Question
adjadj11
FEM_beam_vibration.pdf
The Euler-Bernoulli beam element that is described in the static structural analysis is also used for
the vibration analysis using the Finite Element Method (FEM), as shown in Figure 1. The symbol
^ is used to express the variable in the local coordinate system. In order to generate the motion
equation expressing the transversal vibration of the beam element and determine the natural
frequency, the energy method is used.
Figure 1. The Euler-Bernoulli beam element with the sign convention of forces and moments.
The nodal deflection and rotation of a single element is
2
2
1
1
ˆ
d̂
ˆ
d̂
d̂ y
y
(1)
The deflection of the beam element is described by
d̂Nd̂NNNNx̂v 4321
(2)
where 4321
N,N,N,N are shape functions and they are defined by
223 34
23
33
3223
32
323
31
1 32
1
2 1
32 1
Lx̂Lx̂ L
NLx̂x̂ L
N
Lx̂Lx̂Lx̂ L
NLLx̂x̂ L
N
(3)
The total energy of the beam element constants the kinetic and potential energies of the beam
element. In the case of the free vibration, the potential energy is defined as the strain energy due
to no external force is applied. The strain energy of the homogeneous beam element is expressed
as
L
x̂d x̂
v EIU
2
2
2
2
1 (4)
Expressing the deflection of the beam in the terms of the shape function, the equation (4) yields to
d̂kd̂d̂x̂d x̂
N
x̂
N EId̂U
T
L
T T
2
1
2
1 2
2
2
2
(5)
k is the element stiffness matrix, defined by
L
T
x̂d x̂
N
x̂
N EIk
2
2
2
2
(6)
Using the shape functions in (3) for integrating equation (6), the matrix k is
2
22
3
4
612
264
612612
Lsym
L
LLL
LL
L
EI k (7)
The kinetic energy of the beam element is expressed as
L
x̂d dt
dv AT
2
2
1 (8)
Equation (8) is expressed by the
dt
d̂d m
dt
d̂d
dt
d̂d x̂dNNA
dt
d̂d T
T
L
T
2
1
2
1 2
(9)
with m is the element mass matrix, given by:
L
x̂dNNAm 2
(10)
Calculating m with the shape functions, equation (9) yields:
2
22
4
22156
3134
135422156
420
Lsym
L
LLL
LL
AL m
(11)
The equation of motion for a single beam element is obtained by differentiating the total energy
with respect to time
0 dt
dU
dt
dT
dt
dE
0
2
d̂k dt
d̂d
dt
d̂d m
dt
d̂d TT
0 2
d̂k dt
d̂d m (12)
For the 1D geometry problem, the local and global coordinate systems are coincident, or dd̂ . The motion equations of the entire structure for the free vibration is determined after assembling
all elemental equations (12) in the global coordinate system and yields to this form:
0 dKdM (13)
where M and K are system inertia and stiffness matrices, respectively. Assume that the solution of equation (13) is
tj neDd (14)
where D is the displacement amplitude vector, n is the natural frequencies, and 1j . Substituting solution (14) to equation (13), it becomes
02 DMK n
(15)
In order to solve for non-trivial solution, the condition is set as
02 MK n
(16)
Equation (16) is an eigenvalue problem and it is used to determine the natural frequencies of the
beam problem after applying the boundary conditions.
NOTE: Use the MALAB syntax M,Keige to calculate the eigenvalues 2ne .
Example:
Specify the first three natural frequencies of a simply supported aluminum beam as described in
the following figure and table, then compare to the analytic solution.
Young’s modulus (E) 70 GPa
Poisson’s ration (υ) 0.3
Density (ρ) 2700 kg/m3
Length (2L) 0.2 m
Cross-section (height x width) 0.5 mm x 10 mm
Solution:
Assume that the generated mesh contains two elements. The global motion equation system is
defined as
0
4
612
268
612024
00264
00612612
4
22156
3138
13540312
003134
00135422156
420
3
3
2
2
1
1
2
22
22
3
3
3
2
2
1
1
2
22
22
d
d
d
Lsym
L
LLL
L
LLL
LL
L
EI
d
d
d
Lsym
L
LLL
L
LLL
LL
AL
Boundary conditions are defined as 0 3131 dddd , therefore, the rows and columns
corresponding to these defection at nodes 1 and 3 are removed from M and K . The natural frequencies are specified by equation (16) as
0
4
612
268
612024
00264
00612612
4
22156
3138
13540312
003134
00135422156
420
2
2
22
22
3
2
22
22
n
Lsym
L
LLL
L
LLL
LL
L
EI
Lsym
L
LLL
L
LLL
LL
AL
The eigenvalues of the above equation is determined by using MATLAB syntax M,Keige , and they are
(rad/s) f (Hz)
1 182.053 28.975
2 805.077 128.132
1 2
3
(1) (2)
L L
element 1 element 1 element 2 element 2
3 2023.36 322.070
The analytic solution is defined by
422 LA EIK
f n n
with )e(mod.),e(mod.),e(mod.K
n 388825391879
f1=28.866 (Hz); f2=115.523 (Hz); f1=259.707 (Hz)
Hint: visit the webpage for the analytic calculation:
http://www.amesweb.info/Vibration/Vibration-Calculators.aspx
The FEM solution converges to the analytic results by generating more elements as listed in the
following table
Number of elements f1 (Hz) f2 (Hz) f3 (Hz)
3 28.884 116.807 288.297
4 28.868 115.898 264.492
5 28.864 115.634 261.809
6 28.862 115.536 260.771
7 28.862 115.494 260.310
8 28.861 115.472 260.080
9 28.861 115.462 259.956
10 28.861 115.455 259.885
20 28.607 115.444 259.755
