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Lecture 8 and 9: Two-dimensional elastic problems

Linear Finite-Element Analysis - II

Lecturer: Johan Clausen

Department of Civil Engineering

Lecture 8 – 9: 2D elastic problems

Linear Finite-Element Analysis - II 2

Today

There will be many similarities between this lecture and the lecture

concerning 2D heat conduction.

The main difference is that there will be two degrees of freedom (dof) at each

node. These are the horizontal and the vertical displacements. The main

variable is therefore a vector, u(x,y) compared with the scalar of

temperature analysis, T(x,y).

Instead of heat loads, the loads today will be forces/area (stress), line loads

(force/length) or a single force.

The analogy of the heat fluxes qx and qy [W/m2 = J/(m2s)] are the stresses x,

y and xy [Pa = N/m2].

The analogy of the heat conductivities kxx and kyy [J/(Cms)] are the elasticity

modulus E [Pa] and the Poisson ratio, [-]

We will focus on elements with quadratic shape functions, N, instead of the

linear shape functions from the 2D heat conduction lecture.

We will assume small displacements and linear elasticity. This means that our

calculations are linear.



Two-dimensional elasticity

Plane stress, t is small

Plane strain, t is large

All real bodies are three-dimensional

If the loads, supports and material characteristics are independent of the z-coordinate,

the domain can be approximated with a 2D domain with the thickness t(x,y)

z

z x

y

x

y

x

y

tt

q q q



Discretization into two-dimensional finite elements

The body is discretized with a number of finite elements.

They can be triangular or quadrilateral

Straight or curved element sides

x

y

q q



Today’s elements

The ”quadratic Melosh” element

- Eight nodes

- etype = 14 in the course program

Element side (edge)

Element node at the corner (vertex)

Element interior

(mid-) side node

The isoparametric eight-noded element

(IsoQ8)

- Eight nodes


The six-noded triangular element

(LST, Linear STrain)

- six nodes


The elements of this lecture all have quadratic shape functions.



The element from the 2D heat transfer lecture

The elements from the lecture on 2D heat conduction have all got structural counterparts,

which are included in the course program. They all have linear shape functions

The isoparametric four-noded

element (IsoQ4)

- Four nodes


The three-noded triangular element

(CST, Constant STrain)

- three nodes


The Melosh element

- Four nodes




The Patch Test

The main exercise of today is the patch test.

You will program its main parts and use it to

verify that the elements work properly.

A patch test should include at least one

internal node.

The patch test should be as less constrained

as possible, i.e. the body should be statically

determinate.

Notice that this calls for two more boundary

conditions on the main variable compared to

the heat conduction problem.

The geometry is the same as in the heat

conduction problem.

q

uy

ux

F1



Basic steps of the finite-element method (FEM)

1.Establish strong formulation

Partial differential equation

2.Establish weak formulation

Multiply with arbitrary field and integrate over element

3.Discretize over space

Mesh generation

4.Select shape and weight functions

Galerkin method

5.Compute element stiffness matrix

Local and global system

6.Assemble global system stiffness matrix

7.Apply nodal boundary conditions

Temperature/flux/forces/forced displacement

8.Solve global system of equations

Solve for nodal values of the primary variables (displacements/temperature)

9.Compute temperature/stresses/strains etc. within the element (if needed)

Using nodal values and shape functions



Step 1: Establish strong formulation for 2D linear elasticity (OP pp. 235-258)

nt

x

y

Stress vector

Boundary normal

Boundary traction vector

Strains are given by

[Pa]

x

y

xy

σ

σ

σ

σ

1

x

y

n

nn n

x x x xy y

y xy x y y

t σ n σ n

t σ n σ nt

0

0

2

x

x

xy

y

y

xy

yx

u

x xεuu

εuy y

εuu

y xy x

ε u



nt

x

y

The constitutive relation in matrix notation is:

x

y

xy

σ

σ

σ

σ Dε

x

y

yx

u

x

u

y

uu

y x

ε

where D is the constitutive matrix in plane strain (εz = 0)

12

1 0

1 0(1 )(1 2 )

0 0 (1 2 )

ν νE

ν νν ν

ν

D

or in plane stress (σz = 0)

2

12

1 0

1 01

0 0 (1 )

νE

νν

ν

D

Note: out-of-plane stress given by

Note: out-of-plane strain given by

( ) z x yσ ν σ σ

( )1

z x y

νε ε ε

ν

E is the modulus of elasticity

(Young’s modulus) [Pa]

ν is Poisson’s ratio [-]



nt

x

y

The equilibrium condition leads to: The sum of all surface tractions and body

forces must vanish (equal zero):

A L

t dA t dL b t 0

The first equation is

Gauss’ divergence theorem, (OP p. 73)

x x

A L

b t dA t t dL 0

( )xyx

x x x xy y

L L A

σσt t dL σ n σ n t dL t dA

x y

0 0xy xyx x

x x

A

σ σσ σb t dA b

x y x y

0y xy

y

σ σb

y x

Analogously for the y-direction equations

b is internal

body load

[N/m3]

x x x xy yt σ n σ n

Re-insertion into the first equation leads to

L

A



nt

x

y

To be able to write the equilibrium equations in matrix notation, we define the

following matrices:

T

0

, ,

0

x

x

y

y

xy

σbx y

σb

σy x

σ b

We can then write the strong form of the equilibrium

equations as

T

0

0

0

xyxx

y xy

y

σσb

x y

σ σb

y x

σ b

L

A



nt

x

y

The strong form in the x-direction,

is multiplied by an arbitrary weight function vx(x,y) and

integrated over the domain:

The x-component of the traction vector, t, was given by tx = σxnx + σxyny.

This makes it possible to write the above equation as

0xyxσσ

xx yb

L

A

Step 2: Establish weak formulation for 2D linear elasticity (OP pp. 292-258)

0xyx

x x x x

A A A

σσv t dA v t dA v b t dA

x y

The Green-Gauss

theorem

(OP p. 74)

0

x xx x x x x xy y xy

L A L A

x x

A

v vv σ n t dL σ t dA v σ n t dL σ t dA

x y

v b t dA

0x xx x x xy x x

L A A

v vv t t dL σ σ t dA v b t dA

x y



nt

x

y

Analogous manipulations are carried out for the strong form in the

y-direction, , and hereby the equations for both

directions are given:

Addition of the two equations is still equal to zero:

0y xyσ σ

yy xb

L

A

0

0

x xx x x xy x x

L A A

y y

y y y xy y y

L A A


x y


y x

( ) ( )

0

x x y y x x y y

L A

y yx xx xy y xy

A

v t v t t dL v b v b t dA

v vv vσ σ σ σ t dA

x y y x



nt

x

y

With the following vectors:

the equationL

A

, , ,

and

x

x

x x y

y

y y

xy

yx

x

y

v

xσ

b v vσ

b v yσ

vv

y x

t

t

b v v σ

t

( ) ( )

0

x x y y x x y y

L A

y yx xx xy y xy

A

v t v t t dL v b v b t dA

v vv vσ σ σ σ t dA

x y y x

can be written as the weak form of the equilibrium equations for

a 2D solid in matrix notation:T T T( )

A L A

t dA t dL t dA v σ v t v b

Boundary loads Volume loads (e.g. gravity)



nt

x

y

The weak form of the equilibrium equations is often caled the

principle of virtual displacements or also, in Danish,

“arbejdsligningen”.

The boundary intetral can be split into two terms to reflect

the two different types of bounddary conditions:

A

T T T( )A L A

t dA t dL t dA v σ v t v b :fL t t

:uL u u

Prescribed quantities

Virtual displacements

TT T

f uLL L

t dL t dLt dL v tvt tv

Prescribed traction

(load)

Prescribed displacement

(e.g. a support)



Basic steps of the finite-element method (FEM)

1.Establish strong formulation

Partial differential equation

2.Establish weak formulation

Multiply with arbitrary field and integrate over element

3.Discretize over space

Mesh generation

4.Select shape and weight functions

Galerkin method

5.Compute element stiffness matrix

Local and global system

6.Assemble global system stiffness matrix

7.Apply nodal boundary conditions

Temperature/flux/forces/forced displacement

8.Solve global system of equations

Solve for nodal values of the primary variables (displacements/temperature)

9.Compute temperature/stresses/strains etc. within the element (if needed)

Using nodal values and shape functions



The body is discretized with a number of finite elements.

x

y

Step 3: Discretize over space



As with heat conduction, we choose the Galerkin method.

This means that the main variables u(x,y) (the

displacements) and the weight functions v(x,y) are

interpolated using the same interpolation functions.

In matrix form:

Step 4: Select shape and weight functions

1 2 8

1 2 8

1 2 8

1 2 8

( , ) ( , ) ( , )( , )

( , ) ( , ) ( , )

x x x x

y y y y

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

u N x y u N x y u N x y u

u

1

1

2

1 2 8 2

1 2 8

8

8

0 0 0

0 0 0

x

y

x

x

y

y

x

y

u

u

uu N N N

uu N N N

u

u

( , ) and ( , )x y x y u Na v NV

1

xu 2

xu

3

xu4

xu

5

xu

6

xu

7

xu

8

xu

1

yu 2

yu

3

yu4

yu

5

yu

6

yu

7

yu

8

yu

1 1( , )x y 5 5( , )x y 2 2( , )x y

6 6( , )x y

3 3( , )x y7 7( , )x y4 4( , )x y

8 8( , )x y

Examples:1 1 1 1 2 2 1 7 7( , ) 1, ( , ) 0, ( , ) 0, etc...N x y N x y N x y



With Hooke’s law the weak form may now be written as:

Noticing that V and a are constants (nodal values of the arbitrary field and the nodal

displacements), we can rearrange:

T T T T

T T T T

( )

( )

f u

f u

A L L A

A L L A

t dA t dL t dL t dA

t dA t dL t dL t dA

v σ v t v t v b

v D u v t v t v b

T T T T( ) ) ) )

f uA L L A

t dA t dL t dL t dA NV D Na NV t NV t NV b

T T T T T 0

f uA L L A

t dA t dL t dL t dA

V N D N a N t N t N b

Because V is arbitrary the parenthesis term must vanish (equal zero)

T T T T

T T T T

f u

f u

A L L A

A L L A

t dA t dL t dL t dA

t dA t dL t dL t dA

N D N a N t N t N b 0

B DB a N t N t N b

With B N

σ Dε

ε u

v NV

u Na



In order to be able to write the equations in a compact form the following terms are

introduced:

The stiffness matrixT

T T

T

f u

A

b

L L

l

A

t dA

t dL t dL

t dA

K B DB

f N t N t

f N b

or ,b l b l Ka f f Ka f f f f

1

xu 2

xu

3

xu4

xu

5

xu

6

xu

7

xu

8

xu

1

yu 2

yu

3

yu4

yu

5

yu

6

yu

7

yu

8

yu

1 1( , )x y 5 5( , )x y 2 2( , )x y

6 6( , )x y

3 3( , )x y7 7( , )x y4 4( , )x y

8 8( , )x y

1 1

1 1

2 2

2 2

16 16 16 116 1

8 8

8 8

, ,

x x

y y

x x

y y

x x

y y

u f

u f

u f

u f

u f

u f

K a f

Boundary terms

(boundary load vector)

Internal load vector

The equations can now be written as

Example: The ”Quadratic Melosh” structural

element:

81 2

81 2

8 81 1 2 2

1 2 8

2 161 2 8

3 16

0 0 0

0 0 0

0 0 0

0 0 0

NN N

x x x

NN N

y y y

N NN N N N

y x y x y x

N N N

N N N

N

B

Degrees of

freedom, dof



The definition of the strain was:

0

0

2

x

x

xy

y

y

xy

yx

u

x xεuu

εuy y

εuu

y xy x

ε u

( , ) ( , ) ( )

( , )

x y x y

x y

ε u Na N a Ba

σ Dε DBa

1

xu 2

xu

3

xu4

xu

5

xu

6

xu

7

xu

8

xu

1

yu 2

yu

3

yu4

yu

5

yu

6

yu

7

yu

8

yu

1 1( , )x y 5 5( , )x y 2 2( , )x y

6 6( , )x y

3 3( , )x y7 7( , )x y4 4( , )x y

8 8( , )x y

With the finite element formulation we have

u(x,y) = N(x,y) a. This gives us the strain and

stress within the element as:

81 2

81 2

8 81 1 2 2

1 2 8

2 161 2 8

3 16

0 0 0

0 0 0

0 0 0

0 0 0

NN N

x x x

NN N

y y y

N NN N N N

y x y x y x

N N N

N N N

N

B

B Nx

y

xy

σ

σ

σ

σ

1

1

8

8

x

y

x

y

u

u

u

u

a



2 2

8 51 5 2

2 2

5 62 6 2

2 2

6 73 7 2

2 2

7 84 8 2

( )( ) ( )( )

4 2 2

( )( ) ( )( )

4 2 2

( )( ) ( )( )

4 2 2

( )( ) ( )( )

4 2 2

N Na x b y a x b yN N

ab a b


ab ab


ab a b


ab ab

Example: The ”Quadratic Melosh” element:

Select shape and weight functions

Examples of horizontal displacement of nodes 7 and 3

a

b

a

b

(-a,-b) (a,-b)

(a,b)(-a,b)

8

1 5 2

6

374

y

x

y

y

x

x

N1

N5

1

1

7

xu 3

xu



8 5 5 61 2

6 7 7 83 4

2 2 2 2

5 62 2

2 2 2 2

7 82 2

( )( ) ( )( )

4 2 4 2

( )( ) ( )( )

4 2 4 2

( )( ) ( )( )

2 2

( )( ) ( )( )

2 2

N N N Na x b y a x b yN N

ab ab

N N N Na x b y a x b yN N

ab ab

a x b y a x b yN N

a b ab

a x b y a x b yN N

a b ab

Example: The ”Quadratic Melosh” element:

Notice that there is no overlap or holes

between neighbouring elements:

a

b

a

b

(-a,-b) (a,-b)

(a,b)(-a,b)

8

1 5 2

6

374

y

x

2 2 2 2 2

3 2 2 2

2 2 2 2 2

4 2 2 2

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

4 2 2 2 2

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

4 2 2 2 2

a a b y a a b y a a b y b y yN a y

ab ab a b b

a a b y a a b y a a b y b y yN a y

ab a b ab b



Example: Calculation of the ”Quadratic Melosh” stiffness

matrix using Maple

a

b

a

b

(-a,-b) (a,-b)

(a,b)(-a,b)

8

1 5 2

6

374

y

x

y

y

x

x

N1

N5

1

1

1. Input shape functions.

2. Calculate the strain interpolation matrix, B.

3. Define the constitutive matrix, D.

4. Carry out the matrix multiplication.

5. Integrate each element of the matrix product to obtain K.

6. Program K into the file KQuadMeloshLinElas.m

Assumptions: D and t are constant.

Element width: 2a, height 2b.



Boundary load vector a

b

a

b

8

1 5 2

6

37

4

y

xCan only contain values at nodes. So distributed loads

must be converted into nodal loads:

q

yq

xq

θ

T T

f u

b

L L

t dL t dL f N t N t Prescribed term

Can only contain

nodal valuesReactions

Example: constant load q [N/m2 = Pa], between nodes 1 and 2:

(see the Maple-file ”LineLoadQMelosh.mws”)

T( , )a

b

a

x b t dx

f N q

T

T

( , )

1 1 1 1 2 20 0 0 0 0 0 0 0 0 0 2

6 6 6 6 3 3

ax

b

ya

x y x y x y

qx b t dx

q

q q q q q q at

f N

sin

cos

x

y

q θ

q θ

q

q

Remember: if the load direction is opposite the coordinate axes, it has a negative sign



Internal load vector

Can only contain values at nodes. So distributed loads

must be converted into nodal loads:

T

l

A

t dA f N b

Exercise: Calculate the element load vector for a constant

internal load. It is recommended that you use mathematical

software, e.g. Maple where your starting point can be a

modification of the file ”LineLoadQMelosh.mws”.

3

N

m

x

y

b

b

b

Remember: if the load direction is opposite the coordinate axes, it has a negative sign

a

b

a

b8

1 5 2

6

374

y

xb



Step 6: Assembling

a

b

a

b

(-a,-b) (a,-b)

(a,b)(-a,b)

8

1 5 2

6

374

y

x

y

y

x

x

N1

N5

1

1

1. Determine the local stiffness matrix.

2. Determine the global number of dof corresponding to the

local dof for the element.

3. Add the components of the local stiffness matrix to the

rows and columns of the global stiffness matrix

corresponding to the global dof numbers.

4. Repeat 1-3 until all contributions from all elements have

been added.

The assembling procedure is exactly the same

as in all the other lectures:

In the course MatLab code this is done in

Assemblering.m :

K(gDof,gDof) = K(gDof,gDof) + Ke;



Assembling the load vector

The global matrices are assembled exactly as in

the other lectures.

1 2 3 4 5

87

109

6

11 12 13

1

xf

1

yf

6

yf

9

yf10

yf 11

yf12

yf13

yf

7

yf

2

yf 3

yf4

yf5

yf

8

yf

2

xf3

xf4

xf5

xf

8

xf

13

xf12

xf11

xf

7

xf

10

xf9

xf

6

xf

The figures show a system of two

elements with some applied loads and the

corresponding global nodal load vectors.

1q 2q

3q

4q

b

11

1 2 4

11

1 1 1 2 2

1 12 2 2

12 6

1 1 12 2 2 2

12 6 6

x x

y y

f a bt b a tq

f a bt b a tq a tq

b

b

2a1 2a2

Example:



Exercise: Program and plot the results of a 2D patch test for the ”Quadratic

Melosh” element

t = 0.2 m, E = 1.3 MPa, ν = 0.3

Steps

1. Program the geometry

2. Determine nodal loads fx9, fx

14, fx17, fx

5, fx8,

fx13, fx

16 and fx21.

3. Make the appropriate programming in

Topology.m, Coordinates.m and

BoundaryConditions.m

4. Run the program by hitting F5 in main.m

5. Plot the results by executing the line

”visualize2D”x

y 17

xf 1000Paq

14

xf

9

xf

Hint: Don’t rely on the plot of the deformed structure. It is scaled. You can

display the displacement vector by typing “u” in the command window.

Compare with the analytical solution in plane stress:

( , )

( , )

x

y

qu x y x

E

qu x y ν y

E



Why not use the ”Quadratic Melosh” element always?

The Melosh element must be rectangular and positioned

along the coordinate axes.

It is therefore not very good at approximating boundaries

not aligned with the coordinate axes.

Conclusion: The Melosh element is not very flexible.

The isoparametric eight node element (Iso8)

A quadralilateral element which can be distorted from the

rectangular shape and rotated arbitrarily in the plane.

The element sides can be curved.

Makes use of several important concepts in finite element

theory, such as

Isoparametric coordinates.

Parent and global domain.

The Jacobian matrix and the Jacobian.

Numerical integration by Gauss quadrature.

The Isoparametric forms the basis for nearly all elements in

pratical use



Parent and global domain

In the parent domation the element is always a square

element with the side length 2 with midside nodes.

The isoparametric coordinates ξ and η have their origin

at the element centroid. This means that the nodes

have the coordinates (ξ, η): 1: (-1,-1), 2: (1,-1), 3: (1,1),

4: (-1,1), 5: (0,-1), 6: (1,0), 7: (0,1) and 8: (-1,0).

In the global domain the nodes have the coordinates:

1: (x1,y1), 2: (x2,y2), etc...

The question is ”how do we connect these two sets of

coordinates?”

x

y

1

1 1

1

1

8

47

3

6

25

ξ

η

ξ

η

12

η

1η

12

η

1η

1ξ 12

ξ 1ξ

12

ξ

Parent

domain

Global

domain

Why use the isoparametric formulation?

The overall difficulty is to carry out the stiffness matrix

integration over the element area, A.

The isoparametric formulation is a ”trick” that

fascilitates this integration.T ( , ) ( , )

A

x y x y t dA K B DB



x

y

1

1 1

1

1

8

47

3

6

25

ξ

η

ξ

η

12

η

1η

12

η

1η

1ξ 12

ξ 1ξ

12

ξ

Parent

domain

Global

domain

The Isoparametric coordinate transformation

The idea is that we use the shape functions to

interpolate the coordinates between the nodes.

node node

1 1

( , ) and ( , )i n i n

i i i i

i i

x N ξ η x y N ξ η y

node 8 (for the Iso8-element)n

The Isoparametric eight note element (Iso8) has the

shape functions:

2

8 51 5

2

6 73 6

2

5 62 7

2

7 84 8

(1 )(1 ) (1 )(1 )

4 2 2

(1 )(1 ) (1 )(1 )

2 2

(1 )(1 ) (1 )(1 )

2 2

(1 )(1 ) (1 )(1 )

2 2

N Nξ η ξ ηN N

N Nξ η ξ ηN N

ab

N Nξ η ξ ηN N

ab

N Nξ η ξ ηN N

ab

Notice that these are the shape functions of the ”quadratic

Melosh” element with a = b = 1.



x

y

1

1 1

1

1

8

47

3

6

25

ξ

η

ξ

η

12

η

1η

12

η

1η

1ξ 12

ξ 1ξ

12

ξ

Parent

domain

Global

domain

Displacement and stress interpolation

As usual the displacements within the element are

interpolated from the nodal values

( , )x y u Na 1

1

16 1

8

x

y

y

u

u

u

a

The stresses are still given by:

( , ) ( , )

x

y

xy

σ

x y σ x y

σ

σ DB a

with

81 2

81 2

8 81 1 2 2

3 16

0 0 0

0 0 0

NN N

x x x

NN N

y y y

N NN N N N

y x y x y x

B

Problem: We have to finde the derivatives of the shape

functions with respect to the global (x,y) coordinates, but

the shape functions are expressed in the isoparametric

(ξ,η) coordinates. To overcone this problem we will

define B as a product of three matrices: 1

exp ,expN

B HJ D



x

y

1

1 1

1

1

8

47

3

6

25

ξ

η

ξ

η

12

η

1η

12

η

1η

1ξ 12

ξ 1ξ

12

ξ

Parent

domain

Global

domain

The first matrix, H, relates strains and displacement

derivatives:

1 0 0 0

0 0 0 1

0 1 1 0

H

1

exp ,expN

B HJ D

1 0 0 0

0 0 0 1

2 0 1 1 0

x x

x x

x

yy y

xy

y y

u u

x x

u uε

y yε

u uε

x x

u u

y y

ε H



x

y

1

1 1

1

1

8

47

3

6

25

ξ

η

ξ

η

12

η

1η

12

η

1η

1ξ 12

ξ 1ξ

12

ξ

Parent

domain

Global

domain

The second matrix, Jexp-1, is an expanded form of the

inverse jacobian matrix, J-1 :

1

exp ,expN

B HJ D

1

2 2 2 2

1

2 2 2 2

xx

x x

y y

y y

uu

ξx

u u

y η

u u

x ξ

u u

y η

J 0

0 J

1

1 2 2 2 2

exp 1

2 2 2 2

J 0J

0 J



x

y

1

1 1

1

1

8

47

3

6

25

ξ

η

ξ

η

12

η

1η

12

η

1η

1ξ 12

ξ 1ξ

12

ξ

Parent

domain

Global

domain

The third matrix, DN,exp, is an expanded form of the

matrix, DN :

1

exp ,expN

B HJ D

181

1

281

2

81

8

81 8

0 0

0 0

0 0

0 0

xx

y

x

x

yy

x

y

y

u NNu

ξ ξ ξu

u NNu

η η ηu

u NN

ξ ξ ξu

u NNu

η ηη

,11 ,12 ,18

,21 ,22 ,28

,exp

,11 ,12 ,18

,21 ,22 ,28

0 0 0

0 0 0

0 0 0

0 0 0

N N N

N N N

N

N N N

N N N

D D D

D D D

D D D

D D D

D

Nodal

displacement

vector, a


Linear Finite-Element Analysis - II

,exp ,

x

x

Ny

y

u

ξ

u

η

u

ξ

u

η

D a

38

x

y

1

1 1

1

1

8

47

3

6

25

ξ

η

ξ

η

12

η

1η

12

η

1η

1ξ 12

ξ 1ξ

12

ξ

Parent

domain

Global

domain

The matrix, DN contains the derivatives of the shape

functions with respect to the isoparametric coordinates:

81 2

81 2

N

NN N

ξ ξ ξ

NN N

η η η

D

If we wanted the displacement derivatives in (ξ,η) co-

ordinates we were done now. But we need them in the

(x,y) system in order to determine the stresses. The

transformation of derivatives from the (ξ,η) system into

the (x,y) system is carried out by the inverse of the so-

called ”Jacobian Matrix”, J-1.

,11 ,12 ,18

,21 ,22 ,28

,exp

,11 ,12 ,18

,21 ,22 ,28

0 0 0

0 0 0

0 0 0

0 0 0

N N N

N N N

N

N N N

N N N

D D D

D D D

D D D

D D D

D



x

y

1

1 1

1

1

8

47

3

6

25

ξ

η

ξ

η

12

η

1η

12

η

1η

1ξ 12

ξ 1ξ

12

ξ

Parent

domain

Global

domain

The Jacobian matrix

The matrix J is called ”the Jacobian matrix” and it relates

the displacement derivatives int the two coordinate

systems

and

y yx x

x yx y

u uu u

ξ ξx x

u uu u

yη yη

J J

The elements of J can be found by differentiating the

displacement with respect to (ξ,η) by invoking the chain rule

x x x

x x x

u u ux y

ξ x ξ y ξ

u u ux y

η x η y η

This gives us the elements of J

node node

node node

1 1

1 1

[ ]

i n i n

i ii i

i i

Ni n i n

i ii i

i i

N Nx yx y

ξ ξ ξ ξ

x y N Nx y

η η η η

J D x y

node

node

1

1

( , )

( , )

i n

i i

i

i n

i i

i

x N ξ η x

y N ξ η y



x

y

1

1 1

1

1

8

47

3

6

25

ξ

η

ξ

η

12

η

1η

12

η

1η

1ξ 12

ξ 1ξ

12

ξ

Parent

domain

Global

domain

The Jacobian

The inverse of J, which was needed in the expression for

B, B(x,y) = H Jexp-1 DN,exp , can now be found by

22 121

21 11

1

det

J J

J J

JJ

where the determinant of J, which is often called ”the Jacobian”,

is given by

11 22 21 12det J J J J J

detdx dy dξ dη J

The Jacobian can be regarded as a scale factor that

relates infinitesimal areas in the parent domain and the

global domain, see (OP, pp. 376 – 380):

dA = dξ dη dA = dx dy

dξ

dη dy

dx



x

y

1

1 1

1

1

8

47

3

6

25

ξ

η

ξ

η

12

η

1η

12

η

1η

1ξ 12

ξ 1ξ

12

ξ

Parent

domain

Global

domain

Stiffness matrix for the Iso8 element

The definition of the stiffness matrix was

T T( , ) ( , ) ( , ) ( , )A A

x y t x y dA x y t x y dx dy K B DB B DB

The integral is non-linear and can, in general, not be

solved analytically. Therefore we must use numerical

integration.

As a remark it should be mentioned that the load

distribution along the element sides is the same as for the

”Quadratic Melosh” element, as long as the sides are

straight and have midside nodes. If this is not the case,

the nodal loads must also be found by numerical

integration.

With the Jacobian matrix and its determinant we now have

the tools to calculate the stiffnes smatrix of an isoparametric

quadrilateral element:1 1

1 T 1

exp ,exp exp ,exp

1 1

( ) det( )N Nt dξ dη

K HJ D DHJ D J

1

exp ,expN

B HJ D

detdx dy dξ dη J



f(x)Numerical integration

Our concern is to find the folowing integral numerically

( )b

a

x

x

I f x dx

One way of doing this is to divide the domain into n equal

intervals of width W, and sum up the contribution from n

rectangular areas with the height given by the function value

in the centre of the intervals:

1 1 1

( ) ( )n n n

i i i

i i i

I I f x W W f x

f(x)

f(x)

W W W W

I2I1 I3

I4

f(x1)

f(x2)

f(x3)

f(x4)

xa xb

xa xb

xa xb

x1 x2 x3 x4

x1 x2 x3 x4

f(x1)

f(x2)

f(x3)

f(x4)

Ix

x

x

W1 W2 W3 W4

A second method is analogous to the above but with varying

interval widths Wi :

1 1

( )n n

i i i

i i

I I f x W

Both methods will converge towards the exact integral when

the number of intervals, n, increases.



φ(ξ)

Gauss quadrature

The numerical integration of the stiffness matrices in a

large finite element model is a time consuming task.

Therefore we wish to use as few integration intervals as

possible, while still maintaining a decent accuracy.

-1 1

I

ξ

The position of the evaluation points, xi, and the size of the

widths (also called weitht factors) can be optimized with

respect to the function that is integrated. For integrals of

polynomials in the interval [-1 ; 1] this is known as the so-

called Gauss quadrature. Notice that this is the interval we

are interested in with isoparametric elements as the

isoparametric coordinates are -1 ≤ ξ, η ≤ 1. With Gauss

quadrature we can evaluate integrals as

1

1 11

( ) ( )n n

i i i

i i

I φ ξ dξ I φ ξ W

with the following positons and weights:

φ(ξ)

φ(ξ3)φ(ξ1)φ(ξ2) φ(ξ4)

ξ2ξ1 ξ3 ξ4-1 1

I2I1 I3

I4

W1 W2 W3 W4

ξ



φ(ξ)Gauss points and weights

-1 1

I

ξ

A polynomium of degree 2n – 1 is integrated exact with a

Gauss quadrature of degree n.

φ(ξ)

φ(ξ3)φ(ξ1)φ(ξ2) φ(ξ4)

ξ2ξ1 ξ3 ξ4-1 1

I2I1 I3

I4

W1 W2 W3 W4

ξ

Order n Location of point ξi Weight factor Wi

1 0 2

2 1

3

4

1

3

0.65

9

8

90

3 2 1.2

7

3 2 1.2

7

1 1

2 6 1.2

1 1

2 6 1.2



Gauss quadrature in two dimensions

ξ

For the calculation of the stiffness matrix we need the

Gauss quadrature in two dimensions. This is obtained as

follows:

φ(ξ)

φ(ξ3)φ(ξ1)φ(ξ2) φ(ξ4)

ξ2ξ1 ξ3 ξ4-1 1

I2I1 I3

I4

W1 W2 W3 W4

1

3η

1

3η

1

3ξ 1

3ξ

11

1

1

ξ

η

1 1 1

11 1 1

1 1

1 1

( , ) ( , )

( , )

( , )

i n

i i

i

j n i n

j i j i

j i

j ni n

i j i j

i j

I φ ξ η dξ dη φ ξ η W dη

W φ ξ η W

WW φ ξ η

Example: Numerical integration of φ(ξ,η) with Gauss order

n = 2, which gives 2·2 = 4 points:

1 1 1 1 2 1 2 1 1 2 1 2 2 2 2 2

1 1 1 1 1 1 1 1

3 3 3 3 3 3 3 3

( , ) ( , ) ( , ) ( , )

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

I φ ξ η WW φ ξ η W W φ ξ η WW φ ξ η W W

φ φ φ φ



Exercise: Calculate and program the stiffness matrix of the stuctural

isoparametric eight node element, Iso8

Steps (some may already be implemented):

1. Open the file K_IsoQ8_LinElas.m.

2. Set up a loop over the Gauss points.

3. Initiate the stiffness matrix as zeros.

4. Perform steps 5 to 12 inside a loop over the Gauss points.

5. Calculate and set up the DN-matrix.

6. Calculate the Jacobian matrix.

7. Calculate the determinant of the Jacobian matrix

8. Calculate the inverse Jacobian matrix

9. Calculate the expanded DN-matrix. DN,exp.

10. Calculate the expanded inverse Jacobian matrix, Jexp-1.

11. Calculate the strain distribution matrix, B.

12. Calculate the integral at the current Gauss point and add it to the current value of the stiffness matrix, Ke.

13. A test value with the parameters (x1, x2, x3, x4, x5, x6, x7, x8) = (0, 2, 1.9, -0.1, 0.7, 1.8, 1.1, 0), (y1, y2, y3, y4,

y5, y6, y7, y8) = (0, 0.3, 3.4, 3, 0, 1.9, 3.2, 3), t = 1.3, E = 43, ν = 0.3 and n = 2 (Gauss order) in plane stress

can be seen on the next slide.



K

Does it change the value of K, if you change the number of Gauss points?



Exercise: Program and plot the results of a 2D patch test for the isoparametric

eight node element

t = 0.2 m, E = 1.3 MPa, ν = 0.3

Steps



14, fx17, fx

5, fx8,

fx13, fx

16 and fx21.



BoundaryConditions.m.

4. Run the program by hitting F5 in main.m.


”visualize2D”.x

y 17

xf1000Paq

14

xf

9

xf




( , )

( , )

x

y

qu x y x

E

qu x y ν y

E



Why use triangular elements?

Different elements have different accuracy for different

problems.

This means that the convergence rate to the exact solution

is different for different elements.

Triangular elements can be rotated arbitrarily.

It is relatively simple to make computer code that meshes

an arbitrary area with triangles

Next: The linear strain triangle (LST)

A triangular element with six nodes and quadratic shape

functions.

The sides are straight and have midside nodes.

Gauss quadrature for triangles will be employed.



Preliminary: Natural coordinates (Area coordinates)

Again our main problem is the integration over the element

area necessary to obtain the stiffness matrix.

For general triangular elements it has proven useful to

derive the stiffness matrix using the so-called ”natural

coordinates”, also referred to as ”area coordinates”.

23 3ξ

3 1ξ

13 3ξ 3 0ξ

2 1ξ

22 3ξ

12 3ξ

2 0ξ 1 1ξ

21 3ξ

11 3ξ

1 0ξ

x 1: (x1,y1)

2: (x2,y2)

3: (x3,y3)

y Side 3

Side 1

Side 24

5

6

A1

A2

A3

P

4

2

5

3

6

1

T ( , ) ( , )A

x y x y t dA K B DB

Point P defines three areas. These define the natural

coordinates of the point:

31 21 2 3, ,

AA Aξ ξ ξ

A A A

The folowing constraint applies:

1 2 3 1 2 3 1A A A A ξ ξ ξ

Examples:

Element centroid:

Node 1:

Node 3:

Node 5:

1 1 11 2 3 3 3 3

1 2 3

1 2 3

1 11 2 3 2 2

( , , ) ( , , )

( , , ) (1,0,0)

( , , ) (0,0,1)

( , , ) (0, , )

ξ ξ ξ

ξ ξ ξ

ξ ξ ξ

ξ ξ ξ



The relation between natural and cartesian coordinates is

23 3ξ

3 1ξ

13 3ξ 3 0ξ

2 1ξ

22 3ξ

12 3ξ

2 0ξ 1 1ξ

21 3ξ

11 3ξ

1 0ξ

x 1: (x1,y1)

2: (x2,y2)

3: (x3,y3)

y Side 3

Side 1

Side 24

5

6

A1

A2

A3

P

4

2

5

3

6

1

1 1 2 2 3 3

1 1 2 2 3 3

x x ξ x ξ x ξ

y y ξ y ξ y ξ

In matrix notation:

1 1

1

2 2

3 3

1 1

and

ξ ξ

x ξ ξ x

y ξ ξ y

A A

With the coordinate transformation matrix A:

2 3 3 2 23 32

1

1 2 3 3 1 1 3 31 13

1 2 3 1 2 2 1 12 21

1 1 11

and2

x y x y y x

x x x x y x y y xA

y y y x y x y y x

A A

Here xjk = xj – xk and yjk = yj – yk

The element area can be found by:

21 31 13 122 detA x y x y A



23 3ξ

3 1ξ

13 3ξ 3 0ξ

2 1ξ

22 3ξ

12 3ξ

2 0ξ 1 1ξ

21 3ξ

11 3ξ

1 0ξ

x 1: (x1,y1)

2: (x2,y2)

3: (x3,y3)

y Side 3

Side 1

Side 24

5

6

A1

A2

A3

P

4

2

5

3

6

1

A function Ni (e.g. a shape function) is expressed in

natural coordinates, Ni(ξ1, ξ2, ξ3). The chain rule then

gives us:

31 2

1 2 3

31 2

1 2 3

i i i i

i i i i

ξN N ξ N ξ N

x ξ x ξ x ξ x

ξN N ξ N ξ N

y ξ y ξ y ξ y

With the coordinate transformation:

1 2 3 3 2 23 32

2 3 1 1 3 31 13

3 1 2 2 1 12 21

11

2

ξ x y x y y x

ξ x y x y y x xA

ξ x y x y y x y

xjk = xj – xk

yjk = yj – yk

we get:

23 31 31 2 12

32 13 31 2 21

, ,2 2 2

, ,2 2 2

y y ξξ ξ y

x A x A x A

x x ξξ ξ x

y A y A y A

Differentiation in natural coordinates

A-1



23 3ξ

3 1ξ

13 3ξ 3 0ξ

2 1ξ

22 3ξ

12 3ξ

2 0ξ 1 1ξ

21 3ξ

11 3ξ

1 0ξ

x 1: (x1,y1)

2: (x2,y2)

3: (x3,y3)

y Side 3

Side 1

Side 24

5

6

A1

A2

A3

P

4

2

5

3

6

1

The shape functionsfor the LST-element in natural co-

ordinates are given by (Cook, p. 265):

1 1 1 2 2 2 3 3 3

4 1 2 5 2 3 6 3 1

(2 1), (2 1), (2 1)

4 , 4 , 4

N ξ ξ N ξ ξ N ξ ξ

N ξ ξ N ξ ξ N ξ ξ

The displacement in the element interior is then given by:

1

1

1 2 3 1 6

1 2 3

1 2 3 1 6 6

6

( , , ) 0 0( , , )

( , , ) 0 0

x

y

x

y

x

y

u

uu ξ ξ ξ N N

ξ ξ ξu ξ ξ ξ N N

u

u

u Na

T

A

t dA K B DB

Shape functions for the LST-element

The stiffness matrix is, as always, given by:

0

, 0

x

y

y x

B N with

The constitutive matrix, D, is given

on slide 10.



23 3ξ

3 1ξ

13 3ξ 3 0ξ

2 1ξ

22 3ξ

12 3ξ

2 0ξ 1 1ξ

21 3ξ

11 3ξ

1 0ξ

x 1: (x1,y1)

2: (x2,y2)

3: (x3,y3)

y Side 3

Side 1

Side 24

5

6

A1

A2

A3

P

4

2

5

3

6

1

Example: A Maple look on the product BTDB for the LST-

element.

We choose to carry out the integration with Gauss

quadrature for triangles:

T

A

t dA K B DB

Gauss quadrature for triangles

1 2 3 1 2 3

1 1

1( , , ) det ( , , )

2

n ni i i i i i

i i i i

i iA

I φdA φ JW φ ξ ξ ξ ξ ξ ξ W

J

Te.g.: tB DB

iφiJ

For the LST-element we have a constant Jacobian

(Cook, p. 267):

1 2 3

1det ( , , ) 2

2

i i i

iJ ξ ξ ξ A J

Notice that there is only one counter in triangular Gauss

quadratrure (i), as opposed to the two counters in the

quadrilateral elements (i and j).



23 3ξ

3 1ξ

13 3ξ 3 0ξ

2 1ξ

22 3ξ

12 3ξ

2 0ξ 1 1ξ

21 3ξ

11 3ξ

1 0ξ

4

2

5

3

6

1

Gauss points for triangles

Order n Degree of

precision

Location of

point

Weight factor

Wi

1 1 1

3 2

4 3

1 1 13 3 3, ,

2 1 13 6 6

1 2 16 3 6

1 1 26 6 3

, ,

, ,

, ,

13

13

13

1 1 13 3 3

3 1 15 5 5

31 15 5 5

31 15 5 5

, ,

, ,

, ,

, ,

2748

2548

2548

2548

1 2 3, ,i i iξ ξ ξ






23 3ξ

3 1ξ

13 3ξ 3 0ξ

2 1ξ

22 3ξ

12 3ξ

2 0ξ 1 1ξ

21 3ξ

11 3ξ

1 0ξ

4

2

5

3

6

1


Order n Degree of

precision

Location of

point

Weight factor

Wi

1 1 1

3 2

4 3

1 1 13 3 3, ,

2 1 13 6 6

1 2 16 3 6

1 1 26 6 3

, ,

, ,

, ,

13

13

13

1 1 13 3 3

3 1 15 5 5

31 15 5 5

31 15 5 5

, ,

, ,

, ,

, ,

2748

2548

2548

2548

1 2 3, ,i i iξ ξ ξ






Exercise: Calculate and program the stiffness matrix of the stuctural six node

triangular element (the LST-element)

Steps (some may already be implemented):

1. Open the file K_LST_LinElas.m.

2. Calculate the element area, A.

3. Initiate the stiffness matrix as zeros.

4. Set up a loop over the Gauss points.

5. Perform steps 5 to 12 inside a loop over the Gauss points.

6. Calculate the strain distribution matrix, B.

7. Think about why the element is called the linear strain element (LST).

8. Calculate the integral at the current Gauss point and add it to the current value of the stiffness matrix, Ke.

9. A test value with the parameters (x1, x2, x3, x4, x5, x6) = (0, 2, -0.5, 1, 0.75, -0.25), (y1, y2, y3, y4, y5, y6) = (0,

1, 0.5, 0.5, 0.75, 0.25), t = 1.2, E = 32, ν = 0.2 and n = 3 (Gauss order) in plane stress can be seen on the

next slide.



K

Does it change the value of K, if you change the number of Gauss points?



Exercise: Program and plot the results of a 2D patch test for the six node

triangular element with midside nodes (the LST-element)

t = 0.2 m, E = 1.3 MPa, ν = 0.3

Steps



14, fx17, fx

5, fx8,

fx13, fx

16 and fx21.



BoundaryConditions.m.

4. Run the program by hitting F5 in main.m.


”visualize2D”.x

y 21

xf

1000Paq

16

xf

11

xf




( , )

( , )

x

y

qu x y x

E

qu x y ν y

E



Exercise: Get familiar with the program and examine how many degrees of freedom it

takes to model the bending of a tapered beam using different elements

1. Set atype = 2 in main.m

2. Set etype to any of 2, 4, 8, 14, 15 or 16.

3. Try to run the program. Visualize by executing the line ”Visualize2D”.

4. The number of elements can be controlled by changing the variable nelem_y in Coordinates.m.

5. The vertical displacement of the right-most upper node can be seen by typing

”u(2*KeyParam.RightNodes(1))” in the command window.

6. Compare this displacement for different types of elements and different values of nelem_y.

7. Does it change the result if you change the order of Gauss integration?

8. Try to take a look at how the coordinates and the topology are calculated. This might give you inspiration for

your own program in the semester project.

1.5 m

h(0) = 1.0 m

h(x) = 0.4m-1x2 – 0.93x + 1 m

yq

h(1.5 m) = 0.5 m

ndof

utip

lecture 8 and 9: two-dimensional elastic · pdf filelecture 8 and 9: two-dimensional elastic...

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