lecture n° 3
TRANSCRIPT
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-Element formulation for the different analyses-Modeling techniques
1
Element types and modeling techniques
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Analysis of a 2-D structural problem
2
1: if the element undergoes rigid motion, the strain is zero inside
the elementi
j
k
e
x
y
Properties of the elements
Ue{ } =
ux
uy
ux
uy
ux
uy
Rigid translation
{ } = B[ ] Ue{ }
B[ ] =B11 0 B13 0 B15 0
0 C22
0 C24
0 C26
C11
B22
C13
B24
C15
B26
x= B
11ux+ B
13ux+ B
15ux
B11=
yj yk
2
B13=
yk yi
2
B15=
yi yj
2
x =
yj yk
2ux +
yk yi
2ux +
yi yj
2ux = 0
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Analysis of a 2-D structural problem
3
2: strain evaluated at two adjacent elements is
discontinuous but limited: displacement field has a C0continuity
n
n=
vn
nn = x,y
n
vn
n
vn
finitevalue
Properties of the elements
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Shape functions
4
Summary of the principal properties and conditions that must be
considered in the shape functions definition
Properties:
- the shape function formulated for the node i, assumes the value 1 atnodei andzero at the other element nodes
- for a generic element defined by n nodes the summation of the shapefunction is equal to 1 in each point inside the element (the shape
function is a partition of the unity)
- the number of terms of a shape function must be equal to the numberof conditions that can be imposed at nodes. In the 2D plane analysis
the number of terms is equal to the number of nodes
Ni xj( )= ij =
1 i = j j= 1,...,nnodes
0 i j j= 1,...,nnodes
Ni(x) =1
i
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Shape functions
5
-The shape function must be continuous in the element-The field variable must be continuous in the domain. Therefore, thecontinuity must be assured also at inter-element boundary (the shape
function must be at least ofclass C0)
- In general the shape function must be of class Cm if in the integraldefining the stiffness matrix derivative up to the m+1 order are included
Conditions:
E.g.
In the 2D planestructural problem, first order derivatives of displacement
are present (strain-stresses) inside the integral, the shape function must
therefore have a C0 continuity
In the analysis of beams the function interpolating the displacement must
be of class C1
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Truss
Truss structures
Shell
Shell/plate Axi-
symmetrical3D plate/shell
Beam
Beam structures
SOLID
2D continuum
mechanics3D continuum
mechanics
2D 3D
x
y
x
z
y
Principal element types
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GAP
Contact Pb.
PIPE
Piping and pipe
structures
Mass
Concentrated mass
Spring
Elastic Connecting
Elements
Other element types
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Plane or 3D truss structures / linking / spring ,etc.
Jonly normal actionJ2 nodeJ2 or 3 d.o.f /nodeJonly nodal loadsJgeometrical properties : A (area)
Spar or truss elements /1
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Shape functions
Linear shape function: N11= A11 + B11x
x
y
i
j
For truss structures, links, springs the adopted shape function gives
the exact solution for the internal displacement
Spar or truss elements /2
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These structures are usually
represented as reticular structures
(free rotation at nodes)
The model based on trusses is
acceptable because of:
JLow bending stiffness of connectingelements
JLoose fits between holes and bolts
Installation for petroleum
drilling batteries.
Spar or truss elements /3
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20
1.
5
2
A=900 mm2 A=450 mm2
Upper deck rods Lower deck rods
Connecting
rods
Roof weight = 10 KN/m
Other structures: roof of an industrial building
Spar or truss elements /4
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Spar or truss elements /5
12
Axial force
F.E ModelDeformed shape
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Plane structures
J 2 nodesJ
3 d.o.f /nodeJconcentrated and distributed loadsJgeometry: A (area), J, (moments of inertia)
2D
The x,y plane contains:
J nodesJ applied loadsJ one of the principal inertia
axes
Beam elements /1
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Spatial structures
J 2 (3) nodeJ6 d.o.f /nodeJconcentrated and distributed loadsJgeometry: A, Jzz, Jyy, Jxx,
3D
Beam elements /2
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Theoretical background:
Stress/strain state implicitly involved in the choice of beam elements:
- Strain due to shear loading is neglected
- The only not null stress components are:
y
xx
xy
2D3D
xyx
xz
- xhas a linear variation troughout the cross-section (flexural formula)
y
xx
Beam elements /3
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Beam elements /4
16
vix
viy
y = vyx
x=xi
y
x
y
i
vx y( ) = vix +y = vix vyx
x=xi
y
Beam: the node's d.o.f. represent thedisplacement field of the whole cross
section
Hypothesis of plane cross sections
3 d.o.f. per node
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vjx
vjy j
vix
viyi
Small displacements/deformations
x
y
i jL
Ue{ } =
vix
viy
i
vjx
vjy
j
vx
x( ) = 1x
L
vix +
x
Lvjx = N11vix + N14vjx
vx(x) =f (vix, vjx)
2 conditions for vx(x)
Shape functions linear in x
v x( ){ } =v
x
vy
v x( ){ } = N x( ) U
e{ }3x1 3x6 6x1N
12= N
13= N
15= N
16= 0
Same shape functions as those ofthe truss elements
Beam elements /5
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vy x( ) = A + Bx +Cx2+ Dx
3
= B + 2Cx + 3Dx2
=d vy
dx4 conditions for vy(x)
vy(x) third-degree polynomials in "x" vy(0) = v
iy 0( ) =i
vy(L) = v
jy L( ) =j
vjx
vjyj
vix
viyi
x
yi jL
Ue{ } =
vix
viy
i
vjx
vjy
j
v x( ){ } =v
x
vy
Beam elements /6
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vx
vy
=
N11
0 0 N14
0 0
0 N22
N23
0 N25
N26
0 N32
N33
0 N35
N36
vix
viy
i
vjx
vjy
j
vy= v
iy1 3
x
L
2
+ 2x
L
3
+
ix 2L
x
L
2
+ Lx
L
3
+
+ vjy 3x
L
2
2x
L
3
+j L
x
L
2
+ Lx
L
3
= viy1
L6
x
L
+ 6
x
L
2
+i 1 4
x
L
+ 3
x
L
2
+
+ vjy1
L6
x
L
6
x
L
2
+j 2
x
L
+ 3
x
L
2
Beam elements /7
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The shape function used to represent the beam's deflection is cubic
vy x( ) = A+ Bx +Cx2
+Dx3
The shape functions correctly represent the deflection of the beam's segment only in the
case of constant shear.
In the remaining cases, the representation of displacements, deformations and stresses in
in internal points of the element is approximate. The error decreases with decreasingelement size.
V= constant
V constant
V=d
3vy x( )
dx3
= constant
Beam elements /8
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Beam elements /9
Endtruck wheel base (e1) =5 m
Gauge (S) =20 m
Trolley gauge (Scartamento carrello) = 2.5 m
500
700
200
8
Bridge beam
200
350
5
Endtruck
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Beam elements /10
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FE ModelDeformed shapeShear Z (local axis Z)Bending Moment My (local axis Y)Torsional Moment Mx (load axis X)
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Type of elements adopted to study piping problems in 2D and 3D
Jrectilinear pipe: same as beam elements with appropriate definition ofgeometry (diameters instead of A, J, etc.)
Pipe elements /1
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J Curvilinear pipe: specific definition of the stiffness matrix accounting forthe ratio of curvature radius/pipe diameter
J Special pipes: defined for a correct representation of the stiffness oftypical piping components (T-junctions, valves, etc.)
Pipe elements /2
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Pipe elements /3
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Bicycle frame
Model built with pipe elements
The cross-sectional data (inner and outer
diameter) are input as real constants
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Pipe elements /4
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Deformed shape and
equivalent stress foratypical jump loading
configuration
Deformed shape and equivalent stress fora dynamic biking loading configuration
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Structural, thermal, thermo-mechanical, electromagnetic problems, etc
J4 (3) nodesJ2 d.o.f /nodeClasses of structural problems:
Jplane stressJplane strainJaxisymmetric stress/strain
Plane elements /1
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J One of the principal stresses is zeroJ
Typical for components with a small thickness, if compared withother characteristic dimensions
J Load applied in the mid-plane.
x
y
z
y
Plane elements /2: plane stress
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180
R10
60
The F.E. model is on the X-Y
plane, representing the mid
plane of the body.
Loads can be defined on theentire thickness or per unit of
thickness
Plane elements /3: plane stress
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x
y
z
J one of the principal strains is zeroJ typical for very thick bodies: thickness much bigger than other
characteristics dimensions.
z=0
+-
Plane elements /4: plane strain
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x
y
z
J The model is on the x-y plane and represents a section perpendicular tothe Z-axis of the structure.
J Loads are defined per unit of thickness
Plane elements /5: plane strain
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Plane elements /6: plane strain
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Components characterized by an axis-symmetrical geometry (obtained byrotating a plane section around a fixed axis z)
Axisymmetric loadsz
Notched
bar
undertension
z
r
Cylindrical
vessel with
internal
pressure
z
Jby defining a cylindrical reference system r, , z, the stress/strain are independent of due tothe symmetry, moreover the circumferential displacements () are zero: the problem can be
referred to as plane.
Plane elements /7: axisymmetric problems
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The model represents a section generated by a plane
containing the symmetry axis
Plane elements /8: axisymmetric problems
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x=v
x
x
y=
vy
y
xy= v
x
y+ v
y
x
=vx
x
With respect to the plane
stress condition it is
necessary to define the
circumferential stress/strain
L[ ]=
x
0
0
y
y
x
1
x0
Volumerepresented by
the genericelement
Plane elements /9: axisymmetric problems
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Plane elements /10: application examples
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Geometrically identical models
Plane elements /11: application examples
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Axisymmetric shell elements
Axisymmetric thin walled bodies undergoing to axis-
symmetrical loads
J2 nodesJ3 d.o.f /node(vx, vye qz)
Shell elements /1
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Shell elements /2
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vix
viy
y = vyx
x=xi
y
x
y
i
vx
y( ) = vix +y = vix v
y
x
x=xi
y
Kirchoff-Love hypothesis to
determine the stiffness matrix: aline perpendicular to the mid-plane
remains rectilinear and
perpendicular to the mid-plane after
the deformation
The displacement field can be defined
through the thickness by knowing the
displacement and the rotation of the
mid-plane
Shell elements /3
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Validity of the Kirchoff-Love hypothesis:
Thickness
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As a consequence the stress/strain tensors are defined as follows:
JShear strains are not accounted for
Jnormal stress constant or linearly variable thru the thickness
y
x x
Jnon zero stress tensor components:
X (R)
Y (axial)
Shell elements /5
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The F.E. model represents a section
with a plane containing the axis of
symmetry. The nodes are placed onthe mid-plane.
Shell elements /6
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Thin vessels
Axisymmetric shell
Thick vessels
Plane axisymmetric elements
Shell elements /7
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Example: thin pressure vessel
Shell elements /8
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Shell elements /9
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Shell or plates with
generic geometry:
J4 nodesJ6 d.o.f /node
Kirchoff-Love hypothesis for
determining the stiffness
matrix is valid also for 3D
elements
The displacement field can be
defined through the thickness byknowing the displacement and the
rotation of the mid-plane
Shell elements /10: 3D elements
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Shell elements /11: 3D elements
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Validity of the Kirchoff-Love hypothesis:
Thickness
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x y
z
Stress components :
x, y, xy, xz, yz
Through thickness linear
variation of the normal
stress components
Shell elements /13: 3D elements
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Modelling of a bike frame
Shell elements for the frame, pipe
element for the fork
Shell elements /14: 3D elements
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Results in the jump configuration
Shell elements /15: 3D elements
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S /
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Results for the dynamic byking
Shell elements /16: 3D elements
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3D lid l t /1 b i k
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3D structural, thermal, problems:
J8 nodes: hexahedral., 4 nodes:tetrahedral
J3 d.o.f /node
3D solid elements /1: bricks
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3D lid l t /2 b i k
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Tetrahedral : 4 nodes
Shape function : A+Bx+Cy+Dz
Constant Strain/stress
Hexahedral: 8 nodes
Shape function :
A+Bx+Cy+Dz+Exy+Fyz+Gzx+Hxyz
Linearly variable stress and strain
3D solid elements /2: bricks
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3D lid l t /3 b i k
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Radial stress
Radial stress
3D solid elements /3: bricks
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3D lid l t /4 b d li
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Stress condition depending on local
geometrical parameters (e.g. notch
radius).
70
3D solid elements /4: submodeling
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3D lid l t /5 b d li
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Building up the model is very hard and complex (inclusion of allgeometrical details) and computationally heavy (huge number of d.o.f.)
The analysis requires very small element size to represent the local
geometry very fine mesh (these elements are usually too small to
represent the rest of the component).
3D solid elements /5: submodeling
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3D lid l t /6 b d li
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1 step:
a coarse model is built, not
accounting for the geometrical
details at notches. The external
loads and constraints are
applied.
2 step:a fine model of the local detail
is built up (sub model)
3D solid elements /6: submodeling
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3D solid elements /7: submodeling
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3 Step: The coarse model is used to determine the displacements of
nodes on the sub-models boundary surfaces
Displacement values are accurate if the dimensionsof sub-model are sufficiently higher than those of
the local detail
3D solid elements /7: submodeling
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3D solid elements /8: submodeling
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4 step:
the surface displacements are applied as boundary condition for thesub-model and the problem is solved obtaining an accurate evaluation
of stress and strain at the detail
3D solid elements /8: submodeling
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3D solid elements /9: submodeling
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Sub-modelling can be used with 2D and 3D elements. E.g.
the coarse model can be made of plane or shell elements and
the sub-model of brick elements
3D solid elements /9: submodeling
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3D solid elements /10: submodeling
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Example: Al-alloy suspension arm of a scooter
3D solid elements /10: submodeling
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3D solid elements /11: submodeling
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Full scale testing
Telaio
di prova
Provino
Afferraggio
fisso
Braccio di
flessione
Cuscinetto
assiale orientabile
a semplice effetto
Attuatore idraulico
Cella di carico Zona rottura
3D solid elements /11: submodeling
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3D solid elements /12: submodeling
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Mf
Mt=0.5 Mf
Failure
modes
R=0.1
BendingBending + torsion
3D solid elements /12: submodeling
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3D solid elements /13: submodeling
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Finite element analysis
Submodeling
3D solid elements /13: submodeling
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3D solid elements /14: submodeling
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Bending+torsion
Results Failure localization
Bending
Predicted Effective
3D solid elements /14: submodeling