Etude vibratoire poutre sur appuis simple

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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 rpm24 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 dampersbetween 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 resonanceis 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 motorfrequency. We can consider the beam is on his resonance mode. With this result, the goal of the project should be easier to reach.

MOTOR ECCENTRICITY

Now, we know that the motor works at the same frequency as the resonance of the beam. The second step consists in understanding the origins of those excessive vibrations.
Because the system is a rotating motor, we will solve it as anunbalanced mass problem with an eccentricity will have to calculate with the known data. So, after we know the eccentricity of the rotor, we’ll be able to modify the parameters we are allowed to change such as the shape of the section and the material of the beam or the damping ratio (by including extra dampers). That way, we will reduce the displacement of the beam and so the force transmitted to thebase.

Let’s consider the studied system as the following schematic system.

e

t
m
ms
me

We are in a case of rotating unbalance. According to the book (p.54) the equation of this system is:

X=mreω2(k-Mω2)2+(cω)2

e=X×(k-Mω2)2+(cω)2mrω2

Let’s find c and k :

k=3EILl2×l2=3×2.1011×3,6.10-6×21=4,32.106 N/m

M=mt+me=105+90,92=195,92 kgc=ζcc=ζ×2kM=0,01×24,32.106×195,92=581,85

We now have all the data to find the eccentricity of the motor.

e=32.10-6×(4,32.106-195,92×150,82)2+(581,85×150,8)230×150,82e=7,57.10-6 m |

The eccentricity of the motor is 7,57 µm.

DYNAMIC FORCES

We know the eccentricity of the motor, so we can find the force it exercises and so the force transmitted to the beam by using the transmissibility formula:TR=1+(2ζλ)2(1-λ2)2+(2ζλ)2=1+(2×0,01×1,0155)2(1-1,01552)2+(2×0,01×1,0155)2=26,81 |

The Transmissibility of the system at 24 Hz is 26,81. Now, let’s calculate the force created by the motor in motion.

In this case, we are only interested in the vertical component of the centrifuge force create by the eccentric mass. According to the eccentricity and the frequency, we can find the acceleration it generates...
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