Download - Review 01x
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Summary of the Class before Exam1
Building a FEA Model Ingredients of a FEA software package
Steps in building a FEA model
Modeling considerations
1D Spring/Truss Elements Single 1D spring/truss element
Global stiffness matrix; properties
Boundary conditions
Shape functions and their properties
2D and 3D truss elements; rotation matrix
Beam elements 1D beam elements
2D beam elements
Distributed load
Frames Beam elements with bending and axial loads
3D beam elements with bending, axial loads, and torsion
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Building A FEA Model
Six basic ingredients of a FEA software package
1. Type of analysis
2. Geometry (defined through nodes)
3. Elements
4. Material properties5. Boundary conditions
6. Time functions
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Steps in building a FEA model
1. Discretization of the geometry into an equivalent system of finite elements with
associated nodes and choosing an appropriate element type to closely model the
actual physical behavior.2. Selection of an appropriate displacement (interpolation) function within each
element.
3. Defining the strain/displacement and stress/ strain relationships for deriving the
equations for each finite element
4. Derivation of the element stiffness matrix and equations. The nodal forces arerelated to the nodal displacements using the direct equilibrium method, or energy
methods or the weighted residual method.
5. Assembling the global stiffness matrix from the element stiffness matrices
6. The introduction of the boundary conditions.
7. Solving for the unknown degrees of freedom (or the generalized displacements)8. Solving for the element stresses and strains
9. Interpretation of the results for use in the design/analysis process
Building A FEA Model
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Building An FEA Model
Considerations in building a physical model
1. The nature of real-world problem2. The cost of conducting FEA
Considerations in building an FEA model
1. The accuracy of the FEA model
2. The cost of conducting FEA
The differences between a physical model and an FEA modelDeveloping a physical model is a process that simplifies a real-world problem into a
problem that is suitable for FEA.
Factors that affect the cost:
1. Degree of Freedoms (DOF): p
2. Number of Nodes : N
3. How the nodes are numbered
4. Number of Integration Points in Each Elements
5. Nonlinear Analysis
How to estimate the size (the memory need) of an FEA model?
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Element Stiffness Matrix and Global Stiffness Matrix
Shape funct ions, or interpolation functions:
approximation of displacement field
Nodal force-nodal displacement relationship
Matrix form of above relationship
Element stiffness matrix
Coordinate Transformation
[ ] ][][][ TKTK eTe =Global Stiffness Matrix
Direct stiffness matrix method
Write down global displacement vector
Use this vector as an index for DOFs
Airline tickets
Boundary Conditions
Coordinate rotation for skewed support
Cross out some terms
Solve global stif fness matrix
Solve nodal reaction force (external force)
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Spring/Truss Elements
1 2x
xd1 xf1 xd2 xf2
For the spring to be in equilibrium,
xxx dkdkf 212 +=
xxx dkdkf 211 =
=
x
x
x
x
dd
kkkk
ff
2
1
2
1
[ ] xe
x dkf = [ ]e
k stiffness matrix
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Spring/Truss Elements
1 2 x
1 2
x
xd1 xd2
T T
dx
du= ?=u
xL
dddu xxx
121
+= xx d
L
xd
L
xu 21
1 +
=
[ ]
=
x
x
d
dNNu
2
121
L
x
N
11 = L
x
N
2= Shape functions
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Beam Elements
Mechanics of Beams
2
2
xd
dvEIm=
EI
m
xd
dv=
2
2
Vxd
dm=
( )xV
xd
dvEI
3
3
=
)(
xwxd
dV= ( )xw
xd
dvEI
4
4
=
If ( ) 0 =xw
04
4
=xd
dvEI
m
V
+m
VIn beam theory
+ m
f
+m
f
+
In FEA
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Beam element
yd1
1
yd2
2
Beam Elements
Beam Elements ( )xVxddv
EI 3
3
= In general, ( ) 0 xV
432
23
1)( axaxaxaxv +++=
( ) 410 adxv y ===
( ) 310
axxd
dv===
( ) 322
12 23
aLaLaLxxd
dv++===
( ) 432
2
3
12 aLaLaLadLxv y +++===
Boundary conditions
Why we use a third order polynomial?
Interpolation function for deflection.
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Beam Elements: displacement function
( ) ( )212213112
++= LddLa yy
432
23
1)( axaxaxaxv +++=
( ) ( )212122 213
+= LddLa yy
13 =a yda 13=
[ ]{ }dNv= [ ] [ ]4321 NNNNN = { } [ ]Tyy ddd 2211 =In matrix form
( )32331 321 LLxxLN += ( )3223
32 21
LxLxLxLN +=
( )LxxL
N23
3332
1+= ( )223
34
1LxLx
LN =
Note: ( ) 101 ==xN ( ) 01 ==LxN ( ) 102 ==x
xddN ( ) 0
2 ==Lx
xddN
Beam Elements
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Beam Elements: Stress and Strain
( )2
2
xd
dvEIxm =
( )3
3
xd
dvEIxV =
Beam Elements
[ ]
=
2
2
1
1
2
2
1
1
y
y
e
y
y
d
d
K
m
fm
f
[ ]
=
22
22
3
4626612612
2646
612612
LLLL
LL
LLLL
LL
L
EIK
e
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Torsion in FEA
Consider torsion only
1 2
xm
1
xm
2
x1 x2
L
JL
dGM
=
( )xxxxL
GJmm 1221
==
=
x
x
x
x
L
GJmm
2
1
2
1
1111
x
Frames
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Consider shear, bending, and torsion
1 2
xm1 xm2
x1 x2
L
xzm1 z1zm2 z2
yf1
yd1 y
f2
yd2
=
x
x
x
x
L
GJ
m
m
2
1
2
1
11
11
=
z
y
z
y
z
y
z
y
d
d
LLLLLL
LLLL
LL
L
EI
m
f
m
f
2
2
1
1
22
22
3
2
2
1
1
4626612612
2646
612612
Frames
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Beam In 3D Space
Start from Local Coordinate
x
Consider bending in y zand , and torsion in x
Consider shear in y zand , and axial inx
yy
df22
,
zz df 22,
xx df 22 ,
zzm 22,
xxm 22,
yym 22,
Frames
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Frames
=
z
y
x
z
y
x
z
y
x
z
y
x
z
y
yy
zz
zzz
yyy
yyyy
zzzz
z
y
x
z
y
x
z
y
x
z
y
x
d
d
d
d
d
d
L
EIL
EIL
GJL
EI
L
EIL
EI
L
EIL
AEL
EI
L
EI
L
EIL
EI
L
EI
L
EIL
GJ
L
GJL
EI
L
EI
L
EI
L
EIL
EI
L
EI
L
EI
L
EIL
AE
L
AE
m
m
m
f
f
f
m
m
m
f
f
f
2
2
2
2
2
2
1
1
1
1
1
1
23
23
2
2
2323
2323
2
2
2
2
2
2
1
1
1
1
1
1
4
04
00
06
012
6000
12
00000
2
000
6
0
4
02
06
004
0000000
0
6
0
12
000
6
0
12
6000
120
6000
12
0000000000
Beam In 3D Space, Local Coordinate
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Shape Functions
Shape functions describe the shape of displacement field and are one of
the determinant factors in governing the efficiency and accuracy of FEA.
A shape function takes the value of ONE at the node associated with it,
and takes the value of ZERO at other nodes.
The choice of shape functions depends on:
The number of nodes in the element.
The function should be able to describe rigid body motion, and
constant strain.
L
xN
11 =
L
xN
2=Two nodes truss:
( )32331
321
LLxxL
N += ( )322332 21
LxLxLxL
N +=
( )LxxL
N23
3332
1+= ( )223
34
1LxLx
L
N =
Two nodes
beam bending:
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Coordinate Transformation
Rotation Matrix
=
y
x
y
x
f
f
f
f
1
1
1
1
cossin
sincos
or
=
=
y
x
y
x
y
x
f
f
f
f
f
f
1
1
1
1
1
1
cossin
sincos
cossin
sincos
[ ]{ }fTf = { } [ ] fTf 1=
[ ] [ ]TTT =1
x
y
x
xf1
yf1
xf1
yf1
y
Measured from tocounterclockwisex x
Measured from tocounterclockwise
xx
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=
y
x
y
x
y
x
y
x
f
f
ff
CS
SC
CSSC
f
f
ff
2
2
1
1
2
2
1
1
00
00
0000
=
y
x
y
x
y
x
y
x
d
d
dd
CS
SC
CSSC
d
d
dd
2
2
1
1
2
2
1
1
00
00
0000
xyzzyx
=
y
x
y
x
y
x
y
x
f
f
f
f
CS
SC
CS
SC
f
f
f
f
2
2
1
1
2
2
1
1
00
00
00
00
=
y
x
y
x
y
x
y
x
d
d
d
d
CS
SC
CS
SC
d
d
d
d
2
2
1
1
2
2
1
1
00
00
00
00
zyxxyz
Rotation Matrix
Coordinate Transformation
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Coordinate transformation matrix
x
y
z
xy
z
=
x
y
x
x
y
x
ff
f
ff
f
1
1
1
1
1
1
x
y
x y z
z
( )xx,cos ( )yx,cos ( )zx,cos
( )xy,cos ( )yy,cos ( )zy,cos( )xz,cos ( )yz,cos ( )zz,cos
Coordinate Transformation
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2D and 3D elements: Element stiffness matrix in global coordinate
{ } [ ] { }dTKTfe
T ][][=
[ ] ][][][ TKTK eT=
Coordinate Transformation
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Spring Elements
Global Stiffness matrix: direct stiffness matrix method
i j
Element I
)()(
)()(
Ijj
Iij
IijIii
kk
kk
[ ]
=K
1 i j N
1
i
j
N
)(I
ijk
)(Iijk
)(Iiik
)(I
jjk
N by N matrix
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Global Stiffness Matrix
Global Stiffness matrix: Properties
1. It is symmetric jiij KK =
2. The element on the diagonal of the matrix is always positive. 0>
iiK
3. The product of the i-th row of the global stiffness matrix and the
global displacement matrix gives the external force on the i-th DOF
of the system.
4. Kij is equal to the reaction force on the i-th DOF due to a unit displacement
on thej-th DOF whereas all the other DOFs are fixed.
5. The global stiffness matrix is singular.
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Boundary Conditions
Boundary Conditions
1 3 x
01 =xd
xF1
xd3
2
xd2
2
1k 2k
3P 2P
+
=
x
x
x
x
x
d
d
kkkk
kk
kk
F
F
F
3
2
2121
22
11
3
2
1
0
0
0
+
=
x
x
x
d
d
kkkk
kk
kk
P
P
F
3
2
2121
22
11
3
2
1
0
0
0X X X
X
X
How to find ?xF1 This method is applicable to other elements.
Why we need BC?
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Inclined or Skewed Supports
Sometimes, a boundary condition can be more conveniently applied in a local
coordinate.
In the local coordinate system: 03 =yd
In the global coordinate system:
x
y
x
y
1
2 3
=
y
x
y
x
dd
dd
3
3
3
3
cossinsincos
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[ ] [ ][ ] [ ] [ ] [ ]
[ ][ ] [ ][ ] [ ]
=
3
2
1
3333233
3232212
1211
3
2
1
d
d
d
TKTKT
TKKK
KK
F
F
F
T
T
Frames Inclined or Skewed Supports
=
3
3
3
2
2
2
1
1
1
3
3
3
2
2
2
1
1
1
y
x
y
x
y
x
y
x
y
x
y
x
d
d
d
d
d
d
XXXXXX
XXXXXX
XXXXXXXXXXXXXXX
XXXXXXXXX
XXXXXXXXXXXXXXX
XXXXXX
XXXXXX
M
F
FM
F
FM
F
F
=0
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( ) ( )= L
ddistribute xdxvxwW0
Beam Elements
Distributed loads: Work Equivalence Method
22221111 mdfmdfW yyyydiscrete +++=
( )= L
xdNxwm0
21( )=
L
y xdNxwf0
11
( )= L
y xdNxwf0
32 ( )=
L xdNxwm0
42
Note, x refers to local coordinate, it takes values between 0 and L.