Mechanical Vibration Response of 3 Storey Building

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, and  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

This project is about mechanical vibration response of a 3 storey building. Here in below, there is a model of the system.

Modeling

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

 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