Matakuliah
Tahun
: Dinamika Struktur & Teknik Gempa
: S0774
SINGLE DEGREE OF FREEDOM SYSTEM
Equation of Motion, Problem Statement & Solution
Methods
Pertemuan 19
Systems with two degree of freedom
Recap
Analysis of various 2DOF systems such as:
Linear systems
Torsional systems
Definite and semi-definite systems
Pendulum systems (double pendulum)
String systems
Systems with two degree of freedom
Co-ordinate coupling
Static coupling
Dynamic coupling
Principal co-ordinates
Systems with two degree of freedom
Problem-1
m,J
Obtain the equations of motion of
the system shown in the figure.
The vibration is restricted in
plane of paper
G
K1
a
K2
b
m -mass of the system
J -mass MI of the system
G -centre of gravity
Systems with two degree of freedom
Problem-1
m,J
Static equilibrium
line
G
(x-a)
K2
K1
a
b
b
a
x
G
(x+b)

The system has two generalized
co-ordinates, x and 
Cartesian (x), Polar ()
Systems with two degree of freedom
Problem-1
Eqns. of motion
Static equilibrium
line
(x-a)
The Lagrange’s equation is :
b
a
x
G
(x+b)

d  T  T U

 

 Qi
dt  x i  xi xi
1
1 2
2

T  mx  Jθ
2
2
1
2
U  K 1x - aθ 
2
generalized co-ordinates
x 
xi    
1
2
θ 
 K 2 x  bθ 
2
Dr. S. K. Kudari, Professor, BVB College of Engg. & Tech., Hubli
Systems with two degree of freedom
Problem-1
The Lagrange’s equation is :
d  T  T U

 Q1
 
dt  x  x x
1
1 2
2

T  mx  Jθ
2
2
1
1
2
2
U  K 1x - aθ   K 2 x  bθ 
2
2
x 
xi    
θ 
d  T 
   mx
dt  x 
T
0
x
U
 K1x - aθ(1)
x
 K 2 x  bθ(1)
Dr. S. K. Kudari, Professor, BVB College of Engg. & Tech., Hubli
Systems with two degree of freedom
Problem-1
equations of motion (1st)
U
 K 1 x - aθ (1)  K 2 x  bθ (1)
x
U
 K 1x - K 1aθ  K 2 x  K 2bθ
x
U
 x(K 1  K 2 )  θ(K 1a - K 2b)
x
mx  (K1  K2 )x - (K1a - K2b)θ  0
First Eqn.
motion
of
Systems with two degree of freedom
Problem-1
equations of motion (2nd)
d  T  T U

 Q1
 
dt  θ  θ θ
T
1
1
mx 2  Jθ 2
2
2
1
1
2
2
K 1x - aθ   K 2 x  bθ 
2
2
x 
xi    
θ 
U
d  T 


  Jθ
dt  θ 1 
T
0
θ
U
 K 1 x - aθ ( a)  K 2 x  bθ (b)
θ
U
 K 1xa  K 1a 2θ  K 2 xb  K 2b 2θ
θ
U
 (K 1a  K 2b)x  (K1a 2  K 2b 2 )θ
θ
Systems with two degree of freedom
Problem-1
equations of motion
  (K a  K b)x  (K a2  K b2 )θ  0 Second Eqn. of
Jθ
1
2
1
2
motion
equations of motion are:
mx  (K1  K2 )x - (K1a - K2b)θ  0
First
  (K a  K b)x  (K a2  K b2 )θ  0 Second
Jθ
1
2
1
2
Systems with two degree of freedom
Problem-1
equations of motion
mx  (K1  K2 )x - (K1a - K2b)θ  0
  (K a  K b)x  (K a2  K b2 )θ  0
Jθ
1
2
1
2
Matrix form
m 0 x
 0 J   

 θ 
 (K 1a  K 2b)  x  0
 (K 1  K 2 )
 (K a  K b) (K a 2  K b 2 ) θ   0
1
2
1
2

   
Systems with two degree of freedom
Problem-1
Matrix form
 (K 1a  K 2b)  x  0
m 0 x  (K 1  K 2 )
 0 J     (K a  K b) (K a 2  K b 2 ) θ   0

 θ  
1
2
1
2
   
Stiffness matrix
Mass/inertia matrix
Stiffness matrix shows that co-ordinate x and  are dependent
on each other. Any change in x reflects in change in 
As seen from the matrix, the equations of motion are coupled
with stiffness. This condition is referred as STATIC COUPLING
coupling in mass matrix is referred as DYNAMIC COUPLING
Systems with two degree of freedom
Problem-1
Matrix form
 (K 1a  K 2b)  x  0
m 0 x  (K 1  K 2 )
 0 J     (K a  K b) (K a 2  K b 2 ) θ   0

 θ  
1
2
1
2
   
Mass/inertia matrix
Stiffness matrix
From the above equations, it can be seen that system do not
have dynamic coupling
But, it has static coupling
Systems with two degree of freedom
Problem-1
Matrix form
 (K 1a  K 2b)  x  0
m 0 x  (K 1  K 2 )
 0 J     (K a  K b) (K a 2  K b 2 ) θ   0

 θ  
1
2
1
2
   
Mass/inertia matrix
Stiffness matrix
To have static uncoupling the condition to be satisfied is:
K1a=K2b
Systems with two degree of freedom
Problem-1
Matrix form
0
 x  0
m 0 x (K 1  K 2 )
 
 0 J    
2
2  
0
(K 1a  K 2b ) θ  0

 θ  
The uncoupled Eqns. of motion are
mx  (K1  K2 )x  0 Contains only one coordinate, x
  (K a 2  K b 2 )θ  0 Contains only one coordinate, 
Jθ
1
2
Under such conditions, x and  are referred as
PRINCIPAL COORDINATES
Systems with two degree of freedom
Problem-1
Solution to uncoupled Eqns. of motion:
From Eqn.1:
From Eqn.2:
mx  (K1  K2 )x  0
  (K a 2  K b 2 )θ  0
Jθ
1
2
K  K2 
x   1
x  0
 m 
2
2


K
a

K
b
2
θ   1
θ  0

J


ω1 
K1  K 2
m
ω2 
K1a2  K 2b2
J
Systems with two degree of freedom
Problem-2
m,J
C
G
e
K1
a
K2
b
Obtain the equations of motion of
the system shown in the figure.
The centre of gravity is away from
geometric centre by distance e
The vibration is restricted in plane
of paper
m -mass of the system
J -mass MI of the system
G -centre of gravity
C -centre of geometry
Systems with two degree of freedom
Problem-2
Static equilibrium
line
K1(x-a)
a
x
G
b
x+e
C

K1(x+b)
Due to some eccentricity e, the changes are:
x=x+e
J=J+me2
Substitute in Eqns. of motion of earlier problem having e=0:
Systems with two degree of freedom
Problem-2
equations of motion for system having e=0
mx  (K1  K2 )x - (K1a - K2b)θ  0
  (K a  K b)x  (K a2  K b2 )θ  0
Jθ
1
2
1
2
Substitute x=x+e and J=J+me2=Jn in above Eqns.
)  (K  K )x - (K a - K b)θ  0
m(x  eθ
1
2
1
2
  (K  K )x - (K a - K b)θ  0
mx  me θ
1
2
1
2
  (K a  K b)x  (K a 2  K b 2 )θ  0
Jnθ
1
2
1
2
Systems with two degree of freedom
Problem-2
New equations of motion are
  (K  K )x - (K a - K b)θ  0
mx  me θ
1
2
1
2
  (K a  K b)x  (K a 2  K b 2 )θ  0
Jnθ
1
2
1
2
Matrix form
m me  x
 0 J    
n  θ 

Dynamic
coupling
 (K 1a  K 2b)  x  0
 (K 1  K 2 )
 (K a  K b) (K a 2  K b 2 ) θ   0
1
2
1
2

   
Static coupling
Systems with two degree of freedom
Problem-3
a
a
K
m
a
m
Derive expressions for two
natural frequencies for small
oscillation of pendulum shown in
figure in plane of the paper.
Assume rods are rigid and mass
less
Systems with two degree of freedom
Problem-3
Equilibrium diagram
a
a
1 a
2
a
K
m

J1θ
1
a
m
mg
Ka(2-1)
a

J2θ
2
mg
Systems with two degree of freedom
Problem-3
For first mass
  mg(asinθ )  Ka(θ  θ )(acosθ )  0
J1θ
1
1
2
1
1
1 a
2
a
as  is smaller
  mgaθ  Ka2 (θ  θ )  0
J1θ
1
1
2
1

J1θ
1
mg
Ka(2-1)
  mgaθ  Ka2θ  Ka2θ  0
(ma 2 )θ
1
1
2
1
a

J2θ
2
  (mg  Ka)θ  Kaθ  0
ma θ
1
1
2
First Eqn. of motion
mg
Equilibrium diagram
Systems with two degree of freedom
Problem-3
For second mass
  mg(2asinθ )  Ka(θ  θ )(acosθ )  0
J2θ
2
2
2
1
2
1 a

J1θ
1
mg
2
a
as  is smaller
m(2a) θ
2
Ka(2-1)
1
 2mgaθ2  Ka2θ2  Ka2θ1  0
a
  Kaθ  (2mg  Ka)θ  0
(4ma) θ
1
1
2

J2θ
2
mg
Equilibrium diagram
Second Eqn. of motion
Systems with two degree of freedom
Problem-3
  (mg  Ka)θ  Kaθ  0
ma θ
1
1
2
First Eqn. of motion
  Kaθ  (2mg  Ka)θ  0 Second Eqn. of motion
(4ma) θ
1
1
2
Eqns. of motion in matrix form

0 
ma
θ1 

 0 4ma    


θ2 

 Ka  θ1  0
(mg  Ka)
  
  Ka

(2mg  Ka) θ2  0

For static coupling Ka=0, which is not possible
Systems with two degree of freedom
Problem-3
Equations of motion
  (mg  Ka)θ  Kaθ  0
ma θ
1
1
2
First Eqn. of motion
  Kaθ  (2mg  Ka)θ  0 Second Eqn. of motion
(4ma) θ
1
1
2
Solution to governing eqns.:
Assume SHM
x1  A1sinωt  φ
x2  A 2sin(ωt  φ)
The above equations have to satisfy the governing equations
of motions
Systems with two degree of freedom
String systems
(mg  Ka)  maω A sin(ω  φ)  KaA sin(ω  φ)  0
2
1
2


 KaA1sin(ω  φ)  (2mg  ka)  4maω2 A 2sin(ω  φ)  0
In above equations sin(ωt  φ)  0
Characteristic Eqns.:
(mg  Ka)  maω A  KaA  0
 KaA  (2mg  ka)  4maω A  0
2
1
2
2
1
2
Systems with two degree of freedom
String systems
(mg  Ka)  maω ........  Ka
2
 Ka
(2mg  Ka)  4maω
2
0
The above equation is referred as a characteristic determinant
Solving, we get :
(mg  Ka)  maω (2mg  Ka)  4maω   Ka
2
2
2
Frequency equation
Systems with two degree of freedom
String systems
Solve the frequency Eqn. for Natural frequencies of the
system
ω1....and... ω2
As the system has two natural frequencies, under certain
conditions it may vibrate with first or second frequency,
which are referred as principal modes of vibration
Summary
Co-ordinate coupling
Static coupling
Dynamic coupling
Principal co-ordinates
Thank You