Principles of
Computer

Aided
Design and
Manufacturing
Second Edition 2004
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Author: Prof. Farid. Amirouche
University of Illinois

Chicago
University of Illinois

Chicago
Chapter 10
Dynamic Analysis

A
Finite
–
Element
Approach
Principles of Computer

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Author: Prof. Farid. Amirouche, University of Illinois

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CHAPTER 10
10.1 Introduction
In the preceding chapters we examined the area of finite element associated with the first law
of mechanics. The structure’s response to time variant loads needs to be captured by using
differential equations that govern the motion. Applying Newton’s Second Law or Lagrange
equations usually does this. Up to this point we have discussed only cases in which the load on
the structure is static and therefore the response was also static and proportional to the
structural stiffness and applied loads. It is apparent that the response of a structure will be
different if we account for the inertia forces and the time variant effects due to the kinematics
and forces applied to the system.
This chapter will be used to show how we can use finite element method to formulate and
solve for dynamic problems. The finite element method was used successfully to determine
element stiffness matrices in the study of trusses. The methodology is extended to derive the
mass element matrices and their corresponding global mass matrix for the structure. In most
cases the differential equations for a structure assuming small deformation take the following
form.
F
u
K
u
C
u
M
(10.1)
10.1
Introduction
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CHAPTER 10
If we assume a solution of the form
t
A
u
sin
then the dynamic analysis of structures takes the following form.
F
u
M
K
2
(10.2)
(10.3)
where [K] is the global stiffness matrix, [M] is the global mass,
2
denotes the natural frequency, U
the global displacement array and {F} the external excitation force vector.
One of the finite element’s most popular approaches in vibration is the computation of
the natural frequencies of a mechanical system. Since both [K] and [M] can directly be calculated
from finite element method provided that the geometric properties and the boundary conditions of
the problem are known, we can evaluate [
2
] making use of numerical procedures readily available
in most of the engineering numerical analysis books. In this chapter we will focus on the analysis of
the eigenvalue problem and in the formulation of the mass matrix in particular and the
corresponding stiffness by showing how both the stiffness and mass can be calculated and develop
the necessary steps to compute [
2
].
10.1
Introduction
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CHAPTER 10
10.2 Element Stiffness and Mass Matrices
Once more let us examine a beam under a compressive load, where the compressive force is not
constant but rather a force that is harmonically exciting the beam as function of time. That is the
external force of
t
F
F
sin
0
(10.4)
Let F be applied at the end of the beam as shown in Figure 10.1
In previous chapters we developed the element stiffness matrix when a beam is
discretized into N elements. Essentially each element acts as a spring with a stiffness
value k equal to
l
AE
k
e
Again observe the element shown in Figure 10.2 it is subjected to axial forces at each end
(node), and the actual displacement of the element can be written as function of the node
displacements u1 and u2. The stiffness matrix can be obtained from the energy function
minimization on the basis of equilibrium. This is derived by first developing the potential and
kinetic energy due to axial deformation under the load F.
10.2 Element Stiffness and Mass Matrices
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CHAPTER 10
F=F
0
sin
t
Figure 10.1 A beam excited longitudinally by a harmonic force F
10.2 Element Stiffness and Mass Matrices
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CHAPTER 10
1
2
f
2
u
2
f
1
u
1
Figure 10.2 A beam element with nodes 1 and 2 under compressive load.
10.2 Element Stiffness and Mass Matrices
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CHAPTER 10
First we write the beam displacement in terms of u
1
and u
2
as follows.
t
u
l
x
t
u
l
x
t
x
u
2
1
1
,
Note how the displacement at x = 0 is u
1
and at x =
l
it is u
2.
This is consistent with our assumed
displacement function.
Now we can compute the energy terms needed for our finite element
formulation. The strain energy of the bar is given by the integral
dx
x
t
x
u
EA
t
V
l
0
2
,
2
1
(10.5)
(10.6)
The kinetic energy of the element can be computed from the integral
dx
x
t
x
u
x
A
t
T
2
,
2
1
Where
(x) is the density of the bar.
(10.7)
10.2 Element Stiffness and Mass Matrices
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CHAPTER 10
Differentiating of u (x, t) with respect to x in equation (10.5) and substituting it into (10.6) yield and
explicit expression for the potential energy of the form
)
2
(
2
/
)
(
2
2
2
1
2
1
u
u
u
u
l
EA
t
V
(10.8)
The velocity equation is
t
u
l
x
t
u
l
x
t
t
x
u
2
.
1
.
1
,
Substituting equation (10.9) into (10.7) we obtain the kinetic energy equation in terms of
u
)
(
)
(
)
(
2
1
t
u
t
u
t
u
u
M
t
T
T
2
/
1
)
(
(10.9)
which can be written as
(10.10)
Where M is defined by
2
1
1
2
6
Al
M
(10.11)
10.2 Element Stiffness and Mass Matrices
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CHAPTER 10
10.3 Axial Deformation
10.3.1 One Element equation of motion
The equations of vibration can be obtained from the preceding expressions for the kinetic
energy T( t) and strain energy V (t) by using the variational or Lagrangian approach.
The equation of motion can be calculated from energy in the structure from
n
i
t
f
u
V
u
T
u
T
t
i
i
i
i
,
,
2
,
1
.
Where u
i
is the coordinate of the system, which is assumed to have n degrees of freedom, and where
i
f
is the external force applied at coordinates
u
i
.
(10.13)
Here the energies T and V are the total Kinetic energy and strain energy respectively in the structure.
2
2
.
3
2
1
u
Al
t
T
Analysis of the boundary conditions indicates that the time response at the clamped end must be
zero. Hence the total kinetic energy becomes
(10.14)
10.3 Axial Deformation
Principles of Computer

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CHAPTER 10
2
2
2
1
u
l
EA
t
V
0
3
2
2
..
t
u
l
EA
t
u
Al
(10.15)
Substitution of these two energy expressions into the Lagrange equation
(10.16)
o
n
n
n
u
u
t
u
u
t
u
.
0
1
2
0
.
2
0
2
tan
sin
Which constitutes a simple finite element model of the cantilever bar using only one element.
This finite element model of the bar can now be solved, given a set of initial conditions, for the
nodal displacement u2(t). The solution to the equation is
(10.17)
10.3 Axial Deformation
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CHAPTER 10
10.3.2 Three Elements equation of motion
t
V
t
V
t
V
t
V
3
2
1
4
3
4
3
3
2
3
2
2
2
1
1
1
1
1
1
1
1
0
1
1
1
1
0
2
3
u
u
u
u
u
u
u
u
u
u
l
EA
T
T
(10.18)
=
=
2
4
4
3
2
3
3
2
2
2
2
2
2
2
2
3
u
u
u
u
u
u
u
l
EA
t
u
t
u
t
u
Al
u
T
u
T
u
V
4
..
3
..
2
..
4
.
.
2
2
1
0
1
4
1
0
1
4
18
(10.19)
The vector of derivatives of this total strain energy
indicated in the Lagrangian equation is found to be
10.3 Axial Deformation
T
he total strain energy is the sum of the strain energy associated with each element , that is,
Principles of Computer

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CHAPTER 10
4
.
3
.
4
.
3
.
3
.
2
.
3
.
2
.
2
.
2
.
2
1
1
2
2
1
1
2
0
2
1
1
2
0
36
u
u
u
u
u
u
u
u
u
u
Al
t
T
T
T
T
t
u
t
u
t
u
Al
u
T
u
T
u
T
dt
d
4
..
3
..
2
..
4
.
3
.
.
2
2
1
0
1
4
1
0
1
4
18
In a similar fashion, total kinetic energy can be expressed by adding the contribution of the
kinetic energy from all 3 elements where we write
(10.20)
The first term in the Lagrangian equation, which is responsible for the mass matrix, then becomes
(10.21)
0
)
(
)
(
..
t
Ku
t
u
M
Combining all the Lagrangian terms the equations of motion take the following form
(10.22)
10.3 Axial Deformation
Principles of Computer

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CHAPTER 10
2
1
0
1
4
1
0
1
4
18
Al
M
Where u(t)=[u2 u3 u4]T is the vector of nodal displacements. Here the matrix
(10.23)
1
1
0
1
2
1
0
1
2
3
l
EA
K
is the global mass matrix and the corresponding K
(10.24)
is the global stiffness matrix.
10.3 Axial Deformation
Principles of Computer

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CHAPTER 10
The finite element method allows us to derive both the mass and stiffness of a structure by means of the
energy method. It is obvious that the size of these matrices can expand rapidly if we choose a large
number of elements. When solving a linear problem we are mostly interested in the vibration
characteristics of the problem and not necessarily its time response. Hence we reduce the problem to a
so

called an eigen value problem by first assuming a solution to equations of the form given by (10.22)
that is given by
t
j
Xe
t
x
u
)
,
(
0
2
X
M
w
k
We have essentially reduced the problem to a free

vibration where the right hand side of the
governing equations is set to zero. Substitution of equation (10.25) into (10.22) yields
(10.25)
(10.26)
We can solve for w
2
by setting the following determinant equal to zero.
0
2
M
w
k
Equation (10.26) represents the generalized eigen

value problem.
(10.27)
10.3 Axial Deformation
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CHAPTER 10
Let us consider a beam subject to transverse loading, which causes the beam to bend. The deflection of
the neutral axis at any location x is represented by the variable v, for small deflections we have the
flexure formula which defines the stress, the moment and the vertical displacement measured from the
neutral axis as
10.4 Bending or Transverse Deformation of a Beam
I
My
x
M
dx
v
d
EI
2
2
(10.28)
Or
(10.29)
Where I is the second moment of area. From standard strength of materials the above equation
can be expressed further as
x
w
dx
x
dV
dx
v
d
EI
4
4
(10.30)
10.4 Bending or Transverse Deformation
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CHAPTER 10
i
j
U
i2
U
i1
U
j1
U
j2
Neutral Axis
Figure 10.3 Simply supported beam
10.4 Bending or Transverse Deformation
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CHAPTER 10
The total potential energy of the beam equals the strain energy less the potential energy due to the
external loads. Let us represent the energy equation by
v
u
(10.31)
The strain energy contribution is obtained from the following integral where we can assume that
the energy for a differential element will also have the same equation except the limits on the
integral will be those associated with the element. If we ignore the contribution of the energy due
to the external forces, the energy equation reduces to
dx
dx
v
d
EI
2
2
2
2
(10.31)
The displacement u at each node is expressed in terms of vertical and longitudinal
components as shown in Figure 10.3, hence the deflection can be expressed as
fourth order polynomial
3
2
1
1
dx
cx
x
b
a
v
(10.32)
10.4 Bending or Transverse Deformation
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CHAPTER 10
Which defines a cubic spline. The coefficients can be found by making use of
boundary conditions at both ends. The equation above becomes
4
4
3
3
2
2
1
1
u
N
u
N
u
N
u
N
u
Note how we have substituted u for v in the above equation. We can write equation (10.33) in a
matrix form where we can isolate the shape function and the displacement vector and express u as
4
3
2
1
4
3
2
1
u
u
u
u
N
N
N
N
x
u
(10.34)
(10.33)
Where the shape functions are
3
3
2
2
1
2
3
1
L
x
L
x
N
2
3
2
2
2
L
x
L
x
x
N
(10.35)
3
3
2
3
2
3
L
x
L
x
N
2
3
2
4
L
x
L
x
N
10.4 Bending or Transverse Deformation
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CHAPTER 10
By substituting equation (10.34) into the strain energy equation we obtain
L
u
u
u
u
N
N
N
N
N
N
N
N
EI
u
u
u
u
0
4
3
2
1
"
4
"
3
"
2
"
1
"
4
"
3
"
2
"
1
4
3
2
1
2
1
By taking the second derivatives of the shape functions we get
3
2
"
1
12
6
L
x
L
N
2
"
2
6
4
L
x
L
N
3
2
"
3
12
6
L
x
L
N
2
"
4
6
2
L
x
L
N
(10.36)
(10.37)
By substitution of the above equation into
and recalling that the displacement at the
equilibrium position occurs such that the value of the system total energy is minimum, we get
0
u
(10.38)
10.4 Bending or Transverse Deformation
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CHAPTER 10
4
3
2
1
2
2
2
2
3
4
6
2
6
6
12
6
12
2
6
4
6
6
12
6
12
u
u
u
u
L
L
L
L
L
L
L
L
l
L
L
L
L
EI
Hence, the stiffness matrix for a beam element due to deflection (bending) is
2
2
2
3
3
4
6
2
6
6
12
6
12
2
6
4
6
6
12
6
12
L
L
L
L
L
L
L
L
L
L
L
L
L
EI
k
e
Once again we have denoted by [k
e
] the element stiffness matrix to be consistent with our
formulation of the finite element method. Now we can tackle the formulation of the mass matrix.
(10.39)
(10.40)
10.4 Bending or Transverse Deformation
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CHAPTER 10
Consider the uniform bar shown in Figure 10.5. The element has a velocity
in the x direction. If we
discretize the bar into N elements, we evaluate the kinetic energy of the bar by integrating over the total
length the bar. To obtain the kinetic energy of the element we first define the kinetic energy equation as
follows
.
u
dm
u
T
L
0
.
2
1
dx
A
u
T
.
2
1
Adx
dm
The displacement field can be expressed as a function of basis function or shape functions such that
N
i
i
i
t
q
x
t
x
u
1
,
Where
q
(t) defines the generalized coordinates.
(10.41)
(10.42)
Where
(10.43)
10.4 Bending or Transverse Deformation
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CHAPTER 10
Substituting the above equation into the kinetic energy of the bar yields
L
N
i
i
N
i
i
dx
A
q
q
T
0
1
.
1
.
2
1
(10.44)
which leads to
L
N
i
N
j
j
i
ij
q
q
M
T
0
1
1
.
.
2
1
The mass matrix M
ij
is then defined by the following expression
L
j
i
ij
dx
x
x
x
A
x
M
0
or simply
L
j
i
ij
dx
x
N
x
N
x
A
x
M
0
(10.45)
(10.47)
(10.46)
10.4 Bending or Transverse Deformation
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CHAPTER 10
N
i
& N
j
are the shape functions used in determining the element stiffness k
(e)
matrix, while
i
&
j
are
interpolating functions with the following property: when
is equal to N, M
ij
is referred as the
consistent mass matrix. For the sake of simplicity the shape functions N are used for the evaluation of
the mass matrix.
For a uniform beam subject to bending the mass matrix is found to be
2
2
2
2
4
3
22
13
3
4
13
22
22
13
156
54
13
22
54
156
420
L
L
L
L
L
L
L
L
L
L
L
L
mL
M
m is the total mass of the beam or element and is defined by
A
m
= Mass/unit volume
A: cross section area
L: length of the beam
(10.48)
10.4 Bending or Transverse Deformation
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CHAPTER 10
Example 10.1
Find the coefficients M
11
& M
12
of the mass matrix using the shape function defined in the
evaluation of the element stiffness matrix given by equation (10.35).
Solution:
Using equation (10.47) and substituting both subscripts i and j by 1 we obtain:
L
dx
N
AN
M
0
2
1
11
dx
N
N
A
M
T
1
1
11
Note that
3
2
1
2
3
1
L
x
L
x
N
Factorizing the constants
and A we get
10.4 Bending or Transverse Deformation
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CHAPTER 10
The above integral becomes
L
dx
L
x
L
x
A
M
0
2
3
2
11
2
3
1
=
dx
L
x
L
x
L
x
L
x
A
6
3
2
2
2
4
2
3
1
4
3
1
dx
x
L
x
L
x
L
x
L
x
L
A
L
6
6
5
5
4
4
3
3
2
2
4
12
9
4
6
1
=
420
156
7
4
2
5
9
1
3
6
1
L
L
10.4 Bending or Transverse Deformation
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CHAPTER 10
In a similar fashion we derive the element M
44
L
dx
N
A
M
0
2
4
44
4
6
3
5
3
5
2
4
2
3
2
2
3
2
L
x
L
x
L
x
L
x
L
x
L
x
L
x
L
x
3
3
3
3
4
7
3
6
2
5
0
4
6
3
5
2
4
420
4
7
1
3
1
5
7
1
6
2
5
1
2
L
L
L
L
L
x
L
x
L
x
dx
L
x
L
x
L
x
L
M
44
=
N
4
N
4
10.4 Bending or Transverse Deformation
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CHAPTER 10
10.5 Bernoulli Beam
Consider a beam element whose nodal deformations are described by the displacements
and a rotation at each end as shown in Figure 10.4. We notice that the node displacements are along
the longitudinal and normal axis. Essentially we have two translations at each node plus one
rotation. A beam with both axial and bending deformation is referred as a Bernoulli Beam, which
bears the name of one of the greatest scientists in Beam Theory.
Each joint depicts a longitutidinal and lateral deformation and a rotation. Hence the
element is subject to axial and bending deformation. The transverse shear is neglected. We write the
total strain energy as a combination of both the longitudinal deformation as well the bending
deformation. Therefore, the energy due to the strain deformation can calculated from
L
dx
dx
W
d
EI
dx
dU
EA
U
0
2
2
2
2
2
2
(10.49)
10.5 Bernoulli Beam
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CHAPTER 10
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EA
L
EA
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EA
L
EI
k
e
4
6
0
2
6
0
6
12
0
6
12
0
0
0
0
0
2
6
0
4
6
0
6
12
0
6
12
0
0
0
0
0
2
3
2
3
3
3
3
2
3
3
2
3
(10.50)
Combining the results obtained in equation (10.12) and equation (10.40), we obtain the
elastic stiffness matrix of the beam as
10.5 Bernoulli Beam
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CHAPTER 10
In a similar fashion we obtain the mass element matrix for the beam using equation
(10.11) and (10.48) as follows
105
210
11
0
140
420
13
0
210
11
35
13
0
40
13
70
9
0
0
0
3
1
0
0
6
1
140
1
420
13
0
105
210
11
0
420
13
70
9
0
210
11
35
13
0
0
0
6
1
0
0
3
1
2
2
2
2
L
L
L
L
L
L
L
L
L
L
L
AL
M
e
So we have established a procedure by which a mass and a stiffness element can be computed. In
a similar way we will explore how the load vector is computed so we can complete the governing
dynamical equations. The load vector represents the contribution of the external forces and
moments applied to the beam.
(10.51)
10.5 Bernoulli Beam
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CHAPTER 10
10.6 Force Vector
When the load is distributed over the beam we need to find the generalized force vector that represents the
contribution of the distributed at the degrees of freedom used to denote the beam deformation. In this case the
load will affect the longitudinal direction that is normal to the beam axis and will ultimately affect the bending
moments at the nodes. The contribution along the axial direction along the beam neutral axis will be zero. If we
let the load F (x, t) be w the generalized force vector can be expressed as
L
i
i
dx
x
N
t
x
F
F
0
)
(
,
(10.52)
For the case of a beam subjected to bending at the end points we have a displacement and a rotation, resulting
in both a reaction force and a moment.
12
/
2
/
0
2
/
12
/
0
2
2
wl
wl
wl
wl
F
(10.53)
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CHAPTER 10
Example 10.2
a)
b)
Figure 10.4 Two beams: a) simply supported at the right and fixed at the left end, and
b) fixed at both ends.
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CHAPTER 10
Two distinct uniform beams with different boundary conditions are being evaluated. They are both
of length L, have a cross section area A and a modulus of elasticity E. The boundary conditions
are such that the first beam is simply supported with a pin and roller at one end and fixed at the
other, while the second beam has both ends fixed. a) Derive the mass and stiffness matrix for a 2

element model. For both cases and explain how the boundary conditions alter the global matrix K
and M.
Solution:
1
2
u
2y
u
1x
3
u
3y
u
1y
u
2x
u
3x
Figure 10.5 Bernoulli Beam
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CHAPTER 10
Let the beam be discretized as shown above in Figure 10.5 where the displacement
and rotation at each node are represented by U
ix
, U
iy
and
i
.
(i=1, number of nodes).
Let {U} denote the global displacement array, and then we can determine the size of
the array by simply examining that at each node we have 3 degrees of freedom, hence the
global displacement vector {U} should be of size 9.
a
U
U
U
U
2
1
The global K matrix is defined by summing the stiffness of elements.
2
1
K
K
K
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CHAPTER 10
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EA
L
EA
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EA
L
EI
k
e
4
6
0
2
6
0
6
12
0
6
12
0
0
0
0
0
2
6
0
4
6
0
6
12
0
6
12
0
0
0
0
0
2
3
2
3
3
3
3
2
3
3
2
3
where
This element stiffness for a beam undergoing 2 translations and one rotation at each
joint is given by equation (10.50).
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CHAPTER 10
10.7 Boundary Conditions
From the beam end supports we deduce the following criteria for
a)
The case where the beam is fixed at one end and simply supported at the other
0
1
1
Y
X
U
U
0
3
y
U
b) The case with fixed

fixed conditions at both ends of the beam
U
1x
= U
1y
= 0
U
3x
= U
3y
= 0
Furthermore let the axial deformation U
2x
be negligible, than the global stiffness
matrix reduces for case (a) to
I
AL
L
L
L
L
L
L
L
L
L
L
L
L
EI
K
x
2
2
2
2
2
2
2
2
3
5
5
0
0
0
0
0
4
2
6
0
0
2
8
0
2
0
6
0
24
6
6
0
2
6
4
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CHAPTER 10
Following the outlined procedure above, we develop the global mass matrix by first noting that
for a uniform beam the element mass is
L
L
L
L
L
L
L
L
L
L
L
L
AL
M
e
105
1
210
11
0
140
1
420
13
0
210
11
35
13
0
420
13
10
9
0
0
0
3
1
0
0
6
1
140
420
13
0
105
1
210
11
0
420
13
70
9
0
210
11
35
13
0
0
0
6
1
0
0
3
1
2
2
2
We obtain the global matrix by adding the element mass matrix for each element
2
1
M
M
M
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CHAPTER 10
Applying the boundary conditions for case (a) the mass matrix reduces to
3
1
0
0
0
0
0
4
3
13
0
0
3
8
0
3
0
13
0
312
13
0
0
3
13
4
420
2
2
2
2
2
2
L
L
L
L
L
L
L
L
L
L
L
AL
Combining [M] &[K], and following the assumed solution outlined in equation (10.25) we can
express the characteristic equation for the eigen value problem as the determinant of the
following equation:
0
2
M
w
K
[M] =
The eigen values of which can be computed by determining the determinant of the matrix and
setting it equal to zero. This will result into the natural frequencies of the assumed beam with the
boundary conditions outlined in case a. In a similar fashion we can build the mass and stiffness
matrices for case b. We notice that for case b) the boundary conditions result into 5 displacements to
be zero, hence reducing the size of the global matrices by 5. We should then expect 5 eigen values
for case a) and 4 eigen values for case b).
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CHAPTER 10
10.8 Planar Beam Structure
(1)
(2)
(5)
(3)
(4)
1
2
3
4
X
Y
Figure 10.6 A planar beam structure in space
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CHAPTER 10
u
1y
u
1x
u
2x
u
2y
x
y
X
Y
O
i
j
1
2
Figure10. 7 A simple beam with an orientation
楮ia′䐠獰慣攮
10.8 Planar Beam Structure
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CHAPTER 10
Planar structures are common in engineering or their analysis is vital to design, longevity and safety.
In the previous section we derived the element stiffness or mass of a beam but we assumed the
beam neutral axis to be horizontal coinciding with the x axis of the coordinates X, Y. Obviously if a
beam is positioned in space and we have several of them whose inclinations with respect to the
global frame XY are different, then their mass and stiffness element matrix will contribute
differently to the global stiffness and mass matrix. What follows is derivation of element [M
(e)
] or
[K
(e)
] when its subjected to such orientation in space.
We have previously defined the element stiffness or mass for a uniform beam when the beam is
in a longitudinal position. Let {u
1x
, u
1y
,
1
,u
2x
, u
2y
,
2
} represent the displacement at each node
with respect to a local coordinate system of the beam with x along the neutral axis of the beam.
Then we can transform the local coordinate to a global ones along XY by using a transformation
matrix R such that
2
2
2
1
1
1
2
2
2
1
1
1
1
0
0
0
0
0
0
cos
sin
0
0
0
0
sin
cos
0
0
0
0
0
0
1
0
0
0
0
0
0
cos
sin
0
0
0
0
sin
cos
y
x
y
x
y
x
y
x
u
u
u
u
U
U
U
U
(10.54)
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CHAPTER 10
Or simply
u
R
U
e
The new element mass and stiffness matrices become
e
e
T
e
e
R
M
R
M
e
e
T
e
e
R
K
R
K
Note that the element mass and stiffness matrices on the right hand side of equation (10.56) and
(10.57) are those associated of zero orientation or simply the matrices developed for the element
coinciding with the horizontal (x

axis).
This orientation is justified for the measurement of the orientation angle commencing from the
horizontal line to the neutral axis of the beam element (CCW).
(10.55)
(10.56)
(10.57)
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CHAPTER 10
10.9 Eigen Value Problem
Once the dynamical equations are formulated we need to solve for the eigen values and
eigen vectors of the system. What follows are some common methods used in the solution
of such problems. The actual solution procedure used to solve the eigen problem is
discussed and some considerations are outlined for further study.
A very common eigen problem is encountered in vibration analysis. Examining
equation (10.26) we can generalize the solution of the problem to the following form
M
k
The solution for the n eigen values and corresponding eigen values can be written as
M
k
where the column of
represent the eigen vectors whereas the eigen values are stored in
the matrix
, which is a diagonal matrix listing the eigen values.
The eigen values are the roots of the characteristic polynomial,
)
det(
)
(
M
K
p
(10.58)
(10.59)
(10.60)
10.9 Eigen Value Problem
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CHAPTER 10
This is easily shown in the solution of free vibration problems where the equations of motion
are represented in a matrix as
0
..
KX
X
M
wt
X
x
sin
0
2
X
K
M
0
i
i
X
M
K
2
and X is analogous to
in equation (10.58)
(10.61)
Assuming a solution of the form
(10.62)
we get
(10.63)
or simply
(10.64)
Where
10.9 Eigen Value Problem
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CHAPTER 10
Example 10.3
Consider a linear system composed of two masses and springs interconnected to form an oscillatory
system used to characterize the dynamics of a train

car model. We are interested in the solution of
this problem from the point of view in learning about the frequency response and the system modes
of vibration.
m
1
m
2
x
1
x
2
k
1
k
2
Figure 10.8 A two

degree of freedom mass

spring system
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CHAPTER 10
The equations of motion can easily be derived using Newton’s Law where we denote by x
1
and x
2
the displacements of mass 1 and 2 respectively.
0
x
K
x
M
2
2
2
2
1
k
k
k
k
k
K
and
2
1
0
0
m
m
M
(10.65)
Where
Find:
a) the eigen values
b) the corresponding eigen vectors.
The physical parameters have the following values: m
1
=9 kg, m
2
=1 kg, k
1
=72 N/m and k
2
=4.5 N/m.
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CHAPTER 10
Solution:
For these values the characteristic equation
0
det
2
K
M
0
2
1
2
2
2
1
2
2
1
4
2
1
k
k
k
m
k
m
k
m
m
m
0
4
9
36
13
2
2
2
4
So that
1
2
=9 and
2
2
=4. There are two roots, and each corresponds to two values of the
constant
in the assumed form of the solution:
s
rad
s
rad
/
2
,
/
3
2
1
Now we need to solve for the eigen vectors corresponding to each eigen values.
We proceed by noticing that the equations of motion reduce to
0
1
2
1
u
K
M
Where u
1
denoted the vector associated with the first eigen value
1
. Let such a vector
u
1
be expressed in terms of its components we get
10.9 Eigen Value Problem
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CHAPTER 10
0
12
11
2
2
2
1
2
2
2
1
1
2
1
u
u
k
m
k
k
k
k
m
And for
2
the vector u
2
satisfies
0
2
2
2
u
K
M
Similarly, the two equations and two unknowns are found from the simple relation we have derived
from the assumed solution such as
0
22
21
2
2
2
2
2
2
2
1
1
2
2
u
u
k
m
k
k
k
k
m
These expressions can be solved for the vectors u
1
and u
2
where we can see after substitution
of the assumed values the above equations become:
10.9 Eigen Value Problem
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CHAPTER 10
0
5
.
4
5
.
4
5
.
4
5
.
4
12
11
u
u
0
5
.
5
.
4
5
.
4
5
.
40
22
21
u
u
1
5
.
4
5
.
4
12
11
12
11
u
u
u
u
1
11
u
1
12
u
The solution for the first mode shape vector u1 is as follows:
1
1
1
u
We find the second in a similar fashion to be
9
1
5
.
4
5
.
40
22
21
22
21
u
u
u
u
9
1
21
u
1
22
u
Hence,
and
9
/
1
1
2
u
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CHAPTER 10
Example 10.4
Calculate the displacements of the two masses
–
two springs system in the previous problem
for the given initial conditions :
.
0
0
.
1
20
.
10
.
20
10
x
x
and
x
mm
x
Solution:
The solution of the free vibration problem at hand can be expressed in a matrix form as follows:
)
sin(
)
sin(
2
2
2
1
1
1
22
12
21
11
2
1
t
A
t
A
u
u
u
u
t
x
t
x
)
2
sin(
)
3
sin(
)
2
sin(
9
1
)
3
sin(
2
2
1
1
2
2
1
1
t
A
t
A
t
A
t
A
At t=0 this yields
2
2
1
1
2
2
1
1
sin
sin
sin
9
1
sin
0
1
A
A
A
A
10.9 Eigen Value Problem
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CHAPTER 10
2
2
1
1
sin
9
1
sin
1
A
A
2
2
1
1
sin
sin
0
A
A
Differentiating the above equations and evaluating them at t=0 yields
2
2
1
1
2
2
1
1
2
.
1
.
cos
2
cos
3
cos
9
2
cos
3
0
0
0
0
A
A
A
A
x
x
2
2
1
1
cos
9
2
cos
3
0
A
A
2
2
1
1
cos
2
cos
3
0
A
A
Adding the above two equations yields
0
cos
9
20
2
so that
2
2
the velocity equation reduces to
0
cos
3
1
1
A
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CHAPTER 10
Substituting
2
1
in the above two equation gives
2
1
9
1
1
A
A
And
2
1
0
A
A
which has a solution of
A
1
=

9/10 mm, A
2
= 9/10 mm.
Thus
mm
t
t
t
t
t
x
mm
t
t
t
t
t
x
)
2
cos
3
(cos
9
.
0
)
2
2
sin(
10
9
)
2
3
sin(
10
9
)
(
2
cos
2
.
3
cos
9
.
0
)
2
2
sin(
5
1
)
2
3
sin(
10
9
)
(
2
1
Above is a classic example of a two degrees

of

freedom vibration problem where students can
evaluate the modes shapes and frequencies and deduce the transient solution resulting from the
solution of the free vibration problem. We should also note how the mode shapes and frequencies
contribute to the dynamic solution of lumped mass systems where the magnitude of vibration is
dependent both on the initial conditions as well as the mode shapes or eigen vectors.
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CHAPTER 10
10.10 Modal Analysis
Computing the eigen values and eigen vectors of a mechanical system can become strenuous and
cumbersome if the number of degrees of freedom increases. When a system possesses a large number
of degrees of freedom the determinant of a matrix which results into the characteristic of the equations
can lead to polynomial to the power 2*N. Keep in mind that the roots of such a polynomial need to be
found and the solution of N equations and N unknowns for each eigen value needs to be resolved.
What follows is a modal analysis approach in extracting the eigen values and eigen vectors once the
generalized mass and stiffness matrix are found.
Let the general equation of a free vibration system be given by
0
Kx
x
M
..
And let the initial conditions imposed on the system be such that
0
0
0
0
.
.
x
x
x
x
The steps involved in calculating the eigen values and eigen vectors require that we
make the mass matrix an identity matrix. This is simply obtained by pre multiplying
the equations of motion above by the inverse matrix of M.
(10.66)
10.10 Modal Analysis
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CHAPTER 10
Let us assume that our mass matrix is an identity matrix then it is easy to calculate as a first step
1) Calculate the M

1/2
.
Then we proceed by computing the normalized stiffness matrix following
2
1
2
1
KM
M
K
3) Calculate the symmetric eigen value problem for
~
K
2) Compute the
mass normalized stiffness matrix
to get
2
i
and
v
i
.
4) Normalize
v
i
and form the matrix of eigen vectors
2
1
v
v
P
5) Calculate the matrix of mode shapes
P
M
S
2
1
and
2
1
1
M
P
S
T
6) Calculate the modal initial conditions:
0
1
.
0
1
0
,
0
.
x
r
x
r
S
S
7) Substitute the components of , into equations of solution of modal
equations to get the solutions in the modal coordinates
0
r
)
0
(
.
r
8) Obtain the final solution by
where S is the transformation from
the modal coordinate to the generalized coordinates of the system.
t
S
t
r
x
10.10 Modal Analysis
Principles of Computer

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–
ISBN 0

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064631

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Author: Prof. Farid. Amirouche, University of Illinois

Chicago
CHAPTER 10
Example 10.5
Calculate the solution of the n

degree of freedom system shown for n = 3 by modal analysis. Use the
values m
1
=m
2
=m
3
=1 kg and k
1
=k
2
=k
3
=1 N/m, and the initial condition x
1
(0)=1 m with all other initial
displacements and velocities zero.
Solution
The mass and stiffness matrices for n=3 for the values given become
0
..
kx
x
M
Where
n
m
m
m
m
M
0
0
0
0
0
0
0
0
0
0
0
0
3
2
1
n
n
n
k
k
k
k
k
k
k
k
k
k
k
k
k
k
1
1
4
3
3
3
3
2
2
2
2
1
0
0
0
0
0
0
0
0
10.10 Modal Analysis
Principles of Computer

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ISBN 0

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064631

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Author: Prof. Farid. Amirouche, University of Illinois

Chicago
CHAPTER 10
m
1
m
2
m
3
m
4
m
n
x
1
x
2
x
3
x
4
x
n
Figure 10.9 A Mass

spring system of N Degrees of Freedom
.
10.10 Modal Analysis
Principles of Computer

Aided Design and Manufacturing Second Edition 2004
–
ISBN 0

13

064631

8
Author: Prof. Farid. Amirouche, University of Illinois

Chicago
CHAPTER 10
I
M
1
1
0
1
2
1
0
1
2
K
Following the steps outlined in the modal analysis procedure we get
I
M
2
1
2.
1
1
0
1
2
1
0
1
2
1
1
0
1
2
1
0
1
2
2
1
2
1
I
I
KM
M
K
3.
1
0
2
1
det
0
1
0
1
1
det
1
1
1
1
2
det
2
1
1
0
1
2
1
0
1
2
det
)
det(
k
I
0
1
6
5
1
2
2
1
0
1
1
1
1
2
2
2
3
2
1.
10.10 Modal Analysis
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–
ISBN 0

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064631

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Author: Prof. Farid. Amirouche, University of Illinois

Chicago
CHAPTER 10
The roots of this cubic equation are
2470
.
3
5550
.
1
1981
.
0
3
2
1
Thus the system’s natural frequencies are
8019
.
1
247
.
1
445
.
0
3
2
1
To calculate the first eigenvector, substitute
1
=0.1981 into
0
1
v
I
K
and solve for the vector v
1
=
T
v
v
v
3
2
1
. This yields
0
0
0
1981
.
0
1
1
0
1
1981
.
0
2
1
0
1
1981
.
0
2
31
21
11
v
v
v
Multiplying out this last expression yields three equations, only two of which are independent:
0
)
8019
.
1
(
21
11
v
v
0
8019
.
1
31
21
11
v
v
v
0
)
8019
.
0
(
31
21
v
v
10.10 Modal Analysis
Principles of Computer

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–
ISBN 0

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064631

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Author: Prof. Farid. Amirouche, University of Illinois

Chicago
CHAPTER 10
Solving the first and third equations yields
31
21
31
11
8019
.
0
4450
.
0
v
v
and
v
v
The second equation is dependent and does not yield any new information.
Substituting these values into the vector v
1
yields
1
8019
.
0
445
.
0
31
1
v
v
4. Normalizing the vector yields
1
]
1
)
8019
.
0
(
)
4450
.
0
[(
2
2
2
2
31
1
1
v
v
v
T
Solving for v
31
and substituting back into the expression for v
1
yields the normalized
version of the eigenvector v
1
as
7370
.
0
5910
.
0
3280
.
0
1
v
10.10 Modal Analysis
Principles of Computer

Aided Design and Manufacturing Second Edition 2004
–
ISBN 0

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Author: Prof. Farid. Amirouche, University of Illinois

Chicago
CHAPTER 10
Similarly v
2
and v
3
can be calculated and normalized to be
3280
.
0
7370
.
0
5910
.
0
5910
.
0
3280
.
0
7370
.
0
3
2
v
v
The matrix P is then given by
3280
.
0
5910
.
0
7370
.
0
7370
.
0
3280
.
0
5910
.
0
10
.
0
7370
.
0
3280
.
0
P
5. The matrix
IP
P
M
S
2
1
2
1
or
1640
.
0
2955
.
0
3685
.
0
3685
.
0
1640
.
0
2955
.
0
2955
.
0
3685
.
0
1640
.
0
S
And
6560
.
0
4740
.
1
1820
.
1
1820
.
1
6560
.
0
4740
.
1
4740
.
1
1820
.
1
6560
.
0
2
2
1
1
I
P
M
P
S
T
T
10.10 Modal Analysis
Principles of Computer

Aided Design and Manufacturing Second Edition 2004
–
ISBN 0

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064631

8
Author: Prof. Farid. Amirouche, University of Illinois

Chicago
CHAPTER 10
6. The initial conditions in modal coordinates becomes
0
0
)
0
(
1
0
.
1
.
S
x
S
r
And
1820
.
1
4740
.
1
6560
.
0
0
0
1
6560
.
0
4740
.
1
1820
.
1
1820
.
1
6560
.
0
4740
.
1
4740
.
1
1820
.
1
6560
.
0
)
0
(
0
1
x
S
r
7. The modal solutions are each of the form given by
)
4450
.
0
cos(
6560
.
0
2
4450
.
0
sin
6560
.
0
1
t
t
t
r
)
2470
.
1
cos(
4740
.
1
2
247
.
1
sin
4740
.
1
2
t
t
t
r
t
t
t
r
8019
.
1
cos
1820
.
1
2
8019
.
1
sin
1820
.
1
3
10.10 Modal Analysis
Principles of Computer

Aided Design and Manufacturing Second Edition 2004
–
ISBN 0

13

064631

8
Author: Prof. Farid. Amirouche, University of Illinois

Chicago
CHAPTER 10
8.The solution in physical coordinates is next calculated from
)
8019
.
1
cos(
1820
.
1
)
2470
.
1
cos(
4740
.
1
)
4450
.
0
cos(
6560
.
0
1640
.
0
2955
.
0
3685
.
0
3685
.
0
1640
.
0
2955
.
0
2950
.
0
5
.
0
1640
.
0
t
t
t
)
8019
.
1
cos(
1935
.
0
)
2470
.
1
cos(
4355
.
0
)
4450
.
0
cos(
2417
.
0
)
8019
.
1
cos(
4355
.
0
)
2470
.
1
cos(
2417
.
0
)
4450
.
0
cos(
1938
.
0
)
8019
.
1
cos(
3492
.
0
)
2470
.
1
cos(
5443
.
0
)
4450
.
0
cos(
1075
.
0
3
2
1
t
t
t
t
t
t
t
t
t
t
x
t
x
t
x
This systematic approach is far more convenient than the procedure outlined in the previous
section.
One needs to keep in mind that there are a number of other techniques that can be used in
the solution of eigen value problems.
Students are urged to look up these methods in other textbooks in finite element analysis
and vibration.
x
=S
r
(t)=
10.10 Modal Analysis
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