Mechanical Vibration: 3 Degree of Freedom

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SUMMARY

The discussion focuses on solving a mechanical vibration problem involving a three-degree-of-freedom system using MATLAB. Key equations include the kinetic energy formula T = 1/2 * I * (theta_dot)^2 and potential energy formula V = 1/2 * Kt * theta^2. The user encounters an issue where the first fundamental frequency is calculated as zero, which is typical for positive semi-definite systems. The solution requires a correct setup of equations of motion and a complete formulation of kinetic and potential energy expressions.

PREREQUISITES
  • Understanding of mechanical vibrations and degrees of freedom
  • Proficiency in MATLAB for numerical computations
  • Familiarity with kinetic and potential energy equations in mechanical systems
  • Knowledge of free-free boundary conditions in vibration analysis
NEXT STEPS
  • Review the derivation of equations of motion for multi-degree-of-freedom systems
  • Learn about positive semi-definite systems and their implications in vibration analysis
  • Explore MATLAB functions for eigenvalue problems, specifically the 'eig' function
  • Study the formulation of free body diagrams (FBD) for mechanical systems
USEFUL FOR

Mechanical engineers, students studying dynamics and vibrations, and researchers working on multi-degree-of-freedom systems will benefit from this discussion.

Toni911
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Homework Statement


http://i.imgur.com/wJu0AEs.jpg
http://i.imgur.com/J9sEWgq.jpg?1

Homework Equations


Kinetic Energy = T = 1/2 * I * (theta_dot)^2
Potential Energy = V = 1/2 * Kt * theta^2

The Attempt at a Solution


Matlab Code and Results:
>> I1=3600;
I2=200;
I3=800;
I4=4800;
d1=0.15;
L1=3.5;
d2=0.2;
L2=3;
G=80e9;
J1=(pi/32) *d1^4
J2=(pi/32) *d2^4
Kt1= (J1*G)/L1
Kt2 =(J2*G)/L2
II = [I1 0 0;0 I2+I3 0;0 0 I4]
KK=[Kt1 -Kt1 0;-Kt1 Kt1+Kt2 -Kt2;0 -Kt2 Kt2]

J1 =

4.9701e-05J2 =

1.5708e-04Kt1 =

1.1360e+06Kt2 =

4.1888e+06II =

3600 0 0
0 1000 0
0 0 4800KK =

1.0e+06 *

1.1360 -1.1360 0
-1.1360 5.3248 -4.1888
0 -4.1888 4.1888

>> AA = inv(II)*KK

AA =

1.0e+03 *

0.3156 -0.3156 0
-1.1360 5.3248 -4.1888
0 -0.8727 0.8727

>> [V,D] = eig(AA)

V =

0.0538 0.7956 0.5774
-0.9849 -0.2764 0.5774
0.1648 -0.5391 0.5774D =

1.0e+03 *

6.0878 0 0
0 0.4252 0
0 0 0.0000

The problem is that I am getting lmbda1=0 which means the first fundamental frequency is zero ... and it is mentioned in the second part of the problem that f2=3.89 Hz and In my solution it does not equal that.
Can you please go through my steps and tell me what is wrong.
 
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You will always get one zero natural frequency because this is a positive semi-definite system (also called a free-free system, referring to the BC). This is associated with rigid body rotation; don't worry about it.

I don't think you have this set up correctly. You did not write out your equations of motion (this would really help), but your MatLab code does not look correct. Your kinetic energy expression is incomplete, as is your potential energy expression.

Draw a FBD for each shaft, including the gears mounted on that shaft. Write the equation of motion for each FBD, including a tangential force at the geared connection. Next write the kinematic relation that must exist at the geared connection, and use this to eliminate the tangential force. You should wind up with the equations for one shaft "transformed," that is, expressed as though it were rotating at the same speed as the first shaft.