Slide TSP302 DinamikaStr PengRekGempa TSP 302 P4

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Mata Kuliah
Kode
SKS

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

Single Degree of Freedom System
Arbitrary Force
Pertemuan - 4

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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 mampu menghitung respon sistem struktur sederhana
akibat beban luar sembarang

a home base to excellence


Sub Pokok Bahasan :
 Eksitasi Sembarang

a home base to excellence
Duhamel’s Integral – Undamped System


An impulsive loading is a load which is applied during a short
duration of time.
The corresponding impulse of this type of load is defined as
the product of the force and the time of its duration.
Using Duhamel’s Integral, the
displacement response at time t
due to continuous F(t) is :

F(t)

t

1
u t  
F t  sin n t  t dt

mn 0

F(t)∙dt



t

t

t + dt

(1)

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

Response to Step Forces
p(t)

p(t)

u(t)

po

p(t) = po

m
t

k
SDoF System Without Damping


2t 

u t   u st o 1  cos nt   u st o 1  cos
T
n 


(2)

SDoF System With Damping




n t 
u t   u st o 1  e
cos  D t +
sin  D t 
2



1 



(3)

uo  2u st o

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

Response to Step Force With Finite Rise Time
 p t / t r  t  t r
pt    o
t  tr
 po

p(t)

p(t)

u(t)

po

m

t

k
tr

• If tr < Tn/4 : uo ≈ 2(ust)o
• If tr > 3Tn : uo ≈ 2(ust)o
• If tr/Tn = 1,2,3, … uo = (ust)o

 t sin nt 

u t   u st o  
nt r 
 tr

For t < tr

(4)




1
sin nt  sin n t  tr 
u t   u st o 1 
 nt r


For t > tr

(5)

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 Response to Rectangular Pulse Force
p(t)

p(t)

u(t)

 p t  td

pt    o
0 t  t d

po

m
t

k
td

u t 
2t
 1  cos nt  1  cos
ust o
Tn

For t < td

(6)

u t 
 cos n t  t d   cos nt
ust o

For t > td

(7)

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Response to Rectangular Pulse Force
 The maximum deformation is :
uo  u st o Rd 


(8)

With deformation response factor :
t


2 sin d
Rd  
Tn
 2



po
Rd
k

for t d Tn  1 2

(9)

for t d Tn  1 2

The maximum value of the equivalent static force is :

f So  kuo  po Rd

(10)

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Exercise


A one-story, idealized as a 3,6 m high frame
with two columns hinged at the base and a
rigid beam, has a natural period of 0,5 sec.
Each column is a wide-flange steel section
W8x18. Its properties for bending about its
major axis are Ix = 2,570 cm4, S = Ix/c = 249
cm3; E = 200 GPa. Neglecting damping,
determine the maximum response of this
frame due to a rectangular pulse force of
amplitude 1,800 kgf and duration td = 0,2
sec. The response quantities of interest are
displacement at the top of the frame and
maximum bending stress in the column.

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Response to Half Cycle Sine Pulse Force
p(t)

p(t)

u(t)

 p sin t/t d  t  t d
pt    o
 0 t  td

po

m

t

k
td

For td/Tn ≠ ½

  t  Tn

1
u t 
t 



sin   
sin 2 

2 
ust o 1  Tn 2td    td  2td  Tn 

u t  Tn t d  cost d / Tn    t 1 t d 


sin 2  
2
ust o
Tn 2td   1
  Tn 2 Tn 

For t < td

For t > td

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For td/Tn = ½
u t  1 
2t 2t
2t 

  sin

cos
ust o 2  Tn Tn
Tn 

 t 1
u t  
 cos 2   
ust o 2
 Tn 2 

For t < td

For t > td

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Response to Symmetrical Triangular Pulse Force
p(t)

p(t)
u(t)

po

m
t

k

td/2

td