Slide TSP302 DinamikaStr PengRekGempa TSP 302 P3

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Mata Kuliah
Kode
SKS

: Dinamika Struktur & Pengantar Rekayasa Kegempaan
: TSP – 302
: 3 SKS

Single Degree of Freedom System
Forced Vibration
Pertemuan - 3

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TIU :

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Mahasiswa dapat menjelaskan tentang teori dinamika struktur.
Mahasiswa dapat membuat model matematik dari masalah teknis yang
ada serta mencari solusinya.

TIK :


Mahasiswa mampu menghitung respon struktur dengan eksitasi
harmonik dan eksitasi periodik

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Sub Pokok Bahasan :
 Eksitasi Harmonik
 Eksitasi Periodik

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Undamped SDoF Harmonic Loading



The impressed force p(t) acting on the simple oscillator in
the figure is assumed to be harmonic and equal to Fo sin wt ,
where Fo is the amplitude or maximum value of the force
and its frequency w is called the exciting frequency or forcing
frequency.
u(t)
k

m

p(t) = Fo sin wt

u(t)
fS = ku

fI(t) = mü Fo sin wt

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The differential equation obtained by summing all the forces
in the Free Body Diagram, is :

mu  ku  Fo sin wt

(1)

The solution can be expresses as :

u t   uc t   u p t 

Complementary Solution
uc(t)= A cos wnt + B sin wnt

(3.a)

(2)
Particular Solution
up(t) = U sin wt

(3.b)

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Substituting Eq. (3.b) into Eq. (1) gives :



Or :



 mw 2U  kU  Fo

Fo

Fo k
U

2
k  mw
1 b 2

(4)

Which b represents the ratio of the applied
forced frequency to the natural frequency of
vibration of the system :
b

w
wn

(5)

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Combining Eq. (3.a & b) and (4) with Eq. (2) yields :
u t   A cos wnt  B sin wnt 

Fo k
sin wt
2
1 b

With initial conditions : u  u0

u  u0

(6)

 u 0  Fo k b 
Fo k

u t   u 0  cos wnt  
sin
w
t
sin wt


n
2 
2
1 b 
1 b
 wn
Transient Response

Steady State Response

(7)

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3

u t 

2

Fo / k 

1

0
0

0.5

1

1.5

2

2.5

3

3.5

-1

-2
-3
Total Response

Steady State Response

u 0   0 and u 0   wn Fo / k

w

wn  0 ,20

Transient Response

4

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Steady state response present because of the applied force, no
matter what the initial conditions.
 Transient response depends on the initial displacement and
velocity.
 Transient response exists even if u 0   u 0   0
 In which Eq. (7) specializes to

u t  



Fo k
sin wt  b sin wnt 
2
1 b

(8)

It can be seen that when the forcing frequency is equal to natural
frequency (b = 1), the amplitude of the motion becomes infinitely
large.
 A system acted upon by an external excitation of frequency
coinciding with the natural frequency is said to be at resonance.

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If w = wn (b = 1), the solution of Eq. (1) becomes :
1 Fo
wnt cos wnt  sin wnt 
u t   
2 k

(9)

40
30

u t 

20

Fo / k 

10
0
0
-10
-20
-30
-40

0.5

1

p

1.5

2

2.5

3

3.5

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Damped SDoF Harmonic Loading


Including viscous damping the differential equation governing
the response of SDoF systems to harmonic loading is :

mu  cu  ku  Fo sin wt

(10)

u(t)
c

fD(t) = cú
m

k

u(t)

p(t)
fS(t) = ku

fI(t) = mü

Fo sin wt

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

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The complementary solution of Eq. (9) is :
uc t   e wnt  A cos w D t  B sin w D t 

u p t   C sin wt  D cos wt

(11)

The particular solution of Eq. (9) is :

Where :

Fo
1 b 2
C
k 1  b 2 2  2b 2
D





Fo
 2b
k 1  b 2 2  2b 2





(12)
(13.a)
(13.b)

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The complete solution of Eq. (9) is :

uc t   e wnt  A cos w D t  B sin w D t   C sin wt  D cos wt
Transient Response

(14)

Steady State Response

Response of damped
system to harmonic
force with
b = 0,2,
 = 0,05,
u(0) = 0,
ú(0) = wnFo/k

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
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The total response is shown by the solid line and
the steady state response by the dashed line.
The difference between the two is the transient
response, which decays exponentially with time
at a rate depending on b and .
After awhile, essentially the forced response
remains, and called steady state response
The largest deformation peak may occur before
the system has reached steady state.

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If w = wn (b = 1), the solution of Eq. (10) becomes :


Fo 1  wnt 

e
u t  
cos w D t 
sin w D t   cos wnt 
2


k 2 

1






(15)

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
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Considering only the steady state response, Eq. (12) & Eq.
(13.a, b), can be rewritten as :

u t   U sinwt   

Where :

U

(16)
tan  

1  b   2b 
Fo k
2 2

2

2b
1 b 2

(17)

Ratio of the steady state amplitude, U to the static deflection
ust (=Fo/k) is known as the dynamic magnification factor, D :
U
D

u st

1  b   2b 
1

2 2

2

(18)

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Exercise


The steel frame in the figure
supports a rotating machine
that exerts a horizontal force
at the girder level p(t) = 100
sin 4 t kg. Assuming 5% of
critical damping, determine :
(a)
the
steady-state
amplitude of vibration and
(b) the maximum dynamic
stress in the columns.
Assume the girder is rigid

W = 6,8 tons

EI ∞

WF 250.125
I = 4,050 cm4

4,5 m

(b)
E = 205.000 MPa

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Response To Periodic Excitation
A
periodic function can be separated into its harmonic
components using Fourier Series
pt   a0   a j cos jw0t    b j sin jw0t 

Where :
a0 

1
T0

2
aj 
T0
bj 

2
T0





j 1

j 1

w0 

 pt dt
T0

0

 pt cos jw t dt
T0

0

0

 pt  sin jw t dt
T0

0

0

2p
T0

j  1,2 ,3,...

(19)

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Response to periodic force is given by :

(20)

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