response of mdof systems - chulapioneer.netserv.chula.ac.th/~rchanat/2103433 intro mech...
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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 2mk k k
x1 x2
FBD m 2mkx1 k(x1-x2) kx2
EOM 1211 )( xmxxkkx &&=−−−
2221 2)( xmkxxxk &&=−−
In matrix form, EOM is ⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
−+⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡00
22
200
2
1
2
1
xx
kkkk
xx
mm
&&
&&
EOM -2 (example)
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
−+⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡00
22
200
2
1
2
1
xx
kkkk
xx
mm
&&
&&
x
EOM
M K Fx
)()()()( tttt FKxxCxM =++ &&&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
=21 xx constant
time
x1 x2 x1x2
No phase diff. between x1 and x2
)sin(11 φω += tAx)sin(22 φω += tAx
)(1
φω += tjeA)(
2φω += tjeA
oror
Response of 2DOF system (example-1)
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
−+⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡00
22
200
2
1
2
1
xx
kkkk
xx
mm
&&
&&EOM
Synchronous motion
Sub. into EOM
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
−+⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡
ω−ω−
00
22
200
2
1
2
12
2
xx
kkkk
xx
mm
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡
−−−−
00
222
2
12
2
AA
mkkkmkω
ω
0KxMx =+ω− )()(2 tt
0xMK =ω− )()( 2 t
022
22
2
=ω−−
−ω−mkk
kmk 0)det( 2 =ω− MK Characteristic equation (CHE)
)sin(11 φω += tAx)sin(22 φω += tAx
)(1
φω += tjeA)(
2φω += tjeA
oror
Response of 2DOF system (example-2)
022
22
2
=ω−−
−ω−mkk
kmk0
233
224 =⎟
⎠⎞
⎜⎝⎛+ω⎟
⎠⎞
⎜⎝⎛−ω
mk
mk
mk634.01 =ω
mk366.22 =ω
CHE
Solve the CHE Natural frequencies of the system;
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡
ω−−−ω−
00
222
2
12
2
AA
mkkkmk
kmk
mkk
AA 2
22
1 222
ω−=
ω−=
1ω=ω
731.0)634.0(2
)1(
2
1 =−
=⎟⎟⎠
⎞⎜⎜⎝
⎛
mmkk
kAA
2ω=ω
73.2)366.2(2
)2(
2
1 −=−
=⎟⎟⎠
⎞⎜⎜⎝
⎛
mmkk
kAA
From
Response of 2DOF system (example-3)
1ω=ω
731.0)1(
2
1 =⎟⎟⎠
⎞⎜⎜⎝
⎛AA 73.2
)2(
2
1 −=⎟⎟⎠
⎞⎜⎜⎝
⎛AA
⎭⎬⎫
⎩⎨⎧
=φ1731.0
)(1 x⎭⎬⎫
⎩⎨⎧−
=φ1
73.2)(2 x
Amp. ratio Amp. ratio
The first mode shape The second mode shape
0.731 1
-2.73
1
2ω=ω
same directionOpposite direction
Response of 2DOF system (example-4)
In general, the free vibration contains both modes simultaneously (vibrate at both frequencies simultaneously)
)sin(1
73.2)sin(
1732.0
2221112
1 ψ+ω⎭⎬⎫
⎩⎨⎧−
+ψ+ω⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
tctcxx
2121 ,,, ψψcc are constants (depended on initial conditions)
Initial conditions (1)
)sin(1
73.2)sin(
1732.0
2221112
1 ψ+ω⎭⎬⎫
⎩⎨⎧−
+ψ+ω⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
tctcxx
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
42
)0()0(
2
1
xx
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
00
)0()0(
2
1
xx&
&Initial conditions and
)cos(1
73.2)cos(
1732.0
222211112
1 ψ+ω⎭⎬⎫
⎩⎨⎧−
ω+ψ+ω⎭⎬⎫
⎩⎨⎧
ω=⎭⎬⎫
⎩⎨⎧
tctcxx&
&
Velocity response
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
42
)0()0(
2
1
xx
2211 sin1
73.2sin
1732.0
42
ψ⎭⎬⎫
⎩⎨⎧−
+ψ⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
cc
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
00
)0()0(
2
1
xx&
&222111 cos
173.2
cos1732.0
00
ψ⎭⎬⎫
⎩⎨⎧−
ω+ψ⎭⎬⎫
⎩⎨⎧
ω=⎭⎬⎫
⎩⎨⎧
cc
Initial conditions (2)
2211 sin1
73.2sin
1732.0
42
ψ⎭⎬⎫
⎩⎨⎧−
+ψ⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
cc
222111 cos1
73.2cos
1732.0
00
ψ⎭⎬⎫
⎩⎨⎧−
ω+ψ⎭⎬⎫
⎩⎨⎧
ω=⎭⎬⎫
⎩⎨⎧
cc
4 Eqs., 4 unknowns
Solve for four unknowns,732.31 =c ,268.02 =c 2/21 π=ψ=ψ
)2
sin(1
73.2268.0)
2sin(
1732.0
732.3 212
1 π+ω
⎭⎬⎫
⎩⎨⎧−
+π
+ω⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
ttxx
The response is
ttxx
212
1 cos268.0732.0
cos732.3732.2
ω⎭⎬⎫
⎩⎨⎧−
+ω⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
Initial conditions (3)
)sin(1
73.2)sin(
1732.0
2221112
1 ψ+ω⎭⎬⎫
⎩⎨⎧−
+ψ+ω⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
tctcxx
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
2464.1
)0()0(
2
1
xx
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
00
)0()0(
2
1
xx&
&(a) Initial conditions and
⎭⎬⎫
⎩⎨⎧−
=⎭⎬⎫
⎩⎨⎧
173.2
)0()0(
2
1
xx
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
00
)0()0(
2
1
xx&
&(b) Initial conditions and
Try to do
Summary (Free-undamped) (1)
0KxxM =+ )()( tt&&
The motion is synchronous: constant ω and φ
0KxMx =+ω− )()(2 tt
0xMK =ω− )()( 2 t
0)det( 2 =ω− MKCharacteristics equation
2nω Eigen value
nNnn ω
Eigen value problem
ωω ,,, 21 K N natural freq.
0xMK =ω− ini )( 2
ix Eigen vector
)sin( φ+ω= tAx )( φ+ω= tjeAor
N mode shapesNxxx ,,, 21 K
EOM1
2
3
4 5
Direct Method
Summary (Free-undamped) (2)
Free-undamped response
)sin()sin()sin()( 22221111 NNNN tAtAtAt φ+ω+φ+ω+φ+ω= xxxx K
∑=
φ+ω=N
iiiii tAt
1)sin()( xx
6
where A and φ are from initial condition x(0) and v(0)
Direct Method
Example
1k2k
θx
2l1l
Determine the normal modes of vibration of an automobile simulated by simplified 2-dof system with the following numerical values
lb3220=W
ft5.41 =l lb/ft24001 =k
ft5.52 =l lb/ft26002 =k
ft4=r2rgWJC =
Forced harmonic vibration (1)
Example
EOMt
Fxx
kkkk
xx
mm
ω⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡sin
000 1
2
1
2221
1211
2
1
2
1
&&
&&
System is undamped, the solution can be assumed as
tXX
xx
ω⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡sin
2
1
2
1
Sub. into EOM⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡
ω−ω−
01
2
12
22221
122
111 FXX
mkkkmk
[ ] ⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡ω
0)( 1
2
1 FXX
ZSimpler notation, [ ] ⎥⎦
⎤⎢⎣
⎡ω=⎥
⎦
⎤⎢⎣
⎡ −
0)( 11
2
1 FZ
XX
Forced harmonic vibration (2)
[ ] [ ]⎥⎦
⎤⎢⎣
⎡ωω
=⎥⎦
⎤⎢⎣
⎡ω=⎥
⎦
⎤⎢⎣
⎡ −
0)()(adj
0)( 111
2
1 FZ
ZFZ
XX
))(()( 222
22121 ω−ωω−ω=ω mmZWhere
ω1 and ω2 are natural frequencies
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡
ω−−−ω−
ω=⎥
⎦
⎤⎢⎣
⎡0)(
1 12
11121
122
222
2
1 Fmkkkmk
ZXX
The amplitudes are))((
)(22
222
121
12
2221 ω−ωω−ω
ω−=
mmFmkX
))(( 222
22121
1212 ω−ωω−ω
−=
mmFkX
Forced harmonic vibration (3)
))(()2(
222
221
21
2
1 ω−ωω−ωω−
=m
FmkX
))(( 222
221
21
2 ω−ωω−ω=
mkFX
m mk k k
x1 x2
F1sinωt
tF
xx
kkkk
xx
mm
ω⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
−+⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡sin
022
00 1
2
1
2
1
&&
&&
mk
mk 3, 21 =ω=ω
EOM
Force response of a 2 DOF system
0 1 2 3
012
3
45
-1
-2-3
-4
-5
FXk
1ωω
1
2
FkX
1
1
FkX
1ω=ω 2ω=ω
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
1k 2k
2l1l
mg
θ
xRef.
)( 11 θ− lxk)( 22 θ− lxk
1k 2k1l mg
θ1x
Ref.
11xk)( 12 θ+ lxk
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡θ⎥
⎦
⎤⎢⎣
⎡+−−+
+⎥⎦
⎤⎢⎣
⎡θ⎥
⎦
⎤⎢⎣
⎡00
00
222
2111122
112221 xlklklklklklkkkx
Jm
&&
&&⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡θ⎥
⎦
⎤⎢⎣
⎡ ++⎥
⎦
⎤⎢⎣
⎡θ⎥
⎦
⎤⎢⎣
⎡001
222
2211
11
1 xlklklkkkx
Jmlmlm
&&
&&
Concept of modal analysis
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡θ⎥
⎦
⎤⎢⎣
⎡+−−+
+⎥⎦
⎤⎢⎣
⎡θ⎥
⎦
⎤⎢⎣
⎡0
)(0
0222
2111122
112221 tFxlklklklklklkkkx
Jm
&&
&&)()()( ttt FKxxM =+&&
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡
ωω
+⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡)()(
00
1001
2
1
2
12
2
21
2
1
tNtN
rr
rr
n
n
&&
&& )()()( ttt NΛrr =+&&
EOM in modal coordinate (Independent SDOF equations)
EOM in physical coordinate (Coordinates are coupled)
Solve for )(tr
Transform r(t) back to x(t)
Orthogonality
x = eigen vector (vector of mode shape)
iiiTi M=Mxx
jiiTj ≠= ,0Mxx jii
Tj ≠= ,0Kxx
iiiTi K=Kxx
If M and K are symmetric and then xi and xj are said to be “orthogonal” to each other.
njni ω≠ω
Normalization
u = normalized eigen vector (respect to mass matrix)
1=iTi Muu
jiiTj ≠= ,0Muu
constant is , CC ii xu
0xMK =ω− )()( 2 tFrom eigen value problem
=
0uMK =ω− )()( 2 t
or iii MuKu 2ω=
22ii
Tiii
Ti ω=ω= MuuKuu
Modal matrix
Modal matrix is the matrix that its columns are the mode shape of the system
[ ]nuuuU K21=
Then
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
==
100
010001
K
MOMM
K
K
IMUUT
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
ω
ωω
=
2
22
21
00
0000
nN
n
n
T
K
MOMM
K
K
KUU
Λ (Spectral matrix)
Modal analysis (undamped systems)-1
1. Draw FBD, apply Newton’s law to obtain EOM2. Solve for natural frequencies through CHE3. Determine mode shapes through EVP4. Construct modal matrix (normalized)
Procedures)()()( ttt FKxxM =+&&
0xMK =ω− )()( 2 t0)det( 2 =ω− MK
[ ]nuuuU K21=
IMUU =T
ΛKUU =T
5. Perform a coordinate transformation )()( tt Urx =
)()()( ttt FKxxM =+&& )()()( ttt FKUrrMU =+&&
)()()( ttt TTT FUKUrUrMUU =+&&
)()()( ttt TFUΛrr =+&&
Modal analysis (undamped systems)-2
)()()( ttt TFUΛrr =+&&
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
ω
ωω
+
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
)()()()(
)()()()(
)(
)()(
00
0000
)(
)()(
4
3
2
1
4
3
2
1
21
22221
11211
2
1
2
22
21
2
1
tNtNtNtN
tFtFtFtF
uuu
uuuuuu
tr
trtr
tr
trtr T
NNNN
N
N
NnN
n
n
N K
MOMM
K
K
M
K
MOMM
K
K
&&
M
&&
&&
Independent SDOF equations, can be solve for r(t)
6. Transform the initial conditions to modal coordinates
)()( tt Urx =
IMUU =T
)0()0( Urx =
)0()0( MUrUMxU TT =
)0()0( MxUr T=
From
and )0()0( xMUr && T=
Modal analysis (undamped systems)-3
7. Find the response in modal coordinates8. Transform the response in modal coordinate
back to that in original coordinate )()( tt Urx =
K=)(tr)(tr
)(tx
Example (Modal analysis) -1
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
−+⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡00
33327
1009
2
1
2
1
xx
xx&&
&&
EOM
⎥⎦
⎤⎢⎣
⎡=
01
0x ⎥⎦
⎤⎢⎣
⎡=
00
0vInitial conditions
Example (Modal analysis) -2
2dof string-bead system
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
−+⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡00
2112
00
2
1
2
1
xx
aT
xx
mm
&&
&&
EOM
⎥⎦
⎤⎢⎣
⎡=
00
0x ⎥⎦
⎤⎢⎣
⎡=
10
0vInitial conditions
Example (Modal analysis) -2_2
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. 0=ωn
Rigid-body modes
Compute the solution of the system. Let m1 = 1 kg, m2 = 4 kg and k = 400 N/m.
Initial condition ⎥⎦
⎤⎢⎣
⎡=
001.0
0x ⎥⎦
⎤⎢⎣
⎡=
00
0v
More than two degrees of freedom
Calculate the solution of the n-degree-of-freedom system in the figure for n = 3 by modal analysis. Use the values m1 = m2 = m3 = 4 kg and k1 = k2 = k4 = 4 N/m, and the initial condition x1(0) = 1 m with all other initial displacements and velocities zero.
Modal analysis on damped systems (1)
)()()()( tttt FKxxCxM =++ &&&
CKMKCM 11 −− =
The original modal analysis can be applied to MDOF damped system if and only if
KMC β+α=
α β
However, there are subsets of the above systems where C can be written as a linear combination of M and K.
and are constants. Such system is called “proportionally damped.”
Necessary and sufficient condition
Sufficient but not necessary condition
Such system is called “classically damped”.
EOM
Modal analysis on damped systems (2)
For proportionally damped )()()()( tttt FKxxCxM =++ &&&
)()()()()( tttt FKxxKMxM =+β+α+ &&&
)()( tt Urx = )()()()()( tttt FKUrrUKMrMU =+β+α+ &&&
TU )()()()()( TTTT tttt FUKUrUrUKMUrMUU =+β+α+ &&&
)()()()()()( T ttttt NFUΛrrΛIr ==+β+α+ &&&
)()()(2)( 1121111 tNtrtrtr nn =ω+ζω+ &&&
M2112 nn βω+α=ζω
Let
Premultiply by
Thus, the system when it is written in modal coordinates r(t) can be decoupled into k sets of SDOF equations
where
Modal analysis on damped systems (Ex.)
A belt-driven lathe•bearings are modeled as providing viscous damping •shafts provide stiffness •belt drive provides and applied torque.
/radkg.m10 2321 === JJJ
N.m/rad10321 == kkN.m.s/rad2=c
• Zero initial conditions• Applied moment M(t) is a unit
impulse function
Modal analysis on damped systems (Ex.)