CD-18-Modified Final Project Report

1

MANUAL DESIGN AND ANALYSIS OF

MULTI-STORIED OFFICE BUILDING

A PROJECT REPORT

Submitted by

B.MAHENDER RAO NAYAK (09241A0120)

V.PRAVEEN REDDY (09241A0126)

K.RAJIV (09241A0130)

ABDUL SHABBEER (09241A0139)

VIVEK PATAK (10245A0112)

In fulfilment for the major project of B.TECH in Civil Engineering

GOKARAJU RANGARAJU INSTITUTION OF

ENGINEERING & TECHNOLOGY

HYDERABAD

2

DECLARATION

We here by declare that the project work entitled “MANUAL DESIGN AND

ANALYSIS OF MULTI-STORIED OFFICE BUILDING” is a record of an original

work done by us under the guidance of Mr.V. Mallikarjun Reddy, Associate Prof.,

Department of civil Engineering, GRIET. This project work is done in the fulfilment of the

requirements of the major project. This is a bonafide work carried out by us and the results

provided in this project report have not been copied from any source. The results provided in

this have not been submitted to any other University or Institute for the award of any degree

or diploma.

Date: 15-04-2013

Place: Hyderabad

Civil Engineering Department

GRIET, Hyderabad.

3

ACKNOWLEDGMENT

We express our in deep gratitude to our guide Mr. V.MALLIKARJUN REDDY,

Associate Professor, Department of Civil Engineering (GRIET), for his guidance and extreme

care taken by him in helping us to complete the project work successfully. We are grateful to

him for the extensive care rendered to us right from the initiation of the project to the final

editing of the manuscript. This project would not have been completed without his help who

has given enough exposure in the areas related to the work and as well as provided us with

his ever present help.

We express our sincere thanks to Dr. G.Venkata Ramana, Head of the Department

Civil Engineering, for his support and guidance for doing this project. We are also thankful

to all the teaching and non-teaching staff of Civil Engineering Department.

We are also thankful to each and everyone name by name who helped us directly or

indirectly in execution of this project.

4

GOKARAJU RANGARAJU

INSTITUTE OF ENGINEERING AND TECHNOLOGY , HYDERABAD

CERTIFICATECERTIFICATECERTIFICATECERTIFICATE

This is to certify that the dissertation entitled “MANUAL DESIGN AND

ANALYSIS OF MULTI-STORIED OFFICE BUILDING” is a project work done

under the Guidance of Mr. V. Mallikarjun Reddy (Assosiate professor, civil

engineering department, GRIET).

PROJECT BY TEAM OF

B.MAHENDER RAO NAYAK (09241A0120)

V.PRAVEEN REDDY (09241A0126)

K.RAJIV (09241A0130)

ABDUL SHABBEER (09241A0139)

VIVEK PATAK (10245A0112)

Prof. Dr. G. Venkata Ramana V. Mallikarjun Reddy

HOD, Civil Engineering Project Guide, Civil

Engineering

Signature of the External Examiner

5

ABSTRACT

Due to advancement of technology humans are creating software to make

things easier and time saving. As a result in the civil engineering point of view

the manual design of buildings has lost its importance. It is true that design

using a software is easy and time saving and mostly results are accurate.

On the other hand manual design is a cumbersome job and a time consuming

process, but for a beginner manual design helps to understand the basic

fundamentals that are involved in designing a building. Once a person gains

knowledge in manual design he will be knowing the elements involved in

designing and can easily understand the usage of software.

The main objective of the project is to use the knowledge that we have learnt

during our graduation and learn to deal with practical cases. We wish this

project will fulfill our purpose.

6

CONTENTS

PG.NO

CHAPTER-I

1.INTRODUCTION .......................................................................... 7

CHAPTER-II

2.DESIGN OF SLABS ...................................................................... 10

CHAPTER-III

3. Analysis of frames.............................................................. 45

CHAPTER-IV

4. DESIGN OF BEAMS...................................................................... 115

CHAPTER-V

5. DESIGN OF COLUMNS.................................................. 224

CHAPTER-VI

6. DESIGN OF FOOTINGS.................................................. 275

CHAPTER-VII

7. DESIGN OF STAIR CASE.............................................. 293

CHAPTER-VIII

8. CONCLUSION................................................................. 298

9. REFERENCE.................................................................... 299

7

1. INTRODUCTION

The population explosion and advent of industrial revolution led to the exodus of

people from villages to urban areas. This urbanisation led to a new problem – less space for

housing, work and more people. Because of the demand for land, the land costs got

skyrocketed. So, under the changed circumstances, the vertical growth of buildings i.e.

constructions of multi-storeyed buildings has become inevitable both for residential and as

well as office purposes.

For multi-storeyed buildings, the conventional load bearing structures become

uneconomical as they require larger sections to resist huge moments and loads. But in a

framed structure, the building frame consists of a network of beams and columns which are

built monolithically and rigidly with each other at their joints. Because of this rigidity at the

joints, there will be reduction in moments and also the structure tends to distribute the loads

more uniformly and eliminate the excessive effects of localised loads. Therefore in non-load

bearing framed structures, the moments and forces become less which in turn reduces the

sections of the members. As the walls don’t take any load, they are also of thinner

dimensions. So, the lighter structural components and walls reduce the self weight of the

whole structure which necessitates a cheaper foundation. Also, the lighter walls which can be

easily shifted provide flexibility in space utilisation. In addition to the above mentioned

advantages the framed structure is more effective in resisting wind loads and earth quake

loads.

Work done in this project:

A plot of 369.75 m2 has been selected for the construction of a multi-storeyed office

building. In the office building the functions will be different and it plays a major role

because of different loads acts on different slabs. The frame analysis requires the dimensions

of the members. For the analysis, 6 substitute frames taken in transverse direction and in

longitudinal direction the net moment acting is zero, and this is due to symmetrycity . For the

maximum mid span moment obtained from the analysis, a T-beam has been designed and for

the support moment ‘doubly reinforced’ section has been provided. Stair case has also been

designed. Isolated rectangular sloped footings have been designed to transfer the load to the

ground strata.

8

Analysis of structure:

Kani’s method and substitute frame method is generally used to analyse a multi-

storeyed frame. The substitute frame method requires less computations and easier to carry

out the analysis. Therefore, here substitute frame method has been employed to carry out the

frame analysis and method is discussed in the following paragraphs.

Theoretically, a load applied at any point of the structure cause reactions at all

sections of the frame, but a close study of this aspect has shown that the moments in any

beam or column are mainly due to the loads on spans very close to it. The effect of loads on

distant panels is small. To facilitate the determination of moments in any member of a frame,

it is usual to analyse only a small portion of the frame consisting of adjacent members only.

Such a small portions are termed ‘SUBSTITUTE FRAMES’.

The reactions worked out with the help of substitute frame may sometimes appear to

be lower by about 10% compared with the value obtained from exact analysis. But it may be

mentioned that for analysis all the panels are located at the same time. Such a combination of

loading is highly improbable in practice. Thus the results given by substitute frames are safe

for all practical purposes.

Design concept:

There are three design philosophies to design a reinforced concrete structures. They

are:

1. Working stress method,

2. Ultimate load method and

3. Limit state method.

In the ‘working stress’ method it is seen that the permissible stresses for concrete and

steel are not exceeded anywhere in the structure when it is subjected to the worst

combination of working loads. A linear variation of stress form zero at the neutral axis to

the maximum stress at the extreme fibre is assumed.

Practically, the stress strain curve for concrete is not linear as it was assumed in

working stress method. So, in ‘ultimate load’ design an idealised form of actual stress

strain diagram is used and the working loads are increased by multiplying them with the

load factors.

9

The basis for ‘limit state’ method is a structure with appropriate degrees of reliability

should be able to withstand safely all loads that are liable to act on it throughout its life

and it should also satisfy the serviceability requirements such as limitations on deflection

and cracking.

Limit state method is the most rational method of the three methods. It considers the

actual behaviour of the materials at failure and also it takes serviceability also into

consideration. Therefore, limit state method has been employed in this work.

10

2.DESIGN OF SLABS

Typically we divided the slabs into two types:

i. Roof Slab and

ii. Floor Slab

In case of roof slab the live load obtained is less compared to the floor slab. Therefore we

first design the roof slab and then floor slabs.

We have two types of supports. They are:

1. Ultimate support and

2. Penultimate support

Ultimate support is the end support and the penultimate supports are the intermediate

supports.

Ultimate support tends to have a bending moment of Wu x L

2

10 and the penultimate supports

have Wu x L

2

12

Design of roof slab:

It is a continuous slab on the top of the building which is also known as terrace.

Generally terrace has less live load and it is empty in most of the time except some occasions

in case of any residential building. In case of office buildings it will be empty and live load

act is very less.

According to the end conditions and the dimensions, the slabs are divided into 4 types. They

are Roof S1, Roof S2, Roof S3 and Roof S4.

Slab Dimensions (M x M)

Roof S1 8.62 X 3.05

Roof S2 8.62 X 3.05

Roof S3 5.78 X 3.05

Roof S4 5.78 X 3.05

11

We can observe the slab panels in the above figure and all the slabs are designed as one way

slab for the easy arrangement of the reinforcement and ease of work.

Roof S1 and Roof S2 are the slabs with same dimensions but with different end conditions.

Roof S3 and Roof S4 are also the slabs with the same conditions as mentioned above.

But the point to be noted is that all the Slabs have same shorter span and in the design of one

way slab shorter span is of more importance. Therefore we design any two slabs with

different end conditions and the remaining two slabs also follow the same design.

Design of Roof Slab S1:

Calculation of Depth (D) by using modification factor:

Assume the percentage of the tension reinforcement (Pt) provided is 0.4%

From IS456-2000, P38 Fig4, we get the modification factor (α) = 1.4

Required Depth (D) = L

ra + d

1

12

Where, L

ra =

Span

allowable L

d ratio

d1 = Centre of the reinforcement to the end fibre (= 20mm for slab)

From IS456-2000, P39, Clause 24 & Clause 23.2 for continuous span we have

Span

Effective depth =

L

d = 26

ra = 26 x 1.4 = 36.4

Therefore, D = 3.05 x 10

3

36.4 + 20 = 103.79mm say 110 mm

Effective depth (d) = D – d1 = 110 – 20 = 90mm

Loads:

Dead loads (From IS875 – Part 1):

Terrace water proofing = 2.5 KN/m2

Self weight of the slab = 1 x 1 x D x 25 = 110

1000 x 25

= 2.75 KN/m2

Live loads (From IS875 – Part 2):

Roof = 1.5 KN/m2

Total load (W) = 2.5 +2.75 + 1.5 = 6.75 KN/m2

Ultimate load or limit state load or design load (Wu) = 1.5 x W = 1.5 x 6.75

=10.125 KN/m2

Design moment: (for end panel)

Mu = Wu x L

2

10 =

10.125 x (3.05)2

10 = 9.42 KN-m

13

Calculation of area of steel:

From IS456-2000, P96, Clause G-1.1 (b) we have

Mu = 0.87 x fy x Ast x d x (1 - fy x Ast

fck x b x d )

Or

Ast = 0.5 x fck

fy (1-(1-4.6

Mu

fck x b x d )

1/2) b x d

9.42 x 106 = 0.87 x 415 x Ast x 90 x (1 -

415 x Ast

20 x 1000 x 90 )

Ast = 312.4 mm2

Spacing of 8mm φ bars = ast x 1000

Ast = π4

x 82 x

1000

312.4 = 160.9mm

Therefore, Provide 8mm φ @ 150mm c/c.

Distribution steel:

Ast = 0.12% of Ag

= 0.12

1000 x 110 x 1000 = 132 mm

2.

Spacing of 8mm φ bars = π4 x 8

2 x

1000

132 =380mm


14

Design of Roof slab S2:

Depth D = 110mm

Total load (W) = 6.75 KN/m2

Limit state load (Wu) = 1.5 x 6.75 = 10.125 KN/m2

Design moment: (for intermediate panel)

Mu = Wu x L

2

12 =

10.125 x (3.05)2

12 = 7.85 KN-m



fck x b x d )

7.85 x 106 = 0.87 x 415 x Ast x 90 x (1 -

415 x Ast

20 x 1000 x 90 )

Ast = 256.78 mm2


2 x

1000

256.78 = 195.75mm


Distribution steel:

Ast = 0.12% of Ag

= 0.12

1000 x 110 x 1000 = 132 mm2.


2 x

1000

132 =380mm


15

Design of Roof Slab S3 is same as Roof Slab S1.

Design of Roof Slab S4 is same as Roof Slab S2.

Area of steel at support next to end support:

From IS 456-2000, moment = Wu x L

2

10

Total Load acting on the support (Wu) = 10.125 KN/m

Therefore, Moment = 10.125 x (3.05)

2

10 = 9.42 KN-m

Calculation of Ast:


fck x b x d )

9.42 x 106 = 0.87 x 415 x Ast x 90 x (1 - 415 x Ast

20 x 1000 x 90 )

Ast = 312.4 mm2

Area of steel available by bending up the alternate bars of mid span steel:

From Slab S1 = 1

2 x 312.4 = 160.7 mm2

From Slab S2 = 1

2 x 256.78 = 128.39 mm

2

Total Ast (available) = 160.7 + 128.39 = 284.59 mm2

Therefore, extra bars required for Ast = 312.4 – 284.59 = 27.81 mm2

16

Area of steel at any other interior support:


2

12



2

12 = 7.85 KN-m

Calculation of Ast:


fck x b x d )

7.85 x 106 = 0.87 x 415 x Ast x 90 x (1 - 415 x Ast

20 x 1000 x 90 )

Ast = 256.78 mm2


From Slab S2 = 1

2 x 256.78 = 128.39 mm2

From Slab S2 = 1

2 x 256.78 = 128.39 mm2


Therefore Ast (available) = Ast (required)

No need of providing extra bars.

17

Design of Floor Slab:

It is the slab in which live load is more when compared to the roof slab. In this project

the slab is divided into 9 types according to the end condition and function of slab.

S1 - Toilet and WC’s

S2 - Office

S3 - Office sup dept.

S4 - Assembly hall

S5 (a), S5 (b) - Office chamber and waiting chamber

S6 (a), S6 (b) - Office

S7 - Library

S8 - Secretary Room

S9 (a), S9 (b) - Officers chamber

18

Floor Slab Dimensions

S1 to S5 (b) 8.62 x 3.05

S6 (a) to S9 (b) 5.78 x 3.05

DESIGN OF FLOOR SLAB (S1):

Calculation of Depth (D) by using modification factor:

Assume the percentage of the tension reinforcement (Pt) provided is 0.4%

From IS456-2000, P38 Fig4, we get the modification factor (α) = 1.4

Required Depth (D) = L

ra + d

1

Where, L

ra =

Span

allowable L

d ratio

d1 = Centre of the reinforcement to the end fibre (= 20mm for slab)

From IS456-2000, P39, Clause 24 & Clause 23.2 for continuous span we have

Span

Effective depth =

L

d = 26

ra = 26 x 1.4 = 36.4

Therefore, D = 3.05 x 10

3

36.4 + 20 = 103.79mm say 120 mm


Loads:


Floor finish = 1 KN/m2

Sanitary Blocks including filling = 2.5 KN/m2

19


1000 x 25

= 3 KN/m2


Sanitary blocks public = 3 KN/m2

Corridor = 5 KN/m2

Maximum = 5 KN/m2

For Partition Wall = 1.5 KN/m2

Total load (W) = 1 + 2.5 +3 + 5 + 1.5 = 13 KN/m2

Ultimate load or limit state load or design load (Wu) = 1.5 x W = 1.5 x 13

=19.5 KN/m2


Mu = Wu x L2

10 =

19.5 x (3.05)2

10 = 18.14 KN-m




fck x b x d )

Or

Ast = 0.5 x fck

fy (1-(1-4.6

Mu

fck x b x d )1/2) b x d

18.14 x 106 = 0.87 x 415 x Ast x 100 x (1 -

415 x Ast

20 x 1000 x 100 )

Ast = 569.79 mm2


Ast = π4 x 10

2 x

1000

569.79 = 137.83mm

20


Distribution steel:

Ast = 0.12% of Ag

= 0.12

1000 x 120 x 1000 = 144 mm2.


2 x

1000

144 =349mm



D = 3.05 x 10

3

36.4 + 20 = 103.79mm say 120 mm


Loads:




1000 x 25

= 3 KN/m2


Office = 4 KN/m2

Corridor = 5KN/m2

21

Therefore, Maximum load = 5 KN/m2

For Partition wall = 1.5 KN/m2

Total load (W) = 5 +3 + 1 + 1.5 = 10.5 KN/m2


=15.75 KN/m2

Design moment:

Mu = Wu x L

2

12 =

15.75 x (3.05)2

12 = 12.21 KN-m



fck x b x d )

12.21 x 106 = 0.87 x 415 x Ast x 100 x (1 - 415 x Ast

20 x 1000 x 100 )

Ast = 365.97 mm2

Spacing of 10mm φ bars = π4

x 102 x 1000

365.97 = 214.61mm


Distribution steel:

Ast = 0.12% of Ag

= 0.12

1000 x 120 x 1000 = 144 mm

2.


2 x

1000

144 =349mm


22

Area of steel at support next to end support (between S1 and S2):


2

10

Total Load acting on the support (Wu) = 13

2 +

15.75

2 = 14.375 KN/m

Therefore, Moment = 14.375 x (3.05)2

10 = 13.372 KN-m

Calculation of Ast:


fck x b x d )

13.372 x 106 = 0.87 x 415 x Ast x 100 x (1 -

415 x Ast

20 x 1000 x 100 )

Ast = 404.28 mm2


From Slab S1 = 1

2 x 569.79 = 284.895 mm

2

From Slab S2 = 1

2 x 365.97 = 182.985 mm

2




D = 3.05 x 103

36.4 + 20 = 103.79mm say 120 mm

23


Loads:




1000 x 25

= 3 KN/m2


Private = 2 KN/m2

Corridor = 5 KN/m2

Maximum load = 5 KN/m2


Total load (W) = 5 +3 + 1 + 1.5 = 10.5 KN/m2


=15.75 KN/m2

Design moment:

Mu = Wu x L

2

12 =

15.75 x (3.05)2

12 = 12.21 KN-m



fck x b x d )

12.21 x 106 = 0.87 x 415 x Ast x 100 x (1 - 415 x Ast

20 x 1000 x 100 )

Ast = 365.97 mm2

24


x 102 x 1000

365.97 = 214.61mm


Distribution steel:

Ast = 0.12% of Ag

= 0.12

1000 x 120 x 1000 = 144 mm

2.


2 x

1000

144 =349mm


Area of steel at any other interior support: (Between S2 and S3)

From IS 456-2000, moment = Wu x L2

12

Total Load acting on the support (Wu) = 15.75

2 +

15.75

2 = 15.75 KN/m


12 = 12.21 KN-m

Calculation of Ast:


fck x b x d )

12.21 x 106 = 0.87 x 415 x Ast x 100 x (1 -

415 x Ast

20 x 1000 x 100 )

Ast = 365.97 mm2

25


From Slab S2 = 1

2 x 365.97 = 182.985 mm

2

From Slab S3 = 1

2 x 365.97 = 182.985 mm2





D = 3.05 x 103

36.4 + 20 = 103.79mm say 120 mm


Loads:




1000 x 25

= 3 KN/m2


Assembly = 5 KN/m2

Corridor = 5 KN/m2

Maximum load = 5 KN/m2


Total load (W) = 5 +3 + 1 + 1.5 = 10.5 KN/m2

26


=15.75 KN/m2

Design moment:

Mu = Wu x L2

12 =

15.75 x (3.05)2

12 = 12.21 KN-m



fck x b x d )

12.21 x 106 = 0.87 x 415 x Ast x 100 x (1 -

415 x Ast

20 x 1000 x 100 )

Ast = 365.97 mm2


x 102 x

1000

365.97 = 214.61mm


Distribution steel:

Ast = 0.12% of Ag

= 0.12

1000 x 120 x 1000 = 144 mm2.


2 x

1000

144 =349mm


27

Area of steel at any other interior support: (Between S3 and S4)

Same as between S2 and S3

DESIGN OF FLOOR SLAB (S5 (a)):

D = 3.05 x 103

36.4 + 20 = 103.79mm say 120 mm


Loads:




1000 x 25

= 3 KN/m2


Office chamber = 4 KN/m2

Private = 2 KN/m2


Total load (W) = 2 + 4 +3 + 1 + 1.5 = 11.5 KN/m2


=17.25 KN/m2

Design moment:

Mu = Wu x L

2

12 =

17.25 x (3.05)2

12 = 13.372 KN-m



fck x b x d )

28

13.372 x 106 = 0.87 x 415 x Ast x 100 x (1 - 415 x Ast

20 x 1000 x 100 )

Ast = 404.27 mm2


x 102 x 1000

347.05 = 194.27mm


Distribution steel:

Ast = 0.12% of Ag

= 0.12

1000 x 120 x 1000 = 144 mm

2.


2 x

1000

144 =349mm


Area of steel at any other interior support: (Between S4 and S5 (a))


2

12


2 +

17.25

2 = 16.5 KN/m


12 = 12.79 KN-m

Calculation of Ast:


fck x b x d )

29

12.79 x 106 = 0.87 x 415 x Ast x 100 x (1 - 415 x Ast

20 x 1000 x 100 )

Ast = 385 mm2


From Slab S4 = 1

2 x 365.97 = 182.985 mm

2

From Slab S5 (a) = 1

2 x 404.27 = 202.135 mm2




DESIGN OF FLOOR SLAB (S5 (b)):

D = 3.05 x 103

36.4 + 20 = 103.79mm say 120 mm


Loads:




1000 x 25

= 3 KN/m2



Private = 2 KN/m2


30

Total load (W) = 2 + 4 +3 + 1 + 1.5 = 11.5 KN/m2


=17.25 KN/m2


Mu = Wu x L

2

10 =

17.25 x (3.05)2

10 = 16.047 KN-m



fck x b x d )

16.047 x 106 = 0.87 x 415 x Ast x 100 x (1 - 415 x Ast

20 x 1000 x 100 )

Ast = 495.37 mm2


x 102 x 1000

495.37 = 158.55mm


Distribution steel:

Ast = 0.12% of Ag

= 0.12

1000 x 120 x 1000 = 144 mm

2.

Spacing of 8mm φ bars = π4 x 82 x

1000

144 =349mm


31

Area of steel at support next to end support (between S5 (a) and S5 (b)):


2

10


2 +

17.25

2 = 17.25 KN/m


2

10 = 16.05 KN-m

Calculation of Ast:


fck x b x d )

16.05 x 106 = 0.87 x 415 x Ast x 100 x (1 -

415 x Ast

20 x 1000 x 100 )

Ast = 495.47 mm2



2 x 404.27 = 202.135 mm

2

From Slab S5 (b) = 1

2 x 495.47 = 247.735 mm2




D = 3.05 x 10

3

36.4 + 20 = 103.79mm say 120 mm


Loads:



32


1000 x 25

= 3 KN/m2


Office = 4 KN/m2

Total load (W) = 4 +3 + 1 = 8 KN/m2


=12 KN/m2


Mu = Wu x L2

10 =

12 x (3.05)2

10 = 11.163 KN-m



fck x b x d )

11.163 x 106 = 0.87 x 415 x Ast x 100 x (1 -

415 x Ast

20 x 1000 x 100 )

Ast = 332.06 mm2


x 102 x

1000

332.06 = 236.53mm


Distribution steel:

Ast = 0.12% of Ag

= 0.12

1000 x 120 x 1000 = 144 mm2.


2 x

1000

144 =349mm

33



D = 3.05 x 103

36.4 + 20 = 103.79mm say 120 mm


Loads:




1000 x 25

= 3 KN/m2


Office = 4 KN/m2

Total load (W) = 4 +3 + 1 = 8 KN/m2


=12 KN/m2


Mu = Wu x L

2

12 =

12 x (3.05)2

12 = 9.3025 KN-m

34



fck x b x d )

9.3025 x 106 = 0.87 x 415 x Ast x 100 x (1 - 415 x Ast

20 x 1000 x 100 )

Ast = 273.13 mm2


x 102 x 1000

273.13 = 287.55mm


Distribution steel:

Ast = 0.12% of Ag

= 0.12

1000 x 120 x 1000 = 144 mm

2.


1000

144 =349mm




2

10

Total Load acting on the support (Wu) = 12 KN/m

Therefore, Moment = 12 x (3.05)

2

10 = 11.163 KN-m

35

Calculation of Ast:


fck x b x d )

11.163 x 106 = 0.87 x 415 x Ast x 100 x (1 - 415 x Ast

20 x 1000 x 100 )

Ast = 332.06 mm2



2 x 332.06 = 166.03 mm

2


2 x 273.23 = 136.565 mm2




D = 3.05 x 10

3

36.4 + 20 = 103.79mm say 120 mm


Loads:




1000 x 25

= 3 KN/m2


Library = 10 KN/m2

Total load (W) = 10 +3 + 1 = 14 KN/m2

36


=21 KN/m2


Mu = Wu x L2

10 =

12 x (3.05)2

10 = 19.54 KN-m



fck x b x d )

19.54 x 106 = 0.87 x 415 x Ast x 100 x (1 -

415 x Ast

20 x 1000 x 100 )

Ast = 621.3 mm2


x 102 x

1000

621.3 = 126.41mm


Distribution steel:

Ast = 0.12% of Ag

= 0.12

1000 x 120 x 1000 = 144 mm2.


2 x

1000

144 =349mm


37


D = 3.05 x 10

3

36.4 + 20 = 103.79mm say 120 mm


Loads:




1000 x 25

= 3 KN/m2


Private = 2 KN/m2

Total load (W) = 2 +3 + 1 = 6 KN/m2


= 9 KN/m2

Design moment:

Mu = Wu x L2

12 =

9 x (3.05)2

12 = 6.97 KN-m



fck x b x d )

6.97 x 106 = 0.87 x 415 x Ast x 100 x (1 -

415 x Ast

20 x 1000 x 100 )

Ast = 201.47 mm2


x 102 x

1000

201.47 = 389.8mm

38


Distribution steel:

Ast = 0.12% of Ag

= 0.12

1000 x 120 x 1000 = 144 mm2.


2 x

1000

144 =349mm


Area of steel at support next to end support (between S7and S8):

From IS 456-2000, moment = Wu x L2

10


2 +

9

2 = 15 KN/m

Therefore, Moment = 15 x (3.05)

2

10 = 13.95 KN-m

Calculation of Ast:


fck x b x d )

13.95 x 106 = 0.87 x 415 x Ast x 100 x (1 - 415 x Ast

20 x 1000 x 100 )

Ast = 423.61 mm2


39

From Slab S7 = 1

2 x 621.3 = 310.65 mm2

From Slab S8 = 1

2 x 201.47 = 100.735 mm

2




D = 3.05 x 10

3

36.4 + 20 = 103.79mm say 120 mm


Loads:




1000 x 25

= 3 KN/m2



Total load (W) = 3 +3 + 1 = 7 KN/m2


= 10.5 KN/m2

Design moment:

Mu = Wu x L

2

12 =

10.5 x (3.05)2

12 = 8.14 KN-m

40



fck x b x d )

8.14 x 106 = 0.87 x 415 x Ast x 100 x (1 - 415 x Ast

20 x 1000 x 100 )

Ast = 237.12 mm2


x 102 x 1000

237.12 = 331.23mm


Distribution steel:

Ast = 0.12% of Ag

= 0.12

1000 x 120 x 1000 = 144 mm

2.


1000

144 =349mm


Area of steel at any other interior support: (Between S8 and S9 (a))


2

12


2 +

10.5

2 = 9.75 KN/m


12 = 7.56 KN-m

41

Calculation of Ast:


fck x b x d )

7.56 x 106 = 0.87 x 415 x Ast x 100 x (1 - 415 x Ast

20 x 1000 x 100 )

Ast = 219.38 mm2


From Slab S8 = 1

2 x 201.47 = 100.735 mm

2


2 x 237.12 = 118.56 mm2





D = 3.05 x 10

3

36.4 + 20 = 103.79mm say 120 mm


Loads:




1000 x 25

= 3 KN/m2



42

Total load (W) = 3 +3 + 1 = 7 KN/m2


= 10.5 KN/m2


Mu = Wu x L

2

10 =

10.5 x (3.05)2

10 = 9.77 KN-m



fck x b x d )

9.77 x 106 = 0.87 x 415 x Ast x 100 x (1 - 415 x Ast

20 x 1000 x 100 )

Ast = 287.79 mm2


x 102 x 1000

287.79 = 272.91mm


Distribution steel:

Ast = 0.12% of Ag

= 0.12

1000 x 120 x 1000 = 144 mm

2.


1000

144 =349mm


43



2

10



2

10 = 9.77 KN-m

Calculation of Ast:


fck x b x d )

9.77 x 106 = 0.87 x 415 x Ast x 100 x (1 - 415 x Ast

20 x 1000 x 100 )

Ast = 287.79 mm2



2 x 237.12 = 118.56 mm2


2 x 287.79 = 143.895 mm2



Detail of reinforcement provided in slabs:

Slab Function Main steel Distribution steel End shears

Long

span

(Wu x L

2 )

KN/m

Short

span

(Wu x L

6 )

KN/m

Roof S1 and

Roof S3

Terrace 8mmφ bars @

150mm c/c

8mmφ bars @

300mm c/c

15.441

5.1467

44

Roof S2 and

Roof S4

Terrace 8mmφ bars @

190mm c/c

8mmφ bars @

300mm c/c

15.441

5.1467

Floor S1 Toilet 10mmφ bars @

130mm c/c

8mmφ bars @

300mm c/c

29.738 9.913

Floor S2 Office 10mmφ bars @

210mm c/c

8mmφ bars @

300mm c/c

24.019 8

Floor S3 Office SupDt. 10mmφ bars @

210mm c/c

8mmφ bars @

300mm c/c

24.019 8

Floor S4 Assembly Hall 10mmφ bars @

210mm c/c

8mmφ bars @

300mm c/c

24.019 8

Floor S5 (a) Office chamber

and Waiting space

10mmφ bars @

190mm c/c

8mmφ bars @

300mm c/c

26.306

8.77

Floor S5 (b) Office chamber

and Waiting space

10mmφ bars @

150mm c/c

8mmφ bars @

300mm c/c

26.306

8.77

Floor S6 (a) Office 10mmφ bars @

230mm c/c

8mmφ bars @

300mm c/c

18.3 6.1

Floor S6 (b) Office 10mmφ bars @

280mm c/c

8mmφ bars @

300mm c/c

18.3 6.1

Floor S7 Library 10mmφ bars @

120mm c/c

8mmφ bars @

300mm c/c

32.025 10.675

Floor S8 Secretary room 10mmφ bars @

300mm c/c

8mmφ bars @

300mm c/c

13.725 4.575

Floor S9 (a) Office chamber 10mmφ bars @

300mm c/c

8mmφ bars @

300mm c/c

16.01 5.3375

Floor S9 (b) Office chamber 10mmφ bars @

270mm c/c

8mmφ bars @

300mm c/c

16.01 5.3375

45

3. Analysis of frames

We have many number frames from the plan and the need to be analysed. We have two

different types of frames:

1. Longitudinal direction frame

2. Transverse direction frame

Transverse frame:

The frames are chosen in such a way that the loads vary from one frame to the other and we

have 6 transverse frames.

i. Frame 19-10-01

ii. Frame 20-11-02

iii. Frame 24-15-06

iv. Frame 25-16-07

v. Frame 26-17-08

vi. Frame 27-18-09

In every frame we need to analyse the three types of loading cases and each frame consists of

roof and floor analysis.

Here we assumed the cross sections of beams and columns in advance and with the help of

the assumed dimensions, we calculate the stiffness of the members and there by the

distribution factors for the members especially at the joints.

To analyse the frame we use the substitute frame method and there by applying the moment

distribution method to know the moments carried by the member at the joints.

We take each floor span and we assume the top and bottom storeys are fixed by the substitute

frame principle.

46

Typical building frame:

We have G+5frame with all the details shown in the above figure.

47

Type of

member

b(mm) x

D(mm)

Length

(mm)

M.O.I

(mm4) =

b x D3

12

Multiplying

factor for

flanged

section

Final

M.O.I

(mm4)

K =

final M.O.I

LENGTH

(mm3)

Beam 230 x

300

6000 517.5 x

106

2 1035 x 106 172.5 x 103

230 x

300

8410 517.5 x

106

2 1035 x 106 123.07 x 10

3

Beam 230 x

450

6000 1746.56 x

106

2 3493.125

x 106

582.18 x 103

230 x

450

8410 1746.56 x

106

2 3493.125

x 106

415.35 x 103

Beam 230 x

600

6000 4140 x

106

2 8280 x 106 1.38 x 106

230 x

600

8410 4140 x

106

2 8280 x 106 0.985 x 10

6

Column 230 x

600

3350 4140 x

106

- 4140 x 106 1235.82 x

103

Frame 19-10-01:

Calculation of loads:

Roof:

Slab: 110mm

Dead load = self weight + F.F

= (0.11 x 25) + 2.5 = 5.25 KN/m2

Live load = 1.5 KN/m2

Beam (19-10) (230mm X 300mm)

Dead load:

Self weight = 0.23 x 25 x (0.3-0.11) = 1.1 KN/m

48

Due to Slab = W x L

2 =

5.25 x 3.05

2 = 8 KN/m

250mm thick wall of 1m height = 0.25 x 1 x 20 = 5 KN/m

Live load:

Due to Slab = W x L

2 =

1.5 x 3.05

2 = 2.29KN/m

Therefore, Maximum load = 1.5 x (D.L + L.L) = 1.5 x (1.1 + 8 + 2.29 + 5)

= 24.6 KN/m

Minimum load = 0.9 x D.L = 0.9 x (1.1 + 8 + 5) = 13 KN/m

Beam (10-01) (230mm X 300mm)

Dead load:


Due to Slab = W x L

2 =

5.25 x 3.05

2 = 8 KN/m

250mm thick wall of 1m height = 0.25 x 1 x 20 = 5 KN/m

Live load:

Due to Slab = W x L

2 =

1.5 x 3.05

2 = 2.29KN/m

Therefore, Maximum load = 1.5 x (D.L + L.L) = 1.5 x (1.1 + 8 + 2.29 + 5)

= 24.6 KN/m

Minimum load = 0.9 x D.L = 0.9 x (1.1 + 8 + 5) = 13 KN/m

Case 1:

Maximum load on beam 19-10 and minimum load beam 10-01

49

Calculation of distribution factors:

Joint Member K ΣK D.F

1 Column 1235.82 x 103 1.41 x 10

6 0.88

Beam 172.5 x 103 0.12

2 Beam 172.5 x 103 1.53 x 106 0.11

Column 1235.82 x 103 0.81

Beam 123.07 x 103 0.08

3 Beam 123.07 x 103 1.35 x 106 0.09

column 1235.82 x 103 0.91

Calculation of

moments:

50

Calculation of support reactions:

From beam 1-2, ΣM1=0

64.85 + V2 x 6 = 24.6 x 6 x 6

2 + 78.52

V2 = 76.07 KN

V1 + V2 = 24.6 x 6 = 147.6 KN

V1 = 71.54 KN

Maximum Span moment,

X= 71.54

24.6 = 2.91m from the left support

Therefore, maximum moment = 71.54 x 2.91

2 - 64.85 = 39.24 KN-m


79.93 + V3 x 8.41 = 69.78 + 13 x 8.41 x 8.41

2

V3 = 53.46 KN

V2 + V3 = 13 x 8.41 =109.33 KN

V2 = 55.87 KN

51


X= 53.46

13 = 4.12m from the right support


2 - 69.78 = 40.35 KN-m

Case 2:

Minimum load on beam 19-10 and maximum load on 10-01

Distribution factors are same as calculate in case 1.

Calculation of moments:

52



28.96 + V2 x 6 = 13 x 6 x 6

2 + 53.16

V2 = 42.87 KN

V1 + V2 = 13 x 6 = 78 KN

V1 = 35.13 KN


X= 35.13

13 = 2.7m from the left support


2 - 28.16 = 19.27 KN-m


142.85 + V3 x 8.41 = 136.01 + 24.6 x 8.41 x 8.41

2

V3 = 102.63 KN

V2 + V3 = 24.6 x 8.41 = 206.89 KN

V2 = 104.26 KN


X= 102.63

24.6 = 4.17m from the right support

53


2 - 136.01 = 77.97 KN-m

Case 3:

Max load on beam 19-10 and beam 10-01

Distribution factors are same as calculated in case 1.

Calculations of moments:

54



61.38 + V2 x 6 = 24.6 x 6 x 6

2 + 86.09

V2 = 77.92 KN

V1 + V2 = 24.6 x 6 = 147.6 KN

V1 = 69.68 KN


X= 69.68



2 - 61.38 = 37.22 KN-m


145.76 + V3 x 8.41 = 134.65 + 24.6 x 8.41 x 8.41

2

V3 = 102.05 KN

V2 + V3 = 24.6 x 8.41 = 206.89 KN

V2 = 104.84 KN


X= 102.05



2 - 134.65 = 77.1 KN-m

55

Beam and column bending moments (KN-m):

Joint 1 Joint 2 Joint 3

Case Column Beam Max

span

moment

Beam Column Beam Max

span

moment

Beam Column

1 64.85 64.65 39.24 78.43 1.5 78.93 40.35 69.78 69.86

2 28.96 28.96 29.27 53.16 89.7 142.86 77.97 136.01 135.67

3 61.38 61.38 37.22 86.09 59.67 145.76 77.1 134.64 134.48

Max 64.85 64.65 39.24 86.09 89.7 145.76 77.97 136.01 135.67

Shear or support reaction (KN):

Case no. Joint 4 Joint 5 Joint 6

1 71.54 76.07 55.87 53.46

2 35.13 42.87 104.26 102.63

3 69.68 77.92 104.84 102.05

Maximum shear 71.54 77.92 104.84 102.63

Maximum column load 71.54 182.76 102.63

Floor:

Slab: 120mm


= (0.12 x 25) + 1 = 4 KN/m2

Live load = 4 KN/m2 for Beam 19-10

= 5 KN/m2

for Beam 10-01

Beam (19-10) (230mm X 450mm)

Dead load:


56

Due to Slab = W x L

2 =

4 x 3.05

2 = 6.1 KN/m

250mm thick wall = 0.25 x 20 x (3.35-0.3) = 15.25 KN/m

Live load:

Due to Slab = W x L

2 =

4 x 3.05

2 = 6.1KN/m

Therefore, Maximum load = 1.5 x (D.L + L.L) = 1.5 x (1.9+6.1+15.25+6.1)

= 44.025 KN/m

Minimum load = 0.9 x D.L = 0.9 x (1.9+6.1+15.25) = 21 KN/m

Beam (10-01) (230mm X 450mm)

Dead load:


Due to Slab = W x L

2 =

4 x 3.05

2 = 6.1 KN/m


Live load:

Due to Slab = W x L

2 =

5 x 3.05

2 = 7.625KN/m


= 46.5 KN/m

Minimum load = 0.9 x D.L = 0.9 x (1.9+6.1+15.25) = 21KN/m

57

Case 1:

Maximum load on beam 19-10 and minimum load on beam 10-01



4 Column 1235.82 x 103 3.05 x 10

6 0.4

Beam 582.18 x 103 0.2

Column 1235.82 x 103 0.4

5 Beam 582.18 x 103 3.47 x 10

6 0.17

Column 1235.82 x 103 0.36

Beam 415.35 x 103 0.11

Column 1235.82 x 103 0.36

6 Beam 415.35 x 103 2.89 x 10

6 0.14

Column 1235.82 x 103 0.43

Column 1235.82 x 103 0.43

58




106.54 + V5 x 6 = 44.025 x 6 x 6

2 + 143.19

V5 = 138.18 KN

V4 + V5 = 44.025 x 6 = 264.15 KN

V4 = 125.97 KN

59


X= 125.97



2 - 106.54 = 73.58 KN-m


133.85 + V6 x 8.41 = 105.76 + 21 x 8.41 x 8.41

2

V6 = 84.96 KN

V5 + V6 = 21 x 8.41 = 176.61 KN

V5 = 91.65 KN


X= 84.96



2 - 105.76 = 66.28 KN-m

Case 2:

Minimum load on beam 19-10 and maximum load on beam 10-01

Distribution factors are same as calculated in case-1.

60




34.99 + V5 x 6 = 21 x 6 x 6

2 + 105.91

V5 = 74.81 KN

V4 + V5 = 21 x 6 = 126 KN

V4 = 51.19 KN

61


X= 51.19



2 - 34.99 = 27.72 KN-m


269.10 + V6 x 8.41 = 246.62 + 46.5 x 8.41 x 8.41

2

V6 = 192.86 KN

V5 + V6 = 46.5 x 8.41 = 391.07 KN

V5 = 198.21 KN


X= 192.86



2 - 246.62= 153.57 KN-m

Case 3:

Maximum load on both beams


62




96.02 + V5 x 6 = 44.025 x 6 x 6

2 + 169.53

V5 = 144.33 KN

V4 + V5 = 44.025 x 6 = 264.15 KN

V4 = 119.82 KN


X= 119.82


63


2 - 96.02 = 66.94 KN-m


277.32 + V6 x 8.41 = 242.87 + 46.5 x 8.41 x 8.41

2

V6 = 191.43 KN

V5 + V6 = 46.5 x 8.41 = 391.07 KN

V5 = 199.64 KN


X= 191.43



2 - 242.87= 151.48 KN-m




span

moment


span

moment

Beam Column

1 53.27 106.54 73.58 143.19 4.67 133.84 65.86 105.75 53.13

2 17.5 34.99 27.2 105.91 81.6 269.1 153.57 246.62 122.6

3 48.01 96.02 66.94 169.53 53.9 277.32 151.48 242.87 121.11

Max 53.27 106.54 73.58 169.53 81.6 277.32 153.57 246.62 122.6



1 125.6 138.18 91.65 84.96

2 51.19 74.81 198.21 192.85

3 119.82 144.33 199.64 191.43

Maximum shear 125.6 144.33 199.64 192.85


64

Frame 20-11-02:

Calculation of loads:

Roof:

Slab: 110mm


= (0.11 x 25) + 2.5 = 5.25 KN/m2


Beam (20-11) (230mm X 450mm)

Dead load:


Due to Slab = W x L= 5.25 x 3.05= 16.0125 KN/m

Live load:


Therefore, Maximum load = 1.5 x (D.L + L.L) = 1.5 x (1.2+16.0125+4.6)

= 33 KN/m

Minimum load = 0.9 x D.L = 0.9 x (1.2+16.0125) = 15.49 KN/m

Beam (11-02) (230mm X 450mm)

Dead load:



Live load:


65


= 33 KN/m


Case 1:

Maximum load on beam 20-11 and minimum load beam 11-02



1 Column 1235.82 x 103 1.818 x 10

6 0.68

Beam 582.18 x 103 0.32

2 Beam 582.18 x 103 2.234 x 106 0.26

Column 1235.82 x 103 0.55

Beam 415.35 x 103 0.19

3 Beam 415.35 x 103 1.651 x 106 0.25

column 1235.82 x 103 0.75

66




68.42 + V2 x 6 = 33 x 6 x 6

2 + 111.87

V2 = 106.24 KN

V1 + V2 = 33 x 6 = 198 KN

V1 = 91.76 KN

67


X= 91.76



2 - 68.42 = 59.13 KN-m


105.03 + V3 x 8.41 = 67.38 + 15.49 x 8.41 x 8.41

2

V3 = 60.66 KN

V2 + V3 = 15.49 x 8.41 =130.27 KN

V2 = 69.61 KN


X= 60.66



2 - 67.38 = 51.51 KN-m

Case 2:

Minimum load on beam 20-11 and maximum load on 11-02

Distribution factors are same as calculate in case 1.

68




16.56 + V2 x 6 = 15.49 x 6 x 6

2 + 94.59

V2 = 59.48 KN

V1 + V2 = 15.49 x 6 = 92.94 KN

V1 = 33.46 KN

69


X= 33.46



2 - 16.56 = 19.58 KN-m


188.13 + V3 x 8.41 = 158.87 + 33 x 8.41 x 8.41

2

V3 = 135.28 KN

V2 + V3 = 33 x 8.41 = 277.53 KN

V2 = 142.25 KN


X= 135.28



2 - 158.81 = 118.52 KN-m

Case 3:

Max load on beam 20-11 and beam 11-02


Calculations of moments:

70



57.83 + V2 x 6 = 33 x 6 x 6

2 + 140.51

V2 = 112.78 KN

V1 + V2 = 33 x 6 = 198 KN

V1 = 85.22 KN


X= 85.22


71


2 - 57.83 = 52.1 KN-m


199.52 + V3 x 8.41 = 153.93 + 33 x 8.41 x 8.41

2

V3 = 133.34 KN

V2 + V3 = 33 x 8.41 = 277.53 KN

V2 = 144.19 KN


X= 133.34



2 - 153.93 = 115.42 KN-m




span

moment


span

moment

Beam Column

1 68.42 68.42 59.13 111.87 6.84 105.0 51.51 67.38 68.22

2 16.58 16.58 19.58 94.59 93.54 188.13 118.52 158.81 155.58

3 57.85 57.83 52.1 140.51 59 199.51 115.42 153.93 152.3

Max 68.42 68.42 59.13 140.51 93.54 199.51 118.52 158.81 155.58



1 91.76 106.24 69.61 60.66

2 33.46 59.48 142.25 135.28

3 85.22 112.78 144.11 133.34

Maximum shear 91.76 112.78 144.11 135.28


72

Floor:

Slab: 120mm


= (0.12 x 25) + 1 = 4 KN/m2


= 5 KN/m2

for Beam 11-02

Beam (20-11) (230mm X 600mm)

Dead load:


Due to Slab = W x L= 4 x 3.05= 12.2 KN/m

Live load:



= 41 KN/m


Beam (11-02) (230mm X 600mm)

Dead load:




Live load:

Due to Slab = W x L= 5 x 3.05= 15.25KN/m

73

Therefore, Maximum load = 1.5 x (D.L + L.L) = 1.5x(2.76+12.2+14.5+15.25)

= 67.1 KN/m

Minimum load = 0.9 x D.L = 0.9x(2.76+12.2+14.5) = 26.5 KN/m

Case 1:




4 Column 1235.82 x 103

3.85 x 106

0.32

Beam 1.38 x 106 0.36

Column 1235.82 x 103 0.32

5 Beam 1.38 x 106

4.84 x 106

0.29

Column 1235.82 x 103 0.26

Beam 0.985 x 106 0.19

Column 1235.82 x 103 0.26

6 Beam 0.985 x 106

3.46 x 106

0.28

Column 1235.82 x 103 0.36

Column 1235.82 x 103 0.36

74




75.53 + V5 x 6 = 41 x 6 x 6

2 + 154.22

V5 = 136.12 KN

V4 + V5 = 41 x 6 = 246 KN

V4 = 109.88 KN

75


X= 109.88



2 - 75.53 = 71.71 KN-m


172.08 + V6 x 8.41 = 114.85 + 26.5 x 8.41 x 8.41

2

V6 = 104.63 KN

V5 + V6 = 26.5 x 8.41 = 222.87 KN

V5 = 118.24 KN


X= 104.63



2 - 114.85 = 91.8 KN-m

Case 2:



76




16.29 + V5 x 6 = 13.5 x 6 x 6

2 + 158.17

V5 = 64.15 KN

V4 + V5 = 13.5 x 6 = 81 KN

V4 = 16.85 KN


X= 16.85


77


2 - 12.92 = -2.39 KN-m


375.67 + V6 x 8.41 = 315.93 + 67.1 x 8.41 x 8.41

2

V6 = 275.05 KN

V5 + V6 = 67.1 x 8.41 = 564.31 KN

V5 = 289.26 KN


X= 275.05



2 - 315.93= 247.92 KN-m

Case 3:



78




49.25 + V5 x 6 = 41 x 6 x 6

2 + 228.89

V5 = 152.94 KN

V4 + V5 = 41 x 6 = 246 KN

V4 = 93.06 KN


X= 93.06


79


2 - 49.25 = 56.37 KN-m


393.9 + V6 x 8.41 = 308.24 + 67.1 x 8.41 x 8.41

2

V6 = 271.97 KN

V5 + V6 = 67.1 x 8.41 = 564.31 KN

V5 = 292.34 KN


X= 271.97



2 - 308.24= 242.5 KN-m




span

moment


span

moment

Beam Column

1 37.77 75.53 71.71 154.21 8.9 172.08 91.8 114.85 57.45

2 6.42 12.92 3.092 158.17 108.74 375.67 247.92 315.93 153.37

3 24.66 49.25 56.37 228.89 82.5 393.9 242.5 308.24 150.94

Max 37.77 75.53 71.71 228.89 108.74 393.9 247.92 315.93 153.37



1 109.8 136.12 118.24 104.63

2 16.85 64.15 289.26 275.05

3 93.06 152.94 292.34 271.97

Maximum shear 109.8 152.94 292.34 275.05


80

Frame 24-15-06

Roof:

Same as frame 20-11-02

Floor:

Slab = 120mm


= (0.12 x 25) + 1 = 4 KN/m2

Live load = 10 KN/m2

for Beam 24-15

= 2 KN/m2 for Beam 24-15


Beam (24-15) (230mm X 600mm)

Dead load:



150mm wall of 2.2m height = 0.15 x 20 x 2.2 = 6.6 KN/m

Live load:

Due to Slab = 5 x 3.05

2 +

2 x 3.05

2 = 10.675 KN/m


= 48.5 KN/m

Minimum load = 0.9 x D.L = 0.9 x (2.76+12.2+6.6) = 19.5 KN/m

81

Beam (15-06) (230mm X 600mm)

Dead load:



Live load:

Due to Slab = W x L= 5 x 3.05= 15.25KN/m


= 45.5 KN/m


Case 1:


82



4 Column 1235.82 x 103

3.85 x 106

0.32

Beam 1.38 x 106 0.36

Column 1235.82 x 103 0.32

5 Beam 1.38 x 106

4.84 x 106

0.29

Column 1235.82 x 103 0.26

Beam 0.985 x 106 0.19

Column 1235.82 x 103 0.26

6 Beam 0.985 x 106

3.46 x 106

0.28

Column 1235.82 x 103 0.36

Column 1235.82 x 103 0.36


83



95.08 + V5 x 6 = 48.5 x 6 x 6

2 + 166.11

V5 = 157.34 KN

V4 + V5 = 48.5 x 6 = 291 KN

V4 = 133.66 KN


X= 133.66



2 - 95.08 = 89.37 KN-m


155.12 + V6 x 8.41 = 93.64 + 22.5 x 8.41 x 8.41

2

V6 = 87.3 KN

V5 + V6 = 22.5 x 8.41 = 189.23 KN

V5 = 101.93 KN

84


X= 87.3



2 - 93.64 = 75.72 KN-m

Case 2:




85



14.68 + V5 x 6 = 19.5 x 6 x 6

2 + 133.86

V5 = 78.37 KN

V4 + V5 = 19.5 x 6 = 117 KN

V4 = 38.63 KN


X= 38.63



2 - 14.68 = 23.56 KN-m


261.6 + V6 x 8.41 = 211.34 + 45.5 x 8.41 x 8.41

2

V6 = 185.35 KN

V5 + V6 = 45.5 x 8.41 = 382.66 KN

V5 = 197.31 KN

86


X= 185.35



2 - 211.34= 165.85 KN-m

Case 3:




87



80.19 + V5 x 6 = 44.5 x 6 x 6

2 + 208.43

V5 = 166.87 KN

V4 + V5 = 48.5 x 6 = 291 KN

V4 = 124.13 KN


X= 124.13



2 - 80.19 = 78.7 KN-m


280.82 + V6 x 8.41 = 203.22 + 45.5 x 8.41 x 8.41

2

V6 = 182.1 KN

V5 + V6 = 67.1 x 8.41 = 382.66 KN

V5 = 200.56 KN


X= 182.1

45.5 = 4m from the right support

88

Therefore, maximum moment = 182.1 x 4

2 - 203.22= 160.98 KN-m




span

moment


span

moment

Beam Column

1 47.54 95.08 89.37 166.11 5.5 155.12 75.72 93.64 47.54

2 7.34 14.63 23.56 133.86 63.86 261.6 166.85 211.34 103.08

3 40.11 80.19 78.7 208.43 36.19 280.82 160.98 203.22 100.52

Max 47.54 95.08 89.37 208.43 63.86 280.82 166.85 211.34 103.08



1 133.66 157.34 101.93 87.3

2 38.63 78.37 197.31 185.35

3 124.13 166.87 200.56 182.1

Maximum shear 133.66 166.87 200.56 185.35


Frame 25-16-07

Roof:


Floor:

Slab = 120mm


= (0.12 x 25) + 1 = 4 KN/m2


89


= 6 KN/m2

for Beam 16-07

= 5 KN/m2

for Beam 16-07

Beam (25-16) (230mm X 600mm)

Dead load:




Live load:


2 +

3 x 3.05

2 = 7.625 KN/m


= 44 KN/m


Beam (16-07) (230mm X 600mm)

Dead load:




Live load:


2 +

6 x 3.05

2 = 16.775 KN/m


= 57.51 KN/m

90


Case 1:




4 Column 1235.82 x 103

3.85 x 106

0.32

Beam 1.38 x 106 0.36

Column 1235.82 x 103 0.32

5 Beam 1.38 x 106

4.84 x 106

0.29

Column 1235.82 x 103 0.26

Beam 0.985 x 106 0.19

Column 1235.82 x 103 0.26

6 Beam 0.985 x 106

3.46 x 106

0.28

Column 1235.82 x 103 0.36

Column 1235.82 x 103 0.36

91




85.69 + V5 x 6 = 44 x 6 x 6

2 + 152.33

V5 = 143.11 KN

V4 + V5 = 44 x 6 = 264 KN

V4 = 120.89 KN

92


X= 120.89



2 - 85.69 = 80.45 KN-m


145.58 + V6 x 8.41 = 89.17 + 21.3 x 8.41 x 8.41

2

V6 = 82.86 KN

V5 + V6 = 21.3 x 8.41 = 179.13 KN

V5 = 96.27 KN


X= 82.86



2 - 89.17 = 72 KN-m

Case 2:


93





6.85 + V5 x 6 = 19.5 x 6 x 6

2 + 155.96

V5 = 83.35 KN

V4 + V5 = 19.5 x 6 = 117 KN

V4 = 33.65 KN

94


X= 33.65



2 - 6.85 = 22.26 KN-m


327.23 + V6 x 8.41 = 268.6 + 57.51 x 8.41 x 8.41

2

V6 = 234.86 KN

V5 + V6 = 57.51 x 8.41 = 483.66 KN

V5 = 248.8 KN


X= 234.86



2 - 268.6= 210.5 KN-m

Case 3:



95




62.24 + V5 x 6 = 44 x 6 x 6

2 + 218.95

V5 = 158.11 KN

V4 + V5 = 48.5 x 6 = 264 KN

V4 = 105.89 K

96


X= 105.89



2 - 62.24 = 65.36 KN-m


343.47 + V6 x 8.41 = 257.17 + 57.51 x 8.41 x 8.41

2

V6 = 231.56 KN

V5 + V6 = 67.1 x 8.41 = 483.66 KN

V5 = 252.1 KN


X= 231.56



2 - 257.17= 209.42 KN-m




span

moment


span

moment

Beam Column

1 42.84 85.69 80.45 152.33 3.38 145.56 72 89.17 45.17

2 3.46 6.85 22.26 155.96 85.63 327.23 210.5 268.55 130.75

3 31.15 62.24 65.36 218.95 62.25 343.47 209.42 261.7 128.6

Max 42.84 85.69 80.45 218.95 85.63 343.47 210.5 268.55 130.75

97



1 120.89 143.11 96.27 82.86

2 33.65 83.35 248.8 234.86

3 105.89 158.11 251.1 231.56

Maximum shear 120.89 158.11 251.1 234.86


Frame 26-17-08

Roof:


Floor:

Slab = 120mm


= (0.12 x 25) + 1 = 4 KN/m2



Beam (26-17) (230mm X 600mm)

Dead load:



Live load:

Due to Slab = 3 x 3.05= 9.15 KN/m

Therefore, Maximum load = 1.5 x (D.L + L.L) = 1.5x (2.76+12.2+9.15)

= 36.2 KN/m

98


Beam (17-08) (230mm X 600mm)

Dead load:



Live load:

Due to Slab = 6 x 3.05= 18.3 KN/m

Therefore, Maximum load = 1.5 x (D.L + L.L) = 1.5x (2.76+12.2+18.3)

= 50 KN/m


Case 1:


99



4 Column 1235.82 x 103

3.85 x 106

0.32

Beam 1.38 x 106 0.36

Column 1235.82 x 103 0.32

5 Beam 1.38 x 106

4.84 x 106

0.29

Column 1235.82 x 103 0.26

Beam 0.985 x 106 0.19

Column 1235.82 x 103 0.26

6 Beam 0.985 x 106

3.46 x 106

0.28

Column 1235.82 x 103 0.36

Column 1235.82 x 103 0.36


100



73.1+ V5 x 6 = 36.2 x 6 x 6

2 + 117.93

V5 = 116.07 KN

V4 + V5 = 46.2 x 6 = 217.2 KN

V4 = 101.13 KN


X= 101.13



2 - 73.1 = 68.48 KN-m


97.78 + V6 x 8.41 = 54.2 + 13.5 x 8.41 x 8.41

2

V6 = 51.59KN

V5 + V6 = 13.5 x 8.41 = 113.54 KN

V5 = 61.95 KN


X= 51.59


101


2 - 54.2 = 44.34 KN-m

Case 2:




102



1.84 + V5 x 6 = 13.5 x 6 x 6

2 + 126.71

V5 = 61.31 KN

V4 + V5 = 13.5 x 6 = 81 KN

V4 = 19.69 KN


X= 19.69



2 - 1.84 = 12.54 KN-m


282.21 + V6 x 8.41 = 234.45 + 50 x 8.41 x 8.41

2

V6 = 204.57 KN

V5 + V6 = 50 x 8.41 = 420.5 KN

V5 = 215.93 KN


X= 204.57


103


2 - 234.45= 183.9 KN-m

Case 3:




104



49.5 + V5 x 6 = 36.2 x 6 x 6

2 + 185.08

V5 = 131.2 KN

V4 + V5 = 36.2 x 6 = 217.2 KN

V4 = 86 KN


X= 86


Therefore, maximum moment = 86 x 2.38

2 - 49.5 = 52.84 KN-m


297.26 + V6 x 8.41 = 228.1 + 50 x 8.41 x 8.41

2

V6 = 202.03 KN

V5 + V6 = 50 x 8.41 = 420.5 KN

V5 = 218.47 KN


X= 202.03


105


2 - 228.1= 180 KN-m




span

moment


span

moment

Beam Column

1 36.55 73.1 68.48 117.93 10.07 97.78 44.34 54.2 27.9

2 0.9 1.85 12.54 126.71 77.74 282.21 183.9 234.45 113.98

3 24.76 49.5 52.84 185.08 56.08 297.26 180 228.1 111.97

Max 36.55 73.1 68.48 185.08 77.74 297.26 183.9 234.45 113.98



1 101.13 116.07 61.95 51.59

2 19.69 61.31 215.93 204.57

3 86 131.2 218.47 202.03

Maximum shear 101.13 131.2 218.47 204.57


Frame 27-18-09

Roof:

Same as of roof frame 19-10-01

Floor:

Slab: 120mm


= (0.12 x 25) + 1 = 4 KN/m2


106


Beam (27-18) (230mm X 450mm)

Dead load:


Due to Slab = W x L

2 =

4 x 3.05

2 = 6.1 KN/m


Live load:

Due to Slab = W x L

2 =

3 x 3.05

2 = 4.6KN/m


= 42 KN/m

Minimum load = 0.9 x D.L = 0.9 x (1.9+6.1+15.25) = 21 KN/m

Beam (18-09) (230mm X 450mm)

Dead load:


Due to Slab = W x L

2 =

4 x 3.05

2 = 6.1 KN/m


Live load:

Due to Slab = W x L

2 =

6 x 3.05

2 = 9.15KN/m


= 48.6 KN/m

Minimum load = 0.9 x D.L = 0.9 x (1.9+6.1+15.25) = 21KN/m

107

Case 1:




4 Column 1235.82 x 103 3.05 x 106 0.4

Beam 582.18 x 103 0.2

Column 1235.82 x 103 0.4

5 Beam 582.18 x 103 3.47 x 106 0.17

Column 1235.82 x 103 0.36

Beam 415.35 x 103 0.11

Column 1235.82 x 103 0.36

6 Beam 415.35 x 103 2.89 x 106 0.14

Column 1235.82 x 103 0.43

Column 1235.82 x 103 0.43

108




101.22 + V5 x 6 = 42 x 6 x 6

2 + 137.6

V5 = 132.06 KN

V4 + V5 = 42 x 6 = 252 KN

V4 = 119.94 KN

109


X= 119.94



2 - 101.22 = 70.3 KN-m


133.13 + V6 x 8.41 = 106.52 + 21 x 8.41 x 8.41

2

V6 = 85.14 KN

V5 + V6 = 21 x 8.41 = 176.61 KN

V5 = 91.47 KN


X= 85.14



2 - 106.52 = 65.89 KN-m

Case 2:


110





34.07 + V5 x 6 = 21 x 6 x 6

2 + 108.08

V5 = 75.34 KN

V4 + V5 = 21 x 6 = 126 KN

V4 = 50.66 KN

111


X= 50.66



2 - 34.07 = 26.98 KN-m


280.96 + V6 x 8.41 = 257.96 + 48.6 x 8.41 x 8.41

2

V6 = 201.63 KN

V5 + V6 = 48.6 x 8.41 = 408.73 KN

V5 = 207.1 KN


X= 201.63



2 - 257.96= 160.42 KN-m

Case 3:


112





89.24 + V5 x 6 = 42 x 6 x 6

2 + 166.05

V5 = 138.8 KN

V4 + V5 = 42 x 6 = 252 KN

V4 = 113.2 KN


113

X= 113.2



2 - 89.24 = 63.58 KN-m


288.45 + V6 x 8.41 = 254.55 + 48.6 x 8.41 x 8.41

2

V6 = 200.33 KN

V5 + V6 = 48.6 x 8.41 = 408.73 KN

V5 = 208.4 KN


X= 200.33



2 - 254.55= 158.33 KN-m




span

moment


span

moment

Beam Column

1 50.61 101.22 70.3 137.6 2.24 133.12 65.89 106.09 53.26

2 17.04 34.07 26.98 108.08 86.44 280.96 160.42 257.96 128.22

3 44.62 89.24 63.58 166.06 61.2 288.45 158.33 254.55 126.86

Max 50.61 101.22 70.3 166.06 86.44 288.45 160.42 257.96 128.22


114


1 119.94 132.06 91.47 85.14

2 50.66 75.34 207.28 201.63

3 113.2 138.8 208.4 200.33

Maximum shear 119.94 138.8 208.4 200.33


115

4. DESIGN OF BEAMS

Bending moment diagram for entire building:

116

1. Longitudinal beams

a. Beam 23-24

Roof:

Section: 230mm x 300mm (b x D)

Cover: 25mm

Effective depth (d’) = 300 – 25 = 275mm

Concrete: M20 Steel: Fe415

Loads:

Slab: 110mm thick

Dead load = self weight + floor finish

= 0.11 x 25 +2.5 = 5.25 KN/m2


Acting on beam:

Dead load:

Self weight = 0.23 x 25 x (0.3 – 0.11) = 1.1 KN/m

Slab = 5.25 x 3.05

6 = 2.67 KN/m

Parapet wall (250mm wall including plastering of 1m height)

= 20 x 0.25 x 1 = 5 KN/m

Live load:

Slab = 1.5 x 3.05

6 = 0.76 KN/m

Therefore, Total load (W) = 5 + 2.67 + 1.1 + 0.76 = 9.53 KN/m

Factored load (Wu) = 1.5 x 9.53 = 14.3 KN/m

117

Maximum load on column removing the triangular load = 1.5 (1.1 +5)

= 9.15 KN/m

Shear on column = column load x l

2 =

9.15 x 3.05

2 = 13.96 KN (from each side)

Design moment:

Mu = 14.3 x (3.05)2

12 = 11.09 x 106 N-mm = 11.09 KN-m

Calculation of Ast:

From IS456-2000, P96a, clause G-1.1 (b), we have


fck x b x d )

11.09 x 106 = 0.87 x 415 x Ast x 275 x (1- 415 x Ast

20 x 230 x 275 )

Ast = 116.12 mm2

No. of 10mm diameter bars = 4 x 116.12

π x 102 = 1.5 say 2 bars

Therefore, provided Ast = 2 x π4

x 102 = 157.08 mm

2

Design of shear reinforcement:

Vu = Wu x Lx

2 =

14.3 x 3.05

2 = 21.81KN

Nominal shear stress, v = Vu

b x d =

21.81 x 103

230 x 275 = 0.35 N/mm2

Ast = 157.08 mm2

Consider Pt = 100 xAst

b x d =

100 x 157.08

230 x 275 = 0.25

118

From IS 456-2000, Table 19:

Pt c

0.25 0.36

Therefore, c = 0.36 N/mm2

v approximately equals to c

Therefore, provide minimum shear reinforcement.

Assuming 8mmφ- 2legged stirrups

Asv = 2 x π x 8

2

4 = 100.53mm

2

Min shear reinforcement (IS456-2000, clause 26.5.1.6, P48)

Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 300

0.87 x 415 = 76.44mm2 <100.5mm2 (safe)

Provide 8mm dia stirrups @ 300mm c/c.

Floor:


Cover: 25mm



119

Loads:

Slab: 120mm thick


= 0.12 x 25 +1 = 4 KN/m2

Live load = 10 KN/m2

Acting on beam:

Dead load:


Slab = 4 x 3.05

6 = 2.03 KN/m

Parapet wall (250mm wall including plastering of 2.2m height)

= 20 x 0.25 x (3.35-0.3) = 15.25 KN/m

Live load:

Slab = 10 x 3.05

6 = 5.08 KN/m

Therefore, Total load (W) = 1.9 + 2.03 + 15.25 + 5.08 = 24.26 KN/m


Maximum load on column removing the triangular load = 1.5 (1.9 +5.08)

= 25.725 KN/m


2 =

25.725 x 3.05

2

= 39.23 KN (from each side)

Design moment:

Mu = 36.39 x (3.05)2

12 = 28.21 x 106 N-mm = 28.21 KN-m

120

Calculation of Ast:


fck x b x d )

28.21 x 106 = 0.87 x 415 x Ast x 425 x (1- 415 x Ast

20 x 230 x 425 )

Ast = 191.64 mm2


π x 102 = 2.33 say 3 bars


x 102 = 235.62 mm

2


Vu = Wu x Lx

2 =

36.39 x 3.05

2 = 55.5KN


b x d =

55.5 x 103

230 x 425 = 0.57 N/mm2

Ast = 235.62 mm2


b x d =

100 x 235.62

230 x 425 = 0.24


Pt c

0.15 0.28

0.24 ?

0.25 0.36

Therefore, c = 0.28 + 0.36-0.28

0.25-0.15 x (0.24-0.15) = 0.352 N/mm

2

Therefore, v > c,

121

From IS456-2000(Clause 40.4, P72)

When v exceeds c, shear reinforcement shall be provided

Shear reinforcement shall be provided to carry a shear equal to Vu - cbd. The strength of

shear reinforcement Vus shall be calculated as follows:

Vus = 0.87 x fy x Asv x d

sv

Stirrups are designed for shear Vus = Vu - cbd

= (55.5 x 103) – (0.32 x 230 x 425)

= 24220N


Asv = 2 x π x 8

2

4 = 100.53mm

2

Therefore,

Spacing of stirrups, sv = 0.87 x fy x Asv x d

Vus =

0.87 x 415 x 100.53 x 425

24220

=636.9mm

From IS456-2000, (clause 26.5.1.5, P47)

The maximum spacing of shear reinforcement measured along the axis of the member shall

not be exceed 0.75d for vertical stirrups where d is the effective depth of the section under

consideration. In no case shall the spacing exceed 300mm.

Provide spacing = 300mm

Therefore, Vus = 0.87 x fy x Asv x d

sv =

0.87 x 415 x 100.53 x 425

300 = 51419.83N


Asv

b x Sv ≥

0.4

0.87 x fy

122

Asv = 0.4 x 230 x 300

0.87 x 415 = 76.44mm2 <100.5mm2 (safe)


b. Beam 14-15

Roof:


Cover: 25mm



Loads:

Slab: 110mm thick


= 0.11 x 25 +2.5 = 5.25 KN/m2


Acting on beam:

Dead load:


Slab = 2 x 5.25 x 3.05

6 = 5.34 KN/m

123

Live load:

Slab = 2 x 1.5 x 3.05

6 = 1.53 KN/m

Therefore, Total load (W) = 1.1 + 5.34 + 1.53 = 7.97 KN/m


Maximum load on column removing the triangular load = 1.5 (1.1)

= 1.65 KN/m


2 =

1.65 x 3.05


Design moment:

Mu = 11.96 x (3.05)

2

12 = 9.27 x 10

6 N-mm = 9.27 KN-m

Calculation of Ast:


fck x b x d )

9.27 x 106 = 0.87 x 415 x Ast x 275 x (1-

415 x Ast

20 x 230 x 275 )

Ast = 96.41 mm2


π x 82 = 1.91 say 2 bars


x 82 = 100.53 mm2


Vu = Wu x Lx

2 =

11.96 x 3.05

2 = 18.239KN


b x d =

18.239 x 103

230 x 275 = 0.29 N/mm

2

Ast = 100.53 mm2

124


b x d =

100 x 100.53

230 x 275 = 0.15


Pt c

0.15 0.28





Asv = 2 x π x 82

4 = 100.53mm2



sv =

0.87 x 415 x 100.53 x 275

300 = 33371.66KN


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 300

0.87 x 415 = 76.44mm

2 <100.5mm

2 (safe)


125

Floor:


Cover: 25mm



Loads:

Slab: 120mm thick

Dead load = self weight + floor finish + wall

= 0.12 x 25 +1 + 1.5 = 5.5 KN/m2

Live load = 10 KN/m2 + 5 KN/m2

Acting on beam:

Dead load:


Slab = 5.5 x 3.05

6 +

4 x 3.05

6 = 4.83 KN/m

Internal wall (150mm wall including plastering of 2.2m height)

= 20 x 0.15 x 2.2 = 6.6 KN/m

Live load:

Slab = 15 x 3.05

6 = 7.625 KN/m

Therefore, Total load (W) = 1.9 + 4.83 + 6.6 + 7.625 = 20.955 KN/m



= 12.75 KN/m

126


2 =

12.75 x 3.05

2


Design moment:

Mu = 31.4325 x (3.05)

2

12 = 24.37 x 10

6 N-mm = 24.37 KN-m

Calculation of Ast:


fck x b x d )

24.37 x 106 = 0.87 x 415 x Ast x 425 x (1- 415 x Ast

20 x 230 x 425 )

Ast = 164.5 mm2


π x 102 = 2.1 say 3 bars


x 102 = 235.62 mm

2


Vu = Wu x Lx

2 =

20.955 x 3.05

2 = 31.96KN


b x d =

31.96 x 103

230 x 425 = 0.33 N/mm2

Ast = 235.62 mm2


b x d =

100 x 235.62

230 x 425 = 0.24


Pt c

0.15 0.28

127

0.24 ?

0.25 0.36

Therefore, c = 0.28 + 0.36-0.28

0.25-0.15 x (0.24-0.15) = 0.352 N/mm

2

Therefore, v approximately equals to c,


Asv = 2 x π x 82

4 = 100.53mm2


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 300

0.87 x 415 = 76.44mm

2 <100.5mm

2 (safe)


c. Beam 7-8

Roof:


Cover: 25mm



128

Loads:

Slab: 110mm thick


= 0.11 x 25 +2.5 = 5.25 KN/m2


Acting on beam:

Dead load:


Slab = 5.25 x 3.05

6 = 2.67 KN/m

Parapet wall (250mm wall including plastering of 1m height)

= 20 x 0.25 x 1 = 5 KN/m

Live load:

Slab = 1.5 x 3.05

6 = 0.76 KN/m

Therefore, Total load (W) = 5 + 2.67 + 1.1 + 0.76 = 9.53 KN/m


Maximum load on column removing the triangular load = 1.5 (1.1 +5)

= 9.15 KN/m


2 =

9.15 x 3.05


Design moment:

Mu = 14.3 x (3.05)

2

12 = 11.09 x 10

6 N-mm = 11.09 KN-m

129

Calculation of Ast:

From IS456-2000, P96a, clause G-1.1 (b), we have


fck x b x d )

11.09 x 106 = 0.87 x 415 x Ast x 275 x (1-

415 x Ast

20 x 230 x 275 )

Ast = 116.12 mm2


π x 102 = 1.5 say 2 bars


x 102 = 157.08 mm2


Vu = Wu x Lx

2 =

14.3 x 3.05

2 = 21.81KN


b x d =

21.81 x 103

230 x 275 = 0.35 N/mm

2

Ast = 157.08 mm2


b x d =

100 x 157.08

230 x 275 = 0.25


Pt c

0.25 0.36





130

Asv = 2 x π x 82

4 = 100.53mm2


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 300

0.87 x 415 = 76.44mm

2 <100.5mm

2 (safe)


Floor:


Cover: 25mm



Loads:

Slab: 120mm thick

Dead load = self weight + floor finish + wall

= 0.12 x 25 +1 + 1.5 = 5.5 KN/m2

Live load = 6 KN/m2

Acting on beam:

Dead load:


131

Slab = 5.5 x 3.05

6 = 2.8 KN/m

External wall (250mm wall including plastering)

= 20 x 0.25 x (3.35 – 0.3) = 15.25 KN/m

Live load:

Slab = 6 x 3.05

6 = 3.05 KN/m

Therefore, Total load (W) = 1.9 + 2.8 + 15.25 + 3.05 = 23 KN/m

Factored load (Wu) = 1.5 x 23 = 34.5 KN/m


= 25.73 KN/m


2 =

25.73 x 3.05

2


Design moment:

Mu = 34.5 x (3.05)

2

12 = 26.75 x 10

6 N-mm = 26.75 KN-m

Calculation of Ast:


fck x b x d )

26.75 x 106 = 0.87 x 415 x Ast x 425 x (1- 415 x Ast

20 x 230 x 425 )

Ast = 181.3 mm2


π x 102 = 2.31 say 3 bars


x 102 = 235.62 mm

2

132


Vu = Wu x Lx

2 =

34.5 x 3.05

2 = 52.61KN


b x d =

52.61 x 103

230 x 425 = 0.54 N/mm2

Ast = 235.62 mm2


b x d =

100 x 235.62

230 x 425 = 0.24


Pt c

0.15 0.28

0.24 ?

0.25 0.36

Therefore, c = 0.28 + 0.36-0.28

0.25-0.15 x (0.24-0.15) = 0.352 N/mm

2

Therefore, v > c,


= (52.61 x 103) – (0.35 x 230 x 425)

= 18397.5N


Asv = 2 x π x 8

2

4 = 100.53mm

2

Therefore,


Vus =

0.87 x 415 x 100.53 x 425

18397.5

= 838.48mm

133



sv =

0.87 x 415 x 100.53 x 425

300 = 51419.83N


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 300

0.87 x 415 = 76.44mm2 <100.5mm2 (safe)


134

2. Transverse beams:

a. Beam 19-10 and Beam 10-01

135

ROOF: (section 230mm x 300mm, d' = 25mm)

From analysis of frames the moments acting on the supports are:

At left support (Ast1):

Moment due to external load (Mu) = 64.65 KN-m

From IS456-2000, P96, clause G-1.1 (c), moment of resistance of the section is given by

Mu’Limit = 0.36 x xumax

d x (1- 0.42 x

xumax

d ) x b x d2 x fck

For Fe 415, xumax

d = 0.48

Mu’limit = 0.138 x fck x b x d2

= 0.138 x 20 x 230 x (275)2 = 48 x 10

6 N-mm = 48 KN-m

Mu > Mu’limit , therefore design as doubly reinforced beam.

Consider d'

d =

25

275 = 0.1

Therefore, fsc = 353 N/mm2

Area of steel in compression (Asc):

Asc = Mu - Mu’limit

fsc x (d - d') =

(64.65 - 48) x 106

353 x (275 - 25) =188.67 mm

2

To find Ast1:

Mu’limit = 0.87 x fy x Ast1 x d x (1 - fy x Ast1

fck x b x d )

136

48 x 106 = 0.87 x 415 x Ast1 x 275 x (1 - 415 x Ast1

20 x 230 x 275 )

Ast1 = 602.55 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 188.67

0.87 x 415 =184.46 mm

2

Ast = Ast1 + Ast2 = 602.55 + 184.46 = 787.01 mm2

Between the support:

Span moment = 39.24 KN-m

For T-beam, from IS456-2000, P37, clause 23.1.2 (a)

Effective width of flange (bf) = l06

+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm

Therefore, bf = 4200

6 + (6 x 110) + 230

= 1590 mm

Therefore Mu’limit = 0.36 x fck x bf x Df x(d – 0.42Df)

= 0.36 x 20 x 1590 x 110 x (275 – 0.42 x 110)

= 288.12 KN-m

Mu < < Mu’limit

Ast:

39.24 x 106 = 0.87 x 415 x Ast2 x 275 x (1-

415 x Ast2

20 x 230 x 275 )

Ast2 = 466.65 mm2

137

At right support (Ast3):



= 0.138 x 20 x 230 x (275)2 = 48 x 106 N-mm = 48 KN-m


Consider d'

d =

25

275 = 0.1




fsc x (d - d') =

(86.09 - 48) x 106

353 x (275 - 25) =431.62 mm2

To find Ast1:


fck x b x d )

48 x 106 = 0.87 x 415 x Ast1 x 275 x (1 -

415 x Ast1

20 x 230 x 275 )

Ast1 = 602.55 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 431.62

0.87 x 415 =422 mm2

Ast3 = Ast1 + Ast2 = 602.55 + 422 = 1024.55 mm2

At support (Ast4):



= 0.138 x 20 x 230 x (275)2 = 48 x 10

6 N-mm = 48 KN-m

138


Consider d'

d =

25

275 = 0.1




fsc x (d - d') =

(145.76 - 48) x 106

353 x (275 - 25) =1107.76 mm

2

To find Ast1:


fck x b x d )

48 x 106 = 0.87 x 415 x Ast1 x 275 x (1 -

415 x Ast1

20 x 230 x 275 )

Ast1 = 602.55 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 1107.76

0.87 x 415 =1083.06 mm

2

Ast4 = Ast1 + Ast2 = 602.55 + 1083 = 1685.61 mm2

Between the support (Ast5):



+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 8410 = 5887 mm


6 + (6 x 110) + 230

= 1871.16 mm

139


= 0.36 x 20 x 1871.16 x 110 x (275 – 0.42 x 110)

= 339.07 KN-m

Mu < Mu’limit

Ast:

77.97 x 106 = 0.87 x 415 x Ast5 x 275 x (1-

415 x Ast5

20 x 230 x 275 )

Ast5 = 1411.66 mm2

At right end (Ast6):



= 0.138 x 20 x 230 x (275)2 = 48 x 10

6 N-mm = 48 KN-m


Consider d'

d =

25

275 = 0.1




fsc x (d - d') =

(136.01 - 48) x 106

353 x (275 - 25) =997.28 mm

2

To find Ast1:


fck x b x d )

48 x 106 = 0.87 x 415 x Ast1 x 275 x (1 - 415 x Ast1

20 x 230 x 275 )

Ast1 = 602.55 mm2

140

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 997.28

0.87 x 415 =975.05 mm

2

Ast6 = Ast1 + Ast2 = 602.55 + 975.05 = 1577.6 mm2


At left support:

End shear = (Vu) = 71.54 KN


b x d =

71.54 x 103

230 x 275 = 1.13 N/mm

2

Ast = 787.01 mm2


b x d =

100 x 787.01

230 x 275 = 1.25


Pt c

1.25 0.67


Therefore, v > c,


= (71.54 x 103) – (0.67 x 230 x 275)

= 29162.5N

141


Asv = 2 x π x 8

2

4 = 100.53mm

2

Therefore,


Vus =

0.87 x 415 x 100.53 x 275

29162.5

= 342.27mm



sv =

0.87 x 415 x 100.53 x 275

300 = 33261.73N


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 300

0.87 x 415 = 76.44mm2 <100.5mm2 (safe)

Therefore, provide 8mm dia stirrups @ 300mm c/c.

At middle support:



b x d =

77.92 x 103

230 x 275 = 1.23 N/mm

2

Ast = 1024.55 mm2


b x d =

100 x 1024.55

230 x 275 = 1.62


Pt c

1.5 0.72

142

1.62 ?

1.75 0.75

Therefore, c = 0.72 + 0.75-0.72

1.75-1.5 x (1.62-1.5) = 0.734 N/mm

2

Therefore, v > c,


= (77.92 x 103) – (0.734 x 230 x 275)

= 31494.5N


Asv = 2 x π x 82

4 = 100.53mm2

Therefore,


Vus =

0.87 x 415 x 100.53 x 275

31494.5

= 316.93mm



sv =

0.87 x 415 x 100.53 x 275

300 = 33261.73N


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 300

0.87 x 415 = 76.44mm

2 <100.5mm

2 (safe)


143

At middle support:



b x d =

104.84 x 103

230 x 275 = 1.66 N/mm

2

Ast = 1685.61 mm2


b x d =

100 x 1685.61

230 x 275 = 2.67


Pt c

2.67 0.82


Therefore, v > c,


= (104.84 x 103) – (0.82 x 230 x 275)

= 52975N


Asv = 2 x π x 82

4 = 100.53mm

2

Therefore,


Vus =

0.87 x 415 x 100.53 x 275

52975

= 188.42mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

144

Asv = 0.4 x 230 x 180

0.87 x 415 = 45.87mm2 <100.5mm2 (safe)


At right support:



b x d =

102.63 x 103

230 x 275 = 1.63 N/mm

2

Ast = 1577.6 mm2


b x d =

100 x 1577.6

230 x 275 = 2.5


Pt c

2.5 0.82


Therefore, v > c,


= (102.63 x 103) – (0.82 x 230 x 275)

= 50765N


Asv = 2 x π x 82

4 = 100.53mm2

Therefore,


Vus =

0.87 x 415 x 100.53 x 275

50765

= 196.62mm < 300mm

145


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 190

0.87 x 415 = 48.41mm2 <100.5mm2 (safe)


Reinforcement for roof 19-10-01:

FLOOR: (section: 230mm x 450mm, d’= 25mm)

The moments acting on the beam are:




= 0.138 x 20 x 230 x (425)2 = 114.66 x 106 N-mm = 114.66 KN-m

Mu < Mu’limit , therefore design as singly reinforced beam.

146

To find Ast1:


fck x b x d )

106.54 x 106 = 0.87 x 415 x Ast1 x 425 x (1 - 415 x Ast1

20 x 230 x 425 )

Ast1 = 546 mm2

Between the supports (Ast2):



+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm


6 + (6 x 120) + 230

= 1650 mm

Therefore Mu’limit = 0.36 x fck x bf x Df x (d – 0.42Df)

= 0.36 x 20 x 1650 x 120 x (425 – 0.42 x 120)

= 534.03 KN-m

Mu < Mu’limit

Ast:

73.58 x 106 = 0.87 x 415 x Ast2 x 425 x (1- 415 x Ast2

20 x 230 x 425 )

Ast2 = 852.07 mm2



147


= 0.138 x 20 x 230 x (425)2 = 114.66 x 10

6 N-mm = 114.66 KN-m


Consider d'

d =

25

425 = 0.05




fsc x (d - d') =

(169.53 - 114.66) x 106

353 x (425 - 25) =386.84 mm

2

To find Ast1:


fck x b x d )

114.66 x 106 = 0.87 x 415 x Ast1 x 425 x (1 - 415 x Ast1

20 x 230 x 425 )

Ast1 = 932 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 386.84

0.87 x 415 = 380.36 mm2

Ast3 = Ast1 + Ast2 = 932 + 380.6 = 1312.36 mm2

At support (Ast4):



= 0.138 x 20 x 230 x (425)2 = 114.66 x 106 N-mm = 114.66 KN-m


Consider d'

d =

25

425 = 0.05

148




fsc x (d - d') =

(277.32 -11 4.66) x 106

353 x (425 - 25) =1145.5 mm

2

To find Ast1:


fck x b x d )

114.66 x 106 = 0.87 x 415 x Ast1 x 425 x (1 -

415 x Ast1

20 x 230 x 425 )

Ast1 = 932 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

355 x 1145.5

0.87 x 415 =1126.31 mm

2

Ast4 = Ast1 + Ast2 = 923 + 1126.31 = 2058.31 mm2




+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 8410 = 5887 mm


6 + (6 x 120) + 230

= 1931.67 mm


= 0.36 x 20 x 1871.16 x 120 x (425 – 0.42 x 120)

= 625.2 KN-m

Mu < Mu’limit

149

Ast:

153.57 x 106 = 0.87 x 415 x Ast5 x 425 x (1-

415 x Ast5

20 x 230 x 425 )

Ast5 = 1442.53 mm2




= 0.138 x 20 x 230 x (425)2 = 114.66 x 10

6 N-mm = 114.66 KN-m


Consider d'

d =

25

425 = 0.05




fsc x (d - d') =

(246.62 - 114.66) x 106

355 x (425 - 25) =929.29 mm

2

To find Ast1:


fck x b x d )

114.66 x 106 = 0.87 x 415 x Ast1 x 425 x (1 - 415 x Ast1

20 x 230 x 425 )

Ast1 = 932 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

355 x 929.29

0.87 x 415 =913.72 mm2

Ast6 = Ast1 + Ast2 = 932 + 913.72 = 1845.72 mm2

150


At left support:



b x d =

125.6 x 103

230 x 425 = 1.29 N/mm

2

Ast = 546 mm2


b x d =

100 x 546

230 x 425 = 0.56


Pt c

0.5 0.48

0.56 ?

0.75 0.56

Therefore, c = 0.48 + 0.56-0.48

0.75-0.5 x (0.56-0.5) = 0.5 N/mm

2

Therefore, v > c,


= (125.6 x 103) – (0.5 x 230 x 425)

= 76725N

151


Asv = 2 x π x 8

2

4 = 100.53mm

2

Therefore,


Vus =

0.87 x 415 x 100.53 x 425

76725

= 184.87mm <300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 180

0.87 x 415 = 45.86mm

2 <100.5mm

2 (safe)


At middle support:



b x d =

144.35 x 103

230 x 425 = 1.48 N/mm

2

Ast = 1312.16 mm2


b x d =

100 x 1312.16

230 x 425 = 1.34


Pt c

1.25 0.67

1.34 ?

1.5 0.72

152

Therefore, c = 0.67 + 0.72-0.67

1.5-1.25 x (1.34-1.25) = 0.69 N/mm2

Therefore, v > c,


= (144.32 x 103) – (0.69 x 230 x 425)

= 78827.5N


Asv = 2 x π x 8

2

4 = 100.53mm

2

Therefore,


Vus =

0.87 x 415 x 100.53 x 425

78827.5

= 195.7mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 190

0.87 x 415 = 48.41mm2 <100.5mm2 (safe)


At middle support:



b x d =

199.64 x 103

230 x 425 = 2.04 N/mm

2

Ast = 2058.31 mm2


b x d =

100 x 2058.31

230 x 425 = 2

153


Pt c

2.00 0.79


Therefore, v > c,


= (199.64 x 103) – (0.79 x 230 x 425)

= 122417.5N


Asv = 2 x π x 102

4 = 157.07mm2

Therefore,


Vus =

0.87 x 415 x 157.07 x 425

122417.5

= 196.9mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 190

0.87 x 415 = 48.41mm

2 <157.07mm

2 (safe)


At right support:



b x d =

192.85 x 103

230 x 425 = 1.97 N/mm

2

154

Ast = 1845.72 mm2


b x d =

100 x1845.72

230 x 425 = 1.89


Pt c

1.75 0.75

1.89 ?

2 0.79

Therefore, c = 0.75 + 0.79-0.75

2-1.75 x (1.89-1.75) = 0.77 N/mm

2

Therefore, v > c,


= (192.85 x 103) – (0.77 x 230 x 425)

= 117582.5N


Asv = 2 x π x 102

4 = 157.07mm2

Therefore,


Vus =

0.87 x 415 x 157.07 x 425

117582.5

= 204.98mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 200

0.87 x 415 = 50.96mm

2 <157.07mm

2 (safe)

155


Reinforcement for floor 19-10-01:

156

b. Beam 20-11 and Beam 11-02

157

ROOF (section: 230mm x 450mm, d’ = 25mm)





= 0.138 x 20 x 230 x (425)2 = 114.66 x 106 N-mm = 114.66 KN-m


To find Ast1:


fck x b x d )

68.42 x 106 = 0.87 x 415 x Ast1 x 425 x (1 -

415 x Ast1

20 x 230 x 425 )

Ast1 = 498.67 mm2




+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm


6 + (6 x 110) + 230 = 1590 mm

158


= 0.36 x 20 x 1590 x 110 x (425 – 0.42 x 110)

= 477.02 KN-m

Mu < Mu’limit

Ast:

59.13 x 106 = 0.87 x 415 x Ast2 x 425 x (1-

415 x Ast2

20 x 230 x 425 )

Ast2 = 423.4 mm2




= 0.138 x 20 x 230 x (425)2 = 114.66 x 10

6 N-mm = 114.66 KN-m


Consider d'

d =

25

425 = 0.05




fsc x (d - d') =

(140.51 - 114.66) x 106

353 x (425 - 25) =181.97 mm2

To find Ast1:


fck x b x d )

114.66 x 106 = 0.87 x 415 x Ast1 x 425 x (1 -

415 x Ast1

20 x 230 x 425 )

Ast1 = 932 mm2

159

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

355 x 181.97

0.87 x 415 = 178.92 mm

2

Ast3 = Ast1 + Ast2 = 932 + 178.92 = 1110.4 mm2

At support (Ast4):



= 0.138 x 20 x 230 x (425)2 = 114.66 x 10

6 N-mm = 114.66 KN-m


Consider d'

d =

25

425 = 0.05




fsc x (d - d') =

(199.51 -11 4.66) x 106

353 x (425 - 25) =597.46 mm

2

To find Ast1:


fck x b x d )

114.66 x 106 = 0.87 x 415 x Ast1 x 425 x (1 - 415 x Ast1

20 x 230 x 425 )

Ast1 = 932 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

355 x 597.46

0.87 x 415 =587.45 mm2

Ast4 = Ast1 + Ast2 = 923 + 587.45 = 1518.93 mm2

160




+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 8410 = 5887 mm


6 + (6 x 110) + 230

= 1871.17 mm


= 0.36 x 20 x 1871.16 x 110 x (425 – 0.42 x 110)

= 561.37 KN-m

Mu < Mu’limit

Ast:

118.5 x 106 = 0.87 x 415 x Ast5 x 425 x (1-

415 x Ast5

20 x 230 x 425 )

Ast5 = 973.38 mm2




= 0.138 x 20 x 230 x (425)2 = 114.66 x 106 N-mm = 114.66 KN-m


Consider d'

d =

25

425 = 0.05


161



fsc x (d - d') =

(158.81 - 114.66) x 106

355 x (425 - 25) =310.85 mm

2

To find Ast1:


fck x b x d )

114.66 x 106 = 0.87 x 415 x Ast1 x 425 x (1 - 415 x Ast1

20 x 230 x 425 )

Ast1 = 932 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

355 x 310.85

0.87 x 415 =305.64 mm

2

Ast6 = Ast1 + Ast2 = 932 + 305.64 = 1237.12 mm2


At left support:



b x d =

91.76 x 103

230 x 425 = 0.94 N/mm

2

Ast = 498.67 mm2


b x d =

100 x 498.67

230 x 425 = 0.51

162


Pt c

0.5 0.48

0.51 ?

0.75 0.56

Therefore, c = 0.48 + 0.56-0.48

0.75-0.5 x (0.51-0.5) = 0.483 N/mm

2

Therefore, v > c,


= (91.76 x 103) – (0.483 x 230 x 425)

= 44546.75N


Asv = 2 x π x 82

4 = 100.53mm2

Therefore,


Vus =

0.87 x 415 x 100.53 x 425

44546.75

= 346.29mm >300mm



sv =

0.87 x 415 x 100.53 x 425

300 = 51419.84N


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 300

0.87 x 415 = 76.44mm

2 <100.5mm

2 (safe)

163


At middle support:



b x d =

112.78 x 103

230 x 425 = 1.15 N/mm2

Ast = 1110.4 mm2


b x d =

100 x 1110.4

230 x 425 = 1.14


Pt c

1.00 0.62

1.15 ?

1.25 0.67

Therefore, c = 0.62 + 0.67-0.62

1.25-1 x (1.15-1.00) = 0.65 N/mm

2

Therefore, v > c,


= (112.78 x 103) – (0.65 x 230 x 425)

= 49242.5N


Asv = 2 x π x 8

2

4 = 100.53mm

2

164

Therefore,


Vus =

0.87 x 415 x 100.53 x 425

49242.5

= 313.26mm < 300mm



sv =

0.87 x 415 x 100.53 x 425

300 = 51419.84N


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 300

0.87 x 415 = 76.44mm

2 <100.5mm

2 (safe)


At middle support:



b x d =

144.11 x 103

230 x 425 = 1.47 N/mm

2

Ast = 1598.13 mm2


b x d =

100 x 1598.13

230 x 425 = 1.64


Pt c

1.50 0.72

1.64 ?

1.75 0.75

165

Therefore, c = 0.72 + 0.75-0.72

1.75-1.5 x (1.64 – 1.5) = 0.74 N/mm2

Therefore, v > c,


= (144.11 x 103) – (0.74 x 230 x 425)

= 71775N


Asv = 2 x π x 8

2

4 = 100.53mm

2

Therefore,


Vus =

0.87 x 415 x 100.53 x 425

71775

= 214.92mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 210

0.87 x 415 = 53.51mm2 <100.5mm2 (safe)


At right support:



b x d =

135.28 x 103

230 x 425 = 1.38 N/mm

2

Ast = 1237.12 mm2


b x d =

100 x1237.12

230 x 425 = 1.27

166


Pt c

1.25 0.67

1.27 ?

1.5 0.72

Therefore, c = 0.67 + 0.72-0.67

1.5-1.25 x (1.27-1.25) = 0.674 N/mm

2

Therefore, v > c,


= (135.28 x 103) – (0.674 x 230 x 425)

= 69396.5N


Asv = 2 x π x 82

4 = 100.53mm2

Therefore,


Vus =

0.87 x 415 x 100.53 x 425

69396.5

= 222.28mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 220

0.87 x 415 = 56.06mm

2 <100.5mm

2 (safe)


167


FLOOR (Section: 230mm x 600mm, d’ = 50mm)





= 0.138 x 20 x 230 x (550)2 = 192.027 x 10

6 N-mm

= 192.027 KN-m


To find Ast1:


fck x b x d )

75.53 x 106 = 0.87 x 415 x Ast1 x 550 x (1 - 415 x Ast1

20 x 230 x 550 )

Ast1 = 407.61 mm2

168




+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm


6 + (6 x 120) + 230

= 1650 mm


= 0.36 x 20 x 1590 x 120 x (550 – 0.42 x 120)

= 712.23 KN-m

Mu < Mu’limit

Ast:

71.71 x 106 = 0.87 x 415 x Ast2 x 550 x (1-

415 x Ast2

20 x 230 x 550 )

Ast2 = 385.5 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider d'

d =

50

550 = 0.1


169



fsc x (d - d') =

(228.89 - 192.027) x 106

353 x (550 - 50) =208.86 mm

2

To find Ast1:


fck x b x d )

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 - 415 x Ast1

20 x 230 x 550 )

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 208.86

0.87 x 415 = 201.21 mm

2

Ast3 = Ast1 + Ast2 = 1205.32 + 204.21 = 1409.53 mm2

At support (Ast4):



= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider d'

d =

50

550 = 0.1




fsc x (d - d') =

(393.9 - 192.027) x 106

353 x (550 - 50) =1143.76 mm2

170

To find Ast1:


fck x b x d )

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 - 415 x Ast1

20 x 230 x 550 )

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 1143.76

0.87 x 415 = 1118.26 mm

2

Ast4 = Ast1 + Ast2 = 1205.32 + 1118.26 = 2323.58 mm2




+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 8410 = 5887 mm


6 + (6 x 120) + 230

= 1931.17 mm


= 0.36 x 20 x 1931.7 x 120 x (550 – 0.42 x 120)

= 833.83 KN-m

Mu < Mu’limit

Ast:

247.92 x 106 = 0.87 x 415 x Ast2 x 550 x (1-

415 x Ast2

20 x 230 x 550 )

Ast5 = 1751.94 mm2

171




= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider d'

d =

50

550 = 0.1




fsc x (d - d') =

(315.93 - 192.027) x 106

353 x (550 - 50) =701.76 mm

2

To find Ast1:


fck x b x d )

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -

415 x Ast1

20 x 230 x 550 )

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 701.76

0.87 x 415 = 686.11 mm

2

Ast6 = Ast1 + Ast2 = 1205.32 + 686.11 = 1891.43 mm2

172


At left support:



b x d =

109.8 x 103

230 x 550 = 0.87 N/mm

2

Ast = 407.61 mm2


b x d =

100 x 407.61

230 x 550 = 0.32


Pt c

0.25 0.36

0.32 ?

0.5 0.48

Therefore, c = 0.36 + 0.48-0.36

0.5-0.25 x (0.32-0.25) = 0.4 N/mm2

Therefore, v > c,


= (109.8 x 103) – (0.4 x 230 x 550)

= 59200N

173


Asv = 2 x π x 8

2

4 = 100.53mm

2

Therefore,


Vus =

0.87 x 415 x 100.53 x 550

59200

= 337.21mm >300mm



sv =

0.87 x 415 x 100.53 x 550

300 = 66543.32N


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 300

0.87 x 415 = 76.44mm2 <100.5mm2 (safe)


At middle support:



b x d =

152.94 x 103

230 x 550 = 1.21 N/mm

2

Ast = 1409.33 mm2


b x d =

100 x 1409.33

230 x 550 = 1.12

174


Pt c

1.00 0.62

1.12 ?

1.25 0.67

Therefore, c = 0.62 + 0.67-0.62

1.25-1 x (1.12-1) = 0.65 N/mm

2

Therefore, v > c,


= (152.94 x 103) – (0.65 x 230 x 550)

= 70715N


Asv = 2 x π x 82

4 = 100.53mm2

Therefore,


Vus =

0.87 x 415 x 100.53 x 550

70715

= 282.3mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 280

0.87 x 415 = 71.35mm

2 <100.5mm

2 (safe)


175

At middle support:



b x d =

292.34 x 103

230 x 550 = 2.31 N/mm

2

Ast = 2323.58 mm2


b x d =

100 x 2323.58

230 x 550 = 1.84


Pt c

1.75 0.75

1.84 ?

2.00 0.79

Therefore, c = 0.75 + 0.79-0.75

2-1.75 x (1.84-1.75) = 0.77 N/mm2

Therefore, v > c,


= (292.34 x 103) – (0.77 x 230 x 550)

= 194935N


Asv = 2 x π x 102

4 = 157.07mm2

Therefore,


Vus =

0.87 x 415 x 157.07 x 550

194935

= 160mm < 300mm

176


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 160

0.87 x 415 = 40.77mm2 <157.07mm2 (safe)


At right support:



b x d =

275.05 x 103

230 x 550 = 2.18 N/mm

2

Ast = 1891.43 mm2


b x d =

100 x 1891.43

230 x 550 = 1.5


Pt c

1.5 0.72


Therefore, v > c,


= (275.05 x 103) – (0.72 x 230 x 550)

= 183970N


Asv = 2 x π x 102

4 = 157.07mm2

177

Therefore,


Vus =

0.87 x 415 x 157.07 x 550

183970

= 169.54mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 160

0.87 x 415 = 40.77mm

2 <157.07mm

2 (safe)



178

c. Beam 24-15 and Beam 15-06

179

ROOF – Same as in case of roof 20-11-12

FLOOR:





= 0.138 x 20 x 230 x (550)2 = 192.027 x 10

6 N-mm

= 192.027 KN-m


To find Ast1:


fck x b x d )

95.08 x 106 = 0.87 x 415 x Ast1 x 550 x (1 - 415 x Ast1

20 x 230 x 550 )

Ast1 = 523.81 mm2




+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm

180


6 + (6 x 120) + 230

= 1650 mm


= 0.36 x 20 x 1590 x 120 x (550 – 0.42 x 120)

= 712.23 KN-m

Mu < Mu’limit

Ast:

89.37 x 106 = 0.87 x 415 x Ast2 x 550 x (1-

415 x Ast2

20 x 230 x 550 )

Ast2 = 489.33 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 10

6 N-mm

= 192.027 KN-m


Consider d'

d =

50

550 = 0.1




fsc x (d - d') =

(208.43 - 192.027) x 106

353 x (550 - 50) =92.94 mm

2

To find Ast1:

181


fck x b x d )

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -

415 x Ast1

20 x 230 x 550 )

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 92.94

0.87 x 415 = 90.87 mm2

Ast3 = Ast1 + Ast2 = 1205.32 + 90.87 = 1296.19 mm2

At support (Ast4):



= 0.138 x 20 x 230 x (550)2 = 192.027 x 10

6 N-mm

= 192.027 KN-m


Consider d'

d =

50

550 = 0.1




fsc x (d - d') =

(280.82 - 192.027) x 106

353 x (550 - 50) =503.08 mm

2

To find Ast1:


fck x b x d )

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 - 415 x Ast1

20 x 230 x 550 )

Ast1 = 1205.32 mm2

182

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 503.08

0.87 x 415 = 491.86 mm

2

Ast4 = Ast1 + Ast2 = 1205.32 + 491.86 = 1697.18 mm2




+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 8410 = 5887 mm


6 + (6 x 120) + 230

= 1931.17 mm


= 0.36 x 20 x 1931.7 x 120 x (550 – 0.42 x 120)

= 833.83 KN-m

Mu < Mu’limit

Ast:

166.85 x 106 = 0.87 x 415 x Ast2 x 550 x (1-

415 x Ast2

20 x 230 x 550 )

Ast5 = 1006.35 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 10

6 N-mm

= 192.027 KN-m

183


Consider d'

d =

50

550 = 0.1




fsc x (d - d') =

(211.34 - 192.027) x 106

353 x (550 - 50) =109.42 mm

2

To find Ast1:


fck x b x d )

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -

415 x Ast1

20 x 230 x 550 )

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 109.42

0.87 x 415 = 106.98 mm

2

Ast6 = Ast1 + Ast2 = 1205.32 + 106.98 = 1312.3 mm2


At left support:

184



b x d =

133.66 x 103

230 x 550 = 1.06 N/mm

2

Ast = 523.81 mm2


b x d =

100 x 523.81

230 x 550 = 0.42


Pt c

0.25 0.36

0.42 ?

0.5 0.48

Therefore, c = 0.36 + 0.48-0.36

0.5-0.25 x (0.42-0.25) = 0.44 N/mm

2

Therefore, v > c,


= (133.66 x 103) – (0.44 x 230 x 550)

= 78000N


Asv = 2 x π x 8

2

4 = 100.53mm

2

Therefore,


Vus =

0.87 x 415 x 100.53 x 550

78000

= 255.94mm <300mm


185

Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 250

0.87 x 415 = 63.7mm

2 <100.5mm

2 (safe)


At middle support:



b x d =

166.87 x 103

230 x 550 = 1.32 N/mm

2

Ast = 1296.19 mm2


b x d =

100 x 1296.19

230 x 550 = 1.03


Pt c

1.00 0.62

1.03 ?

1.25 0.67

Therefore, c = 0.62 + 0.67-0.62

1.25-1 x (1.03-1) = 0.63 N/mm

2

Therefore, v > c,


= (166.87 x 103) – (0.63 x 230 x 550)

= 87175N


Asv = 2 x π x 8

2

4 = 100.53mm

2

186

Therefore,


Vus =

0.87 x 415 x 100.53 x 550

87175

= 229mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 220

0.87 x 415 = 56.05mm

2 <100.5mm

2 (safe)


At middle support:



b x d =

200.56 x 103

230 x 550 = 1.59 N/mm2

Ast = 1697.18 mm2


b x d =

100 x 1697.18

230 x 550 = 1.34


Pt c

1.25 0.67

1.34 ?

1.50 0.72

Therefore, c = 0.67 + 0.72-0.67

1.5-1.25 x (1.34-1.25) = 0.69 N/mm2

Therefore, v > c,


187

= (200.56 x 103) – (0.69 x 230 x 550)

= 113275N


Asv = 2 x π x 102

4 = 157.07mm2

Therefore,


Vus =

0.87 x 415 x 157.07 x 550

113275

= 275.35mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 270

0.87 x 415 = 68.8mm2 <157.07mm2 (safe)


At right support:



b x d =

185.35 x 103

230 x 550 = 1.47 N/mm

2

Ast = 1312.3 mm2


b x d =

100 x 1312.3

230 x 550 = 1


Pt c

188

1 0.62


Therefore, v > c,


= (185.35 x 103) – (0.62 x 230 x 550)

= 106920N


Asv = 2 x π x 10

2

4 = 157.07mm

2

Therefore,


Vus =

0.87 x 415 x 157.07 x 550

106920

= 291mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 290

0.87 x 415 = 73.9mm2 <157.07mm2 (safe)



189

d. Beam 25-16 and Beam 16-07

190

ROOF – Same as 20-11-02

191

FLOOR:





= 0.138 x 20 x 230 x (550)2 = 192.027 x 10

6 N-mm

= 192.027 KN-m


To find Ast1:


fck x b x d )

85.69 x 106 = 0.87 x 415 x Ast1 x 550 x (1 - 415 x Ast1

20 x 230 x 550 )

Ast1 = 467.34 mm2




+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm

192


6 + (6 x 120) + 230

= 1650 mm


= 0.36 x 20 x 1590 x 120 x (550 – 0.42 x 120)

= 712.23 KN-m

Mu < Mu’limit

Ast:

80.45 x 106 = 0.87 x 415 x Ast2 x 550 x (1-

415 x Ast2

20 x 230 x 550 )

Ast2 = 436.37 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 10

6 N-mm

= 192.027 KN-m


Consider d'

d =

50

550 = 0.1




fsc x (d - d') =

(218.95 - 192.027) x 106

353 x (550 - 50) =152.54 mm

2

To find Ast1:


fck x b x d )

193

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 - 415 x Ast1

20 x 230 x 550 )

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 152.54

0.87 x 415 = 149.14 mm

2

Ast3 = Ast1 + Ast2 = 1205.32 + 149.14 = 1354.46 mm2

At support (Ast4):



= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


Consider d'

d =

50

550 = 0.1




fsc x (d - d') =

(343.47 - 192.027) x 106

353 x (550 - 50) =858.03 mm2

To find Ast1:


fck x b x d )

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -

415 x Ast1

20 x 230 x 550 )

Ast1 = 1205.32 mm2

194

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 858.03

0.87 x 415 = 838.9 mm

2

Ast4 = Ast1 + Ast2 = 1205.32 + 838.9 = 2044.22 mm2




+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 8410 = 5887 mm


6 + (6 x 120) + 230

= 1931.17 mm


= 0.36 x 20 x 1931.7 x 120 x (550 – 0.42 x 120)

= 833.83 KN-m

Mu < Mu’limit

Ast:

210.5 x 106 = 0.87 x 415 x Ast2 x 550 x (1-

415 x Ast2

20 x 230 x 550 )

Ast5 = 1366.21 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 10

6 N-mm

= 192.027 KN-m

195


Consider d'

d =

50

550 = 0.1




fsc x (d - d') =

(268.55 - 192.027) x 106

353 x (550 - 50) =433.56 mm

2

To find Ast1:


fck x b x d )

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -

415 x Ast1

20 x 230 x 550 )

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 433.56

0.87 x 415 = 423.9 mm

2

Ast6 = Ast1 + Ast2 = 1205.32 + 423.9 = 1629.22 mm2


At left support:


196


b x d =

120.89 x 103

230 x 550 = 0.96 N/mm2

Ast = 467.34 mm2


b x d =

100 x 467.34

230 x 550 = 0.37


Pt c

0.25 0.36

0.37 ?

0.5 0.48

Therefore, c = 0.36 + 0.48-0.36

0.5-0.25 x (0.37-0.25) = 0.42 N/mm

2

Therefore, v > c,


= (120.89 x 103) – (0.42 x 230 x 550)

= 67760N


Asv = 2 x π x 8

2

4 = 100.53mm

2

Therefore,


Vus =

0.87 x 415 x 100.53 x 550

67760

= 294mm <300mm


Asv

b x Sv ≥

0.4

0.87 x fy

197

Asv = 0.4 x 230 x 290

0.87 x 415 = 73.9mm2 <100.5mm2 (safe)


At middle support:



b x d =

158.11 x 103

230 x 550 = 1.25 N/mm

2

Ast = 1354.46 mm2


b x d =

100 x 1354.46

230 x 550 = 1.07


Pt c

1.00 0.62

1.07 ?

1.25 0.67

Therefore, c = 0.62 + 0.67-0.62

1.25-1 x (1.07-1) = 0.634 N/mm2

Therefore, v > c,


= (158.11x 103) – (0.634 x 230 x 550)

= 77909N


Asv = 2 x π x 82

4 = 100.53mm2

198

Therefore,


Vus =

0.87 x 415 x 100.53 x 550

77909

= 256.24mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 250

0.87 x 415 = 63.7mm

2 <100.5mm

2 (safe)


At middle support:



b x d =

251.1 x 103

230 x 550 = 2 N/mm2

Ast = 2044.2 mm2


b x d =

100 x 2044.2

230 x 550 = 1.62


Pt c

1.5 0.72

1.62 ?

1.75 0.75

Therefore, c = 0.72 + 0.75-0.72

1.75-1.5 x (1.62-1.5) = 0.73 N/mm2

Therefore, v > c,

199


= (251.1 x 103) – (0.73 x 230 x 550)

= 158755N


Asv = 2 x π x 10

2

4 = 157.07mm

2

Therefore,


Vus =

0.87 x 415 x 157.07 x 550

158755

= 196.5mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 190

0.87 x 415 = 48.41mm

2 <157.07mm

2 (safe)


At right support:



b x d =

234.86 x 103

230 x 550 = 1.86 N/mm2

Ast = 1629.22 mm2


b x d =

100 x 1629.22

230 x 550 = 1.29


Pt c

1.25 0.67

200

1.29 ?

1.5 0.72

Therefore, c = = 0.67 + 0.72-0.67

1.5-1.25 x (1.29-1.25) = 0.68 N/mm

2

Therefore, v > c,


= (234.86 x 103) – (0.68 x 230 x 550)

= 148840N


Asv = 2 x π x 102

4 = 157.07mm2

Therefore,


Vus =

0.87 x 415 x 157.07 x 550

148840

= 209.56mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 200

0.87 x 415 = 50.96mm2 <157.07mm2 (safe)


201

Reinforcement for roof 25-16-07

e. Beam 26-17 and Beam 17-08

202


203

FLOOR:





= 0.138 x 20 x 230 x (550)2 = 192.027 x 10

6 N-mm

= 192.027 KN-m


To find Ast1:


fck x b x d )

73.1 x 106 = 0.87 x 415 x Ast1 x 550 x (1 - 415 x Ast1

20 x 230 x 550 )

Ast1 = 393.52 mm2




+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm

204


6 + (6 x 120) + 230

= 1650 mm


= 0.36 x 20 x 1590 x 120 x (550 – 0.42 x 120)

= 712.23 KN-m

Mu < Mu’limit

Ast:

68.48 x 106 = 0.87 x 415 x Ast2 x 550 x (1-

415 x Ast2

20 x 230 x 550 )

Ast2 = 366.94 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 10

6 N-mm

= 192.027 KN-m


To find Ast3:


fck x b x d )

185.08 x 106 = 0.87 x 415 x Ast3 x 550 x (1 - 415 x Ast3

20 x 230 x 550 )

Ast3 = 1148.33 mm2

At support (Ast4):


205


= 0.138 x 20 x 230 x (550)2 = 192.027 x 10

6 N-mm

= 192.027 KN-m


Consider d'

d =

50

550 = 0.1




fsc x (d - d') =

(297.26 - 192.027) x 106

353 x (550 - 50) =596.22 mm2

To find Ast1:


fck x b x d )

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -

415 x Ast1

20 x 230 x 550 )

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 596.22

0.87 x 415 = 582.93 mm2

Ast4 = Ast1 + Ast2 = 1205.32 + 582.93 = 1788.25 mm2




+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 8410 = 5887 mm

206


6 + (6 x 120) + 230

= 1931.17 mm


= 0.36 x 20 x 1931.7 x 120 x (550 – 0.42 x 120)

= 833.83 KN-m

Mu < Mu’limit

Ast:

183.9 x 106 = 0.87 x 415 x Ast2 x 550 x (1-

415 x Ast2

20 x 230 x 550 )

Ast5 = 1138.82 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 10

6 N-mm

= 192.027 KN-m


Consider d'

d =

50

550 = 0.1




fsc x (d - d') =

(234.45 - 192.027) x 106

353 x (550 - 50) =240.36 mm

2

To find Ast1:


fck x b x d )

207

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 - 415 x Ast1

20 x 230 x 550 )

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 240.36

0.87 x 415 = 235 mm

2

Ast6 = Ast1 + Ast2 = 1205.32 + 235 = 1440.32 mm2


At left support:



b x d =

101.13 x 103

230 x 550 = 0.8 N/mm

2

Ast = 393.52 mm2


b x d =

100 x 393.52

230 x 550 = 0.31


Pt c

0.25 0.36

0.31 ?

0.5 0.48

208

Therefore, c = 0.36 + 0.48-0.36

0.5-0.25 x (0.31-0.25) = 0.39 N/mm2

Therefore, v > c,


= (101.13 x 103) – (0.39 x 230 x 550)

= 51795N


Asv = 2 x π x 8

2

4 = 100.53mm

2

Therefore,


Vus =

0.87 x 415 x 100.53 x 550

51795

= 385.42mm >300mm



sv =

0.87 x 415 x 100.53 x 550

300 = 66543.32N


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 300

0.87 x 415 = 76.44 mm2 <100.5mm2 (safe)


At middle support:



b x d =

131.2 x 103

230 x 550 = 1.04 N/mm

2

209

Ast = 1148.33 mm2


b x d =

100 x 1148.33

230 x 550 = 0.91


Pt c

0.75 0.56

0.91 ?

1.00 0.62

Therefore, c = 0.56 + 0.62-0.56

1-0.75 x (0.91-0.75) = 0.6 N/mm

2

Therefore, v > c,


= (131.2x 103) – (0.6 x 230 x 550)

= 55300N


Asv = 2 x π x 82

4 = 100.53mm2

Therefore,


Vus =

0.87 x 415 x 100.53 x 550

55300

= 361mm > 300mm



sv =

0.87 x 415 x 100.53 x 550

300 = 66543.32N


210

Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 300

0.87 x 415 = 76.44 mm

2 <100.5mm

2 (safe)


At middle support:



b x d =

218.47 x 103

230 x 550 = 1.73 N/mm

2

Ast = 1788.25 mm2


b x d =

100 x 1788.25

230 x 550 = 1.42


Pt c

1.25 0.67

1.42 ?

1.5 0.72

Therefore, c = 0.67 + 0.72-0.67

1.5-1.25 x (1.42-1.25) = 0.71 N/mm

2

Therefore, v > c,


= (218.47 x 103) – (0.71 x 230 x 550)

= 128655N


Asv = 2 x π x 10

2

4 = 157.07mm

2

211

Therefore,


Vus =

0.87 x 415 x 157.07 x 550

128655

= 242.5mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 240

0.87 x 415 = 61.16mm

2 <157.07mm

2 (safe)


At right support:



b x d =

204.57 x 103

230 x 550 = 1.62 N/mm2

Ast = 1400.32 mm2


b x d =

100 x 1400.32

230 x 550 = 1.11


Pt c

1.00 0.62

1.11 ?

1.25 0.67

Therefore, c = = 0.62 + 0.67-0.62

1.25-1.00 x (1.11-1.00) = 0.64 N/mm2

Therefore, v > c,

212


= (204.57 x 103) – (0.64 x 230 x 550)

= 123610N


Asv = 2 x π x 10

2

4 = 157.07mm

2

Therefore,


Vus =

0.87 x 415 x 157.07 x 550

123610

= 252.33mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 250

0.87 x 415 = 63.7mm

2 <157.07mm

2 (safe)



213

f. Beam 27-18 and Beam 18-09

214


FLOOR:





= 0.138 x 20 x 230 x (550)2 = 192.027 x 106 N-mm

= 192.027 KN-m


To find Ast1:


fck x b x d )

101.22 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -

415 x Ast1

20 x 230 x 550 )

Ast1 = 561.43 mm2




+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 6000 = 4200 mm

215


6 + (6 x 120) + 230

= 1650 mm


= 0.36 x 20 x 1590 x 120 x (550 – 0.42 x 120)

= 712.23 KN-m

Mu < Mu’limit

Ast:

70.3 x 106 = 0.87 x 415 x Ast2 x 550 x (1-

415 x Ast2

20 x 230 x 550 )

Ast2 = 377.38 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 10

6 N-mm

= 192.027 KN-m


To find Ast3:


fck x b x d )

166.06 x 106 = 0.87 x 415 x Ast3 x 550 x (1 - 415 x Ast3

20 x 230 x 550 )

Ast3 = 1000.42 mm2

At support (Ast4):


216


= 0.138 x 20 x 230 x (550)2 = 192.027 x 10

6 N-mm

= 192.027 KN-m


Consider d'

d =

50

550 = 0.1




fsc x (d - d') =

(288.45 - 192.027) x 106

353 x (550 - 50) =546.31 mm2

To find Ast1:


fck x b x d )

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 -

415 x Ast1

20 x 230 x 550 )

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 546.31

0.87 x 415 = 534.13 mm2

Ast4 = Ast1 + Ast2 = 1205.32 + 534.13 = 1739.45 mm2




+ (6 x Df) + bw

l0 = 0.7 x l = 0.7 x 8410 = 5887 mm

217


6 + (6 x 120) + 230

= 1931.17 mm


= 0.36 x 20 x 1931.7 x 120 x (550 – 0.42 x 120)

= 833.83 KN-m

Mu < Mu’limit

Ast:

160.42 x 106 = 0.87 x 415 x Ast2 x 550 x (1-

415 x Ast2

20 x 230 x 550 )

Ast5 = 958.56 mm2




= 0.138 x 20 x 230 x (550)2 = 192.027 x 10

6 N-mm

= 192.027 KN-m


Consider d'

d =

50

550 = 0.1




fsc x (d - d') =

(257.96 - 192.027) x 106

353 x (550 - 50) =373.56 mm

2

To find Ast1:


fck x b x d )

218

192.027 x 106 = 0.87 x 415 x Ast1 x 550 x (1 - 415 x Ast1

20 x 230 x 550 )

Ast1 = 1205.32 mm2

To find Ast2:

Ast2 = fsc x Asc

0.87 x fy =

353 x 373.56

0.87 x 415 = 365.23 mm

2

Ast6 = Ast1 + Ast2 = 1205.32 + 365.23 = 1570.55 mm2


At left support:



b x d =

119.94 x 103

230 x 550 = 0.95 N/mm

2

Ast = 561.43 mm2


b x d =

100 x 561.43

230 x 550 = 0.44


Pt c

0.25 0.36

0.44 ?

0.5 0.48

Therefore, c = 0.36 + 0.48-0.36

0.5-0.25 x (0.44-0.25) = 0.45 N/mm

2

Therefore, v > c,

219


= (119.94 x 103) – (0.45 x 230 x 550)

= 63015N


Asv = 2 x π x 8

2

4 = 100.53mm

2

Therefore,


Vus =

0.87 x 415 x 100.53 x 550

63015

= 316.8mm >300mm



sv =

0.87 x 415 x 100.53 x 550

300 = 66543.32N


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 300

0.87 x 415 = 76.44 mm2 <100.5mm2 (safe)


At middle support:



b x d =

138.8 x 103

230 x 550 = 1.1 N/mm

2

Ast = 1000.42 mm2


b x d =

100 x 1000.42

230 x 550 = 0.79

220


Pt c

0.75 0.56

0.79 ?

1.00 0.62

Therefore, c = 0.56 + 0.62-0.56

1-0.75 x (0.79-0.75) = 0.57 N/mm

2

Therefore, v > c,


= (138.8x 103) – (0.57 x 230 x 550)

= 66695N


Asv = 2 x π x 82

4 = 100.53mm2

Therefore,


Vus =

0.87 x 415 x 100.53 x 550

66695

= 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 300

0.87 x 415 = 76.44 mm

2 <100.5mm

2 (safe)


221

At middle support:



b x d =

208.4 x 103

230 x 550 = 1.65 N/mm

2

Ast = 1739.5mm2


b x d =

100 x 1739.5

230 x 550 = 1.38


Pt c

1.25 0.67

1.38 ?

1.5 0.72

Therefore, c = 0.67 + 0.72-0.67

1.5-1.25 x (1.38-1.25) = 0.7 N/mm2

Therefore, v > c,


= (208.4 x 103) – (0.7 x 230 x 550)

= 119850N


Asv = 2 x π x 102

4 = 157.07mm2

Therefore,


Vus =

0.87 x 415 x 157.07 x 550

119850

= 260.25mm < 300mm

222


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 260

0.87 x 415 = 66.25mm2 <157.07mm2 (safe)


At right support:



b x d =

200.33 x 103

230 x 550 = 1.58 N/mm

2

Ast = 1570.5 mm2


b x d =

100 x 1570.5

230 x 550 = 1.24


Pt c

1.00 0.62

1.24 ?

1.25 0.67

Therefore, c = = 0.62 + 0.67-0.62

1.25-1.00 x (1.24-1.00) = 0.67 N/mm2

Therefore, v > c,


= (200.33 x 103) – (0.67 x 230 x 550)

= 115575N


223

Asv = 2 x π x 102

4 = 157.07mm2

Therefore,


Vus =

0.87 x 415 x 157.07 x 550

115575

= 269.87mm < 300mm


Asv

b x Sv ≥

0.4

0.87 x fy

Asv = 0.4 x 230 x 260

0.87 x 415 = 66.25mm2 <157.07mm2 (safe)



224

5. DESIGN OF COLUMNS

Since the loads and moments in the three columns in a frame are different. Each of the

Column is required to be designed separately. However, when entire building is to be

designed, there will be a number of other columns along with each of the above columns to

form a group.

Since exact values of Pu and Mu are known for all storeys for all columns, the column

section will be designed using exact method using charts and tables. Charts are useful for any

column. It is advisable to have curves plotted of Pu-Mu for standard sections normally used in

building design to avoid calculations.

All the columns are subjected to axial loads and uni-axial bending. They will be

designed to resist Pu and Mu for bending about x- x axis which is the major axis.

For frame 19-10-01:

Columns are C19 at left end, C10 at middle and C01 at right end of the frame.

The moments and axial forces are calculated in analysis of frames and design of

beams. We have transverse frames and in these frames the plinth level transverse beams are

absent and longitudinal beams are present at the plinth level.

In each level the types of loads are:

1. Max shear from transverse beams

2. Shear from longitudinal beams

3. Self weight of columns.

But in plinth level shear from transverse beams is absent. We have only two values at the

plinth level

Section: 230mm x 600mm

Cover (d’) = 50mm

Minimum eccentricity (ex,min) = unsupported length

500 +

lateral dimension

30 or

= 20mm

225

Pu – Load acting on the column

Mux,min = ex,min x Pu

MuD = Maximum (Mu, Mux,min)

fck x b x D = 20 x 230 x 600 = 2760000N = 2760 KN

fck x b x D2 = 20 x 230 x (600)

2 = 1676 x 10

6 N-mm = 1676 KN-m

226

For Column 19:

To find MuD:

Storey Length

(mm)

ex,min (mm) Pu (KN) Mux (KN-

m)

Mux,min

(KN-m)

MuD

Roof – 5 3050 26.1 102.83 64.85 2.68 64.85

5-4 2900 25.8 271.485 53.27 7 53.27

4-3 2900 25.8 440.14 53.27 11.36 53.27

3-2 2900 25.8 608.795 53.27 15.71 53.27

2-1 2900 25.8 777.45 53.27 20.06 53.27

1- Plinth 4900 29.8 946.11 53.27 28.19 53.27

Plinth –

Footing

- 29.8 982.185 0 29.27 29.27

By the help of SP16, chart no:32,

From Pu

fck x b x D and

MuD

fck x b x D2 Plot the P

fck

Where p is the percentage of steel reinforcement

Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.04 0.04 0.015 0.3

5-4 0.1 0.03 0.01 0.2

4-3 0.16 0.03 0.01 0.2

3-2 0.22 0.03 0.01 0.2

2-1 0.28 0.03 0.01 0.2

1- Plinth 0.34 0.03 0.01 0.2

Plinth – Footing 0.36 0.02 0.01 0.2

But from Is 456-2000, P48, Clause 26.5.3.1 (a)

The cross-sectional area of longitudinal reinforcement, shall be not less than

0.8% nor more than 6% of the gross cross-sectional area (b x D) of the column.

227

Therefore, provide P = 0.8%

Ast = 0.8

100 x 230 x 600 = 1104 mm

2

No. of bars of 16mmφ bars = 4 x 1104

π x (16)2 = 5.49 say 6 bars

Therefore provide 6 bars of 16mmφ bars.

Design of lateral ties:

1. Should not be less than 6mm

2. 1

4 x diameter of max main steel bar =

1

4 x 16 = 4mm

Maximum of two values =6mm.

But in the recent construction the usage of 6mm bars is outdated. Therefore consider

minimum 8mm dia bars.

Design of pitch:

1. Least lateral dimension = 230mm

2. 16 x diameter of main steel bar = 16 x 16 = 256

3. 300mm

Minimum of the above values is considered.

Therefore, provide 8mm φ bars @230mm c/c.

No. of bars:

Storey Ast (mm2) No.of bars:

Roof-5 1104 6 bars of 16mm φ

5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ

1-Plinth 1104 6 bars of 16mm φ

Plinth – footing 1104 6 bars of 16mm φ

228

For Column 10:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 3050 26.1 202.61 89.7 5.29 89.70

5-4 2900 25.8 583.35 81.6 15.05 81.60

4-3 2900 25.8 964.09 81.6 24.87 81.60

3-2 2900 25.8 1344.83 81.6 34.70 81.60

2-1 2900 25.8 1725.57 81.6 44.52 81.60

1- Plinth 4900 29.8 2106.31 81.6 62.77 81.60

Plinth –

Footing

-

29.8 2136.1 0 63.66 63.66


From Pu

fck x b x D and

MuD


fck


Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.07 0.05 0.02 0.4

5-4 0.21 0.05 0.02 0.4

4-3 0.35 0.05 0.02 0.4

3-2 0.49 0.05 0.04 0.8

2-1 0.63 0.05 0.08 1.6

1- Plinth 0.76 0.05 0.12 2.4

Plinth – Footing 0.77 0.04 0.12 2.4

But from Is 456-2000, P48, Clause 26.5.3.1 (a)

The cross-sectional area of longitudinal reinforcement, shall be not less than

0.8% nor more than 6% of the gross cross-sectional area (b x D) of the column.

229


Ast = 0.8

100 x 230 x 600 = 1104 mm

2


π x (16)2 = 5.49 say 6 bars


For P = 1.6%

Ast = 1.6

100 x 230 x 600 = 2208 mm

2


π x (20)2 = 7 bars

Therefore provide 8 bars of 20mmφ bars. (Each side 4bars)

For P = 2.4%

Ast = 2.4

100 x 230 x 600 = 3312 mm2


π x (25)2 = 6.75 bars say 8 bars




2. 1


1

4 x 25 = 6.25mm say 8mm


Design of pitch:



3. 300mm

230



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 2208 8 bars of 20mm φ



For Column 01:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 3050 26.1 133.97 136.67 3.50 136.67

5-4 2900 25.8 383.39 122.6 9.89 122.60

4-3 2900 25.8 632.81 122.6 16.33 122.60

3-2 2900 25.8 882.23 122.6 22.76 122.60

2-1 2900 25.8 1131.65 122.6 29.20 122.60

1- Plinth 4900 29.8 1381.07 122.6 41.16 122.60

Plinth –

Footing

-

29.8 1430.66 0 42.63 42.63


From Pu

fck x b x D and

MuD

fck x b x D2 Plot the

P

fck


231

Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.05 0.08 0.02 0.4

5-4 0.14 0.07 0.02 0.4

4-3 0.23 0.07 0.02 0.4

3-2 0.32 0.07 0.04 0.8

2-1 0.41 0.07 0.04 0.8

1- Plinth 0.50 0.07 0.06 1.2

Plinth – Footing 0.52 0.03 0.06 1.2

Provide min P = 0.8%

Ast = 0.8

100 x 230 x 600 = 1104 mm2


π x (16)2 = 5.49 say 6 bars

For P=1.2%

Ast = 1.2

100 x 230 x 600 = 1656 mm

2


π x (20)2 = 5.27 say 6 bars




2. 1


1

4 x 20 = 5mm




Design of pitch:


232


3. 300mm


Therefore, provide 8mm φ bars @230mm c/c

No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ



233

For frame 20-11-02:



Cover (d’) = 50mm


500 +

lateral dimension

30 or

= 20mm

234

fck x b x D = 20 x 230 x 600 = 2760000N = 2760 KN

fck x b x D2 = 20 x 230 x (600)

2 = 1676 x 10

6 N-mm = 1676 KN-m

For Column 20:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 137.01 68.42 3.53 68.42

5-4 2750 25.5 315.59 37.77 8.05 37.77

4-3 2750 25.5 494.17 37.77 12.60 37.77

3-2 2750 25.5 672.75 37.77 17.16 37.77

2-1 2750 25.5 851.33 37.77 21.71 37.77

1- Plinth 4750 29.5 1029.91 37.77 30.38 37.77

Plinth –

Footing

-

29.5 1091.71 0 32.21 32.21

By the help of SP16, chart no: 32,

From Pu

fck x b x D and

MuD


P

fck


Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.05 0.04 0.01 0.2

5-4 0.11 0.02 0.01 0.2

4-3 0.18 0.02 0.01 0.2

3-2 0.24 0.02 0.01 0.2

2-1 0.31 0.02 0.01 0.2

1- Plinth 0.37 0.02 0.01 0.2

Plinth – Footing 0.40 0.02 0.01 0.2

235


Ast = 0.8

100 x 230 x 600 = 1104 mm

2


π x (16)2 = 5.49 say 6 bars




2. 1


1

4 x 16 = 4mm




Design of pitch:



3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ



236

For Column 11:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 279.26 93.54 7.20 93.54

5-4 2750 25.5 806.49 108.74 20.57 108.74

4-3 2750 25.5 1307.98 108.74 33.35 108.74

3-2 2750 25.5 1809.47 108.74 46.14 108.74

2-1 2750 25.5 2310.96 108.74 58.93 108.74

1- Plinth 4750 29.5 2812.45 108.74 82.97 108.74

Plinth –

Footing

- 29.5 2861.68 0 84.42 84.42


From Pu

fck x b x D and

MuD


fck


Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.10 0.06 0.02 0.4

5-4 0.29 0.07 0.02 0.4

4-3 0.47 0.07 0.06 1.2

3-2 0.66 0.07 0.1 2

2-1 0.84 0.07 0.16 3.2

1- Plinth 1.02 0.07 0.2 4

Plinth – Footing 1.04 0.05 0.2 4

For P less than 0.8, Provide P = 0.8%

Ast = 0.8

100 x 230 x 600 = 1104 mm2

237


π x (16)2 = 5.49 say 6 bars


For P = 1.2%

Ast = 1.2

100 x 230 x 600 = 1656 mm

2


π x (20)2 = 5.27, say 6 bars


For P = 2%

Ast = 2

100 x 230 x 600 = 2760 mm

2




For P = 3.2%

Ast = 3.2

100 x 230 x 600 = 4416 mm2


π x (25)2 = 9 bars


For P = 4%

Ast = 4

100 x 230 x 600 = 5520 mm2


π x (25)2 = 11.24 say 12 bars


238



2. 1


1

4 x 25 = 6.25mm say 8mm


Design of pitch:


2. 16 x diameter of main steel bar = 16 x 25 = 400 mm

3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1656 6 bars of 20mm φ

3-2 2760 6 bars of 25mm φ

2-1 4416 10 bars of 25mm φ



239

For Column 02:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 180.53 155.8 4.66 155.80

5-4 2750 25.5 551.39 153.37 14.06 153.37

4-3 2750 25.5 922.25 153.37 23.52 153.37

3-2 2750 25.5 1293.11 153.37 32.97 153.37

2-1 2750 25.5 1663.97 153.37 42.43 153.37

1- Plinth 4750 29.5 2034.83 153.37 60.03 153.37

Plinth –

Footing

-

29.5 2123.66 0 62.65 62.65


From Pu

fck x b x D and

MuD


P

fck


Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.07 0.09 0.04 0.8

5-4 0.20 0.09 0.04 0.8

4-3 0.33 0.09 0.06 1.2

3-2 0.47 0.09 0.08 1.6

2-1 0.60 0.09 0.1 2

1- Plinth 0.74 0.09 0.14 2.8

Plinth – Footing 0.77 0.04 0.16 3.2

For P = 0.8%

Ast = 0.8

100 x 230 x 600 = 1104 mm

2

240


π x (16)2 = 5.49 say 6 bars


For P=1.2%

Ast = 1.2

100 x 230 x 600 = 1656 mm

2


π x (20)2 = 5.27 say 6 bars


For P=1.6%

Ast = 1.6

100 x 230 x 600 = 2208 mm

2


π x (25)2 = 4.5 say 6 bars


For P = 2%

Ast = 2

100 x 230 x 600 = 2760 mm2




For P = 2.8%

Ast = 2.8

100 x 230 x 600 = 3864 mm2




241

For P = 3.2%

Ast = 3.2

100 x 230 x 600 = 4416 mm

2


π x (25)2 = 9 bars




2. 1


1

4 x 25 = 6.25mm say 8mm


Design of pitch:


2. 16 x diameter of main steel bar = 16 x 25 = 400mm

3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1656 6 bars of 20mm φ

3-2 2208 6 bars of 25mm φ

2-1 2760 6 bars of 25mm φ



242

For frame 24-15-06:



Cover (d’) = 50mm


500 +

lateral dimension

30 or

= 20mm

243

fck x b x D = 20 x 230 x 600 = 2760000N = 2760 KN

fck x b x D2 = 20 x 230 x (600)

2 = 1676 x 10

6 N-mm = 1676 KN-m

For Column 24:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 137.01 68.42 3.53 68.42

5-4 2750 25.5 339.45 47.54 8.66 47.54

4-3 2750 25.5 541.89 47.54 13.82 47.54

3-2 2750 25.5 744.33 47.54 18.98 47.54

2-1 2750 25.5 946.77 47.54 24.14 47.54

1- Plinth 4750 29.5 1149.21 47.54 33.90 47.54

Plinth –

Footing

-

29.5 1211.01 0 35.72 35.72


From Pu

fck x b x D and

MuD


P

fck


Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.05 0.04 0.01 0.2

5-4 0.12 0.03 0.01 0.2

4-3 0.20 0.03 0.01 0.2

3-2 0.27 0.03 0.01 0.2

2-1 0.34 0.03 0.01 0.2

1- Plinth 0.42 0.03 0.01 0.2

Plinth – Footing 0.44 0.02 0.01 0.2

244


Ast = 0.8

100 x 230 x 600 = 1104 mm

2


π x (16)2 = 5.49 say 6 bars




2. 1


1

4 x 16 = 4mm




Design of pitch:



3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ



245

For Column 15:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 279.26 93.54 7.20 93.54

5-4 2750 25.5 702.9 63.86 17.92 63.86

4-3 2750 25.5 1126.54 63.86 28.73 63.86

3-2 2750 25.5 1550.18 63.86 39.53 63.86

2-1 2750 25.5 1973.82 63.86 50.33 63.86

1- Plinth 4750 29.5 2397.46 63.86 70.73 70.73

Plinth –

Footing

- 29.5 2446.69 0 72.18 72.18


From Pu

fck x b x D and

MuD


fck


Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.10 0.06 0.01 0.2

5-4 0.25 0.04 0.01 0.2

4-3 0.41 0.04 0.02 0.4

3-2 0.56 0.04 0.06 1.2

2-1 0.72 0.04 0.1 2

1- Plinth 0.87 0.04 0.16 3.2

Plinth – Footing 0.89 0.04 0.16 3.2


Ast = 0.8

100 x 230 x 600 = 1104 mm2

246


π x (16)2 = 5.49 say 6 bars


For P = 1.2%

Ast = 1.2

100 x 230 x 600 = 1656 mm

2


π x (20)2 = 5.27, say 6 bars


For P = 2%

Ast = 2

100 x 230 x 600 = 2760 mm

2




For P = 3.2%

Ast = 3.2

100 x 230 x 600 = 4416 mm2


π x (25)2 = 9 bars




2. 1


1

4 x 25 = 6.25mm say 8mm


247

Design of pitch:



3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1656 6 bars of 20mm φ

2-1 2760 6 bars of 25mm φ



For Column 06:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 180.53 155.8 4.66 155.80

5-4 2750 25.5 461.69 103.08 11.77 103.08

4-3 2750 25.5 742.85 103.08 18.94 103.08

3-2 2750 25.5 1024.01 103.08 26.11 103.08

2-1 2750 25.5 1305.17 103.08 33.28 103.08

1- Plinth 4750 29.5 1586.33 103.08 46.80 103.08

Plinth –

Footing

-

29.5 1675.16 0 49.42 49.42

248


From Pu

fck x b x D and

MuD


P

fck


Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.07 0.09 0.04 0.8

5-4 0.17 0.06 0.02 0.4

4-3 0.27 0.06 0.02 0.4

3-2 0.37 0.06 0.02 0.4

2-1 0.47 0.06 0.04 0.8

1- Plinth 0.57 0.06 0.08 1.6

Plinth – Footing 0.61 0.03 0.08 1.6

For P = 0.8% and P <0.8%

Ast = 0.8

100 x 230 x 600 = 1104 mm

2


π x (16)2 = 5.49 say 6 bars


For P=1.6%

Ast = 1.6

100 x 230 x 600 = 2208 mm

2


π x (25)2 = 4.5 say 6 bars




2. 1


1

4 x 25 = 6.25mm say 8mm

249


Design of pitch:



3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ



250

For frame 25-16-07:



Cover (d’) = 50mm


500 +

lateral dimension

30 or

= 20mm

251

fck x b x D = 20 x 230 x 600 = 2760000N = 2760 KN

fck x b x D2 = 20 x 230 x (600)

2 = 1676 x 10

6 N-mm = 1676 KN-m

For Column 25:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 137.01 68.42 3.53 68.42

5-4 2750 25.5 326.68 42.84 8.33 42.84

4-3 2750 25.5 516.35 42.84 13.17 42.84

3-2 2750 25.5 706.02 42.84 18.00 42.84

2-1 2750 25.5 895.69 42.84 22.84 42.84

1- Plinth 4750 29.5 1085.36 42.84 32.02 42.84

Plinth –

Footing

-

29.5 1147.16 0 33.84 33.84


From Pu

fck x b x D and

MuD


P

fck


Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.05 0.04 0.01 0.2

5-4 0.12 0.03 0.01 0.2

4-3 0.19 0.03 0.01 0.2

3-2 0.26 0.03 0.01 0.2

2-1 0.32 0.03 0.01 0.2

1- Plinth 0.39 0.03 0.02 0.4

Plinth – Footing 0.42 0.02 0.02 0.4

252

For P < 0.8%, Provide P = 0.8%

Ast = 0.8

100 x 230 x 600 = 1104 mm

2


π x (16)2 = 5.49 say 6 bars




2. 1


1

4 x 16 = 4mm




Design of pitch:



3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ



253

For Column 16:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 279.26 68.42 7.20 68.42

5-4 2750 25.5 744.68 42.84 18.99 42.84

4-3 2750 25.5 1210.1 42.84 30.86 42.84

3-2 2750 25.5 1675.52 42.84 42.73 42.84

2-1 2750 25.5 2140.94 42.84 54.59 54.59

1- Plinth 4750 29.5 2606.36 42.84 76.89 76.89

Plinth –

Footing

- 29.5 2655.59 0 78.34 78.34


From Pu

fck x b x D and

MuD


fck


Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.10 0.04 0.01 0.2

5-4 0.27 0.03 0.01 0.2

4-3 0.44 0.03 0.02 0.4

3-2 0.61 0.03 0.08 1.6

2-1 0.78 0.03 0.12 2.4

1- Plinth 0.94 0.05 0.18 3.6

Plinth – Footing 0.96 0.05 0.18 3.6


Ast = 0.8

100 x 230 x 600 = 1104 mm2

254


π x (16)2 = 5.49 say 6 bars


For P = 1.6%

Ast = 1.6

100 x 230 x 600 = 2208 mm

2


π x (20)2 = 4.5 say 6 bars


For P = 2.4%

Ast = 2.4

100 x 230 x 600 = 3312 mm

2




For P = 3.6%

Ast = 3.6

100 x 230 x 600 = 4968 mm2


π x (25)2 = 10.12, say 12 bars




2. 1


1

4 x 25 = 6.25mm say 8mm


255

Design of pitch:



3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 2208 6 bars of 25mm φ

2-1 3312 8 bars of 25mm φ



For Column 07:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 180.53 155.8 4.66 155.80

5-4 2750 25.5 511.2 130.75 13.04 130.75

4-3 2750 25.5 841.87 130.75 21.47 130.75

3-2 2750 25.5 1172.54 130.75 29.90 130.75

2-1 2750 25.5 1503.21 130.75 38.33 130.75

1- Plinth 4750 29.5 1833.88 130.75 54.10 130.75

Plinth –

Footing

-

29.5 1922.71 0 56.72 56.72

256


From Pu

fck x b x D and

MuD


P

fck


Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.07 0.09 0.04 0.8

5-4 0.19 0.08 0.02 0.4

4-3 0.31 0.08 0.04 0.8

3-2 0.42 0.08 0.06 1.2

2-1 0.54 0.08 0.08 1.6

1- Plinth 0.66 0.08 0.12 2.4

Plinth – Footing 0.70 0.03 0.12 2.4

For P < 0.8%, Provide P = 0.8%

Ast = 0.8

100 x 230 x 600 = 1104 mm

2


π x (16)2 = 5.49 say 6 bars


For P=1.2%

Ast = 1.2

100 x 230 x 600 = 1656 mm

2


π x (20)2 = 5.27 say 6 bars


For P=1.6%

Ast = 1.6

100 x 230 x 600 = 2208 mm

2

257


π x (25)2 = 4.5 say 6 bars


For P = 2.4%

Ast = 2.4

100 x 230 x 600 = 3312 mm

2






2. 1


1

4 x 25 = 6.25mm say 8mm


Design of pitch:



3. 300mm



258

No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1656 6 bars of 20mm φ

2-1 2208 6 bars of 25mm φ



259

For frame 26-17-08:



Cover (d’) = 50mm


500 +

lateral dimension

30 or

= 20mm

260

fck x b x D = 20 x 230 x 600 = 2760000N = 2760 KN

fck x b x D2 = 20 x 230 x (600)

2 = 1676 x 10

6 N-mm = 1676 KN-m

For Column 26:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 137.01 68.42 3.53 68.42

5-4 2750 25.5 306.92 36.55 7.83 36.55

4-3 2750 25.5 476.83 36.55 12.16 36.55

3-2 2750 25.5 646.74 36.55 16.49 36.55

2-1 2750 25.5 816.65 36.55 20.82 36.55

1- Plinth 4750 29.5 986.56 36.55 29.10 36.55

Plinth –

Footing

-

29.5 1048.36 0 30.93 30.93


From Pu

fck x b x D and

MuD


P

fck


Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.05 0.04 0.01 0.2

5-4 0.11 0.02 0.01 0.2

4-3 0.17 0.02 0.01 0.2

3-2 0.23 0.02 0.01 0.2

2-1 0.30 0.02 0.01 0.2

1- Plinth 0.36 0.02 0.01 0.2

Plinth – Footing 0.38 0.02 0.01 0.2

261

For P < 0.8%, provide P = 0.8%

Ast = 0.8

100 x 230 x 600 = 1104 mm

2


π x (16)2 = 5.49 say 6 bars




2. 1


1

4 x 16 = 4mm




Design of pitch:



3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ



262

For Column 17:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 279.26 68.42 7.20 68.42

5-4 2750 25.5 685.14 36.55 17.47 36.55

4-3 2750 25.5 1091.02 36.55 27.82 36.55

3-2 2750 25.5 1496.9 36.55 38.17 38.17

2-1 2750 25.5 1902.78 36.55 48.52 48.52

1- Plinth 4750 29.5 2308.66 36.55 68.11 68.11

Plinth –

Footing

- 29.5 2357.89 0 69.56 69.56


From Pu

fck x b x D and

MuD


fck


Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.10 0.04 0.01 0.2

5-4 0.25 0.02 0.01 0.2

4-3 0.40 0.02 0.01 0.2

3-2 0.54 0.02 0.04 0.8

2-1 0.69 0.03 0.1 2

1- Plinth 0.84 0.04 0.14 2.8

Plinth – Footing 0.85 0.04 0.14 2.8


Ast = 0.8

100 x 230 x 600 = 1104 mm2

263


π x (16)2 = 5.49 say 6 bars


For P = 2%

Ast = 2

100 x 230 x 600 = 2760 mm

2




For P = 2.8%

Ast = 2.8

100 x 230 x 600 = 3864 mm

2


π x (25)2 = 7.87, say 8 bars




2. 1


1

4 x 25 = 6.25mm say 8mm


Design of pitch:



3. 300mm



264

No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 2760 6 bars of 25mm φ



For Column 08:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 2900 25.8 180.53 155.8 4.66 155.80

5-4 2750 25.5 480.91 113.98 12.26 113.98

4-3 2750 25.5 781.29 113.98 19.92 113.98

3-2 2750 25.5 1081.67 113.98 27.58 113.98

2-1 2750 25.5 1382.05 113.98 35.24 113.98

1- Plinth 4750 29.5 1682.43 113.98 49.63 113.98

Plinth –

Footing

-

29.5 1771.26 0 52.25 52.25


From Pu

fck x b x D and

MuD


fck


265

Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.07 0.09 0.04 0.8

5-4 0.17 0.07 0.02 0.4

4-3 0.28 0.07 0.02 0.4

3-2 0.39 0.07 0.04 0.8

2-1 0.50 0.07 0.06 1.2

1- Plinth 0.61 0.07 0.1 2

Plinth – Footing 0.64 0.03 0.1 2

For P = 0.8%

Ast = 0.8

100 x 230 x 600 = 1104 mm2


π x (16)2 = 5.49 say 6 bars


For P=1.2%

Ast = 1.2

100 x 230 x 600 = 1656 mm2


π x (20)2 = 5.27 say 6 bars


For P = 2%

Ast = 2

100 x 230 x 600 = 2760 mm2






266

2. 1


1

4 x 25 = 6.25mm say 8mm


Design of pitch:



3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1656 6 bars of 20mm φ



267

For frame 27-18-09:



Cover (d’) = 50mm


500 +

lateral dimension

30 or

= 20mm

268

fck x b x D = 20 x 230 x 600 = 2760000N = 2760 KN

fck x b x D2 = 20 x 230 x (600)

2 = 1676 x 10

6 N-mm = 1676 KN-m

For Column 27:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 3050 26.1 102.83 64.85 2.68 64.85

5-4 2900 25.8 265.825 50.61 6.86 50.61

4-3 2900 25.8 428.82 50.61 11.06 50.61

3-2 2900 25.8 591.815 50.61 15.27 50.61

2-1 2900 25.8 754.81 50.61 19.47 50.61

1- Plinth 4900 29.8 917.805 50.61 27.35 50.61

Plinth –

Footing

-

29.8 953.88 0 28.43 28.43


From Pu

fck x b x D and

MuD


P

fck


Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.04 0.04 0.01 0.2

5-4 0.10 0.03 0.01 0.2

4-3 0.16 0.03 0.01 0.2

3-2 0.21 0.03 0.01 0.2

2-1 0.27 0.03 0.01 0.2

1- Plinth 0.33 0.03 0.01 0.2

Plinth – Footing 0.35 0.02 0.01 0.2

269

For P < 0.8, provide P = 0.8%

Ast = 0.8

100 x 230 x 600 = 1104 mm

2


π x (16)2 = 5.49 say 6 bars




2. 1


1

4 x 16 = 4mm




Design of pitch:



3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ



270

For Column 18:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 3050 26.1 202.61 89.7 5.29 89.70

5-4 2900 25.8 586.58 86.44 15.13 86.44

4-3 2900 25.8 970.55 86.44 25.04 86.44

3-2 2900 25.8 1354.52 86.44 34.95 86.44

2-1 2900 25.8 1738.49 86.44 44.85 86.44

1- Plinth 4900 29.8 2122.46 86.44 63.25 86.44

Plinth –

Footing

- 29.8 2152.25 0 64.14 64.14


From Pu

fck x b x D and

MuD


fck


Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.07 0.05 0.01 0.2

5-4 0.21 0.05 0.01 0.2

4-3 0.35 0.05 0.02 0.4

3-2 0.49 0.05 0.04 0.8

2-1 0.63 0.05 0.08 1.6

1- Plinth 0.77 0.05 0.14 2.8

Plinth – Footing 0.78 0.04 0.14 2.8


Ast = 0.8

100 x 230 x 600 = 1104 mm2

271


π x (16)2 = 5.49 say 6 bars


For P = 1.6%

Ast = 1.6

100 x 230 x 600 = 2208 mm

2


π x (20)2 = 8 bars


For P = 2.8%

Ast = 2.8

100 x 230 x 600 = 3864 mm

2






2. 1


1

4 x 25 = 6.25mm say 8mm


Design of pitch:



3. 300mm



272

No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 2208 8 bars of 20mm φ



For Column 09:

To find MuD:

Storey Length

(mm)


m)

Mux,min

(KN-m)

MuD

Roof – 5 3050 26.1 133.97 136.67 3.50 136.67

5-4 2900 25.8 390.87 128.22 10.08 128.22

4-3 2900 25.8 647.77 128.22 16.71 128.22

3-2 2900 25.8 904.67 128.22 23.34 128.22

2-1 2900 25.8 1161.57 128.22 29.97 128.22

1- Plinth 4900 29.8 1418.47 128.22 42.27 128.22

Plinth –

Footing

-

29.8 1468.06 0 43.75 43.75


From Pu

fck x b x D and

MuD


P

fck


273

Storey Pu

fck x b x D

MuD

fck x b x D2

P

fck From chart

P (%)

Roof – 5 0.05 0.08 0.04 0.8

5-4 0.14 0.08 0.02 0.4

4-3 0.23 0.08 0.02 0.4

3-2 0.33 0.08 0.04 0.8

2-1 0.42 0.08 0.04 0.8

1- Plinth 0.51 0.08 0.06 1.2

Plinth – Footing 0.53 0.03 0.06 1.2

Provide min P = 0.8%

Ast = 0.8

100 x 230 x 600 = 1104 mm2


π x (16)2 = 5.49 say 6 bars

For P=1.2%

Ast = 1.2

100 x 230 x 600 = 1656 mm

2


π x (20)2 = 5.27 say 6 bars




2. 1


1

4 x 20 = 5mm




Design of pitch:


274


3. 300mm



No. of bars:



5-4 1104 6 bars of 16mm φ

4-3 1104 6 bars of 16mm φ

3-2 1104 6 bars of 16mm φ

2-1 1104 6 bars of 16mm φ



275

6. DESIGN OF FOOTINGS

In the design of footing, the loads are known from the column analysis. The working load is

used for the footing design. Footing is a member through which the load of the superstructure

is transferred to the sub soil. Therefore the safe bearing capacity is the main factor in design

of footings. From the data provided by site engineer the Safe Bearing Capacity (SBC) of the

soil is 250 KN/m2. Therefore the footing is isolated rectangular sloped footing. The slope is

provided to decrease the amount of concrete in the construction which results into an

economic construction.

From the analysis of frames and columns, we considered 6 frames and each frame

consists of 3 footings so totally 18 footings are to be designed. But in this project we

designed the most critical 3 footings based on the load they carried are selected . the design

of remaining footings follow the same.

Footing F19:

Steel : Fe415

Concrete : M20

Column : 230mm x 600mm (bmm x Dmm)

Load from column (Pu) : 982.185 KN

Working load (P) : 982.185

1.5 = 654.79 KN

Self weight of footing : 10% of Pu

SBC of soil : 250 KN/m2

Area of footing (Af) : 1.1 x P

SBC of soil =

1.1 x 654.79

250 = 2.88 m

2

Consider, D-b

2 =

600 - 230

2 = 185 mm

Length of footing (Lf) = 185mm + (√(185)2 + 2.88 x 10

6 ) = 1892.11

say 1900mm

276

Breadth of footing (Bf) = Af

Lf =

2.88 x 106

1900 = 1515.79 say 1550 mm

Area of footing provided = Lf x Bf = 2000 x 1520 = 2.95 x 106 = 2.95 m

2

Wu = Pu

Af =

982.185

2.95 = 332.95 KN/m2

X1 = Lf - D

2 =

1900 - 600

2 = 650 mm

Y1 = Bf - b

2 =

1550 - 230

2 = 660 mm

Depth of footing for 2-way bending:

Mux = Wu x Bf x X1

2

2 =

332.94 x 1.550 x (0.65)2

2 = 109.02 KN-m

Muy = Wu x Lf x Y1

2

2 =

332.94 x 1.9 x (0.66)2

2 = 137.78 KN-m

Therefore, max bending moment = Mu = 137.78 KN-m

277

b' = b + 2 x e = 230 + 2 x 50 = 330 mm

D' = D + 2 x e = 600 + 2 x 50 = 700 mm

Clear cover = 50mm

Assume 10mm bars

d'x = 50 + 10

2 = 55mm

d'y = 50 + (3 x 10

2 )= 65mm

278

Depth of foundation:

(Df)x = + d'x = √109.02 x 10

6

0.138 x 20 x 330 + 55 = 400.97 mm

(Df)y = + d'y = √137.78 x 10

6

0.138 x 20 x 700 + 65 = 332.05 mm

Therefore, maximum (Df) = 400.97 mm say 450mm

Therefore dx = 450 – 55 = 395mm

dy = 450 – 65 = 385mm

Provide Df,min = 150mm

df,min ( x-direction) = 150 - 55 = 95mm

df,min ( y-direction) = 150 - 65 = 85mm

279

280

Check for two way shear:

Consider a section ‘ dy

2 ’ from the face of the column.

L2 = D + dy

2 +

dy

2 = 600 + 385 = 985 mm

B2 = b + dy

2 +

dy

2 = 230 + 385 = 615 mm

d2 = Df – (Df – Df,min) x

(dy

2 - 50)

(X1 - 50) – d'y

= 450 – (450 – 150) x

(385

2 - 50)

(650 - 50) – 65 = 313.75 mm

Area of resisting shear (A2) = 2 x(L2 + B2) x d2

= 2 x (985 + 615) x 313.75

= 1004000 mm2

Shear strength of concrete for two way:

τuc2 = ks x τuc

where, τuc = 0.25 x √fck = 0.25 x √20 = 1.12 N/mm2

ks = (0.5 + βc) < 1 else =1

βc = b

D =

230

600 = 0.385

therefore, ks = (0.5 + βc) = 0.5 + 0.385 = 0.885 <1

ks = 0.885

τuc2 = ks x τuc = 0.885 x 1.12 = 0.99 N/mm2

Internal load carrying capacity (Vuc2) = 1004000 x 0.99 = 993960 N

= 993.96 KN

281

External load (VuD2) = Wu x (Lf x Bf – L2 x B2)

= 332.94 x (1.900 x 1.550 – 0.985 x 0.615)

= 778.82 KN

Therefore, Vuc2 > VuD2 (SAFE)

Design of Steel mesh:

Mux = 0.87 x fy x Astx x dx x (1-fy x Astx

fck x b' x dx )

109.02 x 106 = 0.87 x 415 x Astx x 395 x (1-

415 x Astx

20 x 330 x 395 )

Astx = 890.72 mm2

No. of 10mmφ bars = 4 x 890.74

π x 102 = 11.34 say 12 bars

Muy = 0.87 x fy x Asty x dy x (1-fy x Astx

fck x D' x dy )

137.78 x 106 = 0.87 x 415 x Astx x 385 x (1-

415 x Astx

20 x 700 x 385 )

Asty = 1081.2 mm2

No. of 10mmφ bars = 4 x 1081.2

π x 102 = 13.77 say 14 bars

Check for one way shear:

Consider a section at distance‘d’ from the face of column

d1 = Df – (Df – Df,min) x (dy- 50)

(Y1 - 50) – d'y

= 450 – (450 – 150) x (385- 50)

(660 - 50) – 65 = 220.24 mm

L1 = D + 2 x dy = 600 + 2 x 385 = 1370 mm

282

Area resisting shear (A1) = (Lf x Dy,min) + (d1 – Dy,min) x ( Lf + L1

2 )

= (1900 x 150) + (220.24 – 150) x ( 1900 + 1370

2 )

= 399842.4 mm2

Asty = 14 bars of 10mm dia = 1099.56 mm2

Pt = 100 x Asty

A1 =

100 x 1099.56

399842.4 = 0.28

From IS456-2000, P73, Table 19

P τc (N/mm2)

0.25 0.36

0.28 ?

0.50 0.48

τc = 0.36 + (0.48 - 0.36)

(0.5 - 0.25) x (0.28 – 0.25) = 0.38 N/mm

2

Internal shear resisting capacity = 0.38 x 399842.4

1000 = 151.94 KN

External shear = Wu x Lf x (Y1 – dy)

= 332.94 x 1.900 x (0.660 – 385)

= 173.96 KN

Internal shear resisting capacity < External Shear (NOT SAFE)

Increase the depth.

Shear = load

area

0.38 = 173.96 x 10

3

A1

283

A1 =457789.47 mm2

To calculate d1:

A1 = (Lf x Dy,min) + (d1 – Dy,min) x ( Lf + L1

2 )

457789.47 = (1900 x 150) + (d1 – 150) x ( 1900 + 1370

2 )

d1 = 255.68 mm

Footing F11:



1.5 = 1907.79 KN




SBC of soil =

1.1 x 1907.79

250 = 8.4 m

2

Consider, D-b

2 =

600 - 230

2 = 185 mm

Length of footing (Lf) = 185mm + (√(185)2 + 8.4 x 10

6 ) = 3089.17

say 3100mm


Lf =

8.4 x 106

3100 = 2709.67 say 2725 mm


2

Wu = Pu

Af =

2861.68

8.45 = 338.66 KN/m

2

X1 = Lf - D

2 =

3100 - 600

2 = 1250 mm

Y1 = Bf - b

2 =

2725 - 230

2 = 1247.5 mm

284


Mux = Wu x Bf x X1

2

2 =

338.66 x 2.725 x (1.25)2

2 = 720.98 KN-m

Muy = Wu x Lf x Y1

2

2 =

338.66 x 3.1 x (1.2475)2

2 = 816.92 KN-m


b' = b + 2 x e = 230 + 2 x 50 = 330 mm

D' = D + 2 x e = 600 + 2 x 50 = 700 mm

Clear cover = 50mm

Assume 16mm bars

d'x = 50 + 16

2 = 58 mm

d'y = 50 + (3 x16

2 ) = 74 mm


(Df)x = + d'x = √720.98 x 10

6

0.138 x 20 x 330 + 58 = 947.71 mm

(Df)y = + d'y = √816.92 x 10

6

0.138 x 20 x 700 + 74 = 724.26 mm


Therefore dx = 1000 – 58 = 942 mm

dy = 1000 – 74 = 926 mm


df,min ( x-direction) = 150 - 58 = 92 mm

df,min ( y-direction) = 150 - 74 = 76 mm

285




L2 = D + dy

2 +

dy

2 = 600 + 926 = 1526 mm

B2 = b + dy

2 +

dy

2 = 230 + 926 = 1156 mm


(dy

2 - 50)

(X1 - 50) – d'y

= 1000 – (1000 – 150) x

(926

2 - 50)

(1250 - 50) – 74 = 633.45 mm


= 2 x (1526 + 1156) x 633.45

= 3.4 x 106 mm2


τuc2 = ks x τuc


ks = (0.5 + βc) < 1 else =1

βc = b

D =

230

600 = 0.385

Therefore, ks = (0.5 + βc) = 0.5 + 0.385 = 0.885 <1

ks = 0.885

τuc2 = ks x τuc = 0.885 x 1.12 = 0.99 N/mm2

286

Internal load carrying capacity (Vuc2) = 3.4 x 106 x 0.99 = 3.37 x 106 N

= 3370 KN


= 338.66 x (3.1 x 2.725 – 1.526 x 1.156)

= 2263.41 KN




fck x b' x dx )

720.98 x 106 = 0.87 x 415 x Astx x 942 x (1-

415 x Astx

20 x 330 x 942 )

Astx = 2555.91 mm2

No. of 16mmφ bars = 4 x 2555.91

π x 162 = 12.71 say 13 bars


fck x D' x dy )

816.92 x 106 = 0.87 x 415 x Astx x 926 x (1-

415 x Astx

20 x 700 x 926 )

Asty = 2671.98 mm2

No. of 16mmφ bars = 4 x 2671.98

π x 162 = 13.29 say 14 bars



d1 = Df – (Df – Df,min) x (dy- 50)

(Y1 - 50) – d'y

= 1000 – (1000 – 150) x (926- 50)

(1247.5 - 50) – 74 = 304.21 mm

287

L1 = D + 2 x dy = 600 + 2 x 926 = 2452 mm


2 )

= (3100 x 150) + (304.21 – 150) x ( 3100 + 2452

2 )

= 893086 mm2


Pt = 100 x Asty

A1 =

100 x 2814.87

893086 = 0.32


P τc (N/mm2)

0.25 0.36

0.32 ?

0.5 0.48

τc = 0.36 + (0.48 - 0.36)

(0.5 - 0.25) x (0.32 – 0.25) = 0.4 N/mm

2

Internal shear resisting capacity = 0.4 x 893086

1000 = 357.23 KN


= 338.66 x 3.1 x (1.2475 – 0.926)

= 337.53 KN

Internal shear resisting capacity > External Shear (SAFE)

Footing F07:



1.5 = 1281.81 KN

288




SBC of soil =

1.1 x 1281.81

250 = 5.64 m

2

Consider, D-b

2 =

600 - 230

2 = 185 mm

Length of footing (Lf) = 185mm + (√(185)2 + 5.64 x 106 ) = 2567.06

say 2575mm


Lf =

5.64 x 106

2575 = 2190.29 say 2200 mm


2

Wu = Pu

Af =

1922.71

5.665 = 339.4 KN/m2

X1 = Lf - D

2 =

2575 - 600

2 = 987.5 mm

Y1 = Bf - b

2 =

2200 - 230

2 = 985 mm


Mux = Wu x Bf x X1

2

2 =

339.4 x 2.2 x (0.9875)2

2 = 364.06 KN-m

Muy = Wu x Lf x Y1

2

2 =

339.4 x 2.575 x (0.985)2

2 = 423.97 KN-m


b' = b + 2 x e = 230 + 2 x 50 = 330 mm

D' = D + 2 x e = 600 + 2 x 50 = 700 mm

Clear cover = 50mm

Assume 12mm bars

289

d'x = 50 + 12

2 = 56 mm

d'y = 50 + (3 x12

2 ) = 68 mm


(Df)x = + d'x = √364.06 x 106

0.138 x 20 x 330 + 56 = 688.23 mm

(Df)y = + d'y = √423.97 x 10

6

0.138 x 20 x 700 + 68 = 536.45 mm


Therefore dx = 700 – 56 = 644 mm

dy = 700 – 68 = 632 mm


df,min ( x-direction) = 150 - 56 = 94 mm

df,min ( y-direction) = 150 - 68 = 82 mm




L2 = D + dy

2 +

dy

2 = 600 + 632 = 1232 mm

B2 = b + dy

2 +

dy

2 = 230 + 632 = 862 mm


(dy

2 - 50)

(X1 - 50) – d'y

= 700 – (700 – 150) x

(632

2 - 50)

(987.5 - 50) – 68 = 475.95 mm

290


= 2 x (1232 + 862) x 475.95

= 1993278.6 mm2


τuc2 = ks x τuc


ks = (0.5 + βc) < 1 else =1

βc = b

D =

230

600 = 0.385

therefore, ks = (0.5 + βc) = 0.5 + 0.385 = 0.885 <1

ks = 0.885

τuc2 = ks x τuc = 0.885 x 1.12 = 0.99 N/mm2

Internal load carrying capacity (Vuc2) = 1993278.6x 0.99 = 1973345.814N

= 1973.35 KN


= 339.4 x (2.575 x 2.2 – 1.232 x 0.862)

= 1562.27 KN




fck x b' x dx )

364.06 x 106 = 0.87 x 415 x Astx x 644 x (1- 415 x Astx

20 x 330 x 644 )

291

Astx = 1929.08 mm2

No. of 12mmφ bars = 4 x 1929.08

π x 122 = 17.1 say 18 bars


fck x D' x dy )

423.06 x 106 = 0.87 x 415 x Astx x 632 x (1-

415 x Astx

20 x 700 x 632 )

Asty = 2051.42 mm2

No. of 12mmφ bars = 4 x 2051.42

π x 122 = 18.13 say 19 bars



d1 = Df – (Df – Df,min) x (dy- 50)

(Y1 - 50) – d'y

= 700 – (700 – 150) x (632- 50)

(985 - 50) – 68 = 289.65 mm

L1 = D + 2 x dy = 600 + 2 x 632 = 1864 mm


2 )

= (2575 x 150) + (289.65 – 150) x ( 2575 + 1864

2 )

= 696203.175 mm2


Pt = 100 x Asty

A1 =

100 x 2148.84

696203.175 = 0.31

292


P τc (N/mm2)

0.25 0.36

0.31 ?

0.5 0.48

τc = 0.36 + (0.48 - 0.36)

(0.5 - 0.25) x (0.31 – 0.25) = 0.39 N/mm2

Internal shear resisting capacity = 0.39 x 696203.175

1000 = 271.52 KN


= 339.4 x 2.575 x (0.985 – 0.632)

= 308.51 KN

Internal shear resisting capacity < External Shear (NOT SAFE)

Increase the depth.

Shear = load

area

0.39 = 308.51 x 10

3

A1

A1 =791051.28 mm2

To calculate d1:

A1 = (Lf x Dy,min) + (d1 – Dy,min) x ( Lf + L1

2 )

791051.28 = (2575 x 150) + (d1 – 150) x ( 2575 +1864

2 )

d1 = 332.38 mm

293

7. DESIGN OF STAIR CASE

Stairs consist of steps arranged in a series for purpose of giving access to different floors of a

building. Since a stair is often the only means of communication between the various floors

of a building, the location of the stair requires good and careful consideration. In a residential

house, the staircase may be provided near the main entrance. In a public building, the stairs

must be from the main entrance itself and located centrally, to provide quick accessibility to

the principal apartments. All staircases should be adequately lighted and properly ventilated.

Various types of Staircases

Straight stairs

Dog-legged stairs

Open newel stair

Geometrical stair

RCC design of a Dog-legged staircase

In this type of staircase, the succeeding flights rise in opposite directions. The two

flights in plan are not separated by a well. A landing is provided corresponding to the level at

which the direction of the flight changes.

Dimensions:

B x L x H = 2820mm x 6000mm x 3350mm

Assume, Rise = 150mm

Tread = 300mm

sec θ = √(150)2+(300)2

300 = 1.12

294

No. of risers = 3350

150 = 22.33 say 22

Provide 11 + 11.

For flight-1: 11 and for Flight-2: 11

Going = 11 x treads = 11 x 300 = 3300mm

Total width of landings = 6000 – 3300 = 2700 mm

Therefore, width of landing at each end = 1350 mm

295

Design of Flight -1:

Type: One way single span simply supported inclined slab

Span (L): 3300 + 1350 = 4650mm

Trail Depth:

From IS456-2000, P39, Clause 24 & Clause 23.2 for Simply supported span we have

Span

Effective depth =

L

d = 20

ra = 20 x 1.4 = 28

Therefore, D = 4750

28 + 20 = 189.65mm say 200 mm


296

Loads:

Self weight of slab (inclined) = 25 x D x secθ

= 25 x 0.2 x 1.12

= 5.6 KN/m2

Weight of steps = 25 x rise x secθ

2 =

25 x 0.15 x 1.1

2

= 2.09 KN/m2

Live load = 5 KN/m2

Floor Finish = 1 KN/m2

Total Load (W) = 13.69 KN/m2

Ultimate load (Wu) = 1.5 x 13.69 = 20.54 KN/m2

Design moment (Mu) = Wu x L

2

10 =

20.54 x (4.65)2

10 = 44.41 KN-m

Mu,limit = 0.138 x fck x b x d2 = 0.138 x 20 x 1000 x (180)

2

= 89.42 x 106 N-mm = 89.42 KN-m

Mu < Mu,limit

Main steel:



fck x b x d )

44.7 x 106 = 0.87 x 415 x Ast x 180 x (1 -

415 x Ast

20 x 1000 x 180 )

Ast = 753.21 mm2


Ast = π4 x 12

2 x

1000

753.21 = 150.15mm

297


Distribution steel:

Ast = 0.12% of Ag

= 0.12

1000 x 200 x 1000 = 240 mm2.


2 x

1000

240 = 209.44mm


Design of flight-2:

Same as design of flight-1.

298

8. CONCLUSION

The complete details and the necessity of the Multi-Storeyed building have been

explained in the introduction part. Live loads and dead loads are taken into consideration and

analysis is done manually. Manual analysis comprises of load distribution of slabs on to

beams and calculation of bending moment and shear force by any approximate method.

Analysis of frames is done by the substitute frame method and is furnished in the

introduction part. With the help of moment distribution method the moments carried at each

joints are known.

The design method adopted is limit state method. The bending moments obtained from

chapter 4 are used in calculating the area of steel in each frame section. The percentage of

steel obtained by limit state method is minimum as mentioned in limit state method of IS

456-2000.

Design of columns and footings are done depending on the load acting on them.

This project explains the basic concept behind the software. With the help of any software

we can do this analysis within no time but the basic thing is to know the concept behind it.

The same result can be achieved using the software like “Staad-Pro”. The wind load and

seismic analysis can be included for the future extension of the project.

299

9. REFERENCE

I. Design of Reinforced Concrete Structures by A. K. Jain

II. Illustrated design of reinforced concrete buildings by Dr.V.L. Shah & Dr.S.R.Karve

III. Basic Principles of Analysis and Design of an RCC Framed Structures by Prof. H. R.

Surya Prakash S. Krishna Murthy

IV. Design of R.C.C structural elements by S.S. Bhavikatti

V. Design of R.C.C slabs by K.C.Jain

VI. R.C.C Design and Drawing by Neelam Sharma

VII. Treasure of R.C.C Designs by Sushil kumar

VIII. Design of R.C.C structures by prof.N.Krishna Raju

IX. Design of R.C.C structures by prof.S.Ramamrutham X. R.C.C design by S.Unnikrishina pillai & Devdas Menon

XI. The code books referred for this project are:

● IS 456:2000 (reinforced concrete for general building construction)

● IS 875, part 1, 1987(dead loads for building and structures)

● IS 875, part 2, 1987(imposed loads for buildings and structures)

● SP 16 (design aids for IS 456)

Documents

CD-18-Modified Final Project Report