76
3 Forced Vibration of Single Degree of Freedom System
Y
y s0 k
1 +
cω
k
2
mp 2
1 −
ω
p
2
2
+
2 ·
n
p
·
ω
p
2
or
Y
y s0
=
1 + ( 2η ζ ) 2
( 1 − η 2 ) 2 + ( 2η ζ ) 2
In this case, Y/y s0 the ratio of the mass displacement to support displacement is known as the transmissibility. Comparing Eq. (3.37) with Eq. (3.33), the
transmissibility for both force and motion is identical.
Example 3.6 A vertical single-cylinder diesel engine, of 500 kg mass, is mounted
on springs with k =200 kN/m and dampers with ζ = 0.2. The rotating parts are well
balanced. The mass of the equivalent reciprocating parts is 10 kg, and the stroke is
200 mm. Find the dynamic amplitude of the vertical motion, the transmissibility and
the force transmitted to the foundation, if engine is operated at 200 rpm.
The magnitude of the unbalanced force is F 0 = meω
2
m = 10 kg, e = 100 mm = 0.1 m and ω =
200 × 2π
60
=
20π
3
rad/s
Therefore, F 0 = 10 × 0.1 ×
20π
3
2 = 438.65 N
k = 200,000 N/m
c = 2 × 500 × 0.2 ×
2,000,000
500
= 4000 Nm/s
The dynamic amplitude is
x max =
438.65
200,000 − 10 ×
20π
3
2
2 +
4000 ×
20π
3
2
= 2.06 × 10
− 3 m
Natural frequency of the system is
p =
200,000
500
= 20 rad/s
η =
ω
p
=
20π
3 × 20
= 1.047