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TECHNICAL REPORT
MODELLING EFFECT OF EARTHQUAKE ON
DOUBLE-STOREY BUILDING USING SECOND ORDER
ORDINARY DIFFERENTIAL EQUATIONS (ODE)
NORA FARAHIN BINTI MUHAMMAD
2009351351 TR12/27
NOR ATIQAH BINTI DOLLAH SANI2009161531 TR12/27
NURUL FAJRIYAH BINTI HAMID
2009944889 TR12/27
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UNIVERSITI TEKNOLOGI MARA
TECHNICAL REPORT
MODELLING EFFECT OF EARTHQUAKE ON DOUBLE-
STOREY BUILDING USING SECOND ORDER ORDINARY
DIFFERENTIAL EQUATIONS (ODE)
NORA FARAHIN BINTI MUHAMMAD
2009351351 TR12/27
NOR ATIQAH BINTI DOLLAH SANI
2009161531 TR12/27
NURUL FAJRIYAH BINTI HAMID
2009944889 TR12/27
Report submitted in partial fulfillment of the requirement
for the degree of
Bachelor of Science (Hons.) (Computational Mathematics)
Center for Mathematics
Faculty of Computer and Mathematical Sciences
Jan 2012
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ACKNOWLEDGEMENTS
ALL PRAISE TO ALLAH OF AL MIGHTY, THE MOST GRACIOUS, MOST
MERCIFUL AND MOST BENEVOLENT.
Firstly, I am grateful to Allah S.W.T for giving us the strength, the opportunity and
guidance in every decision that we made and the barriers that we faced. Without
this Grace, it would be virtually impossible for us to have the courage, endurance
and the strength to do the research and complete it. With His help, this effort would
be senseless.
First and foremost, we would like to take this opportunity to express our sincere anddeepest gratitude to our supervisor Puan Liza Nurul Fazila Dr. O K Rahmat who
has continuously and patiently provide us with guidance and spent time to supervise
us in order to complete the research paper.
Thousands of thanks are also dedicated to all of our friends and families for sharing
prayers, views, information and many things that are valuable for our research
paper. Dear friends and families, only Allah S.W.T can pay all your cooperation.
We really appreciate your support.
Last but not least, once again, name and people that we mentioned above, there is
no word that can describe how thankful we are to have all of you in our life. The
entire thing we have done together during this research will we treasured in our
heart. All burdens that we carried became easy and simple with all your helps.
Last words, thank you so much and may Allah bless all of you always.
Thank You.
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TABLE OF CONTENTS
ACKNOWLEDGEMENTS..
TABLE OF CONTENTS ....
LIST OF TABLE .....
LIST OF FIGURES .
ABSTRACT.....
1. INTRODUCTION..2. METHODOLOGY.....
2.1 Learning and understanding the model.....2.2
Solving the equation...
2.3 Project testing .2.4 Collecting Data...
3. IMPLEMENTATION4. RESULTS AND DISCUSSION ..5. CONCLUSIONS AND RECOMMENDATIONSREFERENCES..
APPENDIX A
APPENDIX B ...
ii
iii
iv
v
vi
1
4
4
611
12
13
15
32
33
35
36
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LIST OF TABLES
Table 1 : The collecting data of mass, stiffness and magnitude ........ 12
Table 2 : tu1 and tu2 when 21 mm , as 22431 !m , 22801 !m , 3001 !k and
2002 !k ...17
Table 3 : tu1 and tu2 when 21 mm " , as 22801 !m , 22431 !m , 3001 !k and
2002 !k .
19
Table 4 : )(1 tu and )(2 tu when 21 mm ! as 22801 !m and 22802 !m ,
3001 !k and 2002 !k ....21
Table 5 : )(1tu
and )(2tu
when 21kk
as 2001!k
and 3002!k
, 22801!
m and
22432 !m ...23
Table 6 : )(1 tu and )(2 tu when 21 kk " as 3001 !k and 2002 !k , 22801 !m
and 22432 !m ....25
Table 7 : )(1 tu and )(2 tu when 21 kk ! as 3001 !k and 3002 !k , 22801 !m
and 22432 !m ....27
Table 8 : tu1 and tu2 when 1m = 2m = 2280 and 21 kk ! = 300 . 29
Table 9 : The value of1[
and 2[ in different conditions . 31
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LIST OF FIGURES
Figure 1 : Point of fracture or slippage, termed focus or hypercenter .. 1
Figure 2 : Model of a single-storey building . 3
Figure 3 : Model of a spring-mass-damper system ... 3
Figure 4 : Model of a spring-mass-damper system with two masses ... 4
Figure 5 : A building consisting of floors of mass m connected by stiff but
flexible vertical walls (unforced)... 8
Figure 6 : The model of double storey-building system at equilibrium 11
Figure 7 : The Maple 12 coding for the double-storey building model 13
Figure 8 : The floors moved at the same direction of the equilibrium point 15Figure 9 : The first floor moved to left and second floor moved to right of the
equilibrium point .. 16
Figure 10 : The first floor moved to right and second floor moved to left of the
equilibrium point .. 16
Figure 11 : Graph of tu1 and tu2 when 21 mm , as 22431 !m , 22801 !m ,
3001 !k and 2002 !k .18
Figure 12 : Graph of tu1 and tu2 when 21 mm " , as 22801 !m , 22431 !m ,
3001 !k and 2002 !k .....20
Figure 13 : Graph of )(1 tu and )(2 tu when 21 mm ! as 22801 !m and
22802 !m , 3001 !k and 2002 !k ..22
Figure 14 : Graph of )(1 tu and )(2 tu when 21 kk as 2001 !k and 3002 !k ,
22801 !m and 22432 !m ..24
Figure 15 : Graph of )(1 tu and )(2 tu when 21 kk " as 3001 !k and 2002 !k ,
22801 !m and 22432 !m ...... 26
Figure 16 : Graph of )(1 tu and )(2 tu when 21 kk ! as 3001 !k and 3002 !k ,
22801 !m and 22432 !m .....................28
Figure 17 : Graph of tu1 and tu2 when 1m = 2m = 2280 and 21 kk ! = 300 . 30
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ABSTRACT
Logically, a double-storey building will collapse, depending on the strength of an
earthquake. In mathematics the strength of the earthquake can be interpreted as the
frequency of the vibration and external force that affect the building due to the
earthquake. Hence, from our study, we will present reasons and conditions for a building
to survive or collapse during earthquake. It is our hope that the ability to make
predictions could help scientist, engineers and builders to evaluate, plan and make safer
double-storey buildings. The model of forced vibrations using second order differential
equation on double-storey building is used to study the effect of earthquake on double-
storey building. This model is used to calculate natural frequencies, [ and time, t of the
vibration of the building. The analysis is carried out using Maple 12 and graphs are
presented and discussed.
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1. INTRODUCTIONThroughout the years, earthquake happened without prior warning, destroying houses,
buildings, farmlands and general human activities. According to the US GeologicalSurvey, the earth sustains around 500,000 earthquakes a year, with 100 causing
significant damage. Records from Malaysian Meteorological Department, in year 2000 to
2009, a total of 69 earthquakes have been reported in Malaysia, respectively 31 times in
the peninsular and Sabah, while another seven is in Sarawak. During the period of two
years from 2007 to 2009, there were 40 earthquake recorded, and the highest ever
recorded in Malaysia is 3.5 on the Richter scale. However, according to the Malaysian
Meteorological Department, Peninsular is stable in seism tectonic. The latest earthquake
happened near Malaysia was reported on 14 October 2011 at Papua Guinea, Indonesia
with 6.7 Richter scale. Luckily, Malaysian was not affected by this earthquake.
An earthquake is a natural process that occurs suddenly like volcanic eruption and
meteorite impact. However, an earthquake happens due to sudden release of accumulated
stress in the long run of the earth surface. This pressure may be cause by plate movement
along the crack tectonic and glacial. The earths crust bends and folded. When this
happens, the hard rock can be broken into blocks along the weak zone. When the two
parts differed miles horizontally and vertically, the movement is called faulting.
Faulting can occur on a small scale or large scale. There are three types of fault which
are normal fault, thrust fault and strike-slip fault.
Figure 1: Point of fracture or slippage,
termed focus or hypocenter
fault
epicenter
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The strength of an earthquake is measured by the Richter scale. The Richter scale was
developed in 1935 by American seismologist Charles Richter (1891-1989) as a way of
quantifying the magnitude or strength of the earthquakes.
Since the seventies, Malaysian has started building multi-storey dwelling, single-
storey terrace and double-storey houses. For example today, in Kuala Lumpur we haveKuala Lumpur City Centre (KLCC), Tabung Haji Tower, Putra World Trade Centre
(PWTC), the Dayabumi as well as many multi-storey flats and also condominiums.
However, we are still considered lucky because of the earthquake that occurred in
Malaysia is small at about 3 to 4 on the Richter scale and did not caused much casualties,
like the one that happened in Sabah with 4.5 Richter scale in September 2004. Most of
devastating earthquake often involves destruction, typically occurs at 7 and above on the
Richter scale. That is what happened in Haiti, Peru and recently in Italy, all so horrible
and involves the destruction of countless possession and cause a huge loss. Hence, the
safety of dwellers is of utmost importance to make sure that there is no death due to the
earthquake and reduce the number of destruction in Malaysia.
Although the earthquake that occurred in Malaysia did not caused much
casualties, but this is a big problem to take into consideration since Malaysia have built
many multi-storey building. Therefore, by this reason, we are doing this research because
we are concern with the possibilities of the number of building that may collapses, due to
future earthquake. However, the scope of the project is on the double-storey concrete
buildings in Malaysia. Hence, by using second order differential equation, we may
identify the relation between floors with the movement of each floor. The movements to
be considered are lateral and two-dimensional, that is, either each floor moves parallel or
right to left movements..
Marchand and McDevitt (1999), defined the model of the vibrations of a single-
storey building and multi-storey building using initial value problem of second order
ordinary differential equation. Their research required the understanding of the behavior
at a building during an earthquake. This is important in knowing the reaction of the
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building to various initial conditions. Figure 2 shows the model of a single-story building
and the idealized with a spring-mass-damper system in Figure 3. For this project we
ignore damping effects, c as shownin figure 2 and 3.
Figure 2 : Model of a single-storey buildingSource: Marchand and McDevitt (1999)
Figure 3 : Model of a spring-mass-damper systemSource: Marchand and McDevitt (1999)
where
tu : Displacement of the roof on building at time t.
We are going to consider several linear dynamical systems in which each mathematical
model is a second order differential equations with constant coefficients along with initial
conditions specified at a time that we shall take to be 0!t .
2. METHODOLOGY
2.1 Learning and understanding the model
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We can derive the equation of motion using Hookes law and Newtons second law of
motion in equation (1). Equation (1) is for a single-storey building with forcing and
without damping with the initial displacement and the give initial velocity at time t.
)(//
tFkumu ! 0)0( uu ! 1/ )0( uu ! (1)
where
u : Function of displacement of the block from equilibrium.
/u : First order derivative ofu with respect to t.
//u : Second order derivative ofu with respect to t.
F(t) : Time-dependent applied external force.
m : The total mass of the roof.
k : The overall stiffness of the walls in (Ib/ft).
u(0) : Initial displacement at time t= 0, 0u
u(0) : Initial velocity at time t= 0,1u
For the double-storey building, we model the structural dynamics of a building consisting
of two floors as shown in Figure 4.
Figure 4: Model of a spring-mass-damper system with two masses
Source: Marchand and McDevitt (1999)
We assume that the floors have equal and homogeneous mass and we have two spring
stiffness constant k that attached to the first floor and a weight of mass1m is attached to
the second floor end of this spring. To this weight, a the second spring of stiffness is
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attached having spring constant ck . To the second of this second spring, a weight of
mass2m is attached to the system appears as illustrated in Figure 5. We also noted that
kkc
. As we did with the single spring mass system, we also ignore the damping
effect, c. We will make the same assumption that the springs obeys Hookes law and
exert a linear restoring force on the mass given by xkFs (! . Then the force will be
acting as
)(
)(
123
122
11
uukF
uukF
kuF
c
c
!
!
!
(2)
where force1F acting on its left side of first mass due to one spring and a force 2F acting
on its right side due to the second spring. The 12 uu is the net displacement of the
second spring from its natural length and the force 3F acting at the second mass.
2.2 Solving the equation
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To understand the model for double-story building, we need to solve the Ordinary
Differential Equation (ODE) relevant to the single story building. For the purpose, we
focus on equation (1) which is using the second order ODE model of simple motion with
forcing and without damping. From equation (1), we let
tFtF Pcos0! (3)
where 0F is non-negative frictional force andK
is angular / circular frequency. Substitute
equation (3) in equation (1) will be
tFkumu Pcos0//
! (4)
Hence, by dividing equation (4) with m will give
m
tFu
m
ku
Pcos0//! (5)
We let
m
k![ (6)
where [ is frequency of the earthquake (Nur Hamiza, 2010). Hence we get
m
tFuu
P[
cos02//! (7)
To solve equation (7), we used the method of undetermined coefficients. We assume that
tututu ph ! where tuh is general solution to the corresponding homogeneous
equation and tup is the particular solution of undetermined coefficients. We solve the
equation of tuh first. We consider the general solution of equation (7) will be
02// ! uu [ (8)
In this case,dt
duis proportional to )(tu where )(tu is displacement of a point on the floor
of the building at time t. Then )(tudt
du
w or pudt
du! wherep is constant.
The generalized solution of the form pteu ! , which we differentiate to give,pt
peu !/
andptepu
2//! . Substitute of this function into equation (8) will becomes,
022
!ptpt
eep [ (9)
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The equation can be factorized to
0)(22
! [pept
(10)
where 0{pte for all t. The auxiliary equation of 022
! [p , which has two complex
conjugate roots [is . Hence, the general solution is
tCtCtuh [[ sincos)( 21 ! . (11)
where1C and 2C are magnitude of the earthquake. With simple harmonic motion, we can
express )(tuh in the form
J[ ! tAtuh sin)( (12)
with 0uA by letting Jsin1 AC ! and Jcos2 AC ! , then
tCtC
tAtAtA
[[
J[J[J[
sincos
cossinsincossin
21!
!
(13)
where A is magnitude of earthquake (Nur Hamiza, 2010). and J is distance correction
factor of earthquake (Nur Hamiza, 2010). Hence in order to determine particular solution
to equation (7), the equation
tm
Ftf [cos0! (14)
will providedmk!![K , and tf is eternal forcing function. To determined pu , by
using the method of undetermined coefficients, we know that
ttAttAtup [[ sincos 21 ! (15)
tAttAtAttAtu
tAttAtAttAtu
p
p
[[[[[[[[
[[[[[[
cos2sinsin2cos
sincoscossin
2
2
21
2
1
//
2211
/
!
!
(16)
where1A and 2A are constants. By substituting equation (15) and (16) in (5)
tm
FttAttA
tAttAtAttA
[[[[[
[[[[[[[[
cossincos
cos2sinsin2cos
02
2
2
1
22
212
1
!
(17)
Comparing coefficients in equation (17) gives
01 !A ,[m
A2
12 !
(18)
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Hence, substitute (18) in equation (15)
ttm
Ftup [
[
sin2
0! (19)
Lastly, the general solution for equation (5) by adding tuh and tup
ttm
FtAtu [
[
J[ sin2
sin)( 0! (20)
This is a general solution for undamped resonant case which is [K ! . When [K ! , it
means frequency of earthquake is equal with the natural frequency. To find the general
solution for double-storey building, we apply Newtons second law and Hookes law
which is an unforced vibration of a double-storey building. By Hookes law, its extension
or compression u and forceFor reaction are related by the formula xkF (! . Figure 5
shows and illustration of a building idealized as a collection of double floors, each of
mass m, connected together by vertical walls (Nur Hamiza, 2010).
Figure 5 : A building consisting of floors of mass m connected by stiff but flexible
vertical walls (unforced)
Source : Nur Hamiza Adenan (2010)
Then, by the application of Newtons law maF! to the double storey building yields
the equation of motion. The equation of motion for the first floors and the second floors
are
tFukukkum 1022121//
11 cos [! (21)
and
tFukukum 202212//
22 cos [! (22)
tu1
tu2
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Transforming this equation in the matrix form by considering the general solution of
equation (21) and (22) are the homogeneous equation will be
-
!
-
-
-
-
0
0
0
0
2
1
22
221
//
2
//
1
2
1
u
u
kk
kkk
u
u
m
m(23)
In terms of the displacement, the mass matrix can be form as
M =
-
2
1
0
0
m
m(24)
and stiffness matrix
K=
-
22
221
kk
kkk(25)
the system can be written as
Mu// + Ku = 0 (26)
To obtain equation (26), we take the same step as in equation (4) in the form of matrix.
We will let the matrix of =M
Kwhich is the frequency of the earthquake (Nur Hamiza
A., 2010), and in the form matrix,
=
-
2
2
1
21
0
0
m
k
m
kk
,
u// + u = 0 (27)
Then, taking the same step in equation (9) and (10), the auxiliary equation of 0P ! 22 ,
which has two complex conjugate roots is . Since matrix P =
-
2
1
p
pand =
-
2
1
[
[where
1
211
m
kk ![ and
2
22
m
k![ . Hence, the general solution is
tCtCtCtCtu 242312111 sincossincos [[[[ ! (28)
where 321 ,, CCC and 4C are the magnitude of the earthquake. To obtain a formula for tu2
we use equation (28) to express tu2 in term of tu1 by substitution. Upon simplifying,
we get
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tCtCtCtCtu 242312112 sincossincos [[[[ ! (29)
We are considering the equation (14) to (18) to get the particular solution for the equation
(21) and (22). Thus the particular solution for ttm
Ftu p 1
11
01 sin
2[
[
!
and
ttm
Ftu p 2
22
02 sin
2[
[
!
. Then the final general solution will get
ttm
FtCtCtCtCtu 1
11
0242312111 sin
2sincossincos [
[
[[[[ !
(30)
and
ttm
FtCtCtCtCtu 2
22
0242312112 sin
2sincossincos [
[
[[[[ !
(31)
The equation (28), tu1 refer to the first floor and the equation (29) which is tu2 refer to
the second floor of the double-storey building. Those two equations are the general
solution of the double-storey building. Since we have the general solution for double-
storey building, then we can set our system some initial conditions to solve. We assume
that we the initial displacement tu1 and tu2 to be 0 and the initial velocity, tu/
1 and
tu /2 will be released from rests that are equal to zero.
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2.3 Project testing
We consider testing this project for the two-storey bungalows as in Figure 6.
Figure 6 : The model of double storey-building system at equilibrium
This model is similar to the situations of coupled springs connected to the two masses, m1
and m2 that are attached to the fixed wall of the stiffness k1 and k2 and the values of
displacement of the block from equilibrium, u1 and u2. The equilibrium points are
considered at the center. Our assumption based on the direction of the floor either to the
left or to the right for both of the floor. When the earthquakes happened, we will consider
the external forceF0, natural frequency, 1[ and 2[ which is based on the magnitude of the
earthquake. For model of double storey building, we assume the width of the building is
the 24 ft or 7.3152 m. However, we assume the equilibrium point is in the center of the
building. Therefore, the displacement is between mtum 6576.36576.3 (Arkireka
Architect Department, 2011)to the left and to the right of the equilibrium point. Here we
assume that the value of 0)0(1 !u , 0)0(2 !u , 0)0(
/
1 !u , 0)0(/
2 !u , 1.40 !f .
k1
k2
m1
m2
u1
u2
0 +u2-u2
+u1-u1 0
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2.4 Collecting Data
Table 1 is the data that we use for our project. The basic data for mass of the building
denoted as m is 2280 and 2433 (Arkireka Architect Department, 2011). For the stiffness
as denotes by k, we use values 300 and 200 respectively(). For the magnitude, 0F we
fixed the number to 4.1 (Malaysian Meteorology Department,2010). We fix the
magnitude because it is the latest number of force from the earthquake that happened in
Malaysia. At the time we take the data on the building, the building is static. Hence,
001 !u , 00
2 !u . For the velocity of the building 00/1 !u , 002 !u because we
assume the building is not moving at the time we take the data.
Table 1 : Mass, Stiffness and Magnitude
Condition 1m (kg) 2m (kg) 1k (N/m) 2k (N/m) 0F
1m < 2m 2243 2280 300 200 4.1
1m > 2m 2280 2243 300 200 4.1
1m = 2m 2280 2280 300 200 4.1
1k < 2k 2280 2243 200 300 4.1
1k > 2k 2280 2243 300 200 4.1
1k = 2k 2280 2243 300 300 4.1
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3. IMPLEMENTATION
In this section, we solve the Ordinary Differential Equation (ODE) for the system of
double storey building by using the Maple 12 software to get the solution and graph.
The basic step to solve the equation (21) and (22) are by inserting the initial value for
two floors of the buildings. To explain about our Maple 12 Programming, we take the
simple value of1m , 2m , 1k , 2k , 0F , 1[ , 2[ and the initial value of tu1 , tu
/
1, tu2 , tu
/
2
when 0!t .
Step 1: Determine the value of1m , 2m , 1k , 2k , 0F , 1[ , 2[
Step 2: Implement the equation of double storey building which is the equation (21) andequation (22).
Step 3: Determine the initial value of tu1 , tu/
1, tu2 , tu
/
2
Step 4: Applied ODE solvers.
Step 5: Solution of tu1 and tu2 .
Step 6: Visualize the result by plotting the graph.
We get the all values of the1
m ,2m,
1
k ,2
k ,0
F ,1
[ ,2
[ , tu1
, tu
/
1
, tu2
, tu
/
2 based on
the data of the building and the earthquakes.
Figure 7 : The Maple 12 coding for the double-storey building model
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where
m1, m2 : mass of the building in the first floor and second floor in kg
(kilogram).
k1 , k2 : the stiffness in the building between first floor and second
floor inN/m (Newton per meter).
F0 : the magnitude of the earthquake.
1, 2 : the natural frequency which is based on the value of
stiffness and the mass of the building
u1(0) , u2(0) : the initial displacement of the building at the time t = 0
second (s) in meter(m).
D[1](u1(0)) ,
D[1](u2(0))
: the initial velocity when the earthquake happen at the time
t = 0second (s) in meter per second( 1ms )
The result will show the graph of the implemented equation in the first floor which is in
the red line and the second floor in the blue line based on time the earthquake attract the
building.
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4. RESULTS AND DISCUSSION
In this section, the summary for result and analysis for the model that have been used will
be shown and discuss.
4.1 Tables and Figures
We analyze the equation by using MAPLE. We developed a program to calculate and will
be shown in the graph. The equation will be calculated by using a constant value of )(1 tu
and )(2 tu at initial displacement and initial velocity which is t=0 from the equilibrium
point with mtum 6576.36576.3 ee (Arkireka Architect Department, 2011) and with
external forces, 1.40 !F (Malaysian Meteorology Department, 2010). We are using the
same value of two masses and stiffness which are 2,1m = 2280 kg, 2243 kg and 2,1k = 300
N/m, 200 N/m. We calculate the length of the movement of the floors by adding the
displacement of first and second floor when the floors moved at different direction of the
equilibrium point and when the floors moved at the same direction of the equilibrium
point, we will subtract that to get the value of that movement.
Figure 8: The floors moved at the same direction of the equilibrium point
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Figure 9: The first floor moved to left and
second floor moved to right of the
equilibrium point
Figure 10 : The first floor moved to right
and second floor moved to left of the
equilibrium point
All figures below show the type of movement when ground of building shaken. This
physical interpretation is that an earthquake wave of the proper frequency, having time
duration sufficiently long, can crack a floor and hence collapse the entire building. The
amplitude of the earthquake wave does not have to be large, once the floors move of a
centimeter might be enough to start the oscillation of the floor.
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4.1.1 Estimated value points when using the different value of mass and fixed the
value of stiffness and external force in the three conditions.
Table 2 : tu1 and tu2 when 21 mm , as 22431 !m , 22802 !m , 3001 !k and 2002 !k
t (second) 0 5 10 15 20 25 30
tu1 0 0.0039 0.0435 -0.0714 -0.0257 0.0469 0.0587
tu2 0 -0.0472 0.0667 -0.0089 0.0113 -0.0481 0.0333
)()( 12 tutu 0 0.0511 0.0232 0.0625 0.0370 0.0950 0.0254
Table 2 shows the tu1 and tu2 when 21 mm , as 22431 !m , 22802 !m , while
3001 !k and 2002 !k have different values when the value of t increases. At 0 second,
there is no displacement recorded. We can see the motions of the floors of different
direction from the equilibrium point recorded at 5, 20 and 25 seconds. At 10, 15
and 30 seconds, both of the floors moved at the same direction of the equilibrium point.
From table above, the highest difference in displacement with 0.0950m between first and
second floor is at time 25 seconds.
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Figure 11: Graph of tu1 and tu2 when 21 mm , as 22431 !m , 22802 !m , 3001 !k
and 2002 !k
The graph of tu as shown in Figure 11visualizes the motion more clearly. The highest
peak or maximum value of tu is amplitude or maximum displacement. From the graph,
we conclude that the building moved at different and same direction and it is possible that
the building may collapse. In this condition, the value of natural frequency are
472139532.0112152243
101 !![
and 296174438.028557
12 !![
.
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Table 3: tu1 and tu2 when 21 mm " , as 22801 !m , 22432 !m , 3001 !k and 2002 !k
t (second) 0 5 10 15 20 25 30
tu1 0 0.0026 0.0468 -0.0642 -0.0363 0.0328 0.0669
tu2 0 -0.0484 0.0633 -0.0050 0.0158 -0.0460 0.0254
)()( 12 tutu 0 0.0510 0.0165 0.0592 0.0521 0.0788 0.0415
Table 3 shows the tu1 and tu2 when 21 mm " , as 22801 !m , 22432 !m , 3001 !k
and 2002 !k have different values when the value of t increases. At 0 second, there is no
displacement recorded. We can see the motions of the floors of different direction from
the equilibrium point recorded at 5, 20 and 25 seconds. At 10, 15 and 30 seconds,
both of the floors moved at the same direction of the equilibrium point. From table above,
the highest difference in displacement with 0.0788m between first and second floor is at
time 25 seconds.
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Figure 12: Graph of tu1 and tu2 when 21 mm " , as 22801 !m , 22432 !m , 3001 !k
and 2002 !k
The graph of tu as shown in Figure 12 visualizes the motion more clearly. The highest
peak or maximum value of tu is amplitude or maximum displacement. From the graph,we conclude that the building moved at different and same direction and it is possible that
the building may collapse. In this condition, the value of natural frequency are
468292905.0114114
51 !![
and 298607259.044862243
102 !![
.
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Table 4: )(1 tu and )(2 tu when 21 mm ! as 22801 !m and 22802 !m , 3001 !k and
2002 !k
t (second) 0 5 10 15 20 25 30
tu1 0 0.0035 0.0476 -0.0675 -0.0368 0.0355 0.0710
tu2 0 -0.0474 0.0638 -0.0101 0.0190 -0.0428 0.0269
)()( 12 tutu 0 0.0509 0.0162 0.054 0.0558 0.0783 0.0441
Table 4 shows the displacement of )(1 tu and )(2 tu when 21 mm ! as 22801 !m and
22802 !m , 3001 !k and 2002 !k as the time t increases. At 0 second, there is no
displacement recorded. We can see the motions of the floors of different direction from
the equilibrium point recorded at 5, 20 and 25 seconds. At 10, 15 and 30 seconds,
both of the floors moved at the same direction of the equilibrium point. From table above,
the highest difference in displacement with 0.0783m between first and second floor is at
time 25 seconds.
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Figure 13: Graph of )(1 tu and )(2 tu when 21 mm ! as 22801 !m and 22802 !m ,
3001 !k and 2002 !k
The graph of tu as shown in Figure 13 visualizes the motion more clearly. The highest
peak or maximum value of tu is amplitude or maximum displacement. . From the
graph, we conclude that the building moved at different and same direction and it is
possible that the building may collapse. In this condition, the value of natural frequency
are 468292905.0114114
51 !![
and 296174438.028557
12 !![
.
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4.1.2 Estimated value points when using the different value of stiffness and fixed
the value of mass and external force in the three conditions.
Table 5 : )(1 tu and )(2 tu when 21 kk as 2001 !k and 3002 !k , 22801 !m and
22432 !m
t (second) 0 5 10 15 20 25 30
tu1 0 -0.0137 -0.0019 -0.0213 0.0204 -0.0001 0.0385
tu2 0 -0.0357 0.0219 -0.0009 0.0219 -0.0093 0.0036
)()( 12 tutu 0 0.0220 0.0238 0.0204 0.0015 0.0092 0.0349
Table 5 shows the displacement of )(1 tu and )(2 tu when 21 kk as 2001 !k and
3002 !k , 22801 !m and 22432 !m have the different value when the value of time
increases. At 0 second, there is no displacement recorded. We can see the motion of the
floors of different direction from the equilibrium point recorded at 10 seconds. At 5,
15, 20, 25 and 30 seconds, both of the floors moved at the same direction of the
equilibrium point. From table above, the highest difference in displacement with 0.0349mbetween first and second floor is at time 30 seconds.
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Figure 14: Graph of )(1 tu and )(2 tu when 21 kk as 2001 !k and 3002 !k , 22801 !m
and 22432 !m
The graph of tu as shown in Figure 14 visualizes the motion more clearly. The highest
peak or maximum value of tu is amplitude or maximum displacement. From the graph,
we conclude that the building moved at different and same direction and it is possible that
the building may collapse. In this condition, the value of natural frequency are
468292905.0114114
51 !![
and 365717709.067292243
102 !![
.
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Table 6: )(1 tu and )(2 tu when 21 kk " as 3001 !k and 2002 !k , 22801 !m and
22432 !m
t (second) 0 5 10 15 20 25 30
tu1 0 -0.0142 0.0594 -0.0226 -0.0805 -0.0110 0.1458
tu2 0 -0.0484 0.0633 -0.0050 0.0158 -0.0460 0.0254
)()( 12 tutu 0 0.0342 0.0039 0.0176 0.0963 0.0350 0.1204
Table 6 shows the )(1 tu and )(2 tu when 21 kk " as 3001 !k and 2002 !k , 22801 !m and
22432 !m have the different value when the value of time increases. At 0 second, there is
no displacement recorded. We can see the motion of the floors of different direction from
the equilibrium point recorded at 20 seconds. At 5, 10, 15, 25 and 30 seconds,
both of the floors moved at the same direction of the equilibrium point. From table above,
the highest difference in displacement with 0.1204m between first and second floor is at
time 30 seconds.
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Figure 15: Graph of )(1 tu and )(2 tu when 21 kk " as 3001 !k and 2002 !k , 22801 !m
and 22432 !m
The graph of tu as shown in Figure 15 visualizes the motion more clearly. The highest
peak or maximum value of tu is amplitude or maximum displacement. From the graph,
we conclude that the building moved at different and same direction and it is possible that
the building may collapse. In this condition, the value of natural frequency are
468292905.0114114
51 !![
and 298607259.044862243
102 !![
.
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Table 7: )(1 tu and )(2 tu when 21 kk ! as 3001 !k and 3002 !k , 22801 !m and
22432 !m
t (second) 0 5 10 15 20 25 30
tu1 0 -0.0141 -0.0211 0.0138 0.0075 0.0133 -0.0052
tu2
0 -0.0342 0.0268 -0.0207 0.0029 0.0275 0.0068
)()( 12 tutu 0 0.0201 0.0479 0.0345 0.0046 0.0142 0.0120
Table 7 shows the )(1 tu and )(2 tu when 21 kk ! as 3001 !k and 3002 !k , 22801 !m and
22432 !m have the different value when the value of time increases. . At 0 second, there
is no displacement recorded. We can see the motion of the floors of different direction
from the equilibrium point recorded at 10, 15 and 30 seconds. At 5, 20, 1and 25
seconds, both of the floors moved at the same direction of the equilibrium point. From
table above, the highest difference in displacement with 0.0479m between first and
second floor is at time 10 seconds.
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Figure 16: Graph of )(1 tu and )(2 tu when 21 kk ! as 3001 !k and 3002 !k , 22801 !m
and 22432 !m
The graph of tu as shown in Figure 16 visualizes the motion more clearly. The highest
peak or maximum value of tu is amplitude or maximum displacement. From the graph,
we conclude that the building moved at different and same direction and it is possible that
the building may collapse. In this condition, the value of natural frequency are
512989176.09519
11 !![
and 365717709.067292243
102 !![
.
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4.1.3 Estimated value points when mass and the stiffness are same value.
Table 8: tu1 and tu2 when 228021 !! mm and 30021 !! kk
t (second) 0 5 10 15 20 25 30
tu1 0 -0.0135 -0.0202 0.0152 0.0111 0.0120 -0.0039
tu2 0 -0.0344 0.0285 -0.0198 0.0063 0.0275 0.0070
)()( 12 tutu 0 0.0209 0.0487 0.0350 0.0048 0.0155 0.0109
Table 8 shows the tu1 and tu2 when 228021 !! mm and 30021 !! kk have the
different value when the value of time increases. At 0 second, there is no displacement
recorded. We can see the motion of the floors of different direction from the equilibrium
point recorded at 10, 15 and 30 seconds. At 5, 20 and 25 seconds, both of the
floors moved at the same direction of the equilibrium point. From table above, the highest
difference in displacement with 0.0487m between first and second floor is at time 10
seconds.
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Figure 17: Graph of tu1 and tu2 when 228021 !! mm and 30021 !! kk
The graph of tu as shown in Figure 17 visualizes the motion more clearly. The highest
peak or maximum value of tu is the amplitude or maximum displacement. From the
graph, we conclude that the building moved at different and same direction and it is
possible that the building may collapse. In this condition, the value of natural frequency
are 512989176.09519
11 !![
and 362738125.019038
12 !![
.
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Table 9 : The value of1[
and2[in different conditions.
Conditions 1[ 2[
21mm 0.472139532 0.296174438
21mm " 0.468292905 0.298607259
21mm ! 0.468292905 0.296174438
21kk 0.468292905 0.365717709
21 kk " 0.468292905 0.298607259
21 kk ! 0.512989176 0.365717709
21mm ! and 21 kk ! 0.512989176 0.362738125
Table 9 shows the value of1[
and2[in the different condition. The value of
1[for the
conditions of21
mm " ,21
mm ! ,21
kk and 21 kk " shown the same value which is
0.468292905 and the condition of21 kk ! and the combination of 21 mm ! and 21 kk !
give the value 0.512989176. The value of2[for the conditions
21mm and
21mm ! is
0.296174438, for the conditions21
mm " and 21 kk " is 0.298607259 and for the
conditions21
kk and is 0.365717709.
Therefore, the effect of the earthquake depends on the natural frequencies of oscillation of
the floors and the displacement of the building motion. The displacement of building
motion increase in )(tu , causing the floors to take an excursion that is too large to
maintain the structural the integrity of the floor. Hence, we conclude that the highest
difference in displacement and natural frequency of oscillation of the floors may cause
severe damage. For example in this research, the condition when 21 kk " has the highest
displacement with 0.1204m and its natural frequency about 0.298607259 to 0.468292905
will collapse the entire building at time 30 seconds.
21 kk !
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5. CONCLUSIONS AND RECOMMENDATIONS
In this project, the study is to identify the movement between each floor in order to help
minimize destruction and to help in construction of vibration proof buildings. We use
Ordinary Differential Equation (ODE), modified for double-storey building with the
collecting data that we have already collected and assume that the initial displacement
and the initial condition for two of the floors are equal to zero. That means there are no
movement of the floors before earthquake happens. Then we solve the equation using
Maple 12 that will show the graph of seven conditions depending on the data. Then we
identify the movement of the building by the result shows in Maple.
As we know, earthquakes are a major problem of mankind. But we can say that the
buildings are not usually collapsed depending on the magnitudes of earthquake. The
structure of the building, especially the balancing of the mass and stiffness of each floor
are important to make the building last for a long time.
For this research, we have limited our scope to double-storey building and using second
order ordinary equation with force and without damping. For future research, we
recommended that the study can be extended using equations with damping and multi
storey building. Another area to study with regards to the effect of the earthquake on
buildings is resonance. We may also use numerical method with real data.
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REFERENCES
A.K. Chopra. (1995). Dynamics o f Structures, Theory and Applications to Eartquake
Engineering.New Jersey: Prentice Hall, Inc., Upper Saddle.
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Benaroya, H. (1998). Mechanical Vibration: Analysi s, Uncertainty, and Control. New
Jersey: Prentice-Hall.
Blanchard, P., Devaney, R.L., & Hall, G.R. (2002).Differential Equations. Pacific Grove:
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Boyce, W.E., & DiPrima, R.C. (2004). Elementary Differential Equations. New York:
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Chakraverty, S., Gupt, P., & Sharma, S. (2009). Neural network-based simulation for
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Dhang, N. Structural Dynamics: An overview. Department of Civil Engineering.
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Gagen, Alex & Larson, Sean (2000). CoupledOscillators.
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Kelly, S.G. (1993). Fundamental o f Mechanical Vibrations. New York: McGraw-Hill,
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M. C. M. Bakker, J. G. M. Kerstens & J. Weerheijm (2008).Dynamic response of high-
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Marchand, R., & McDevitt, T.J. (1999). Learning Differential Equations by Exploing
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Nagle, R.K., Saff, E.B., & Snider, A.D. (2004).Fundamentals of Differential Equations
andBoundary Value Problem.New York: Pearson Education, Inc.
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Nur Hamiza A. (2010),Mathematical Modelling o f Earthquake Induced Vibration on
Multistory Buildings. Malaysia: Universiti Teknologi Malaysia(UTM).
Zappler, G., & Long, L.T. (2002).Popular Science 2. Danbury: Grolier Incorporated.
Zill, D.G., & Cullen, M.R. (2009). Differential Equations with Boundary-Value Problem.
Canada: Brooks/Cole, Cengage Learning.
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APPENDIX A
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APPENDIX B
The Maple 12 coding for the double-storey building model based on the condition.
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