (8U 8v 8W)
CTz == (1 + v)(1 _ 2v) v 8x + v 8y + (1 - v) 8z

T xy == G ( 8U + 8V) T yZ == (8V + 8W) ; T 1:-< == (8U + 8W)
8y 8x; G 8z 8y G 8z 8x