Systems with Translational and Rotational
Displacements
Free-body diagram
kx2
kx1

L/4
G
L
x1
x
x2
Assume the center of mass G moves upwards by x and the rod
rotates counter clockwise around it by .
The equations of motion for x and  are
JG=ml2/12
l
l
J G  kx1  kx2
4
2
mx  kx1  kx2  k ( x1  x2 )
The geometric relationship between displacements is
x1  x 
l

4
x2  x 
l

2
The final equations of motion become
m

 0

0   x  2k
ml 2     
  kl



12
4
kl 
4  x  0

5kl 2  


 
16 
frequencies and modes ?
1
Vibration of an object off-centre mounted
k
K
G
O
m
eg
k
K
m J
x
K

m JG
e
G
Mass center
Center of flexure
k
e
c

y (z)
w
x
C
Vertical displacement
w  x  z  x  y  e
dF
y
z
Vertical acceleration
  x  z  x  y  e
w
dz (dy)
: density
The initial force on the infinitesimal element is
2
A: area
  Adyw

dF  (dm)w
Total vertical inertia force
F   dF A w dz  A w dy  m( x  e)
z
y
Total moment of inertia about C
 dz  A ( y  e) w
 dy  ( J G  me 2 )  mex
M G    zdF   A zw
z
y
Equations of motion
m( x  e)  kx  0
kx
( J G  me2 )  mex  K  0
K
Jc
Matrix equation of motion
 m me  x k 0   x 
me J    0 K     0
C  

 

The system is mass-coupled.
Any object attached at a point, which is not its center of mass, vibrates in
both translation and rotation. One motion can affect the other. This is the
basic mechanism for flutter type of dynamic instability to occur. Onedegree-of-freedom systems will not produce flutter.
3
Force Vibration of Two-Degrees-of-Freedom Systems
k1
f1 sint
m1
m1 0   x1  k1  k2  k2   x1   f1 sin t 
 0 m  x     k
x    0 
k

2  2  
2
2  2  

The solution of this undamped system is
k2
 x1 (t )   X 1 

    sin  t
 x2 (t )  X 2 
m2
One gets
k1  k 2  m1 2

 k2

  X 1   f1 
 
2 
k 2  m2    X 2   0 
 k2
It can be found that
( k 2  m2 2 ) f 1
X1 
m1 m2 (12   2 )( 22   2 )
X2 
k 2 f1
m1 m2 (12   2 )( 22   2 )
where 1 and 2 are the two natural frequencies of the system.
It can be seen from the above expression that when the excitation
frequency  equals either of 1 and 2 , resonance takes place.
4
Example:
k
k
Let m1  m2  m, k1  k 2  k . This gives 1  0.618 m , 2  1.618 m .
The forced response for this system is shown below
kX/f
30
20
10
for X2
0
0
1
2
3
4
for X1
/ 1
-10
-20
-30
Vibration Absorber
(k 2  m2 2 ) f1
X1 
m1m2 (12   2 )(22   2 )
If m2 and k 2 are chosen such that
the first mass does not vibrate at all.
5
k2
  , then X 1  0 , that is,
m2
So if a primary system is a one-degree-of-freedom mass-spring
( m1  k1 ) system, which is subjected to an oscillatory force of
excitation frequency  , then by attaching it with another massspring ( m2  k2 ) system with
k2
  , the primary system will not
m2
vibrate under this excitation. Therefore, the attached mass-spring
system serves as a vibration absorber.
k
k
1
2
Introduce 11  m and  22  m . Re-draw the response curve in
1
2
terms of a new frequency ratio below.
Indeed, a zero response is produced at  / 22  1 .
If damping is present in the primary system or the attached system,
a zero response cannot be achieved at  / 22  1 . However, a low
response is produced for a range of frequencies around  / 22  1 .
kX1 / f1
20
15
undamped absorber
original system
10
10% damped absorber
5
 / 22
0
0
1
2
6
3
Solving Equations of Motion Using Laplace Transforms
One-Degree-of-Freedom Systems
2 x(t )  ax (t )  4 x(t )  sin t ; x(0)  2 , x (0)  1 (where a is a constant)
Laplace transform
1
2( s 2 X  2s  1)  a( sX  2)  4 X  2
s 1
Work out the X(s) as
4 s  2  2a
1
X (s)  2
 2
2s  as  4 (2s  as  4)( s 2  1)
a
a2  4
a
2
(2 
)
s

1

a

s

4  a2
8  2a 2  4  a 2
4  a2

2
a
s
1
2
s  s2
2
nonhomogeneous part
forced vibration
homogeneous part
free vibration
Let us look at two cases where a  2 and a  2 .
(1) a  2 :
2.25s  3  0.25s  0.25
X ( s)  2

s s2
s2 1
2.25( s  0.5)  3  2.25  0.5 0.25s 0.25

 2  2
s 1 s 1
( s  0.5) 2  ( 1.75 ) 2
 2.25
s  0.5
1.875
0.25s 0.25


 2
2
( s  0.5) 2  ( 1.75 ) 2 ( s  0.5) 2  ( 1.75 ) 2 s  1 s  1
Obviously the system is stable.
Inverse Laplace transforms gives
x(t )  2.25 exp( 0.5t ) cos1.323 t  1.417 exp( 0.5t ) sin 1.323 t  0.25(sin t  cos t )
homogeneous part
free vibration
nonhomogeneous part
forced vibration
7
Obviously the response is bounded, and the above solution satisfies both
the original ordinary differential equation and the initial conditions.
(2) a  2 :
1.75s  1 0.25s  0.25

s2  s  2
s2 1
1.75( s  0.5)  1  1.75  0.5 0.25s 0.25

 2  2
s 1 s 1
( s  0.5) 2  ( 1.75 ) 2
X ( s) 
 1.75
s  0.5
0.125
0.25s 0.25


 2
2
( s  0.5) 2  ( 1.75 ) 2 ( s  0.5) 2  ( 1.75 ) 2 s  1 s  1
Obviously the system is unstable.
Inverse Laplace transforms gives
x(t )  1.75 exp( 0.5t ) cos1.323 t  0.0945 exp( 0.5t ) sin 1.323 t  0.25(sin t  cos t )
homogeneous part
free vibration
nonhomogeneous part
forced vibration
Obviously the response is unbounded, and the above solution satisfies both
the original ordinary differential equation and the initial conditions.
Two-Degrees-of-Freedom Systems
k1
A mass-spring system:
f1
m1
k2
f2
m2
The equations of motion are
m1x1  (k1  k2 ) x1  k2 x2  f1
m2 x2  k2 x1  (k2  k3 ) x2  f 2
The Laplace transforms are
(m1s 2  k1  k2 ) X 1  k2 X 2  F1
 k2 X 1  (m2 s 2  k2  k3 ) X 2  F2
Determine X1 (s) and X 2 (s) first and then get x1 (t ) and x2 (t ) .
8
k3