tech report without damping!!! (final draft) (1)

8/3/2019 Tech Report Without Damping!!! (Final Draft) (1)

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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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ii

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


468292905.0114114

51 !![

and 298607259.044862243

102 !![

.


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21

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,



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


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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23

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





468292905.0114114

51 !![

and 365717709.067292243

102 !![

.


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25

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,



time 30 seconds.


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26

Figure 15: Graph of )(1 tu and )(2 tu when 21 kk " as 3001 !k and 2002 !k , 22801 !m

and 22432 !m





468292905.0114114

51 !![

and 298607259.044862243

102 !![

.


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27

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





512989176.09519

11 !![

and 365717709.067292243

102 !![

.


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29

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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30

Figure 17: Graph of tu1 and tu2 when 228021 !! mm and 30021 !! kk


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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31

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.

Arkireka Architect Department (2011).

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:

Thomson Learning, Inc.

Boyce, W.E., & DiPrima, R.C. (2004). Elementary Differential Equations. New York:

John Wiley & Sons, Inc.

Chakraverty, S., Gupt, P., & Sharma, S. (2009). Neural network-based simulation for

response identification of two-storey shear building subject to earthquake motion.

Journal of neural computing an

dappl

icat

ion. 19(3). Doi: 10.1007/s00521-009-

0279-6.

Dhang, N. Structural Dynamics: An overview. Department of Civil Engineering.

IIT,Kharagpur.

Earthquake Hazards Program. (2009). Last earthquake in Malaysia. Retrieve from

http://earthquake.usgs.gov/earthquakes/eqarchives/last_event/world/world_malays

ia.php

Gagen, Alex & Larson, Sean (2000). CoupledOscillators.


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Kelly, S.G. (1993). Fundamental o f Mechanical Vibrations. New York: McGraw-Hill,

Inc.

M. C. M. Bakker, J. G. M. Kerstens & J. Weerheijm (2008).Dynamic response of high-

ri se building structuresto blast loading. Netherland: Eindhoven University of

Technology.

Magnitude Earthquake (2011). Earthquake hits papua new guinea. Retrive from

http://www.irishweatheronline.com/news/earth-science/geology/6-7-magnitude-

earthquake-hits-papua-new-guinea/41757.html.

Malaysian Meteorological Department (2010). Retrieve from

http://www.met.gov.my/index.php?option=com_content&task=view&id=1717&It

emid=1635

Marchand, R., & McDevitt, T.J. (1999). Learning Differential Equations by Exploing

Earthquake Induced Structural Vibrations, 15(6), 477-485.

Nagle, R.K., Saff, E.B., & Snider, A.D. (2004).Fundamentals of Differential Equations

andBoundary Value Problem.New York: Pearson Education, Inc.

New Straits Time (2011). Is Malaysia in Danger. Retrieve from

http://www.nst.com.my/nst/articles/IsMalaysiaindanger/Article/art_print

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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tech report without damping!!! (final draft) (1)

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