lecture n° 3

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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


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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.


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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


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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


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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


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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


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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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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


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Modelling of a bike frame

Shell elements for the frame, pipe

element for the fork


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Results in the jump configuration


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S /

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Results for the dynamic byking


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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


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3D lid l t /3 b i k

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Radial stress

Radial stress


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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).

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3D solid elements /4: submodeling

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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).


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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)


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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


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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


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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


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Example: Al-alloy suspension arm of a scooter


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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


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Mf

Mt=0.5 Mf

Failure

modes

R=0.1

BendingBending + torsion


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Finite element analysis

Submodeling


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Bending+torsion

Results Failure localization

Bending

Predicted Effective


lecture n° 3

Documents

Transcript of lecture n° 3