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 (Cont..)
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Rotating Vector Presentation: Damped System
The harmonic vibration response of undamped system is given by: u static =
po sin Ωt k
u steady = ρ sin (Ωt − φ )
ρ=
po k
1
(1 − β ) + (2ζβ ) 2 2
2
,− − tan φ =
2ζβ 1− β 2
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Force Balance Diagram: Damped System p = po sin Ωt
u steady = ρ sin (Ωt − φ ),
u& steady = ρΩ cos(Ωt − φ )
u&&steady = − ρΩ 2 sin (Ωt − φ ) f I = mu&& = − mΩ 2 ρ sin (Ωt − φ )
f D = cu& = cΩρ cos(Ωt − φ )
f S = ku = kρ sin (Ωt − φ )
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Dynamic Response Factors
The steady state harmonic response is:
u (t ) =
po k
1
(1 − β ) + (2ζβ ) 2 2
2
sin (Ωt − φ )
u (t ) = Rd sin (Ωt − φ ) po / k Rd =
Dynamics of Structure by A. K. Chopra p = po sin Ωt
1
(1 − β ) + (2ζβ ) 2 2
u& (t ) = po / km
2
m Rd Ω cos(Ωt − φ ) = Rv cos(Ωt − φ ) k
u&&(t ) m = − Rd Ω 2 sin (Ωt − φ ) = − Ra sin (Ωt − φ ) po / m k Ra
β
u steady = ρ sin (Ωt − φ ),
= Rv = β Rd Dynamics of Structures
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Four-way Logarithmic Plot
Dynamics of Structure by A. K. Chopra Dynamics of Structures
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Resonant Frequencies and Responses
The frequency at which largest response amplitude occurs is called resonant frequency The steady state displacement, velocity and acceleration response of a dynamic system are: u (t ) =
u&&(t ) = −
po k
po k
1
sin (Ωt − φ ),
(1 − β ) + (2ζβ ) 2 2
2
Ω2
(1 − β ) + (2ζβ ) 2 2
2
u& (t ) =
po k
Ω
(1 − β ) + (2ζβ ) 2 2
2
cos(Ωt − φ )
sin (Ωt − φ ),
By setting the first derivatives of u, u& and u&& w.r.t β equal to zero we can determine the resonant frequencies. z z z
Displacement resonant frequency ratio : Velocity resonant frequency ratio : Acceleration resonant frequency ratio :
β d = 1 − 2ζ 2
βv = 1 β a = 1 / 1 − 2ζ 2
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Resonant Frequencies and Responses (Cont..)
The three dynamic response factors are: Rd = Rv =
Ra =
1
2
⇒ Rd =
2
⇒ Rv =
(1 − β ) + (2ζβ ) 2 2
β
(1 − β ) + (2ζβ ) 2 2
β2
(1 − β ) + (2ζβ ) 2 2
2
⇒ Ra =
1 2ζ 1 − ζ 2 1 2ζ 1 2ζ 1 − ζ 2
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Half-Power Bandwidth The difference of frequency ratios on either side of the resonant frequency at which the amplitude is 1/√2 times the resonant amplitude. For small damping:
Rd =
1
(1 − β ) + (2ζβ ) 2 2
2
Rd ,resonant =
,
Dynamics of Structure by A. K. Chopra
1 2ζ
For half-power bandwidth:
1
(1 − β ) + (2ζβ ) 2 2
(β )
2
(
2 2
=
1 1 ⇒ 1− β 2 2 2ζ
(
) + (2ζβ ) 2
2
= 8ζ 2
)( )
− 2 1 − 2ζ 2 β 2 + 1 − 8ζ 2 = 0
2 2 2 β 2 = (1 − 2ζ 2 ) ± (1 − 2ζ 2 ) − (1 − 8ζ 2 ) = (1 − 2ζ ) ± 2ζ 1 + ζ
For small damping:
β = (1 − 2ζ 2 ) − 2ζ _ and _ β b2 = (1 − 2ζ 2 ) + 2ζ 2 a
β b − β a = 2ζ ⇒ ζ =
βb − β a 2
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Rotating Machinery/Eccentric Mass Vibrator In case of rotating machinery:
sin
p (t ) = mo eΩ sin Ωt 2
Here po is moeΩ2 the response is therefore:
u (t ) = ρ sin (Ωt − φ )
ρ=
mo eΩ 2 k
(1 − β ) + (2ζβ )
tan φ =
2 2
2
=
e
Ω 2 mo k/m m
(1 − β ) + (2ζβ ) 2 2
2
=
eβ 2
mo m
(1 − β ) + (2ζβ ) 2 2
2
2ζβ 1− β 2
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Example 6.2
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Example 6.2
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Example 6.2
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Transmissibility: Support Movement
Let the support move according to the equation:
⇒
u g = G sin Ωt
u&&g = −GΩ 2 sin Ωt
ut
Equation of motion is:
⇒
mu&&t + cu& + ku = 0
m(u&& + u&&g ) + cu& + ku = 0
mu&& + cu& + ku = −mu&&g = mGΩ 2 sin Ωt
2 Here p(t ) = mGΩ sin Ωt
Therefore u (t ) =
⇒
mGΩ 2 k
u (t ) = G
1
(1 − β ) + (2ζβ ) 2 2
2
β2
(1 − β ) + (2ζβ ) 2 2
2
sin (Ωt − φ )
sin (Ωt − φ ),
Dynamics of Structures
tan φ =
2ζβ 1− β 2
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Transmissibility: Support Movement (Cont..)
The total displacement is: β2
u t = u g + u = G sin Ωt + G ⇒ u t = χ sin (Ωt − γ ),
(1 − β ) + (2ζβ ) 2 2
2
where χ = G
sin (Ωt − φ )
1 + (2ζβ )
2
, 2
(1 − β ) + (2ζβ ) 2 2
and
tan γ =
2ζβ 3
(1 − β ) + (2ζβ ) 2 2
2
The ratio of amplitude of total displacement (χ) to amplitude of ground displacement (G) is called transmissibility (TR) TR =
χ G
1 + (2ζβ )
2
=
(1 − β ) + (2ζβ ) 2 2
2
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Example 6.3
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Example 6.3
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Force Transmission
The total force transmitted to the base is: fT = f D + f S = cu& + ku
The displacement and velocity response to harmonic force is: u (t ) =
po k
1
(1 − β ) + (2ζβ ) 2 2
2
sin (Ωt − φ ),
po k
Ω
(1 − β ) + (2ζβ ) 2 2
2
cos(Ωt − φ )
Therefore: f T = po
1
(1 − β ) + (2ζβ ) 2 2
2
c ⎧ ⎫ ⎨sin (Ωt − φ ) + Ω cos(Ωt − φ )⎬ k ⎩ ⎭
1 + (2ζβ )
2
⇒ fT = po
u& (t ) =
(1 − β ) + (2ζβ ) 2 2
2
sin (Ωt − γ )
Transmissibility is also the ratio amplitudes of force transferred to the base to that of applied force Dynamics of Structures
TR =
fT = po
1 + (2ζβ )
2
(1 − β ) + (2ζβ ) 2 2
2
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