Mechanical Vibration Response of 3 Storey Building

vibration waves
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YILDIZ TECHNICAL UNIVERSITY
Department of Mechatronics Engineering

Lecture:
MKT-4201 MECHANICAL VIBRATIONS

 

Year/Semester: 

2017 – 2018/ Fall

 

Lecturer and Asistant:

Prof. Dr. Faruk YİĞİT   

 

Name Surname:Tunahan Vatansever-Alparslan Sipahi-Çağcan Koç

Contents:

Summary   

Project Defination   

Modeling   

Analysis  

Matlab Code   

Result and Conclusion   

 

Summary

 In this Project, we examined the vibration response of 3 storey building. We examined vibration response of each floor and vibration response of chandelier on the 3.floor.

First, we examined which method would be easier to extract the motion equation of the system.

  • Newton 2. Law Principle
  • D’Alembert Principle
  • Conservation of Energy Principle
  • Langrange Principle

   We decided that the Langrange method would be easier. Then we set the degree of freedom.
We solved our equations according to Langrange method and degree of freedom. Since our degree of freedom is 4, we obtained 4 different equations in the motion equation.  We performed our analyzes according to these 4 equations.

   We used our equation of motion to extract the state space form of our system. We found matrices A, B, C and D of state space. Then we used these matrices in the matlab to plot to see the changes of x1, x2, x3 and ∅ . To plot graphics in matlab, we have assumed the masses of the building, the damping ration and the spring ratio.

 

Problem Defination

  In this Project, we examined mechanical vibration response of 3 storey building. We examined vibration response of each floor and vibration response of chandelier on the 3.floor.

 

Modeling

We used langrange equation method to find out equation of motion of this system.

Langrange Equations
State Space Form
State Space Form

Matlab Code

 

function xdot=deprem(u)
x=u(1:6);
w=u(7);
wdot=u(8);

m1=35000;

m2=38000;

m3=42000;

m4=2;

c1=12000;

c2=13000;

c3=15000;

k1=600000;

k2=750000;

k3=800000; %System parameters;

A=[0 0 0 0 1 0 0 0;

   0 0 0 0 0 1 0 0;

   0 0 0 0 0 0 1 0;

   0 0 0 0 0 0 0 1;

-k1/m1 0 0 0 -c1/m1 0 0 0;

 k2/m2 -k2/m2 0 0 c2/m2 -c2/m2 0 0;

   0 (k2/m2)+m4+(m3/(m3+m4)) -k3/(m3+m4) 0 0 c3/(m3+m4) -c3/(m3+m4) 0

   0 0 0 0 0 0 0 0];

Bw=[0;0;0;0;1/m1;0;0;0];

Bw(:,2:8)=0;

Bwdot=[0;0;0;c1/m1;0;0;0;0];

Bwdot(:,2:8)=0;

Bwteta=[0;0;0;0;0;0;1;0];

Bwteta(:,2:8)=0;

Bwtetadot=[0;0;0;0;0;0;0;1];

Bwtetadot(:,2:8)=0;

C=[1 0 0 0 0 0 0 0;0 1 0 0 0 0 0 0;0 0 1 0 0 0 0 0;0 0 0 1 0 0 0 0];

D=[0;0;0;0];

xdot=(A+Bw+Bwdot+Bwteta+Bwtetadot);

plot(xdot);

 

Result and Conclusion

Response Graphic

2 Replies to “Mechanical Vibration Response of 3 Storey Building

  1. We are a bunch of volunteers and starting a brand new scheme in our community. Your website provided us with helpful information to work on. You’ve performed an impressive activity and our whole group might be grateful to you.

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