lecture6 FE modeling topics Part I - PADT, Inc. · 2011 Alex Grishin MAE 323 Lecture 6 FE Modeling Topics: Part 1 7 •2D Continuum Elements s r r s •All the above isoparametric

Modeling Topics: Part IMAE 323: Lecture 6

2011 Alex Grishin 1MAE 323 Lecture 6 FE Modeling Topics: Part 1

Modeling Topics



•Common element types for structural analyis:

oPlane stress/strain, Axisymmetric

oBeam, truss,spring

oPlate/shell elements

o3D solid

oSpecial: Usually used for contact or other constraints

•What you need to know before selecting an element:

oWhat is the dominant mechanism by which work is dissipated?

oWhat DoF’s and derived quantities does the element support?

oWhat special inputs (if any are required)

oWhat parent shape(s) does the element support?

•These considerations make some elements more suitable for certain problems

than others



•Beams and Shells

x

z

y1

2

3

1 2

v1,y

x

θ2θ1

θy1

θx1θz1

ux1

uy1

uz1

θx2

θy2

θz2 ux2

uy2

uz2

X-X

y

z

Cross

section

X-X

•The user must input cross

sectional properties.

v2

u1u2

•The traditional Euler-Bernoulli beam element has

two nodes, with a third optional orientation node in

three dimensions. This is a reduced continuum

element. Supported nodal DoF’s are: displacement,

and slope. The displacements usually include

longitudinal extension. Element DoF’s are bending

stress, strain,

•The element dissipates work by bending, torsion

and linear extension exclusively. This means that

only bending, torsion and compressive/tensile

stresses and strains are calculated for derived

quantities



•Beam and Shells

•Traditional (Mindlin) shell elements come in rectangular (4-node) shapes, as well as

triangular (3-node) shapes. These elements are also dominated by bending deformation.

Nodal and element DoF’s are the same as beams but with an additional spatial dimension

•Again, the user must supply cross-sectional information

z

x

y

2

3

4

rst

θx1

uy1

θy1

ux1

θz1

uz1

θx1

uy1

θy1

ux1

θz

1

uz

1

θx1

uy1

θy1

ux1

θz1

uz1

θx1

uy1

θy1

ux1

θz1

uz1

1

12

3

r

s

t



•Beam and Shells

z,uz

x,ux

z,uz

x,ux

∆x

C

P P

C

z,uz

x,ux

z,uz

x,ux

∆x

C

PP

C

•Kirchoff Assumption:

Straight lines normal to

the mid-plane before

bending remain straight

and normal to midplane

after bending. Suitable

for thin shells

•Mindlin Assumption:

straight lines normal to

midplane before

bending remain straight

BUT not necessarily

normal after bending.

This is due to transverse

shear. Suitable for thick

shells

θy

γxz



•Beam and Shells

•Today, some commercial FE software products offer fully parametric beam and

shell elements like those shown below. The advantage of these types of elements

over traditional small-deflection beams and shells is that they carry transverse

shear components (and so are better at modeling thick beams and shells), and they

more accurately represent large rotations and large curvatures

•They also offer a fully integrated solution through the cross-section. This is helpful

when modeling composites or assessing the affects of cross section warping.

•The disadvantage of these element types is that they may experience shear locking

and/or localized spurious deflections. Also, care must be taken to reproduce Euler-

Bernoulli (predominantly bending) behavior

t

rs

1

2

3

4

2

3

4

rst

1



•2D Continuum Elements

rs rs

•All the above isoparametric elements (whose shape functions we have already

seen) can be used to model plane stress, plane strain, and axisymmetric problems in

structural mechanics . The work dissipation mechanism is governed by the elastic

internal strain energy relationship (Chapter 4, equation 3). Nodal DoF’s are

displacement x and y displacement. Element DoF’s are stress and strain

•At this point, the reader should notice a relationship between the number of nodes

an element has and the number of shape functions. When we solve the algebraic

system equations, the solution vector represents shape function coefficients, which

in turn represents solution values at nodal locations. Because of this, an element is

said to have “nodal degrees of freedom”.



•3D Continuum Elements

•All the above isoparametric elements (whose shape functions are obtained the

same way as the 2D elements) may be used to model structural problems over any

3D continuum. Nodal and element DoF’s are the same as for plane stress and strain,

but with an additional spatial dimension

•The energy dissipation mechanism is again supplied by the elastic strain energy

density relationship (Hooke’s Law)

r

s

t

r

s

t



•Spring Elements

•We have already seen spring elements (Chapter 2). They may be defined in two

spatial dimensions or three. The longitudinal spring is usually fully defined by two

nodal locations and a longitudinal stiffness, k (as in Chapter 2). The second type of

spring (the Coincident-Node spring) is fully defined by two stiffness components in

2D (three in 3D) and two coincident nodal locations. This type of spring may also have

either translational OR rotational DoF’s (the latter are identical to the former except

that the stiffness values are defined in terms of force x distance divided by degree or

radian)

•The energy dissipation mechanism is supplied by Hooke’s Law.

x

y

z

ux1

uy1

uz1

1

2 ux2

Uy2

uz2

1 2

ux1

uy1

uz1

ux2

uy2

uz2

Type 1: Longitudinal Spring Type 2: Coincident-Node Spring



•Element vs Nodal Degrees of Freedom

•In Chapter 2, the student (hopefully) got a feel for what nodal degrees of freedom

are when we spoke of spring, truss and beam elements. In that chapter, we saw that

they are simply the primary variable (the one we want to solve for) at nodal

locations. If we were to count the total number of nodal DoF’s an element has, we

would see a pattern that looks something m*n , where m is the spatial dimension

and n is the number of nodes. For example, if we have a two-dimensional spring

element, it appears that we should have 2 x 2 = 4 nodal DoF’s. However, in the

element coordinate system, this reduces to 1 x 2 =2 nodal DoF’s (spring elements

don’t use the transverse, or y-element direction). Furthermore, a two-dimensional

beam element appears to have (with extension) 6 DoF’s instead of 4!

•So, it seems that the DoF count per node for an element depends on :

- coordinate reference

-degree and form of the governing equation(s)

Canonical DoF’s and Numerical DoF’s




•In an attempt to make this all a little less bewildering, we’ll introduce some general

principles. It helps to make a distinction between what we’ll call canonical DoF’s and

numerical DoF’s. Numerical DoF’s will refer to the total number of DoF’s per element

that we use in constructing the algebraic system. This will usually determine the

matrix system size during solution. Canonical DoF’s will refer to the theoretical

minimum number of DoF’s needed to solve the problem, usually for a single

characteristic differential equation in a “preferred” coordinate system (this will be an

important idea if you’re ever asked to solve a matrix structural problem “manually”).

This preferred coorinate system is usually obvious in 1 dimension (along the domain),

but can get more complicated in two or more dimensions (more on this later)

•The number of canonical DoF’s of an element solving N differential equations may

be found by:


1

Nc

i

i

nDoF n p=

= ∑

Where pi is the order of the ith differential equation divided by 2 and the index i

ranges over the number of independent differential equations defined over the

element and n is the number of nodes.

(1)





•The number of canonical degrees of freedom, nDoFc is important in physics and

relates to the idea of generalized coordinates. This notion helps us to understand

what an element can do when it is used to approximate a structure.

And it has two nodes, n=2

•Example 1:

A bar/truss element utilizes a single differential equation:

So, p1= 2/2=1

So, by equation (1), nDoFc=1*2=2

02

2

=+ xbdx

udEA





And it has two nodes, n=2

•Example 2:

A canonical beam element utilizes a single differential equation:

So, p1= 4/2=2

So, by equation (1), nDoFc=2*2=4

•In the previous two examples, it is instructive to compare this number to m*n (the number of spatial

dimensions times the number of nodes), because this is the number of total free-particle DoF’s (the

cumulative number of DoFs each node would have in space individually). In example 1, It is less

because the element can only dissipate energy in the direction of its length. In example 2, it is equal to

m*n simply by accident (the order of the governing differential equation).

•The governing differential equation places restrictions on whether a node can move in a given spatial

direction, thus implying a preferential coordinate orientation - a set of coordinates in which motion is

prescribed by a unique differential equation along each axis. In structural mechanics, this coordinate

system will always coincide with principle coordinates

04

4

=dx

vdEI




Element name Dimension # DOFs per nodewhere these elements are

foundwhen these elements are used

Solid ( 2D elastic continuum --pl.

stress, strain, axisymmetry)2D 2 (x and y translation) everywhere

applications involving fully elastic domains in

only two spatial dimensions

Solid (2D elastic continuum) 3D 3 (x,y,z translations) everywhere Applications involving fully elastic domains

Plates/Shells 2D Canonical3 (1 transverse deflection and 2 in-

plane rotations)

textbooks and specialized

code

Thin structural members under transvers load

or moments with two distinct principle

curvatures

Plates/Shells 3D general

6 (x, y, z translations and x,y,z

rotations. Extension and torsion

added to canonical)

commercial code

thin structural members with two distinct

principle curvatures, but also extension and

surface-normal torsion

Beams 1D canonical2 (1 transverse translation and 1

rotation)

textbooks and specialized

code

long-hand calculations involving structural

bending problems of thin members (transverse

loads and moments only)

Beams 2D general

3 (x,y translations and 1 rotation.

Usually, extension is added to

canonical behavior)

textbooks and commercial

code

hand calcs and commercial applications

involving structural bending problems of thin

structures

Beams 3D general

6 (x, y, z translations and x,y,z

rotations. Extension and Torsion

added to canonical)

commercial code

commercial applications involving thin

members subjected to bending, extension, and

torsion

Springs/bars/trusses 1D canonical 1 (axial translation)textbooks and specialized

code

hand calcs involving structural members with

extension only

Springs/bars/trusses 2D general2 (x and y translation. Same

behavior as canonical)

textbooks and commercial

code

hand calcs and commercial applications

involving extension in multiple directions

Springs/bars/trusses 3D general3 (x ,y, z translations. Same

behavior as canonical)commercial code

commercial applications involving extension in

multiple directions





•Thus, this simple assessment of nDoFc tells us something very important about

truss/bar elements. Without any further calculations or evidence, we can

immediately deduce what will happen in the following situation:

•We already know from the previous example that this element cannot

dissipate energy in the y-direction (only along it’s length). So, what does this

mean in terms of the finite element solution?

ux=0

uy=0

1 2

Fy

A truss element fixed at node

1 with a transverse load at

node 2x

y





•To reinforce this line of reasoning, we can always check it by actually solving the

finite element system!

•We utilize equation 12b of Chapter 2:

2 211

2 211

2 222

2 222

x

y

x

y

Fuc cs c cs

Fvcs s cs sEA

FuL c cs c cs

Fvcs s cs s

− −

− − = − − − −





•Using the technique we learned there (slide 12 of Chapter 2) of striking out DoF’s

that are set to zero:

2 211

2 211

2 2

2

2 2

2

0

x

y

y

Fuc cs c cs

Fvcs s cs sEA

uL c cs c cs

vcs s s Fc s

− −

− − = − − − −

2

2

0

y

c csEA

L cs s F

=

•But recall that c and s are the direction cosines the element makes with the

coordinate axes:





•So, we have:

01 0

0 0 y

A

L F

E =

•This equation has no solution (k is singular)!

•On the other hand, if we had used a beam element (slide 28 of Chapter 2) and set

rotation at node 1 equal to zero, we would get a solution. So, if we need a line

element that carries a transverse load, use beams (this is the difference between

trusses and frames)! This is a major theme of ROM (Reduced Order Modeling -

using reduced continuum elements)

•We’ll leave the student with a final thought (and exercise). Count the DoF’s of the

beam of slide 28 in Chapter 2 and compare to equation (1). Then compare this

number to the rigid-body DoF’s m*n and give an explanation




•So far, we have discussed nodal DoF’s, which usually represent the primary

variable(s) at nodal locations. However, just as the slope of the transverse deflection

of beam elements may constitute yet another DoF, quantities involving other

deflection derivatives (often called derived quantities) may also constitute DoF’s of

their own. The most notable examples of such quantities are stress and strain.

However, in continuum elements, numerical integration results in an odd

phenomenon.

•Recall that in the FEM, we are actually solving integral equations and the integrals

are performed numerically with quadrature rules. This produces some error. This

error gets amplified when we take derivatives*, which we must do to calculate

stresses and strains (see slides 15 and 16 of Chapter 4). This error is not great (and

diminishes with increased order of shape functions used), but is enough to give

stress and strain contours unphysical and spurious patterns

*This is usually the case when estimating derivatives approximately




•As an example, consider a two-element (2nd order Serendipity) plane stress model

with an applied uniform shear stress at the free end:

x

yv=-1

•Now, let’s look at the internal shear stress. Before we continue, students should be

able to predict what the analytical result would be

r

s

r

s




•But look at the shear stress contours that result if one simply plots the stresses as

calculated naively from equation (8) of Chapter 5 (with second degree

isoparametric shape functions):

0

0

r

s

s r

∂ ∂

∂ = ⋅ ⋅ ⋅ ∂

∂ ∂ ∂ ∂

N 0 uσ C

0 N v

rsσ →




•Specifically, let’s focus on the shear stress distribution at

( 1 / 3)xy

sσ = →

1/ 3s =

•The location 1/ 3s =

happens to coincide with

one of the Gauss points of a

two-point rule

•The figure marks the x-locations of the same Gauss rule for both elements.

Note that the average stress at Gauss point locations is closer to the analytical

result. As we refine the elements over this domain, these points would

converge faster to the analytical result than other points.

r




•The fact that derived quantity averages are closer to the analytical solution at Gauss

points reflects the superconvergent properties of such sites. Even though taking

derivatives of approximate quantities amplifies their error, it turns out that Gauss

points of a rule 1 order less than that used to perform the integration have the same

error as the displacements!*

•Most commercial FE code calculates and stores stress and strains at these Gauss

point locations OR at element centroids. Such locations may be thought of as

providing the element DoF’s (and are often referred to as such)

•And because these sites provide such increased accuracy for derived quantities, the

value of these quantities at other locations is usually extrapolated from these sites

*R. D. Cook, D. S. Malkus, and M. E. Plesha, Concepts and Applications of Finite

Element Analysis, 3rd ed. New York, NY, USA: John Wiley & Sons, 1989.




•Thus, to obtain smoother, more accurate stress and strain contour plots, most

modern FE codes extrapolate these quantities according to a scheme like the one

shown below for 2nd order isoparametric elements

•Here, r and s are the usual

isoparametric coordinates,

such that Gauss Points 1, thru

4 lie at , 1/ 3r s = ±

•To make things a little easier,

let a new parametrization, p, q

range from ±1 over the Gauss

points

p

q

r

s

1 2

34

A B

CD

•Thus, the two coordinate systems are related by a scale factor of sqrt(3). In other

words: 3

3

p r

q s

=

=

1 2

34




•To find element stresses anywhere within the domain, use:

•So, if we wanted to calculate σx (say) at point A (node 1: p=q=-sqrt(3)), we would

have:

4

1

p p

i i

i

Nσ σ=

=∑

Where σi is any stress component at the four Gauss point locations, and the Ni are

now bilinear shape functions in p and q given by:

( ) ( )( )( )( ) ( )( )( )

1 1

1 11( , )

1 14

1 1

p q

p qp q

p q

p q

− −

+ − =

+ + − +

N

1 2 3 41.866 0.500 0.134 0.500

xA x x x xσ σ σ σ σ= − + −




•Similar extrapolation formulas can be used for triangular, as well three dimensional

tetrahedral and hexahedral elements

•Once the extrapolation is performed for each element, coincident stress/strain

values at corner nodes are averaged. In commercial systems, such plots are often

referred to as averaged, or nodal stresses (unmodified stress values at Gauss points

are referred to as element stresses).

•With schemes such as this, only corner node stresses are calculated and stored.

Below is an averaged shear stress plot of the problem on slide 17

•Even with just two

elements, the stress plot is

now much closer to the

analytical value. This plot at

least now starts to make

physical sense




•Unaveraged element stresses and strains also provide a good measure of mesh

quality. This usually manifests itself as elevated stress or strain levels in poorly

shaped elements, as shown below

Ω2Ω3

Ω4 Ω1

Poorly

shaped

element…

…Leads to

spurious

strain result



•The Patch Test and Energy Error

•In the last lecture, we learned about the element shape metrics. Most

commercial FE software comes with some such metric to alert the user

when element aspect ratios, lengths, or angles become so distorted that

the solution accuracy may suffer. There are also ways the user may check

element quality directly

•One way is to conduct a “patch test”. This is done by selecting a group of

suspect elements and applying either a constant strain or constant stress

condition to them. If the element aspect ratios are within acceptable

limits (and the underlying formulation is sound), the elements should

respond by re-producing the constant strain or stress field exactly.

Wherever this is not the case, the user can conclude that the elements

may be excessively distorted.




ux=

0uy=

0

p

x

y Example showing a test

for constant σy



•Implementing the patch test in Mechanical APDL

ux=0

uy=0

p

•Below is an example of apply a constant σy condition on an

element “patch”

x

y




•Results of the constant stress patch test:

x

y

Problem

elements




•Below is an example of applying a constant strain condition to

the element patch. This is done by applying a bi-linear

displacement field to all nodes

x

y

Y

X 0.25 0.6

-0.4 0.1 0.2

0.45 0.2 0.2

Displacement

Gradient




•The resulting strain in the x-direction…

x

y




•Most commercial FE codes also have the capability of plotting the

strain energy or strain energy density. This is usually done on an

unaveraged (or element) basis. This corresponds to element strain

energy, which is proportional to the square of the stress. Because

of this, it is easy to see differences between adjacent elements. If

the differences are large, that may indicate element problems



•Structural Symmetry And Modeling Boundary Conditions

•We have already encountered two types of symmetry found in structural elastostatic

problems: planar and axisymmetry. Both of these types of symmetry exploit the fact that

a domain’s loads, boundary conditions, and geometry admit a planar section in which all

other parallel planar sections yield an identical structural response. Other types of

symmetry one can encounter in structural problems involve boundary conditions. These

are:

•Symmetry Boundary Condition

•Anti-Symmetry Boundary Condition

•Cyclic Symmetry Boundary Condition



•Structural Symmetry

oSymmetry Boundary Condition

•A symmetry boundary condition is a surface on which primary solution variables

(displacement in structural models) are set to zero normal to the surface. Such a

boundary condition usually reflects a theoretical dividing line through a domain – both

sides of which behave identically. The figure below shows a planar model (one type of

symmetry), which is identically loaded at opposite ends. The geometry and loading admit

a plane of symmetry which can be used to “cut” the model in half

Half-symmetry model

x

y

ux=0




oAnti-Symmetry Boundary Condition

•An anti-symmetry boundary condition is a surface on which primary solution variables

(displacement in structural models) are set to zero within the plane of the surface. Such a

boundary condition also reflects a hypothetical dividing line through a domain, but this

time, it usually represents a half of structure – the other half of which deforms in the

opposite direction along the symmety plane. The figure below shows a planar model

whose loading suggests a plane of symmetry parallel to it

x

yHalf-anti- symmetry model

ux=0




oCyclic Symmetry Boundary Condition

•A cyclic symmetry boundary condition can be imposed on a structure if it has repeated or

identical features (or parts) around a central axis of revolution.

x

y

r

θ S1

S2

1 2S Sθ θ=



More About Shell/Plate

Elements



DoF’s at each node:

ux,uy,uz,θx, θy, θz

Solid (tet)

elements Shell elements

A tubular section

modeled with shell

elements

•Shells are a planar extension

of beam elements. They are

used to model thin structures

whose primary energy

dissipation mechanism involves

bending in more than 1

dimension



•ANSYS also supports a special kind of shell (Solsh190) that actually has

the same geometry (node configuration) as a linear hexahedral or prism

element It is referred to as a “solid shell element”. This works by

extrapolating displacement and rotation DoF results from the midplane

(or neutral axis) to 8 (or six) corner nodes

•The advantage of this type of element is that it eliminates the need for

the analyst to extract mid-plane geometry from solid CAD data in order to

define the shell/plate surfaces. It is still a shell element (dominated by

bending), but it has full 3D solid element geometry.

DoF’s at each node:

ux,uy,uz

Solid Shell Elements



•Below is an example of the type of model for which the solid

shell option makes sense. It has several thin members in bonded

contact…

A typical member



•This special solid shell is implemented in Workbench by first

inserting a sweep meshing method. Then select “Automatic

Thin” next to “Src/Trg Selection”

Solid shell elements

Sweep Method

Automatic Thin

Documents

lecture6 FE modeling topics Part I - PADT, Inc. · 2011 Alex Grishin MAE 323 Lecture 6 FE Modeling Topics: Part 1 7 •2D Continuum Elements s r r s •All the above isoparametric