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Vibration 3
2DOF and Multiple DOF systems
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Chapter 4 Multiple Degree of Freedom Systems
Extending the first 3 chapters to more then one degree of freedom
Rao’s textbook: Chap 5
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Two Degrees of Freedom (4.1)
x1 x2
k1m1 m2
k2
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Free-Body Diagram
x1 x2
m1 m2k1 x1
k2(x2 -x1)
5
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Equations of Motion
( )( )
0)()()(0)()()()(
:gRearrangin)()()(
)()()()(
221222
2212111
12222
1221111
=+−=−++
−−=−+−=
txktxktxmtxktxkktxm
txtxktxmtxtxktxktxm
&&
&&
&&
&&
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Initial Conditions
• Two coupled, second -order, ordinary differential equations with constant coefficients
• Needs 4 constants of integration to solve• Thus 4 initial conditions on positions and
velocities
202202101101 )0(,)0(,)0(,)0( xxxxxxxx &&&& ====
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Solution by Matrix Methods
0KxxM
KM
xxx
=+
⎥⎦
⎤⎢⎣
⎡−
−+=⎥
⎦
⎤⎢⎣
⎡=
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡=
&&
&&
&&&&
&
&&
22
221
2
1
2
1
2
1
2
1
,0
0
)()(
)(,)()(
)(,)()(
)(
kkkkk
mm
txtx
ttxtx
ttxtx
t
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Initial Conditions
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡=
20
10
20
10 )0( ,)0(xx
xx
&
&&xx
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Solution:
Let x(t) = ue jωt
j = −1, u / = 0, ω unknown
⇒ -ω2M + K( )ue jωt = 0
⇒ -ω2M + K( )u = 0
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Changes ODE into algebraic equation
-ω2M + K( )u = 0 ⇒
two algebraic equation in 3 uknowns
u =u1
u2
⎡
⎣ ⎢ ⎤
⎦ ⎥ , and ω
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Condition for Solution:
inv -ω2 M + K( ) exists ⇒ u = 0
Require u / = 0 ⇒ -ω2 M + K( )−1 does not exist
or det -ω2M + K( )= 0
One equation in one unknown w
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Back to our specific system: the characteristic equation
det -ω2M + K( )= 0 ⇒
det−ω2m1 + k1 + k2 −k2
−k2 −ω2m2 + k2
⎡
⎣ ⎢ ⎤
⎦ ⎥ = 0 ⇒
m1m2ω4 − (m1k2 + m2k1 + m2k2)ω2 + k1k2 = 0
Quadratic in ω2 so four solutionsω2
1 and ω22 or +ω1 and + ω2
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Calculating the corresponding vectors u1 and u2
(−ω12M + K)u1 = 0
(−ω22M + K)u2 = 0
A vector equation for each square frequency
And:
4 equations in the 4 unknowns (eachvector has 2 components, but...
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Numerical examples
• m1=9 kg,m2=1kg, k1=24 N/m and k2=3 N/m• Characteristic equation becomes
ω4-6ω2+8=(ω2-2)(ω2-4)=0ω2 =2 and ω2 =4 or
ω1,3 = ± 2 rad/s, ω2,4 = ±2 rad/s
Each value of ω2 yields an expression or u:
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Computing the vectors u
For ω12 = 2, let u1 =
u11
u12
⎡ ⎣ ⎢
⎤ ⎦ ⎥ then we have
(-ω12M + K)u = 0 ⇒
27 − 9(2) −3−3 3 − (2)
⎡ ⎣ ⎢
⎤ ⎦ ⎥
u11
u12
⎡
⎣ ⎢ ⎤
⎦ ⎥ =00
⎡ ⎣ ⎢
⎤ ⎦ ⎥
⇒
9u11 − 3u12 = 0 and − 3u11 + u12 = 0
2 equations, 2 unknowns but DEPENDENT!
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0u0u
u0u
u
=+−⇔=+−
=+−
=+−
=⇒=
1211
21
1121
1
21
121112
11
)()(
:arbitrary , does so ,)(
satisfies Suppose arbitrary. is magnitude The.0)(det :because is This
!determined becan magnitude not the direction, only the
:equationsboth from 31
31
KMaKM
aaKM
KM
uuuu
ωω
ω
ω
continued
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Likewise for the second value of ω2:
For ω22 = 4, let u2 =
u21
u22
⎡
⎣ ⎢ ⎤
⎦ ⎥ then we have
(-ω12M + K)u = 0 ⇒
27 − 9(4) −3−3 3 − (4)
⎡ ⎣ ⎢
⎤ ⎦ ⎥
u21
u22
⎡
⎣ ⎢ ⎤
⎦ ⎥ =00
⎡ ⎣ ⎢
⎤ ⎦ ⎥
⇒
−9u21 − 3u22 = 0 or u21 = −13
u22
Note that the other equation is the same
18-
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What to do about the magnitude!
⎥⎦
⎤⎢⎣
⎡−=⇒=
⎥⎦
⎤⎢⎣
⎡=⇒=
11
11
31
222
31
112
u
u
u
u
Several possibilities, here we just fix one element:
Choose:
Choose:
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Thus the solution to the algebraic matrix equation is:
ω1,3 = ± 2, u1 =13
1⎡ ⎣ ⎢
⎤ ⎦ ⎥
ω2,4 = ±2, u2 =−13
1⎡ ⎣ ⎢
⎤ ⎦ ⎥
20
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Return now to the time response:
x(t) = u1e− jω1t ,u1e
jω1t ,u2e− jω2t ,u2e
jω2t ⇒
x(t) = au1e− jω1t + bu1e
jω1t + cu2e− jω2t + du2e
jω2t
⇒ x(t) = ae− jω1t + be jω1t( )u1 + ce− jω2t + de jω2t( )u2
= A1 sin(ω1t + φ1)u1 + A2 sin(ω2t + φ2 )u2
where A1, A2 ,φ1, and φ2 are constants of integration
We have four solutions:
Since linear we can combine as:
determined by initial conditions
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Physical interpretation of all that math!
• Each of the TWO masses is oscillating at TWO natural frequencies ω1 and ω2
• The relative magnitude of each sine term, and hence of the magnitude of oscillation of m1 and m2 is determined by the value ofA1u1 and A2u2
• The vectors u1 and u2 are calledmode shapes
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What is a mode shape?
• First note that A1,A2, φ1 and φ2 are determined by the initial conditions
• Choose them so that A2 = φ1 = φ2 =0• Then:
• Thus each mass oscillates at (one) frequency ω1with magnitudes proportional to u1 the 1st mode shape
x(t) =x1(t)x2 (t)
⎡ ⎣ ⎢
⎤ ⎦ ⎥
= A1
u11
u12
⎡ ⎣ ⎢
⎤ ⎦ ⎥ sinω1t
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Mode shapes:
Mode 1:
Mode 2:
1/3
-1/3
1
1
m1
m2
m2
m2
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Example (continue on)
( ) ( )( ) ( )
( ) ( )( ) ( ) ⎥
⎥
⎦
⎤
⎢⎢
⎣
⎡
+++
+−+=⎥
⎦
⎤⎢⎣
⎡
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
+++
+−+=⎥
⎦
⎤⎢⎣
⎡
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡
2211
22
11
2
1
2211
22
11
2
1
2cos22cos2
2cos23
2cos23)(
)(
2sin2sin
2sin3
2sin3)(
)(
00
)0( mm, 01
=(0)consider
φφ
φφ
φφ
φφ
tAtA
tAtA
txtx
tAtA
tAtA
txtx
&
&
&xx
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( ) ( )( ) ( )
( ) ( )( ) ( ) ⎥
⎥
⎦
⎤
⎢⎢
⎣
⎡
+
−=⎥
⎦
⎤⎢⎣
⎡
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
+
−=⎥⎦
⎤⎢⎣
⎡
2211
22
11
2211
22
11
sin2sin2
sin3
2sin230
0
sinsin
sin3
sin30
mm 1
φφ
φφ
φφ
φφ
AA
AA
AA
AA
At t=0 we have
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3 = A1 sin φ1( )− A2 sin φ2( )0 = A1 sin φ1( )+ A2 sin φ2( )0 = A1 2 cos φ1( )− A2 2cos φ2( )0 = A1 2 cos φ1( )+ A2 2cos φ2( )
A1 = 1.5 mm, A2 = −1.5 mm,φ1 = φ2 =π2
rad
4 equations in 4 unknowns:
Yields:
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Solution:
x1(t) = 0.5cos 2t + 0.5cos2tx2 (t) = 1.5cos 2t −1.5cos2t
0 5 10 15 20
4
2
2
4
x1 t
x2 t
t
mm
seconds
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Solution as a sum of modes
x(t) = u1 cosω1t + u2 cosω2t
Determines how the first frequency contributes to theresponse
Determines how the second frequency contributes to theresponse
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Things to note
• Two degrees of freedom implies twonatural frequencies
• Each mass oscillates at with these two frequencies present in the response
• Frequencies are not those of two component systems
ω1 = 2 ≠k1
m1
= 1.63 , ω 2 = 2 ≠k 2
m 2
= 1.732
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Motor-pump system on springs
Figs.5.1
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Packaging of an instrument (portable electronics)
Figs.5.2
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• As is evident from the systems shown in Figs.5.1 and 5.2, the configuration of a system can be specified by a set of independent coordinates termed as generalized coordinates, such as length, angle, or some other physical parameters.
• Principle coordinates is defined as any set of coordinates that leads a coupled equation of motion to an uncoupledsystem of equations.
Principle coordinates
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)2.5()()()1.5()()(
2232122321222
1221212212111
FxkkxkxccxcxmFxkxkkxcxccxm
=++−++−=−++−++
&&&&
&&&&
)3.5( )()(][)(][)(][ tFtxktxctxmrr&r&&r =++
Equations of Motion for 2DOF System
where [m], [c], and [k] are called the mass, damping, and stiffness matrices, respectively, and are given by
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⎥⎦
⎤⎢⎣
⎡+−
−+=⎥
⎦
⎤⎢⎣
⎡=
322
221
2
1
][ 0
0 ][
kkkkkk
km
mm
][][],[][ kkmm TT ==
where the superscript T denotes the transpose of the matrix.
Matrices [m] and [k] are symmetric:
Properties of M & K Matrices
35
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Solution: For the given data, the eigenvalueproblem, Eq.(5.8), becomes
Ex 5.3 Free Vibration Response of a Two Degree of Freedom System
).0()0()0( ,1)0( 2211 xxxx && ===
Find the free vibration response of the system shown in Fig.5.3(a) with k1 = 30, k2 = 5, k3 = 0, m1= 10, m2 = 1 and c1 = c2 = c3 = 0 for the initial conditions
(E.1)00
55- 5 3510
00
2
12
2
2
1
322
22
2212
1
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+−
−+−
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
++−−
−++−
XX
XX
kkmk
kkkm
ω
ω
ω
ω
or
36
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from which the natural frequencies can be found as
By setting the determinant of the coefficient matrix in Eq.(E.1) to zero, we obtain the frequency equation,
Example 5.3 Solution
(E.2)01508510 24 =+− ωω
E.3)(4495.2,5811.10.6,5.2
21
22
21
====
ωωωω
The normal modes (or eigenvectors) are given by
E.5)(5
1
E.4)(21
)2(1)2(
2
)2(1)2(
)1(1)1(
2
)1(1)1(
XX
XX
XX
XX
⎭⎬⎫
⎩⎨⎧−
=⎪⎭
⎪⎬⎫
⎪⎩
⎪
37
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By using the given initial conditions in Eqs.(E.6) and (E.7), we obtain
The free vibration responses of the masses m1and m2 are given by (see Eq.5.15):
(E.7))4495.2cos(5)5811.1cos(2)(
(E.6))4495.2cos()5811.1cos()(
2)2(
11)1(
12
2)2(
11)1(
11
φφ
φφ
+−+=
+++=
tXtXtx
tXtXtx
(E.11)sin2475.121622.3)0(
(E.10)sin4495.2sin5811.10)0(
(E.9)cos5cos20)0(
(E.8)coscos1)0(
2)2(
1)1(
12
2)2(
11)1(
11
2)2(
11)1(
12
2)2(
11)1(
11
φ
φφ
φφ
φφ
XXtx
XXtx
XXtx
XXtx
+−==
−−===
−===
+===
&
&
Example 5.3 Solution
38
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while the solution of Eqs.(E.10) and (E.11) leads to
The solution of Eqs.(E.8) and (E.9) yields
(E.12)72cos;
75cos 2
)2(11
)1(1 == φφ XX
(E.13)0sin,0sin 2)2(
11)1(
1 == φφ XX
Equations (E.12) and (E.13) give
(E.14)0,0,72,
75
21)2(
1)1(
1 ==== φφXX
Example 5.3 Solution
39
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Thus the free vibration responses of m1 and m2are given by
(E.16)4495.2cos7
105811.1cos7
10)(
(E.15)4495.2cos725811.1cos
75)(
2
1
tttx
tttx
−=
+=
Example 5.3 Solution
40
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which upon rearrangement become
5.4 Torsional System
22312222
11221111
)(
)(
ttt
ttt
MkkJ
MkkJ
+−−−=
+−+−=
θθθθ
θθθθ&&
&&
Consider a torsional system as shown in Fig.5.6. The differential equations of rotational motion for the discs can be derived as
)19.5()(
)(
22321222
12212111
tttt
tttt
MkkkJ
MkkkJ
=++−
=−++
θθθ
θθθ&&
&&
For the free vibration analysis of the system, Eq.(5.19) reduces to
41
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From
dynami
)20.5(0)(
0)(
2321222
2212111
=++−
=−++
θθθ
θθθ
ttt
ttt
kkkJ
kkkJ&&
&&
Figure 5.6: Torsional system with discs mounted on a shaft
5.4 Torsional System
42
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Find the natural frequencies and mode shapes for the torsional system shown in Fig.5.7 for J1 = J0 , J2 = 2J0 and kt1 = kt2 = kt .
Solution: The differential equations of motion, Eq.(5.20), reduce to (with kt3 = 0, kt1 = kt2 = kt, J1 = J0 and J2 = 2J0):
Ex 5.4 Natural Frequencies of a Torsional System
(E.1) 02
02
2120
2110
=+−
=−+
θθθ
θθθ
tt
tt
kkJ
kkJ&&
&&Fig.5.7:Torsional system
43
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The solution of Eq.(E.3) gives the natural frequencies
gives the frequency equation:
Example 5.4 Solution
(E.2)2,1);cos()( =+Θ= itt ii φωθ
(E.3)052 20
220
4 =+− tt kkJJ ωω
0t0t
02
01
/k5102.1/k0.4682
(E.4))175(4
and)175(4
JJ
Jk
Jk tt
==
+=−= ωω
Rearranging and substituting the harmonic solution:
44
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Equations (E.4) and (E.5) can also be obtained by substituting the following in Eqs.(5.10) and (5.11).
The amplitude ratios are given by
Example 5.4 Solution
(E.5)2808.04
)175(2
7808.14
)175(2
)2(1
)2(2
2
)1(1
)1(2
1
−=+
−=ΘΘ
=
=−
−=ΘΘ
=
r
r
0and2,,,
3022011
2211
=========
kJJmJJmkkkkkk tttt
45
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Generalized coordinates are sets of n coordinates used to describe the configuration of the system.Equations of motion for a lathe Using x(t) and θ(t)
5.5 Coordinate Coupling and Principal Coordinates
46
-
and the moment equation about C.G. can be expressed as
)21.5()()( 2211 θθ lxklxkxm +−−−=&&
From the free-body diagram shown in Fig.5.10a, with the positive values of the motion variables as indicated, the force equilibrium equation in the vertical direction can be written as
)22.5()()( 2221110 llxkllxkJ θθθ +−−=&&
Eqs.(5.21) and (5.22) can be rearranged and written in matrix form as
5.5 Coordinate Coupling and Principal Coordinates
47
-
The lathe rotates in the vertical plane and has vertical motion as well, unless k1l1 = k2l2. This is known as elastic or static coupling.
From Fig.5.10b, the equations of motion for translation and rotation can be written as
)23.5(00
)( )(
)( )(
00
22
212211
221121
0 21 ⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
+−−
−−++
⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡θθx
lklklklk
lklkkkxJ
m&&
&&
•Equations of motion Using y(t) and θ(t).
θθθ &&&& melyklykym −′+−′−−= )()( 2211
5.5 Coordinate Coupling and Principal Coordinates
48
-
These equations can be rearranged and written in matrix form as
)24.5()()( 222111 ymellykllykJ P &&&& −′′+−′′−= θθθ
)25.5(00
)()(
)()(
22
2112211
112221
2 ⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡′+′′+′−
′−′++
⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡θθy
lklklklk
lklkkkyJmemem
P&&
&&
If , the system will have dynamic or inertiacoupling only.
2211 lklk ′=′
Note the following characteristics of these systems:
5.5 Coordinate Coupling and Principal Coordinates
49
-
1. In the most general case, a viscously damped 2DOF system has the equations of motions in the form:
)26.5(00
2
1
2221
1211
2
1
2221
1211
2
1
2221
1211
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡+
⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡+
⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡xx
kkkk
xx
cccc
xx
mmmm
&
&
&&
&&
2.The system vibrates in its own natural way regardless of the coordinates used. The choice of the coordinates is a mere convenience.
3.Principal or natural coordinates are defined as system of coordinates which give equations of motion that are uncoupled both statically and dynamically.
5.5 Coordinate Coupling and Principal Coordinates
50
-
Ex 5.6 Principal Coordinates of Spring-Mass System
Determine the principal coordinates for the spring-mass system shown in Fig.5.4.
51
-
We define a new set of coordinates such that
Approach: Define two independent solutions as principal coordinates and express them in terms of the solutions x1(t) and x2(t).
The general motion of the system shown is
Example 5.6 Solution
(E.1)3coscos)(
3coscos)(
22112
22111
⎟⎟⎠
⎞⎜⎜⎝
⎛+−⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
⎟⎟⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
φφ
φφ
tmkBt
mkBtx
tmkBt
mkBtx
52
-
Since the coordinates are harmonic functions, their corresponding equations of motion can be written as
Example 5.6 Solution
(E.2)3cos)(
cos)(
222
111
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
φ
φ
tmkBtq
tmkBtq
(E.3)03
0
22
11
=⎟⎠⎞
⎜⎝⎛+
=⎟⎠⎞
⎜⎝⎛+
qmkq
qmkq
&&
&&
53
-
The solution of Eqs.(E.4) gives the principal coordinates:
From Eqs.(E.1) and (E.2), we can write
Example 5.6 Solution
(E.4))()()()()()(
212
211
tqtqtxtqtqtx
−=+=
(E.5))]()([21)(
)]()([21)(
212
211
txtxtq
txtxtq
−=
+=
54
-
The equations of motion of a general 2DOF system under external forces can be written as
Consider the external forces to be harmonic:
Forced Vibration Analysis
)27.5(
2
1
2
1
2221
1211
2
1
2221
1211
2
1
2212
1211
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡+
⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡+
⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡FF
xx
kkkk
xx
cccc
xx
mmmm
&
&
&&
&&
)28.5(2,1,)( 0 == jeFtF tijj
ω
where ω is the forcing frequency. We can write the steady-state solutions as
)29.5(2,1,)( == jeXtx tijj
ω
55
-
Substitution of Eqs.(5.28) and (5.29) into Eq.(5.27) leads to
Forced Vibration Analysis
)30.5(
)()(
)()(
20
10
2
1
22222221212122
1212122
1111112
⎭⎬⎫
⎩⎨⎧
=
⎭⎬⎫
⎩⎨⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
++−++−
++−++−
FF
XX
kcimkcim
kcimkcim
ωωωω
ωωωω
We defined as in Section 3.5 the mechanical impedance Zre(iω) as
)31.5(2,1,,)( 2 =++−= srkcimiZ rsrsrsrs ωωω
56
-
And write Eq.(5.30) as:
[ ] )32.5()( 0FXiZrr
=ωwhere
[ ]
⎭⎬⎫
⎩⎨⎧
=
⎭⎬⎫
⎩⎨⎧
=
=⎥⎦
⎤⎢⎣
⎡=
20
100
2
1
2212
1211 matrix Impedance)( )()( )(
)(
FF
F
XX
X
iZiZiZiZ
iZ
r
r
ωωωω
ω
Eq.(5.32) can be solved to obtain:
Forced Vibration Analysis
57
-
where the inverse of the impedance matrix is given
Forced Vibration Analysis
[ ] )33.5()( 01 FiZXrr −= ω
[ ] )34.5()( )()( )(
)()()(1)(
1112
12222
122211
1⎥⎦
⎤⎢⎣
⎡−−
=−
ωωωω
ωωωω
iZiZi-ZiZ
iZiZiZiZ
Eqs.(5.33) and (5.34) lead to the solution
)35.5()()()(
)()()(
)()()()()()(
2122211
201110122
2122211
201210221
ωωωωωω
ωωωωωω
iZiZiZFiZFiZiX
iZiZiZFiZFiZiX
−+−
=
−−
=
58
-
Find the steady-state response of system shown in Fig.5.13 when the mass m1 is excited by the force F1(t) = F10 cos ωt. Also, plot its frequency response curve.
Ex 5.8 Steady-State Response of Spring-Mass System
59
-
The equations of motion of the system can be expressed as
Example 5.8 Solution
(E.1)0
cos2
2 0
0 10
2
1
2
1
⎭⎬⎫
⎩⎨⎧
=⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡+
⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡ tFxx
k-k-kk
xx
mm ω
&&
&&
E.2)(2,1;cos)( == jtXtx jj ω
We assume the solution to be as follows.
Eq.(5.31) gives(E.3))(,2)()( 12
22211 kZkmZZ −=+−== ωωωω
60
-
Example 5.8 Solution
(E.5)))(3()2(
)(
(E.4)))(3(
)2()2(
)2()(
2210
22210
2
2210
2
22210
2
1
kmkmkF
kkmkFX
kmkmFkm
kkmFkmX
+−+−=
−+−=
+−+−+−
=−+−
+−=
ωωωω
ωωω
ωωω
Eqs.(E.4) and (E.5) can be expressed as
Hence,
E.6)(
1
2
)(2
1
2
1
2
1
2
10
2
1
1
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎟⎟⎠
⎞⎜⎜⎝
⎛−
=
ωω
ωω
ωω
ωω
ω
k
F
X
61
-
Fig.5.14: Frequency response curves
Example 5.8 Solution
E.7)(
1
)(2
1
2
1
2
1
2
102
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
ωω
ωω
ωω
ω
k
FX
62
-
Semidefinite Systems
)36.5(0)(0)(
1222
2111
=−+=−+
xxkxmxxkxm
&&
&&
Semidefinite systems are also known as unrestrained or degenerate systems. Two examples of such systems are shown in Fig.5.15. For Fig.5.15a, the equations of motion can be written as
For free vibration, we assume the motion to be harmonic:
)37.5(2,1),cos()( =+= jtXtx jjj φω
63
-
Semidefinite Systems
)38.5(0)(
0)(
22
21
212
1
=+−+−
=−−−
XkmkX
kXXkm
ω
ω
Substituting Eq.(5.37) into Eq.(5.36) gives
Fig.5.15: Semidefinite Systems
64
-
Semidefinite Systems
From which the natural frequencies can be obtained:
Such systems, which have one of the natural frequencies equal to zero, are called semidefinitesystems.
)40.5()(and 021
2121 mm
mmk +== ωω
We obtain the frequency equation as)39.5(0)]([ 21
221
2 =+− mmkmm ωω
65
-
)10.5())((4
)()(21
)()(21,
2/1
21
223221
2
21
132221
21
13222122
21
⎭⎬⎫
⎩⎨⎧ −++
−
⎢⎢⎣
⎡
⎭⎬⎫
⎩⎨⎧ +++
⎭⎬⎫
⎩⎨⎧ +++
=
mmkkkkk
mmmkkmkk
mmmkkmkk
m
ωω
{ }{ } )9.5(0))((
)()()(223221
1322214
21
=−+++
+++−
kkkkk
mkkmkkmm ωω
Frequency (characteristic) equation for 2DOF mass-spring system
