Method of Finite Elements I...Method of Finite Elements I Chapter 4 . Isoparametric Elements. ... If polynomials of higher order are to be used, further nodes should be typically introduced

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Method of Finite Elements I

Chapter 4

Isoparametric Elements


Method of Finite Elements I30-Apr-10

Today’s Lecture Contents

• The Shape Function: Reminder from the previous lecture

• Looking for a uniform mapping across element types

• Isoparametric elements

• 1D Demonstration: Bar elements



In the previous lecture, we defined an approximation for the displacement field:

This also defined the first derivative of the approximation (strain)

Strain field

Displacement field

The Galerkin Method



{ } [ ]{ }( )u N x d=

where u1,u2 are so far random coefficients. Instead, we can choose to write the same relationship using a different basis N(x):

Vector of degrees of freedom at the

element nodes

Shape Function Matrix

Displacement field

In this case we express u in terms of the degrees of freedom, i.e., the displacements at the ends of the bar:

Substituting in our initial displacement approximation we obtain:

𝑑𝑑 = 𝑢𝑢(𝑥𝑥 = 0)𝑢𝑢(𝑥𝑥 = 𝐿𝐿) =

𝑢𝑢0𝑢𝑢𝐿𝐿 = 1 0

1 𝐿𝐿𝑢𝑢1𝑢𝑢2 ⇔

𝑢𝑢1𝑢𝑢2 = 1 0

1 𝐿𝐿

−1 𝑢𝑢0𝑢𝑢𝐿𝐿

𝑢𝑢 = 1 𝑥𝑥 1 01 𝐿𝐿

−1 𝑢𝑢0𝑢𝑢𝐿𝐿 = 𝐿𝐿 − 𝑥𝑥

𝐿𝐿𝑥𝑥𝐿𝐿

𝑢𝑢0𝑢𝑢𝐿𝐿 ⇒ 𝑁𝑁(𝑥𝑥) = 𝐿𝐿 − 𝑥𝑥

𝐿𝐿𝑥𝑥𝐿𝐿

The Galerkin Method



[ ]{ } [ ]{ }( ) d N x

d B ddx

ε = =

Then:

The weak form also involves the first derivative of the approximation

Strain field

[ ] [ ] [ ]( ) 1 1 1d N x

Bdx L

= = −where

Strain Displacement Matrix

Displacement field

[ ]{ }( )u N x d=Vector of degrees of freedom at the

element nodes

The Galerkin Method



Therefore if we return to the weak form :

The following FUNDAMENTAL FEM expression is derived

and set:

or even betterWhy??

The Galerkin Method



EA has to do only with material and cross-sectional properties

We call The Finite Element stiffness Matrix

Ιf E is a function of {d}

Ιf [B] is a function of {d}

Material Nonlinearity

Geometrical Nonlinearity

The Galerkin Method



The Galerkin Method

[ ] [ ] [ ]

[ ]

[ ] [ ]

[ ]

[ ]

00

2

11 1 1 11

11 1 1 1 11

1 11 1

LL T k EA dxk B EA B dx L L

B EA LL L L

EAkL

− = − ⇒ = ⇒

−− = = − ⇒

− = −

∫∫

Indeed, if we use the proposed formulation for [N], [B]:

bar stiffness!



Therefore: Shape functions will be defined as interpolation functions which relate the variables in the finite element with their values in the element nodes. The latter are obtained through solving the problem using finite element procedures.

In general we may write:

where: is the function under investigation (for example: displacement field)is the shape functions matrix

is the vector of unknowns in the nodes (for example: displacements)

The Shape Function

( )( ) i ii

u x N x u u= =∑ N

( )u x

( )xN

[ ]1 2T

Nu u u u=



Regardless of the dimension of the element used, we have to bear in mind that Shape Functions need to satisfy the following constraints:

• in node i has a value of 1 and in all other nodesassumes a value of 0.

• Furthermore we have to satisfy the continuity between the adjoining elements.

Example: rectangular element with 6 nodes

The Shape Function

( )i xN

H4 H5

Why??



Usually, polynomial functions are used as interpolation functions, for example:

( )0

Ni

n ii

P x a x=

=∑

where n is the order of the polynomial; is equal to the number of unknowns in the nodes (degrees of freedom).

In the MFE we use three different polynomials: Lagrange Serendipity and Hermitian polynomials.

The Shape Function



Lagrange Polynomials

The Shape Function

A function φ(x) can be approximated by a polynomial of the order mand the values of φ(x) in those m+1 points:

( ) ( ) ( )1

1

mi

ii

x x L xφ φ φ+

=

≈ =∑where:

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

11 2 1

1 1 2 1

mj m

ij i j i i i mi j

x x x x x x x xL x

x x x x x x x x

++

= +≠

− − − −= =

− − − −∏




The Shape Function

A function φ(x) can be approximated by a polynomial of the order mand the values of φ(x) in those m+1 points:



Lagrange Polynomials – 2D Example

( ) ( )( ) ( )

1 2 1 21 2 1 1 2 2

4 3 4 33 4 1 3 2 4

L x L x

L x L x

φ φ φ

φ φ φ

− −−

− −−

= +

= +( )1 11 ,x y

( )2 22 ,x y

( )4 44 ,x y ( )3 33 ,x y

Let us break the process into steps:

1. Interpolate along borders 1-2, 4-3:

11L 1

2L

x

y

( ) ( )

( ) ( )

1 2 1 22 11 2

2 1 2 1

4 3 4 3 341 2

4 3 4 3

,

,

x x x xL x L xx x x x

x xx xL x L xx x x x

− −

− −

− −= =

− −−−

= =− −

Therefore:




( )1 11 ,x y

( )2 22 ,x y

( )4 44 ,x y ( )3 33 ,x y

11L 1

2L

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

1 1 2 2 3 4

1 4 1 2 2 3 1 2 2 3 4 3 1 4 4 31 1 1 1 2 2 2 1 3 2 2 4

L y L y

L y L x L y L x L y L x L y L x

φ φ φ

φ φ φ φ− −

− − − − − − − −

= + =

= + + +

2. Interpolate along borders 1-2, 4-3: (one possible way is below)

x

y

2 11 2 1 2

2 1 2 1

343 4 3 4

4 3 4 3

x x x xx x x x

x xx xx x x x

φ φ φ

φ φ φ

−

−

− −= +

− −−−

= +− −




1 11: u ,v

2 22 : u ,v

4 44 : u ,v3 33 : u ,v

To interpolate the displacement field for this 2D element, using the horizontal (ui) and vertical displacements (vi) at each one of the 4 nodes (i) of this element

x

y

we can use the following approximation:

( ) ( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

2 4 1 3 4 2 3 11 2 3 4

2 1 4 1 2 1 3 2 4 3 3 2 4 3 4 1

,x x y y x x y y x x y y x x y y

u x y u u u ux x y y x x y y x x y y x x y y

− − − − − − − −= + + +

− − − − − − − −

( ) ( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

2 4 1 3 4 2 3 11 2 3 4

2 1 4 1 2 1 3 2 4 3 3 2 4 3 4 1

,x x y y x x y y x x y y x x y y

x y v v v vx x y y x x y y x x y y x x y y

v− − − − − − − −

= + + +− − − − − − − −




1 11: u ,v

2 22 : u ,v

4 44 : u ,v3 33 : u ,v

The polynomials we used in this simple example are of order 1

x

y

If polynomials of higher order are to be used, further nodes should be typically introduced within the elements.

In fact, in the 2D domain a simple geometrical rule, based on Pascal’s triangle defines how many nodes are required for the representation of displacement fields of any order, according to the type of element used.



Pascal’s Triangle – 2D Example

©PM Moite, IITK

For triangular elements:

https://www.iitk.ac.in/tkic/workshop/FEM/ppt/Families%20of%20Shape%20Functions%20_PM%20Mohite.pdf




©PM Moite, IITK

For square elements & Lagrange Polynomials (which are complete:




Serendipity PolynomialsThese functions are similar to Lagrangian polynomials, but for theirincompleteness (missing terms). Due to this fact we do not need to introduceadditional inner nodes, as for Lagrangian polynomials of higher order.

Serendipity Polynomials





©PM Moite, IITK

For square elements & serendipity Polynomials (which are incomplete:




Hermitian Polynomials

Lagrangian polynomials and serendipity functions provide a C0

continuity. If we additionally need continuity of the first derivativesbetween the finite elements we use Hermitian polynomials.A Hermitian polynomial of the order n, Hn(x), is a 2n+1 orderpolynomial. For example a Hermitian polynomial of the first order isactually a third order polynomial.Let us consider a bar element with nodes on its ends. Unknowns arevalues of the function φ in the nodes 1 and 2, φ1 and φ2, and firstderivatives of φ in respect to x , φ1,x and φ2,x

Remember: The 1st derivative of displacement, corresponds to rotation11 ( )x

( )xϕ

12 ( )x

( ) ( )1 2

1, 2,,x xx x x x

d x d xdx dxϕ ϕ

ϕ ϕ= =

= =

2ϕ1ϕ

1,xϕ2,xϕ



Hermitian PolynomialsHermitian shape functions relate not only the displacements at nodes to displacements within the elements but also to the first order derivatives (e.g. rotational DOFs for a beam element).

( ) ( )2

0 11

( )( ) ii i i

i

u xu x N x u N xx=

∂= +

∂∑

( )( )( )( )

0

0

1

1

1 at node i and 0 at other nodes

0 at all nodes

0 at all nodes

1 at node i and 0 at other nodes

i

i

i

i

N x

N x

N x

N x

=

′ =

=

′ =

Shape function ofthe derivative u´(x)

Shape functionof u(x)

01N 02N

11N12N



In general, we would like to be able to represent any element in a standardized manner – introducing a transformation between a set of standardized (natural) coordinates and the real (global) coordinatesDifferent schemes exist for establishing such transformations:

1 sub parametric representations (less nodes for geometric than for displacement representation)

2 isoparametric representations (same nodes for both geometry and displacement representation)

3 super parametric representations (more nodes for geometric than for displacement representation)

Isoparametric ElementsH

ere

we

will

use

this

one



Displacement fields as well as the geometrical representation of the finite elements are approximated using the same approximating functions – shape functions

x

y

r

s

1, -1

1, 1-1, 1

-1, -1

1 1ˆ ˆ,u v

2 2ˆ ˆ,u v

3 3ˆ ˆ,u v

4 4ˆ ˆ,u v

1

4

2

3

This transformation allows us to refer to similar elements (eg. truss, beams, 2D elements, in a standard manner, using the natural coordinates (r,s), without every time referring to the specific global coordinate system (x,y).




Isoparametric elements can be one-, two- or three-dimensional:

The principle is to assure that the value of the shape function hi is equal to one in node i and equals zero in other nodes.

1 1 1; ;

n n n

i i i i i ii i i

x h x y h y z h z= = =

= = =∑ ∑ ∑

1 1 1

ˆ ˆ ˆ; ; n n n

i i i i i ii i i

u h u v h v w h w= = =

= = =∑ ∑ ∑


same approximating functions – shape

functions for

Geometric interpolation

Displacement Interpolation



In order to establish the stiffness matrixes we must differentiate the displacements with respect to the coordinates (x, y, z).

Since the shape functions are usually defined in natural coordinates we must introduce the necessary coordinate transformation between natural and global or local coordinate system. For example in the 3D domain we can express the derivative of a function which is expressed in terms of coordinates (x,y,z), with respect to another set ofcoordinates (r,s,t) as:

x y zr x r y r z r

x y zs x s y s z s

x y zt x t y t z t

φ φ φ φ

φ φ φ φ

φ φ φ φ

∂ ∂ ∂ ∂ ∂ ∂ ∂= + +

∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂

= + +∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂

= + +∂ ∂ ∂ ∂ ∂ ∂ ∂

Chain rule of differentiation!


𝜙𝜙,



Written in matrix notation:

x y zr x r y r z r

x y zs x s y s y s

x y zt x t y t z t

φ φ φ φ

φ φ φ φ

φ φ φ φ

∂ ∂ ∂ ∂ ∂ ∂ ∂= + +

∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂

= + + ⇒∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂ ∂

= + +∂ ∂ ∂ ∂ ∂ ∂ ∂

x y zr r r r x

x y zs s s s y

x y zt t t t z

φ φ

φ φ

φ φ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂




We recall the Jacobian operator J:

1 −∂ ∂= ⇒ =

∂ ∂J J

r x x r∂ ∂∂ ∂

x y zr r r r x

x y zs s s s y

x y zt t t t z

φ φ

φ φ

φ φ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂


J



Consider the bar element with two end nodes at points x1, x2 defined on the Cartesian axis x.

and now consider the following standardtruss element defined on the natural axis r

Is there a function that can map each point on the x axis to a point on the r axis, so that:

1.2.

Bar Element

r=-1 r=1

r

r=0



The relation between the x-coordinate and the r-coordinate is given as:

The relation between the displacement u and the nodal displacements are defined in the same way:

1u 2u

y

x

1x2x

0r =1r = − 1r =

( )

( )

1 2

2

1

1 1ˆ ˆ(1 ) (1 )2 2

ˆi ii

u r r u r u

N r u=

= − + +

=∑

( )

1 2

2

1

1 1(1 ) (1 )2 2

i ii

x r x r x

N r x=

= − + +

=∑

Bar Element



Let us now consider the derivation of the stiffness matrix K. Firstly, we write for the strain matrix (using matrix B):

and then we can write up the integrals for calculating the stiffness matrix:

ˆ=ε Bu

det

T

V

T

V

EA dV

EA dr ds dt

=

=

∫

∫

K B B

B B J




We need to be able to establish the strains – meaning we need to be able to take the derivatives of the displacement field in regard to the x-coordinate

du du drdx dr dx

ε = =

( )

[ ]

1 2 2 1

1 2 2 1

22 1 2 1

12 1

1 1 1ˆ ˆ ˆ ˆ(1 ) (1 )2 2 21 1 1(1 ) (1 ) ( )2 2 2 2

ˆˆ ˆ ˆ ˆ 1 1 1ˆ

du d r u r u u udr drdx d Lr x r x x x Jdr dr

uu u u udu du drudx dr dx x x L L

ε

= − + + = − = − + + = − = =

⇓

− −= = = = = − −

1u 2u

y

x

1x2x

0r =1r = − 1r =

1 2dx Jdr L

−= =

Bar Element

B



[ ]

( ) ( ) [ ] [ ]

2

1

21

ˆ1 1 1 since ˆ

or this also occurs as

1 11 1 = 1 12

LJ

uuL

N x dN r dr d r r Jx dr ds dx L

ε

=−

= − =

∂ = = = − + − ∂

B B

B

Bar ElementThe strain-displacement matrix then becomes:

[ ]1

21

1

21

11 1 ,

1 2

1 1 1 1

1 1 1 12

AE dx LdrL dr

AE L AErL L

−

−

− = − = = ⇒

− − = ⇒ = − −

∫K J J

K K

and the stiffness matrix is calculated as:



Quadrature

How can we calculate such integrals in an automated manner for higher order elements and higher dimensions?

Demo following next

[ ]1

21

11 1

1AE drL −

− = −

∫K J

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Method of Finite Elements I...Method of Finite Elements I Chapter 4 . Isoparametric Elements. ... If polynomials of higher order are to be used, further nodes should be typically introduced