MECH 375 – Mechanical Vibrations Lab Report # 6 (Characteristics o
a Multi Degree-of-Freedom System Concordia University
Title of Experiment:
Characteristics of a Multi Degree-of-Freedom System
Lab Section: AL-X
Amardip Singh Ghuman
Ian Llamas
Kevin Balie
Objective:
,The objective of this experiment is to determine the fundamental natural frequencies of three rotor system as
well as to find the natural frequencies and natural modes of a four rotor system.
Introduction:
Today, mechanical systems composed by multi degrees of freedom are extremely common in industry.
Usually, the fundamental frequency for these types of systems is more important than its higher natural
frequencies because, sometimes, the forced response can be even higher.
The system presented in the following figure shows us a schematic of a three degree of freedom system
composed by three rotors. The systems consists of three discs with different mass moments of inertia (J 1,
J2 and J3) coupled by two torsion shafts with stiffness K2 and K3 respectively.
[1]
For the schematic above, the free body diagram can be schematized according to the next figure.
Leading us to an equation which the roots represent the natural frequencies:
K
¿
(¿ 3−ω2 J ¿)=0
3
−K 2 ¿
(K 1 + K −ω
2
2
J 1
)[ ( K1 + K 3 −ω2 J 2
)( K 3 −ω2 J 3 ) −K 23] + K ¿2
Where:
Kx Stiffness of Shaft x
θ1 is the angular rotation for the disc 1
θ2 is the angular rotation for the disc 2
θ3 is the angular rotation for the disc 3
J 1 is the mass moment of inertia for disc 1 J
J3
2 is the mass moment of inertia for disc 2
is the mass moment of inertia for disc 3
Solving the previous equation will give the value of all natural frequencies. It is clear to see that as the
number of degrees increases, the complexity also increases. The use of Matlab® software can be useful to
solve these equations quickly.
According to Dunkerley´s method no shaft can ever be perfectly straight or perfectly balanced; moreover, a
good estimation for the natural frequency can be given by:
n
1 1
≅
ω2
∑ω 2
1 i=1 ii
Where:
ω=
ii √ K1
J1
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