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1
FFinite Element Methodinite Element Method
FEM FOR BEAMS
for readers of all backgroundsfor readers of all backgrounds
G. R. Liu and S. S. Quek
CHAPTER 5:
2Finite Element Method by G. R. Liu and S. S. Quek
CONTENTSCONTENTS INTRODUCTION FEM EQUATIONS
– Shape functions construction– Strain matrix– Element matrices – Remarks
EXAMPLE– Remarks
3Finite Element Method by G. R. Liu and S. S. Quek
INTRODUCTIONINTRODUCTION
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.
4Finite Element Method by G. R. Liu and S. S. Quek
INTRODUCTIONINTRODUCTION
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.
5Finite Element Method by G. R. Liu and S. S. Quek
FEM EQUATIONSFEM EQUATIONS
Shape functions constructionStrain matrixElement matrices
6Finite Element Method by G. R. Liu and S. S. Quek
Shape functions constructionShape functions construction
Consider a beam element d1 = v1
d2 = 1
d3 = v2
d4 = 2
x = - a = 1
x = a = 1
x,
2a
0
Natural coordinate system: a
x
7Finite Element Method by G. R. Liu and S. S. Quek
Shape functions constructionShape functions construction
33
2210)( vAssume that
In matrix form:
3
2
1
0
321)(
v or αp )()( Tv
)32(11 2
321
a
v
ax
v
x
v
8Finite Element Method by G. R. Liu and S. S. Quek
Shape functions constructionShape functions construction
1
11
(1) ( 1)
d(2)
d
v v
v
x
To obtain constant coefficients – four conditions
2
21
(3) (1)
d(4)
d
v v
v
x
At x= a or = 1
At x= a or = 1
9Finite Element Method by G. R. Liu and S. S. Quek
Shape functions constructionShape functions construction
4
3
2
1
321
321
2
2
1
1
0
1111
0
1111
aaa
aaa
v
v
or αAd ee
aa
aa
aa
aa
e
11
00
33
22
4
11A or ee dAα 1
10Finite Element Method by G. R. Liu and S. S. Quek
Shape functions constructionShape functions construction
Therefore,
ev dN )(
where
)()()()()( 43211 NNNNe PAN
)1()(
)32()(
)1()(
)32()(
3244
341
3
3242
341
1
a
a
N
N
N
Nin which
11Finite Element Method by G. R. Liu and S. S. Quek
Strain matrixStrain matrix
yLvx
vy
x
uxx
2
2
Therefore,
NNNNB
22
2
22
2
a
y
a
y
xyyL
4321 NNNN Nwhere
)31(2
,2
3
)31(2
,2
3
43
21
aN N
a
N N(Second derivative of shape functions)
12Finite Element Method by G. R. Liu and S. S. Quek
Element matricesElement matrices
dd][][1
d)()(dd
1
132
2
2
2
4
1
1
2
2
2
22
NNNN
NNBcBk
TzTz
Ta
aA
T
V
e
a
EIa
aEI
xxx
AyEV
x
NNNNNNNN
NNNNNNNN
NNNNNNNN
NNNNNNNN
a
EI ze d
41342414
41332313
42322212
41312111
1
13
k
2
22
3
4.
33
234
3333
2
asy
a
aaa
aa
a
EI zek
Evaluate integrals
13Finite Element Method by G. R. Liu and S. S. Quek
Element matricesElement matrices
x
NNNNNNNN
NNNNNNNN
NNNNNNNN
NNNNNNNN
Aa
aAxAV TTa
aA
T
V
e
d
dddd
44342414
43332313
42322212
41312111
1
1
1
1
NNNNNNm
2
22
8.
2278
6138
13272278
105
asy
a
aaa
aa
Aae
m
Evaluate integrals
14Finite Element Method by G. R. Liu and S. S. Quek
Element matricesElement matrices
13
2
13
1
1
2
1
1
1
1
4
3
2
1
2
2
ddd
s
af
sy
s
af
sy
s
s
s
s
yf
S
sT
V
bT
e
m
faf
m
faf
m
f
m
f
N
N
N
N
afSfVf
y
y
f
NNf
eeeee fdmdk
15Finite Element Method by G. R. Liu and S. S. Quek
RemarksRemarks
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.
16Finite Element Method by G. R. Liu and S. S. Quek
EXAMPLEEXAMPLEConsider 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
0.5 m
0.06 m
0.1 m
E=69 GPa =0.33
17Finite Element Method by G. R. Liu and S. S. Quek
EXAMPLEEXAMPLE Step 1: Element matrices
33 6 41 10.1 0.06 1.8 10 m
12 12zI bh
9 6
3
6 -2
3 0.75 3 0.7569 10 1.8 10 0.75 0.25 0.75 0.125
3 0.75 3 0.752 0.25
0.75 0.125 0.75 0.25
3 0.75 3 0.75
0.75 0.25 0.75 0.1253,974 10 Nm
3 0.75 3 0.75
0.75 0.125 0.75 0.25
e
K k
18Finite Element Method by G. R. Liu and S. S. Quek
EXAMPLEEXAMPLE
Step 1 (Cont’d):
1 1
1 16
2 2
2 2
?3 0.75 3 0.75 unknown reaction shear force
?0.75 0.25 0.75 0.125 unknow3,974 10
3 0.75 3 0.75
00.75 0.125 0.75 0.25
v Q
M
v Q P
M
DK F
n reaction moment
Step 2: Boundary conditions 1 1 0v
1 1
1 16
2 2
2 2
03 0.75 3 0.75
00.75 0.25 0.75 0.1253.974 10
3 0.75 3 0.75
00.75 0.125 0.75 0.25
v Q
M
v Q P
M
19Finite Element Method by G. R. Liu and S. S. Quek
EXAMPLEEXAMPLE Step 2 (Cont’d):
6 -23 0.753.974 10 Nm
0.75 0.25
K
Therefore, Kd=F where
dT = [ v2 2] , 1000
0
F
20Finite Element Method by G. R. Liu and S. S. Quek
EXAMPLEEXAMPLE
Step 3: Solving FE equation– Two simultaneous equations
v2 = -3.355 x 10-4 m
2 = -1.007 x 10-3 rad 6
1 2 2
6 4 3
3.974 10 ( 3 0.75 )
3.974 10 [ 3 ( 3.355 10 ) 0.75 ( 1.007 10 )]
998.47N
Q v
61 2 2
6 4 3
3.974 10 ( 0.75 0.125 )
3.974 10 [ 0.75 ( 3.355 10 ) 0.125 ( 1.007 10 )]
499.73Nm
M v
Substitute back into first two equations of Kd=F
21Finite Element Method by G. R. Liu and S. S. Quek
RemarksRemarks
FE solution is the same as analytical solution Analytical solution to beam is third order
polynomial (same as shape functions used) Reproduction property
22Finite Element Method by G. R. Liu and S. S. Quek
CASE STUDYCASE STUDY
Resonant frequencies of micro resonant transducer
Membrane
Bridge
23Finite Element Method by G. R. Liu and S. S. Quek
CASE STUDYCASE STUDYNumber of 2-
node beam elements
Natural Frequency (Hz)
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
24Finite Element Method by G. R. Liu and S. S. Quek
CASE STUDYCASE STUDY
Mode 1 (0.44 MHz)
0
0.2
0.4
0.6
0.8
1
1.2
0 20 40 60 80 100
x (um)
Dy
(um
)
25Finite Element Method by G. R. Liu and S. S. Quek
CASE STUDYCASE STUDYMode 2 (1.21MHz)
-1.5
-1
-0.5
0
0.5
1
1.5
0 20 40 60 80 100
x (um)
Dy
(um
)
26Finite Element Method by G. R. Liu and S. S. Quek
CASE STUDYCASE STUDY
Mode 3 (2.38 MHz)
-1.5
-1
-0.5
0
0.5
1
1.5
0 20 40 60 80 100
x (um)
Dy
(um
)