field consistency

7/29/2019 field consistency

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FIELD CONSISTENCY

-M.SURENDRAN

(QHS 016)


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3D problems 2D problems Plane stress (dimension small in normal direction)

Plane strain (one dimension is very long)

Axisymmetric (Involving solids of revolution)

Sufficient accuracy Linear 3-noded triangle

Bilinear 4-noded quadrilateral

Higher accuracy by higher order elements

Quadratic 6-noded triangle Quadratic 8-noded quadrilateral

Quadratic 9-noded quadrilateral


3/26

Problems with current approach

Plane stress modelling of beam flexure

Poissons ratio 0.5 (nearly incompressiblematerials and materials undergoing plasticdeformation)

Parasitic shear Similar to shear locking in 1D Timoshenko beam

element

Error increases with increase in aspect ratio

Solved by Reduced integration

Addition of bubble modes for 4-nodedrectangular elements


4/26

8-noded and 9-noded elements No locking with reduced integration

Show severe linear and quadratic shear stress

oscillations

Shear locking, membrane locking, parasitic

shear field inconsistency

Adding drilling degrees of freedom (rotational

DOF normal to the plane of element)

Why does the technique work?


5/26

Problems due to inconsistency

Locking, poor convergence and violent stressoscillations in c0 elements formulation are dueto lack of a consistent definition of critical

strain field (strain field constrained in penaltyregime)

Eg. Shear strain in Timoshenko beam andMindlin plates and shells, membrane strain incurved beams and shells, volumetric strain inincompressible elasticity


6/26

The error in a timoshenko beam increases as

the beam becomes thin even if shape

functions are chosen to satisfy completeness

and continuity

As the beam becomes thin the shear strains

should vanish and automatically enforce

kirchhoffs codition


7/26

Symptoms vs Causes

Locking FEM solution vanishes quickly with increase inpenalty multipliers

Locking is linked to the non-singularity of stiffness matrix

Eg. Timoshenko beam Shear stiffness becomes large

as depth decreases

So high rank and non-singularity

But these are the outcomes of assumptions in

formulation. Unexpected requirements are unsatisfied

(symptoms)

Not the reason that fem does not

Cause???


8/26

FIELD CONSISTENCY

So the penalty linked strain fields should be

discretized in a consistent way

Only physically meaningful constraints

The requirement that a certain strain field

interpolation may have to be defined in a

manner that only physically realistic constraint

conditions will emerge in constrained physical

regimes


9/26

BEAM THEORY

Transverse deflection of thin cantilever beam

Length = L

Load = q N/m A

A

A


10/26

By Ritz approximation

A

A

A This is best-fit or least-squares fit

This seeks the best approximation


11/26

THE TIMOSHENKO BEAM THEORY

General formulation of beam flexure with

transverse shear deformation

Total strain energy 2 independent functions

w(x) and (x)

A

A


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2-TERM RITZ APPROXIMATION

(Inconsistent)

approx = a1x

u = b1x

approx= a1x b1

a1= -3qL2/(12EI + L2)

b1= -qL/2 1.5qL3/(12EI + L2)

As ; a1, b1 0 [, introduces a

penalty condition on the shear strain]

a10 imposes zero bending strain [spurious]


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(consistent)

approx = a1x

u = b1x + b2x2

approx

= -b1 + (a1 2b2)x So if shear strain 0 , (a1 2b2) 0. No

spurious constraint on bending strain (approx )

BM is constant approximation of the quadraticvariation

SF has linear variation even if


14/26


(Inconsistent)

approx = a1x + a2x2

u = b1x + b2x2

approx= -b1 + (a1 2b2)x + a2x

2

; shear strain 0 , a2 0

This means constant bending strain(approx,x)

So constant BM

Same as 3-term approximation

Quadratic shear oscillations. But they vanish at = 1/3. (which correspond to 2 point Gaussianintegration rule)


15/26


(Consistent)

Assume such that shear strain is consistent

Inconsistent term is quadratic term in a2

Replace x2 by (Lx L2

/6) approx

= (-b1 - a2L2/6) + (a1 + a2L 2b2)x

Yield physically meaningful constraints

But we see that approx solution should bejudged correctly


16/26

Explained easily by shear flexible beamelement

Using reduced integration produced accurate

elements It introduced errors that compensate the

other constraining errors

Functional reconstitution procedure toderive errors resulting from use ofinconsistent strain field interpolations


17/26

PLANE STRESS MODEL OF

BEAM FLEXURE

Length = L

Depth = T Thickness = t

DOF = u,v

Strain energyfunctional consists of

energy from normal

strain (UE) and shearstrain (UG)`

X,U2l

2t

L

x,u

y,v

Y,V

T


18/26

a

Timoshenko beam theory is obtained bysubstituting u = y

For slender beams shear strains=0 (Classical

Euler Bernoulli beam theory)

But shear stress, shear force is finite

This is the condition that causes difficulties


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4-NODED RECTANGULAR ELEMENT

4-nodes

8-DOF u1 u4, v1 v4 Field variables are interpolated as

u = a0 + a1x + a2y + a3xy v = b0 + b1x + b2y + b3xy

Problem is associated with shear strain as it willbe constrained and will become vanishingly smallwhen flexural action of thin beam is modelled

= u,y + v,x = (a2 +b1) + a3x + b3y


20/26

Shear strain energy within element will be

E

Since shear strain = 0 in thin beam a2 + b1 0

a3 0

b3 0

First constraint imposes conditions on both u and v

Second and third impose conditions separately

These are undesirable stiffening effect parasitic shear


21/26

For slender beams (l > t) l2 >>t2

If shear strain energy approaches zero in thin

beams, then a3 0 will be enforced more

rapidly than b3 0

The spurious energy generated from these

terms will be in similar proportions

Since shear force at a section along length is

constant shear strain will also be constant


22/26

EQ

Constant term averaged shear strain

Linear oscillating termrelated to the spurious

constraint This oscillation is self-equilibriating over the

element.

But it contributes a finite energy in equation In slender beams it becomes major and

dominates behavior of beam


23/26

The discretized strain energy functional of a

beam portion from to can be re-constituted

as

A very simple trick is to use a one point

integration rule for the shear strain energy.

e= e is ratio of actual a3 and

consistent a3


24/26

By comparing the displacement eqns with

actual displacements we get a3,consis=M0/EI

average shear-stress representation in a 4-

node field-inconsistent plane stress element

as

For long element (slender beam)

It also produces shear oscillations


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8-NODED RECTANGULAR ELEMENT

Quadratic shear stress oscillations appear

The performance of this element can be

improved by reduced integration of the shear

strain energy

A

a

=


26/26

Oscillating term is

If a6

0 it will create inconsistency the quadratic function is an oscillating

function with zero-points at the Gauss points

associated with the 2-point integration rule

FET Piece wise Ritz method stresses

taken at gauss points so accurate

field consistency

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