Finite Element Method
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Transcript Finite Element Method
The Finite Element Method
A Practical Course
CHAPTER 3:
FEM FOR BEAMS
CONTENTS
INTRODUCTION
FEM EQUATIONS
– Shape functions construction
– Strain matrix
– Element matrices
– Remarks
EXAMPLE AND CASE STUDY
– Remarks
INTRODUCTION
The element developed is often known as
a beam element.
A beam element is a straight bar of an
arbitrary cross-section.
Beams are subjected to transverse forces
and moments.
Deform
only
in
the
directions
perpendicular to its axis of the beam.
INTRODUCTION
In beam structures, the beams are joined
together by welding (not by pins or
hinges).
Uniform cross-section is assumed.
FE matrices for beams with varying
cross-sectional area can also be developed
without difficulty.
FEM EQUATIONS
Shape functions construction
Strain matrix
Element matrices
Shape functions construction
Consider a beam element
d1 = v1
d3 = v2
d4 = 2
d2 = 1
x,
0
2a
x= - a
x= a
= 1
=1
x
Natural coordinate system:
a
Shape functions construction
Assume that
v( ) 0 1 2 3
In matrix form: v( ) 1
2
2
0
3 1
2
3
or
3
v( ) p T ( )α
v v 1 v 1
(1 2 2 3 3 2 )
x x a a
Shape functions construction
d1 = v1
d3 = v2
d4 = 2
d2 = 1
x,
0
2a
x= - a
x= a
= 1
=1
To obtain constant coefficients – four conditions
(1)
v(1) v1
(2)
dv
1
dx 1
(3)
v(1) v2
(4)
dv
2
dx 1
At x= a or = 1
At x= a or = 1
Shape functions construction
v1 1 1
0 1
1
a
v2 1 1
2 0 1a
1
2
a
1
2
a
1 1
3
a 2
or d e A eα
1 3
3
a 4
a
2 a
2
3 a 3 a
1
A e1
a
4 0 a 0
a 1 a
1
or α A e1d e
Shape functions construction
Therefore,
v N( )d e
where
N( ) PA e1 N1 ( ) N 2 ( ) N 3 ( ) N 4 ( )
in which
N1 ( ) 14 (2 3 3 )
N 2 ( ) a4 (1 2 3 )
N 3 ( ) 14 (2 3 3 )
N 4 ( ) a4 (1 2 3 )
Strain matrix
u
2v
xx
y
yLv
2
x
x
Eq. (2-47)
Therefore,
2
y 2
y
B yLN y 2 N 2
N
N
2
2
x
a
a
where N N1
N 2
3
a
N1 , N 2 (1 3 )
2
2
3
a
N 3 , N 4 (1 3 )
2
2
N 3
N 4
(Second derivative of
shape functions)
Element matrices
k e B cBdV E y dA
T
2
a
a
V
A
1
EI z
1
EI
k e 3z
a
1
1
2
2
T
( 2 N) ( 2 N)dx
x
x
2
EI z
1 2
T
[ 2 N] [ 2 N]ad 3
4
a
a
N1N1
N N
2 1
N3N1
N 4N1
N1N 2 N1N3
N 2N 2 N 2N 3
N3N 2 N3N3
N 4N 2 N 4N 3
Evaluate
integrals
1
1
N T N d
N1N 4
N 2N 4
d
N 3N 4
N 4N 4
3 3a
4a 2
EI z
ke 3
2a
sy.
3 3a
3a 2a 2
3
3a
4a 2
Element matrices
m e N NdV dA
T
a
a
V
N Ndx A
T
1
1
NT Nad
A
Aa
1
1
N1 N1
N N
2 1
N 3 N1
N 4 N1
N1 N 2
N1 N 3
N2 N2
N3 N 2
N4 N2
N 2 N3
N3 N3
N 4 N3
Evaluate
integrals
N1 N 4
N 2 N 4
d
N3 N 4
N 4 N 4
78 22a 27 13a
8a 2 13a 6a 2
Aa
me
78 22a
105
2
sy
.
8
a
Element matrices
f s1 f y a f s1
N1
f y a2
1 N2
ms1 3 ms1
T
T
fe N fb dV N f s dS f f y a
d
1 N
f
a
f
f
s2
3
V
Sf
s2 y
f y a2
m
N
4
s 2 3 ms 2
f
k ede med
e
e
Remarks
Theoretically, coordinate transformation can also be used
to transform the beam element matrices from the local
coordinate system to the global coordinate system.
The transformation is necessary only if there is more than
one beam element in the beam structure, and of which
there are at least two beam elements of different
orientations.
A beam structure with at least two beam elements of
different orientations is termed a frame or framework.
EXAMPLE
Consider the cantilever beam as shown in the figure. The beam is fixed
at one end and it has a uniform cross-sectional area as shown. The beam
undergoes static deflection by a downward load of P=1000N applied at
the free end. The dimensions and properties of the beam are shown in
the figure.
P=1000 N
0.1 m
E=69 GPa
=0.33
0.5 m
0.06
m
EXAMPLE
Exact solution:
4v
EI y 4 f y 0
x
fy 0
Eq. (2.59)
Step 1: Element matrices
1 3 1
3
I z bh 0.1 0.06 1.8 106 m 4
12
12
69 10 1.8 10
9
K ke
6
2 0.253
0.75
3
0.75 0.25
3.974 106
3 0.75
0.75 0.125
0.75
3
0.75
3
0.75 0.25 0.75 0.125
3 0.75
3
0.75
0.75
0.125
0.75
0.25
3
0.75
0.75 0.125
Nm -2
3
0.75
0.75 0.25
Px 2
v( x)
(3L x)
6 EI y
PL3
v( x L)
3EI y
= -3.355E10-4 m
P=1000 N
E=69 GPa
=0.33
0.5 m
P=1000 N
EXAMPLE
v1 1 0
E=69 GPa
Step 1 (Cont’d):
=0.33
0.5 m
0.75
3
0.75 v1 Q1 ? unknown reaction shear force
3
0.25 0.75 0.125 1 M 1 ? unknown reaction moment
6 0.75
3.974 10
3 0.75
3
0.75 v2 Q2 P
0.75
0.125
0.75
0.25
2 M 2 0
D
K
F
Step 2: Boundary conditions
v1 1 0
0.75
3
0.75 v1 0 Q1
3
0.75 0.25 0.75 0.125 0 M
1
1
3.974 106
3 0.75
3
0.75 v2 Q2 P
0.75
0.125
0.75
0.25
2 M 2 0
EXAMPLE
Step 2 (Cont’d):
0.75
3
-2
K 3.974 10
Nm
0.75
0.25
6
Therefore, K d = F,
dT
where
1000
= [ v2 2] , F
0
Step 3: Solving FE equation
(Two simultaneous equations)
v2 = -3.355 x 10-4 m
2 = -1.007 x 10-3 rad
EXAMPLE
0.75
3
0.75 v1 Q1 ?
3
0.25 0.75 0.125 1 M 1 ?
6 0.75
3.974 10
3 0.75
3
0.75 v2 Q2 P
0.75
0.125
0.75
0.25
2 M 2 0
D
K
F
Step 4: Stress recovering
v2 = -3.355 x 10-4 m
2 = -1.007 x 10-3 rad
Q1 3.974 106 (3v2 0.75 2 )
3.974 106 [3 (3.355 104 ) 0.75 (1.007 103 )]
998.47N
M 1 3.974 106 (0.75v2 0.125 2 )
3.974 106 [0.75 (3.355 104 ) 0.125 (1.007 103 )]
499.73Nm
Substitute
back into
first two
equations
Remarks
FE solution is the same as analytical solution
Analytical solution to beam is third order
polynomial (same as shape functions used)
Reproduction property
CASE STUDY
Resonant frequencies of micro resonant transducer
Bridge
Membrane
CASE STUDY
Natural Frequency (Hz)
Number of 2node beam
elements
Mode 1
Mode 2
Mode 3
10
4.4058 x 105
1.2148 x 106
2.3832 x 106
20
4.4057 x 105
1.2145 x 106
2.3809 x 106
40
4.4056 x 105
1.2144 x 106
2.3808 x 106
60
4.4056 x 105
1.2144 x 106
2.3808 x 106
Analytical
Calculations
4.4051 x 105
1.2143 x 106
2.3805 x 106
[ K lM ]f = 0
Section 3.6, pg. 58
CASE STUDY
Mode 1 (0.44 MHz)
1.2
Dy (um)
1
0.8
0.6
0.4
0.2
0
0
20
40
60
x (um)
80
100
CASE STUDY
Mode 2 (1.21MHz)
1.5
Dy (um)
1
0.5
0
-0.5
0
20
40
60
-1
-1.5
x (um)
80
100
CASE STUDY
Mode 3 (2.38 MHz)
1.5
Dy (um)
1
0.5
0
-0.5
0
20
40
60
-1
-1.5
x (um)
80
100