Free Vibration -...
Transcript of Free Vibration -...
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Pertemuan - 2
Single Degree of Freedom System Free Vibration
Mata Kuliah : Dinamika Struktur & Pengantar Rekayasa Kegempaan
Kode : CIV – 308
SKS : 3 SKS
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TIU : Mahasiswa dapat menjelaskan tentang teori dinamika struktur.
Mahasiswa dapat membuat model matematik dari masalah teknis yang
ada serta mencari solusinya.
TIK : Mahasiswa dapat memformulasikan persamaan gerak sistem struktur
berderajat kebebasan tunggal yang bergetar bebas
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Sub Pokok Bahasan :
Getaran Bebas SDoF Tak Teredam
Frekuensi dan Periode
Getaran Bebas SDoF Dengan Redaman
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The differential equation governing free vibration of the systems
without damping (c = 0 & p(t) = 0) is :
This equation called second order homogeneous differential equation
Undamped SDoF Free-Vibration A structure is said to be undergoing free vibration when it is
disturbed from its static equilibrium position and then
allowed to vibrate without any external dynamic excitation.
0 tkutum (1)
k m
u(t)
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To find the solution of Eq. (1), assuming a trial
solution given by :
Where A and B are constants depending on
the initiation of the motion while w is a
quantity denoting a physical characteristic of
the system.
tcosAu nw tsinBu nw(2.a) (2.b)
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The substitution eq.(2.a&b) into eq. (1) gives :
If eq.(3) is to be satisfied at any time, the factor
in parentheses must be equal to zero, or :
wn is known as the natural circular frequency of
the system (rad/s)
02
tcosAkm nn ww (3)
m
kn
2w
m
kn w (4)
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Since either eq.(2.a) or (2.b) is a solution of eq.(1), and since
the differential equation is linear, the superposition of these
two solution is also solution :
The expression for velocity, ú, is :
Constant A and B, will be determined based on initial
condition (at t = 0) :
tsinBtcosAu nn ww
tcosBtsinAu nnnn wwww
(5)
(6)
00 uuuu
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Subject to these initial conditions, the solution to the eq. (5)
will be :
tsinu
tcosutu n
n
n ww
w0
0
(7)
2
2
0
00
n
uuu
w
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Frequency and Period
The time required for the undamped system to
complete one cycle of free vibration is the natural
period of vibration, Tn (in second)
A system executes 1/Tn cycles in 1 sec. This natural
cyclic frequency of vibration (in Hz) is denoted by :
n
nTw
2
w
2
nnf (9)
(8)
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Exercise Determine natural frequency and natural period from each
system below
k
k
E, I, L
W
(a)
40x40 cm2
E= 23.500 MPa
(b)
EI ∞
W = 2,5 tons
3 m
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Assignment 1 If system (b) have initial condition
a) Determine the displacement at t = 2 sec
b) Plot a time history of displacement response of the
system, for t = 0 s until t = 5 s.
cm/s 20 0 cm 30 u&u
40x40 cm2
E= 23.500 MPa
(b)
EI ∞
W = 2,5 tons
3 m
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Damped SDoF Free-Vibration If damping is present in the system, the differential equation
governing free vibration of the systems is :
The trial solution to satisfy Eq. (14) is
Subtitute Eq. (11) to Eq. (10) yield the characteristic eq. :
0 tkutuctum (10)
ptCeu (11)
02 kcpmp (12) c
k
m
u(t)
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The roots of the quadratic Eq. (12) are :
Thus the general solution of Eq. (10) is :
Three distinct cases may occur depends on
the sign under the radical in Eq. (13)
m
k
m
c
m
cp ,
2
2122
tptpeCeCtu 21
21
(13)
(14)
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1. Critically Damped System
For critical damping :
The expression under the radical in Eq. (13) is equal to zero,
and the roots are equal, they are :
The general solution of Eq. (14) is :
kmmcc ncr 22 w (15)
ncr
m
cpp w
221
(16)
tneBtAtuw
(17)
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1. Critically Damped System
Constant A and B, will be determined based on initial
condition (at t = 0) :
Substitute the initial condition, yield :
tuuuetu n
tn 000 ww
00 uuuu
(18)
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2. Overdamped System
In an overdamped system, the damping coefficient is greater
that the value for critical damping (c > ccr)
EoM of an overdamped system due to initial condition is :
Where :
ξ is the damping ratio (%)
ttt DDn BeAeetuwww
D
n
D
n
uu
B
uu
A
w
w
w
w
2
010
2
010
2
2
r
nD
c
c
ww 12
(19)
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3. Underdamped System
When the value of the damping coefficient is less than the
critical value (c < ccr), the roots of the characteristic eq. (12)
are complex conjugates, so that :
Regarding the Euler’s equations which relate exponential and
trigonometric functions :
2
2122
m
c
m
ki
m
cp , 1i
xsinixcose
xsinixcose
ix
ix
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3. Underdamped System
The substitution of the roots p1 and p2 into Eq. (14) together
with the use of Euler equation, gives :
Where
Using initial condition,
tsinBtcosAetu DD
tn www
(20)
21 ww D
00 uuuu
tsin
uutcosuetu D
D
nD
tn ww
www 00
0
(21)
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Many types of structures, such as buildings, bridges, dams, nuclear power plants, offshore
structures, etc., all fall into underdamped system (c < ccr), because typically their
damping ratio is less than 0,10.
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Logarithmic Decrement
The ratio ui/ui+1 of successive peaks (maxima) is :
The natural logarithm of this ratio, called the logarithmic
decrement,
If z is small, this gives an approximate equation : d ≈ 2z
2
1 1
2
exp
u
u
i
i
21 1
2ln
d
i
i
u
u
(22)
(23)
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Over j cycles the motion decreases from u1 to uj+1. This ratio
is given by :
Therefore,
dj
j
j
j
eu
u...
u
u
u
u
u
u
u
u
14
3
3
2
2
1
1
1
d 21
1
1 ju
uln
j
(24)
(25)
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Exercise A one-story building is idealized as a rigid girder supported by
weightless columns. In order to evaluate the dynamic properties of this
structure, a free-vibration test is made, in which the roof system (rigid
girder) is displaced laterally by a hydraulic jack and then suddenly
released.
During the jacking operation, it is observed that a force of 10 kg is
required to displace the girder 0.508 cm.
After the instantaneous release of this initial displacement, the maximum
displacement on the first return swing is only 0.406 cm and the period
of this displacement cycle is T = 1.40 sec.
k/2 k/2
c
Weight, W = mg
Find :
• W
• f and w
• d, , c, wD
• u6
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Assignment 2 If system in the figure have initial condition
Plot a time history of displacement response of the system,
for t = 0 s until t = 5 s, if :
a. ξ = 5%
b. ξ = 100%
c. ξ = 125%
cm/s 20 0 cm 30 u&u
40x40 cm2
E= 23.500 MPa
(b)
EI ∞
W = 2,5 tons
3 m