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Institute of Structural Engineering

Method of Finite Elements I

Chapter 4

Isoparametric Elements

[O] V1/Ch3/77-95[F] Ch16


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

Note: this is just a vector of thecoefficients of the linear approximation function.

u1, u2 do not correspond to displacements at the nodes



{ } [ ]{ }( )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 (DOFs) 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:

𝑑𝑑 = 𝑢𝑢(𝑥𝑥 = 0)𝑢𝑢(𝑥𝑥 = 𝐿𝐿) =

𝑢𝑢0𝑢𝑢𝐿𝐿 = 1 0

1 𝐿𝐿𝑢𝑢1𝑢𝑢2 ⇔

𝑢𝑢1𝑢𝑢2 = 1 0

1 𝐿𝐿−1 𝑢𝑢0

𝑢𝑢𝐿𝐿

The Galerkin Method

u0, uL correspond to the displacements at the nodes (nodal DOFs), which are contained in vector {d}. So here, we expressed the coefficients in terms of the nodal DOFs.



{ } [ ]{ }( )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 𝑢𝑢0

𝑢𝑢𝐿𝐿

𝑢𝑢 = 1 𝑥𝑥 1 01 𝐿𝐿

−1 𝑢𝑢0𝑢𝑢𝐿𝐿 = 𝐿𝐿 − 𝑥𝑥

𝐿𝐿𝑥𝑥𝐿𝐿

𝑢𝑢0𝑢𝑢𝐿𝐿 ⇒ 𝑁𝑁(𝑥𝑥) = 𝐿𝐿 − 𝑥𝑥

𝐿𝐿𝑥𝑥𝐿𝐿

The Galerkin Method

shape functions



The Shape Functions for the generic bar (truss) Element

Truss Element Shape Functions

(in global coordinates)

( ) [ ] 11 2

2

uu x N N

u

=



[ ]{ } [ ]{ }( )

d N xd B d

dxε = =

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



Strain-Displacement relation

defineThe truss element

stress displacement matrix

Constant Variation of strains along the element’s length



EA depends on material and cross-sectional properties

We call the nodal force vector for distributed loads

The Galerkin Method

{ } [ ] ( )∫L

T

0

f = N -αx dx

This is defined as: { } ( )( )

0f xf x L

= =

f = for concentrated loads* at the nodes

The following FUNDAMENTAL FEM expression is derived

*Notice the placement of the concentrated load on the corresponding degree of freedom where it applies, i.e., x = 0 → Node 1, x = L → Node 2



where

The Galerkin Method

[ ]{ } { }fK d =

The FEM equation to solve is (as in the DSM):

for a bar element with only two nodes(2-noded bar element)[ ] 1 1

1 1EAKL

− = −

We are however not restricted to only using 2-noded elements!

EduApp1

EduApp2

https://eduapp-app1.ethz.ch/

https://eduapp-app1.ethz.ch/



Generalized ElementsWe are not restricted to only using 2-noded elements!

In order to generalize to

• elements of higher order (more nodes) and

• generalized geometries

we need to generalize the shape functions

Higher order

Why?

[O] Ch3§2, Fig. 3.2



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)

Revisiting 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 order n and the values of φi in those n+1 points:

( ) ( ) ( )1

1

ni

ii

x x L xϕ ϕ ϕ+

=

≈ =∑where the normalized Lagrange polynomial of order n-1 isdefined by n points, as follows:

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

1 2

1 1 2

nj nn

ij i j i i i ni j

x x x x x x x xL x

x x x x x x x x=≠

− − − −= =

− − − −∏

Note that ( ) ( )0 & 0 for n ni i i jL x L x i j≠ = ≠

niL



Note that

( ) ( )0 & 0 for n ni i i jL x L x i j≠ = ≠

The Shape FunctionLagrange Polynomials




The Shape Function

The shape function Ni of a Lagrange element with n nodes coincideswith the normalized Lagrange polynomial of degree n-1, i.e.

( ) ( )ni iN x L x=

This explains why C0 continuous 1D elements are also called Lagrange elements.



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 yLet us break the process into steps:

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:

Step 1: Interpolate along x




( )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

Step 2: Interpolate along y

Step 1: Interpolate along x

( ) ( )( ) ( )

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

− −−

− −−

= +

= +

Interpolate along borders 1-2, 4-3:




1 11: u ,v

2 22 : u ,v

4 44 : u ,v3 33 : u ,v

To interpolate the displacement field for this 2D element, both• the horizontal u(x,y) and • vertical displacements v(x,y)should be interpolated

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− − − − − − − −

= + + +− − − − − − − −

ui, vi indicate the horizontal/vertical displacements at each one of the 4 nodes (i) of this element




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.

the nine-node biquadraticQuadrilateral (Lagrangian) [F

] Ch1

6§8



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

Serendipity Polynomials




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

Serendipity PolynomialsLagrange Polynomials

the nine-node biquadraticQuadrilateral (Lagrangian)

the eight-node “serendipity” quadrilateral.

[F] Ch16§8



Hermitian PolynomialsLagrangian polynomials and serendipity functions provide a C0

continuity.

If we additionally need continuity of the first derivatives between thefinite elements we use Hermitian polynomials.A Hermitian polynomial of the order n, Hn(x), is a 2n+1 order polynomial.For example a Hermitian polynomial of the first order is actually a thirdorder polynomial.The first few Hermitian polynomials:

( )( )( )( )( )

0

1

22

33

4 24

1

2

4 2

8 12

16 48 12...

x

x x

x x

x x x

x x x

Η =

Η =

Η = −

Η = −

Η = − +



Hermitian Polynomials

Let us consider a bar element with nodes on its ends. Unknowns are values ofthe function φ at the end nodes 1 and 2, φ1 and φ2, and the first derivatives ofφ with respect to x , φ1,x and φ2,x

Remember: The 1st derivative of displacement, corresponds to rotation

( ) ( )1 2

1, 2,,x xx x x x

d x d xdx dxϕ ϕ

ϕ ϕ= =

= =

11 ( )x

( )xϕ

12 ( )x

2ϕ1ϕ

1,xϕ2,xϕ

Question: What well-known type of element will need to be represented by Hermitian Polynomials?



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



next lecture & lab

So far we discussed the form of Shape Functions using a formulation of the element in global coordinates

This element formulation may be extended to any type of element however, but suffers from the following limitations:

1 The construction of shape functions for high order elements with curved geometry is highly complicated.

2 The evaluation of integrals for element stiffness matrix and force vector is not possible in closed form.

These limitations may be overcome by combining isoparametric representation & numerical quadrature.


current lecture




Natural coordinates

Shapefunctions

Geometrydescription

Displacement interpolation

,x yu u

,x y

( , , )iN r s t, ,r s t

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) coordinates

[F] Ch16§2



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 Ni 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 N r s t x y N r s t y z N r s t z= = =

= = =∑ ∑ ∑

1 1 1( , , ) ; ( , , ) ; ( , , )

n n n

i i i i i ii i i

u N r s t u v N r s t v w N r s t w= = =

= = =∑ ∑ ∑


same approximating functions – shape

functions for

Geometric interpolation

Displacement Interpolation



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

Bar Element


(in global coordinates)

( ) [ ] 11 2

2

uu x N N

u

=



Strain-Displacement relation

define The truss element stress displacement matrix




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

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

Bar Element

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

ξ = -1 ξ = 1

ξ

ξ = 0

( )( )

1

2

1. @ 0 1

2. @ 1

x x

x L x

ξ

ξ

= = = −

= = =

MM



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

1u 2u

y

x

1x2x

0r =1r = − 1r =

( ) ( )2

1 21

1 1ˆ ˆ ˆ(1 ) (1 )2 2 i i

iu u u N uξ ξ ξ ξ

=

= − + + =∑

( )2

1 21

1 1(1 ) (1 )2 2 i i

ix x x N xξ ξ ξ

=

= − + + =∑

Bar Element

ξ = -1 ξ = 1ξ = 0

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



Finite Element Idealization – Isoparametric Coordinates


(in isoparametric “natural” coordinates)

, 2A l =ξ

1ξ = − 1ξ =0ξ =

( ) ( )11 12

N ξ ξ= −

-1 1

( ) ( )21 12

N ξ ξ= +

-1 1

( ) ( ) ( ) 11 2

2

uu N N

uξ ξ ξ

=



1 2

1 1 1(1 ) (1 )2 12 2

1

2 22

x x x x xd dx dx L Jx dx d d

d Jdx L

ξ ξξ ξξ ξ

ξ

−= − + +

−

−∂= = → = = = ∂

⇒ = =

Why?

Isoparametric formulation:


( ) ( ) ( ) 11 2

2

uu N N

uξ ξ ξ

=

( ) ( )11 12

N ξ ξ= − ( ) ( )21 12

N ξ ξ= +

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

B Jx x L

ξ ξξξ ξ

−∂ ∂ ∂∂= = = = −

∂ ∂ ∂ ∂

, 2A l =ξ

1ξ = − 1ξ =0ξ =

Chair rule of differentiation

This is termed the Jacobian, J



Why?

Isoparametric formulation:



( ) ( ) ( ) 11 2

2

uu N N

uξ ξ ξ

=

( ) ( )11 12

N ξ ξ= − ( ) ( )21 12

N ξ ξ= +

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

B Jx x L

ξ ξξξ ξ

−∂ ∂ ∂∂= = = = −

∂ ∂ ∂ ∂

, 2A l =ξ

1ξ = − 1ξ =0ξ =

Chair rule of differentiation



Let us now consider the derivation of the stiffness matrix K, using the iso-parametric expression. *The strain-displacement matrix is the same for the case of the bar element for both global and isoparametric coordinates:

We write up the integral for calculating the stiffness matrix:

Why? det T T

L L

EA dx EA dξ= =∫ ∫K B B B B J


Because dxdx d dd

ξ ξξ

= = J

[ ] 1

2

ˆ1 1 1 with ˆuuL

ε

= − =

B B



[ ] 1

2

ˆ1 1 1 with ˆuuL

ε

= − =

B B

Bar ElementThe strain-displacement matrix in this case remains unchanged:

[ ]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:



As pointed out, in order to evaluate the stiffness matrix we must differentiate the displacements with respect to the coordinates (x, y, z).

The shape functions will be expressed in natural coordinates. Therefore the coordinate transformation (Jacobian) between natural and cartesiancoordinate system should be used.

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!

Isoparametric Elements in 3D

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

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

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




This is the Jacobian operator J:

1

−

∂= = ⇒

∂∂

⇒ =∂

x Jr x r x

Jx 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

which relates the natural coordinate derivatives to the Cartesian coordinate derivatives.




T

V

dV= ∫K B CB

The expression of element stiffness matrix

The strain-displacement matrix B will be expressed in terms of the natural coordinates (r,s,t) and the volume differential dV refers to the Cartesian coordinate system.

Therefore: detdV dr ds dt= J

*The Jacobian determinant is used when making a change of variables and evaluating a multiple integral of a function over a region within its domain. To accommodate for the change of coordinates the magnitude of the Jacobian determinant arises as a multiplicative factor within that integral.




T

V

dV= ∫K B CB

The expression of element stiffness matrix

The strain-displacement matrix B will be expressed in terms of the natural coordinates and the volume differential dV refers to the Cartesian coordinate system.


detT

V

dr ds dt= ∫K B CB J




T

V

q dV= ∫f N

The expression of the equivalent nodal force vector is

The strain-displacement matrix N will be expressed in terms of the natural coordinates (r,s,t) and the volume differential dV refers to the Cartesian coordinate system.


detT

V

q dr ds dt= ∫f N J



Example: Quadratic Bar Element with 3 nodes

Based on what we so far discussed, how can we define the equations for a quadratic bar element with 3 nodes?Define the higher order element by using the Lagrange Polynomial for 3 nodes, in isoparametric coordinates ξ

( ) ( ) ( )

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

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

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

1 2

3

1 2

3

1 2

3

3

1, 02 31 11

1 2 1 3

1, 01 3 22 21

2 1 2 3

1, 01 23 31

3 1 3 2

1 12

1

1 12

ni i iN L L

N N

N N

N N

ξ ξξ

ξ ξξ

ξ ξξ

ξ ξ ξ

ξ ξ ξ ξξ ξ ξ ξ

ξ ξ ξ ξ

ξ ξ ξ ξξ ξ ξ

ξ ξ ξ ξ

ξ ξ ξ ξξ ξ ξ ξ

ξ ξ ξ ξ

=− ==

=− ==

=− ==

= = ⇒

− −= → = −

− −

− −= → = −

− −

− −= → = +

− −



Example: Quadratic Bar Element with 3 nodes

Self Study TaskDerive the stiffness matrix and the equivalent nodal force vector of the quadratic 3-noded bar element!

Solution: [O] Ch3§3.3

Stiffness Matrix

Nodal force vector



Higher order bar Elements

[O] Ch3§2, Fig. 3.2

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