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Analysis and Design of Inclined Columns

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  • PSZ 19:16 (Pind. 1/07)

    DECLARATION OF THESIS / UNDERGRADUATE PROJECT PAPER AND COPYRIGHT

    Authors full name :

    Date of birth : Title :

    Academic Session:

    I declare that this thesis is classified as :

    I acknowledged that Universiti Teknologi Malaysia reserves the right as follows :

    1. The thesis is the property of Universiti Teknologi Malaysia. 2. The Library of Universiti Teknologi Malaysia has the right to make copies for the purpose

    of research only. 3. The Library has the right to make copies of the thesis for academic exchange.

    Certified by :

    SIGNATURE SIGNATURE OF SUPERVISOR

    (NEW IC NO. /PASSPORT NO.) NAME OF SUPERVISOR

    Date : 25 APRIL 2008 Date : 25 APRIL 2008

    2007/2008

    21 NOVEMBER 1985

    SHAFIAH BINTI DOLHAKIM

    ANALYSIS AND DESIGN OF INCLINED COLUMN

    NOTES : * If the thesis is CONFIDENTIAL or RESTRICTED, please attach with the letter from the organisation with period and reasons for confidentiality or restriction.

    UNIVERSITI TEKNOLOGI MALAYSIA

    CONFIDENTIAL (Contains confidential information under the Official Secret Act 1972)*

    RESTRICTED (Contains restricted information as specified by the

    organisation where research was done)* OPEN ACCESS I agree that my thesis to be published as online open access

    (full text)

    IR AZHAR AHMAD 851121-13-5362

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  • I hereby declare that I have read this report and in my opinion this report is

    sufficient in terms of scope and quality for the award of the degree of Bachelor of Civil Engineering

    Signature : _____________________________ Name of Supervisor : IR AZHAR AHMAD

    Date : 25 APRIL 2008

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  • i

    ANALYSIS AND DESIGN OF INCLINED COLUMN

    SHAFIAH BINTI DOLHAKIM

    A report submitted in partial fulfillment of the

    requirements for the award of the degree of Bachelor of Civil Engineering

    Faculty of Civil Engineering Universiti Teknologi Malaysia

    25 APRIL, 2008

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  • ii

    ANALISIS DAN REKABENTUK TIANG CONDONG

    SHAFIAH BINTI DOLHAKIM

    Tesis ini dikemukakan sebagai memenuhi

    sebahagian daripada syarat penganugerahan Ijazah Sarjana Muda Kejuruteraan Awam

    Fakulti Kejuruteraan Awam Universiti Teknologi Malaysia

    25 APRIL, 2008

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  • iii

    I declare that this report entitled Analysis and Design of Inclined Column is the result of my own study except as cited in the references. The report has not been

    accepted for any degree and is not concurrently submitted in candidature of any other degree.

    Signature : ________________________ Name : SHAFIAH BINTI DOLHAKIM

    Date : 25 APRIL 2008

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  • iv

    Specially For;

    My Parent

    My sisters and brothers Nephews and nieces

    For giving me support, love, advices and encourage me all the times

    And not forgotten to all my friends and lecturers

    Wish we are success in our own future and God Bless us all

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  • v

    ACKNOWLEDGEMENTS

    I would like to take this opportunity to extend my outmost gratitude to Allah SWT for giving me good health, enjoyable life and guiding me in completing this study.

    I am also grateful for the invaluable guidance to my supervisor, Mr Ir Azhar

    Ahmad for his continuous guidance, support, encouragement and patience throughout my research work. His suggestions and comments have given me the courage and confidence to handle this research work and formed a valuable part of this thesis.

    I would also like to express my gratitude to my parent for their unreserved support and concern for me to accomplish this study. Not forgotten to my family

    members, group members, classmates, schoolmates and friends for giving me brilliant ideas and advices and for supporting me on all the decisions that I have made.

    Thank you so much..

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  • vi

    ABSTRAK

    Kajian ini dijalankan bagi menentukan prosedur, kaedah dan formula dalam merekabentuk tiang condong. Disebabkan pembangunan yang pesat pada masa kini, elemen struktur seperti tiang boleh direkabentuk secara unik bagi

    memperlihatkan nilai-nilai kesenian senibina sesuatu pembangunan. Kini, tiang bukan sahaja boleh direkabentuk secara tegak tetapi juga condong. Rekabentuk ini adalah berdasarkan spesifikasi yang terdapat dalam BS8110 dan juga merujuk kepada kajian kes Cadangan Pembangunan Taman Ekologi, Hutan Bandar, Johor Bahru yang telah diubahsuai. Tiang condong yang direkabentuk bergantung kepada jenis struktur kerangka yang mana tiang ini menanggung beban graviti dan dikenakan pada jasad tegar atau sesetengah kerangka dirembat. Satu set keputusan diperolehi daripada kajian ini dan keputusan tersebut dibandingkan dengan keputusan yang diperolehi daripada tiang biasa. Daripada analisis dan rekabentuk tiang condong, dapat disimpulkan bahawa momen yang wujud daripada tiang condong adalah lebih besar daripada momen tiang tegak. Ini adalah disebabkan oleh wujudnya kesipian dan akibat pertambahan beban.

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  • vii

    ABSTRACT

    Nowadays, people prefer to construct buildings with sophisticated design due to the rapid development. The structural elements such as column are also design uniquely to show its aesthetic value. For instance, column is not longer build vertically but it can be inclined. This research done to determine the procedure,

    method and formula of designing an inclined column based on BS8110 and a case study of Cadangan Pembangunan Taman Ekologi, Hutan Bandar, Johor Bahru that

    had been modified. Inclined columns results from the type of structural framing which are gravity load only columns and apply to both rigid and some braced frames. A set of results have been obtained from this research and these results are compared with the results in vertical column. Analysis and designing the inclined column leads

    to the conclusions that the moment in the inclined column were larger than those in the normal one due to the existing of eccentricity and the increasing of load.

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  • viii

    TABLE OF CONTENTS

    CHAPTER TITLE PAGE

    TITLE PAGE i DECLARATION iii DEDICATION iv ACKNOWLEDGEMENTS v ABSTRACT vi ABSTRAK vii TABLE OF CONTENTS viii LIST OF FIGURES xii LIST OF APPENDICES xiii

    1

    INTRODUCTION

    1

    1.1 Background 1

    1.2 Statement of Problem 1

    1.3 Objectives 2

    1.4 Scope of the Study 1.5 Significance of the Study

    2

    3

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  • ix

    CHAPTER

    TITLE

    PAGE

    2

    LITERATURE REVIEW

    4

    2.1 Introduction to the Column 2.2 Inclined/ Leaning Column

    2.2.1 Definition

    2.2.2 Leaner Column / Inclined Column 2.3 Analysis

    2.4 Types of Column 2.4.1 Braced and Unbraced Column

    2.4.2 Short and Slender Column 2.5 Effective Height 2.6 End Condition 2.7 Mode of Failure 2.8 Load and Deflection 2.9 Design of Short column 2.9.1 Braced Short Column 2.9.2 Short Braced Columns Supporting an Approximately Symmetrical Arrangement of Beams (3.8.4.4) 2.9.3 Short Braced Columns Support Both Moment and Axial Load ( 3.8.4.5) 2.10 Design of Slender Column 2.10.1 Braced Slender Column

    2.10.2 Unbraced Slender Column 2.11 Design of Inclined Column 2.12 Frame Analysis

    2.13 Main Reinforcement

    4

    5 5 5 6 7

    7

    9 10

    10 11

    11

    13

    14

    14

    14

    15 16 17

    17

    19 21

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  • x

    CHAPTER

    TITLE

    PAGE

    3

    RESEARCH METHODOLOGY

    22

    3.1 Introduction

    3.2 Method use to analyze structure 3.3 Case study 3.4 Design process

    22

    24

    24

    25

    4

    STRUCTURAL ANALYSIS 27

    4.1 Introduction 4.2 Frame analysis

    4.3 Column analysis

    4.3.1 Inclined column 4.3.2 Vertical column

    27

    43 29 29 32

    CHAPTER TITLE PAGE

    5 RESULTS AND DISCUSSIONS 35

    5.1 Introduction 5.2 Results from analysis and design 5.3 Comparison between inclined column and

    vertical column.

    35 35 37

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  • xi

    CHAPTER TITLE PAGE

    6 CONCLUSIONS AND RECOMMENDATIONS

    38

    6.1 Conclusions 6.2 Recommendations

    38 39

    REFERENCES

    40

    Appendices A-F 43

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  • xii

    LIST OF FIGURES

    FIGURE TITLE PAGE

    2.2.2a

    2.4.1a

    2.4.1b 2.4.1c

    2.8a

    2.8b

    2.11

    2.12.1a

    2.12.1b 2.12.1c

    3.1 3.3 4.2a

    4.2b

    Inclined column. The building are braced in both direction.

    The building are unbraced in both direction. The building is braced in y direction but unbraced in x direction. Inclined column subjected to eccentric axial load. Simply supported column subjected to axial load. Relationship between load and bending moment

    in inclined column. Type 1 braced frame. Type 2 braced frame. Type 3 braced frame. Simplified research methodology in this study. Front view of the building (case study). Roof Second Floor First Floor.

    First Floor Ground Floor -Foundation

    6 8

    8 9

    12

    13

    18

    19 20 21

    23 25 28

    28

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  • xiii

    LIST OF APPENDICES

    APPENDIX.

    TITLE

    PAGE

    A

    B

    C D

    E

    F

    Architecture drawing of case study Frame Analysis - Braced Frame

    Column Analysis Inclined column Column Analysis Vertical column Chart no 28 and chart no 29 Detailing of the column reinforcement

    43 44

    46 53 61 63

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  • CHAPTER 1

    INTRODUCTION

    1.1 BACKGROUND

    In structural engineering, a column is a vertical structural element that

    transmits the weight of the structure above to other structural elements below through compression. Columns can be either compounded of parts or made as a single piece and frequently used to support beams, arches and slabs on which the

    upper parts of walls or ceilings rest. Column refers specifically to such a structural element that also has certain proportional and decorative features.

    Design of columns is influenced by the ultimate limit state which is the deflections and cracking during service conditions are not usually a problem, besides correct detailing of the reinforcement and adequate cover are important.

    1.2 STATEMENT OF PROBLEM

    Nowadays, people prefer to construct buildings with sophisticated design due to the rapid development. The structural elements such as column are also design

    uniquely to show its aesthetic value. For instance, column is not longer build vertically but it can be inclined. This study done to determine the procedure, method

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  • 2

    and formula of designing inclined column based on BS8110 and a case study of Cadangan Pembangunan Taman Ekologi Hutan Bandar, Johor Bahru focusing on the analysis and designing a column only. However, the limitations of references make

    the analysis of this research difficult. Thus, the specific solution cannot be obtained. Furthermore, the design consideration in BS8110 is only specifically for vertical

    column.

    1.3 OBJECTIVES

    The objectives of this study are:

    i. To analyze the equilibrium forces in inclined column. ii. To design an inclined column based on BS8110.

    iii. To find out the comparison between inclined column and vertical column due to the method used and factors influence the designing.

    iv. As references for future uses.

    1.4 SCOPE OF THE STUDY

    The scopes of this study include:

    i) Literature review for analyzing and designing an inclined column in concrete structures.

    ii) Literature review for analyzing and designing an inclined column based on specification in British Standard, BS 8110.

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  • 3

    1.5 SIGNIFICANCE OF THE STUDY

    This study is more focusing on analyzing and designing an inclined column due to the specification given in BS 8110. The result of analyzing and designing this

    type of column will be compared with another type of column, the vertical column.

    One of the rational reasons this study were conducted is to determine the method and formula used in designing an inclined column. With the information at hand, standard method and formula to design an inclined column could be publicized

    and more extensive researches could be planned for the future.

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  • CHAPTER 2

    LITERATURE REVIEW

    2.1 INTRODUCTION

    As the axial load on a perfectly straight slender column is increased in magnitude, this column will passes through three states: stable equilibrium, neutral equilibrium, and instability. The straight column under load is in stable equilibrium if a lateral force is applied between the two ends of the column and produces a small

    lateral deflection which disappears and this column will returns to its straight form when the lateral force is removed. If the column load is gradually increased, a

    condition is reached in which the straight form of equilibrium becomes neutral equilibrium where a small lateral force will produce a deflection that does not disappear and the column remains in this slightly bent form when the lateral force is removed (cl 3.8.3.2). The load at which neutral equilibrium of a column is reached is called the critical or buckling load. The state of instability is reached when a slight increase of the column load causes uncontrollably growing lateral deflections leading to complete collapse.

    A reinforced concrete column is extended by having the steel reinforcing bars stick out a few inches or feet above the top of the concrete, then placing the next

    level of reinforcing bars to overlap, and pouring the concrete of the next level. Steel reinforcement in concrete columns provides compressive capacity, but its most important role is in controlling the mode of failure. When taken beyond yield point, concrete will suddenly fails abruptly-even explosively and therefore, reinforcement

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  • 5

    is used even in columns subject only to axial compression as small steel columns and even though encased in concrete, their slenderness ratio must be limited with horizontal ties.

    2.2 INCLINED/LEANING COLUMN

    2.2.1 Definition

    Inclined or leaning means departing or being caused to depart from the true

    vertical or horizontal or the property depend on a line or surface that departs from the vertical or move away from a vertical position.

    2.2.2 Leaner column/ inclined column

    The general stability of a structure must be provided as it relates to the each individual column. Therefore, the consideration must be given to the load effects resulting from the deflected shape of the structure. The stability of a column not involved with the lateral bracing system is therefore dependent on the rigidity of the

    columns associated with the lateral bracing system or rigid bents. The columns that are dependent on the rigid frame columns are referred to as "Leaner Columns". Leaner columns are gravity load only columns and apply to both rigid and some

    braced frames as shown in Figure 2.2.2a.

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  • 6

    Figure 2.2.2a Inclined column

    2.3 ANALYSIS

    The analysis of an engineering structure ideally involves a complete evaluation of structural behavior such as loading conditions. Fortunately, by a process of rational elimination it is usually possible to reduce the problem to manageable proportions.

    The stage of designing a column representative loading cases and method of

    analysis are chosen is a vital one. The calculation may usually be simplified by making assumptions. Subject to this provision methods of analysis should be as simple as possible and more complex techniques should only be utilized when;

    Inclined column

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  • 7

    i) The low calculated strength of a component is referring to the accuracy of the assumptions.

    ii) More exact analysis is required to justify weight saving.

    The strength design method requires service loads or related internal

    moments and forces to be increased by specified load factors (required strength) and computed nominal strengths to be reduced by specified strength reduction factors, (design strength). Therefore, the equations in designing inclined column are based on the equations in designing vertical column. BS 8110 has classified column as a

    compression members with cross-sectional dimension does not exceed four times its smaller dimension, h 4b but if, h > 4b it will be classified as a wall.

    2.4 TYPES OF COLUMN

    There are two types of column which are braced and unbraced column. For braced column the calculation only considered on dead load and imposed load while

    for unbraced column the consideration taken are dead load, imposed load and wind load.

    2.4.1 Braced and Unbraced Column

    From clause 3.8.1.5 BS8110, a column may be considered as a braced column if lateral stability to the whole structure is provided by walls or bracing to resist all lateral forces. Therefore, braced column will only carry vertical load as

    shown in figure 2.4.1a and figure 2.4.1c. With a braced column the axial forces and moments are caused by the dead and imposed load only. If one of the braced members is not provided, the effect of the vertical load in any direction will be support together by column and beam.

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  • 8

    In the other word, this column is design to carry both vertical and horizontal load and this is known as unbraced column as shown in figure 2.4.1b and figure

    2.4.1c. With an unbraced column the loading arrangements which include the effects of the lateral loads must also be considered.

    For a braced column the critical arrangement of the ultimate load is usually causes by the largest moment in the column, together with a large axial load. When the moments in columns are large and particularly with unbraced columns, it may also be necessary to check the case of maximum moment combined with the

    minimum axial load.

    Figure 2.4.1a The building are braced in both direction

    Figure 2.4.1b The building are unbraced in both direction

    Column

    Column

    Load Bearing Wall

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  • 9

    Figure 2.4.1c The building is braced in y direction but unbraced in x direction

    The axial forces due to the vertical loading may be calculated as though the beams and slabs are simply supported. In some structures it is unlikely that all the floors of a building will carry the full imposed load at the same instant, therefore, a reduction is usually allowed in the total imposed load when designing columns and

    foundations in buildings which are two or more storey high.

    2.4.2 Short and Slender Column

    Both columns will be classified into short column and slender column. A

    column may be considered as a short column when the ratios of hlex / and bley / are

    less than 15 (braced) and 10 (unbraced). On the other hand, slender column may be considered when the ratios of hlex / and bley / are more than the limit above. Both

    exl and eyl are effective height of a column respectively with x and y axis. According

    to Clements (1981), in designing a column the value of h which is the side of the column that is parallel to the deflection direction have to be considered.

    Column

    Load Bearing Wall

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  • 10

    2.5 EFFECTIVE HEIGHT

    The effective height, el of a column in a given plane may be obtained from equation

    below in BS 8110 (3.8.1.6):

    el ol

    The values of are given in table 3.19 BS8110: Part 1 for braced column and table

    3.20 for unbraced column as a function of the end conditions of the column while ol

    is the clear height between end of the column and restraints.

    2.6 END CONDITIONS

    The four end conditions are: (3.8.1.6.2)

    a) Condition 1 The end of the column is connected monolithically to beams on either side which are at least as deep as the overall dimension of the column in the plane considered. Where the column is connected to a foundation structure, this should be of a form specifically designed to carry moment.

    h beam > h column

    b) Condition 2 The end of the column is connected monolithically to beams or slabs on either side which are shallower than the overall dimension of the column in the plane considered.

    h beam < h column

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  • 11

    c) Condition 3 The end of the column is connected to members which, while not

    specifically designed to provide restraint to rotation of the column will, nevertheless, provide some nominal restraint.

    d) Condition 4 The end of the column is unrestrained against both lateral movement and rotation.

    2.7 MODE OF FAILURE

    The mode of failure of a column can be one of the following:

    a) Material failures with negligible lateral deflection, which usually occurs with short columns but can, also occur when there are large end moments on a column with an intermediate slenderness ratio.

    b) Material failures are caused by the lateral deflection and the additional moment. This type of failure is typical of indeterminate columns.

    c) Instability failure which occurs with slender columns and is liable to be preceded by excessive deflections.

    2.8 LOAD AND DEFLECTION

    A column with a cross section that lacks symmetry may face torsion buckling (sudden twisting) lateral buckling. Eccentricity, e of the load or defects such as initial bent will decreased the column strength. If the axial load on the column is not

    concentric, that is, its line of action is not parallel with the centrically axis of the

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  • 12

    column; the column is characterized as eccentrically loaded (3.8.2.4) as shown in figure 2.8a and figure 2.8b. The eccentricity, e of the load, or an initial curvature,

    subjects the column to immediate bending. The increased stresses due to the combined axial-plus-flexural stresses result in a reduced load-carrying ability.

    For an inclined column, these changes happened to the forces in the structural members which is bending moment and shear force. The changes are depending on the magnitude and the inclination direction.

    Figure 2.8a Inclined column subjected to eccentric axial load

    P

    M

    e

    H

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  • 13

    Figure 2.8b Simply supported column subjected to eccentric load

    2.9 DESIGN OF SHORT COLUMN

    A short column usually failed in compression because the effect of bending is smaller. Short columns usually need to only be designed for the maximum design moment about the one critical axis (3.8.4.3). The maximum axial load that column can support,

    uzN is calculated based on the ultimate capacity of the concrete and

    reinforced which is:

    scyccuuz AfAfN 95.045.0

    Where, uzN = ultimate axial load

    cA = nett cross section area of the column scA = vertical reinforced cross section area cuf = concrete strength yf = reinforced strength

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  • 14

    2.9.1 Braced Short Column

    With considering eccentricity, the ultimate load can be calculated using equation below after reduction of 10 percent:

    yscccu fAAfN 8.04.0

    2.9.2 Short Braced Columns Supporting an Approximately Symmetrical Arrangement of Beams (3.8.4.4)

    In this type of column, the bending moment is small because of the non-symmetrically arrangement of the live load at the both side of the column in a

    direction.

    yscccu fAAfN 70.035.0

    This equation only can be used when,

    a) The beam spans do not differ by more than 15 percent of the longer. b) The beams are designed to support uniformly distributed loads.

    2.9.3 Short Braced Columns Support Both Moment and Axial Load (3.8.4.5)

    The reinforced area for this type of column can be obtained by using analysis

    of section (3.4.4.1) and design charts (3.4.4.2). Symmetrically-reinforced rectangular sections may be designed to withstand an increased moment about one axis given by:

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  • 15

    '' bM

    hM yx , yxx Mb

    hMM'

    '

    '

    '' bM

    hM yx , xyy Mh

    bMM'

    '

    '

    2.10 DESIGN OF SLENDER COLUMN

    A slender column must be designed for an additional moment caused by its curvature at ultimate conditions. The expression given in BS8110 for the additional moments was derived by studying the moments curvature behaviors for a member subject to bending plus axial load. The equations for calculating the design moments are only applicable to columns of a rectangular or circular section and with symmetrical reinforcement.

    A slender column should be designed for an ultimate axial load (N) plus an increased moment given by,

    addMMM it

    ui NaM

    Where, iM is the initial moment in the column.

    addM is the moment caused by the deflection of the column.

    ua is the deflection of the column.

    The deflection of a rectangular or circular column is given by,

    Kha au

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  • 16

    The coefficient 2

    '20001

    ble

    a

    with 'b being generally the smaller dimension of the column section except when

    biaxial bending is considered. The coefficient K is a reduction factor to allow for the

    fact that the deflection must be less when there is a large proportion of the column section in compression. The value for K is given by the equation:

    0.1

    baluz

    uz

    NNNNK

    Where uzN is the ultimate axial load such that,

    scyccuuz AfAfN 87.045.0

    And balN is the axial load at balanced failure and may be taken as,

    ccubal AfN

    In order to calculate K, the area scA of the columns reinforcement must be

    known and hence a trial and error approach is necessary, taking an initial conservative value of K=1.0. Values of K are also marked on the column design charts.

    2.10.1 Braced Slender Column

    The maximum additional moment addM occurs near the mid-height of the

    column and at this location the initial moment is taken as,

    iM = 14.0 M + 22 4.06.0 MM

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  • 17

    Where 1M the smaller initial end moment due to the design ultimate loads and 2M is

    the corresponding larger initial end moment.

    For the usual case with double curvature of a braced column, 1M should be

    taken as negative and 2M as positive. From figure 3.2.1, the final design moment

    should never be taken as less than,

    2M addi MM 2/addi MM Or mineN

    2.10.2 Unbraced Slender Column

    The sway of an unbraced structure causes larger additional moments in the columns.

    2.11 DESIGN INCLINED COLUMN

    In analyzing leaner columns, the members are considered as simple pinned end columns, laterally supported at their ends. Lateral stability provided by rigid or braced frames must be properly sized to provide restraint for all loading within the

    structure. Figure 2.10a shows the relationship between load and bending moment in an inclined column.

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  • 18

    L\

    Figure 2.11 Relationship between load and bending moment in inclined column

    The value of w is a total load which is imposed to a beam and will be transferred to a column in vertical axis. The axial force P can be obtained by using equation:

    2wLP

    The relationship between load and moment at the mid span can be showed as:

    PeM

    While, the eccentricity can be derived from the following equation:

    cos2L

    e

    2L 2/2L

    2/2L

    F

    w

    Y

    X

    L

    e P

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  • 19

    2.12 Frame Analysis

    After the slab analysis for determine the design load that will be carried by beam and column, the beam and column then will be design as a fixed frame to

    determine the critical shear forces, axial forces and moments acting at a column. The frame analyses are divided into two types:

    a) braced frame which are considering the live load and dead load only (3.2.1.2). b) unbraced frame which are considering the live load, dead load and wind load

    (3.2.1.3).

    By the way, in this research the wind load is not considered because it can be stated that the inclined column is braced element due to its inclined design. Therefore, the braced frame is used to determine the forces and the moments. There are three types of braced frame:

    a) Simplification into sub-frames (3.2.1.2.1)

    All beams design together with column. The ends of the column are assumed to be fixed or pinned. The load arrangement is as showed:

    Figure 2.12.1a Type 1 braced frame

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  • 20

    b) Alternative simplification for individual beams and associated columns (3.2.1.2.3)

    The moments and forces in each individual beam can be obtained by considering a simplified sub-frame consisting only of that beam, the columns

    attached to the ends of the beam. The ends of the column and beam are considering fixed or pinned. The stiffness value of the beams considered should be taken as half their actual values if they are taken to be fixed at the ends. The moments in the column can also be obtain if the sub-frame has its

    central beam span longer than both side beam spans.

    Figure 2.12.1b Type 2 braced frame

    c) Continuous beam simplification (3.2.1.2.4)

    For conservative choice, the moments and forces in beams also can be obtain by considering the beam as a continuous beam over supports providing no

    restraint to rotation.

    Beam A-B

    K K/2 A B

    Beam B-C

    K/2 K K/2 B C

    Beam C-D K/2 K C D

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  • 21

    Figure 2.12.1c Type 3 braced frame

    2.13 Main Reinforcement

    The minimum area of longitudinal reinforcement in a column for various condition is 0.4% bh (Table 3.25) and the minimum bar should be four in rectangular columns and six in circular column. The size of bar should not be less than 12mm (3.12.5.3).

    The maximum area of longitudinal reinforcement should not exceed 6% bh (3.12.6.2)

    K/2

    A

    K/2 K/2

    B&C

    K/2

    D

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  • CHAPTER 3

    METHODOLOGY

    3.1 INTRODUCTION

    This chapter will discuss the method used to analyze and design inclined column. The basic step in designing vertical column is then used to analyze and designing inclined column. The manual calculation will be used by referring to the project plans of the case study. The outcome from the analysis and the calculation of the case study will be therefore used to compare with vertical column. All of the specification and design criteria are accordance to British Standard 8110. Figure 3.1

    shows the flow chart of a simplified overview of the project flow from start to its completion.

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  • 23

    Figure 3.1: Simplified research methodology for this study

    SCOPE START

    OBJECTIVES:

    i) To analyze the equilibrium forces in inclined column.

    ii) To design an inclined column based on BS8110.

    iii) To find out comparison between inclined column and vertical column.

    iv) As references for future uses.

    CASE STUDY

    ANALYSIS & COLLECTING

    DATA

    DESIGN

    RESULTS

    CONCLUSIONS PREPARING FOR

    COMPLETE THESIS REPORT

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  • 24

    3.2 METHOD USE TO ANALYZE STRUCTURE

    Load analysis is important to determine the equilibrium forces and moment act to the column. The analysis must begin with an evaluation of all the loads carried

    by the structure, including its own weight. Many of the loads are variable in magnitude and position, and all possible critical arrangements of loads must be considered. First the structure itself is rationalized into simplified forms that represent the load-carrying action of the prototype. Then, the forces in each member can be determined by one of the following methods:

    i) Applying moment and shear coefficients ii) Manual calculations

    iii) Computer methods

    In this research, the manual calculation is used to determine the moments and forces react in the column. A braced frame analysis and column analysis will be applied and

    the calculation is based on the plan of a case study.

    3.3 CASE STUDY

    A plan of Development Proposal of Taman Ekologi Hutan Bandar, Johor Bahru is adopted as a case study. This project involved three storeys building which consist of inclined column, ring beam and ring slab. Besides that, this building also construct with ranking piles which resist horizontal forces. Figure 3.3 shows the front

    view of the building. While, the architectures drawing is shows in appendix.

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  • 25

    Figure 3.3 Front view of the building

    3.4 DESIGN PROCESS

    The design process for reinforced concrete column structures consists of the following steps:

    1. Determine design data: design loads, design criteria and specifications. Specify material properties.

    2. Make a first estimate of member sizes.

    3. Calculate member cross sectional properties; perform frame analysis to obtain internal forces: moment, shear forces and axial force.

    Inclined Column

    Ring beam and Ring Slab

    Vertical Column

    Ranking Piles

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  • 26

    4. Calculate the required reinforcement based on moment, shear forces and axial force demands.

    5. If members do not satisfy the specification, modify the design and make changes at step 2.

    6. Detail reinforcement. Develop design drawings.

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  • CHAPTER 4

    STRUCTURAL ANALYSIS

    4.1 INTRODUCTION

    The objectives of structural analysis are to determine shear forces, axial force and moments at the whole structural element. The analysis begins with the calculation of many types of load that act at the structure. Usually, at the beginning stage, all sizes used are been estimate first, and then the suitability will be checked.

    The analysis can be done with several methods such as:

    a) Calculation using moment distributed method. b) Calculation using moment and shear coefficient in BS8110.

    In this study, the structural analysis is divided into two parts which are the

    frame analysis and column analysis. The frame analysis is used to determine the critical shear forces, axial forces and moments in a column while column analysis is used to design the column.

    4.2 Frame Analysis

    In this study, the frame analysis involved are based on 3.2.1.2.1 BS 8110 which is concern each sub-frame may be taken to consist of the beams at one level together

    with the column above and below. The ends of the column are assumed to be fixed.

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  • 28

    The moments and the shear forces will be obtain in this analysis. The procedures in frame analysis invoved:

    a) Determine the stiffness value of the column and beam. b) Calculate the maximum load and minimum load. c) Do the frame analysis.

    59.5 9.31

    w = 11.89 kN/m

    29.8 59.5 9.31

    Figure 4.2a Roof Second Floor First Floor (only one case involved - maximum)

    59.5 9.31 cL

    w 1 w 2

    24.5 29.8

    208.8 32.6

    Figure 4.2b First floor Ground floor Foundation

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  • 29

    Case w 1 (kN/m) w 2 (kN/m) 1 max, max, max 11.26 4.02

    2 max, min, max 11.26 2.00 3 min, max, min 5.20 4.02

    Table 4.2 Maximum and minimum load

    The calculations are shown in appendix.

    4.3 Column Analysis

    They are two column analysis involved which are:

    4.3.1 Inclined column

    The procedures of designing inclined column are:

    a) Determine the types of column by calculating the effective length and decides the end condition.

    b) Compare the three moments obtained and take the biggest value of moment.

    c) Determine the reinforcement.

    Sample Calculations of inclined column

    Effective Length:

    30805003580: oeyex lll

    End Condition:

    Top : Beam (500) > Column (400) .. Condition 1

    Bottom : Beam (500) > Column (400) . Condition 1

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  • 30

    From Table 3.19, 75.0

    Therefore;

    oeyex lll = 308075.0 = 2310mm

    Checking

    8.5400

    2310

    8.5400

    2310

    blh

    l

    ey

    ex

    Both values are less than 15, so the column is short column.

    Column 1

    kNPkNN

    11.30767.323

    1

    1

    Moment analysis

    kNmM analysis 09.13

    min05.0 NhM column = 0.05 ( 323.67) (0.4) = 6.47 kNm < analysisM

    ePM inclined . = 307.11 ( 0.372)

    = 114.24kNm > analysisM

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  • 31

    Reinforcement

    22

    6

    2

    23

    /785.1400400

    1024.114

    /02.2400400

    1067.323

    mmNbhM

    mmNbhN

    mmd 36522025400 , 91.0

    400365

    hd

    Using Chart No.29,

    2656100

    40040041.0

    41.0100

    mmA

    bhA

    sc

    sc

    Use 4T16 ( 2804mmAs )

    Link

    Diameter minimum mm41641

    . R6 mm

    Clear Distance Maximum mm1921612 .150mm

    Therefore, use R6 150.

    Other calculations are in Appendix.

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  • 32

    4.3.2 Vertical column

    a) Determine the types of column by calculating the effective length and decides the end condition.

    b) Compare the two moments obtained and take the biggest moment.

    c) Determine the reinforcement.

    Sample Calculations of vertical column

    Effective Length:

    30005003500: oeyex lll

    End Condition:

    Top : Beam (500) > Column (250) .. Condition 1

    Bottom : Beam (500) > Column (250) . Condition 1

    From Table 3.19, 75.0 Therefore;

    oeyex lll = 300075.0 = 2250mm

    Checking

    92502250

    92502250

    blh

    l

    ey

    ex

    Both values are less than 15, so the column is short column.

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  • 33

    Column 1

    From frame analysis, moment at roof is,

    kNmM analysis 29.5

    1N kN67.207

    Moment Analysis

    min05.0 NhM column

    kNmkNm

    okh74.660.2

    !......205.1225005.005.0........25.067.20705.0

    Therefore, use kNmM analysis 29.5

    Reinforcement

    mmd 2192

    1225250

    So, 85.088.0250219

    hd

    Use Chart No. 28 with 22 /460,/30 mmNfmmNf ycu and 85.0hd

    32.3250250

    1067.207

    34.02502501029.5

    3

    2

    6

    2

    bhNbhM

    kNmM analysis 29.5

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  • 34

    From the chart, use the minimum reinforcement for column which is:

    4T12 with 2453mmAs

    Other calculations are in Appendix.

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  • CHAPTER 5

    RESULTS AND DISCUSSIONS

    5.1 INTRODUCTION

    From the analysis and design, the moments, axial load and reaction force which is the vertical and horizontal force increase due to increasing of load. The results are summarized between the inclined column and vertical column based on the size of column use, the eccentricity, loads, moments and reinforcement obtained.

    The results then compared with each other where this two column have same structural criteria but different in the loadings and the moments.

    5.2 Results from analysis and design

    For inclined column, the value of moments from frame analysis is quite small, but because of the existing of eccentricity the value of moment taken became bigger. Column below needs more reinforcement because it supports large number of

    loading which is from the roof until the ground floor. For these inclined columns, the

    minimum reinforcement area is 640mm 2 while the maximum reinforcement area is

    9600mm 2 .

    On the other hand, for vertical column, the value of moment taken is small

    and the minimum of reinforcement has to be used in almost of the vertical column except for Column 4. The moment in Column 4 is quite big because it supports all

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  • 36

    the loading from the roof to the ground floor. The minimum reinforcement of these

    vertical columns is 453 mm 2 and the maximum reinforcement is 3750 mm 2 .

    INCLINED COLUMN ( e = 0.372m ) Column size 400 400 mm

    VERTICAL COLUMN ( e = 0.05h ) Column size 250 250 mm

    Column 1

    N = 323.67 kN

    M = 114.24 kNm

    4T16 (A s = 804 mm 2 )

    R6 150

    Column 1

    N = 207.67 kN

    M = 5.29 kNm

    4T12 (A s = 453 mm 2 )

    R6 125

    Column 2

    N = 761.07 kN

    M = 253.66 kNm

    4T25 + 4T16 ( A s = 2767mm 2 )

    R8 150

    Column 2

    N = 440.9 kN

    M = 21.14 kNm

    4T12 (A s = 453 mm 2 )

    R6 125

    Column 3

    N = 1212.99 kN

    M = 415.60 kNm

    4T32 + 4T25 (A s = 5180 mm 2 )

    R8 300

    Column 3

    N = 666.15 kN

    M = 8.91 kNm

    4T12 (A s = 453 mm 2 )

    R6 125

    Column 4

    N = 1765.54 kN

    M = 600.50 kNm

    6T40 + 4T25 (As= 9503 mm 2 )

    R10 300

    Column 4

    N = 1017.34 kN

    M = 32.43 kNm

    4T20 + 4T10 (A s = 1571 mm 2 )

    R8 100

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  • 37

    5.3 Comparison between Inclined column and Vertical column

    Inclined Column

    Vertical Column

    Frame Analysis

    a) Stiffness value considered the inclined length.

    a) Stiffness value considered the vertical length.

    Column Design

    a) The effective length take the inclined length, el .

    b) Considered the eccentricity.

    a) The effective length take the vertical length, el .

    b) Considered only a small eccentricity, 0.05h min or

    zero eccentricity.

    Result

    a) Has a large value of moment due to the eccentricity and

    load. b) Big size of column. c) Need more reinforcement. d) Have horizontal force to

    support the column.

    a) Has small value of moment.

    b) Small size of column. c) Averagely used minimum

    reinforcement.

    d) No horizontal force exists.

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  • CHAPTER 6

    CONCLUSIONS AND RECOMMENDATIONS

    6.1 CONCLUSIONS

    From the analysis and design, it can be concluded that:

    i) The equilibrium forces (moments, axial load and reaction force) act in a column is important to get correct detailing of reinforcement.

    ii) The equations, methods and formula used in designing inclined column are similar with designing vertical column only in inclined column the existing of eccentricity have to be considered..

    iii) The moment increase due to the increasing of load.

    iv) The existing of the eccentricity will increase the moment.

    v) The increasing in inclination of the column will increase the eccentricity.

    vi) The horizontal force increase when the inclination of the column increases.

    vii) Manual calculation is quite complicated compare to software application.

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  • 39

    6.2 RECOMMENDATIONS

    Stated here are some recommendations that can be organised for future studies:

    a) Use different materials in designing the structure such as steel and wood.

    b) Do analysis and designing inclined column based on other design standard.

    c) Do the analysis by using finite element. ( if can )

    d) Do the analysis and designing inclined column for slender column and unbraced frame.

    e) Use different shape of column. For example; U column or round column.

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  • REFERENCES

    1. Structural Design (A Practical Guide For Architecture) by Rod Underwood, Michele Chiuini.

    2. Reinforced Concrete Designer Handbook by Charles E Reynolds.

    3. Reinforced Concrete Design, 3rd Edition by W.H. Mosley and J.H. Bungey.

    4. Structural Engineering and Construction Management by Mohd Salleh Jaafar, Salihuddin Hasim, Mohd Razali Abd Kadir, Izian Abd Karim.

    5. Introduction to Stress Analysis: Engineering Officers Course.

    6. Effects of Axial Load on Shear Behavior of Short RC Columns under Cyclic Lateral Deformations.

    7. Structural Engineering Reference Manual.

    8. British Standard, BS8110.

    9. Principles of Structural Design, edited by W.F. Chen and E.M. Lui.

    10. Nota Kuliah SAB 4333 Rekabentuk Konkrit Bertetulang 2, berpandukan BS 8110 1997, disediakan oleh Prof. Madya Dr. Ramli Abdullah.

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  • 41

    11. Mohd. Azrul Helmi bin Yusof (2004), Carta Rekabentuk untuk Keratan Tiang Konkrit Tidak Seragam. Projek Sarjana Muda. Universiti Teknologi Malaysia.

    12. Nazarina bt. Munjiyat (2007), Analisis Tiang Condong. Projek Sarjana Muda. Universiti Teknologi Malaysia.

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  • APPENDICES

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  • APPENDIX B FRAME ANALYSIS

    Roof - Second Floor - First Floor

    26.18 -10.57 0.18 -0.03 0.6

    -0.29 5.98 -0.93

    19.42

    -9.32 0.4 0.2 0.616 0.192

    0.4 0.192 -48.55

    48.55 19.42 9.71 -29.9 -9.32

    -14.95 4.86 5.98 2.99 -2.99 -0.93

    -1.5 1.5 0.6 0.3 0.92 -0.29

    0.46 0.15 0.18 0.09 0.09 -0.03

    26.18 -52.37 21.16 -10.57

    Table 1

    First Floor - Ground Floor - Foundation

    Case 1 - maximum, maximum,maximum

    15.41 -6.28 0.09 -0.04 1.56 -0.33

    13.76 -5.91 0.203 0.084 0.301 0.115 0.183 0.713 0.401

    -67.79 65.79 -16.42 48.33 5.69 -15.46 -20.6 -9.4

    -7.73 2.85 0 5.51 0.65 -0.86 -1.14 -0.63

    -0.43

    0.33 0 0.31 0.04 -0.1 -0.13 -0.07

    54.15 -74.56 54.55 -

    21.87 -26.52

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  • Case 2 - maximum, minimum,maximum

    15.66 -7.23 0.08

    -0.04 1.82 -0.33

    13.76

    -6.86 0.203 0.084 0.301 0.115 0.183

    0.713 0.401 -67.79 67.79 -8.17

    48.33 6.91 -17.95 -

    23.91 -10.91 -8.98 2.85 0

    6.4 0.75 -0.86 -1.14 -0.52 -0.43 0.38 0 0.31 0.04 -0.11 -0.15 -0.07

    55.04 70.72 52.1 -25.2 -19.67

    Case 3 - minimum, maximum, minimum

    6.85

    -1.87 0.04 -0.01

    0.45

    -0.15 6.36 -1.71

    0.203 0.102 0.301 0.115 0.183 0.713 0.401 -31.31 31.31 -16.42 22.32 2.63 -4.48 -5.97 -2.72 -2.24 1.32 0

    1.6 0.19 -0.4 -0.53 -0.24 0.2 0.1 0 0.14 0.02

    -0.03 -0.04 -0.02 24.06 -30.91 27.82 -6.54 -19.4

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  • APPENDIX C

    COLUMN ANALYSIS

    Inclined column

    Effective Length:

    30805003580: oeyex lll

    End Condition:

    Top : Beam (500) > Column (400) .. Condition 1

    Bottom : Beam (500) > Column (400) . Condition 1

    From Table 3.19, 75.0

    Therefore;

    oeyex lll = 308075.0 = 2310mm

    Checking

    8.5400

    2310

    8.5400

    2310

    blh

    l

    ey

    ex

    Both values are less than 15, so the column is short column.

    Column 1

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  • kNPkNN

    11.30767.323

    1

    1

    kNmM analysis 09.13

    min05.0 NhM column = 0.05 ( 323.67) (0.4) = 6.47 kNm < analysisM

    ePM inclined . = 307.11 ( 0.372) = 114.24kNm > analysisM

    22

    6

    2

    23

    /785.1400400

    1024.114

    /02.2400400

    1067.323

    mmNbhM

    mmNbhN

    mmd 36522025400 , 91.0

    400365

    hd

    Using Chart No.29,

    2656100

    40040041.0

    41.0100

    mmA

    bhA

    sc

    sc

    Use 4T16 ( 2804mmAs )

    Link

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  • Diameter minimum mm41641

    . R6 mm

    Clear Distance Maximum mm1921612 .150mm

    Therefore, use R6 150.

    Column 2

    kNPkNN

    88.68107.761

    2

    2

    kNmM analysis 36.52

    min05.0 NhM column = 0.05( 761.07) (0.4) = 15.22 kNm < analysisM

    ePM inclined . = 681.88 ( 0.372) = 253.66kNm > analysisM

    22

    6

    2

    23

    /96.3400400

    1066.253

    /76.4400400

    1007.761

    mmNbhM

    mmNbhN

    mmd 36522025400 , 91.0

    400365

    hd

    Using Chart No.29,

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  • 22400100

    4004005.1

    5.1100

    mmA

    bhA

    sc

    sc

    Use 4T25 + 4T16 ( 22767mmAs )

    Link

    Diameter minimum mm3.62541

    . R8 mm

    Clear Distance Maximum mm1921612 .150mm

    Therefore, use R8 150.

    Column 3

    kNPkNN

    24.11799.1212

    3

    3

    kNmM analysis 92.20

    min05.0 NhM column = 0.05 (1212.99) (0.4) = 24.26 kNm > analysisM

    ePM inclined . = 1117.24 ( 0.372) = 415.6kNm > analysisM

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  • 22

    6

    2

    23

    /49.6400400

    106.415

    /58.7400400

    1099.1212

    mmNbhM

    mmNbhN

    mmd 36522025400 , 91.0

    400365

    hd

    Using Chart No.29,

    24960100

    4004001.3

    1.3100

    mmA

    bhA

    sc

    sc

    Use 4T32 + 4T25 ( 25180mmAs )

    Link

    Diameter minimum mm83241

    . R8 mm

    Clear Distance Maximum mm3002512 .300mm

    Therefore, use R8 300

    Column 4

    Effective Length:

    mmlll oeyex 5205001020:

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  • End Condition:

    Top : Beam (500) > Column (400) .. Condition 1

    Bottom : Beam (500) > Column (400) . Condition 1

    From Table 3.19, 75.0

    Therefore;

    oeyex lll = 52075.0 = 390mm

    Checking

    98.0400390

    98.0400390

    blh

    l

    ey

    ex

    Both values are less than 15, so the column is short column.

    kNPkNN

    26.161454.1765

    1

    1

    kNmM analysis 39.69

    min05.0 NhM column = 0.05 (1765.54) (0.4) = 35.31 kNm < analysisM

    ePM inclined . = 1614.26 (0.372) = 600.5kNm > analysisM

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  • 22

    6

    2

    23

    /38.9400400

    105.600

    /03.11400400

    1054.1765

    mmNbhM

    mmNbhN

    mmd 36522025400 , 91.0

    400365

    hd

    Using Chart No.29,

    28800100

    4004005.5

    5.5100

    mmA

    bhA

    sc

    sc

    Use 6T40 + 4T25 ( 29503mmAs )

    Link

    Diameter minimum mm104041

    . R10 mm

    Clear Distance Maximum mm3002512 .300mm

    Therefore, use R10 300.

    Minimum Reinforcement

    0.4 % bh = 2640400400100

    4.0mm

    Maximum Reinforcement

    6% bh = 29600400400100

    6mm

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  • APPENDIX D

    COLUMN ANALYSIS

    Vertical Column

    Effective Length:

    30005003500: oeyex lll

    End Condition:

    Top : Beam (500) > Column (250) .. Condition 1

    Bottom : Beam (500) > Column (250) . Condition 1

    From Table 3.19, 75.0

    Therefore;

    oeyex lll = 300075.0 = 2250mm

    Checking

    92502250

    92502250

    blh

    l

    ey

    ex

    Both values are less than 15, so the column is short column.

    Column 1

    From frame analysis, moment at roof is,

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  • kNmM analysis 29.5

    1N kN67.207

    Moment Analysis

    min05.0 NhM column

    kNmkNm

    okh74.660.2

    !......205.1225005.005.0..................25.067.20705.0

    Therefore, use kNmM analysis 29.5

    Reinforcement

    mmd 2192

    1225250

    So, 85.088.0250219

    hd

    Use Chart No. 28 with 22 /460,/30 mmNfmmNf ycu and 85.0

    hd

    32.3250250

    1067.207

    34.02502501029.5

    3

    2

    6

    2

    bhNbhM

    From the chart, use the minimum reinforcement for column which is:

    4T12 with 2453mmAs

    kNmM analysis 29.5

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  • Column 2

    From frame analysis, moment at second floor is,

    kNmM analysis 14.21

    2N kN9.440

    Moment Analysis

    kNmM analysis 14.21

    min05.0 NhM column

    kNmkNm

    okh14.2151.5

    !......205.1225005.005.0..................25.09.44005.0

    Therefore, use kNmM analysis 14.21

    Reinforcement

    mmd 2192

    1225250

    So, 85.088.0250219

    hd

    Use Chart No. 28 with 22 /460,/30 mmNfmmNf ycu and 85.0hd

    05.7250250109.440

    35.1250250

    1014.21

    3

    2

    6

    2

    bhNbhM

    From the chart, use the minimum reinforcement for column which is:

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  • 4T12 with 2453mmAs

    Column 3

    From frame analysis, moment at second floor is,

    kNmM analysis 91.8

    3N kN15.666

    Moment Analysis

    kNmM analysis 91.8

    min05.0 NhM column

    kNmkNm

    okh95.833.8

    !......205.1225005.005.0..................25.015.66605.0

    Therefore, use kNmM analysis 95.8

    Reinforcement

    mmd 2192

    1225250

    So, 85.088.0250219

    hd

    Use Chart No. 28 with 22 /460,/30 mmNfmmNf ycu and 85.0

    hd

    66.10250250

    1015.666

    57.02502501091.8

    3

    2

    6

    2

    bhNbhM

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  • From the chart, use the minimum reinforcement for column which is:

    4T12 with 2453mmAs

    Link

    Diameter minimum = 31241

    mm.. use R8

    Clear Maximum Distance = mm1441212

    So, use R8 125

    Column 4

    From frame analysis, moment at second floor is,

    kNmM analysis 43.32

    4N kN34.1017

    Effective Length:

    5005001000: oeyex lll

    End Condition:

    Top : Beam (500) > Column (250) .. Condition 1

    Bottom : Beam (500) > Column (250) . Condition 1

    From Table 3.19, 75.0

    Therefore;

    oeyex lll = 50075.0 = 375mm

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  • Checking

    5.1250375

    5.1250375

    blh

    l

    ey

    ex

    Both values are less than 15, so the column is short column.

    Moment Analysis

    kNmM analysis 43.32

    min05.0 NhM column

    kNmkNm

    okh79.3272.12

    !......205.1225005.005.0................25.034.101705.0

    Therefore, use kNmM analysis 79.32

    Reinforcement

    mmd 21522025250

    So, 85.086.0250215

    hd

    Use Chart No. 28 with 22 /460,/30 mmNfmmNf ycu and 85.0

    hd

    3.16250250

    1034.1017

    08.2250250

    1043.32

    3

    2

    6

    2

    bhNbhM

    From the chart,

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  • 21375100

    2502502.2

    2.2100

    mmA

    bhA

    sc

    sc

    Therefore, use 4T20 + 4T10 with 21571mmAs

    Link

    Diameter minimum = 52041

    . Use R8

    Clear Distance Maximum = mm1201012

    So, use R8 100

    Minimum Reinforcement

    0.4 % bh = 2250250250100

    4.0mm

    But in BS8110 stated that, minimum area of reinforcement in a column at least 0.4%

    bh (Table 3.25) and the minimum bar should be 4 in rectangular columns and the size of bar should not be less than 12mm. So, in this research the minimum

    reinfoecement used is 4T12 ( sA = 453mm 2 ).

    Maximum Reinforcement

    6% bh = 23750250250100

    6mm

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  • pdfMachine A pdf writer that produces quality PDF files with ease!

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  • FOOTING PLAN GROUND FLOOR PLANFIRST FLOOR PLAN

    SECOND FLOOR PLANROOF PLAN

    DATA-DATA TERLIBAT

    1. fcu = 30N/mm22. fy = 460 @ 250 N/mm2, fy(slab)= 250 N/mm23. Ruang Pameran (Exhibition)4. Dinding Kaca = 26.7kN/m35. Bearing Capacity = 150kN/m26. qk(w/tah) = 3.kKN/m2, gk(finishes) = 1.0kN/m27. Tebal papak = 150 mm8. Saiz rasuk(mm) = 200x5009. Saiz tiang(mm) = 250x25010. Saiz asas(foundation) = 2500x250011. Penutup=25 mm

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