Slide CIV 308 DinamikaStr PengRekGempa CIV 308 P4

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Mata Kuliah
Kode
SKS

: Dinamika Struktur & Pengantar Rekayasa Kegempaan
: CIV – 308
: 3 SKS

Single Degree of Freedom System
General Dynamic Loading
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

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

Sub Pokok Bahasan :
 Eksitasi Sembarang

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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.
F(t)

Using Duhamel’s Integral, the
displacement response at time t
due to continuous F(t) is :

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

F(t)∙dt



t

t + dt

t

(1)

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

Response to Step Forces
p(t)

p(t)
u(t)

po

p(t) = po

m
t

k

u st 0 

SDoF System Without Damping

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

uo  2u st o

Maximum displacement :

(2)

SDoF System With Damping





n t 
u t   u st o 1  e
cos  D t +
sin  D t 
2




1 



po
k

(3)

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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

Maximum displacement :

m
t

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

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

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

For t < tr

(4)

For t > tr

(5)

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Example


The elevated water tank in Fig
weighs 45,5 tons when full with
water.
 The tower has a lateral stiffness
of 145 kg/mm.
 Treating the water tower as an
SDoF system, estimate the
maximum lateral displacement
due to each of the two dynamic
forces shown without any
“exact” dynamic analysis.
Neglect damping.

P(t)

W

k

P(t)
25 tons

P(t)
25 tons

t (dt)

0,2

4

t (dt)

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Response to Pulse Excitations


We next consider an important class of excitations
that consist of essentially a single pulse
 Air pressure generated on a structure due to
aboveground blasts or explosions is an example of
pulse force.
Force

Time

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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

2t
u t 
 1  cos n t  1  cos
u st o
Tn
u t 
 cos n t  t d   cos n t
u st o

For t < td

(6)

For t > td

(7)

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





po
Rd
k

(8)

With deformation response factor :
t

2 sin d
Rd  
Tn
 2

for t d Tn  1 2

for t d Tn  1 2

(9)

The maximum value of the equivalent static force is :

fSo  kuo  po Rd

(10)

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Example


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

p(t)

p(t)
u(t)

po

m
t

k
td

For td/Tn ≠ ½

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

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

For t < td

For t > td

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For td/Tn = ½

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

 t 1
u t  
 cos 2   
u st 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