Model Project for replacement of fine aggregate

ANALYSIS AND DESIGN OF MULTICOMPLEX

BUILDING AT TIRUNELVELI

A DESIGN PROJECT REPORT

Submitted by

NAME REG . NO

NAME REG . NO

NAME REG . NO

NAME REG . NO

in partial fulfilment for the award of the degree

of

BACHELOR OF ENGINEERING

IN

CIVIL ENGINEERING

XXXXX COLLEGE OF ENGINEERING,

CHENNAI - 6

ANNA UNIVERSITY: CHENNAI 6

ANNA UNIVERSITY: CHENNAI

DOWNLOADED FROM http://annacivil.blogspot.com

http://annacivil.blogspot.in

BONAFIDE CERTIFICATE

Certified that this project report DESIGN OF MULTICOMPLEX

BUILDING is the bonafide work of NAMES who carried out the project under my supervision.

SIGNATURE SIGNATURE

HEAD OF THE DEPARTMENT SUPERVISOR

Civil Engineering Civil Engineering

XXXXXXXXXXX XXXXXXXXXXXX

College of Engineering College of Engineering

Submitted to the viva voce held at XXXXXXXXXXXXX COLLEGE OF

ENGINEERING on..............................

INTERNAL EXAMINER EXTERNAL EXAMINER



CONTENTS

PAGE NO

1 INTRODUCTION

1.1 Objectives 1.2 Analysis of framed structure 1.3 Design of structures 1.4 Slab 1.5 Beam 1.6 Column 1.7 Footing

2 PLAN, ELEVATION AND SECTION

2.1 Ground floor plan 2.2 First floor plan 2.3 Second floor plan 2.4 Third floor plan 2.5 Fourth floor plan 2.6 Fifth floor plan 2.7 Sixth floor plan 2.8 Seventh floor plan

3 ANALYSIS OF MULTISTOREY FRAME

3.1 Frame analysis

4 DESIGN

4.1 Design of slab

4.2 Design of beams

4.3 Design of column

4.4 Design of footing

5 CONCLUSION

6 REFERENCE

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 19 35 40 54 61 62



LIST OF DRAWING

I. Plan II. Elevation III. Section IV. Slab detailing V. Beam detailing VI. Column detailing VII. Footing detailing

LIST OF TABLES

I. Beam maximum bending moments

II. Beam maximum shear forces

III. Column maximum axial forces

IV. Column maximum bending moments



ACKNOWLEDGEMENT

We thank the almighty for his kind guidance all throughout the tenure of this work

We take this opportunity to express our thanks to our beloved principal

Dr.S.Joseph Sekhar M.E., Ph.D and our correspondent Rev.Fr.Jesu Marian for

their co-operative guidance and encouragement.

We are extremely grateful to our beloved H.O.D, civil Engg, and Dr.S.Carmel Jawahar M.E., Ph.D for his help and encouragement.

We owe greatest debt of gratitude to our beloved guide Er.J.Sherin Nisha M.E

for her technical idea for the well completion of this project.

We sincerely acknowledgement thanks to our entire teaching and non-

teaching staff members.



ABSTRACT

This report discusses on the Design of a Multicomplex building. This project

mainly includes the analysis and design of the building. The load conditions as

per code IS456 are assumed. The slab and beams are designed for bending,

shear and deflection limit.

The plan, Elevation, Section and Reinforcement retails are drawn in AutoCAD.

The analysis made in STADD Pro. Designs are made manually by using IS 456:

2000.

Design is carried out manually and it is based on IS 456- 2000. And detailing

was done for the structural elements as per SP 3.

Concrete grade of M20 and steel HYSD bars of grade Fe415 are used. Safe

Bearing Capacity of soil is taken as 180KN/m2. Footing is designed as Circular.

Plan and detailing are enclosed in this report.



SYMBOLS

A - Total area

Ast - Area of tensile reinforcement

Asc - Area of compression reinforcement

D - Overall depth of beam or slab

d - Effective depth of beam or slab

Leff - Effective span of beam, slab or column

B.M - Bending moment

R.M - Resisting moment

m - Modular ratio



fy - Characteristic strength of steel

V - Shear force

Z - Modules of section

cbc - Permissble stress in concrete in bending compression

st - Permissble stress in tensile steel

fck - Chracteristic strength of concrete

B - Breadth of footing

L - Length of footing

Pu - Ultimate load

W - Total load

v - Normal shear stress

c - Permissble shear stress

Ac - Area of concrete

d - Diameter of steel

DL - Dead load

LL - Live load

WL - Wind load



1

CHAPTER 1

INTRODUCTION

1.1. GENERAL Soils having low bearing capacity are found in several parts of the world.

In Tamilnadu also low bearing capacity soil occur in many parts. The

basic problems associated with these types of deposits are low shear

strength and high compressibility. Whenever poor soil conditions are at

site such as, loose sand, soft clay, highly organic deposit or dumped

heterogeneous material are encountered the following are the alternatives

to overcome the problems.

i. Avoid the site by relocating the proposed structure at a site with

better soil conditions.

ii. Adopt a foundation system that recognizes the inadequacies of the

soil and transfers load to it in a manner that the inadequacies are

unable to cause any harm to the superstructure

iii. Adopt a foundation system that by-passes the poor soil and

transfers the load from the superstructure to better soil located from

below the poor soil.

iv. Improve or modify the properties of the soil either by excavating

the poor quality soil and replacing it with soil having better

engineering properties or by in-situ treatment without excavation

these processes are known as ground improvement or ground

modification.

Due to growth of population and scarcity of land, construction of

structures in poor soil is unavoidable.



2

1.2. GROUND IMPROVEMENT AND MODIFICATION

The terms ground improvement and modification refer to the

improvement or modification to the engineering properties of the soil so

that soil posses properties that are acceptable to us for the proposed civil

engineering activity. The improved or modified soil exhibits satisfactory

performance when foundations, earth retaining structures are constructed

on within or using the soil.

The term Ground improvement means a permanent or long term

improvement and the term Ground modification means a temporary or

short term modification effected for the construction stage only.

Improvements in soil behaviour required are listed below.

i. An increase in the bearing capacity.

ii. A reduction in the amount of settlement and on the time in which it

occurs.

iii. An increase in the capacity to retard seepage.

iv. Acceleration in the rate at which drainage occurs.

v. Elimination of the possibility of liquefaction.

vi. An increase in the suitability of a slope or a vertical cut or an

underground opening.

1.3. IMPROVEMEND BY MIXING WITH ADDITIVES There are various methods of modification that are possible to

achieve improvement in soil properties. Of all the available

methods, improving the properties by mixing with additives such

as



3

1.1 OBJECTIVES

The objectives of project are

To prepare drawings for the building.

To analyse and Design a multi-storeyed building of eight storey.



4

1.2 ANALYSIS OF FRAMED STRUCTURE

The method by which multicomplex building frames resist horizontal

or lateral forces depends on how the structures has be laid down or planned to

bear these loads.

1.2.1 METHODS OF ANALYSIS

Analysis of the whole frame by using STADD Pro.



5

1.2.1 a. MAXIMUM BENDING MOMENTS IN BEAMS

The magnitude of bending moments in beams and columns respectively depend upon their relative rigidity. Generally the beams are made of the same

dimensions in all the floors, while the dimensions of column vary from storey to

storey. Columns have smallest dimensions at the top and largest dimensions at

bottom, due to this the ratio of the rigidity of the beam to that of the column is

larger in the upper floors than in lower floors. Beam in all floors are made of the

same dimensions and provided with same amount of steel, only one substitute

frame may be sufficient when placed in a position in the structure for which the

B.M is largest.

1.2.1 b. MAXIMUM BENDING MOMENT IN COLUMNS

The bending moment in columns increases with increase in their rigidity. Hence they are largest in the lower storeys, and smallest in upper

storey. The maximum compressive stress in columns is found by combining

maximum vertical loads with the maximum bending moments.



6

1.3. DESIGN OF RCC STRUCTURES

Reinforced cement concrete members can be designed by one of the following methods.

a) Working Stress method

b) Limit state method

1.3.1 LIMIT STATE DESIGN

Limit state method of design is based on the plastic theory.

Partial safety factors are used in this method to determine the design

loads and design strength of materials from their characteristic values.

The design aids to IS: 456 published by the Bureau of Indian Standards

made by the design by limit state method very simple and hence this

method is being widely used in practice.

This method gives economical results when compared with the

conventional working stress method.

1.3.2. WORKING STRESS METHOD

This is conventional method adopted in the past in the design of

R.C structures.

It is based on the elastic theory in which the materials, concrete and

steel, are assumed to be stressed well above their elastic limit under

the load.



7

1.4. SLABS

Slabs are the primary members of a structure, which support the imposed loads directly on them and transfer the same safely to the

supporting elements such as beams, walls, columns etc.

A slab is a thin flexural member used in floors and roofs of structures to support the imposed loads.

CLASSFICATION OF SLAB

Solid slab

Hollow slab or voided slab

Ribbed slabs



8

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.

The size of a beam is designed considering the maximum B.M in it and generally kept uniform throughout its length.

IS 456:2000 recommends that minimum grade of concrete should not be less than M20 in R.C works.

1.5.1. BREADTH OF BEAM

It shall not exceed the size of the supports. Generally the breadth of a beam is kept as 1/3 to 2/3 of its depth.

1.5.2. DEPTH OF BEAMS

The depth of the beam is to be designed to satisfy the strength and stiffness requirement.

It also satisfy sufficient M.R. and deflection check as recommended in IS456:2000

For preliminary analysis purpose over all depth of beam may assumed 1/10 to 1/12 of clear span for simply support 1/7 to 1/5

for continuous and cantilever beam.



9

1.6. COLUMN

Members in compression are called column and struts.

The term column is reserved for members who transfer loads to the ground. Classification of column, depending slenderness

a. Short column

b. Slender column

a. Short column

IS 456:2000 classifying rectangular columns as short when ratio of

the effective length (Le) to the least dimensions less than 12. This

ratio is called slenderness ratio of the column.

b. Slender column

The ratio of Le to the least dimensions is greater than 12 is called

slender column.

End conditions Effective length factor

1. Both end fixed - 0.65L

2. One end fixed-one end hinged - 0.80L

3. Both end hinged - 1.00L

4. One end fixed one end free - 2.00L



10

1.7. FOOTING

Foundation is the bottom most but the most important component

of a structure.

It should be well planned and carefully to ensure the safety and

stability of the structure.

Foundation provide for R.C columns are called column base.

1.7.1. BASIC REQUIREMENT OF FOOTING

a. It should withstand applied load moments and induced reactions.

b. Sufficient area should be providing according to soil pressure.

Footings carry two different loads then it should plan carefully.

1.7.2. TYPES OF FOOTING

a. Isolated base/footing

b. Combined footing

c. Strap footing

d. Solid raft footing

e. Annular raft footing



11

2.1. GROUND FLOOR PLAN



12

2.2. FIRST FLOOR PLAN



13

2.3. SECOND FLOOR PLAN



14

2.4. THIRD FLOOR PLAN



15

2.5. FOURTH FLOOR PLAN



16

2.6. FIFTH FLOOR PLAN



17

2.7. SIXTH FLOOR PLAN



18

2.8. SEVENTH FLOOR PLAN



19

ELEVATION



20

SECTION X-X



21

ANALYSIS OF MULTISTOREY FRAME

3.1 FRAME ANALYSIS

STAAD.Pro Report

To: From:

Copy to:

Date: 02/11/2011 12:11:00

Ref: ca/ Document1

Job Information

Engineer Checked Approved

Name: Date: 23-Oct-11

Structure Type SPACE FRAME

Number of Nodes 1674 Highest Node 1676

Number of Elements 4158 Highest Beam 4158

Number of Basic Load Cases 4

Number of Combination Load Cases 10



22

Included in this printout are data for: Included in this printout are results for load cases:

Type L/C Name

Combination 5 combination load case 5 Basic Load Cases

Number Name

1 seismic load 2 wind load 3 dead load 4 live load

Combination Load Cases

Comb. Combination L/C Name Primary Primary L/C Name Factor

5 combination load case 5 1 seismic load 1.50 dead load 1.50 live load 1.50 1.50 1.50 1.50 1.50 1.50 1.50 1.50 1.50



23



24

My 889.644kN-m:1m Mz 889.644kN-m:1m 1 SEISMIC LOAD



25

BEAM Beam Maximum Moments Distances to maxima are given from beam end A.

Beam Node A Length

(m) L/C d

(m) Max My (kN-m)

d (m)

Max Mz (kN-m)

3930 1491 10.000 5:COMBINATION LOAD CASE 5

Max -ve 0.000 1.979 10.000 425.281

Max +ve 10.000 -1.615 5.000 -215.180


Max -ve 0.000 2.319 10.000 436.108

Max +ve 10.000 -2.414 5.000 -199.390


Max -ve 0.000 0.346 10.000 407.302

Max +ve 10.000 -0.484 5.000 -200.438


Max -ve 0.000 1.062 10.000 413.841

Max +ve 10.000 -0.936 5.000 -200.359


Max -ve 0.000 1.461 10.000 409.856

Max +ve 10.000 -1.350 5.000 -200.002


Max -ve 0.000 1.781 10.000 406.231

Max +ve 10.000 -1.755 5.000 -200.051


Max -ve 0.000 1.502 10.000 401.913

Max +ve 10.000 -1.548 5.000 -200.180



26


Max -ve 0.000 1.301 0.000 399.014

Max +ve 10.000 -1.305 5.000 -200.352


Max -ve 0.000 1.429 0.000 403.629

Max +ve 10.000 -1.627 5.000 -214.118



27

Beam Maximum Shear Forces Distances to maxima are given from beam end A.

Beam Node A Length

(m) L/C d

(m) Max Fz

(kN) d

(m) Max Fy

(kN)


Max -ve 0.000 230.857

Max +ve 0.000 -0.359 10.000 -247.742


Max -ve 0.000 231.850

Max +ve 0.000 -0.473 10.000 -246.749


Max -ve 0.000 237.406

Max +ve 0.000 -0.083 10.000 -241.198


Max -ve 0.000 236.109

Max +ve 0.000 -0.200 10.000 -242.490


Max -ve 0.000 236.978

Max +ve 0.000 -0.281 10.000 -241.621


Max -ve 0.000 237.693

Max +ve 0.000 -0.354 10.000 -240.906


Max -ve 0.000 238.530

Max +ve 0.000 -0.305 10.000 -240.069



28


Max -ve 0.000 239.523

Max +ve 0.000 -0.261 10.000 -239.081


Max -ve 0.000 243.199

Max +ve 0.000 -0.306 10.000 -235.400



29

COLUMN

Beam Maximum Moments Distances to maxima are given from beam end A .

Beam Node A Length

(m) L/C d

(m) Max My (kN-m)

d (m)

Max Mz (kN-m)


Max -ve 3.500 45.798 3.500 187.455

Max +ve 0.000 -25.159


Max -ve 3.500 39.379 0.000 97.025

Max +ve 0.000 -21.942 3.500 -4.218


Max -ve 3.500 1.359 0.000 107.504

Max +ve 0.000 -3.188 3.500 -5.854


Max -ve 3.500 53.052 3.500 168.181

Max +ve 0.000 -53.491 0.000 -152.967


Max -ve 3.500 35.748 0.000 73.210

Max +ve 0.000 -46.405 3.500 -39.973


Max -ve 3.500 2.081 0.000 79.087

Max +ve 0.000 -1.714 3.500 -51.438


Max -ve 3.500 56.325 3.500 166.002



30

Max +ve 0.000 -54.858 0.000 -169.227


Max -ve 3.500 27.518 0.000 32.105

Max +ve 0.000 -24.766 3.500 -31.219


Max -ve 3.500 1.473 0.000 57.030

Max +ve 0.000 -0.394 3.500 -59.893


Max -ve 3.500 27.518 0.000 32.105

Max +ve 0.000 -24.766 3.500 -31.219


Max -ve 3.500 59.396 3.500 166.500

Max +ve 0.000 -58.365 0.000 -177.914


Max -ve 3.500 32.583 0.000 20.869

Max +ve 0.000 -28.549 3.500 -32.916


Max -ve 0.000 0.740 0.000 46.108

Max +ve 3.500 -58.906


Max -ve 3.500 32.583 0.000 20.869

Max +ve 0.000 -28.549 3.500 -32.916


Max -ve 3.500 61.939 3.500 168.501

Max +ve 0.000 -60.918 0.000 -186.126



31


Max -ve 3.500 37.176 0.000 14.355

Max +ve 0.000 -34.141 3.500 -32.551


Max -ve 0.000 1.547 0.000 36.501

Max +ve 3.500 -0.418 3.500 -55.096


Max -ve 3.500 37.176 0.000 14.355

Max +ve 0.000 -34.141 3.500 -32.551


Max -ve 3.500 64.558 3.500 174.454

Max +ve 0.000 -62.687 0.000 -194.580


Max -ve 3.500 41.438 0.000 7.323

Max +ve 0.000 -38.002 3.500 -26.356


Max -ve 0.000 2.644 0.000 24.854

Max +ve 3.500 -0.398 3.500 -44.675


Max -ve 3.500 41.438 0.000 7.323

Max +ve 0.000 -38.002 3.500 -26.356


Max -ve 3.500 64.685 3.500 184.871

Max +ve 0.000 -62.594 0.000 -199.406


Max -ve 3.500 47.700 0.000 5.545



32

Max +ve 0.000 -39.571 3.500 -14.102


Max -ve 0.000 6.338 0.000 18.375

Max +ve 3.500 -1.343 3.500 -26.546


Max -ve 3.500 47.700 0.000 5.545

Max +ve 0.000 -39.571 3.500 -14.102


Max -ve 3.500 103.950 3.500 340.859

Max +ve 0.000 -62.252 0.000 -193.960


Max -ve 3.500 72.529 0.000 14.721

Max +ve 0.000 -34.631 3.500 -24.154


Max -ve 0.000 13.873 0.000 30.317

Max +ve 3.500 -14.354 3.500 -44.532


Max -ve 3.500 72.529 0.000 14.721

Max +ve 0.000 -34.631 3.500 -24.154



33

Beam Maximum Axial Forces Distances to maxima are given from beam end A.

Beam Node A Length (m) L/C d

(m) Max Fx

(kN)

1 1 3.500 5:COMBINATION LOAD CASE 5 Max -ve 0.000 4.44E 3

Max +ve


Max +ve


Max +ve


Max +ve


Max +ve


Max +ve


Max +ve


Max +ve




34

Max +ve


Max +ve


Max +ve


Max +ve


Max +ve


Max +ve


Max +ve


Max +ve


Max +ve


Max +ve




35

Max +ve


Max +ve


Max +ve


Max +ve


Max +ve


Max +ve


Max +ve


Max +ve

3639 1305 3.500 5:COMBINATION LOAD CASE 5 Max -ve 0.000 551.286

Max +ve


Max +ve




36

Max +ve


Max +ve



37

DESIGN

4.1. DESIGN OF SLAB

DATA:

SLAB - B C1, C12,C B2

Size of floor = 5m x 10m

Materials M20 concrete and Fe415 HYSD bars

PERMISSIBLE STRESSES

cbc = 7 N/mm2 (from IS456:2000 table 21)

st = 230 N/mm2(from table 22)

For M20 concrete;

Q = 0.91

j = 0.90

m = 13.33

TYPE OF SLAB

LY / Lx = 10 / 5 = 2

Two-way slab



38

DEPTH OF SLAB

(from IS456:2000 page no.39 cl.24.1)

Span / overall depth = L/D = 35 x 0.8 = 28

5000 / 28 = 178 mm

Adopt Overall depth = 180 mm

D = 180 mm

Effective depth, d = D C /2

Adopt , d = 155 mm

EFFECTIVE SPAN

Effective span is the least of;

(i) Centre to centre of supports = 5 + 0.23 = 5.23m

(ii) Clear span + effective depth = 5 + 0.155 = 5.155 m

(iii) Le = 5.155 m

LOAD CALCULATION

Self weight = 0.18 x 1 x 25 = 4.5 KN/m2

Live load = 1.5 KN/m2

Floor finishes = 0.75 KN/m2

Total load = 6.75 KN/m2



39

BENDING MOMENT

Ly/Lx = 10 / 5 = 2

x = 0.118

y = 0.029 (from IS456:2000 table 27)

Mx = x x W x lx2

= 0.118 x 6.75 x 5.1552

= 21.13 KN.m

My = y x W x lx2

= 0.029 x 6.75 x 5.1552

= 5.19 KN.m

CHECK FOR DEPTH

Effective depth required, de = M/Qb

= 21.13 x 106 / 0.91 x 1000

= 152 mm

Depth provided Depth required

Hence safe



40

REINFORCEMENT

Reinforcement in shorter span;

Ast = Mx / st jd

= 21.13 x 106 / 230 x 0.9 x 155

= 658.56 mm2

Spacing = 1000 x ast / Ast

= 1000 x (10/2) / 658.56

= 119 mm

Adopt 10mm dia. Bars @ 120 mm spacing

Reinforcement in longer span

Ast = My / st jd

= 5.19 x 106 / 230 x 0.9 x 155

= 161.75 mm2

Spacing = 1000 x ast / Ast

= 1000 x (10/2) / 161.75

= 485 mm

Adopt 10mm dia. Bars @ 300 mm spacing



41

Reinforcement detailing



42

4.2. DESIGN OF BEAMS

DATA:

BEAM - A 12 =10 m

Effective Span = 10m

fy = 415 N/mm2

fck = 20N/mm2

Thickness of Beam = 450mm

STAAD OUTPUT

Maximum positive BM Mu(+ive) = 215 KNm

Maximum negative BM Mu(-ive) = 425 KNm

Maximum Shear force Vu = 247 KN

DEPTH

Mu lim = 0.138fckb.d2

d = 425106(0.13825300)

d = 585 600mm

D = 600+50 =650mm

REINFORCEMENT AT TOP



43

Mu = 0.87 Ast fy d 1 styck 425 106 = 0.87 Ast 415 600 1 st 41545060020 425 106 = 216630Ast 16.65 Ast2

Ast=2407mm2

Using 2 layers of 5 nos of 25mm dia bars on compression face

Ast pro = 2418.1mm2

REINFORCEMENT AT BOTTOM

Mu = 0.87 Astfy d 1 styck 215 106 = 0.87 Ast 415 600 1 st 415 45060020 215 106 =216630Ast 16.65 Ast2

Ast=1082.5mm2

Using 4nos of 25mm dia bars on the Tension face

Ast pro = 1081.2mm2



44

SHEAR REINFORCEMENT

CHECK FOR SHEAR STRESS

v =

= 247103 450600 = 0.91 N/mm2

Pt = 100Ast = 1002418

450600 = 0.89N/mm2

Ref. table 19 from IS456

c = 0.59

c < v

Then shear reinforcement are to be designed to resist the balance shear

compute as

Vus = (v c) bd

Vus = (0.91 0.59)45060010-3



45

Vus = 86.4 KN

Using 8 mm dia 2 legged stirrups

Sv =0.87

Sv = 251mm

Adopt 8 mm dia 2 legged stirrups at 251mm spacing

CHECK FOR DEFLECTION CONTROL

max =

basicKtKcKf

fs = 0.58fy

fs = 0.584152407

2418.1 = 239.6 Pt = 0.89 , fs =239.6, Fig 4 IS 456-2000 kt= 1.0

max = 26111 = 26

actual = 5000

600 = 8.3



46

max >

actual

Hence safe

REINFORCEMENT DETAILING



47

DESIGN OF SECONDARY BEAM

DATA:

BEAM AB1 = 5m

Effective Span = 5m

fy = 415 N/mm2

fck = 20N/mm2

Thickness of Beam = 450mm

STAAD OUTPUT

Maximum positive BM Mu(+ive) = 215 KNm

Maximum negative BM Mu(-ive) = 425 KNm

Maximum Shear force Vu = 247 KN

DEPTH

Mu lim = 0.138fckb.d2

d= 425106(0.13825300)

d = 585 600mm

D = 600+50 =650mm

REINFORCEMENT AT TOP



48

Mu = 0.87 Ast fy d 1 styck 425 106 = 0.87 Ast 415 600 1 st 41545060020 425 106 =216630Ast 16.65 Ast2

Ast=2407mm2

Using 2 layers of 5 nos of 25mm dia bars on compression face

Ast pro = 2418.1mm2

REINFORCEMENT AT BOTTOM

Mu = 0.87 Astfy d 1 styck 215 106 = 0.87 Ast 415 600 1 st 415 45060020 215 106 =216630Ast 16.65 Ast2

Ast=1082.5mm2

Using 4nos of 25mm dia bars on the Tension face

Ast pro = 1081.2mm2

SHEAR REINFORCEMENT


v =



49

= 247103 450600 = 0.91 N/mm2

Pt = 100Ast = 1002418

450600 = 0.89N/mm2

Ref. table 19 from IS456

c = 0.59

c < v

Then shear reinforcement are to be designed to resist the balance shear

compute as

Vus = (v c) bd

Vus = (0.91 0.59)45060010-3

Vus = 86.4 KN

Using 8 mm dia 2 legged stirrups

Sv = 0.87 Sv = 251mm



50

Adopt 8 mm dia 2 legged stirrups at 251mm spacing

CHECK FOR DEFLECTION CONTROL

max =

basicKtKcKf

fs = 0.58fy

fs = 0.584152407

2418.1 = 239.6

Pt=0.89, fs =239.6, Fig 4 IS 456-2000 Kt= 1.0

max = 26111 = 26

actual= 10000

600 = 16.67

max >

actual

Hence safe



51




52

4.3. DESIGN OF COLUMN

DATA:

COLUMN A1

Pu = 915.875 KN

fck = 20 N/mm2

fy = 415 N/mm2

Strength of compression members with helical reinforcement is 1.05 times the

strength similar members with lateral ties (IS456:2000 cl. 39.4)

Pu = 915.875 / 1.05 = 873 KN

MINIMUM ECCENTRICITY

Assume; emin = 20mm ; (IS456:2000 CL 25.4)

emin < 0.05D

20mm < 40mm

Pu = 0.4 fck + 0.67 fy Asc (IS456:2000 cl 39.3)

Assume 1 % of steel

Asc = 1/100 x Ag

Pu = 0.4 x 20 x(Ag Asc) + 0.67 x 415 x Ag / 100

Ag = 85591mm2



53

CHECK FOR ECCENTRICITY

0.05D = 0.05 x 800 = 40 mm

emin < 0.05D

Hence ok

REINFORCEMENT

Provide 18 mm dia. Bars

Asc = Ag / 100 = 85591 / 100 = 855.91mm2

No. Of bars = 855.91 / x (18 / 2)2 = 3.4

Adopt 5 nos. of 18mm dia. Bars

HELICAL REINFORCEMENT

Use 8 mm dia bars mild steel for helical reinforcement

fy = 250 N/mm2

Core dia. = Dc = 800 (2 X29.5) = 740mm

Area of core = Ac = x (Dc)2 / 4 = x 7402 / 4 = 430084mm2

0.36(Ag/Ac 1 )fck / fy = 0.36 ((800/2)2 1/430084 - 1) x 20/250= 4.859 x 10-3



54

dn = 800 (2 x 33.5) = 733mm

Let, S = Pitch of spiral

Vu = dn / S x ( x s / 4)

Vu = 115750.7204 S

Vc = Ac x 1 = 430084 x 1 = 430084

115750.7204S /430084 = 4.8596x 10-3

S = 55mm

PITCH SCHEDULE

(i) < 75mm

(ii) 1/ 6 Dc = 1 /6 x 740 = 123.33mm

(iii) >3s = 3 x 8 =24mm

(iv) S = 55mm

Hence ok



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56

4.4. DESIGN OF FOOTING

DATA:

Pu = 915.875 KN

D = 800mm

P = 180 KN/m2

fck= 20 N/mm2

fy = 415 N/mm2

pu = (1.5 x 180) = 270 KN/m2

DIMENSIONS OF FOOTING

Load on column = 915.875 KN

Self weight of footing = (10%) = 91.6 KN

Total load on soil = wu = 1007.5 KN

Let, Df = diameter of the circular footing

Af = area of the footing

Af = ( Df2 / 4) = (wu / pu) = (1007.5 / 800) = 1.26 m2

Df = (4 x 1.26) / = 1.27 m

Adopt diameter of footing = Df = 2m



57

Upward soil pressure = pu = (915.875 x 4 / x 22) = 291.5 KN/m2 < 300 KN/m2

Hence, the diameter of the footing is adequate to resist the loads.

Rx = 0.6 [R2 + r2 + Rr / R +r ] = 0.6[10002 + 4002 +(1000 x 400) /(1000 + 400)]

= 668mm

Upward load on area (b b c c) is expressed as Wq and computed as,

Wq =[ (1 0.42) 291.5 / 4 ] = 192 KN

BENDING MOMENT

Maximum bending moment at the face of the column quadrant is computed as,

Mu = 192 (0.668 0.4) = 51.5 KN.m

Breadth of footing = rc / 2 = x 0.4 / 2 = 0.63m = 630 mm

Depth of footing = d = Mu / 0.138 fck b =1.72 m

Depths required from shear considerations will nearly 1.5 times that for moment

computations.

Hence adopt effective depth = d = 250 mm = and overall depth = 300 mm

REINFORCEMENTS

Mu = [0.87fyAstd][1-{(Ast fy)/(bdfck)}]

= [0.87 x 415 Ast x 250][1 {(Ast 415) / (172 x 250 x 20]



58

= 1100mm2

Ast min = (0.0012 x 172 x 300) = 61.92 mm2

Provide 10 mm diameter bars at 150 mm spacing.


Ultimate shear force at a distance of 0.25m from the face of column is given by,

Vu = 291.5(22 1.352) (/4) = 498.5 KN

Shear per metre width of perimeter = (498.5 / x 1.35) = 110 KN

v = (Vu / bd ) = (110 x 103 / 1000 x 250 ) = 0.44 N/mm2

(100 Ast / bd) = (100 x 1100 / 1000 x 250) = 0.44 N/mm2

Refer table 19 of IS 456:2000 and read out the permissible shear stress in concrete

(ks c) = (1 x 0.45) = 0.45 N/mm2 > 0.44 N/mm2

Hence, the shear stresses are within safe permissible limits.



59




60

ESTIMATION

ITEM

NO

PARTICULARS

NO LENGTH m

BREADTH m

HEIGHT OR DEPTH

M

QTY

1 EARTH WORK EXCAVATION

190 3 3 1.5 2565m3

2 PLAIN CEMENT

CONCRETE

190 3 3 0.05 85.5m3

3 RCC FOR FOUNDATION

190 3 3 0.6 1026m3

4 BACKFILL 1453.5m3

5 DISPOSAL 1111.5m3

6 RCC FOR COLUMN

190 DIA=0.8 AREA=0.5027m2

30.5 2913.14m2

7 RCC FOR BEAM1

1022

5 0.45 0.6 1379.7m3

8 RCC FOR BEAM2

1106

10 0.45 0.6 2986.2m3

9 RCC FOR SLAB 1008

10 5 0.18 9072m3

10 BRICK WORK1 4 75.6 0.2 28 1693.44m3

11 BRICKWORK2 2 82.8 0.2 28 927.36m3

12 BRICKWORK3 10 36.8 0.2 28 2060.8m3

13 BRICKWORK4 2 65.2 0.2 28 730.24m3

14 PLASTERING 2 1020 28 57120m2

15 WHITE WASH 2 1020 28 57120m2

16 PAINTING 2 1020 28 57120m2

17 REINFORCEMENT FOR

FOOTING(10mmDIA)

11 3m 33X190=6270m 3870.kg


FOOTING(6mm DIA)

16 3m 48X190=9120m 2026.7kg



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19 REINFORCEME

NT FOR COLUMN(18m

m DIA)

5 30.5m 152.5X190=28975m

57950kg


COLUMN(10mm DIA)

102 0.8m 81.6X190=15504m

9570.4kg


BEAM1(25mm DIA)

8 10m 80X1022=81760m

315432.02kg

22 REINFORCEMNT FOR

BEAM1(8mm DIA)

34 2.1m 71.4X1022=72970.8m

28828kg


BEAM2(25mm DIA)

8 5m 40X1106=44240m

170679.0kg


BEAM2(8mm DIA)

17 2.1m 35.7X1106=39484.2m

15598.7kg


SLAB(10mm)

11 5m 55X1008=55440m

34222.2kg


SLAB(10mm DIA)

17 10m 170X1008=17136m

10577.7kg



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ITEM NO PARTICULARS QUANTITY RATE AMOUNT

1 EARTH WORK EXCAVATION

2565m3 175.00 448875 /-

2 PLAIN CEMENT CONCRETE

85.5m3 3000 256500 /-

3 RCC FOR FOUNDATION

1026m3 5500 5643000 /-

4 BACKFILL 1453.5m3 50 72675 / - 5 DISPOSAL 1111.5m3 50 55575 / - 6 RCC FOR COLUMN 2913.14m2 5500 16022270 / -

7 RCC FOR BEAM1 1379.7m3 5500 7588350 / -

8 RCC FOR BEAM2 2986.2m3 5500 16424100 / -

9 RCC FOR SLAB 9072m3 5500 49896000 / -

10 BRICK WORK1 1693.44m3 2750 4656960 / -

11 BRICKWORK2 927.36m3 2750 2550240 / - 12 BRICKWORK3 2060.8m3 2750 5667200 /- 13 BRICKWORK4 730.24m3 2750 2008160 /- 14 PLASTERING 57120m2 135 7711200 /- 15 WHITE WASH 57120m2 10 571200 /- 16 PAINTING 57120m2 115 6568800/- 17 REINFORCEMENT 648754.72kg 51 33086490 / - 18 ELECTRIFICATION 500000/-

TOTAL 309727595/-



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CONCLUSION

The multicomplex building of eight storey height is designed to be

located at Tirunelveli. The limit state and working stress method was

adopted. The design for slabs, beams, columns, footing has been

done.

It is concluded that the design by manual method satisfies the

entire requirement and it would be sufficient for the construction of

building.

In this work a eight storey multicomplex with all the facilities for

a comfortable working has been properly analysed and designed as

per IS456-2000.



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REFERENCES

1. P.C.Varghese, Limit state design of reinforced concrete

2. N.Krishna Raju, R.N.Pranesh, Reinforced concrete design

3. B.C.Punmia, Theory of structures

4. B.N.Dutta, Estimating and Costing in Civil Engg.

5. Code IS456:2000

6. Code Sp:16

7. Code IS 875:1986



DESGIN AND ANALYSIS OF MULTICOMPLEX BUILDING.pdfDATA:STAAD OUTPUTDEPTHREINFORCEMENT AT TOPREINFORCEMENT AT BOTTOMSHEAR REINFORCEMENTCHECK FOR SHEAR STRESSCHECK FOR DEFLECTION CONTROLDATA:STAAD OUTPUTDEPTHREINFORCEMENT AT TOPREINFORCEMENT AT BOTTOMSHEAR REINFORCEMENTCHECK FOR SHEAR STRESSCHECK FOR DEFLECTION CONTROL

Model Project for replacement of fine aggregate

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