single degree-of-freedom forced...
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SDOF Forced Vibrations 3.6 Response Under the Harmonic Motion of Base
)64.3(0)()( =−+−+ yxkyxcxm
tYty ωsin)( =
)65.3()sin(cossin
αωωωω
−=+=+=++
tAtYctkYyckykxxcxm
−=+= −
kcckYA ωαω 122 tan and )(
If ,
From the figure, the equation of motion is
where
SDOF Forced Vibrations REMEMBER
0( ) ( ) ( ) cosmx t cx t kx t F tω+ + =
The system with equation of motion:
the particular solution is :
( ) cos( ) px t X tω φ= −
( )0
2 2 2 22 2 2 (1 ) (2 )stFX
r rk m c
δζω ω
= =− +− +
1 12 2
2tan tan1
c rk m r
ω ζϕω
− − = = − −
SDOF Forced Vibrations
[ ] )66.3()sin()()(
)()( 12/1222
22
αφωωω
ω−−
+−
+= t
cmkckY
txp
−= −
21
1 tanω
ωφmk
c
)67.3()sin()( φω −= tXtxp
)68.3()2()1(
)2(1)()(
)(2/1
222
22/1
22
22
+−
+=
+−
+=
rrr
cmkck
YX
ζζ
ωωω
The steady-state response of the system
can be expressed as
where
or
where
and )69.3()14(1
2tan)()(
tan 22
31
22
31
−+
=
+−
= −−
rr
cmkkmc
ζζ
ωωωφ
3.6 Response Under the Harmonic Motion of Base
sin( )mx cx kx A tω α+ + = −
SDOF Forced Vibrations
The variations of displacement transmissibility is shown in the figure below.
3.6 Response Under the Harmonic Motion of Base
222
2
)2()1()2(1
rrr
YXTd ζ
ζ+−
+==
−+
= −22
31
)14(12tan
rr
ζζφ
SDOF Forced Vibrations
1. The value of Td is unity at r = 0 and close to unity for small values of r.
2. For undamped system (ζ = 0), Td →∞ at resonance (r = 1).
3. The value of Td is less than unity (Td < 1) for values of r >√2 (for any amount of damping ζ ).
4. The value of Td = 1 for all values of ζ at r =√2.
5. For r <√2, smaller damping ratios lead to larger values of Td. On the other hand, for r >√2, smaller values of damping ratio lead to smaller values of Td.
6. The displacement transmissibility, Td, attains a maximum for 0 < ζ < 1 at the frequency ratio r = rm < 1 given by:
The following aspects of Td can be noted from the figure: 3.6 Response Under the Harmonic Motion of Base
18121 2 −+= ζζmr
SDOF Forced Vibrations 3.6 Response Under the Harmonic Motion of Base
2
( ) ( ) (3.72)Xsin( ) sin( )T
F k x y c x y mxm t F tω ω ϕ ω ϕ
= − + − = −
= − = −
•Force transmitted: (force transmitted to base from spring and damper)
The force transmissibility is given by:
Disturbing force
22
2 2 2
1 (2 )(1 ) (2 )
rrr r
ζζ
+=
− +T
fo
FTF
= =
kY=
2
2 2 2
1 (2 )(1 ) (2 )
X rY r r
ζζ
+=
− +
SDOF Forced Vibrations
2 sin (3.75)mz cz kz my m Y tω ω+ + = − =
The equation of motion can be written as
•Relative Motion:
3.6 Response Under the Harmonic Motion of Base
( ) ( ) 0
For :
mx c x y k x y
z x y
+ − + − =
= −
SDOF Forced Vibrations
1( ) sin( ) (3.76)z t Z tω ϕ= −
)75.3(sin2 tYmymkzzczm ωω=−=++
)77.3()2()1()()( 222
2
222
2
rrrY
cmkYmZ
ζωωω
+−=
+−=
−
=
−= −−
21
21
1 12tantan
rr
mkc ζω
ωφ
For
The steady-state solution is given by:
where, the amplitude Z:
•Relative Motion:
And phase angle
3.6 Response Under the Harmonic Motion of Base
SDOF Forced Vibrations Example 3.3 Vehicle Moving on a Rough Road
The figure shows a simple model of a motor vehicle that can vibrate in the vertical direction while traveling over a rough road. The vehicle has a mass of 1200kg. The suspension system has a spring constant of 400 kN/m and a damping ratio of ζ = 0.5. If the vehicle speed is 20 km/hr, determine the displacement amplitude of the vehicle. The road surface varies sinusoidally with an amplitude of Y = 0.05m and a wavelength of 6m.
SDOF Forced Vibrations
rad/s 290889.061
3600100022 vvf =
×
== ππω
rad/s 2574.181200
104002/13
=
×==
mk
nω
318653.02574.18
81778.5===
n
rωω
The frequency can be found by
For v = 20 km/hr, ω = 5.81778 rad/s. The natural frequency is given by,
Hence, the frequency ratio is
Example 3.3 Solution
SDOF Forced Vibrations
469237.1)318653.05.02()318653.01(
)318653.05.02(1)2()1(
)2(12/1
22
22/1
222
2
=
××+−××+
=
+−+
=rr
rYX
ζζ
m 073462.0)05.0(469237.1469237.1 === YX
Thus, the displacement amplitude of the vehicle is given by
The amplitude ratio can be found from Eq.(3.68):
This indicates that a 5cm bump in the road is transmitted as a 7.3cm bump to the chassis and the passengers of the car.
Example 3.3 Solution