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Beams and frames
• Beams are slender members used for supporting transverse loading.
• Beams with cross sections symmetric with respect to loading are considered.
EIMdxvdE
yI
M
///
22 =
∈=
−=
σ
σ
Potential energy approach
Strain energy in an element of length dx is
∫
∫
∫
⎟⎟⎠
⎞⎜⎜⎝
⎛=
=
A
A
A
IinertiaofmomenttheisdAy
dxdAyEIM
dAdxdU
2
22
2
21
21 σε
The total strain energy for the beam is given by-
( )∫=L
dxdxvdEIU0
22 /21
( ) ∑∑∫∫ −−−=Πk
kkm
mm
LL
vMvppvdxdxdxvdEI '
00
22 /21
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 Approachp
dx
V
V+dV
M M+dM
•Here we start from equilibriumof an elemental length.
dV/dx = pdM/dx =V
EIMdxvd // 22 =
( ) 022
2
2
=− pdxvdEI
dxd
( )∫ =Φ⎥⎦⎤
⎢⎣⎡ −
Ldxpdx
vdEIdxd
0 22
2 0
For approximate solution by Galerkin’s approach-
Φ is an arbitrary function using same basic functions as v
Integrating the first term by parts and splitting the interval 0 to Lto (0 to xm), (xm to xk) and (xk to L) we get-
02
2
02
2
2
2
02
2
002
2
2
2
=Φ
−Φ
−Φ⎟⎟⎠
⎞⎜⎜⎝
⎛+
Φ⎟⎟⎠
⎞⎜⎜⎝
⎛+Φ−
Φ∫∫
L
x
xL
x
xlL
k
k
m
m
dxd
dxvdEI
dxd
dxvdEI
dxvdEI
dxd
dxvdEI
dxddxpdx
dxd
dxvdEI
∑∑∫∫ =Φ−Φ−Φ−Φ
kkk
mmm
LL
Mpdxpdxdxd
dxvdEI 0'
002
2
2
2
Further simplifying-
Φ and M are zero at support..at xm shear force is pm and at xkBending moment is -Mk
•Beam is divided in to elements…each node has two degrees offreedom.•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
Q1 Q3 Q5 Q7 Q9
Q2 Q8Q6Q4 Q10
e1 e3e2 e4
[ ]TQQQQQ 10321 ,, K=
Q is the global displacement vector.
e
q1
q4q2
q3
Local coordinates-
],,,[
],,,['22
'11
4321
vvvv
qqqqq T
=
=
•The shape functions for interpolating v on an element are definedin terms of ζ from –1 to 1.•The shape functions for beam elements differ from those definedearlier. Therefore, we define ‘Hermite Shape Functions’
H1
H4
H3
H21
1
Slope=0
Slope=0
Slope=0
Slope=0
Slope=0
Slope=1
Slope=1
Slope=0
4,3,2,132 =+++= idcbaH iiiii Kζζζ
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
( ) ( )
( ) ( )
( ) ( )
( ) ( )1141
2141
1141
2141
24
23
22
21
−+=
++=
+−=
+−=
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-
2433
1211)( ⎟⎟
⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
ζζζ
ddvHvH
ddvHvHv
The coordinates transform by relationship-
ζ
ζ
ζζ
dldx
xxxx
xxx
e
2
22
21
21
1221
21
=
−+
+=
++
−=
(x2-x1) is the length of element le
⎥⎦⎤
⎢⎣⎡=
=
+++=
=
4321
44332211
2,,
2,
22)(
,2
HlHHlHH
whereHqv
qHlqHqHlqHv
Thereforedxdvl
ddv
ee
ee
e
ζ
ζ
( )
qd
Hdd
Hdl
qdx
vd
equationaboveinngsubstitutid
vdldx
vdandddv
ldxdv
dxdxvdEIU
T
e
T
ee
ee
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
==
= ∫
2
2
2
2
42
2
2
2
2
2
22
16
42
/21
ζζ
ζζ
⎥⎦⎤
⎢⎣⎡ +−+−
=⎟⎟⎠
⎞⎜⎜⎝
⎛22
31,23,
2231,
23
2
2ee ll
dHd ζζζζζ
Where-
∫−
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎟⎠⎞
⎜⎝⎛ +
+−
+−+−
−⎟⎠⎞
⎜⎝⎛ +−
+−+−
=1
1
22
2
22
22
22
3
431
)31(834916
91)31(83
431
)31(8349)31(8349
821 qd
l
lsymmetric
lll
ll
lEIqU
e
e
eee
ee
e
Te ζ
ζ
ζζζ
ζζζζ
ζζζζζζ
qkqU
ddd
eT
e 21
2032 1
1
1
1
1
1
2
=
=== ∫∫∫−−−
ζζζζζ
Note that-
This result can be written as-
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
−
−−−−
−
=
22
22
3
4626
6126122646
612612
eeee
ee
eeee
ee
ee
llll
llllll
ll
lEIk
Where Ke 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.
qHdplpvdx e
le⎟⎟⎠
⎞⎜⎜⎝
⎛= ∫∫
−
1
12ζ
Substituting the value of H we get-
T
eeeee
e
l
plplplplf
where
qfpvdxT
e
⎥⎦
⎤⎢⎣
⎡−=
=∫
12,
2,
12,
2
22
This is equivalent to the element shown below-
1 2le
p
21 e
Ple/2 Ple/2Ple
2/12 -Ple2/12
The point loads Pm and Mk are readily taken care of byintroducing the nodes at the point of application.
Introducing local-global correspondence from potential energy approach we get-
.
0
21
vectorntdisplacemevirtualadmissiblewhere
FKQ
QFKQQ
TT
T
=Ψ=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.
CCa
Dof =(2i-1)
CCa
Dof = 2i
Shear Force and Bending Moment-
HqvanddxdMV
dxvdEIM === 2
2
We have,
⎪⎪
⎭
⎪⎪
⎬
⎫
⎪⎪
⎩
⎪⎪
⎨
⎧
−−
−
+
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
−
−−−−
−
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
12
212
2
4626
6126122646
612612
2
2
4
3
2
1
22
22
3
4
3
2
1
e
e
e
e
eeee
ee
eeee
ee
e
pl
plpl
pl
qqqq
llll
llllll
ll
lEI
RRRR
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
21
21
0
2
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
−−−
−
−
=
= ∑
22
22
422313
221561354313422
135422156
420
21
eeee
ee
eeee
ee
ese
se
e
se
T
llll
llllll
ll
slk
foundationelasticformatrixstiffnessiskwhere
kq
PLANE FRAMES•Plane structure with rigidly connected members.
•Similar to beams except that axial loads and deformationsare present.
•We have 2 displacements and 1 rotation at each node.•3 dof at each node.
q =[q1, q2, q3, q4, q5, q6]T
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)
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.
q’ = Lq
Where-
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−
=
1000000000
000000010000000000
lmml
lmml
L
q’2, q’3,q’5 and q’6 are beam element dof while q’1andq’4are like rod element dof.
Combining two stiffness and rearranging at proper locations weget element stiffness matrix as-
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−−−
−
−
−
−
=
eeee
eeee
ee
eeee
eeee
ee
e
lEI
lEI
lEI
lEI
lEI
lEI
lEI
lEI
lEA
lEA
lEI
lEI
lEI
lEI
lEI
lEI
lEI
lEI
lEA
lEA
k
460260
61206120
0000
260460
61206120
0000
22
2323
22
2323
'
LkLK
LqkL
qkWapproachsGalerkinby
LqkLq
qkqU
eTe
eTT
eTe
eTT
eTe
'
'
'
'
'
'','
21
''21
=
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 onmember, we have-
FKQformglobalIn
fLf
plplplplf
wherefLqfq
T
T
eeee
TTT
=
=
⎥⎦
⎤⎢⎣
⎡−=
=
,'
12,
2,0,
12,
2,0'
'''
22