Seismic Analysis and Design of Hospital Building-libre

SEISMIC ANALYSIS AND DESIGN OF MULTI STOREYED HOSPITAL BUILDING


Submitted at the th e end of final final semester by K.Aslam [Reg.No.11609103006]

In partial fulfillment for the award of the degree of

BACHELOR OF ENGINEERING IN CIVIL ENGINEERING

SRI VENKATESHWARA COLLEGE OF ENGINEERING & TECHNOLOGY THIRUVALLUR

ANNA UNIVERSITY: CHENNAI 600 025 APRIL 2012

1


CERTIFICATE This is to certify that the project report entitled “SEISMIC

ANALYSIS

he BUILDING” BUILDING” is t he

AND

DESIGN

OF

MULTI

STROREYED

HOSPITAL

bonafied project work done by K.ASLAM (Reg.No.11609103006) in

partial fulfillment of the requirement for the award of Bachelor of Engineering Degree in Civil Engineering under Sri Venkateswara College of Engineering & Technology during the year 2011-2012.

Internal Guide

S.SANTHANA RAMAN,M.E,C.A.R(CSU)USA Dean& Project Guide Department of Civil Engineering. S.V.C.E.T.

.

Mrs.GAYATRI PADHY,B.E Head of the Department Department of Civil Engineering S.V.C.E.T.

External Guide

Dr.S.JUSTIN Engineering Manager SDDH, EDRC, Larsen & Toubro, Chennai.

Mr.UP.VIJAY Assistant Engineering Manager, EDRC, Larsen & Toubro, Chennai.

Mr.M.RAMPRATHAP Assistant Engineering Manager, EDRC, Larsen & Toubro, Chennai.

2


ACKNOWLEDGEMENT

I take this oppurtunity to thank Mr.Bikram K.Naik [HR Manager, EDRC Division] and Mr.Mukesh Kumar Singh [HR Manager P&OD] for giving me the oppurtunity to work with Larsen & Toubro. I thank Dr.S.Justin [Engineer Manager (SDDH),EDRC], Mr.UP.Vijay [Assitant Engineering Manager] and Mr.M.Ramprathap [Assistant Engineering Manager] without them this project would not have been possible. Their feedback, comments and suggestions were helpful throughout the entire work. I also thank all other Engineers of the Health and Leisure Division of B&F, L&T for guiding me throughtout the program. I thank Dr.Suresh Mohan Kumar M.Tech,Ph.D our Principal, for giving me the oppurtunity to work with Larsen & Toubro Limited Limited and I thank Mr.S.Santhanaraman Mr.S.Santhanaraman M.E,C.A.R (CSU) USA and Mrs.Gayatri Padhy,B.E. without them this project would not have been possible throughout the entire work. Last but not the least I thank my friends and family for their whole hearted support and cooperation.

K.ASLAM

3


ABSTRACT

Industrial training is an essential in the development of the practical and professional skills required of an Engineer and as an aid to prospective employment undertaking the training in a reputed firm adds to the advantage. The training was done with Larsen & Toubro, Chennai. The work allotted during the Indutrial Training period was the Seismic Analysis and Design of multistoryed New Teaching Hospital block at Agartala one of the projects undertaken by L&T. Earthquake Engineering was developed a lot from the early days and seismically analysing the structures requires specialized explicit finite element analysis software,which divides the element into very small slices and models the actual physics. The seismic analysis of the proposed building was done in the software ETABS, version- 9.7, which is one of the most advanced software in the structural design field. The loads applied on the structure was based on IS:875(part I)-1987[dead load],IS:875(part II)-1987[live load], IS:875(part III)-1987[wind load], IS:1893-2000 [Earthquake load]. Scale factor is calculated from the design base shear (Vb) to the base shear calculated using fundamental time period (T a). Once the analysis was completed all the structural components were designed according to Indian standard code IS:456-2000. Footing, columns, beams, slab, staircase and shear wall were designed. Ductile detailing of the structural elements were done as per code IS:139201993.

4


LIST OF CONTENTS

PAGE NO.

1.0 INTRODUCTION

1

1.1 New Teaching Hospital Building

1

1.2 Larsen & Toubro

2

1.2.1 Vision and Mission

2

2.0 FEATURES OF PROJECT

2

3.0 INTRODUCTION TO ETABS

3

3.1 Modeling Features

3

3.2 Analysis Features

4

4.0 LOAD CALCULATION

5

4.1 General data

5

4.2 Dead load

5

4.2.1 Typical Floor slab

5

4.2.2 Terrace Floor slab

5

4.3 Live load

5

4.4 Wall load

6

4.4.1 Member load due to 230mm thick exterior wall

6

4.4.2 Member load due to 100mm thick interior wall

6

4.5 Static check for load combintion

6

4.5.1 Plinth level

6

4.5.2 Floor level

7

4.5.3 Static check for dead and live load combination.

8

4.6 Wind load calculation

9

4.6.1 Basic wind speed

9

4.6.2 Design wind speed

9

4.6.3 Design wind pressure

9

4.6.4 Wind force

10

4.7 Seismic load calculation

13

4.7.1 Zone factor

14

5


4.7.2 Response reduction factor

14

4.7.3 Importance factor

14

4.7.4 Seismic weight

14

4.7.5 Soil classification

15

4.7.6 Seismic base shear calculation

16

4.7.8 Analysis

17

4.7.9 Scale factor calculation

18

5.0 MODELLING AND ANALYSIS OF THE BUILDING

23

5.1 Basic Grid system

23

5.2 Define Geometry

23

5.3 Define material Property

24

5.4 Define frame section

24

5.5 Define Wall or Slab section

25

5.6 Define Diaphram

26

5.7 Define Response Spectrum Function

26

5.8 Response Spectrum Function

26

5.9 Define Static load cases

26

5.10 Define load combination

28

5.11 Output

30

6.0 DESIGN OF FOUNDATION

36

6.1 General

36

6.2 Design of Isolated footing

36

6.3 Design of typical isolated footing

37

6.3.1 Design Parameters

37

6.3.2 Bending moment in X direction

38

6.3.3 Bending moment in Y direction

39

6.3.4 Check for one way shear

39

6.3.5 Check for two way shear

39

6.3.6 Transfer of forces at column base

40

6.4 Foundation detailing

40

6


7.0 COLUMN DESIGN

41

7.1 General

41

7.2 Column design

42


42

7.2.2 Design of bending moment

42

7.2.3 Reinforcement provided

43


43

7.2.5 Design of bending moment

44

7.2.6 Calculation of reinforcement

44

7.3 Column detailing

45

8.0 BEAM DESIGN

46

8.1 General

46

8.2 Beam design

47


47

8.2.2 Bending moment calculation

47

8.2.3 Shear Reinforcement

48

8.2.4 Shear Reinforcement at mid span

48

8.2.5 Design of Hanger bars

49

8.2.6 Development Length

49

8.2.7 Check for deflection

49

8.3 Beam Detailing

50

9.0 DESIGN OF SLAB

51

9.1 Design of one way and two way slab

51

9.2 Design of typical two way slab

53

9.2.1 Design parameters

53

9.2.2 Design of moments

53

9.2.3 Reinforcement calculation

54

9.2.4 Check for depth

54

9.2.5 Check for shear

54

9.2.6 Check for deflection

55

9.3 Reinforcement details

55

9.4 Slab detailing

56

10. DESIGN OF STAIRCASE

57

10.1 General

57 7


10.2 Design parameters

59

10.3 Load acting on stairs

59

10.4 Design of waist slab type staircase

59

10.5 Reinforcement provided

60

10.6 Staircase detailing

60

11. DESIGN OF SHEARWALL

61

11.1 General

61

11.2 Wall dimensions

61

11.3 Check for boundary element

62

11.4 Fixing boundary element

62

11.5 Determination of minimum steel

62

11.6 Shear reinforcement

63

11.7 Flexural strength

63

11.8 Transverse reinforcement

64

11.9 Shear wall detailing

64

12. Conclusion

65

13. References

66

8


LIST OF SYMBOLS

Asc

=

Area of Compression Steel

Ast

=

Area of tension steel

A sv

=

Area of shear reinforcement

Ag

=

Gross area of concrete

b

=

Breadth of the section

B.M

=

Bending Moment

C/C

=

Center to Center

d

=

Effective depth

d’

=

Depth of compression reinforcement.

DL

=

Dead load

D

=

Over all depth

Df

=

Depth of flange

EL

=

Earthquake load

f ck

=

Characteristic Strength of concrete

f y

=

Characteristic Strength of steel

I

=

Importance Factor (IS: 1893-1987)

L

=

Clear Span

L.L

=

Live Load

k 1

=

Modification Factor (IS: 456-2000)

K1

=

Risk Factor (IS 875:-1987) part III

K2

=

Topography factor (IS 875-1987) part III

Ld

=

Development Length

Leff

=

Effective length

Lx

=

Length of span in x direction

Ly

=

Length of span in x direction

M.F

=

Modification factor 9


M.R

=

Moment of Resistance

Mu

=

Ultimate Moment

Mu limit

=

Limiting moment of reinforcement of a section

Pu

=

Axial load on a compression member

Pt

=

Percentage of reinforcement

R.C.C

=

Reinforcement cement concrete

R

=

Response Reduction factor (IS: 1893-1987) part-III

S

=

Spacing of reinforcement

S.F

=

Shear force

T

=

Torsion moment

Vu

=

Shear force

WL

=

Wind load

Zp

=

Plastic section modulus (IS: 456-2000)

Z

=

Zone factor (IS: 1893-1987)-part III

τc

=

Critical shear stress in concrete

τv

=

Nominal shear stress in concrete

φ

=

Nominal diameter of bars

10


LIST OF TABLES

PAGE NO.

1. Value of Topography Factor K 2

11

2. Wind load calculation

11

3. Wind forces from Analysis (X Direction)

12

4. Wind forces from Analysis (Y Direction)

12

5. Seismic zone

14

6. Base shear in X-direction using ETABS model

17

7. Base shear in Y-direction using ETABS model

18

8. Base shear in X-direction using ETABS model

19

9. Base shear in Y-direction using ETABS model

20

10. Period Vs Sa/g

20

11. Seismic weight output

22

12. Load combination

28

13. Moment coefficients

54

14. Reinforcement provided

54

11


LIST OF FIGURES

PAGE NO.

1. Location map

01

2. Panel drawing

07

3. Axial load from Analysis

09

4.

Plan & Elevation

10

5. Soil classification

15

6. Response spectra for soft soil

22

7. The new model initialization form.

23

8. Building Plan Grid System and Storey Data Definition

23

9. Material property data form

24

10. Define Frame Properties form

24

11. Section properties and Reinforcement details .

25

12. Define wall or slab section

25

13. Define Diaphragms

26

14. Define response spectrum function.

26

15. Response spectrum function graph

27

16. Define static load case

27

17. Define load combinations

28

18. Model output

30

19. Bending moment diagram from Analysis

31

20. Shear force diagram from Analysis

32

21. Diaphragm

33

22. Plan showing slab details

34

12


23. Plan showing beam and column ID

35

24. Types of footing

36

25. Pressure diagram

38

26. Footing detailing

40

27. Types of column

41

28. Column detailing

45

29. Beam detailing

50

30. Types of Slab

51

31. Load distribution in two way slab

53

32. R.C Details of Slabs

55

33. Slab detailing

56

34. Types of staircase

58

35. Detailing of staircase

60

36. Shear wall diagram

61

37. Shear wall detailing.

64

13


1.0 INTRODUCTION 1.1 NEW TEACHING HOSPITAL BUILDING :

Figure1-location map The New teaching hospital block located at Agartala. The total build up area of the hospital building is 12596.67 square meter and having five floors (G+5). The Hospital building consists of various divisions like Ortho ward,Orthopedic ward, Opthamal ward, ENT ward, major and minor operation theaters, out patient ward, seminar halls for medical students, scanning and X-ray centre and medicine store room etc. The building located at seismic prone zone (zone factor V). The building has designed according to the Earthquake resistant considerations.

14


1.2 LARSEN & TOUBRO LIMITED 1.2.1 Vision and Mission

The design and execution of New teaching hospital block were awarded to Larsen & Toubro Limited (L&T) is an Indian multinational conglomerate company headquarted in Mumbai, India. The company has four main sectors; Technology, Engineering, Construction and Manufacturing. L&T has an international presence with a global spread of office and factories further, supplemented by a comprehensive marketing and distribution network. The firm has more than 60 units in 25 countries. Domestic business with in India dominates, but

the company steadily growing it’s global operations with a focus on China and Middle East. The company was founded in Mumbai in 1938 by Danish Engineers, Mr.Henning HlockLarsen and Seren Kristian Toubro. In 1944 ECC was incorporated by the partners; presently ECC(Engineering Construction & Contracts Division of L&T) is the largest contruction organization. L&T covers various disciplines of construction- Civil, Mechanical, Instrumentation and Electrical. The design wing of L&T, EDRC division provides consultancy design and total Engineering solutions to customers. It carries out both residential and commercial projects. The training work was carried out in the EDRC division of L&T. 2.0 FEATURES OF PROJECT

The project consists of Seismic Analysis and Design of New Teaching Hospital block (NTH) located at Agartala. The architectural drawings of NTH were done in the Auto CAD 2010 and the structural modeling was done by using ETABS software. The concrete mix used for all the structural member is M 30 and steel is Fe 500. The load combination were taken to obtain the maximum design loads, bending moments and shear forces. The structural element designs were carried as per IS:456-2000 for the load combinations. Earthquake resistant design detailing of the structure was done as per IS:13920-1993.

15


3.0 INTRODUCTION TO ETABS:

ETABS is sophisticated software for analysis and design program developed specifically for buildings systems. ETABS version-9.7 features an in intuitive and powerful graphical interface coupled with unmatched modeling, analytical, and design procedures, all integrated using common database. Although quick and easy for simple structures, ETABS can also handle the largest and most complex building models, including a wide range of nonlinear behaviors, making it the tool of choice for structural engineers in the building industry. 3.1 MODELING FEATURES

The ETABS building is idealized as an assemblage of area, line and point objects. Those objects are used to represent wall, floor, column, beam, brace and link / spring physical members. The basic frame geometry is defined with reference to a simple three-dimensional grid system. With relatively simple modeling techniques, very complex framing situations may be considered. The building may be unsymmetrical and non-regulator in plan, Torsional behavior of the floors and understory compatibility of the floors are accurately reflected in the results. The solution enforces complete three-dimensional displacement compatibility, making it possible to capture tubular effects associated with the behavior of tall structures having relatively closely spaced columns. Semi-rigid floor diaphragms may be modeled to capture the effects of in plane floor deformations. Floor objective may span between adjacent levels to create sloped floors (ramps), which can be useful for modeling parking garage structures. 3.2 ANALYSIS FEATURES

Static analysis for user specified vertical and lateral floor on story loads are possible. If floor elements with plate bending capability are modeled, vertical uniform loads on the floor are transferred to the beams and columns through bending of the floor elements. The program can automatically generate lateral wind and seismic load patterns to meet the requirements of various building codes. Three dimensional mode shapes and frequencies, model participation factors, direction factors and participating mass percentage are evaluated

16


using Eigen vector or Ritz-vector analysis-Delta analysis effects may be included with static or dynamic analysis. Response spectrum analysis, linear time history analysis, nonlinear analysis and static nonlinear analysis are possible. The static nonlinear capabilities also allow you to perform incremental construction analysis, so that forces that arise as a result of construction sequence are included. Results from the various static load cases may be combined with each other or with the results from the dynamic response dynamic response spectrum or time history method. Output may be viewed graphically, displayed in tabular output, the types of output include reactions, member forces, mode shapes, participation factors, static and dynamic story displacements and story shears inter story drifts and joint displacements, time history traces and more.

17


4.0 LOAD CALCULATIONS 4.1 GENERAL DATA

Structure

=G+5

Floor height ( First Floor to Fifth Floor)

= 4.0 m

Grade of concrete (for all structural elements)

= M 30

Unit weight of concrete

= 25kN/m

Unit weight of cement mortar

= 24kN/m

Unit weight of water

= 10kN/m

Unit weight of Brick

= 20kN/m

3

3

3

3

4.2 DEAD LOAD (As per IS 875 part I)

4.2.1 TYPICAL FLOOR SLAB: 2

Self-weight of slab (150mm thick)

= 3.75 kN/m

Floor Finish

= 1.5 kN/m

Total Load

= 5.25 kN/m

2

2

4.2.2 TERRACE FLOOR SLAB: 2

Self-weight of slab (150mm thick)

= 3.75 kN/m

Terrace (Roof finish + Water proofing)

= 3.20 kN/m

Total Load

= 6.95 kN/ m

2

2

4.3 LIVE LOAD (As per IS 875 part II)

Living room, Toilet & Bath room

= 3.00 kN/m

2

Recreation Room and Pantry

= 3.00 kN/m

2

Store and Laundry

= 4.00 kN/m

2

18


2

Corridors & stair case

= 4.00 kN/m

Terrace Floor

= 1.5 kN/m

ICU

= 3.00 kN/m

2

Lounge

= 3.00 kN/m

2

Nurse station

= 3.00 kN/m

2

Operation Theatres

= 3.00 kN/m

2

Dining room

= 4.00 kN/m

2

Waiting room

= 4.00 kN/m

2

2

4.4 WALL LOAD

4.4.1 Member load due to 230mm thick exterior wall Wall thickness

= 230 mm

Floor Height

= 4000 mm

Beam Depth

= 750 mm

(0.23 × (4-0.75) × 20)

= 15kN/m

4.4.2 Member load due to 230mm thick interior wall Wall thickness

= 115 mm

Floor Height

= 4000 mm

Beam Depth

= 750 mm

(0.115 × (4-0.75) × 20

= 7.5 kN/m

4.5 STATIC CHECK FOR LOAD COMBINATION (D.L + L.L)

4.5.1 PLINTH LEVEL Beam size

= 300×750mm

Self-weight of beam (B1)

= 25×0.75 ×0.3×3.42 19


= 19.24kN/m Self-weight of beam (B2)

= 25× 0.75 ×0.3×4.28 = 24.08kN/m


= 25× 0.75 ×0.3×3.60 = 20.25kN/m


= 25× 0.75 ×0.3×4.95 = 27.85kN/m

Wall load on beam (B2, B3 & B4)

= (0.115×(4-0.75)×20 = 7.5kN/m

Column size

= 750×750mm

Self-weight of column C1

=25× (4-0.75) ×0.75×0.75 = 45.70kN

Total Load at Plinth level

= 159.62 kN/m

. Figure 2-Panel drawing

4.5.2 FLOOR LEVEL Beam size

= 300×750mm 20



= 25× 0.75 ×0.3×3.42 = 19.24kN/m


= 25× 0.75×0.3×4.28 = 24.08kN/m


= 25× 0.75 ×0.3×3.60 = 20.25kN/m


= 25× 0.75 ×0.3×4.95 = 27.85kN/m

Self-weight of Slab

= 25× 0.15×9.23×7.02 = 242.98kN/m

Column size

= 750×750mm

Self-weight of column

=25× (4-0.75) ×0.75×0.75 = 45.70kN

Wall load on beam (B2, B3 & B4)

= 0.115×(4-0.75) ×20 = 7.5kN/m

Live load on Slab

= 5.5×9.23×7.02 = 356.37kN/m

2

4.5.3 STATIC CHECK FOR DEAD AND LIVE LOAD COMBINATION Total dead load

= 2417.72kN/m

Total dead Imposed load

= 157.50kN/m

Total live load (Floors+ Terrace)

= 2138.22kN/m

Total load

= 4713.44 kN/m

Axial load from Analysis

= 4737.58 kN/m 21


Figure 3- Axial load from software 4.6 WIND LOAD CALCULATION

Wind is air in motion relative to the surface of the earth. The primary cause of wind is traced to earth’s

rotation and differences in terrestrial radiation. The radiation effects are primarily

responsible for convection either upwards or downwards. The wind generally blows horizontal to the ground at high wind speeds. Since vertical components of atmospheric

motion are relatively small, the term ‘wind’ denotes almost exclusively the horizontal wind, vertical winds are always identified as such. The wind speed is assessed with the aid of anemometers or anemographs. 4.6.1BASIC WIND SPEED:

Basic wind speed is based on peak gust velocity averaged over a short time interval of about three seconds and corresponds to mean heights above ground level in an open terrain. Basic wind speed given in figure1 in IS 875(part):1987. 4.6.2 DESIGN WIND SPEED :

The basic wind speed for any site is obtained from figure and shall be modified to i nclude the following effects to get design wind velocity at any height shall be modified to include the following effects to get design wind velocity at any height(Vz) for the chosen structure: Vz = Vb× k 1×k 2× k 3 22


The suggested life period to be assumed in design and the corresponding k 1 factors for different structures for different class of structures given in table 1, IS 875(part):1987. k 2 is the terrain, Height and structure size factor given in table 2, IS 875(part):1987. k 3 is the topography factor given in table 3, IS 875(part):1987. 4.6.3 DESIGN WIND PRESSURE :

The design wind pressure at any height above mean ground level shall be obtained by the following relationship between wind pressure and wind velocity 2

Pz = 0.6×V z 4.6.4 WIND FORCES:

The value of force coefficient apply to the building or structure as a whole and multiplied by effective frontal area of the building by design wind pressure, P z gives the total wind load on that particular building or structure. The force coefficients are given in two mutually perpendicular directions relative to reference axis of the structural member. They are designed as C pn and Cpt, give the normal and transverse, respectively to the reference plane. Fn = Cpn× Pz×l×b Ft= Cpt × Pz× l×b Basic wind speed at Agartala (As per IS 875-1987)

= 50km/s (Vb)

Terrain category (clause 5.3.2.1)

=2

Building class (B=36m, W=53.0) (clause 5.3.2.2)

=C

Risk coefficient (K1) (Assume 100 years Life period)

= 1.08

Topography factor (K3)(clause 5.3.2.3)

=1

a=53.0m b=36m

a=53.0m h=28.0m

Fig4.A- Plan

Fig 4.B- Elevation 23


In X-direction (a/b) = 1.47

In Y-direction (b/a)

= 0.68

In X-direction (h/b) = 0.78

In Y-direction (h/a)

= 0.53

Width of building in X-direction

= 53.0m

Width of building in Y-direction

= 36m

1. Value of Topography Factor (k 2) - Table 1

Height(m)

K2

10

0.93

15

0.97

20

1.00

30

1.04

50

1.10

Force coefficient in X-direction

= 1.075(from code IS 875-1987(part3))

Force coefficient in Y-direction

= 1.20(from code IS 875-1987(part 3))

Design speed Vz

= Vb× K1 × K2 × K3

2. Wind load calculation-Table 2 Load Story

Height(m)

K2

Vz(km/s)

X

2

in

Pz=0.6Vz (kN) direction(kN)

Load

in

direction(kN)

FIRST

4.0

0.930

50.22

1.513

351.32

577.36

SECOND

8.0

0.930

50.22

1.513

234.21

384.91

THIRD

12.0

0.950

51.30

1.579

244.43

401.69

FOURTH

16.0

0.979

52.87

1.677

259.60

426.63

FIFTH

20.0

1.000

54.00

1.749

270.75

444.95

TERRACE

24.0

1.016

54.86

1.805

139.71

229.60

Total

1500.02

2465.14

24

Y


3. Wind forces from Analysis – Table 3 Story

Point

Load

FX

FY

FZ

MX

MY

MZ

(kN)

(kN)

(kN)

(kN-m)

(kN-m)

(kN-m)

BASE

1

WINDPX

-3.94

-0.36

-11.44

0.304

-15.199

0.071

BASE

3

WINDPX

-5.17

0.35

1.63

-1.013

-16.898

-0.036

BASE

58

WINDPX

-0.22

-7.95

-30.13

0.341

-0.544

0.041

BASE

59

WINDPX

-0.35

10.82

42.97

-0.494

-0.649

0.05

BASE

96

WINDPX

-6.44

-0.43

119.41

1.174

-18.245

0.121

BASE

115

WINDPX

-6.22

0.15

-13.46

-0.426

-19.163

-0.216

BASE

379

WINDPX

-27.28

-0.03

-107.35

0.097

-2.789

0

BASE

380

WINDPX

-27.92

0

-33.24

0.012

-2.75

0

BASE

184

WINDPX

-4.39

-0.13

-1.19

0.179

-0.241

-0.001

BASE

191

WINDPX

3.41

-0.67

-16.6

0.358

-0.051

0.007

BASE

200

WINDPX

-3.12

0.06

-2.57

0.054

-0.102

0.002

BASE

201

WINDPX

4.55

-0.78

23.06

0.423

-0.029

-0.007

WINDPX

1500.02

FX

FY

FZ

MX

MY

MZ

(kN)

(kN)

(kN)

(kN-m)

(kN-m)

(kN-m)

Summation

4. Wind forces from software-Table 4 Story

Point

Load

BASE

1

WINDPY

-1.01

-6.36

-33.52

23.774

-3.106

-0.198

BASE

3

WINDPY

-1.36

-9.92

-34.13

31.666

-3.66

-0.205

BASE

5

WINDPY

-1.94

-6.61

-39.98

25.27

-4.641

-0.171

BASE

73

WINDPY

-18.29

-0.3

-62.8

0.695

-0.976

-0.097

BASE

74

WINDPY

5.39

-65.76

-255.22

6.363

-0.504

0.21

BASE

117

WINDPY

-0.6

-2.1

-224.58

0

0

0

BASE

119

WINDPY

0.75

-2.38

-229.76

0

0

0

BASE

374

WINDPY

20.31

-0.44

94.16

1.313

1.667

0

BASE

375

WINDPY

20.38

-0.46

105.38

1.36

1.582

0.001

BASE

197

WINDPY

2.35

0.3

-4.6

0.196

0.157

-0.002

BASE

199

WINDPY

-19.47

-1.77

31.28

0.924

-0.613

0.034

BASE

200

WINDPY

-3.02

0.32

-5.15

0.212

-0.177

0.003

BASE

201

WINDPY

14.03

-2.17

77.89

1.349

0.306

-0.02

Summation

WINDPY

2465.14 25


4.7 SEISMIC LOAD CALCULATION (Based on code IS 1893-2002)

During an earthquake, ground motions develop in a random manner both horizontally and vertically in all directions radiating from the epicenter. The ground motions develop vibrations in the structure inducing inertial forces on them. Hence structures located in seismic zones should be suitably designed and detailed to ensure strength, serviceability and stability with acceptable levels of safety under seismic forces. The satisfactory performance of a large number of reinforced concrete structures subject to severe earthquake in various parts of the world has demonstrated that it is possible to design structures to successfully withstand the destructive effects of major earthquakes. The Indian standard codes IS: 1893-1984 and IS: 13920-1993 have specified the minimum design requirements of earthquake resistant design probability of occurrence of earthquakes, the characteristics of the structure and the foundation and the acceptable magnitude of damage. Determination of design earthquake forces is computed by the following methods, 1) Equivalent static lateral loading. 2) Dynamic Analysis. In the first method, different partial safety factors are applied to dead, live, wind earthquake forces to arrive at the design ultimate load. In the IS: 456-2000 code, while considering earthquake effects, wind loads assuming that both severe wind and earthquake do not act simultaneously. The American and Australian code recommendations are similar but with different partial safety factors. The dynamic analysis involves the rigorous analysis of the structural system by studying the dynamic response of the structure by considering the total response in terms of component modal responses. 4.7.1 ZONE FACTOR (Z): 

The values of peak ground accelerat ion considered earthquake.

26

given in units ‘g’ for the ma ximum




The value of (Z/2) corresponds to design basis earthquake damage control in limit state.



Based on history of seismic activities seism tectonic understanding the entire country has been divided in to four zones. The zone factor from table 2(IS 1893:2002)

Seismic zone

II

III

IV

V

Seismic Intensity

Low

Moderate

Severe

Very severe

Z

0.10

0.16

0.24

0.36

Zone factor values- Table 5 4.7.2 RESPONSE REDUCTION FACTOR (R): 

R is the response reduction factor and controls the permitted damage in design basis earthquake.



The minimum value of R is 3 and maximum is 5 however to use higher values of R special ductile detailing requirements are must and the designer is accepting more damages but in the controlled manner. The Response reduction factor from table 7(IS 1893:2002)

4.7.3 IMPORTANCE FACTOR (I): 

I is the importance factor and permitted damage could be reduced by setting the value

of I more than ‘1’. 

For the buildings like ‘HOSPITALS’, communication and community buildings the value is 1.5 from table 6 (IS 1893:2002).

4.7.4 SEISMIC WEIGHT (W): 

Seismic weight of the building is measured in Newton. Seismic weight includes the dead loads (that of floor, slabs, finishes, columns, beams, water tanks, permanent machines etc.



Seismic weight includes only a part of Imposed loads, for example 25% to 50% of imposed loads for buildings from table 8 (IS 1893:2002).

27


4.7.5 SOIL CLASSIFICATION: 

2

Sa/g is the lateral acceleration to be established in m/s . For 5 present of damping three different types of curves are recommended in IS 1893:2002 for different stiffness of supporting media-Rock, Medium soil and Soft soil.



The classification of soil is based on the average shear velocity for 30m o f rock or soil layers or based on average Standard Penetration Test (SPT) values for top 30m.

3 ) g / a 2.5 S ( t n e i c i f f 2 e o C n o 1.5 i t a r e l e c c 1 A l a r t c e 0.5 p S

Type I (Rock, or Hard Soil) Type II (Medium Soil) Type III (Soft Soil)

0 0

0.5

1

1.5

2

2.5

3

3.5

4

Period (s)

Fi

Figure 5- soil classification graph

Zone factor of the building

(Z)

= 0.36 (Zone V)

Importance of the building

(I)

= 1.5 (Post Earth quake service needed)

Response reduction factor

(R)

= 5 (Ductile shear walls with SMRF system)

Soil type

= Soft soil

Width of building in X-direction

= 53.0m

Width of building in Y-direction

= 36m

Height of the building

= 28m

Seismic weight of the building

= 177309kN

28


4.7.6 SEISMIC BASESHEAR CALCULATION:

The design base shear is the sum of lateral forces applied at all levels that are finally transferred to the ground. Vb

= A h× W

Ah

= (ZI/2R)(Sa/ g)

Fundamental Natural Time period (without infill) 0.75

Ta

= 0.075(h)

Ta (X-Direction)

= 0.913

Ta (Y-Direction)

= 0.913

Sa/g

= 1.67/T = 1.83

Ah (X-Direction)

= 0.099

Ah (Y-Direction)

= 0.099

Fundamental Natural Time period (with infill) Ta Ta (X-Direction) Sa/g Ah (X-Direction)

= (0.09×h)/√d = 0.420 & Ta (Y-Direction) = 0.346 = 2.5 = 0.135 &Ah (Y-Direction)

= 0.135

Base shear (without infill) Base shear (X direction) (A h×W)

= 17554kN

Base shear (Y direction) (A h×W)

= 17554kN

Base shear (with infill) Base shear (X direction) (A h×W)

= 23937kN

Base shear (Y direction)(A h×W)

= 23937kN

29


Average base shear Base shear in X-direction using Code

= 20746kN

Base shear in Y-direction using Code

= 20746kN

Base shear in X-direction using ETABS model

= 6760kN

Base shear in Y-direction using ETABS model

= 6173.94kN

4.7.7 ANALYSIS

Analysis options are set before the analysis, the analysis is performed with a scale factor 1. The number of modes is initially set as 1.5 after anlysis. If the cumulative mass participation factor is less than 95 percentage, then it is modified accordingly with base shear values obtained for the earth quake load case the new scale factor is calculated and again the model is analysed for the new scale factor. It can be observed that the base shear value calculated from the code and by the software with the new scale factor are the same. 6. Base shear in X-direction using ETABS model- Table 6 Story

Point

Load

FX

FY

FZ

MX

MY

MZ

(kN)

(kN)

(kN)

(kN-m)

(kN-m)

(kN-m)

BASE

1

SPECX 21.12

6.06

51.23

23.359

78.266

0.911

BASE

3

SPECX 26.97

8.54

30.56

25.089

86.532

1.435

BASE

5

SPECX 29.62

3.71

63.14

11.368

90.891

1.676

BASE

7

SPECX 161.37

34.86

772.61

1.607

14.971

0.587

BASE

8

SPECX 190.16

4.52

19.77

0.66

20.901

0.464

BASE

9

SPECX 161.54

29.35

759.54

1.607

15.034

0.604

BASE

62

SPECX 36.09

4.51

13.19

12.545

100.706

1.788

BASE

66

SPECX 10.92

1.08

12.92

0

0

0

BASE

115

SPECX 26.77

4.34

63.2

12.201

80.561

2.515

BASE

1591

SPECX 0.11

6.7

0.93

0

0

0

BASE

1592

SPECX 0.82

6.79

5.12

0

0

0

BASE

1594

SPECX 2.23

7.86

15.27

0

0

0

BASE

1595

SPECX 0.08

5.91

11.84

0

0

0

Sum

6760

30


7.Base shear in Y-direction using ETABS model- Table 7 Story

Point

Load

FX

FY

FZ

MX

MY

MZ

(kN)

(kN)

(kN)

(kN-m)

(kN-m)

(kN-m)

BASE

1

SPECY 5.67

14.41

91.79

52.89

17.692

1.018

BASE

3

SPECY 6.92

25.76

92.27

77.876

19.551

2.137

BASE

5

SPECY 8.2

18.87

118.39

65.448

21.629

1.569

BASE

7

SPECY 66.6

69.75

520.95

6.717

4.599

0.125

BASE

159

SPECY 0.34

39.45

118.78

5.311

0.793

0.009

BASE

160

SPECY 0.15

68.93

307.44

6.871

0.433

0.001

BASE

172

SPECY

11.22

0.16

77.79

0

0

0

BASE

174

SPECY

60.13

0.3

402.37

0

0

0

BASE

1587

SPECY

0.19

105.15

76.18

0

0

0

BASE

1588

SPECY

0.1

107.69

219.67

0

0

0

BASE

1591

SPECY

0.01

54.74

65.87

0

0

0

BASE

1592

SPECY

0.2

65.62

205.77

0

0

0

BASE

1594

SPECY

0.47

98.15

299.42

0

0

0

BASE

1595

SPECY

0.03

46.88

222.45

0

0

0

Sum

6173.94

4.7.8 SCALE FACTOR CALCULATION

Summation of support reaction gives the seismic weight of the building. Base shear in X direction calculated as per the code. ETABS did the dynamic analysis and total base shear in X and Y direction are found from results. The ratio of base shear obtained from code to that obtained from dynamic analysis gives the shear force in X and Y direction. Since the spectra value for different time period is calculated manually the base shear values might not be the same values obtained from manual calculations. Thus the new scale factor is calculated and the structure reanalyzed to obtain the new base shear values. Base shear (from Analysis) Base shear in X-direction using Code

= 20746kN


= 20746kN

31



= 6760kN


= 6173.94kN

Scale Factor (Vbx/ Vbx')

= 3.07

Scale Factor (Vby/ Vby')

= 3.36

8. Base shear in X-direction using ETABS model- Table 8 Story

Point

Load

FX

FY

FZ

MX

MY

MZ

(kN)

(kN)

(kN)

(kN-m)

(kN-m)

(kN-m)

BASE

1

SPECX 64.85

18.61

157.27

71.713

240.277 2.796

BASE

3

SPECX 82.81

26.2

93.83

77.024

265.653 4.404

BASE

5

SPECX 90.92

11.39

193.85

34.901

279.036 5.146

BASE

7

SPECX 495.42

107.03

2371.91 4.934

45.96

BASE

115

SPECX 82.19

13.33

194.02

37.458

247.321 7.72

BASE

117

SPECX 21.32

6.37

714.21

0

0

0

BASE

119

SPECX 23.94

6.78

724.44

0

0

0

BASE

121

SPECX

5.94

40.5

0

0

0

BASE

442

SPECX 340.62

9.65

1323.44 8.414

34.379

0.113

BASE

1587

SPECX 0.81

99.15

997.9

0

0

0

BASE

1588

SPECX 0.68

87.3

1162.35 0

0

0

BASE

1591

SPECX

0.32

20.57

2.86

0

0

0

BASE

1592

SPECX 2.52

20.84

15.73

0

0

0

BASE

1594

SPECX 6.84

24.14

46.89

0

0

0

BASE

1595

SPECX 0.26

18.13

36.35

0

0

0

Sum

13.81

1.803

20753.07

9. Base shear in Y-direction using ETABS model- Table 9 Story

Point

Load

FX

FY

FZ

MX

MY

MZ

(kN)

(kN)

(kN)

(kN-m)

(kN-m)

(kN-m)

BASE

1

SPECY

19.05

48.43

308.4

177.711

59.445

3.419

BASE

3

SPECY

23.25

86.57

310.02

261.662

65.691

7.179

BASE

5

SPECY

27.56

63.42

397.8

219.907

72.672

5.271

32


BASE

7

SPECY

223.79

234.38

1750.39

22.57

15.453

0.422

BASE

119

SPECY

10.08

20.21

2062.4

0

0

0

BASE

121

SPECY

4.34

22.2

24.92

0

0

0

BASE

123

SPECY

2.91

62.47

11.13

156.838

5.821

0.665

BASE

440

SPECY

210.17

7.31

1056.86

14.678

17.001

0.01

BASE

442

SPECY

216.26

17.22

1617.36

21.921

16.802

0.136

BASE

1592

SPECY

0.67

220.5

691.38

0

0

0

BASE

1594

SPECY

1.58

329.78

1006.05

0

0

0

BASE

1595

SPECY

0.09

157.52

747.42

0

0

0

Sum

20744.37

Base shear in X-direction using Code

= 20746 kN


= 20746 kN


= 20753.07 kN


= 20744.37 kN

10. Period Vs Sa/g- Table 10 Period

Acceleration 2

(seconds)

Sa /g

(m /s)

0

1.000

0.530

0.1

2.500

1.324

0.2

2.500

1.324

0.3

2.500

1.324

0.4

4.175

2.212

0.5

3.340

1.769

0.6

2.783

1.474

0.7

2.386

1.264

0.8

2.088

1.106

0.9

1.856

0.983

1

1.670

0.885

1.1

1.518

0.804

1.2

1.392

0.737

33


1.3

1.285

0.681

1.4

1.193

0.632

1.5

1.113

0.590

1.6

1.044

0.553

1.7

0.982

0.520

1.8

0.928

0.491

1.9

0.879

0.466

2

0.835

0.442

2.1

0.795

0.421

2.2

0.759

0.402

2.3

0.726

0.385

2.4

0.696

0.369

2.5

0.668

0.354

2.6

0.642

0.340

2.7

0.619

0.328

2.8

0.596

0.316

2.9

0.576

0.305

3

0.557

0.295

3.1

0.539

0.285

3.2

0.522

0.276

3.3

0.506

0.268

3.4

0.491

0.260

3.5

0.477

0.253

3.6

0.464

0.246

3.7

0.451

0.239

3.8

0.439

0.233

3.9

0.428

0.227

4

0.418

0.221

34


Figure 6-Response spectra for Soft soil sites

11. Seismic weight output- Table 11 Story

Point

Load

FX

FY

FZ

MX

MY

MZ

(kN)

(kN)

(kN)

(kN-m)

(kN-m)

(kN-m)

BASE

1

SISWT

27.65

21.32

2497.5

-34.428

42.726

0.476

BASE

3

SISWT

6.31

19.1

4190.33

-33.679

12.518

-0.08

BASE

5

SISWT

0.11

19.9

3577.2

-33.288

3.005

0.052

BASE

72

SISWT

-17.8

-13.2

2293.75

19.13

-29.645

-0.359

BASE

73

SISWT

55.84

-1.31

243.8

1.129

2.512

-0.173

BASE

74

SISWT

-57.53

62.27

545.42

-4.294

-3.983

0.017

BASE

148

SISWT

-13.74

-31.27

3478.73

54.117

-21.508

0.292

BASE

149

SISWT

-15.04

-14.49

1985.17

20.963

-23.614

0.588

BASE

410

SISWT

21.85

53.21

477.07

-2.793

2.787

-0.171

BASE

416

SISWT

-0.1

22.94

277.04

1.069

0.413

-0.059

BASE

198

SISWT

34.07

-1.58

56.5

0.839

1.065

-0.028

BASE

199

SISWT

-33.99

-1.54

56.36

0.832

-1.057

0.028

BASE

200

SISWT

-7.72

0.22

-8.62

0.142

-0.455

0

Sum

177309.1

35


5.0 MODELING AND ANALYSIS OF THE BUILDING

The primary purpose of structural analysis in building structure i s to establish the distribution of internal forces and moment over the whole or part of the structure and to identify the critical design conditions at all sections the geometry is commonly idealised by considering the structure to make up of linear elements. Here the structural analysis is carried out by ETABS 5.1 BASIC GRID SYSTEM:

Begin creating the grid system by clicking the File menu > New model command or the New model icon. The form shown in figure below wil display. Select the No option on that form the next figure will display.

Figure 7-The new model initialization form. 5.2 DEFINE GEOMETRY:

The Building Plan Grid System and Storey Data form is used to specify horizontal and vertical grid line spacing, storey data, storey elevation and units. They automatically add the structural objects with appropriate properties to the model.

Figure 8- Building Plan Grid System and Storey Data Definition

36


5.3 DEFINE MATERIAL PROPERTY:

The material properties of each object in the model is specified in the appropriate form. The material used is concrete, the grade of concrete, the properties of concrete such as Mass per unit volume, Modulus of Elasticity of concrete,Poisson ratio are specified and for steel yield strength is specified.

Figure 9- Material pr operty data form. 5.4 DEFINE FRAME SECTION:

Assign the frame section such as Column and Beam. Select the section property as Rectangle and define the depth,width and reinfprcement details,cover provisions.similarly for various sections like circular,pipe, steel joist sections aiso assigned with suitable data.

Figure 10- Define Frame Properties form.

37


Figure 11- Section properties a nd Reinforcement details . 5.5 DEFINE WALL OR SLAB SECTION:

Assign the slab or wall section then assign the section name,thickness,material used,type and reinforcement details.

Figure 12- Define wall or slab section

38


5.6 DEFINE DIAPHRAM:

The lateral loads can be in the form of wind or seismic loads, the loads are automatically calculated from the dimensions and properties of the structure based on built-in options for a variety of building codes. For Rigid diaphragm systems, the wind loads applied at the geometric centers of each rigid floor diaphragm.

Figure 13-Define Diaphragms 5.7 DEFINE RESPONSE SPECTRUM FUNCTION:

Functions are defined to describe how a load varies as a function of period, time or frequency.

Figure 14- Define response spectrum function. 5.8 RESPONSE SPECTRUM FUNCTIONS

Response spectrum functions are pseudo spectral acceleration versus period functions for use in response spectrum analysis. In this program the acceleration values to be normalized; that 39


is, the functions themselves are not assumed to have units. Instead, the units are associated with a scale factor that multiplies the function and is specified when we define the response spectrum case.

Figure 15- Response spectrum function graph 5.9 STATIC LOAD CASES:

Loads represent the actions upon the structure, such as force, pressure, support displacement, thermal effects and others. A spatial distribution of loads upon the structure is called static load case. Define as many load cases as needed. Typically separate load case definitions would be used for dead load, live load, static earthquake load, wind load, snow load, and Thermal load.

Figure 16- Define static load case 40


5.9 DEFINE LOAD COMBINATIONS :

Define the load combinations in the appropriate form; Select the Add new combo option, then enter the load combination name, assign the loads with suitable scale factor. When the combination is defined, it applies to the results for every object in the model.

Figure 17- Define load combinations

12. Load combination - Table 12 COMB

DEAD LOAD

LIVELOAD<3

LIVELOAD>3

COMB01

1.5 DEADLOAD +

1.5LIVELOAD +

1.5 LIVELOAD

COMB02

1.2 DEADLOAD +

0.6LIVELOAD +

0.3 LIVELOAD

+1.2 E.Q.LOADX

COMB03

1.2 DEADLOAD +

0.6 LIVELOAD +

0.3 LIVELOAD

+1.2 E.Q.LOADY

COMB04

1.5 DEADLOAD +

+1.5 E.Q.LOADX

COMB05

1.5 DEADLOAD +

+1.5 E.Q.LOADY

COMB06

0.9 DEADLOAD +

-1.5 E.Q.LOADX

COMB07

0.9 DEADLOAD +

-1.5 E.Q.LOADY

COMB08

1.2 DEADLOAD +

1.2 LIVELOAD +

1.2 LIVELOAD

+1.2WINDLOADX

COMB09

1.2 DEADLOAD +

1.2 LIVELOAD +

1.2 LIVELOAD

-1.2WINDLOADX

COMB10

1.2 DEADLOAD +

1.2 LIVELOAD +

1.2 LIVELOAD

+1.2WINDLOADY

COMB11

1.2 DEADLOAD +

1.2 LIVELOAD +

1.2 LIVELOAD

-1.2WINDLOADY

COMB12

1.5 DEADLOAD +

+1.5WINDLOADX

COMB13

1.5 DEADLOAD +

-1.5 WINDLOADX

COMB14

1.5 DEADLOAD +

+1.5WINDLOADY 41

WIND/ E.QLOAD


COMB15

1.5 DEADLOAD +

-1.5 WINDLOADY

COMB16

0.9 DEADLOAD +

+1.5WINDLOADX

COMB17

0.9 DEADLOAD +

-1.5 WINDLOADX

COMB18

0.9 DEADLOAD +

+1.5WINDLOADY

COMB19

0.9 DEADLOAD +

-1.5 WINDLOADY

UNFACTORED COMBINATION COMB 20

1 DEADLOAD +

1 LIVELOAD +

1 LIVELOAD

COMB 21

1 DEADLOAD +

0.5 LIVELOAD +

0.25LIVELOAD

+1 E.Q.LOADX

COMB 22

1 DEADLOAD +

0.5 LIVELOAD +

0.25LIVELOAD

+1 E.Q.LOADY

COMB 23

1 DEADLOAD +

-1 E.Q.LOADX

COMB 24

1 DEADLOAD +

-1 E.Q.LOADY

COMB 25

1 DEADLOAD +

1 LIVELOAD +

1 LIVELOAD

+1 WINDLOADX

COMB 26

1 DEADLOAD +

1 LIVELOAD +

1 LIVELOAD

-1 WINDLOADX

COMB 27

1 DEADLOAD +

1 LIVELOAD +

1 LIVELOAD

+1 WINDLOADY

COMB 28

1 DEADLOAD +

1 LIVELOAD +

1 LIVELOAD

-1 WINDLOADY

COMB 29

1 DEADLOAD +

+1 WINDLOADX

COMB 30

1 DEADLOAD +

-1 WINDLOADX

COMB 31

1 DEADLOAD +

+1 WINDLOADY

COMB 32

1 DEADLOAD +

-1 WINDLOADY

42


5.11 MODEL OUTPUT A) OUTPUT: 3D-MODEL OF A RC FRAME

Figure.18-3D model of RC frame 43


B) BENDING MOMENT DIAGRAM FROM ANALYSIS

Figure 19- Bending Moment from analysis.

44


C) SHEAR FORCE FROM ANALYSIS

Figure 20- Shear Force from analysis. 45


D) DIAPHRAGM

Figure 21-Diaphram output from analysis.

46


E) PLAN SHOWING SLAB ID

Figure 22- Plan showing slab ID

47


E) PLAN SHOWING BEAM ID

Figure 23- Plan showing Beam ID

48


6.0 DESIGN OF FOUNDATION 6.1GENERAL

In a typical structure built on a ground, that part of structure which is located above the ground is generally referred to as the superstructure and the part which lies below ground is referred to as substructure or foundation. The purpose of the foundation is to effectively support the superstructure by 

Transmitting the applied load effects to the soil below without exceeding the safe bearing capacity of soil.



Ensuring that the settlement of the structure is within the tolerable limits, and as uniform as possible.

Further the foundation should provide adequate safety against possible instability due to overturning or sliding and possible pullout. Footings belong to the category of shallow foundation, types of footings are Isolated footing, Combined footing and Wall footing. In some cases it may be inconvenient to provide separate isolated footings for columns (or walls) on account of inadequate areas available in plan. This may occur when two or more columns are located close to each other or if they are relatively heavily loaded and rest on soil with low safe bearing capacity resulting in an overlap of areas if isolated footings are attempted. In such cases it is advantageous to provide a single footing for the column.

Isolated footing

Figure 24-Types of footing 6.2 DESIGN OF ISOLATED FOOTING

For ordinary structures located on reasonably firm soil, it usually suffices to provide a separate footing is also called an isolated column. It is generally square or rectangular in plan other shapes are resorted to under special circumstances. The footing basically comprises a thick slab which may be flat, stepped or sloped. 49


6.3 DESIGN OF TYPICAL ISOLATED FOOTING 6.3.1 DESIGN PARAMETERS : 2

Grade of concrete using

(f ck )

= 30N/mm

Grade of steel using

(fy)

= 500 N/mm

2

Load combination considered (COMBO 5)

= 1.5D.L+1.5E.L

Axial load

= 1584.89 kN

Moment in X direction (Mux)

= 18.26 kNm

Moment in X direction (Muy)

= 16.13 kNm

Safe bearing capacity of soil

= 100 kN/ m

Unit weight of concrete

= 25kN/m

Shape of the footing

= Square footing

Type of the footing

= Flat

Assume self-weight of footing as 15%

= 1.15×1584.89 kN

Total weight

= 1822.62 kN(unfactored)

[P/A] ± [Mx /Z] ± [My]

≤ q (SBC of soil)

2

3

2

2

2

[1822.62/B.L] + [18.26×6/B.L ] + [18.26×6/B.L ]

= 100 kN/ m

Assume L/B ratio

=1 (L=B)

Solve the above equation, Size of the footing

= 4.4m×4.4 m

[P/A] + [Mx /Z] + [My]

P1

= 96.01 kN/ m

2

[P/A] +[Mx /Z] - [My]

P2

= 93.73 kN/ m

2

[P/A] - [Mx /Z] + [My]

P3

= 94.55 kN/ m

2

[P/A] - [Mx /Z] - [My]

P4

= 92.27 kN/ m

2

Assume depth of footing

= 700 mm

Downward pressure due to base slab (0.7×25)

= 17.5 kN/ m

50

2


Figure 25-Pressure diagram 6.3.2 BENDING MOMENT IN X DIRECTION

Cantilever projection along X

= 1.825 m

2

2

Mx [94.22×0.5×1.825 ]+[.5×1.825×1.945×.67×1.825]-[17.5×0.5×1.825 ] =129.91kNm Steel required along X direction Mu lim

= 0.138×f ck ×b×d

Depth required

= 178mm

Depth provided [700-50-(25/2)]

= 637mm

2

6

2

Mu/b.d = [1.5×129.91×10 ] / [1000×637 ]

= 0.48

Percentage of steel

= 0.117

Provide percentage of steel

= 0.75 2

Provide 25mm diameter bars and spacing [.785×25 ×1000]/4400 =111.50 Provide 25mm diameter bars at spacing of 100mm c/c.

51

2


6.3.3 BENDING MOMENT IN Y DIRECTION:

Cantilever projection along Y

= 1.825 m

2

2

My[94.22×0.5×1.825 ]+[.5×1.825×1.945×.67×1.825]-[17.5×0.5×1.825 ]=129.91kNm Steel required along Y direction Mu lim

= 0.138×f

Depth required

= 178mm

Depth provided [700-50-(25/2)]

= 637mm

2

6

2

Mu/b.d = [1.5×129.91×10 ] / [1000×637 ]

= 0.48

Percentage of steel

= 0.117

Provide percentage of steel

= 0.75

ck ×b×d

2

Provide 25mm diameter bars and spacing [0.785×25 ×1000]/4400 =111.50 Provide 25mm diameter bars at spacing of 100mm c/c.

6.3.4 CHECK FOR ONE WAY SHEAR:

Critical section located a distance of

‘d’ from the face of the column.

Average pressure (95.187+94.215)/2

= 94.70 kN/ m

Shear force at critical section (94.70-15)×(1.825-.637)

= 94.60

Nominal shear stress τv = (Vu×1.5)/b.d

= 0.22N/ mm

2

From IS: 456-2000, table 19 Percentage of reinforcement

= 0.59N/ mm

2

τv

Safe in one way shear.

2

< τc

6.3.5 CHECK FOR TWO WAY SHEAR : 2

Average pressure (94.345+94.14)/2

= 94.24 kN/ m

Vu = 94.24× [(4.4×4.4)-(0.75+0.637)(0.75+0.637)]

=1563.06kN

Nominal shear stress τv = (Vu×1.5)/b.d

= 0.66 N/ mm

2

= 1.37 N/ mm

2

τc = 0.25×√(f ck ) τv < τc Hence safe in two way shear.

52

2


6.3.6 TRANSFER OF FORCES AT COLUMN BASE:

As some bars are in tension, no transfer of forces is possible through bearing at the column footing interface. So, these bars may be extended into the footing. Development length required for 32mm diameter bars (In tension) Ld = (φ×fy×0.87)/ (4×1.6×1.8)

= 38φ

Ld = 38 ×32

= 1216mm

Length available = 750-50-25-16+(8×32)

= 915mm

The balance length = 301mm adopt

= 350mm

Need to be provided beyond the bend point.

6.4 DETAILING OF FOOTING

Figure 26-RC detailing of footing detailing

Y 25 AT 100 mm C/C = 25mm dia bars at 100mm c/c Y 25 AT 100 mm C/C = 25mm dia bars at 100mm c/c

53


7.0 COLUMN DESIGN 7.1 GENERAL

The column is a compression member which is subjected (predominantly) to axial forces. The

IS 456:2000 code (clause 25.1.1) defines the column as a ‘compression member’ the effective length of which exceeds three times the least lateral dimensio n. The term ‘pedestal’ is used to describe a vertical compression member whose effective length is less than three times to least lateral dimension. Classification of columns Based on type of Reinforcement: 1)

Tied columns

2)

Spiral columns

3)

Composite columns

Figure 27-Types of column

Based on type of loading 1)

Axially loaded column

2)

Columns with uniaxial eccentric loading

3)

Columns with biaxial eccentric loading

Based on Slenderness Ratio 1)

Short columns

2)

Slender or long columns

54


7.2 COLUMN DESIGN 7.2.1 DEIGN PARAMETERS :

Terrace level: C-1 2


(f ck )

= 30N/mm


(fy)

= 500 N/mm

2


= 1.5D.L+1.5E.L

Axial load

= 316.44kN

Moment in X direction (Mux)

= 235.84kNm

Moment in X direction (Muy)

= 128.0 kNm

Size of the column

= 750×750mm

d'

= 60mm

Ratio of d’/D

= 60/750= 0.08

Pt / f ck

= 1.0/ 30= 0.03 3

Pu / f ck bd =(316×10 )/(30×750×750)

= 0.018

2

Mu / f ck bd [From the code IS 456- 1978 chart 48]

= 0.08

Mux1

=0.08×30×750×750

2

=1012.50kNm

Ratio of d’/D Pt / f ck

=60/750= 0.08 = 1.0/ 30

=0.03

3

Pu / f ck bd =(1586×10 )/(30×750×750)

= 0.018

2

Mu / f ck bd [From the code IS 456- 1978 chart 48]

= 0.08

Muy1

=0.08×30×750×750 =1012.50kNm

Percentage of reinforcement

= 1.0

Puz / Ag [From the code IS 456- 1978 chart 63]

= 17 = 17×750×750 = 9562 kN/m

2

7.2.2 BENDING MOMENT CALCULATION: αn

[Mux / Mux1] + [Muy / Muy1]

αn

≤1

[Pu /Puz]

= (316.44/ 9562) = 0.03

[Pu /Puz]≤0.2 so

αn

[235.84/ 1012.50] +[128/1012.50]

=1 = 0.36

55

2


0.36

[Muy / Muy1]

αn

≤1 = 0.80 > 0.0.13

[From the code IS 456- 1978 chart 64]

7.2.3 REINFORCEMENT PROVIDED :

Pt

= 1.0

Ast

= (1.0×750× 750)/ 100

Ast

=5625mm

2

2

Provide 20 numbers of 20 mm diameter bars. (Ast=6280mm ) Transverse reinforcement Tie diameter

φt

= φ long/4

25/4

= 6.25mm

Provide 8 mm diameter tie bars Tie spacing

St < or = 16

φ

=16×25= 517mm Spacing of ties should not greater than 300mm. As per IS13920:1993 Spacing of hoops should not greater than = (1/4 of least lateral dimension) (1/4)×750=187.5mm Provide 8 mm bars at spacing 190mm c/c as a transverse reinforcement 7.2.4 DESIGN PARAMETERS :

Fifth floor level 2


(f ck )

=30N/mm


(fy)

= 500 N/mm

2


= 1.5D.L+1.5E.L

Axial load

= 659.88kN/mm

Mux

= 122.65kNm

Muy

= 59.10kNm

Size of the column

= 750×750mm

d'

= 60mm

Ratio of d’/D

= 60/750= 0.08

Pt / f ck

= 1.0/ 30= 0.03

Pu / f ck bd 2

Mu / f ck bd

2

3

(659.88×10 )/(30×750×750) = 0.04 [From the code IS 456- 1978 chart 48] 56

= 0.08


Mux1

=0.08×30×750×750

2

=1012.50kNm

Ratio of d’/D

=60/75= 0.08

Pt / f ck

= 1.0/ 30=0.03 3

Pu / f ck bd

(659.88×10 )/(30×750×750) = 0.04

2

Mu / f ck bd

[From the code IS 456- 1978 chart 48]

= 0.08

Muy1

=0.08×30×750×750

2

=1012.50kNm Percentage of reinforcement

= 1.0

Puz / Ag [From the code IS 456- 1978 chart 63]

= 17 = 17×750×750 = 9562 kN/m

2

7.2.5 BENDING MOMENT CALCULATION: αn

[Mux / Mux1] + [Muy / Muy1]

αn

≤1

[Pu /Puz] = (659.88 / 9562.5)

= 0.07

[Pu /Puz]≤0.2 so

αn

[122.16/ 1012.5] + [59.10/1012.5]

=1 = 0.18

0.18 αn

[Muy / Muy1] [From the code IS 456- 1978 chart 64]

≤1 =0.80> 0.05(calculated).

7.2.6 REINFORCEMENT DETAILS:

Pt

= 1.0

Ast

= (1.0×750× 750)/ 100

Ast

=5625mm 2

Provide 20 numbers of 20 mm diameter bars 4 on each face (6280 mm )

57

2


7.3 COLUMN DETAILING:

Figure 28.a-Longitudinal section

Figure 28.b-Cross section

58


8.0 BEAM DESIGN 8.1 GENERAL: 

A reinforced concrete flexure member should be able to resist tensile, compressive and shear stresses induced in it by loads acting on the member.



Concrete is fairly strong in compression but week in tension.



Thus the tensile weakness of concrete is overcome by the provision of reinforcing steel in the tension zone.



A flexure member may be Beam, slab, Wall or Component of foundation.

There are three types of reinforced concrete beams: 1. Singly reinforced beams 2. Doubly reinforced beams 3. Singly or doubly reinforced flanged beams Singly Reinforced Beams. 

In Singly reinforced simply supported beams or slabs reinforcing steel bars are placed near the bottom of beam or slabs where they are most effective in resisting the tensile stresses.



In the case of cantilever beams or slabs reinforcing steel bars are placed near the top of the beam or slabs for the same reason.

Doubly Reinforced Beams. A doubly reinforced concrete section is reinforced in both compression and tension regions. The section of the beam or slab may be rectangle, T and L section. The necessity of using steel in compression region arises due to two main reasons: 

When depth of the section is restricted, the strength available from singly reinforced section is in adequate.



At a support of a continuous beam or slab where bending moment changes sign. Such a situation may also arise in the design of a beam circular in plan.

59


Flanged beams: When a reinforced concrete slab is cast monolithically with the beam as in the case of beam supported floor slab system, the beams can be considered as flanged beams with slab acting as an effective flange on compression side. 

Minimum reinforcement – The minimum area of tension reinforcement shall not be less than that given by the following: Ast min=0.85bd/ fy



The maximum area of compression reinforcement shall not exceed 0.04 bD. Compression reinforcement shall be enclosed by stirrups for effective restraint. The anchorage length of straight bars in compression shall be equal to the development length of bars in compression.



Where the depth of the web in a beam exceeds 750 mm side face reinforcement shall be provided along the two faces. The total area of such reinforcement shall be not less than 0.1 percent of the web area and shall be distributed equally on two faces at a spacing not exceeding 300 mm or web thickness whichever is less.

8.2 DESIGN OF TYPICAL BEAM 8.2.1 DESIGN PARAMETERS 2


(f ck )

= 30N/mm


(fy)

= 500 N/mm

2

Length of the beam

= 8.20m

Size of the beam

= 300×750mm

Load combination considered Envelope of factored loads. 8.2.2 BENDING MOMENT CALCULATION :

Maximum positive moment [Mmax] 2

6

Mu / b.d = (124.61×10 )/ 300×690

= 124.61kNm 2

= 0.87

Mulim = 3.99 > 0.87, Beam section is designed as singly reinforced. [From table 4 of SP:16] Pt

= 0.215

Ast = (Pt×bd)/100

= 438.84mm

2

60


Minimum Ast = (0.85/f y)= Ast/bd

= 360.16 mm

2

Provide 3 numbers of 16mm dia bars Maximum negative moment [M max] 2

6

Mu / b.d = (300.20×10 )/ 300×690

= 300.20kNm

2

= 2.10 2

[From table 4 of SP:16] Pt

= 1097.10mm

Ast = (Pt×bd)/100

= 1097.10mm

2

3 numbers of 16mm dia bars, 2 numbers of 12mm dia bars. 8.2.3 SHEAR REINFORCEMENT AT SUPPORT:

τv = Vu/ b.d Pt

3

=134×10 /300×690

= 0.64

(100×1097.10/690×300)

= 0.50

τc

= 0.50

τc

< τv

Hence the beam is unsafe against shear so shear reinforcement is required. = Vu- (τc × b×d)

Vs -3

(134-(0.5×300×690×10 )

= 30.5kN 3

Sv = (0.87×500×2×113×690)/ (30.5×10 ) As per ductile detailing should be greater than (d/4) = (690/4)

= 222mm =172.50mm

Adopt stirrup spacing 200mm c/c for a distance 2d = 2×690 =1.5m from the face of the support. 8.2.4 SHEAR REINFORCEMENT AT MID SPAN:

τv = Vu/ b.d

3

= 70.5×10 /300×690 Pt

= 0.34 (100×1097.10/750×300)

τc

= 0.5 = 0.5

τc

> τv

Minimum shear reinforcement is to be provided. Asv / b.Sv

= 0.4/ 0.87× f y

2

0.785×8 / 300×Sv Sv

= 0.4/ 300×500 = 182mm

< 0.75×d

= 517.5mm

< Least lateral dimension

= 300 mm

61


< 300mm So provide 8mm dia 2 legged stirrups at 100 mm c/c for a distance of L-4d = 8.20- 2.76 = 5.44m 8.2.5 DESIGN OF HANGER BARS:

Maximum shear due to secondary beam Vu

= 61.19kN

For single bar or single group of parallel bars all bent up to the same cross section. = 0.87×f y×Asv ×Sin α×2

Vus 2

0.87×500×.785×12 ×Sin60 ×2

= 170.34kN

So provide 2 numbers of bars 12 mm diameters at an angle of 60’ 8.2.6 DEVELOPMENT LENGTH

Ld = (φ×fy×0.87)/ (4×1.6×1.8)

= 1.2

For ductile detailing Development length = L d + 10d-

allowance for 90’ bent.

1.2+(10×.032×)-(8×.032)

= 1.7m

8.2.7 CHECK FOR DEFLECTION:

Deflection actual

= 8250 / 690 = 11.96 mm

Deflection allowable for continuous beam

= 26 mm

Stress in steel f s = 0.58×fy×[Ast required/ Ast provided]

=0.58(438.82/452.16)500

f s (Permissible stress in steel)

= 281 N/mm

2

Find k1 (From IS 456-2000figure 4 modification factor chart) Pt =(100×452.16)/ bd

= 0.22

k1

= 0.95

Find k2

Pt =(100×1205.86)/ bd

= 0.58

K2

= 0.9

Deflection allowable

= k1×k2×26


= 0.95×0.9×26 = 22.23mm 11.26 < 22.23 Safe in deflection

62


8.3 BEAM DETAILING

Figure 29-Longitudinal section of beam

3a -Y16 = 3 numbers of 16mm dia bars (top layer-1) 2b -Y12 = 2 numbers of 12mm dia bars (top layer-2) 3c- Y16 = 3 numbers of 16mm dia bars (bottom layer-1) Shear reinforcement =2×d = 1500mm stirrup spacing 200mm c/c for a distance 2d = 2×750 =1.5m from the face of the support.

63


9.0 DESIGN OF SLAB 9.1 ONE WAY AND TWO WAY SLAB

The most common type of structural element used to cover floors and roofs of buildings are reinforced concrete slabs of different types. One way slabs are those supported on the two opposite sides so that the loads are carried along one direction only. A common example of one way slab is the verandah slab spanning in the shorter direction with main reinforcements and distribution reinforcements in the transverse direction. ONE -WAY SLAB Reinforced concrete slabs supported on two opposite sides with their longer dimension exceeding two times the shorter dimension are referred to as one- way slabs.[ Ly /Lx >2 ] TWO WAY SLAB Reinforced concrete slabs supported on all the four sides with their effective span in the longer direction not exceeding two times the effective span in the shorter direction are designed as two way slabs. Two- way slabs bending moments are maximum at the centre of the slab and the larger moment invariably develops along the short span. [ Ly /Lx ≤2]

Figure 30-Types of slab

64




Shorter of the two spans should be used for calculating the span to effective depth ratios.



For two way slab of shorter span- up to 3.5m with fy=250, the span to overall depth ratios given below & vertical deflection limits up to 3 kN per square meter. 1. f y = 500, values given below should be multiplied to 0.8. 2. Simply Supported slabs = 35 3. Continuous slabs =40



Span to effective depth ratios for spans up to 10m i. Cantilever =7 ii. Simply supported =20 iii. Continuous = 26



Spans above 10m- value may be multiplied by b y 10/span in m.

Effective span 

Simply Supported slab & Continuous slab(width of the support is less than 1/12 of clear span) the least of, 1. Clear span + effective Depth 2. Centre to Centre of supports 3. Continuous slab-(width slab-(width of the support are wider than 1/12 of clear span or 600mm whichever is less)

clear span bet’ support.



One end fixed and other continuous-



One end free and other continuouscontinuous- the least one, one, 1. clear span+1/2x eff. Depth of slab 2. Clear span+1/2x width of the discontinuous support.



Roller or rocket bearing-

distance bet’ the Centre of bearing.

i. Cantilever-Face of the support +1/2 eff. Depth. ii. Frame : (continuous frame )c/c distance 

Slabs spanning in two directions at right a ngles 1. The most commonly used limit state of collapse method is based on yield-line theory.

65


Figure 3131- load distribution in two way slab

9.2 DESIGN OF TYPICAL TWOWAY SLAB 9.2.1 DESIGN PARAMETERS :

Length of slab in X-direction (L×)

= 3.60m

Length of slab in Y direction (Ly)

=4.55m


(f ck ck )

=30N/mm


(fy)

= 500 N/mm

Ly /L×

2 2

= 4.55/3.60 = 1.2 < 2 (two way slab)

Basic value of L/D ratio

= 3600/28

Effective depth

= 130mm

Overall depth

= 150mm

Effective span

=clearspan+eff.depth =clearspan+eff.depth = 3.6+0.130 = 3.73m

Self-weight of the slab

= 1× 25× 0.150 = 3.75kN/m

2

2

Live load

= 4kN/m

Floor Finish

= 24×1×0.06 24×1×0.06 = 1.5kN/m

Total Load

=9.25kN/m

Ultimate load

= 9.25×1.5

2

2

=13.875kN/m 9.2.2 DESIGN OF BENDING MOMENT:

Two adjacent edges discontinuous (from code IS456:2000 table 26) Ly /L×

= 1.2

Mxe

= α×e× W × L×2 66

2


13. Moment calculation Table-13 Moment coefficients

M= α×e×W× L× (kNm)

Ultimate load(W) in Length (Lx) in m kN

α×e=0.065 (Edge)

13.875

3.73

12.54

α×m=0.049 (Span)

13.875

3.73

9.45

αye =0.047 (Edge)

13.875

3.73

9.07

αym=0.035 (Span)

13.875

3.73

6.75

9.2.3 REINFORCEMENT CALCULATION: TABLE-14

Moment

Ast calculation

Ast

No .of bars with

(kN-m) 12.54

Spacing

diameter. 87×500 × A st ×130(1 – 230 mm

Provide Provide 10mm dia bars

300mm 300mm


300mm 300mm


300mm 300mm


300mm 300mm

500×Ast /30×1000 ×130) ×130) 9.45

87×500 × A st ×130(1 – 171mm 171mm 500×Ast /30×1000 ×130) ×130)

9.07

87×500 × A st ×130(1 – 164mm 164mm 500×Ast /30×1000 ×130) ×130)

6.75

87×500 × A st ×130(1 – 122mm 122mm 500×Ast /30×1000 ×130) ×130)

9.2.4 CHECK FOR DEPTH: 2

Max. B.M in the slab

= 12.68 kN/m

M.R

= 0.138 × f ck × b × d

12.68 kNm

= 0.138 ×30×1000× d

Depth

= 55.34 < 130 mm

2 2

9.2.5 CHECK FOR SHEAR:

V ux

= 0.5×13.875×3.73 0.5×13.875×3.7 3 = 25.88kN 3

Max. S.F. of support, V u Nominal shear stress

=(25.88×10 )/(1000×150)

τv

= 0.17 N/mm

100Ast / bd Critical shear

2

= 0.15

τc

= 0.29 N/mm

τc

Hence the slab is safe against shear.

67

>

τv

2


9.2.6 CHECK FOR DEFLECTION:

Deflection actual

= 3730/ 130 = 28mm

Stress in steel f s

= 0.58 × (230/304) × 500

f s(From IS 456-2000 figure 4 ) modification factor)

= 220 N/mm

Modification factor (k 1)

= 1.8


= 1.8×20×1×1 = 36mm

2

[28mm < 36mm] The slab having the sufficient stiffness and the deflection will be within the permissible value. 9.3 REINFORCEMENT DETAILS

Figure 32- R.C.Slab details



Tension reinforcement;(lower part of the slab) 1. 0.25L-continuous edge 2. 0.15L-discontinuous edge



Continuous edges of middle strip, the tension reinforcement. Shall extend in the upper part of the slab a distance 0.15L from the support and at least 50 % shall extend a distance of 0.3L.

1. Negative moments at discontinuous edge- 0.1L 2. Torsion reinforcement shall be provided at any corner 

Minimum distance- 1/5 x shorter span



Area of reinforcement-3/4 x area required for the max. Mid-span meters in the slab.

68




Spacing 1. main steel- ≠ > 3d or 300mm whichever is smaller 2. Distribution steel- ≠> 5d or 450mm whichever is smaller



Minimum reinforcement: ≠< 0.15 % of the total cross sectional area for f y=250 and 0.12% when f y=415 bars are used



Maximum diameter of bars : ≠> 1/8 x D



Cover: should not be less then15mm nor < dia. of bar whichever is higher

Figure 33-Slab detailing

1- 10mm dia bars at 300mm spacing

3,4-10mm dia bars at 300mm spacing



69


10.0 DESIGN OF STAIRCASE 10.1 GENERAL:

Functionally the staircase is an important component of building, and often the only means of access between the various floors in the building. It consists of a flight of steps, usually with one or more intermediate landings provided between floor levels. Following are the structural components of staircase. A) Thread: The horizontal portion of a step where the foot rests is referred to as tread. The typical dimension of a tread is 250mm to 300mm. B) Riser: Riser is the vertical distance between the adjacent treads or the vertical projection of one step with value of the step 150 to 190mm depending upon the type of building. The width of stairs is generally 1 to 1.5m and in case not less than 850mm. Public buildings should be provided with larger widths to f acilitate free passage to users and prevent overcrowding. C) Going: Going is the horizontal projection (plan) of an inclined flight of steps between the first and last riser. A typical flight comprises two landings and one going. To break the monotony of climbing, the number of steps in a flight should not exceed 10 to12. The tread riser combination can be provided in conjunction with 1) Waist slab type 2) Tread riser type 3) Isolated cantilever type 4) Double cantilever precast tread slab with a central inclined beam.

70


Figure 34-Types of stair cases

Types of staircases The various types of staircases adopted in different types of buildings can be grouped under geometrical and structural classifications depending upon their shape and plan pattern and their structural behavior under loads. Types of staircases based on geometrical classification. 1) Straight stairs (with or without landing) 2) Quarter-turn stairs 3) Dog-legged stairs

71


4) Open well stairs 5) Spiral stairs 6) Helical stairs 10.2 DESIGN PARAMTERS

Type of staircase

= Waist slab type

Number of steps in flight

= 12

Tread (T)

=300mm

Riser (R)

=150mm

Width of landing beams

=600mm 2


(f ck )

=30N/mm


(fy)

= 500 N/mm

Effective span (L) = [(12×300) +600]

= 4200mm

Thickness of waist slab

= [4200/20]

2

= 210mm 10.3 LOAD ACTING ON STAIRS

Dead load of slab (on slope) (Ws) = (0.21×1×25)

= 5.25kN/m

Dead load of slab on hor izontal span [Ws√R +T / T]

= 5.59kN/m

Dead of one step (0.5×0.15×0.3×25)

= 0.56kN/m

Load of steps/m length = [0.56×(1000/300)]

=1.86kN/m

Floor finishes

= 1.5kN/m

Total dead

= 8.95kN/m

Live load (overcrowding) per meter

= 5kN/m

Total service load

=13.95kN/m

Total ultimate load

= (1.5×13.95)

2

2

=20.93kN/m 10.4 DESIGN OF WAIST SLAB TYPE STAIRCASE 2

Maximum bending moment (0.125×20.93×4.2 )

= 46.15kNm

Check for depth of waist slab(d)=√ 46.15×106 /(0.138×30×1000)

= 106mm

Assume a cover of 20mm and using 12mm dia bars [210-20-6]

= 184mm

Effective depth provided is greater than the required depth

(184 >106)

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10.5 REINFORCEMENT PROVIDED

Using table 2 from SP: 16 2

2

[Mu/ b×d ] = [46.15×106 /1000×184 ]

= 1.35

Pt = [Ast×100/b×d]

=0.329

Ast

=606mm

2

2

Provide 12mm diameter bars at 200mm spacing (Ast=678mm ) as main reinforcement. Distribution reinforcement (0.0012×1000×210)=252mm

2 2

Provide 8mm diameter bars at 200mm spacing (Ast=252mm ) as distribution reinforcement.

10.6 STAIRCASE DETAILING

Figure 35-Detailing of staircases

Longitudinal reinforcement: Y 12 AT 200 c/c Distribution reinforcement: Y 8 at 200 c/c

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11.0 DESIGN OF SHEAR WALL 11.1 GENERAL

Lateral force resisting system in the building is a dual system consisting of SMRF and shear walls. In general, the shear walls will resist all the lateral force being a relatively stiff element. The design of shear wall is based on the assumption that it will be the part of the lateral force resisting system of the structure. The shear wall is provided in between the middle two columns of the exterior frames. These columns will act as a flange element or boundary elements for the shear wall. Therefore, there is no need for further thickening of shear wall at the end or boundary regions. Calculated reinforcement in horizontal and vertical direction is greater than the minimum prescribed reinforcement provided reinforcement is uniformly distributed in both the directions.

Figure 36-Shear Wall 11.2 WALL DIMENSIONS:

Length of the wall

= 7.2m

Thickness of wall

= 230mm

Height of the wall

= 4.0m

Load combinations considered

= 1.5DL + 1.5 EQL = 0.9DL + 1.5 EQL

P1 Axial load (1.5DL + 1.5 EQL)

= 3349 kN

Shear force

= 7532.98 kN

Bending moment

= 33781.95

P2 Axial load

= 2197 kN

Shear force

= 7529.02 kN 74


Bending moment

= 33756 kNm 2


(f ck )

=30N/mm


(fy)

= 500 N/mm

2

11.3 CHECK FOR BOUNDARY ELEMENT: 3

3

Moment of inertia I= (t ×L )/ 12

= (230 ×7.2 )/ 12 12

4

= 7.15 ×10 mm Area= t×L

= 230×7200 3

2

=1656×10 mm f c= (P/A)±( Mt ×L/ I)

=2.02 ± 17.03 = 19.05 and -15.01kN/ m

2

11.4 FIXING BOUNDARY ELEMENT:

Assume width of flange (Wf)

= 230 mm

Length of flange (Lf) = (0.1×L)

= 0.1×7200 = 720mm

Provide 750×250mm boundary element. Lw = 7200-(2×750)

= 5700

c/c = 7200-750

= 6450

Wall thickness is 230mm so provide two layers of steel.

11.5 DETERMINATION OF MINIMUM STEEL:

1) VERTICAL REINFORCEMENT

= 0.0025×Ag

0.0025×1000×230

= 575 mm

2

Provide area of steel is 1% of gross area. That is 0.5% in each layer. Area of steel in each layer (Ast)

= (0.575×1000×230)/100 = 1322.5mm

2

Diameter of bar required should not be greater than (1/10 th) of wall thickness. (1×230)/10 = 23 adopt 16mm dia bars. 2

Spacing of bars = (1000×0.785× 16 )/ 1322.5

= 150mm

Spacing should not be greater than Lw/5 = 5700/5

= 1140mm

3×t w = 3× 230

= 690mm

Provide 16mm diameter bar at 350mm spacing. 2

Area of steel in each layer (Ast)

= (0.785×16 ×1000)/100 = 2000.60mm

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2


11.6 SHEAR REINFORCEMENT:

τv

3

=7532.98 ×10 /230×(0.8×7.2)

= 5.68 N/mm

Percentage of steel =(100×Ast)/ (b×t)

= 1.70

From IS 456:2000(table 19 & 20)

τc

= 0.7925N/mm

τcmax

= 3.5N/mm

2

2

Shear reinforcement to be designed for shear reinforcement Vus Vs

= Vu- (τc × t w ×dw)

Vs

=6483.73kN

HORIZONTAL REINFORCEMENT A sv (Minimum reinforcement)

= 0.25% of Ag = 575mm

3

Sv = (0.87×500×575×0.8×7200)/ (6483.73×10 )

2

= 222.20 mm c/c

Provide 12mm diameter bar at 230mm spacing. 2)

= (1000×0.785×22 / 230 = 491.48mm

2

A sv required < A st provided. 11.7 FLEXURAL STRENGTH:

Load on web (Pw)

=P1[Lw.tw /(Lw.tw+2.Bf .Lf )]

=2844.78kN.

λ = Pw / (f ck ×tw×Lw)

=0.06

φ= [(0.87× f y)/ f ck ]×[Ast× tw]

= 0.39

β=(0.87× f y)/(έc×E)

= 0.62

xu / L= (λ +φ) /0.36+2 φ

=0.39

xu’/ L=0.0035/(0.0035+[(0.87× f y)/Es])

=0.616

xu / L < xu’/ L 2

/ φ)×(0.5-(0-4116(xu / L)]-[( xu’/ L)×(0.168+( β /3)]} Muv= f ck ×t×Lw× φ{[1+( λ Muv

= 29741.70kNm

M1= Mu- Muv

=33781-29741.70 =4039.30kNm

f

=[Lf .B /(L f w.tw+2.Bf .Lf )]

f

=0.04 76


Pc= (M1 /c)+f.P2 =(4039.30/6.45)+(0.04×7529.30)

=927.42kN

Pt= (M1 /c)-f.P1 =(4039.30/6.45)+(0.04×7532.31)

=324.42kN

Ast= [Pc-(0.4× Bf .Lf × f ck )]/ [(0.67× f y)-(0.4 f ck )]

=920mm

Area of steel [Ast min] =0.008×250×750

=1500 mm

2

2

Provide 20mm diameter bar at 170mm spacing.(1884mm ) 11.8 TRANSVERSE REINFORCEMENT:

Diameter of tie bars ≥ φ /4 =(20/4)

=5mm

Provide 6mm diameter bar as tie bars. Spacing of reinforcement minimum of 16

φ

=320mm

Provide 6mm diameter bar at a spacing of 300mm.

11.8 SHEAR WALL DETAILING

Figure 37-Detailing of shear wall 1-16mm diameter bars at 350mm spacing 2-12mm diameter bars at 230mm spacing.

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2


CONCLUSION The New teaching hospital building was designed with the earthquake resistant design consideration. Seismic analysis and design were done by using ETABS software and verified manually as per IS 1893-2002 the provision of shear wall in the staircase and lift region have the ultimate shear resistance, the total base shear produced by the earth quake for that maximum percentage of the shear resistance produced by the shear wall and the remaining shear resistance produced by the columns. The detailing of the structural elements were done as per IS 13920-1993(Ductile detailing for Earthquake resistant structures). To conclude a complete design involving several parameters so as to result the earthquake has been done.

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Seismic Analysis and Design of Hospital Building-libre

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