Slide CIV 308 DinamikaStr PengRekGempa CIV 308 P3

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Mata Kuliah
Kode
SKS

: Dinamika Struktur & Pengantar Rekayasa Kegempaan
: CIV - 308
: 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
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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) = F o sin wt

u(t)
fS = ku

fI(t) = mü F o sin wt

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

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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
1 b 2
k  mw

(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   Acos wn t  B sin wn t 

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 wn t  
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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Assignment 3
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

If system in the figure have initial condition
u 0 to
 3 harmonic
cm & u 0 loading
20 cm/s p(t) = 5000 sin (wt)
And subjected
W =of
2,5the
tons system,
Plot a time history of displacement response
for t = 0 s until t = 5 s, if ξ = 0%, and :
b = 0,1 ; 0,25 ; 0,60 ; 0,90; 1,00; 1,25 & 1,75EI ∞
40x40 cm2



3m

(b)
E= 23.500 MPa

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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

F o sin wt

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

A

Fo
2b
k 1  b 2 2  2b 2












Fo
b 2 2  1  b 2
B
k  1  b 2 2  2b 2  1   2









Steady State Response



0 ,5

(14)

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

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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 wn t 
2


k 2 

1






(15)

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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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Assignment 4




If system in the figure have initial condition u 0  3 cm & u 0  20 cm/s
And subjected to harmonic loading p(t) = 5000 sin (wt)
Plot a time history of displacement response of the system,
for t = 0 s until t = 5 s, if : ξ = 5%
W = 2,5 tons

b = 0,1 ; 0,25 ; 0,60 ; 0,90; 1,00; 1,25 40x40 cm2
& 1,75

EI ∞
3m

(b)
E= 23.500 MPa