1. DYNAMIC CONSIDERATIONS
Static analysis holds when the loads are slowly applied. When
the loads are suddenly applied, or when the loads are of a
variable nature, the mass and acceleration effects come into
the picture. If a solid body, such as an engineering structure,
is deformed elastically and suddenly released, it tends to
vibrate about its equilibrium position. This periodic motion due
to the restoring strain energy is called free vibration.
1
2. The number of cycles per unit time is called frequency.
The maximum displacement from the equilibrium position is
the amplitude.
FORMULATION
We define the Lagrangean by
L = T - П (11.1)
where T is the kinetic energy and П is the potential energy.
2
3. 2
1
t
t
I Ldt
1 2 , 1 2
, ,..., , ,...,
n n
q q q q q q i i
q = dq /dt,
0 1
i i
d L L
i to n
dt q q
Hamilton’s principle For an arbitrary time interval from t1 to t2,
the state of motion of a body extremizes the functional
If L can be expressed in terms of the generalized variables
where then the equations of
motion are given by
(11.3)
(11.2)
3
4. Example 11.1
Consider the spring-mass system in Fig. E11.1. The kinetic
and potential energies are given by
2 2
1 1 2 2
2
2
1 1 2 2 1
1 1
2 2
1 1
2 2
T m x m x
k x k x x
4
5. Using L = T – П, we obtain the equations of motion
1 1 1 1 2 2 1
1 1
2 2 2 2 1
2 2
0
0
d L L
m x k x k x x
dt x x
d L L
m x k x x
dt x x
which can be written in the form
1 2 2
1 1 1
2 2 2
2 2
0
0
0
k k k
m x x
m x x
k k
0
Mx kx
x
which is of the form
where M is the mass matrix, K is the stiffness matrix, and
and x are vectors representing accelerations and displacements.
(11.4)
5
6. Solid Body with Distributed Mass
Consider a solid body with distributed mass (Fig.11.1). The
potential-energy term has already been considered in Chapter 1.
The kinetic energy is given by
1
2
T
V
T u u dV
(11.5)
6
7.
, ,
T
u u v w
, , .
u v and w
u Nq
where ρ is the density (mass per unit volume) of the material
and
is the velocity vector of the point at x, with components
In the finite element method, we dive the body into elements,
and in each element, we express u in terms of the nodal
displacements q, using shape functions N.
u = Nq (11.7)
In dynamic analysis, the elements of q are dependent on time,
while N represents (spatial) shape functions defined on a
master element. The velocity vector is then given by
(11.8)
(11.6)
7
8. 1
2
T T
e
T q N N dV q
e T
e
m N N dV
1 1
2 2
T e T
e
e e
T T q m q Q MQ
When Eq.11.8 is substituted into Eq.11.5, the kinetic energy Te
in element e is
where the bracketed expression is the element mass matrix
This mass matrix is consistent with the shape functions
chosen and is called the consistent mass matrix. On summing
over all the elements, we get
(11.11)
(11.9)
(11.10)
8
9. 1
2
T T
Q KQ Q F
MQ KQ F
0
MQ KQ
The potential energy is given by
Using the Lagrangean L = T – П, we obtain the equation of
motion:
For free vibrations the force F is zero. Thus,
(11.14)
(11.12)
(11.13)
For the steady-state conditions, starting from the equilibrium
state, we get
sin
Q U t
(11.15)
9
10. 2
KU MU
KU MU
where U is the vector of modal amplitudes of vibration and ω
(rad/s) is the circular frequency (2πf, f = cycles/s or Hz).
Introducing Eq. 11.15 into Eq.11.14, we have
This is the generalized eigen value problem
(11.17)
(11.16)
ELEMENT MASS MATRICES
Treating the material density ρ to be constant over the
element, we have, from Eq. 11.10,
e T
e
m N N dV
(11.18)
10
11.
1 2
1 2
T
q q q
N N N
1 2
1 1
2 2
N N
1
1
2
e T T
e e
e
A
m N NA dx N N d
One-dimensional bar element For the bar element shown in
Fig. 11.2,
where
(11.19)
11
13. 2 1
1 2
6
e e e
A
m
1 2 3 4
1 2
1 2
,
0 0
0 0
T
T
u u v
q q q q q
N N
N
N N
On carrying out the integration of each term in NTN, we find that
Truss element For the truss element shown in Fig. 11.3,
(11.21)
(11.20)
where
1 2
1 1
2 2
N N
13
14. in which ξ is defined from -1 to +1. Then,
2 0 1 0
0 2 0 1
1 0 2 0
6
0 1 0 2
e e e
A
m
(11.22)
FIGURE 11.3 Truss element 14
15. CST element For the plane stress and plane strain conditions
for the CST element shown in Fig. 11.4, we have, from
Chapter 5,
1 2 6
1 2 3
1 2 3
...
0 0 0
0 0 0
T
T
u u v
q q q q
N N N
N
N N N
(11.23)
The element mass matrix is then given by
e T
e
e
m t N N dA
Noting that 2
1 1 2
1 1
, ,
6 12
e e
e e
N dA A N N dA A
etc., we have
15
16. 2 0 1 0 1 0
2 0 1 0 1
2 0 1 0
2 0 1
12
2 0
2
e e e
t A
m
Symmetric
(11.24)
Axisymmetric triangular element For the axisymmetric
triangular element, we have
uT = [ u w]
16
18. where u and w are the radial and axial displacements,
respectively. The vectors q and N are similar to those for the
triangular element given in Eq.11.23. We have
2
e T T
e e
m N N dV N N r dA
(11.25)
1 1 2 2 3 3
2
e T
e
m N r N r N r N N dA
3 2
1 1 2 1 2 3
2 2 2
, , , .
20 60 120
e e e
e e e
A A A
N dA N N dA N N N dA etc
Since r = N1r1 + N2r2 + N3r3, we have
Noting that
18
19. we get
3 2
1
3 2
1
1
2
1
2
3
3
4
2 0 2 0 2 0
3 3 3
4
2 0 2 0 2
3 3 3
4
2 0 2 0
3 3
4
2 0 2
3 3
10
4
2 0
3
4
2
3
e
e
r r
r r r r
r r
r r r r
r
r r r
r
A
r r r
m
Symmetric r r
r r
(11.26)
19
20. 1 2 3
3
r r r
r
1 2 8
1 2 3 4
1 2 3 4
...
0 0 0 0
0 0 0 0
T
T
u u v
q q q q
N N N N
N
N N N N
where
Quadrilateral element For the quadrilateral element for plane
stress and plane strain,
(11.28)
(11.27)
20
21. The mass matrix is then given by
1 1
1 1
det
e T
e
m t N N J d d
This integral needs to be evaluated by numerical integration.
(11.29)
21
22. Beam element For the beam element shown in Fig. 11.5, we
use the Hermite shape functions given in Chapter 8. We have
v = Hq
1
1
2
e T e
e
m H H A d
(11.30)
On integrating, we get
2 2
2
156 22 54 13
4 13 3
156 22
420
4
e e
e e e
e e e
e
e
A
m
Symmetric
(11.31)
22
23. Frame element We refer to Fig.8.9, showing the frame
element. In the body coordinate system x’, y’, the mass matrix
can be seen as a combination of bar element and beam
element. Thus, the mass matrix in the prime system is given
by
2
2 2
'
2
2 0 0 0 0
156 22 0 54 13
4 0 13 3
2 0 0
156 22
4
e e
e e e e
e
e
a a
b b b b
b b b
m
a
Symmetric b b
b
(11.32)
23
24. where
6 420
e e e e
A A
a and b
Using the transformation matrix L is given by Eq. 8.48, we
now obtain the mass matrix me in the global system:
me = LTme’L (11.33)
Tetrahedral element For the tetrahedral element presented in
Chapter 9,
1 2 3 4
1 2 3 4
1 2 3 4
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
T
u u v w
N N N N
N N N N N
N N N N
(11.34)
24
26. Lumped mass matrices Practicing engineers also use
lumped mass techniques, where the total element mass in
each direction is distributed equally to the nodes of the
element, and the masses are associated with translational
degrees of freedom only. For the truss element, the lumped
mass approach gives a mass matrix of
1 0 0 0
1 0 0
1 0
2
1
e e e
A
m
Symmetric
(11.36)
26
27. For the beam element, the lumped element mass matrix is
1 0 0 0
0 0 0
1 0
2
0
e e e
A
m
Symmetric
(11.37)
EVALUATION OF EIGENVALUES AND EIGENVECTORS
KU = λMU (11.38)
We observe here that K and M are symmetric matrices.
Further, K is positive definite for properly constrained
problems.
27
28. Properties of Eigenvectors
For a positive definite symmetric stiffness matrix of size n,
there are n real eigenvalues and corresponding
eigenvectors satisfying Eq.11.38. The eigenvalues may be
arranged in ascending order:
0 ≤ λ1 ≤ λ2 ≤ …≤ λn (11.39)
If U1, U2…Un are the corresponding eigenvectors, we have
KUi = λiMUi (11.40)
28
29. 0
T
i j
U MU if i j
0
T
i j
U KU if i j
1
T
i i
U MU
T
i i i
U KU
The eigenvectors possess the property of being orthogonal
with respect to both the stiffness and mass matrices
(11.41a)
The lengths of eigenvectors are generally normalized so that
The foregoing normalization of the eigenvectors leads to
the relation
(11.42b)
(11.41b)
(11.42a)
29
30. Eigenvalue – Eigenvector Evaluation
The eigenvalue-eigenvector evaluation procedures fall into the
following basic categories:
1.Characteristic polynomial technique
2.Vector iteration methods
3.Transformation methods
Characteristic polynomial From Eq. 11.38, we have
(K - λM)U = 0 (11.43)
30
31. If the eigenvector is to be nontrivial, the required condition is
det(K - λM)= 0 (11.44)
This represents the characteristic polynomial in λ.
Example 11.2
Determine the eigenvalues and eigenvectors for the stepped
bar shown in Fig. E11.2a.
Solution Gathering the stiffness and mass values
corresponding to the degrees of freedom Q2 and Q3, we get
the eigenvalue problem
31
32.
1 2 2
2 1 1 2 2 2 2 2
1 2 2
3 3
2 2 2 2
2 2
2 2
2
6 2
A A A
U A L A L A L U
L L L
E
U U
A L A L
A A
L L
4 2 4
0.283
7.324 10 / .
32.2 12
f
lbs in
g
2 2
6 4
3 3
0.2 0.1 25 2.5
30 10 1.22 10
0.1 0.1 2.5 5
U U
U U
We note here that the density is
Substituting the values, we get
32
35. Note that λ = ω2, where ω is the circular frequency given by
2πf and f = frequency in hertz (cycles/s).
These frequencies are
f1 = 4115 Hz
f2 = 13884 Hz
The eigenvector for λ1 is found from
(K – λ1M)U1 = 0
which gives
2
6
3 1
3.96 3.024
10 0
3.204 2.592
U
U
35
36. The two previous equations are not independent, since the
determinant of the matrix is zero. This gives
3.96 U2 = 3.204U3
Thus,
1 2 2
,1.236
T
U U U
1 1 1
T
U MU
1 14.527 17.956
T
U
2 11.572 37.45
T
U
For normalization, we set
On substituting for U1, we get
The eigenvector corresponding to the second eigenvalue is
similarly found to be
The mode shapes are shown in Fig. E11.2b. 36
