SDOF Forced Vibrations

MDPN471
3
CHAPTER
Mechanical Vibrations
Single degree-of-freedom
Forced Vibrations
SDOF Forced Vibrations
3.6 Response Under the Harmonic Motion of Base
From the figure, the equation of motion is
mx + c( x − y ) + k ( x − y ) = 0 (3.64)
If y (t ) = Y sin ωt,
mx + cx + kx = ky + cy = kY sin ωt + cωY cos ωt
= A sin(ωt − α ) (3.65)
 cω 
where A = Y k 2 + (cω ) 2 and α = tan −1 − 
 k 
REMEMBER
The system with equation of motion:
F0 cos ωt
mx(t ) + cx (t ) + kx (t ) =
the particular solution is :
x p (t ) X cos (ωt − φ )
=
F0 δ st
=X =
( k − mω ) ω
2 2
+ c 2 2 (1 − r 2 ) 2 + (2ζ r ) 2
 cω  −1  2ζ r 
=ϕ tan
=  2 
−1
tan  2 
 k − mω   1 − r 
The steady-state response of the system
kx A sin(ωt − α )
mx + cx +=
can be expressed as
Y k 2 + (cω ) 2
x p (t ) = sin(ωt − φ1 − α )
[(k − mω ) ]
(3.66)
2 2
+ (cω ) 2 1/ 2
 cω 
where φ1 = tan 
−1
2 
 k − mω 
or x p (t ) = X sin(ωt − φ ) (3.67)
where X  k 2 + (cω ) 2 
1/ 2
 1 + (2ζr ) 2 
1/ 2
=  = 2
(3.68)
Y  (k − mω 2 ) + (cω ) 2   (1 − r 2 2
) + ( 2ζ r ) 
 mcω 3  −1  2ζr 3 
and φ = tan 
−1
2
= tan  2
(3.69)
 k (k − mω ) + (ωc)  1 + (4ζ − 1)r 
2 2
The variations of displacement transmissibility is
shown in the figure below.
1 + ( 2ζ r ) 2
 2ζ r 3

Td = X = φ = tan 
−1
2
Y (1 − r 2 ) 2 + (2ζr ) 2 1 + ( 4ζ 2
− 1) r 
The following aspects of Td can be noted from the figure:
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:
rm = 1 1 + 8ζ 2 − 1
2ζ
•Force transmitted: (force transmitted to base from spring and damper)
k ( x − y ) + c( x − y ) =
F= −mx (3.72)
= mω 2 X sin(ωt=
− ϕ ) FT sin(ωt − ϕ )
X 1 + (2ζ r ) 2
=
Y (1 − r 2 ) 2 + (2ζ r ) 2
The force transmissibility

is given by:
FT 1 + (2 ζ r ) 2
T= = = r 2
(1 − r 2 ) 2 + (2ζ r ) 2
f
Fo
Disturbing force = kY
•Relative Motion:
mx + c( x − y ) + k ( x − y ) =
0
For z= x − y :
The equation of motion can be written as
mz + cz + kz =−my =mω 2Y sin ωt (3.75)

•Relative Motion:
For mz + cz + kz = −my = mω 2Y sin ωt (3.75)

The steady-state solution is given by:
z (t ) Z sin(ωt − ϕ1 )
= (3.76)
where, the amplitude Z:

mω 2Y r2
Z= =Y (3.77)
(k − mω ) + (cω )
2 2 2
(1 − r ) + (2ζr )
2 2 2
cω  −1  2ζ r 
And phase angle φ1 = tan −1  2 
= tan  2 
 k − mω   1 − r 
r2
Z =Y
(1 − r 2 ) 2 + (2ζr ) 2
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.
Example 3.3 Solution
The frequency can be found by
 v × 1000  1
ω = 2πf = 2π   = 0.290889v rad/s
 3600  6
For v = 20 km/hr, ω = 5.81778 rad/s. The natural

frequency is given by,
1/ 2
k  400 ×10 3

ωn = =   = 18.2574 rad/s
m  1200 
Hence, the frequency ratio is

ω 5.81778
r= = = 0.318653
ωn 18.2574
Example 3.3 Solution
The amplitude ratio can be found from Eq.(3.68):
1/ 2 1/ 2
X  1 + (2ζr ) 2
  1 + (2 × 0.5 × 0.318653) 2

=  = 2
Y  (1 − r 2 ) 2 + (2ζr ) 2   (1 − 0. 318653 ) 2
+ ( 2 × 0 . 5 × 0.318653) 
= 1.469237
Thus, the displacement amplitude of the vehicle is given by
X = 1.469237Y = 1.469237(0.05) = 0.073462 m
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.
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MDPN471

the particular solution is :

The force transmissibility

mz + cz + kz =−my =mω 2Y sin ωt (3.75)

For mz + cz + kz = −my = mω 2Y sin ωt (3.75)

where, the amplitude Z:

For v = 20 km/hr, ω = 5.81778 rad/s. The natural

Hence, the frequency ratio is

Thus, the displacement amplitude of the vehicle is given by

X = 1.469237Y = 1.469237(0.05) = 0.073462 m

This indicates that a 5cm bump in the road is transmitted

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