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Finite Element Method - WordPress.com · Finite Element Method FEM FOR TRUSSES . 2 CONTENTS ......
Transcript of Finite Element Method - WordPress.com · Finite Element Method FEM FOR TRUSSES . 2 CONTENTS ......
1
Finite Element Method
FEM FOR
TRUSSES
2
CONTENTS
INTRODUCTION
FEM EQUATIONS
– Shape functions construction
– Strain matrix
– Element matrices in local coordinate system
– Element matrices in global coordinate system
– Boundary conditions
– Recovering stress and strain
EXAMPLE – Remarks
HIGHER ORDER ELEMENTS
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INTRODUCTION
Truss members are for the analysis of skeletal type systems – planar trusses and space trusses.
A truss element is a straight bar of an arbitrary cross-section, which can deform only in its axis direction when it is subjected to axial forces.
Truss elements are also termed as bar elements.
In planar trusses, there are two components in the x and y directions for the displacement as well as forces at a node.
For space trusses, there will be three components in the x, y and z directions for both displacement and forces at a node.
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INTRODUCTION
In trusses, the truss or bar members are joined
together by pins or hinges (not by welding), so
that there are only forces (not moments)
transmitted between bars.
It is assumed that the element has a uniform cross-
section.
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Example of a truss structure
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FEM EQUATIONS
Shape functions construction
Strain matrix
Element matrices in local coordinate system
Element matrices in global coordinate
system
Boundary conditions
Recovering stress and strain
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Shape functions construction
Consider a truss element
D3i - 1
D3i - 2
D3i
D3j - 1
D3j - 2
D3j
le
x
u1
u2
u(x)
fs1
fx
global node j
local node 2
global node i
local node 1
fs2
X
Y
Z
o
0
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Shape functions construction
0
0 1
1
( ) 1h T
T
u x x x
p α
pα
Let
Note: Number of terms of basis function, xn
determined by n = nd - 1
At x = 0, u(x=0) = u1
At x = le, u(x=le) = u2
1 0
2 1
1 0
1 e
u
lu
0 1
1 2
1 0
1 1
e e
u
ul l
1 2
1 1
2 2
( ) ( )
1 0
( ) 1 1 ( )1 1
( )
h T
e
e e
e eN x N x
e
u ux xu x x x
u ul ll l
x
P α N d
d
N
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Shape functions construction
)()()( 21 xNxNx N
1
2
( ) 1
( )
e
e
xN x
l
xN x
l
N1 N2
x
le 0
1 1
1 2
2 11 1 2 2 1( ) ( ) ( )
e
u uu x N x u N x u u x
l
(Linear element)
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Strain matrix
2 11 1 2 2 1( ) ( ) ( )
e
u uu x N x u N x u u x
l
2 1x
e
u uu
x l
or
eex Lx
uBdNd
1 11
e e e e
x xL
x l l l l
B Nwhere
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Element Matrices in the Local Coordinate
System
0
1
1 11 1d d
1 1 1
e
e
l eT
e
e e eV
e
l AEV A E x
l l l
l
k B cB
Note: ke is symmetrical
Proof: BcB][BcBB]cBTTTTTTT [
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Element Matrices in the Local Coordinate
System
1 1 1 2
02 1 2 2
2 1d d
1 26
e
e
lT e
e e
V
N N N N A lV A l x
N N N N
m N N
Note: me is symmetrical too
111
022
1
2d d d
2
e
e e
x es
l sT T
e b s x
s x eV Ss
f lf
fNf V f S f x
fN f lf
f N N
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Element matrices in global coordinate
system
Perform coordinate transformation
Truss in space (spatial truss) and truss in
plane (planar truss)
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Element matrices in global coordinate
system
Spatial truss
ee TDd (Relationship between local
DOFs and global DOFs)
where
eijijij
ijijij
nml
nml
000
000T
j
j
j
i
i
i
e
D
D
D
D
D
D
3
13
23
3
13
23
D,
cos( , )
cos( , )
cos( , )
j i
ij
e
j i
ij
e
j i
ij
e
X Xl x X
l
Y Ym x Y
l
Z Zn x Z
l
Direction cosines
(2x1)
(6x1)
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Element matrices in global coordinate
system
Spatial truss (Cont’d)
2 2 2( ) ( ) ( )e j i j i j il X X Y Y Z Z
Transformation applies to
force vector as well:
ee TFf where
j
j
j
i
i
i
e
F
F
F
F
F
F
3
13
23
3
13
23
F
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Element matrices in global coordinate
system
Spatial truss (Cont’d)
ee TDd
eeeee fdmdk eeeee fDTmTDk
e
T
ee
T
ee
TfTDTmTDTkT )()(
eeeee FDMDK
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Element matrices in global coordinate
system
Spatial truss (Cont’d)
2 2
2 2
2 2
2 2
2 2
T
e e
ij ij ij ij ij ij ij ij ij ij
ij ij ij ij ij ij ij ij ij ij
ij ij ij ij ij ij ij ij ij ij
ij ij ij ij ij ij ij ij ij ije
ij ij ij ij ij ij ij ij ij ij
ij
l l m l n l l m l n
l m m m n l m m m n
l n m n n l n m n nAE
l l m l n l l m l nl
l m m m n l m m m n
l n
K T k T
2 2
ij ij ij ij ij ij ij ij ijm n n l n m n n
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Element matrices in global coordinate
system
Spatial truss (Cont’d)
2 2
2 2
2 2
2 2
2 2
2 2 2
2 2 2
2 2 2
2 2 26
2 2 2
T
e e
ij ij ij ij ij ij ij ij ij ij
ij ij ij ij ij ij ij ij ij ij
ij ij ij ij ij ij ij ij ij ije
ij ij ij ij ij ij ij ij ij ij
ij ij ij ij ij ij ij ij ij ij
ij
l l m l n l l m l n
l m m m n l m m m n
l n m n n l n m n nA l
l l m l n l l m l n
l m m m n l m m m n
l n
M T m T
2 22 2 2ij ij ij ij ij ij ij ij ijm n n l n m n n
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Element matrices in global coordinate
system
Spatial truss (Cont’d)
1
1
1
1
1
1
( )2
( )2
( )2
( )2
( )2
( )2
x es ij
x es ij
x es ij
T
e ey e
s ij
y e
s ij
y e
s ij
f lf l
f lf m
f lf n
f lf l
f lf m
f lf n
F T f Note: 1
1
2
2
x es
e
x es
f lf
f lf
f
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Element matrices in global coordinate
system
Planar truss
ee TDd
where
ijij
ijij
ml
ml
00
00T ,
j
j
i
i
e
D
D
D
D
2
12
2
12
D
j
j
i
i
e
F
F
F
F
2
12
2
12
FSimilarly (4x1)
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Element matrices in global coordinate
system
Planar truss (Cont’d)
2 2
2 2
2 2
2 2
ij ij ij ij ij ij
ij ij ij ij ij ijT
e e
ij ij ij ij ij ije
ij ij ij ij ij ij
l l m l l m
l m m l m mAE
l l m l l ml
l m m l m m
K T k T
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Element matrices in global coordinate
system
Planar truss (Cont’d)
2 2
2 2
2 2
2 2
2 2
2 2
2 26
2 2
ij ij ij ij ij ij
ij ij ij ij ij ijT ee e
ij ij ij ij ij ij
ij ij ij ij ij ij
l l m l l m
l m m l m mA l
l l m l l m
l m m l m m
M T m T
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Boundary conditions
Singular K matrix rigid body movement
Constrained by supports
Impose boundary conditions cancellation
of rows and columns in stiffness matrix,
hence K becomes SPD
Recovering stress and strain
x e eE E Bd BTD (Hooke’s law)
x
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EXAMPLE
Consider a bar of uniform cross-sectional area shown in the figure. The
bar is fixed at one end and is subjected to a horizontal load of P at the
free end. The dimensions of the bar are shown in the figure and the
beam is made of an isotropic material with Young’s modulus E.
P
l
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EXAMPLE
Exact solution of 2
20x
uE f
x
( )
Pu x x
EA , stress:
x
P
A :
FEM:
(1 truss element) 1 1
1 1e
AE
l
K = k
1 1
2 2
?1 1
1 1
u FAE
u F Pl
1 1
2 2
?1 1
1 1
u FAE
u F Pl
2
Plu
AE
1
2
0
( ) ( ) 1 1e
ux x x x Pu x x xPl
ul l l l EAEA
N d
2
01 1x e
PE E
ul l A
Bd
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Remarks
FE approximation = exact solution in example
Exact solution for axial deformation is a first order
polynomial (same as shape functions used)
Hamilton’s principle – best possible solution
Reproduction property
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HIGHER ORDER ELEMENTS
1 2 3 4 1 2 3
Quadratic element Cubic element
1
2
3
1( ) (1 )
2
1( ) (1 )
2
( ) (1 )(1 )
N
N
N
2
1
2
2
2
3
2
4
1( ) (1 )(1 9 )
16
1( ) (1 )(1 9 )
16
9( ) (1 3 )(1 )
16
9( ) (1 3 )(1 )
16
N
N
N
N