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Mechanical vibrations

Teacher:
Mr.XX xxxx Students:
Mr. XX xxxxx
Mr. LE SANT Dimitri

PROJECT ONE: VIBRATION SUPPRESSION

L w h l Beam :
L = 2m w = 0,20 m h = 0,06 m
Density : ρ = 7800 kg/m3
Young Modulus E = 2 x 1011 Pa
Effective mass me = 0,4857 x mb

Motor :
Mass motor mm = 30 kg
Mass support ms = 75 kg
Total mass mt = 105 kg
Rotational speed = 1440 rpm 24 Hz

Equivalent damping and mesured vibration : ζ = 0,01
Displacement amplitude of the beam Db = 32 µm

Requirement:
1: Reducing the displacement amplitude to less than 3 µm
2: Reducing the dynamic force transmitted to the base to less than 10 % of the original

SuMMARY I. BEAM RESONANCE FREQUENCY 2 II. MOTOR ECCENTRICITY 3 III. DYNAMIC FORCES 4 IV. SOLUTIONS 5 V. ANNEX: matlab code 2

BEAM RESONANCE FREQUENCY

The first step of this study is to find the frequency of resonance of the beam in those conditions. If it is close to the motor frequency, we’ll easily figure out a solution to change the resonance frequency of the beam by changing the material or / and the shape of the section. We’ll also be able to include dampers between the motor support and the beam to reduce its vibrations.
If it’s far from the resonance frequency, the method will be the same but should be harder.

To find the resonance of the beam, we consider the motor and its support as a point mass upon the middle of the beam

mt=105 kg
L=2m
l=1m

According to the book Theory and Applications of Mechanical Vibration (p.25) the resonance is defined in that way:

ωn=48EI(mt+me)L3
Where:
-E = 2 x 1011 Pa
-mt = 105 kg
-me = 0,4857 x 7800 x 0,06 x 0,2 x 2 = 90,92 kg
-L = 2m
-I = wh312=0,20×0,06312 = 3,6.10-6

ωn=48EI(mt+me)L3=48×2.1011×3,6.10-6(105+90,92)×23=148,49 rad/s |

Fn=ωn2π=148,49 2π=23,63 Hz |

The natural frequency of the beam in those conditions of material, dimension and mass is very close to the motor