# Mechanical Vibration: 3 Degree of Freedom

## 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-05

J2 =

1.5708e-04

Kt1 =

1.1360e+06

Kt2 =

4.1888e+06

II =

3600 0 0
0 1000 0
0 0 4800

KK =

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 0.5774

D =

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.

## Answers and Replies

OldEngr63
Gold Member
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.