Free Vibrations of TDOF Systems: Analysis
The oscillating beams illustrated in Figure Q1 each have a moment of inertia of J
about their pivots. Derive the differential equations of motion and write them in
matrix notation.
Further, solve for the natural mode frequencies and the
corresponding amplitude ratios.
2l
k
l
2
k
1
l
2l
k
Figure Q1
Solution
The free body diagram of the lower beam is illustrated below. Applying Newton’s
second law of motion to the lower beam leads to the following equation:
J1  kl 2 1   2   4kl 2 1 


 5kl 2 1  kl 2  2
Thus, the equation of motion for the lower beam is as follows:
J1  5kl 2 1  kl 2  2  0
(1)
Free Vibrations of TDOF Systems: Analysis
kl 1  2 
1
l
2kl1
2l
Free Body Diagram of the Lower Beam
The free body diagram of the upper beam is also shown below.
2l
l
2kl2
2
kl 1  2 
Free Body Diagram of the Upper Beam
Applying Newton’s second law of motion to the upper beam leads to the following
equation:
J2  kl 2 1   2   4kl 2  2 


 kl 2 1  5kl 2  2
Thus, the equation of motion for the lower beam is as follows:
J 2  kl 2 1  5kl 2  2  0
(2)
The equations of motion, written in matrix notation are the following:
J
0

0  1  5kl 2
  
J   2   kl 2
kl 2   1  0
    
5kl 2   2  0
(3)
In the natural modes, the solutions to the equations of motion may be represented as
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Free Vibrations of TDOF Systems: Analysis
follows:
i (t )  Ai sin n t

 (t )   2 A sin  t 
i
n
i
n 
(4)
Substituting equations (4) into equations (1) and (2) leads to the following:
5kl

2
 n J A1  kl 2 A2  0

2
kl 2 A1  5kl 2  n J A2  0
2


(5)
Equations (5) may be re-written as a single matrix equation, as follows:

 5kl 2  n 2 J

kl 2


  A1  0
 
2 
5kl 2  n J   A2  0

kl 2

(6)
Application of Cramer’s rule to the above equation leads to the following:
5kl
 k l
2
 0

4
2
2
2
2 4
J n  10 Jkl n  24k l  0
2
2
 n J
2 4
(7)
Equation (7) can now be solved for the two natural frequencies, as follows:
10 Jkl 2  100 J 2 k 2l 4  96 J 2 k 2l 4 
n1 

2J 2

2
2
2

10 Jkl  2 Jkl
6kl



J
2J 2

2

6kl

n1 

J
2
(8)
Similarly:
n 2
2
n 2
10 Jkl 2  2 Jkl 2 4kl 2 



J 
2J 2

4kl 2



J
(9)
To obtain the corresponding ratios of amplitudes, the expressions of natural mode
frequencies can be substituted into either of the equations (5). Thus:
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Free Vibrations of TDOF Systems: Analysis
5kl
Similarly:
5kl
2
2

 6kl 2 A11   kl 2 A21



A11  kl 2 
u1 

 1
A21  kl 2 
(10)
 4kl 2 A12   kl 2 A22



A12  kl 2
u2 
 2  1
A22
kl

(11)
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