The Finite Element Method
A Practical Course
CHAPTER 3:
FEM FOR BEAMS
CONTENTS

INTRODUCTION
 FEM EQUATIONS
– Shape functions construction
– Strain matrix
– Element matrices
– Remarks

EXAMPLE AND CASE STUDY
– Remarks
INTRODUCTION

The element developed is often known as
a beam element.
 A beam element is a straight bar of an
arbitrary cross-section.
 Beams are subjected to transverse forces
and moments.
 Deform
only
in
the
directions
perpendicular to its axis of the beam.
INTRODUCTION

In beam structures, the beams are joined
together by welding (not by pins or
hinges).
 Uniform cross-section is assumed.
 FE matrices for beams with varying
cross-sectional area can also be developed
without difficulty.
FEM EQUATIONS

Shape functions construction
 Strain matrix
 Element matrices
Shape functions construction

Consider a beam element
d1 = v1
d3 = v2
d4 = 2
d2 = 1
x, 
0
2a
x= - a
x= a
 = 1
=1
x
Natural coordinate system:  
a
Shape functions construction
Assume that
v( )   0  1   2   3
In matrix form: v( )  1 
2
2
 0 
 
 
3  1
 2 
 3 

or
3
v( )  p T ( )α
v v  1 v 1
 

 (1  2 2  3 3 2 )
x  x a  a
Shape functions construction
d1 = v1
d3 = v2
d4 = 2
d2 = 1
x, 
0
2a
x= - a
x= a
 = 1
=1
To obtain constant coefficients – four conditions
(1)
v(1)  v1
(2)
dv
 1
dx  1
(3)
v(1)  v2
(4)
dv
 2
dx  1
At x= a or  = 1
At x= a or  = 1
Shape functions construction
 v1  1  1
  0 1
 1 
a

 
v2  1 1
 2  0 1a
1
2
a
1
2
a
 1 1 

3 

a  2 
  or d e  A eα
1   3 
3 

a   4 

a
2  a
2
 3  a 3  a 
1

A e1  
a 
4  0 a 0


a 1 a 
1
or α  A e1d e
Shape functions construction
Therefore,
v  N( )d e
where

N( )  PA e1  N1 ( ) N 2 ( ) N 3 ( ) N 4 ( )
in which
N1 ( )  14 (2  3   3 )
N 2 ( )  a4 (1     2   3 )
N 3 ( )  14 (2  3   3 )
N 4 ( )  a4 (1     2   3 )

Strain matrix
u
 2v
 xx 
 y
  yLv
2
 x
 x
Eq. (2-47)
Therefore,
2
y 2
y
B   yLN   y 2 N   2
N


N
2
2
x
a 
a
where N  N1
N 2
3
a
N1   , N 2  (1  3 )
2
2
3
a




N 3    , N 4  (1  3 )
2
2
N 3
N 4 
(Second derivative of
shape functions)
Element matrices
k e   B cBdV  E  y dA
T
2
a
a
V
A
1
 EI z 
1
EI
k e  3z
a

1
1
2
2
T 
( 2 N) ( 2 N)dx
x
x
2
EI z
1 2
T 
[ 2 N] [ 2 N]ad  3
4
a 

a
 N1N1
 N N 
 2 1
 N3N1

 N 4N1
N1N 2 N1N3
N 2N 2 N 2N 3
N3N 2 N3N3
N 4N 2 N 4N 3
Evaluate
integrals
1

1
N T N d
N1N 4 
N 2N 4 
d
N 3N 4 

N 4N 4 
 3 3a

4a 2
EI z 
ke  3
2a 

 sy.
 3 3a 
 3a 2a 2 
3
 3a 

4a 2 
Element matrices
m e   N NdV    dA
T
a
a
V
N Ndx   A
T
1
1
NT Nad
A
  Aa 
1
1
 N1 N1
N N
 2 1
 N 3 N1

 N 4 N1
N1 N 2
N1 N 3
N2 N2
N3 N 2
N4 N2
N 2 N3
N3 N3
N 4 N3
Evaluate
integrals
N1 N 4 
N 2 N 4 
d
N3 N 4 

N 4 N 4 
 78 22a 27  13a 

8a 2 13a  6a 2 
Aa 
me 
78  22a 
105 

2 
sy
.
8
a


Element matrices
 f s1   f y a  f s1 
 N1 
 f y a2





1 N2
 ms1   3  ms1 
T
T


fe   N fb dV   N f s dS f  f y a 
d     

1  N 
f
a

f
f
s2 
3
V
Sf
 s2   y
 
f y a2



m
N
 4 
 s 2   3  ms 2 
  f
k ede  med
e
e
Remarks

Theoretically, coordinate transformation can also be used
to transform the beam element matrices from the local
coordinate system to the global coordinate system.

The transformation is necessary only if there is more than
one beam element in the beam structure, and of which
there are at least two beam elements of different
orientations.

A beam structure with at least two beam elements of
different orientations is termed a frame or framework.
EXAMPLE
Consider the cantilever beam as shown in the figure. The beam is fixed
at one end and it has a uniform cross-sectional area as shown. The beam
undergoes static deflection by a downward load of P=1000N applied at
the free end. The dimensions and properties of the beam are shown in
the figure.
P=1000 N
0.1 m
E=69 GPa
=0.33
0.5 m
0.06
m
EXAMPLE
Exact solution:
 4v
EI y 4  f y  0
x
fy  0
Eq. (2.59)

Step 1: Element matrices
1 3 1
3
I z  bh   0.1 0.06   1.8 106 m 4
12
12
69 10 1.8 10 


9
K  ke
6
2  0.253
0.75
 3
0.75 0.25
 3.974 106 
 3 0.75

0.75 0.125
0.75
3
0.75 
 3
0.75 0.25 0.75 0.125 


 3 0.75
3
0.75


0.75
0.125

0.75
0.25


3
0.75 
0.75 0.125 
Nm -2
3
0.75

0.75 0.25 
Px 2
v( x) 
(3L  x)
6 EI y
PL3
v( x  L) 
3EI y
= -3.355E10-4 m
P=1000 N
E=69 GPa
=0.33
0.5 m
P=1000 N
EXAMPLE

v1  1  0
E=69 GPa
Step 1 (Cont’d):
=0.33
0.5 m
0.75
3
0.75   v1   Q1  ?   unknown reaction shear force
 3

0.25 0.75 0.125  1   M 1  ?   unknown reaction moment
6  0.75
3.974 10
 

 3 0.75
3
0.75  v2   Q2  P 

   
0.75
0.125

0.75
0.25

  2   M 2  0
D
K
F

Step 2: Boundary conditions
v1  1  0
0.75
3
0.75   v1  0   Q1 
 3
0.75 0.25 0.75 0.125    0  M 
 
1 
  1
3.974 106 

 

 3 0.75
3
0.75  v2   Q2  P 

 
0.75
0.125

0.75
0.25

   2   M 2  0
EXAMPLE

Step 2 (Cont’d):
0.75
 3
-2
K  3.974 10 
Nm


0.75
0.25


6
Therefore, K d = F,
dT

where
1000
= [ v2 2] , F  

0


Step 3: Solving FE equation
(Two simultaneous equations)
v2 = -3.355 x 10-4 m
2 = -1.007 x 10-3 rad
EXAMPLE

0.75
3
0.75   v1   Q1  ? 
 3

0.25 0.75 0.125  1   M 1  ? 
6  0.75
3.974 10
 

 3 0.75
3
0.75  v2   Q2  P 

   
0.75
0.125

0.75
0.25

  2   M 2  0 
D
K
F
Step 4: Stress recovering
v2 = -3.355 x 10-4 m
2 = -1.007 x 10-3 rad
Q1  3.974  106 (3v2  0.75 2 )
 3.974  106 [3  (3.355  104 )  0.75  (1.007  103 )]
 998.47N
M 1  3.974  106 (0.75v2  0.125 2 )
 3.974  106 [0.75  (3.355  104 )  0.125  (1.007  103 )]
 499.73Nm
Substitute
back into
first two
equations
Remarks

FE solution is the same as analytical solution
 Analytical solution to beam is third order
polynomial (same as shape functions used)
 Reproduction property
CASE STUDY

Resonant frequencies of micro resonant transducer
Bridge
Membrane










CASE STUDY
Natural Frequency (Hz)
Number of 2node beam
elements
Mode 1
Mode 2
Mode 3
10
4.4058 x 105
1.2148 x 106
2.3832 x 106
20
4.4057 x 105
1.2145 x 106
2.3809 x 106
40
4.4056 x 105
1.2144 x 106
2.3808 x 106
60
4.4056 x 105
1.2144 x 106
2.3808 x 106
Analytical
Calculations
4.4051 x 105
1.2143 x 106
2.3805 x 106
[ K  lM ]f = 0
Section 3.6, pg. 58
CASE STUDY
Mode 1 (0.44 MHz)
1.2
Dy (um)
1
0.8
0.6
0.4
0.2
0
0
20
40
60
x (um)
80
100
CASE STUDY
Mode 2 (1.21MHz)
1.5
Dy (um)
1
0.5
0
-0.5
0
20
40
60
-1
-1.5
x (um)
80
100
CASE STUDY
Mode 3 (2.38 MHz)
1.5
Dy (um)
1
0.5
0
-0.5
0
20
40
60
-1
-1.5
x (um)
80
100