Finite Element Theory

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Transcript Finite Element Theory

FEA Simulations
• Usually based on energy minimum or virtual work
• Component of interest is divided into small parts
– 1D elements for beam or truss structures
– 2D elements for plate or shell structures
– 3D elements for solids
• Boundary conditions are applied
– Force or stress (i.e., pressure or shear)
– Displacement
– Multi-point constraints
FEA Simulations (Contintued)
• Solution of governing equations
–
–
–
–
Static: solution of simulataneous equations
Vibrations: eigenvalue analysis
Transient: Numerically step through time
Nonlinear: includes buckling uses an iterative solution
• Evaluation of stress and strain
• Post-processing: e.g., contour or history plots
Principal of Virtual Work

Change in energy for a “virtual” displacement,u , in
the structure of volume, V , with surface, S, is
 
 
 
 
where E   e  dV   u  b dV   u  fdS u  P
V
E

u



e

b
f

P
V
S
is the energy change
is the virtual displacement
is the internal stress


u
is the virtual strain due to
are the body forces (e.g., gravity, centrifugal)
are surface tractions(e.g., pressure, friction)
are point loads
Principal of Virtual Work (continued)
The principal of virtual work must hold for all possible

virtual displacements. We must convert u at the nodes

to e in the elements.
For example if we have a beam we can take
  x  x    x  x1  
u1  
u2 and
u   2
 x2  x1 
 x2  x1 

 1    1  

u

u1  
u2
e 
 
x
 x2  x1 
 x2  x1 
or


u

  1 1 1

e   ,      B u
 L L  u2 
B is the strain displacement matrix.
Energy Minimization
Let   U  V where U is the internal energy and V is
the potential energy due to the loads and are given by
T 
T 
T 
T 
U  1/ 2  e  dV and V   u  b dV   u  fdS u  P.
V
S
T  VT T


Recall e  B u then e    u B D B u and
1 T T

T 
T 
T 
   u B D Bu dV   u  b dV   u  fdS u  P
2V
V
S
and the variation becomes 
T T

T
T 
T 
   u B D BudV   u  b dV   u  fdS   u  P  0 .
V
S

  V
T
T T
Since DBu  De   , and e  u B
T 
T 
T 
T 
   e dV   u  b dV   u  fdS   u  P  0 ,
V
V
is the same as virtual work.
S
Finite Elements
• Assume displacements inside
an element is a linear (or quadratic)
function of the displacements of
the nodes of each element.
• Assume the function outside
each element is zero.
• Add the displacements of each
element to get the displacements
of the nodes.
Finite Elements (continued)
For example:
For element
For element
Hence
u1 
 1 1  
e
1:u1   L1 , L1 ,0 u2  x1  x  x2 otherwise, u1e  0
u 
 3
u1 

1 1  
e
2: u2  0, L , L 2  u2  x2  x  x3 otherwise, u2e  0

 u 
2
 3
 1 1 
   , ,0 
   L1 L1 
e 
0, 1 , 1 

L2 L2 

x1  x  x2  u1 
 
u2 
x2  x  x3 u3 

Interpolation Function
• Isoparametric elements have
same function for displacements
and coordinates
• Transform to an element with
coordinates at nodes
Node
x
h
1
-1
-1
2
1
-1
3
1
1
4
-1
1
Interpolation Function (continued)
• 2D isoparametric element
1
x  1  x 1  h x1  1  x 1  h x2  1  x 1  h x3  1  x 1  h x4 
4
1
y  1  x 1  h  y1  1  x 1  h  y2  1  x 1  h  y3  1  x 1  h  y4 
4
1
u  1  x 1  h u1  1  x 1  h u2  1  x 1  h u3  1  x 1  h u4 
4
1
v  1  x 1  h v1  1  x 1  h v2  1  x 1  h v3  1  x 1  h v4 
4
• 1D isoparametric element
1
1
1
1
x  1  x x1  1  x x2 , u  1  x u1  1  x u2
2
2
2
2
• 3D isoparametric element in a similar manner
Volume Integration
• For the parallelepiped the


the shaded area is dr2  dr3
 

• The volume is dr1  dr2  dr3 
- i.e. a determinant
• For a rectangular
parallelepiped dx  dy  dz
dx
• The volume is
  
dx  dy  dz   0
0
0 0 1 0 0
dy 0  0 1 0 dxdydz
0 dz 0 0 1
Volume Integration (continued)
• Using the isoparametric coordinates
x
x
x
dx 
dx 
dh 
d
x
h

• Use similar expressions for dy, dz
x
• The volume becomes
x
*Determinant is Jacobian y
x
of coordinates
z
x
x
h
y
h
z
h
x

y
dxdhd

z

Gaussian Quadrature
• Numerical integration is more efficient if both the
multiplying factor and the location of the integration
points are specified by the integration rule.
• The rule for integration along x can be expanded to
include y and z.
b
• For example, in one dimension, let    f x dx
a
• Transform to isoparametric coordinates
a b ba
1  x  1  x 
ba
x
a

b


x
and
dx

 




dx
2
 2   2 
 2 
 2 
Gaussian Quadrature (continued)
• Then the integral becomes
b  a 1 b  a b  a  


 f 
x dx
 2 1  2
 2  
• We can write the integral as
b a ba 
ba n

A
f

 k 
xk 
 2  k 1
 2  
 2
Gaussian Quadrature (continued)
n
1
2
3
k
1
1
2
1
2
3
Exact for polynomial
Ak x k
of degree
2
0
1
1  1/ 3
3
1 1/ 3
5/9  3 / 5
5
8/9 0
5/9 3 / 5
Governing Equations
• Use virtual work or energy minimization
• Sum over each element since each element
has no influence outside its boundary



• Let uk  NU k where U k is the vector of nodal
displacements for element k


• Then the strains are ek  B U k
Governing Equations (continued)
• Then





T
   Bk D Bk dV U k    b dV    fdV  P
elementsV
elementsS
elementsV

• Let
K
• And

T
B
 k D Bk dV
elem entsV

F
• Hence


 b dV 
elem entsV
 
KU  F
 
elem ents
S


fdV  P
Boundary Conditions
• Surface forces have been included
• So far the FE model is not restrained from rigid body
motion
• Hence displacement boundary conditions are needed
1 T  T 
• Recall   U KU  U F
2
 
• Let the displacement constraints be C U  A  0
• Develop an augmented energy
 
1  T   T  T
  U K U  U F   C U  A
2
Boundary Conditions (continued)
• The energy minimum

 0
U
• Leads to
K

 C
T 

C  U   F 
      
0      A 
Summary
• The governing equations are based on virtual work or the
minimization of energy.
• Displacement functions are zero outside elements.
• Isoparametric elements use the same function for the
coordinates and displacements inside elements
• A Jacobian is required to complete integrations over the
internal coordinates.
• Gaussian quadrature is typically used for integrations.
Hence stresses and strains are calculated at integration
points.
• Enforced displacements can be included through the use
of Lagrange multipliers.