Fundamental-Finite-Element-Analysis-and-Applications

SPACE FRAMES 283

-applied only at the element ends (no applied distributed twisting moment along the span),
'the problem can be described in terms of the following differential equation:

GJd21jJ = 0
dt 2

where ljJ(t) is the rotation of the section about its taxis, G is the shear modulus, and J is the
torsional constant. From the similarity of this equation to the one for the axial deformation
problem, it is easy to see that a linear solution in terms of nodal degrees of freedom d and

4

d 10 is as follows:

The twisting moment per unit length is related to the angle of twist by the following equa-
tion:

Mr(t) = GJ~~

By combining axial deformations, bending in r-t and s-t planes, and effects due to twisting
actions, the total strain energy in a frarne element can be written as follows:

2V)2 (dd2t2W)2dt

ei,(ddt2 dt
t' t't'U Jto'.GJ(d7lfjtJ)2dt
= 2I' Jo EA(ddUt )2dt + 21' Jo + 21' Jo m, + 2I'

It is convenient to express this in matrix form as follows:

u=~ JorL(f!Edt. EA o0 du

2 <!'!!.) 0 tu, 0 dt
o EI~
dt 0 00 0] 2

[o o ddr2v

~j ~;d2w dt
dt

Using the different displacement interpolations given above, the required derivatives can
be written as follows:

du oo

dt

d2 v

odr2

-r1 0 0 00 0 r1 0 0 00 0

BT= 0 l1!2r-IJ6 0 0 0 I6Jr - 4 0 I6J - l1!21 0 0 0 I6Jr - 2

L L

0 0 l1!2r - I6J 0 4 I6Jr 0 '0 0 I6J r t12:1 0 2 I6J1 0

L- L-

00 0 -r1 0 00 0 0 1 0 0

t.

284 TRUSSES, BEAMS, AND FRAMES

Thus the strain energy can be written as follows:

where

e-- 1][EA 0 000st, 0

0 EIs

000

LLk= BeBT dt

Carrying out matrix multiplications and integration, we get the following space frame stiff-
ness matrix in its local coordinates:

EA 0 0 00 0 -TEA 0 0 00 0

T 12EI, ~ _12EI, ~
L3 L2 L3 L2
0 0 0 0 0 0 0 0

11£/, 6El, 0 0 _12EI, 0 -L~2
L3 L3
- I FIIIIIII 0 0 0 0 0

11111" GJ 00 0 -TGJ 0
00 0 0 0 0
T
t..,~~; -L~2 ~ ~ 0 -2ELl'
'0 0 0 0 L 0 00 0
U!h~ 0 0 L2
6El, 0 0 0 4El, 0 -L~2 0 0 3§.!L
"It,,, k= -TEA L 0 0
IF 0 EA L
II,·tll 0 0 0
T 00 0
.",1.-"'

0 -~L2 0
"."II 0 _12EI, 12El, 0
L3 L3
-~:11' 12~/,
- 71,U, 0 L
0 0 00 L
0
00 12EI, 6El, 0 0 0 ~ 0
L2
'lIIit. 0 -TGJ 0 ./ 0 GJ L2

00 0 0 0 T 0 0

0 0 -~L2 0 2E1, 0 0 0 6EI, 0 ~ 0'
0 0 0 0 -L~2 L2 L
~L L 2EI, ~
0 L
0 L 0 0

In a similar manner, the equivalent nodal load vector can be established by considering the
work done by the distributed applied forces along the local coordinate directions qr and qs
as follows:

Writing the displacement expressions in a matrix form, we have

0 +2t3 3t2 1 0 0 0 t3 2t2 + t J[ ~]
( vet) ) = [ d12
L3 - L2 lJ L

wet) 0 0 T2rJJ - 3t2 + 1 0 -ltJ3 + T2t2 - t 0
L2

=NTd

SPACE FRAMES 285

Using these, the work done can be written as follows:

where rq is the equivalent load vector. If all distributed forces are assumed constant over
an element, then the integration can easily be carried out to give the following equivalent

load vector:

{orT = qsL q,L 0 _ q,Lz qsLZ 0 qsL qrL 0 q,.Lz _ q,L Z }

q ' 2 ' 2 " 12' 12' , 2' 2' , 12' 12

4.9.2 local-to-Global Transformation

The stiffness matrix and the equivalent load vector derived so far have been in terms of
a local coordinate system. Since different elements in a space frame will generally have
different local axes, before assembly these matrices must be transformed to a global co-
ordinate system that is common to all elements in the frame. In the global x, y, and z
coordinate system the nodal displacements are identified as follows:

Ll1, v]> wI x, y, and z displacements at node 1
ex p ey]> ezl Rotations about x, y, z axes at node 1
x, y, and z displacements at node 2
LiZ' vz' Wz Rotations about x, y, z axes at node 2

e e exz' yZ' zz

The corresponding nodal forces and moments in the global x, y, and z coordinate system
are as follows:

Fxl' F;,I' Ftl Applied forces in the global x, y, and z at node 1

»:M.d' lvIY1' Applied moments about the global x, y, and z at node 1

FtzF.tZ' FyZ' Applied forces in the global x, y, and z at node 2
Applied moments about the global x, y, and z at node 2
lvIxz' lvIyZ' lvIa

The local-to-global transformation matrix is developed by considering three components
of displacements and rotations at each node as vector quantities. Thus the complete trans-
formation matrix is a 12 x 12 matrix consisting offour identical 3 x 3 rotation matrices as
follows:

where H is a 3 x 3 three-dimensional rotation matrix and 0 is a 3 x 3 zero matrix. The rota-
tion matrix H transforms a vector quantity from the local to the global coordinate system.

286 TRUSSES, BEAMS,AND FRAMES

Within the small-displacement assumption, the nodal displacements, rotations, forces, and

moments are all vector quantities and can all be transformed using this H matrix. The com-
ponents of a vector along the local s, t, and r coordinates are simply the sum of projections
of its x, Y, and z components along the local axes. In matrix form the transformation can be

written as follows:

where 1( is the cosine of the angle between the t and x axes (direction cosine). The other
terms have a similar meaning. Thus nine direction cosines are needed to establish the H
matrix for each element in a space frame. Explicit expressions for these direction cosines
are given later in the section.

Using this transformation matrix, the element equations in the local coordinate system
can be related to those in the global coordinate system as follows:

Multiplying both sides by TT, we get

r,Noting that TT is the transformation of applied loads from the local to the global coordi-

nates, we have the following element equations in terms of global degrees of freedom and

applied nodal loads in the global directions:

kd =r

I

where

and

Three-Node Method for Calculating Direction Cosines Just knowing the ele-
ment end coordinates is not enough to establish all nine direction cosines in the rotation
matrix H. We need additional information about one of the local principal axes. The sim-
plest method is the so-called three-node method. Nodes I and 2 are at the element end
points, as in the case of a plane frame element. A third node is used to define either the
local r-t or the s-t plane. In the following development it is assumed that the third node
is used to define the local r-t plane, as shown in Figure 4.44. It is important to note that
the third node is used for defining the element orientation alone. It does not determine the
element length or any other parameter used in the element equations. In a given frame any
suitable node can be used to define the third node as long as this node, together with ele-
ment nodes I and 2, determines the r-t local plane for the element. The third node cannot
be collinear with element nodes I and 2 because in this case the three nodes do not form a
plane.

SPACE FRAMES 287

The coordinates of the three nodes for an element are denoted by (xl' Yl' z.), (X2•Y2' Z2)'
and (x3•Y3' Z3)' Using these COOrdinates, the nine direction cosines can be computed as
follows.

The local t axis is defined first by a vector V12 pointing from node 1 to node 2 of the
element: .

where i,j, k are unit vectors along the global coordinate directions. The length of the vector

=V12 is the element length L ~ (x2 - x 1)2 + (Y2 - Yl)2 + (Z2 - Zl)2. A unit vector along the

local t axis is therefore given by .

The three direction cosines defining the local t axis are thus

=1 X2 -xl. =m Y2 -Yl. =111 _Z2_-Z_1

I L' I L' L

The local s axis is normal to the plane defined by nodes 1, 2, and 3. A vector along the

local s axis is obtained by taking the cross product of the vector V12 with vector V13 that
goes from node 1 to node 3. Thus we have

ij
Vs = Vl3 X V12 = det [ x3 - xl Y3 - YI

X2-XI Y2-YI

Xs = Y3(Z2 - Zl) + Y2(~1 - Z3) + YI(Z3 - Z2)

=Y, X3(z, - Z2) + Xl (Z2 - Z3) + X2(Z3 - Zl)
=Z, X3(Y2 - Yl)'± X2(YI - Y3) + Xl (Y3 - Y2)

The length of this vector Vs is

Therefore a unit vector along the local s axis is given by

The three direction cosines defining the local s axis are thus

y
nt =~.
L '
S
s

288 TRUSSES, BEAMS,AND FRAMES

Finally, the local r axis is normal to the s-t plane. It can be defined by taking the cross
product of the unit vector along the s axis with the one along the t axis

where

Example 4.13 An f-shaped frame element is oriented in such a way that its longitudinal
axis is along the global z axis and its minor moment of inertia axis makes an angle of 30°

with the global x axis. The element is 100 em long. Establish the 3 X 3 rotation matrix H

for the element.
Assuming node 1 is at (0, 0, 0), node 2 should be at (0, 0, 100). According to the conven-

tion adopted here, the third node should define the major moment of inertia axis. The angle
between the global x axis and the major moment of inertia axis will be 60°. Thus, to define
the orientation of this element, we choose a third point located at 100(- cos(60), sin(60), 0).
This will establish the t - s - r coordinate system for the element, as shown in Figure 4.47.
The computations of direction cosines using the three-node method are as follows:

Nodal coordinates:

x yz

10 o0
20 o 100.
3 /-50
50·..j3· 0

Vector from node 1 to 2, VIZ = (O, 0,100.)

=Element length, L 100.

z,t

Figure 4.47. Local and global axes for a space frame element

SPACE FRAMES 289

" Unit vector, VI =vlzlL =(O, 0,1.)
. Vector from node I to 3, Vl3 = {- 50, 50{3, 0)
Vector, Vs = VI3 X VIZ = (8660.25, 5000., 0)

Length, L, = 10000.

Unit vector, Vs =V/Ls = (O. 866025,0.5, 0)
Unit vector, Vr =VI X Vs = (-0.5,0.866025,0.)

Thus the rotation matrix is as follows:

H =[ ~.866025 ~.5 ~'J
-0.5 0.866025 O.

4.9.3 Element Solution

For computing element solutions, the global degrees of freedom for each element are first
'transformed into the local degrees of freedom by multiplying them by the T matrix for the
element. These local degrees of freedom are then used with the axial deformation and beam
interpolation functions to get complete displacement solutions. Finally the axial forces,
bending moments, and shear forces are computed by appropriate differentiation. In terms
of notation used in this section the appropriate expressions are as follows:

Global to local transformation:

Axial displacement:

I)(~~); O:5.t:5.L

i)

Axial force:

=F(t) EA duCt) .
dt '
Transverse displacement in the local s direction:

290 TRUSSES,BEAMS.AND FRAMES
Transverse displacement in the local r direction:

=(wet) +1 - 3(2 2(3 -(t _ 2(2 + ..t.) O.s ts L
L2 L3 L L2

Bending moments:

(~:~Mr(t) =EIr ) ; 0 :5, t :5, L

Ms(t) =-EIs ( ~:n; 0:5, t :5, L

Shear forces:

III

iii

d~ ~:~= ='" Vs(t) r EIr ( ) ; 0 :5, t :5, L

~I
'~I ~:t;)V,(t) =- d;s =EIs (

; 0 :5, t :5, L

Twisting moment:

! Lt ) ( ddl4O) .
'
i:if!(t) = ( 1-

dif!
M(t) = GJ dt

Example 4.14 Analyze the one-story three-dimensional frame shown in Figure 4.48. The
height of the columns is 12ft and the length of the beams is 10 ft. Each beam is subjected
to a uniformly distributed load of 2 kip/ft in the downward direction. I-shape sections are
used for both columns and beams with the arrangement shown in the figure. The columns
are connected to the foundation through simple connections that do not resist moments. The
material is steel with E = 29000kip/irr' and G = 11200kip/irr', The section properties are
as follows:

Beams: J=43in4; Imax = I, = 450 in"; = =Imin Is 32 in2
Columns: J = 60in4; = =Imin Is 54 in2
= =I max I, 650 in";

SPACE FRAMES 291

3 3
22

y

x
Flgure 4.48. One-story space frame

Taking advantage of symmetry, we model a quarter of the frame using three elements.
Because of symmetry, the boundary conditions at nodes 3 and 4 are as follows:

Node 3: £1=0; ex = 0;
Node 4:
v=O;

The distributed load is applied to the elements in their local coordinates. Therefore, to
assign proper direction and sign to the distributed loads, we must carefully establish the
local coordinates for the elements as follows:

Element 1: Nodes 1,2, and 4
=> t axis along global z; s axis along global x; r axis along global y

Element 2: Nodes 2, 3, and 4 ,

=> t axis along global x; s axis along global -z; r axis along global y

fz=Distributed ~oad: q, ":' 0; q, kip/in

Element 3: Nodes 2, 4, and 3

=> t axis along global y; s axis along global z; r axis along global x

fzDistributed load:' qr = 0; qs = - kip/in

Matrices for this example are too large for printing. A complete solution can seen on the
book website. Solving the final system of global equations, we get

{ex! =0.000398634, By!.= ,-0.00015917, ez! = 0, Uz = 0.000575402, Vz = 0.000117025,

Wz = -0.0248276, exz = -0.000799706, eyz = 0.000330328, ezz =0, "s = 0.000117025,

w3 = -0.041634, eX3 = -0.000799706, £14 =0.000575402,:w4 = -0.0557153,
By4 = 0.000330328}

292 TRUSSES, BEAMS, AND FRAMES

The solution for element 1 is as follows:
Nodal values in global coordinates,

dT = (0,0,0,0.000398634, -0.00015917, 0, 0.000575402, 0.000117025,

-0.0248276, -0.000799706, 0.000330328, 0)

Nodal values in local coordinates,

d, =Td = (0.,0.,0.,0.,0.000398634, -0.00015917, -0.0248276,0.000575402,

0.000117025,0., -0.000799706, 0.000330328)

Axial effects:

=Interpolation functions, N~ (1. - 0.00694444t, 0.00694444t)

N~ ~~Axial displacement, u(t) = ( ) =-0.000172414t

Axialforce, EAdu(t)ldt =-20.

Torsional effects:

Twist angle, ljJ(t) =N~ ( ~:) =0

Twisting moment, GJ dljJ(t)/dt =O.

Bending about r axis:

N'[, = ( 6.69796 X 10-7t3 - 0.000144676t2 + 1, 0.0000482253t3 - 0.0138889t2 + t,

0.000144676t2 - 6.69796 x 1O-7t3, 0.0000482253t3 - 0.00694444t2 )

N~ ~:] ~d ;'
7.86875 x 1O-9t3 - 0.00015917t
vet) = [

d12

= =Bending moment, Mr. E1r d2v(t)/dt 2 0.889955t
Shear force, V. =dM/dt =0.889955

Bending about saxis:

Nrv = (6.69796 x 1O-7t3 - 0.000144676t2 + 1, -0.0000482253t3 + 0.0138889t2 - t,

0.000144676t2 - 6.69796 x 1O-7t3, 0.00694444t2 - 0.0000482253t3 )

~wet) =Nrv [ =] -1.94202x 1O-8t3 + 3.38615 x 1O-8t2 + 0.000398634t

dll

Bending moment, M, = -E1, d2w(t)/dt 2 = -0. 180999t
Shear force, Vr = -dM/ dt = 0.180999

FRAMES IN MULTISTORY BUILDINGS 293

Solutions over the remaining elements can be obtained in a similar manner:

Forces and Moments at Element Ends

x y Z Axial Force V. Vr 'M r Ms Mt
0.180999
00 0 -20. 0.889955 0.180999 0 0 0
128.154 -26.0639 0
0 0 144. ~20. 0.889955 0
0 128.154 0 0
2 0 0 144. -0.889955 -10. -171.846 0 0
60. 0 144. -0.889955 0 0
0 -26.0639 0 0
3 0 0 144. -0.180999 10. 273.936 0 0
0 60. 144. -0.180999 0

~ MathematicafMATLAB Implementation 4.5 on the Book Web Site:
Analysis of space frames

4.10 FRAMES IN MULTISTORY BUILDINGS

From a structural point of view most modem multistory buildings consist of a grid of
structural frames. The floor system, consisting of metal decks and concrete slabs, span
across the frames and tie everything together to create a three-dimensional structure, as
shown in Figure 4.49. Because of large dimensions, the floor system within its own plane
is essentially rigid and is known as a rigid diaphragm. Thus individual frames in a building
do not behave as their isolated counterparts. We must analyze the entire three-dimensional
system.

With the assumption of an in-plane rigid-floor system, each story has only three inde-
pendent in-plane degrees of freedom: the two displacements in the x and y directions (£I and

Figure 4.49. Multistory building

294 TRUSSES, BEAMS, AND FRAMES

Figure 4.50. Rigid zone at beam-column connections

v) and a rotation about the z axis (8). Thus, to analyze a building, we use standard frame
element equations and assemble contributions from all frames in the building in the usual
manner. For each story we define constraints that relate all in-plane degrees of freedom to
one node that we call the master node for that story. Identifying the master node degrees
of freedom by subscript m, the constraints for degrees of freedom at any other node in the
story are expressed as follows:

l

'1.

'Ii,

i = 1,2, ...

where Xi and Yi are the coordinates of node i.
Joints in a building frame also require special modeling care. The column dimensions

in a typical building are of the order 14 to 20 in while the beams may vary in depth from
21 in to 30 in. When creating frame models using centerline dimensions, the joint zone
with these large member sizes is fairly large. Very little deformations are expected within
this jointzone. Thus rigid joint zones are typically created when analyzing building frames,
as shown in Figure 4.50. The nodes are placed outside of the joint region.

The following constraints are defined between the degrees of freedom at these nodes to
create the effect of the rigid joint zone:

After incorporating these constraints, the rest of the analysis follows the standard proce-
dures used in other examples in this c h a p t e r . -

PROBLEMS 295

PROBLEMS

Plane Trusses
4.1 Determine joint displacements and axial forces in the three-bar pin-jointed structure
shown in Figure 4.51. All members have the same cross-sectional area and are of

= = =the sarne material, A 1 in2 and E 30 X 106 lb/irr'. The load P 15,000 lb. The

dimensions in inches are shown in the figure.

72

o • )I--------------+~-p

(in)

o 108

Figure 4.51.

4.2 Determine joint displacements and axial forces in the truss shown in Figure 4.51 if
the support where the load is applied settles down by ~ in.

4.3 Taking advantage of symmetry, determine joint displacements and axial forces in the

three-bar truss shown iIi Figure 4.52. All members have the same cross-sectional

area and are of the same material, A = 0.001 m2 and E = 2000Pa. The load

P = 20kN. The dimensions in meters are shown in the figure. One of the goals for

this problem is for you to learn the correct use of symmetry. Thus no credit will be
given if you do not-use symmetry even'if your solution is correct.

3

o o (m)
4
-4 Figure 4.52.

296 TRUSSES, BEAMS, AND FRAMES

4.4 Determine joint displacements and axial forces in the truss shown in Figure 4.53. All

members have the same cross-sectional area and are of the same material, A = 2 in2

= =and E 30 x 106lb/in2. The load P 30,000 lb. The dimensions in feet are shown

in the figure.

6

o p
o 30

(ft)
8

Figure 4.53.

4.5 Determine joint displacements and axial forces in the truss shown in Figure 4.53 if
in addition to the applied force the lower left support moves horizontally toward the
right by ~ in.

4.6 Determine joint displacements and axial forces in the truss shown in Figure 4.54.
Note the diagonals are not connected to each other at their crossing. The cross-
sectional area of vertical and horizontal members is 30 x 10-4 m2 and that for the
diagonals is 10 x 10-4 m2. All members are made of steel with E = 210 GPa. The

=load P 20 kN. The dimensions.in meters are shown in the figure.

6 (}-------.,----{)---p

o (m)
o
8

Figure 4.54.

Space Trusses

4.7 Determine joint displacements and axial forces in the space truss shown in Figure
4.55. The cross-sectional area of members is 15 x 10-4 m2. All members are made

PROBLEMS 297

p

3P

Figure 4.55.

of steel with E = 210 GPa. The load P = 10 leN is applied in the -z direction and

3P in the x direction. The nodal coordinates in meters are as follows:

x yz

1 O. O. 10.

2 O. 8. O.

3 6.9282 -4. O.

4 -6.9282 -4. O.

4.8 Determine joint displacements and axial forces in the space truss shown in Figure
4.56. The cross-sectional area of members is 15 x 10-4 m2. All members are made

p

Figure 4.56.

298 TRUSSES, BEAMS, AND FRAMES

1I1I~ = =of aluminum with E 70 GPa. The load P 10 kN is applied in the -z direction.

'" The nodal coordinates in meters are as follows:
x yZ
I1I1.;
1 O. 4. O.
II, 2 -3. 2. 5.
3 -3. O. O.
" 4 3. O. O.
5 O. O. 5.

Thermal and Initial Strains

4.9 A copper rod 1.4 in in diameter is placed in an aluminum sleeve with inside diameter
1.42 in and wall thickness 0.2 in, as shown in Figure 4.57. The rod is 0.005 in longer

=than the sleeve. A load P 60,000 lb is applied to the assembly through a large

bearing plate that can be considered rigid. Determine stresses in the rod and the

sleeve. The modulus of elasticity for copper is 17 x 106 psi and that for aluminum

is 10 X 106 psi.

Rigid plate

p

Gap=0.005

Figure 4.57.

4.10 Determine joint displacements and axial forces in the truss shown in Figure 4.51 if
instead of load P the diagonal member experiences a temperature decrease of 70"F.

=The coefficient of thermal expansion is a 6 x 10-6j"F.

4.11 Determine joint displacements and axial forces in the truss shown in Figure 4.53.

!During construction the diagonal member was fabricated in too short and was

forced to fit into the assembly through heat treatment.

4.12 Determine joint displacements and axial forces in the space truss shown in Figure

4.56 if instead of load P the member between nodes 1 lind 2 experiences a temper-
ature rise of 50"C. The coefficient of thermal expansion is a = 23 x 1O-6j"C.

Spring Elements

4.13 Determine support reactions and forces in the springs for the assembly shown in
Figure 4.58.

Figure 4.58.

PROBLEMS 299

4;14 Determine support reactions and forces in the springs when the assembly is forced
to close the gap g, as shown in the Figure 4.59.

Gap,g

Figure 4.59.

Beam Bending

4.15 A fixed-end beam is subjected to a concentrated load at the midspan P.= 200 Ib, as

shown in Figure 4.60. The beam has a rectangular cross section with width = 12 in '
and height = 1 in. The length of the beam is L =200 in and its modulus of elasticity

is E = 107 lb/in'', Taking advantage of symmetry, use only one beam element to
determine the maximum bending and shear stresses in the beam.

p

,.

Figure 4.60. Fixed-end beam

4.16 Two cantilever beams are joined together through a simple pin, as shown in Figure

=4.61. The concentrated load at the midspan is P 200 lb. The beam has a rectan-
gular cross section with width = 12 in and height = 1 in. The length of the beam
is L = 200 in and its modulus of elasticity is E = 107 lb/irr'. Taking advantage of

symmetry, use only one beam element to determine maximum bending and shear
stresses in the beam.

p

I" --~I>I""f---- L/2 ------<J>j
Figure 4.61. Beams connected through' a pin

300 TRUSSES, BEAMS, AND FRAMES

4.17 Tho cantilever beams are joined together through a simple pin, as shown in Figure

=4.62. The E1 107 for the right beam and is twice that for the left beam. The
=concentrated load at the hinge location is P 200 lb. Determine the displacement

and bending moment distribution over the beams.
p

- - L/4 ~;------- 3L/4 "I
Figure 4.62. Beams connected through a pin

4.18 A two-span beam is subjected to a moment, as shown in the Figure 4.63. Find the re-

sulting displacement, shear force, and bending moments. Compute shear and bend-

ing stresses at the middle support in the beam. Use the following numerical data
(1 k = 1000 lb).
.

L= 180 in; E = 29,000 k/in2; M= 1000k·in

=Beam cross section: Standard l-shape (W16 x 26) with 1 301 in" and dimensions

as shown in the figure.

I

L
~~oo;~J~ I+I- L -I

h =15.69 :;
br =5.5

-be-'

Figure 4.63.

4.19 Immediately after construction the right support of the beam shown in Figure 4.64
undergoes a settlement equal to A.Find the resulting displacement, shear force, and
bending moments. Compute shear and bending stresses at the middle support in the
beam. Use the following numerical data:

L=8m; E =200 GPa; A= lOmm

Beam cross section: Standard I-shape with 1 = 125.3 X 10- 6 m" and dimensions (in

min) as shown in the figure.

PROBLEMS 301

L ~

r +h =399

IT1br =140
L

tw =6.3
tr =8.8

.... tr

......b r

Figure 4.64.

4.20 Find displacements and draw shear force and bending moment diagrams for the

beam shown in Figure 4.65. Assume E = 210 GPa, I = 4 X 10-4 m", L = 2 m,
P = 10 leN,M =20 leN· m.

Figure 4.65. Beam with variable section

4.21 Two uniform beams are connected together through a spring as shown in Figure
4.66. Determine deflections, bending moment, and shear forces in the beams. What
is the force in the spring? Tile numerical values are

=E 104ksi; L = 100in; P = 100 kips; k =2000 kip/in

Use kip . in units in calculations.

k

- + - -1L/jco-- L

Figure 4.66. Two beams connected througha spring

302 TRUSSES, BE~MS, AND FRAMES

4.22 Consider a uniform beam simply supported at the ends and spring supported in the
middle, as shown in Figure 4.67. Taking advantage of symmetry, analyze the beam
using only one element. Assume the following numerical data:

L=4m; E =70 GPa; =I 2 X 104 rnm"; k= 400N/mm; P = 20kN

p

.&~ =-.' .-~·t-o-~~o,c,%
EI k EI

I - - - L ·1 L - - - j

Figure 4.67. Beam supported by a spring

4.23 The left half of a beam is subjected to a uniformly distributed load q = 1 lb/in, as
shown in Figure 4.68. The beam has a rectangular cross section with width = 12 in
= =and height 1in. The length of the beam is L 200 in and its modulus of elasticity
is E = 1071b/in2• Create the simplest possible finite element model for this beam.

Using this model, determine the bending and shear stresses in the beam at U 4 from
the left end.

q

1;:;::LJ.L::I,~.J-J::l~::J.-lJ::::::::::,::;::;,,·::::;;:;;;, ,,c"":~"-:::I

I- L/2 -I- L/2 -I

Figure 4.68. Fixed-end beam
;'

4.24 Using only one element, find the midspan displacement, bending moment, and shear
force for the beam shown in Figure 4.69. Note that the loading is large enough to
cause the gap to close. Use the following data: E = 104 ksi,I = 1000 in", L = 100 in,

q = 1 k/in, gap = 0.25 in.

q

Gap'A~

I· L ~I

Figure 4.69. Single span beam with a gap

PROBLEMS 303

4.25 Compute stresses in the two-span beam shown in Figure 4.70. There is a small gap
between the middle support and the beam:

=L 6m; q = lOkN/m; E =200 GPa; Gap D. = 5 mm

Beam cross section: rectangular with width = 120 mID and height = 300 mrn.

Figure 4.70. Two-spanbeam

4.26 A beam with an overhang is subjected to a uniformly distributed load q = 27 Ib/in
over the right span and a moment M = ~ lb . in at the free end, as shown in Figure
4.71. The beam has a rectangular cross section with width = 12in and height = 1in,
giving A = 12 in2 and 1= 1 in", The length of the beam is L = 1 in and its modulus
of elasticity is E = 11b/in2•

(a) Create the simplest possible finite element model for this beam. Write element
equations, assemble to form global equations, and obtain the reduced system
of equations, taking boundary conditions into consideration.

(b) After a analysis of this beam it is found that the rotation at the free end is -&.
rad and that at the leftsupport is --&. rad. Determine the vertical displacement

and bending moment at point a in the beam.

q

1""'1--- L/3 ---ll>l<oI--- L/3 - - -.i. 1-!>1<*>l - - - L/3 ---!13>l

Figure 4.71.

4.27 Using the simplest possible finite element model, determine the maximum bending
and shear stress in a uniform beam simply supported at the left end and spring
supported at the right end, as shown in Figure 4.72. Assume the following numerical
data:

a = 5 kN/m; L = 5 m; E = 200 GPa;

304 TRUSSES, BEAMS, AND FRAMES

y
q(x) = a xfL

EI

I-- L/3·-I...- - - 2L/3 --~~I

Figure 4.72. Uniform beam subjected to a linearly increasing load

4.28 Two overhanging beams are joined together through a simple pin connection as
shown in Figure 4.73. Determine displacement, shear force, and bending moments.
Use the following numerical data:

L= 8m; E =200 GPa; q =3kN/m

Beam cross section: Standard l-shape with 1= 125.3 X 10- 6 m" and dimensions (in
rom) as shown in the figure:

q

. ..A··-··-·"_· ~WLUjl.;Z£;~ ~ '=Z>.

I---- L ---'-Ooo,..I.t--.7L--!-:3L.-tI-.. - L ---<>j

h =399 =tw 6.3

bf;;' 140 tf = 8.8

Figure 4.73.

Plane Frames
4.29 A 300-rom-wide and 100-rom-thick bar is supported and loaded as shown in Figure

=4.74. Determine displacement, shear force, and bending moments. Assume E

10 GPa.

PROBLEMS 305

4
8
3

o

- - - - - - - - - (m)

o 23

Figure 4.74.

4.30 Determine displacements, bending moments, and shear forces in the plane frame
shown in Figure 4.75. Draw free-body diagrams for each element clearly showing

= = =all element end forces and moments. Assume q 10 kNlm, L 2 m, E 210 GPa,

A =4 X 10-2 m2, and I = 4 X 10-4 m". Take advantage of symmetry.

q

I

L

1

Figure 4.75.

4.31 Determine displacements, bending moments, arid shear forces in the plane frame
shown in Figure 4.76. Draw free-body diagrams for each element clearly showing

= = =all element end forces and momentS. Assume q 10kN(m, L 2 m, E 210 GPa,

A = 4 X 10-2 m2, and I = 4 X 10-4 m". Take advantage of symmetry. Use kN . m

units.

306 TRUSSES, BEAMS,AND FRAMES

I

L

t

L

1

Figure 4.76.

4.32 The element stiffness matrices for vertical and horizontal elements of a plane frame
have already been computed and assembled into a 9 x 9 global matrix as follows:

1875. 0 -3750. -1875. 0 -3750. 0 0 0
0 25000. 0 0 -25000. 0 0 0 0
0 0 0
-3750. 0 10000. 3750. 0 5000. -50000. 0 0
-1875. 0 3750. 51875. 0 3750. 0 -15000. 15000.
-25000. 40000. 15000. 0 -15000. 10000.
0 0 0 0 15000. 30000. 50000. 0 0
-3750. 0 5000. 3750. 0 0 15000.
0 -50000. -15000. 0 0 -15000. ~15000.
0 0 0 15000. -15000.
0 0 0 10000. 20000.
0 0 0

It is decided to brace the frame by an inclined truss member as shown in Figure 4.77.
Write down the stiffness matrix for the inclined truss member. Assemble it into the
global matrix. Determine nodal displacements due to a downward concentrated load

y 3
P

5
42
3
2

4
0

x
0 2345

Figure 4.77.

PROBLEMS 307

of P = 50 kN applied at the corner. For all members, E = 210 GPa, I = 10-4 m",

=and A 10-2 m2 .

4.33 Determine displacement, shear force, and bending moments in the plane frarne
shown in Figure 4.78. Use the following numerical data:

P = l2kN; E = 200 GPa; q = 3lcN/m
Cross-section: Standard l-shape with A = 4.95 X 10-3 m2 and I = 125.3 X 10-6 m",

p
6

3

o 6 (m)
o Figure 4.78. 9

4.34 Two 200 mm x 200 mm square bars are joined through a pin and loaded as shown in
Figure 4.79. Determine displacement, shear force, and bending moments. Assume
E = 10 GPa and P = 20lcN e .

6

3 4 6 (m)
1.5 Figure 4.79. 8

o
o

308 TRUSSES, BEAMS,AND FRAMES

4.35 The lower portion of an aluminum step bracket, shown in Figure 4.80, is subjected
to a uniform pressure of 20 Nzrnnr'. The left end is fixed and the right end is free.

The dimensions of the bracket are thickness = 3 mm, L = 150 mm, b = 10 mm,
and h = 24 mm. The material properties are E = 70,000 Nzrnm? and v = 0.3.

For a preliminary analysis determine stresses in the bracket using a simple model
consisting of three plane frame elements.

\

L
Free \
f-- h/2 - f - - h/2---\
Figure 4.80. Step bracket

4.36 The top surface of an S-shaped aluminum block, shown in Figure 4.81, is subjected
to a uniform pressure of 20 N/mm2. The bottom is fixed. The dimensions of the

block are t = 3 mm, L = 15 mm, b = 10 mm, and h = 24 mm. The material prop-
erties are E = 70,000 N/mm2 and v = 0.3. For a preliminary analysis determine

stresses in the block using a simple model consisting of four plane frame elements.

/

Figure 4.81. S-shaped aluminum block Figure 4.82. Concrete dam

4.37 The cross section of a concrete dam is shown in Figure 4.82. Using a simple plane
frame model, estimate stresses in the-darn due to self-weight of the dam and the
water pressure. Use a reasonable number of elements and assign average section
properties to each element. Assume unit thickness perpendicular to the plane. The
depth of the water behind the dam is 18 m, The density of concrete is 2400 kg/m''
and that of water is 1000 kg/m". The modulus of elasticity of concrete is 30 GPa.

PROBLEMS 309

4.38 The side view of a pry bar is shown in Figure 4.83. The cross section of the bar

is rectangular with thickriess = ~ in and width (perpendicular to the plane of
paper) = 1.2 in. The other dimensions (in inches) are shown in the figure. The
material properties are E = 29 X 106 psi and v = 0.3. A load of P = 200 lb is ap-

plied at the center of the handle. Using a suitable number of plane frame elements,
determine maximum deflection and stress in the bar.

p

Figure 4.83. Pry bar

Space Frames

4.39 A cantilever beam is supported by a beam fixed at both ends, as shown in Figure
4.84. Analyze the system when it is subjected to loading as shown. Use the following
numerical data:

L=4m; b =1 m; q =100N/m; E = 200 GPa; G = 100GPa

Beams, W360 x 44.6: =A 5.71 X 10-3 m2 ; J =0.1729 X 10- 6 m";
Beams, W310 x 38.7: I min = 8.16 X 10-6 rn"
Imax = 121.1 X 10-6 m";
J = 0.1421 X 10- 6 m";
A = 4.94 X 10-3 m2 ; l min = 7.2 X 10-6 m"
I max = 84.9 X 10-6 m";

Figure 4.84.

310 TRUSSES, BEAMS, AND FRAMES

4.40 Two beam cantilever out of an Hsframe, as shown in Figure 4.85. The columns are
fixed at the top and the bottom. Analyze the frame when it is subjected to loading
as shown. Use the following numerical data.

L = 20 ft; h = 15 ft; b = 20 ft; P = 30 kips; q = 2kips/ft; w = 0.5kip/ft

Beams, W16 x 40: =A 11.8 in2; J =0.851 in";
=l min 28.9 in"
=Imax 518 in";
J = 3.062 in":
Columns, W12 x 72: A =21.1 in2; =Imin 195 in"

=Imax 597 in";

E =29,000 ksi; G = 11,200 ksi

/ Figure 4.85.

CHAPTER FIVE

h i -& "dHll . , PFHH s iH FW ;a 54'

TWO-DIMENS~ONALELEMENTS

In the previous chapters the basic finite element concepts were demonstrated through
simple one-dimensional elements. For most of those problems either the exact solution is
known or there are several other numerical methods available that may be more convenient
to use than the finite element method. This is not the case for two- and three-dimensional
problems. The exact solutions are rare for these problems. Most other numerical meth-
ods are not as widely applicable as the finite element method. One of the key advantages
of the finite element method is that the fundamental concepts extend easily from one-
dimensional to higher dimensional problems. The only complication is that the required
integration and differentiation become more difficult, but these are calculus-related prob-
lems and have nothing to do with the finite element concepts. A review of vector calculus,
especially integration over two- and three-dimensional domains, will make the transition
from one-dimensional problems to two- and three-dimensional problems a lot smoother.

In this chapter the basic finite element concepts are illustrated with reference to the
following partial differential equation defined over an arbitrary two-dimensional region,
such as the one shown in Figure 5.1:

=-aax (ka.t-aux) + -aay (kY-aauy) + pu + q 0

where kxCx, y), kyCx, y), p(x, y), and q(x,y) are known functions defined over the area. The
unlrnownsolution is u(x, y). The equation can easily be recognized as a generalization of the
one-dimensional boundary value problem considered in Chapter 3. Steady-state heat flow,
variety of fluid flow, and torsion of planar sections are some of the common engineering
applications that are governed by the differential equations which are special cases of this
general boundary value problem.

311

--

312 TWO-DIMENSIONAL ELEMENTS

y
Normal, n

x

Figure 5.1. Two-dimensional solution domain

The differential equation is second order and therefore, based on the discussion in Chap-

.ter 2, the boundary conditions involving u are essential and those involving its derivatives

are natural boundary conditions. The area of the solution domain is denoted by A and its
boundary by C. The boundary is defined in terms of a coordinate e that runs along the
boundary and an outer unit normal to the boundary n, as shown in Figure 5.1. The x and y
components of the normal vector are denoted by nx and l1y:

The formulation presented in this chapter can handle boundary conditions of the following
types:

(i) Essential boundary conditioy(~pecified on portion of the boundary indicated by Ce:

uCe) specified on C,
where e is a coordinate along the boundary.
(ii) Natural boundary condition specified on portion of the boundary indicated by Cn :

=au au

lex ax I1x + ley ay ny aCc)u(e) + f3(c) on CIl

where aCe) andf3(c) are specified parameters along the boundary. When lex = ley == k,
the left-hand side is the derivative of u in the direction of the outer normal to the
boundary, and the boundary condition is expressed as

au au) au
k ( ax nx + ay ny == k an = au + f3

Observe carefully that this expression does not allow specification of any arbitrary first
derivative of u. Only the normal derivative can be specified. For example, on a boundary

SELECTED APPLICATIONS OF THE 2D BVP 313

that is perpendicular to the x axis, the components of the unit normal to the boundary will

ben, = 1 and ny = 0 and thus' only au/ax can be specified by this form. The reason

for choosing the normal derivative is because it gives rise to a convenient weak form.
Furthermore, for most physical problems it is natural to specify the normal derivative or it
is possible to use mathematical manipulations to express the derivative in this form. Thus
it is not a severe limitation for practical problems.

A few selected applications that are governed by the differential equation of this form
are presented in the first section. The second section presents a general finite element for-
mulation for the problem. Rectangular and triangular elements for solution of this differen-
tial equation are presented in the last two sections in this chapter. These elements are used
to obtain approximate solutions for a variety of problems.

5.1 SELECTED APPLICATIONS OF THE 2D BVP

5.1.1 Two-Dimensional Potential Flow

The problem of two-dimensional frictionless and incompressible fluid flow, lmown as po-
tential flow, is governed by the following two equations:

1. Continuity equation: ~ + ~ = 0

t; -2. Irrotational flow condition: ~ =0

where u(x, y) and v(x, y) are the fluid velocities in the x and y directions. Two different
formulations are used to find solutions of these equations and are known as stream function
and potential function formulations:

In the stream function formulation it is assumed that a scalar function, called the stream
function ifJ(x, y), exists that is related to the fluid velocities in the x and y directions as
follows:

u =aa-ifyJ; v = -ai-fJ
ax

With this assumption the continuity equation is satisfied exactly while the irrotational flow
condition yields the following second-order partial differential equation:

In the potentialfunction formulation it is assumed that a scalar function, called the potential
function ¢(x, y), exists that is related to the fluid velocities in the x and y directions as
follows:

v =aa-¢y

314 TWO·DIMENSIONAL ELEMENTS
y

x

Figure 5.2. Flow around an object

With this assumption the irrotational flow condition is satisfied exactly while the continuity

equation yields the following second-order partial differential equation: .

Thus both formulations are governed by the differential equation of the same type with
kx == ky == I and p == q == O. However, the two formulations differ in the way the bound-
ary conditions are imposed. To illustrate this difference, consider a typical potential flow
problem to determine fluid flow around a symmetric object, as shown in Figure 5.2. Away
from the object the fluid is flowing in the x direction at a known velocityzq, In the vicinity
of the object the velocities change and are unknown. At a sufficient distance downstream
from the object, once again we have a constant flow in the x direction.

The first task is to es.t.a.blish the boI undaries of the solution domain. On all sides of the
object, we must extend the solution domain far enough so that the assumption of uniform
flow in the x direction is valid (indicated by horizontal streamlines). Since the computa-
tional effort will depend on the size of the solution domain, extending the domain too far
would be inefficient. However, if the domain is not large enough, the results will be inac-
curate. Thus few trials may be necessary before settling on a proper size for the solution
domain. It is generally recommended to extend the computational domain in all directions
to about four times the size of the object.

The next task is to define appropriate boundary conditions on all sides. Taking advantage
of symmetry, we need to model only a quarter of the solution domain, as shown in Figure

5.3. For the stream function formulation If!, we can take any arbitrary reference value for
the base because the velocities depend only on the derivatives of If!. We choose If! == 0 along

x == 0 and come up with the following boundary conditions:

Boundary Conditions for Stream Function.Formulation

1. On the bottom and along the obstruction

If!==O

SELECTED APPLICATIONS OF THE 2D BVP 315

y
ifi=Huo

r ..-

V' .,.

.- ...V' ...

V" ~... .(
V' .. _

ifi=O

ifi=O
Figure 5.3. Boundary conditions in terms of stream function for flow around an object

2. On the left side

Integrating this expression, we have

Thus 1/t varies linearly with height on this side.
3. On the top: From linear variation of 1/t on the left side with y == H, we have

4. On the right side, because of Symmetry,

81jJ == 0
8x
Note that the original form of the boundary condition 81/t/8y == Uo on the left side is not in
the form that can be handled by the natural boundary condition defined for the differential
equation since the normal derivative for this side is

==> -81/t ==8-1-/t

8n 8x
Thus we can specify 81/t/8x on the ends but not 81/t/8y. On the right side 81/t/8n == 81/t/8x,
and thus we can use this boundary condition directly. .

With the potential function formulation the boundary conditions on a quarter of the
solution domain are as shown in Figure 5.4. On the bottom, -top, and left side we have
natural boundary conditions. The right vertical symmetry line must be a potential line and
therefore we must specify that 1> == constant. The actual value does not matter.

316 TWO·DIMENSIONAL ELEMENTS

y
8¢!8y=O

poo-.,... f"" f i/>=Constant
"... "J'
P'" ~ .- if

8¢!8x=uo " if

.-" of
~

8i/>/8n=O

x

8¢!8y=O

Figure 5.4. Boundary, conditions in termsof potentialfunction for flow aroundan object

Boundary Conditions for PotentialFunctionFormulation
1. On the top and bottom sides

aa¢y =0

2. Along the obstruction

3. On the left end

4. On the right end

¢ = constant

5.1.2 Steady-State Heat Flow

T(x,Consider the problem of finding the steady-state temperature distribution y) in long

chimneylike structures. Assuming no temperature gradient in the longitudinal direction,
we can take a unit slice of such a structure and model it as a two-dimensional problem.
Using conservation of energy on a differential volume, the following governing differential
equation can easily be established:

aax (kax Tax) + aay (ky aaTy ) + Q = 0

where kx and ky are thermal conductivities in the x and y directions and Q(x, y) is specified
heat generation per unit volume. Typical units for k are W/m- °C or Btu/hr- ft- of and those
for Q are'W/m3 or BtuIhr· fil. The possible boundary conditions are as follows:

SELECTED APPLICATIONS OF THE 2D BVP 317

~ (i) Known temperature along a boundary:

T(e) = To specified

(ii) Specified heat flux along a boundary:

The sign convention for heat flow is that heat flowing into a body is positive and

that out of the body is negative. Comparing with the general form this boundary

condition implies a = 0 and f3 = ~qo' On an insulated boundary or across a line of
symmetry there is no heat flow and thus qo = O.

(iii) Convection heat loss along a boundary:

oT =h(T(e) - T'" )
- kon-

where h is the convection coefficient, T(e) is the unknown temperature at the

boundary, and Too is the temperature of the surrounding fluid. Typical units for h
are W/m2 • °C and Btn/hr -ft2. "F, Comparing with the general form, this boundary

= =condition implies a -h andf3 hToo '

5.1.3 Bars Subjected to Torsion

The problem of determining stress distribution in bars of arbitrary cross section subjected
to twist can be formulated in terms ?f a stress function ¢ defined as follows:

The total shear stress can be computed by taking the vector sum as follows:

All other stress components on the cross section are assumed to be zero. (See Chapter 7 for
a review of basic stress and strain concepts and derivation of stress equilibrium equations.)
With this assumption the stress equilibrium equations are automatically satisfied. The strain
compatibility conditions lead to the following differential equation:

ewhere G is the shear modulus and is the angle of twist per unit length. The boundary

condition is that ¢ must be constant on the boundary. Usually ¢ = 0 is chosen on the
boundary.

318 TWO-DIMENSIONAL ELEMENTS
The total torque is related to the stress function as follows:

Using these equations, the finite element formulation can be employed to analyze thefol-
lowing two situations.

Determining Stresses in a Bar Subjected to a Given Torque T The governing
differential equation needs 0'8, the angle of twist per unit length. The torque does not show
up in the governing differential equation and thus we cannot proceed directly as in other
common stress analysis situations. Recognizing that the stress function ¢ depends linearly
on OB, we use the following procedure:

(i) Assume arty convenient value of OBa• To clearly identify that this value is assumed,
we have used the subscript a.

(ii) Find the stress function solution ¢a from the governing differential equation cor-
responding to OBa • Compute the total torque Ta by integrating over the bar cross
section using

(iii) Use ratios and proportions to determine the actual values of the angle of twist and,.

¢ as follows: ,/.

Once ¢ is known, the two nonzero stress components can be determined by simple
differentiation. All other stress components are zero.

Determining Torsional Constant An important use of the stress function approach
is to determine the torsional constant J for a given section. The torsional constant is a
property of a cross section and is related to the torque as follows:

T =JOB

=Thus, if we set OB l , then the total torque computed is the torsional constant of the cross

section:

SELECTED APPLICATIONS OF THE 20 BVP 319

Figure 5.5. Rectangular waveguide

(i) Set GB= 1. "

(ii) Find the stress function solution ¢ by solving the governing differential equation.
Compute the total torque T by integrating over the bar cross section using

IIT=]=2 ¢dA

A

Recall from Chapter 4 that] values are required for space frame members. The equations
given there for some simple cross sections are based on the closed-form solutions of ¢
available for those sections. For complicated sections, closed-form solutions are not pos-
sible. However, one can use the finite element formulation presented here to determine a
numerical value of] for virtually any cross section.

5.1.4 Waveguides in Electromagnetics

Waveguides are hollow conductors that are employed to efficiently transmit electrical sig-
nals at microwave frequencies (in the GHz range). A waveguide with a rectangular cross
section is shown in Figure 5.5. Electromagnetic waves inside a waveguide experience mul-
tiple reflections from the enclosing walls and produce discrete modes that depend on the
shape and size of the waveguide, the medium within the waveguide, and the operating fre-
quency. The two types of modes supported by a waveguide are identified as the transverse
magnetic (TM) andthe transverse electric (TE). These modes can propagate only when the
operating frequency is higher than a certain frequency known as the cutofffrequency, Thus
determination of the cutoff frequency is a critical parameter in the design of waveguides.

The governing differential equation for determination of the cutoff frequency is the

Helmholtz equation, written in terms of a scalar potential if! as follows:

where lee is the unlmown cutoff frequency. The potential is related to the transverse electric
field components as follows:

320 TWO·DIMENSIONAL ELEMENTS

For TM modes:

For TE modes:

where 20 is the characteristic wave impedance for TM modes. For the TM modes IjJ is set
to 0 along the boundary. For the TE modes the normal derivative of the potential (8i/J18n)
is 0 on the boundary.

This differential equation is of the same form as the general BVP with kx = ky = 1,
p = 12;;, and q = O. However, there is one key difference in this application, and that

is the cutoff frequency kc is not known. The situation is similar to the buckling problem
considered in Chapter 3. Proceeding with the standard finite element formulation in the
usual manner will give rise to an eigenvalue problem. The eigenvalues of the system will
correspond to the cutoff frequencies and the corresponding eigenvectors will be the wave
propagation modes.

5.2 INTEGRATION BY PARTS IN HIGHER DIMENSIONS

In order to derive the weak form for two- and three-dimensional boundary value problems,
we need a formula equivalent to the integration by parts for one-dimensional problems.
This is accomplished by using the well-known Gauss and Green-Gauss theorems from
vector calculus. These theorems are presented here in the context of a two-dimensional
problem. Their extension to the three-dimensional boundary value problems is obvious.

Gauss's Divergence Theorem Consider a vector of functions F = (F1(x,y), F2(x, y)l.

The divergence of F is defined as follows:

=divP 8F1 + 8F2

8x 8y

According to Gauss's divergence theorem, the area integral of div P is equal to line integral
of its dot product with the unit outer normal 11. That is,

ff i=divP dA p T 11de

A

The dA = dx dy is the differential area. All terms in the boundary integral must be ex-

pressed in terms of the boundary coordinate e. The de in the boundary integral is the length
of a differential segment on the boundary:

de= ~d~+di

INTEGRATION BY PARTS IN HIGHERDIMENSIONS 321

Written explicitly,

where 11.< and l1y are the components of the unit normal to the boundary,

Green-Gauss Theorem Given another function g(x, y), the integral of the divergence
of the product gF can be written as follows:

Using the product rule, the divergence of the product gF can be written as follows:

di ( F) - a(agxFI ) + a(gF2) _ aF1 + F ag + aF2 F ag
IV g - ay g ax I ax g ay + -? ay
-

Rearranging terms,

dI' V(gF) -_g (a-F1+a-F2 ) + (a- g

ax ay ax

Noting that the row vector in the second term is the transpose of the gradient vector of g,

ag)T
ay

we have

div(gF) = g(div F) + CilglF

Thus the integral of div(gF) can be expressed as follows: .

II IIg(divF)dA + C'lglFdA = 1 gFTndc

AA

Rearranging terms, we get the following relationship, commonly known as the Green-
Gauss theorem:

IIg(diVF)dA= 19FTl1dC- II('i1~lFdA

A

322 TWO-DIMENSIONAL ELEMENTS

Green-Gauss Theorem as Integration by Parts in Two Dimensions The

Green-Gauss theorem can be written in a form very similar to the usual one-dimensional

= =integration by parts by taking the vector of functions with F1
F = (f(x, y), O)T. The divergence of this function is
f and' F2 0, i.e.,

divF = af
ax
and thus applying the Green-Gauss theorem, we have

fJ agf dA
ax .
A

=Similarly, by taking the vector of functions with F1 0 and F2 == f, we can see that

Note that, if the derivative on function f(x, y) is with respect to x, then the boundary in-
tegral involves nx and vice versa. Similar to the one-dimensional case, this form of the
Green-Gauss theorem gives an area integral in which the derivative is switched from one
function to the other. We also get a boundary integral involving the product of the functions
and the component of the unit normal. These forms, and their obvious extensions to three
dimensions, will be used in the following section and the remaining chapters to obtain
weak forms for given boundary value problems.

,/

Example 5.1 A two-dimensional boundary value problem is stated as follows:

a a2u 2u + (x + y)u + x- = 0; 1 < x < 3; 1<y<2
3ay-2
2ax-2 - y

=u(x, 1) == sin(x); u(1, y) sin(y); au au3

a/x, 2) ==x ;
. ax (3, y) =u(3, y) + 1

Comparing the problem with the general form, identify the terms kx' ky, p, ... ,fJ for this

problem. Classify the boundary conditions into essential and natural types. Obtain a weak
form for the problem.

The solution domain is a rectangle, as shown in Figure 5.6. The coefficients· in the

differential equation are as follows:

P =x+y; x

q==-

y

The boundary conditions are as follows:

INTEGRATION BY PARTS IN HIGHERDIMENSIONS 323

y Side 3
2 Side 2 .

Side 4

Side 1

,-0----------- x

3

Figure 5.6.

Essential: On side 1, u = sin(x); on side 4, u = sin(y).

Natural: On side 3

au =;2.
ay ,

au au au and [3 = -3;2

ky ay12y = -3 ay :=:} kyay 12y = -3;2 :=:} a = 0

Natural: On side 2

au = u(3,y) + 1; 12x = 1, 12y = O
ax

au au au and [3=2
kxax 12x = 2 ax :=:} kxa/Ix = 2u(3, y) + 2 :=:} a = 2

Multiplying the differential equation by the weighting function Nj , the Galerkin weighted

residual is as follows:

I f ( », x)2a-2?u - 3-? + (x +y)u +- NjdA =0; ,i =1,2, ...
Bx: ay- y

A

Using integration by parts on the first two terms,

For the rectangular domain for this example, by substituting the normal vector components,
the expressions can be written more explicitly:

324 TWO-DIMENSIONAL ELEMENTS

First term:

3 l. 2 2-a--2:xuNi dx dy = l3 2a-uNi(x, 1) x (0) dx + l2 2a-u Ni(3, y) x (1) dy
x=1 aX aX
x=1 y=1 ye l
ax-l

(3 au (2 au

+ JX=1 2 ax Ni(x, 2) x (0)dx + J 2 ax Ni(1, y) x (-l)dy
y=1

- 3 l 2 2a-ua-Ni dx dY
l x=1 ye l ax ax .

ax-3 l 2 2a~2Nuidxd.y= l2 2a-uNp,y)dy
l x=1 ye I ye l ax

- 2 2a-uN/l,y)dy- l3l2 2a-u -aaN' dxdy
l y=1 ax xe l y=1 aX X

Second term:

3 l 2 - 3aa2y-2uNi dx dy = l3 - 3a-u Ni(x, 1) x (-1)dx
aY
l xe l y=1 x=1

+ l 2 -3 au x (O)dy + l3 -3 au 2) x (l)dx

Fl 7YJNP,y) _1 7YJNi(x,

+ 2 -3a-uNp,y) X (O)dy+ l3l2 3a-u-aaN'dxdy
l y=1 ay x=1 y=1 aX X

3 l 2 - 3aa2y-2uNidxdy= l3 3faJ-Yu Ni(x,l)dx- l3 3a-u Ni(x.2)dx
aY
l xe I ye l x=1 x=1

+ 3 l 2 3-aua-Ni dx dY
l xe l y=1 ax ax

Thus the weighted residual is

l 2 2a-uNp,y)dy- l2 2a-uN/l,y)dy+ l3 3aa-YuNi(x,l)dx- l3 3a-u Ni(x,2)dx
y=1 ax aX aY
ye l _1 x=1

+ 3 l 2 (-2au-a-N'+3a-ua-N'+ ((x+y)u+X-) N )dxdy=O
l x=1 y=1 ax ax ay ay y,

Consider boundary terms:

First term:

l~ au y=1 2- Ni(3,y)dy
ax

FINITE ELEMENTEQUATIONS USINGTHE GALERKIN METHOD 325

=. au u(3, y) + 1; .
Natural
SIde 2: ax

I:I~ 2(u(3, y) + 1)Ni(3,y) dy

Second term:

L a2 au
- 2-Ni(l , y) dy
ye I X .

=Side 4: u sin(y); Essential

=Must satisfy explicitly: - u
(2 2 aa Ni( l , y) dy 0
Jy=l x

Third term:

L3 au

x=1 37YJNJx,1)dx
Side 1: u = sin(x); Essential

Must satisfy explicitly: J<(3=I 3 :uNi(x, 1) dx =0
y

Fourth term:

L3 au

- -,=1 37JyNi(x,2)dx

Side 3: :~ = x3; Natural

1:1~ -3x3 NJx, 2) dx

Thus the wealeform is

1:1 1:11)~i~3,2(u(3,y) +
y) dy + -3x3Ni(x, 2)dx

1 L2 (+. 3 2a-ua N . + 3a-u a-Na.1 ((x+y)u+ -X) N,)dxdy
-a- + =0
aX X 1 aY Y

-,=1 ye l Y

Admissible solutions must satisfy the EBe on sides 1 and 4.

5.3 FINITE ELEMENT EQUATIONS USING THE GALERKIN METHOD

The solution domain is discretized into elements. For simplicity of notation we do not in-
troduceany super- or subscriptsand assume that A and C now refer to an arbitrary element
area and its boundary. For the general derivation considered in this section, no specific

326 TWO·DIMENSIONAL ELEMENTS

shape of the element is assumed. Explicit equations for rectangular and triangular are de-
veloped in later sections.

The general form of finite element equations for the problem can be written using ex-
actly the same steps used for the one-dimensional problems. Recall from Chapter 2 that for
one-dimensional problems a weak form is written by using standard steps of writing the
weighted residual, integration by parts, and incorporating the natural boundary conditions.
Furthermore, in the finite element form the weighting functions are the same as the inter-
polation functions N, used to define an assumed solution over a finite element. The same
general procedure works for two-dimensional problems.

The Galerkin weighted residual for the two-dimensional boundary value problem is

ff( aax (kax aux) + aay (kyaaUy ) + pu + q.) N,dA =0; i = 1,2•...

A

Using the Green-Gauss theorem on the first two terms in the weighted residual, we have

Thus the weighted residual is

f( ff(J
c
kx aauxnx + kyaau/y) Nide + -kx aaux &aNi - kyaauy IaNiii + puNi+ qNi)dA = 0

A.
(

Splitting the boundary integral into two parts, the one over which the essential boundary

condition is specified (Ce) and the other where the natural boundary condition is specified

(C ), and substituting the natural boundary condition, we get
II

As was the case for one-dimensional problems, requiring the assumed solution to satisfy
the essential boundary conditions results in weighting functions that are zero over the Ce
boundary, Thus with the admissible assumed solutions the weak-form equivalent to the
given boundary value problem is as follows:

i = 1,2•...

FINITE ELEMENT EQUATIONS USING THE GALERKIN METHOD 327

The assumed solution over an element is written as follows:

N,1(X'Y))[~~~]u(x,y) = (N1(x,y) N2(x,y) ... =NTd

un

=811 (8NI 8N2

8x 8x 8x

=au (8NI 8N2

8y 8y 8y

where up u2' ••• are the unknown solutions at the element nodes. Over the boundary of
the element the assumed solution must be written in terms of boundary coordinate e in the

following form:

NIl(e))[~] N~d=u(e) (NI(e) N2(e) ... =

ull

Substituting these into the weak form, we have

L (aNi(e)N~d +[3N)de i = 1,2, ...

c;

II+ (-kx ~~iB;-d-k)'~~iB~d +pNiNTd + qNi)dA = 0;

II

Writing all n equations, with Ni' i = 1, 2, ... , n, together in a matrix form yields

or,

L,,(aNcN~d +[3Nc)de + II(-k. J1./i[d - kjJ)'B~d ~ pNNTd + qN)dA =0

II

328 TWO-DIMENSIONAL ELEMENTS

Rearranging terms by keeping quantities involving d on the left-hand side, we get the
element equations as follows:

The first two terms inside the area integral can be arranged in a more compact form by
using matrices as follows:

where

~Jax and C =(J ~)

~ay

Thus the finite element equations are as follows:

Define n x n matrices

IIkk = BCBT dA;

A

Define n X 1 vectors

IIrq = qNdA;

A

Thus the element equations are

=(kk + kp + lea)d rq + rfJ

The quantities lea and rfJ are a result of applied natural boundary conditions and will affect

only those elements that have a boundary on which an NBC is specified. For the interior

elements these terms will be zero. .

In order to actually evaluate the integrals defining the element equations, we need to

assume an element shape and develop appropriate interpolation functions. Any element

shape can be created as long as it is useful in modeling the geometry of practical solution

domains and it is possible to carry out required integrations and differentiations. Some

typical elements are presented in the remaining sections and the following chapter.

RECTANGULAR FINITE ELEMENTS 329

5.4 RECTANGULAR FINITE ELEMENTS

A rectangular element is the simplest element that can be developed for two-dimensional
problems. The assumed solutions can easily be written by using the Lagrange interpolation
formula in the x and y directions. The area and boundary integrals are also straightforward
to evaluate over rectangular domains. However, from a practical point of view a rectangular
element is not very useful sinceit is difficult to model irregular geometries using rectan-
gular elements alone. The rectangular elements are developed here primarily to show the
procedure without integration issues complicating the discussion.

5.4.1 Four-Node Rectangular Element

A typical rectangulai element with dimensions 2a x 2b is shown in Figure 5.7. The bound-
ary is defined by four sides of the element. The outer normals and boundary coordinate
c for each side are also shown in the figure. The computations are easier if we choose a
local coordinate system (s, t) with the origin at the element center. For each element these
local coordinates are related to the global coordinates by a simple translation. Denoting the
global coordinates of the element centroid by (xc'Yc)' we have the following relationship
between the local and the global coordinates:

t =Y - Yc

Note that with tins change of variables dt = dy

ds =dx;

and therefore the derivatives with respect to x and y are exactly the same as those with
respect to sand t. In the local coordinate system, the nodal coordinates are (-a, -b), (a, -b),
(a, b), and (-a, b). To evaluate boundary terms, for each side a boundary coordinate c is
defined with the origin at the center of each side. The positive directions of these boundary
coordinates are consistent with the requirement of moving counterclockwise around the
boundary of the element.

yy
n

n

---------- x n

---:--------x

Figure 5.7. Four-node rectangular element

330 TWO·DIMENSIONAL ELEMENTS
Assumed Solution Assuming unknown solutions at the four corners as nodal degrees
of freedom, we have a total of four degrees of freedom. The assumed finite element solu-
tions are needed in the following form:

=Over A: U(S, t) (NI (s, t) Nz(s, t)

Over C:

Over the area the assumed solution can be written starting from a polynomial in two di-

mensions with four coefficients. Following the discussion in Chapter 2, we must use com-

plete polynomials as much as possible. A complete linear polynomial in two dimensions

has three terms Co + CIS + czt. For the fourth term we must choose from one of the three
quadratic terms c3st + c4sz + cst z. For an element that is appropriate for any general appli-

cation the assumed solution should be symmetric; otherwise the element will give different

results depending upon its orientation. Based on this consideration, we choose the st term

as the fourth term, and thus a suitable polynomial solution for a rectangular element is as

follows: .

The coefficients co' c1' •.. can be expressed in terms of nodal degrees of freedom by eval-
uating the polynomial at the nodes as follows:

["'] [1 -a -b
Uz _ 1 a -b -oabb]['c"IJ
u- 1 a b ab Cz

3

u4 1 -a b -ab c3

Inverting the 4 x 4 matrix, we get

I III
4 4 [~:J4 4
u3
-4cI i I 4Ici. -4cI i

m" -4iIj 4ci II
I 4ij 4ij
- 4 iI j I I u4

I

4ab - 4ab 4ab - 4ab

RECTANGULAR FINITEELEMENTS 331

Substituting into the polynomial, we have

I III

4 444

I I -4Ili
["']-4Ili u2
4li 4li u3
u(s, t) = (l s t st)
-4Iij
-4iIj
II
4ij 4ij

I II I U4
4ab - 4ab 4ab - 4ab

Carrying out matrix multiplication,

U(s t) = ( (a-s)(b-t) (a+s)(b-t) (a+s)(b+tl
----;jQb ----;jQb
, 4ab

N) UI] = NTd

U2
[4 U3

U4

(a - s)(b - t) N. _ (a + s)(b - t) .
2- 4ab'
NI = 4ab ;

N _ (a + s)(b + t) . N _ (a - s)(b + t)

3- 4ab ' 4 - 4ab

where N1, •• . , N4 are the required iriterpolation functions. Note that each of the interpola-
tion functions N, is 1 at the ith node of the element and 0 at every other node.

The solutions along the elementsides are written by substituting appropriate values of
sand t for each side. Expressing boundary interpolation functions in terms of e for each

side, we have

b ==*NTe,1-2 a+e
For side 1: = =- =s e, t (a2-Qe 2Q 0 0); -a::;; c s a
For side 2: -b s; c s b
For side 3: = = =s b-e b+e 0); -a::;; c s a
For side 4: a, t e ==*N{2-3 (0 -b::;; c s b
2b 2b

a );= = =s -e, t b ==*N{3-4 (0 0 a-e
2Q 2:c

= = =( os -a, t -e ==*N{4-1 b2+be 0 b-e ).
2b '

Element Equations The element equations involve the following matrices and vec-

tors:

Ifkk = BeBT dA-' IIkp =- T dA;
pNN

A A

I Ir, = qNdA; iliJ = c,f3Ne de

A

332 TWO-DIMENSIONAL ELEMENTS

~l (~ ~)i!!a!:sJ..as
~at
i!!a!:tJ..
and c =

For simplicity in integrations, it will be assumed that k", ky, p, q, a, and f3 are all constant

over an element. Thus these terms can be taken out of the integrals. Substituting the as-

sumed solution and carrying out integrations, the matrices and vectors needed for element

equations can easily be written:

NT _ ( (a-s)(b-t) (a+s)(b-t) (a+s)(b+t) (a-s)(b+t) )
- 4ab 4ab 4ab 4ab

C-b b-t b+t _b4+abt)
4ab 4ab
BT = 4ab
a+s a+s a-s
s-a
4ab - 4ab 4ab 4ab

k,(a-s)2+kx(b-tl' k,.<a2 - s ' ) - k ky(?-a2)+k,(t2_b2) kx(b2-t2)-k,.<a-s)'
16a2b2 16a'b' 16a2b2
x(b-t)2
ky(a2-s2)-k/b-t)2 16a2b'
16a2b2
ky(a+s)2+k,(b-t)2 k,(b'-t')-k,(a+s)' k,.<s2_ a2)+kx(t2 - b ' )
k/s2-a2)+k,(t2-b') 16a2b' 16a2b2
16a2b2 16a'b'

k,,(b2-t2)-k,.<a_s)2 kx( 2-t2)-k,.<a+s)2 k,(a+sl'+k/b+t)2 k,.<a2 - ?)-k
16a'b' x(b+t)2
16a'b' b 16a2b2

16a'b'

ky(s'-a2)+k,,(t2_b2) k,(a 2-s2)-k,(b+t)' k,.<a-s)2+k,,(b+t)2
16a'b' 16a2b2
16a'b'

!'t:. + fkcbi !'t:. _ kxb _!'t:._~ ~_!'t:.
6b 3a 6b 6a 6a 3b
3b

kya k,b k,a + kxb ~_!'t:. _!'t:._~
6a 3b 6b 6a
6iJ-3li 3b 3a

_!'t:. _ k,b ~_!'t:. !'t:. ~ !'t:._~
6a 3b
;6b 6a _!'t:._~ 3b+3a 6b 3a
6b 6a
k6ab - k,a !'t:. _ kxb !'t:. k,b

3ii 6b 3a 3b+3a

(a_s)2(b_t)2 (a-s)(a+s)(b-t)2 (a-s)(a+s)(b-t)(b+t) (a-s)2(b-t)(b+t)
16a2b' 16a2b' 16a'b' 16a2 b2

(a-s)(a+s)(b-t)2 (a+s)2(b_t)2 (a+s)'(b-t)(b+t) (a2_?)(b'_t2)
16a2b2 16a2b2 16a'b' 16a'lb2

(a-s)(a+s)(b-t)(b+t) (a+s)2(b-t)(b+t) (a+s)'(b+t)' (a-s)(a+s)(b+t)2
16a2b' 16a2b2 16a'b' 16a2b2

(a-s)2(b-t)(b+t) (a'-s')(b2-t') (a-s)(a+s)(b+t)' (a-s)'(b+d
16a'b2 16a2b' 16a'b' 16a2b2

-~abp -~abp -~abp -~abp

I" r =», = -~abp -~abp -~abp -~abp
-pNNT dsdt
La Lb -~abp -~abp -~abp -~abp

-~abp -~abp -~abp -~abp,

1: 1:r~ = qNT ds dt =(abq abq abq abq)