Method of Finite Elements I - Homepage | ETH Zürich...Constant Strain Triangle The Constant Strain...

Institute of Structural Engineering

Method of Finite Elements I

Chapter 6

2D Elements

*slides are prepared in collaboration with Dr. S. Triantafyllou, Assistant Professor at the University of Nottingham


Method of Finite Elements I30-Apr-10

Today’s Lecture Contents

• Continuum Elements– Plane Stress

– Plane Strain

• Structural Elements– Plate Elements

Extending to higher Dimensions



FE Classification

`



2D vs. 3D Formulations

Three-dimensional elasticity problems are very difficult to solve. Thus we will first develop governing equations for two-dimensional problems, and will explore two basic theories:

- Plane Strain- Plane Stress

The basic theories of plane strain and plane stress represent the fundamentalplane problem in elasticity. While these two theories apply to significantlydifferent types of two-dimensional bodies, their formulations yield verysimilar field equations.

Since all real elastic structures are three-dimensional, theories set forthhere will be approximate models. The nature and accuracy of theapproximation will depend on problem and loading geometry



Plane StrainConsider an infinitely long cylindrical (prismatic) body as shown. If the bodyforces and tractions on lateral boundaries are independent of the z-coordinateand have no z-component, then the deformation field can be taken in thereduced form

x

y

z R



Examples of Plane Strain

x

y

z

x

y

z

P

Long CylindersUnder Uniform Loading

Semi-Infinite Regions Under Uniform Loadings



Plane Strain Equations

0,2)()(

2)(,2)(

===

+=+=

++=++=

yzxzxyxy

yxyxz

yyxyxyxx

eee

eeeeee

ττµτ

σσνλσ

µλσµλσ

Equilibrium Equations

0

0

=+∂σ∂

+∂τ∂

=+∂τ∂

+∂σ∂

yyxy

xxyx

Fyx

Fyx

Strain Compatibility

yxe

xe

ye xyyx

∂∂

∂=

∂

∂+

∂∂ 2

2

2

2

2

2

Strain Displacement RelationsStrains vs. Stresses

021,,

===

∂∂

+∂∂

=∂∂

=∂∂

=

xzyzz

xyyx

eeexv

yue

yve

xue



Plane StressConsider a where one dimension, eg. along z, is small in comparison to theother dimensions in the problem. Since the region is thin in the z-direction,there can be little variation in the stress components

through the thickness, and thus they will be approximately zerothroughout the entire domain. Finally, since the region is thin in the z-direction it can be argued that the other non-zero stresses will have littlevariation with z. Under these assumptions, the stress field can be taken as

0),(),(),(

===

=

==

yzxzz

xyxy

yy

xx

yxyxyx

ττσ

ττ

σσσσ

x

y

z R

2h

yzxzz ττσ ,,



Examples of Plane Stress Problems

Thin Plate WithCentral Hole

Circular Plate UnderEdge Loadings



Plane Stress Equations

Strain Displacement Relations

Equilibrium Equations

0

0

=+∂σ∂

+∂τ∂

=+∂τ∂

+∂σ∂

yyxy

xxyx

Fyx

Fyx

Strain Compatibility

yxe

xe

ye xyyx

∂∂∂

=∂∂

+∂∂ 2

2

2

2

2

2

0,1

)(1

)(

)(1,)(1

==τν+

=

+ν−

ν−=σ+σ

ν−=

νσ−σ=νσ−σ=

yzxzxyxy

yxyxz

xyyyxx

eeE

e

eeE

e

Ee

Ee

021,0

21

21,,,

=

∂∂

+∂∂

==

∂∂

+∂∂

=

∂∂

+∂∂

=∂∂

=∂∂

=∂∂

=

xw

zue

yw

zve

xv

yue

zwe

yve

xue

xzyz

xyzyx

Strains vs. Stresses



Plane Stress/ Strain Elasticity Plane Strain

Premise 1:

Premise 2: Loads are applied only within the plane Premise 3: The applied loads are independent of zPremise 4: No load is applied on the boundary surfaces normal to the

Conclusion:

Premise 5: The edge surfaces are rigid

Stress Tensor

Strain Tensor

Constitutive Matrix



Plane Stress/ Strain Elasticity Plane Stress

Premise 1:

Premise 2: Loads are applied only within the plane Premise 3: The applied loads are independent of Premise 4: No load is applied on the boundary surfaces normal to the

Conclusion:

Stress Tensor

Strain Tensor

Constitutive Matrix


Method of Finite Elements I

What are the types of Finite Elements usedin plane stress/strain formulations?



Constant Strain Triangle The Constant Strain Triangle element is historically the first finite element ever used in engineering practice (Argyris, 1960, Turner, 1956) for the evaluation of stress distribution in wing panels.

The CST Element

Plane Elementi. The displacement field varies within the x-y plane

Constant strain fieldi. The strain does not vary within the element

Elastic Material Behaviour

1 2

3

We number the nodes in counter-clockwise order. In this way, the normal vector to the 123 surface will point towards the positive z axis and the resulting area of the triangle will be a positive quantity.



The CST Element

Remember that the deformation is the first derivative of the displacement field Compatibility equations in

the 2D plane

Therefore if we ask for the strain field to be constant

The displacement field has to be a linear function of



The CST Element

1 2

3

Therefore, the following “candidate” displacement field approximation is considered:



The CST ElementTherefore the shape functions of the CST element are readily derived as

where the shape function matrix assumes the following form

and



CST Shape Functionsand if we plot these shape functions over the surface of the element:

1

2

3 1

3

2

1

2

3



The CST element is fairly accurate within zones of small variation of stresses. However a fine mesh is required in every other case in order for FEA to converge to an accurate solution.

As already discussed during the Galerkin lectures, the FEA solution is refined either by increasing the number of finite elements or by increasing the order of the interpolating (test) functions used in the FE formulation. Thus, there is a trade-off between required mesh-size and interpolation complexity for the same degree of accuracy.

Based on that rationale, the Quadrilateral 4-node (Q4) finite element was introduced in an effort to effectively model the stress variations of plane elasticity problems.

The Q4 Element



The Q4 Element

1 2

34

Plane Elementi. The displacement field varies only within the x-y plane


Formulation Assumptions



The Q4 Element

We number the nodes in counter-clockwise order. In this way, the normal vector to the 1234 surface will point towards the positive axis and the resulting area of the triangle will be a positive quantity.

1 2

34

Plane Elementi. The displacement field varies only within the x-y plane


Formulation Assumptions



The Q4 Element Formulation Assumptions

The following bilinear “candidate” displacement field approximation is considered:

1 2

34



The Q4 Element

The following bilinear “candidate” displacement field approximation is considered:1 2

34

The formulation is indifferent to the coordinate system.

So why not make things easier?



The Q4 Element

The arbitrary nodal displacement values are introduced at the r.h.s. of the interpolation equation:

Exactly the same procedure as in the CST element!



The Q4 ElementFollowing the standard procedure, the arbitrary nodal displacement values are introduced at the r.h.s. of the interpolation equation:

Therefore the Shape Function Matrix is derived form the following:

Shape Functions



The Q4 Element



The Q4 ElementThe compatibility relations are again expressed in matrix form

and therefore by substituting the interpolation equation into the r.h.s.

where now



Q4 Stiffness MatrixThe stiffness matrix is derived as

and given that the element is rectangular

and if the thickness is constant

The evaluation of the stiffness matrix can be performed analytically. Integration involves only linear expressions of and

/2 /2

/2/ 2x a= −

/2 /2

/2/ 2x a= −



Consistent nodal force vectorConsider the case of distributed loading along the element’s 12 side

1 2

34

Equivalent nodal vector due to a traction load

In this case

For example, the component along the 12 side is

External Work due to surface tractions




1 2

34


In this case


In matrix form




1 2

34


In this case


Consistent Load Vector



Taxonomy of Finite Elements



The Euler/Bernoulli beam theory

Uniaxial Elementi. The longitudinal direction is sufficiently larger than the

other two Prismatic Element

i. The cross-section of the element does not change along the element’s length

Euler/ Bernoulli assumptioni. Upon deformation, plane sections remain plane AND

perpendicular to the beam axis

Assumptions

But what happens if two dimensions are sufficiently larger that the third one?



The plate problem

• The slab thickness is sufficiently smaller that the two leading dimensions

• Consequently, the three dimensional problem is reduced to a two-dimensional one and the plate problem is examined at the mid-surface.

• If the thickness of the slab is then the mid-surface is located at a distance from each lateral surface



The plate problemTwo possible loading states

Case 1: Loads applied within the plane Case 2: Loads applied perpendicular to the mid-surface

Plane stress (or membrane) problem Bending



The plate problemTwo possible loading states

Case 1: Loads applied within the plane Case 2: Loads applied perpendicular to the mid-surface

Plane stress (or membrane) problem Bending

i. Plate theory is only concerned with the response of the body due to bending loads

ii. The combined response under membrane and bending conditions of plane (or curved) surfaces is treated under the framework of shell theory



Plate Theories

thick thin very thin

L/t 5-10 5-100 >100

characteristicstransvereseshear deformations

no transverse shear deformations

Geometrically nonlinear

Plate theory Reissner,Mindlin Kirchhoff Von Karman

Beam Theory Timoshenko Euler, Bernoulli



Kirchhoff-Love Plate TheoryAssumptions• The plate is thin in the sense that the thickness

is small compared to the leading dimensions, but not so thin that the lateral deflection becomes comparable to .

• The plate thickness is either uniform or varies slowly so that three-dimensional stress effects are ignored.

• The plate is symmetric in fabrication about the mid-surface.

• Applied transverse loads are distributed over plate surface areas of dimension or greater.

• The support conditions are such that no significant extension of the mid-surface develops. Augustus Edward Hough Love

(1863-1940)

Gustav Robert Kirchhoff(1824-1887)



Kirchhoff-Love Plate TheoryAssumptions

From (i), since is very small, the variation of with respect to z can be neglected. Therefore:

Additionally, the following kinematic assumption is introduced:

“Planes perpendicular to the mid-surface will remain plane and perpendicular to the deformed

mid-surface”

This is the two-dimensional equivalent of the Euler-

Bernoulli kinematic assumption for beams!!

Augustus Edward Hough Love(1863-1940)

Gustav Robert Kirchhoff(1824-1887)



Kirchhoff-Love Plate TheoryRemember in the one-dimensional beam problem...

Point A displacement(that’s because the section remains plane)

(that’s because the plane remains perpendicular to the neutral axis)



Kirchhoff-Love Plate TheorySimilarly in the two-dimensional plate problem – Bending with respect to axis

Point A displacement(that’s because the section remains plane)


Positive rotation with respect to results in positive displacements

along



Kirchhoff-Love Plate TheorySimilarly in the two-dimensional plate problem – Bending with respect to axis

Point A horizontaldisplacement(that’s because the section remains plane)


Positive rotation with respect to results in negative displacements

along



Kirchhoff-Love Plate TheoryCompatibility Relations (Strain-Displacement)



Kirchhoff-Love Plate TheoryStress-strain relations (Considering elastic isotropic material)

and

1st RemarkThe constitutive relation results when we substitute the zero deformation terms (derived in the previous slide) to the three-dimensional elastic stress-strain relations. The derived relation is identical to the plane-strain case. Thus, according to the Kirchhoff-Love assumptions, every infinitesimal particle within the plate is in a plane-strain condition

2nd RemarkAn immediate consequence is that the shear stress components and vanish



Kirchhoff-Love Plate TheoryStress-strain relations (Considering elastic isotropic material)

and

Substituting for the strain components with respect to the displacement

As expected, all the stress components are linear functions of

(I)



Kirchhoff-Love Plate TheoryStress Resultants

Since the normal stress distribution is not uniform they give rise to a moment vector (created from the couple of tension and compression forces.

Since the shear stress distribution is not uniform they also give rise to a moment vector (created from the couple of tension and compression forces.



Kirchhoff-Love Plate Theory

Stress Resultants

However, are not the only stress-resultants. For these moments to be in equilibrium, a pair of shear forces must exist. Considering an infinitesimal mid-surface element the positive moments are defined as:




If a distributed load is applied onto the element, equilibrium with respect to the vertical axis z results in:




Similarly, the following equations are derived, considering moment equilibrium with respect to axes



Kirchhoff-Love Plate TheoryDifferential Form with respect to

Substituting the stress-displacement relations (I) into the definition of the moment components the following relations are derived

the bending rigidity of the plate



Kirchhoff-Love Plate Theory

Square simply supported plate subject to uniform distributed loading

bending in both directions

pay attention to the way the corner regions deform….



Kirchhoff-Love Plate TheoryThis is the effect of the twisting moment

bending in both directions

darker areas denote larger values of twisting moments

and this is why specifications make sure that an additional amount of reinforcement is provided for corner areas of slabs



Kirchhoff-Love Plate TheoryThe Kirchhoff-Love plate theory • extends the Euler/Bernoulli beam assumptions to the two-dimensional

case• Based on that, every significant measure of rotation, force, moment is

evaluated with respect to the vertical deflection

Assumptions:

The main kinematic assumption is that “Plane surfaces remain plane and perpendicular to the mid-surface of the plate”.

General Remarks

The Kirchhoff-Love theory predicts a zero distribution of shear stresses along the z direction. Thus, it can only be applied in problems where the variation of such stresses is expected to be small and their mean value does not deviate from 0. Such can be considered the case of thin plates.

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