3.2 Response of Damped Systems to Harmonic Loading
65
Fig. 3.5 Example 3.1
Stiffness for each column =
3E I
L 3
Total stiffness =
2 × 3 × E I
L 3
=
2 × 3 × 21 × 10
10
× 1500 × 10
− 7
4 3
= 2,953,125 N/m
m = 5000 kg.
The natural frequency is given by
p =
k
m
=
2,953,125
5000
= 24.3 rad/s
η =
ω
p
=
11
24.3
= 0.453
ζ = 0.04
F 0 = 50, 000 N
Therefore,
x max =
F 0
k
(1 − η 2 ) + ( 2ηζ ) 2
=
50,000
2,953,125
( 1 − 0.453 2 ) 2 + ( 2 × 0.04 × 0.453) 2
= 2.13 × 10
− 2 m = 21.3 mm
Example 3.2 Determine the magnification factor of forced vibration produced by
an oscillator, fixed at the middle of the beam at a speed of 600 rpm. The weight
concentrated at the middle of the beam is W = 5000 N and produces a static deflection
of the beam equal to δ st = 0.025 cm. Neglect the weight of the beam and assume that
65
Fig. 3.5 Example 3.1
Stiffness for each column =
3E I
L 3
Total stiffness =
2 × 3 × E I
L 3
=
2 × 3 × 21 × 10
10
× 1500 × 10
− 7
4 3
= 2,953,125 N/m
m = 5000 kg.
The natural frequency is given by
p =
k
m
=
2,953,125
5000
= 24.3 rad/s
η =
ω
p
=
11
24.3
= 0.453
ζ = 0.04
F 0 = 50, 000 N
Therefore,
x max =
F 0
k
(1 − η 2 ) + ( 2ηζ ) 2
=
50,000
2,953,125
( 1 − 0.453 2 ) 2 + ( 2 × 0.04 × 0.453) 2
= 2.13 × 10
− 2 m = 21.3 mm
Example 3.2 Determine the magnification factor of forced vibration produced by
an oscillator, fixed at the middle of the beam at a speed of 600 rpm. The weight
concentrated at the middle of the beam is W = 5000 N and produces a static deflection
of the beam equal to δ st = 0.025 cm. Neglect the weight of the beam and assume that