Post on 13-Apr-2018
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 1/29
Beams and frames
• Beams are slender members used forsupporting transverse loading.
• Beams with cross sections symmetric with
respect to loading are considered.
EI dxvd
E
y
I
M
//
/
22 =
∈=
−=
σ
σ
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 2/29
Potential energy approach
Strain energy in an element of length dx is
∫∫
∫
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ =
=
A
A
A
I inertiaof moment theisdA y
dxdA y
EI
M
dAdxdU
2
2
2
2
2
1
2
1 σε
The total strain energy for the beam is given by-
( )∫= L
dxdxvd EI U 0
22 /2
1
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 3/29
( ) ∑∑∫∫
−−−=Πk
k k
m
mm
L L
v M v p pvdxdxdxvd EI '
00
22 /2
1
Potential energy of the beam is then given by-
Where-
-p is the distributed load per unit length-pm is the point load at point m.
-Mk is the moment of couple applied at point k
-vm
is the deflection at point m
-v’k is the slope at point k.
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 4/29
Galerkin’s Approach p
dx
V
V+dV
MM+dM
•Here we start from equilibrium
of an elemental length.
dV/dx = pdM/dx =V
EI M dxvd // 22 =
( ) 02
2
2
2=− p
dxvd EI
dx
d
( )∫ =Φ⎥⎦⎤
⎢⎣⎡ −
L
dx pdx
vd EI dx
d 0
2
2
20
For approximate solution by Galerkin’s approach-
Φ is an arbitrary function using same basic functions as v
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 5/29
Integrating the first term by parts and splitting the interval 0 to L
to (0 to xm), (xm to xk ) and (xk to L) we get-
02
2
0
2
2
2
2
0
2
2
00
2
2
2
2
=Φ
−Φ
−Φ⎟⎟
⎠
⎞⎜⎜
⎝
⎛ +
Φ⎟⎟ ⎠
⎞
⎜⎜⎝
⎛
+Φ−
Φ
∫∫ L
x
x L
x
xl L
k
k
m
m
dx
d
dx
vd EI
dx
d
dx
vd EI
dx
vd EI
dx
d
dx
vd
EI dx
d
dx pdxdx
d
dx
vd
EI
∑∑∫∫ =Φ−Φ−Φ−
Φ
k k k
mmm
L L
M pdx pdxdx
d
dx
vd
EI 0
'
002
2
2
2
Further simplifying-
Φ and M are zero at support..at xm shear force is pm and at xk
Bending moment is -Mk
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 6/29
•Beam is divided in to elements…each node has two degrees of
freedom.
•Degree of freedom of node j are Q2j-1 and Q2j
• Q2j-1 is transverse displacement and Q2j is slope or rotation.
FINITE ELEMENT FORMULATION
Q1Q3
Q5 Q7 Q9
Q2Q8Q6Q4 Q10
e1 e3e2 e4
[ ]T
QQQQQ 10321 ,, K
=
Q is the global displacement vector.
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 7/29
e
q1
q4q2
q3
Local coordinates-
],,,[
],,,[
'
22
'
11
4321
vvvv
qqqqq T
=
=
•The shape functions for interpolating v on an element are defined
in terms of ζ from –1 to 1.
•The shape functions for beam elements differ from those defined
earlier. Therefore, we define ‘Hermite Shape Functions’
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 8/29
H1
H4
H3
H2
1
1
Slope=0
Slope=0
Slope=0
Slope=0
Slope=0
Slope=1
Slope=1
Slope=0
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 9/29
4,3,2,132 =+++= id cba H iiiii Kζ ζ ζ
Each Hermite shape function is of cubic order represented by-
The condition given in following table must be satisfied.
10010000ζ=1
00001001ζ=-1
H’4H4H’3H3H’2H2H’1H1
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 10/29
( ) ( )
( ) ( )
( ) ( )
( ) ( )1141
214
1
114
1
214
1
24
2
3
2
2
2
1
−+=
++=
+−=
+−=
ζ ζ
ζ ζ
ζ ζ
ζ ζ
H
H
H
H
Finding out values of coefficients and simplifying,
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 11/29
Hermite functions can be used to write v in the form-
2
433
1
211)( ⎟⎟ ⎠
⎞⎜⎜⎝
⎛ ++⎟⎟
⎠
⎞⎜⎜⎝
⎛ +=
ζ ζ ζ
d
dv H v H
d
dv H v H v
The coordinates transform by relationship-
ζ
ζ
ζ ζ
d l
dx
x x x x
x x x
e
2
22
21
21
1221
21
=
−+
+=
++−=
(x2-x1) is the length of
element le
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 12/29
⎥
⎦
⎤⎢
⎣
⎡=
=
+++=
=
4321
44332211
2
,,
2
,
22)(
,
2
H l
H H l
H H
where
Hqv
q H l
q H q H l
q H v
Therefore
dx
dvl
d
dv
ee
ee
e
ζ
ζ
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 13/29
( )
qd
H d
d
H d
lq
dx
vd
equationaboveinngsubstituti
d
vd
ldx
vd
and d
dv
ldx
dv
dxdxvd EI U
T
e
T
ee
e
e
⎟⎟ ⎠
⎞⎜⎜⎝
⎛ ⎟⎟ ⎠
⎞⎜⎜⎝
⎛ =
==
= ∫
2
2
2
2
42
2
2
2
2
2
22
16
42
/2
1
ζ ζ
ζ ζ
⎥⎦
⎤⎢⎣
⎡ +−+−=⎟⎟
⎠
⎞⎜⎜⎝
⎛
22
31,
2
3,
22
31,
2
32
2
ee ll
d
H d ζ ζ
ζ ζ
ζ
Where-
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 14/29
∫−
⎥⎥⎥⎥
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎢⎢⎢⎢
⎣
⎡
⎟ ⎠
⎞⎜⎝
⎛ +
+−
+−+−
−⎟ ⎠
⎞⎜⎝
⎛ +−
+−+−
=1
1
2
2
2
22
2
2
22
3
4
31
)31(8349
16
91)31(
8
3
4
31
)31(8349)31(8349
8
2
1qd
l
lsymmetric
lll
ll
l
EI qU
e
e
eee
ee
e
T
e ζ
ζ
ζ ζ ζ
ζ ζ ζ
ζ
ζ ζ ζ ζ ζ ζ
qk qU
d d d
e
T
e2
1
203
21
1
1
1
1
1
2
=
=== ∫∫∫−−−
ζ ζ ζ ζ ζ
Note that-
This result can be written as-
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 15/29
⎥⎥⎥⎥
⎥
⎦
⎤
⎢⎢⎢⎢
⎢
⎣
⎡
−
−−−
−
−
=
22
22
3
4626
612612
2646
612612
eeee
ee
eeee
ee
e
e
llll
ll
llll
ll
l EI k
Where K e is element stiffness matrix given by
It can be seen that it is a symmetric matrix.
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 16/29
Load vector-
•We assume the uniformly distributed load p over the element.
q Hd
pl
pvdx
e
le ⎟⎟ ⎠
⎞
⎜⎜⎝
⎛ =
∫∫ −
1
12 ζ
Substituting the value of H we get-
T
eeeee
e
l
pl pl pl pl f
where
q f pvdxT
e
⎥⎦
⎤⎢⎣
⎡−=
=∫
12,
2,
12,
2
22
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 17/29
This is equivalent to the element shown below-
1 2le
p
21e
Ple/2 Ple/2
Ple2/12 -Ple2/12
The point loads Pm and Mk are readily taken care of by
introducing the nodes at the point of application.
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 18/29
Introducing local-global correspondence from potential energy
approach we get-
.
0
2
1
vector
nt displacemevirtualadmissiblewhere
F KQ
QF KQQ
T T
T
=Ψ
=Ψ−Ψ
−=Π
And from Galerkin’s approach we get-
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 19/29
BOUNDARY CONSIDERATIONS
•Let Qr = a…….single point BC
•Following Penalty approach, add 1/2C(Qr -a)2 to Π
•C represents stiffness which is large in comparison
with beam stiffness terms.
•C is added to K rr and Ca is added to Fr to get-
KQ = F
•These equations are solved to get nodal displacements.
CCa
Dof =(2i-1)
CCa
Dof = 2i
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 20/29
Shear Force and Bending Moment-
Hqvand
dx
dM V
dx
vd EI M ===
2
2
We have,
⎪⎪
⎭
⎪⎪
⎬
⎫
⎪⎪
⎩
⎪⎪
⎨
⎧
−
−
−
+⎪⎪
⎭
⎪⎪
⎬
⎫
⎪⎪
⎩
⎪⎪
⎨
⎧
⎥⎥⎥
⎥⎥
⎦
⎤
⎢⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
=⎪⎪
⎭
⎪⎪
⎬
⎫
⎪⎪
⎩
⎪⎪
⎨
⎧
12
2
12
2
4626
612612
2646
612612
2
2
4
3
2
1
22
22
3
4
3
2
1
e
e
e
e
eeee
ee
eeee
ee
e
pl
pl
pl
pl
q
q
q
q
llll
ll
llll
ll
l
EI
R
R
R
R
V1 = R 1 V2 = -R 3 M1 = -R 2 M2 = R 4
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 21/29
Beams on elastic support
•Shafts supported on ball, roller, journal bearings
•Large beams supported on elastic walls.
•Beam supported on soil (Winkler foundation).
•Stiffness of support contributes towards PE.
•Let ‘s’ be the stiffness of support per unit length.
additional term =
∑ ∫
∫
=
=
e e
T T
l
qdx H H sq
Hqv
dxsv
2
1
2
10
2
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 22/29
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
−−−
−−
=
= ∑
22
22
422313221561354
313422135422156
420
2
1
eeee
ee
eeee
ee
es
e
se
e
s
e
T
llllll
llll
ll
slk
foundationelastic
for matrixstiffnessisk where
k q
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 23/29
PLANE FRAMES
•Plane structure with rigidly connected members.
•Similar to beams except that axial loads and deformations
are present.
•We have 2 displacements and 1 rotation at each node.
•3 dof at each node.
q =[q1, q2, q3, q4, q5, q6]T
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 24/29
X
Yq1
q2
q4
q5
q3 (q’3)
X ’ Y ’
1
2
Global and local coordinate systems
θ
q’4q’5
q’1q’2
q6 (q’6)
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 25/29
q’ =[q’1, q’2, q’3, q’4, q’5, q’6]
l,m are the direction cosines of local coordinate system. X’Y’
• l = cos(θ)
• m =sin(θ)
We can see from the figure that-
•q3 = q’3
•q3
= q’6which are rotations with respect to body.
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 26/29
q’ = Lq
Where-
⎥⎥⎥⎥
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎢⎢⎢⎢
⎣
⎡
−
−
=
100000
00000000
000100
0000
0000
lmml
lm
ml
L
q’2, q’3,q’5 and q’6 are beam element dof while q’1andq’4
are like rod element dof.
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 27/29
Combining two stiffness and rearranging at proper locations we
get element stiffness matrix as-
⎥⎥⎥⎥
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎥
⎦
⎤
⎢⎢⎢⎢
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎢
⎣
⎡
−
−−−
−
−
−
−
=
eeee
eeee
ee
eeee
eeee
ee
e
l
EI
l
EI
l
EI
l
EI
l
EI
l
EI
l
EI
l
EI
l
EA
l
EAl
EI
l
EI
l
EI
l
EI
l
EI
l
EI
l
EI
l
EI
l
EA
l
EA
k
460
260
612
0
612
0
0000
260
460
6120
6120
0000
22
2323
22
2323
'
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 28/29
Lk LK
Lqk L
qk W
approachsGalerkinby
Lqk Lq
qk qU
eT e
eT T
eT
e
eT T
eT
e
'
'
'
'
'
''
,'2
1
''2
1
=
Ψ=
Ψ=
=
=
Strain energy of an element is given by-
Element stiffness matrix in global form can be written as-
7/27/2019 Herimite Shape Function for Beam
http://slidepdf.com/reader/full/herimite-shape-function-for-beam 29/29
X
Y
p
Y’
X’
If there is distributed load on
member, we have-
F KQ
formglobal In f L f
pl pl pl pl f
where
f Lq f q
T
T
eeee
T T T
=
=
⎥⎦
⎤⎢⎣
⎡−=
=
,'
12,
2,0,
12,
2,0'
'''
22