1. GANDHINAGAR INSTITUTE
OF TECHNOLOGY
Subject: Dynamics of machine
Topic:-Two degree of freedom system
Branch:-Mechanical
Sem.:-6
Div:-D
Prepared By:- Guided By:-
Patel Yash S. (150123119039) Prof. Samir Raval
2. 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.
3. 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
4. Equations of Motion for Forced Vibration
)2.5()()(
)1.5()()(
2232122321222
1221212212111
Fxkkxkxccxcxm
Fxkxkkxcxccxm
)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
5. Equations of Motion for Forced Vibration
322
221
322
221
2
1
][
][
0
0
][
kkk
kkk
k
ccc
ccc
c
m
m
m
)(
)(
)(
)(
)(
)(
2
1
2
1
tF
tF
tF
tx
tx
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.
6. 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).
x
x x
x
7. Free Vibration Analysis of an Undamped System
)5.5(0)()()()(
)4.5(0)()()()(
2321222
2212111
txkktxktxm
txktxkktxm
)6.5()cos()(
)cos()(
22
11
tXtx
tXtx
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:
8. Free Vibration Analysis of an Undamped System
)7.5(0)cos()(
0)cos()(
232
2
212
22121
2
1
tXkkmXk
tXkXkkm
)8.5(0)(
0)(
232
2
212
22121
2
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),
9. Free Vibration Analysis of an Undamped System
0
)(
)(
det
21
2
12
221
2
1
kkmk
kkkm
)9.5(0))((
)()()(
2
23221
132221
4
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:
11. Free Vibration Analysis of an Undamped System
)11.5(
)(
)(
)(
)(
32
2
22
2
2
21
2
21
)2(
1
)2(
2
2
32
2
12
2
2
21
2
11
)1(
1
)1(
2
1
kkm
k
k
kkm
X
X
r
kkm
k
k
kkm
X
X
r
)12.5(and )2(
12
)2(
1
)2(
2
)2(
1)2(
)1(
11
)1(
1
)1(
2
)1(
1)1(
Xr
X
X
X
X
Xr
X
X
X
X
The normal modes of vibration corresponding to ω1
2
and ω2
2 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.
12. Free Vibration Analysis of an Undamped System
(5.17)modesecond
)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
tx
tx
tXr
tX
tx
tx
tx
0)0(,)0(
,0)0(constant,some)0(
2
)(
12
1
)(
11
txXrtx
txXtx
i
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
13. 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:
14. Free Vibration Analysis of an Undamped System
)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
)(
)0()0(
sin,
)(
)0()0(
sin
)0()0(
cos,
)0()0(
cos
122
211
2
)2(
1
121
212
1
)1(
1
12
211
2
)2(
1
12
212
1
)1(
1
rr
xxr
X
rr
xxr
X
rr
xxr
X
rr
xxr
X
Substituting into Eq.(5.15) leads to
The solution can be expressed as
19. Solution
(E.12)
7
2
cos;
7
5
cos 2
)2(
11
)1(
1 XX
(E.13)0sin,0sin 2
)2(
11
)1(
1 XX
(E.14)0,0,
7
2
,
7
5
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
21. 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
23. Coordinate Coupling and Principal Coordinates
Generalized coordinates are sets of n coordinates used
to describe the configuration of the system.
•Equations of motion Using x(t) and θ(t).
24. 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
25. Coordinate Coupling and Principal Coordinates
)23.5(
0
0
)()(
)()(
0
0
2
2
2
12211
221121
0 21
x
lklklklk
lklkkkx
J
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).
26. Coordinate Coupling and Principal Coordinates
)24.5()()( 222111 ymellykllykJP
)25.5(
0
0
)()(
)()(
2
2
2
112211
112221
2
y
lklklklk
lklkkky
Jme
mem
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:
27. Coordinate Coupling and Principal Coordinates
)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
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.
28. Example :Principal Coordinates of Spring-Mass System
Determine the principal coordinates for the spring-mass
system shown in Fig.