Dynamics of Machines - Part II - IFS.pdf

DYNAMICS OF MACHINES 41614
PART I – INTRODUCTION TO THE BASIC TOOLS OF
MODELLING, SIMULATION, ANALYSIS & EXPERIMENTAL VALIDATION
(a) Amplitude [m/s 2 ]
(b) Amplitude [m/s 2 ]
x 10−4
3
2
1
0
−1
−2
Signal (a) in Time Domain − (b) in Frequency Domain
0 5 10 15
time [s]
20 25 30
x 10−5
2
1
0
0 5 10 15 20 25
frequency [Hz]
6
4
2
0
−2
−4
−6
1
x 10 −4
−4
1
x 10 −3
4
3
2
1
0
−1
−2
−3
0.5
x 10 −7
0.5
Angular Velocity: 0 rpm Mode: 2 Nat. Freq.: 14.7288 Hz
0
0
−0.5
−0.5
−1 0
−1 0
2
Angular Speed: 1 Mode: 1 Nat. Freq.: 15.6815 Hz
2
Kompendie 1034 – February 2004
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Dr.-Ing. Ilmar Ferreira Santos, Associate Professor
Department of Mechanical Engineering
Technical University of Denmark
Building 358, Room 159
2800 Lyngby
Denmark
Phone: +45 45 25 62 69
Fax: +45 45 88 14 51
E-Mail: ifs@mek.dtu.dk

Contents
1 Introduction to Dynamical Modelling of Machines and Structures and Experimental
Analysis of Mechanical Vibrations based on the Human Senses 3
1.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Description of the Test Facilities . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Data of the Mechanical System . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4 Calculating Equivalent Stiffness Coefficients – Beam Theory . . . . . . . . . . . 5
1.5 Calculating Stiffness Matrices – Beam Theory . . . . . . . . . . . . . . . . . . . 7
1.6 Mechanical Systems with 1 D.O.F. . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.6.1 Physical System and Mechanical Model . . . . . . . . . . . . . . . . . . . 9
1.6.2 Mathematical Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.6.3 Analytical and Numerical Solution of the Equation of Motion . . . . . . . 10
1.6.4 Analytical and Numerical Solution of Equation of Motion using Matlab . 15
1.6.5 Comparison between the Analytical and Numerical Solutions of Equation
of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
1.6.6 Homogeneous Solution or Free-Vibrations or Transient Response - Experimental
Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.6.7 Natural Frequency – ωn [rad/s] or fn [Hz] . . . . . . . . . . . . . . . . . . 19
1.6.8 Damping Factor ξ or Logarithmic Decrement β . . . . . . . . . . . . . . . 19
1.6.9 Forced Vibrations or Steady-State Response . . . . . . . . . . . . . . . . 24
1.6.10 Resonance and Phase . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
1.6.11 Superposition of Transient and Forced Vibrations in Time Domain (Simulation)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
1.6.12 Resonance – Experimental Analysis in Time Domain . . . . . . . . . . . . 28
1.7 Mechanical Systems with 2 D.O.F. . . . . . . . . . . . . . . . . . . . . . . . . . . 31
1.7.1 Physical System and Mechanical Model . . . . . . . . . . . . . . . . . . . 31
1.7.2 Mathematical Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
1.7.3 Analytical and Numerical Solution of System of Differential Linear Equations
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
1.7.4 Modal Analysis using Matlab eig-function [u, w] = eig(−B, A) . . . . . . . 37
1.7.5 Analytical and Numerical Solutions of Equation of Motion using Matlab . 40
1.7.6 Analytical and Numerical Results of the System of Equations of Motion . 41
1.7.7 Programming in Matlab – Frequency Response Analysis . . . . . . . . . . 42
1.7.8 Understanding Resonances and Mode Shapes using your Eyes and Fingers 43
1.7.9 Resonance – Experimental Analysis in Time Domain . . . . . . . . . . . . 48
1.8 Mechanical Systems with 3 D.O.F. . . . . . . . . . . . . . . . . . . . . . . . . . . 49
1.8.1 Physical System and Mechanical Model . . . . . . . . . . . . . . . . . . . 49
1.8.2 Mathematical Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
1.8.3 Programming in Matlab – Theoretical Parameter Studies and Experimental
Validation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
1.8.4 Theoretical Frequency Response Function . . . . . . . . . . . . . . . . . . 55
1.8.5 Experimental – Natural Frequencies . . . . . . . . . . . . . . . . . . . . . 56
1.8.6 Experimental – Resonances and Mode Shapes . . . . . . . . . . . . . . . . 56
1.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
1.10 Project 0 – Identification of Model Parameters (An Example) . . . . . . . . . . . 61
1.11 Project 1 – Modal Analysis & Validation of Models . . . . . . . . . . . . . . . . . 66
2

1 Introduction to Dynamical Modelling of Machines and Structures
and Experimental Analysis of Mechanical Vibrations
based on the Human Senses
1.1 Summary
The aim of this study is to show some theoretical and experimental examples to facilitate the
understanding of the physical meaning of the main topics and definitions used in relation to
vibrations in machines. Theoretical and experimental studies are led side-by-side, clarifying
the definitions of stiffness, flexibility, natural frequency, damping factor, logarithmic decrement,
resonance, phase, beating, unbalance, natural mode shape, modal node and so on. The experimental
investigations are always carried out in low frequency ranges, aiming at making easy
the visualization of the movements by the human eyes and the understanding of the mechanical
vibrations without necessarily having to use sensors and electronic devices.
1.2 Description of the Test Facilities
Figures 1 and 2 show the simple elements used during the experimental investigations: a flexible
beam (ruler), concentrated masses or magnets, a support, an accelerometer, a signal amplifier, a
shaker and a signal analyzer. The beam or ruler already has a scale, enabling it to easily achieve
the information about the position (length) where the magnets will be mounted. At each position
more than one magnet can be mounted, allowing changes in the values of the concentrated
masses. The masses or magnets can be easily moved and attached to different positions along
the ruler, aiming at investigating changes in the natural frequencies of the magnet-ruler system
(mass-spring system). The ruler is very flexible in one plane only due to the characteristic of
its cross-section (moment of inertia of area). The beam can easily be mounted with different
boundary conditions, as free-free, clamped-free, simply supported in both ends, etc. allowing an
investigation of the influence of the boundary conditions on its flexibility, and consequently on
the natural frequencies of the system.
Regarding rotating machines some analogies can be made between the mass-spring system presented
here and a centrifugal compressor, while comparing the ruler (or flexible beam) to the
flexible shaft, and the magnets (or concentrated masses) to the impellers or rigid discs. Changes
in the montage of shaft into the bearings (boundary conditions) or changes in the positioning
of the impellers along the shaft will lead to different critical speeds and mode shapes.
As mentioned above, the mass-spring system was designed to have a very flexible behavior with
very low natural frequencies. This allows the detection of natural frequencies, modes shapes
and resonance using the human eyes as sensors (or sighting senses). Moreover, it is also made
possible to excite the structure with human fingers, aiming at understanding the 90 degrees
phase between displacement and excitation (while operating around resonance conditions) by
means of tactile senses. Accelerometer, amplifier, shaker and signal analyzer are used aiming at
confirming what the human senses detect.
3

By understanding the topics related to mechanical vibration in low frequency ranges (high
flexibility and slow motions detectable by human senses), it gets easier to understand mechanical
vibration in high frequency ranges where one has faster motions with small amplitudes, just
detectable by sensors and electronic devices.
Figure 1: Experimental investigation of a mechanical continuous system with concentrated
masses modelled as equivalent spring-mass systems with 1, 2 and 3 degrees of freedom (D.O.F.).
Figure 2: Signal analyzer and shaker used for inducing and measuring mechanical vibrations
while analyzing the behavior of the spring-mass systems with 1 D.O.F., 2 D.O.F. and 3 D.O.F.
4

1.3 Data of the Mechanical System
ρ material density of the beam 7, 800 Kg/m 3
E elasticity modulus 2 × 10 11 N/m 2
L total length of the beam 0.600 m
b width of the beam 0.030 m
h thickness 0.0012 m
mi concentrated mass (i = 1, ...,6) 0.191 Kg
Table 1: Data of the mass-spring system ”A”.
ρ material density of the beam 7, 800 Kg/m 3
E elasticity modulus 2 × 10 11 N/m 2
L total length of the beam 0.300 m
b width of the beam 0.025 m
h thickness 0.0010 m
ρ material density (steel) 7, 800 Kg/m 3
mi concentrated mass (i = 1, ...,6) 0.191 Kg
Table 2: Data of the mass-spring system ”B”.
1.4 Calculating Equivalent Stiffness Coefficients – Beam Theory
(a) (b)
Figure 3: (a) Flexible beam – clamped-free boundary condition case with force applied to the end
L; (b) clamped-free boundary condition case with force applied to a general position L ∗ ;
By applying a vertical force F at the end of the beam as shown in figure 3(a) and using Beam
Theory, one can write:
EI d4y(x) = 0 (1)
dx4 Integrating in X once, one has:
EI d3 y(x)
dx 3 = F(x) = C1 (2)
5

Integrating twice in X, it gives:
EI d2 y(x)
dx 2 = M(x) = C1x + C2 (3)
Integrating again in X, one gets the rotation angle of the beam:
EI dy(x)
dx
x
= EIΘ(x) = C1
2
2 + C2x + C3
And finally, integrating the last time in X, one achieves the equation responsible for describing
the deflexion of the beam:
x
EI y(x) = C1
3
6
x
+ C2
2
2 + C3x + C4
The boundary conditions for the clamped-free beam are:
•y(x = 0) = 0
•Θ(x = 0) = 0
•M(x = L) = 0
•F(x = L) = −F (reaction)
⎫
⎪⎬
⎪⎭
After calculating the constants Ci using the boundary condition for the clamped-free beam case,
one gets
y(x) = − F
E I
dy(x)
dx
x 3
6
= Θ(x) = − F
E I
L x2
−
2
x 2
2
− L x
Using the relationship between applied force F and the induced deflexion at a given point along
the beam length, x = L for instance, one gets the equivalent stiffness as:
K = F
y(L)
= 3 EI
L 3
Suggestion (I): Change the beam boundary conditions, for example bi-supported at both ends,
and calculate the equivalent stiffness in the new case.
Suggestion (II): Change the beam boundary conditions, for example clamped-clamped at both
ends, and calculate the equivalent stiffness in the new case.
6
(4)
(5)
(6)
(7)
(8)
(9)

1.5 Calculating Stiffness Matrices – Beam Theory
Two Different Lengths for Applying Forces – To facilitate the understanding of steps
which will be presented, one can introduce the follow nomenclature (see figure 3(b)):
• L ∗ = L1 or L ∗ = L2 – length where the force F is applied.
• x = L1 or x = L2 – length where the displacement is measured.
Taking into account two different points for applying the forces and measuring the displacements
of the beam, one works with the following set of equations
x [0, L ∗ ]
and
x [L ∗ , L]
⎧
⎪⎨
⎪⎩
⎧
⎪⎨
⎪⎩
y(x) = − F
E I
dy(x)
dx
= − F
E I
x3 6 − L∗ x2
2
x2 2 − L∗
x
y(x) = y(L ∗ ) + dy(x)
dx
dy(x)
dx
= dy(x)
dx
x=L ∗
x=L∗ · (x − L∗ )
which are responsible for describing the deflection of the beam, considering the loading on
different coordinates.
Let us introduce an example of a system with two points of force application. Assuming in case
(I) the force F is applied to the first coordinate L ∗ = L1. One can measure and/or calculate
the beam deflection at the coordinates x = L1 and x = L2 through equations (10) and (11):
y1 = y(L1) = F · L3 1
3 · EI
and
y2 = y(L2) = F
6EI · (2L3 1 + 3L 2 1(L2 − L1)) (13)
Assuming in case (II) that the force F is applied to the second coordinate L ∗ = L2, one can
measure and/or calculate the follow beam displacements at the coordinates x = L1 and x = L2
through the equations (10) and (11):
y1 = y(L1) = F
6EI · (2L3 1 + 3L 2 1(L2 − L1)) (14)
and
y2 = y(L2) = F · L3 2
3 · EI
7
(10)
(11)
(12)
(15)

In order to calculate the stiffness matrix (k11, k12, k21, k22) of the 2 d.o.f. system,
Ky = f (16)
where
K =
k11 k12
k21 k22
y = { y1 y2 } T
f = { f1 f2 } T
one has to solve the following system of equations for case (I) and case (II):
k11 k12
k21 k22
y1
·
y2
=
Case (I): f = { F 0 } T
k11 k12
k21 k22
·
Case (II): f = { 0 F } T
k11 k12
k21 k22
f1
f2
F ·L 3 1
3·EI
F
6EI · (2L3 1 + 3L2 1 (L2 − L1))
F
6EI
·
· (2L31 + 3L21 (L2 − L1))
F ·L3 2
3·EI
=
=
F
0
0
F
It leads to a system of 4 equations, which can be written in a matrix form:
⎡
⎢
⎣
F ·L 3 1
3·EI
F
6EI · (2L31 + 3L 2 0
1(L2 − L1))
0
0
F ·L
0
3 1
3·EI
F
6EI · (2L31 + 3L 2 F
6EI
1(L2 − L1))
· (2L31 + 3L 2 1(L2 − L1))
F ·L 3 0
2
3·EI
0
0
F
6EI
0
· (2L31 + 3L 2 1(L2 − L1))
F ·L 3 ⎤
⎥
⎦
2
3·EI
·
⎧
⎪⎨
·
⎪⎩
⎫ ⎧
⎪⎬ ⎪⎨
F
0
=
⎪⎭ ⎪⎩
0
F
⎫
⎪⎬
⎪⎭
(21)
k11
k12
k21
k22
Solving this matrix system of order 4 by using the software MATHEMATICA, or by using
Cramer’s rule, one achieves the stiffness matrix:
EI
K =
(4L2 − L1)(L1 − L2) 2
⎡
12(L2/L1)
⎣
3 ⎤
6(L1 − 3L2)/L1
⎦ (22)
6(L1 − 3L2)/L1 12
Similar procedure can be made in order to get the stiffness matrix of the 3 d.o.f system. This is
the motivation of an exercise later on. The results (stiffness matrix with 9 stiffness coefficient)
will be presented in the section describing mechanical systems with 3 d.o.f.
8
(17)
(18)
(19)
(20)

1.6 Mechanical Systems with 1 D.O.F.
1.6.1 Physical System and Mechanical Model
(a)
(b)
(c)
Figure 4: (a) Real mechanical system composed of a turbine attached to an airplane flexible wing;
(b) Laboratory prototype built by a lumped mass attached to a flexible beam); (c) Equivalent
mechanical model with 1 D.O.F. for a lumped mass attached to a flexible beam.
1.6.2 Mathematical Model
It is important to point out, that the equations of motion in Dynamics of Machinery will frequently
have the form of second order differential equations: ¨y(t) = F(y(t), ˙y(t)). Such equations
can generally be linearized around an operational position of a physical system, leading to second
order linear differential equations. It means that the coefficients which are multiplying the
variables ¨y(t) , ˙y(t) , y(t) (co-ordinate chosen for describing the motion of the physical system)
do not depend on the variables themselves. In the mechanical model presented in figure 4 these
coefficients are constants: m1, d1 and k1. One of the aims of the course Dynamics of Machinery
is to help the students to properly find these coefficients so that the equations of motion can
really describe the movement of the physical system. The coefficients can be predicted using
theoretical or experimental approaches.
9

After having created the mechanical model for the physical system, the next step is to derive
the equation of motion based on the mechanical model. The mechanical model is composed of
a lumped mass m1 (assumption !!!), spring with equivalent stiffness coefficient k1 (calculated
using beam theory) and damper with equivalent viscous coefficient d (obtained experimentally).
While creating the mechanical model and assuming that the mass is a particle, the equation of
motion can be derived, for example, using Newton’s second law:
m1¨y1(t) + d1 ˙y1(t) + k1y1(t) = f1(t) (23)
¨y1(t) + d1
˙y1(t) +
m1
k1
y1(t) =
m1
f1
(t) (24)
m1
¨y1(t) + 2ξωn ˙y1(t) + ω 2 ny1(t) = f1
(t) (25)
m1
1.6.3 Analytical and Numerical Solution of the Equation of Motion
After having created the mechanical model (step 1) and derived the mathematical model (step
2) for this mechanical model, the next step is to solve the equation of motion (step 3), aiming
at analyzing (step 4) the dynamical behavior of the physical system. Here the analytical and
numerical solutions of linear differential equations are presented and compared. Later on you
can choose the most convenient way to solve the equations and analyze the dynamical behavior
of physical systems. It is important to mention that the numerical procedure is very simple,
an integrator of first-order. Other integrators can be used depending on the characteristics of
the mechanical models, for example, Runge-Kutta of higher order (third, fourth, etc.) among
others.
Homogeneous Solution or Transient Solution and Transient Analysis – The homogeneous
solution of a linear differential equation is also called transient solution. The homogeneous
differential equation is achieved when the right side of the equation is set zero (see eq.(26), or in
other words, when no force acts on the system. All analyzes and conclusions obtained from the
homogeneous solution are called transient analyzes, and provide information about the dynamical
behavior of the system while perturbations of displacement and velocities (in case of second
order differential equations of motion) are introduced into the system.
¨y(t) + 2ξωn ˙y(t) + ω 2 ny(t) = 0 (26)
yh(t) = Ce λt , (assumption) (27)
˙yh(t) = λCe λt
¨yh(t) = λ 2 Ce λt
The assumption (27) and its derivatives are introduced into the differential equation (26), leading
to
⇕
λ 2 Ce λt + 2ξωnλCe λt + ω 2 nCe λt = 0
(λ 2 + 2ξωnλ + ω 2 n)Ce λt = 0 (28)
10

Demanding (λ 2 + 2ξωnλ + ω 2 n) = 0, because Ce λt = 0 in eq.(28), one gets two values of λ, or
two roots for the equation (λ 2 + 2ξωnλ + ω 2 n) = 0:
λ1 = −ξωn − ωn 1 − ξ2 · i
λ2 = −ξωn + ωn 1 − ξ2 · i
It is important to highlight that all analyzes carried out here will be concentrated
in cases where ξ < 1 (sub-critically damped system). Cases where ξ = 1 or ξ > 1
are called critically or super-critically damped systems respectively. In these cases
no problem related to amplification of vibration amplitudes can be found while
crossing resonances. Using the roots λ1 and λ2 the homogenous solution is:
yh(t) = C1e λ1t + C2e λ2t
where C1 and C2 are defined as a function of the initial condition of the movement when t = 0:
• initial displacement y(0) = yini
• initial movement ˙y(0) = vini
It is important to point out that these constants have to be calculated by using the general
solution, which will be presented later.
Permanent Solution and Steady-State Analysis – The permanent solution of a linear
differential equation is also called steady-state solution. The complete differential equation is
achieved when the right side of the equation is completed with the excitation (see eq.(30)), or
in other words, when static or dynamic forces act on the system. All analyzes and conclusions
obtained from the permanent solution are called steady-state analyzes, and provide information
about the dynamical behavior of the system while excitation forces are introduced into the
system. Let us introduce an excitation force which value oscillates in time with frequency ω
[rad/s]:
¨y(t) + 2ξωn ˙y(t) + ω 2 ny(t) = f
m eiωt
(29)
(30)
yh(t) = Ae iωt , (assumption) (31)
˙yh(t) = jωAe iωt
¨yh(t) = (jω) 2 Ae iωt = −(ω) 2 Ae iωt
The assumption (31) and its derivatives are introduced into the differential equation (30), leading
to:
− (ω) 2 Ae iωt + 2ξωn(jωAe jωt ) + ω 2 nAe iωt = f
m eiωt
⇕
(−ω 2 + ω 2 n + i2ξωnω)A = f
m
11
(32)

The amplitude A eq.(33) and the particular solution yp(t) eq.(34) are derived by eq.(32) and
(31):
A =
f/m
−ω 2 + ω 2 n + i2ξωnω
yp(t) = A · e iωt
=
f/m
−ω 2 + ω 2 n + i2ξωnω
· e iωt
General Solution = Transient Solution + Steady-State Solution – The general solution
of a linear differential equation is achieved by adding the homogenous and the permanent
solutions, and sequentially by defining the initial conditions of the movement. This solution
will provide information about the transient and steady-state response of the mechanical model.
Considering that the order of the mechanical model is correct (in this case, one degree-of-freedom
system), the solution of the linear differential equation will be useful for predicting the dynamical
behavior of the physical system, if the coefficients of the differential equation (m, d and k,
or ωn and ξ) are properly chosen, using either the theoretical or experimental information or a
combination of both. The general solution of the differential equation of motion is given by:
y(t) = C1e λ1t + C2e λ2t + Ae iωt
where y(t) is the displacement of the mass-damping-spring system.
For achieving the velocity of the mass-damping-spring system, eq.(35) has to be differentiated
in time:
˙y(t) = λ1C1e λ1t + λ2C2e λ2t + iωAe iωt
Introducing the initial conditions of displacement and velocity into eq.(35) and (36), when t = 0,
one gets:
y(0) =C1e λ10 + C2e λ20 + Ae iω0 ⇒ yini = C1 + C2 + A (37)
˙y(0) =λ1C1e λ10 + λ2C2e λ20 + iωAe iω0 ⇒ vini = λ1C1 + λ2C2 + iωA (38)
Rewriting eq.(37) and eq.(38) in matrix form, one gets:
1 1
λ1 λ2
C1
C2
=
yini − A
vini − jωA
Solving the linear system eq.(39) using Cramer’s rule, the constants C1 and C2 are calculated:
C1 =
yini − A 1
det
vini − A λ2
=
1 1
det
λ2(yini − A) − (vini − iωA)
λ2 − λ1
λ1 λ2
12
(33)
(34)
(35)
(36)
(39)

1 yini − A
det
λ1 vini − A
C2 = =
1 1
det
−λ1(yini − A) + (vini − iωA)
λ2 − λ1
λ1 λ2
Summarizing, below is the analytical solution of second order differential equation, which is
responsible for describing the movements of the mass-damping-spring system in time domain,
as a function of the excitation force and initial condition of displacement and velocity:
y(t) = C1e λ1t + C2e λ2t + Ae iωt
where
λ1 = −ξωn − ωn 1 − ξ2 · i
λ2 = −ξωn + ωn 1 − ξ2 · i
A =
f/m
−ω 2 + ω 2 n + i2ξωnω
C1 = λ1(yini − A) − (vini − iωA)
λ2 − λ1
C2 = −λ1(yini − A) + (vini − iωA)
λ2 − λ1
Numerical Solution According to Taylor’s expansion, an equation can be approximated by:
f(t) ⋍ f(t0) + df
dt
t=t0
(t − t0) + d2 f
dt 2
t=t0
(t − t0)... + dn f
dt n
t=t0
(40)
(t − t0) (41)
Assuming a very small time step t −t0 ≪ 1, the higher order terms of eq.(42) can be neglected.
It turns:
f(t) ⋍ f(t0) + df
dt
t=t0
(t − t0) (42)
Knowing the initial conditions of the movement when t = 0,
y(0) = y0 = yini
˙y(0) = ˙y0 = vini
and the equation of motion, which has to be solved,
¨y(t) = − 2ξωn ˙y(t) − ω 2 ny(t) + f
m eiωt
13
(43)

one can get the initial acceleration, when t = 0, on the basis of initial conditions:
t0 = 0
˙y0
y0
⎫
⎬
⎭ ⇒ ¨y0 = −2ξωn ˙y0 − ω 2 ny0 + f
m eiωt0
The first predicted values of displacement, velocity and acceleration in time t1 = ∆t , using the
approximation given by eq.(42), are:
t1 = ∆t
˙y1 = ˙y0 + ¨y0∆t
y1 = y0 + ˙y1∆t
¨y1 = −2ξωn ˙y1 − ω 2 ny1 + f
m eiωt1
The second predicted values of displacement, velocity and acceleration in time t2 = t1 + ∆t ,
using the approximation given by eq.(42), are:
t2 = 2∆t
˙y2 = ˙y1 + ¨y1∆t
y2 = y1 + ˙y2∆t
¨y2 = −2ξωn ˙y2 − ω 2 ny2 + f
m eiωt2
The N-th predicted values of displacement, velocity and acceleration in time tN = tN−1 + ∆t ,
using the approximation given by eq.(42), are:
tN = N∆t
˙yN = ˙yN−1 + ¨yN−1∆t
yN = yN−1 + ˙yN∆t
¨yN = −2ξωn ˙yN − ω 2 nyN + f
m eiωtN (44)
Plotting the points [y1, y2, y3, ..., yN] versus [t1, t2, t3, ..., tN], one can observe the numerical
solution of the differential equation, which describes the displacement of the mass-dampingspring
system in time domain. Plotting the points [ ˙y1, ˙y2, ˙y3, ..., ˙yN] versus [t1, t2, t3, ..., tN] or
[¨y1, ¨y2, ¨y3, ..., ¨yN] versus [t1, t2, t3, ..., tN] one can also observe velocity and acceleration of the
mass-damping-spring system in time domain. The analytical and numerical solutions eq.(43) of
the second order differential equation are illustrated using a Matlab code.
14

1.6.4 Analytical and Numerical Solution of Equation of Motion using Matlab
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% DYNAMICS OF MACHINERY LECTURES (41035) %
% MEK - DEPARTMENT OF MECHANICAL ENGINEERING %
% DTU - TECHNICAL UNIVERSITY OF DENMARK %
% %
% Copenhagen, October 30th, 2003 %
% %
% IFS %
% %
% 1 D.O.F. SYSTEM - EXACT AND NUMERICAL SOLUTION %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
clear all;
close all;
%Concentred Masses
m1= 0.191; %[Kg]
m2= 0.191; %[Kg]
m3= 0.191; %[Kg]
m4= 0.191; %[Kg]
m5= 0.191; %[Kg]
m6= 0.191; %[Kg]
%Elastic Properties of the Beam of 600 [mm]
E = 2e11; %elasticity modulus [N/m^2]
b = 0.030 ; %width [m]
h = 0.0012 ; %thickness [m]
Iz= (b*h^3)/12; %area moment of inertia [m^4]
%Mass-Spring-Damping System Properties
L=0.610; %beam length
K= 3*E*Iz/L^3; %stiffness coefficient
M=m1+m2; %mass coefficient
xi=0.005; %damping factor [no-dimension]
D=2*xi*sqrt(K*M); %damping coefficient
wn=sqrt(K/M); %natural frequency [rad/s]
fn=wn/2/pi %natural frequency [Hz]
fnexp=0.875; %measured natural frequency [Hz]
dif=(fn-fnexp)/fnexp;%error between calculated
%and measured frequencies
%_____________________________________________________
%Initial Condition
y_ini= -0.000 % beam initial deflection [m]
v_ini= -0.000 % beam initial velocity [m/s]
freq_exc=0.95 % excitation frequency [Hz]
force=-0.100 % excitation force [N]
time_max=30.0; % integration time [s]
%_____________________________________________________
15
%_____________________________________________________
%EXACT SOLUTION % EQUATION (40)
n=600; % number of points for plotting
j=sqrt(-1); % complex number
w=2*pi*freq_exc; % excitation frequency [rad/s]
lambda1=-xi*wn+j*wn*sqrt(1-xi*xi);
lambda2=-xi*wn-j*wn*sqrt(1-xi*xi); AA=(force/M)/(wn*wn-w*w +
j*2*xi*wn*w); C1=(
lambda2*(y_ini-AA)-(v_ini-j*w*AA))/(lambda2-lambda1);
C2=(-lambda1*(y_ini-AA)+(v_ini-j*w*AA))/(lambda2-lambda1);
for i=1:n,
t(i)=(i-1)/n*time_max;
y_exact(i)=C1*exp(lambda1*t(i)) + ...
C2*exp(lambda2*t(i)) + ...
AA*exp(j*w*t(i));
end
%_____________________________________________________
%NUMERICAL SOLUTION % EQUATION (44)
% trying different time steps to observe convergence
% deltaT=0.3605; % time step [s]
% deltaT=0.3; % time step [s]
% deltaT=0.1; % time step [s]
% deltaT=0.05; % time step [s]
deltaT=0.01; % time step [s]
n_integ=time_max/deltaT; % number of points (integration)
% Initial Conditions
y_approx(1) = y_ini; % beam initial deflection [m]
yp_approx(1) = v_ini; % beam initial velocity [m/s]
for i=1:n_integ,
t_integ(i)=(i-1)*deltaT;
ypp_approx(i) =-(wn*wn)*y_approx(i) ...
-(2*xi*wn)*yp_approx(i) ...
+(force/M)*exp(j*w*t_integ(i));
yp_approx(i+1)= yp_approx(i) + ypp_approx(i)*deltaT;
y_approx(i+1) = y_approx(i) + yp_approx(i+1)*deltaT;
end
%_____________________________________________________
%Graphical Results
title(’Simulation of 1 D.O.F System in Time Domain’)
subplot(3,1,1), plot(t,real(y_exact),’b’)
title(’(a) Exact Solution - (b) Numerical Solution
(delta T = 0.01 s) - (c) Comparison’)
xlabel(’time [s]’)
ylabel(’(a) y(t) [m]’)
grid
subplot(3,1,2),
plot(t_integ(1:n_integ),real(y_approx(1:n_integ)),’r’)
xlabel(’time [s]’)
ylabel(’(b) y(t) [m]’)
grid
subplot(3,1,3), plot(t,real(y_exact),’b’,t_integ(1:n_integ),
real(y_approx(1:n_integ)),’r’)
xlabel(’time [s]’)
ylabel(’(c) y(t) [m]’)
grid

1.6.5 Comparison between the Analytical and Numerical Solutions of Equation of
Motion
(a) y(t) [m]
0.01
0.005
0
−0.005
(a) Exact Solution − (b) Numerical Solution (delta T = 0.3605 s) − (c) Comparison
−0.01
0 1 2 3 4 5
time [s]
6 7 8 9 10
0.4
(b) y(t) [m]
(c) y(t) [m]
0.2
0
−0.2
−0.4
0 1 2 3 4 5
time [s]
6 7 8 9 10
0.4
0.2
0
−0.2
−0.4
0 1 2 3 4 5
time [s]
6 7 8 9 10
Figure 5: Comparison between the analytical and numerical solutions when a time step of
0.3605 [s] is used – Divergence and numerical instability.
(a) y(t) [m]
(b) y(t) [m]
(c) y(t) [m]
0.01
0.005
0
−0.005
(a) Exact Solution − (b) Numerical Solution (delta T = 0.3 s) − (c) Comparison
−0.01
0 1 2 3 4 5
time [s]
6 7 8 9 10
0.02
0.01
0
−0.01
−0.02
0 1 2 3 4 5
time [s]
6 7 8 9 10
0.02
0.01
0
−0.01
−0.02
0 1 2 3 4 5
time [s]
6 7 8 9 10
Figure 6: Comparison between the analytical and numerical solutions when a time step of 0.3 [s]
is used – Divergence between the numerical and analytical solution due to a roof time step.
16

(a) y(t) [m]
(b) y(t) [m]
(c) y(t) [m]
0.01
0.005
0
−0.005
(a) Exact Solution − (b) Numerical Solution (delta T = 0.1 s) − (c) Comparison
−0.01
0 1 2 3 4 5
time [s]
6 7 8 9 10
0.01
0.005
0
−0.005
−0.01
0 1 2 3 4 5
time [s]
6 7 8 9 10
0.01
0.005
0
−0.005
−0.01
0 1 2 3 4 5
time [s]
6 7 8 9 10
Figure 7: Comparison between the analytical and numerical solutions when a time step of 0.1 [s]
is used – Accumulation of errors with the number of iterations and divergence between the numerical
and analytical solutions.
(a) y(t) [m]
(b) y(t) [m]
(c) y(t) [m]
0.01
0.005
0
−0.005
(a) Exact Solution − (b) Numerical Solution (delta T = 0.05 s) − (c) Comparison
−0.01
0 1 2 3 4 5
time [s]
6 7 8 9 10
0.01
0.005
0
−0.005
−0.01
0 1 2 3 4 5
time [s]
6 7 8 9 10
0.01
0.005
0
−0.005
−0.01
0 1 2 3 4 5
time [s]
6 7 8 9 10
Figure 8: Comparison between the analytical and numerical solutions when a time step of 0.05 [s]
is used – Accumulation of errors with the number of iterations and divergence between the numerical
and analytical solutions.
17

(a) y(t) [m]
(b) y(t) [m]
(c) y(t) [m]
0.01
0.005
0
−0.005
(a) Exact Solution − (b) Numerical Solution (delta T = 0.01 s) − (c) Comparison
−0.01
0 1 2 3 4 5
time [s]
6 7 8 9 10
0.01
0.005
0
−0.005
−0.01
0 1 2 3 4 5
time [s]
6 7 8 9 10
0.01
0.005
0
−0.005
−0.01
0 1 2 3 4 5
time [s]
6 7 8 9 10
Figure 9: Comparison between the analytical and numerical solutions when a time step of 0.01 [s]
is used – Good agreement between the numerical and analytical solutions in the total range of
time.
(a) y(t) [m]
(b) y(t) [m]
(c) y(t) [m]
1
0.5
0
−0.5
(a) Exact Solution − (b) Numerical Solution (delta T = 0.01 s) − (c) Comparison
−1
0 5 10 15
time [s]
20 25 30
1
0.5
0
−0.5
−1
0 5 10 15
time [s]
20 25 30
1
0.5
0
−0.5
−1
0 5 10 15
time [s]
20 25 30
Figure 10: Comparison between the analytical and numerical solutions when a time step of
0.01 [s] is used – Good agreement between the numerical and analytical solutions in the total
range of time. System behavior when excited by a harmonic force with a frequency near the
natural frequency.
18

1.6.6 Homogeneous Solution or Free-Vibrations or Transient Response - Experimental
Analysis
¨y1(t) + 2ξωn ˙y1(t) + ω 2 ny1(t) = 0 (45)
1.6.7 Natural Frequency – ωn [rad/s] or fn [Hz]
fn = 1
k1
2π m1
= 1
2π
3EI
L3 1
mi
[Hz]
number Length fn fexp fexp
of masses [m] [Hz] [Hz] [Hz]
(theor.) (*) (**)
1 0.610 1.23 12/10 = 1.2 1.250
2 0.610 0.87 8.5/10 = 0.85 0.875
3 0.610 0.71 7/10 = 0.7 0.705
4 0.610 0.61 6/10 = 0.6 0.625
Table 3: Measuring the natural frequency of the mass-spring system ”A” with 1. d.o.f, (*) using
the human eyes and a watch, and (**) using an accelerometer attached to the mass, and making
a comparison to the theoretical mathematical model.
number Length fexp
of masses [m] [Hz]
(*)
2 0.610 0.875
2 0.310 1.875
Table 4: Measuring the natural frequency of the mass-spring system ”A” with 1. d.o.f, (*) using
the human eyes and a watch, when the equivalent stiffness of the system is changed, by modifying
the position (length) where the lumped mass is attached to the beam.
1.6.8 Damping Factor ξ or Logarithmic Decrement β
• Experimental identification without using sensors (OBS: Experiment carried out using a
watch, light and shadow, a calculator, and the mass-spring system oscillating from an
initial condition of displacement yo = yini until half the initial amplitude yN = yini/2.)
1
2πN
ξ =
ln
yo
yN
1 1 + 2πN ln
yo
yN
2
or β = 2π ξ
19

Using the information of figure 11 (signal in time domain), where
yo = 3.0 · 10 −5 [m/s 2 ] (first peak of signal in time domain, figure 11)
yN = y6 = 2.0 · 10 −5 [m/s 2 ] (after 6 peaks)
N = 6,
one gets
• Damping Factor of the system ”A” (ξ)
ξ =
• Log Dec
1 +
1
2π·6 ln
1
2π·6 ln
3.0·10 −5
2.0·10 −5
3.0·10 −5
2.0·10 −5
β = 2π ξ = 2π 0.010 = 0.068
• Equivalent Viscous Damping (d)
=
2 0.010755
≈ 0.010
1.000058
d = 2 · ξ · ωn · m = 2 · 0.010 · (0.87 · 2 · π) · 0.382 ≈ 0.016 [N · s/m]
(a) Amplitude [m/s 2 ]
(b) Amplitude [m/s 2 ]
2
0
−2
x Signal 10−5
4
(a) in Time Domain − (b) in Frequency Domain
−4
0 5 10 15
time [s]
20 25 30
0.8
0.6
0.4
0.2
x 10−5
1
0
0 5 10 15 20 25
frequency [Hz]
Figure 11: Transient Vibration – Acceleration of the clamped-free flexible beam when two concentrated
masses m = m1 +m2 = 0.382 Kg are attached at its free end (L = 0.610 m) – Natural
frequency of the mass-spring system ”A”: 0.87 Hz.
20

(a) Amplitude [m/s 2 ]
x Signal 10−5
5
0
(a) in Time Domain − (b) in Frequency Domain
−5
0 5 10 15
time [s]
20 25 30
(b) Amplitude [m/s 2 ]
x 10−5
2
1
0
0 5 10 15 20 25
frequency [Hz]
Figure 12: Transient Vibration – Acceleration of the clamped-free flexible beam when two concentrated
masses m = m1 +m2 = 0.382 Kg are attached at its free end (L = 0.310 m) – Natural
frequency of the mass-spring system ”B”: 1.75 Hz.
• Damping Factor of the system ”B” – From fig.12, one gets: yo = 4.6 · 10 −5 [m/s 2 ], yN =
y54 = 1.0 · 10 −5 [m/s 2 ] and N = 54:
ξ =
1 +
1
2π·54 ln
1
2π·54 ln
4.6·10 −5
1.0·10 −5
4.6·10 −5
1.0·10 −5
• Equivalent Viscous Damping (d)
=
2 0.004498
≈ 0.005 (46)
1.000010
d = 2 · ξ · ωn · m = 2 · 0.005 · (1.75 · 2 · π) · 0.382 ≈ 0.04 [N · s/m]
21

(a) Amplitude [m/s 2 ]
(b) Amplitude [m/s 2 ]
0.5
0
−0.5
x Signal 10−4
1
(a) in Time Domain − (b) in Frequency Domain
−1
0 5 10 15
time [s]
20 25 30
x 10−5
2.5
2
1.5
1
0.5
0
0 5 10 15 20 25
frequency [Hz]
Figure 13: Free vibration – Spring-mass systems with 1 D.O.F. Two masses m = m1 + m2 =
0.382 Kg fixed at the middle of the beam L1 = 0.155 m, resulting in a system ”B” natural
frequency of 3.81 Hz
• Damping Factor of the system ”B” – From fig.13, one gets: yo = 0.95 · 10 −4 [m/s 2 ],
yN = y34 = 0.50 · 10 −4 [m/s 2 ] and N = 34:
ξ =
1 +
1
2π·34 ln
1
2π·34 ln
0.95·10 −4
0.50·10 −4
0.95·10 −4
0.50·10 −4
• Equivalent Viscous Damping (d)
=
2 0.003029
≈ 0.003
1.000005
d = 2 · ξ · ωn · m = 2 · 0.003 · (3.81 · 2 · π) · 0.382 ≈ 0.05 [N · s/m]
22

(a) Amplitude [m/s 2 ]
(b) Amplitude [m/s 2 ]
1
0.5
0
−0.5
−1
x 10 −4 Signal (a) in Time Domain − (b) in Frequency Domain
0 5 10 15
time [s]
20 25 30
0.8
0.6
0.4
0.2
x 10−5
1
0
0 5 10 15 20 25
frequency [Hz]
Figure 14: Free vibration – Spring-mass systems with 1 D.O.F. One mass m = m1 = 0.191 Kg
fixed at the middle of the beam L1 = 0.155 m, resulting in a system ”B” natural frequency of
4.94 Hz
• Damping Factor of the system ”B” – From fig.13, one gets: yo = 0.95 · 10 −4 [m/s 2 ],
yN = y14 = 0.50 · 10 −4 [m/s 2 ] and N = 14:
ξ =
1 +
1
2π·14 ln
1
2π·14 ln
0.95·10 −4
0.50·10 −4
0.95·10 −4
0.50·10 −4
• Equivalent Viscous Damping (d)
=
2 0.007296
≈ 0.007
1.000026
d = 2 · ξ · ωn · m = 2 · 0.007 · (4.94 · 2 · π) · 0.191 ≈ 0.08 [N · s/m]
IMPORTANT CONCLUSION: THE DAMPING FACTOR IS A CHARACTER-
ISTIC OF THE GLOBAL MECHANICAL SYSTEM AND SIMULTANEOUSLY DE-
PENDS ON MASS m, STIFFNESS k AND DAMPING d COEFFICIENTS, NOT
ONLY ON THE DAMPING COEFFICIENT, AS YOU CAN SEE IN THE DEFINI-
TION:
• ξ = d
2·m·ωn
= d1
2· √ m·k
IT IS POSSIBLE TO INCREASE THE DAMPING FACTOR OF A MECHANICAL
SYSTEM EITHER BY DECREASING THE MASS m, OR BY DECREASING THE
STIFFNESS k OR BY INCREASING THE DAMPING COEFFICIENT d. THE
DAMPING FACTOR IS A VERY USEFUL PARAMETER FOR DEFINING THE
RESERVE OF STABILITY IN MACHINERY DYNAMICS.
23

1.6.9 Forced Vibrations or Steady-State Response
The two most frequent ways of representing the frequency response function of mechanical
systems are presented in figure 15: (a) and (b) real and imaginary parts as a function of the
excitation frequency; (c) and (d) magnitude and phase as a function of the excitation frequency.
Other alternative ways are presented in figures 16 and 17.
(a) Frequency Response Function (Real Part)
5
ξ=0.005
ξ=0.05
Real(A(ω)) [m/N]
0
−5
0 0.5 1 1.5 2
Frequency [Hz]
(b) Frequency Response Function (Imaginary Part)
0
ξ=0.005
−2
ξ=0.05
Imag(A(ω)) [m/N]
−4
−6
−8
−10
0 0.5 1 1.5 2
Frequency [Hz]
Phase(A(ω)) [ o]
(c) Frequency Response Function (Amplitude)
10
ξ=0.005
8
ξ=0.05
||A(ω)|| [m/N]
6
4
2
−100
−150
0
0 0.5 1 1.5 2
Frequency [Hz]
(d) Frequency Response Function (Phase)
0
−50
ξ=0.005
ξ=0.05
−200
0 0.5 1 1.5 2
Frequency [Hz]
Figure 15: Steady state response in the frequency domain or Frequency Response Function
f/m
(FRF): (a) and (b) illustrate the real and imaginary part of A = −ω2 +ω2 n +i2ξωnω; (c) and (d)
illustrate the magnitude and phase of the complex function A =
1.6.10 Resonance and Phase
f/m
−ω 2 +ω 2 n+i2ξωnω .
• In order to understand the ”90 degree phase while in resonance” use the tactile senses –
Remember the experiments in the classroom using the mass-beam system and forces applied
by your finger, and outside building 404, using a car and a tree and the forces applied by
your hands (synchronization).
• Complex Vector Diagram of Resonance and Phase
24

Imag(A(ω)) [m/N]
0
−1
−2
−3
−4
−5
−6
−7
−8
Frequency Response Function (Real Part)
ξ=0.005
ξ=0.05
−9
−5 −4 −3 −2 −1 0 1 2 3 4 5
Real(A(ω)) [m/N]
Figure 16: Steady state response in the frequency domain or Frequency Response Function (FRF)
illustrated as the real versus the imaginary part of A =
Imag(A(ω)) [m/N]
0
−0.5
−1
−1.5
−2
1
0.5
Real(A(ω)) [m/N]
0
Frequency Response Function
−0.5
0
0.5
f/m
−ω 2 +ω 2 n +i2ξωnω.
1
1.5
Frequency [Hz]
2
ξ=0.05
Figure 17: Steady state response in the frequency domain or Frequency Response Function (FRF)
f/m
illustrated in a 3D-plot: the real and imaginary parts of A = −ω2 +ω2 as a function of the
n+i2ξωnω
frequency.
25

(a)
(b)
(c)
Figure 18: Complex vector diagram using an excitation force of constant magnitude: (a) the
frequency of the excitation force is lower than the natural frequency; (b) the frequency of the
excitation force is coincident with the natural frequency, characterizing a resonance case where
the phase between the force and the displacement is 90 degrees, i.e. the phase between the force
and the velocity is 0 degrees, meaning a synchronization between force and velocity; (c) the
frequency of the excitation force is bigger than the natural frequency.
26

1.6.11 Superposition of Transient and Forced Vibrations in Time Domain (Simulation)
(a)
(c)
(e)
y(t) [m]
y(t) [m]
2.5
2
1.5
1
0.5
0
−0.5
−1
−1.5
−2
(a) Excitation Frequency : 0.1 Hz
−2.5
0 5 10 15
time [s]
20 25 30
−2.5
0 5 10 15
time [s]
20 25 30
y(t) [m]
2.5
2
1.5
1
0.5
0
−0.5
−1
−1.5
−2
2.5
2
1.5
1
0.5
0
−0.5
−1
−1.5
−2
(a) Excitation Frequency : 0.8 Hz
(a) Excitation Frequency : 0.9 Hz
−2.5
0 5 10 15
time [s]
20 25 30
(b)
(d)
(f)
y(t) [m]
y(t) [m]
y(t) [m]
2.5
2
1.5
1
0.5
0
−0.5
−1
−1.5
−2
(a) Excitation Frequency : 0.7 Hz
−2.5
0 5 10 15
time [s]
20 25 30
2.5
2
1.5
1
0.5
0
−0.5
−1
−1.5
−2
(a) Excitation Frequency : 0.8702 Hz
−2.5
0 5 10 15
time [s]
20 25 30
2.5
2
1.5
1
0.5
0
−0.5
−1
−1.5
−2
(a) Excitation Frequency : 1.0 Hz
−2.5
0 5 10 15
time [s]
20 25 30
Figure 19: Time response of the system with 1 D.O.F. excited by forces with different frequencies
(a) ω = 0.1 Hz (before resonance); (b) ω = 0.7 Hz; (before resonance); (c) ω = 0.8 Hz (before
resonance but close – beating); (d) ω = 0.8702 Hz (at resonance); (e) ω = 0.9 Hz (after
resonance but close – beating); (f) ω = 1.0 Hz (after resonance).
27

1.6.12 Resonance – Experimental Analysis in Time Domain
(a) Amplitude [m/s 2 ]
(b) Amplitude [m/s 2 ]
3
2
1
0
−1
−2
−3
x 10 −5 Signal (a) in Time Domain − (b) in Frequency Domain
0 5 10 15
time [s]
20 25 30
x 10−5
1.5
1
0.5
0
0 5 10 15 20 25
frequency [Hz]
(a) Amplitude [m/s 2 ]
(b) Amplitude [m/s 2 ]
(a) Amplitude [m/s 2 ]
(b) Amplitude [m/s 2 ]
3
2
1
0
−1
−2
−3
0.5
x 10 −5 Signal (a) in Time Domain − (b) in Frequency Domain
0 5 10 15
time [s]
20 25 30
x 10−5
1.5
1
0
0 5 10 15 20 25
frequency [Hz]
3
2
1
0
−1
−2
−3
0.5
x 10 −5 Signal (a) in Time Domain − (b) in Frequency Domain
0 5 10 15
time [s]
20 25 30
x 10−5
1.5
1
0
0 5 10 15 20 25
frequency [Hz]
Figure 20: Resonance phenomena due to force excitations with frequency around the natural
frequency of the mass-spring system: 1 D.O.F. system with natural frequency of 0.87 Hz, excited
by the shaker with frequencies of 0.80 Hz, 0.87 Hz and 0.90 Hz – Spring-mass system (A) with
two masses m = m1 + m2 = 0.382 Kg fixed at the beam (A) length L1 = 0.600 m, resulting in a
natural frequency of 0.87 Hz.
28

(a) Amplitude [m/s 2 ]
(b) Amplitude [m/s 2 ]
x Signal 10−5
5
0
(a) in Time Domain − (b) in Frequency Domain
−5
0 5 10 15
time [s]
20 25 30
x 10−5
2.5
2
1.5
1
0.5
0
0 5 10 15 20 25
frequency [Hz]
Figure 21: Beating phenomena with two transient responses – Two spring-mass systems with 1
D.O.F. each, vibrating with very similar natural frequencies (transient responses), resulting in
beating phenomena: Spring-mass system ”A” with three masses m = m1 +m2 +m3 = 0.576 Kg
fixed at the beam length L1 = 0.285 m, resulting in a natural frequency near 1.75 Hz; Springmass
system ”B” with masses m = m4 + m5 = 0.382 Kg fixed at the end of the beam L1 =
0.310 m, resulting in a natural frequency of 1.75 Hz
29

(a) Amplitude [m/s 2 ]
(b) Amplitude [m/s 2 ]
6
4
2
0
−2
−4
−6
x 10 −6 Signal (a) in Time Domain − (b) in Frequency Domain
−8
0 5 10 15
time [s]
20 25 30
x 10−6
2.5
2
1.5
1
0.5
0
0 5 10 15 20 25
frequency [Hz]
(a) Amplitude [m/s 2 ]
x Signal 10−5
1.5
1
0.5
0
−0.5
−1
(a) in Time Domain − (b) in Frequency Domain
−1.5
0 5 10 15
time [s]
20 25 30
(b) Amplitude [m/s 2 ]
x 10−6
8
6
4
2
0
0 5 10 15 20 25
frequency [Hz]
Figure 22: Beating phenomena due to transient (low damping factor) and forced vibrations
with similar frequencies: 1 D.O.F. system ”A” with natural frequency of 0.87 Hz, excited by
a shaker with frequencies of 0.80 Hz and 0.90 Hz - Spring-mass system ”A” with two masses
m = m1+m2 = 0.382 Kg fixed at the beam length L1 = 0.600 m, resulting in a natural frequency
of 0.87 Hz.
30

1.7 Mechanical Systems with 2 D.O.F.
1.7.1 Physical System and Mechanical Model
(a)
(b)
(c)
Figure 23: (a) Real mechanical system built by two turbines attached to an airplane flexible wing;
(b) Laboratory prototype built by two lumped masses attached to a flexible beam); (c) Equivalent
mechanical model with 2 D.O.F. for the two lumped masses attached to the flexible beam.
1.7.2 Mathematical Model
M¨y(t) + D˙y(t) + Ky(t) = f(t) (47)
m11 m12
m21 m22
¨y1
¨y2
d11 d12
+
d21 d22
˙y1
˙y2
k11 k12
+
k21 k22
31
y1
y2
=
f1
f2
(48)

where
m11 = m1 + m2
m12 = 0
m21 = 0
m22 = m3 + m4
d11 = 2ξ k11m11
d12 = 0
d21 = 0
d22 = 2ξ k22m22
(49)
⎫
⎪⎬
(Approximation of damping coefficients using the exp. damping factor!)
⎪⎭
12EI
3
k11 =
(4L2 − L1)(L1 − L2) 2(L2/L1)
6EI
k12 =
(4L2 − L1)(L1 − L2) 2(L1 − 3L2)/L1
6EI
k21 =
(4L2 − L1)(L1 − L2) 2(L1 − 3L2)/L1
12EI
k22 =
(4L2 − L1)(L1 − L2) 2
The coefficients of the mass matrix can easily be achieved. Each of the single masses has
m1 = m2 = m3 = m4 = m5 = m6 = 0.191 Kg. The stiffness coefficients kij were calculated
in the section 1.5, using the geometry of the beam (I, L1 , L2) and its material properties
(E). The damping matrix can be approximated by using proportional damping, for example,
D = αM + βK. The coefficients α and β can be chosen, so that the damping factor ξ of the
first resonance is of the same order as the one in the previous section. Please, note that this
is just an approximation which be verified using Modal Analysis Techniques. Another way of
approximating the damping coefficients is given by eq.(50).
1.7.3 Analytical and Numerical Solution of System of Differential Linear Equations
System of Equation of motion
m11
m21
m12 ¨y1 d11
+
m22 ¨y2 d21
d12 ˙y1
d22 ˙y2
M¨y + D ˙y + Ky = ¯fe jωt
+
k11 k12 y1
k21 k22
Differential Equations 2nd order → 1st order
M D ¨y
0 M ˙y
+
0 K ˙y
−M 0 y
=
¯f
0
A˙z(t) + Bz(t) = fe jωt
32
y2
f1
=
f2
e jωt
e jωt
(50)
(51)
(52)
(53)

⎧ ⎫
⎪⎨
˙y1(t)
⎪⎬
˙y(t) ˙y2(t)
z(t) = =
y(t) ⎪⎩
y1(t) ⎪⎭
y2(t)
→ velocity
→ velocity
→ displacement
→ displacement
The analytical solution can be divided into three steps: (I) homogeneous solution (transient analysis);
(II) permanent solution (steady-state analysis) and (III) general solution (homogeneous
+ permanent).
Homogeneous Solution and Transient Analysis – The homogeneous differential equation
is achieved when the right side of the equation is set zero (see eq.(55)), or in other words, when
no force is acting on the system.
A˙zh(t) + Bzh(t) = 0 (55)
(54)
zh(t) = ue λt , (assumption) (56)
˙zh(t) = λue λt
The assumption (56) and its derivative are introduced into the differential equation (55), leading
to an eigenvalue-eigenvector problem:
[λA + B]ue λt = 0 ⇒ [λA + B]u = 0 (57)
• Eigenvalues λi can be calculated by using eq.(58):
determinant(λA + B) = 0 ⇒ λ1 , λ2 , λ3 , λ4 (58)
• Eigenvectors ui can be calculated by using eq.(59):
λ1Au = −Bu ⇒ u1
λ2Au = −Bu ⇒ u2
λ3Au = −Bu ⇒ u3
λ4Au = −Bu ⇒ u4
• The homogeneous solution can be written as:
zh(t) = C1u1e λ1t + C2u2e λ2t + C3u3e λ3t + C4u4e λ4t
where C1, C2, C3 and C4 are constants depending on the initial displacement and velocities of
the coordinates y1 and y2, when t = 0.
33
(59)

Permanent Solution and Steady-State Analysis – The permanent solution takes into
account the right side of the differential equation (see eq.(60)), or in other words, the effect of
the force on the system. In case of harmonic excitation, one can write:
A˙zp(t) + Bzp(t) = fe jωt
(60)
zp(t) = Ae jωt , (assumption!) (61)
˙zp(t) = jωAe jωt
The assumption adopted in eq.(61) and its derivative are introduced into the differential equation
(60), leading to:
[jωA + B]Ae jωt = fe jωt ⇒ A = [jωA + B] −1 f (62)
The permanent solution of the equation of motion is given by:
zp(t) = Ae jωt ⇒ zp(t) = [jωA + B] −1 fe jωt
General Solution – The general solution of a linear differential equation is achieved by adding
the homogenous and the permanent solutions, and by sequentially defining the initial conditions
of the movement. This solution will provide information about the transient and steady-state
response of the mechanical model. Considering that the order of the mechanical model is correct
(in this case, the two degree-of-freedom system), the solution of the linear differential equation
will be useful for predicting the dynamical behavior of the physical system, if the coefficients
of the differential equation (M, D and K, or A and B) are properly chosen, either by using
theoretical or experimental information or both. The general solution of the differential equation
of motion is given by:
z(t) = C1u1e λ1t + C2u2e λ2t + C3u3e λ3t + C4u4e λ4t + Ae iωt
where z(t) gives information about the displacement and velocity of the coordinates y1 and y2.
Introducing the initial conditions of displacement and velocity
zini = { v1ini
v2ini y1ini y2ini }T
into eq.(64), when t = 0, one obtains
z(0) = zini = C1u1e λ10 + C2u2e λ20 + C3u3e λ30 + C4u4e λ40 + Ae iω0
or
⎧
⎪⎨
zini = C1u1 + C2u2 + C3u3 + C4u4 + A = [ u1 u2 u3 u4 ]
⎪⎩
C1
C2
C3
C4
(63)
(64)
(65)
(66)
⎫
⎪⎬
+ A (67)
⎪⎭
Solving the linear system by inverting the modal matrix U = [ u1 u2 u3 u4 ] one achieves the
vector c = { C1 C2 C3 C4 } T :
34

zini = U c + A ⇒ c = U −1 {(zini − A)} (68)
Summarizing, below is the analytical solution of a second order differential equation, which is
responsible for describing the displacements and velocities of the coordinates y1(t) and y2(t) in
time domain, as a function of the excitation force and the initial condition of displacement and
velocity:
z(t) = C1u1e λ1t + C2u2e λ2t + C3u3e λ3t + C4u4e λ4t + Ae iωt
where
λ1 = −ξ1ωn1 − ωn1 1 − ξ2 1 · i
and u1
λ2 = −ξ1ωn1 + ωn1 1 − ξ2 1 · i
and u2
λ3 = −ξ2ωn2 − ωn2 1 − ξ2 2 · i
and u3
λ4 = −ξ2ωn2 + ωn2 1 − ξ2 2 · i and u4
⎧
⎪⎨
⎪⎩
C1
C2
C3
C4
A = [jωA + B] −1 f
⎫
⎪⎬
= [ u1 u2 u3 u4 ]
⎪⎭
−1 { zini − A}
Numerical Solution – The numerical solution of the system of differential equations can be
found by using the approximation according to Taylor’s expansion. Thus, one equation can be
approximated by:
f(t) ⋍ f(t0) + df
dt
t=t0
(t − t0) + d2 f
dt 2
t=t0
(t − t0)... + dn f
dt n
t=t0
(69)
(t − t0) (70)
Assuming a very small time step t −t0 ≪ 1, the higher order terms of eq.(71) can be neglected.
It turns:
f(t) ⋍ f(t0) + df
dt
t=t0
(t − t0) (71)
Knowing the initial conditions of the movement, when t = t0 = 0,
y(0) = y0
y1(0)
y2(0)
=
y1ini
y1ini
35
(72)

and
˙y(0) = ˙y0
˙y1(0)
˙y2(0)
=
v1ini
v2ini
and the equation of motion eq.(52), which has to be solved, one can calculate the acceleration,
when t = t0 = 0:
¨y0 = −M −1 D ˙y0 + Ky0 − ¯ jωt0 fe
The first predicted values of displacement, velocity and acceleration in time t1 = ∆t , using the
approximation given by eq.(71), are:
t1 = ∆t
˙y1 = ˙y0 + ¨y0∆t
y1 = y0 + ˙y1∆t
¨y1 = −M −1 D ˙y1 + Ky1 − ¯ jωt1 fe
The second predicted values of displacement, velocity and acceleration in time t2 = t1 + ∆t ,
using the approximation given by eq.(71), are:
t2 = 2∆t
˙y2 = ˙y1 + ¨y1∆t
y2 = y1 + ˙y2∆t
¨y2 = −M −1 D ˙y2 + Ky2 − ¯ jωt2 fe
The N-th predicted values of displacement, velocity and acceleration in time tN = tN−1 + ∆t ,
using the approximation given by eq.(74), are:
tN = N∆t
˙yN = ˙yN−1 + ¨yN−1∆t
yN = yN−1 + ˙yN∆t
¨yN = −M −1 D ˙yN + KyN − ¯ jωtN fe
Plotting the points [y1,y2,y3, ...,yN] versus [t1, t2, t3, ..., tN], one can observe the numerical
solution of the differential equation, which describes the displacements of the mass-dampingspring
system in time domain. Plotting the points [ ˙y1, ˙y2, ˙y3, ..., ˙yN] versus [t1, t2, t3, ..., tN] or
[¨y1, ¨y2, ¨y3, ..., ¨yN] versus [t1, t2, t3, ..., tN] one can also observe the velocity and acceleration of
the mass-damping-spring system in time domain. The analytical and numerical solutions of the
second order differential equation, eq.(52), are illustrated using a Matlab code.
36
(74)
(73)

1.7.4 Modal Analysis using Matlab eig-function [u, w] = eig(−B, A)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% MACHINERY DYNAMICS LECTURES (41614) %
% MEK - DEPARTMENT OF MECHANICAL ENGINEERING %
% DTU - TECHNICAL UNIVERSITY OF DENMARK %
% %
% Copenhagen, February 11th, 2000 %
% IFS %
% %
% MODAL ANALYSIS %
% %
% 2 D.O.F. SYSTEMS - MODAL ANALYSIS - 3 EXPERIMENTAL CASES %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
clear all; close all;
%Concentred Masses
m1= 0.191; %[Kg]
m2= 0.191; %[Kg]
m3= 0.191; %[Kg]
m4= 0.191; %[Kg]
m5= 0.191; %[Kg]
m6= 0.191; %[Kg]
%Elastic Properties of the Beam of 600 mm
E= 2e11; %elasticity modulus [N/m^2]
b= 0.030 ; %width [m]
h= 0.0012 ; %thickness [m]
I= (b*h^3)/12; %area moment of inertia [m^4]
% (1.CASE) Data for the mass-spring system
%__________________________________________________
M1=m1; %concentrated mass [Kg] |
M2=m2; %concentrated mass [Kg] |
L1= 0.310; %length for positioning M1 [m] |
L2= 0.610; %length for positioning M2 [m] |
%__________________________________________________|
% Coefficients of the Stiffness Matrix
LL=(L1-4*L2)*(L1-L2)^2;
K11= -12*(E*I/LL)*L2^3/L1^3; %equivalent Stiffness [N/m]
K12= -6*(E*I/LL)*(L1-3*L2)/L1; %equivalent Stiffness [N/m]
K21= -6*(E*I/LL)*(L1-3*L2)/L1; %equivalent Stiffness [N/m]
K22= -12*(E*I/LL); %equivalent Stiffness [N/m]
%Mass Matrix
M= [M1 0; 0 M2];
%Stiffness Matrix
K= [K11 K12; K21 K22];
%Damping Matrix
D= [0 0; 0 0];
%State Matrices
A= [ M D ;
zeros(size(M)) M ] ;
B= [ zeros(size(M)) K ;
-M zeros(size(M))];
%Dynamical Properties of the Mass-Spring System
[u,w]=eig(-B,A); % Eigenvectors u
% Eigenvalues w [rad/s]
w
u
pause;
w_vector=diag(w);
[natfreq,indice]=sort(w_vector); % Sorting the natural Frequencies
% Natural Frequencies (Hz)
w1=abs(imag(w_vector(indice(1))))/2/pi % First natural frequency
w2=abs(imag(w_vector(indice(3))))/2/pi % Second natural frequency
mass_position(1) = 0; mass_position(2) = L1; mass_position(3) =
L2;
% first mode shape
uu1(1)=0; uu1(2)=u(1,indice(1)); uu1(3)=u(2,indice(1)); figure(1)
plot(uu1,mass_position,’r*-.’,-uu1,mass_position,’r*-.’,’LineWidth’,1.5)
grid f1=num2str(w1); set(gca,’FontSize’,12,’FontAngle’,’oblique’);
title([’First Mode Shape - Freq.: ’,f1,’ Hz’])
37
% second mode shape
uu2(1)=0; uu2(2)=u(1,indice(3)); uu2(3)=u(2,indice(3)); figure(2)
plot(uu2,mass_position,’r*-.’,-uu2,mass_position,’r*-.’,’LineWidth’,1.5)
grid f2=num2str(w2); set(gca,’FontSize’,12,’FontAngle’,’oblique’);
title([’Second Mode Shape - Freq.: ’,f2,’ Hz’]) pause;
% (2.CASE) Increasing the Mass Values
%__________________________________________________
M1=m1+m2; %concentrated mass [Kg] |
M2=m3+m4; %concentrated mass [Kg] |
L1= 0.310; %length for positioning M1 [m] |
L2= 0.610; %length for positioning M2 [m] |
%__________________________________________________|
% Coefficients of the Stiffness Matrix
LL=(L1-4*L2)*(L1-L2)^2;
K11= -12*(E*I/LL)*L2^3/L1^3; %equivalent Stiffness [N/m]
K12= -6*(E*I/LL)*(L1-3*L2)/L1; %equivalent Stiffness [N/m]
K21= -6*(E*I/LL)*(L1-3*L2)/L1; %equivalent Stiffness [N/m]
K22= -12*(E*I/LL); %equivalent Stiffness [N/m]
%Mass Matrix
M= [M1 0; 0 M2];
%Stiffness Matrix
K= [K11 K12; K21 K22];
%Damping Matrix
D= [0 0; 0 0];
%State Matrices
A= [ M D ;
zeros(size(M)) M ] ;
B= [ zeros(size(M)) K ;
-M zeros(size(M))];
%Dynamical Properties of the Mass-Spring System
[u,w]=eig(-B,A); % Eigenvectors u
% Eigenvalues w [rad/s]
w
u
pause;
w_vector=diag(w);
[natfreq,indice]=sort(w_vector); % Sorting the natural Frequencies
% Natural Frequencies (Hz)
w1=abs(imag(w_vector(indice(1))))/2/pi % First natural frequency
w2=abs(imag(w_vector(indice(3))))/2/pi % Second natural frequency
mass_position(1) = 0; mass_position(2) = L1; mass_position(3) =
L2;
% first mode shape
uu1(1)=0; uu1(2)=u(1,indice(1)); uu1(3)=u(2,indice(1)); figure(3)
plot(uu1,mass_position,’b*-.’,-uu1,mass_position,’b*-.’,’LineWidth’,1.5)
grid f1=num2str(w1); set(gca,’FontSize’,12,’FontAngle’,’oblique’);
title([’First Mode Shape - Freq.: ’,f1,’ Hz’])
% second mode shape
uu2(1)=0; uu2(2)=u(1,indice(3)); uu2(3)=u(2,indice(3)); figure(4)
plot(uu2,mass_position,’b*-.’,-uu2,mass_position,’b*-.’,’LineWidth’,1.5)
grid f2=num2str(w2); set(gca,’FontSize’,12,’FontAngle’,’oblique’);
title([’Second Mode Shape - Freq.: ’,f2,’ Hz’]) pause;
% (3.CASE) Increasing the Mass Values
%__________________________________________________
M1=m1+m2+m3; %concentrated mass [Kg] |
M2=m4+m5+m6; %concentrated mass [Kg] |
L1= 0.310; %length for positioning M1 [m] |
L2= 0.610; %length for positioning M2 [m] |
%__________________________________________________|
...

(1.CASE)
w =
%Dynamical Properties of the Mass-Spring System
[u,w]=eig(-B,A); % Eigenvectors u
% Eigenvalues w [rad/s]
0 +48.8527i 0 0 0
0 0 -48.8527i 0 0
0 0 0 + 7.3517i 0
0 0 0 0 - 7.3517i
u =
1.0000 1.0000 0.3300 0.3300
-0.3300 -0.3300 1.0000 1.0000
0 - 0.0355i 0 + 0.0355i 0 - 0.0778i 0 + 0.0778i
0 + 0.0117i 0 - 0.0117i 0 - 0.2356i 0 + 0.2356i
(2.CASE)
w =
0 +34.5441i 0 0 0
0 0 -34.5441i 0 0
0 0 0 + 5.1985i 0
0 0 0 0 - 5.1985i
u =
1.0000 1.0000 -0.3300 -0.3300
-0.3300 -0.3300 -1.0000 -1.0000
0 - 0.0289i 0 + 0.0289i 0 + 0.0635i 0 - 0.0635i
0 + 0.0096i 0 - 0.0096i 0 + 0.1924i 0 - 0.1924i
(3.CASE)
u =
1.0000 1.0000 -0.3300 -0.3300
-0.3300 -0.3300 -1.0000 -1.0000
0 - 0.0205i 0 + 0.0205i 0 + 0.0449i 0 - 0.0449i
0 + 0.0068i 0 - 0.0068i 0 + 0.1360i 0 - 0.1360i
w =
0 +28.2051i 0 0 0
0 0 -28.2051i 0 0
0 0 0 + 4.2445i 0
0 0 0 0 - 4.2445i
38

0.7
0.6
0.5
0.4
0.3
0.2
0.1
First Mode Shape − Freq.: 1.1701 Hz
0
(a) −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
(b)
(c)
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
0.7
0.6
0.5
0.4
0.3
0.2
0.1
First Mode Shape − Freq.: 0.82736 Hz
First Mode Shape − Freq.: 0.67554 Hz
0
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Second Mode Shape − Freq.: 7.7751 Hz
0
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Second Mode Shape − Freq.: 5.4979 Hz
0
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Second Mode Shape − Freq.: 4.489 Hz
0
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
Figure 24: First and second mode shapes of the mechanical system modelled with 2 D.O.F. (a)
(1.CASE) – one mass attached to each of the two coordinates; (b) (2.CASE) – two masses
attached to each one of the two coordinates; (c) (3.CASE) – three masses attached to each one
of the two coordinates.
39

1.7.5 Analytical and Numerical Solutions of Equation of Motion using Matlab
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% DYNAMICS OF MACHINERY LECTURES (72213) %
% MEK - DEPARTMENT OF MECHANICAL ENGINEERING %
% DTU - TECHNICAL UNIVERSITY OF DENMARK %
% %
% Copenhagen, February 11th, 2000 %
% %
% IFS %
% %
% 2 D.O.F - EXACT AND NUMERICAL SOLUTION %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
clear all;
close all;
%Concentred Masses
m1= 0.191; %[Kg]
m2= 0.191; %[Kg]
m3= 0.191; %[Kg]
m4= 0.191; %[Kg]
m5= 0.191; %[Kg]
m6= 0.191; %[Kg]
%Elastic Properties of the Beam of 600 [mm]
E = 2e11; %elasticity modulus [N/m^2]
b = 0.030 ; %width [m]
h = 0.0012 ; %thickness [m]
Iz= (b*h^3)/12; %area moment of inertia [m^4]
% (1.CASE) Data for the mass-spring system
%__________________________________________________
M1=m1+m2; %concentrated mass [Kg] |
M2=m3+m4; %concentrated mass [Kg] |
L1= 0.310; %length for positioning M1 [m] |
L2= 0.610; %length for positioning M2 [m] |
%__________________________________________________|
% Coefficients of the Stiffness Matrix
LL=(L1-4*L2)*(L1-L2)^2;
K11= -12*(E*Iz/LL)*L2^3/L1^3; % Stiffness coeff.[N/m]
K12= -6*(E*Iz/LL)*(L1-3*L2)/L1; % Stiffness coeff.[N/m]
K21= -6*(E*Iz/LL)*(L1-3*L2)/L1; % Stiffness coeff.[N/m]
K22= -12*(E*Iz/LL); % Stiffness coeff.[N/m]
% Coefficients of the Damping Matrix
% (damping factor xi=0.005)
D11= 2*0.005*sqrt(M1/K11); % Damping coeff.[Ns/m]
D12= 0; % Damping coeff.[Ns/m]
D21= 0; % Damping coeff.[Ns/m]
D22= 2*0.005*sqrt(M2/K22); % Damping coeff.[Ns/m]
%Mass Matrix
M= [M1 0; 0 M2];
%Damping Matrix
D=[D11 D12; D21 D22];
%Stiffness Matrix
K= [K11 K12; K21 K22];
%State Matrices A & B % EQUATION (52)
A= [ M D ;
zeros(size(M)) M ] ;
B= [ zeros(size(M)) K ;
-M zeros(size(M))];
%Dynamical Properties of the Mass-Spring System
[u,w]=eig(-B,A); %eigenvectors u
%eigenvalues w
w1=abs(imag(w(3,3)))/2/pi; %first natural freq.[Hz]
w2=abs(imag(w(1,1)))/2/pi; %second natural freq.[Hz]
wexp1=0.78125; %measured natural freq.[Hz]
wexp2=5.563; %measured natural freq.[Hz]
dif1=(w1-wexp1)/wexp1; %error calculated and measured freq.
dif2=(w2-wexp2)/wexp2; %error calculated and measured freq.
%_____________________________________________________
%Initial Condition
y1_ini = -0.000 % beam initial deflection [m]
y2_ini = -0.000 % beam initial deflection [m]
v1_ini = -0.001 % beam initial velocity [m/s]
v2_ini = -0.000 % beam initial velocity [m/s]
freq_exc = 0.000 % excitation frequency [Hz]
force1 = -0.000 % excitation force 1 [N]
force2 = -0.000 % excitation force 2 [N]
time_max = 30.0; % integration time [s]
%_____________________________________________________
40
%_____________________________________________________
%EXACT SOLUTION % EQUATION (68)
n=2000; % number of points for plotting
j=sqrt(-1); % complex number
w_exc=2*pi*freq_exc; % excitation frequency [rad/s]
z_ini = [v1_ini v2_ini y1_ini y2_ini]’;
force_exc = [force1 force2 0 0 ]’;
vec_aux = z_ini - inv((j*w_exc*A + B))*force_exc;
lambda1=w(1,1);
lambda2=w(2,2);
lambda3=w(3,3);
lambda4=w(4,4);
u1=u(1:4,1);
u2=u(1:4,2);
u3=u(1:4,3);
u4=u(1:4,4);
C=inv(u)*(vec_aux);
c1=C(1);
c2=C(2);
c3=C(3);
c4=C(4);
for i=1:n,
t(i)=(i-1)/n*time_max;
y_exact=c1*u1*exp(lambda1*t(i)) + ...
c2*u2*exp(lambda2*t(i)) + ...
c3*u3*exp(lambda3*t(i)) + ...
c4*u4*exp(lambda4*t(i)) + ...
inv((j*w_exc*A + B))*force_exc*exp(j*w_exc*t(i));
end
y1_exact(i) = y_exact(3);
y2_exact(i) = y_exact(4);
figure(1)
title(’Simulation of 2 D.O.F System in Time Domain’)
subplot(2,1,1), plot(t,real(y1_exact),’b’)
title(’Exact Solution ’)
xlabel(’time [s]’)
ylabel(’ y1_{exact}(t) [m]’)
grid
subplot(2,1,2), plot(t,real(y2_exact),’b’)
xlabel(’time [s]’)
ylabel(’y2_{exact}(t) [m]’)
grid
pause;
%_____________________________________________________
%NUMERICAL SOLUTION % EQUATION (73)
% deltaT=0.3605; % time step [s]
% deltaT=0.3; % time step [s]
% deltaT=0.1; % time step [s]
% deltaT=0.05; % time step [s]
% deltaT=0.01; % time step [s]
deltaT=0.005; % time step [s]
n_integ=time_max/deltaT; % number of points (integration)
% Initial Conditions
y1_approx(1) = y1_ini; % beam initial deflection [m]
y2_approx(1) = y2_ini; % beam initial deflection [m]
yp1_approx(1) = v1_ini; % beam initial velocity [m/s]
yp2_approx(1) = v2_ini; % beam initial velocity [m/s]
for i=1:n_integ,
t_integ(i)=(i-1)*deltaT;
ypp1_approx(i)=-1/M1*(K11*y1_approx(i)+K12*y2_approx(i) ...
+ D11*yp1_approx(i)+D12*yp2_approx(i) ...
-(force1)*exp(j*w_exc*t_integ(i)));
ypp2_approx(i)=-1/M2*(K21*y1_approx(i)+K22*y2_approx(i) ...
+D21*yp1_approx(i)+D22*yp2_approx(i) ...
-(force2)*exp(j*w_exc*t_integ(i)));
yp1_approx(i+1)=yp1_approx(i) + ypp1_approx(i)*deltaT;
yp2_approx(i+1)=yp2_approx(i) + ypp2_approx(i)*deltaT;
y1_approx(i+1)=y1_approx(i)+yp1_approx(i+1)*deltaT;
y2_approx(i+1)=y2_approx(i)+yp2_approx(i+1)*deltaT;
end
%_____________________________________________________

%_____________________________________________________
%Graphical Results
figure(2)
title(’Simulation of 2 D.O.F System in Time Domain’)
subplot(2,1,1), plot(t_integ(1:n_integ),
real(y1_approx(1:n_integ)),’r’)
title(’Numerical Solution (delta T = 0.005 s)’)
xlabel(’time [s]’)
ylabel(’y1_{approx}(t) [m]’)
grid
subplot(2,1,2), plot(t_integ(1:n_integ),
real(y2_approx(1:n_integ)),’r’)
xlabel(’time [s]’)
ylabel(’y2_{approx}(t) [m]’)
grid
%_____________________________________________________
%Graphical Results (Comparison Exact vs. Numerical)
figure(3)
subplot(2,1,1), plot(t,real(y1_exact),’b’,
t_integ(1:n_integ),real(y1_approx(1:n_integ)),’r’)
title(’Simulation of 2 D.O.F System in Time Domain -
Exact Solution vs. Numerical Solution
(delta T = 0.005 s)’)
xlabel(’time [s]’)
ylabel(’y1_{approx}(t) [m]’)
grid
subplot(2,1,2), plot(t,real(y2_exact),’b’,
t_integ(1:n_integ),real(y2_approx(1:n_integ)),’r’)
xlabel(’time [s]’)
ylabel(’y2_{approx}(t) [m]’)
grid
1.7.6 Analytical and Numerical Results of the System of Equations of Motion
y1 exact (t) [m]
y2 exact (t) [m]
y1 approx (t) [m]
y2 approx (t) [m]
2
0
−2
−4
x Exact 10−5
4
Solution
−6
0 5 10 15
time [s]
20 25 30
4
2
0
−2
−4
−6
x 10−5
6
−8
0 5 10 15
time [s]
20 25 30
x Numerical 10−5
5
0
Solution (delta T = 0.005 s)
−5
0 5 10 15
time [s]
20 25 30
4
2
0
−2
−4
−6
x 10−5
6
−8
0 5 10 15
time [s]
20 25 30
Figure 25: Analytical and Numerical Solutions – (a) Analytical solution with initial velocity
condition at ˙y1(0) = 1 mm/s, ˙y2(0) = 0 mm/s, y1(0) = 0 mm and y2(0) = 0 mm; (b) Numerical
solution (time step of 0.005 [s]) with the same initial conditions – Transient Analysis.
41

1.7.7 Programming in Matlab – Frequency Response Analysis
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% MACHINERY DYNAMICS LECTURES (72213) %
% IKS - DEPARTMENT OF CONTROL ENGINEERING DESIGN %
% DTU - TECHNICAL UNIVERSITY OF DENMARK %
% %
% Copenhagen, February 11th, 2000 %
% IFS %
% %
% 2 D.O.F. SYSTEMS - FRF (FREQUENCY RESPONSE FUNCTION) %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Concentred Masses
m1= 0.191; %[Kg]
m2= 0.191; %[Kg]
m3= 0.191; %[Kg]
m4= 0.191; %[Kg]
m5= 0.191; %[Kg]
m6= 0.191; %[Kg]
%Elastic Properties of the Beam of 600 [mm]
E= 2e11; %elasticity modulus [N/m^2]
b= 0.030 ; %width [m]
h= 0.0012 ; %thickness [m]
I= (b*h^3)/12; %area moment of inertia [m^4]
% (1.CASE) Data for the mass-spring system
%__________________________________________________
M1=m1+m2+m5; %concentrated mass [Kg] |
M2=m3+m4+m6; %concentrated mass [Kg] |
L1= 0.310; %length for positioning M1 [m] |
L2= 0.610; %length for positioning M2 [m] |
%__________________________________________________|
% Coefficients of the Stiffness Matrix
LL=(L1-4*L2)*(L1-L2)^2;
K11= -12*(E*I/LL)*L2^3/L1^3; %equivalent Stiffness [N/m]
K12= -6*(E*I/LL)*(L1-3*L2)/L1; %equivalent Stiffness [N/m]
K21= -6*(E*I/LL)*(L1-3*L2)/L1; %equivalent Stiffness [N/m]
K22= -12*(E*I/LL); %equivalent Stiffness [N/m]
% Coefficients of the Damping Matrix
D11= 2*0.01*(2*pi*5.0)*M1; %equivalent Damping [N/m]
D12= 0; %equivalent Damping [N/m]
D21= 0; %equivalent Damping [N/m]
D22= 2*0.01*(2*pi*1.0)*M2; %equivalent Damping [N/m]
%Mass Matrix
M= [M1 0; 0 M2];
%Damping Matrix
D=[D11 D12; D21 D22];
%Stiffness Matrix
K= [K11 K12; K21 K22];
%State Matrices
A= [ M D ;
zeros(size(M)) M ] ;
B= [ zeros(size(M)) K ;
-M zeros(size(M))];
%Dynamical Properties of the Mass-Spring System
[u,w]=eig(-B,A); %natural frequency [rad/s]
%Dynamical Properties of the Mass-Spring System
w=sort(diag(abs(w)))/2/pi %natural frequency [rad/s]
w1=w(1) %first natural frequency [Hz]
w2=w(3) %second natural frequency [Hz]
wexp1=0.625 %measured natural frequency [Hz]
wexp2=4.405 %measured natural frequency [Hz]
dif1=(w1-wexp1)/wexp1 %error between calculated and measured freq.
dif2=(w2-wexp2)/wexp2 %error between calculated and measured freq.
42
% FRF -- FREQUENCY RESPONSE FUNCTION
N = 800 ; % number of points for plotting (see HP-Analyzer)
N_factor = 100 ;
% Given Excitation Function acting on the the point 1
fo1 = 1 ; % [N] force amplitude acting on the point 1 of the beam
fo2 = 0 ; % [N] force amplitude acting on the point 2 of the beam
for i=1:N;
F1(i)= fo1;
F2(i)= fo2;
end;
% Calculation of the displacement, velocity and acceleration responses
for i=1:N;
w(i) = 2*pi*i/N_factor;
j = sqrt(-1);
AA = [(-M*w(i)*w(i)+K)+D*w(i)*j]; % Dynamical Stiffness Matrix
x = inv(AA)*[F1(i) F2(i)]’; % Displacement (complex)
x11(i) = x(1); % rail vibration displacement
x21(i) = x(2); % sleeves displacement
end;
% Given Excitation Function acting on the the point 2
fo1 = 0 ; % [N] force amplitude acting on the point 1 of the beam
fo2 = 1 ; % [N] force amplitude acting on the point 2 of the beam
for i=1:N;
F1(i)= fo1;
F2(i)= fo2;
end;
% Calculation of the displacement, velocity and acceleration responses
for i=1:N;
w(i) = 2*pi*i/N_factor;
j = sqrt(-1);
AA = [(-M*w(i)*w(i)+K)+D*w(i)*j]; % Dynamical Stiffness Matrix
x = inv(AA)*[F1(i) F2(i)]’; % Displacement (complex)
x12(i) = x(1); % rail vibration displacement
x22(i) = x(2); % sleeves displacement
end;
% Plotting the results
figure(1)
subplot(2,2,1),plot(w/2/pi,abs(x11))
title(’Excitation on Point 1 and Response of Point 1’)
xlabel(’Frequency [Hz]’)
ylabel(’(a) y11 [m/N]’)
grid on
subplot(2,2,2),plot(w/2/pi,abs(x12))
title(’Excitation on Point 2and Response of Point 1’)
xlabel(’Frequency [Hz]’)
ylabel(’(b) y12[m/N]’)
grid on
subplot(2,2,3),plot(w/2/pi,abs(x21))
title(’Excitation on Point 1 and Response of Point 2’)
xlabel(’Frequency [Hz]’)
ylabel(’(c) y21 [m/N]’)
grid on
subplot(2,2,4),plot(w/2/pi,abs(x22))
title(’Excitation on Point 2 and Response of Point 2’)
xlabel(’Frequency [Hz]’)
ylabel(’(d) y22 [m/N]’)
grid on

||y 1 (ω)|| [m/N]
Phase [ o]
Excitation on Point 1 and Response of Point 1
0.25
0.2
0.15
0.1
0.05
−100
−150
−200
−250
−300
0
0 2 4
Frequency [Hz]
6 8
0
−50
−350
0 2 4
Frequency [Hz]
6 8
||y 2 (ω)|| [m/N]
Phase [ o]
Excitation on Point 1 and Response of Point 2
0.8
0.6
0.4
0.2
−100
−150
−200
−250
−300
0
0 2 4
Frequency [Hz]
6 8
0
−50
−350
0 2 4
Frequency [Hz]
6 8
Figure 26: Forced Vibration – Theoretical Frequency Response Function (FRF) of the clampedfree
flexible beam when two concentrated masses m = m1 + m2 = 0.382 Kg are attached at its
free end (L = 0.610 m) and two additional masses m = m1 + m2 = 0.382 Kg are attached at its
middle (L = 0.310 m) – Natural frequencies of the 2. D.O.F. mass-spring system ”A”: 0.81 Hz
and 5.56 Hz.
1.7.8 Understanding Resonances and Mode Shapes using your Eyes and Fingers
• Understanding the ”90 Degree Phase between excitation force and displacement response
while in Resonance” using tactile senses. In other words, understanding ”Zero Degree Phase
between the excitation force and velocity response while in resonance” using tactile senses.
• Visualization of the participation of modes shapes in the transient response – Visualization
using your eyes! Transient motion of the physical system excited with different initial
conditions by using your fingers!
43

||y 1 (ω)|| [m/N]
Phase [ o]
Excitation on Point 2 and Response of Point 1
0.8
0.6
0.4
0.2
−100
−150
−200
−250
−300
0
0 2 4
Frequency [Hz]
6 8
0
−50
−350
0 2 4
Frequency [Hz]
6 8
||y 2 (ω)|| [m/N]
Phase [ o]
Excitation on Point 2 and Response of Point 2
2.5
2
1.5
1
0.5
−100
−150
−200
−250
−300
0
0 2 4
Frequency [Hz]
6 8
0
−50
−350
0 2 4
Frequency [Hz]
6 8
Figure 27: Forced Vibration – Theoretical Frequency Response Function (FRF) of the clampedfree
flexible beam when two concentrated masses m = m1 + m2 = 0.382 Kg are attached at its
free end (L = 0.610 m) and two additional masses m = m1 + m2 = 0.382 Kg are attached at its
middle (L = 0.310 m) – Natural frequencies of the 2. D.O.F. mass-spring system ”A”: 0.81 Hz
and 5.56 Hz.
44

||y i (ω)|| [m/N]
Phase [ o]
0.8
0.6
0.4
0.2
−100
−150
−200
−250
−300
Excitation on Point 1
0
0 2 4
Frequency [Hz]
6 8
0
−50
point 1
point 2
point 1
point 2
−350
0 2 4
Frequency [Hz]
6 8
||y i (ω)|| [m/N]
Phase [ o]
2.5
2
1.5
1
0.5
−100
−150
−200
−250
−300
Excitation on Point 2
0
0 2 4
Frequency [Hz]
6 8
0
−50
point 1
point 2
point 1
point 2
−350
0 2 4
Frequency [Hz]
6 8
Figure 28: Forced Vibration – Theoretical Frequency Response Function (FRF) of the clampedfree
flexible beam when two concentrated masses m = m1 + m2 = 0.382 Kg are attached at its
free end (L = 0.610 m) and two additional masses m = m1 + m2 = 0.382 Kg are attached at its
middle (L = 0.310 m) – Natural frequencies of the 2. D.O.F. mass-spring system ”A”: 0.81 Hz
and 5.56 Hz.
45

Imag(y i (ω)/f 1 (ω)) (i=1,2) [m/N]
0.1
0
−0.1
−0.2
−0.3
−0.4
−0.5
−0.6
FRF − Excitation on Point 1
point 1
point 2
−0.7
−0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4
Real(y (ω)/f (ω)) (i=1,2) [m/N]
i 1
Figure 29: Forced Vibration – Theoretical Frequency Response Function (FRF) of the clampedfree
flexible beam when two concentrated masses m = m1 + m2 = 0.382 Kg are attached at its
free end (L = 0.610 m) and two additional masses m = m1 + m2 = 0.382 Kg are attached at its
middle (L = 0.310 m) – Natural frequencies of the 2. D.O.F. mass-spring system ”A”: 0.81 Hz
and 5.56 Hz.
46

Imag(y i (ω)/f 1 (ω)) (i=1,2) [m/N]
0.5
0
−0.5
−1
−1.5
1
0.5
Real(y i (ω)/f 1 (ω)) (i=1,2) [m/N]
0
FRF − Excitation on Point 1
−0.5
0
2
4
6
Frequency [Hz]
8
point 1
point 2
Figure 30: Forced Vibration – Theoretical Frequency Response Function (FRF) of the clampedfree
flexible beam when two concentrated masses m = m1 + m2 = 0.382 Kg are attached at its
free end (L = 0.610 m) and two additional masses m = m1 + m2 = 0.382 Kg are attached at its
middle (L = 0.310 m) – Natural frequencies of the 2. D.O.F. mass-spring system ”A”: 0.81 Hz
and 5.56 Hz.
47

1.7.9 Resonance – Experimental Analysis in Time Domain
(a) Amplitude [m/s 2 ]
1
0
−1
x Signal 10−5
2
(a) in Time Domain − (b) in Frequency Domain
−2
0 5 10 15
time [s]
20 25 30
(b) Amplitude [m/s 2 ]
4
3
2
1
x 10 −6
0
0 5 10 15 20 25
frequency [Hz]
(a) Amplitude [m/s 2 ]
1
0
−1
x Signal 10−5
2
(a) in Time Domain − (b) in Frequency Domain
−2
0 5 10 15
time [s]
20 25 30
(b) Amplitude [m/s 2 ]
8
6
4
2
x 10 −6
0
0 5 10 15 20 25
frequency [Hz]
Figure 31: Resonance phenomena due to the excitation force with frequency around the natural
frequency of the mass-spring system: 2 D.O.F. system with the natural frequencies of 0.62 Hz
and 4.59, excited by the shaker – Spring-mass system ”A” with three masses m = m1+m2+m3 =
0.573 Kg fixed at the beam length L2 = 0.610 m and three additional masses m = m4+m5+m6 =
0.573 Kg fixed at the middle L1 = 0.310 m resulting in two natural frequencies of 0.62 Hz and
4.60 Hz.
48

1.8 Mechanical Systems with 3 D.O.F.
1.8.1 Physical System and Mechanical Model
(a)
(b)
(c)
Figure 32: (a) Real mechanical system built by three turbines attached to an airplane flexible
wing; (b) Laboratory prototype built by three lumped masses attached to a flexible beam); (c)
Equivalent mechanical model with 3 D.O.F. for the three lumped masses attached to a flexible
beam.
1.8.2 Mathematical Model
It is important to point out again, that the equations of motion in Dynamics of Machinery will
frequently have the form of a second order differential equation: ¨y(t) = F(y(t), ˙y(t)). Such
equations can generally be linearized around an operational position of the physical system,
leading to second order linear differential equations. It means that the coefficients which are
multiplying the variables ¨y1(t) , ˙y1(t) , y1(t) , ¨y2(t) , ˙y2(t) , y2(t) , ¨y3(t) , ˙y3(t) , y3(t) (coordinates
chosen to describe the motion of the physical system) do not depend on the variables
themselves. In the case of the mechanical model presented in figure 32, these coefficients are
constants: m1, m2 and m3 are related to the masses; d11, d12, d13, d21, d22, d23, d31, d32 and
d33 are related to the equivalent viscous damping; and k11, k12, k13, k21, k22, k23, k31, k32 and
k33 are related to equivalent stiffness. One of the goals of the course (Dynamics of Machines) is
to present theoretical or experimental approaches to properly find these coefficients, so that the
49

equations of motion can really describe the movement of the physical system.
After creating the mechanical model for the physical system, the next step is to derive the
equation of motion based on the mechanical model. The mechanical model is built by lumped
masses m1, m2, m1 (assumption !!!), springs with equivalent stiffness coefficient (calculated
using Beam Theory) and dampers with equivalent viscous coefficient (obtained experimentally).
While creating the mechanical model and assuming that the mass is a particle, the equation of
motion can be derived using Newton’s or Lagrange axioms. For the 3 D.O.F system one can
write:
M¨y(t) + D˙y(t) + Ky(t) = f(t) (75)
or
⎡
⎣
m11 m12 m13
m21 m21 m23
m31 m32 m33
The mass coefficients
⎫
m11 = m1 + m2
m12 = 0
m13 = 0
m21 = 0
m22 = m3 + m4
m23 = 0
m31 = 0
m32 = 0
m33 = m5 + m6
⎤⎧
⎨ ¨y1
⎦ ¨y2
⎩
¨y3
⎪⎬
⎪⎭
⎫
⎬
⎭ +
⎡
⎣
d11 d12 d13
d21 d22 d23
d31 d32 d33
⎡
+ ⎣
⎤⎧
⎨
⎦
⎩
˙y1
˙y2
˙y3
k11 k12 k13
k21 k22 k23
k31 k32 k33
⎫
⎬
⎭ +
⎤⎧
⎨
⎦
⎩
y1
y2
y3
⎫
⎬
⎭ =
⎧
⎨
⎩
f1
f2
f3
⎫
⎬
⎭ ejωt
can easily be achieved either by measuring the masses or by having the material density and
mass dimensions.
The equivalent damping coefficients can be approximated by
d11 = 2ξ k11m11
d12 = 0
d13 = 0
d21 = 0
d22 = 2ξ k22m22
d23 = 0
d31 = 0
d32 = 0
d33 = 2ξ ⎫
⎪⎬
(Approximation!!!) (78)
⎪⎭
k33m33
or by assuming, for example, proportional damping D = αM + βK. The coefficients α and β
can be chosen, so that the damping factor ξ of the first resonance is of the same order as the
damping factor achieved in the previous section. Please, note that this is just an approximation
50
(76)
(77)

which could be verified using Modal Analysis Techniques. Another way of approximating the
damping coefficients is given by eq.(78), which should also be verified through experiments!
The stiffness coefficients
k11 =
3EIL 3 2 (L2 − 4L3)
L 3 1 (L1 − L2) 2 (2L1L2 + L 2 2 + L1L3 − 4L2L3)
k12 = −3EI(−3L2(L2 − 2L3)L3 + L1(L 2 2 − 2L2L3 − 2L 2 3 ))
L1(L1 − L2) 2 (L2 − L3)(2L1L2 + L 2 2 + L1L3 − 4L2L3)
k13 =
−9EIL 2 2
L1(L1 − L2)(L2 − L3)(2L1L2 + L 2 2 + L1L3 − 4L2L3)
k21 = −3EI(−3L2(L2 − 2L3)L3 + L1(L 2 2 − 2L2L3 − 2L 2 3 ))
L1(L1 − L2) 2 (L2 − L3)(2L1L2 + L 2 2 + L1L3 − 4L2L3)
k22 =
k23 =
k31 =
k32 =
k33 =
3EI(L1 − 4L3)(L1 − L3) 2
(L1 − L2) 2 (L2 − L3) 2 (2L1L2 + L 2 2 + L1L3 − 4L2L3)
−3EI(L 2 1 − 2L1L2 − 2L 2 2 − 3L1L3 + 6L2L3)
(L1 − L2)(L2 − L3) 2 (2L1L2 + L 2 2 + L1L3 − 4L2L3)
−9EIL2 2
L1(L1 − L2)(L2 − L3)(2L1L2 + L 2 2 + L1L3 − 4L2L3)
−3EI(L 2 1 − 2L1L2 − 2L 2 2 − 3L1L3 + 6L2L3)
(L1 − L2)(L2 − L3) 2 (2L1L2 + L 2 2 + L1L3 − 4L2L3)
3EI(L1 − 4L2)
(L2 − L3) 2 (2L1L2 + L 2 2 + L1L3 − 4L2L3)
can be calculated using the geometry of the beam (I, L1 , L2, L3) and its material properties (E),
according to section 1.3. You can try to get these coefficients using the information presented
in section 1.5.
Differential Equations 2nd order → 1st order
M D ¨y
0 M ˙y
+
0 K ˙y
−M 0 y
=
¯f
0
A˙z(t) + Bz(t) = fe jωt
z(t) =
˙y(t)
y(t)
⎧
⎪⎨
=
⎪⎩
˙y1(t)
˙y2(t)
˙y3(t)
y1(t)
y2(t)
y3(t)
⎫
⎪⎬
⎪⎭
→ velocity
→ velocity
→ velocity
→ displacement
→ displacement
→ displacement
51
⎫
⎪⎬
⎪⎭
e jωt
(79)
(80)
(81)

The analytical solution can be divided into three steps: (I) homogeneous solution (transient analysis);
(II) permanent solution (steady-state analysis) and (III) general solution (homogeneous
+ permanent), as mentioned in section 1.7.3. Introducing the initial conditions of displacement
and velocity
zini = { v1ini
one gets
v2ini v3ini y1ini y2ini y3ini }T
z(t) = C1u1e λ1t + C2u2e λ2t + C3u3e λ3t + C4u4e λ4t + C5u5e λ5t + C6u6e λ6t + Ae iωt
⎧
⎪⎨
⎪⎩
C1
C2
C3
C4
C5
C6
where
λ1 = −ξ1ωn1 − ωn1 1 − ξ2 1 · i
and u1
λ2 = −ξ1ωn1 + ωn1 1 − ξ2 1 · i
and u2
λ3 = −ξ2ωn2 − ωn2 1 − ξ2 2 · i
and u3
λ4 = −ξ2ωn2 + ωn2 1 − ξ2 2 · i
and u4
λ5 = −ξ3ωn3 − ωn3 1 − ξ2 3 · i
and u5
λ6 = −ξ3ωn3 + ωn3 1 − ξ2 3 · i and u6
⎫
A = [jωA + B] −1 f
⎪⎬
= [ u1 u2 u3 u4 u5 u6 ] −1 { zini − A}
⎪⎭
1.8.3 Programming in Matlab – Theoretical Parameter Studies and Experimental
Validation
52
(83)
(82)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% MACHINERY DYNAMICS LECTURES (72213) %
% IKS - DEPARTMENT OF CONTROL ENGINEERING DESIGN %
% DTU - TECHNICAL UNIVERSITY OF DENMARK %
% %
% Copenhagen, February 11th, 2000 %
% IFS %
% %
% 3 D.O.F. SYSTEMS - 4 DIFFERENT EXPERIMENTAL CASES %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Concentred Masses Values
m1= 0.191; %[Kg]
m2= 0.191; %[Kg]
m3= 0.191; %[Kg]
m4= 0.191; %[Kg]
m5= 0.191; %[Kg]
m6= 0.191; %[Kg]
%Elastic Properties of the Beam of 600 [mm]
E= 2.07e11; %elasticity modulus [N/m^2]
b= 0.030 ; %width [m]
h= 0.0012 ; %thickness [m]
Iz= (b*h^3)/12; %area moment of inertia [m^4]
% (1.CASE) Data for the mass-spring system
%__________________________________________________
M1=m1; %concentrated mass [Kg] |
M2=m2; %concentrated mass [Kg] |
M3=m3; %concentrated mass [Kg] |
L1= 0.203; %length for positioning M1 [m] |
L2= 0.406; %length for positioning M2 [m] |
L3= 0.610; %length for positioning M3 [m] |
%__________________________________________________|
% Coefficients of the Stiffness Matrix [N/m]
K11= (3*E*Iz*L2^3*(L2 - 4*L3))/(L1^3*(L1 - L2)^2*( ...
2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K12= (-3*E*Iz*(-3*L2*(L2 - 2*L3)*L3 + L1*(L2^2 - ...
2*L2*L3 - 2*L3^2)))/(L1*(L1 - L2)^2*(L2 - ...
L3)*(2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K13= (-9*E*Iz*L2^2)/(L1*(L1 - L2)*(L2 - L3)*(...
2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K21= (-3*E*Iz*(-3*L2*(L2 - 2*L3)*L3 + L1*(L2^2 - ...
2*L2*L3 - 2*L3^2)))/(L1*(L1 - L2)^2*(L2 - ...
L3)*(2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K22= (3*E*Iz*(L1 - 4*L3)*(L1 - L3)^2)/((L1 - ...
L2)^2*(L2 - L3)^2*(2*L1*L2 + L2^2 + ...
L1*L3 - 4*L2*L3));
K23= (-3*E*Iz*(L1^2 - 2*L1*L2 - 2*L2^2 - 3*L1*L3 + ...
6*L2*L3))/((L1 - L2)*(L2 - L3)^2*(2*L1*L2 + ...
L2^2 + L1*L3 - 4*L2*L3));
K31= (-9*E*Iz*L2^2)/(L1*(L1 - L2)*(L2 - L3)*(2*L1*L2 ...
+ L2^2 + L1*L3 - 4*L2*L3));
K32= (-3*E*Iz*(L1^2 - 2*L1*L2 - 2*L2^2 - 3*L1*L3 + ...
6*L2*L3))/((L1 - L2)*(L2 - L3)^2*(2*L1*L2 + ...
L2^2 + L1*L3 - 4*L2*L3));
K33= (3*E*Iz*(L1 - 4*L2))/((L2 - L3)^2*(2*L1*L2 + ...
L2^2 + L1*L3 - 4*L2*L3));
%Mass Matrix
M= [M1 0 0; 0 M2 0; 0 0 M3];
%Stiffness Matrix
K= [K11 K12 K13; K21 K22 K23; K31 K32 K33];
%Damping Matrix
D= [0 0 0; 0 0 0; 0 0 0];
%State Matrices
A= [ M D ;
zeros(size(M)) M ] ;
B= [ zeros(size(M)) K ;
-M zeros(size(M))];
%Dynamical Properties of the Mass-Spring System
[u,w]=eig(-B,A); %natural frequency [rad/s]
%Dynamical Properties of the Mass-Spring System
w=sort(diag(abs(w)))/2/pi %natural frequency [rad/s]
w1=w(1); %first natural frequency [Hz]
w2=w(3); %second natural frequency [Hz]
w3=w(5); %third natural frequency [Hz]
53
wexp1=1.031 %measured natural frequency [Hz]
%IMPORTANT: Freq resolution 400 lines
wexp2=7.000 %measured natural frequency [Hz]
%IMPORTANT: Freq resolution 400 lines
wexp3=19.312 %measured natural frequency [Hz]
%IMPORTANT: Freq resolution 400 lines
dif1=(w1-wexp1)/wexp1 %error between calculated and measured freq.
dif2=(w2-wexp2)/wexp2 %error between calculated and measured freq.
dif3=(w3-wexp3)/wexp3 %error between calculated and measured freq.
pause;
% (2.CASE) Increasing the Mass Values
% Data for the mass-spring system
%__________________________________________________
M1=m1+m4; %concentrated mass [Kg] |
M2=m2+m5; %concentrated mass [Kg] |
M3=m3+m6; %concentrated mass [Kg] |
L1= 0.203; %length for positioning M1 [m] |
L2= 0.406; %length for positioning M2 [m] |
L3= 0.610; %length for positioning M3 [m] |
%__________________________________________________|
% Coefficients of the Stiffness Matrix [N/m]
K11= (3*E*Iz*L2^3*(L2 - 4*L3))/(L1^3*(L1 - L2)^2*( ...
2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K12= (-3*E*Iz*(-3*L2*(L2 - 2*L3)*L3 + L1*(L2^2 - ...
2*L2*L3 - 2*L3^2)))/(L1*(L1 - L2)^2*(L2 - ...
L3)*(2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K13= (-9*E*Iz*L2^2)/(L1*(L1 - L2)*(L2 - L3)*(...
2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K21= (-3*E*Iz*(-3*L2*(L2 - 2*L3)*L3 + L1*(L2^2 - ...
2*L2*L3 - 2*L3^2)))/(L1*(L1 - L2)^2*(L2 - ...
L3)*(2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K22= (3*E*Iz*(L1 - 4*L3)*(L1 - L3)^2)/((L1 - ...
L2)^2*(L2 - L3)^2*(2*L1*L2 + L2^2 + ...
L1*L3 - 4*L2*L3));
K23= (-3*E*Iz*(L1^2 - 2*L1*L2 - 2*L2^2 - 3*L1*L3 + ...
6*L2*L3))/((L1 - L2)*(L2 - L3)^2*(2*L1*L2 + ...
L2^2 + L1*L3 - 4*L2*L3));
K31= (-9*E*Iz*L2^2)/(L1*(L1 - L2)*(L2 - L3)*(2*L1*L2 ...
+ L2^2 + L1*L3 - 4*L2*L3));
K32= (-3*E*Iz*(L1^2 - 2*L1*L2 - 2*L2^2 - 3*L1*L3 + ...
6*L2*L3))/((L1 - L2)*(L2 - L3)^2*(2*L1*L2 + ...
L2^2 + L1*L3 - 4*L2*L3));
K33= (3*E*Iz*(L1 - 4*L2))/((L2 - L3)^2*(2*L1*L2 + ...
L2^2 + L1*L3 - 4*L2*L3));
%Mass Matrix
M= [M1 0 0; 0 M2 0; 0 0 M3];
%Stiffness Matrix
K= [K11 K12 K13; K21 K22 K23; K31 K32 K33];
%Damping Matrix
D= [0 0 0; 0 0 0; 0 0 0];
%State Matrices
A= [ M D ;
zeros(size(M)) M ] ;
B= [ zeros(size(M)) K ;
-M zeros(size(M))];
%Dynamical Properties of the Mass-Spring System
[u,w]=eig(-B,A); %natural frequency [rad/s]
%Dynamical Properties of the Mass-Spring System
w=sort(diag(abs(w)))/2/pi %natural frequency [rad/s]
w1=w(1); %first natural frequency [Hz]
w2=w(3); %second natural frequency [Hz]
w3=w(5); %third natural frequency [Hz]
wexp1=0.71875 %measured natural frequency [Hz]
%IMPORTANT: Freq resolution 400 lines
wexp2=5.125 %measured natural frequency [Hz]
%IMPORTANT: Freq resolution 400 lines
wexp3=14.312 %measured natural frequency [Hz]
%IMPORTANT: Freq resolution 400 lines
dif1=(w1-wexp1)/wexp1 %error between calculated and measured freq.
dif2=(w2-wexp2)/wexp2 %error between calculated and measured freq.
dif3=(w3-wexp3)/wexp3 %error between calculated and measured freq.
pause;

% (3.CASE) Changing the Position of the Concentrated Masses
% Data for the mass-spring system
%__________________________________________________
M1=m1+m2; %concentrated mass [Kg] |
M2=m3+m4; %concentrated mass [Kg] |
M3=m5+m6; %concentrated mass [Kg] |
L1= 0.150; %length for positioning M1 [m] |
L2= 0.300; %length for positioning M2 [m] |
L3= 0.450; %length for positioning M3 [m] |
%__________________________________________________|
% Coefficients of the Stiffness Matrix [N/m]
K11= (3*E*Iz*L2^3*(L2 - 4*L3))/(L1^3*(L1 - L2)^2*( ...
2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K12= (-3*E*Iz*(-3*L2*(L2 - 2*L3)*L3 + L1*(L2^2 - ...
2*L2*L3 - 2*L3^2)))/(L1*(L1 - L2)^2*(L2 - ...
L3)*(2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K13= (-9*E*Iz*L2^2)/(L1*(L1 - L2)*(L2 - L3)*(...
2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K21= (-3*E*Iz*(-3*L2*(L2 - 2*L3)*L3 + L1*(L2^2 - ...
2*L2*L3 - 2*L3^2)))/(L1*(L1 - L2)^2*(L2 - ...
L3)*(2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K22= (3*E*Iz*(L1 - 4*L3)*(L1 - L3)^2)/((L1 - ...
L2)^2*(L2 - L3)^2*(2*L1*L2 + L2^2 + ...
L1*L3 - 4*L2*L3));
K23= (-3*E*Iz*(L1^2 - 2*L1*L2 - 2*L2^2 - 3*L1*L3 + ...
6*L2*L3))/((L1 - L2)*(L2 - L3)^2*(2*L1*L2 + ...
L2^2 + L1*L3 - 4*L2*L3));
K31= (-9*E*Iz*L2^2)/(L1*(L1 - L2)*(L2 - L3)*(2*L1*L2 ...
+ L2^2 + L1*L3 - 4*L2*L3));
K32= (-3*E*Iz*(L1^2 - 2*L1*L2 - 2*L2^2 - 3*L1*L3 + ...
6*L2*L3))/((L1 - L2)*(L2 - L3)^2*(2*L1*L2 + ...
L2^2 + L1*L3 - 4*L2*L3));
K33= (3*E*Iz*(L1 - 4*L2))/((L2 - L3)^2*(2*L1*L2 + ...
L2^2 + L1*L3 - 4*L2*L3));
%Mass Matrix
M= [M1 0 0; 0 M2 0; 0 0 M3];
%Stiffness Matrix
K= [K11 K12 K13; K21 K22 K23; K31 K32 K33];
%Damping Matrix
D= [0 0 0; 0 0 0; 0 0 0];
%State Matrices
A= [ M D ;
zeros(size(M)) M ] ;
B= [ zeros(size(M)) K ;
-M zeros(size(M))];
%Dynamical Properties of the Mass-Spring System
[u,w]=eig(-B,A); %natural frequency [rad/s]
%Dynamical Properties of the Mass-Spring System
w=sort(diag(abs(w)))/2/pi %natural frequency [rad/s]
w1=w(1); %first natural frequency [Hz]
w2=w(3); %second natural frequency [Hz]
w3=w(5); %third natural frequency [Hz]
wexp1=1.094 %measured natural frequency [Hz]
%IMPORTANT: Freq resolution 400 lines
wexp2=7.188 %measured natural frequency [Hz]
%IMPORTANT: Freq resolution 400 lines
wexp3=20.25 %measured natural frequency [Hz]
%IMPORTANT: Freq resolution 400 lines
dif1=(w1-wexp1)/wexp1 %error between calculated and measured freq.
dif2=(w2-wexp2)/wexp2 %error between calculated and measured freq.
dif3=(w3-wexp3)/wexp3 %error between calculated and measured freq.
pause;
54
% (4.CASE) Changing the Position and the Values of the Concentrated Masses
% Data for the mass-spring system
%__________________________________________________
M1=m1+m4+m5; %concentrated mass [Kg] |
M2=m2+m6; %concentrated mass [Kg] |
M3=m3; %concentrated mass [Kg] |
L1= 0.150; %length for positioning M1 [m] |
L2= 0.300; %length for positioning M2 [m] |
L3= 0.450; %length for positioning M3 [m] |
%__________________________________________________|
% Coefficients of the Stiffness Matrix [N/m]
K11= (3*E*Iz*L2^3*(L2 - 4*L3))/(L1^3*(L1 - L2)^2*( ...
2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K12= (-3*E*Iz*(-3*L2*(L2 - 2*L3)*L3 + L1*(L2^2 - ...
2*L2*L3 - 2*L3^2)))/(L1*(L1 - L2)^2*(L2 - ...
L3)*(2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K13= (-9*E*Iz*L2^2)/(L1*(L1 - L2)*(L2 - L3)*(...
2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K21= (-3*E*Iz*(-3*L2*(L2 - 2*L3)*L3 + L1*(L2^2 - ...
2*L2*L3 - 2*L3^2)))/(L1*(L1 - L2)^2*(L2 - ...
L3)*(2*L1*L2 + L2^2 + L1*L3 - 4*L2*L3));
K22= (3*E*Iz*(L1 - 4*L3)*(L1 - L3)^2)/((L1 - ...
L2)^2*(L2 - L3)^2*(2*L1*L2 + L2^2 + ...
L1*L3 - 4*L2*L3));
K23= (-3*E*Iz*(L1^2 - 2*L1*L2 - 2*L2^2 - 3*L1*L3 + ...
6*L2*L3))/((L1 - L2)*(L2 - L3)^2*(2*L1*L2 + ...
L2^2 + L1*L3 - 4*L2*L3));
K31= (-9*E*Iz*L2^2)/(L1*(L1 - L2)*(L2 - L3)*(2*L1*L2 ...
+ L2^2 + L1*L3 - 4*L2*L3));
K32= (-3*E*Iz*(L1^2 - 2*L1*L2 - 2*L2^2 - 3*L1*L3 + ...
6*L2*L3))/((L1 - L2)*(L2 - L3)^2*(2*L1*L2 + ...
L2^2 + L1*L3 - 4*L2*L3));
K33= (3*E*Iz*(L1 - 4*L2))/((L2 - L3)^2*(2*L1*L2 + ...
L2^2 + L1*L3 - 4*L2*L3));
%Mass Matrix
M= [M1 0 0; 0 M2 0; 0 0 M3];
%Stiffness Matrix
K= [K11 K12 K13; K21 K22 K23; K31 K32 K33];
%Damping Matrix
D= [0 0 0; 0 0 0; 0 0 0];
%State Matrices
A= [ M D ;
zeros(size(M)) M ] ;
B= [ zeros(size(M)) K ;
-M zeros(size(M))];
%Dynamical Properties of the Mass-Spring System
[u,w]=eig(-B,A); %natural frequency [rad/s]
%Dynamical Properties of the Mass-Spring System
w=sort(diag(abs(w)))/2/pi %natural frequency [rad/s]
w1=w(1); %first natural frequency [Hz]
w2=w(3); %second natural frequency [Hz]
w3=w(5); %third natural frequency [Hz]
exp1=1.312 %measured natural frequency [Hz]
%IMPORTANT: Freq resolution 400 lines
wexp2=7.219 %measured natural frequency [Hz]
%IMPORTANT: Freq resolution 400 lines
wexp3=18.000 %measured natural frequency [Hz]
%IMPORTANT: Freq resolution 400 lines
dif1=(w1-wexp1)/wexp1 %error between calculated and measured freq.
dif2=(w2-wexp2)/wexp2 %error between calculated and measured freq.
dif3=(w3-wexp3)/wexp3 %error between calculated and measured freq.
pause;

1.8.4 Theoretical Frequency Response Function
(a) y11 [m/N]
(d) y21 [m/N]
(c) y31 [m/N]
0.1
0.05
Excitation on Point 1
0
0 10 20 30
Frequency [Hz]
0.4
0.3
0.2
0.1
0
0 10 20 30
Frequency [Hz]
0.8
0.6
0.4
0.2
0
0 10 20 30
Frequency [Hz]
(b) y12 [m/N]
(a) y22 [m/N]
(d) y32 [m/N]
0.4
0.3
0.2
0.1
Excitation on Point 2
0
0 10 20 30
Frequency [Hz]
0.8
0.6
0.4
0.2
0
0 10 20 30
Frequency [Hz]
2
1.5
1
0.5
0
0 10 20 30
Frequency [Hz]
(c) y13 [m/N]
(b) y23 [m/N]
(d) y33 [m/N]
0.8
0.6
0.4
0.2
Excitation on Point 3
0
0 10 20 30
Frequency [Hz]
2
1.5
1
0.5
0
0 10 20 30
Frequency [Hz]
3
2
1
0
0 10 20 30
Frequency [Hz]
Figure 33: Forced Vibration – Theoretical Frequency Response Function (FRF) of the clampedfree
flexible beam when two concentrated masses m33 = m1 + m2 = 0.382 Kg are attached at
its free end (L = 0.610 m), two additional masses m22 = m3 + m4 = 0.382 Kg are attached
at L = 0.410 m and two additional masses m11 = m5 + m6 = 0.382 Kg are attached at
L = 0.210 m – – Natural frequencies of the 3 D.O.F. mass-spring system ”A”: 1.03 Hz,
7.00 Hz and 19.31 Hz.
55

1.8.5 Experimental – Natural Frequencies
(a) Amplitude [m/s 2 ]
(b) Amplitude [m/s 2 ]
2
1
0
−1
−2
x Signal 10−4
3
1
(a) in Time Domain − (b) in Frequency Domain
0 5 10 15
time [s]
20 25 30
x 10−5
2
0
0 5 10 15 20 25
frequency [Hz]
Figure 34: Transient Vibration – Acceleration of the clamped-free flexible beam when two concentrated
masses m = m1 + m2 = 0.382 Kg are attached at its free end (L3 = 0.610 m), two
additional masses m = m1 + m2 = 0.382 Kg are attached at its length (L2 = 0.410 m) and two
additional masses m = m1 +m2 = 0.382 Kg are attached at its length (L1 = 0.210 m) – Natural
frequencies of the 3 D.O.F. mass-spring system ”A”: 1.03 Hz, 7.00 Hz and 19.31 Hz.
1.8.6 Experimental – Resonances and Mode Shapes
• Visualization of the participation of modes shapes in the transient response – Visualization
using your eyes! Transient motion of the physical system excited with different initial
conditions by using your fingers!
• Applying an oscillatory excitation by using your finger at the 3 different points of the
physical system (co-ordinates of the mechanical model) and detecting the participation of
the mode shapes in the permanent solution or steady-state response by using your eyes.
56

(a) Amplitude [m/s 2 ]
(b) Amplitude [m/s 2 ]
(a) Amplitude [m/s 2 ]
(b) Amplitude [m/s 2 ]
(a) Amplitude [m/s 2 ]
(b) Amplitude [m/s 2 ]
2
1
0
−1
−2
x 10 −5 Signal (a) in Time Domain − (b) in Frequency Domain
0 5 10 15
time [s]
20 25 30
1.5
1
0.5
x 10 −5
0
0 5 10 15 20 25
frequency [Hz]
2
1
0
−1
−2
1.5
0.5
x 10 −5 Signal (a) in Time Domain − (b) in Frequency Domain
0 5 10 15
time [s]
20 25 30
1
x 10 −5
0
0 5 10 15 20 25
frequency [Hz]
2
1
0
−1
−2
1.5
0.5
x 10 −5 Signal (a) in Time Domain − (b) in Frequency Domain
0 5 10 15
time [s]
20 25 30
1
x 10 −5
0
0 5 10 15 20 25
frequency [Hz]
Figure 35: Resonance phenomena due to the excitation force with frequencies around the natural
frequencies of the mass-spring system: 3 D.O.F. system with the natural frequencies of 0.75 Hz,
5.12 Hz and 14.68 Hz, excited by the shaker – Spring-mass system ”A” with two masses m =
m1 + m2 = 0.382 Kg fixed at the beam length L3 = 0.610 m, two additional masses fixed at
L2 = 0.410 m and two more at L1 = 0.210 m.
57

1.9 Exercises
• Answer the questions using the Matlab program dof1-integration.m:
1. Vary the cross section parameters of the beam (b,h) while exciting the mass-spring system
with just an initial velocity (initial displacement and excitation force are set zero). (a)
Explain what happens with the natural frequency of the mass-spring system; (b) What
happens with the maximum vibration amplitude of the system?
2. Vary the beam length (L) while exciting the mass-spring system with just an initial velocity
(initial displacement and excitation force are set zero). (a) Explain what happens with
the natural frequency of the mass-spring system; (b) What happens with the maximum
vibration amplitude of the system?
3. Vary the number of masses attached to the beam while exciting the mass-spring system with
just an initial velocity (initial displacement and excitation force are set zero). (a) Explain
what happens with the natural frequency of the mass-spring system; (b) What happens
with the maximum vibration amplitude of the system?
4. Vary the damping factor ξ while exciting the mass-spring-damping system with just an
initial velocity (initial displacement and excitation force are set zero). (a) Explain what
happens with the natural frequency wn of the mass-spring system and the damped natural
frequency wd; (b) What happens with the maximum vibration amplitude of the system?
5. Set the damping factor ξ = 0.005 while exciting the mass-spring-damping system with just
an excitation force of f = 0.1 · e j·w·t [N] and initial velocity (initial displacement is set
zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies
when: (a) w = 10%wn; (b) w = 50%wn; (c) w = 90%wn; (d) w = wn; (e) w = 110%wn; (f)
w = 150%wn and (g) w = 200%wn.
6. Set the damping factor at 10 times more than before, ξ = 0.05, while exciting the massspring-damping
system with just an excitation force of f = 0.1·e j·w·t [N] and initial velocity
(initial displacement is set zero). Explain the vibration behavior of the system in terms of
amplitudes and frequencies when: (a) w = 10%wn; (b) w = 50%wn; (c) w = 90%wn; (d)
w = wn; (e) w = 110%wn; (f) w = 150%wn and (g) w = 200%wn.
7. Explain how this parameters variation could be useful in the case of a real machine?
• Answer the questions using the Matlab program dof2-integration.m:
1. Excite the mass-spring system just with an initial velocity at the first coordinate ( ˙y1ini )
(initial displacements and excitation forces are set zero). Describe the vibration behavior
of points y1 and y2.
2. Excite the mass-spring system just with an initial velocity at the second coordinate ( ˙y2ini )
(initial displacements and excitation forces are set zero). Describe the vibration behavior
of points y1 and y2.
3. Compare the two last simulations. Why is the transient behavior so different when the
system is perturbed with initial velocity at point y1 or at point y2?
58

4. Vary the number of masses attached to the first coordinate y1 the beam while exciting
the mass-spring system with just an initial velocity at the first coordinate ( ˙y1ini ) (initial
displacements and excitation forces are set zero). (a) Explain what happens with the natural
frequencies of the system; (b) How many natural frequencies change when you change the
mass in just one point of the structure? Explain.
5. Vary the number of masses attached to the second coordinate of the beam, y2, while exciting
the mass-spring system with just an initial velocity at the first coordinate ( ˙y1ini ) (initial
displacements and excitation forces are set zero). (a) Explain what happens with the natural
frequencies of the system; (b) How many natural frequencies change when you change the
mass in just one point of the structure? Explain.
6. Set the damping factor ξ = 0.005, while exciting the mass-spring-damping system with just
an excitation force of f1 = 0.1 · e j·w·t [N] (initial velocities and initial displacements are set
zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies,
when: (a) w = 10%wn1; (b) w = 50%wn1; (c) w = 90%wn1; (d) w = wn1; (e) w = 110%wn1;
(f) w = 90%wn2; (g) w = wn2; (h) w = 110%wn2; (i) w = 200%wn2.
7. Set the damping factor ξ = 0.05, while exciting the mass-spring-damping system with just
an excitation force of f1 = 0.1 · e j·w·t [N] (initial velocities and initial displacements are set
zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies,
when: (a) w = 10%wn1; (b) w = 50%wn1; (c) w = 90%wn1; (d) w = wn1; (e) w = 110%wn1;
(f) w = 90%wn2; (g) w = wn2; (h) w = 110%wn2; (i) w = 200%wn2.
8. Set the damping factor ξ = 0.005, while exciting the mass-spring-damping system with just
an excitation force of f2 = 0.1 · e j·w·t [N] (initial velocities and initial displacements are set
zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies,
when: (a) w = 10%wn1; (b) w = 50%wn1; (c) w = 90%wn1; (d) w = wn1; (e) w = 110%wn1;
(f) w = 90%wn2; (g) w = wn2; (h) w = 110%wn2; (i) w = 200%wn2.
9. Set the damping factor ξ = 0.05, while exciting the mass-spring-damping system with just
an excitation force of f2 = 0.1 · e j·w·t [N] (initial velocities and initial displacements are set
zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies,
when: (a) w = 10%wn1; (b) w = 50%wn1; (c) w = 90%wn1; (d) w = wn1; (e) w = 110%wn1;
(f) w = 90%wn2; (g) w = wn2; (h) w = 110%wn2; (i) w = 200%wn2;
10. Explain how the variation of such parameters could be useful in a case with a real machine?
• Create a program dof3-integration.m based on dof2-integration.m and answer the following
questions:
1. Make use of the beam theory, and show how to get the 9 stiffness coefficients k11, k12, k13,
k21, k22, k23, k31, k32 and k33.
2. Excite the mass-spring system with just an initial velocity at the first coordinate ( ˙y1ini )
(initial displacements and excitation forces are set zero). Describe the vibration behavior
of the points y1, y2 and y3.
3. Excite the mass-spring system with just an initial velocity at the second coordinate ( ˙y2ini )
(initial displacements and excitation forces are set zero). Describe the vibration behavior
of the points y1, y2 and y3.
59

4. Excite the mass-spring system with just an initial velocity at the third coordinate ( ˙y3ini )
(initial displacements and excitation forces are set zero). Describe the vibration behavior
of the points y1, y2 and y3.
5. Compare the three last simulations. Why is the transient behavior so different when the
system is perturbed with initial velocity at point y1, y2 or y3.
6. Vary the number of masses attached to the first coordinate of the beam, y1, while exciting
the mass-spring system with just an initial velocity at the first coordinate ( ˙y1ini ) (initial
displacements and excitation forces are set zero). (a) Explain what happens with the natural
frequencies of the system; (b) How many natural frequencies change when you change the
mass in just one point of the structure? Explain.
7. Vary the number of masses attached to the second coordinate of the beam, y2, while exciting
the mass-spring system with just an initial velocity at the first coordinate ( ˙y1ini ) (initial
displacements and excitation forces are set zero). (a) Explain what happens with the natural
frequencies of the system; (b) How many natural frequencies change when you change the
mass in just one point of the structure? Explain.
8. Set the damping factor ξ = 0.005, while exciting the mass-spring-damping system with just
an excitation force of f1 = 0.1 · e j·w·t [N] (initial velocities and initial displacements are set
zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies,
when: (a) w = 10%wn1; (b) w = 50%wn1; (c) w = 90%wn1; (d) w = wn1; (e) w = 110%wn1;
(f) w = 90%wn2; (g) w = wn2; (h) w = 110%wn2; (i) w = 200%wn2; (j) w = 90%wn3; (k)
w = wn3; (l) w = 110%wn3; (m) w = 200%wn3.
9. Set the damping factor ξ = 0.05, while exciting the mass-spring-damping system with just
an excitation force of f1 = 0.1 · e j·w·t [N] (initial velocities and initial displacements are set
zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies,
when: (a) w = 10%wn1; (b) w = 50%wn1; (c) w = 90%wn1; (d) w = wn1; (e) w = 110%wn1;
(f) w = 90%wn2; (g) w = wn2; (h) w = 110%wn2; (i) w = 200%wn2; (j) w = 90%wn3; (k)
w = wn3; (l) w = 110%wn3; (m) w = 200%wn3.
10. Set the damping factor ξ = 0.005, while exciting the mass-spring-damping system with just
an excitation force of f3 = 0.1 · e j·w·t [N] (initial velocities and initial displacements are set
zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies,
when: (a) w = 10%wn1; (b) w = 50%wn1; (c) w = 90%wn1; (d) w = wn1; (e) w = 110%wn1;
(f) w = 90%wn2; (g) w = wn2; (h) w = 110%wn2; (i) w = 200%wn2; (j) w = 90%wn3; (k)
w = wn3; (l) w = 110%wn3; (m) w = 200%wn3.
11. Set the damping factor ξ = 0.05, while exciting the mass-spring-damping system with just
an excitation force of f3 = 0.1 · e j·w·t [N] (initial velocities and initial displacements are set
zero). Explain the vibration behavior of the system in terms of amplitudes and frequencies,
when: (a) w = 10%wn1; (b) w = 50%wn1; (c) w = 90%wn1; (d) w = wn1; (e) w = 110%wn1;
(f) w = 90%wn2; (g) w = wn2; (h) w = 110%wn2; (i) w = 200%wn2; (j) w = 90%wn3; (k)
w = wn3; (l) w = 110%wn3; (m) w = 200%wn3;
12. Explain how such a variation of parameters could be useful in a case with a real machine?
60

1.10 Project 0 – Identification of Model Parameters (An Example)
GOAL – With the first project the student will face a practical problem of the real life: how
to properly choose the coefficients of linear differential equations of second order, aiming at
achieving a reliable mathematical model, which can predict the machine dynamics?
(a) (b)
Figure 36: (a) Offshore platform http : //www.civl.port.ac.uk/comp−prog/offshore−platforms;
(b) Laboratory prototype composed of one concentrated mass (foundation and rotor) attached
to four flexible beams – An equivalent model of 1 D.O.F. system for analyzing the platform’s
linear vibration in the horizontal direction.
To represent the 2D-movements of the offshore platform shown in figure 36(a) a laboratory
prototype was built, as it can be seen in figure 36(b). This simplified test rig is composed of one
concentrated mass (foundation and rotor) attached to four flexible beams. An equivalent model
of 1 D.O.F. system can be created with the purpose of analyzing the platform’s linear vibration
in the horizontal direction.
m0 2.108 [kg] platform mass
L0 0.205 [m] beam length
b0 0.025 [m] beam width
h0 0.001 [m] beam thickness
E 1.9 · 10 11 [N/m 2 ] steel elastic modulus
Table 5: Main parameters of the test rig (platform)
1. Create a mechanical model of one-degree-of-freedom for describing the horizontal vibration
of the test rig. Use Newton’s law and equivalent coefficients of mass m [Kg], viscous
damping d [N/(m/s)] and linear stiffness k [N/m].
2. There are two different ways of experimentally obtaining the forced vibration response
of the platform in the frequency domain, i.e. its frequency response functions FRF(ω),
namely by means of H1(ω) and H2(ω) functions. Detail about how to experimentally
obtain H1(ω) and H2(ω) will be given in the second part of manuscript. Anyway, for now,
it is important to relate such experimental functions to the frequency response functions
61

presented in section 1.6.9, to learn how to deal with experimental data and how to extract
important information. H1(ω) as well as H2(ω) are obtained when an excitation force
is horizontally acting on the platform mass and its value is measured by using a force
transducer, and simultaneously the acceleration response of the platform is measured using
a acceleration sensor. The experimental setup can be seen in figure 36(b). The most
common ways of representing the frequency response functions H1(ω) and H2(ω) are in
the form of amplitude and phase, or real and imaginary. In table 6 such values are presented
in the real and imaginary forms and in figure 37 they are plotted. Please, plot H1(ω) and
H2(ω) in terms of magnitude and phase.
Frequency - H1 - H2 COHERENCE
[Hz] [(m/s 2 )/N] [(m/s 2 )/N]
1.6250 0.0490 - 0.0040i 0.0595 - 0.0049i 0.8236
1.7500 0.0610 - 0.0051i 0.0698 - 0.0058i 0.8745
1.8750 0.0750 - 0.0071i 0.0855 - 0.0080i 0.8776
2.0000 0.0826 - 0.0068i 0.0945 - 0.0078i 0.8744
2.1250 0.0974 - 0.0025i 0.1093 - 0.0028i 0.8907
2.2500 0.1124 + 0.0015i 0.1204 + 0.0016i 0.9335
2.3750 0.1310 - 0.0018i 0.1393 - 0.0019i 0.9403
2.5000 0.1411 - 0.0100i 0.1477 - 0.0105i 0.9555
2.6250 0.1540 - 0.0112i 0.1601 - 0.0116i 0.9621
2.7500 0.1817 - 0.0040i 0.1889 - 0.0041i 0.9619
2.8750 0.2113 - 0.0145i 0.2184 - 0.0150i 0.9677
3.0000 0.2474 - 0.0172i 0.2520 - 0.0176i 0.9819
3.1250 0.2676 - 0.0230i 0.2726 - 0.0234i 0.9817
3.2500 0.3016 - 0.0364i 0.3080 - 0.0371i 0.9792
3.3750 0.3460 - 0.0487i 0.3536 - 0.0497i 0.9787
3.5000 0.4072 - 0.0467i 0.4173 - 0.0479i 0.9758
3.6250 0.4601 - 0.0673i 0.4697 - 0.0687i 0.9795
3.7500 0.5107 - 0.0903i 0.5165 - 0.0914i 0.9887
3.8750 0.5894 - 0.1239i 0.5974 - 0.1256i 0.9866
4.0000 0.6862 - 0.1427i 0.6994 - 0.1454i 0.9812
4.1250 0.8110 - 0.1959i 0.8228 - 0.1987i 0.9857
4.2500 0.9381 - 0.2824i 0.9539 - 0.2872i 0.9834
4.3750 1.1113 - 0.4002i 1.1429 - 0.4116i 0.9723
4.5000 1.3290 - 0.5385i 1.3711 - 0.5556i 0.9693
4.6250 1.5586 - 0.8030i 1.6312 - 0.8405i 0.9555
4.7500 1.8764 - 1.4666i 1.9844 - 1.5509i 0.9456
4.8750 1.8832 - 2.0905i 2.0664 - 2.2938i 0.9114
5.0000 1.2159 - 3.1027i 1.3216 - 3.3723i 0.9201
5.1250 0.2098 - 3.6778i 0.2413 - 4.2301i 0.8694
5.2500 -1.6165 - 3.2734i -1.7564 - 3.5568i 0.9203
5.3750 -2.2154 - 2.3597i -2.3326 - 2.4844i 0.9498
5.5000 -2.1657 - 1.4778i -2.2476 - 1.5337i 0.9635
5.6250 -1.9289 - 1.1157i -1.9659 - 1.1371i 0.9812
5.7500 -1.7058 - 0.8104i -1.7278 - 0.8208i 0.9873
5.8750 -1.5511 - 0.6322i -1.5665 - 0.6384i 0.9902
6.0000 -1.4389 - 0.5016i -1.4438 - 0.5033i 0.9966
6.1250 -1.3520 - 0.4295i -1.3559 - 0.4307i 0.9971
6.2500 -1.2591 - 0.3428i -1.2621 - 0.3436i 0.9976
6.3750 -1.1843 - 0.3000i -1.1866 - 0.3006i 0.9980
6.5000 -1.1241 - 0.2668i -1.1264 - 0.2673i 0.9980
6.6250 -1.0716 - 0.2302i -1.0725 - 0.2304i 0.9991
6.7500 -1.0206 - 0.2121i -1.0216 - 0.2123i 0.9990
6.8750 -0.9715 - 0.1980i -0.9722 - 0.1981i 0.9993
7.0000 -0.9376 - 0.1766i -0.9384 - 0.1767i 0.9992
7.1250 -0.9073 - 0.1613i -0.9077 - 0.1614i 0.9995
7.2500 -0.8793 - 0.1528i -0.8797 - 0.1529i 0.9995
7.3750 -0.8516 - 0.1459i -0.8519 - 0.1459i 0.9996
7.5000 -0.8271 - 0.1377i -0.8273 - 0.1377i 0.9998
7.6250 -0.8107 - 0.1259i -0.8109 - 0.1260i 0.9997
7.7500 -0.7879 - 0.1204i -0.7881 - 0.1204i 0.9997
7.8750 -0.7695 - 0.1170i -0.7696 - 0.1170i 0.9998
8.0000 -0.7513 - 0.1098i -0.7515 - 0.1098i 0.9998
8.1250 -0.7374 - 0.1058i -0.7376 - 0.1059i 0.9998
8.2500 -0.7240 - 0.1038i -0.7242 - 0.1038i 0.9998
8.3750 -0.7122 - 0.0994i -0.7123 - 0.0994i 0.9998
8.5000 -0.6963 - 0.0949i -0.6964 - 0.0950i 0.9998
8.6250 -0.6861 - 0.0930i -0.6862 - 0.0930i 0.9999
8.7500 -0.6792 - 0.0897i -0.6793 - 0.0897i 0.9998
8.8750 -0.6705 - 0.0886i -0.6706 - 0.0886i 0.9998
9.0000 -0.6606 - 0.0847i -0.6607 - 0.0847i 0.9999
9.1250 -0.6544 - 0.0815i -0.6545 - 0.0815i 0.9998
9.2500 -0.6463 - 0.0793i -0.6465 - 0.0794i 0.9997
9.3750 -0.6365 - 0.0765i -0.6367 - 0.0765i 0.9997
9.5000 -0.6287 - 0.0774i -0.6289 - 0.0774i 0.9997
9.6250 -0.6250 - 0.0762i -0.6253 - 0.0762i 0.9996
9.7500 -0.6207 - 0.0722i -0.6209 - 0.0722i 0.9997
9.8750 -0.6160 - 0.0711i -0.6162 - 0.0711i 0.9997
10.0000 -0.6081 - 0.0741i -0.6083 - 0.0741i 0.9998
Table 6: Two experimental frequency response functions H1(ω) and H2(ω) of the test rig.
3. Identification of model parameters – Suppose you have no access to the values of mass
62

(a)
(b)
REAL(Acc/force) [(m/s 2 )/N]
REAL(Acc/force) [(m/s 2 )/N]
15
10
5
0
−5
−10
Experimental Frequency Response Function (with damper)
measured
−15
0 1 2 3 4 5
Frequency [Hz]
6 7 8 9 10
IMAG(Acc/force) [(m/s 2 )/N]
0
−1
−2
−3
−4
15
10
5
0
−5
−10
measured
0 1 2 3 4 5
Frequency [Hz]
6 7 8 9 10
Experimental Frequency Response Function (with damper)
measured
−15
0 1 2 3 4 5
Frequency [Hz]
6 7 8 9 10
IMAG(Acc/force) [(m/s 2 )/N]
0
−1
−2
−3
−4
measured
0 1 2 3 4 5
Frequency [Hz]
6 7 8 9 10
Figure 37: Two experimental frequency response function FRF(ω) of the test rig obtained by
means of H1(ω) and H2(ω) functions.
63

m [kg], damping d [N/(m/s)] and stiffness k [N/m] of the platform and then can not be
calculated. However, you can measure the forces applied to the platform and its acceleration
response (steady-state response) and are able to experimentally obtain the frequency response
function presented in table 1 6 and plotted in figure 37. Remember that the frequency
response function is acceleration/force in this case, or
FRF(ω) =
−ω 2 /m
(−ω 2 + ω 2 n + 2 · ξ · ωn · ω · i) =
−ω 2
(−m · ω 2 + d · ω · i + k)
Elaborate a simple parameter identification procedures based on the Least-Square Method,
assuming that the frequency response functions FRF(ω) are known, and identify simultaneously
the three parameter, i.e. mass, stiffness and damping:
⎡
⎢
⎣
−ω 2 1 1
−ω 2 2 1
−ω 2 3 1
... ...
−ω 2 N 1
⎤
⎥
⎥ m
⎥
⎦ k
⎧
⎪⎩
REAL
REAL
⎪⎨
= REAL
...
REAL
ω2 1
FRF(ω1)
ω2 2
FRF(ω2)
ω2 3
FRF(ω3)
ω 2 N
FRF(ωN)
⎫
⎪⎬
⎪⎭
(84)
=⇒ A · x = b (85)
x = A T · A −1 · A T · b (86)
⎡
⎢
⎣
ω1
ω2
ω3
...
ωN
⎤
⎧
IMAG
IMAG
⎥ ⎪⎨ ⎥
⎥ d = IMAG
⎦
...
⎪⎩ IMAG
ω2 1
FRF(ω1)
ω2 2
FRF(ω2)
ω2 3
FRF(ω3)
ω 2 N
FRF(ωN)
⎫
⎪⎬
⎪⎭
=⇒ Ā · ¯x = ¯ b (87)
¯x = Ā T · Ā −1 · Ā T · ¯ b (88)
Implement the identification procedure using MAPLE, or MATHEMATICA or MATLAB or
MATCAD or another software. Use H1(ω) as well as H2(ω) for identifying the coefficients
of mass m [kg], stiffness k [N/m] and damping d [N/(m/s)].
4. Find theoretical and experimental ways of checking the identified values of mass, stiffness
and damping, in order to assure that such values are really the correct mass, stiffness and
damping coefficients of the real system. The more ”checking procedures” you can find, the
better! Explain them all in details.
5. Model application – On the platform top a rotating machine is mounted, as it can be seen in
figure.36(b). Its characteristics are delivered by the manufacturer. The maximum angular
velocity is 1, 200 [rpm] (20 [Hz]). It is also known that the machine unbalance (m unb · ε) is
0.00012 [Kg · m]. By using your mathematical model, plot the vibration amplitude of the
platform as a function of the machine angular velocity. Determine the maximum vibration
amplitude of the platform.
1 It is important to notice that the values of H1(ω) and H2(ω) presented in table 6 are −H1(ω) and −H2(ω).
When using the values of such functions to identify the model parameters, they have to be multiplied by -1.
64

6. Model application – As explained in the last item, the motor characteristics are delivered
by the manufacturer. The maximum angular velocity is 1, 200 [rpm] (20 [Hz]). It is also
known that the machine unbalance (m unb · ε) is 0.00012 [Kg · m]. Based on the dynamic
equilibrium between motor torque, and torques associated to inertia, losses and external
loading, the motor start-up curve shows a linear variation of angular velocity from 0 to
1, 200 [rpm] in a period of 40 [s]. Based on your mathematical model, please simulate
computationally the vibration behavior of the platform during the period of 40 [s], knowing
that, when the motor starts, the platform is already deformed due to a constant lateral
wind. The platform static deformation is 0.001 [m]. Plot the platform displacement in
the time domain, considering two cases: (a) considering the static wind force acting on
the platform the whole time; (b) considering the lateral wind force suddenly disappears 30
[s] after the motor start-up. Analyze and discuss the behavior of the plots. What is the
maximum vibration amplitude of the platform in both cases?
7. Question – Explain why the test rig can be modelled as an one-degree-of-freedom system in
the range of 0 to 10 Hz, if one knows that a rigid body in the space (platform mass of the
test rig) should be represented by a mechanical model of six-degrees-of-freedom, i.e. three
linear and three angular displacements.
8. Write your final conclusions.
(No Technical report!)
65

1.11 Project 1 – Modal Analysis & Validation of Models
GOAL – To get familiar with the dynamic interaction between machine and structure, the elaboration
of mechanical and mathematical models for representing rotor-structure vibrations in
2D, implementation and vibration analysis using Matlab, visualization of natural frequencies and
mode shapes. To test the accuracy of an analytical mathematical model proposed for describing
the system dynamic behavior, i.e. natural frequencies and mode shapes. Remember, if the measured
frequencies and mode shapes agree with those predicted by the analytical mathematical
model, the model is verified and can be useful for design proposes and vibration predictions with
some confidence. Otherwise, the analytical models are useless.
(a) (b)
Figure 38: Machine-structure dynamical interaction – (a) Offshore platform http :
//www.oil−gas.uwa.edu.au/Troll−A−Graphics.htm; (b) Equivalent laboratory prototype composed
of four concentrated masses attached to four flexible beams: 1- mass on the first floor, 2mass
on the second floor, 3- mass on the third floor, 4- mass on the fourth floor, 5- motor-disk
with unbalanced masses for simulating a rotating machine with unbalance problems, 6- acceleration
sensor attached to the second mass, 7- acceleration sensor attached to the third mass,
8- magnetic actuator for simulating wave excitation, 9- magnetic actuator for simulating waves
excitation.
Figures 38(a) and (b) illustrate an offshore platform and an equivalent laboratory prototype,
where the students can carry on measurements and vibration analyzes. The laboratory prototype
is composed of four concentrated masses attached to four flexible beams. Elements 1,2,3 and
4 are the four masses connected by means of flexible beams. Element 5 is a motor-disk with a
66

changeable unbalanced mass for simulating rotating machines (for example compressors, turbines
or pumps) with an unbalance problem. Elements 6 and 7 are two acceleration sensors attached
to the second and third masses. Elements 8 and 9 are magnetic actuators built to apply forces
with different dynamic characteristics (oscillatory, random, pulse etc.) to the structure (first
floor). In that way it is possible to simulate the wave forces coming from the ocean by means
of the magnetic actuators.
The motor-disk with unbalanced masses is mounted at the top of the platform (fourth floor). The
motor has variable angular velocity from 0 to 40 Hz (2400 rpm). Due to the unbalanced masses
strong vibration amplitudes can be detected on the second and third floor of the platform.
To represent the dynamical behavior of the system a mechanical model has to be created.
Considering the range of frequencies between 0 and 40 Hz, all rotor-structure movements happen
in a vertical plane (2D motion), and an appropriate mechanical model would be the one presented
in figure 39(b). For the suggested mechanical model:
(a) (b)
Figure 39: Machine-structure dynamical interaction – (a) Laboratory prototype; (b) Mechanical
model composed of four lumped masses attached to four flexible beams, an equivalent model of a
4 D.O.F. system for analyzing the linear vibrations of the platform in the horizontal direction
due to the interaction with a machine and ocean waves.
1. MODELLING – The main information about the geometric properties of the structure
presented in figure 39(b) is given below:
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% Lumped Masses
m1 = 1.95 % [kg] lowest mass
m2 = 1.72 % [kg] above lowest mass
m3 = 1.95 % [kg] below highest mass
m4 = 3.04 % [kg] highest mass with motor
% Beam Properties
E=2.0e11 % [N/m^2] elasticity modulus
b=0.0291 % [m] width
h=0.0011 % [m] thickness
I=(b*h^3)/12 % [m^4] area moment of inertia
% Position of the Platforms
L1=0.122 % [m] Length to Platform 1
L2=0.123 % [m] Length from Platform 1 to Platform 2
L3=0.149 % [m] Length from Platform 2 to Platform 3
L4=0.127 % [m] Length from Platform 3 to Platform 4
• Write the mathematical model (equations of motion) based on Newton’s laws, achieving
the mass M, stiffness K and damping D matrices.
• How is the structure of the mass matrix M? How to properly calculate the mass
coefficients?
• How is the structure of the stiffness matrix K? Find two different ways of obtaining
the stiffness coefficients.
• How is the structure of the damping matrix D? Find three different ways of obtaining
the damping coefficients, using proportional damping,
D1 = α · M.
D2 = β · K.
D3 = α · M + β · K.
Remember that when you do not know how to exactly model and achieve the damping
coefficients of the matrix D, assumptions have to be made. A realistic approximation
of the structural damping factor ξ is 0.005, according to the experimental results
obtained in equation (46). Try to adjust the proportionality factors α and β so that
the damping factor ξ1 related to the first mode shape of the structure is 0.005. Feel
free to do your own assumptions regarding damping, if you want!
2. NUMERICAL IMPLEMENTATION – Write a program in Matlab dof4-integration.m, or
use the program of your preference. Use as reference the dof2-integration.m, if you want.
Calculate the natural frequencies ωi (i = 1, ...,4) of the mechanical model and the damping
ratios ξi (i = 1, ...,4) related to the 4 modes shapes, considering the three different damping
matrices D1, D2 and D3.
%State Matrices A and B
A= [ M D ;
zeros(size(M)) M ] ;
B= [ zeros(size(M)) K ;
-M zeros(size(M))];
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%Modal Matrix u with mode shapes
%Matrix w with damping factors and damped natural frequencies
[u,w]=eig(-B,A);
3. ANALYSIS – Neglecting the damping, write the modal matrix with help of your program
dof4-integration.m. Based on the information contained in such a matrix describe the mode
shapes of the structure with some drawings;
4. ANALYSIS – Without neglecting the damping, write the modal matrix with help of your
program dof4-integration.m. Based on the information contained in such a matrix describe
the mode shapes of the structure and try to explain the physical meaning of the complex
numbers in the modal matrix.
5. EXPERIMENTAL – Try to predominately excite the first mode shape of the building using
an appropriate initial condition of displacement. Capture the acceleration signal in time
domain and plot it. Based on the logarithmic decrement try to establish the damping
factor ξ1 associated to the first mode shape of the building. Please, download the file
yyy4−trans.txt from campus net in order to rebuild figure 40. After obtaining the damping
ratio ξ1, compare with your assumption of 0.005.
acc [m/s 2 ]
3
2
1
0
−1
−2
−3
Acceleration of Mass 4
−4
0 1 2 3 4
time [s]
5 6 7 8
Figure 40: Experimental transient vibration response (acceleration) of mass 4 after initial condition
of displacement, which excites only the first mode shape of the structure.
6. EXPERIMENTAL – Obtain 4 frequency response functions in the range of 0 − 40 Hz,
when the building is excited on the first mass by magnetic forces. Please, download the
files: frf−general.m, xxx1.txt, xxx2.txt, xxx3.txt, xxx4.txt, yyy1.txt, yyy2.txt, yyy3.txt,
yyy4.txt (campus net)
7. EXPERIMENTAL – Experimental Modal Analysis (EMA) deals with the determination of
natural frequencies, modes shapes, and damping ratios from experimental measurements.
The fundamental idea behind modal testing is the resonance. If a structure is excited at
resonance, its response exhibits two distinct phenomena: (a) as the excitation frequency
69

approaches the natural frequency of the structure, the magnitude at resonance rapidly
approaches a sharp maximum value, provided that the damping ratio is less than about
0.5; (b) the phase of the response between excitation force and displacement shift by 180 o
as the frequency sweeps through resonance, with the value of the phase at resonance being
90 o . This physical phenomenon is used to determine the natural frequency of a structure
from measurements of the magnitude and phase of the force response of the structure as
the driving frequency is swept through a wide range of values.
Identify 4 different 1-DOF systems around each one of the 4 natural frequencies of the
building. Use your frequency domain identification procedure, which was already developed
in the project 1 and is based on the Least Square Method, and obtain the experimental
natural frequencies and the experimental damping factors of each one of the 4 mode shapes
of the building.
8. MODEL VALIDATION (VERIFICATION) – Compare the theoretical and experimental
frequency response functions of the first, second, third and fourth floors, when the structure
is excited by the magnetic forces on the first floor. Justify the discrepancies between
theoretical and experimental results.
9. APPLICATION OF THE MODEL – The values of the unbalance mass and eccentricity
are:
% Disk Unbalance
m=0.045 % [kg] unbalance mass
e=0.040 % [m] eccentricity
Consider the 5 different angular velocities, close to the resonances of the building and among
them, as following:
• 225 rpm (3,75 Hz),
• 495 rpm (8,25 Hz),
• 615 rpm (10,25 Hz),
• 900 rpm (15,00 Hz) and
• 975 rpm (16,25 Hz).
Use your mathematical model to predict the vibration amplitude of the top mass,
i.e. acceleration of the top mass, when the rotor-disk operates unbalanced at
the 5 different angular velocities. Check your results comparing with the experimental
results. Explain the discrepancies between the results obtained with
help of your mathematical model and the experiments. Please, download the
files yyy4−unbal−3−75−HZ.txt, yyy4−unbal−8−25−HZ.txt, yyy4−unbal−10−25−HZ.txt,
yyy4−unbal−15−00−HZ.txt, yyy4−unbal−16−25−HZ.txt and acc−in−time−domain.m to
rebuild figure 41.
10. MODEL ADJUSTMENT – Try to adjust the natural frequencies ωi (i = 1, ...,4) of the
analytical model using the experimental natural frequencies obtained via Experimental
Modal Analysis. Remember that the parameter
% Beam Properties
E=(2.0 +- 0.1)e11 % [N/m^2] elasticity modulus
70

acc [m/s 2 ]
acc [m/s 2 ]
0.8
0.6
0.4
0.2
−0.2
−0.4
−0.6
−0.8
0 1 2 3 4
time [s]
5 6 7 8
5
4
3
2
1
0
−1
−2
−3
−4
1
0
(b) Acceleration of Mass 4
(d) Acceleration of Mass 4
−5
0 1 2 3 4
time [s]
5 6 7 8
acc [m/s 2 ]
2.5
2
1.5
1
0.5
0
−0.5
−1
−1.5
−2
(a) Acceleration of Mass 4
−2.5
0 1 2 3 4
time [s]
5 6 7 8
acc [m/s 2 ]
acc [m/s 2 ]
6
4
2
0
−2
−4
(c) Acceleration of Mass 4
−6
0 1 2 3 4
time [s]
5 6 7 8
10
8
6
4
2
0
−2
−4
−6
−8
(e) Acceleration of Mass 4
−10
0 1 2 3 4
time [s]
5 6 7 8
Figure 41: Experimental forced vibration response (acceleration) of mass 4 due to a disk operating
with an unbalance mass m = 0.045 Kg with an eccentricity radius e = 0.040 m at 5 rotational
speeds: (a) 225 rpm (3,75 Hz); (b) 495 rpm (8,25 Hz); (c) 615 rpm (10,25 Hz); (d) 900 rpm
(15,00 Hz) and (e) 975 rpm (16,25 Hz).
71

=(0.0291 +- 0.0001) % [m] width
h=(0.0011 +- 0.0001) % [m] thickness
I=(b*h^3)/12 % [m^4] area moment of inertia
% Position of the Platforms
L1=(0.122 +- 0.001) % [m] Length to Platform 1
L2=(0.123 +- 0.001) % [m] Length from Platform 1 to Platform 2
L3=(0.149 +- 0.001) % [m] Length from Platform 2 to Platform 3
L4=(0.127 +- 0.001) % [m] Length from Platform 3 to Platform 4
can be varied based on the geometry and assembly tolerances, as well as on material properties.
Changes of such parameters lead to changes in the stiffness matrix K.
11. MODEL ADJUSTMENT – After adjustment of the natural frequencies, try to adjust the
damping ratios ξi (i = 1, ...,4) of the analytical model using the experimental damping
ratios obtained via Experimental Modal Analysis. Remember that alpna and β are the
parameters to be varied, if you have proportional damping matrices D1, D2 and D3.
12. APPLICATION OF THE ADJUSTED MODEL – Consider the 5 different angular velocities,
close to the resonances of the building and among them, as following:
• 225 rpm (3,75 Hz),
• 495 rpm (8,25 Hz),
• 615 rpm (10,25 Hz),
• 900 rpm (15,00 Hz) and
• 975 rpm (16,25 Hz).
Use your adjusted analytical model to predict the vibration amplitude of the top mass, i.e.
acceleration of the top mass, when the rotor-disk operates unbalanced at the 5 different
angular velocities. Check your theoretical results comparing with the experimental results.
Explain the discrepancies between the results obtained with help of your mathematical
model and the experiments.
13. Write about your conclusions!
(Technical report until 01/04/2005)
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