Seismic Analysis and Design of Hospital Building-libre

8/20/2019 Seismic Analysis and Design of Hospital Building-libre

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SEISMIC ANALYSIS AND DESIGN OF MULTI STOREYED HOSPITAL BUILDING

1

SEISMIC ANALYSIS AND DESIGN OF MULTI STOREYED

HOSPITAL BUILDING

Submitted at the end of 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


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CERTIFICATE

This is to certify that the project report entitled

“SEISMIC ANALYSIS AND DESIGN OF MULTI STROREYED HOSPITAL

BUILDING” is t he 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 Mrs.GAYATRI PADHY,B.E

Dean& Project Guide Head of the Department Department of Civil Engineering. Department of Civil Engineering

S.V.C.E.T. . S.V.C.E.T.

External Guide

Dr.S.JUSTIN

Engineering Manager

SDDH,

EDRC, Larsen & Toubro,

Chennai.

Mr.UP.VIJAY Mr.M.RAMPRATHAP

Assistant Engineering Manager, Assistant Engineering Manager,

EDRC, Larsen & Toubro, EDRC, Larsen & Toubro,

Chennai. Chennai.


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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 and I thank 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


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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 (Ta).

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 wallwere designed. Ductile detailing of the structural elements were done as per code IS:13920-

1993.


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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


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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


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7.0 COLUMN DESIGN 41

7.1 General 41

7.2 Column design 42


7.2.2 Design of bending moment 42

7.2.3 Reinforcement provided 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


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


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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


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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


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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


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LIST OF TABLES PAGE NO.

1. Value of Topography Factor K2 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


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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


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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


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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.


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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 Hlock-

Larsen 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.


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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. Thesolution 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


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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.


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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/m3

Unit weight of cement mortar = 24kN/m3

Unit weight of water = 10kN/m3

Unit weight of Brick = 20kN/m3

4.2 DEAD LOAD (As per IS 875 part I)

4.2.1 TYPICAL FLOOR SLAB:

Self-weight of slab (150mm thick) = 3.75 kN/m2

Floor Finish = 1.5 kN/m2

Total Load = 5.25 kN/m2

4.2.2 TERRACE FLOOR SLAB:

Self-weight of slab (150mm thick) = 3.75 kN/m2

Terrace (Roof finish + Water proofing) = 3.20 kN/m2

Total Load = 6.95 kN/ m2

4.3 LIVE LOAD (As per IS 875 part II)

Living room, Toilet & Bath room = 3.00 kN/m2

Recreation Room and Pantry = 3.00 kN/m2

Store and Laundry = 4.00 kN/m2


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Corridors & stair case = 4.00 kN/m2

Terrace Floor = 1.5 kN/m2

ICU = 3.00 kN/m2

Lounge = 3.00 kN/m2

Nurse station = 3.00 kN/m2

Operation Theatres = 3.00 kN/m2

Dining room = 4.00 kN/m2

Waiting room = 4.00 kN/m2

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


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= 19.24kN/m

Self-weight of beam (B2) = 25× 0.75 ×0.3×4.28

= 24.08kN/m


= 20.25kN/m


= 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


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= 19.24kN/m

Self-weight of beam (B2) = 25× 0.75×0.3×4.28

= 24.08kN/m


= 20.25kN/m


= 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/m2

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


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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 include 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


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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

Pz = 0.6×Vz2

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, Pz 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 Cpn 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 a=53.0m

b=36m h=28.0m

Fig4.A- Plan Fig 4.B- Elevation


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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

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

Story Height(m) K 2 Vz(km/s) Pz=0.6Vz2(kN)

Load X in

direction(kN)

Load in Y

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

Height(m) K2

10 0.93

15 0.9720 1.00

30 1.04

50 1.10


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3. Wind forces from Analysis – Table 3

Story Point Load FX

(kN)

FY

(kN)

FZ

(kN)

MX

(kN-m)

MY

(kN-m)

MZ

(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

Summation WINDPX 1500.02

4. Wind forces from software-Table 4

Story Point Load FX

(kN)

FY

(kN)

FZ

(kN)

MX

(kN-m)

MY

(kN-m)

MZ

(kN-m)

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


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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 acceleration given in units ‘g’ for the maximum

considered earthquake.


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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).


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4.7.5 SOIL CLASSIFICATION:

Sa/g is the lateral acceleration to be established in m/s2. 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 of rock or soil

layers or based on average Standard Penetration Test (SPT) values for top 30m.

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

0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2 2.5 3 3.5 4

Period (s)

S p e c t r a l A c c e l e r a t i o n C o e f f i c i

e n t ( S a / g )

Type I (Rock, or Hard Soil)

Type II (Medium Soil)

Type III (Soft Soil)


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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 = Ah× W

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

Fundamental Natural Time period (without infill)

Ta = 0.075(h) 0.75

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 = (0.09×h)/√d

Ta (X-Direction) = 0.420 & Ta (Y-Direction) = 0.346

Sa/g = 2.5

Ah (X-Direction) = 0.135 &Ah (Y-Direction) = 0.135

Base shear (without infill)

Base shear (X direction) (Ah×W) = 17554kN

Base shear (Y direction) (Ah×W) = 17554kN

Base shear (with infill)

Base shear (X direction) (Ah×W) = 23937kN

Base shear (Y direction)(Ah×W) = 23937kN


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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

(kN)

FY

(kN)

FZ

(kN)

MX

(kN-m)

MY

(kN-m)

MZ

(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


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

Story Point Load FX

(kN)

FY

(kN)

FZ

(kN)

MX

(kN-m)

MY

(kN-m)

MZ

(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 thespectra 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

Base shear in Y-direction using Code = 20746kN


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Base shear in X-direction using ETABS model = 6760kN

Base shear in Y-direction using ETABS model = 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

(kN)

FY

(kN)

FZ

(kN)

MX

(kN-m)

MY

(kN-m)

MZ

(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 1.803

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 13.81 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 20753.07

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

Story Point Load FX

(kN)

FY

(kN)

FZ

(kN)

MX

(kN-m)

MY

(kN-m)

MZ

(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


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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

Base shear in Y-direction using Code = 20746 kN

Base shear in X-direction using ETABS model = 20753.07 kN

Base shear in Y-direction using ETABS model = 20744.37 kN

10. Period Vs Sa/g- Table 10

Period

(seconds) Sa /g

Acceleration

(m2 /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


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1.3 1.285 0.681

1.4 1.193 0.632

1.5 1.113 0.590

1.6 1.044 0.5531.7 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 0.3852.4 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 0.2953.1 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 0.2393.8 0.439 0.233

3.9 0.428 0.227

4 0.418 0.221


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Figure 6-Response spectra for Soft soil sites

11. Seismic weight output-Table 11

Story Point Load FX

(kN)

FY

(kN)

FZ

(kN)

MX

(kN-m)

MY

(kN-m)

MZ

(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


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5.0 MODELING AND ANALYSIS OF THE BUILDING

The primary purpose of structural analysis in building structure is 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


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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 property 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.


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Figure 11- Section properties and 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


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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 usein response spectrum analysis. In this program the acceleration values to be normalized; that


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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


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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 LIVELOAD3 WIND/ E.QLOAD

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


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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 WINDLOADXCOMB 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


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5.11 MODEL OUTPUT

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

Figure.18-3D model of RC frame


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B) BENDING MOMENT DIAGRAM FROM ANALYSIS

Figure 19- Bending Moment from analysis.


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C) SHEAR FORCE FROM ANALYSIS

Figure 20- Shear Force from analysis.


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D) DIAPHRAGM

Figure 21-Diaphram output from analysis.


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E) PLAN SHOWING SLAB ID

Figure 22- Plan showing slab ID


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E) PLAN SHOWING BEAM ID

Figure 23- Plan showing Beam ID


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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.


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6.3 DESIGN OF TYPICAL ISOLATED FOOTING

6.3.1 DESIGN PARAMETERS:

Grade of concrete using (f ck ) = 30N/mm2

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/ m2

Unit weight of concrete = 25kN/m3

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)

[1822.62/B.L] + [18.26×6/B.L2

] + [18.26×6/B.L2

] = 100 kN/ m2

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/ m2

[P/A] +[Mx /Z] - [My] P2 = 93.73 kN/ m2

[P/A] - [Mx /Z] + [My] P3 = 94.55 kN/ m2

[P/A] - [Mx /Z] - [My] P4 = 92.27 kN/ m2

Assume depth of footing = 700 mm

Downward pressure due to base slab (0.7×25) = 17.5 kN/ m2


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Figure 25-Pressure diagram

6.3.2 BENDING MOMENT IN X DIRECTION

Cantilever projection along X = 1.825 m

Mx [94.22×0.5×1.8252]+[.5×1.825×1.945×.67×1.825]-[17.5×0.5×1.825

2] =129.91kNm

Steel required along X direction

Mu lim = 0.138×f ck ×b×d2

Depth required = 178mm

Depth provided [700-50-(25/2)] = 637mm

Mu/b.d2 = [1.5×129.91×10

6] / [1000×637

2] = 0.48

Percentage of steel = 0.117

Provide percentage of steel = 0.75

Provide 25mm diameter bars and spacing [.785×252×1000]/4400 =111.50

Provide 25mm diameter bars at spacing of 100mm c/c.


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6.3.3 BENDING MOMENT IN Y DIRECTION:

Cantilever projection along Y = 1.825 m

My[94.22×0.5×1.8252]+[.5×1.825×1.945×.67×1.825]-[17.5×0.5×1.825

2]=129.91kNm

Steel required along Y direction

Mu lim = 0.138×f ck ×b×d2

Depth required = 178mm

Depth provided [700-50-(25/2)] = 637mm

Mu/b.d2 = [1.5×129.91×10

6] / [1000×637

2] = 0.48

Percentage of steel = 0.117

Provide percentage of steel = 0.75

Provide 25mm diameter bars and spacing [0.785×252×1000]/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/ m2

Shear force at critical section (94.70-15)×(1.825-.637) = 94.60

Nominal shear stress τv = (Vu×1.5)/b.d = 0.22N/ mm2

From IS: 456-2000, table 19 Percentage of reinforcement = 0.59N/ mm2

Safe in one way shear. τv < τc

6.3.5 CHECK FOR TWO WAY SHEAR:

Average pressure (94.345+94.14)/2 = 94.24 kN/ m2

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/ mm2

τc = 0.25×√(f ck ) = 1.37 N/ mm2

τv < τc Hence safe in two way shear.


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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


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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 dimension. 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


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7.2 COLUMN DESIGN

7.2.1 DEIGN PARAMETERS:

Terrace level: C-1


Grade of steel using (fy) = 500 N/mm2


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

Pu / f ck bd =(316×103)/(30×750×750) = 0.018

Mu / f ck bd2[From the code IS 456- 1978 chart 48] = 0.08

Mux1 =0.08×30×750×7502

=1012.50kNm

Ratio of d’/D =60/750= 0.08

Pt / f ck = 1.0/ 30 =0.03

Pu / f ck bd =(1586×103)/(30×750×750) = 0.018

Mu / f ck bd2[From the code IS 456- 1978 chart 48] = 0.08

Muy1 =0.08×30×750×7502

=1012.50kNm

Percentage of reinforcement = 1.0

Puz / Ag [From the code IS 456- 1978 chart 63] = 17

= 17×750×750

= 9562 kN/m2

7.2.2 BENDING MOMENT CALCULATION:

[Mux / Mux1] αn

+ [Muy / Muy1] αn≤ 1

[Pu /Puz] = (316.44/ 9562)

= 0.03

[Pu /Puz]≤0.2 so αn =1

[235.84/ 1012.50] +[128/1012.50] = 0.36


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0.36 ≤ 1

[Muy / Muy1] αn

[From the code IS 456- 1978 chart 64] = 0.80 > 0.0.13

7.2.3 REINFORCEMENT PROVIDED:Pt = 1.0

Ast = (1.0×750× 750)/ 100

Ast =5625mm2

Provide 20 numbers of 20 mm diameter bars. (Ast=6280mm2)

Transverse reinforcement

Tie diameter φt = φ long/4

25/4 = 6.25mmProvide 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=187.5mmProvide 8 mm bars at spacing 190mm c/c as a transverse reinforcement


Fifth floor level

Grade of concrete using (f ck ) =30N/mm2



Axial load = 659.88kN/mm2

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 (659.88×103)/(30×750×750) = 0.04

Mu / f ck bd2 [From the code IS 456- 1978 chart 48] = 0.08


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Mux1 =0.08×30×750×7502

=1012.50kNm

Ratio of d’/D =60/75= 0.08

Pt / f ck = 1.0/ 30=0.03

Pu / f ck bd (659.88×103)/(30×750×750) = 0.04

Mu / f ck bd2 [From the code IS 456- 1978 chart 48] = 0.08

Muy1 =0.08×30×750×7502

=1012.50kNm

Percentage of reinforcement = 1.0

Puz / Ag [From the code IS 456- 1978 chart 63] = 17

= 17×750×750

= 9562 kN/m2


[Mux / Mux1] αn

+ [Muy / Muy1] αn≤ 1

[Pu /Puz] = (659.88 / 9562.5) = 0.07

[Pu /Puz]≤0.2 so αn =1

[122.16/ 1012.5] + [59.10/1012.5] = 0.18

0.18 ≤ 1

[Muy / Muy1] αn

[From the code IS 456- 1978 chart 64] =0.80> 0.05(calculated).

7.2.6 REINFORCEMENT DETAILS:

Pt = 1.0

Ast = (1.0×750× 750)/ 100

Ast =5625mm2

Provide 20 numbers of 20 mm diameter bars 4 on each face (6280 mm2)


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7.3 COLUMN DETAILING:

Figure 28.a-Longitudinal section

Figure 28.b-Cross section


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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 usingsteel 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.


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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



Length of the beam = 8.20m

Size of the beam = 300×750mm

Load combination considered Envelope of factored loads.


Maximum positive moment [Mmax] = 124.61kNm

Mu / b.d2 = (124.61×10

6)/ 300×690

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 = 438.84mm2


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Minimum Ast = (0.85/f y)= Ast/bd = 360.16 mm2

Provide 3 numbers of 16mm dia bars

Maximum negative moment [Mmax] = 300.20kNm

Mu / b.d2= (300.20×10

6)/ 300×690

2 = 2.10

[From table 4 of SP:16] Pt = 1097.10mm2

Ast = (Pt×bd)/100 = 1097.10mm2

3 numbers of 16mm dia bars, 2 numbers of 12mm dia bars.

8.2.3 SHEAR REINFORCEMENT AT SUPPORT:τv = Vu/ b.d =134×10

3 /300×690 = 0.64

Pt (100×1097.10/690×300) = 0.50

τc = 0.50

τc < τv

Hence the beam is unsafe against shear so shear reinforcement is required.

Vs = Vu- (τc × b×d)

(134-(0.5×300×690×10-3

) = 30.5kN

Sv = (0.87×500×2×113×690)/ (30.5×103) = 222mm

As per ductile detailing should be greater than (d/4) = (690/4) =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 = 70.5×103 /300×690 = 0.34

Pt (100×1097.10/750×300) = 0.5

τc = 0.5

τc > τv

Minimum shear reinforcement is to be provided.

Asv / b.Sv = 0.4/ 0.87× f y

0.785×82 / 300×Sv = 0.4/ 300×500

Sv = 182mm

< 0.75×d = 517.5mm

< Least lateral dimension = 300 mm


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< 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.

Vus = 0.87×f y×Asv ×Sin α×2

0.87×500×.785×122×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 = Ld + 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/mm2

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

Deflection allowable = 0.95×0.9×26

= 22.23mm

11.26 < 22.23 Safe in deflection


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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.


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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


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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 squaremeter.

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 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 of the support are wider than 1/12 of clear

span or 600mm whichever is less)

One end fixed and other continuous- clear span bet’ support.

One end free and other continuous- the least 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 angles

1. The most commonly used limit state of collapse method is based on

yield-line theory.


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Figure 31- load distribution in two way slab

9.2 DESIGN OF TYPICAL TWOWAY SLAB


Length of slab in X-direction (L×) = 3.60m

Length of slab in Y direction (Ly) =4.55m



Ly /L× = 4.55/3.60

= 1.2 < 2 (two way slab)

Basic value of L/D ratio = 3600/28

Effective depth = 130mmOverall depth = 150mm

Effective span =clearspan+eff.depth

= 3.6+0.130

= 3.73m

Self-weight of the slab = 1× 25× 0.150

= 3.75kN/m2

Live load = 4kN/m2

Floor Finish = 24×1×0.06 = 1.5kN/m2

Total Load =9.25kN/m2

Ultimate load = 9.25×1.5

=13.875kN/m2

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


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13. Moment calculation Table-13

Moment coefficients Ultimate load(W) in

kN

Length (Lx) in m M= α×e×W× L× (kNm)

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

(kN-m)

Ast calculation Ast No .of bars with

diameter.

Spacing

12.54 87×500 × Ast ×130(1 –

500×Ast /30×1000 ×130)

230 mm Provide 10mm dia bars 300mm

9.45 87×500 × Ast ×130(1 –

500×Ast /30×1000 ×130)

171mm Provide 10mm dia bars 300mm

9.07 87×500 × Ast ×130(1 –

500×Ast /30×1000 ×130)


6.75 87×500 × Ast ×130(1 –

500×Ast /30×1000 ×130)


9.2.4 CHECK FOR DEPTH:

Max. B.M in the slab = 12.68 kN/m2

M.R = 0.138 × f ck × b × d2

12.68 kNm = 0.138 ×30×1000× d2

Depth = 55.34 < 130 mm

9.2.5 CHECK FOR SHEAR:

V ux = 0.5×13.875×3.73

= 25.88kN

Max. S.F. of support, Vu =(25.88×103)/(1000×150)

Nominal shear stress τv = 0.17 N/mm2

100Ast / bd = 0.15

Critical shear τc = 0.29 N/mm2

Hence the slab is safe against shear. τc > τv


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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/mm2

Modification factor (k 1) = 1.8

Deflection allowable = 1.8×20×1×1 = 36mm

[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.


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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

2,3-10mm dia bars at 300mm spacing 5,6-10mm dia bars at 300mm spacing


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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 facilitate 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.


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Figure 34-Types of staircases

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


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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



Effective span (L) = [(12×300) +600] = 4200mm

Thickness of waist slab = [4200/20]

= 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 horizontal span [Ws√R

2

+T

2

/ T] = 5.59kN/mDead 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)

=20.93kN/m

10.4 DESIGN OF WAIST SLAB TYPE STAIRCASE

Maximum bending moment (0.125×20.93×4.22) = 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

[Mu/ b×d2] = [46.15×106 /1000×184

2] = 1.35

Pt = [Ast×100/b×d] =0.329

Ast =606mm2

Provide 12mm diameter bars at 200mm spacing (Ast=678mm2) as main reinforcement.

Distribution reinforcement (0.0012×1000×210)=252mm2

Provide 8mm diameter bars at 200mm spacing (Ast=252mm2) 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


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Bending moment = 33756 kNm



11.3 CHECK FOR BOUNDARY ELEMENT:

Moment of inertia I= (t ×L3)/ 12 = (230 ×7.2

3)/ 12

= 7.15 ×1012

mm4

Area= t×L = 230×7200

=1656×103 mm

2

f c= (P/A)±( Mt ×L/ I) =2.02 ± 17.03

= 19.05 and -15.01kN/ m2

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 mm2

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

= 1322.5mm2

Diameter of b

Seismic Analysis and Design of Hospital Building-libre

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