CE-5113: DYNAMICS OF STRUCTURES By: Dr. Mohammad Ashraf (engineerashraf@yahoo.com) Office: CE: B109
Department of Civil Engineering, University of Engineering and Technology, Peshawar
Module-4 Single Degree of Freedom System: Forced Harmonic Vibration
Dynamics of Structures
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1
Forced Vibration
The force vibration response of system is governed by: mu&& + cu& + ku = p (t )
The exciting force may be of: z z
Short duration Long duration
The response of a system subjected to a short duration exciting force is also of short duration called transient response. Damping in the system causes the vibration to decay and the system returns to rest after a while. This type of response may be critical because of low cycle fatigue. The response to long duration force has two components: z
z
Transient component is due to the initial conditions and present only at the beginning and decays due to damping Steady state component lasts as long as the exciting force acts.
The failure may be due to fatigue. Dynamics of Structures
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Harmonic Forced Vibration
A harmonic force is given by: p(t ) = po sin Ωt _ or _ po cos Ωt
Where po is the amplitude of the force and Ω is the frequency of vibrating force. pocosΩt
Example: Unbalanced rotating Machine The study of harmonic force vibration, while useful by itself, also provide an insight into the nature of force vibration of a more general type. Dynamics of Structures
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2
Undamped Harmonic Vibration
The undamped harmonic vibration of a dynamic system is governed by: mu&& + ku = po sin Ωt
The solution of the equation of motion is made with two parts: z
Complimentary solution: obtained by putting p(t) equal to zero and is equivalent to free vibration response of undamped system and is called transient response. uc (t ) = A cos ωt + B sin ωt
z
Particular solution: giving the steady state response of the system. The particular response is of the form: u p (t ) = G sin Ωt
z
The value of G is obtained by satisfying the equation of motion: po G (k − mΩ 2 )sin Ωt = po sin Ωt ⇒ G = k − mΩ 2 The complete solution is therefore: po u (t ) = uc (t ) + u p (t ) = A cos ωt + B sin ωt + sin Ωt k − mΩ 2 Dynamics of Structures
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Undamped Harmonic Vibration (Cont..)
Constant A and B are determined from the initial conditions u (0) = uo _ and _ u& (0) = vo
Steady State
po ⎛ Ω ⎞ u (t ) = u0 cos ωt + sin ωt + ⎜ sin Ωt − sin ωt ⎟ k − mΩ 2 ⎝ ω ω ⎠ vo po 1 (sin Ωt − β sin ωt ) u (t ) = u0 cos ωt + sin ωt + ω k 1− β 2
Steady State Response Complete Response
Transient
vo
β = 0.2
po (0) = A k − mΩ 2 po po Ω vo = Aω (0 ) + Bω (1) + Ω(1) = Bω + 2 k − mΩ k − mΩ 2 p0 ⎞ po ⎛v Ω u (t ) = u0 cos ωt + ⎜ o − sin ωt + sin Ωt 2 ⎟ k − mΩ 2 ⎝ ω ω k − mΩ ⎠
u0 = A(1) + B(0) +
Where β = Ω/ω is known as the frequency ratio. Dynamics of Structures
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3
Undamped Harmonic Vibration (Cont..) Dynamic Load Factor, D is the ratio of maximum dynamic deflection (steady state) to static deflection up = D=
po 1 sin Ωt k 1− β 2
up u static
=
4
po k
3 2
1 sin Ωt 1− β 2
1
The amplitude of dynamic load factor AD called dynamic magnification factor, is given by: A = 1 D
u static =
AD
2
0 -1 0.0
0.5
1.0
1.5
2.0
2.5
3.0
-2 -3
1− β 2
-4
β
For small value of β, AD is approaching 1.0 and for large values AD is approaching zero For β =1.0 the response is infinitely large called resonance. Dynamics of Structures
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Undamped Harmonic Vibration (Cont..) For β <1.0, AD is +ve and the system is said to be vibrating in phase with the applied force and for β <1.0, AD is –ve i.e. out-of-phase 4 β = 0.5 β = 2.0 deflection
0
static -4 0
0.5
1
1.5
2
time Dynamics of Structures
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4
Resonance Response: Undamped System
The harmonic vibration response of undamped system is given by: v p 1 (sin Ωt − β sin ωt ) u (t ) = u0 cos ωt + o sin ωt + o ω k 1− β 2
For a simple case of uo = vo = 0 1 p (sin Ωt − β sin ωt ) = po ⎛⎜⎜ sin βωt − β2 sin ωt ⎞⎟⎟ u (t ) = o 1− β k 1− β 2 k ⎝ ⎠
For β = 1.0, the numerator and the denominator are both zero and the displacement becomes indeterminate. In the limiting case, the problem can be solved by L’Hospital’s rule lim
β →1
u (t ) =
u (t ) =
po k
lim
β →1
⎛ sin βωt − β sin ωt ⎞ po ⎜⎜ ⎟⎟ = 1− β 2 ⎝ ⎠ k
lim
β →1
⎛ ωt cos βω t − sin ωt ⎞ ⎜⎜ ⎟⎟ − 2β ⎝ ⎠
po ⎛ ωt cos ωt − sin ωt ⎞ po (sin ωt − ωt cos ωt ) ⎟= ⎜ k ⎝ −2 ⎠ 2k Dynamics of Structures
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Resonance Response: Undamped System (Cont..)
The response is periodic with period of T = 2π/ω. Peak response is obtained by differentiating: p 10 u = o (sin ωt − ωt cos ωt ) 2k p u& = o {ω cos ωt − (ω cos ωt + ωt (− ω sin ωt ))} 2k p u& = o ω 2t sin ωt 2k u& = 0 ⇒ ωt = nπ − − − − n = 0,1,2,3......
uk 1 = (sin ωt − ωt cos ωt ) po 2
5 0 -5
The difference between two successive -10 peaks is given by
0
5
10
15
20
p ⎛ 2π nπ ⎞ ⎛ nπ ⎞ p0 u⎜ 2π cos nπ = ± 0 π + ⎟= ⎟ − u⎜ k ω ⎠ ⎝ ω ⎠ 2k ⎝ω Dynamics of Structures
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5
Damped Harmonic Vibration
Damped Harmonic Vibration is governed by: mu&& + cu& + ku = po sin Ωt − − − − − − − − − (1)
The particular solution (steady state response) is of the form: u = G1 cos Ωt + G2 sin Ωt − − − − − − − − − (2)
On putting in eq.1 we get values of G1 and G2 the particular solution is thus
u=
{(
)
}
po 1 1 − β 2 sin Ωt − 2ζβ cos Ωt − − − − − (3) k 1 − β 2 2 + (2ζβ )2
(
)
The complimentary solution (transient response) is: utrans = e −ζωt ( A cos ωd t + B sin ωd t ) − − − − − − − − − (4) Dynamics of Structures
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Damped Harmonic Vibration (Cont..)
The complete solution is thus:
u = e −ζωt ( A cos ωd t + B sin ωd t ) + G1 sin Ωt + G2 cos Ωt − − − − − −(5) Transient
Steady State
where A and B are arbitrary constants determined from the initial conditions: A=
2ζβ po + uo k (1 − β 2 )2 + (2ζβ )2
B=
(
)
po ω 2 βζ 2 − β 1 − β 2 vo + uoωζ + − − − − − − − (7 ) k ωd 1 − β 2 2 + (2ζβ )2 ωd
(
)
The transient response is given by:
⎞ ⎛ v + u ωζ utrans = e −ζωt ⎜⎜ uo cos ωd t + o o sin ωd t ⎟⎟ ωd ⎠ ⎝ −ζωt ⎡ ⎤ p e ω + o 2βζ 2 − β 1 − β 2 sin ωd t ⎥ − − − − − −(8) ⎢2ζβ cos ωd t + k 1 − β 2 2 + (2ζβ )2 ⎣ ωd ⎦
(
)
{
Dynamics of Structures
(
)}
12
6
Damped Harmonic Vibration (Cont..) 4
Total Response Steady State Response
Undamped Harmonic Vibration z
The transient response does not decay with time
0
-4 0
2
4
Damped Harmonic Vibration z
1
The transient response decay with time
3
4
5
6
4
5
6
Total Response Steady State Response
0
-4 0
1
2
3
Dynamics of Structures
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Damped Harmonic Vibration (Cont..)
The steady state response (Eq.3) of a damped harmonic vibration can be written in the alternative form as:
u = ρ sin (Ωt − φ ) − − − − − − − − − − − (9)
Where ρ is the amplitude of steady state response and Φ is the phase angle by which the response lags the exciting force.
ρ=
po k
1
(1 − β ) + (2ζβ ) 2 2
2
,− − tan φ =
Dynamics of Structures
2ζβ 1− β 2
14
7
Damped Harmonic Vibration : Magnification Dynamic load factor, i.e. the ratio of dynamic to static displacement, is given by: D (t ) =
2 2
2
3
sin (Ωt − φ )
2.5
The amplitude of dynamic load factor called the dynamic magnification factor is: AD =
1
(1 − β ) + (2ζβ ) 1
Magnification
0.5
1.5
0.7 1
1
2
0 0
The plot between β and AD shows that AD is not maximum at β =1.0. The value of β to maximize AD is obtained as: dAD = 0 ⇒ β = 1 − 2ζ 2 dβ
0.2
0.5
(1 − β ) + (2ζβ ) 2 2
0
2
( AD )max =
1 2 Frequency Ratio
3
1 2ζ 1 − ζ 2
Dynamics of Structures
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Damped Harmonic Vibration : Phase Angle
At resonance, i.e. β =1.0, all the curves pass through a single point (Φ = 90). For β < 1.0 the phase angle is less than 90o. For β > 1.0 the phase angle is between 90o and 180o. For undamped system the steady state response is in phase (Φ = 0) with the exciting force when β < 1.0. For undamped system the steady state response is in out-of-phase (Φ = 180) with the exciting force when β > 1.0.
tan φ =
2ζβ 1− β 2
180
Phase Angle
0
90
0.2 0.5 0.7 1
0 0
Dynamics of Structures
1 2 Frequency Ratio
3
16
8
Resonance Response: Damped System At resonance (β =1.0 )the steady state and transient responses are: p u steady = − o cos Ωt 2ζk
6
⎡ ⎤ p e ωζ = o sin ωd t ⎥ ⎢cos ωd t + k 2ζ ⎣ ωd ⎦ −ζωt
utrans
In the initial phase the transient response will minimize the total response. When the transient response diminish, the system vibrates with the exciting force but lagging by 90o. For small damping, the total response is:
utotal
(
Transient
12
0 -6 -12
Steady State
12 6 0 -6 -12
)
p 1 −ζωt e = o − 1 cos ωt k 2ζ Dynamics of Structures
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Resonance Response: Damped System (Cont..) Envelope curve
Steady-state amplitude (po/2ζk)
12 6 0 -6 -12 0.5
1
1.5
2
50 25
0.01% 0.05% 0.10%
0 -25
2.5 1.2 Amplitude Ratio
0
3
1 0.8 0.6 0.02 0.05 0.1 0.2
0.4 0.2 0
-50 0
1
2
3
4
5
6
Dynamics of Structures
0
2
4
6
8 10 12 14 16 18 20 No. of Cycles 18
9