42
TWO DEGREE OF FREEDOM SYSTEM
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TWO DEGREE OF FREEDOM SYSTEM
INTRODUCTION Systems that require two independent coordinates
to describe their motion; Two masses in the system X two possible types of
motion of each mass. Example: motor pump system.
There are two equations of motion for a 2DOF system, one for each mass (more precisely, for each DOF).
They are generally in the form of couple differential equation that is, each equation involves all the coordinates.
EQUATION OF MOTION FOR FORCED VIBRATION
Consider a viscously damped two degree of freedom spring-mass system, shown in Fig.5.3.
Figure 5.3: A two degree of freedom spring-mass-damper system
4
Both equations can be written in matrix form as
The application of Newton’s second law of motion to each of the masses gives the equations of motion:
EQUATIONS OF MOTION FOR FORCED VIBRATION
)2.5()()(
)1.5()()(
2232122321222
1221212212111
Fxkkxkxccxcxm
Fxkxkkxcxccxm
)3.5( )()(][)(][)(][ tFtxktxctxm
where [m], [c], and [k] are called the mass, damping, and stiffness matrices, respectively, and are given by
5
And the displacement and force vectors are given respectively:
EQUATIONS OF MOTION FOR FORCED VIBRATION
322
221
322
221
2
1
][
][
0
0 ][
kkk
kkkk
ccc
cccc
m
mm
)(
)()(
)(
)()(
2
1
2
1
tF
tFtF
tx
txtx
It can be seen that the matrices [m], [c], and [k] are all 2 x 2 matrices whose elements are known masses, damping coefficient and stiffnesses of the system, respectively.
6
where the superscript T denotes the transpose of the matrix.
EQUATIONS OF MOTION FOR FORCED VIBRATION
oThe solution of Eqs.(5.1) and (5.2) involves four constants of integration (two for each equation). Usually the initial displacements and velocities of the two masses are specified as
][][],[][],[][ kkccmm TTT
oFurther, these matrices can be seen to be symmetric, so that,
x1(t = 0) = x1(0) and 1( t = 0) = 1(0), x2(t = 0) = x2(0) and 2 (t = 0) = 2(0).
xx x
x
7
Assuming that it is possible to have harmonic motion of m1 and m2 at the same frequency ω and the same phase angle Φ, we take the solutions as
FREE VIBRATION ANALYSIS OF AN UNDAMPED SYSTEM
)5.5(0)()()()(
)4.5(0)()()()(
2321222
2212111
txkktxktxm
txktxkktxm
By setting F1(t) = F2(t) = 0, and damping disregarded, i.e., c1 = c2 = c3 = 0, and the equation of motion is reduced to:
)6.5()cos()(
)cos()(
22
11
tXtx
tXtx
8
Since Eq.(5.7)must be satisfied for all values of the time t, the terms between brackets must be zero. Thus,
FREE VIBRATION ANALYSIS OF AN UNDAMPED SYSTEM
)7.5( 0)cos()(
0)cos()(
2322
212
221212
1
tXkkmXk
tXkXkkm
Substituting into Eqs.(5.4) and (5.5),
)8.5(0)(
0)(
2322
212
221212
1
XkkmXk
XkXkkm
9
or
0
)(
)(det
212
12
2212
1
kkmk
kkkm
which represent two simultaneous homogenous algebraic equations in the unknown X1 and X2. For trivial solution, i.e., X1 = X2 = 0, there is no solution. For a nontrivial solution, the determinant of the coefficients of X1 and X2 must be zero:
)9.5(0))((
)()()(223221
1322214
21
kkkkk
mkkmkkmm
FREE VIBRATION ANALYSIS OF AN UNDAMPED SYSTEM
10
The roots are called natural frequencies of the system.
which is called the frequency or characteristic equation. Hence the roots are:
)10.5())((
4
)()(
2
1
)()(
2
1,
2/1
21
223221
2
21
132221
21
13222122
21
mm
kkkkk
mm
mkkmkk
mm
mkkmkk
FREE VIBRATION ANALYSIS OF AN UNDAMPED SYSTEM
11
The normal modes of vibration corresponding to ω1
2 and ω22 can be
expressed, respectively, as
)11.5()(
)(
)(
)(
32222
2
2
21221
)2(1
)2(2
2
32212
2
2
21211
)1(1
)1(2
1
kkm
k
k
kkm
X
Xr
kkm
k
k
kkm
X
Xr
To determine the values of X1 and X2, given ratio
)12.5( and )2(
12
)2(1
)2(2
)2(1)2(
)1(11
)1(1
)1(2
)1(1)1(
Xr
X
X
XX
Xr
X
X
XX
which are known as the modal vectors of the system.
FREE VIBRATION ANALYSIS OF AN UNDAMPED SYSTEM
12
Where the constants , , and are determined by the initial conditions. The initial conditions are
(5.17)mode second)cos(
)cos(
)(
)()(
modefirst )cos(
)cos(
)(
)()(
22)2(
12
22)2(
1
)2(2
)2(1)2(
11)1(
11
11)1(
1
)1(2
)1(1)1(
tXr
tX
tx
txtx
tXr
tX
tx
txtx
The free vibration solution or the motion in time can be expressed itself as
0)0(,)0(
,0)0(constant, some )0(
2)(
12
1)(
11
txXrtx
txXtxi
i
i
FREE VIBRATION ANALYSIS OF AN UNDAMPED SYSTEM
)1(1X
)2(1X 1
2
13
Thus the components of the vector can be expressed as
)14.5()()()( 2211 txctxctx
The resulting motion can be obtained by a linear superposition of the two normal modes, Eq.(5.13)
)15.5()cos()cos(
)()()(
)cos()cos()()()(
22)2(
1211)1(
11
)2(2
)1(22
22)2(
111)1(
1)2(
1)1(
11
tXrtXr
txtxtx
tXtXtxtxtx
where the unknown constants can be determined from the initial conditions:
FREE VIBRATION ANALYSIS OF AN UNDAMPED SYSTEM
14
Substituting into Eq.(5.15) leads to
)16.5()0()0(),0()0(
),0()0(),0()0(
2222
1111
xtxxtx
xtxxtx
)17.5(sinsin)0(
coscos)0(
sinsin)0(
coscos)0(
2)2(
1221)1(
1112
2)2(
121)1(
112
2)2(
121)1(
111
2)2(
11)1(
11
XrXrx
XrXrx
XXx
XXx
The solution can be expressed as
)(
)0()0(sin,
)(
)0()0(sin
)0()0(cos,
)0()0(cos
122
2112
)2(1
121
2121
)1(1
12
2112
)2(1
12
2121
)1(1
rr
xxrX
rr
xxrX
rr
xxrX
rr
xxrX
FREE VIBRATION ANALYSIS OF AN UNDAMPED SYSTEM
15
from which we obtain the desired solution
)18.5()0()0([
)0()0(tan
cos
sintan
)0()0([
)0()0(tan
cos
sintan
)0()0()0()0(
)(
1
sincos
)0()0()0()0(
)(
1
sincos
2112
2111
2)2(
1
2)2(
112
2121
2121
1)1(
1
1)1(
111
2/1
22
22112
21112
2/12
2)2(
1
2
2)2(
1)2(
1
2/1
21
22122
21212
2/12
1)1(
1
2
1)1(
1)1(
1
xxr
xxr
X
X
xxr
xxr
X
X
xxrxxr
rr
XXX
xxrxxr
rr
XXX
FREE VIBRATION ANALYSIS OF AN UNDAMPED SYSTEM
16
Solution: For the given data, the eigenvalue problem, Eq.(5.8), becomes
EXAMPLE 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)0
0
55-
5 3510
0
0
2
1
2
2
2
1
322
22
2212
1
X
X
X
X
kkmk
kkkm
or
17
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
E.3)(4495.2,5811.1
0.6,5.2
21
22
21
The normal modes (or eigenvectors) are given by
E.5)(5
1
E.4)(2
1
)2(1)2(
2
)2(1)2(
)1(1)1(
2
)1(1)1(
XX
XX
XX
XX
18
By using the given initial conditions in Eqs.(E.6) and (E.7), we obtain
The free vibration responses of the masses m1 and 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
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
XXtx
XXtx
XXtx
XXtx
EXAMPLE 5.3 SOLUTION
19
while the solution of Eqs.(E.10) and (E.11) leads to
The solution of Eqs.(E.8) and (E.9) yields
(E.12)7
2cos;
7
5cos 2
)2(11
)1(1 XX
(E.13)0sin,0sin 2)2(
11)1(
1 XX
Equations (E.12) and (E.13) give
(E.14)0,0,7
2,
7
521
)2(1
)1(1 XX
EXAMPLE 5.3 SOLUTION
20
EXAMPLE 5.3 SOLUTION
Thus the free vibration responses of m1 and m2 are given by
(E.16)4495.2cos7
105811.1cos
7
10)(
(E.15)4495.2cos7
25811.1cos
7
5)(
2
1
tttx
tttx
21
Figure 5.6: Torsional system with discs mounted on a shaft
TORSIONAL SYSTEM
Consider a torsional system as shown in Fig.5.6. The differential equations of rotational motion for the discs can be derived as
22
which upon rearrangement become
TORSIONAL SYSTEM
22312222
11221111
)(
)(
ttt
ttt
MkkJ
MkkJ
)19.5()(
)(
22321222
12212111
tttt
tttt
MkkkJ
MkkkJ
For the free vibration analysis of the system, Eq.(5.19) reduces to
)20.5(0)(
0)(
2321222
2212111
ttt
ttt
kkkJ
kkkJ
23
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):
EXAMPLE 5.4:NATURAL FREQUENCIES OF TORSIONAL SYSTEM
(E.1) 02
02
2120
2110
tt
tt
kkJ
kkJ
Fig.5.7:
Torsional system
24
The solution of Eq.(E.3) gives the natural frequencies
gives the frequency equation:
EXAMPLE 5.4 SOLUTION
(E.2)2,1);cos()( itt ii
(E.3)052 20
220
4 tt kkJJ
(E.4))175(4
and)175(4 0
20
1 J
k
J
k tt
Rearranging and substituting the harmonic solution:
25
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
(E.5)4
)175(2
4
)175(2
)2(1
)2(2
2
)1(1
)1(2
1
r
r
0and2,
,,
3022011
2211
kJJmJJm
kkkkkk tttt
EXAMPLE 5.4 SOLUTION
26
Generalized coordinates are sets of n coordinates used to describe the configuration of the system.•Equations of motion Using x(t) and θ(t).
COORDINATE COUPLING AND PRINCIPAL COORDINATES
27
and the moment equation about C.G. can be expressed as
)21.5()()( 2211 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
Eqs.(5.21) and (5.22) can be rearranged and written in matrix form as
COORDINATE COUPLING AND PRINCIPAL COORDINATES
28
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(0
0
)( )(
)( )(
0
0 2
22
12211
221121
0 21
x
lklklklk
lklkkkx
J
m
•Equations of motion Using y(t) and θ(t).
melyklykym )()( 2211
COORDINATE COUPLING AND PRINCIPAL COORDINATES
29
These equations can be rearranged and written in matrix form as
)24.5()()( 222111 ymellykllykJP
)25.5(0
0
)()(
)()(
2
22
112211
112221
2
y
lklklklk
lklkkky
Jme
mem
P
If , the system will have dynamic or inertia coupling only.
2211 lklk
Note the following characteristics of these systems:
COORDINATE COUPLING AND PRINCIPAL COORDINATES
30
1.In the most general case, a viscously damped two degree of freedom system has the equations of motions in the form:
)26.5(0
0
2
1
2221
1211
2
1
2221
1211
2
1
2221
1211
x
x
kk
kk
x
x
cc
cc
x
x
mm
mm
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.
COORDINATE COUPLING AND PRINCIPAL COORDINATES
31
EXAMPLE 5.6:PRINCIPAL COORDINATES OF SPRING-MASS SYSTEM
Determine the principal coordinates for the spring-mass system shown in Fig.5.4.
32
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)3
coscos)(
3coscos)(
22112
22111
tm
kBt
m
kBtx
tm
kBt
m
kBtx
33
Since the coordinates are harmonic functions, their corresponding equations of motion can be written as
(E.2)3
cos)(
cos)(
222
111
tm
kBtq
tm
kBtq
(E.3)03
0
22
11
qm
kq
qm
kq
EXAMPLE 5.6 SOLUTION
34
The solution of Eqs.(E.4) gives the principal coordinates:
From Eqs.(E.1) and (E.2), we can write
(E.4))()()(
)()()(
212
211
tqtqtx
tqtqtx
(E.5))]()([2
1)(
)]()([2
1)(
212
211
txtxtq
txtxtq
EXAMPLE 5.6 SOLUTION
35
The equations of motion of a general two degree of freedom 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
F
F
x
x
kk
kk
x
x
cc
cc
x
x
mm
mm
)28.5(2,1,)( 0 jeFtF tijj
where ω is the forcing frequency. We can write the steady-state solutions as
)29.5(2,1,)( jeXtx tijj
36
Substitution of Eqs.(5.28) and (5.29) into Eq.(5.27) leads to
)30.5(
)()(
)()(
20
10
2
1
22222221212122
1212122
1111112
F
F
X
X
kcimkcim
kcimkcim
We defined as in section 3.5 the mechanical impedance Zre(iω) as
)31.5(2,1,,)( 2 srkcimiZ rsrsrsrs
FORCED VIBRATION ANALYSIS
37
And write Eq.(5.30) as:
)32.5()( 0FXiZ
Where,
20
100
2
1
2212
1211 matrix Impedance)( )(
)( )()(
F
FF
X
XX
iZiZ
iZiZiZ
FORCED VIBRATION ANALYSIS
38
where the inverse of the impedance matrix is given
)33.5()( 01FiZX
)34.5()( )(
)( )(
)()()(
1)(
1112
1222
2122211
1
iZiZ
i-ZiZ
iZiZiZiZ
Eqs.(5.33) and (5.34) lead to the solution
)35.5()()()(
)()()(
)()()(
)()()(
2122211
201110122
2122211
201210221
iZiZiZ
FiZFiZiX
iZiZiZ
FiZFiZiX
FORCED VIBRATION ANALYSIS
Eq.(5.32) can be solved to obtain:
39
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.
EXAMPLE 5.8:STEADY-STATE RESPONSE OF SPRING-MASS SYSTEM
40
The equations of motion of the system can be expressed as
EXAMPLE 5.8 SOLUTION
(E.1)0
cos
2
2
0
0 10
2
1
2
1
tF
x
x
k-k
-kk
x
x
m
m
E.2)(2,1;cos)( jtXtx jj
We assume the solution to be as follows.
Eq.(5.31) gives
(E.3))(,2)()( 122
2211 kZkmZZ
41
(E.5)))(3()2(
)(
(E.4)))(3(
)2(
)2(
)2()(
2210
22210
2
2210
2
22210
2
1
kmkm
kF
kkm
kFX
kmkm
Fkm
kkm
FkmX
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
k
F
X
EXAMPLE 5.8 SOLUTION
42Fig.5.14: Frequency response curves
E.7)(
1
)(2
1
2
1
2
1
2
102
k
FX
EXAMPLE 5.8 SOLUTION
