Transcript Damped single degree of freedom system
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Mechanical Vibrations
Single degree of freedom system with viscous damping:
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System free body diagram C is the damping constant or coefficient
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Mathematical model (governing equation of motion):
..
.
m x
c x
kx
0
c
.
x
F d
is the damping force.
It is proportion al to velocity
Solution: Assume x(t) = B e st, substitute into the equation of motion :
ms
2
cs
k
0
s
1 , 2
c
c
2 4
mk
2
m s
1 , 2
c
2
m
c
2
m
2
k m
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s
1 , Define 2 C 2
m
n
2 , then the 1
n
roots , where is of the solution known as the can be written
damping ra tio
.
as :
C c
C
, where
C c
is the
critical d amping coe fficient C c
2
m
n
.
In the face of the value of the square root shown above three possible solutions are exist.
These solutions are as follows :
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Solution # 1: under-damped vibration: When ζ<1 The roots S as:
s
1 , 2
1,2
of the characteristic equation can now be written
1 2
n
The solution for this case becomes :
x
e
n t
OR
x
B
1 cos 1
Xe
n t
2
n t
cos 1
B
2 sin 2
n t
1 2
n t
Define d
x
1
Ae
2
n t
n t
as the cos d
t
damped nat ural frequ
ency
, then
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The solution (above) for the under-damped vibration case (
<1
consists of a harmonic motion of frequency
d and an amplitude
Ae
n t
Note that:
A
B
1 2
B
2 2 and tan 1
B
2
B
1 is the phase angle.
A
and can be determined from the inatial conditions of the motion.
The amplitde for this case is exponantia lly decaying with time as shown.
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Frequency of damped vibration:
d
1 2
n
Under-damped vibration
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Critical damping (c c )
The critical damping
c c
is defined as the value of the damping constant for which the radical in s-equation becomes zero:
c
2
c c m
2
k m
0
c c
2
m k m
2
m
n
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Solution # 2: critically damped vibration (ζ=1): For this case the roots of the characteristic equation become:
s
1 , 2
n
Therefor t
x
e
n t
he solution
B
1
B
2
t
can be written as :
B
1 and
B
2 are constants that can be determined from inatial conditions .
The motion is no longer harmonic as shown in the figure.
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-The system returns to the equilibrium position in short time -The shape of the curve depends on initial conditions as shown -The moving parts of many electrical meters and instruments are critically damped to avoid overshoot and oscillations .
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Solution # 3: over-damped vibration (ζ>1) For this case the roots of the characteristic equation become:
x s
1 , 2
Therefore,
the
2 1
solution
n
can be written
B
1
e
2 1
n t
B
2
e
2 1
n t
as,
B
1
and
B
2
are constants to be determined from knowing the initial conditions of the motion.
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Graphical representation of the motions of the damped systems
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Logarithmic decrement The logarithmic decrement represents the rate at which the amplitude of a free-damped vibration decreases. It is defined as the natural logarithm of the ratio of any two successive amplitudes.
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Logarithmic decrement:
x
1 (
t
)
x
2 (
t
)
So
Ae
n t
1
Ae
n t
2 cos cos
d d t
1
t
2
x
ln 1 (
t
)
x
2 (
t
)
Assume: δ is the logarithmic decrement
x
1 (
t
)
x
2 (
t
)
e e
n t
1
n
t
1
d
n
d
n
But
e
n
d t
2 cos 2 1 2
n
t
1
d
t
2
d
d
cos cos 2
d
2
d
2 1 2
t
1
d
t
1
Logarithmic decrement: is dimensionless
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Logarithmic decrement
2 2
For small damping ; ζ <<1
2
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Generally, when the amplitude after a number of cycles “n” is known, logarithmic decrement can be obtained as follows:
1 ln
n x
1 (
t
)
x n
(
t
) 2 1 2 where
x
1 (t) is the first known amplitude,
x n
(t) is the other known amplitude after n cycles of the decayed motion.
Example 2.6
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An under-damped shock absorber is to be designed for a motorcycle of mass 200 kg (Fig.(a)).When the shock absorber is subjected to an initial vertical velocity due to a road bump, the resulting displacement-time curve is to be as indicated in Fig.(b).
Requirements: 1.
Find the necessary stiffness and damping constants of the shock absorber if the damped period of vibration is to be 2s and the amplitude x 1 /4 ).
x 1 is to be reduced to one-fourth in one half cycle (i.e. x 1.5
=
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Example 2.6
Note that this system is under-damped system
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Solution: Finding k and c
Here,
n
x
0 .
5 and
x n
1 4, then the logarithmi 1
n
ln
x
1
x n
0 .
4037 1 0 .
5 ln 2 .
7726 2 1 2 c decrement is :
d
2 2
d
n
2 1 2
n
n
2 1 0 .
4037 2 3 .
4338
rad
/
s
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The critical damping can be found as:
c c
2
m
n c
c c k
m
n
2 2 3 .
4338 1373 .
54 N.sec/m
Damping coefficient C and stiffness K can be found as:
0 .
4037 1373 .
54 554 .
4981 N.sec/m...
....(Ans.) 3 .
4338 2 2358 .
2652 N/m.......
.(Ans.)
S t a t i c s