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..)
Dynamics of Structures
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1
Transmissibility: Support Movement
Let the support move according to the equation:
u&&g = −GΩ 2 sin Ωt
⇒
u g = G sin Ωt
ut
Equation of motion is:
m(u&& + u&&g ) + cu& + ku = 0
⇒
mu&&t + 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 ),
tan φ =
2ζβ 1− β 2
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3
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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2
Example 6.4 (2nd Edition)
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Example 6.3 (Cont..)
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3
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
TR =
fT = po
Dynamics of Structures
1 + (2ζβ )
2
(1 − β ) + (2ζβ ) 2 2
2
7
Vibration Measuring Instruments
Vibration measurement is very important in structural engineering, e.g. measurement of ground motion, building motion during earthquake and laboratory testing Although today’s vibration measuring instruments also called seismic instruments, are highly developed and intricate, the underlying principle is simple, i.e. spring-mass-damper system When subjected to motion, the mass moves and the motion is recorded after suitable magnifications Different instruments are required for seismological observation and earthquake engineering measurements
Dynamics of Structures
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4
Measurement of Acceleration
Let the support move with a harmonic acceleration given by:
The measured displacement and phase angle of the motion of mass are given by: u (t ) = ρ sin (Ωt − φ )
Where
u&&g = A sin Ωt
tan φ = ρ=
1
(1 − β ) + (2ζβ ) 2 2
2
=
A
ω2
1
(1 − β ) + (2ζβ ) 2 2
2
Now the ratio of the amplitude of measured displacement and the amplitude of ground acceleration is given by: ρ A
mA k
2ζβ 1− β 2
=
1
ω2
1
(1 − β ) + (2ζβ ) 2 2
2
The ratio is dependent on the frequency ratio, i.e the frequency of ground acceleration Dynamics of Structures
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Measurement of Acceleration (Cont..)
Since the ground motion due to earthquake is having different frequency contents, the measured response should be independent of frequency of ground motion, i.e. ρω2/A should not vary with β From the table and figure, ρω2/A remains almost constant for ζ = 0.7 for β varying between 0 and 0.6, if the error is not to exceed 5% Since the amplitude ratio (ρ/A) is inversely proportional to frequency of the system (ω), a high instrument frequency will result in low magnification and therefore must be magnified through some other means
Dynamics of Structures
3 Magnification
2.5
0.0
2
0.5
1.5
0.7
0.2
1.0
1 0.5 0 0
β 0.0 0.1 0.2 0.3 0.4 0.5 0.6
0.5
1 1.5 2 Frequency Ratio
ρω2/A ζ=0 ζ=0.7 1.00 1.000 1.01 1.000 1.04 1.000 1.10 0.998 1.19 0.991 1.33 0.975 1.56 0.947
ζ=0 0 0 0 0 0 0 0
2.5
3
Φ/β ζ=0.7 1.400 1.405 1.419 1.442 1.470 1.502 1.533 10
5
Measurement of Acceleration (Cont..) 180
The time shift between the measure response and ground motion is given by:
135 Phase Angle
φ
1 2ζβ 1 2ζβ t s = = tan −1 = tan −1 Ω Ω 1 − β 2 βω 1− β 2 ts 1 2ζβ == tan −1 T 2πβ 1− β 2
0.0 0.2 0.5
45
0.7 1.0
0 0
When the motion being measured is harmonic, the shift is of no particular significance When the motion have several harmonic components as in the case of earthquake, this shift will distort the motion of instrument For ζ=0.7 the shift is practically constant, i.e. phase angle is a linear function of β and Ω
ts /T
90
0.5
1 1.5 2 Frequency Ratio
100 90 80 70 60 50 40 30 20 10 0
2.5
3
0.0 0.2 0.5 0.7 1.0
0
0.5
1 1.5 2 Frequency Ratio
2.5
Dynamics of Structures
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11
Measurement of Displacement
Let the support move with a harmonic displacement given by: u g = G sin Ωt
The measured displacement and phase angle of the motion of mass are given by: u (t ) = ρ sin (Ωt − φ )
Where
tan φ = ρ=
mGΩ 2 k
1
(1 − β ) + (2ζβ ) 2 2
2
=G
β2
(1 − β ) + (2ζβ ) 2 2
2
Now the ratio of the amplitude of measured displacement and the amplitude of ground displacement is given by: ρ G
2ζβ 1− β 2
=
β2
(1 − β ) + (2ζβ ) 2 2
2
The measured motion is dependent on frequency ratio Dynamics of Structures
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6
Measurement of Displacement (Cont..)
Since the ground motion due to earthquake is having different frequency contents, the measured response should be independent of frequency of ground motion, i.e. ρ/G should not vary with β From the figure, ρ/G remains almost constant for high values of β Thus the frequency of displacement measuring instrument shall be very low in comparison with forcing frequency. Such instrument is unwieldy because of the heavy mass and soft spring For high value of β, ρ/G is almost unity and phase angle is 180o, for any value of damping therefore: u (t ) = −G sin Ωt Dynamics of Structures
3 0.0 0.2 0.5 0.7 1.0
2.5 2 ρ/G
1.5 1 0.5 0 0
0.5
1 1.5 2 Frequency Ratio
2.5
3
13
Example 6.7 (2nd Edition)
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7
Example 6.7 (Cont..)
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Example 6.7 (Cont..)
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8
Energy in Viscous Damping
The energy input per cycle to the system due to the applied force is given by:
E I = ∫ p(t )du =
2π / Ω
∫(p
= πρpo sin φ
o
sin Ωt )(u&dt ) =
0
2π / Ω
∫(p
sin Ωt )(ρΩ cos(Ωt − φ )dt )
o
0
The energy per cycle dissipated in the system is given by:
ED = ∫ f D du =
2π / Ω
2π / Ω
0
0
∫ (cu& )(u&dt ) = ∫ c(ρΩ cos(Ωt − φ )) dt 2
= cπΩρ 2
2ζβ
From the figure: sin φ =
=
2ζβk ρ po
(1 − β ) + (2ζβ ) ⎛ 2ζβ k ⎞ Therefore: E I = πρpo ⎜⎜ ρ ⎟⎟ = 2πρ 2ζβ k = cπΩρ 2 p ⎝ o ⎠ Hence:
2 2
2
tan φ =
2ζβ 1− β 2
2ζβ
φ 1− β 2
EI = ED Dynamics of Structures
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Energy in Viscous Damping (Cont..)
The Potential energy per cycle due to spring force is given by: ES = ∫ f S du =
2π / Ω
0
0
∫ (ku )(u&dt ) = ∫ k (ρ sin (Ωt − φ ))(ρΩ cos(Ωt − φ ))dt = 0
The Kinetic energy per cycle due to inertia force is given by: EK = ∫ f I du =
2π / Ω
2π / Ω
2π / Ω
0
0
∫ (mu&&)(u&dt ) = ∫ k (− ρΩ
For β = 1.0, ( E I = (πpo )ρ
)
sin (Ωt − φ ) (ρΩ cos(Ωt − φ ))dt = 0
sin φ = 1 )
ED = (2πζk )ρ 2
For EI = ED πpo ρ = (2πζk )ρ 2 ⇒ ρ =
2
po 2ζk
Which is the same relation which was already derived using other method Dynamics of Structures
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9
Energy in Viscous Damping (Cont..)
The spring, inertia and damping forces are given by:
f S = ku
f D = cu& = cΩρ cos(Ωt − φ )
f I = − mΩ 2u
fD
f D = cΩ ρ 2 − ρ 2 sin 2 Ωt = cΩ ρ 2 − u 2 2
cΩ ρ
2
⎛u⎞ ⎛ f ⎞ ⇒ ⎜⎜ ⎟⎟ + ⎜⎜ D ⎟⎟ = 1.0 ⎝ ρ ⎠ ⎝ cΩρ ⎠
u ρ
Which is the equation of ellipse. The area enclosed by the ellipse is equal to dissipated energy The fD-u curve is not a single value function but a loop called hysteresis loop The total resisting force (elastic + damping) is:
f S + f D = ku + cΩ ρ − u 2
f S = ku
fD + fS
kρ
cΩρ
u
2
ρ
The plot of fS + fD is given in the figure which is the same but rotated ellipse Dynamics of Structures
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Hysteretic Damping
The equation of motion for forced vibration under hysteretic damping is of the form: mu&& + f t (u ) = p (t ) S
The general solution for calculating the dissipated energy is quite complex However for steady state harmonic motion, hysteretic damping can be accounted for expressing the total spring force as the sum of two components; average spring force and damping force given by:
f St = f S + f D = ku +
Ω
u&
The equation of motion thus becomes: mu&& + ku +
ηk
ηk
u& = po sin Ωt Ω The solution of this equation is:
u = ρ h sin (Ωt − φh )
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10
Hysteretic Damping (Cont..)
Where:
ρh =
po k
1
(1 − β )
2 2
+η
,− − tan φh =
2
η 1− β 2
The general shapes of curve are similar to that for viscous damping However the amplitude ratio ρh/(po/k) is maximum at β=1.0 unlike the case of viscous damping At β=0, the phase angle is given by tan φh = η which is in the case of viscous damping equal to zero.
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Hysteretic Damping (Cont..)
As
u = ρ h sin (Ωt − φh ) − − − − − (1)
And
u& = ρ h Ω cos(Ωt − φh )
t Therefore f S = ku +
ηk Ω
u& = ku + ηkρ h cos(Ωt − φh )
f St − ku = cos(Ωt − φh ) − − − − − (2) ηkρ h
Squaring and adding (1) and (2) 2
⎛ f St − ku ⎞ ⎛ u ⎞ ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ = 1 − − − (3) ⎝ ηkρ h ⎠ ⎝ ρ h ⎠
This is the equation of a skew ellipse 2 The area enclosed by ellipse is, ηkπρ h Dynamics of Structures
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Hysteretic Damping (Cont..)
From (3) it is obvious that the dissipated energy is independent of exciting frequency Thus the hysteresis loop can be obtained experimentally by carrying out quasistatic cyclic test Horizontal intercept “d” and consequently η (for light damping) can be obtained by:
d = ρh
η 1+η
2
− ⇒ −η =
d
ρ −d 2 h
2
≈
d
ρh
2
⎛ f St − ku ⎞ ⎛ u ⎞ ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ = 1 − − − (3) ⎝ ηkρ h ⎠ ⎝ ρ h ⎠
ηkπρ h2 Dynamics of Structures
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12