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ANALYSIS AND DESIGN OF THREE STOREY FRAMED BUILDING. Data · May 2012 DOI: 10.13140/RG.2.1.34 10.13140/RG.2.1.3437.5528 37.5528

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PROJECT REPORT REPORT SUBMITTED IN PARTIAL FUL FILL FILLMENT MENT FOR THE THE DEGREE OF BACHELOR OF TECHNOLOGY IN CIVIL ENGINEERING ON

ANAL A NALYSIS YSIS AND A ND DESIGN OF THREE THREE STOREY FRAMED BUILDING BY UMER FAROOQ

MUZAMIL YOUSUF

ROLL NO 01-CE-2008

ROLL NO 22-CE-2008

Department of civ il enginee engi neeri ring ng Baba Ghul Ghul am Shah Badshah Universit y, Rajouri Rajouri

1

PROJECT REPORT REPORT SUBMITTED IN PARTIAL FUL FILL FILLMENT MENT FOR THE THE DEGREE OF BACHELOR OF TECHNOLOGY IN CIVIL ENGINEERING ON

ANAL A NALYSIS YSIS AND A ND DESIGN OF THREE THREE STOREY FRAMED BUILDING BY UMER FAROOQ

MUZAMIL YOUSUF

ROLL NO 01-CE-2008

ROLL NO 22-CE-2008

Underr the Unde t he guidance of MR. VASEEM AHMAD SHAHNAZ

Hea ead d of Department Department o f civil ci vil enginee engineeri ring ng Baba Ghul Ghul am Shah Badshah Universi ty, Rajouri Rajouri

2

ACKNOWLEGDGEMENTS At the outset, I express my deepest thanks to our Lord, the most Gracious, the most merciful, the cherisher and sustainer of the worlds. I am thankful to my parents who continuously look after me right from my birth, provided me with the best facilities that they could provide me and continuously encouraged me to perform well in my studies and my daily life in general. All these factors were the favorable conditions that made me to strive for the better future life. May the almighty forgive me, my parents and all the believers on the Day of Reckoning I express my deepest gratitude to my teacher Mr. Mir Aijaz Ahmad who as a mentor and Asstt. Professor in the Department of Civil Engineering for his invaluable guidance and inputs. I thank him for allowing me to work on this project and for his constant help and support throughout my endeavor. I consider myself fortunate to have worked under his supervision. It was a matter of great pleasure to have him as a guide in the preparation of this project report. I take this great opportunity to express my profound gratitude to HOD Civil Engineering Mr Vaseem Shahnaz for his superlative flow of ideas and guidance as well as moving inspirations all throughout the classes. I owe a lot to him for his kind gesture in serving our teacher. I also thank him for checking the report before submission.

Umer Farooq

Muzamil Yousuf

Roll no. 01-CE-2008

Roll no. 22-CE-2008

3

DEPARTMENT OF CIVIL ENGINEERING BABA GHULAM SHAH BADSHAH UNIVERSITY

RAJOURI J&K- 185131 (INDIA)

CERTIFICATE This is to certify that this report entitled “ ANALYSIS AND DESIGN OF THREE STOREY FRAMED BUILDING ” submitted by UMER FAROOQ (01-CE-2008), and MUZAMIL YOUSUF (22-CE-2008) in partial fulfillments

for the requirements for the award of Bachelor of Technology Degree in Civil Engineering at College of Engineering & Technology, BGSB University, Rajouri (J&K) is an authentic work carried out by them under my supervision and guidance. To the best of my knowledge, the matter embodied in this report has not been submitted to any other University/Institute for the award of any Degree or Diploma.

Date: Mr. Vaseem Ahmad Shahnaz

HOD

External Examiner

Department of Civil Engineering

4

CONTENTS

S. No.

Description

Page No.

1.

Abstract

6

2.

History

7

3.

Introduction

4.

12

Analysis under Vertical loads

15

5.

Design of Slabs

25

6.

Design of Beams

38

7.

Design of Columns

47

8.

Design of Footing

53

9.

Design of Stairs

71

References

74

10.

5

ABSTRACT

Structural design is the primary aspect of civil engineering. The foremost basic in structural engineering is the design of simple basic components and members of a building viz., Slabs, Beams, Columns and Footings. In order to design them, it is important to first obtain the plan of the particular building. Thereby depending on the suitability; plan layout of beams and the position of columns are fixed. Thereafter, the vertical loads are calculated namely the dead load and live load. Once the loads are obtained, the component takes the load first i.e the slabs can be designed. Designing of slabs depends upon whether it is a one-way or a two-way slab, the end conditions and the loading. From the slabs, the loads are transferred to the beam. The loads coming from the slabs onto the beam may be trapezoidal or triangular. Depending on this, the beam may be designed. Thereafter, the loads (mainly shear) from the beams are taken by the columns. For designing columns, it is necessary to know the moments they are subjected to. For this purpose, frame analysis is done by Moment Distribution Method. After this, the designing of columns is taken up depending on end conditions, moments, eccentricity and if it is a short or slender column. Most of the columns designed in this mini project were considered to be axially loaded with uniaxial bending. Finally, the footings are designed based on the loading from the column and also the soil bearing capacity value for that particular area. Most importantly, the sections must be checked for all the four components with regard to strength and serviceability.

6

CHAPTER 1 HISTORY

7

HISTORY STRUCTURAL ANAL YSIS A structure refers to a system of two or more connected parts use to support a load. It is an assemblage of two or more basic components connected to each other so that they serve the user and carry the loads developing due to the self and super-imposed loads safely without causing any serviceability failure. Once a preliminary design of a structure is fixed, the structure then must be analyzed to make sure that it has its required strength and rigidity. To analyze a structure a structure correctly, certain idealizations are to be made as to how the members are supported and connected together. The loadings are supposed to be taken from respective design codes and local specifications, if any. The forces in the members and the displacements of the joints are found using the theory of structural analysis. The whole structural system and its loading conditions might be of complex nature so to make the analysis simpler, we use certain simplifying assumptions related to the quality of material, member geometry, nature of applied loads, their distribution, the type of connections at the joints and the support conditions. This shall help making the process of structural analysis simpler to quite an extent.

Methods of structural analysis When the number of unknown reactions or the number of internal forces exceeds the number of equilibrium equations available for the purpose of analysis, the structure is called as a statically indeterminate structure. Most of the structures designed today are statically indeterminate. This indeterminacy may develop as a result of added supports or extra members, or by the general form of the structure. While analyzing any indeterminate structure, it is essential to satisfy equilibrium, compatibility, and force-displacement requisites for the structure. When the reactive forces hold the structure at rest, equilibrium is satisfied and 8

compatibility is said to be satisfied when various segments of a structure fit together without intentional breaks or overlaps. Two fundamental methods to analyze the statically indeterminate structures are discussed below.

Force methodsOriginally developed by James Clerk Maxwell in 1864, later developed by Otto Mohr and Heinrich Muller-Breslau, the force method was one of the first methods available for analysis of statically indeterminate structures. As compatibility is the basis for this method, it is sometimes also called as compatibility method or the method of consistent displacements. In this method, equations are formed that satisfy the compatibility and forcedisplacement requirements for the given structure in order to determine the redundant forces. Once these forces are determined, the remaining reactive forces on the given structure are found out by satisfying the equilibrium requirements.

Displacement methodsThe displacement method works the opposite way. In these methods, we first write load-displacement relations for the members of the structure and then satisfy the equilibrium requirements for the same. In here, the unknowns in the equations are displacements. Unknown displacements are written in terms of the loads (i.e. forces) by using the load-displacement relations and then these equations are solved to determine the displacements. As the displacements are determined, the loads are found out from the compatibility and load- displacement equations. Some classical techniques used to apply the displacement method are discussed. Slope deflection methodThis method was first devised by Heinrich Manderla and Otto Mohr to study the secondary stresses in trusses and was further developed by G. A. Maney extend its application to analyze indeterminate beams and framed structures. 9

The basic assumption of this method is to consider the deformations caused only by bending moments. It’s assumed that the effects of shear force or axial force deformations are negligible in indeterminate beams or frames. The fundamental slope-deflection equation expresses the moment at the end of a member as the superposition of the end moments caused due to the external loads on the member, while the ends being assumed as restrained, and the end moments caused by the displacements and actual end rotations. A structure comprises of several members, slope-deflection equations are applied to each of the member. Using appropriate equations of equilibrium for the joints along with the slope-deflection equations of each member we can obtain a set of simultaneous equations with unknowns as the displacements. Once we get the values of these unknowns i.e. the displacements we can easily determine the end moments using the slope-deflection equations.

Moment distribution methodThis method of analyzing beams and multi-storey frames using moment distribution was introduced by Prof. Hardy Cross in 1930, and is also sometimes referred to as Hardy Cross method. It is an iterative method in which one goes on carrying on the cycle to reach to a desired degree of accuracy. To start off with this method, initially all the joints are temporarily restrained against rotation and fixed end moments for all the members are written down. Each joint is then released one by one in succession and the unbalanced moment is distributed to the ends of the members, meeting at the same joint, in the ratio of their distribution factors. These distributed moments are then carried over to the far ends of the joints. Again the joint is temporarily restrained before moving on to the next joint. Same set of operations are performed at each joints till all the joints are completed and the results obtained are up to desired accuracy. The method does not involve solving a number of simultaneous equations, which may get quite complicated while applying large structures, and is therefore preferred over the slope-deflection method. 10

Kani’s methodThis method was first developed by Prof. Gasper Kani of Germany in the year 1947. The method is named after him. This is an indirect extension of slope deflection method. This is an efficient method due to simplicity of moment distribution. The method offers an iterative scheme for applying slope deflection method of structural analysis. Whereas the moment distribution method reduces the number of linear simultaneous equations and such equations needed are equal to the number of translator displacements, the number of equations needed is zero in case of the Kani’s method. This method may be considered as a further simplification of moment distribution method wherein the problems involving sway were attempted in a tabular form thrice (for double story frames) and two shear coefficients had to be determined which when inserted in end moments gave us the final end moments. All this effort can be cut short very considerably by using this method. Advan tag es of Kani ’s meth od :

 All the computations are carried out in a single line diagram of the structure.

 The effects of joint rotations and sway are considered in each cycle of iteration. Henceforth, no need to derive and solve the simultaneous equations. This method thus becomes very effective and easy to use especially in case of multistory building frames.

 The method is self correcting, that is, the error, if any, in a cycle is corrected automatically in the subsequent cycles. The checking is easier as only the last cycle is required to be checked.

 The convergence is generally fast. It leads to the solutions in just a few cycles of iterations.

11

CHAPTER 2 INTRODUCTION

12

Introduction Structural analysis is the backbone of civil engineering. During recent years, there has been a growing emphasis on using computer aided softwares and tools to analyze the structures. There has also been advancement in finite element analysis of structures using Finite Element Analysis methods or matrix analysis. These developments are most welcome, as they relieve the engineer of the often lengthy calculations and procedures required to be followed while large or complicated structures are analyzed using classical methods. But not all the time such detailed analysis are necessary to be performed i.e. sometimes, just approximate analysis could suffice our requirements as in case of preparing the rough estimates and participating in the bidding process for a tender. It may even happen that sometimes the analysis software or tool is not available at hand? Or the worst case, the computer itself is not available?? Then in such cases, accurate analysis of such large and complicated structures involving so many calculations is almost impossible. Now-a-days, high rise buildings and multi-bay-multi-storey buildings are very common in metropolitan cities. The analysis of frames of multistoreyed buildings proves to be rather cumbersome as the frames have a large number of joints which are free to move. Even if the commonly used Moment distribution method is applied to all the joints, the work involved shall be tremendous. However, with certain assumptions, applying the substitute analysis methods like substitute frame method, portal method, cantilever method or factor method, the structures can be analyzed approximately.

Substitute frame method By considering any floor of the frame called substitute frame, the moments can be calculated and results can be obtained in good 13

agreement with the results from rigorous analysis. The moments carried from floor to floor through columns are very small as compared to the beam moments; therefore, the moments in one floor have negligible effect on the moments on the floors above and below. Therefore, in this method, the analysis of the multi-storeyed frames is carried out by taking one floor at a time. Each floor is taken with columns above and below fixed at far ends, and the moments and shears are calculated in beams and columns. The method is very effective in analyzing any framed structure under vertical loadings. This work is focused to check its applicability and efficacy under the lateral loading conditions Objectives • To manually analyze the problem frame, using Kani’s method under vertical loading conditions. • To perform the same analysis using standard analysis software Staad.Pro • Perform substitute frame analysis for the loading cases • Compare the accuracy of the substitute frame analysis with manual and Staad.Pro analysis and check its validity in lateral loading cases. • Optimize the substitute frame method to further lessen the calculations so as to get the final results within permissible limit of errors. • Design the Structural members of the multistory.

14

CHAPTER 3 ANALYSIS UNDER VERTICAL LOADS

15

Ap proximate analysis vertical l oads:

of

the

frame

for

The analysis is made by considering the following assumptions: 1) Beams do not receive axial forces. 2) For any beam a point of contra flexure occurs at a distance of 0.1 L form each end of the span, where L is the span of beam

Dead load on the beams = 25 kN/m Live load on the beams = 15 kN/m Span of the beams between AB, EF, IJ, CD, GH, KL = 6.1 m Span of the beams between BC, FG, JK = 2.1 m Assume points of contra flexure at 0.1 L from each end i.e. 0.1 x 6.1 = 0.61 m from each end for the long beams. Distance between points of contra flexure, L 1 = 6.1 - (2x0.61) = 4.88 m 16

Maximum positive B.M =

 

2 x 4.88 = 119.1 kN-m

Shear at each point of contra flexure = (15+25)

. = 97.6 kN 

Maximum negative moment at each end of the beam =

- [97.6 x 0.61+

∗.∗.] 

= - 67 kN

Maximum shear force for a beam occurs at each end of the beam and is . equal to = 122 kN

∗  

Analysis of columns: Consider the end column AEMN Axial force in AE = 40 x 3.05 = 122 kN Axial force in EI = 122+122 = 244 kN Axial force in IM = 244+122 = 366 kN Moment at the upper end of column = Moment at the end of beam = 67 kN-m Assuming equal stiffness for column moment transmitted to each column by a beam at E or I = 67/2 = 33.5 kN-m Now consider any interior column. There will be no moment transmission to the interior column. Consider the column BFJN Axial force in BF = 40 x 6.1 = 244 kN, Axial force in FJ = 244+244 = 488 kN Axial force in JN = 488+244 =732 Kn

17

Method of s ubstitute frames :

Analysis of multi-storey building frames involves lot of complications and tedious calculations by using conventional methods. To carry out exact analysis is a time consuming task. Substitute frame method for analysis of multistory frame can be handy in approximate and quick analysis. This method has been applied only for vertical loading conditions. The method assumes that the moments in the beams of any floor are influenced by loading on that floor alone. The influence of loading on the lower or upper floors is ignored altogether. The process involves the division of multi-storied structure into smaller frames. These sub frames are known as equivalent frames or substitute frames. The sub frames are usually analyzed by the moment distribution method, using only Two cycle of distribution. It is only necessary to consider the loads on the two nearest spans on eac h side of the point . The substitute frames are formed by the beams at the floor level under consideration, together with the columns above and below with their far ends fixed. 18

The distributed B.M are not carried over far ends of the columns in this process; the moments in the columns are computed at each floor level independently and retained at that floor irrespective of further analysis.

Assumptions – Slab thickness = 0.15m

floor finish thickness = 0.05m

Beam section = 0.35mx0.45m

column section = 0.35mx0.35m

Density of concrete used = 25 Kn/m 3 live load for educational building = 3kN/m2 Clockwise moment positive and vice-versa

LoadingSlab dead load = 0.15x1x25 = 3.75 kN/m 2

Floor finish = 1.25 kN/m ,

2

live load = 3kN/m

2

Beam self weight = 0.35x0.45x25 = 3.9 kN/m Total vertical load per metre length of beam = (3.75+1.25+3)+3.9 = 12kN/m Fixed end moments induced 19

∗∗ = - 36 kN-m  ∗∗ = 36kN-m = M =  ∗∗ = - 4kN-m  ∗∗ = 4kN-m 

Mab = Mef = Mij =Mcd = Mgh = Mkl = Mba = Mfe = M ji =Mdc = Mhg Mbc = Mfg = M jk = Mbc = Mgf = Mkj = Moment

lk

of Length

Relative

Distribution

Rotation

3

stiffness

factors

Factor =

= K/∑k

-(1/2)( K/∑k)

Span inertia (mm ) (L) x10

3

metres

(K)

AB

443

6

73.8

0.34

-0.17

AE

417

3

139

0.65

-0.325

BA

443

6

73.8

0.08

-0.04

BF

417

3

139

0.15

-0.075

BC

1328

2

664

0.75

-0.375

CB

1328

2

664

0.75

-0.375

CG

417

3

139

0.15

-0.075

CD

443

6

73.8

0.08

-0.04

DC

443

6

73.8

0.34

-0.17

DH

417

3

139

0.65

-0.325

EA

417

3

139

0.39

-0.195

EF

443

6

73.8

0.21

-0.105

EI

417

3

139

0.39

-0.195

FE

443

6

73.8

0.07

-0.035

FB

417

3

139

0.14

-0.07

FG

1328

2

664

0.65

-0.325

FJ

417

3

139

0.14

-0.07

GF

1328

2

664

0.65

-0.325

GC

417

3

139

0.14

-0.07

GH

443

6

73.8

0.07

-0.035

GK

417

3

139

0.14

-0.07

HD

417

3

139

0.39

-0.195

HL

417

3

139

0.39

-0.195

HG

443

6

73.8

0.21

-0.105

20

IE

417

3

139

0.48

-0.24

IJ

443

6

73.8

0.25

-0.125

II’

313

4

78.3

0.27

-0.135

JI

443

6

73.8

0.08

-0.04

JF

417

3

139

0.14

-0.07

JK

1328

2

664

0.70

-0.35

JJ’

313

4

78.3

0.08

-0.04

KJ

1328

2

664

0.70

-0.35

KG

417

3

139

0.14

-0.07

KL

443

6

73.8

0.08

-0.04

KK’

313

4

78.3

0.08

-0.04

LH

417

3

139

0.48

-0.24

LL’

313

4

78.3

0.27

-0.135

LK

443

6

73.8

0.25

-0.125

Spacing of the frame = 5.4 m 2

D.L of Slab = 5 kN/m , L.L of Slab = 3 kN/m

2

D.L of floor per meter run of girder = (5 x 5.4 ) + 8.1 = 35.1 kN/m L.L of floor per meter run of girder = 3 x 5.4 = 16.2 kN/m Fixed end moments: The fixed end moments due to dead load and live load areMember

D.L per metre L.L per metre F.E.M due to F.E.M (kN/m)

(kN/m)

D.L (kN-m)

EF

35.1

16.2

105.3

153.9

FG

35.1

16.2

11.7

17.1

GH

35.1

16.2

105.3

153.9

21

due

D.L+L.L(kN-m)

to

Frame will be analyzed be two cycle moment distribution: Bending Moment Moment in columns: Loading: D.L on EFGH, L.L on EF and GH GH JOINT Column distribution factors

E

F

G

Just above floor

0.39

0.14

0.14

0.39

Just below floor

0.39

0.14

0.14

0.39

Horizontal members Distribution factors

EF

FE

0.21

0.07

FG

FEM due to D.L

GF

GH

HG

0.65

0.65

0.07

0.21

-11.7

11.7

F.E.M due to total load

-153.9

153.9

Distribution +carry over

-4.977

16.16

Summation

-158.877

170.06 -104.13

-92.43

Distribution to 22

H

-153.9

153.9

92.43

-16.16

4.977

104.13

-170.06

158.87

columns Just above floor

61.95

9.23

-9.23

61.95

Just below floor

61.95

9.23

-9.23

61.95

join joints ts

E

F

G

H

Membe Memberrs

EF

FE

FG

GF

GH

HG

Distribution

0.21

0.07

0.65

0.65

0.07

0.21

FEM,s FEM,s

-153.9

153.9

-17.1 -17.1

17.1

-153.9

153.9

Distri Distr ibution

32.32

-9.57

-88.92 -88.92

88.92

-9.57

-32.32

Car Carry over

-4.78

16.16

-49.24 -49.24

-4.78 -4.78

-16.16

4.8


1.0

-2.31

21.47

25.79

2.77

1.08

Car Carry over

-1.15

0.5

12.89

10.73

0.54

1.38


0.24 0.24

-0.57

8.25

6.62

0.71

0.28

-12 -126.3

158.1

-11 -112.7

128.6

-15 -156.5

124.8

factors

Tota Totall moment

23

Membe Memberrs

Staad Staad Pro Pr o

Kani’s Method

End moments End Mom Moments ents

Substitu ubstitute te fram

Kani’s Method

end moment moments s

Vs S/ F Method Method

EF

-129.23

-133.37

-126.3

5.5%

FE

156.52

160.43

158.1

1.47%

FG

108.57

-113.51 -113.51

-112.7 -112.7

0.7%

GF

120.6 120.6

128.6 128.6

128.6 128.6

0

GH

-150.42 -150.42

-159.76

-156.5 -156.5

2.08%

HG

127.60

125

124.8

0.2%

The inference made made from the table is that in case of vertical loading, the difference between the Kani’s analysis and substitute frame method (S/F method) is very less.

24

Chapter 4 Design of Slab

25

Design of sl ab:

Plan of Ground Floor and First Floor 26

Design of s lab: S1 (Two-way slab) Size of slab = 5.4 m x 6.3 m Edge conditions = two adjacent edges discontinuous Materials used = M-20 grade of concrete and Fe-415 HYSD bars Depth of slab: 6 inch = 152.4mm Adopt the effective depth (d) = 135mm Loads: Self weight of slab = (25 x 0 .22) = 5.5 kN/m 2 Imposed load = 3 kN/m

2

Weight of flooring 50 mm thick = (0.05 x 24) = 1.2 kN/m 2 Total working load = w = 9.7 kN/m

2

Therefore design ultimate load = w u = (1.5 x 9.7) = 14.55kN/m 2 Ultimate design mo ments The moment coefficients for (L y/Lx) = (6.3/5.4) =1.16 Short span moment coefficients: a) –ve moment coefficient = α x = 0.053 b) +ve moment coefficient = α x = 0.040 Long span moment coefficients: a) – ve moment coefficient = α y = 0.042 b) +ve moment coefficient= α y = 0.035 2

2

Mux(-ve) = (α x wu Lx ) = (0.053 x 14.55 x 5.4 ) = 22.9 kN-m Mux(+ve)= (αx wu Lx2) = (0.040 x 14.55 x 5.4 2) = 16.97 kN-m 2

2

Muy(-ve) = (α y wu Lx ) = (0.042 x 14.55 x 5.4 ) = 17.82 kN-m 2

2

Muy(+ve) = (α y wu Lx ) = (0.035 x 14.55 x 5.4 ) = 14.85 kN-m Vu = 0.5 wu Lx = 0.5 x 14.55 x 5.4 = 39.3 kN 27

Check for d epth: Mu,lim = 0.138 f ck b d

2

√

d = (22.9 x 106)/(0.138 x 20 x 1000) = 91mm < 135mm hence the effective depth selected is sufficient to resist the design ultimate moment Ast,min =(0.12% b d) = 0.0012 x 1000 x 153 = 183.6 mm 2 Reinforcements along short and long s pan directions the area of reinforcement is calculated using the relation, Mu = 0.87 f y Ast d {1- Ast f y/ b d f ck} Spacing of the selected bars are computed using the relation, Spacing = S = (Area of 1 bar/ total area) x1000 such that A st(provided) ≥ Ast(min) In addition, the spacing should be the least Of three times the effective de pth or 300 mm. using 10 mm diameter bars for long span, d = 135 mm & for short span, d = 125 mm. The detail of reinforcements provided in the two-way slab is compiled in the table below: Location

Ast (required)

Spacing of 10 mm ϕ bars

1) Short span 2

140 mm c/c

2

190 mm c/c

2

200 mm c/c

2

240 mm c/c

a) –ve B.M(top of supports)

560 mm

b) +ve B.M(centre of span)

403 mm

2) Long span a) –ve B.M (top of supports)

388 mm


320 mm

Torsion Reinforcement at corners Area of torsional steel in each 4 layers = (0.75 x 403) = 302.25 mm

28

2

Distance over which the torsion reinforcement is provided = (1/5 short span) = (0.2 x 5400) = 1080 mm. Provide 6 mm dia meter bars at 75 mm c/c for a length of 1080 mm at all 4 corners in 4 layers

Design of slab: S2 (Two-way Slab) Size of slab = 5.4 m x 6.3 m Edge conditions = one edge discontinuous Materials used = M-20 grade of concrete and Fe-415 HYSD bars Depth of slab: 6 inch = 152.4mm Adopt the effective depth (d) = 135mm Loads: Self weight of slab = (25 x 0 .22) = 5.5 kN/m 2 Imposed load = 3 kN/m

2

Weight of flooring 50 mm thick = (0.05 x 24) = 1.2 kN/m Total working load = w = 9.7 kN/m

2

2

Therefore design ultimate load = w u = (1.5 x 9.7) = 14.55 kN/m Ultimate design moments The moment coefficients for (L y/Lx) = (6.3/5.4) =1.16 Short span moment coefficients: c) – ve moment coefficient = α x = 0.048 d) +ve moment coefficient = α x = 0.036 Long span moment coefficients: c) – ve moment coefficient = α y = 0.037 d) +ve moment coefficient= α y = 0.028 2

2

Mux (-ve) = (α x wu Lx ) = (0.048 x 14.55 x 5.4 ) = 20.36 kN-m 29

2

2

2

Mux (+ve) = (α x wu Lx ) = (0.036 x 14.55 x 5.4 ) = 15.27 kN-m 2

2

Muy (-ve) = (αy wu Lx ) = (0.037 x 14.55 x 5.4 ) = 15.69 kN-m Muy (+ve) = (α y wu Lx2) = (0.028 x 14.55 x 5.4 2) = 11.88 kN-m Vu = 0.5 wu Lx = 0.5 x 14.55 x 5.4 = 39.3 kN Check for depth: 2

Mu, lim = 0.138 f ck b d

√

d = (20.36 x 106) / (0.138 x 20 x 1000) = 85.88mm < 135mm Hence the effective depth selected is s ufficient to resist the design ultimate moment Ast, min = (0.12% b d) = 0.0012 x 1000 x 153 = 183.6 mm 2 Reinforcements along short and long s pan directions The area of reinforcement is calculated using the relation, Mu = 0.87 f y Ast d {1- Ast f y/ b d f ck} Spacing of the selected bars are computed using the relation, Spacing = S = (Area of 1 bar/ total area) x1000, such that A st (provided) ≥ Ast (min) In addition, the spacing should be the least Of three times the effective depth or 300 mm. using 10 mm diameter bars for long span, d = 135 mm & for short span, d = 125 mm. The detail of reinforcements provided in the two-way slab is compiled in the table below:

30

Location

Ast (required)


3) Short span c) –ve B.M(top of supports)

490 mm

2

160 mm c/c

d) +ve B.M(centre of span)

360 mm2

215 mm c/c

c) –ve B.M (top of supports)

340 mm2

230 mm c/c


254 mm

2

305 mm c/c

4) Long span

Torsion Reinforcement at corners Area of torsional steel in each 4 layers = (0.75 x 360) = 270 mm 2 Distance over which the torsion reinforcement is provided = (1/5 short span) = (0.2 x 5400) = 1080 mm. Provide 6 mm diameter bars at 75 mm c/c for a length of 1080 mm at all 4 corners in 4 layers (Same for Design of other slabs having same area and end conditions)

Design o f sl ab S3 (One-way Slab) Size of slab = 5.4 m x 2.1m Edge conditions = one edge discontinuous Materials used = M-20 grade of concrete and Fe-415 HYSD bars Depth of slab: 6 inch = 152.4mm Adopt the effective depth (d) = 135mm Loads: Self weight of slab = (25 x 0 .22) = 5.5 kN/m Imposed load = 3 kN/m

2

2

Weight of flooring 50 mm thick = (0.05 x 24) = 1.2 kN/m 31

2

Total working load = w = 9.7 kN/m

2

Therefore design ultimate load = w u = (1.5 x 9.7) = 14.55 kN/m

2

Ultimate design mo ments The moment coefficients for (L y/Lx) = (6.3/5.4) =1.16 Short span moment coefficients: e) – ve moment coefficient = α x = 0.048 f) +ve moment coefficient = α x = 0.036 Long span moment coefficients: e) – ve moment coefficient = α y = 0.037 f) +ve moment coefficient= α y = 0.028 2

2

Mux (-ve) = (α x wu Lx ) = (0.048 x 14.55 x 5.4 ) = 20.36 kN-m 2

2

Mux (+ve) = (α x wu Lx ) = (0.036 x 14.55 x 5.4 ) = 15.27 kN-m 2

2

Muy (-ve) = (αy wu Lx ) = (0.037 x 14.55 x 5.4 ) = 15.69 kN-m 2

2

Muy (+ve) = (α y wu Lx ) = (0.028 x 14.55 x 5.4 ) = 11.88 kN-m Vu = 0.5 wu Lx = 0.5 x 14.55 x 5.4 = 39.3 kN Check for depth: 2

Mu, lim = 0.138 f ck b d

√

6

d = (20.36 x 10 ) / (0.138 x 20 x 1000) = 85.88mm < 135mm Hence the effective depth selected is s ufficient to resist the design ultimate moment Ast, min = (0.12% b d) = 0.0012 x 1000 x 153 = 183.6 mm

2

Reinforcements along short and long s pan directions The area of reinforcement is calculated using the relation, Mu = 0.87 f y Ast d {1- Ast f y/ b d f ck} Spacing of the selected bars are computed using the relation, 32

Spacing = S = (Area of 1 bar/ total area) x1000, such that A st (provided) ≥ Ast (min) In addition, the spacing should be the least Of three times the effective depth or 300 mm. using 10 mm diameter bars for long span, d = 135 mm & for short span, d = 125 mm. The detail of reinforcements provided in the two-way slab is compiled in the table below: Location

Ast (required)


5) Short span a) –ve B.M(top of supports)

490 mm2

160 mm c/c


360 mm

2

215 mm c/c

2

230 mm c/c

2

305 mm c/c

Long span c) –ve B.M (top of supports)

340 mm


254 mm

Torsion Reinforcement at corners Area of torsional steel in each 4 layers = (0.75 x 360) = 270 mm

2

Distance over which the torsion reinforcement is provided = (1/5 short span) = (0.2 x 5400) = 1080 mm. Provide 6 mm diameter bars at 75 mm c/c for a length of 1080 mm at all 4 corners in 4 layers (Same Design for other slabs having same area and end conditions)

Design o f sl ab: S4 (One-way Slab) Size of slab = 5.4 m x 2.1 m Edge conditions = one edges discontinuous Materials used = M-20 grade of concrete and Fe-415 HYSD bars 33

Depth of slab: 6 inch = 152.4mm Adopt the effective depth (d) = 135mm Effective span: The least of i)

(Clear span + eff depth ) = 2.1 + 0.135 = 2.235

ii)

(Center to center supports) = 2.1+0.23 = 2.33

Loads: Self weight of slab = (25 x 0 .22) = 5.5 kN/m

2

Imposed load = 3 kN/m 2 Weight of flooring 50 mm thick = (0.05 x 24) = 1.2 kN/m

2

Total working load = w = 9.7 kN/m2 Therefore design ultimate load = w u = (1.5 x 9.7) = 14.55kN/m

2

Ultimate moments and shear forc es, 2

2

Mu = (0.125wuL ) = (0.125x14.55x2.235 ) =9.09 kN-m Vu = (0.5wuL) = (0.5 x 14.55 x2.235) = 16.25 KN Limiting moment of resistance Mu, lim = 0.138 f ck b d

2

=(0.138x20x1000x 135 2) = 50 KN-m Mu < Mu, lim. Section is under reinforced. Tension Reinforcements Mu = 0.87 f y Ast d {1- Ast f y/ b d f ck} 6

9.09x10 =( 0.87x415xAst135) {1- Ast 415/1000x135x20} 2

Solving Ast = 192 mm > Ast, min = 160 mm

2

Using 10 mm diameter of bars, the spaci ng of the selected bars are computed using the relation,

34

Spacing = S = (Area of 1 bar/ total area) x1000 such that A st(provided) ≥ Ast(min) S = (1000x78.5 / 193 ) = 400mm Adopt spacing of 400 with alternative bars are bent up at supports Distribution bars Ast = 0.12 percent = (0.12x1000x155) = 186mm 2 Providing 8mm bars at 270 c/c Check for shear stress 3

tv = (vu / bxd) = (16.25x10 /1000x135) = 0.120 N/mm

2

pt = (100Ast / bxd ) =(100x192x0.5 / 1000x135) = 0.142 Permissible shear stress slab (as IS- 456 ) is calculated as 2

ktc = 0.36 N/mm > tv Hence slab is safe. Check for deflectio n contro l (L/d)max = (L/d)basic x kt x kc x kf For pt = s(100x192 / 1000x135) = 0.33 perc ent ,k c = 1, kf = 1. (L/d)max = ( 26x1.4x1x1) = 36.4 (L/d) provided = 13.8 Hence deflection criteria is satisfied

Design o f sl ab: S5 (One-way Slab) Size of slab = 5.4 m x 2.1 m Edge conditions = No one edges discontinuous Materials used = M-20 grade of concrete and Fe-415 HYSD bars Depth of slab: 6 inch = 152.4mm 35

Adopt the effective depth (d) = 135mm Effective span: The least of (Clear span + eff. depth) = 2.1 + 0.135 = 2.2 35 (Center to center supports) = 2.1+0.23 = 2.33 Loads: Self weight of slab = (25 x 0 .22) = 5.5 kN/m Imposed load = 3 kN/m

2

2

Weight of flooring 50 mm thick = (0.05 x 24) = 1.2 kN/m 2 Total working load = w = 9.7 kN/m

2

Therefore design ultimate load = w u = (1.5 x 9.7) = 14.55kN/m 2 Ultimate moments and shear forc es, Mu = (0.125wuL2) = (0.125x14.55x2.235 2) = 9.09 kN-m Vu = (0.5wuL) = (0.5 x 14.55 x2.235) = 16.25 KN Limiting moment of resistance Mu, lim = 0.138 f ck b d2 2

= (0.138x20x1000x 135 ) = 50 KN-m Mu < Mu, lim. Section is under reinforced. Tension Reinforcements Mu = 0.87 f y Ast d {1- Ast f y/ b d f ck} 6

9.09x10 = ( 0.87x415xA st135) {1- A st 415/1000x135x20} 2

Solving Ast = 192 mm > Ast, min = 160 mm

2

Using 10 mm diameter of bars, the spaci ng of the selected bars are computed using the relation, Spacing = S = (Area of 1 bar/ total area) x1000 such that A st(provided) ≥ Ast(min) 36

S = (1000x78.5 / 193 ) = 400mm Adopt spacing of 400 with alternative bars are bent up at supports Distribution bars Ast = 0.12 percent = (0.12x1000x155) = 186mm

2

Providing 8mm bars at 270 c/c Check for shear stress 3

tv = (vu / bxd) = (16.25x10 /1000x135) = 0.120 N/mm

2

pt = (100Ast / bxd ) =(100x192x0.5 / 1000x135) = 0.142 Permissible shear stress slab (as IS- 456 ) is calculated as ktc = 0.36 N/mm2> tv Hence slab is safe. Check for deflectio n contro l (L/d)max = (L/d)basic x kt x kc x kf For pt = s (100x192 / 1000x135) = 0.33 percent, kc = 1, k f = 1. (L/d)max = ( 26x1.4x1x1) = 36 (L/d) provided = 13.7 Hence deflection criteria is satisfied.

37

CHAPTER 5 Design of Beams

38

DESIGN OF BEAM B1: 350mm x 450mm (Long beam) Calculation of loads Self weight of beam = 3.9 kN/m , le ngth of beam = 6.1 m Weight of the 9” wall above Beam = 0.15m x 3 x 20 = 9 KN/m Area of load control of slab under beam = 9.18 m

2

Weight of slab on beam = 9.18 x 0.1524 x 25 = 5.7 kN/m Total weight = 18.6 KN/m f ck = 20 N/mm

2

f y = 415 N/mm

2

Assuming 25 mm effective cover, b = 350 mm, d = 425 mm Fixed end moment at one end = -126.3 kN-m Fixed end moment at another end = 158.1 kN-m Design end moment at mid span = 2

2

= (1.5 x 18.6 x 6.1 )/8 – (18.6 x 6.1 )/2 = 129.76 – 346.05 = -216.29 kNm Location

Mu (kNm)

Mu/bd

Pt (%)

Area of

From SP-16

steel mm

Ist Support

-126.3

1.78

0.563

837

IInd Support

158.1

2.23

0.731

1087

Mid Span

-216.29

3.05

0.955

1504

Design for shear reinforcement Maximum shear force = 56.73 kN Max steel percentage = 0.955 % tc = 0.62 N/mm 2 (I.S-456:2000) Vus = Vu –tc b d = 56.73 – 0.62 x 0.35 x 0.45 = 56.63 kN Shear resisted by stirrups = 56.63 kN Vus / d = 56.63/425 = 0.13 , Provide 2L -8mm shear stirrups Provide 2L-8mm dia bars @ 75 mm c/c

39

2

DESIGN OF BEAM B2: 350mm x 450mm (shor t beam) Calculation of loads Self weight of beam = 3.9 kN/m , length of beam = 2.8 m Weight of the 9” wall above Beam = 0.15m x 3 x 20 = 9 KN/m Area of load control of slab under beam = 9.18 m

2


2

f y = 415 N/mm

2


2

= (1.5 x 18.6 x 2.8 )/8 – (18.6 x 2.8 )/2 = 27.34 – 72.91 = -45.57 kNm Location

Mu (kNm)

Mu/bd

Pt (%)

Area of

From SP-16

steel mm

Ist Support

-112.7

1.59

0.493

777

IInd Support

128.6

1.81

0.564

889

Mid Span

-45.57

0.64

0.186

293

Design for shear reinforcement Maximum shear force = 26.04 kN Max steel percentage = 0.56 % 2

tc = 0.5 N/mm (I.S-456:2000) Vus = Vu –tc b d = 26.04 – 0.5 x 0.35 x 0.45 = 25.96 kN Shear resisted by stirrups = 25.96 kN Vus / d = 25.96/425 = 0.06, Provide 2L-8mm shear stirrups Provide 2L-8mm dia bars @ 80 mm c/c

40

2

DESIGN OF BEAM B3: 350mm x 450mm (interm ediate beam) Calculation of loads Self weight of beam = 3.9 kN/m , length of beam = 5.4m Weight of the 9” wall above Beam = 0.15m x 3 x 20 = 9 KN/m Area of load control of slab under beam = 7.29 m

2


2

f y = 415 N/mm

2


2

= (1.5 x 18.04 x 5.4 )/8 – (18.04 x 5.4 )/2 = 98.63 – 263.02 = -164.39 kNm Location

Mu (kNm)

Mu/bd

Pt (%)

Area of

From SP-16

steel mm

Ist Support

-112.7

1.59

0.493

777

IInd Support

128.6

1.81

0.564

889

Mid Span

-164.39

2.31

0.758

1194


tc = 0.57 N/mm (I.S-456:2000) Vus = Vu – tc b d = 48.71 – 0.57 x 0.35 x 0.45 = 48.62 kN Shear resisted by stirrups = 48.62 kN Vus / d = 48.62/425 = 0.11, Provide 2L-8mm shear stirrups Provide 2L-8mm dia bars @ 80 mm c/c

41

2

DESIGN OF BEAM B4: 350mm x 450mm Calculation of loads Self weight of beam = 3.9 kN/m , length of beam = 6.1m Weight of the 9” wall above Beam = 0.15m x 3 x 20 = 9 KN/m Area of load control of slab under beam = 18.38 m

2


2

f y = 415 N/mm

2


2

= (1.5 x 24.36 x 6.1 )/8 – (24.36 x 6.1 )/2 = 169.95 – 453.21 = -283.26 kNm Location

Mu (kNm)

Mu/bd

Pt (%)

Area of

From SP-16

steel mm

Ist Support

-126.3

1.78

0.493

777

IInd Support

158.1

2.23

0.564

889

Mid Span

-283.26

3.99

0.955

1504


tc = 0.62 N/mm (I.S-456:2000) Vus = Vu –tc b d = 453.21 – 0.62 x 0.35 x 0.45 = 453.11 kN Shear resisted by stirrups = 453.11 kN Vus / d = 453.11/425 = 1.06, Provide 2L-8mm shear stirrups. Provide 2L-8mm dia bars @ 80 mm c/c

42

2


2


2

f y = 415 N/mm

2


2

= (1.5 x 24.36 x 6.1 )/8 – (24.36 x 6.1 )/2 = 169.95 – 453.21 = -283.26 kNm Location

Mu (kNm)

Mu/bd

2

Pt (%)

Area of

From SP-16

steel mm

Ist Support

-126.3

1.78

0.493

777

IInd Support

158.1

2.23

0.564

889

Mid Span

-283.26

3.99

0.955

1504

Design for shear reinforcement Maximum shear force = 453.21 kN Max steel percentage = 0.955 % tc = 0.62 N/mm 2 (I.S-456:2000) Vus = Vu –tc b d = 453.21 – 0.62 x 0.35 x 0.45 = 453.11 kN Shear resisted by stirrups = 453.11 kN Vus / d = 453.11/425 = 1.06 , Provide 2L -8mm shear stirrups Provide 2L-8mm dia bars @ 80 mm c/c 43

2


2


2

f y = 415 N/mm

2


2

= (1.5 x 23.18 x 5.4 )/8 – (23.18 x 5.4 )/2 = 187.76 – 337.96 = -150.20 kNm Location

Mu (kNm)

Mu/bd

Pt (%)

Area of

From SP-16

steel mm

Ist Support

-126.3

1.78

0.493

777

IInd Support

158.1

2.23

0.564

889

Mid Span

-150.20

2.12

0.685

1079


tc = 0.53 N/mm (I.S-456:2000) Vus = Vu –tc b d = 62.58 – 0.53 x 0.35 x 0.45 = 62.49 kN Shear resisted by stirrups = 62.49 kN Vus / d = 62.49/425 = 0.14, Provide 2L-8mm shear stirrups Provide 2L-8mm dia bars @ 80 mm c/c

44

2

3-D model of Building

3-storey frame in Staad Pro

45

Shear Diagram of beam B1

Deflection profile for beam b1

Reinforcement details for Beam B1 in Staad

46

CHAPTER 6 Design of Columns

47

LOADS 1) Floor height = 10 ft = 3.048 m 2) Weight of 6” thick slab (0.1524m) = 25 x 0.1524 = 3.8 kN/m 3) Weight of flooring 50 mm thick = 24 x 0.05 = 1.2 KN/m Total D.L =

2

2

5.0kN/m 2

4) Imposed floor loads for class rooms and lecture rooms = 3 KN/m 2 5) Self weight of column (350 mm x 350 mm) of 10 m height 2

= 0.35 x10x25 6) Loads from roof (truss) = 1kN/m

=

30.6 kN

2

7) Load of 9” thick wall(228.6mm) for 10m height = 0.2286x10x20

=

45.72kN/m

8) For each floor height of 3.048m, loads of 9” wall on the beam = 0.2286x3.048x20

=

13.93kN/m

9) Wind loads: Design wind pressure is Px = 0.6V x2 Where P x = design wind pressure in N/mm2 at a height Z and Vx = design wind velocity in m/s at a height Z Column A1 Area under column = 2.7 m x 3.16 m = 8.53 m

2

Total dead weight(weight of slab, flooring, finishes) = 8.53 x 5 = 42.65 kN For three strorey building, total dead load carried = 42.65 x 3 =128 kN Weight of 9” thick wall = 45.72(3.16+2.7-0.35) = 251.9 kN 2

Self weight of column (350mm x 350mm) of 10m height = 0.35 x 10 x 25 = 30.6 kN Total dead load = 128+251.9+30.6 =410.5 kN Live load = 3 x 8.53 x 3 = 76.7 kN Load from Roof = 1 x 8.5 = 8.5 kN Grand total: 410.5+76.7+8.53 = 495.73 kN Loading is same for column A4 Column A2 Area of load = 2.7 x 4.36 = 11.7m

2

Dead load(weight of slab, flooring, finishes) =11.7 x 5 = 58.5 kN 48

For three storey building, DL = 58.5 x 3 = 175.5 kN Weight of 9” thick wall = 45.72(4.36+2.7-0.35) = 306.78 kN 2

Self weight of column (350mm x 350mm) of 10m height = 0.35 x 10 x 25 = 30.6 kN Total D.L = 175.5+306.78+30.6 = 512.88 kN L.L = 3 x 11.7 x 3 = 105.3 kN Load from roof = 1 x 11.7 = 11.7 kN Grand total: 629.88 kN Loading is similar for the column A3 Column B1 Area of load control under column B1 = 5.4 x 3.16 = 17.06m 2 Dead load(weight of slab, flooring, finishes) =17.06 x 5 = 85.3 kN For three storey building, DL = 85.3 x 3 = 256 kN Weight of 9” thick wall = 45.72(5.4+3.16-0.35) = 375.36 kN 2

Self weight of column (350mm x 350mm) of 10m height = 0.35 x 10 x 25 = 30.6 kN Total D.L = 256+375.36+3.6 = 747.26 kN L.L = 3 x 17.06 x 3 = 153 kN Load from roof = 1 x 17.06 = 17.06 kN Grand total: 832.02 kN Loading is similar for the column B4,C1,C4,D1,D4,E1,E4,F1,F4,G1,G4,H1,H4 Column B2: Area of load control under column B2 = 5.4 x 4.36 = 23.5m 2 Dead load(weight of slab, flooring, finishes) =23.5 x 5 = 117.5 kN For three storey building, DL = 117.5 x 3 = 352.5 kN Weight of 9” thick wall = 45.72(5.4+4.36) = 446.2 kN Self weight of column (350mm x 350mm) of 10m height = 2

0.35 x 10 x 25 = 30.6 kN 49

Total D.L = 256+375.36+3.6 = 747.26 kN L.L = 3 x 23.5 x 3 = 211.5 kN Load from roof = 1 x 23.5 = 23.5 kN Grand total: 1064.3 kN Loading is similar for the column B3,C2,C3,E2,D1,D2,E3,F2,F3,G2,G3,H2,H3 Column J1 Area of load control under column J1 = 3.16 x 1.25 = 3.95 m

2

Dead load(weight of slab, flooring, finishes) =3.95 x 5 = 19.75 kN For three storey building, DL = 19.75 x 3 = 59.25 kN Weight of 9” thick wall = 45.72(1.25+3.16-0.35) = 185.62 kN Self weight of column (350mm x 350mm) of 10m height = 0.35 2 x 10 x 25 = 30.6 kN Total D.L = 59.25+185.62+30.6= 275.47 kN L.L = 3 x 3.95 x 3 = 35.55 kN Load from roof = 1 x 3.95 = 3.95 kN Grand total: 314.97 kN Loading is similar for the column J4 Column J2 Area of load control under column J2 = 1.25 x 4.36 = 5.45 m

2

Dead load(weight of slab, flooring, finishes) =5.45 x 5 = 27.25 kN For three storey building, DL = 27.75 x 3 = 83.25 kN Weight of 9” thick wall = 45.72(4.36+1.25-0.35) = 240.48 kN 2

Self weight of column (350mm x 350mm) of 10m height = 0.35 x 10 x 25 = 30.6 kN Total D.L = 83.45+240.48+30.6= 354.53 kN L.L = 3 x 5.45 x 3 = 49.05 kN Load from roof = 1 x 5.45 = 5.45 kN Grand total: 409.03 kN 50

Loading is similar for the column J3 Colum n I2 Area of load control under column I2 = 3.95 x 4.36 = 17.22 m

2

Dead load(weight of slab, flooring, finishes) = 17.22 x 5 = 86.1 kN For three storey building, DL = 86.1 x 3 = 258.3 kN Weight of 9” thick wall = 45.72(4.36+3.95) = 379.93 kN 2

Self weight of column (350mm x 350mm) of 10m height = 0.35 x 10 x 25 = 30.6 kN Total D.L = 258.3+379.93+30.6= 668.83 kN L.L = 3 x 17.22 x 3 = 154.98 kN Load from roof = 1 x 17.22 = 17.22 kN Grand total: 841.03 kN Loading is similar for the column I3 Colum n I1 Area of load control under column I2 = 3.325 x 3.05 = 10.14 m

2

Dead load(weight of slab, flooring, finishes) = 10.14 x 5 x 0.1524 = 7.72 kN For three storey building, DL = 7.72 x 3 = 23.18 kN Weight of 9” thick wall = 45.72(3.325+3.05) = 291.46 kN 2

Self weight of column (350mm x 350mm) of 10m height = 0.35 x 10 x 25 = 30.6 kN Total D.L = 152.1+291.46+30.6= 345.24 kN L.L = 3 x 10.14 x 3 = 91.26 kN Load from roof = 1 x 10.14 = 10.14 kN Grand total: 575.56 kN, Loading is similar for the column I4

51

Bending Shear Profile for Column C1

Deflection profile for Column C1

Reinforcement details for column C1

52

Chapter 7 Design of f ooting

53

Design of fo otin g 1 (F1) 1.) Data Axial load of column=p= 496KN Size of column =

350 x 350 mm

2

Use M20 & fe-415HYSD bars 2.) Size of footing; Load on column = 496KN Wt of footing & backfill at 10%= 49.6 Total load = 545.6KN Area of footing = (545.6/200) = 2.728m2 Size of footing =L=B = √2.728 = 1.651m Adopt 1.7 by 1.7m square footing. Net soil pressure at ultimate loads with a load factor of 1.5 is given by Qu (496x1.5/1.7x1.7) = 257KN/m 2 =

0.26 N/mm

2

3.) One way shear. The critical section is at a d istance “d” from the column face . Factored shear force = V u1 = (0.26x1700)(1000-d) =442(1000-d) Assuming percentage of reinforcement in the footing pt = 0.25 percent for M20 grade concrete Now from IS-456 code permissible shear stress as =tc =0.36 N/mm One way shear resistance Vc1 =(tcx1700xd). = (612d) N Vu1 < Vc1 442(1000-d) < 612d . d > 420 mm 54

2

4.) Two way shear, Assuming the effective depth of slab 430mm and computing the two way shear resistance at critical section (d/2) from face of column, We have the relation 2

2

Vc2 = 0.26[1700 -(350+d) ] =597246 N Two way shear resistance is calculated as Vc2= ksx tc[4(350+d)d] Where ks=1, tc=0.25 = 1.118 N/mm

2

2

=1x1.118[1400d+4d ] Vu2 < vc2 , 2

597246 < (1565.2d + 4.472d ). Solving, d > 230.18 mm Hence, one way shear is more critical, Adopt eff depth =d = 425mm, and overall depth = 450mm. 5.) Design of reinforcements. Ultimate moment at col face is computed as Mu=(0.5xbxs.p)

=(260x0.675x0.5)

=87.75 KNm/m 2

6

3

2

(Mu / bd ) = (87.75x10 /10 x425 ) =0.4858 Refer table 2 SP-16, and interpolate the percentage of reinforcement as p t =0.140 which is less than 0.25 percent assumed for one- way shear Therefore, Ast= (ptbd/100) =(0.25x1000x425/100) 2

=1062.5mm /m Using 20mm dia bars, Spacing of bars is S = (1000x314/1063) =295.39mm c/c 55

Adopt 20mm dia bars at 290mm centres in both directions

Design of footin g 2 (F2) 1. Data Axial load of column=p= 630KN Size of column =

350 x 350 mm

2

Use M20 & fe-415HYSD bars 2.) Size of footing; Load on column = 630KN Wt of footing & backfill at 10%= 63 Total load = 693KN Area of footing = (693/200) = 3.465m

2

Size of footing =L=B = √3.465 = 1.86m Adopt 1.9 by 1.9m square footing. Net soil pressure at ultimate loads with a load factor of 1.5 is given by Qu (630x1.5/1.9x1.9) = 261.77KN/m =

0.26 N/mm

2

2

3. One way shear. The critical section is at a d istance “d” from the column face . Factored shear force = V u1 = (0.26x1900)(1000-d) =498(1000-d) Assuming percentage of reinforcement in the footing pt = 0.25 percent for M20 grade concrete Now from IS-456 code permissible shear stress as =tc =0.36 N/mm One way shear resistance Vc1 =(tcx1900xd). = (684d) N Vu1 < Vc1 56

2

498(1000-d) < 684d . d > 421 mm 4. Two way shear, Assuming the effective depth of slab 430mm and computing the two way shear resistance at critical section (d/2) from face of column, We have the relation 2

2

Vc2 = 0.26[1900 -(350+d) ] =790079.46N Two way shear resistance is calculated as Vc2= ksx tc[4(350+d)d] Where ks=1, tc=0.25 = 1.118 N/mm 2 2

=1x1.118[1400d+4d ] Vu2 < vc2 , 2

790079.46 < (1565.2d + 4.472d ). Solving, d > 280.3 mm Hence, one way shear is more critical, Adopt eff depth =d = 425mm, and overall depth = 450mm. 5. Design of reinforcements. Ultimate moment at col face is computed as Mu=(0.5xbxs.p)

=(0.5x0.75x260)

=97.5 KNm/m 2

6

3

2

(Mu / bd ) = (97.5x10 /10 x425 ) =0.539 Refer table 2 SP-16, and interpolate the percentage of reinforcement as p t =0.155 which is less than 0.25 percent assumed for one- way shear Therefore, Ast= (ptbd/100) =(0.25x1000x425/100) =1062.5mm2/m Using 20mm dia bars, 57

Spacing of bars is S = (1000x314/1063) =295.39mm c/c Adopt 20mm dia bars at 290mm centres in both directions Desig n of fo oti ng 3 (F3) 1. Data Axial load of column=p=833KN Size of column =

350 x 350 mm

2

Use M20 & fe-415HYSD bars 2.) Size of footing; Load on column = 833KN Wt of footing & backfill at 10%= 83.3 Total load = 916.3KN Area of footing = (916.3/200) = 4.58m

2

Size of footing =L=B = √4.58 = 2.14m Adopt 2.2 by 2.2 m square footing. Net soil pressure at ultimate loads with a load factor of 1.5 is given by Qu (833x1.5/2.2x2.2) = 258.16KN/m

2

=

0.26 N/mm2

3. One way shear. The critical section is at a d istance “d” from the column face . Factored shear force = V u1 = (0.26x2200)(1000-d) =572(1000-d) Assuming percentage of reinforcement in the footing pt = 0.25 percent for M20 grade concrete Now from IS-456 code permissible shear stress as =tc =0.36 N/mm 2 One way shear resistance Vc1 =(tcx2200xd). = (792d) N 58

Vu1 < Vc1 572(1000-d) < 792 d . d > 420 mm 4. Two way shear, Assuming the effective depth of slab 430mm and computing the two way shear resistance at critical section (d/2) from face of column, We have the relation 2

2


2

2

=1x1.118[1400d+4d ] Vu2 < vc2 , 2

1104246< (1565.2d + 4.472d ). Solving, d >315.82 mm Hence, one way shear is more critical, Adopt eff depth =d = 425mm, and overall depth = 450mm. 5. Design of reinforcements. Ultimate moment at col face is computed as Mu=(0.5xbxs.p)

=(0.5x0.925x260)

=120.25 KNm/m 2

6

3

2


=1062.5mm /m 59

Using 20mm dia bars, Spacing of bars is S = (1000x314/1063) =295.39mm c/c Adopt 20mm dia bars at 290mm centres in both directions Design of footin g 4 (F4) 1. Data Axial load of column=p=1065KN Size of column =

350 x 350 mm

2

Use M20 & fe-415HYSD bars 2.) Size of footing; Load on column = 1065KN Wt of footing & backfill at 10%=106.5 Total load =1171. KN Area of footing = (1171.5/200) = 5.86m2 Size of footing =L=B = √5.86 = 2.42m Adopt 2.5 by 2.5m square footing. Net soil pressure at ultimate loads with a load factor of 1.5 is given by Qu (1065x1.5/2.5x2.5) = 255.6KN/m 2 =

0.26 N/mm2

3. One way shear. The critical section is at a d istance “d” from the column face . Factored shear force = V u1 = (0.26x2500)(1000-d) =650(1000-d) Assuming percentage of reinforcement in the footing pt = 0.25 percent for M20 grade concrete Now from IS-456 code permissible shear stress as =tc =0.36 N/mm One way shear resistance 60

2

Vc1 =(tcx2500xd). = (900d) N Vu1 < Vc1 650(1000-d) < 900d . d > 420 mm 4. Two way shear, Assuming the effective depth of slab 420mm and computing the two way shear resistance at critical section (d/2) from face of column, We have the relation 2

2


2

2

=1x1.118[1400d+4d ] Vu2 < vc2 , 1470846 < (1565.2d + 4.472d 2). Solving, d > 425 mm Hence, one way shear is more critical, Adopt eff depth =d = 425mm, and overall depth = 450mm. 5. Design of reinforcements. Ultimate moment at col face is computed as Mu=(0.5xbxs.p)

=(0.5x1.075x260)

=139.75 KNm/m 2

6

3

2

(Mu / bd ) = (140x10 /10 x425 ) =0.78 Refer table 2 SP-16, and interpolate the percentage of reinforcement as p t =0.224 which is less than 0.25 percent assumed for one- way shear Therefore, Ast= (ptbd/100) =(0.25x1000x425/100) 61

2

=1062.5mm /m Using 20mm dia bars, Spacing of bars is S = (1000x314/1063) =295.39mm c/c Adopt 20mm dia bars at 290mm centres in both directions Design of footin g 5 (F5) 1. Data Axial load of column=p= 315KN Size of column =

350 x 350 mm

2


2

Size of footing =L=B = √1.73 = 1.316m Adopt 1.3 by 1.3m square footing. Net soil pressure at ultimate loads with a load factor of 1.5 is given by Qu (315x1.5/1.3x1.3) = 279.58KN/m 2 =

0.28 N/mm

2

3. One way shear. The critical section is at a d istance “d” from the column face Factored shear force = V u1 = (0.28x1300)(1000-d) =364(1000-d) Assuming percentage of reinforcement in the footing pt = 0.25 percent for M20 grade concrete Now from IS-456 code permissible shear stress as =tc =0.36 N/mm 62

2

One way shear resistance Vc1 =(tcx1300xd). = (468d) N Vu1 < Vc1 364(1000-d) < 468d . d > 437.5 mm 4. Two way shear, Assuming the effective depth of slab 438mm and computing the two way shear resistance at critical section (d/2) from face of column, We have the relation 2

2

Vc2 = 0.28[1300 -(350+d) ] =299335.65 N Two way shear resistance is calculated as Vc2= ksx tc[4(350+d)d] Where ks=1, tc=0.25 = 1.118 N/mm

2

2

=1x1.118[1400d+4d ] Vu2 < vc2 , 299335.65 < (1565.2d + 4.472d 2). Solving, d > 138 mm Hence, one way shear is more critical, Adopt eff depth =d = 440mm, and overall depth = 465mm. 5. Design of reinforcements. Ultimate moment at col face is computed as Mu=(0.5xbxs.p)

=(0.5x0.475x280)

=66.5 KNm/m 2

6

3

2

(Mu / bd ) = (66.5x10 /10 x425 ) =0.368 Refer table 2 SP-16, and interpolate the percentage of reinforcement as p t =0.102 which is less than 0.25 percent assumed for one- way shear 63

Therefore, Ast= (ptbd/100) =(0.25x1000x425/100) 2

=1062.5mm /m Using 20mm dia bars, Spacing of bars is S = (1000x314/1063) =295.39mm c/c Adopt 20mm dia bars at 290mm centres in both directions. Design of fo otin g 6 (F6) 1. Data Axial load of column=p= 410KN Size of column =

350 x 350 mm2

Use M20 & fe-415HYSD bars 2.) Size of footing; Load on column = 410KN Wt of footing & backfill at 10%= 41 Total load = 451kN Area of footing = (451/200) = 2.25m

2


0.27 N/mm

2

2

3. One way shear. The critical section is at a d istance “d” from the column face . Factored shear force = V u1 = (0.26x1500)(1000-d) =390(1000-d) Assuming percentage of reinforcement in the footing pt = 0.25 percent for M20 grade concrete 64

Now from IS-456 code permissible shear stress as =tc =0.36 N/mm

2

One way shear resistance Vc1 =(tcx1500xd). = (540d) N Vu1 < Vc1 390(1000-d) < 540d . d > 295.45 mm 4. Two way shear, Assuming the effective depth of slab 300mm and computing the two way shear resistance at critical section (d/2) from face of column, We have the relation Vc2 = 0.27[1500 2-(350+d)2] =493425 N Two way shear resistance is calculated as Vc2= ksx tc[4(350+d)d] Where ks=1, tc=0.25 = 1.118 N/mm

2

=1x1.118[1400d+4d 2] Vu2 < vc2 , 2

493425 < (1565.2d + 4.472d ). Solving, d > 200.44 mm Hence, one way shear is more critical, Adopt eff depth =d = 200mm, and overall depth = 225mm. 5. Design of reinforcements. Ultimate moment at col face is computed as Mu=(0.5xbxs.p)

=(273x0.575x0.5)

=78.48 KNm/m (Mu / bd2) = (78.48x10 6/103x4252) =1.55 65

Refer table 2 SP-16, and interpolate the percentage of reinforcement as p t =0.477 which is greater than 0.25 percent assumed for one- way shear Therefore, Ast= (ptbd/100) =(0.477x1000x200/100) 2

=954 mm /m Using 16mm dia bars, Spacing of bars is S = (1000x154/954) =161.42mm c/c Adopt 16 mm dia bars at 160mm centres in both directions

Desig n of fo otin g 7 (F7) 1. Data Axial load of column=p= 842 KN Size of column =

350 x 350 mm

2

Use M20 & fe-415HYSD bars 2.) Size of footing; Load on column = 842 KN Wt of footing & backfill at 10%= 84.2 Total load = 926.2KN Area of footing = (926.2/200) = 4.631m

2


0.286 N/mm

2

2

3. One way shear. The critical section is at a d istance “d” from the column face . ( refer fig 7) 66

Factored shear force = V u1 = (0.286x2100)(1000-d) =600(1000-d) Assuming percentage of reinforcement in the footing pt = 0.25 percent for M20 grade concrete Now from IS-456 code permissible shear stress as =tc =0.36 N/mm

2

One way shear resistance Vc1 =(tcx2100xd). = (756d) N Vu1 < Vc1 600(1000-d) < 756d . d > 442 mm 4. Two way shear, Assuming the effective depth of slab 442mm and computing the two way shear resistance at critical section (d/2) from face of column, We have the relation Vc2 = 0.286[2100 2-(350+d)2] =1081862.5 N Two way shear resistance is calculated as Vc2= ksx tc[4(350+d)d] Where ks=1, tc=0.25 = 1.118 N/mm

2

=1x1.118[1400d+4d 2] Vu2 < vc2 , 2

1081862.5 < (1565.2d + 4.472d ). Solving, d > 347 mm Hence, one way shear is more critical, Adopt eff depth =d = 445mm, and overall depth = 470mm. 5. Design of reinforcements. 67

Ultimate moment at col face is computed as Mu=(0.5xbxs.p)

=(286x0.675x0.5)

=96.52 KNm/m 2

6

3

2


=1112.5mm /m Using 20mm dia bars, Spacing of bars is S = (1000x314/1113) =282.12mm c/c Adopt 20mm dia bars at 280mm centre’s in both directions Desig n of fo oti ng 8 (F8) 1. Data Axial load of colum,P = 576KN Size of column =

350 x 350 mm2


2

Size of footing =L=B = √3.168 = 1.779m Adopt 1.8 by 1.8m square footing. Net soil pressure at ultimate loads with a load factor of 1.5 is given by Qu (576x1.5/1.8x1.8) = 266.6KN/m

2 68

=

2

0.266N/mm

3. One way shear. The critical section is at a d istance “d” from the column face . Factored shear force = V u1 = (0.266x1800) (1000-d) =478.8(1000-d) Assuming percentage of reinforcement in the footing pt = 0.25 percent for M20 grade concrete Now from IS-456 code permissible shear stress as =tc =0.36 N/mm

2

One way shear resistance Vc1 =(tcx1800xd). = (648d) N Vu1 < Vc1 479(1000-d) < 648d . d > 425 mm 4.

Two way shear,

Assuming the effective depth of slab 430mm and computing the two way shear resistance at critical section (d/2) from face of column, We have the relation Vc2 = 0.266[1800 2-(350+d)2] =702073.75 N Two way shear resistance is calculated as Vc2= ksx tc[4(350+d)d] Where ks=1, tc=0.25 = 1.118 N/mm 2 =1x1.118[1400d+4d 2] Vu2 < vc2 , 2

702073.7 < (1565.2d + 4.472d ). Solving, d > 258.15mm Hence, one way shear is more critical, Adopt eff depth =d = 425mm, and overall depth = 450mm. 69

5. Design of reinforcements. Ultimate moment at col face is computed as Mu = (0.5xbxs.p) = (266x0.725x0.5) = 96.42 KNm/m 2

6

3

2

(Mu / bd ) = (96.42x10 /10 x425 ) =0.533 Refer table 2 SP-16, and interpolate the percentage of reinforcement as p t =0.152 which is less than 0.25 percent assumed for one- way shear 2

Therefore, A st= (ptbd/100) =(0.25x1000x425/100) =1062.5mm /m Adopt 20mm dia bars at 290mm centres in both directions.

Column

Total

Footing

Footing 2

Loading(KN)

size (m )

Depth o f footin g (mm)

A1, A4

496

F1

1.7x1.7

450

A2, A3

630

F2

1.9x1.9

425

B1,B4,C1,C4,D1,D4

833

F3

2.2x2.2

425

1034

F4

2.5x2.5

425

J1, J4

315

F5

1.3x1.3

465

J2, J3

375

F6

1.5x1.5

225

I2, I3

842

F7

2.1x2.1

470

I1, I4

576

F8

1.8x1.8

450

E1,E4,F1,F4,G1,G4 H1,H4 B2,B3,C2,C3,D2,D3 E2,E3,F2,F3,G2,G3

70

CHAPTER 8 Design of Stairs

71

Design of Stair Dimensions of Stair = 3.6 m x 6.1 m Height of the floor is 10ft = 3.04 m Live load on the stair = 2 kN/m

2

Since as per the architectural drawing stair has two flights therefore the height of each flight is 1.52 m Let the Risers provided = 10, therefore the height of each riser = 152mm Number of Treads 10 -1 = 9 Let the width of stairs be 1600 mm Let the treads of steps be 270 mm Design of flight AB: Let the bearing of flight be 150 mm Effective horizontal span = 3+1.6 + (0.15/2) = 4.675 m Let the thickness of waist be 220 mm Loads: D.L of 220 mm waist = 25 x 220 = 5500 N/m

2

Ceiling finish (12.5 mm) = 24 x 12.5 = 300 N/m

2 2

2

Corresponding load per sq meter on plan = √ (R +T )/T x 5500 = 2

2

= √ (152 +270 )/270 x 5500 = 6300 N/m

2

Hence the actual load per sq meter of the plan area will consist of waist and ceiling finish = 6300 N/m

2

D.L of steps (152/2 mm average) = 76 x 25 = 1900 N/m Top finish (12.5 mm) = 12.5 x 24 = 300 N/m

2

2

Live load = 2000 N/m2 Therefore total load = 10500 N/m2 = 10.5 kN/m 2 Maximum B.M per meter width of stairs, M = (10.5 x 4.675 2)/8 = 27.4kN/m 2 Ultimate bending moment per meter width of stairs, M u = 1.5 x 27.4 = 41.1 kN/m 2

2

3

2

0.138f ckbd = 0.138 x 20 x 1000 d = 41.1 x 10 , thus d = 122 mm Providing 10 mm diameter bars, effective cover = 20 mm 72

Overall depth required is 122 + 20 = 142 mm Provide an overall depth of 220 mm Therefore effective depth = d eff = 220 – 20 = 180 mm 2

6

2

Mu/bd = (41.1 x 10 )/ (1000 x 200 ) = 1.03

 (.∗.) /  } Percentage of steel, p = 50{ /  t

Ast = 0.47/100(1000 x 200) = 940 mm

= 0.47%

2

Spacing of 10 mm diameter bars = (79 x 1000)/ 940 = 84 mm c/c Provide 10 mm ϕ bars @ 84 mm c/c Distribution steel = 0.12/100(1000 x 200) = 240 mm

2

Spacing of 8 mm bars = (50 x 1000)/240 = 208 mm c/c Provide 8mm ϕ bars @ 208 mm c/c

73

Major Project Report

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