Analysis and Design of Inclined Columns

PSZ 19:16 (Pind. 1/07)

UNIVERSITI TEKNOLOGI MALAYSIA DECLARATION OF THESIS / UNDERGRADUATE PROJECT PAPER AND COPYRIGHT

SHAFIAH BINTI DOLHAKIM Author’s full name :

21 NOVEMBER 1985 Date of birth

:

Title

:

ANALYSIS AND DESIGN OF INCLINED COLUMN

2007/2008 Academic Session: I declare that this thesis is classified as :

√

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)

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

851121-13-5362

IR AZHAR AHMAD

(NEW IC NO. /PASSPORT NO.)

NAME OF SUPERVISOR

Date : 25 APRIL 2008

NOTES :

SIGNATURE OF SUPERVISOR

*

Date : 25 APRIL 2008

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

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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

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_____________________________

Name of Supervisor

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IR AZHAR AHMAD

Date

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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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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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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

1

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

INTRODUCTION

1

1.1

Background

1

1.2

Statement of Problem

1

1.3

Objectives

2

1.4

Scope of the Study

2

1.5

Significance of the Study

3

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ix

CHAPTER

2

TITLE

PAGE

LITERATURE REVIEW

4

2.1

Introduction to the Column

4

2.2

Inclined/ Leaning Column

5

2.2.1 Definition

5

2.2.2 Leaner Column / Inclined Column

5

2.3

Analysis

6

2.4

Types of Column

7

2.4.1 Braced and Unbraced Column

7

2.4.2 Short and Slender Column

9

2.5

Effective Height

10

2.6

End Condition

10

2.7

Mode of Failure

11

2.8

Load and Deflection

11

2.9

Design of Short column

13

2.9.1 Braced Short Column

14

2.9.2 Short Braced Columns Supporting

14

an Approximately Symmetrical Arrangement of Beams (3.8.4.4) 2.9.3 Short Braced Columns Support Both

14

Moment and Axial Load ( 3.8.4.5) 2.10

Design of Slender Column

15

2.10.1 Braced Slender Column

16

2.10.2 Unbraced Slender Column

17

2.11

Design of Inclined Column

17

2.12

Frame Analysis

19

2.13

Main Reinforcement

21

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x

CHAPTER

3

4

TITLE

RESEARCH METHODOLOGY

22

3.1

Introduction

22

3.2

Method use to analyze structure

24

3.3

Case study

24

3.4

Design process

25

STRUCTURAL ANALYSIS

27

4.1

Introduction

27

4.2

Frame analysis

43

4.3

Column analysis

29

4.3.1

Inclined column

29

4.3.2

Vertical column

32

CHAPTER

5

PAGE

TITLE

PAGE

RESULTS AND DISCUSSIONS

35

5.1

Introduction

35

5.2

Results from analysis and design

35

5.3

Comparison between inclined column and 37 vertical column.

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xi

CHAPTER

6

TITLE

CONCLUSIONS AND RECOMMENDATIONS

PAGE

38

6.1

Conclusions

38

6.2

Recommendations

39

REFERENCES

40

Appendices A-F

43

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xii

LIST OF FIGURES

FIGURE

TITLE

PAGE

2.2.2a

Inclined column.

6

2.4.1a

The building are braced in both direction.

8

2.4.1b

The building are unbraced in both direction.

8

2.4.1c

The building is braced in y direction but

9

unbraced in x direction. 2.8a

Inclined column subjected to eccentric axial

12

load. 2.8b

Simply supported column subjected to axial

13

load. 2.11

Relationship between load and bending moment

18

in inclined column. 2.12.1a

Type 1 braced frame.

19

2.12.1b

Type 2 braced frame.

20

2.12.1c

Type 3 braced frame.

21

3.1

Simplified research methodology in this study.

23

3.3

Front view of the building (case study).

25

4.2a

Roof – Second Floor – First Floor.

28

4.2b

First Floor –Ground Floor -Foundation

28

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xiii

LIST OF APPENDICES

APPENDIX.

TITLE

PAGE

A

Architecture drawing of case study

43

B

Frame Analysis - Braced Frame

44

C

Column Analysis – Inclined column

46

D

Column Analysis – Vertical column

53

E

Chart no 28 and chart no 29

61

F

Detailing of the column reinforcement

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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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

Inclined column

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;

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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, ø (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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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.

Load Bearing Wall

Column

Figure 2.4.1a The building are braced in both direction

Column

Figure 2.4.1b The building are unbraced in both direction pdfMachine A pdf writer that produces quality PDF files with ease! Produce quality PDF files in seconds and preserve the integrity of your original documents. Compatible across nearly all Windows platforms, simply open the document you want to convert, click “print”, select the “Broadgun pdfMachine printer” and that’s it! Get yours now!

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Column

Load Bearing Wall

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

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2.5 EFFECTIVE HEIGHT

The effective height, l e of a column in a given plane may be obtained from equation below in BS 8110 (3.8.1.6):

le   lo

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 l o 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.

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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 pdfMachine A pdf writer that produces quality PDF files with ease! Produce quality PDF files in seconds and preserve the integrity of your original documents. Compatible across nearly all Windows platforms, simply open the document you want to convert, click “print”, select the “Broadgun pdfMachine printer” and that’s it! Get yours now!

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.

P

M e

H

Figure 2.8a Inclined column subjected to eccentric axial load

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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, N uz is calculated based on the ultimate capacity of the concrete and reinforced which is:

N uz  0.45 f cu Ac  0.95 f y Asc

Where, N uz Ac Asc f cu fy

= ultimate axial load = nett cross section area of the column = vertical reinforced cross section area = concrete strength = reinforced strength

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2.9.1 Braced Short Column

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

N  0.4 f cu Ac  0.8 Asc f y

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 nonsymmetrically arrangement of the live load at the both side of the column in a direction.

N  0.35 f cu Ac  0.70 Asc f y

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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Mx My  ' h' b

,

Mx  Mx 

Mx My  ' h' b

,

My  My 

'

'

h' My b'

b' Mx h'

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,

M t  M i  M add

 M i  Na u

Where, M i is the initial moment in the column.

M add is the moment caused by the deflection of the column. au is the deflection of the column.

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

a u   a Kh

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1  le  The coefficient  a    2000  b ' 

2

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:

K

N uz  N  1. 0 N uz  N bal

Where N uz is the ultimate axial load such that,

Nuz  0.45 f cu Ac  0.87 f y Asc

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

N bal  f cu Ac

In order to calculate K, the area Asc 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 M add occurs near the mid-height of the column and at this location the initial moment is taken as,

M i = 0.4 M 1 + 0.6 M 2  0.4 M 2 pdfMachine A pdf writer that produces quality PDF files with ease! Produce quality PDF files in seconds and preserve the integrity of your original documents. Compatible across nearly all Windows platforms, simply open the document you want to convert, click “print”, select the “Broadgun pdfMachine printer” and that’s it! Get yours now!

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

For the usual case with double curvature of a braced column, M 1 should be taken as negative and M 2 as positive. From figure 3.2.1, the final design moment should never be taken as less than, M2 M i  M add

M i  M add / 2 Or

N  emin

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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L

w

L\ L2

L2 / 2

F e

L2 / 2

P Y

X

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:

P

wL 2

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

M  Pe

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

e

L cos  2

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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 pdfMachine A pdf writer that produces quality PDF files with ease! Produce quality PDF files in seconds and preserve the integrity of your original documents. Compatible across nearly all Windows platforms, simply open the document you want to convert, click “print”, select the “Broadgun pdfMachine printer” and that’s it! Get yours now!

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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.

Beam A-B K A

K/2 B Beam B-C

K/2

K B

K/2 C

Beam C-D K/2

K C

D

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.

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K/2 A K/2

K/2 B&C

K/2 D

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)

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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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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

Figure 3.1: Simplified research methodology for this study

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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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Vertical Column Inclined Column

Ring beam and Ring Slab

Ranking Piles

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.

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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. pdfMachine A pdf writer that produces quality PDF files with ease! Produce quality PDF files in seconds and preserve the integrity of your original documents. Compatible across nearly all Windows platforms, simply open the document you want to convert, click “print”, select the “Broadgun pdfMachine printer” and that’s it! Get yours now!

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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

w1 24.5 208.8

cL

w2 29.8 32.6

Figure 4.2b First floor – Ground floor – Foundation

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Case 1 – max, max, max 2 – max, min, max 3 – min, max, min

w 1 (kN/m) 11.26 11.26 5.20

w 2 (kN/m) 4.02 2.00 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:

l ex  l ey : l o  3580  500  3080

End Condition: Top

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

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From Table 3.19,   0.75 Therefore; l ex  ley   l o

= 0.75  3080 = 2310mm

Checking

l ex 2310   5.8 h 400 l ey 2310   5.8 b 400 Both values are less than 15, so the column is short column.

Column 1

N 1  323.67 kN P1  307.11kN

Moment analysis

M analysis  13.09kNm

M column  0.05 Nhmin = 0.05 ( 323.67) (0.4) = 6.47 kNm < M analysis M inclined  P.e = 307.11 ( 0.372) = 114.24kNm > M analysis

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Reinforcement

N 323.67  10 3   2.02 N / mm 2 bh 400  400 M 114.24  10 6   1.785 N / mm 2 bh 2 400  400 2

d  400  25 

20  365mm , 2

d 365   0.91 h 400

Using Chart No.29, 100 Asc  0.41 bh

Asc 

0.41  400  400  656mm 2 100

Use 4T16 ( As  804mm 2 )

Link 1  16  4mm ……. R6 mm 4 Clear Distance Maximum  12  16  192mm …….150mm

Diameter minimum 

Therefore, use R6 – 150. Other calculations are in Appendix.

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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:

l ex  l ey : l o  3500  500  3000

End Condition:

Top

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

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

From Table 3.19,   0.75 Therefore; l ex  ley   l o

= 0.75  3000 = 2250mm

Checking l ex 2250  9 h 250 l ey 2250  9 b 250 Both values are less than 15, so the column is short column.

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Column 1

From frame analysis, moment at roof is,

M analysis  5.29kNm

N 1  207.67kN

Moment Analysis

M analysis  5.29 kNm

M column  0.05 Nhmin  0 . 05 207 . 67 0 . 25 ........ 0 . 05 h  0 . 05  250  12 . 5  20 ...... ok !  2 . 60 kNm  6 . 74 kNm

Therefore, use

M analysis  5.29kNm

Reinforcement

d  250  25 

So,

12  219mm 2

d 219   0.88  0.85 h 250

Use Chart No. 28 with

f cu  30 N / mm 2 , f y  460 N / mm 2 and

d  0.85 h

M 5.29  10 6   0.34 bh 2 250  250 2 N 207.67  10 3   3.32 bh 250  250

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

4T12 with As  453mm 2

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 pdfMachine A pdf writer that produces quality PDF files with ease! Produce quality PDF files in seconds and preserve the integrity of your original documents. Compatible across nearly all Windows platforms, simply open the document you want to convert, click “print”, select the “Broadgun pdfMachine printer” and that’s it! Get yours now!

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 Column 1

VERTICAL COLUMN ( e = 0.05h ) Column size 250  250 mm Column 1

N = 323.67 kN

N = 207.67 kN

M = 114.24 kNm

M = 5.29 kNm

4T16 (A s = 804 mm 2 )

4T12 (A s = 453 mm 2 )

R6 – 150

R6 – 125

Column 2

Column 2

N = 761.07 kN

N = 440.9 kN

M = 253.66 kNm

M = 21.14 kNm

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

4T12 (A s = 453 mm 2 )

R8 – 150

R6 – 125

Column 3

Column 3

N = 1212.99 kN

N = 666.15 kN

M = 415.60 kNm

M = 8.91 kNm

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

4T12 (A s = 453 mm 2 )

R8 – 300

R6 – 125

Column 4

Column 4

N = 1765.54 kN

N = 1017.34 kN

M = 600.50 kNm

M = 32.43 kNm

6T40 + 4T25 (A s = 9503 mm 2 )

4T20 + 4T10 (A s = 1571 2

mm ) R10 – 300 R8 – 100 pdfMachine A pdf writer that produces quality PDF files with ease! Produce quality PDF files in seconds and preserve the integrity of your original documents. Compatible across nearly all Windows platforms, simply open the document you want to convert, click “print”, select the “Broadgun pdfMachine printer” and that’s it! Get yours now!

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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, l e . b) Considered the eccentricity.

a) The effective length take the vertical length, l e . 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

a) Has small value of moment. b) Small size of column. c) Averagely used minimum reinforcement. d) No horizontal force exists.

support the column.

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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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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 Officer’s 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 0.18 0.6 5.98 19.42 0.4 0.4 19.42 5.98 0.6 0.18 26.18

0.2 -48.55 9.71 -14.95 2.99 -1.5 0.3 0.46 0.09 -52.37

-10.57 -0.03 -0.29 -0.93 -9.32 0.192 0.192

0.616 48.55 -29.9 4.86 -2.99 1.5 0.92 0.15 0.09 21.16

-9.32 -0.93 -0.29 -0.03 -10.57

Table 1 First Floor - Ground Floor - Foundation

Case 1 - maximum, maximum,maximum 15.41 0.09 1.56 13.76 0.203 0.713

0.084

0.31

-67.79 5.69 -7.73 0.65 -0.43 0.04

54.15

-74.56

48.33 5.51

0.301 65.79 -15.46 2.85 -0.86 0.33 -0.1

-6.28 -0.04 -0.33 -5.91 0.115 0.401 -20.6 -1.14

-0.13 54.55 21.87

0.183 -16.42 -9.4 0 -0.63 0 -0.07 -26.52

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Case 2 - maximum, minimum,maximum 15.66 0.08 1.82 13.76 0.203 0.713

0.084 -67.79

48.33 6.4 0.31 55.04

6.91 -8.98 0.75 -0.43 0.04 70.72

0.301

-7.23 -0.04 -0.33 -6.86 0.115 0.401

67.79

0.183 -8.17

-17.95 23.91 2.85 -0.86 -1.14 0.38 -0.11 -0.15 52.1 -25.2

-10.91 0 -0.52 0 -0.07 -19.67

Case 3 - minimum, maximum, minimum 6.85 0.04 0.45 6.36 0.203 0.713 22.32 1.6 0.14 24.06

0.102 -31.31 2.63 -2.24 0.19 0.2 0.02 -30.91

0.301 31.31 -4.48 1.32 -0.4 0.1 -0.03 27.82

-1.87 -0.01 -0.15 -1.71 0.115 0.401 -5.97 -0.53 -0.04 -6.54

0.183 -16.42 -2.72 0 -0.24 0 -0.02 -19.4

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APPENDIX C

COLUMN ANALYSIS

Inclined column

Effective Length:

l ex  l ey : l o  3580  500  3080

End Condition:

Top

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

Bottom : Beam (500) > Column (400) …. Condition 1 From Table 3.19,   0.75 Therefore; l ex  l ey   l o

= 0.75  3080 = 2310mm

Checking l ex 2310   5.8 h 400 l ey 2310   5. 8 b 400

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

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N 1  323.67 kN P1  307.11kN

M analysis  13.09kNm

M column  0.05 Nhmin = 0.05 ( 323.67) (0.4) = 6.47 kNm < M analysis M inclined  P.e = 307.11 ( 0.372) = 114.24kNm > M analysis

N 323.67  10 3   2.02 N / mm 2 bh 400  400 M 114.24  10 6   1.785 N / mm 2 bh 2 400  400 2

d  400  25 

20  365mm , 2

d 365   0.91 h 400

……Using Chart No.29, 100 Asc  0.41 bh

Asc 

0.41  400  400  656mm 2 100

Use 4T16 ( As  804mm 2 )

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1  16  4mm ……. R6 mm 4 Clear Distance Maximum  12  16  192mm …….150mm Diameter minimum 

Therefore, use R6 – 150.

Column 2

N 2  761.07 kN P2  681.88kN

M analysis  52.36kNm

M column  0.05 Nhmin = 0.05( 761.07) (0.4) = 15.22 kNm < M analysis

M inclined  P.e = 681.88 ( 0.372) = 253.66kNm > M analysis

N 761.07  10 3   4.76 N / mm 2 bh 400  400 M 253.66  10 6   3.96 N / mm 2 bh 2 400  400 2

d  400  25 

20  365mm , 2

d 365   0.91 h 400

…Using Chart No.29,

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100 Asc  1. 5 bh

Asc 

1.5  400  400  2400mm 2 100

Use 4T25 + 4T16 ( As  2767mm 2 )

Link

1  25  6.3mm ……. R8 mm 4 Clear Distance Maximum  12  16  192mm …….150mm Diameter minimum 

Therefore, use R8 – 150.

Column 3

N 3  1212.99kN P3  117.24kN

M analysis  20.92kNm

M column  0.05 Nhmin = 0.05 (1212.99) (0.4) = 24.26 kNm > M analysis M inclined  P.e = 1117.24 ( 0.372) = 415.6kNm > M analysis

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N 1212.99  10 3   7.58 N / mm 2 bh 400  400 M 415.6  10 6   6.49 N / mm 2 2 2 bh 400  400

d  400  25 

20  365mm , 2

d 365   0.91 h 400

Using Chart No.29, 100 Asc  3.1 bh

Asc 

3.1  400  400  4960mm 2 100

Use 4T32 + 4T25 ( As  5180mm 2 )

Link 1  32  8mm ……. R8 mm 4 Clear Distance Maximum  12  25  300mm …….300mm

Diameter minimum 

Therefore, use R8 – 300

Column 4

Effective Length:

l ex  l ey : l o  1020  500  520mm

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End Condition:

Top

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

Bottom : Beam (500) > Column (400) …. Condition 1 From Table 3.19,   0.75 Therefore; l ex  l ey   l o

= 0.75  520 = 390mm

Checking l ex 390   0.98 h 400 l ey 390   0.98 b 400

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

N1  1765.54kN P1  1614.26kN

M analysis  69.39kNm

M column  0.05 Nhmin = 0.05 (1765.54) (0.4) = 35.31 kNm < M analysis M inclined  P.e = 1614.26 (0.372) = 600.5kNm > M analysis

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N 1765.54  10 3   11.03 N / mm 2 bh 400  400 M 600.5  10 6   9.38 N / mm 2 2 2 bh 400  400

d  400  25 

20  365mm , 2

d 365   0.91 h 400

Using Chart No.29,

100 Asc  5.5 bh

Asc 

5.5  400  400  8800mm 2 100

Use 6T40 + 4T25 ( As  9503mm 2 )

Link

1  40  10mm ……. R10 mm 4 Clear Distance Maximum  12  25  300mm …….300mm

Diameter minimum 

Therefore, use R10 – 300. Minimum Reinforcement 0.4 % bh =

0.4  400  400  640mm 2 100

Maximum Reinforcement 6% bh =

6  400  400  9600mm 2 100

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APPENDIX D

COLUMN ANALYSIS

Vertical Column

Effective Length:

l ex  l ey : l o  3500  500  3000

End Condition:

Top

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

Bottom : Beam (500) > Column (250) …. Condition 1 From Table 3.19,   0.75 Therefore; l ex  l ey   l o

= 0.75  3000 = 2250mm

Checking l ex 2250  9 h 250 l ey 2250  9 b 250 Both values are less than 15, so the column is short column.

Column 1

From frame analysis, moment at roof is, pdfMachine A pdf writer that produces quality PDF files with ease! Produce quality PDF files in seconds and preserve the integrity of your original documents. Compatible across nearly all Windows platforms, simply open the document you want to convert, click “print”, select the “Broadgun pdfMachine printer” and that’s it! Get yours now!

M analysis  5.29kNm N 1  207.67kN

Moment Analysis M analysis  5.29 kNm

M column  0.05 Nhmin  0.05207.67 0.25..................0.05h  0.05  250  12.5  20......ok!  2.60kNm  6.74kNm

Therefore, use

M analysis  5.29kNm

Reinforcement

d  250  25 

So,

12  219mm 2

d 219   0.88  0.85 h 250

Use Chart No. 28 with

f cu  30 N / mm 2 , f y  460 N / mm 2 and

d  0.85 h

M 5.29  10 6   0.34 bh 2 250  250 2 N 207.67  10 3   3.32 bh 250  250 From the chart, use the minimum reinforcement for column which is: 4T12 with As  453mm 2

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Column 2

From frame analysis, moment at second floor is, M analysis  21.14kNm

N 2  440.9kN

Moment Analysis M analysis  21.14kNm

M column  0.05 Nhmin  0.05440.9 0.25..................0.05h  0.05  250  12.5  20......ok!  5.51kNm  21.14kNm Therefore, use

M analysis  21.14kNm

Reinforcement

d  250  25 

So,

12  219mm 2

d 219   0.88  0.85 h 250

Use Chart No. 28 with

f cu  30 N / mm 2 , f y  460 N / mm 2 and

d  0.85 h

M 21.14  10 6   1.35 bh 2 250  250 2 N 440.9  10 3   7.05 bh 250  250

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

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4T12 with As  453mm 2

Column 3

From frame analysis, moment at second floor is, M analysis  8.91kNm

N 3  666.15kN

Moment Analysis M analysis  8.91kNm

M column  0.05 Nhmin  0.05666.150.25..................0.05h  0.05  250  12.5  20......ok!  8.33kNm  8.95kNm

Therefore, use

M analysis  8.95kNm

Reinforcement

d  250  25  So,

12  219mm 2

d 219   0.88  0.85 h 250 Use Chart No. 28 with

f cu  30 N / mm 2 , f y  460 N / mm 2 and

d  0.85 h

M 8.91  10 6   0.57 bh 2 250  250 2 N 666.15  10 3   10.66 bh 250  250

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From the chart, use the minimum reinforcement for column which is: 4T12 with As  453mm 2

Link

Diameter minimum =

1  12  3 mm……………….. use R8 4

Clear Maximum Distance = 12  12  144mm So, use R8 – 125

Column 4

From frame analysis, moment at second floor is, M analysis  32.43kNm

N 4  1017.34kN

Effective Length: l ex  ley : l o  1000  500  500

End Condition: Top

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

Bottom : Beam (500) > Column (250) …. Condition 1 From Table 3.19,   0.75 Therefore; l ex  l ey   l o

= 0.75  500 = 375mm pdfMachine A pdf writer that produces quality PDF files with ease! Produce quality PDF files in seconds and preserve the integrity of your original documents. Compatible across nearly all Windows platforms, simply open the document you want to convert, click “print”, select the “Broadgun pdfMachine printer” and that’s it! Get yours now!

Checking l ex 375   1.5 h 250 l ey 375   1.5 b 250

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

Moment Analysis

M analysis  32.43kNm

M column  0.05 Nhmin  0.051017.34 0.25................0.05h  0.05  250  12.5  20......ok!  12.72kNm  32.79kNm

Therefore, use

M analysis  32.79kNm

Reinforcement d  250  25 

So,

20  215mm 2

d 215   0.86  0.85 h 250

Use Chart No. 28 with

f cu  30 N / mm 2 , f y  460 N / mm 2 and

d  0.85 h

M 32.43  10 6   2.08 bh 2 250  250 2 N 1017.34  10 3   16.3 bh 250  250 From the chart, pdfMachine A pdf writer that produces quality PDF files with ease! Produce quality PDF files in seconds and preserve the integrity of your original documents. Compatible across nearly all Windows platforms, simply open the document you want to convert, click “print”, select the “Broadgun pdfMachine printer” and that’s it! Get yours now!

100 Asc  2.2 bh

Asc 

2.2  250  250  1375mm 2 100

Therefore, use 4T20 + 4T10 with As  1571mm 2 Link

Diameter minimum =

1  20  5 …………………. Use R8 4

Clear Distance Maximum = 12  10  120mm So, use R8 – 100

Minimum Reinforcement 0.4 % bh =

0.4  250  250  250mm 2 100

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 ( As = 453mm 2 ).

Maximum Reinforcement 6% bh =

6  250  250  3750mm 2 100

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FIRST FLOOR PLAN FOOTING PLAN

GROUND FLOOR PLAN

DATA-DATA TERLIBAT 1. fcu = 30N/mm2

2. fy = 460 @ 250 N/mm2, fy(slab)= 250 N/mm2 3. Ruang Pameran (Exhibition) 4. Dinding Kaca = 26.7kN/m3 5. Bearing Capacity = 150kN/m2 6. qk(w/tah) = 3.kKN/m2, gk(finishes) = 1.0kN/m2 7. Tebal papak = 150 mm 8. Saiz rasuk(mm) = 200x500 9. Saiz tiang(mm) = 250x250 10. Saiz asas(foundation) = 2500x2500 11. Penutup=25 mm ROOF PLAN SECOND FLOOR PLAN

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