Response of MDOF systems
Degree of freedom (DOF): The minimum number of
independent coordinates required to determine completely the positions of all parts of a system at any instant of time.
Two DOF systems Three DOF systems
The normal mode analysis (EOM-1)
Example: Response of 2 DOF system
m 2m
k k k
x1 x2
FBD kx1 m k(x1-x2) 2m kx2
EOM kx1 k(x1 x2) mx1 2 2
2
1 ) 2
(x x kx mx
k
In matrix form, EOM is
0 0 2
2 2
0
0
2 1 2
1
x x k
k
k k
x x m m
EOM -2 (example)
0 0 2
2 2
0
0
2 1 2
1
x x k
k
k k
x x m
m
EOM
M x¨ K x F
) ( )
( )
( )
(t Cx t Kx t F t x
M In general form
M is the inertia of mass matrix (n x n) C is the damping matrix (n x n)
K is the stiffness matrix (n x n)
F is the external force vector (n x 1) x is the position vector (n x 1)
Synchronous motion
From observations, free vibration of undamped MDOF system is a synchronous motion.
• All coordinates pass the equilibrium points at the same time
• All coordinates reach extreme positions at the same time
• Relative shape does not change with time
2
1 x
x constant
time
x1 x2 x1
x2
No phase diff. between x1 and x2
)
1sin(
1 A t x
)
2 sin(
2 A t
x
) ( 1
Aej t
) ( 2
A ej t or
or
Response of 2DOF system (example-1)
0 0 2
2 2
0
0
2 1 2
1
x x k
k
k k
x x m m
EOM
Synchronous motion
Sub. into EOM
0 0 2
2 2
0
0
2 1 2
1 2
2
x x k
k
k k
x x m m
0 0 2
2 2
2 1 2
2
A A m
k k
k m
k
0 Kx
Mx
2 (t) (t)
0 x
M
K ) ( )
( 2 t
2 0 2
2
2 2
m k
k
k m
k det(K 2M) 0 Characteristic equation (CHE)
)
1sin(
1 A t x
)
2 sin(
2 A t
x
) ( 1
Aej t
) ( 2
A ej t or
or
Response of 2DOF system (example-2)
2 0 2
2
2
2
m k
k
k m
k 0
2 3 3
2 2
4
m
k m
k
m 634 k .
1 0
m
366 k .
2 2
CHE
Solve the CHE Natural frequencies
of the system
;
0 0 2
2 2
2 1 2
2
A A m
k k
k m
k
k
m k
m k
k A
A 2
2 2
1 2 2
2
1
731 . 0 )
634 . 0 ( 2
) 1 (
2
1
m m k k
k A
A
2
73 . 2 )
366 . 2 ( 2
) 2 (
2
1
m m k k
k A
A From
Response of 2DOF system (example-3)
1
731 . 0
) 1 (
2
1
A
A 2.73
) 2 (
2
1
A
Amp. ratio Amp. ratio A
1
731 .
) 0
1(x
1
73 . ) 2
2(x
The first mode shape The second mode shape
0.731 1 1
2
same direction Opposite
direction
Response of 2DOF system (example-4)
In general, the free vibration contains both modes
simultaneously (vibrate at both frequencies simultaneously)
) 1 sin(
73 . ) 2
1 sin(
732 . 0
2 2
2 1
1 1
2
1
c t c t
x x
2 1
2
1, c , ,
c are constants (depended on initial conditions)
Initial conditions (1)
) 1 sin(
73 . ) 2
1 sin(
732 . 0
2 2
2 1
1 1
2
1
c t c t
x x
4 2 )
0 (
) 0 (
2 1
x x
0 0 )
0 (
) 0 (
2 1
x x
Initial conditions and
) 1 cos(
73 . ) 2
1 cos(
732 . 0
2 2
2 2 1
1 1
1 2
1
c t c t
x x
Velocity response
4 2 )
0 (
) 0 (
2 1
x x
2 2
1
1 sin
1 73 . sin 2
1 732 . 0 4
2
c c
0 0 )
0 (
) 0
1( x x
2 2
2 1
1
1 cos
1 73 . cos 2
1 732 . 0 0
0
c c
Initial conditions (2)
2 2
1
1 sin
1 73 . sin 2
1 732 . 0 4
2
c c
2 2
2 1
1
1 cos
1 73 . cos 2
1 732 . 0 0
0
c c
4 Eqs.,
4 unknowns
Solve for four unknowns ,
732 .
1 3
c c2 0.268, 1 2 /2
2) 1 sin(
73 . 268 2
. 0 2)
1 sin(
732 . 732 0 .
3 1 2
2
1
t t
x x
The response is
t x t
x
2 1
2
1 cos
268 . 0
732 . cos 0
732 . 3
732 .
2
Initial conditions (3)
) 1 sin(
73 . ) 2
1 sin(
732 . 0
2 2
2 1
1 1
2
1
c t c t
x x
2 464 . 1 )
0 (
) 0 (
2 1
x x
0 0 )
0 (
) 0 (
2 1
x x
(a) Initial conditions and
1 73 . 2 )
0 (
) 0
1( x
x
0 0 )
0 (
) 0
1( x
x
(b) Initial conditions and
Try to do
Summary (Free-undamped) (1)
0 Kx
x
M(t) (t)
The motion is synchronous:
constant and
0 Kx
Mx
2 (t) (t) 0 x
M
K ) ( )
( 2 t
Eigen value problem
0 )
det(K 2M
Characteristics equation
2
n Eigen value
nN n
n
1, 2,, N natural freq.
0 x
M
K ni ) i
( 2
xi Eigen vector )
sin(
A t
x or Aej(t)
N mode shapes
xN
x
x1, 2,, 1 EOM
2
3
4 5
Direct Method
Summary (Free-undamped) (2)
Free-undamped response
) sin(
) sin(
) sin(
)
(t x1A1 1t 1 x2A2 2t 2 xN AN Nt N
x
N
i
i i
i
iA t
t
1
) sin(
)
( x
x 6
where A and are from initial condition x(0) and v(0)
Direct Method
Example
k1
k2
x
l2
l1
Determine the normal modes of vibration of an automobile simulated by simplified 2-dof system with the following numerical values
lb
3220 W
ft 5 .
1 4
l k1 2400 lb/ft
ft 5 .
2 5
l k2 2600 lb/ft
ft
4
2 r g r JC W
Rigid body mode
• Rigid body mode is the mode that the system moves as a rigid body.
• The system moves as a whole without any relative motion among masses.
• There is no oscillation.
n 0
Forced harmonic vibration (1)
Example
EOM F t
x x k
k
k k
x x m
m
sin
0 0
0 1
2 1 22
21
12 11
2 1 2
1
System is undamped, the solution can be assumed as
X t X x
x
sin
2 1 2
1
Sub. into EOM
0
1 2
1 2
2 22
21
12 2
1
11 F
X X m
k k
k m
k
) 0
( 1
2
1 F
X Z X
Simpler notation,
) 0
( 1 1
2
1 F
X Z X
Forced harmonic vibration (2)
) 0 (
) ( adj ) 0
( 1 1 1
2
1 F
Z F Z
X Z X
) )(
( )
( m1m2 12 2 22 2 Where Z
1 and 2 are natural frequencies
) 0 (
1 1
2 1 11
21
12 2
2 22
2
1 F
m k
k
k m
k X Z
X
The amplitudes are
) )(
(
) (
2 2
2 2
2 1 2 1
1 2 2 22
1
m m
F m
X k
) )(
( 2 2 2 2
1 21
2
m m
F X k
Forced harmonic vibration (3)
) )(
(
) 2
(
2 2
2 2 2
1 2
1 2
1
m
F m
X k
) )(
( 12 2 22 2
2
1
2
m X kF
m m
k k k
x1 x2
F1sint
F t x
x k k
k k
x x m
m
sin
0 2
2 0
0 1
2 1 2
1
m k m
k 3
, 2
1
EOM
Force response of a 2 DOF system
0 1 2 3
0 1 2 3 4 5
-1 -2 -3 -4 -5
F Xk
1
1 2
F k X
1 1
F k X
1
2
Same direction
Opposite direction
Solving methods
Modal analysis
• is a method for solving for both transient and steady state responses of free and forced MDOF systems through
analytical approaches.
• Uses the orthogonality property of the modes to
“decouple” the EOM breaking EOM into independent SDOF equations, which can be solved for response separately.
Introduction
Coordinate coupling
k1 k2
l2
l1
mg
x Ref.
)
( 1
1 xl k
)
( 2
2 xl k
k1 l1 mg k2
x1 Ref.
1 1x k
) ( 1
2 x l k
0 0 0
0
2 2
1 1 2 2 2
1 x
l k l k l k l k
l k l k k
k x
J m
0
1 0
2 2 2
1 1
1 x
l k l
k
l k k
k x
J ml
ml m
Concept of modal analysis
0 ) ( 0
0
2 2 2 2 1 1 1 1 2 2
1 1 2 2 2
1 x F t
l k l k l k l k
l k l k k
k x
J m
Mx(t)Kx(t) F(t)
) (
) ( 0
0 1
0 0 1
2 1 2
1 2
2 2
1 2
1
t N
t N r
r r
r
n n
r(t) Λr(t) N(t)
EOM in physical coordinate (Coordinates are coupled)
EOM in modal coordinate (Independent SDOF equations)
Solve for r(t)
Transform r(t) back to x(t)
Orthogonality
x = eigen vector (vector of mode shape)
TMx M
x
j
i i
T
jMx 0,
x xTjKxi 0, i j
TKx K
x
If M and K are symmetric and then xi and xj are said to be “orthogonal” to each other. ni nj
Normalization
u = normalized eigen vector (respect to mass matrix)
1
i T
i Mu
u
j
i i
T
jMu 0, u
0 x
M
K ) ( )
( 2 t
From eigen value problem
constant is
, C
C i
i x
u
0 u
M
K ) ( )
( 2 t
or Kui i2Mui
2 2
i i
T i i i
T
i Ku u Mu
u
Modal matrix
Modal matrix is the matrix that its columns are the mode shape of the system
u u un
U 1 2 Then
1 0
0
0 1
0
0 0
1
I
MU UT
2 2
2 2
1
0 0
0 0
0 0
nN n
n T
KU
U
Λ (Spectral matrix)
Modal analysis (undamped systems)-1
1. Draw FBD, apply Newton’s law to obtain EOM 2. Solve for natural frequencies through CHE 3. Determine mode shapes through EVP
4. Construct modal matrix (normalized)
Procedures
) ( )
( )
(t Kx t F t x
M 0 x
M
K ) ( )
( 2 t
0 ) det(K 2M
u u un
U 1 2 I
MU UT
Λ KU
UT
5. Perform a coordinate transformation x(t) Ur(t) )
( )
( )
(t Kx t F t x
M MUr(t)KUr(t) F(t)
) ( )
( )
(t T t T t
TMUr U KUr U F
U
) ( )
( )
(t Λr t UTF t r
