Beams and frames

sublimefrontΠολεοδομικά Έργα

15 Νοε 2013 (πριν από 3 χρόνια και 6 μήνες)

188 εμφανίσεις

Beams and frames
•Beams are slender members used for
supporting transverse loading.
•Beams with cross sections symmetric with
respect to loading are considered.
E
I
M
dxvd
E
y
I
M
//
/
22
=
∈=
−=
σ
σ
Potential energy approach
Strain energy in an element of length dx is











=
=
A
A
A
IinertiaofmomenttheisdAy
dxdAy
EI
M
dAdxdU
2
2
2
2
2
1
2
1
σε
The total strain energy for the beam is given by-
()

=
L
dxdxvdEIU
0
22
/
2
1
()
∑∑
∫∫
−−−=Π
k
kk
m
mm
LL
vMvppvdxdxdxvdEI
'
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.
Galerkin’s Approach
p
dx
V
V+dV
M
M+dM
•Here we start from equilibrium
of an elemental length.
dV/dx = p
dM/dx =V
EIMdxvd//
22
=
(
)
0
2
2
2
2
=−p
dx
vd
EI
dx
d
(
)









L
dxp
dx
vd
EI
dx
d
0
2
2
2
0
For approximate solution by Galerkin’s approach-
Φis an arbitrary function using same basic functions as v
Integrating the first term by parts and splitting the interval 0to L
to (0 to xm), (xm
to xk) and (xk
to L) we get-
0
2
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
x
lL
k
k
m
m
dx
d
dx
vd
EI
dx
d
dx
vd
EI
dx
vd
EI
dx
d
dx
vd
EI
dx
d
dxpdx
dx
d
dx
vd
EI
∑∑
∫∫
=Φ−Φ−Φ−
Φ
k
kk
m
mm
LL
Mpdxpdx
dx
d
dx
vd
EI0
'
00
2
2
2
2
Further simplifying-
Φand M are zero at support..at xm
shear force is pm
and at xk
Bending moment is -Mk
•Beam is divided in to elements…each node has two degrees of
freedom.
•Degree of freedom of node j are Q2j-1
and Q
2j
•Q
2j-1
is transverse displacement and Q2j
is slope or rotation.
FINITE ELEMENT FORMULATION
Q1
Q3
Q5
Q7
Q9
Q2
Q8
Q6
Q4
Q10
e1
e3
e2
e4
[
]
T
QQQQQ
10321
,,K=
Q is the global displacement vector.
e
q1
q4
q2
q3
Local coordinates-
],,,[
],,,[
'
22
'
11
4321
vvvv
qqqqq
T
=
=
•The shape functions for interpolating v on an element are defined
in terms of ζfrom –1 to 1.
•The shape functions for beam elements differ from those defined
earlier. Therefore, we define ‘Hermite Shape Functions’
H1
H4
H3
H2
1
1
Slope=0
Slope=0
Slope=0
Slope=0
Slope=0
Slope=1
Slope=1
Slope=0
4,3,2,1
32
=+++=idcbaH
iiiii
K
ζζζ
Each Hermite shape function is of cubic order represented by-
The condition given in following table must be satisfied.
10010000ζ=1
00001001ζ=-1
H’4
H4
H’3
H3
H’2
H2
H’1
H1
()()
()()
()()
()()
11
4
1
21
4
1
11
4
1
21
4
1
2
4
2
3
2
2
2
1
−+=
++=
+−=
+−=
ζζ
ζζ
ζζ
ζζ
H
H
H
H
Finding out values of coefficients and simplifying,
Hermite functions can be used to write v in the form-
2
433
1
211
)(








++








+=
ζζ
ζ
d
dv
HvH
d
dv
HvHv
The coordinates transform by relationship-
ζ
ζ
ζ
ζ
d
l
dx
xxxx
xxx
e
2
22
2
1
2
1
1221
21
=

+
+
=
+
+

=
(x2-x1) is the length of
element le






=
=
+++=
=
4321
44332211
2
,,
2
,
22
)(
,
2
H
l
HH
l
HH
where
Hqv
qH
l
qHqH
l
qHv
Therefore
dx
dv
l
d
dv
ee
ee
e
ζ
ζ
()
q
d
Hd
d
Hd
l
q
dx
vd
equationaboveinngsubstituti
d
vd
ldx
vd
and
d
dv
ldx
dv
dxdxvdEIU
T
e
T
ee
e
e
















=
==
=

2
2
2
2
42
2
2
2
2
2
22
16
42
/
2
1
ζζ
ζζ






+−+−
=








22
31
,
2
3
,
22
31
,
2
3
2
2
ee
ll
d
Hd
ζ
ζ
ζ
ζ
ζ
Where-




























+
+−
+−
+−







+−
+−+−
=
1
1
2
2
2
2
2
2
2
22
3
4
31
)31(8349
16
91
)31(
8
3
4
31
)31(8349)31(8349
8
2
1
qd
l
lsymmetric
lll
ll
l
EI
qU
e
e
eee
ee
e
T
e
ζ
ζ
ζζζ
ζ
ζζ
ζ
ζζζζζζ
qkqU
ddd
e
T
e
2
1
20
3
2
1
1
1
1
1
1
2
=
===
∫∫∫
−−−
ζζζζζ
Note that-
This result can be written as-















−−−


=
22
22
3
4626
612612
2646
612612
eeee
ee
eeee
ee
e
e
llll
ll
llll
ll
l
EI
k
WhereK
e
is element stiffness matrix given by
It can be seen that it is a symmetric matrix.
Load vector-
•We assume the uniformly distributed load p over the element.
qHd
pl
pvdx
e
le








=
∫∫

1
1
2
ζ
Substituting the value of H we get-
T
eeee
e
e
l
plplplpl
f
where
qfpvdx
T
e






−=
=

12
,
2
,
12
,
2
22
This is equivalent to the element shown below-
1
2
le
p
21
e
Ple/2
Ple/2
Ple
2/12
-Ple
2/12
The point loads P
m
and Mk
are readily taken care of by
introducing the nodes at the point of application.
Introducing local-global correspondence from potential energy
approach we get-
.
0
2
1
vector
ntdisplacemevirtualadmissiblewhere
FKQ
QFKQQ
TT
T

=Ψ−Ψ
−=Π
And from Galerkin’s approach we get-
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 Krr
and Ca is added to Fr
to get-
KQ = F
•These equations are solved to get nodal displacements.
C
Ca
Dof =(2i-1)
C
Ca
Dof = 2i
Shear Force and Bending Moment-
Hqvand
dx
dM
V
dx
vd
EIM===
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
= R1
V2
= -R3
M1
= -R2
M2
= R4
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
TT
l
qdxHHsq
Hqv
dxsv
2
1
2
1
0
2














−−−


=
=

22
22
422313
221561354
313422
135422156
420
2
1
eeee
ee
eeee
ee
e
s
e
s
e
e
s
e
T
llll
ll
llll
ll
sl
k
foundationelastic
formatrixstiffnessiskwhere
kq
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, q
5, q6
]T
X
Y
q1
q2
q4
q5
q3 (q’3)
X

Y

1
2
Global and local coordinate systems
θ
q’4
q’5
q’1
q’2
q6 (q’6)
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’6
which are rotations with respect to body.
q’ = Lq
Where-






















=
100000
0000
0000
000100
0000
0000
lm
ml
lm
ml
L
q’2, q’3,q’5 and q’6 are beam element dof whileq’
1andq’4
are like rod element dof.
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
EA
l
EI
l
EI
l
EI
l
EI
l
EI
l
EI
l
EI
l
EI
l
EA
l
EA
k
46
0
26
0
612
0
612
0
0000
26
0
46
0
612
0
612
0
0000
22
2323
22
2323
'
LkLK
LqkL
qkW
approachsGalerkinby
LqkLq
qkqU
eTe
eTT
eT
e
eTT
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-
X
Y
p
Y’
X’
If there is distributed load on
member, we have-
FKQ
formglobalIn
fLf
plplplpl
f
where
fLqfq
T
T
eeee
TTT
=
=






−=
=
,
'
12
,
2
,0,
12
,
2
,0'
'''
22