Post on 03-Apr-2018
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Beams and Frames
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beam theory can be used to solve simple
beamscomplex beams with many cross section
many 2-d and 3-d frame structures are
better modeled by beam theory
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One Dimensional Problems
x
y
P
p
u(x)
y
px
StretchContract
u(x)
kx + P(x=l)=0dudx
The geometry of the problem is three dimensional, but the
variation of variable is one dimensional
x v(x)v(x)
kx + P(x)=0d4v
dx4
Variable can be scalar field liketemperature, or vector field like
displacement.
For dynamic loading, the variable can
be time dependent
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Element Formulation assume the displacement w is a cubic polynomial in
L = Length
a1, a2, a3, a4 are the undetermined coefficients
2 3
1 2 3 4v(x) a a x a x a x= + + +
the cross sectional areaE = Modulus of Elsaticity
v = v(x) deflection of the
neutral axis
= dv/dx slope of theelastic curve (rotation of
the section
F = F(x) = shear force
M= M(x) = Bending
moment about Z-axis
i j
L, EI
viFi,Fj,vj
Mi,i
Mj,j
j vjivi
y
x
x
v(x)
(x)i
j
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`
1 1
x 0
dvx 0, v(0) v ;
dx
dv
=
= = =
{ } { }
1 1
2 22 3 2
3 3
4 4
a a
a av(x) 1 x x x ; (x) 0 1 2x 3x
a a
a a
= =
2 2x L
x , v v ;dx =
= = =
i 1
i 2
2 3j 3
2j 4
v 1 0 0 0 a
0 1 0 0 a
v a1 L L L
a0 1 2L 3L
=
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Applying these boundary conditions, we get
1
1 1 2 1
{d} [P(x)]{a}
{a} [P(x)] {d}
a v ; a
1a 3v 2L 3v L
=
=
= =
= +
Substituting coefficients ai back into the original equation
for v(x) and rearranging terms gives
L
{ }
1
22 3
3
4
a
av(x) 1 x x x
a
a
=
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The interpolation function or shape function is given by
2 3 2 3
1 12 3 2
2 3 2 3
3x 2x 2x xv(x) (1 )v (x )
L L L L
3x 2x x x( )v ( )
= + + +
+ + +
N1=1
dN1dx
(x=0) = 0
N1(x=L) =1
dN1dx
(x=L) = 0
N2=1
[ ]
1
1
1 2 3 4
2
2
v
Lv N (x) N (x) N (x) N (x) [N]{d}
vL
= =
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strain for a beam in bending is defined by the curvature, so
Hence
N A
dv
dx
y
x
dvdx
u(x) = y2 2
2 2
du d v d [N]y {d} y[B]{d}
dx dx dx = = = =
3 2 3 2 2 3 3 2
12x 6 6x 4 6 12x 6x 2[B]
L L L L L L L h
=
{ } { }
{ } [ ]{ }
{ } [ ]{ }
{ } [ ] { }
{ } [ ] [ ][ ]{ }
e
e
e
Te
v
Te
v
T Te 2
v
Internal virtual energy U = dv
substitute E in above eqn.
U = E dv
= y B d
U = d B E B d y dv
=
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e eFrom virtual work principle U W =
{ } { } { } [ ] { }
{ } { } { } [ ] { }
{ } { } { }
e e
T T Te
b y
v v
T T Te
s y
s s
TTe e e
c
External virtual workdue to body force
w = d(x) b dv d N b dv
External virtual work due to surface force
w = d(x) p dv d N p ds
External virtual work due to nodal forces
w d P , P
=
=
=
{ }yi i yj= P , M , P ,....
{ } [ ] [ ][ ] { } { } [ ] { } [ ] { } { }
{ } { }
[ ] [ ][ ]
{ } [ ] { } [ ] { } { }
T TT T T2 ed ( B E B y dv d d N b dv N p dv Py y
e e sv v
eK U Fe e
where
T 2K B D B y dv Element stiffness matrixe
ev
T T eF N b dv N p ds P Total nodal force vectore y y
e sv
= + +
=
= =
= + + =
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the stiffness matrix [k] is defined
dAy
To compute equivalent nodal force vector
( )L
T 2 T
VA 0
2 2
3
2 2
[k] [B] E[B]dV dAy E [B] [B]dx
12 6L 12 6L
6L 4L 6L 2LEI
12 6L 12 6LL
6L 2L 6L 4L
= =
=
for the loading shown
{ } [ ] { }
[ ]
T
e y
s
y
y
2 3 2 3 2 3 2 3
2 3 2 2 3 2
F N p ds
From similar triangles
p w w; p x; ds = 1 dxx L L
3x 2x 2x x 3x 2x x xN (1 ) (x ) ( ) ( )
L L L L L L L L
=
= =
= + + +
w
x
py
L
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{ } [ ] { }
{ }
T
e y
s
2 3
2 3
22 3
2
e 2 3L
2 3
F N p ds
3wL3x 2x(1 )
20L L
wL2x x(x )
wx 30L LF dx
7wLL3x 2x( )
20L L
=
+
+
= =
+ve directions
vi vj
i j
22 3
2wLx x( )20L L
+
w
Equivalent nodal force due toUniformly distributed load w
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v1 v2v3
1 23
v4
v3
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1
1
2 5
2
3
3
V 12 6 -12 6
M 6 4 -6 2
V -12 -6 12+12 -6+6 -12 68x10
M 6 2 -6+6 4+4 -6 2
-12 -6V
M
=
1
1
2
2
3
3
1 1 2 3
v
v
12 -6 v
6 2 -6 4
Boundary condition
v , , v , v 0
=
2 3
5
M 1000; M 1000
8 28x10
2
= =
{ } [ ]{ }
2
3
42
5 43
1000
4 1000.0
4 -2 1000 2.679x101
-2 8 1000.028*8x10 4.464x10
Final member end forces
f k d {FEMS}
=
= =
= +
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Find slope at joint 2 and deflection at
joint 3. Also find member end forces
v
vv
EI EI
20 KN 20kN/m Guided Support
2m 2m 6m
EI=2 x 10
4
kN-m
2
Global coordinates
20 KN 20kN/m
m mm
m
m
m
m
Fixed end reactions (FERs)
Action/loads at global
coordinates
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1 14
1 1
32 2
2 2
1 1 2 2
For element 1
f v12 24 -12 24
m 24 64 -24 321X10
f -12 -24 12 -24 v4
24 32 -24 64m
v v
For el
=
1 1
41 1
ement 2
f v12 36 -12 36
m 36 144 -36 721X10
=2 2
2 2
2 2 3 3
f -12 -3 12 -3 v6
36 72 -36 144m
v v
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1
1
2
2
3
3
F 1875 3750 -1875 3750
M 3750 10000 -3750 5000
F -1875 -3750 1875+555.56 -3750+1666.67 -555.56 1666.67
M
F
M
=
3750 5000 -3750+1666.67 10000+6666.67 -1666.67 3333.33
-555.56 -1666.67 555.56 -1666.67
1666.67
1
1
2
2
3
3
1 1 2 3
v
v
v
3333.332 -1666.67 6666.67
Boundary condition
v , , v , 0
=
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2 3
M 50; F 60
16666.67 -1666.67
-1666.67 555.56
= =
{ } [ ]{ }
2
3
2
3
50
v 60
555.56 1666.67 50 0.0197141
v 1666.67 16666.67 60 0.167146481481.5
Final member end forces
f k d {FEMS}
=
= =
= +
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x
y
q3
q2
q1
q5
q6
q4
displacement in local coordinates
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3 3 3 3
3 3 3 3
AE AE0 0 0 0
L L
12EI 6EI 12EI 6EI0 0
L L L L6EI 4EI 6EI 2EI
0 0L L L L
[k]AE AE
0 0 0 0L L
=
{ } [ ]{ }
If f ' member end forces in local coordinates then
f' k ' q '=
3 3 3 3
3 3 3 3
12EI 6EI 12EI 6EI
0 0L L L L
6EI 2EI 6EI 4EI0 0
L L L L
{ }q {q q q q q q }
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'
1 1 2
'
2 1 2
'
At node i
q q cos q sin
q q sin q cos
= +
= +
{ } 1 2 3 4 5 6q {q ,q ,q ,q ,q ,q }
are forces in global coordinate direction
=
3 3q q
l cos ; m sin
=
= =
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{ } { }
[ ] [ ] [ ][ ]T
using conditions q' [L]{q}; and f' [L]{f}
Stiffness matrix for an arbitrarily oriented beam element is given by
k L k ' L
= =
=
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Grid Elements
a
a
f = GJ/l
qxi fi qxj fj
xi i
xj j
JG JGq fL L
q fJG JG
L L
=
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3 3 3 3
3 3 3 3
GJ GJ0 0 0 0L L
12EI 6EI 12EI 6EI0 0
L L L L
6EI 4EI 6EI 2EI0 0
L L L LGJ GJ
0 0 0 0L L
12EI 6EI 12EI 6EI
{ } [ ]{ }
If f ' member end forces in local coordinates then
f' k ' q '=
3 3 3 3
3 3 3 3
L L L L
6EI 2EI 6EI 4EI0 0
L L L L
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[ ]
C 0 -s 0 0 0
0 1 0 0 0 0
-s 0 c 0 0 0L
0 0 0 c 0 -s
=
0 0 0 0 1 0
0 0 0 --s 0 c
[ ] [ ] [ ][ ]T
k L k ' L
=
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Beam element for 3D analysis
x
y
q2
q1
q5q7
q8
q11
z
q3
q6q4
displacement in local coordinatesq9
q10
q12
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if axial load is tensile, results from beam
elements are higher than actual
resultsare conservative
than actual size of error is small until load is about 25% of
Euler buckling load
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for 2-d, can use rotation matrices to get
stiffness matrix for beams in anyorientation
,
add capability for torsional loads about theaxis of the element, and flexural loading in
x-z plane
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to derive the 3-d beam element, set up the
beam with thex axis along its length, andyandz axes as lateral directions
of simple strength of materials solution
JG
L
JG
LJG
L
JG
L
T
Txi
xj
i
j
=
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J = torsional moment aboutx axis
G = shear modulusL = length
xi, xj are nodal degrees of freedom ofangle of twist at each end
Ti, Tj are torques about thex axis at each
end
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flexure in x-z plane adds another stiffness
matrix like the first one derivedsuperposition of all these matrices gives a
to orient a beam element in 3-d, use 3-d
rotation matrices
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for beams long compared to their crosssection, displacement is almost all due toflexure of beam
for short beams there is an additional lateral
displacement due to transverse shearsome FE programs take this into account,
but you then need to input a shear
deformation constant (value associated withgeometry of cross section)
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limitations:
same assumptions as in conventional beam andtorsion theories
no better than beam anal sis
axial load capability allows frame analysis, butformulation does not couple axial and lateral
loading which are coupled nonlinearly
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Finite Element Model
Element formulation exact for beam spans
with no intermediate loads need only 1 element to model any such
member that has constant cross section
for distributed load, subdivide into severalelements
need a node everywhere a point load isapplied
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need nodes where frame members connect,
where they change direction, or where thecross section properties change
,
have the same linear and rotationaldisplacement
boundary conditions can be restraints onlinear displacements or rotation
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simple supports restrain only linear
displacementsbuilt in supports restrain rotation also
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restrain vertical and horizontal displacements
of nodes 1 and 3
no restraint on rotation of nodes 1 and 3
need a restraint inx direction to revent ri id
body motion, even if all forces are inydirection
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cantilever beam
has x and y linear displacements and rotation of node 1
fixed
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point loads are idealized loads
structure away from area of applicationbehaves as though point loads are applied
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only an exact formulation when there are no
loads along the span for distributed loads, can get exact solution
ever where else b re lacin the distributed
load by equivalent loads and moments at thenodes
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C t I t A i t
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Computer Input Assistance
preprocessor will usually have the same
capabilities as for trussesa beam element consists of two node
physical properties
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material properties:
modulus of elasticity
if dynamic or thermal analysis, mass densityand thermal coefficient of expansion
physical properties: cross sectional area
2 area moments of inertia
torsion constant location of stress calculation point
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boundary conditions:
specify node numbers and displacementcomponents that are restrained
specify by node number and load components
most commercial FE programs allows
application of distributed loads but they use
and equivalent load/moment set internally
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Output Processing and Evaluation
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Output Processing and Evaluation
graphical output of deformed shape usually
uses only straight lines to representmembers
constraints on the deformed shape of eachmember
to check these, subdivide each member andredo the analysis
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most FE codes do not make graphical
presentations of beam stress results
user must calculate some of these from the stress
values returned
for 2- beams, you get a norma stress norma to
the cross section and a transverse shear acting on
the face of the cross section
normal stress has 2 components
axial stress
bending stress due to moment
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3-d beams
normal stress is combination of axial stress,flexural stress from localy- andz- moments
stress due to moment is linear across a section
the combination is usually highest at theextreme corners of the cross section
may also have to include the effects of torsion
get a 2-d stress state which must be evaluated
also need to check for column buckling