Multi-degree of Freedom Systems
• Motivation: Many systems are too complex
to be represented by a single degree of
freedom model.
• Objective of this chapter: Understand
vibration of systems with more than one
degree of freedom.
Free-vibration of undamped twodegree of freedom system
• We learn how to analyze free vibration by
considering an example
x1
x2
k3
k2
k1
m1
m2
Deriving equations of motion
k1x1
k2(x1-x2)
m1
k2(x2-x1)
m2
k3x2
m1x1  ( k1  k2 ) x1  k2 x2  0
m2 x2  k2 x1  ( k2  k3 ) x2  0
Interpretation of coefficients
First equation: k1+k2 is the force on m1 needed to move it slowly by one unit while m2 is
held stationary. –k2 is the force on m1 need to hold it steady if m2 is displaced slowly by one
unit.
Second equation: k1+k2 is the force on m2 needed to move it slowly by one unit while m1 is
held stationary. –k2 is the force on m2 need to hold is steady if m1 is displaced slowly by
one unit.
Maxwell reciprocity theorem: the force on m1 need to hold it steady if m2 is displaced slowly
by one unit=force on m2 need to hold is steady if m1 is displaced slowly by one unit.
Special case
• Let m1=m2=m and k1=k2=k3=k. Then:
mx1  2kx1  kx2  0
mx2  kx1  2kx2  0
Solution of equations of motion
We know from experience that:
x1(t )  A sin( t )
x2 (t )  B sin( t )
Substituting the above equation to eq. of motion, we
obtain two eqs w.r.t. A, B :
(  mA 2  2kA  kB) sin t  0
(  mB 2  2kB  kA) sin t  0
The above two equations are satisfied for every t
 mA 2  2kA  kB  0
 mB 2  2kB  kA  0
Trivial solution A, B=0. In order to have a nontrivial solution:
2k  m 2
det 
  k


0
2
2k  m 
k
k
3k
,
m
m
First natural frequency:

k
 AB
m
The displacement for the first natural frequency is:
 A
 
 A
This vector is called mode shape. Constant cannot be determined.
Usually assume first entry is 1. Therefore, mode shape is: 1
1
Thus, both masses move in phase and the have
the same amplitudes.
Second natural frequency:

3k
 A  B
m
Mode shape for above frequency:
 A 



A


Usually assume first entry is 1. Therefore, mode shape is:
 1
 
 1
The two masses move in opposite directions.
Displacements of two masses are sums of displacements
in the two modes:
 x1(t ) 
1
1
k
3k
t 1)  c2   sin(
t  2 )
  c1  sin(

x
(
t
)
1

1
m
m
 
 
 2 
General expression for vibration of the
two-degree of freedom system
 x1(t ) 
 A1 
 A2 
  c1  sin( 1t 1)  c2   sin( 2t  2 )

 x2 (t ) 
 B1 
 B2 
Observations:
• System motion is superposition of two
harmonic (sinusoidal) motions with
frequencies 1 and 2.
• Participation of each mode depends on
initial conditions.
• Four unknowns (c1, c2, 1, 2) can be
found using four initial conditions.
• Can find specific initial conditions so that
only one mode is excited.
How to solve a free vibration problem
involving a two degree of freedom system
1) Write equations of motion for free vibration (no external force or moment)
2) Assume displacements are sinusoidal waves, and plug in equations of
motion: Obtain equation:
 A 0
[C ( )] *     
 B  0
where
[C( )]  [K ]   2[ M ]
3) Solve det[C(ω)]=0, obtain two natural frequencies, ω1, ω2.
4) Solve,
A  0
[C(1)]*     
 B  0
Assume A=1, and obtain first mode shape. Obtain second mode shape in a
similar manner.
Free vibration response is:
 x1(t ) 
 A1 
 A2 
  c1  sin( 1t 1)  c2   sin( 2t  2 )

 x2 (t ) 
 B1 
 B2 
5) Find constants from initial conditions.