SDOF Free Vibrations
Free Vibration with Viscous Damping
• Torsional systems with Viscous Damping:
Consider a single degree of freedom torsional system with a
viscous damper, as shown in figure (a). The viscous damping
torque is given by:
T = −ctθ
(2.101)
The equation of motion can be derived as:
J 0θ + ctθ + ktθ = 0
(2.102)
where J0 = mass moment of inertia of disc
kt = spring constant of system
θ = angular displacement of disc
SDOF Free Vibrations
Free Vibration with Viscous Damping
In the underdamped case, the frequency of damped vibration is
given by:
ωd = 1 − ζ 2 ωn
(2.103)
where
ωn =
and
kt
J0
ct
ct
ct
=
=
ζ =
ctc 2 J 0ωn 2 kt J 0
where ctc is the critical torsional damping constant
(2.104)
(2.105)
SDOF Free Vibrations
Example 2.11 Shock Absorber for a Motorcycle
An underdamped shock absorber is to be
designed for a motorcycle of mass 200kg
(shown in Fig.(a)). When the shock absorber
is subjected to an initial vertical velocity due
to a road bump, the resulting displacementtime curve is to be as indicated in Fig.(b).
Find the necessary stiffness and damping
constants of the shock absorber if the damped
period of vibration is to be 2 s and the
amplitude x1 is to be reduced to one-fourth in
one half cycle (i.e., x1.5 = x1/4). Also find the
minimum initial velocity that leads to a
maximum displacement of 250 mm.
SDOF Free Vibrations
Example 2.11 Shock Absorber for a Motorcycle
Since
x1.5 = x1 / 4,
x2 = x1.5 / 4 = x1 / 16
,
Hence the logarithmic decrement becomes
x 
2πζ
δ = ln 1  = ln(16) = 2.7726 =
1− ζ 2
 x2 
(E.1)
From which ζ can be found as 0.4037. The damped period of
vibration given by 2 s. Hence,
2 =τd =
ωn =
2π
ωd
=
2π
ωn 1 − ζ 2
2π
2 1 − (0.4037)
2
= 3.4338 rad/s
SDOF Free Vibrations
Example 2.11 Shock Absorber for a Motorcycle
The critical damping constant can be obtained:
cc = 2mωn = 2(200)(3.4338) = 1.373.54 N - s/m
Thus the damping constant is given by:
c = ζcc = (0.4037)(1373.54) = 554.4981 N - s/m
and the stiffness by:
k = mωn2 = (200)(3.4338) 2 = 2358.2652 N/m
The displacement of the mass will attain its max value at time t1:
(
x(t ) = X 0 e −ζω nt sin 1 − ζ 2 ωnt + φ
sin ωd t1 = 1 − ζ 2
This gives:
or
Try to prove
sin ωd t1 = 1 − (0.4037) 2 = 0.9149
t1 = 0.3678 sec
)
(2.70)
SDOF Free Vibrations
Example 2.11 Shock Absorber for a Motorcycle
The envelope passing through the max points is:
x = 1 − ζ 2 Xe −ζω nt
(E.2)
Since x = 250mm,
0.25 = 1 − (0.4037) 2 Xe − ( 0.4037 )(3.4338)( 0.3678)
X = 0.4550 m
The velocity of mass can be obtained by differentiating the
displacement:
x(t ) = Xe −ζω nt sin ωd t
as
x (t ) = Xe −ζω nt (−ζω n sin ωd t + ωd cos ωd t )
(E.3)
When t = 0,
x (t = 0) = x0 = Xωd = Xωn 1 − ζ 2 = (0.4550)(3.4338) 1 − (0.4037) 2
= 1.4294 m/s
SDOF Free Vibrations
QUIZ