FEM SOLID MECHANICS II - Simulación y Tecnología Part 1 Linear and nonlinear problems in solid mechanics Fundamental equations: Kinematics The stress tensor Equilibrium 3 …

FEM in Solid MechanicsPart II

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Part II

Eduardo Dvorkin

Contents

Part 1

� Linear and nonlinear problems in solid mechanics� Fundamental equations:

� Kinematics� The stress tensor� Equilibrium

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� Equilibrium� Constitutive relations

� The principle of virtual work

Part 2

� FEM in solid mechanics� Elasto-plasticity� Structural elements� Nonlinear problems- Collapse� Dynamic problems

FEM in Solid Mechanics

P(t)

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The solution : ui=ui(x,y,z)Has to fulfill:• Compatibility with rigid boundary conditions• Internal equilibrium• Equilibrium between applied loads and stresses• Constitutive equations

FEM in Solid MechanicsAny continuous set of functions ui=ui(x,y,z) that satisfies the rigid boundary conditions can be our solution.

Approximate solutions:

We seek among a finite set of functions and we pick the one that best fits the solution.

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If we want to improve a solution we just increase the number offunctions in the trial set.

It can be shown that the requirements on the functions are:

Continuity of the functions (not necessarily of their derivatives).Continuity of the functions (not necessarily of their derivatives).Exact representation of rigid body movements.Exact representation of rigid body movements.Exact representations of constant strain cases.Exact representations of constant strain cases.


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We can calculate the above integrals with functionsthat are only C0 continuous


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We can calculate the above integrals with functionsthat are only C0 continuous

FEM in Solid MechanicsElements and nodes

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The interpolation functions


1

2

r=1

s=1

2D example Natural coordinate systeminside each element (r, s)

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3

4

r=-1

s=-1


2D four-nodes element

r

s

12

h1+1

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r

3 4



s

12

h2+1

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r

3 4



s

12

h3

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r

3 4

+1



s

12

h4

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r

3

+1

FEM in Solid Mechanics1D elements

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From Bathe, Finite Element Procedures

Pop Quiz # 4

and why?

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to be able to represent rigid translations

Show at home that it can also represent rigid rotations and constant strain states

Pop Quiz # 5 Are these meshes acceptable?

YES

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NOYES

YES

YES

Pop Quiz # 5 Isoparametric elements

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1

2

12

Local 1Local 2Local 1

Local 2

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4

5

6

Local 3 Local 4

Local 3

Local 4

Vector of unknown nodal variables

X

Y

12

45

6

1 2

Local 1

Local 2

Local 3

Local 4

Local 1

Local 2

Local 3

Local 4

Interpolación Elemento 1

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Interpolación Elemento 2

FEM in Solid MechanicsInside each element

And calculating the derivatives via the Jacobian Matrix as shown above

x

u

∂

∂

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UB

z

u

x

w

y

w

z

v

x

v

y

uz

w

y

vx

zx

yz

xy

zz

yy

xx

=

∂

∂+

∂

∂∂

∂+

∂

∂∂

∂+

∂

∂∂

∂∂

∂∂

=

=

ε

ε

ε

ε

ε

ε

ε

2

2

2


Matrix 3D 2D

Plane stress

2D

Plane strain

2D

Axisym.

ε 6 x 1 3 x 1 3 x 1 4 x 1

B 6 x 3*NNOD 3 x 2*NNOD 3 x 2*NNOD 4 x 2*NNOD

U 3*NNOD 2*NNOD 2*NNOD 2*NNOD

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U 3*NNOD 2*NNOD 2*NNOD 2*NNOD

zx

yz

xy

zz

yy

xx

ε

ε

ε

ε

ε

ε

2

2

2

yz

zz

yy

ε

ε

ε

2

yz

zz

yy

ε

ε

ε

2

θθε

ε

ε

ε

yz

zz

yy

2

FEM in Solid Mechanicsεθθ in axisymmetric problems

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oy ty= oy+v

y

z

( )y

v

y

yvy00

00

2

22=

−+=

π

ππεθθ

FEM in Solid Mechanicsεθθ in axisymmetric problems

( )y

v

y

yvy00

00

2

22=

−+=

π

ππεθθ

........................

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U

yh

h

yh

h

yh

h

n

k

k

NNOD

n

k

k

n

k

k

yz

zz

yy

=

∑∑∑0......00

........................

........................

........................

221

θθε

ε

ε

ε

FEM in Solid MechanicsFor linear elastic materials

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FEM in Solid MechanicsFor an isotropic linear elastic material:

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FEM in Solid MechanicsFor an elastic-plastic material we have to define an incremental relation:

From the references

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From the references


For a rigid-viscoplastic material:

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Linear elasticity

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[ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ]

[ ] [ ] [ ]dStHUdSut

dvbHUdvub

dvBUdvdv

T

S

S

T

i

S

i

V

TT

i

V

i

T

V

T

V V

T

ijij

∫∫

∫∫

∫∫ ∫

=

=

==

δδ

δδ

σδσδεδεσ


Linear elasticity

[ ] [ ] [ ] [ ] [ ] [ ]

[ ] [ ]ext

T

S

S

V

TT

V

RF

dStHdvbHdvB

=

+= ∫∫∫

int

σ

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The nodes are always in equilibriumThe nodes are always in equilibrium


Linear elasticity

[ ] [ ][ ] [ ] [ ]RUdvBCB ext

T

V

∫ =

Stiffness matrix

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[ ] [ ] [ ][ ]dvBCBKT

V

∫=

The stiffness matrix is symmetric


Linear elasticity

In natural coordinates

[ ] [ ] [ ][ ] dtdsdrJBCBKT

∫ ∫ ∫− − −

=1 1 1

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[ ] [ ] [ ] [ ] [ ] dsdrJtHdtdsdrJBR S

T

S

T

ext ∫ ∫∫ ∫ ∫

∫ ∫ ∫

− −− − −

− − −

+=1

1

1

1

1

1

1

1

1

1

1 1 1

σ

Numerical Numerical integration is integration is used (e.g. Gauss)used (e.g. Gauss)

FEM in Solid MechanicsGetting to know the elements

1

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10

Pure bending using a 4-node element

Plane stress; E=2.1E6; ν=0.3; thickness= 0.1


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Mesh


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Vertical displacements


Vertical displacement at the tip

Analytical= ML2/2EI 2.86

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FEM 0.072

FEM/Analytical 0.025

What is going on?


Let’s examine the predicted stresses

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Shear stressesShould be zero … but are only zero at the element center


Standard solution:

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Let’s use more elements!!


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New mesh


New mesh

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FEM 1.93

FEM/Analytical 0.67


Let’s examine the predicted stresses

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Shear stressesShould be zero … but are only zero at the element center


The 4N displacement interpolation cannot model

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a constant bending situation


1

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10

Pure bending using a 9-node element

Plane stress; E=2.1E6; ν=0.3; thickness= 0.1


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Mesh


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FEM 2.86

FEM/Analytical 1.00



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Shear stresses

Zero everywhere!!


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Axial stresses

Exact result!!


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Transverse stresses

Zero everywhere!!


The 9N element exactly representsa constant bending situation

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Always?


Distorted 9N – det(J)‡const

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Mesh


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FEM 0.74

FEM/Analytical 0.26


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Shear stresses

Should be zero … but are only zero at the element center


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Axial stresses

Bad results


Locking in elasto-plastic problemsThe mixed elements

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The mixed elements


p

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p

R=10

R=20

E=2.1E6ν=0.3σy=2100ET=0.0


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Mesh


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Lambda

Displacement Based Element


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Mixed Element


The displacement based element shows an improper hardeningThe mixed element shows the correct flat response

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HOWEVERHOWEVER

The pressure predictions clearly show whatdoes locking mean


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Displacement based: locking


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Mixed: non-locking

Structural Elements

Beam elements (Bernoulli and Timoshenko)Plate / Shell elements

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Plate / Shell elements

Structural ElementsThe Bernoulli (Hermitian) beam element (2 nodes only)

It is the classical beam we know from strengthof materials

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of materialsNo shear deformationNo stresses through the thickness

Cannot be used in conjunction with modern Cannot be used in conjunction with modern plate/shell elementsplate/shell elements

Structural ElementsThe Bernoulli (Hermitian) beam element: d.o.f.

12

6 6 d.o.fd.o.f. / node. / node

v

θy

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u

w

vθx

θz

Structural ElementsThe Bernoulli (Hermitian) beam element: d.o.f.

No shear deformation means:

v∂=θ

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x

w

x

v

y

z

∂

∂−=

∂

∂=

θ

θ

Structural ElementsTimoshenko (isoparametric) beam element(2,3 and 4 nodes to describe curved beams)

Includes shear deformationNo stresses through the thickness

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No stresses through the thickness

Can be used in conjunction with modern Can be used in conjunction with modern plate/shell elementsplate/shell elements

Structural ElementsTimoshenko (isoparametric) beam element

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6 6 d.o.fd.o.f. / node. / node


In order to avoid locking use:

Nodes Integration points

2 1

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2 1

3 2

4 3


Timoshenko beam element: locking Section

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M 1

1100

Structural ElementsTimoshenko beam element: locking

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Structural ElementsTimoshenko beam element: locking

Tip displacement

2-points integration along 0.74266E-3

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2-points integration along length

0.74266E-3

1-point integration along length(default in ADINA)

2.85714

Analytical 2.85714

Structural ElementsThe isoparametric shell element: geometric definition

5 d.o.f. / node

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Structural ElementsThe isoparametric shell element: locking

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Use nonUse non--locking elements (e.g. MITC4)locking elements (e.g. MITC4)

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Structural ElementsThe isoparametric shell element: b.c.

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6 d.o.f. at intersection nodes

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6 d.o.f. at intersection nodes

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Nonlinear ProblemsIterative methods

PVW

σσσ += ∆− ttt

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σσσ += ∆− ttt

hence,


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Nomenclature:

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Hence, for equilibrium:


Since the problem is nonlinear (plasticity, contacts, etc.) we must iterate to get the solution:

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Iterate until:


We have plenty of freedom to select the matrix

because we just have to get,

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independently of the iteration path.

Iterative methods: full Newton, modified Newton, BFGS.Iterative methods: full Newton, modified Newton, BFGS.Combine with line searches.Combine with line searches.


Iteration tolerances

A fundamental decision:

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A fundamental decision:If too restrictive we may not get a solutionIf too ample we may get a very bad solution

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Examples:

P

Softening problem

Alone is not enough

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Alone is not enough

P

Stiffening problem

Alone is not enough


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Cannot iterate controlling loadUse LDC (Collapse analysis in ADINA)

Nonlinear ProblemsIterative methods: LDC

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Dynamic Problems

)(tRUKUCUM =++•••

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)(tRUKUCUM =++

Dynamic ProblemsExplicit Integration

Conditionally stableCheapM: lumped

Tt <∆

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limTt <∆


Simple stability example 1

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lim2.0 Tt =∆ lim6.0 Tt =∆ lim9.0 Tt =∆


Simple stability example 1

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lim001.1 Tt =∆


Simple stability analysis 2

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lim02.0 Tt =∆ lim6.0 Tt =∆



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lim001.1 Tt =∆

Dynamic ProblemsImplicit integration

Uconditionally stable

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Dynamic ProblemsNonlinear Analysis

Implicit No iterationsMay give incorrect solution

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May give incorrect solutionCareful

Explicit integration Iterate at every stepMay not convergeCareful

Documents

FEM SOLID MECHANICS II - Simulación y Tecnología Part 1 Linear and nonlinear problems in solid mechanics Fundamental equations: Kinematics The stress tensor Equilibrium 3 …