the finite element method fem for beams a practical course chapter 3:
TRANSCRIPT
The FThe Finite Element inite Element MethodMethod
FEM FOR BEAMS
A Practical CourseA Practical Course
CHAPTER 3:
CONTENTSCONTENTS INTRODUCTION FEM EQUATIONS
– Shape functions construction– Strain matrix– Element matrices – Remarks
EXAMPLE AND CASE STUDY– Remarks
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.
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.
Shape functions constructionShape functions construction
Consider a beam element
Natural coordinate system: a
x
d1 = v1
d2 = 1
d3 = v2
d4 = 2
x = - a = 1
x = a = 1
x,
2a
0
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
Shape functions Shape functions constructionconstruction
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
d1 = v1
d2 = 1
d3 = v2
d4 = 2
x = - a = 1
x = a = 1
x,
2a
0
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
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
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)
Eq. (2-47)
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
1 1 1 2 1 3 1 4
1 2 1 2 2 2 3 2 4
3 13 1 3 2 3 3 3 4
4 1 4 2 4 3 4 4
dze
N N N N N N N N
N N N N N N N NEI
N N N N N N N Na
N N N N N N N N
k
2
22
3
4.
33
234
3333
2
asy
a
aaa
aa
a
EI zek
Evaluate integrals
Element Element matricesmatrices
1
1
1 1 1 2 1 3 1 4
1 2 1 2 2 2 3 2 4
13 1 3 2 3 3 3 4
4 1 4 2 4 3 4 4
d d d d
d
aT T Te a
V A
V A x A a
N N N N N N N N
N N N N N N N NAa
N N N N N N N N
N N N N N N N N
m N N N N N N
2
22
8.
2278
6138
13272278
105
asy
a
aaa
aa
Aae
m
Evaluate integrals
Element matricesElement matrices
2
2
111
1 112 3
1223
2423
d d dy
f
y
y ss
f a
ssT Te b s f y
y ssV S
f as
s
f a ffN
mmNf V f S f a
f a ffN
mN m
f N N
eeeee fdmdk
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.
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
EXAMPLEXAMPLEE
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
Exact solution:4
40y y
vEI f
x
2
( ) (3 )6 y
Pxv x L x
EI
0yf
P=1000 N
0.5 m
E=69 GPa =0.33
3
( )3 y
PLv x L
EI
= -3.355E10-4 m
Eq. (2.59)
EXAMPLEXAMPLEE
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 conditions1 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
P=1000 N
0.5 m
E=69 GPa =0.33
1 1 0v
EXAMPLEXAMPLEE Step 2 (Cont’d):
6 -23 0.753.974 10 Nm
0.75 0.25
K
Therefore, K d = F, where
dT = [ v2 2] , 1000
0
F
Step 3: Solving FE equation(Two simultaneous equations) v2 = -3.355 x 10-4 m
2 = -1.007 x 10-3 rad
EXAMPLEXAMPLEE
Step 4: Stress recovering
v2 = -3.355 x 10-4 m
2 = -1.007 x 10-3 rad
61 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
1 1
1 16
2 2
2 2
?3 0.75 3 0.75
?0.75 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
DK F
RemarksRemarks
FE solution is the same as analytical solution Analytical solution to beam is third order
polynomial (same as shape functions used) Reproduction property
CASE CASE STUDYSTUDY
Number 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
[ K M ]= 0 Section 3.6, pg. 58