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# Chapter 8 Dynamic Finite Element Methods (FEM)

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## Chapter 8 Dynamic Finite Element Methods (FEM)

A method of discretizing continuous systems, such as Rayleigh-Ritz

Also, and more importantly, a method of modeling

### complex systems

Numerous large commercial codes are available based on

these ideas

Here we examine the fundamentals

(2)

## Section 8.1 Example: The Bar

Recall the longitudinal vibration of a bar from section 6.3

A two node FEM is presented

(3)

Consider the Static Case

### Two approximations 1) element/ mesh size

2) static solution

Here c1 and c2 are constants of integration But we treat them as functions of time

c1(t) and c2 (t)

2

2

1 2

( ) 0

( ) d u x EA dx

u x c x c

  

(4)

## Nodal Displacements

u1(t) and u2(t) are called nodal displacements

They are functions of time

Eventually we will solve for these

The solution is approximated by the spatial solution u(x) and the two temporal functions u1(t) and u2(t) matched at the nodes

(5)

1 2

1 1 2

2 1

2 1 2

1

1 2 1

( )

(0) ( ) (0)

( ) ( ) ( ) ( )

( )

l

[ ( ) ( )]

u x c x c

u u t c c

c t u t

u l u t c l c

c t u t u t

 

  

 

  

  

Solving for the functions of integration

Recall:

From the sketch:

(6)

## The approximate solution is

### Here we do not yet know the values

of u1(t) and u2(t). If we did, we would be done!

(7)

## Finding an Equation of Motion for u

i

(t)

Write the strain energy and kinetic energy

### and use the Lagrangian approach.

The strain energy is

2 0

1 ( , )

( ) 2

l u x t

V t EA dx

x

Substitute in the approximate value of u(x,t):

 

 

2

1 2

2 0

2 2

1 1 2 2

( ) 1 ( ) ( )

2

2 2

l EA

V t u t u t dx

l

EA u u u u

l

(8)

Recognize the Quadratic Form

12 1 2 22

1 1

2 2

2 2

1 1

1 1

1 1

2 2

T

T

EA u u u u l

u EA u

u l u K

  

      

      

    u u

Where:

1 2

( ) 1 1

( ) , 1 1

u t EA

u t K l

    

          

u

(9)

## Calculate the Kinetic Energy

2

0

1 2

1 2

1 ( , )

( ) ( )

2

( , ) 1 ( ) ( )

( , ) 1 ( ) ( )

l

t

u x t

T t A x dx

t

x x

u x t u t u t

l l

x x

u x t u t u t

l l

 

 

    

 

    

 

 

     

 

(10)

Substitute and Expand to get:

12 1 2 22

( ) 1

2 3 1

2

T

T t Al u u u u

M

   

 u u

1 2

2 1

,

1 2

6

u Al

u M

   

     

 

u  

(11)

## Apply the boundary condition:

1

2 2

2 2

( ) 0

1 1

( ) , ( )

2 2 3

u t

EA Al

V t u T t u

l

  

Apply the Lagrangian:

i i i i( )

T T V

t u u u f t

   

   

 

  

 

 

2

( )

2

( ) 0

3

Al EA

u t u t l

 
(12)

Now we can solve to get:

2

( ) sin( 1 3 )

( , ) sin( 1 3 )

u t A E t

l

x E

u x t A t

l l

 

 

 

  

(13)

## Comparison Between Exact and FEM

The frequency for FEM is: FEM

1.732 E

l

The exact frequencies are: 1 1.57l E ,2 4.712l E

First mode shape

(14)

8.2 Three Element Bar

Next we expand the one element solution to three (or more)

Use the same strain energy computations for each element

Change only the coordinates and length of the element

(15)

## Strain Energy for Each Element

1 1

1

2 2

2 2

2

3 3

3 3

3

4 4

Replace

3

1 1

( ) 3

1 1 2

1 1

( ) 3

1 1 2

1 1

( ) 3

1 1 2

T

T

T

l l

u u

V t EA

u u

l

u u

V t EA

u u

l

u u

V t EA

u u

l

      

             

      

             

      

             

(16)

Expand to Compute the Total Strain Energy

1 1

2 2

1

3 3

4 4

1 1

2 2

2

3 3

4 4

1 2 3

3 4

1 1 0 0 1 1 0 0 3

0 0 0 0 2

0 0 0 0 0 0 0 0 0 1 1 0 3

0 1 1 0 2

0 0 0 0 0 0 0 0 0 0 0 0 3

0 0 1 1

2

T

T

T

u u

u u

V EA

u u

l

u u

u u

u u

V EA

u u

l

u u

u EA u

V l u

u

   

   

   

   

   

   

   

   

   

   

   

   

  

 

 

  

1 2 3

0 0 1 1 4

u u u u

 

 

 

 

 

  

1 1

2 2

3 3

4 4

1 1 0 0

1 2 1 0

3

0 1 2 1

2

0 0 1 1

T

T

u u

u u

V EA

u u

l

u u

   

   

   

   

   

   

(17)

## Apply the Boundary Condition

1 1

2 2

3 3

4 4

2 2

3 3

4 4

1 1 0 0

1 2 1 0

3

0 1 2 1

2

0 0 1 1

2 1 0

3 1 2 1

2 0 1 1

( ) 1

2

T

T

T

T T

u u

u u

V EA

u u

l

u u

u u

EA u u

l u u

V t K

   

   

   

   

   

   

     

     

     

     

     

u u Defines the Global

Stiffness Matrix K

(18)

Now Compute the Kinetic Energy

From Eq. (8.10):

2 1 ( ) 12 1 2

Al T

T t

u u

Let: 3

l l and get: 1

2 1

( ) 36 1 2

T

i i

T t Al

u u

This defines the kinetic energy per element.

Next write these for each element and,

Assemble the elements to get the global kinetic energy.

(19)

## Elemental Kinetic Energy:

1 1

1

2 2

2 2

2

3 3

3 3

3

4 4

2 1

1 2

36

2 1

1 2

36

2 1

1 2

36

T

T

T

u u

T Al

u u

u u

T Al

u u

u u

T Al

u u

     

      

 

   

     

      

 

   

     

      

 

   

(20)

1 1

2 2

3 3

4 4

2 1 0 0 1 4 1 0 1

0 1 4 1 2 18

0 0 1 2

T

T

u u

u u

T Al

u u

u u

   

   

   

   

   

   

Expand and Assemble Global KE

1 1

2 2

1

3 3

4 4

1 1

2 2

2

3 3

4 4

1 2 3

3 4

2 1 0 0 1 2 0 0 0 0 0 0 36

0 0 0 0 0 0 0 0 0 2 1 0 0 1 2 0 36

0 0 0 0

36

T

T

u u

u u

T Al

u u

u u

u u

u u

T Al

u u

u u

u Al u

T u

u

   

   

   

   

   

   

   

   

   

   

   

   

  

 

 

1 2 3 4

0 0 0 0 0 0 0 0 0 0 2 1 0 0 1 2

T u

u u u



 

 

 

 

 

  

(21)

Apply the Clamped BC

1 1

2 2

3 3

4 4

2 2

3 3

4 4

2 1 0 0 1 4 1 0 1

0 1 4 1 2 18

0 0 1 2 4 1 0

( ) 1 1 4 1

2 18 0 1 2

1 2

T

T

T

T

T T

u u

u u

T Al

u u

u u

u u

T t Al u u

u u

T M

     

     

     

      

     

 

   

     

     

      

     

     

u u

(22)

Use Lagrange Implementation for EOM

T T

1 1

2

and

2

0 where all gen. coords

, , 0,

T M V K

d T T V

dt

T d T T V

M M K

dt

M K

  

  

   

   

 

 

   

 

 

 

     

 

  

u u u u

u u u u

u u u

u u u u

u u

(23)

## The Equation of Motion is:

2 2

3 3

4 4

4 1 0 2 1 0

1 4 1 3 1 2 1

18 0 1 2 0 1 1

u u

Al EA

u u

u l u

                     

       

        

       

### 0

(24)

Comparison with Analytical Solution

Consider a 1 meter Al bar, E =7.0x1010 N/m2

And ρ = 2700 kg/m3.

1 1

2 2

3 3

FEM-3 Elements Exact (% error)

8092 rad/s 7998 rad/s (0.55%) 26, 458 rad/s 23, 994 rad/s (9.64%) 47, 997 rad/s 39, 900 rad/s (19.3%)

 

 

 

(25)

## System-Level Assembly

How can we assemble the system stiffness matrix directly?

First, define a system DOF vector containing all the degrees of freedom

2 3 4

list of all DOFs in system

ndof = 3 here u

u u

u

  

   

  

(26)

Global Stiffness Matrix

In this case, each “element” attaches to two coordinates

Positions in the global matrix are determined by the coordinate vector

2 2

3 3

4 4

2 3 4

since

u u

K u u

u u

u u u

 

 

    u

(27)

## Element Stiffness Matrix

Note that for the ith element,

   

th

T

1 2 1 1

2 1 2 2

1 1

stiffness matrix for i element

1 1 1

1 1 2 2

2 2

0 0 0 0

0 0

0 0 0 0

element 1 element 2

i i i i e T e e

i i i i i i i

i i i i

u k k u

V k u u u k u

u k k u

k k k k

K k k k k

k k

 

 

   

 

 

1

1 1 2 2

2 2

0

0

k k k k

k k

(28)

Basic Elements of a FE Code (I)

Mesh generation (1D)

Choose length of beam;

Choose number of nodes or elements (each implies the other) and create uniform-length elements (W.L.O.G.)

Define material and mass properties Thickness, width, modulus, density,

Define loads (spatial information)

(29)

## MATLAB Suggestions (I)

Presize matrices

ndof = 3; % number of nodes in this problem M=zeros(ndof,ndof); K=zeros(ndof,ndof); etc

Define element stiffness matrices

k = [k1 k2];% list of element (spring) stiffnesses kelem = k(j) * [1 -1 ; -1 1]; % for jth element

Use pointers to locate rows and columns

point = [j j+1]; % for jth element

K(point,point) = K(point,point) + kelem ;

(30)

MATLAB Suggestions (II)

Use pointers again to enforce boundary conditions. This point to rows and cols

active = [2 3];% the active DOFs in our problem

- Can create a reduced-size matrix

Mr = M(active,active); % Mr is a 2x2 Kr = K(active,active); % Kr is a 2x2

[V,D] = eig(Kr,Mr);% the direct EV problem

- Can just point to the active DOFs. This is equivalent to above; no new matrices needed

[V,D] = eig(K(active,active), M(active,active));

(31)

## Sample FE Code (I)

% Simple spring-mass FE code. EMA 11/17/99 clear all

num_elem = 2; % define number of elements (springs)

% write your code so that any num_elem will work!

num_nodes = num_elem + 1;

ndof = num_nodes; % number of DOFs

nodal_forces = zeros(ndof,1); nodal_forces(ndof) = 1;% put a unit force on the last DOF nodal_masses = ones(ndof,1); % make all mass values = 1.0

spring_stiffnesses = 100 * ones(ndof,1); % all k's = 100.0

% presize the system-level matrices M = zeros(ndof,ndof);

K = zeros(ndof,ndof);

F = zeros(ndof,1);

(32)

## Sample FE Code (II)

% loop on elements and assemble the system matrices for j = 1:num_elem

point = [j j+1]; % point to the global DOFs for this element kelem = spring_stiffnesses(j) * [1 -1 ; -1 1]; % element k K(point,point) = K(point,point) + kelem;

end % looping on elements

% Apply the discrete masses and forces to appropriate DOF for i = 1:ndof

M(i,i) = M(i,i) + nodal_masses(i);

F(i) = F(i) + nodal_forces(i);

end

(33)

## Sample FE Code (III)

% Enforce boundary conditions by defining the active DOFs as

% everything but the clamped (first) DOF active = 2:ndof;

% What is the static displaced shape?

% ur is "reduced" length, i.e., ndof - number of BC

ur = K(active,active)\F(active); % inv(K)*F in MATLAB

% recreate the full-size solution vector u_static = zeros(ndof,1);

u_static(active) = ur; % just that simple!

Just treat like any other MDOF M, K, and F matrices

(34)

## 8.3 Beam Elements

We just did the bar element: i.e. motion in the longitudinal direction

Next we will do a beam element to capture motion in the transverse direction

Note from Chapter 6 that there are other elements to consider: plate, shell, etc.

(35)

The Bending Beam Element

Note the 4 coordinates to approximate u(x,t)

The transverse static displacement is:

2 2

2 2

( , ) u x t 0 x EI x

 

 

 

  

 

(36)

## Approximate Bending Solution

### Integrating the static deflection expression yields:

3 2

1 2 3 4

( , ) ( ) ( ) ( ) ( )

u x t  c t x  c t x  c t x c t 

This is subject to the following end conditions:

1 2

3 4

(0, ) ( ) (0, ) ( )

( , ) ( ) ( , ) ( )

x x

u t u t u t u t

u l t u t u l t u t

 

 

(37)

## Use the End Conditions to find the “Constants” of

Integration

(38)

Following the Energy Method as Before Yields the Beam Element Energy, Mass and Stiffness:

2 2

2 2

2 2

3

2 2

156 22 54 13

22 4 13 3

54 13 156 22

420

13 3 22 4

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

l l

l l l l

M Al

l l

l l l l

l l

l l l l

K EI

l l

l

l l l l

(39)

Example 8.3.1: Simply supported Beam

3 4 3

3 4 420

4 2 2 4 M Al

K EI

l

Using a single element

Note from the sketch that u1(t) and u3(t) are fixed

Deleting these from the mass and stiffness expressions yields:

(40)

## Equation of Motion and Frequencies

2 2

4

4 4

1 4 4

2 4 4

( ) ( )

4 3 840 2 1

( ) ( )

3 4 1 2

10.95445 (9.86965 )

50.199605 (39.47860 )

u t EI u t

u t Al u t

EI EI

Al Al

EI EI

Al Al

  

  

    

   

 

   

    

       

### 0

(41)

Example 8.3.3

Derive the global mass and stiffness matrices for a clamped free beam using 3 nodes and two

elements.

Obtain global matrices directly from the beam FEM of equations (8.53) and (8.56)

Change l for l/2, write in global coordinates and apply the clamped constraint

(42)

The First Element Changing l for l/2 yields:

(43)

3 4 2

5 6

u u u u

  

 

  

  u

The Second Element Changing l for l/2 yields:

2 2

2

2 2

156 11 54 13

2

13 3

11 2 4

13

840 54 156 11

2

13 3

2 4 11

l l

l l l l

M Al

l l

l l l l

2 2

2 3

2 2

12 3 12 3

3 3 0.5

8

12 3 12 3

3 0.5 3

l l

l l l l

K EI

l l

l

l l l l

(44)

3 4 5 6

u u u u

  

 

  

  u

M and K

u  u

2

2 2

1 2

2 2

312 0 54 13

2

13 3

0 2

2 4

840 13

54 156 11

2

13 3

2 4 11

l

l l l

M M M Al

l l

l l l l

2 2

1 2 3

2 2

24 0 12 3

0 2 3 0.5

8

12 3 12 3

3 0.5 3

l

l l l

K K K EI

l l

l

l l l l

(45)

## 8.5 Trusses

As an example of using FEM for structures that do not have analytical solutions consider a simple truss.

The ability of FEM to model very complex structures and

(46)

3 3 4

4 5 6

( ) cos sin

( ) cos sin

u t U U

u t U U

## A Simple Frame

Considered to be

### formed of bar elements

Local coordinates

pointing in different directions

But system can vibrate in the x-y plane

Local coordinates: ui(t) Global coordinates: Ui(t)

Global to local transformation for element 2:

(47)

## Developing the Stiffness For Bar 2

(48)

Element 2 Potential In Global Coordinates

2 2

2 2

(2) 2 2

2 2

cos 0

sin 0 1 1 cos sin 0 0

0 cos 1 1 0 0 cos sin

0 sin

cos sin cos cos sin cos

sin cos sin sin cos sin

cos sin cos cos sin cos

sin cos sin sin cos sin

T e

K EA

l

K EA

l

  

  

     

     

     

     

 

      

 

         

 

 

  

 

  

 

 

 

 

 

 

(49)

2 2

2 2

(1) 2 2

2 2

1 2 1

5 6

cos sin cos cos sin cos

sin cos sin sin cos sin

cos sin cos cos sin cos

sin cos sin sin cos sin

( ) ( )

( ) ( ) K EA

l

U t U t U t U t

U

Element 1 Stiffness

(50)

## Expand to Complete Element Stiffness in Global Coordinates

2 2

2 2

(1)

2 2

2 2

cos sin cos 0 0 cos sin cos

sin cos sin 0 0 sin cos sin

0 0 0 0 0 0

0 0 0 0 0 0

cos sin cos 0 0 cos sin cos

sin cos sin 0 0 sin cos sin

K EA

l

 

1 2 3 4 5 6

T U U U U U U

U

(51)

## For θ =30° add Expanded local to get Global Stiffness

1 2 3 4 5 6

0.75 0.4330 0 0 0.75 0.4330

0.4330 0.25 0 0 0.4330 0.25

0 0 0.75 0.4330 0.75 0.4330

0 0 0.4330 0.25 0.4330 0.75

0.75 0.4330 0.75 0.4330 1.5 0

0.4330 0.25 0.4330 0.25 0 0.5

K

U U EA U

l U

U U

 

  

  

U

(52)

Apply the BC at Nodes 1 & 3

U1 U2 U3 U4 0

5 6

1.5 0

,

0 0.5

EA U

K l U

   

      

  U  

(53)

Compute the Global Mass Matrix

1 0 0 1

M

Al

 

1 2

1 0.5 1 1.5

E , E

l l

 

 

 

(54)

## Several Other Features

Variable Grid Sizes

Inconsistent Mass Matrices

Model Reduction

Damping

(55)

Variable Grid Size: Example 8.2.2

(56)

## Comparison

3 elements 2 non uniform Exact Elements

1 2

8092, 8367, 7998 26458, 26460, 23994

rad/s

### Variable grid size can accommodate non uniformities and works to increase

accuracy and or reduce size

(57)

## 8.4 Lumped Mass Matrix

An alternative way to compute the mass matrix that avoids

having to compute integrals off diagonal elements

The energy approach we have used is call the “consistent mass matrix”

The simplified version we are about to compute is also called the “inconsistent mass matrix

(58)

## Bar Element

For two elements, lump half the mass at each element, for 3 elements, a third

2 elements

1 0 0 1 2

MAl

3 elements

1 0 0 0 1 0 3 0 0 1 M Al

(59)

Beam Element

### Divide mass among TRANSVERSE coordinates

1 0 0 0 0 0 0 0 0 0 1 0 2

0 0 0 0 MAl

A singular matrix!

(60)

## 8.6 Model Reduction

To increase accuracy, the order of the system often makes computation (simulation or modal computations) prohibitive.

Thus many techniques have been developed to eliminate coordinates in meaningful ways.

(61)

Removing “insignificant” terms via partitioning:

(62)

## Energy Expressions:

1 11 12 1

2 21 22 2

1 11 12 1

2 21 22 2

1 2 1 2

T

T

K K

V K K

M M

T M M

     

      

     

     

      

     

u u

u u

u u

u u

Next assume f2 is zero, which

implies that the variation of V with u2 is Zero (actually an assumption)

(63)

Constraint that V does not change with respect to u

2

yields:

1 11 1 1 12 2 2 21 1 2 22 2

2

1

2 22 21 1

0

T T T T

V K K K K

K K

 

u u u u u u u u

u

u u

Allows us to remove u2 in terms of u1 and suggests the Transformation: 22

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