Finite Element Method II 1 - Aalborg Universitet beam element 1 Beam in three dimensions. Finite Element Method II Structural elements 3D beam element 2 Lecture plan. Finite Element

Finite Element Method II

Structural elements

3D beam element

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Beam in three dimensions


Structural elements

3D beam element

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Lecture plan


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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 displacements

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

Using nodal values and shape functions


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Structural elements, n degrees-of-freedom (ndof) in each

node

degrees-of-freedom are displacement or rotation components in

cartesian coordinate system and these are the so-called primary

variables we solve for

A 3D beam has 6dof in each node:

2 nodes, one at each end (in this case)

3 deformation components

3 rotation components

node 1 node 2


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Beam assumptions, (Cook: section 2.3-2.5 p24-36),

(OP:chapter 17, p311-334)

Small deformations

axial deformation, bending and twist can be decoupled and looked at

seperately

Bernoulli-Euler beam theory for bending

Plane sections normal to the beam axis remain plane and normal to the

beam axis during the deformation.

Twist is considered free

Saint-Venant torsion


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Deformation figures: displacements

Node 1

Node 2


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Deformation figures: rotations

Node 1

Node 2


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step 1: Strong formulation for axial deformation, (Cook:

section 2.2 p20-21), (OP: p52-53)


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equlibrium equation, see slide 20-21 lecture 1

Sum of all forces are equal to zero

The force in terms of normal stress

Material property or constitutive relation (Hooks law)

Kinematic relation or geometric relation

axial deformation equation

¡N + bdx+N + dN = 0 )dN

dx+ b = 0

N =A¾


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Second order differential equation needs two boundary conditions

Possible boundary conditions: displacement (kinematic) or

displacement gradient (boundary force,static)

This is the strong formulation axial deformation


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Step 2: Establish weak formulation

Strong form

Multiply with an arbitrary function v(x) (weight function) and integrate

over the pertinent region


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Use integration by parts of the first term to obtain the same

derivative of the weight function and primary variable u

Weak formulation of axial deformation

boundary conditionsdistributed load


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Step 3: Discretize over space

Discretized problem. Define: nodes, unknown (degree-of-freedom dof)

numbering, element numbering

Nodes

Elements

dof

coordinateNode number


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Step 4: Select shape and weight functions (Cook: section

3.2-3.3 p83-91), (OP: chapter 7, 98-106)

Assuming nodal values to be known

Linear variation of deformation allows a constant deformation

gradient (strain)

Simplest one-dimensional element (p98-99)

Matrix notation

shape functions

nodal values (dof)


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Step 5: Compute element stiffness matrix

If the weak formulation holds for the entire field, it also holds for part

of the field, i.e. integration is done over one element

Insert the displacement field and arbitrary field (Galerkin approach,

equal to the approximation of the displacement field) into the weak

formulation

a and c are constants, i.e. they can be taken outside the integrals


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In compact form


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Exercise: Determine the stiffness matrix for the axial

deformation

Solve the integral on slide 17

Enter the stiffness matrix into the local stiffness matrix in

K_3d_beam.m

The local element dof are u1 and u7, i.e. the stiffness should be added to

rows and columns 1 and 7. This is easily done by the following way Kel([1 7],[1 7]) = [ - - ; - - ]


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Exercise: Enter the shape functions in shape_3d_beam.m

Enter N1 and N2 from slide 17 into the function

Test the shape function in the program beam_shape_test.m

Try to run the program with different displacements

u1=1 u7=1


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Beam bending problem in the xz-plane, (Cook: section 2.3

p24-27), (OP: chapter 17, p 311-334)

u3=1 u5=1 u9=1 u11=1


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step 1: strong formulation, (OP: p.311-318)

Infinitely small part of the beam

Vertical equilibrium

Moment equilibrium around left end, counter clockwise

second order terms are disregarded


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Combining the equilibrium equations

Kinematic relations (rotation and curvature)

Constitutive relation, uniaxial state of stress

Moment-displacement relation (homogeneous material)


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Strong formulation in terms of displacements

4th order differential equation, 4 boundary conditions (two at each

end)

Free end

Simple support

Fixed support


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Assumptions in the formulation, (Cook: p28-29),

(OP: p315-317)

rotation is taken as first derivative of displacement. This is only

approximately true if the displacements are small

The shear strain is assumed equal to zeros which gives zeros shear

stress. This is not true but comes out of simplifying a 3D problem to

2D. We will not be concerned about this inconsistency

The beam axis is located at the so-called neutral axis where an

evenly distribution of normal stresses don't introduce a moment.

This gives that the axial and bending problem decouples and can be

considered separately.


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step 2: weak formulation, (OP: p.318-319)

Multiply with arbitrary field and integrate over element

Integrate by parts


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Integrate by parts again

boundary conditionsdistributed load


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Step 3: Discretize over space

Discretized problem. Define: nodes, unknown (degree-of-freedom dof)

numbering, element numbering


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Step 4: Select shape and weight functions,

(Cook: sektion 3.2-3.3 p83-91), (OP: p.323-328)

Assuming nodal values to be known

The approximation for the deflection must be able to produce a

constant deflection and curvature, i.e. it should at least be twice

differentiable.

The second derivative of the displacements enters the formulation

hence the first derivative should be continuous over element

boundaries (C1-continuity) , or the second derivative will be infinite.

We have four node values available, i.e. four shape functions giving

the deformation shape

C0 continuity C1 continuity


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Shape functions

test u3=1, u5=u9=u11=0


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Including all dof


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Exercise: Enter the shape functions in shape_3d_beam.m

Enter N3 - N6 from slide 28-29 into the function

Test the shape function in the program beam3D_example.m, make

sure the signs are correct!

Try to run the program with different displacements

u3=1 u11=1


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step 5: compute the element stiffness matrix

Weak form

FE approximation


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In compact form

natural boundary

conditions, cancels

between elements.

Only at supports they

have a value

(reactions).

consistent load


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Exercise: solve the bending in the xz-plane part of the

stiffness matrix and enter into K_3d_beam.m

The local element dof are u3, u5, u9 and u11, i.e. the stiffness should be added

to rows and columns 3, 5, 9 and 11. This is easily done by

Kel([3 5 9 11],[3 5 9 11]) =


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xy-plane, what changes? (Cook: p27-28)

index for dof

Signs on shape functions for rotation

u6=1 u5=1 u12=1 u11=1


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xy-plane


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displacement field


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Free Torsion, Saint-Venant, (Cook: p27-28), (OP: chapter 14,

p261-281)

u4=1 u10=1


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Torsional stiffness


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Torsional moment of inertia

Thin walled sections (statik 4, 5th semester, 4th lecture)

Open sections Closed sections (Bredts equation)

"Teknisk STÅBI", steel sections

From bending moment of inertia


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FE formulation is identical with the axial deformation

See slide 9

Torsion

mx is a distributed twisting load

A linear approximation (shape function) for the torsion between the

nodes, see slide 14


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FE-formulation for torsion in compact form


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The full displacement field for a 3D beam


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Stiffness matrix for torsion is identical with the one for axial

deformation with AE replaced with GIx , se exercise slide 8

The dofs are u4 and u10 , i.e. the stiffness matrix should enter the

corresponding rows and columns


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Exercise: Type in the stiffness matrix and shape functions and test

your beam element (Teknisk STÅBI)


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Transformation, (Cook: section 2.4 p29-32)

Why do we need to do a transformation?


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Local beam axis


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Combined structure including the global dofs is defined in global

coordinate system (xg,yg,zg)

Stiffness are evaluated in local coordinate system

E, G, A, Ix, Iy, Iz, L (all values are independent of how the beam is

located in space)


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Assume at first that there exists a relation between a vector in the

global coordinate system and local coordinate system (at first we

only consider the xz-plane)

This relation is valid for any vector, e.g. displacement, rotation, force

or moment vector.

transformation matrix


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Until now everything has been described in a local coordinate

system where the beam axis is located at the local x-axis. I.e. the

element stiffness matrix Ke is described in a local system and the

compact form (slide 16, 33 and 42) is only valid in a local system

a includes both displacements and rotations and f includes both

forces and moments

Property of a transformation matrix T

The global form of the system

Global element stiffness matrix,

this is used to assemble the

global system of equations


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We have already identified the local element stiffness matrix Ke, all

we need is to determine the transformation matrix T

If we want to describe the components of a vector given in one

coordinate system (xg,yg) in another coordinate system (x,y), we can

multiply the vector with the unit vectors spanning the (x,y) system

This corresponds to rotating the vector - equal the angle between

the two systems


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Vectors defined in global system

V defined in local system

The transformation matrix is an orthogonal set of unit vectors placed

in the columns. This also holds in 3D


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In the beam case we need the y- and z-coordinates equal zero (the

beam axis is equal the x-axis). The y- and z-axis are the axis around

which the moments of inertia, Iy and Iz, are defined.

the plane spanned by node 1, node 2 and node 3 defines the xy-

plane

the xz-plane is orthogonal to the xy-plane

I.e. the beam has two nodes but we need three nodes to define the

location in space (this is the only thing the 3rd node is used for)


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How do we find the unit vectors describing the xyz-system?


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Exercise: Include the transformation in the program

Update transformation.m according to the previous slide.

Make the full transformation matrix (12x12) from T (3x3) in

K_3d_beam.m and multiply the local element stiffness matrix with

the transformation to obtain the global element stiffness matrix

hint introduce a matrix null = zeros(3,3)

The cross product V3xV1 in matlab: cross(V3,V1);

The transposed: Tg'


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Test the transformation function

The local y- and z-components should all be zero


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Exersice: Solve the test case from "Teknisk STÅBI"

supports are type 1 boundary conditions

loads are type 2 boundary conditions

where a type 1 BC has been defined a reaction is determined

where a type 2 BC has been defined a displacement is determined

A BC should be defined in all nodes in all dofs. Where nothing is

defined a load equal 0 is assumed

Below infinite axial stiffness is assumed. How do we get that?


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Field values: displacement, strain, stress, section forces

All field values are evaluated in local coordinate system, i.e. nodal

dofs needs to be rotated via the transformation matrix [12x12]

We need to identify the numbers of the12 dofs for the beam of interest

done in the matrix ElemDof defined in calc_globdof.m

Normal force, slide 9

Bending moments, slide 22 (index just change when considering Mz

and Vz)

Torsional moment, slide 39

N = EA@ux

@x


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Derivative of the field values are taken as derivative of the shape

functions multiplied by the nodal values, se slide 43 (used in the

visualization part)


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Exercise: 3Dframe

Create the indicated geometry


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Exercise: Varying section height

Divide the beam into 5 elements with varying height


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Thank you for your attention

Finite Element Method II 1 - Aalborg Universitet beam element 1 Beam in three dimensions. Finite Element Method II Structural elements 3D beam element 2 Lecture plan. Finite Element

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