# Chapt 11 Finite Element Analysis by 7390T04

VIEWS: 71 PAGES: 39

• pg 1
```									                     William J. Palm III : Mechanical Vibration

11. DYNAMIC FINITE ELEMENT ANALYSIS

 Bar element
 Model with multiple bar elements
 Consistent and lumped-mass matrices
 Analysis of trusses
 Torsional elements
 Beam elements
 MATLAB applications
 Chapter review

Outline
This chapter introduces the finite element method that provides a more
accurate system description than that used to develop a lumped-parameter
model but that for complex systems is easier to solve than partial differential
equations and more accurate than their finite-difference approximations.

Pukyong National University                                           Intelligent Mechanics Lab.
Introduction
 Finite element method (FEM) or sometime called finite element analysis (FEA)
particularly useful for irregular geometries such as bars with variable cross sections
and for systems such as trusses that are made up of several bars
 In FEM, system is modeled as consisting of several pieces (elements)
Each piece is treated as a continuous element (finite element)
 Structure is divided into a number of simple structure parts such as bars, beams
or plates, whose equations of motion (EOM) are easily derived
 Equations obtained for each element are assembled and combined according to
how elements are connected in the system

Discretization of frame and plane stress problems

Pukyong National University                                            Intelligent Mechanics Lab.
Introduction
 Connection points : joints or nodes
Resulting collection of finite element and nodes : mesh
 To solve EOM for each element, approximate solution with low-order polynomials
 These solutions are then assembled to obtain mass and stiffness matrices
for entire structure as a whole
 Once equations have been assembled, modal analysis technique are used to
compute natural frequencies and mode shapes
 Displacements obtained are displacements of nodes in the mesh

Different 1D, 2D and 3D basic elements
Pukyong National University                                       Intelligent Mechanics Lab.
Introduction
 Here treated bar and beam elements only without plate and shell elements
 For modeling structural element, approximation needs to vibration mode shape
 When only one finite element is used between structural joints or corners,
usually obtain an accurate result only for the lowest mode
 Then to estimate higher modes, several elements must use between structural
joints. In most practical problems, a large number of elements is used
 Fortunately, FEM has now widely implemented in many powerful commercially
available computer programs
 Programs provide a graphical interface for specifying locations of nodes,
assemble equations and often provide graphical displays of results
 Here, provide a simple introduction with basic concept and terminology
 FEM is a very strong tool for vibration analysis
 Undoubtedly need a deeper understanding of FEM

Pukyong National University                                          Intelligent Mechanics Lab.
Bar Elements
Stiffness Matrix
 Force-deflection relation :                             E, A
EA                 EA                                             f
f     u  ku,     k                                  L
L                  L
 When forces are applied at both ends :
 Stiffness matrix :
 f1   k11 k12   u1      f1  k11u1  k12u2
 f   k                                             k       k  EA  1 1
f 2  k21u1  k22u2       K             
 2   21 k22  u2 
 k     k  L 1 1 
    


 Determine elements of matrix :
First column contains endpoint forces
when u1 = 1, u2 = 0, i.e. f1 = ku1, f2 = ku1
Second column contains endpoint forces
when u1 = 0, u2 = 1, i.e. f1 = ku2, f2 = ku2

f1  k u1  k u2 ,   f 2  k u1  k u2

 f1   k      k   u1  EA  1 1  u1 
 f    k               
k  u2  L  1 1  u2 
 2                             

Pukyong National University                                              Intelligent Mechanics Lab.
Bar Element

Mass Matrix
d 2u ( x )
 Static displacement of bar : EA             0 (0  x  L)
dx 2
 Solution : u( x)  a  bx

 Taking this static mode shape to be an approximation of dynamic mode shape
u ( x, t )  a(t )  b(t ) x

 Displacements of two nodes :u1 ( x), u2 ( x)
at x = 0: u (0, t )  u1 (t ) , x = L: u ( L, t )  u2 (t )
u2 (t )  u1 (t )
where a(t )  u1 (t ), b(t ) 
L
 Mode shape :
 x                 x
u ( x, t )  1   u1 (t )  u2 (t )
 L                 L
 S1 ( x) u1 (t )  S2 ( x) u2 (t )

 Shape functions :
   x              x
S1 ( x)  1   , S 2 ( x) 
 L                L
An approximate solution of EOM of bar

Pukyong National University                                          Intelligent Mechanics Lab.
Bar Element
Mass Matrix
 Kinetic energy of bar using assumed mode shape :
1  u 
2
1
 A   dx   A  S1 ( x)u1 (t )  S2 ( x)u2 (t ) dx
L                      L
KE  
2
0   2  t      2    0

1               L                                L                 1             L
      Au12 (t )  S12 ( x)dx   Au1 (t )u2 (t )  S1 ( x)S2 ( x)dx   Au2 (t )  S22 ( x)dx
2

2              0                                0                  2            0

 Evaluation of integral :

1                                            1
KE         AL[u12 (t )  u1 (t )u2 (t )  u2 (t )]  uT (t )Mu(t )
2

6                                            2

 Velocity vector :
 u (t ) 
u(t )   1 
u2 (t ) 

 Mass matrix of bar element :
 AL 2 1 
M
6 1 2 
    

This result is dependent on the assumed mode shape                                 Full Ship Model

Pukyong National University                                                                 Intelligent Mechanics Lab.
Bar Element

Potential Energy Approach :
 Similar approach to derive stiffness matrix K from potential energy expression :
1  u 
2                                                      2
L                     1       L  dS ( x)          dS ( x)          
PE   EA   dx  EA  1 u1 (t )  2 u2 (t )  dx
0 2
 x            2      0
 dx                  dx            
1 EA L                             1 EA 2
2 0
          [u2 (t )  u1 (t )]2 dx         [u1 (t )  2u1 (t )u2 (t )  u2 (t )]
2

2 L                                2 L
1
 uT (t ) K u(t )
2

EA  1 1
K
L 1 1 
     

 Shape function used are equivalent to linear force-deflection relation used to derive
stiffness matrix K
 So, derivation of K based on potential energy should give same result

Pukyong National University                                                                    Intelligent Mechanics Lab.
Bar Element

Fixed-Free Bar (Cantilever Bar)
 Fixed end (left-hand) of bar : u1 (t )  0
 Kinetic and potential energy of bar :
1               1 EA 2
KE   ALu2 PE 
2
u2
6               2 L
 From conservation of mechanical energy : KE  PE  constant
 AL          EA              A     E
u2 u 2     u2 u2  0 or    u2  u2  0  Dynamic mode of bar element
3            L               3     L

 Natural frequency :
1 3E 1.7321 E
n        
L      L   

 Given initial condition : u2 (0), u2 (0)

   x          x
u ( x, t )  1   u1 (t )  u2 (t )
 L            L
x
 u2 (t )
L

Pukyong National University                                      Intelligent Mechanics Lab.
Bar Element

Comparison with Other Models
 Natural frequency of a concentrated mass attached to a spring element :
k                                           EA    EA1 3E
n                               If mc  0, n                    , ms   AL)
( k
mc  ms / 3                                   AL2 / 3 L 
L
d 2u E d 2u
 Compare results with solution of partial differential equation,     
dt 2  dx 2
 Natural frequencies and mode shapes :
(2n  1)       E                        (2n  1) x
n                        , Fn ( x)  An sin                 (n  1, 2,3, )
2L                                       2L

 First natural frequency and mode shape :
1st mode
      E       1.5708 E                    x
1                           , F1 ( x)  A1 sin
2L               L                       2L
 Second natural frequency and mode shape :
3 E 4.7124 E                    3 x
2                  , F2 ( x)  A2 sin
2L     L                        2L

 FEM predicts 63% lower than that by 2nd mode of
distributed-parameter model                                                                     Node
2nd mode
 2nd mode shape is quite different with that FEM used
 Higher modes all have higher frequencies,
more complex mode shapes

Pukyong National University                                                                 Intelligent Mechanics Lab.
Bar Element

Applying Boundary Conditions

 Dynamic model of bar element with standard matrix form :
 AL  2 1   u1  EA  1    1  u1  0
                        
6 1 2 u2  L  1        1  u2  0
Mu + Ku = 0
                          

 If left-hand of element is fixed, u1 (t )  u1 (t )  0
first equation is not applicable, immediately
 AL          EA
u2       u2  0
3            L

This means 1st row and 1st column of M, K can strike out

Striking out row and column corresponding to the displacement of a fixed node
is a procedure often used as a quick way to reduce the number of equations

 If bar has an axial force f(t) applied at its free end, EOM :
 AL          EA
u2       u2  f (t )
3            L

Pukyong National University                                                       Intelligent Mechanics Lab.
Models with Multiple Bar Elements
 Next step id to decide how many elements to use
One element model give inaccurate results if higher modes are excited
 To present higher modes, more elements must be used to model entire bar
 If multiple elements is used, equations for all elements must be assembled into
a model of entire structure as a whole

One element

Two elements

Three elements

Pukyong National University                                          Intelligent Mechanics Lab.
Models with Multiple Bar Elements
Example : Two-element Bar Model
 Demonstrate how to develop a finite element model
 A1L1 2 1               A2 L2 2 1      EA  1 1         EA  1 1
M1           1 2  , M 2            1 2  , K1  1          , K2  2 
6                       6               L1  1 1 
        L2  1 1 


 Dynamic model of each element with standard matrix form :
M i ui + K i ui = 0 (i  1, 2)

Pukyong National University                                               Intelligent Mechanics Lab.
Multiple Bar Elements

Example : Two-element Bar Model

 For first element,
m1  2 1   u1        1 1  u1  0     m1
 k1                         u2  k1u2  0 (1)         u1 (t )  u1 (t )  0)
6 1 2 u2             1 1  u2  0
(
                                 3

 For second element,                             m2
(2u2  u3 )  k2 (u2  u3 )  0 (2)
m2    2 1  u2        1 1 u2  0     6
1 2  u     k2              
6          3         1 1  u3  0
          m2
(u2  2u3 )  k2 (u3  u2 )  0 (3)
6

 Three equations, two unknown u2, u3 , reduce to just two equations by adding Eqs. (1), (2)
 m1 m2       m2
       u2     u3  (k1  k2 )u2  k2u3  0 (4)
 3  3        6

 From Eq. (3) and (4),

1  2(m1  m2 ) m2  u2  (k1  k2 ) k2  u2  0
                      
6    m2       2m2  u3   k2
                k2  u3  0
   

Finite element model of bar with two elements

Pukyong National University                                                     Intelligent Mechanics Lab.
Multiple Bar Elements

Example : Two-element Bar Model

 Solve frequencies and mode shapes for case with constant cross section and length
A1  A2  A, L1  L2  L / 2, m1  m2  m / 2, k1  k2  k  2 EA / L

m  4 1  u2     2 1 u2  0      2 1 u2      4 1  u2     m 2  L2 2 
12 1 2 u3 
k            or  1 1  u    1 2 u     12k  24 E 
            1 1  u3  0           3           3                   

 Characteristic equations : 7 2  10  1  0,   0.1082, 1.3204

24EA 1.611 E 5.629 E                                         (1.611-1.5708)/1.5708
 Natural frequencies : i               ,                                                = 2.559%
 L2   L   L   L   L                                         (5.629-4.7124)/4.7124
= 19.451%
 Natural frequencies and mode shapes for exact solution :
(2n  1)   E                  E       1.5708 E        3 E 4.7124 E
i                    , 1                           , 2       
2L                   2L               L           2L     L   

FEM predicts 2.6%, 19% higher for 1st and 2nd modes than distributed-parameter model

 Compare mode shapes :
x                    3 x        F1 ( L)     sin  / 2             F2 ( L)     sin 3 / 2
F1 ( x)  A1 sin    , F2 ( x)  A2 sin                                   1.414,                           1.414
2L                     2L        F1 ( L / 2) sin  / 4             F2 ( L / 2) sin 3 / 4

Both models predict same mode shapes for first two modes

Pukyong National University                                                               Intelligent Mechanics Lab.
Multiple Bar Elements

Extension to Multiple Element

 Consider fixed left-hand and free right-hand
 k1  k2       k2                                     2m1  2m2      m2                                   
 k          k2  k3    k3                            m           2m2  2m3       m3                      
      2                                              1
2                                              
K              k3      k3  k4    k4           , M                  m3        2m3  2m4      m4          
                                                     6                                                     
                         k4     k4  k5    k5                                    m4       2m4  2m5   m5 

                                  k5       k5       
                                         m5       m5 

Ei Ai
ki          , mi   Ai Li
Li

Banded structure due to that a given element is affected only by its adjacent elements
A main diagonal, one upper sub-diagonal, one lower sub-diagonal

 Eigenvalue problem : M-1Ku   2u

Pukyong National University                                                              Intelligent Mechanics Lab.
Consistent and Lumped-Mass Matrices
 Mass matrix derived from kinetic energy of element, assuming that velocity is time
derivative of displacement function specified by shape function
 Functions used specify static deflection shape, and thus only approximation
 This is a consistent mass matrix because using deflection shape function
 Stiffness matrix also derived from these shape function, however accurate
because stiffness is associated with static deflection
 An easy way of calculating a mass matrix is to place an appropriate mass value
at each node as a lumped mass
 If total mass is uniformly distributed throughout the system and if kinematics are
simple, then an appropriate mass value for a given node would be proportional
to dimension of element
 For bar element of length L, density , area A, total mass is AL
 Placing one-half of total mass at each node, lumped-mass matrix gives :
 AL 1 0            AL/ 2         AL              AL/ 2
M
2 0 1 
    

Advantage : diagonal matrix, easily inverted to compute M-1
Disadvantage : not always obvious how mass should be apportioned to each node
For beam elements with coordinates equal to slope of beam deflection curve, with no mass
lumped for that coordinate, mass matrix will have some zero elements along it diagonal

Pukyong National University                                                Intelligent Mechanics Lab.
Consistent and Lumped-Mass Matrices

Example : Fixed-Fixed Bar
Compare the results obtained from consistent mass matrix, lumped-mass matrix
and distributed-parameter model (exact model)
EA  1 1
 Stiffness matrix :   K
L 1 1 
       
 AL 2 1             AL 1 0
 Consistent mass matrix and lumped-mass matrix : MC                 ML 
6 1 2
                2 0 1 
    
 Assembled matrices :
1     1    0         1 1 0            2     1      0        2 1 0
 AL                    AL 
1              
EA 
1 (1  1) 1 
EA 
K                               1 2 1 MC         1 (2  2)             1 4 1 
L                 L                    6                     6
0
      1    1        0 1 1 
                   0
      1      2
        0 1 2
      
1    0    0        1 0 0 
 AL              AL       
ML       0 (1  1) 0   2 0 2 0 
2
0
     0    1       0 0 2 
      

 Boundary conditions : u1  u3  0
2EA
 Stiffness matrix : K 
L

Pukyong National University                                       Intelligent Mechanics Lab.
Consistent and Lumped-Mass Matrices

Example : Fixed-Fixed Bar
4  AL
 Consistent mass matrix : M C 
6              L       E
 Equation of motion with consistent mass matrix :    u2  u2  0
3       L
1 3E 1.73 E
 Frequency :                                      (1.73  1.57)/1.57
L        L                        = +10.19%

 Lumped-mass matrix : M L   AL
2 EA
 Equation of motion with lumped-mass matrix :  ALu2             u2  0
L
1 2 E 1.41 E
 Frequency :                                           (1.41 – 1.57)/1.57
L     L                               = 10.19%

1.57    E
 Lowest frequency for distributed-parameter model :  
L     

Lumped-mass approximation :
 Leads to a diagonal mass matrix
 More convenient in models having a large number of elements
 More easily inverted without excessive numerical error
 Introduce modeling error
 Choice between use of consistent versus lumped-mass matrices is usually not clear

Pukyong National University                                                  Intelligent Mechanics Lab.
Torsional Elements
 Finite element model of a rod in torsion

Stiffness Matrix
GJ
 Force-deflection relation of a uniform rod having a net applied torque T : T       k
L
GJ           EA
 Torsional stiffness : k            k    (Longitudinal vibration)
L            L
 Torques are applied at both end of bar :           Stiffness matrix :         GJ    1 1
K         1 1 
T1   k11 k12  1                                                      L         
T    k       
 2   21 k22   2                          Only GJ replaces EA
 Determine elements of matrix :
First column contains endpoint torques
when 1 = 1, 2 = 0, i.e. f1 = k1, f2 = k1
Second column contains endpoint forces
when 1 = 0, 2 = 1, i.e. f1 = k2, f2 = k2

T1   k     k  1  GJ    1 1 1 
T    k            
k   2  L     1 1   
 2                              2

Pukyong National University                                            Intelligent Mechanics Lab.
Torsional Elements

Mass Matrix
d 2 ( x)
 Static displacement of rod : GJ            0 (0  x  L)
dx 2
 Solution :  ( x)  a  bx

 Taking this static mode shape to be an approximation of dynamic mode shape
 ( x, t )  a(t )  b(t ) x

 Displacement of two nodes : 1 ( x),  2 ( x)
at x = 0:  (0, t )  1 (t ) , x = L:  ( L, t )   2 (t )
 (t )  1 (t )
where a(t )  1 (t ), b(t )  2
L

 Mode shape :  ( x, t )  1   1 (t )   2 (t )  S1 ( x) 1 (t )  S 2 ( x)  2 (t )
x           x
     
 L            L

   x              x
 Shape functions : S1 ( x)  1   , S 2 ( x) 
 L                L

 An approximate solution of equation of motion of bar
 Shape function is same with longitudinal vibration

Pukyong National University                                                                   Intelligent Mechanics Lab.
Torsional Elements
Mass Matrix
 Kinetic energy of bar using assumed mode shape:
1   
2
L                1      L
KE            J   dx   J   S1 ( x)1 (t )  S2 ( x)2 (t )  dx
2

0    2  t      2     0                                 

 Evaluation of integral :
1                                          1     12 (t ) 1 (t )2 (t ) 22 (t )  1 T
KE   JL[1 (t )  1 (t ) 2 (t )   2 (t )]   JL 
2                         2
                         θ (t )Mθ(t )
6                                          2        3           3           3  2
 Velocity vector :
 (t ) 
θ(t )   1 
 2 (t ) 
 Mass matrix of bar element :
 JL  2 1 
M
6 1 2
   
 Equation of motion for torsional element :
Mθ  Kθ  0

 This result is dependent on the assumed mode shape
 Only J replaces A

Pukyong National University                                                              Intelligent Mechanics Lab.
Torsional Elements

Example : Fixed-Free Bar in Torsion
 Equation of motion :
 L  2 1  1  G  1 1 1  0
                  
6 1 2  2  L  1 1   2  0
                      

 Fixed end (left-hand) of bar : 1 (t )  1 (t )  0
 First equation is not applicable
L     G
2  2  0
3      L
1 3G 1.7321 G
 Natural frequency : n            
L            L        
 Given initial condition :  2 (0),  2 (0)
     x          x            x
 ( x, t )  1   1 (t )   2 (t )   2 (t )
 L            L          L

 Natural frequency of a concentrated inertia attached to end of rod :
k
n 
Ic  Ir / 3
GJ       1 3G                GJ
If I c  0, n                               ( k      , I r   JL)
 JL2 / 3 L                    L

Pukyong National University                                                            Intelligent Mechanics Lab.
Bar Element

Example : Fixed-Free Bar in Torsion
d 2 G d 2
 Compare results with solution of distributed-parameter model,     
dt 2  dx 2
 Natural frequencies and mode shapes :
(2n  1)     G                        (2n  1) x
n                      , Fn ( x)  An sin                 (n  1, 2,3, )
2L                                     2L

 First natural frequency and mode shape :
    G       1.5708 G                    x
1                          , F1 ( x)  A1 sin
2L             L                       2L                      1st mode :
 Second natural frequency and mode shape :                                  (1.7321  1.5708)/1.5708
= +10.27%
3 G 4.7124 G                    3 x
2                  , F2 ( x)  A2 sin
2L     L                        2L

 FEM predicts 10% higher than that by 1st mode of distributed-parameter model

Pukyong National University                                                             Intelligent Mechanics Lab.
Beam Elements
 Element equation for a beam element (call Euler-Bernoulli beam element) are
derived in a fashion similar to that used for longitidinal vibration of a bar
 Beam displacements normal to its length is denoted : v( x, t )
Translational displacements of endpoints : v1 , v2
Rotational displacement of endpoints : 1 ,  2
 Displacement vector of beam element :
 v1 
 
v   1
 v2 
 
 2 

 Boundary conditions :
v (0, t )  v1 (t )
v (0, t )
 1 (t )
x
v ( L, t )  v2 (t )
v ( L, t )
  2 (t )
x

Pukyong National University                                               Intelligent Mechanics Lab.
Beam Elements

Shape Functions
 Solution of distributed-parameter model approximately :
v( x, t )  a(t )  b(t ) x  c(t ) x 2  d (t ) x 3

 Applying boundary conditions :
1
a(t )  v1 (t ), b(t )  1 (t ), c(t )           [ 3v1 (t )  2 L1 (t )  3v2 (t )  L 2 (t )]
L2
1
d (t )       [2v1 (t )  L1 (t )  2v2 (t )  L 2 (t )]
L3
 Solution : v( x, t )  S1 ( x)v1 (t )  S2 ( x)1 (t )  S3 ( x)v2 (t )  S 4 ( x) 2 (t )

 Shape function : S i ( x )
2           3
x     x
S1 ( x)  1  3    2  
L     L
2           3
x     x
S2 ( x)  x  2 L    L  
L     L
2               3
x     x
S3 ( x )  3    2  
L     L
2               3
x     x
S4 ( x)   L    L  
L     L

Pukyong National University                                                                              Intelligent Mechanics Lab.
Beam Elements
Mass Matrix
 Kinetic energy of beam element using assumed mode shape:
1  v 
2
L
KE           A   dx
0    2  t 
1      L
      A [ S1 ( x)v1 (t )  S2 ( x)1 (t )  S3 ( x)v2 (t )  S 4 ( x) 2 (t )]2 dx
2     0

 AL
 Evaluation of integral : KE                      vT Mv
420
 Velocity vector :          v1 
 
1
v 
 v2 
 
 2 
 

 Mass matrix of beam element :
 156 22 L                    54  13L 
                           13L 3L2 
 AL  22 L 4 L
2

M                                                               This result is dependent on
420  54    13L                   156 22 L                   the assumed mode shape S(x)
                                      
 13L 3L                  22 L 4 L2 
2

Pukyong National University                                                                         Intelligent Mechanics Lab.
Beam Elements

Stiffness Matrix
 Similar approach to derive stiffness matrix K from potential energy expression :
2
L  v 
2
1
PE  EI   2  dx  vT Kv
2    0
 x 
I : Area moment of inertia of the section

 Stiffness matrix of beam element :

 12  6 L 12 6 L 
                 2 
EL  6 L 4 L 6 L 2 L 
2

K 3
L  12 6 L 12 6 L 
                 2 
 6 L 2 L 6 L 4 L 
2

 Equation of motion of beam element :
Mv  Kv  0

Pukyong National University                                           Intelligent Mechanics Lab.
Beam Elements

Matrix Representation
 For future reference, when treat beams having multiple elements
Ma       Mb             K a   Kb 
M T                     K T
M b      Mc 
            K b   Kc 


 AL  156 22L           AL  54 13L          AL  156 22L 
Ma                      , Mb                   , Mc 
420  22L
       4L2 
        420 13L 3L2 
                 420 22L 4L2 
          

EL 12 6L         EL  12 6L        EL  12 6 L 
Ka                 , Kb  3           , Kc  3 
L3 6L 4L2 
               L 6L 2L2 
        L 6L 4L2 

Pukyong National University                                           Intelligent Mechanics Lab.
Beam Elements

Example : Pinned-Pinned Beam
 Displacement vector of beam element :        v1 
 
v   1    v1 1 v2  2 
T
 Boundary conditions :
 v2 
v1  v2  0, 1  0,  2  0              
 2 
 Then strike out first and third columns and rows of M, K
 AL  4 L2   3L2         EL  4 L2      2 L2 
M                     ,     K 3  2              
420  3L2   4 L2          L 2L         4 L2 

 Equation of motion :
 AL  4 L2   3L2  1  EI    4 L2   2 L2  1  0
                       2               
420  3L2   4 L2  2  L3   2L      4 L2  2  0

 4 3 1  840 EI      2 1  1  0
 3 4      AL4     1 2    0
       2                  2  

Pukyong National University                                                      Intelligent Mechanics Lab.
Beam Elements

Example : Pinned-Pinned Beam


840 EI
,  j   j eit        (2  4 2 )1  (3 2   )2  0
 AL4
(3 2   )1  (2  4 2 )2  0

 Frequency equation :
7 4  22 2  3 2  0
 2  0.142857 ,  2  3

10.9544 EI        50.9544 EI
 Natural frequency : 1                  , 2 
L2   A           L2   A
 For distributed-parameter model :

 n 
2
EI
n   
L          A                                    1st mode :
(10.9544  9.8696)/ 9.8696
9.8696 EI        39.4784 EI
1              , 2                                   = 10.99%
L2   A           L2   A
2nd mode :
(50.9544  39.4784)/ 39.4784
= 29.07%

Pukyong National University                                                    Intelligent Mechanics Lab.
Beam Elements

Example : Two Elements
 For each element : i  1, 2

 156        11Li      54     13Li / 2                12   3Li   12     3Li 
                    13Li / 2 3L2 / 4                            3Li   L2 / 2
 Ai Li  11Li        L2                              8Ei I i  3Li  L2                 
Mi                         i                 i
Ki  3               i           i

420  54           13Li / 2  156      11Li           Li  12 3Li        12    3Li 
                                                                              
 13Li / 2 3L2 / 4 11Li
i                L2 
i                      3Li Li / 2 3Li
2
L2 
i

 Displacement vector of first element :
 v1 
 
 1    v  v  T
 v2      1 1 2 2

 
 2 
 Applying boundary conditions :
Fixed condition : v1  1  0

 Equation of motion :
v           v  0 
M c1  2   K c1  2    
 2         2  0

Pukyong National University                                                 Intelligent Mechanics Lab.
Beam Elements

Example : Two Elements
 Displacement vector of second element : (v2  2 v3 3 )T
 Adding equations that share common elements :
 v2                           v2  0
M c1  M a 2   Mb 2   2  K c1  K a 2
                     K b 2  2  0
   
 MT                   v    KT                   v   0
      b2       Mc 2  3            b2       K c2  3
                                
3                           3  0

 Result for mass and stiffness matrices :

156  156 11L  11L   54    13L / 2        312         0       54    13L / 2
                    13L / 2 3L2 / 4   AL  0                 13L / 2 3L2 / 4 
 AL 11L  11L L  L
2     2
                 2 L2                    
M
840  54         13L / 2   156     11L  840  54           13L / 2   156     11L 
                                                                               
 13L / 2  3L2 / 4  11L     L2            13L / 2 3L2 / 4 11L      L2 

 12  12 3L  3L 12 3L                          24    0  12 3L 
                                                 0   2 L2 3L L2 / 2
8EI 3L  3L L  L 3L L / 2 8EI
2     2     2

K 3                                                                       
L  12       3Li    12 3L  L3                     12 3Li 12 3L 
                                                                    
 3L       L2 / 2 3L  L2                         3L L / 2 3L
2
L2 

Because M, K are (44) matrices, 4 frequencies and 4 mode shapes

Pukyong National University                                                     Intelligent Mechanics Lab.
Beam Elements

Example : Three Elements
 Displacement vector of total element :     (v1 1 v2  2 v3 3 )T
 Applying boundary conditions at fixed condition : v1  1  0
(v1 1 v2  2 v3 3 )T
 General structure of equations of motion :        v2   v2  0 
       
 2    2  0
 v3   v3  0 
M K    
3   3  0 
 Result for mass and stiffness matrices :         v    v4  0 
 4      
M c1  M a 2       Mb2       0             4   4  0 
 
                                            
M   MT 2   b
M c 2  M a3 Mb3  
 0
              Mb3 T
M c3 
K c1  K a 2     K b2     0 
             K c 2  K a3 K b3 
K   K T2 b                        
 0

T
K b3         K c3 


Pukyong National University                                                  Intelligent Mechanics Lab.
Beam Elements

Example : Concentrated Forces on Fixed-free Beam
 Mass and stiffness matrices :
 312         0      54    13L / 2            24   0   12 3L 
                  13L / 2 3L2 / 4               2 L2 3L L2 / 2
 AL  0          2 L2                         8EI  0                 
M                                                 K 3
840  54        13L / 2  156     11L           L  12 3Li 12 3L 
                                                                
 13L / 2 3L / 4 11L                         3L L / 2 3L
2
L2                     2
L2 

 Equations of motion :
 Forcing terms : f 2  kv2 , f3   P, T2  T3  0
 v2      v2   f 2   0 
         T   0 
M  2
K  2   2   
 v3      v3   f3    P 
             
3      3   T3   0 

 Modified stiffness matrices :
       kL3                   
 24          0    12 3L 

8 EI 
8 EI                   
K 3          0        2L 2
3L L / 2 
2
L                                  
    12       3Li 12    3L 
              L2 / 2 3L  L2 
     3L                        

Pukyong National University                                                Intelligent Mechanics Lab.
Beam Elements
Distributed Beam Forces
 When a distributed force acts on a beam : f ( x, t )
L
 Virtual work done by force :  W (t )   f ( x, t ) v( x, t )dx   vT F(t )
0

 Force vector of forces and torques :                                 F1   f1 
F  T 
F   2   1 
 F3   f 2 
   
 F4  T2 
 Using shape function : v( x, t )  S1 ( x)v1 (t )  S2 ( x)1 (t )  S3 ( x)v2 (t )  S 4 ( x) 2 (t )
L                                 L                                 L                                 L
 W (t )   f ( x, t ) S1 ( x) v1 (t )dx   f ( x, t ) S 2 ( x)1 (t )dx   f ( x, t ) S3 ( x) v2 (t )dx   f ( x, t ) S 4 ( x) 2 (t )dx
0                                 0                                0                                  0

 Force or torque at node i :
L
Fi (t )   Si ( x) f ( x, t )dx (i  1, 2,3, 4)
0
L
F1 (t )  f1 (t )   S1 ( x) f ( x, t ) dx
0
L
F2 (t )  T1 (t )   S2 ( x) f ( x, t ) dx
0
L
F3 (t )  f 2 (t )   S3 ( x) f ( x, t ) dx
0
L
F4 (t )  T2 (t )   S4 ( x) f ( x, t ) dx
0

Pukyong National University                                                                               Intelligent Mechanics Lab.
Beam Elements
Example : Distributed Forces on Fixed-Free Beam

 Applying boundary conditions at fixed condition : v1  1  0

 AL  156  22 L      EL  12 6 L 
M                        , K 3 
420  22 L 4 L2 
                   L  6 L 4 L2 


 Force : f ( x, t )  w
L       x 2 x 
3
wL
f 2    w 3    2    dx  
0
 L
        L         2
L     x 2    x 
3
wL2
T2   w   L    L    dx 
0
 L
           L      12

 Equations of motion :
 wL 

 AL  156 22 L   v2  EI  12 6 L   v2   2 
                            
420  22 L 4 L2   2  L3  6 L 4 L2   2   wL2 
                                
 12 
     

Need not evaluate integrals for f1, T1 since they do not appear in equations of
motion because of particular boundary conditions

Pukyong National University                                            Intelligent Mechanics Lab.
Example : Marine & Aircraft Engine

Finite element model   Natural vibration frequencies

Pukyong National University                             Intelligent Mechanics Lab.
Chapter Review

You should be able to do the following:
 Understand basic concept of finite element analysis
 Write equations for the bar, torsional and beam elements
 Assemble equations using given boundary conditions for a system consisting
of multiple elements
 Solve assembled equations to determine mode shapes and mode frequencies

Pukyong National University                                         Intelligent Mechanics Lab.

```
To top