two degree of freddom system
TRANSCRIPT
GANDHINAGAR INSTITUTE OF TECHNOLOGY
Subject: Dynamics of machineTopic:-Two degree of freedom system
Branch:-MechanicalSem.:-6Div:-D
Prepared By:- Guided By:-Patel Yash S. (150123119039) Prof. Samir Raval
INTRODUCTION• The System which require two independent co-
ordinates to specify its motion at configuration at instant is called two degree of freedom system.– 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.
Figure. A two degree of freedom spring-mass-damper system
Equations of Motion for Forced Vibration
)2.5()()()1.5()()(
2232122321222
1221212212111
FxkkxkxccxcxmFxkxkkxcxccxm
)3.5( )()(][)(][)(][ tFtxktxctxm
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:
where [m], [c], and [k] are called the mass, damping, and stiffness matrices, respectively, and are given by
Equations of Motion for Forced Vibration
322
221
322
221
2
1
][
][ 0
0 ][
kkkkkk
k
cccccc
cm
mm
)()(
)()()(
)(2
1
2
1
tFtF
tFtxtx
tx
And the displacement and force vectors are given respectively:
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.
Equations of Motion for Forced Vibration
][][],[][],[][ kkccmm TTT
6
where the superscript T denotes the transpose of the matrix.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
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
Free Vibration Analysis of an Undamped System
)5.5(0)()()()()4.5(0)()()()(
2321222
2212111
txkktxktxmtxktxkktxm
)6.5()cos()()cos()(
22
11
tXtxtXtx
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
By setting F1(t) = F2(t) = 0, and damping disregarded, i.e., c1 = c2 = c3 = 0, and the equation of motion is reduced to:
Free Vibration Analysis of an Undamped System
)7.5( 0)cos()(
0)cos()(
2322
212
221212
1
tXkkmXk
tXkXkkm
)8.5(0)(
0)(
2322
212
221212
1
XkkmXk
XkXkkm
Since Eq.(5.7)must be satisfied for all values of the time t, the terms between brackets must be zero. Thus,
Substituting into Eqs.(5.4) and (5.5),
Free Vibration Analysis of an Undamped System
0
)(
)(det
212
12
2212
1
kkmk
kkkm
)9.5(0))((
)()()(223221
1322214
21
kkkkk
mkkmkkmm
or
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:
)10.5())((4
)()(21
)()(21,
2/1
21
223221
2
21
132221
21
13222122
21
mmkkkkk
mmmkkmkk
mmmkkmkk
The roots are called natural frequencies of the system.
which is called the frequency or characteristic equation. Hence the roots are:
FREE VIBRATION ANALYSIS OF AN UNDAMPED SYSTEM
Free Vibration Analysis of an Undamped System
)11.5()(
)(
)()(
32222
2
2
21221
)2(1
)2(2
2
32212
2
2
21211
)1(1
)1(2
1
kkmk
kkkm
XXr
kkmk
kkkm
XXr
)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
The normal modes of vibration corresponding to ω12
and ω22 can be expressed, respectively, as
To determine the values of X1 and X2, given ratio
which are known as the modal vectors of the system.
Free Vibration Analysis of an Undamped System
(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
0)0(,)0(
,0)0(constant, some )0(
2)(
12
1)(
11
txXrtx
txXtxi
i
i
Where the constants , , and are determined by the initial conditions. The initial conditions are
The free vibration solution or the motion in time can be expressed itself as
)1(1X
)2(1X 1 2
Free Vibration Analysis of an Undamped System
)14.5()()()( 2211 txctxctx
)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
Thus the components of the vector can be expressed as
The resulting motion can be obtained by a linear superposition of the two normal modes, Eq.(5.13)
where the unknown constants can be determined from the initial conditions:
Free Vibration Analysis of an Undamped System
)16.5()0()0(),0()0(),0()0(),0()0(
2222
1111
xtxxtxxtxxtx
)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
)()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
rrxxrX
rrxxrX
rrxxrX
rrxxrX
Substituting into Eq.(5.15) leads to
The solution can be expressed as
Free Vibration Analysis of an Undamped System
)18.5()0()0([
)0()0(tancossintan
)0()0([)0()0(tan
cossintan
)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
2/122
)2(1
22
)2(1
)2(1
2/1
21
22122
21212
2/121
)1(1
21
)1(1
)1(1
xxrxxr
XX
xxrxxr
XX
xxrxxrrr
XXX
xxrxxrrr
XXX
from which we obtain the desired solution
Example :Free Vibration Response of a Two Degree of Freedom System
).0()0()0( ,1)0( 2211 xxxx
(E.1)00
55-
5 3510
00
2
1
2
2
2
1
322
22
2212
1
XX
XX
kkmk
kkkm
Solution: For the given data, the eigenvalue problem, Eq.(5.8), becomes
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
or
Solution
(E.2)01508510 24
E.3)(4495.2,5811.10.6,5.2
21
22
21
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
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,
The normal modes (or eigenvectors) are given by
Solution
(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
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):
Solution
(E.12)72cos;
75cos 2
)2(11
)1(1 XX
(E.13)0sin,0sin 2)2(
11)1(
1 XX
(E.14)0,0,72,
75
21)2(
1)1(
1 XX
while the solution of Eqs.(E.10) and (E.11) leads to
The solution of Eqs.(E.8) and (E.9) yields
Equations (E.12) and (E.13) give
Solution
(E.16)4495.2cos7
105811.1cos7
10)(
(E.15)4495.2cos725811.1cos
75)(
2
1
tttx
tttx
Thus the free vibration responses of m1 and m2 are given by
Torsional System
Figure : Torsional system with discs mounted on a shaft
Consider a torsional system as shown in Fig. The differential equations of rotational motion for the discs can be derived as
Torsional System
22312222
11221111
)(
)(
ttt
ttt
MkkJ
MkkJ
)19.5()(
)(
22321222
12212111
tttt
tttt
MkkkJ
MkkkJ
which upon rearrangement become
For the free vibration analysis of the system, Eq.(5.19) reduces to
)20.5(0)(
0)(
2321222
2212111
ttt
ttt
kkkJ
kkkJ
Coordinate Coupling and Principal CoordinatesGeneralized 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
)21.5()()( 2211 lxklxkxm
)22.5()()( 2221110 llxkllxkJ
and the moment equation about C.G. can be expressed as
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
Eqs.(5.21) and (5.22) can be rearranged and written in matrix form as
Coordinate Coupling and Principal Coordinates
)23.5(00
)( )(
)( )(
00
22
212211
221121
0 21
x
lklklklk
lklkkkxJ
m
melyklykym )()( 2211
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
•Equations of motion Using y(t) and θ(t).
Coordinate Coupling and Principal Coordinates
)24.5()()( 222111 ymellykllykJP
)25.5(00
)()(
)()(
22
2112211
112221
2
y
lklklklk
lklkkkyJmemem
P
2211 lklk
These equations can be rearranged and written in matrix form as
If , the system will have dynamic or inertia coupling only.
Note the following characteristics of these systems:
Coordinate Coupling and Principal Coordinates
)26.5(00
2
1
2221
1211
2
1
2221
1211
2
1
2221
1211
xx
kkkk
xx
cccc
xx
mmmm
1. In the most general case, a viscously damped two degree of freedom system has the equations of motions in the form:
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.
Example :Principal Coordinates of Spring-Mass SystemDetermine the principal coordinates for the spring-mass system shown in Fig.
Solution
(E.1)3coscos)(
3coscos)(
22112
22111
tmkBt
mkBtx
tmkBt
mkBtx
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
Solution
(E.2)3cos)(
cos)(
222
111
tmkBtq
tmkBtq
(E.3)03
0
22
11
qmkq
qmkq
Since the coordinates are harmonic functions, their corresponding equations of motion can be written as
Solution
(E.4))()()()()()(
212
211
tqtqtxtqtqtx
(E.5))]()([21)(
)]()([21)(
212
211
txtxtq
txtxtq
The solution of Eqs.(E.4) gives the principal coordinates:
From Eqs.(E.1) and (E.2), we can write
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