Chapter 8 Dynamic Finite Element Methods (FEM)

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

Section 8.1 Example: The Bar

• Recall the longitudinal vibration of a bar from section 6.3

• A two node FEM is presented

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



  

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

1 2

1 1 2

2 1

2 1 2

1

1 2 1

( )

(0) ( ) (0)

( ) ( ) ( ) ( )

( )

[ ( ) ( )]

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:

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!

Finding an Equation of Motion for u

(t)

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

  

  



Recognize the Quadratic Form





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

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

 



 

    

 

    

 

 

     

 



Substitute and Expand to get:





( ) 1

2 3 1

2

T

T t Al u u u u

M

   

 u u

Where

1 2

2 1

,

1 2

6

u Al

u M



   

     

 

u  

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:

T T V

t u u u f t

   

   

 

  

 

 

2

( )

( ) 0

3

Al EA

u t u t l





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

 

 

 

  

An approximate solution to the bar using one finite element!

Comparison Between Exact and FEM

• The frequency for FEM is: ^FEM

1.732 E

 l

 

• The exact frequencies are: ^{1  1.57}_l ^E_ ^{,2  4.712}_l ^E_

First mode shape

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

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



      

             

      

             

      

             

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

      

      

     

        

      

   

Now add these as they are all in the same coordinate system:

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

Now Compute the Kinetic Energy

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.

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







     

      

 

   

     

      

 

   

     

      

 

   

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















 

 

 

 

 

 

   

   

   

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

Use Lagrange Implementation for EOM

T T

1 1

2

and

0 where all gen. coords

, , 0,

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

Just what we expected!

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

Comparison with Analytical Solution

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%)

 

 

 

 

 

 

Rule of thumb: need at least twice as many elements as you want accurate frequencies

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

  

   

  

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

Element Stiffness Matrix

• Note that for the i^th 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

 

   

 

  

 

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)

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 ;

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));

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);

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

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

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.

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

 

 

 

  

 

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

 

 

Use the End Conditions to find the “Constants” of

Integration

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



  

  

 

   

   

 

  

  

 

    

  

 

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:

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

Example 8.3.3

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

The First Element Changing l for l/2 yields:

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

  

  

 

    

  

 

3 4 5 6

u u u u

  

  

  

  u

Expand to Global Coordinates and Add to get Global

In this case the global coordinates are:

u  u

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

  

  

 

  

   

  

 

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

hence to provide solutions where analytical solutions do not exist makes the method a key engineering tool.

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: u_i(t) Global coordinates: Ui(t)

Global to local transformation for element 2:

Developing the Stiffness For Bar 2

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



  

  



     

     

     

     

 

      

 

         

 

 

  

 

  

 

 

 

 

 

 

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

Corresponding to:

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

     

     

     

     

   

  

 

 

   

 

   

 

  

 

 





T  U U U U U U

U

Compute a similar expression for element 2.

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







Apply the BC at Nodes 1 & 3

5 6

1.5 0

,

0 0.5

EA U

K l U

   

      

  U  

Compute the Global Mass Matrix

1 0 0 1



 

Compute the natural frequencies:

1 2

1 0.5 1 1.5

E , E

l l

 

 

 

Several Other Features

• Variable Grid Sizes

• Inconsistent Mass Matrices

• Model Reduction

• Damping

Variable Grid Size: Example 8.2.2

Can use a variable grid size to increase accuracy without increasing the size of the model.

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

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

Bar Element

2 elements

1 0 0 1 2

M Al  

  

 

3 elements

1 0 0 0 1 0 3 0 0 1 M _ Al ^_ ^_

 

 

 

Beam Element

Divide mass among TRANSVERSE coordinates

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

0 0 0 0 M Al

 

 

 

  

 

 

A singular matrix!

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.

Removing “insignificant” terms via partitioning:

Re-Arranged so that those with low mass or thought to be insignificant are placed last in the model

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)

Constraint that V does not change with respect to u

yields:





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