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ANALYSIS, DESIGN AND ESTIMATION OF TWO

STOREYED RESIDENTIAL BUILDING

A PROJECT REPORT

Submitted by

In partial fulfillment for the award of the degree

of

BACHELOR OF ENGINEERING

in

CIVIL ENGINEERING

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BONAFIDE CERTIFICATE

Certified that this project report "ANALYSIS, DESIGN ANDESTIMATION OF TWO STOREYED RESIDENTIAL BUILDING" is

the bonafide work of

.who carried out the project work under my supervision.

SIGNATURE SIGNATURE

. ..

HEAD OF THE DEPARTMENT SUPERVISOR

Civil Engineering Civil Engineering

Submitted for the viva-voce held at . on

Internal Examiner External Examiner

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ABSTRACT

The Project reports on design of a new two Storeyed Residential

Building at . near m. This project mainly includes the

Analysis, Design and Estimation. The next site area of field is

687.82 Sq.m and the plinth area of the building is 958.53 Sq.m.

The analysis was done by Kani's Methods, using this method the

moment of each beam and column was calculated.

Limit state method of design is used for the design purpose. The

load condition is taken as per IS:875. Manual estimation was done and the

cost the building Rs. 3129.8 Per sq.m.

The plan, section, elevation and reinforcement details are drawn

using AUTO CAD.

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ACKNOWLEDGEMENT

We record our sincere thanks and gratitude to our supervisor

.. Lecturer, Department of Civil Engineering,

. College of Engineering, .., who evinced keen interest and helped us

to a lot every now and then to complete and project.

We express our sincere thanks to Head of the Department of

Civil Engineering, for his valuable suggestions guidance throughout the course

of the study.

Further we express our sincere thanks to ..Principal, .

College of Engineering, for giving encouraging suggestions for the successful

completion of the project.

We express our sincere thanks to correspondent of ourcollege for his valuable suggestion garden throughout the ware about of this

study.

We express our sincere thanks to the . Lecturer of our

Department for his valuable suggestion garden throughout the ware of this

study.

We also thank all the faculty members of the Department of

Civil Engineering, . College of Engineering, for their constant support

and inspiration for the successful completion of this project .

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CONTENTS

Page No

Abstract iii

List of Tables viii

List of figures ix

List of Symbols xi

I. INTRODUCTION 1

1.1. General 1

1.1.1 The Concept of Design 11.1.2 objectives 2

1.1.3. Components of RC Structures 2

1.1.3.1 Limit state design 3

1.1.4 Slabs 3

1.1.5 Beam 3

1.1.6 Columns 4

1.1.6.1. Short Column 4

1.1.6.2. Slender Column 5

1.1.7 Footing 5

1.1.7.1 Types of column base 5

1.1.8 Staircase 5

1.1.8.1 Classification of Stairs 6

1.1.9. Summary 6

1.2. Methodology 12

1.2.1 Analysis 12

1.2.2. Design 12

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1.2.3. Estimation 12

1.2.4. Limit state design 12

1.2.5. Partial safety factor 12

1.3. Analysis 18

1.3.1. Introduction 18

1.3.2. Kanis Method 18

1.3.3. Primary Structure

2. Designs 85

2.1. Design of slab 85

2.1.1. Design Procedure for one way slab 86

2.1.2. Design Procedure for two way slab 88

2.1.3. Design of One Way Slab 90

2.1.4. Design of two way Slabs 95

2.2. Design of Beams 101

2.2.1.Types of Beams 101

2.2.2. Design procedure for beams 102

2.2.3. Design of singly Reinforced Beam 104

2.2.4.Design of doubly reinforced Beam 108

2.3. Design of curved Beam 103

2.4. Design of Plinth Beam 121

2.5. Design of lintels 125

2.6.Design of Lintel Cum Sunshade 126

2.7.Design of staircase With single beam 133

2.8. Design of column 139

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2.8.1. Column 139

2.8.2. Types of columns 139

2.8.3. Design Procedure for Column 140

2.8.4. Load calculation of Column 142

2.8.5. Square Column with Biaxial Bending

(Intermediate Column) 144

2.8.6. Square Column with Biaxial Bending

(Corner Column) 149

2.9.Design of footing 154

2.9.1. Footing 154

2.9.2. Types of footing 154

2.9.3. Design Procedure 155

2.9.4. Design of footing 161

2.10 Estimation and costing 166

2.10.1. Abstract Estimation 175

3. CONCLUSION 176

4. REFERENCE 177

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LIST OF TABLES

Table No. Description Page. no

1. Moment for each frame 83

2. Reinforcement Detail about all beams 119

3. Estimation and costing 166

4. Abstract Estimate 175

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LIST OF FIGURES

Fig. No. Description Page.no

1.1.1. Site Plan 7

1.1.2 Ground floor plan 8

1.1.3 First floor plan 9

1.1.4. Cross section of AB 10

1.1.5. Elevation 11

1.2.1. Analysis of frame AA' 13

1.2.2. Analysis of frame BB' 13

1.2.3. Analysis of frame CC' 14

1.2.4. Analysis of frame DD' 14

1.2.5. Analysis of frame EE' 15

1.2.6. Analysis of frame FF' 15

1.2.7. Analysis of frame GG' 16

1.2.8. Analysis of frame HH' 16

1.2.9. Analysis of frame II' 17

1.2.10 Analysis of frame JJ' 17

1.3.1. Deformed shape of the member 18

1.3.2. Final end moment 19

1.3.3. Multistoreyed frame 21

1.3.4. Rotation end moment at A 22

1.3.5. Relative stiffness 24

2.1.3.1. Reinforcement Details for one way slab 94

2.1.4.1. Reinforcement Details for two way slab 100

2.2.3.1. Reinforcement Details for Singly reinforcement beam 107

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2.2.4.1. Reinforcement Details for Doubly reinforcement beam 112

2.3.1 Reinforcement Details for curved beam 118

2.4.1 Reinforcement Details for Plinth beam 124

2.6.1 Reinforcement Details for Lintel cum beam 132

2.7.1.1. Reinforcement Details for Stair Case 138

2.8.5.1. Reinforcement Details for Column

(Intermediate Column)

148

2.8.6.1. Reinforcement Details for Column

(Corner Column)

153

2.9.4.1. Reinforcement Details for Footing 165

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LIST OF SYMBOLS

Mx - Moment in shorter direction

My - Moment in shorter direction

d - Effective depth

D - Overall depth

Ast - Area of Steel

P - Load

Wu (or) Pu - Design load

Mu - Design moment

Asc - Area of concrete

Fy - Characteristic strength of steel

Fck - Characteristic strength of concrete

B.M - Bending Moment

b - Breadth of Beam

D - Overall depth

Vus - Strength of shear reinforcement

L - Clear span

Le - Effective span

N.A - Neutral Axis

MF - Modification factor

Q - Angle of repose of soil

M - Modular of rupture

c - Permissible shear stress in concrete

v - Nominal shear stress

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CHAPTER I

INTRODUCTION

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CHAPTER - 1

INTRODUCTION

1.1. General

The main motive of a Civil Engineer is to design a structure,

which is to be safe, serviceable and economical. Safety means that the

structure should not fail under loads unless exceeds by a given margin.

Serviceability means that structures must perform well throughout their service

life in both appearance and comfort to their clients. This requirement includes

cracking and deflection under working loads. Economy means that structures

must be designed in such a way as to minimize the quantities of materials used

in them. Although safety and serviceability are the basic requirements, the test

of an acceptable structural design is economy.

1.1.1 The Concept of Design

In the design of reinforced concrete structures, all critical sections

are checked for the effect of forces acting on them. Sizing of columns, beams

and spacing of frames will affect the economy as well as the stability of a

framed building. Framed structure mode of construction is more suitable for a

commercial type of construction. Basically in the framed structures the wallsare not the load carrying structures, so that the size of the walls can be

decreased. By decreasing the size of the walls the need for building material

would get reduced and the floor space of the building would also get

increased. With the reduction in the building material, a reduced amount of

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load will be transmitted to the footing. The modern method of Design, which

is adopted recently, is the Limit State Design. Limit state method includes

consideration of structures at both working and ultimate load levels with a

view to satisfy the requirements of safety and serviceability. The aim of the

limit stage design is to ensure that a reinforced concrete section does not reach

any of the limit states to which it may be subjected. The usual approach is to

design for a limit state which is likely to govern it and them to check it for the

remaining limit states.

In this project "Analysis, Design and Estimation of a two storied

residential building" the structure is completely analyzed by Kanis method

and as per IS 456:2000 code.

1.1.2 Objectives

1. To analyse the soil condition of the site.

2. To analyse the frames in the building using Kanis method.

3. To design the structural components of the two storey building

4. To prepare the detailed drawing for the design carried out.

5. To analyse the construction cost of building.

1.1.3 Component of RC Structures

Reinforced cement concrete members can be designed by the

following methods.

1. Working stress method

2. Limit state method

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1.1.3.1 Limit state design

Limit state method of design is based on elastic theory.

Partial safety factors are used in this method to determine thedesign loads and design strength of materials from their

characteristics values.

The design aids to IS:456, published by the bureau of Indian

standards. The design of limit state method is very simple and

hence widely used in practice.

This method gives economical results when compared with the

conventional working stress method.

1.1.4 Slabs

Slabs are primary members of a structure, which support the imposed

load directly on them and transfer the same safety to the supporting

elements such as beams, walls, columns etc.

A slab is a thin flexural member used in floor and roof of a structure

to support the imposed loads.

The Slabs are classified as Solid Slab, Hollow slab and Ribbed slab

based on their construction.

1.1.5 Beam

A beam has to be generally designed for the actions such as bending

moments, shear forces and twisting moments developed by the

lateral loads.

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The size of the beam is designed considering the maximum moment

in it and generally kept uniform throughout its length.

IS:456:2000 recommends that the minimum grade of concrete should

not be less than M20 in RC works.

When there is a Reinforced concrete slab over a concrete beam, then

the beam and the slab can be constructed in such a way that they act

together.

The combined beam and slab are called as flanged beams. It may be

'I' or 'L' beams. Here both T-beams and L-beams are designed.

1.1.6 Column

Vertical members in compression are called as columns and struts.

The term column is reserved for members which transfer load to the

ground. Classification of column, depending upon slenderness ratio

is, Short Column and Slender column.

Columns are classified as axially loaded column, Eccentrically

loaded column & column subjected to axial load and moment

depending the action of load.

1.1.6.1 Short Column

IS:456:2000 classifies rectangular column as short when the ratio

of effective length (Le) to the least lateral dimension is less than 12. This ratio

is called slenderness ratio of the column.

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1.1.6.2 Slender Column

The ratio of effective length (Le) to the least lateral dimension is

grater than 12 are called as slender column.

1.1.7 Footing

Foundation is the most important component of a structure.

It should be well planned and carefully designed to ensure the

safety and stability of the structure.

Foundation provided for RCC columns are called as column base.

1.1.7.1 Types of column base

1. Isolated footing

2. Combined footing

3. Strap footing

4. Solid raft foundation5. Annular raft foundation

1.1.8 Staircase

A staircase is a flight of steps leading from one floor to another. It

is provided to afford the means of ascent and descent between various floors of

the building. It should be suitably located in a building. In a domestic building

the stair should be centrally located to provide easy access to all rooms. In

public buildings stairs should be located near the entrance. In big building

there can be more than one stairs. Fire protection to stairs is important too.

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Stairs are constructed using timber, bricks, stone, steel or reinforced cement

concrete.

1.1.8.1 Classification of Stairs

1. Single flight stairs

2. Quarter turn stairs

3. Dog legged stairs

4. Open well type stairs

5. Bifurcated stairs

6. Circular stairs

7. Spiral stairs

1.1.9. Summary

This project deals with the design of two storeyed building using

limit state method of one way and two way solids slabs are designed. Singly

and doubly reinforced beams and plinth beam, are designed as per

IS 456 / 2000. Columns with biaxial loaded column base of isolated column

footing are design. Single beam staircase designed with working stress

method.

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Fig : 1.1.1. Site plan

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Fig : 1.1.2. Ground Floor Plan

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Fig : 1.1.3. First Floor Plan

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Fig : 1.1.4. Cross Section at AB

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Fig : 1.1.5. Elevation

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1.2 Methodology

1.2.1. Analysis

Various methods are available for the analysis of a structure.

Kanis Method is adopted in this project.

1.2.2. Design

Various methods are available for the design of a structure. Limit

sate method is adopted in this project.

1.2.3. Estimation

Various methods are available for the estimation of a structure.

Centre line method is adopted in this project.

1.2.4. Limit state design

The acceptable limit for safety and serviceability requirement

before failure occur is called limit state. The aim of design is to achieve

acceptable probabilities that the structure will not become unfit for use. All

relevant limit state shall be considered in the design to ensure adequate degree

of safety and serviceability.

1.2.5. Partial safety factor

The value of load which has a 95% probability of a structure of

structural member for the limit state of collapse the following values of partial

safety factor is applied for limit state of collapse.

Ym = 1.5 for concrete

Ym = 15 for steel

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Fig : 1.2.1. Analysis of Frame AA'

Fig : 1.2.2. Analysis of Frame BB'

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Fig : 1.2.3. Analysis of Frame CC'

Fig : 1.2.4. Analysis of Frame DD'

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Fig : 1.2.5. Analysis of Frame EE'

Fig : 1.2.6. Analysis of Frame FF'

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Fig : 1.2.7. Analysis of Frame GG'

Fig : 1.2.8. Analysis of Frame HH'

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Fig : 1.2.9. Analysis of Frame II'

Fig : Analysis of Frame JJ'

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1.3 Analysis

1.3.1. Introduction

A multistoried frame is a complicated statically indeterminate

structure. The analysis by moment distribution method is very lengthy and

difficult. Hence Rotation Contribution Method (Kanis method) is adopted for

better and easier calculation.

1.3.2 Kanis Method

This is a method developed by Gasper Kanis of Germany. This

method is an excellent extension of the slope deflection method. Since this

new method has been recognized as a very useful method, it is proposed to

discuss this method in detail.

Let AB represent one of the spans of a framed or continuous

structure. Let the span carry a loading.

Fig : 1.3.1. Deformed shape of the member

Fig 1.3.1. shows the deformed shape of the member AB. Let the

ends A and B undergo rotations a and b respectively. Let us assume that

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lateral displacements of the ends do not occur. Let Mab and Mba represent the

end moments for the span AB. Let us follow the following sign conventions

regarding end moments and rotations.

i) Clockwise end moments are positive.

ii) Clockwise rotations at ends are positive.

In this method the final end moments are split up in to certain

components. By an orderly successive approximation process these

components are computed.

The components to which the final end moments Mab and Mba can

be split up are shown in Fig.1.3.2.

Fig: 1.3.2. Final end moments

The moments are determined by going through the followingstages :

1. The ends A and B of the member are first regarded as fixed.

Corresponding to this condition the fixed end moments abM and baM at

B are determined.

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2. Now maintaining the fixity of the end B, the end A is rotated through an

angle a by the application of a moment 2 M 'ab at A. For this condition a

moment of M'ab is induced at the end B. The moment M'ab is called the

rotation contribution of the end A.

3. In this stage, the end A is considered as fixed, and the end B is rotated

through an angle b by the application of a moment 2M'ba at B. For this

condition a moment of M 'ba is induced at the end A. The moment M'ba is

called the rotation contribution of the end B.

Thus, the final moments Mab and Mba can be expressed as follows.

Mab = abM + 2M'ab + baM

and

Mba =baM + 2M

'ba +

abM ....................(i)

For the member AB, when we refer to the final moment Mab at A,

the end a may be referred to as the near end and the end B as the far end.

Similarly, when we refer to the final moment Mba at B, the end B

may be referred so as the near end and the end A as the far end.Relation shown in eq. (i) above may be worded as follows:

Moment at the near end of a member = sum of

a) The fixed end moment at the near end due to the loading on the member.

b) Twice the rotation contribution of the near end,

c) The rotation contribution of the far end.

Figure 1.3.3. shows a multistoreyed frame. If no lateral joint displacements

occur for the members, the equations of the type obtained above are applicable

to all the members.

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Fig 1.3.3. Multistoreyed frame

Consider the various members meeting at the joint A. The end

moments at A for the members meeting at A are therefore given by

Mab = abM + 2M'ab + M

'ab

Ma-10 = 10aM + 2M'a-10 + M

'10-a

Ma-5 = 5aM + 2M'a-5 + M

'5-a

Ma-2 = 2aM + 2M'a-2 + M'2-a

For the equilibrium of the joint A, the sum of all the end

moments at A must be equal to zero. i.e.

Mab = 0

Mab = abM + 2M'ab + M'ba = 0 ....................(iii)

Where,

abM = algebraic sum of the fixed end moments at A for all the

members meeting at A.

M'ab = algebraic sum of the rotation contributions at A for all

the members meeting at A.

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M'ba = algebraic sum of the rotation contributions at the far end

joints with respect to joint A.

For the frame shown in Fig.1.3.3.

abM = abM + 10aM + 5aM + 2aM

M'ab = M'ab+ M'a-10+ M'a-5+ M'a-2

M'ba = M'ba+ M'10-a+ M'5-a+ M'2-a

From eq. (iii), we have, M'ab = { }baab MM'

2

1+

....................(i)

But know for the member AB, with the end B fixed, the moment

required at A so as to rotate the joint A by a is given by

Fig : 1.3.4. Rotational end moment at A

2M'ab =ab

ab

l

El4

a = 4EKaba

M'ab = 2EKaba

Where, Kab =ab

ab

l

l

and E = Young's modulus

Now consider the members meeting at the joint A. The end

rotation at A of each member meeting at A equals a. Assuming that the

young's moduli to be the same for all the members, we have,

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M'ab = 2EaKab

ab

ab

M

M

'

'

= abab

K

K

M'ab =

ab

ab

K

K abM' ..................(v)

From equations (iv) and (v), we get,

M'ab =

abab

K

K

2

1 { }baab MM

'+

The ratio

abab

K

K

2

1is called the rotation factor for the member AB at the

joint A. Let Uab be the rotation factor at the joint A for the member AB.

M'ab = Uab { }baab MM'+ ................... (vi)

Now consider equation (vi). The summation abM can be

computed. This is therefore a known quantity. If trial values be assumed for

the far end rotation contributions the approximate value of the near end

rotation contribution can be computed from equation (vi).

By successive application of equation (vi) to the various joints the

rotation contributions can be determined.

For instance as a first approximation assuming that the rotation

contributions of the far ends of the members meeting at A to be zero, the

rotation contribution at A for the member AB is given by

M'ab = Uab abM

By a similar approximation the rotation contributions at other

joints are also determined. With the approximate values of the rotationcontributions computed, it is possible to again determine a more correct value

of the rotation contribution at A for the member AB using the equation.

M'ab = Uab { }baab MM'+

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The process mentioned above is repeated so that more and more

accurate values of the rotation contributions are obtained. After attaining the

desired extent of accuracy in the values of the rotation contributions the final

moments can be easily computed from the relations like.

Mab = abM +2M'ab+M

'ba

Some important points to be remembered

a) If an end of a member is fixed the rotation at the end being zero, the

rotation contribution at the end is also zero.

b) If an end of a number is hinged or pinned, it will be convenient to

consider the end as fixed and to take the relative stiffness as .43ll

Fig : 1.3.5. Relative Stiffness

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Load Calculation on Analysis of frame

Roof

Live load =73.4

6.52+

73.4

8.1

= 2.75 kN/m

Parapet = 2.25 kN/m

D.L (Roof. beam) = 3.75 + 2.58

= 6.34 kN/m

Floor terrazo = 0.89 kN/m

Plastering = 0.3 kN/m

Finishing = 0.47 kN/m

Total load W = 13 kN/m

Factored load Wu = 1.5 x 13

= 19.5 kN/m

I Floor

Live load = 2.75 kN/m

Dead load(Roof, beam) = 6.34 kN/m

Brick wall (230mm)

of 3m height = 13 kN/m

Stone Floor = 0.71 kN/m

Plastering = 0.3 kN/m

Finishing = 0.9 kN/m

Total load W = 24 kN/m

Factored load Wu = 1.5 x 24

= 36 kN/m

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Frame @ AA'

Fixed End Moment

BDM =12

2wl

=12

)88.4(36 2

= -71.44 kN.m

DB = 71.44 kN.m

CFM =12

)1.6(5.19 2

= -60.46 kN.m

FCM = 60.46 kN.m

DGM =12

)1.6(36 2

= -111.63 kN.m

GDM = 111.63 kN.m

GIM =

12

)96.3(36 2

= -47.04 kN.m

IGM = 47.04 kN.m

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Joint Member R.S T.R.S D.F R.F

BBA I/3

0.538 I0.62 -0.31

BD I/4.88 0.38 -0.19

CCD I/3

0.49I0.68 -0.34

CF I/6.10 0.32 -0.16

D

DC I/3

1.03 I

0.32 -0.16

DB I/4.88 0.20 -0.1

DG I/6.1 0.16 -0.08

DE I/3 0.32 -0.16

FFC I/6.1

0.49I0.32 -0.16

FG I/3 0.68 -0.34

G

GF I/3

1.08 I

0.31 -0.155

GD I/6.1 0.15 -0.175

GI I/3.96 0.23 -0.115

GH I/3 0.31 -0.155

IIG I/3.96

0.58 I0.43 -0.215

IJ I/3 0.57 -0.285

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41

CF

-60.46 -0.

16

-0.34

-60.46 3.75

-0.

16

60.46

-0.34

8.99

11.16

11.19

11.23

11.24

-11.11

-10.39

-10.61

-10.64

-10.6519.11

23.7223.77

23.86

23.88

0.850.872

0.908

1.8

4.25

-0.16

-0.

1

-40.19 -0.

08

D-111.63 111.63

2.13

0.9

0.45

0.43

0.42

-3.23

-2.57

-2.49

-2.48

-2.48

-5.13-5.13

-5.15

-5.32

-6.68

-23.61

22.07

-22.55-22.62

-22.63

4.25

1.8

0.9080.872

0.85

-0.155-0.

075

64.59

-0.155

-0.

115

0.16

-4.95

-3.95

-3.82

-3.80

-3.80

-47.04

B -71.44

-0.31

-0.

19

-71.44 71.44

13.57

13.06

13.3613.46

13.47

2.66

1.12

0.56

0.54

0.5322.1421.32

24.71

21.97

21.97

HEA

47.04

-0.285

-0.

12

-11.99-12.28

-12.31

-12.32-12.32

J

47.04

-9.04

-9.26

-9.29-9.29

-9.29

-0.

215

-6.68

-5.32

-5.15-5.13

-5.13

G


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42

B

C-60.46 60.46

11.24

11.24-10.65

-48.63

-10.65

-10.6511.24

50. 4

25.58

23.88

0.850.85

-111.63 111.63

0. 42

0. 42

-2 . 48

-113.27

-2. 48

-2. 48

0. 42

-113.27

0.85

0.85

1. 7

F

-22.63

-22.63

-5.13

50. 4

-32.87

-5.12

-5.12

-22.63

-47.04G 47.04

-3.80

-3.80

-9.29

-48.63

-9.29

-9.29

-3.80

24.66

-5. 13

-5. 13

-10.26

I

-12. 3

-12. 3

-24.6

D

0. 53

0. 53

13 .47

85. 97

13.47

13.47

0.53

-43.97

21.97

21.97

43.94

-71.44 71.44

A E H J

43.94

0.85 -5.13 -12


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43

B

C-48.63 50.4

-113.27 107.09

F

- 63.93G 24.66I

D-43.97 85.97

A E H J

- 50.448.63

43.94

25.58

1.7

-32.87

- 10.26- 24.6

43.94 0.85 -5.13 -12.32


44/197

Frame @ BB'

Fixed End Moment

MBC =12

)88.4(36 2

= -71.44 kNm

MCB = 71.44 kN.m

MEF =12

)88.4(5.19 2

= -38.69 kNm

MFE = 38.69 KN.M

MFG =12

)1.6(5.19 2

= -60.46 KNM

MGF = 60.46 KNM

MGH =12

)96.3(5.19 2

= -25.48 kNm

MHG = 25.48 kNm

MIJ =12

)96.3(362

= -47.04 kNm

MJI = 47.04 KN.M

44


45/197


B

BA I/3

0.87 I

0.38 -0.14

BC I/4.88 0.24 -0.12BE I/3 0.38 -0.14

C

CB I/4.88

0.87 I

0.24 -0.12

CD I/3 0.38 -0.19

CF I/6.10 0.38 -0.19

EEB I/3

0.54 I0.62 -0.31

EF I/4.88 0.38 -0.19

FFE I/4.88

0.70 I0.29 -0.145

FG I/6.1 0.23 -0.115

FC I/3 0.48 -0.24

G

GF I/6.1

0.75 I

0.22 -0.11

GH I/3.96 0.33 -0.165

GI I/3 0.45 -0.225

HHG I/3.96

0.58 I0.44 -0.22

HJ I/3 0.56 -0.28

I

IG I/3

0.92 I

0.36 -0.18

IJ I/3.96 0.28 -0.14

IK I/3 0.36 -0.18

J

JH I/3

0.92 I

0.36 -0.18

JI I/3.96 0.27 -0.135

JL I/3 0.37 -0.185

45


46/197

46

FG

-21.78 -0.

115

-0.24

-60.46 60.46

-0.

11

34.98

-0.225

3.37

3.90

4.014.02

-4.22

-5.17

-5.45-5.40

-16.89-16.84

-16.53

-14.93

-0.19

-0.

12

-7.44 -0.

08

D

11.78

11.78

11.77

10.22

-8.63

-10.56

-11.16

-11.05

-14.93

-16.54

-16.84-16.89

-0.18

-47.04

-0.18

0.

14

0.19

7.79

9.15

9.169.16

-47.04 47.04

-7.80

-7.25

-7.38

-7.30

47.04-0.

135

-0.185

I

-10.69

-9.94

-10.12-10.01

B -71.44

-0.19

-0.

120

-71.44 71.44

7.13

8.84

9.09

9.13

-9.43

-10.47

-10.64

-10.6711.29

14.0114.4

14.46

LKDA

-0.

145

-0.

165

-6.33

-7.75

-8.18

-8.14

-1.93

-1.15

-1.64

-1.03

-25.48 25.480.

22

25.28

-0.28

-2.45

-1.47-2.08

-1.31 -9.74

-9.85

-9.67

-10.40J

-0.18

10.22

11.7711.78

11.78

I

C

-0.19

14.46

14.40

14.0111.29

12.00

7.176.125.96

7.35

4.39

3.75

3.65

4.25

4.93

5.06

5.07

-38.69

-0.31

-0.

19

-38.69 38.69

E

7.05

8.158.368.40


47/197

47

B

F-60.46 60.46

4.024.02

-5.40

-57.82

-5.40-5.40

4.02

53.68

-25.38

8.40

-16.89

-16.89

-16.89

-16.89

-33.78

11.78

11.78

23.56

-8.17

-8.17

-1.03

-42.85

12.5

-11.05

11.78

11.78

-47.04G 47.04

9.16

9.16

-7.30

-36.02

-7.30

-7.30

-9.16

41.60

J

-10.01

-10.01

-20.02

C

-10.67

-10.67

9.13

59.23

9.13

9.13

-10.67

-63.85

14.4614.46

28.92

-71.44 71.44

A D K L

14.64

-16.89 11.78 -10.01

G -25.48 25.48H

8.40

-16.89

-16.89

I

- 1.03- 1.03

-8.17

-15.25

- 1.31

- 1.31

-9.74

-12.36

-11.05

11.78

11.7810.32 -20.79

- 1.31- 9.74

-9.74

38.69-38.69

E

3.653.65

5.07

-26.32

5.075.07

3.65

52.48

5.965.96

14.46

26.38

34.88

5.96

14.46

14.46


48/197

48

B

F-57.82 53.65

36.02I 41.60J

C-63.85 59.23

A D K L

14.64

-16.89 11.78 -10.01

G -42.85 15.25H

52.48-26.32

E

26.38 -0.09-10.32 -12.36

34.88

28.92 -33.78 -2356

-25.38

12.51

-20.79

-20.02


49/197

Frame @ CC'

Fixed End Moment

CDM =12

2wl

=12

)27.4(5.19 2

= -29.62 kN.m

DCM = 29.62 kN.m

DGM =

12

)1.6(5.19 2

= -60.46 kN.m

GDM = 60.46 kN.m

BEM =12

)27.4(36 2

= -54.69 kN.m

EBM = 54.69 kN.m

EHM =12

)1.6(36 2

= -111.63 kN.m

HEM = 111.63 kN.m

49


50/197


B

BA I/3

0.9 I

0.37 -0.185

BE I/4.27 0.26 -0.130

BC I/3 0.37 -0.185

CCB I/3

0.56 I0.59 -0.295

CD I/4.27 0.41 -0.205

D

DC I/4.27

0.73 I

0.32 -0.16

DG I/6.10 0.22 -0.11

DE I/3 0.46 -0.23

E

ED I/3

1.06 I

0.31 -0.15

EB I/4.27 0.22 -0.11

EH I/6.10 0.15 -0.08

EF I/3 0.31 -0.15

GGD I/6.10

0.497 I0.33 -0.165

GH I/3 0.67 -0.335

H

HG I/3

0.83 I

0.40 -0.2

HE I/6.10 0.20 -0.1

HI I/3 0.40 -0.2

50


51/197

51

D G

-30.84 -0.

11

-0.23

-60.46 60.46-0.

165

60.46

-0.335

2.95

3.43

2.95

2.87

2.86

-10.46

-7.44

-7.13-7.09

-7.09

8.31

8.31

8.278.01

6.55

-0.15

-0.

12

-56.94 -0.

08

D

-20.33-20.33

-20.31

-20.15

-18.77

6.55

8.01

8.27

8.318.31

-18.77

-20.15

-20.31

-20.33-20.33

-0.2

111.63

-0.2

0.

1

0.15

B -54.69

-0.185

-0.

130

-54.69 54.69

7.10

5.73

5.68

5.68

5.65

4.80

5.87

6.066.09

6.10

10.118.16

8.09

8.098.05

KFA

-0.

16

C

-0.185

8.05

8.09

8.098.16

10.11

5.75

5.06

4.875.10

5.13

3.99

3.51

3.393.53

3.56

4.29

4.99

4.304.18

4.16

-29.62

-0.295

-0.

205

-29.62 29.62

C

-111.63 111.63

3.49

4.27

4.41

4.43

4.43

-9.38

-10.07

-10.15

-10.16

-10.16

-21.24

-15.11

-14.49-14.41

-14.40

6.17

7.186.186.01

6.01


52/197

52

5.13

5.13

8.05

18.31

B

60.46

2.86

2.86

-7.09

-61.83

- 4. 43

4. 43

-10. 16

-111.83

12.5

-10.16-10.16

4.43

95.74

J

-20. 33

-20. 33

-40.66

6. 106. 10

5. 65

59.23

5. 65

5. 65

6. 10

-37.31

8.318.31

16.62

-54.69

A F I

- 14.40- 14.40

-20.33

-49. 13

-29.62

21.23

5.13

8.05

8.05

EB H

D GC

54.69

3. 56

3. 56

4. 16

-18.34

4. 16

4. 16

3. 56

41. 5

29.62 - 60.46

8.316.01

6.01

53.68

-7.09

-7.09

2.86

49.14

-55.06

-14.40

-20.33

-20.33

- 111.63 111.63

6.01

8.31

8.31

8.058.05

16.10

8.05 8.31 -20.3


53/197

53

B

D-61.83

22.63

E95.74

H-37.31 72.54

A D I

49.14G

-18.34E 41.5

-111.83

-55.0

16.62 -40.6

21.23

16.1

8.05 8.31 -20.


54/197

Frame @ DD'

Fixed End Moment

ABM =12

2wl

=12

)73.4(5.19 2

= -36.35 kN.m

BAM = 36.35 kN.m

BCM =

12

)52.1(5.19 2

= -3.75 kN.m

CBM = 3.75 kN.m

DEM =12

)73.4(36 2

= -67.12 kN.m

EDM = 67.12 kN.m

EFM =

12

)52.1(36 2

= -6.93 kN.m

FEM = 6.93 kN.m

54


55/197

FGM =12

)96.3(36 2

= -47.04 kN.m

GFM = 47.04 kN.m


AAB 0.211 I

0.544 I0.39 -0.195

AD 0.333 I 0.61 -0.305

B

BA 0.211 I

1.202 I

0.17 -0.085

BC 0.658 I 0.55 -0.275

BE 0.333 I 0.28 -0.14

C

CB 0.658 I

0.991 I

0.66 -0.33

CF 0.333 I 0.34 -0.17

D

DA 0.333 I

0.877 I

0.38 -0.19

DH 0.333 I 0.38 -0.19

DE 0.211 I 0.24 -0.12

E

EB 0.333 I

1.535 I

0.22 -0.11

ED 0.211 I 0.14 -0.07

EF 0.658 I 0.42 -0.21

EI 0.333 I 0.22 -0.11

F

FC 0.333 I

1.576 I

0.21 -0.105

FE 0.658 I 0.42 -0.21FJ 0.333 I 0.21 -0.105

FG 0.252 I 0.16 -0.08

GGF 0.252 I

0.585 I0.43 -0.215

GK 0.333 I 0.57 -0.285

55


56/197

56

BC

32.6 -0.

275

-0.14

-3.75 3.75

-0.

33

3.75

-0.17

-10.91

-9.32

-8.21

-7.82

-7.8

2.36

0.043

-0.83

-0.95

-1.0-5.56

-4.74-4.18

-3.98

-3.9

-8.62-8.61

-8.53

-8.12

-6.75-0.11

-0.

07

60.19 -0.

21

E-6.93 6.93

-12.88

-15.50

-16.29

-16.43

-16.45

10.87

13.98

14.29

14.34

14.35

7.177.17

7.15

6.99

5.44

1.22

0.022

-0.43-0.49

-0.5

-6.75

-8.12

-8.53-8.61

-8.62

5.446.99

7.15

7.177.17

-0.105-0.

21

-40.11

-0.105

0.

08

0.11

4.14

5.33

5.45

5.46

5.46

-47.04 47.04-11.00

-11.26

-11.28

-11.28

-11.28

47.04

-0215

-0.285

G

-14.59

-14.93

-14.96

-14.96-14.96

D -67.12

-0.19

-0.

12-67.12 67.12

6.72

7.50

7.687.75

7.74

-4.29

-5.17

-5.43

-5.48

-5.4810.6411.88

12.15

12.27

12.26

KJIH

A

-36.35

-0.305

11.098.87

8.34

8.15

8.10

7.095.67

5.33

5.21

5.27

-3.37-2.86

-2.54

-2.42

-2.4

-0.

088

12.2612.27

12.1511.88

10.64

-0.19

-0.

195

-36.35 36.35

F

12.26-8.62 7.17

-14.9


57/197

57

-21.14

E

-67.12

-36.35A

-6.93

8.1

12.26

12.26

12.26

12.2624.52

-0.5

-0.5

7.17

6.17

-47.04F

6.93

36.35 -3.75 3.75

-1. 0

-1. 0

-7. 8

-6.05

-7. 8

-7. 3

-1. 0

-20.35

-2.4

-2.4

5.2

36.75

5. 2

5. 2

-2 .4

-28.35

8. 1

8. 1

12.26

28.46

32.62

67.12

7. 74

7. 74

-5.48

-57.12

-5.48

-5.48

7.74

63 .9

-16.45

-16.4514.35

- 25.48

14.35

14.35

-16.45

19.18

-8. 62

-8. 62

-17.24

47.04

5.46

5.46

-11.28

25.48

-11.28

-11. 28

5.46

29.94

-14

-14.9-29

H I J K

BC

-3.90

-8.62

-8.62

-3.9-3.9

-8.6

-16.40

G

7.17

7.17

-17.24

12.26-8.62 7.17

-14.9


58/197

58

-57.12

-28.35A

-25.48

24.52

-47.4E

19.18

36.75 -20.35 -6.05

63.9 29.94

12.26 -8.62 7.17-14.96

BC

-16.4

32.62

D

E

6.17

13.84

G

14.34-17.24

28.46

H I JK

-29.92


59/197

Frame @ EE'

Fixed End Moment

ABM =12

2wl

=12

)96.3(5.19 2

= -25.48 kN.m

BAM = 25.48 kN.m

CDM =12

)98.1(36 2

= -11.76 kN.m

DCM = 11.76 kN.m

DEM =12

)96.3(36 2

= -47.04 kN.m

EDM = 47.04 kN.m

EFM =12

)27.4(362

= -54.69 kN.m

FEM = 54.69 kN.m


59


60/197

CCD 0.505 I

0.838 I0.60 -0.3

CG 0.333 I 0.40 -0.2

D

DE 0.253 I

1.406 I

0.18 -0.09

DH 0.333 I 0.24 -0.12

DC 0.505 I 0.34 -0.17

DA 0.333 I 0.24 -0.12

AAB 0.253 I

0.586 I0.43 -0.215

AD 0.333 I 0.57 -0.285

BBA 0.253 I

0.586 I0.43 -0.215

BE 0.333 I 0.57 -0.285

E

EF 0.234 I

1.153 I

0.20 -0.1

ED 0.253 I 0.22 -0.11

EI 0.333 I 0.29 -0.145

EB 0.333 I 0.29 -0.145

FFE 0.234 I

0.567 I0.40 -0.2

FI 0.333 I 0.60 -0.3

60


61/197

61

AB

-25.48 -0.

215

-0.285

-25.48 25.48-0.

215

25.48

-0.285

4.66

6.25

6.47

6.58

6.59

-6.48

-7.24

-7.69

-7.74

-7.75

6.17

8.288.58

8.73

8.75

-0.12

-0.

17

-35.28 -0.

09

D

-47.04 47.04

2.86

2.17

1.96

1.92

1.90

1.47

2.89

3.01

3.02

3.02

3.98

3.98

3.96

3.811.94

-8.59

-9.59-10.19

-10.26

-10.27

3.812.89

2.62

2.55

2.53

1.94

3.81

3.963.98

3.98

-0.145-0.

11

-7.65

-0.145

0.

1

0.12

1.34

2.63

2.73

2.75

2.75

-54.69 54.69

-11.21

-11.46

-11.48

-11.49

-11.49

54.69-02

-0.3

F

-16.81

-17.19

-17.23-17.23

-17.23

C

-11.76

-0.2

-0.

3-11.76 11.76

3.528

1.912.3

2.4

2.4

5.39

4.09

3.71

3.62

3.592.352

1.27

1.53

1.61

1.62

JIHG

-0.19

2.53

2.55

2.62

2.893.81 E


62/197

62

B

A-25.48 25.48

6.59

6.59-7. 75

-20.05

-7.75

-7.756.59

16.57

13.818.75

2.53

2.53

-47.04 47.04

1. 90

1. 90

3. 02

-40.22

3.02

3.02

1.90

54.98

2.532.53

5.06

B

-10.27

-10.27

3.98

-16.56

-5.12

-5.12

-22.63

-54.69E 54.69

2. 75

2. 75

-11.49

-60.68

-11.49

-11.49

2.75

34.46

3.983.98

7.96

F

-17.2

-17.2

-34.4

D

3. 59

3. 59

2. 4

21.34

2.4

2.43.59

-3.37

1.62

1.623.24

-11.76 11.76

G H I J

1.62 2.53 -3.98 -17.2

8.75

8.75

2.53

20.03

13.81 -2.31

-10.27

3.983.98


63/197

63

C

A-20.05 16.57

-40.22 54.98

B

-60.68E 34.46

F

D-3.37 21.34

G H I J

1.62 2.53 -3.98 -17.23

13.81 2.31

7.963.24 5.06

-34.46

20.03 -16.57


64/197

Frame @ FF'

Fixed End Moment

CDM =12

2wl

=12

)96.3(5.19 2

= --25.48 kN.m

DCM = 25.48 kN.m

BEM =12

)96.3(36

2

= -47.04 kN.m

BBM = 47.04 kN.m

EGM =

12

)27.4(36 2

= -54.69 kN.m

GEM

= 54.69 kN.m

64


65/197

Joint Member R.S (I/l) T.R.S D.F R.F

B

BA I/3

0.92 I

0.36 -0.18

BC I/3 0.36 -0.18

BE I/3.96 0.28 -0.14

CCB I/3

0.58 I0.57 -0.285

CD I/3.96 0.43 -0.215

D DC I/3.96 0.58 I 0.43 -0.215

DE D/3 0.57 -0.285

E

ED I/3

1.15 I

0.29 -0.145

EB I/3.96 0.22 -0.11

EG I/4.27 0.20 -0.10

EF I/3 0.29 -0.145

GGE I/4.27

0.56 I0.41 -0.205

GH I/3 0.59 -0.295

65


66/197

D

-0.

16

60.46

-0.34

-8.30-9.13

-9.17

-9.84

-9.86

1.36

3.13

3.44

3.48

3.49

-0.145

-0.

11

-7.69

-0.145

0.

10

0.94

2.16

2.37

2.40

2.40

-47.04

-11.40

-11.65

-11.69-11.70

-11.70

54.69

-0205

-0.295

G

-16.41

-16.77

-16.83

-16.84

-16.84

B -47.04

-0.18

-0.

14 -47.04

6.58

5.76

5.28

5.19

5.178.46

7.40

6.796.68

6.65

HFA

C

-25.48

-0.285

6.65

6.68

6.79

7.40

8.46

3.65

5.32

5.49

5.61

5.64

4.85

6.93

7.29

7.44

7.48

-0.18

-0.

215

-25.48 25.48

-6.26

-6.89

-7.33

-7.42

-7.43

47.04

1.03

2.38

2.61

2.64

2.64

E

3.493.48

3.44

-3.13

1.36


67/197

B

C -25.48 25.48

7. 48

7. 486. 65

21.61

5. 64

5. 64-7. 43

-21.63 -16.23

2. 64

2. 64

5. 17

57.49

3.49

3.49

6.98

-54.69 54.69

2. 40

2. 40

-11 70

-61.59-16.84

-16.84

-33.68

G

5. 17

5. 17

2. 64

-34.06

6.65

6.65

13.3

-47.04 47.04

A F H

6.65 3.49 -16.84

-7.43

-7.43

5.64

16.26

E

-11.70

-11.70

2.40

33.69

20.78

7.48

6.65

6.65

-9.86

-9.863.47

-9.86

3.47

3.47

-2.88


68/197

B

C -21.63 16.26

-61.59 33.69G

-34.06 57.49

A F H

6.65 3.49 -16.84

E

-16.23

-2.58

D

33.686.98

21.61

-20.78

13.3


69/197

Frame @ GG'

Fixed End Moment

CDM =12

2wl

=( )

12

98.15.192

= -6.37 kN.m

DCM = 6.37 kN

.m

BEM =( )12

98.1362

= -11.76 kN.m

EBM = 11.76 kN.m


70/197


71/197

@ C

Sum of F.E.M. @ C = -6.37

Rotation Contributions of far ends,

@ B = 0@ D = 0

-6.37

M'CD = Ucd cdc MM'

+

= -0.3 [-6.37 + 0]

= 1.91 kNm

M'CB = Ucb cbc MM'

+

= -0.2 [-6.37 + 0]

= 1.27 kNm

@ B

Sum of F.E.M. @ B = -11.76

Rotation Contributions of far ends,

@ C = 1.27

@ A = 0.00

@ E = 0.00

-16.37

M'BE = -10.47 x -0.22

= 2.3 kNm

M'BC = -10.47 x -0.14

= 1.46 kNm

M'BA = -10.47 x -0.14

= 1.46 kNm

4-1

-1

-1

-1

-1

47

-0

-14

-14

-14

-14

-14

4-1

-1

-1

-1

-1

47

-0

-14

-14

-14

-14

-14


72/197


73/197

4-1

-1

-1

-1

-1

47

-0

-14

-14

-14

-14

-14

4-1

-1

-1

-1

-1

47

-0

-14

-14

-14

-14

-14

CD

EB

-

3.

4

5

A

-6.37

1. 46

1 .46

0. 00-3.45

6.37

0.000.00

1.46

7.83

0.00

0.97

0.971.51

3.45

3.99

0.971.51

1.510.00

-11.76 11.76

2. 37

2. 37

0. 00

-7.02

0. 00

0. 00

2. 37

14.13

0. 00

1. 51

1.51

0. 00

3.02

1.51

1.51

0.00

0.00

0.00


74/197

C -3.457.83 D - 7.83 3.45

3.99 -7.02 14.13 - 14.13 7.02

G

E

H

- 3.99

- 1.511.51

B

3.02 - 3.02

A F I


75/197


76/197


B

BA I/3

0.99 I

0.34 -0.17

BC I/1.52 0.66 -0.33

C

CB I/1.52

1.24 I

0.53 -0.265

CD I/3 0.217 -0.135

CE I/3.96 0.2 -0.1

E

EC I/3.96

0.58 I

0.43 -0.215

EF I/3 0.57 -0.285


77/197

CB

-3.75-0.

33

-0.17

-3.75 3.75

1.24

-0.55-1.23

-1.3

-1.3

5.43

7.477.69

7.72

7.72

-0.

265

-21.73-0.

1

-0.135

-25.48 25.48

2.049

2.822.90

2.91

2.91

-5.92

-6.08-6.10

-6.1

-6.1

E

25.48-0.

215

-0.285

-7.85

-8.06

-8.08

-8.09

-8.09

2. 766

3.81

3.92

3.93

3.93

0.64

-0.28

-0.63

-0.67

-0.67

A D F


78/197

@ B

Sum of F.E.M = -3.75

@ C = 0.00@ A = 0.00

-3.75

-3.75 x -0.33 = 1.24

-3.75 x -0.17 = 0.64

@ C

Sum of F.E.M = -21.73

1.24

0.00

0.00

-20.49

-20.49 x -0.265 = 5.43

-20.49 x -0.1 = 2.049

-20.4 x -0.135 = 2.766

@ E

Sum of F.E.M =

2.48

@ C = 2.049

@ F = 0.000

27.529

27.53 x -0.215 = -5.92

27.53 x -0.285 = -7.85


79/197

@ B

-3.75

@ A = 0.00

@ C = 5.43

1.68

1.68 x -0.33 = -0.55

1.68 x -0.17 = -0.28

@ C

-21.73

@ B = -0.55

@ E = -5.92

@ D = 0.00

-28.20

-28.2 x -0.265 = 7.47

-28.2 x -0.135 = 3.81

-28.2 x -0.1 = 2.82

@ E

2548

@ C = 2.82

@ F = 0.0028.30

28.3 x -0.215 = -6.08

28.3 x -2.85 = -8.06

@ B

-3.75

@ C = 7.47

@ A = 0.003.72

3.72 x -0.33 = -1.23

3.72 x -0.17 = -0.63


80/197

@ C

-21.73

@ B = -1.23

@ D = 0.00

@ E = -6.08-29.04

-29.04 x -0.265 = 7.69

-29.04 x -0.135 = 3.92

-29.04 x -0.1 = 2.9

@ E

25.48

@ C = 2.90@ F = 0.00

28.38

28.38 x -0.215 = -6.10

28.38 x -0.285 = -8.08

@ B

-3.75

@ C = 7.69

@ A = 0.003.94

3.94 x -0.17 = -0.67

3.94 x -0.33 = -1.3

@ C

Sum of F.E.M = -21.73

@ B = -1.3

@ E = -6.10

@ D = 0.00-29.13

-29.13 x -0.265 = 7.72

-29.13 x -0.135 = 3.93

--29.13 x -0.1 = 2.913


81/197

@ E

Sum of F.E.M = 25.48

2.91

0.00

28.3928.39 x -0.215 = -6.1

28.39 x -0.285 = -8.09

@ B

-3.75

7.72

0.00

3.973.97 x -0.17 = -0.67

3.97 x -0.33 = -1.3

@ C

-21.73

-1.3

-6.1

-29.13

-29.13 x -0.265 = 7.72

-29.13 x -0.135 = 3.93

-29.13 x -0.1 = 2.913

@ E

25.48

2.91

0.00

28.3928.39 x -0.215 = -6.10

28.39 x -0.285 = -8.09


82/197

H

- 1.51

A

B

-0.67

-0.67

-1.34

-3.753.75 C

-1.33

-1.37

7.72

1.37

7.72

7.72

-1.3

17.89

3.93

3.93

0

0

0

3.93

3.93

0.0

7.86

-6.1

-6.1

2.91

16.11

-8.09

-8.09

0.00

16.18

-8.09

-8.09

0

0

0

0.67

-0.67

0.00

0.00

0.00

D F

H

- 1.51

A

B -3.753.75 C

-1.33

-1.37

7.72

1.37

7.72

7.72

-1.3

17.89

3.93

3.93

0

0

0

3.93

3.93

0.0

7.86

-6.1

-6.1

2.91

16.11

-8.09

-8.09

0.00

16.18

-8.09

-8.09

0

0

0

0.67

-0.67

0.00

0.00

0.00

D F

-25.76

-6.1

2.91

2.91-25.48

25.48


83/197

E

A

B

-1.34

C

1.3717.89

3.93

-25.76

16.18

-8.09-0.67

D F

16.11

7.86


84/197

Frame @ II'

Fixed End Moment

BDM =12

2wl

=( )

12

61.0362

= -1.116 kN.m

DBM = 1.116 kN.m

CFM =( )12

27.45.192

= -29.62 kN.m

FCM = 29.62 kN.m

DGM =

( )

12

27.4362

= -54.69 kN.m

GDM = 54.69 kN.m



85/197

B

BA I/3

1.97 I

0.17 -0.085

BD I/0.61 0.83 -0.415

C

CD I/3

0.56 I

0.59 -0.295

CF I/4.27 0.41 -0.205

D

DB I/0.61

2.54 I

0.64 -0.32

DC I/3 0.13 -0.065

DG I/4.27 0.1 -0.05

DE I/3 0.13 -0.065

F

FC I/4.27

0.56 I

0.41 -0.205

FG I/3 0.59 -0.295

G

GF I/3

0.9 I

0.37 -0.185

GD I/4.27 0.26 -0.13

GH I/3 0.37 -0.185


86/197

CF

-29.62 -0.

205

-0.295

-29.62 29.62 -0.

205

29.62

-0.295

5.36

6.93

6.65

6.61

6.61

-7.17

-5.70

-5.58

-5.57

-5.572.91

2.92

2.88

2.94

3.45

-0.065

-0.

32

-53.57 -0.

05

D

-54.69 54.69

2.653

2.26

2.21

2.24

2.24

-6.11

-6.33

-6.35-6.35

-6.36

-9.05

-9.03

-9.01

-8.70

-10.31

-8.21

-8.04-802

-8.02

3.45

2.942.88

2.92

2.91

-8.70

-9.01

-903

-9.04

-9.05

-0.185-0.

13

54.69

-0.185

0.

1

0.065

B

-1.116

-0.085

-0.

415

-1.116 1.116

0.46

-6.58

-5.56

-5.52

-5.50

16.99

14.52

14.1914.38

14.360.094

-1.34

-1.13-1.11

-1.12

HEA

-0.19

7.72

9.98

9.56

9.52

9.52

G

-1.12 2.91-9.05


87/197

F 29.62-29.62 D - 7.83

54.69 -54.69 1.116

E

H

H

G

-5.57

-5.57

6.61

25.09

6.61

6.61

-5.57

-21.97

9.52

9.52

2.91

21.95

15.34

9.52

2.91

2.91

-26.12

-8.02

-9.05

-9.05

-6.36

-6.362.24

44.21

-9.05

-9.05

-18.1

2.24

-2.24-6.36

-56.57

D

14.36

14.365.50

35.30

-5.50

-5.5014.36

2.244

-1.116

-1.1

-1.12.24

2.91

-2.91-5.82

E A

-8.02

-8.09-9.05

-25.09


88/197

F25.09 -21.97

C

21.95

44.21 -56.57 35.33B

H

G

D

2.244

E A

-25.09

-26.12

-18.1-5.82

-2.24

-9.052.91

-1.12


89/197

Frame @ JJ'

Fixed End Moment

CDM =

12

2wl

=( )

12

73.45.192

= -36.35 kN.m

DCM = 36.35 kN.m

BEM =( )

12

73.4362

= -67.12 kN.m

EBM = 67.12 kN.m


76


90/197

B

BA I/3

1.29 I

0.43 -0.215

BE I/9.46 0.14 -0.07

BC I/3 0.43 -0.215

CCB I/3

2.27 I0.76 -0.38

CD I/9.46 0.24 -0.12

77

-36.35 -0.

12

-0.38

-36.35

4.36

2.99

2.87

2.8713.819.46

9.11

9.08

CD

E

12.47

12.39

11.46

-67.12

-0.215

-0

.07

-0.215

B -67.12

3.73

4.04

4.06

4.0611.46

12.39

12.47

12.47

A


91/197

78

CD

EB

-

3.

4

5

A

-36.35

2.87

2.870. 00

-30.61

0.00

9.089.08

12.47

30.63

34.029.08

12.4712.47

0.00

0. 00

12.47

12.47

0. 00

24.94

12.41

12.47

0.00

0.000.00

-67.12

4.06

4.06

-59.00


92/197

79

C

B

-

3.

45

A

-30.61

-59.00 59.00

-30.61

-30.63

-34.02

-30.63

34.02

-24.9424.94

12.47 - 12.47

D

E


93/197


94/197


95/197

-36.35

12.47

-23.88

-23.88 x -0.12 = 2.87-23.88 x -0.38 = 9.08

@ E

-67.12

9.08

-58.04

-58.04 x -0.07 = 4.06

-58.04 x -0215 = 12.48

82


96/197

Table: 1

Moment for each frame

Frame MemberMax. B.M

Design MomentPositive

(wl2/8)

Joint Moment

Negative

AA'

A1 A2 90.69 48.63 50.4 48.17 50.4

A3 A4 107.16 43.97 85.97 42.19 85.97

A4 A5 167.44 113.27 107.09 57.26 113.27

A5 A6 70.67 63.93 24.66 26.37 63.93

BB'

B1 B2 58.04 26.32 52.48 18.64 52.48

B2 B3 90.69 57.82 53.68 34.94 57.82

B3 B4 38.22 42.85 15.25 9.17 42.85

B4 B5 107.16 63.85 59.23 45.62 63.85

B5 B6 107.16 63.85 59.23 45.62 63.85

B7 B8 70.56 36.02 41.60 31.75 41.60

CC'

C1 C2 44.44 18.34 41.5 14.52 41.50

C2 C3 90.69 61.83 49.14 35.20 61.83

C4 C5 82.04 37.31 72.54 27.11 72.54

C5 C6 167.44 111.83 95.74 63.65 111.83

DD'

D1 D2 54.53 28.35 36.75 21.98 36.75

D2 D3 5.63 20.35 6.05 -7.57 20.35

D4 D5 100.67 57.12 63.9 40.51 63.9

D5 D6 10.39 25.48 19.18 -11.94 25.48

D6 D7 70.56 47.4 29.94 31.89 47.4

83


97/197

EE'

E1 E2 38.22 20.05 16.57 19.91 20.05

E3 E4 17.64 3.37 21.34 5.28 21.34

E4 E5 70.56 40.22 54.98 22.96 54.98

E5 E6 82.04 60.68 34.46 34.47 60.68

FF'

F1 F2 38.22 21.63 16.26 19.27 21.63

F3 F4 70.56 34.06 57.49 24.78 57.49

F4 F5 82.04 61.59 33.69 34.4 61.59

GG'

G1 G2 9.55 3.45 7.83 3.91 7.83

G2G3 9.55 7.83 3.45 3.91 7.83

G4 G5 17.64 7.02 14.13 7.06 14.13

A2 B2 17.64 14.13 7.02 7.06 14.13

HH'H1 H2 5.63 1.37 17.89 -4.00 17.89

H2 H3 38.22 25.76 16.11 17.28 25.76

II

'

I1 I2 44.44 25.09 21.97 20.91 25.09

I3 I4 0.90 44.21 56.57 -49.49 56.57I4 I5 82.04 35.33 22.44 63.25 63.25

JJ'J1 J2 54.53 30.61 30.61 6.69 30.61

J3 J4 100.67 59.00 59.00 -17.33 59.00

84


98/197

CHAPTER IIDESIGNS

85


99/197

CHAPTER - 2

DESIGN

2.1 Design of slab

Slabs

The most common type of structural element used to cover floors

and roofs of building are reinforced concrete slabs of different types. One-

way slabs are those supported on the two opposite sides so that the loads are

carried along one direction only. Two-way slabs are supported on all four

sides with such dimensions such that the loads are carried to the supports

along both directions.

If Ly / Lx < 2, then the slab is designed as two way slab

If Ly / Lx > 2, then the slab is designed as one way slab

Where,

Ly = Longer span dimension of the slab

Lx = Shorter span dimension of slab

Restrained slabs are referred to as slabs whose corners are prevented form

lifting. They may be supported on continuous or discontinuous edges.

1. Types of Slabs

i) One way slab

ii) Two way slab

86


100/197

i) If 2>panshortersspanlonger

the slab is designed as one way slab.

ii) If 2 v

Hence safe the stress is with in permissible limit.

93


107/197

Check for deflection control

%of reinforcement Pt =bd

Ast100

= 1201000

600100

= 0.5%

Ast (pro) Sv =Ast

As1000

150 =Ast

1000104 2

Ast (pro) = 523.5mm2

Kt = 1.1Kc = 1

Kf = 1

d

lmax = 1.1x1x1x20

= 22

d

lactual =

120

1620= 13.5

= 22 > 13.5

Hence safe against deflection

94


108/197

95


109/197

Fig : 2.1.3.1. Reinforcement Details for one way slab

2.1.4. Design of two way Slabs

1. Master Bed Room (Two adjacent sides are discontinuous)

lx = 3.96 m

ly = 4.87 m

fy = 415 N/mm2

fck = 20 N/mm2

x

y

l

l

= 96.3

87.4

= 1.23 < 2

It is designed as two way slab

Live load = 3 kN/m2

Floor finish = 1.06 kN/m2

Depth =20

span

=20

3960

= 158 mm

Over all depth = 120mm

Effective depth = 100mm

Effective Span

i) c/c distance = 3.96 + 0.23 = 4.19m

ii) Span + Effective depth = 3.96 + 0.1 = 4.06m

Effective span = 4.060m

Total service load = 7.00 kN/m2

Design load = 1.5x7.00,

96


110/197

Wu = 10.50 kN/m2

Ultimate moment

Mux = 2xux l

Muy =2

xwy l

x = 1.2 0.060

1.3 0.065

x = 0.060 + )2.13.1()060.0065.0(

(1.23 - 1.2)

x = 0.0615

y = 0.047

Mx = 0.0615 x 10.5 x 4.062

= 10.64 kNm

My = 0.047 x 10.5 x 4.062

= 8.13 kNm

Shear Force Vux =2

xulw

=2

060.4505.10

= 21.31 kN

Check for depth

d =bf

Mu

ck138.0

=100020138.0

1050.106

d = 65.74 mm < 100mm

Hence safe

97


111/197

Tension Reinforcement

Reinforcement of Shorter Span

Mu = 0.87 fy Ast d

bdf

Astf

ck

y1

10.64x106 = 0.87x415x100

100100020

4151

Ast

10.64x106 = 36105 Ast - 7.49 Ast2

Ast = 310.86 mm2

Spacing =Ast

ast1000

=

08.277

4

101000

2x

= 252.5 mm

Provide 10 mm # Bars @ 200 mm c/c distance

(Astpro = 392.69mm2)

Reinforcement at Longer Span

9.12x106 = 0.87x415xAstx 120

120100020

4151

Ast

9.12x106

= Ast - 7.49Ast2

Ast = 209.35mm2

Min Ast =100

12.0x1000x150

= 180 mm2

Spacing =Ast

Ast1000

Assume 10mm # bars.

Spacing =35.209

41000

2

= 240 mm

Provide 10mm # bars @ 250mm c/c distance

98


112/197

Reinforcement in edge strip

Astmin = 1501000100

12.0

= 180mm2

Provide 10mm # bars @ 300 c/c distance

Check for shear stress

v =bd

Vu

=1001000

1034.21 3

= 0.18 N/mm2

% of tension reinforcement Pt =bd

Ast100

=1001000

69.392100

= 0.39%

From table. 19 Page. 73, IS 456/2000

1.25 0.36

1.50 0.48

= 0.36 + )25.039.0()25.050.0(

36.048.0

= 0.42

kc = 1.3x0.42

= 0.5555 N/mm2 > 0.18

Check for deflection

Pt = 0.225

d

lbasic = 28

99


113/197

fs = 0.58x415x96.392

86.310

= 190.41

Kt = 2.0

Kc = 1

Kf = 1

d

lmax = 28x2x1x1

= 48

d

lactual =

100

3960

= 39 < 48 Hence safe

Check for crack control

1) Reinforcement provided in more than minimum % of steel.

=100

12.0x1000x120

= 144 mm2 < 392.96

2) Spacing of main reinforcement >3d

= 3x120

= 360mm

3) Diameter of reinforcement should be less than d/8

=8

920

= 15 mm

100


114/197

Fig : 2.1.4.1. Reinforcement Details for Two way slab

101


115/197

102


116/197

2.2. Design of Beams

Beams are defined as structural members subjected to transverse

load that caused bending moment and shear force along the length. The plane

of transverse loads is parallel to the plane of symmetry of the cross section of

the beam and it passes through the shear center so that the simple bending of

beams occurs. The bending moments and shear forces produced by the

transverse loads are called as internal forces.

2.2.1.Types of Beams

Depending upon the supports and end conditions, beams are

classified as below.

1. Simply supported beams

2. Over hanging beams

3. Cantilever beam

4. Fixed beam

The reinforced concrete beams, in which the steel reinforced is

placed only on tension side, are known as singly reinforced beams, the

tension developed due to bending moment is mainly resisted by steel

reinforcement and compression by concrete.

When a singly reinforced beams needs considerable depth to

exist large bending moment, then the beam is also reinforced in the

compression zone. The beams having reinforcement in compression and

tension zone is called as doubly reinforced beam.

103


117/197

2.2.2.Design procedure for beams

1. Find the cross sectional dimensions of the beam

Over all depth D =20

Span

Effective depth = Overall depth - Effective cover

2. Effective span of the beam

i. Clear Span + effective span

ii. c/c distance between two supports

The effective span of the beam is taken from the above

conditions which ever is less than that is taken as effective span of the beam.

3. Find the moment of the sections

4. Refer IS: 456 / 2000, Page No: 96, Annex G. 1.1(5)

Limiting moment Mu limit = 0.138 fck bd2

If Mu < Mu limit the section is under reinforced since the beam is

designed as singly reinforced beam If Mu > Mu limit the section is over

reinforced. Since the beam is designed as Doubly reinforced beam.5. Tension reinforcement

Refer IS : 456 / 2000, page No : 96, Annex G. 1.1(5)

Mu = 0.87 fy Ast d

bdfck

Astfy1

6. Check for shear stress

Refer IS : 456 / 2000, Page No : 72, Clause 40.1

Nominal shear stressbd

Vuv=

7. Find the percentage of steel Pt =bd

Astpro

100

Refer IS : 456 / 2000, Page No : 73, table - 19

104


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The value cis taken as based on the percentage of steel and grade of

concrete.

If c < v, minimum shear reinforcement is to be provide.

Refer IS : 456 / 2000, Page No : 48, Clause 26.5.1.6

fybs

A

v

st

87.0

4.0

If c < v, shear reinforcement is to be provide. in following method.

IS :456 / 2000 page no : 73, Clause 40.4

Strength of shear reinforcement Vus [vu - c bd]

for vertical stirrups Vus = v

sv

s

dAfy87.0

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2.2.3. Design of singly Reinforced Beam

Frame DD'

Span = 4.73 m

Width = 230 mm

fck = 20 N/mm2

fy = 415 N/mm2

Cross sectional dimensional

Effective depth d =20

Span

=

20

4730

= 236.5 mm

Provide overall depth D = 450mm

Effective depth d = 400 mm

Moment taken form the analysis,

Mu = 63.9 KNm

Check for depth

d =bfck

Mu

138.0

=23020138.0

109.63 6

xx

x

= 317.27 mm < 400 mm

Hence safe

Area of tension Reinforcement

Mu = 0.87 fy Astd

bdfck

Afy st1

63.9 x 106 = 0.87 x 415 x Ast x 400

40023020

4151

Ast

63.9 x 106 = 144420 Ast - 32.57 Ast2

106


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Ast = 498.50 mm2

Provide 3 nos of 16mm # bars

Astpro = 3 x4

102

= 603.18mm2

Check for shear stress

v =bd

Vu.

Ultimate load Wu = 13.50

73.4

9.63

Shear load Vu = 2

lwu

= 31.95 KN

v =400230

1095.31 3

= 0.34 N/mm2

Percentage of steel Pt = bd

Ast100

= 40023018.603100

= 0.65%

0.50 0.48

0.75 0.56

= 0.48 + ( )50.065.050.075.0

48.056.0

c = 0.53 N/mm2

CV < Since minimum shear reinforcement is to be provided

Sv =b

Asvfy

4.0

87.0

Assume 8 mm # 2 legged vertical stirrups

107


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Sv =2304.0

4

8241587.0

2

= 394. 52 mm

Provide 8mm # 2 legged vertical stirrups at 300 mm c/c distance.

Size of beam = 230 x 450 mm

Mu = 36.75 KNm

Tension Reinforcement

Mu = 0.87 fy Ast d

bdfck

Astfy1

36.75 x 106 = 0.87 x 415 x Ast x 400

40023020

4151 st

A

36.75 x 106 = 144420 Ast - 32.57 Ast2

Ast = 271.03 mm2

Provide 3 nos of 12mm # bars

Provide 8mm # 2 legged vertical stirrups at 300 mm c/c

108


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109


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124/197

2.2.4.Design of doubly reinforced Beam

Frame AA'

L = 6.1 m

fck = 20 N/mm2

fy = 415 N/mm2

Cross section dimension

Effective depth =20

span

= mm30520

6100=

Over all depth D = 450mm


Breadth b = 230 mm

Moment taken from the Analysis

Mu = 113.27 KN.m

Check for depth

d =bfck

Mu

..138.0

=23020138.0

1027.113 6

= 422 mm < 450mm

Hence ok

Area of tension Steel

Mu = 0.87 fy.Ast d

dbf

fA

ck

yst

..

.1

113.27 x 106 = 0.87 x 415 Ast (400)

400230204151

xxAst

113.27x106 = 144420 Ast - 32.57 Ast2

Ast = 1018.04 mm2

Provide 4 nos of 20 mm # bars

111


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126/197

Assume 8mm # 2 legged stirrups

Sv =( )

2304.0

84

241587.0 2

= 394.mm 300mm

Provide 8mm # 2 legged stirrups300 mm c/c distance

Frame AA' @ top

fck = 20 N/mm2

fy = 415 N/mm2

Cross section dimension

Effective depth =20

Span

= mm30520

6100=

Over all depth D = 450mm


Breadth b = 230 mm

Moment taken from the analysis

Mu = 50.4 KN.m

Check for depth

d =bfck

Mu

.138.0

=23020138.0

104.506

= 281.77 < 450 mm

Tension reinforcement

Mu = 0.87 fy.Ast.d

dbfck

fyAst

..

.1

113


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50.4x106 = 0.87 x 415 Ast (400)

40023020

4151

Ast

50.4x106 = 144420 Ast - 32.57 Ast2

Ast = 381.86 mm2

Provide 4 nos of 12mm # bars.

Provide 8mm # 2 legged stirrups of 300 mm

114


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Fig : 2.2.4.1. Reinforcement Details for doubly reinforced Beam

115


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116


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2.3. Design of curved Beam

Thickness of slab = 120 mm

Radious R0 = 2.43 m

L.L = 3 KN/ m2

fck = 415 N/mm2

fy = 415 N/mm2

Live load = 3 KN/m

Inner radious R = 1.97 m

Weight of slab = 0.12 x 25

= 3 KN/m2

Total load = 3+3

= 6 KN/m2

Total load from Slab per meter =R

wR so

2

2

=21.22

643.2 2

= 8.01 KN/m

Assume size of Beam = 230 x 450 mm

Self weight of beam = 0.23 x 0.45 x 25

= 2.58 KN/m

Total load = 8.01 + 2.58

= 10.59

Ultimate load Wu = 10.59 x 1.5= 15.88 KN

q =2

2

6

5.3

b

h

=2

2

23.06

4.05.3

117


131/197

= 1.76

C =

4= 1.27

Critical moment at mid span

at = 0 the mid Span

Moc = WR2 ( cos -1)

WR2 = 15.88 x 1.972

= 61.63 KN.m

Moc = 61.63 (1-27-1)

= 16.64 KNm

At = 900 at the support

Mof = WR2 (cos -1)

= - 61.63 KNm

Mof = WR2 (c sin -1)

= 61.63 (1.27 - 1.57)

= -18.48 KNm

M = 61.63 KNmT = 18.48 KNm

Design of section

Max. Bm Mc = m + M1

Mt =

+

7.1

/1 bDu

= 18.48

+

7.1

23.0

45.01

= 32.14 KNm

Me = 61.63 + 32.14

= 93.77 KNm

118


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= 1 KNm

Check for depth d =23020138.0

77.93 6

xx

x

= 384.34 < 400 mmHence Safe

The areas of the bottom and top reinforcement

Mu = 0.87 fy Ast d

bdfck

Astfy1

16.64 x 106 = 0.87 x 415 x Ast x 400

40023020

4151

Ast

16.64 x 106 = 144420 Ast - 32.57 Ast2

Ast = 118 mm2

Minimum Ast = 0.12% cross sectional area

= 230450100

12.0xx

= 124.2 mm2

Provide 2 nos of 10 mm # bar at bottom

Mu = 0.87 f y Ast d

bdfAf

ck

sty1

93.77x106 = 0.87 x 415 x Ast x 400

40023020

4151 st

A

93.77x106 = 144420 Ast - 32.57Ast2

Ast = 790mm2

Provide 4 nos of 16mm # bars.

Ast pro = 804mm2

119


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Design for Shear

Pc =bd

Ast100

= 400230804100

= 0.87%

0.75 0.56

1.0 0.62

= 0.56 +)75.01(

)56.062.0(

(0.87-0.75)

c = 0.59 N/mm2

Shear Capacity of the section Ve = c bd

= 0.59 x 230 x 400

= 54.28x103N

Critical Shear force

V = w(R-d)

= 15.88 (1.9) ( )2/ - 400

= 19.65x103N

Tersional force at the section which should be converted in to

equivalent shear.

Ve = V + 1.6b

T

= (19.65x103) + 1.6

30.2

1048.18 6

= 19.77x106N

Nominal Shear Stress,

v =bd

ve

120


134/197

=3

3

10230

1077.19

v = 0.24 N/mm2

Minimum shear reinforcement is to be provided,

sv

sv

B

A= 87.0

4.0

yf

Sv =b

fy

4.0

87.0

Assume 10mm # of two legged vertical stirups

Asv = 24

102

= 157.08mm2

Sv =2304.0

08.15741587.0

= 616.45mm

Provide 10mm # bars at 300mm c/c distance.

121


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Fig : 2.3.1. Reinforcement Details for curved Beam

122


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137/197

EE'

E1 E2 2-16 # - 8# @ 300mm c/c

E3 E4 2-16 # - 8# @ 300mm c/c

E4 E5 3-16 # 1-16 # 8# @ 300mm c/c

E5 E6 3-16 # 1-16 # 8# @ 300mm c/c

FF'

F1 F2 2-16 # - 8# @ 300mm c/c

F3 F4 3-16 # 1-16 # 8# @ 300mm c/c

F4 F5 3-16 # 1-16 # 8# @ 300mm c/c

GG'

G1 G2 2-12 # - 8# @ 300mm c/c

G2G3 2-12 # - 8# @ 300mm c/c

G4 G5 2-12 # - 8# @ 300mm c/c

A2 B2 2-12 # - 8# @ 300mm c/c

HH'H1 H2 2-12 # - 8# @ 300mm c/c

H2 H3 3-12 # 1-16 # 8# @ 300mm c/c

II'

I1 I2 3-16 # 1-16 # 8# @ 300mm c/c

I3 I4 3-16 # 1-16 # 8# @ 300mm c/c

I4 I5 3-16 # 1-16 # 8# @ 300mmc/c

JJ'J1 J2 2-16 # - 8# @ 300mm c/c

J3 J4 2-16 # - 8# @ 300mm c/c

124


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2.4. Design of Plinth Beam

1. Span = 6.09m

Deed load = 15.18 KN/m

Effective depth =20

6090

= 304.5mm

Overall depth D = 300 mm

Effective. depth d = 250mm

Effective span :-

i). c/c distance = 6.09 + 0.23 = 6.32m

ii) Span + Effective. depth = 6.09 + 0.25 = 6.34m

Effective. Span = 6.32 m

Load Calculations

Self weight of Beam = 0.23 x 0.3 x 25

= 1.725 KN

Dead Load = 15.18 KN/m

Total Load = 16.905 KN/m

Max .ultimate load = 25.35 KN/m

Max. ultimate moment =8

2Wl

=8

32.635.25 2x

= 126.60 KNm

Mulimit = 0.138 f ckbd2

= 0.138 x 20 x 230 x 4002

= 101.56

Mu > Mu limit

Balanced Moment = 126.60 - 101.56

125


139/197

= 86.93 KNM = 25.04

fsc = 361

ASc =250

04.25

= 198.17 mm2

Provide 6 nos

Ast2 = fyfAscsc

87.0

.=

41587.0

361198

= 197.97

Ast1 = fyitxbfck u

87.0

)lim(36.0

=41587.0

)40048.0(2302036.0

x

= 880.63mm2

Total Ast = 1078 mm2

Provide 6 nos of 16 mm # bars

Shear reinforcement

V = bd

Vu

Vu =2

Wul

=2

32.635.25 x

= 80.106 KN

=400230

10106.803

x

x

= 0.87 N/mm2

Pt =bd

Ast100 =400230

1206100

=1.3

1.25 -> 0.67

126


140/197

1.5 -> 0.72

= 0.67 + )25.13.1(25.15.1

62.072.0

= 0.69 N/mm2

Vus = (80.106 x 103 - 0.69 x 230 x 400)

= 16.626 x 103

Spacing = 310626.16

40010141587.0

= 877 mm

Provide 8mm # 2 legged vertical stirrups at 300mm c/c

Deflection

actuald

l

=400

6090= 15.22

Safe against deflection

127


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128


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Fig : 2.4.1. Reinforcement Details for Plinth Beam

2.5.Design of lintels

Lintels are horizontal structural elements provided over the

opening on walls (doors, window etc.,) to carry the masonry over them. When

they are very near to the roof level, they have to carry the loads transmitted

from the roof also.

They are designed, as small rectangular beams of width always

equal to the thickness of wall in which they are provided. Since lintels are of

minor structural importance, the minimum requirements of reinforcement

specified for beams need not be strictly complied with in lintels.

A minimum of 20mm nominal cover may be provide for the

main reinforcement bars of lintels. The lintels shall have a bearing of at least

150mm on walls at their ends.

129


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2.6.Design of Lintel Cum Sunshade

step1. Design Constant and limiting depth of N.A

fy = 415 N/mm2

fck = 20 N/mm2

for Fe 415 steel

d

xu max = 0.479

Ru = 0.36

fck =d

xu max (1-0.416d

xu max )

= 0.36 x 20 x 0.479 (1-.416 x 0.479)

= 2.761

Design of sunshade

Live Load = 1 KN/m

Dead Load = 1 x 1 x 0.1 x 25 = 2.5 KN/m

Total = 3.5 KN/m

Factored load = 1.5 x 3.5 = 5.25 KN/m

Mu =2

2lWu

=2

)45.0(25.5 2

= 0.531 kNm

Vu = 5.25 x 0.45 = 2.36 KNm

d =1000761.210531.0

6

= 13.86mm

d = 80 mm

D = 100 mm

130


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Mu = 0.87 f y Ast.d

dbf

fA

ck

yts

..

.1

0.538 x 106 = 0.87 x 415 Ast (80)

80100020

4151 st

A

0.538 x 106 = 28884 Ast - 7.491 Ast2

Ast = 18.71mm2

Ast (min) = 10012.0 x 1000 x 100 = 120 mm2

Spacing =st

st

A

a1000=

120

)(4

10002

d

= 418

Provide 8 mm # bars @ 250 mm c/c

Ast (Pro.) = 201 .06 mm2

Check for shear

V =bd

Vu

=8010

1036.2 3

= 0.0029 N/mm2

Pt =bd

Ast100

=801000

120100

= 0.15 %

c = 0.28 N/mm2

K = 1.3

K. c = 0.195

K. c > n

Hence safe

131


145/197

Check for development length

Ld = 47

= 47 x 8

= 376 mm

xu =bf

Af

ck

sty

.36.0

87.0

=10002026.0

06.20141587.0

= 10.08mm

Mu = 0.87 fyAst (d - 0.416 xu)

= 0.87 x 415 x 201.06 (80 - 0.416 x 10.08)

= 5.50 x 106 Nmm

Provided 900 bend and clear cover 35mm

L0 =2

sL - x1 + 3

=2

300- 35 + 3 x 8

= 139mm

1.3.u

u

V

M+ L0 = 3

6

1036.2

1050.53.1

= 3029.66 > Ld

Hence Safe

Design of Lintel Beam

Self weight of slab = 0.12 x 1x 2.5 = 3 kN/m

B.w. Load = (1.8 x 0.23 x 0.5) 25

= 4.65 kN/m

Self wt. of lintel beam = 0.23 x 0.45 x 1

= 0.1035 kN/m

132


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Dead load of sunshade = 1 x 1 x 0.1 x 25

= 2.5 kN/m

Finish Load = 1 x 1x 0.025 x 2

= 0.55 kN/m

Total load = 10.80 kN/m

Factored Load = 16.2 kN/m

Mu =8

2lWu

=8

)8.1(2.16 2

= 6.56 kNm

d =Fasd

Mu

=230761.2

1056.6 6

= 101.63mm

d = 120mm

D = 150mm

Main Reinforcement

6.56 x 106 = 0.87 x 415 Ast (120)

120100020

4151 st

A

= 43326 Ast - 7.491Ast2

Ast = 155.59mm2

Assume 10mm# bar, use

Check for deflection

Pt =bd

Ast100

=120230

59.155100

133


147/197

= 0.56%

Modification factor = 1.05

d =05.120

L

= 85.714

= 85.714 < 120mm

Check for shear

Vud =2

.LWu -

+ 261.0

2

30.0uW

=2

)8.1(102.16 3- (16.2x103) x 0.411

= 7921.8 N

v =bd

VuD

=120230

8.7921

= 0.028 N/mm2

Pt =bd

Ast100

=120230

57.155100

= 0.56%

c = 0.499 N/mm2

Providing 8mm legged stirrups

Sv =b

fA ysv.175.2

=230

4155.100175.2

= 394mm

Max

134


148/197

Sv = lesser of 0.75 d

Hence provide 8mm # 2 legged stirrups @ 150mm c/c

Check for development length

xu =bf

Astf

ck

y

.36.087.0

=2302036.0

59.15541587.0

= 33.92mm

Mu = 0.87 fy Ast (d-0.416 ux )

= 0.87 x 415 x 155.59 (120-0.416x33.92)

= 5.9 x 106

Nmm

Vu =2

.Lu

=2

8.12.16

= 14.58 kN

Use 10mm dia. bar,

Ld = 47

= 47 x 10= 470mm

L0 =2

sL - x1

=2

300-30

= 120mm

1.3u

u

V

M

+L0 = 3

6

1058.14

1094.53.1

+120

= 649.62mm > Ld

Hence ok.

135


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Fig : 2.6.1. Reinforcement Details for Lintel Cum Sunshade

136


150/197

2.7.Design of staircase With single beam

1. Assume the rise and tread of the step; normally Rise of the steps is 150

to 250 mm. Tread is 250 to 350mm.

2

b.tech civil pyuhkjroject1

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