Lecture 19: Intro to Isoparametric Formulations.mech420/Lecture19.pdf · 2007-05-07 · Lecture 19: Intro to Isoparametric Formulations. Chapter #8 Development of the Linear Strain

MECH 420: Finite Element Applications

Lecture 19: Intro to Isoparametric Formulations.

Chapter #8 Development of the Linear Strain Triangle Equations.

Our look at Chapter #8 will be brief.In Chapter #10 we will use isoparametric formulations (natural coordinates) to form the plane element stiffness Chapter #8 serves as our background/motivation to the material of Chapter #10.

The LST element uses a linear strain approximation.The displacement field must be a multivariate polynomial with second order terms.How does a second order displacement function change the element derivation that we looked at for the CST element?



§8.1. Derivation of the Linear-Strain Triangular Element Stiffness Matrix and Equations.

Writing out the assumed displacement field:

12 unknown coefficients to solve for.12 coefficients need to be calculated. Found by applying constraints on the displacement field at each of the 6 node points.Another way to interpret this is that we need 6 node points on our LST element to capture the LST element “state.”

2 21 2 3 4 5 6

2 27 8 9 10 11 12

ˆ( , )ˆ( , )u x y a a x a y a x a xy a y

uv x y a a x a y a x a xy a y

ψ⎧ ⎫+ + + + +⎧ ⎫

≈ = =⎨ ⎬ ⎨ ⎬+ + + + +⎩ ⎭ ⎩ ⎭



The extra 3 nodes above the CST element maintain compatibility along element boundaries.The element boundaries are now curved due to the second order displacement field

3

2

8

9

7

4

By adding a control point (an intermediate node point) at the midpoint of each edge we can

form a parabolic edge and ensure it is common to elements

1 and 2.

1

2

44

4

ud

v⎧ ⎫⎨ ⎬⎩ ⎭



Pascal’s Triangle is a tabulation of how plane stress elements increment in complexity (pg. #346).



ANSYS #3 works with a 8 noded quadrilateral element that is addressed in §10.6.We will see shortly why we turn to an isoparametric formulation for that element.

How would the choice of a linear displacement field affect the modelling of the arc?

There is conflict between the desire for simple stress strain measures (constant) and the straight line edges of the CST element.



Follow the same steps as the CST derivation.Using a global reference frame to define the assumed displacement field.

Set up 12 constraint equations in terms of a1, a2, …, a12.

2 21 2 3 4 5 6

2 27 8 9 10 11 12

1 12 2

2 22 2

12 12

1

ˆ( , )ˆ( , )

1 0 0 0 0 0 00 0 0 0 0 0 1

u x y a a x a y a x a xy a yu

v x y a a x a y a x a xy a y

a aa ax y x xy y

x y x xy ya

M

a

ψ

ψ

⎧ ⎫+ + + + +⎧ ⎫≈ = =⎨ ⎬ ⎨ ⎬

+ + + + +⎩ ⎭ ⎩ ⎭⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎡ ⎤ ⎪ ⎪ ⎪ ⎪= =⎨ ⎬ ⎨ ⎬⎢ ⎥

⎣ ⎦ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

2 21 2 3 4 5 6

2 27 8 9 10 11 12

ˆ for 1..6

î

i ii

x xy y

u a a x a y a x a xy a yi

v a a x a y a x a xy a y ==

⎧ ⎫+ + + + +⎧ ⎫= =⎨ ⎬ ⎨ ⎬

+ + + + +⎩ ⎭ ⎩ ⎭



This produces a set of 12 equations that defines the 12 coefficients…

The approximate displacement field, ψ , is given by:

2 21 11 1 1 1 1 1

2 22 11 1 1 1 1 1

2 211 1212 12 12 12 12 12

2 212 1212 12 12 12 12 12

1

1 0 0 0 0 0 00 0 0 0 0 0 1

1 0 0 0 0 0 00 0 0 0 0 0 1

ˆ

a ux y x x y ya vx y x x y y

a ux y x x y ya vx y x x y y

a X d−

⎡ ⎤⎧ ⎫ ⎧ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎣ ⎦

=

[ ]11

1

ˆ( , ) ˆ ˆˆ( , )u x y

a Xv

M N dM dx y

ψ −⎧ ⎫⎡ ⎤= = = =⎨ ⎬ ⎣ ⎦

⎩ ⎭

d̂



The element strains (which now vary over the element) can be calculated from the assumed displacement field.

ˆˆˆˆˆ ˆˆ ˆ

x

y

xy

udxv

dyu v

dy dx

ε

ε

γ

∂=

∂=

∂ ∂= +

ˆBdε =[ ] û N dψ≈ =

Given that N is quadratic in x and y then the B matrix must now be linear in x and y. That

is, the entries in B vary as one moves through the element domain.

Actual Definition of Stress

Approximate displacement field Approximate strains



Using the total potential energy expression:

where,

we would apply the variational principle (stationary PE at equilibrium) …

P B S PUπ = + Ω +Ω +Ω

{ }( ) ( ) ( )

1 1 1ˆ ˆ ˆ ˆ2 2 2

TT T T

e e eV V A

U dV DBd BddV td B DB dA dσ ε⎡ ⎤

= = = ⋅⎢ ⎥⎢ ⎥⎣ ⎦

∫∫∫ ∫∫∫ ∫∫

{ }0

ˆP

d

π∂=

∂ ( ) ( ) ( )

ˆ ˆ ˆ0 T T T TS

e e eA A S

t B DB dA d t N b dA N T dA f d⎡ ⎤

= ⋅ − ⋅ − ⋅ −⎢ ⎥⎢ ⎥⎣ ⎦∫∫ ∫∫ ∫∫

We have to integrate the 12x12 matrix expression.



We need to alleviate some of the difficulty within the LST element formulation.

1 44

4

ˆû

dv⎧ ⎫

= ⎨ ⎬⎩ ⎭

[ ]

[ ]

1

2 11 1 12 12 12 12 12

12

1 2 3 4 5 6

ˆ( , ) ˆˆ( , )

ˆ ˆ

aau x y

M M X dv x y

a

Bd B B B B B B d

ψ

ε

−× ××

⎧ ⎫⎪ ⎪⎧ ⎫ ⎪ ⎪= = =⎨ ⎬ ⎨ ⎬

⎩ ⎭ ⎪ ⎪⎪ ⎪⎩ ⎭

∴ = =

00m

m m

m m

Bβ

γγ β

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

Each entry is a linear (binomial or trinomial) function of x and y.



§8.2. Example LST Stiffness Derivation.Generates the B matrix for a triangular element with given node points.“The explicit expression for the 12x12 stiffness matrix, being extremely cumbersome to obtain, is not given here.”

Why use an LST over a CST element?We get a linear approximation to the strain in the element domain.Better approximation to true deformations experienced.Expect faster convergence to a consistent single solution.

But there are more state variables with each LST element so there have to significantly less elements (less than 1/2).

§8.3. Comparison of Elements.Looks at the convergence of both elements for a sample case.



More motivation to persevere through Chapter #10:In your ANSYS #3 question you are using 8 node quadrilateral elements (presented in §10.6). Still must have parabolic

sides. The extra 2 nodes over the LST element allow for two more second order terms to be placed in the

assumed displacement field.



We end up again with a B matrix containing terms that are linear in x and y.For this 8 node planar element, we have to integrate 16x16 matrix expressions in order to get the stiffness matrix.This is after getting expressions for entries of the shape function matrix by solving a 16x16 linear system of equations.

2 2 2 21 2 3 4 5 6 7 8

2 2 2 29 10 11 12 13 14 15 16

2 2 2 21 2 3 4 5 6 7 8

2 2 29 10 11 12 13 14 15

ˆ( , )ˆ( , )

ˆˆ

i

i

u x y a a x a y a x a xy a y a x y a xyu

v x y a a x a y a x a xy a y a x y a xy

u a a x a y a x a xy a y a x y a xyv a a x a y a x a xy a y a x y

ψ⎧ ⎫+ + + + + + +⎧ ⎫

≈ = =⎨ ⎬ ⎨ ⎬+ + + + + + +⎩ ⎭ ⎩ ⎭

+ + + + + + +⎧ ⎫=⎨ ⎬

+ + + + + +⎩ ⎭2

16

for 1..8ii

x xy y

ia xy =

=

⎧ ⎫=⎨ ⎬

+⎩ ⎭



Isoparametric formulations avoid the hassle associated with using Cartesian coordinates to form the interpolating polynomials.

We change from working in terms of x and y to working in terms of ‘natural’ or ‘intrinsic’ coordinates.We define a ‘mapping’ between the coordinate systems.No matter what the actual shape of the planar element, the natural variables always vary between ‘-1’ and ‘+1’.In the intrinsic coordinate plane our element will look like a square.This means that our volume and area integrals are always over the same bounds and that the integrals will be much easier to form.



If we follow the strict use of global cartesian reference frames:

( ) ( ) ( )

ˆ ˆ ˆ0 T T T TS

e e eA A S


= ⋅ + ⋅ + ⋅ +⎢ ⎥⎢ ⎥⎣ ⎦∫∫ ∫∫ ∫∫

We have to integrate the 12x12 matrix expression over an area.

1s = − 1s = +

1t = +

1t = −

s

t

It is pretty easy to set up this strip if we can

integrate in a s-t plane



In cylindrical coordinates:R defines a family of concentric cylindersz defines a family of elevated horizontal planes.

θ defines a family of rays.At any point in 3D space defined by there is an intrinsic cartesian frame.

this frame defines how subsequent motions in R, θ, and z will move you in space

{ }TR zθ



0.0η =ξ

0.00.666

ηξ

=⎧ ⎫⎨ ⎬= −⎩ ⎭

The curvilinear coordinates in an isoparametric formulation create an instrinsic or natural coordinate system.

Natural coordinate system: one that is defined by knowledge of the element’s nominal shape.

L1 is the name of a line in the plane.



The idea behind isoparametric formulations is to replace the x, y, and z cartesian coordinates with the curvilinear coordinates.For any problem the curvilinear coordinates are always bounded by fixed limits (‘-1’ to ‘+1’ for MECH 420):

1.0 1.0ξ− ≤ ≤ +

xy Rz

ξζη

⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

{ }* * * Tξ ζ η

{ }* * * Tx y z

Mapping (mapping

generated by the shape functions)



In the context of how we get the element equations…

We will consider the application of isoparametric forms for the bar element (simple case).Consider the use of quadrature in evaluating the element equation integral statements.Quadrature allows us to automate the numerical evaluation of the integral terms.

( ) ( ) ( )

ˆ ˆ ˆ0 T T T TS

e e eA A S


= ⋅ + ⋅ + ⋅ +⎢ ⎥⎢ ⎥⎣ ⎦∫∫ ∫∫ ∫∫

Change the variable (and the limits) of integration. Will always have integrals that are from -1..+1.

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Lecture 19: Intro to Isoparametric Formulations.mech420/Lecture19.pdf · 2007-05-07 · Lecture 19: Intro to Isoparametric Formulations. Chapter #8 Development of the Linear Strain