Planning It Park

PLANNING, ANALYSING AND DESIGNING OF AN IT PARK WITH GREEN HOUSE CONCEPT

A DESIGN PROJECT Submitted by Project members

Reg. no

B.ARIVUNIDHI

51909103002

P.VIJAYAKUMAR

51909103020

Under The Guidance Of Mr.V.VENGAI VENTHAN. ME (structural engg) In partial fulfillment for the award of the degree

Of BACHELOR OF ENGINEERING

In CIVIL ENGINEERING

BHARATHIDASAN ENGINEERING COLLEGE, NATTRAMPALLI

ANNAUNIVERSITY CHENNAI 600 025 NOV-DEC 2012

BONAFIDE CERTIFICATE This is to certify that this project report “PLANNING AND ANALYSING AND DESIGNING OF AN IT BUIDING WITH GREEN HOUSE CONCEPT” is bonafide work of B.ARIVUNIDHI

51909103002

P.VIJAYAKUMAR

51909103020

of final year, civil engineering branch, in partial fulfillment of the requirement for the award of Bachelor Of Engineering in Civil Engineering by Anna University, Chennai.

SIGNATURE

SIGNATURE

Mr. PRIYANKA

Mr. V.VENGAI VENTHAN

HEAD OF THE DEPARTMENT

LECTURER

Civil Engineering Department

Civil Engineering Department

Bharathidasan Engineering College

Bharathidasan Engineering College

Nattrampalli 635 852

Nattrampalli 635852

Submitted for the VIVA - VOICE examination held on.............................. At BHARATHIDASAN ENGINEERING COLLEGE, NATTRAMPALLI - 635852.

------------------------

------------------------

(Internal Examiner)

(External Examiner)

ACKNOWLEDGEMENT It is our great pleasure and privilege to have an opportunity to take this project work during the course in Engineering during the year NOV/DEC 2012 We gratefully acknowledge our sincere thanks Dr. SENTHIL KUMAR M.E, PhD the Principal of our college for granting us permission to do the mini project work in our college. We also express our sincere thanks with a sense of gratitude to our respectful Head of the Department, Mr. PRIYANKA M.E,for their interest and encouragement shown in our project. We are immensely grateful to our lecturer, Mr.V.VENGAI VENTHAN, M.E for guiding our project to a great success. Special thanks to our entire department faculty members for helping us in solving our problem and doubts that we encountered in our project. We are also indebted to thank our parents, without whose constant support we would not have made a career in this field. Last but not the least, we thank our friends who have encouraged and helped us during the course of this project.

SYNOPSIS This project work deals with the Design of IT BUILDING and Study of Green Building Concept..The vertical and horizontal loads are calculated using IS456-2000 code . The design has been done according to the Limit State Method of design and conforming to Indian standard code IS 456-2000. The slab is designed according to the edge condition by limit state method. The maximum bending moment is taken and the beams are designed, by using the Axial, Uniaxial and Biaxial moments and the vertical loads of the columns and footings are designed accordingly. Column strap Footing is provided. Dog-legged staircase is provided for all the floors. The concept and significance of “GREEN BUILDING” and incorporation of certain “GREEN MATERIALS” for the deigned “IT BUILDING” is detailed.

LIST OF SYMBOLS The following symbols carrying the meanings noted against them are used in this volume A

= Area

Ast

= Area of the steel reinforcement

Asc

=Area of compression steel

BM

= Bending Moment

B

= Breadth of the beam, slab

D

= Overall depth of beam or slab

b

= Breadth of column

d

= Effective depth of beam or slab

fy

= Characteristic strength of steel

fck

= Characteristic compressive strength of concrete

l

= Length of the beam

lx

= Length of shorter span of slab

ly

= Length of the longer span of slab

lex

= Effective length of the slab along shorter span

ley

= Effective length of the slab along longer span

Mx, My

= Moments on the strip of unit width spanning ly and lx

Mux, Muy

= Moments about x and y axes due to design loads

Mux1, Muy1

= Maximum uniaxial moment capacity for an axial load of Pu

MOR

= Moment of Resistance

Pu

= Axial load on compression member

Sv

= Spacing of stirrups

V

= Shear force

Vs

= Design shear force

W

= Total load

ax

= Bending moment co-efficient along shorter span

ay

= Bending moment co-efficient along longer span

τv

= Permissible shear stress in concrete

τc

= Maximum shear stress in concrete

o

= Diameter of bars

N

= Newton

KN

= Kilo Newton

mm

= Millimeter

m

= Meter

c/c

= Center to center

Fe415

= High yield strength deformed bars

M20

= Grade of concrete

M45

= Grade of concrete

FEM

= Fixed End Moment

COM

= Carry Over Moment

LL

= Live Load

DL

= Dead Load

WL

= Wind Load

LIST OF TABLES 1. BENEFITS OF GREEN BUILDING. 2. IGBC AND LEED RATING SYSTEM. 3. REQUIREMENTS THAT ARE ESSENTIAL IN AN IT BUILDING. 4. REINFORCEMENT DETAILS 4.1Reinforcement details of slabs 4.2Reinforcement details of beams 4.3Reinforcement details of columns 4.4 Reinforcement details of footings 5. COMPARISSION OF NORMAL AND WUL’S

LIST OF ABBREVATIONS IGBC

- Indian Green Building Council

LEED

- Leadership in Energy and Environmental Design Certification

WLU

- Water less Urinals

ACP

- Aluminum Composite Panel

VOC

- Volatile Organic Compounds

HVAC

- Heating. Ventilation and Air Conditioning

GDP

- Growth Development Period

EPA

- Environmental Protection Agency

SGG

- Self Cleaning Glass

INDEX 1. INTRODUCTION

1

1.1 General 1.2 Information Technology Park 1.3 Advantages of IT-Park 1.4 Green Building Concept 1.5 Need for study 1.6 Objectives 2. REVIEW OF LITERATURE

8

2.1 General 2.2 Types of Analysis 3. STUDY AREA AND DATA COLLECTION

9

3.1 Location 3.2 Data Collected 4. METHODOLOGY

10

4.1 Description 5. PLANNING 5.1 Key plan

12 14

5.2 Plan 6. ANALYSIS AND DESIGN WITH REINFORCEMENT DETAILS 6.1 Design of Slab

16 17

6.2 Design of Beams

30

6.3 Design of lintel

39

6.4 Design of Columns

44

6.5 Design of Footing

52

6.6 Design of Staircase

56

7. GREEN MATERIALS

60

7.1 Water less urinals 7.1.1 Economic payback 7.1.2 Green building credits 7.2 Aluminum Composite Panel (ACP) 7.2.1 Economic payback 7.2.2 Green building credits 7.3 Glass 8.3.1 Economic payback 8.3.2 Green building credits\ 7.4 Paint 7.4.1 Economic payback 7.4.2 Green building credits 7.5 Carpet 7.5.1 Economic payback 7.5.2 Green building credits 7.6Other Proposal 7.6.1 HVAC Ventilation CO2 sensor 7.6.2 Chillers 8. REFERENCE

69

CHAPTER-1 INTRODUCTION 1.1

GENERAL Information Technology as it denotes to technology, IT spans a wide variety of areas that include but are not limited to things such as Processes, Computer Software, Computer Hardware, Programming Languages, and Data Constructs. In short, anything that renders Data, Information or perceived Knowledge, in any visual format whatsoever, via any multimedia distribution mechanism, is considered to be a part of the domain space known as Information Technology. As IT is applicable to organizations within enterprises, it represents an operational group that helps solve problems as those related to data, information and knowledge capture, persistence, processing, brokering, discovery and rendering. Information Technology (IT) Parks, which are also referred to as Techno Parks, Cyber Parks and Science Parks, have been established to facilitate the development of IT industries that foster new business development and technological innovation by emerging ideas within a cluster environment.

1.2

INFORMATION TECHNOLOGY PARK IN INDIA The Indian Information Technology industry accounts for a 5.19% of the country's GDP and export earnings as of 2009, while providing employment to a significant number of its tertiary sector workforce. More than 2.3 million people are employed in the sector either directly or indirectly, making it one of the biggest job creators in India and a mainstay of the national economy. India's IT Services industry was born in Mumbai in 1967 with the establishment of Tata Group in partnership with Burroughs. The first software export zone SEEPZ was set up here way back in 1973, the old avatar of the modern day IT park. More than 80 percent of the country's software exports happened out of SEEPZ, Mumbai in 80s.Today Bangalore is considered as the IT hub.

1

`1. 3

ADVANTAGES OF IT-PARK IT Parks provide infrastructure and support services for businesses, particularly high-quality (high-capacity) communications, real estate and office space. Frequently, IT Parks have links to research institutions providing a pipe line of both innovations personnel and advice. This serves as the technology transfer function. Today, IT Parks are present throughout the developing world, including India, China, Malaysia, the Philippines, Costa Rica and the Dominican Republic. To recruit and co-locate new and established knowledge-based companies, Promote innovation based on "smart" technologies, Provide an interface or shared research environment for research organizations and private industry.

1.4

GREEN BUILDING CONCEPT Green building (also known as green construction or sustainable building) is the practice of creating structures and using processes that are environmentally responsible and resource-efficient throughout a building's life-cycle: from siting to design, construction, operation, maintenance, renovation, and deconstruction. This practice expands and complements the classical building design concerns of economy, utility, durability, and comfort.

Some of the salient features of a Green Building are: 1.

Use of Recycled and Environmental Friendly Building Materials.

2.

Use of Non-Toxic and recycled/recyclable Materials.

3.

Efficient use of Water and Water Recycling.

4.

Use of Energy Efficient and Eco-Friendly Equipment.

5.

Use of Renewable Energy.

6.

Quality of Indoor Air Quality for Human Safety and Comfort.

7.

Effective Controls and Building Management Systems.

2

Purpose of Green Building 1.

Green buildings are designed to reduce the overall impact of the built environment on human health and the natural environment.

2.

Efficiently using energy, water, and other resources.

3.

Protecting occupant health and improving employee productivity.

4.

Reducing waste, pollution and environmental degradation. Evaluation bodies of Green building

1.

Indian Green Building Council (IGBC) o (Launched by Confederation of India Industry (CII) in 2001)

2.

Pioneering Green Building movement in India o Leadership in energy and environmental design certification (LEED)

Benefits

Tangible Benefits

Intangible Benefits

Electricity saving-40% to 50%

Daylight and Views

Water saving-20%-30%

Health and productivity

Tangible Cost Premiums 1.

LEED Registration and Application Costs

2.

Commissioning Agent Costs

3.

Energy Modeling Costs

4.

Improved HVAC equipment and controls

5.

Improved lighting and controls

6.

Improved building envelope or glazing

7.

Storm water quantity or quality improvements

8.

Showers, bike racks, etc. 3

1.

Plumbing fixture premiums

2.

Green Power contracts

3.

Green Kiosks

4.

Vegetative Roofing materials

Intangible Cost Premiums 1.

LEED Documentation Costs

2.

Waste Hauling Premiums and additional on site labor

3.

Construction cleanliness to meet IAQ Measures Additional babysitting on site

4.

Material premiums

Advantages of Green Buildings Meets high standards of energy efficiency and environmental responsibility. Focus is mainly on resources such as energy, water and materials and attains efficiency of these resources. It is said to reduce the energy bills and offer a healthier and more comfortable living environment. Reduce the effect of environmental hazards and ease its effects on human health and environment too. It is said that natural daylight design reduces a building's electricity needs, and improves people's health and productivity. It is the use of eco-friendly materials that highlights the concept of green building. (Purchase eco-friendly products to build a green home or a sustainable environmental building). Enhance the energy efficiency of our building.Put to use environmental friendly technology and see how energy efficient your building can get. Passive solar design can dramatically lower down the heating and cooling costs of a building, as with high levels of insulation and energy-efficient windows. It usually requires a systemic attention to the full life cycle impacts of resources which is embodied in such green building and to the resource consumption and pollution emissions over the building's complete life cycle. Green buildings are also popularly known as eco-homes or sustainable buildings. It is generally agreed that green buildings 4

are structures which are designed, sited, built, renovated and operated to energy-efficient guidelines, and that they will have a positive environmental, economic and social impact over their life cycle.Make use of low energy appliances, energy efficient lighting and renewable energy technologies which uses solar panels and wind turbines. 1.4.1 IGBC AND LEED RATING SYSTEM

IGBC Green Building Certification Levels Certification Level Points Certified 26-32 Silver 33-38 Gold 39-51 Platinum 52 or more The following are the basic pre-req. for any building to attain the green rating: (No points will be given for these requisites since they are the eligibility criteria for the certification) 1. Erosion and sedimentation control. 2. Fundamental building system commissioning. 3. Minimum energy performance. 4. CFC reduction in HVAC and R equipments. 5. Storage and collection of recyclables. 6. Minimum IAQ performance. 7. Environmental tobacco smoke control.

5

REQUIREMENTS THAT ARE ESSENTIAL IN IT-PARK: S.NO

CRITERIA

DESCRIPTIONS

1

Daylight

Indoor environmental 1 quality

2

Lighting

Controllability of system

7

POINTS/CREDITS

1

3

Chemical and pollutant source

Controllability of System

4

Composite wood,agrifiber products Low emitting Materials

1

5

Paints

Low emitting Materials

1

6

Carpet

Low emitting Materials

1

points

1

Materials and reuse

13

8

Optimize energy performance

Energy Atmosphere

And 10

9

Onsite renewable energy

Energy Atmosphere

And 3

10

Water use reduction 20%

Water efficiency

1

11

Water use reduction

Water efficiency

1

12

Development density and community connectivity

Sustainable sites

6

1

1.5 NEED FOR STUDY Due to the complexity of design and implementation, IT Parks can often take a number of years to mature and become fully sustainable, as well as requiring a significant investments in infrastructure.

1.5.1 GREEN RATED BUILDINGS IN TAMILNADU(CHENNAI) 1.

Motorola Manufacturing Facility-silver.

2.

TCS Techno park-Gold.

3.

Wipro Chennai Development Center- SEZ-Silver.

4.

Cognizant Green Campus-Coimbatore-Gold.

5.

Olympia Technology Park in Chennai.

6.

Tamilnadu Legislative Assembly in Chennai-Gold.

1.6

OBJECTIVES 1.

To plan the GREEN TECHNO-PARK building.

2.

To design the Green Techno-Park building based on edge condition.  To design the beam and column.  To design the foundation.

3.

To analyse the structure using software STAAD Pro.

4.

To study GREEN BUILDING CONCEPT and it's importance. Properties of certain GREEN BUILDING MATERIALS.

5.

To give an estimate comparison between conventional and green materials in the proposed building.

6.

To give some salient features based on GREEN BUILDING CONCEPT in our project.

7

CHAPTER 2 LITREATURE REVIEW 2.1

GENERAL In the case of Multi-storey frames, the degree of indeterminacy is very high and hence solution by consistent deformation, slope deflection, moment distribution or column analogy methods is almost ruled out. Kani's method, however, may be employed, but it needs more computational efforts. For quicker solution, design engineers use the following approximate methods of analysis

2.2

TYPES OF ANALYSIS Vertical loads are taken from the appropriate code (IS 875) Any one of the following methods for horizontal loads:

1.

Portal method

2.

Cantilever method

3.

Factor method

4.

STAAD Pro

5.

ANSYS

8

CHAPTER 3 STUDY AREA AND DATA COLLECTION 3.1

LOCATION The study area of the project is located at Kelambakkam High Road in Chennai district, Tamilnadu. The plot area of the site is 9600 m 2, the plinth area of the building is 710 m2 .The Google map of study area is given figure 1 below.

3.2

DATA COLLECTED Type of soil

: Coarse sand

Ultimate bearing capacity of soil

: 440 kN/m2

Figure 1 Google Image of Study Area

9

CHAPTER 4 METHODOLOGY 4.1DESCRIPTION First of all the plan for the building should be drafted. According to the needs and the usage of the building the loads are calculated and the slabs are then design for the loads based on its edge conditions specified. The size of the beam, area of steel reinforcement and the spacing of stirrups are calculated. The columns are designed according to the loads transmitted from the floors, from the design the size of the column, area of steel reinforcement and the spacing of lateral ties required are determined. According to the bearing capacity of the soil and the load from the super-structure the type of foundation suitable is chosen and it is designed. Based on the space, requirement and the usage of the building the type of staircase suitable is chosen and it is designed. The reinforcement details of the various structural elements should be drafted. The schematic representation of methodology of Design, Estimation & Analysis of IT-PARK is presented in figure 2.

10

Data Collection

Drafting Plan

Design Of Staircase

Software Analysis

Design Of slabs

Design of Beams, Lintels & Columns Design of Footing

Figure 2 Shows Schematic representation of the methodology of the Design, Estimation and Analysis of multi-storey building

11

PLANNING SPECIFICATION REPORT The project work has been proposed as an IT building. Each floor consists of working cabins having enough spacing provision. Each room has minimum dimension as per rules laid by government. The parking area has been provide at ground floor. Foundation The earth work excavation for foundation shall be carried out up to hard dense soil which is available at 2.4 meters. The safe bearing capacity of soil at 2.4 m is taken as 440KN/m2. Concrete R.C.C column footing shall be designed and constructed with M30 concrete, Fe415 steel, shape, dimension are followed as per detailed drawing. Filling in basement The basement of ground floor will be filled with sand, crushed dust and it is consolidated with watering and ramping. Anti termite treatment Pre construction anti termite treatment to the foundation bed, floor wall, floor junction, perimeter of the building has to be carried out through pest control of termites. Super structures Super structures elements like column & beam shall be reinforced cement concrete of M25 concrete and Fe415 steel and the main walls to super structures shall be brick work in cement mortar using best quality of first class bricks. Floor slab The floor slab of building shall be R.C.C of M25 grade concrete and Fe415 steel. The depth of slab are designed and detailed as shown in drawings. Roof slab The roof slab of building shall be RCC of M25 grade concrete and Fe415 steel. The depth of slab and reinforcement details is designed as per code provision and details are shown in diagrams. The top of floor slab exposed to weather is finished with the best qualiy of hydraulic pressed mangalore clay tiles over the weathering course of surkhi concrete and tiles are laid over CM 1:3 and pointed with CM 1:3.

12

Plastering All walls are to be plastered with CM 1:5 , 12 mm thick externally and internally. All exposed RCC works are plastered with CM 1:3, 10 mm thick. An approved water proofing compound is to be mixed with mortar for external work. Flooring Mabonite tiled flooring is laid over PCC 1:4:8 with CM 1:3. All toilets shall be finished with ceramic tiles att floor and finished with glazed tiles in all four sides. Finishes All wood works are painted with best quality synthetic enamel paint to get even shade and 3 coats over primer coat. All external and internal surfaces shall be painted with 3 coats of best quality cement paint of approved colors. Electrification Concealed wiring is carried out for all electrical lights, special fittings etc. suitable switch board & MCB are provided in each room, to have control over the electrification. A main line supply is available at outside. Sanitary and drainage arrangements For every toilet sanitary arrangements are carried out with necessary provision made for closets, urinals and wash basins and proper disposal lines are provided and connected with septic tank. All around the building, concrete pavement is provided with plinth protection from rain water and proper drains are provided to collect rain water.

13

KEY PLAN

14

DESIGN OF SLAB

15

DESIGN OF SLAB Introduction The Reinforced concrete slabs may be designed in two ways. One way Slabs Two way Slabs One way Slab The reinforced concrete Slabs may be supported on two parallel long edges only and free on any support along the parallel short edges. The Structural action of the Slab is essentially one way. The Slab in the Direction perpendicular to the supporting beams or walls, carries the loads. Such Slabs are called slab spanning in one direction. These are known as one-way slabs. Conditions for a One way Slab Consider long span as Ly and Short span as Lx. If the ratio between the Ly and Lx is greater than 2, then it is said to be a one-way Slab. Two way Slab The reinforced concrete slabs supported on its four sides on beams or walls having the ratio of a long span less than or equal to two are called as slabs spanning in two directions or two way slabs. The structural action in such slabs is in two way. The loads are carried by the slab along both short span and long span. The bending moments and deflection in two way slabs are considerably lesser than those in one way slab for similar loading and similar support conditions. The deflected surfaces of such slabs have double curvature. Conditions for a Two way slab Consider long span as Ly and Short span as Lx. If the ratio between the Ly and Lx is lesser than 2, then it is said to be a two-way Slab. Most of the Slabs in our project are two-way Slabs.

16

DESIGN OF TWO WAY SLAB EDGE CONDITION-TWO ADJACENT EDGES DISCONTINUOUS Step 1: Fy Fck Step 2: ly lx

Design constants2 = 415N/mm2 = 25 N/mm slenderness ratio = 5.23m = 4.23m

= = 1.23 < 2 (therefore two way slab) Step 3:Depth of Slab Effective depth = (span/depth) = 40 x 0.8 = 32 D = = 132.8 mm Adopt d = 140 mm D= d + d' = 140 + 15 = 155 mm Overall depth, D=155mm. Effective span Step 4: a. Clear span + effective depth = 4.23 + 0.14 = 4.37 m b. Clear span + wall thickness = 4.23 + 0.23 = 4.46 m Thus, effective span lex = 4.37 m Load Calculations Step 5: Self weight of the slab = = =

b x D x 25 1 x 0.155x 25 3.875 KN/m2

Floor Finishes

=

1 KN/m2

Ceiling plastering

= = =

21 x 0.02 x21 0.42 Kn/m 5 KN/m2 (from IS875)

Live load Total Load, W Ultimate load, Wu

-----------------------2 10.295 KN/m ----------------------=1.5 x W = 1.5 x 10.295 =15.44 KN/m2 =

17

Step 6:Bending moment and Shear Force End condition: Two adjacent edges discontinuous

Shorter span Negative Positive

longer span

0.061 0.046

0.047 0.035

Shorter span : Mx = xWuL2xe Mx = 0.061 x 15.44 x 4.372 = 17.98 kn.m Mx = 0.046 x 15.44 x 4.372 = 13.56 kn.m

-ve +ve Longer span : -ve +ve Vu Step 7:

My = yWuL2xe My = 0.047 x 15.44 x 4.372 = 13.85 kn.m My = 0.035 x 15.44 x 4.372 = 10.31 kn.m =0.5 x wu x lxe =0.5 x 15.44 x 4.37 = 33.73 KN Check for Effective depth Mu 6 = 0.138 Fckbd2 17.98X10 =0.138X23X1000Xd 2

D = 72.19 mm < 140 mm Hence the provided depth is safe. Step 8: Main Reinforcement For shorter span: Mu = 0.87 fy Ast d [ 1 17.98 x 106

]

= 0.87 x 415 x Ast x 140 [ 1 -

5.99 Ast2 - 50547 Ast + 17.98 x 106 = 0 Ast = 375mm 2 Provide 12mm  bars 18

]

Spacing S = = = 301.57 mm Hence , 12 mm  bars provided @300mm c/c in shorter span . Therefore, actual Ast = x1000 = 376 mm2. For Longer span: Effective depth = short span eff. depth = 140 -

= 130 mm

Mu = 0.87 fy Ast d [1 -

]

13.35 x 106= 0.87 x 415 x Ast x 130 [1 -

]

5.99 Ast2 - 46936 Ast + 13.35 x 106 = 0 Ast = 296 mm2 Adopt, Ast = 300 mm2 provide 12mm  bars Spacing S = = x 1000 = 350 mm Hence, 12 mm  bars provided @350 mm c/c in longer span. Therefore, actual Ast = x1000 = 320 mm2 Step 9: Minimum Reinforcement Ast (min) = 0.12 % b D = 0.0012 x 1000 x155 =186 mm2 Ast (min) < Ast, Hence provided reinforcement is safe. Step 10: v

Check for shear stress = =

Pt

= 0.24 N/mm2

= =

= 0.26

From page no 73 IS456:2000 K c = 1.26 x 0.36 = 0.45N/mm2 K c > v Hence the slab is safe against shear stress. 19

Step 11:

Check for deflection control

(

) provided =

(

) max kt fs pt Kt

(

33 ) max

= = = =

( ) basic x kt x kc fs & pt relationship from page no 38 IS456:2000 0.58 x fy = 0.58 x 415 = 240.7

= = = = > >

= 0.26 1.70 (from graph 38) 20 x 1.65 x1 33 31.2 ( )provided

Hence it is safe.

20

DESIGN OF TWO WAY SLAB EDGE CONDITION- ONE LONG EDGE DISCONTINUOUS Step 1: Fy Fck Step 2: ly lx

Design constants2 = 415N/mm2 = 25 N/mm slenderness ratio = 5.23m = 4.23m

= = 1.23 < 2 (therefore two way slab) Step 3:Depth of Slab Effective depth = (span/depth) = 40 x 0.8 = 32 D= = 132.8 mm Adopt d = 140 mm D= d + d' = 140 + 15 = 155 mm Overall depth, D=155mm. Step 4: Effective span a. Clear span + effective depth = 4.23 + 0.14 = 4.37 m b. Clear span + wall thickness = 4.23 + 0.23 = 4.46 m Thus, effective span lex = 4.37 m Load Calculations Step 5: Self weight of the slab = = = Floor Finishes

=

Ceiling plastering

= = =

Live load Total Load, W Ultimate load, Wu

b x D x 25 1 x 0.155x 25 3.875 KN/m2 1 KN/m2

21 x 0.02 x21 0.42 Kn/m 5 KN/m2 (from IS875) ----------------------= 10.295 KN/m2 ----------------------=1.5 x W = 1.5 x 10.295 =15.44 KN/m2

21

Step 6:Bending moment and Shear Force End condition: ONE LONG EDGE DISCONTINUOUS

Shorter span Negative Positive

longer span

0.053 0.040

0.037 0.028

Shorter span : Mx = xWuL2xe Mx = 0.053 x 15.44 x 4.372 = 15.62 kn.m Mx = 0.040 x 15.44 x 4.372 = 11.79 kn.m

-ve +ve Longer span :

My = yWuL2xe My = 0.037 x 15.44 x 4.372 = 10.90 kn.m My = 0.028 x 15.44 x 4.372 = 8.25 kn.m

-ve +ve Vu

= =

Step 7:

0.5 x wu x lxe 0.5 x 15.44 x 4.37 = 33.73 KN Check for Effective depth

Mu 6 = 0.138 Fckbd2 15.62 x 10 = 0.138 x 25 x 1000 x d 2 d = 67.28 mm < 140 mm Hence the provided depth is safe. Step 8:

Main Reinforcement For shorter span: Mu = 0.87 fy Ast d [ 1 -

15.62 x 106

]

= 0.87 x 415 x Ast x 140 [ 1 -

5.99 Ast2 - 50547 Ast + 15.62 x 106 = 0 Ast = 322mm 22 Adopt Ast = 325 mm Provide 12mm  bars 22

]

Spacing S = = = 347.9 mm 340mm Hence , 12 mm  bars provided @340mm c/c in shorter span . Therefore, actual Ast = x1000 = 333 mm2. For Longer span: Effective depth = short span eff. depth = 140 Mu

= 130 mm

= 0.87 fy Ast d [1 -

]

10.90 x 106= 0.87 x 415 x Ast x 130 [1 -

]

5.99 Ast2 - 46936 Ast + 13.35 x 106 = 0 Ast = 240 mm2 provide 12mm  bars Spacing S = = = 471.2 mm 450 mm Hence, 12 mm  bars provided @450 mm c/c in longer span. Therefore, actual Ast = x1000 = 251 mm2 Step 9: Minimum Reinforcement Ast (min) = 0.12 % b D = 0.0012 x 1000 x155 =186 mm2 Ast (min) < Ast, Hence provided reinforcement is safe. Step 10: Check for shear stress v = = Pt

= 0.24 N/mm2

=

= = 0.23 From page no 73 IS456:2000 K c = 1.26 x 0.36 = 0.45N/mm2 K c > v Hence the slab is safe against shear stress. 23

Step 11: (

Check for deflection control

) provided (

) max kt fs pt

= = = = =

= Kt = = = 33 > ( )max > Hence it is safe.

( ) basic x kt x kc fs & pt relationship from page no 38 IS456:2000 0.58 x fy = 0.58 x 415 = 240.7 = 0.23 1.70 (from graph 38) 20 x 1.65 x1 33 31.2 ( )provided

24

DESIGN OF TWO WAY SLAB EDGE CONDITION- INTERIOR PANEL Step 1: Fy Fck Step 2: ly lx

Design constants2 = 415N/mm2 = 25 N/mm slenderness ratio = 5.23m = 4.23m

= = 1.23 < 2 (therefore two way slab) Step 3:Depth of Slab Effective depth = (span/depth) = 40 x 0.8 = 32 d=

= 132.8 mm

Adopt d = 140 mm D= d + d' = 140 + 15 = 155 mm Overall depth, D=155mm. Step 4: Effective span a. Clear span + effective depth = 4.23 + 0.14 = 4.37 m b. Clear span + wall thickness = 4.23 + 0.23 = 4.46 m Thus, effective span lex = 4.37 m Step 5: Load Calculations Self weight of the slab

= = =

b x D x 25 1 x 0.155x 25 3.875 KN/m2

Floor Finishes

=

1 KN/m2

Ceiling plastering

= = =

Live load Total Load, W Ultimate load, Wu

21 x 0.02 x21 0.42 Kn/m 5 KN/m2 (from IS875) ----------------------= 10.295 KN/m2 ----------------------=1.5 x W = 1.5 x 10.295 =15.44 KN/m2

25

Step 6:Bending moment and Shear Force End condition: INTERIOR PANEL

Shorter span Negative Positive

longer span

0.044 0.033

0.032 0.024

Shorter span : Mx = xWuL2xe Mx = 0.044 x 15.44 x 4.372 = 12.97 kn.m Mx = 0.033 x 15.44 x 4.372 = 9.7 kn.m

-ve +ve Longer span :

My = yWuL2xe My = 0.032 x 15.44 x 4.372 = 9.43 kn.m My = 0.024 x 15.44 x 4.372 = 7.07 kn.m

-ve +ve Vu Step 7:

= =

0.5 x wu x lxe 0.5 x 15.44 x 4.37 = 33.73 KN Check for Effective depth

Mu 6 = 0.138 Fckbd2 12.9X10 =0.138X23X1000Xd2 D = 61.3 mm < 140 mm Hence the provided depth is safe. Step 8: Main Reinforcement For shorter span: Mu = 0.87 fy Ast d [ 1 12.97 x 106

]

= 0.87 x 415 x Ast x 140 [ 1 -

5.99 Ast2 - 50547 Ast + 12.97 x 106 = 0 Ast = 265 mm 2 Adopt Ast = 280 mm2 26

]

Provide 12mm  bars Spacing S = = = 403.8 mm 380mm Hence , 12 mm  bars provided @380mm c/c in shorter span . Therefore, actual Ast = x1000 = 297 mm2. For Longer span: Effective depth = short span eff. depth = 140 Mu

= 130 mm

= 0.87 fy Ast d [1 -

]

9.43 x 1026= 0.87 x 415 x Ast x 130 [1 5.99 Ast - 46936 Ast + 9.432x 106 = 0 Ast = 206.342mm Adopt Ast=210mm

]

provide 12mm  bars Spacing S = = = 538.52 mm 500 mm Hence, 12 mm  bars provided @500 mm c/c in longer span. Therefore, actual Ast = x1000 =226 mm2 Step 9: Minimum Reinforcement Ast (min) = 0.12 % b D = 0.0012 x 1000 x155 =186 mm2 Ast (min) < Ast, Hence provided reinforcement is safe. Step 10: Check for shear stress v = = Pt

= 0.24 N/mm2

= = =

0.2

From page no 73 IS456:2000 K c = 1.26 x 0.29 = 0.36N/mm2 K c > v Hence the slab is safe against shear stress. 27

Step 11: (

Check for deflection control

) provided (

=

) max kt fs pt

= = = =

= Kt = = = 34 > ( ) max > Hence it is safe.

( ) basic x kt x kc fs & pt relationship from page no 38 IS456:2000 0.58 x fy = 0.58 x 415 = 240.7 = 0.2 1.70 (from graph 38) 20 x 1.7 x1 34 31.2 ( )provided

REINFORCEMENT DETAILS OF SLABS Table 4 Reinforcement details of slabs This table shows Reinforcement details of slabs Diameter of bars(mm)

Spacing (mm)

Edge conditions

Type of slab

Ast (mm2)

1

Two Adjacent edges discontinuous

Two way

376

12

300

2

One long edge discontinuous

Two way

333

12

340

3

Interior panel

Two way

297

12

380

Slab

28

DESIGN OF BEAM

29

DESIGN OF BEAM Considering a beam, bending moment and shearing stresses are greater than those of slabs. Therefore, the depth of the beam is governed by the bending moment criteria while deflection criteria normally get satisfied. A beam simply supported at its ends carrying a uniformly distributed loads bends with a concavity upwards. It is subjected to a maximum sagging or positive bending moment at its mid span and zero at its supports. Sometimes the beam will be subjected to maximum negative or hogging bending moment. The point where the curvature changes from sagging to hogging is called at the point of contra flexure. While designing a beam, according to the bending moment we design as a flanged section or rectangular section or a Tee Beam or L Beam. At the time of load calculation, end shear from slab, load due to the wall carried by the beam, self weight of the beam, concentrated load transferred by secondary beam resting on the main beam are considered.

30

DESIGN OF T BEAM Step 1: Fy Fck L Step 2:

Data = = =

415N/mm22 25 N/mm 5.23m

Dimension Effective depth, d Overall depth, D

= = (5230/12) = 436 mm Say, d = 450 mm = 450 + 50 = 500 mm

Depth of the rib Width of the rib, bw Effective flange width bf

= 500 - 155 = 345 mm = 230 mm = [ +bw +6Df] =[

Step 3:

Load Calculations Self weight of the slab= = Floor Finishes = Self weight of the rib = = Live load = = Plaster finishes = Total Load, W =

+ 230 + (6x 155) ] = 2032 mm < 4230 mm

0.155 x 4.23 x 25 16.39 KN/m (0.6 x 4.23) = 2.538 KN/m 0.23 x 0.345 x 25 2.04 KN/m 4 x 4.23 16.92 KN/m 0.45 KN/m 38.33 KN/m

Ultimate load, Wu = (1.5 x W) = 57.49KN/m Step 4:Bending moment and Shear Force Mu = 0.125WuL2 = 0.125 x 57.49x 5.232 = 196.56 KNm Vu = 0.5 x wu x lx = 0.5 x 57.49 x 5.23 = 150.33 KN Step 5:Main Reinforcement xu = 0.48 d = 0.48 x 450 = 216 mm > Df = 155 mm Assuming the neutral axis depth, xu = 155 mm Mur

=

0.36 fck bf Df (d-0.42xu)

=

0.36 x 25 x 2032 x 155 (450 - 0.42 x 155)

=

1091.05 KNm > 196.56 KNm

Mu

= 0.87 fy Ast d [1 -

]

196.56 x 106 = 0.87 x 415 x Ast x 450 [1 -

] 31

2.94 Ast2 - 162472.5 Ast + 196.56 x 106 = 0 Ast = 1240 mm2 Provide 4 nos. of 20 mm  bars. Min. Reinforcement: Ast(min) =

=

= 1048 mm2

Ast(min) < Ast Hence, the provided Ast is safe. Step 6:Check for shear stress v = =

= 1.45 N/mm2

Two bars of 20 mm  are bent up near the supports, the two remaining bars provides an area, Ast = 2 x x 2 = 628.31 mm2 Pt

= =

= 0.607

From table 19 (IS456:2000) = 0.52 N/mm2 c = 1.27 x 0.52 = 0.66 K c v < K c (but here Kc less than v ) since, v >c shear reinforcement are to be designed. Shear carried by bent-up bars is given by : Vus

= = =

0.87fyAsv 2 x sin 0.87 x 415 x 314 x 2 x sin 45 160.64 KN > 150.33 KN

Provide nominal shear reinforcement using 8 mm  bars two legged stirrups, the spacing of stirrups is given: Sv

=

=

= 397.5 mm = 350 mm.

Use 8 mm  bars two legged stirrups throughout the span @350 mm c/c.

32

Step 7:Check for deflection control Ignoring the contribution of flanges (conservative) Pt =

(

) max

= =

1.21 & fy = 240.7

= =

( ) basic x kt x kc 20 x 0.95 x 1 = = 11.62

(

) provided

=

(

) max

>

(

19

) provided

Hence it is safe.

REINFORCEMENT DETAILS OF BEAMS This table shows Reinforcement details of T beam Beam

Span (m)

Ast (mm2)

Reinforcement

All floors

5.3

1240

4-20mm 

33

Beam size 230mm x 500mm

Spacing of stirrups 350mm c/c

DESIGN OF L BEAM Step 1:Data Fy Fck L

= = =

Step 3:

Bending moment and Shear Force

415N/mm2 2

25 N/mm 5.23 m Effective depth, d = = 450 mm Overall depth, D = 500 mm Depth of the rib = 345 mm Width of the rib, bw = 230 mm Step 2:Load Calculations Self weight of the slab = 0.155 x 0.5 x 4.23 x 25 = 8.19 KN/m Floor Finishes = 0.6 x 0.5 x 4.23 = 1.26 KN/m Self weight of the rib = 0.23 x 0.34 x 25 = 2.07 KN/m Live load = 4 x 0.5 x 4.23 = 8.46 KN/m Plaster finishes = 0.975 KN/m Total Load, W = 20.955 KN/m Ultimate load, Wu = 31.43 KN/m

Bending moment at2support section

Mu

= =

WuL /12 31.43 x 5.232/12 = 71.64 KNm

Bending moment at2center of the section Mu = WuL /24 = 31.43 x 5.232/24 = 35.82 KNm Vu = 0.5 x wu x lx = 0.5 x 31.43 x 5.23 = 82.18 KN Step 4:Effective flange width 1. bf = [(L0/12) +bw +3Df] = [ (5230/12) + 230 + (3 x 155) ] = 1131 mm 2. bf = bw + 0.5 ( spacing b/w ribs) = 230 + 0.5 ( 4000) = 2230 mm Therefore effective flange width, bf = 1131 mm Step 5:Torsional moment at support section Total self weight of the rib = 20.955 - 2.07 =18.88 KN/m 34

Total ultimate load on L beam = 1.5(18.88 x 5.23) = 148.19 KN Factored shear force, Vu = 0.5 x 148.19 = 74.09KN The distance between the centroid of the shear force from the center line of the beam is given by D = bf /2 - d/2 = (1131/ 2) - (230/2) = 450.5 mm Torsional moment, Tu = 74.09x 0.4505 = 33.37 KNm STEP 6:

EQUIVALENT BENDING MOMENT AND SHEAR FORCE

Equivalent B.M.

Mu

= Tu [

Mel

= 33.37[ ] = 62.30 KNm = Mu + Mt = 71.64 +62.30 = 133.94KNm

Equivalent shear force, Ve

]

= Vu + 0.6(

= 82.18+ 0.6( STEP 7: MAIN REINFORCEMENT At the support section: Mu = 0.87 fy Ast d [ 1 -

) ) = 314.31 KN

]

133.94 x 2 106 = 0.87 x 415 x Ast x 450 [ 1 26.05 Ast - 162472.5 Ast 2+ 133.94 x 106 = 0 Ast = 978 mm Thus provide 4 nos. of 20 mm  bars2 Actual Ast = 1256 mm At centre of span section: Mu = 0.87 fy Ast d [ 1 ]

]

35.82x 106 ] = 0.87 x 415 x Ast x 450 [ 1 2 Ast = 222 mm Minimum reinforcement as per clause 26.5.1.1 IS456:2000 Ast min = = = 216.9 mm2 < 222 mm2 section.

Hence provide 2 nos. of 12mm  bars on the tension side at the centre of the span

35

Step 7: Shear reinforcements =  v

Pt



= = = = = =

3.03N/mm2

<

1.21 0.68 N/mm2 (From Table 19 of IS456:2000) v shear reinforcement are required

=

[

c

Since c

Asv

] not less than [

]

Using 10 mm diameter two legged stirrups with 50 mm cover, = b - cover = 200 mm b1 = d - cover = 410 mm d1 = 2x = 157 mm2 Asv Sv

=

[

[[

Sv

= = = =

]

] = 117.16 mm ]

104.87 mm

Therefore Provide 10 mm  bars @ 100 mm c/c near supports gradually increasing to 230 mm c/c towards the canter of the span

36

This table shows Reinforcement details of L beam Beam

Span (m)

Ast (mm2)

Reinforcement

All floor

5.3

1256

4-20mm 

37

Beam size

Spacing of stirrups

230mm x 500mm

100mm c/c

DESIGN OF LINTEL

38

DESIGN OF LINTEL STEP 1:

DATA

Door size = 4m X 3m Assume size of the lintel = 230 x 150 mm Using 8 mm diameter with a nominal cover of 20 mm Effective depth available = 150 - (20-8/2) =126 mm ~ 130mm STEP 2: EFFECTIVE SPAN 1. Centre to centre of bearings = 4000 + 0.230 = 4230 mm 2. Centre of opening + effective depth = 4000 + 130 = 4130 mm Effective span = 4130mm STEP 3: LOAD CALCULATIONS Height of equilateral triangle

STEP 4:

= 0.866 x L = 0.8766 x 4.130 = 3.57 m = ½ x 4.13 x 3.57 x 0.230 x 20 W1 = 33.91KN Impose load = 4.13 x 0.6 x 750 = 1.85KN Self weight of lintel = 0.230 x 0.15 x 4.13 x 25 = 3.56 KN Total u.d.l w2 = 33.92 + 1.85 + 3.56 = 39.32 KN MAXIMUM BENDING MOMENT AT MID SPAN M

= =

STEP 5:

= 43.66KN/m = 1.5 x 43.66= 65.49 KNm Mu = 39.32 x 1.5 Vu = 58.98KN MAIN REINFORCEMENT ] Mu = 0.87 fy Ast d [ 1 -

65.49 X 106 Ast

= 0.87 X 415 X 130 [ 1= 902.6 mm2

]

Provide 8 nos. of 12 mm  bars.

39

STEP 5: v

SHEAR REINFORCEMENT = = = 1.97 N/mm2

Pt

=

= = 3.01 = 0.92N/mm2 c c X K = 0.92 X 1.30 = 1.19 Sv

= =

= 394 mm

Spacing of the stirrups is the least of 1. Sv = 390 mm 2. 0.75d = 0.75 x 130 = 97.5 mm ~ 100 mm Hence provide 8mm  two legged stirrups @ 100 mm c/c.

40

DESIGN OF LINTEL STEP 1:

DATA

Door size = 1.2m X 1.2m Assume size of the lintel = 230 x 150 mm Using 8 mm diameter with a nominal cover of 20 mm Effective depth available = 150 - (20-8/2) =126 mm ~ 130mm STEP 2: EFFECTIVE SPAN 3. Centre to centre of bearings = 1200 + 0.230 = 1430 mm 4. Centre of opening + effective depth = 1200 + 130 = 1330 mm Effective span = 1330mm STEP 3: LOAD CALCULATIONS Height of equilateral triangle

STEP 4:

= 0.866 x L = 0.8766 x 1.330 = 1.151 m = ½ x 1.33 x 1.151x 0.230 x 20 W1 = 3.52KN Impose load = 1.33 x 0.6 x 0.75 = 0.59KN Self weight of lintel = 0.230 x 0.15 x 1.33 x 25 = 1.14KN Total u.d.l w2 = 3.52 + 0.591 + 1.14 = 5.25 KN MAXIMUM BENDING MOMENT AT MID SPAN M

STEP 5:

=

= = 1.65KN/m = 1.5 x 1.65= 2.475 KNm Mu = 5.25x 1.5 Vu = 7.87KN MAIN REINFORCEMENT ] Mu = 0.87 fy Ast d [ 1 -

2.47 X 106 Ast

= 0.87 X 415 X 130 [ 1= 55mm2

]

Provide 2 nos. of 8 mm  bars.

41

STEP 5: v

SHEAR REINFORCEMENT = = = 0.26 N/mm2

Pt

= =

= 0.33 = 0.36N/mm2 c c X K = 0.36X 1.30 = 0.46 0.26 0.46 Hence it is safe.

42

DESIGN OF COLUMNS

43

DESIGN OF COLUMN (Axial loaded column) GROUND FLOOR: STEP 1: DATA Factored load, Pu 2 = 3100 KN Fck = 30 N/mm F y =415N/mm Unsupported length of the column = 4.42 m STEP 2: GROSS SECTIONAL AREA Assume percentage of reinforcement, P = 1 Using the chart 25, Required gross sectional area of column, Ag = 2025 cm2 Thus provide a section 45 cm x 45 cm. S TEP3 :

MINIMUM ECCENTRICITY = + = + = 0.884 + 1.5= 2.384 cm

emin

= = 0.05 < 0.05D =

+ =

emin STEP 4: As

=

+

= 0.884 + 1.5= 2.384 cm

= 0.05 < 0.05D

AREA OF STEEL = = = 2025 mm2

Provide 8 nos. of 20mm  bars.

44

STEP 5: LATERAL TIES TRANSVERSE REINFORCEMENT The diameter of the transverse links shall not be less than  ¼ th the column of the largest longitudinal bar  6 mm o D / 4 = 20 / 4 = 5mm o 6 mm. SPACING OF TRANSEVERSE LINKS: This shall not exceed the least of the following: The least lateral dimension of the column 1. 16 times the dia of the smallest longitudinal reinforcing rod in the column 2. 48 times the dia of the transverse reinforcement 3. Here spacing a)450mm b)16 X 20=320 mm c)48 X 6 =288mm Provide 6mm dia of lateral ties @300mm c/c.

45

DESIGN OF COLUMN (Axial loaded column with uniaxial bending) STEP 1: DATA Factored load, Pu MU Fck Fy STEP 2:

= 1750 KN = 196.56 KN = 30 N/mm2 2 = 415 N/mm

MINIMUM ECCENTRICITY = + = + = 0.884 + 1.5= 2.384 cm

emin

= = 0.05 < 0.05D =

+ =

emin

=

+

= 0.884 + 1.5= 2.384 cm

= 0.05 < 0.05D

STEP 3: REINFORCEMENTS Using 20 mm  of bars with 40 mm cover = = 0.11 Adopt,

= 0.15 = = 0.28 =

= 0.07 Using the chart 44, = 0.04 P

= 0.04 x 30 =1.2

46

STEP 4: As

AREA OF STEEL =

= = 2430 mm2 Provide 8 nos. of 20mm  bars. STEP 5: LATERAL TIES TRANSVERSE REINFORCEMENT The diameter of the transverse links shall not be less than  ¼ th the column of the largest longitudinal bar  6 mm o D / 4 = 20 / 4 = 5mm o 6 mm. SPACING OF TRANSEVERSE LINKS: This shall not exceed the least of the following: The least lateral dimension of the column 1. 16 times the dia of the smallest longitudinal reinforcing rod in the column 2. 48 times the dia of the transverse reinforcement 3. Here spacing a)450mm b)16 X 20=320 mm2 c)48 X 6 =288mm Provide 6mm dia of lateral ties @300mm c/c.

47

DESIGN OF COLUMN (Axial loaded column with biaxial bending) STEP 1: DATA Factored load, Pu = 1000 KN Mux = 71.64 KN Muy = 35.82KN D = 450 mm b = 450 mm Fck = 30 N/mm 2 Fy = 415 N/mm STEP 2: MINIMUM ECCENTRICITY = + = + = 0.884 + 1.5= 2.384 cm emin

= = 0.05 < 0.05D =

+ =

emin

=

+

= 0.884 + 1.5= 2.384 cm

= 0.05 < 0.05D

STEP 3: REINFORCEMENTS Reinforcement distributed equally on four sides Assume percentage of reinforcement, p = 1.2 = = 0.11 Adopt,

= 0.15 = = 0.16 =

= 0.04

Using Sp 16 chart 44, = 0.08 Mux

= 0.08 X 30 X 450 X 4502 =218.7 KN.M

Puz Puz

= Axial load capacity of the section under pure axial load. = 0.45 fck Ac + 0.75 fy Asc 48

Asc Ac A g

Puz

= 1.2 % Ag = 0.012 x 450 x 450 = 2430 mm2 = Ag - Asc = (450 X 450) = 202500 mm2 = 202500 - 2430 = 200070 mm2 = (0.45 x 30 x 200070) + (0.75 x 415 x 2430) = 3457.2 KN =

= 0.28 IS 456:2000 recommends a simplified procedure based on bresler's empirical formulation involving the salient design parameters, Mux & Muy are moments about X and Y axes due to design loads Mux1 & Muy1 are maximum uniaxial moment capacity for an axial load Pu bending about X and Y axis respectively. an is an exponent whose value depends on the ratio (Pu/Puz) The values of an varies linearly from 1 to 2. For (Pu/Puz) less than 0.2, an = 1. For (Pu/Puz) greater than 0.2, an = 2.

0.13 < 1.0 Hence the assumed area of reinforcement satisfies. STEP 4: AREA OF STEEL As

= = = 2430 mm2

Provide 8 nos. of 20mm  bars. STEP 5: LATERAL TIES TRANSVERSE REINFORCEMENT The diameter of the transverse links shall not be less than  ¼ th the column of the largest longitudinal bar  6 mm o D / 4 = 20 / 4 = 5mm o 6 mm.

48

SPACING OF TRANSEVERSE LINKS: This shall not exceed the least of the following: The least lateral dimension of the column  16 times the dia of the smallest longitudinal reinforcing rod in the column  48 times the dia of the transverse reinforcement  Here spacing  a)450mm  b)16 X 20=320 mm2 c)48 X 6 =288mm 2 Provide 6mm dia of lateral ties @300mm c/c.

REINFORCEMENT DETAILS OF COLUMN This table shows Reinforcement details of columns Size Of column 450mm x

Main reinforcement diameter (mm) 20

Lateral ties Diameter (mm)

Spacing of lateral tie (mm)

Floor

6

300

Ground

450mm 450mm x

Floor 20

6

300

450mm

First to fourth floor

50

DESIGN OF FOOTING

51

Design of Footing Dimensions: Size of column = 450mmX450mm Factored load = 3200KN. Safe bearing capacity = 440 KN/mm2 fcK = 30 N/mm2 fy = 415 N/mm2 a) Loads on footing: Total load

= 3200 KN

Self wt.

=10% = 320 KN

Total load

=3520 KN

b) Size of footing: Area

= 3520/440 = 23.46 m2

Size of footing = = 2.85 m Adopt size of the footing

= 2.85m X 2.85m

Adopt width of strap beam,b = 450mm = 0.45m. c) Design of footing: Soil pressure ,Pu = 3200/(2.85X2.85) = 408.16 KN/m2 440 KN/m2 Factored soil pressure = 1.5 X 408.16 =0612.24KN/m Cantilever projections from the short side face of the column = 0.5(2.85 - 0.45) = 1.2 m Cantilever projections from the long side face of the column = 0.5(2.85 - 0.45) = 1.2 m Ultimate design moment M

= mu . = 0.5 Pul2

=0.5X612.24X1.22

52

= 440.81 KN m.

d) Effective depth of footing: d=

(Mu/0.138 fcK b)

=

(440.81X106/0.138X30X103)  350 mm. = 326.3 mm= Depth (d) based on shear consideration will be double than that due to moment consideration. Effective depth = d = 2 X 350 mm = 700 mm. Overall depth = d = 720mm 40mm D = d+d' = 740mm e)Reinforcement: Mu 440.81X106

= {0.87X fy X AstXd} [1- {Ast X fy/ bd fcK}] = (0.87X415XAstX700) [1- 415X Ast/103X30X700] = 1810 mm2

Ast

Adopt 20mm diameter bars (8 nos) @ 200 mm c/c Ast provided = 2513 mm2 f) Check for shear stress : Shear stress at a distance = Vu. ( equal to effective depth) Vu = (1.2-0.70) 612.24 = 306.12 KN = Vu/bd = 306.12X103/103X700 τv = 0.43 N/mm2 100Ast/bd

= {100X2513/103X700} = 0.28 τc = 0.45 N/mm2

Permissible shear stress = Ks τc = 1X τc =0.45 N/mm2 (Ks τc)> τv Safe factor shear permissible limit (within safety limit)

REINFORCEMENT DETAILS OF FOOTING This table shows Reinforcement details of footing Size of footing

Main Reinforcement (one direction)

Main Reinforcement (other direction)

2.85m x2.85m

8 Nos of

8 Nos of

20mm 

20mm 

54

Spacing of bar (mm ) 200

Depth of footing (m) 0.74

DESIGN OF STAIRCASE

55

DESIGN OF STAIRCASE Stair cases are generally provided connecting successive floors of a building and in small buildings they are the only means of access between the floors. The staircases comprise of flight of steps generally with one or more intermediate landings provided between the floor levels. A flight of steps consists of two landings and one going with 10 to 12 steps. The structural component of a flight of stairs consists of: (a) Tread which forms the horizontal portion of the step. The tread is usually 250 to 300 mm wide depending upon the type of the building. (b) Riser is the vertical distance between the adjacent treads or the vertical projection of the step, generally in the range of 150 to 190 mm depending upon the type of the building. The width of the stairs varies in the range of 1 to 1.5m with a minimum value of 850mm. (c) Going forms the horizontal plan projection of an inclined flight of steps between the first and the last riser.

56

DESIGN OF STAIRCASE (DOG LEGGED STAIRCASE WITH WAIST SLAB) STEP 1:

DATA

Tread, T = 250 mm Rise, R =200mm Width of landing =900mm Width of passage = 850 mm =2250 mm Width of each flight =2000 mm Height of each flight = 2000/200 = 10 Nos No of Rises in each flight = 9 Nos No of Tread STEP 2: EFFECTIVE SPAN Effective span = (230/2) + 850 + 2250 + 900 + (230/2) = 4230 mm Thickness of waist slab = = = 211.5 mm ~ 215 mm Effective depth =215 - 25 = 190 mm STEP 3: LOAD CALCULATION Dead load of the slab = 0.215 x 1 x 25 = 5.375 KN/m Dead load of the slab on horizontal span /1m, W

=[

]

=[

Dead load of one step /m Finishes

= = =

= 6.88 KN/m = 0.5 x 0.20 x 0.25 x 25 0.625 KN = 2.5 KN/m 1 KN/m

Live load Total load Ultimate load, Wu

= = =

5 KN/m 15.38 KN/m 23.07 KN/m

Dead load of one step

STEP 4: Mu = Vu

=

STEP 5: Mu d

(liable to overcrowding)

BENDING MOMENT AND SHEAR FORCE 0.125wuL2 = 0.125 x 23.07 x 4.232 = 51.59 KNm 0.5 wuL =0.5 x 23.07 x 4.23 =48.79 KN CHECK FOR EFF2ECTIVE DEPTH = 0.138 fck b d =

= 122.28 mm

122.28 < 215 Hence it is safe. 57

]

STEP 6:

MAIN REINFORCEMENT Mu

= 0.87 fy Ast d [1 -

]

51.59 x 106 = 0.87 x 415 x Ast x 190 [1 = 810 mm2 Ast Assume 12mm  bars Spacing S = = 139mm Adopt S = 130mm Thus provide 12 mm  bars @130mm c/c. STEP 7: DISTRIBUTION REINFORCEMENT Ast (min) = 0.12 % b D = 0.0012 x 1000 x 215 =258 mm2 Provide 8mm  bars Spacing S = = 180 mm Thus provide 8 mm  bars @180 mm c/c. STEP 8: CHECK FOR SHEAR v = Pt K c K c

=

= 0.25 N/mm2

= =

= 0.426

= >

1.2 x 0.36 = 0.43 N/mm2 v

58

]

Hence the design of doglegged staircase is safe.

59

GREEN MATERIALS 7.1 WATER LESS URINALS As urine is about 96% liquid, no additional water is really needed to wash it down the drain. The waterless, urinal, looking much like its conventional counterpart, takes advantage of this concept with generally positive results. Waterless urinals do away with the requirement of water for flushing and result in saving of between 56,800 litres to 170,000 litres of water per urinal per year. To achieve odour control in waterless urinals, odour traps using sealant liquid, microbial control, rubber membrane and curtain valve have been developed across the world. The waterless urinal appears and works like a conventional urinal, except that it does not flush and, therefore, requires no water. Like their traditional counterparts, waterless urinals are made of fibreglass or vitreous china, and are offered in white as well as various custom colours.

ENVIRONMENTALLY FRIENDLY Waterless urinals contribute positively to the environment. First, the absence of water for flushing reduces the demand for water, an increasingly scarce commodity in some areas. Also, since no water goes down the drain, additional wastewater requiring treatment is not generated. Next, the special drain cartridges and inserts used in some models are recyclable. Finally, the sealant liquid composed of natural oils, is biodegradable.

60

COMPARISSION OF NORMAL AND WATERLESS URINALS S.NO

NORMAL URINALS

WATERLESS URINALS

1

Maintenance cost is high

High cost in initialization but can be paid back during it's life time

2

Labour cost (maintenance and

Labour cost (maintenance and cleaning)-less

cleaning)-High 3

Wet operation in flush provide

Dry operation makes hostile conditions for

favourable conditions for

bacteria and viruses

gems.(unhygienic) 4

High water consumption

5

Chances for Communicable

No water consumption

diseases

No handle(Touch-free) thus reducing the spreading of communicable diseases

7.1.1 ECONOMIC PAYBACK OF WATERLESS URINALS 1.

Waterless urinals can be installed in both URBAN and RURAL areas.

2.

Conserve water and energy, Reduces the waste water generation.

3.

Collected urinals can be used in productive industrial and agricultural applications (New Paradigm in waste water management).

1.

Rebates and Incentives-Some water utility companies offer rebates and incentive payments to owners installing waterless urinals. Payments range from a partial to full reimbursement for the cost of no-flush urinals.

2.

Repair of flush valves due to use, failure, or vandalism will not be required. Neither is cleanup resulting from backed up drains and overflows

3.

Therefore, labor costs for no-flush urinals should be less than for flush-types. The costs of replacement cartridges can outweigh the maintenance labour saving.

7.1.2 "GREEN BUILDING" CREDITS Many new construction projects nowadays are earning certification as "green buildings" under the LEED program developed by the U.S. Green

Building Council.

Installation of waterless urinals helps gain water conservation points.

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7.2 ALUMINIUM COMPOSITE PANEL Aluminium Composite Panel (ACP) or Aluminium Composite Material (ACM) is a widely-used term describing flat panels that consist of a non-aluminium core bonded between two aluminium sheets. Aluminium sheets can be coated with Polyester paint. ACP is very rigid and strong despite its light weight. Aluminium can be painted in any kind of colour, and ACPs are produced in a wide range of metallic and non-metallic colours as well as patterns that imitate other materials, such as wood or marble. Applications of ACPs are not limited to external building cladding, but can also be used in any form of cladding such as partitions, false ceilings etc. Aluminium Composite Panels are also widely used within the signage industry as an alternative to heavier, more expensive substrates. Features Excellent fireproof property. Super peeling strength. Perfect cold resistance performance. Excellent surface flatness and smoothness. Superior weather,corrosion,pollutant resistance. Superior impact resistance. Lightweight and easy to process. Easy to maintain. 7.2.1 ECONOMIC PAYBACK OF (ACP) Less maintenance. Aesthetic look for longer period of time. Reduces health hazards.

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7.2.3"GREEN BUILDING" CREDITS Many new construction projects nowadays are earning certification as "green buildings" under the LEED program developed by the U.S. Green Building Council. Installation of Aluminium Composite Panel (ACP) helps gain indoor environment quality points. 7.3 GLASS Glass is completely recyclable and non-toxic in nature. It satisfies all the ecological parameters of being the most sought after "green" building material in Green Buildings Green building design criteria emphasizes the energy-efficient performance of fenestration materials and maximum use of natural daylight. Given this background, Glass is an indispensable material for green building. It has a wide range of functional benefits. Its transparency allows day-lighting of the interiors and integrates the interiors with the exteriors. Studies have proven time and again that this substantially improves the productivity and health of the occupants of the building. "GREEN BENEFITS" of GLASS  Day-lighting - The use of glass brings in lot of light that helps in giving a high amount of natural day lighting instead of depending solely on artificial lighting thus reducing considerably electricity consumption.  Blending interiors with exteriors (Views) - Glass facades give a spectacular view of the outside world from the cozy interiors.  Recyclability - Glass being recyclable satisfies the important parameter of being a "Green" building material.  Achieving energy efficiency - High performance glass helps in controlling the solar and thermal heat in the interiors and helps to maintain the temperature at its minimum best and in turn helps to tone down the air-conditioning expenses. Innovative application - Being very flexible building material glass helps to satisfy and capture an architect's utmost imagination in its shape and form.

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 Controls noise: Double glazed glass facades help in achieving a high degree of acoustic comfort by keeping away noise penetrating from the exteriors to the interiors thus ensuring a calmer atmosphere inside.  Self Cleaning: The future belongs to self-cleaning glass which keeps itself clean on its own and brings out an ever sparkling effect.

7.4 PAINTS Today, most interior and exterior paints are found to have high levels of VOCs, which help them to dry faster. But, these VOCs emit smog-forming chemicals into the air and thus become the major contributor to ground-level ozone pollution. These release low-level emissions into the air for years after the application. And the major source of these toxins is a variety of volatile organic compounds, which, until recently, were essential to the performance of the paint. Volatile organic compounds are substances that evaporate from paint allowing it to dry and are very toxic to humans. Even after the paint has dried, VOCs can continue to be released from the paint for years, thus, harming the occupants. The easiest solution is to use paints that do not contain VOCs and instead contain a non-harmful drying agent. Low VOC paints are the ones which use water as a carrier instead of petroleumbased solvents. They contain reduced levels of volatile organic compounds (VOCs), which emit smog producing pollutants into the air. The new environmental regulations have resulted in alternative solutions - Low VOC and Zero VOC paints. Lower VOC paints preserve both indoor and outdoor air quality and reduce the incidence of eye or respiratory irritation from exposure to VOC fumes. Types of non-VOC paints: 1. Natural Paints and Finishes - These are paints made from natural raw ingredients such as water, plant oils and resins, plant dyes and essential oils,natural minerals such as clay, chalk and talcum; milk casein, natural latex, bees wax, earth and mineral dyes. Water based natural paints give off almost no smell. The oil based natural paints usually 64

have a pleasant fragrance of citrus or essential oils. Allergies or sensitivities to these paints are uncommon. These paints are the safest for one's health and also for the environment. 2. Zero VOC paints - According to the EPA (Environmental Protection Agency) standard, any paint in the range of 5 grams/litre or less can be called 'Zero VOC' paint. Adding a colour tint usually brings the VOC level up to 10 grams/litre, which is still quite low. 3. Low VOC paints - As described above, the level of harmful emissions are lower than solvent-borne surface coatings, as they carry water as a carrier instead of petroleum base solvents. These certified coatings also contain no, or very low levels, of heavy metals and formaldehyde. The amount of VOCs in paints should not exceed 200 grams/litre and in varnishes, it should not exceed 300 grams/litre. Low VOC paints tend to emit odour until dry. To avoid this, one should buy paints that contain VOCs less than 25 grams/litre. Advantages: 1.

Environment friendly, as there are lower levels of ozone pollution.

2.

Fewer emissions of smog-forming chemicals.

3.

Better indoor and outdoor air quality.

4.

Allergies or sensitivities to these paints are uncommon.

5.

Ideal for commercial applications, and offer excellent scrub ability.

6.

Quick Drying.

7.

Low Odour.

8.

Non-yellowing.

9.

Increased UV resistance, flexibility. In order to meet EPA (Environmental Protection Agency) standards, paints and stains must not contain VOC's in excess of 200 grams per litre and varnishes must not contain VOC's in excess of 300 grams per liter.

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"GREEN BUILDING" CREDITS Many new construction projects nowadays are earning certification as "green buildings" under the LEED program developed by the U.S. Green Building Council. Using LOW VOC PAINTS helps gain Ozone safety and Innovative design points.

7.5 CARPETS A carpet is a textile floor covering consisting of an upper layer of "pile" attached to a backing. The pile is generally either made from wool or a manmade fibre such as polypropylene, and usually consists of twisted tufts which are often heat-treated to maintain their structure. WOOL CARPET MATERIAL Wool carpet has no synthetic dyes. Wool carpeting is usually dyed with natural pigments so you can avoid synthetic dyes. This is important for people allergies or chemical sensitivities. Natural wool offers amazing insulating properties. Wool is naturally fire-resistant; Wool carpet comes in endless design options. Carpet made from synthetic fibers often contains petroleum products, and petroleum is a non-renewable resource. Wool, on the other hand, renews itself quite rapidly. The Benefits of Choosing Carpet The benefits derived from selecting good quality carpets in any type of application are manifold: what other type of floor covering will provide sound insulation, energy savings, underfoot comfort, a safe, non-slip floor, and be easy to clean and install, with good wear and non-allergenic properties. ECONOMIC PAYBACK OF CARPET USE Sound Insulation The usage of carpet gives excellent noise proof thus providing a quiter environment for the workers.

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Energy Saving Carpets can make a measurable contribution to retaining the warmth in a room and therefore saving energy. Carpets have low heat conduction and are natural thermal insulators creating a heat barrier. As a result, as much as 10% of the heat in a room which would be dissipated with smooth floor covering is retained in the room by the carpet. This, combined with the outstanding underfoot comfort of a carpet, can result in a considerable reduction in the use of heating in the transition from a warm to a cold season. It has been estimated that up to 30 days heating can be saved, resulting in an energy saving of around 4-6% and a consequent reduction in heating bills. Safety Safety covers two aspects - reduced slippage, especially when wet, and a decrease in stress on joints The soft resilient fibres of a carpet provide a cushioning effect and excellent orthopedic properties. The underfoot safety of carpets is an important feature, both in the home and the workplace. Infants and the elderly in particular benefit from the non-slip aspects of carpets and, in the event of a fall, the soft resilience of the carpet lowers the risk of injury. Health With their superb dust-trapping properties, carpets are conducive to a healthy living environment. Dust settles quickly and is then securely held by the pile fibres of the carpet until it is vacuumed again. Cleaning and Maintenance Cleaning and maintenance of carpet is quick, easy and low cost.

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"GREEN BUILDING" CREDITS Many new construction projects nowadays are earning certification as "green buildings" under the LEED program developed by the U.S. Green Building Council. Provision of WOOL CARPET helps gain Energy saving and Innovative design points.

7.6 OTHER PROPOSALS HVAC ventilation co2 sensors HVAC (Heating, Ventilating, and Air Conditioning) refers to technology of indoor or automotive environmental comfort. HVAC system design is a major sub discipline of mechanical engineering, based on the principles of thermodynamics, fluid mechanics, and heat transfer. CHILLER A chiller is a machine that removes heat from a liquid via a vapor-compression or absorption refrigeration cycle. This liquid can then be circulated through a heat exchanger to cool air or equipment as required. In air conditioning systems, chilled water is typically distributed to heat exchangers, or coils, in air handling units, or other type of terminal devices(Vents) which cool the air in its respective space(s), and then the chilled water is re-circulated back to the chiller to be cooled again. These cooling coils transfer sensible heat and latent heat from the air to the chilled water, thus cooling and usually dehumidifying the air stresam.

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REFERENCE: 1. "Advanced Reinforced Concrete Structures", by Dr. Krishna Raju. 2. "R.C.C DESIGN", by Dr.S. Ramamrutham. 3. INDIAN STANDARD code of practice for plain and reinforced concrete, IS456:2000 1. Code of practice for load calculation IS 875:1987. 2. Code of practice for load calculation SP 16. 3. "Limit state Design of reinforced concrete" by P.C. Varghese . 4. Illustrated design of R.C Buildings" by K.L Shah and S.R Karve. 5. Structural design and drawings" by Dr. KrishnaRaju. 6. Soil mechanics and foundation engineering" by B.C Punmia.

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