A04 - Introduction to the Finite Element Method
1
Introduction to the Finite Element Method
Background Material
 AERO 306 notes and Introduction to Aerospace Structural Analysis,
Allen and Haisler
 http://ceaspub.eas.asu.edu/structures/FiniteElementAnalysis.htm
 http://www.myb2o.net/myb2ous/Analysis/Features/10547.htm
 http://www.myb2o.net/myb2ous/Analysis/Tools/Process/10407.htm
 http://larcpubs.larc.nasa.gov/randt/1993/RandT/SectionG/G11.html
Assumptions
It is assumed that you are familiar with basic FEM theory (AERO 306)
and with applications to truss or beam elements and structures, and:
 know FEM theory based on an energy or variational formulation,
 know what a stiffness matrix is,
 know how to assemble element stiffness and force matrices into
global (structural) stiffness and force matrices,
 know how to solve the resulting equilibrium equations
[ K ]{q}  {Q} for displacements, and
 know how to determine resulting strains and stresses.
A04 - Introduction to the Finite Element Method
2
Structures are often analyzed using complex finite element analysis
methods. These tools have evolved over the past decades (since
early 1960's) to be the basis of most structural design tasks. A
candidate structure is analyzed subject to the predicted loads and
the finite element program predicts deflections, stresses, strains,
and even buckling of the many elements. The designed can then
resize components to reduce weight or prevent failure. In recent
years, structural optimization has been combined with finite
element analysis to determine component gauges that may
minimize weight subject to a number of constraints. Such tools are
becoming very useful and there are many examples of substantial
weight reduction using these methods. Surprisingly, however, it
appears that modern methods do not do a better job of predicting
failure of the resulting designs, as shown by the figure below,
constructed from recent Air Force data. The data compares static
test failures of wing, fuselage, vertical tail, horizontal tail, landing
gear and other components for New Aircraft (designed with finite
element methods) and Older aircraft (designed without FEM).
A04 - Introduction to the Finite Element Method
3
Moral of the story as presented by the chart above? Don't believe that
just because you are using modern, sophisticated finite element tools that
the analysis will in some way be better or safer. It still takes engineering
judgment, know-how and experience.
A04 - Introduction to the Finite Element Method
Aircraft have many main structural components in the wings, fuselage,
tail section, landing gear, etc. as shown below:
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A04 - Introduction to the Finite Element Method
Structural members shown above may be variously modeled as
beams, thin plates, membranes, shells, etc. In some cases, it may
be necessary to perform a full 3-D stress analysis (an elasticity
type analysis as opposed to an approximate one like plate theory).
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A04 - Introduction to the Finite Element Method
6
The webs in wing ribs and floor beams, and wing and fuselage
skins are typically thin members that might be considered as being
in a state of plane stress.
Plane stress describes a three-dimensional geometry wherein the
non-zero stresses all occur in a single plane. For a thin plate, web
or skin lying in the x-y plane, the only non-zero stress components
are  xx , yy , xy .
A04 - Introduction to the Finite Element Method
7
We now consider the development of a plane stress finite element.
Suppose we have a geometry like that shown below where the
thickness is small compared to the other two dimensions:
The only major stress components are  xx , yy , xy . Consider
another example of a thin bracket in tension and the corresponding
finite element mesh. The boundary at the hole is welded and
therefore fixed. A static tensile load of 5,500 lbs is applied to one
end. The dimensions are 25" x 10" x 0.5" thick.
A04 - Introduction to the Finite Element Method
t = 0.5 in.
Bracket Geometry & FEM Mesh, Loading and BC
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A04 - Introduction to the Finite Element Method
9
Notice that the geometry has been divided up into a number of
rectangular regions (elements) - these are called Quad elements.
We could also use triangular elements. We will demonstrate the
development of the stiffness matrix and load vector for a triangular
element as shown
below. We assume that
plane stress occurs in the
x-y plane and define
displacement
components u ( x, y ) and
v( x, y ) . We define
nodal displacements at
the three corners as:
Node 1: u1 and v1
Node 2: u2 and v2
Node 3: u3 and v3
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A04 - Introduction to the Finite Element Method
The process of developing the equilibrium equations for a given
element requires that we utilize an energy or variational principle;
for example, the principle of minimum potential energy given by:
 (U  V )  0
(1.1)
where U is the internal strain energy and V is the external potential
energy.
For a plane stress state, the internal (potential) energy is given by
U  12  { }T { }dV
(1.2)
 
 
xx


 xx 
{ }   yy  and { }   yy 


 

 xy 
 xy 
(1.3)
V
where
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A04 - Introduction to the Finite Element Method
Assuming a linear elastic material, the constitutive equation may
be written as
(1.4)
{ }  [ D]{ }
where [D] for plane stress is given by:
0
1 

E 

[ D] 

1
0

1  2 
 0 0 (1  ) / 2 
Note that [D] is symmetric.
(1.5)
The infinitesimal strains are given:
 xx
u
 ,
x
 yy
v
 ,
y
 xy
u v


y x
where u ( x, y ) and v( x, y ) are the displacement fields.
(1.6)
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A04 - Introduction to the Finite Element Method
Substituting (1.4) into (1.2) gives the strain energy as
U  12  { }T [ D]{ }dV
V
(1.7)
where use has been made of the symmetry of [D]. Note that if we
substitute (1.6) into (1.7), U is now in terms of the displacement
fields u ( x, y ) and v( x, y ) .
The external potential V can be evaluated once the external
tractions and body forces are specified. In general, V will have the
form of
V    pxu  p y v)ds   ( Fxu  Fy v )dV
(1.8)
S
V
where px and p y are boundary tractions, S is the element
boundary surface, Fx and Fy are body forces. Note that V is in
terms of displacements u ( x, y ) and v( x, y ) .
A04 - Introduction to the Finite Element Method
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We now have U and V in terms of u ( x, y ) and v( x, y ) ; and the
equilibrium equations for are an element are defined by the energy
principle  (U  V )  0 (1.1). However, we can't apply (1.1) just
yet. Why? Because u ( x, y ) and v( x, y ) must be in terms of
discrete variables (nodal displacements) but u ( x, y ) and v( x, y ) are
continuous functions.
Constant Strain Triangle
For any element, the displacement components u ( x, y ) and v( x, y )
are unknown. Following a Rayleigh-Ritz type solution, we assume
a solution for each. The simplest assumption that can be made in
this case is to assume that the displacement varies linearly over the
element. Hence, we assume:
u ( x, y )  1   2 x  3 y
v( x, y )  1  2  x  3 y
where the  ' s and  's are constants. These constants can be
related to nodal displacements for the triangular element:
(1.9)
A04 - Introduction to the Finite Element Method
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Assume the corners of
the triangle (nodes) are
numbered CCW, and
have coordinates
( x1, y1 ) , etc. as shown.
At each node (i=1,2,3),
assume the nodal
displacements are given
by (ui , vi ) . We can now
write 6 "boundary
conditions" as follows:
For u(x,y):
At node 1: u1  u ( x1, y1 )  1   2 x1  3 y1
At node 2: u2  u ( x2 , y2 )  1   2 x2  3 y2
At node 3: u3  u ( x3 , y3 )  1   2 x3  3 y3
(1.10)
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A04 - Introduction to the Finite Element Method
For v(x,y):
At node 1: v1  v( x1, y1 )  1   2 x1  3 y1
At node 2: v2  v( x2 , y2 )  1   2 x2  3 y2
At node 3: v3  v( x3 , y3 )  1   2 x3  3 y3
(1.11)
We can now solve for the constants in terms of nodal
displacements. Eqs. (1.10) can be written in matrix form as
1 x1
1 x
2

1 x3
Solution is:
y1  1   u1 
   

y2  2   u2 

y3  3  u3 
1  (a1u1  a2u2  a3u3 ) /(2 A)
 2  (b1u1  b2u2  b3u3 ) /(2 A)
 3  (c1u1  c2u2  c3u3 ) /(2 A)
(1.12)
(1.13)
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A04 - Introduction to the Finite Element Method
where
a1  x2 y3  x3 y2 , a2  x3 y1  x1 y3 , a3  x1 y2  x2 y1
b1  y2  y3 ,
b2  y3  y1 ,
b3  y1  y2
c1  x3  x2 ,
c 2  x1  x3 ,
c3  x2  x1
(1.14)
and
1 x1
2 A  1 x2
1 x3
y1
y2  2(area of triangle)
y3
Substituting (1.13) into (1.9) and rearranging, u(x,y) can be written
1
u ( x, y ) 
[(a1  b1x  c1 y )u1  (a2  b2 x  c2 y )u2
2A
 ( a3  b3 x  c3 y )u3 ]
(1.15)
Note that the a's, b's and c's are constants and depend only upon the
nodal coordinates (x,y) of the 3 corner nodes.
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A04 - Introduction to the Finite Element Method
Defining the coefficients of ui as Ni , equation (1.15) becomes:
3
u ( x, y )   Ni ui
(1.16)
1
N i ( x, y ) 
(ai  bi x  ci y )
2A
(1.17)
i 1
where
A similar result is obtained for v(x,y):
3
v( x, y )   Ni vi
(1.18)
i 1
The quantities Ni ( x, y ) are called shape functions. Note that the
same shape functions apply for both u ( x, y ) and v( x, y ) .
We can now obtain the strains by substituting displacement
functions (1.16) and (1.18) into strain expressions (1.6) to obtain:
A04 - Introduction to the Finite Element Method
 xx
3
b
u 3 Ni


ui   i ui
x i 1 x
i 1 2 A
 yy
3
ci
v 3 Ni


vi   vi
y i 1 y
i 1 2 A
 xy
3
3
3
Ni
ci
bi
u v 3 Ni

 
ui  
vi   ui   vi
y x i 1 y
i 1 x
i 1 2 A
i 1 2 A
18
(1.19)
The last 3 equations for strains can be put into matrix notation as:
 u1 
v 
1
 

b
0
b
0
b
0


2
3
 xx  1  1
u2 

(1.20)
0 c1 0 c2 0 c3  
 yy  


  2 A c b c b c b   v2 
 1 1 2 2 3 3  u 
 xy 
3
 
 v3 
A04 - Introduction to the Finite Element Method
Or, more compactly as (for any element "e"):
{ e }  [ Be ]{qe }
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(1.21)
b1 0 b2 0 b3 0 
1 
e
where
(1.22)
[B ] 
0 c1 0 c2 0 c3 


2A
c1 b1 c2 b2 c3 b3 
 u1 
v 
 1
u2 
e
and
(1.23)
{q }   
 v2 
u3 
 
 v3 
Since the terms in [ B e ] are constant for an element, the strains
{ e } are constant within an element; hence the name "constant
strain triangle" or CST.
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A04 - Introduction to the Finite Element Method
We can now evaluate the internal strain energy U. Substituting
(1.21) into (1.7) gives:
Ue 

e T
e T
e
e
e
1
{
q
}
[
B
]
[
D
][
B
]{
q
}dV
2 V
= 12 {q e }T   [ B e ]T [ D e ][ B e ]dV {q e }
 V

(1.24)
The quantity in parentheses can be identified as the element
stiffness matrix [ k e ] and (1.24) can be written as:
U e  12 {q e }T [k e ]{q e }
(1.25)
where the element stiffness matrix [ k e ] is defined by:
[k e ]   [ Be ]T [ De ][ Be ]dV
V
(1.26)
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A04 - Introduction to the Finite Element Method
e
If the element has a constant thickness t , then dV=tdA. Assuming
that E is constant over the element and noting that the terms in B
are constants, then
[k e ]  t e Ae [ B e ]T [ D e ][ B e ]
(1.27)
Note that the element stiffness matrix [ k e ] is a 6x6 matrix, i.e., we
have a 6 degree-of-freedom (dof) element.
Note that the general form for the strain energy (1.25) can be
written in index notation also:
6
6
U  12 {q } [k ]{q }  12  kije qie q ej
e
e T
e
e
(1.28)
i 1 j 1
Because [ D e ] is symmetric, the stiffness matrix [ k e ] defined by
either (1.26) or (1.27) is a symmetric matrix (always the case).
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A04 - Introduction to the Finite Element Method
The stiffness matrix for the CST defined by (1.27) can be written
in sub-matrix notation as:
 k11 k12
[k e ]   k21 k22

(6 x 6)
 k31 k32
k13 
k23 

k33 
(1.29)
where each of the kij is a (2x2) sub-matrix defined by
(bi b j D11  ci c j D33 ) (bi c j D12  ci b j D33 ) 

[kij ]  41A 
 (1.30)
(ci b j D21  bi c j D33 ) (ci c j D22  bi b j D33 ) 
(2 x 2)
where the Dij are material properties ( E , ) defined by (1.5) and
the bi and ci are geometry parameters (x-y coordinates of nodes)
defined by (1.14).
A04 - Introduction to the Finite Element Method
To define the external potential
energy V, we have to define the
external load. Suppose we have a
uniform traction (pressure) p applied
on the element edge defined by
nodes 1 and 2. The external
potential then becomes:
L
V    [u ( s) p cos  v( s) p sin  ]tds
e
0
Note that p cos is the component of p in the x direction.
Displacements u and v on boundary 1-2 must be written as
functions of position s on the boundary:
u12 ( s )  (1  s / L)u1  ( s / L)u2
v12 ( s)  (1  s / L)v1  ( s / L)v2
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A04 - Introduction to the Finite Element Method
Substituting u(s) and v(s) into V, and integrating over the
boundary, gives:
V  

 12 ptL cos  u1   12 ptL sin  v1   12 ptL cos  u2   12 ptL sin  v2 
The last result can be written in matrix notation as
6
V e  {q e }T {F e }   Fie qie
(1.31)
 1 pt e L cos 
2

 1 pt e L sin  
2

 1 pt e L cos 
e
{F }   2

 1 pt e L sin  
2



0


0


(1.32)
i 1
where
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A04 - Introduction to the Finite Element Method
The matrix {F} represents the equivalent generalized nodal force
vector due to pressure load on boundary 1-2, i.e., we have replaced
the pressure p on boundary 1-2 by the nodal forces {F} at nodes 1
and 2.
=
Note that the total force due to p on boundary 1-2 is (ptL) and
divides equally between nodes 1 and 2.
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A04 - Introduction to the Finite Element Method
Another set of forces exists on the boundary of any element.
These are due to surrounding elements that apply forces due to
contact with the element in question, i.e., surrounding elements are
being deformed and hence they try to deform the element in
question and thereby put forces on this element. Additionally,
where a node is at a support or "fixed," there will be a reaction
force on the element node. Call these reaction forces {S }.
S4
2
S3
S2
S6 1
S1
S5
3
Si  reactions from adjacent elements
The ext. pot. energy due to reactions is V e  {qe }T {S e }
(1.33)
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A04 - Introduction to the Finite Element Method
We can determine the equations of equilibrium for the element.
Using (1.1) and noting that U e and V e are functions of nodal
displacements qie , i  1,...,6 , we have
6
 (U e  V e )
i 1
qie
 (U e  V e )  0  
 qie
(1.34)
Since  qi  0 , then
 (U e  V e )
qie
0
for i  1, 2,...,6
(1.35)
Substituting U (1.28) and V [(1.31) and (1.33)] into (1.35) gives
the equilibrium equation for any element.
[k e ]{qe }  {F e }  {S e }
Note that [ K e ] is (6x6) and {F e } & {S e } are (6x1) matrices.
(1.36)
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A04 - Introduction to the Finite Element Method
Equations (1.34)-(1.36) provide the equilibrium equation for a
single element. Suppose we look at a collection of elements (i.e., a
complete structure). Then the total energy of the structure is given
by the sum of internal and potential energy of all the elements
( N el ):
N el
N el
U str  U  
e
e1
e1

 6 6 e e e
1 {q } [ k ]{q } 
1
kij qi q j  (1.37)



2
2


e1  i 1 j 1

e T
e
e

N el

N el
and

N el
N el
e1
e1
N el  6


Vstr  V e   {q e }T {F e }   {q e }T {S e }
   
e1  i 1
e e
Fi qi  
e1

e e
   Si qi 
 e1  i 1

N el
6
The principle of minimum potential energy for the structure
requires that
(1.38)
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A04 - Introduction to the Finite Element Method
M
 (U str  Vstr )  0  
 (U str  Vstr )
qi
i 1
 qi
(1.39)
where {q} contains the M degrees of freedom for the structure
(NOT dof for each element). For  qi  0 , the last equation
requires that
 (U str  Vstr )
(1.40)
 0 for i  1, 2,..., M
qi
Substituting U str (1.37) and Vstr (1.38) into (1.40) gives
 N el


 e1

1 {q e }T [ k e ]{q e }
2

{q } {S } 

 0
   {q } {F }   
N el
e T
e 1
qi
e
N el
e 1
e T
e

for i  1, 2,..., M
(1.41)
Problem: The energy terms for each element are in terms of the
element dof, but in order to obtain the equations of equilibrium for
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A04 - Introduction to the Finite Element Method
the structure (above equation), we have to take the partial
derivatives with respect to the global structural dof. In order to
complete the above, the element degrees of freedom {q e } must be
written in terms of the M global structural degrees of freedom {q}.
For any element, we can write a transformation between element
local and global dof (called the local-global transformation):
{q e }  [T e ] {q}
(6 x1)
y
(1.42)
(6 xM ) ( Mx1)
The transformation will be nothing more then 1's and 0's. As an
example, suppose we have the following element and structural
node numbering:
q2
q6
q8
q4
1
2
3
4
1 q1
2 q3
3 q5
4 q7
q10
1
5
9
5
2 6
6 7
10 8
p
5 q9
6
7
8
q18
3
7
11
q24
q17
q23
4
8
12
9
x
10
11
12
y
9
x
10
11
12
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A04 - Introduction to the Finite Element Method
Consider element 7. Suppose we place element node 1 at global
node 6.
q12
q14
q2e
q6e
q11 7
e
q1e
q13
3
q
6
5
1
7
7
q22
q4e
q21
q3e
11
2
Element nodes
and local dofs
Structural nodes
and global dofs
We see that for element 7, there is a correspondence between the 6
element local dofs at element nodes 1, 2 and 3, and the 6 structural
global dofs at nodes 6, 11and 7. We see that local (element) node
1 corresponds to global node 6, local (element) node 2 corresponds
to global node 11, and local node 3 corresponds to global node 7.
We can write this local to global transformation {qe }  [T e ]{q} as:
A04 - Introduction to the Finite Element Method
32
 q1,2 
 q 
1 2 3 4
8 9 10 11 12  3,4 
 q5,6 



  q7,8 
1 0 
[0]
[0] [0] [0]
[0]
[0]
[0] [0] [0] [0] [0] 0 1 


e 7  


q1,2
  q9,10 
 7  
  q11,12 
1 0 
[0]
[0]
[0] [0] [0] 
 q3,4   [0] [0] [0] [0] [0]

 [0]  q
0
1


  13,14 
 7  
  q15,16 
 q5,6  
1 0 
[0] [0] [0] [0] [0]

[0]
[0]
[0] 
0 1  [0] [0] [0]
q
17,18




 
 q19,20 


q
 21,22 
q23,24 


global node #
5
6
7
Each [0] is a (2x2). The above says that for element 7, local
(element) node 1 corresponds to global node 6, i.e., local dofs 1,2
correspond to global dofs 11,12; local node 2 corresponds to global
node 11, i.e., local dofs 3,4 correspond to global dofs 21,22, etc.
33
A04 - Introduction to the Finite Element Method
Now transform U e and V e from local to global dof by substituting
(1.42) into (1.37) and (1.38) to obtain
U e  12 {q e }T [k e ]{q e }  12 {q}T [T e ]T [ k e ][T e ]{q}  12 {q}T [ K ge ]{q}
V  {q} [T ] {F }  {q} [T ] {S }  {q}
e
T
e T
e
T
e T
e
T
{Fge }  {q}T {S ge }
(1.43)
Now we can define the following element matrices in global dof
(instead of local element dof):
[ K ge ]  [T e ]T [k e ] [T e ]
( MxM )
( Mx 6) (6 x 6) (6 xM )
{Fge }  [T e ]T {F e }
( Mx1)
(1.44)
( Mx 6) (6 x1)
{S ge }  [T e ]T {S e }
To see what an element stiffness and force matrix written in global
dof looks like, consider element 7 again. We obtain for [ K g7 ] and
{Fg7 }:
6
A04 - Introduction to the Finite Element Method
7
6  k11

7
11  k21
 7
7  k31
Element 7
Each block is a (2x2) sub-matrix
1 2 3 4 5
1
2
3
4
5
[ K g7 ]  6
7
8
9
10
11
12
6
7
11
7
k12
7
k22
7
k32
7
7 
k13

7
k23

7 
k33 

 F17 
 
7
 F2 
 7
 F3 
34
8 9 10 11 12
7
7
k13
k11
7
k12
7
7
k31
k33
7
k32
7
7
k23
k21
7
k22
{Fg7 } 
F37
35
A04 - Introduction to the Finite Element Method
Now the internal and external potential energy is given by
N el
N el
N el  M M

e
T e
e
1
1
U str  U   2 {q} [k g ]{q}    2  k gij qi q j  (1.45)


e1
e1
e1  i 1 j 1




N el
N el
e1
e1
N el  M

N el


Vstr  V e   {q}T {Fge }   {q}T {S ge }
   
e1  i 1
Fgie qi
e1
N el  M


e
     S gi qi 
 e1  i 1

(1.46)
Now we can substitute (1.45) and (1.46) into (1.40) to obtain:
 N el


 e1

   {q}
1 {q}T [ k e ]{q}
g
2
N el
e1
qi
T
{Fge }
   {q}
N el
e1
T
{S ge }



 0
for i  1, 2,..., M
(1.47)
36
A04 - Introduction to the Finite Element Method
which gives a system of M equations in terms of the structural
displacements:

N el
e1

N el
N el
[k ge ]{q}  {Fge }  {S ge }  {0}
e1
(1.48)
e1
or
N el
N el
 Nel e 
e
e
[
k
]
{
q
}

{
F
}

{
S
 g 
 g  g}
 e1

e1
e1
(1.49)
When all the element contributions have been summed, we simply
write
(1.50)
[ K ]{q}  {Q}  {S}
Note that when the element stiffness and force matrices are written
in terms of structural displacements (using local to global
transformation), they become additive [see eq. (1.49)]; i.e., to get
the structural stiffness matrix [K], we sum the contributions for all
elements.
37
A04 - Introduction to the Finite Element Method
Assemblage of Elements
A single element by itself is useless. We must determine the
equilibrium equations for an assemblage of elements that comprise
the entire structures.
Consider the following structure (only a few elements are taken to
simplify the discussion) with a uniformly pressure p on the right
boundary and fixed on the left boundary (assume a constant
thickness t).
1
We number the structural nodes
from 1 to 12 as shown. We also
number the elements from 1 to
12 as shown (in any order).
5
y
2
1
4
9
x
2
3
6
3
5
8
10
6
7
7
4
9
12
11
For each global node of the structure, we can specify the (x,y)
coordinates: xi , yi , i=1, 2, …, 12.
10 8
11
12
p
A04 - Introduction to the Finite Element Method
38
q6
q8
q4
Each node of the structure will q2
q3
q5
1 q1
2
3
4 q7
have two degrees of freedom
q10
(dof). We label these
5 q9
6
7
8
structural (global) degrees of
q18
q24
q17
q23
freedom in order as shown to
y
9
the right. Note that the
10
11
12
x
structural nodal displacements
are written without the superscript "e." The nodal displacement
vector is written as {q} and is (24x1) for this problem.
We note that the left side is fixed (nodes 1, 5 and 9). Hence,
displacement boundary conditions will require that
q1  q2  q9  q10  q17  q18  0 .
Note: we do not have to number the dof consistently and in
sequence with the structural nodes. However, this makes the
bookkeeping much, much simpler!
For each element, we can construct a table called the element
connectivity that specifies which structural (global) nodes are
A04 - Introduction to the Finite Element Method
39
connected by an element. Hence, for the problem above, we have
the following element connectivity table:
Element No.
Element Node 1 Element Node 2 Element Node 3
1
1
5
2
2
5
6
2
3
5
10
6
4
5
9
10
5
2
6
3
6
6
7
3
7
6
11
7
8
6
10
11
9
3
7
4
10
7
8
4
11
7
12
8
12
7
11
12
Note that for the CST, element nodes MUST be given as CCW.
Element node 1 can be attached with any global node of the
element.
A04 - Introduction to the Finite Element Method
40
Note that if we are careful in numbering the nodes and choosing
the element connectivity in a "systematic" manner, there will be a
pattern to the element connectivity table (see above). An
automatic mesh generator, like the one in FEMAP, tries to follow
this pattern.
Note that the global node numbers for the structure are somewhat
arbitrary, i.e., we could number them in any order. However, it
will turn out that there are optimum ways to number nodes (for a
given structure and mesh) in order to reduce the bandwidth of the
structural stiffness matrix [K] - this saves time solving the
equations. For the mesh above, it would be optimum to number
downward and left-to-right, as opposed to left-to-right and
downward. We'll discuss that later. Likewise, the element
numbering is arbitrary, but again there may be optimum
approaches. An automatic mesh generator tries to do the
numbering in an optimum fashion.
41
A04 - Introduction to the Finite Element Method
Note that for this structure, we have 12 global nodes. There are 2
degrees of freedom (dof) at each node (u and v). Hence, the
structure has 24 dof and the structural stiffness matrix [K] will be
(24x24). The structural equilibrium equations can be written as:
[ K ] {q}  {Q}  {S}
(24 x 24) (24 x1)
(1.51)
(24 x1)
where [K]=structural stiffness matrix,
{Q}=structural forces matrix (due to applied tractions
and body forces)
{S}=structure reaction forces due to boundary conditions
Lets see how each element contributes to global matrices. Take
element 1 to start with. Note that we can use sub-matrix notation
to divide the element matrices as following. Use a superscript of 1
on the k terms to indicate element 1.
42
A04 - Introduction to the Finite Element Method
1
1
 k11
k12
 (2 x 2)
 1
1
1
[k ]   k21 k22
(6 x 6)  1
1
k
k
32
 31

1 
k13


1
k23  ,
1 
k33 

 F11 
(2 x1) 
 1 
1
{F }   F2 
 1
 F3 


We now look at element 1 and note that element node numbers 1, 2, 3
correspond to global node numbers 1, 5, 2 (from the drawing of the
mesh, or from the element connectivity table). We can indicate this
information on the stiffness and force matrices as follows:
1
5
1
1
 k11
k
12
1 
(2 x 2)
 1
1
1
[k ]  5  k21 k22
(6 x 6)
 1
1
k
k
32
2  31

2
1 
k13


1
k23  ,
1 
k33 

 F11 
1 (2 x1) 
 1 
{F 1}  5  F2 
(6 x1)
 1
2  F3 


43
A04 - Introduction to the Finite Element Method
Hence, we see that element 1 contributes stiffness and forces to
global nodes 1, 5 and 2. Placing these contributions into the global
stiffness matrix gives:
Element 1 only
K
1
2
3
4
5
6
7
8
9
10
11
12
1
2
3
4
5
6
7
8
9 10 11 12
{q}
Q
1
1
k13
k11
1
k12
q1,2
F11
1
1
k31
k33
1
k32
q3,4
F31
1
k22
q5,6
q7,8
q9,10
F21
1
1
k23
k21
=
q23,24
** remember, each block is a (2x2) sub-matrix
6
A04 - Introduction to the Finite Element Method
7
6  k11

7
k

11 21
 7
7  k31
Now take element 7.
Element 7 only
K 1
1
2
3
4
5
6
7
8
9
10
11
12
2
3
4
5
6
7
8
11
7
k12
7
k22
7
k32
7
7 
k13

7
k23 
7 
k33 

9 10 11 12
 F17 
 
7
F
 2
 7
F
 3 
{q}
Q
q1,2
q3,4
q5,6
q7,8
q9,10
=
q23,24
** remember, each block is a (2x2) sub-matrix
44
A04 - Introduction to the Finite Element Method
45
Note that the distributed pressure load p is applied only to the right
boundary of elements 10 and 11. Hence {F} for all elements
except 10 and 11 will be zero. For elements 10 and 11, we will
have
0
 0 


 0 7

7
0




1 ptL
1 ptL



48 
812 
2
2

10
11
{F }  
 8 {F }  
 12
0
 0 


 1 ptL48 
 1 ptL812 
2
4
2
8
0
 0 


where L48 is the length between global nodes 4 and 8, etc.
If we assemble all element stiffness matrices [k] and forces
matrices {F} to the global equilibrium equations, we have the
following result:
A04 - Introduction to the Finite Element Method
46
Structural Equations of Equilibrium
K
1
2
3
4
5
6
7
8
9
10
11
12
1 2 3 4 5 6 7 8 9 10 11 12
{q}
Q
q1,2
X X
X
q3,4
X X X
X X
q5,6
X X X
X X
q7,8
X X
X X
X
q9,10
X X
X X
X X
X X
X X X
X X
=
X X
X X X
X X
X
X X
X
X
X
X X
X X
X X X
X X
X X X
q23,24
X X
X X
X
X means that one or more elements have contributed here
** remember, each block is a (2x2) sub-matrix
Note that [K] is symmetric; also it is banded (semi-bandwidth=12).
47
A04 - Introduction to the Finite Element Method
In the previous page, each X
1
2
3
means that one or more elements
3 3
1
3 31
1
5
have contributed to that (2x2)
sub-matrix. For example, we note 5 2 1 2 2 2 1 6 2
7
1 1 3 36 1 7
that node 2 will have stiffness
from elements 1, 2 and 5. Hence,
22 8
2
2 4
3
3
the 2,2 position of the global
9
10
11
stiffness matrix will be equal to
(note: you have to refer to the element connectivity to see which
element node for each element corresponds to global node 2):
1
2
5
[ K 22 ]  [k33
]  [k33
]  [k11
]
each sub-matrix is (2x2)
The global node 2-6 coupling term [ K 26 ] will have contributions
from elements 2 and 5 since only these elements share the
2
5
boundary between nodes 2 and 6: [ K 26 ]  [k32
]  [k12
].
Global node 6 will have stiffness contributions from elements 2, 3,
2
3
5
6
7
8
5, 6, 7, and 8: [ K66 ]  [k22
]  [k33
]  [k22
]  [k11
]  [k11
]  [k11
].
48
A04 - Introduction to the Finite Element Method
Question? What happened to the reactions {S} for each element?
Why don't they show up in the structural stiffness matrix?
Simple. It is equilibrium. Recall that when we make a free-body,
in this case take a single finite element as the free-body, we will
have equal and opposite reactions where the cut is made though the
body. Consider elements 1 and 2 below:
S41
2
1
1
2
3
S21
S31
S 42
S11
S61 1
S51
S62
3
2
2
S52
S32
S22
S12
1
At the boundary between elements 1 and 2, the reactions are equal
and opposite. Hence, we add them up we have: S11  S32  0 ,
S21  S42  0, S51  S52  0, and S61  S62  0 . Hence, all the reactions
49
A04 - Introduction to the Finite Element Method
between elements sum to zero and do not have to be put into the
structural equilibrium equations.
OK, but what about the boundary where there are supports? What
happens to the reactions there? For example, the cantilever plate
example above:
1
2
3
4
1
They don't disappear and should
5
2
be included in the structural
3
4
stiffness matrix.
y
9
We know that there will be
x
unknown reactions at global
nodes 1, 5 and 9. We could call R2 1 R1
these reactions R1, R2 , R9 , R10 ,
R10 1
5 R9 2
R17 and R18 (consistent with
R18
3
global displacements). So we
4
y
have the free body of the
9 R17
structure:
x
6
5
8
6
7
9
7
12
10 8
11
10
11
12
2
3
4
6
5
8
10
6
7
7
9
12
11
10 8
11
12
p
p
50
A04 - Introduction to the Finite Element Method
Structural Equations of Equilibrium with Support Reactions
K
1
2
3
4
5
6
7
8
9
10
11
12
1 2 3 4 5 6 7
X X
X
X X X
X X
X X X
X X
X X
X
X X
X X
X X
X X X
X X
X X
X
X
X
X X
X X
X
8
9 10 11 12
X
X X
X X
X
X X
X
X
X X
X X X
X X X
X
X X
{q}
q1,2
Q
R1,2
q3,4
q5,6
q7,8
q9,10
X
R9,10
=
X
R17,18
q23,24
X means that one or more elements have contributed here
** remember, each block is a (2x2) sub-matrix
X
A04 - Introduction to the Finite Element Method
51
OK, now one last step. We have to apply displacement boundary
conditions. The structure is fixed at nodes 1, 5 and 9; thus,
q1  q2  q9  q10  q17  q18  0 . The easiest way to apply
boundary conditions to any system of equations is as follows:
1. Zero out the row and column on the left side matrix (the [K]
matrix) corresponding to each B.C., and zero out the row of the
right side (the {Q} matrix) corresponding to each B.C.
2. Place a 1 on the diagonal of the left side matrix (the [K] matrix)
corresponding to each B.C.
 You will notice that every dof that has a B.C. also corresponds to
a dof where a support reaction (R) occurs. Applying B.C. as
described above will thus eliminate the reactions from the
equilibrium equations.
 A theoretical reason why we don’t have to worry about reactions
in structural equations of equilibrium? Because these support
reactions R do no work (displacement is zero at support) and
hence do not affect equilibrium of the structure!!!
52
A04 - Introduction to the Finite Element Method
Structural Equations of Equilibrium with B.C. Applied
K
1
2
3
4
5
6
7
8
9
10
11
12
1
2
1 
 1


0 0
0 0


0 0
0 0


0 0
0 0


3 4
X X
X X X
X X
0 0
0 0


X X
X X
X
5
0
0

0
0

0
0 
0
0 
1 
 1


0 0
0 0


0
0

0
0

0
0 
0
0 
6
7 8
9
10 11 12
0 0
0 0


X X
X X X
X X
X
X X
X X
Q
q1,2
0 
 
0 
q3,4
X
X X
X X
0 0
0 0


{q}
0 0
0 0


q5,6
q7,8
q9,10
X X
X X
X
1 
 1


0 0
0 0


X
0 
 
0 
=
X
0 0
0 0


X X
X X X
X X
0 
 
0 
q23,24
X means that one or more elements have contributed here
** remember, each block is a (2x2) sub-matrix
X
53
A04 - Introduction to the Finite Element Method
The structural equations with B.C. may now be solved for the
unknown displacements. Note that when we solve the system of
equations, the solution will give q1  q2  q9  q10  q17  q18  0 ,
i.e, the 1st equation simply says (1)q1  0, etc.
Element Strains and Stresses
Now we are ready to solve for the element strains and stresses. For
each element, we can substitute the 6 global displacements
corresponding to that element into (1.21):
{ e }  [ B e ]{q e }
(3 x1)
e=1, 2, …, no. of elements
(3 x 6) (6 x1)
The stresses for each element can then be obtained by substituting
the strains for that element into (1.4):
{ e }  [ De ]{ e }
(3 x1)
(3 x3) (3 x1)
e=1, 2, …, no. of elements
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54
Evaluation of stress results based on stress components in the
Cartesian coordinates directions ( xx , yy , xy , etc.) leaves
something to be desired. Why? Stresses in these directions may
not necessarily represent the largest stresses and we need these in
order to consider yielding or failure. You already know that you
can calculate principal stresses and maximum shear stress using
stress transformation equations or Mohr's Circle. Hence, stress
results (stress components) are often represented in two additional
ways:
 Principal stresses and maximum shear stress, and
 von Mises stress.
Principal stresses can, as noted above, be obtained by either stress
transformation equations or through the use of Mohr's Circle. An
alternate approach to define principal stresses is to write:
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A04 - Introduction to the Finite Element Method
 xx   p
 xy
 xy
 yx
 yy   p
 yz
 zx
 zy
 zz   p
0
(1.52)
Expansion of the determinant provides a cubic equation that can be
solved for the three principal stresses  p . Comparing principal
stresses to a tensile yield stress provides some measure of
evaluation; however, one has to keep in mind that comparing the
principal stress (obtained from a three-dimensional stress state) to
a yield stress obtained from a uniaxial tension test is risky at best.
The von Mises stress provides a means to extrapolate uniaxial
tensile test data (for yield stress) to a three-dimensional stress state.
In effect, the von Mises stress provides an "equivalent" uniaxial
stress approximation to the three-dimensional stress state in a body
through the following equation:
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A04 - Introduction to the Finite Element Method
 VM
1
22
( xx   yy )  ( yy   zz )  ( zz   xx )
 12 
 (1.53)
 6 xy  6 yz  6 zx


2
2
or
 VM 
1
2
1
22
( p1   p 2 )2  ( p 2   p 3 ) 2  ( p 3   p1 )
(1.54)


where ( p1, p 2 , p3 ) are the principal stresses. Given the stress
components ( xx , yy , xy , etc.) or principal stresses, one can
compute the von Mises stress.
This representation has been used quite successfully to model the
onset of yielding in ductile metals and collaborates well with
experiment. It is widely used in industry. For a material to remain
elastic,
 VM   y (for no yielding)
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57
Equation (1.54) forms an ellipsoid in 3-D (ellipse in 2-D) when the
stresses are plotted in principal stress space. As long as the stress
state represented by the principal stresses is inside the ellipse (the
yield surface), the material is elastic.
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58
Element Libraries
(or, choose the right element for a structural component and
loading, in order to maximize potential for correct results with the
least amount of computation)
Many, many finite elements have been developed for use in
modern FEM software. Choosing the correct element for a
particular structural is paramount. For example,
 if a structural member behaves like a beam in bending, we
should choose a beam element to model it,
 if a structural member behaves like a thin plate in plane stress,
we should choose an appropriate element to model it,
 if a structural member looks like a shell of revolution, we
should use a thin shell of revolution element,
 if a structural member will experience a three-dimensional
stress state, we have to choose an element that models that
behavior,
 etc.
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A04 - Introduction to the Finite Element Method
Here are some examples of the types of elements available:
 Truss element (2-D and 3-D)
 Beam bending element (2-D and 3-D; straight and curved)
dof at each node
 Membrane element (no bending; flat and curved)
 Triangular, Quad (both straight and curved sides)
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60
 Planes Stress and Plane Strain elements
 Triangular and Quadrilateral shapes (both straight and curved
sides).
Plane stress requires that the only non-zero stresses occur in
the plane of the element (however, strain does occur normal
to plane). Generally applicable to thin geometries. Two
displacement dof per node (NO rotational dof).
Plane strain requires that the only non-zero strains occur in
the plane of the element (strain is zero normal to plane, but
stress is not zero). Long constrained geometries (for
example, a long pipe, a dam). Elements with curved
boundaries will always have 3 or more nodes per edge.
A04 - Introduction to the Finite Element Method
 Plate and shell bending elements (bending and in-plane
stresses; flat and curved elements)
 Triangular, Quad (both straight and curved sides)
Plate and shell bending
elements are characterized
as being thin compared to
other dimensions, and
having no stress normal to
the plate (similar to plane
stress).
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A04 - Introduction to the Finite Element Method
62
Plate and shell bending elements will have in-plane and normal
displacements ( u, v, w) and rotations ( x , y ) about the two axes
in the plane of the plate/shell. No stiffness about the normal
axes.
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63
 Axisymmetric shell bending element (for shell of revolution)
The shell of revolution can be
described geometrically by a curved
(or straight) line that is revolved
about the axis of symmetry. Top
two figures are examples of thin
shells of revolution. Degrees of
freedom and stress state for a shell
of revolution are similar to plate
and shell bending elements, i.e.,
displacements parallel and
perpendicular to the shell surface,
and rotations about the two axes
that lie in the plane of the shell. No
stress normal to surface.
Lower figure (multicell tube) is NOT a shell of revolution, but
would require plate or shell elements.
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64
 Axisymmetric body of revolution element (3-D stress analysis)
Body is solid and axisymmetric about some axis of revolution.
Stress state is fully three-dimensional (includes all stress
components). Note that the element forms a triangular-shaped
ring, i.e., the element itself is a body of revolution.
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65
 General solid element (for full 3-D stress analysis)
 Tetrahedron, Brick shapes (both straight and curved sides)
Stress state is fully three-dimensional (includes all stress
components). Three-dimensional solid elements have 3
displacements per node (NO rotational dofs). Elements with
curved boundaries will always have 3 or more nodes per
edge.
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66
Mesh Generation, Pre and Post-processing, Solvers
In order to generate the structural stiffness matrix, we must specify
the nodal coordinates (x,y,z) of each global node and element
connectivity for each element. For a large problem with complex
geometry, this is a monumental task if done by hand. Mesh
generators are software tools that automate this process.
Mesh Generation and Preprocessing
In general, mesh generation starts by specifying the coordinates of
a few key locations that will sufficiently define the outer boundary
t = 0.5 in.
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67
of a structural component and other distinguishing features such as
holes, fillets, etc. Consider the bracket shown above. We could
first define the corners of a 10" x 25" rectangle. To specify the
rounded edges on the right boundary, we could specify that a 4"
radius fillet is to be placed at each right corner. Finally, to specify
the location of the hole, we would define a 2" radius circle whose
center is located 5" from the right boundary.
To specify the mesh (how many elements are to be used), we
would specify so many elements in the x direction and so many
elements in the y direction.
Since the structure is 2-D and in plane stress, we would specify the
type of element (triangular, quad, etc.) to be used in defining the
mesh.
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68
The mesh generator would then automatically determine the
coordinates of all global nodes and determine the element
connectivity for all elements.
As a part of the mesh generation, we would also specify the
thickness to be used for each element. Generally, the mesh
generator would allow for a linear thickness variation within the
region being meshed. For a variable thickness case, one would
have to specify the thickness at several key locations and mesh
generator would determine the thickness for all elements in the
region.
Also, the material properties would be specified for each element
in the region. Most mesh generators only provide for one material
set to be used within a region (i.e., same material for all elements).
Next, the mesh generator (also called the preprocessor, because it
processes all the required input) would require the type of loading
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69
that is applied to elements or element boundaries (point forces,
body forces, surface tractions). For a thermal stress analysis, the
preprocessor would also require information on thermal loading.
For a dynamic (transient) analysis, we would have to specify mass
properties for each element and the time history of the loading.
Lastly, the displacement boundary conditions must be specified.
Given all this information, the mesh generator/preprocessor will
have generated a model something like that shown below:
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70
With this information, the preprocessor would create a data set (an
output file) in a format suitable to be read and processed by the
finite element analysis program (also called the solver).
Solver
Once the solver has created the structural stiffness and force
matrices, solved for global nodal displacements, and solved for
stresses and strains (plus a few other things appropriate to each
element type), it generates an output file and we are now ready to
examine and interpret the results.
Post-processing
Clearly, for a complicated structure with many nodes and
elements, the examination and evaluation of all results is an
enormous task because of the shear volume of data (displacements,
strains, stresses, etc). The post-processor now takes over. Its
purpose is to provide output such as the deformed geometry,
contour plots of principal stress components, contour plots of von
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71
Mises stress, contour plots of strain components, etc. (what ever
you decide is important). What, and how, the postprocessor can
display information depends on the type of elements being used.
Below are outputs of the deformed geometry and vonMises stress.
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72
Often the pre and post-processor are combined into one software
package (like FEMAP or PATRAN). Most preprocessor programs
will create output files in formats acceptable to various solvers
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73
(like CAEFEM, NASTRAN, ANSYS, ABACUS, etc.). Likewise,
most postprocessor programs will accept the output file of most
solvers and display a variety of data results generated by the
solver.
Last Thoughts on Mesh Generators, Pre/Post-Processors & Solvers
 They will do a lot the tedious work for you.
 They are not a substitute for good engineering know-how,
judgement and experience.
 In the end, YOU are responsible for the model you have
generated for the structure, and for the correct interpretation and
usage of the analysis results.
 They aid you in the design process; but in the end, YOU are the
designer.