FEMModeling

8/8/2019 FEMModeling

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MANE 4240 & CIVL 4240

Introduction to Finite Elements

Practical considerations

in FEM modeling

Prof. Suvranu De


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Reading assignment:

Logan Chap 7 + Lecture notes

Summary:

Aspect ratio and element shapes Use of symmetry

Natural subdivisions at discontinuities

Stress equilibrium in FEM solutions


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2002 Brooks/Cole Publishing / Thomson Learning

Figure 7-1a (a) Beam with

loading: effects of the aspect

ratio (AR) illustrated by the

five cases with different

aspect ratios

Aspect ratio = longest

dimension/ shortest

dimension

Aspect ratio and element shapes


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Figure 7-1b (b) Inaccuracy of solution as a function of the aspect

ratio (numbers in parentheses correspond to the cases listed in

Table 7-1)


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Figure 7-2 Elements with poor shapes


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Avoid abrupt changes in element sizes

Abrupt change in

element size

Gradual change in

element size


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Examples of how NOT to connect elements


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Figure 7-3 Use of symmetry applied to a soil mass subjected to

foundation loading (number of nodes = 66, number of elements

= 50) (2.54 cm = 1 in., 4.445 N = 1 lb)

Use of symmetry in modeling


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Figure 7-4 Use of symmetry applied to a uniaxially loaded

member with a fillet


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Figure 7-5 Problem reduction

using axes of symmetry applied

to a plate with a hole subjected

to tensile force


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Natural subdivisions at discontinuities


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Look before you leap!

1. Check the model that you have developed: Boundary conditions

Loadings

Symmetry?

Element aspect ratios/shapes

Mesh gradation

2. Check the results

Eyeball

Anything funny (nonzero displacements where they should be zero?)

Are stress concentrations in places that you expect?

Comparison with known analytical solution/literature

3. If you remesh the same problem and analyze, do the solutions converge?

(specifically check for convergence in strain energy)


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Stress equilibrium in FEM analysis

Example:

80cm

1 22

( ) 140

x A x sqcm

!

Consider a linear elastic bar with varying cross section

P=3E/80

x

The governing differential (equilibrium) equation

( ) 0 (0,80)d du

E A x for xdx dx

Boundary conditions

80

( 0) 0

3

80 x cm

u x

du

A Pdx !

! !

! !

Analytical solution3 1

( ) 12

1

40

exactu x

x

!

Eq(1)E: Youngs modulus


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Lets us discretize the bar using a 2-noded (linear) bar element. The finite element

approximation within the bar is

1 1 2 2( ) ( ) ( )FEM x xu x N x u N x u!

where the shape functions

1

2

( ) 180

( )80

xx

xx

!

!

If we incorporate the boundary condition at x=0

2 2 2( ) ( )

80

FEM

x x

xu x x u u! !

Does this solution satisfy the equilibrium equation (Eq 1)?


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2

2

22

1

40( ) 1 ??80 80 40

FEM

xx

x

ud du d d xE A x E udx dx dx dx

! ! !

Conclusion: The FEM displacement field does NOT satisfy the equilibrium

equations at every point inside the elements.

However, the solution gets better as the mesh is refined.


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Stress equilibrium in FEM analysis

To obtain exact solution of the mathematical model in solidmechanics we need to satisfy

1. Compatibility

2. Stress-strain law

3. Stress-equilibriumat every point in the computational domain.

In a FE model one satisfies the first 2 conditions exactly.

But stress-equilibrium is NOT satisfied point wise.Question: Then what is satisfied?


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Let us compute the FEM solution using a bar element

The stiffness matrix is

80

0

2

( ) 1 1

1 180

1 113

1 1240

xE A x dx

K

E

!

! -

!

-

The system equations to solve are

1 1

2

1 113

1 1240

x x

x

u fE

u P

! -

With u1x=0; we solve for

2

240 240 3 9

13 13 80 13x

Eu c

E E

! ! !

(Note that the exact solution for the displacement at node 2 is 1cm!!)


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Let us now compute the nodal forces due to element stresses using the formula

f K d !

1 113

1 1240

EK

! -

0

9

13

d

!

30

1 113 809

1 1 3240 1380

E

Ef d

E

! ! !

-


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1. Element equilibrium

2. Nodal equilibrium

P=3E/80

1

3

80x

Ef ! 2

3

80x

Ef !

Two observations


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The following two properties are ALWAYS satisfied by the

FEM solution using a coarse or a fine mesh

Property 1: Nodal point equilibriumProperty 2. Element equilibrium

El #4 El #3

El #1El #2

P

PROPERTY 1: (Nodal point equilibrium) At any node the

sum of the element nodal point forces is in equilibrium with the

externally applied loads (including all effects due to body forces,

surface tractions, initial stresses, concentrated loads, inertia,

damping and reaction)


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How to compute the nodal reaction forces for a given finite element?

!

!

!!

e

e

e

V

T

V

T

V

T

dVB

dVdBDB

ddVBDBdkf

X

! eVT

dVBf X

Once we have computed the element stress, we may obtain the

nodal reaction forces as


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El #4 El #3

El #1El #2

P

This is equal in

magnitude and in the

same direction as P

Sum of forces equal externally applied load (=0 at this node)

Nodal point equilibrium implies:


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PROPERTY 2: (Element equilibrium) Each element is in

equilibrium under its forcesf

i.e., each element is underforce and moment equilibrium

e.g.,

F3x

F2xF1x

F4x

F1yF2y

F3yF4y

Define ? AT

d 01010101!

00 !!!

!

!

dB

dVBd

dVBdfd

e

e

V

TT

V

TTT

I

X

X

3

1 2 3 4

T

x x x xd f F F F F @ !

But

since this is a rigid

body

displacement, the

strains are zero

Hence

1 2 3 4 0 x x x x F F F F !

as a rigid body displacement in x-direction


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Example (Finite Element Procedures, Bathe 1996)


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NOTE: In a finite element analysis

1. Stress equilibrium violated inside each element

2. Stresses are discontinuous across elements

3. Stresses are not in equilibrium with the applied traction


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Figure7-10 Example 6.2, illustrating violation of equilibrium of

a differential element and along the diagonal edge between two

elements (the coarseness of the mesh amplifies the violation of

equilibrium)


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Figure 7-11 Convergence of a finite element solution based on the

compatible displacement formulation


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Hence a finite element analysis can be interpreted as a process in

which

1. The structure or continuum is idealized as an assemblage of

elements connected at nodes pertaining to the elements.

2. The externally applied forces are lumped to these nodes to

obtain the equivalent nodal load vectors

3. The equivalent nodal loads are equilibriated by the nodal point

forces that are equivalent to the element internal stresses.

4. Compatibility and stress-strain relationships are exactlysatisfied, but instead of force equilibrium at the differential level,

only global equilibrium for the complete structure, of the nodal

points and ofeach element under its nodal point forces is

satisfied.

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