ANALYSIS AND DESIGN OF WORKING WOMENS’

HOSTEL WITH DAYCARE CENTER

A DESIGN PROJECT REPORT

Submitted by

NAGARATHINAM.M (910715103045)
NIVETHA.J (910715103048)
RIVETHA.S (910715103058)
SIVANI.M (910715103068)

In partial fulfillment for the award of the degree

of

BACHELOR OF ENGINEERING

IN

CIVIL ENGINEERING

K.L.N. COLLEGE OF INFORMATION TECHNOLOGY, SIVAGANGAI

ANNA UNIVERSITY: CHENNAI 600 025

October 2018

ANNA UNIVERSITY: CHENNAI 600 025

1
 BONAFIDE CERTIFICATE
This is to certify that the dissertation entitle “ANALYSIS AND DESIGN OF
WORKING WOMENS’ HOSTEL WITH DAYCARE CENTER ”
is the bonafide work of M.NAGARATINAM (910715103045), J.NIVETHA
(910715103048), S.RIVETHA (910715103058), M.SIVANI(910715103068), who
carried out the project work under my Supervision.

SIGNATURE SIGNATURE

Dr. A.S.S.SEKAR M.E., Ph.D Dr. K.R.YOGANATHAN

M.E.,Ph.D

HEAD OF THE DEPARTMENT SUPERVISOR

Professor and Head Professor, Professor,

Department of Civil Engineering, Department of Civil Engineering,

K.L.N. College of Information Technology, K.L.N. College of Information
Technology,
Sivagangai-630612. Sivagangai-630612.

Submitted for the Viva – Voce examination held at K.L.N.COLLEGE OF

INFORMATION TECHNOLOGY, Sivagangai on ………………

Internal Examiner External Examiner

2
 ACKNOWLEGDGEMENT

Firstly and fore mostly , we would like to praise and thank the
Almighty lord showering . His blessings towards the successful completion
of this project.
We would like to express our sincere thanks to our Principal
Dr.N.Balaji M.E., Ph.D., for his standard support and encouragement in
this project.
We would also like to express our sincere thanks to our Professor &
Head of Civil Engineering Department Dr.A.S.S.Sekar M.E.,Ph.D., for his
constant encouragement throughout this project.
We would also like to thank Dr.K.R.YOGANATHAN M.E.,Ph.D..,
Professor, Department of Civil Engineering for being our Project
Supervisor. His guidance throughout this project made us to complete the
project within the stipulated time.
We extend our deep sense of gratitude to Mr.K.M. BASANTH BABU
B.E,M.E,(Ph.D)., Professor, Department of Civil Engineering and
Mrs.G.R.POORNIMA M.E,M.C.A., Assistant Professor, Department of Civil
Engineering coordinators of design project for the completion of project.We
would also like to thank all our teaching and non-teaching faculty
members who directly or indirectly extended their support in completing
this project.
We would also like to thank all our classmate and friends who
helped in documenting this project report.
Last but not the least we would like to thank our parents for their
untiring and moral love and support throughout this project.

NAGARATHINAM.M
NIVETHA.J
RIVETHA.S
SIVANI.M

3
 TABLE OF CONTENTS

CHAPTER TITLE PAGE NO

ABSTRACT

LIST OF SYMBOLS

1. INTRODUCTION 1

2. PLANNING AND SPECIFICATION 2

3. ANALYSIS AND DESIGN 12

5. DESIGN OF STRUCTURAL MEMBERS

DESIGN OF SLAB 18

DESIGN OF BEAM 25

DESIGN OF COLUMN 34

DESIGN OF FOOTING 40

DESIGN OF STAIRCASE 49

6. CONCLUSION 58

7. REFERENCES 59

4
 ABSTRACT

The progress of scientific knowledge and means by which development has

been equally impressive. Knowledge of options that could be adopted to
accept changes of life style provides for growth and development that is
truly sustainable.

The sustainable development is defined as the “development that

meets the needs of the present without compromising the ability of future
generations to meet their own needs”.

Keeping in mind the necessity and importance of development

activities, we planned to do the project as “ANALYSIS AND
DESIGN OF WORKING WOMENS’ HOSTEL WITH
DAYCARE CENTER”, in karadikal, Madurai which is possessing the
matter of aesthetics as well , it has utilitarian value to serve also .

The project consist of phases of planning, analysis and design of

structural elements . The detailing of reinforcements is also done .

It is framed structure and the structural elements such as footings ,

columns ,slabs and beams are analysed and designed with respect to the
latest trends in technologies STAAD PRO software and with reference to
IS 456:2000, IS 875: 1987 (Part 1,2,3) and SP-16 for load calculation and
drawings are prepared using AutoCADD.

5
 LIST OF SYMBOLS

AC =Area of Concrete

Ast =Area of tension reinforcement

Asc = Area of compressive reinforcement

B.M =Bending moment

b = Breadth of the section

C/C = Center to center distance

D = Overall depth of the section

DL = Dead load

d = Effective depth of section

fck = Characteristic compressive strength of concrete

fy = Characteristic strength of steel

LL = Live load

Leff =Effective length

Mu = Moment of resistance

S = Spacing of reinforcement

Vu =Ultimate shear force

Ø = Diameter of steel reinforcement

τc = Permissible shear stress of concrete

τv =Nominal shear stress in concrete

𝝋 =Angle of shear resistance of soil

𝜸 =Unit weight of soil

6
 CHAPTER 1

1.1 INTRODUCTION

This project contains the following:

 The Working Womens’ Hostel Building with Daycare Centre is designed

(G+2)Good foundation is provided at the sufficient depth in this building.
 Suitable electrical arrangements, water supply and drainage facilities are
provided.
 The Hostel Building is planned with good ventilation and lighting by
providing sufficient number of doors and windows.
 Sick room, Dorm room, Recreation room, Reading room, Service room,
Store room, Gym, Warden room, kitchen, Dining hall are provided.

The project incorporates planning and designing, drafting of all essential

drawings including structural drawings.

The structural designing procedure and calculation have been done in

accordance with IS 456-200 and design aids by SP-16 by limit state method
of designing.

1.2 OBJECTIVES OF PROJECT

• The ultimate aim of hostel with day care centre is to give a space for
working mother to stay along with their children.
• The primary purpose is to provide safe & secure accommodation for
women.
• Also provide facilities required for pre-school children up to 8 years.

7
 CHAPTER 2

PLANNING AND BRIEF SPECIFICATIONS

2.1 PLANNING - INTRODUCTION :

Planning comprises mainly two aspects, one is to provide more
comfortable accommodation and the other is structural planning however,
two play simultaneous role and the Engineer is expected to plan and design
for both of these aspects.

2.2 PLANNING FOR COMFORT:

The hostel building is designed with G+2 floors and there are 67 rooms
in hostel. Ground floor consists of 15 rooms, First floor and second floor consist
of 25 rooms Out of 67 rooms and remaining rooms are provides for the facilities
there is a dining hall and kitchen in the ground floor along with the warden and
guest rooms and utility

2.3 STRUCTURAL PLANNING:

Columns are designed as rectangular column so that the maximum
dimension of column can be placed along the maximum span of slab,
secondary beams are provided whenever necessary to reduce the span of the
slab so that deflection at the centre of the slab gets reduced.
The layout of the column of the building is determined by shapes,
size and the height of the occupancy of the building. It is observed that
compared to long, narrow and irregular shaped building. Hence such a
buildings are to be sectionalized in the independent part so that each part
becomes rectangular with length not exceeding approximately the three
times the width. It economical and advantageous from the view point of
concrete placement to use high strength deformed bars, the effectiveness of
a reinforcement in strengthening in concrete member in proportional to its
yield stress which is higher in case of deformed bars, thus resulting in
savings in the quality of reduction in width of the member, thereby reducing
the dead loads which further reduces the depth of the beam providing more
head room. Reduction in cross section member owing to the reduction of
dead loads not only saves the construction materials, but also handling and
transportation charges.
8
 2.4 DESCRIPTION OF THE PLAN:
The building consists of three floors,
 Ground floor
 First floor
 Second floor

The ground floor consists of

 Dining hall
 Kitchen
 Store room
 Wash basin
 Gym
 Sick room
 Warden room
The first floor consists of
 Recreation room
 Dorm room
The second floor consists of

Plinth area of proposed Hostel with Daycare center is 821m2

2.5 SPECIFICATION

2.5.1 SITE CLEARANCE

The proposed area is to be cleaned off all stones, plants, trees,

rubbish etc.,

2.5.2 MARKING OF CENTER LINE

As per the centre line the border line of the footing are marked on
the site for doing earthwork excavation.

2.5.3 EARTH WORK EXCAVATION

The side of the pits should be vertical and the depth of excavation of
the footing varies in accordance to the depth and width as per the design
below the ground level. Suitable temporary fencing is to be provided around
the site of excavation to avoid any accidental fall into the pits.

9
2.5.4 FOUNDATION CONCRETE

After the excavation on the earth, the foundation concrete should be laid
by using P.C.C 1:5:10 mix. Over this P.C.C of the footing concrete M40 for
the superstructure is laid.

2.5.5 EARTH FILLING

After the laying of foundation and footing the open area is to be filled by
the excavated earth and consolidated as per the requirements.

2.5.6 SAND FILLING

The sand filling is done with river sand for following thickness and with
uniform compaction below the levelling coarse.

2.5.7 FLOORING CONCRETE AND FLOOR FINISH

The flooring concrete is of mix 1:2:4 and of grade M20 is provided and
well compacted river sand filling is provided in basement of all rooms in
ground floor.

2.5.8 BRICK WORK IN SUPERSTRUCTURE

Brick work in superstructure is done in all rooms for a height of 3.2m

using C:M 1:5.

2.5.9 COLUMN

The columns are planned with R.C.C work in M20 grade taken up to
parapet with reinforcement as per design and drawing. The size of column is
taken as 0.3m X 1.2m.

2.5.10 BEAMS

The beams are taken as R.C.C work in M40 grade concrete using 40mm
aggregate with reinforcement as per design and drawing. The size of beam is
taken as 0.23m X 0.23m.

2.5.11 SLAB

In corridor and room, R.C.C roof slab has to be laid in M20 grade with
sufficient depth of 0.15m.
10
2.5.12 MATERIALS ADOPTED

 Reinforced cement concrete of grade M20.

 Steel reinforcement Fe415.
 Brick compressive strength 7.5N/mm

2.5.13 STAIRCASE

In approach area the dog-legged staircase are to be provided with M20

grade, the flight slab thickness is 185mm. The rise and tread in the staircase is
150mm and 300mm respectively. The steps are constructed with brick work.

2.5.14 LINTELS

Lintels are signed as spread over the opening in M20 grade concrete
with sufficient depth as per design.

2.5.15 SUNSHADE

The lintel cum sunshade are to be projected beyond the wall and at the
free end.

2.5.16 WOOD WORK

The doors, windows and ventilators used are to be well seasoned wood
with better quality.

2.5.17 WEATHERING COURSE

Weathering course is to be provided in brick jelly lime concrete for

average thickness.

2.5.18 PLASTERING

Exposed walls are to be plastered with C:M 1:3 12mm thick, ceiling
plastering with C:M 1:3 10mm thick.

11
GROUND FLOOR

12
FIRST FLOOR

13
SECOND FLOOR

14
COLUMN –FOOTING – CENTER LINE LAYOUT

15
ELEVATION

16
SECTION ON A-A

17
 CHAPTER 3

ANALYSIS AND DESIGN

3.1 METHODOLOGY

3.1.1 LOADS CONSIDERED

Dead load(self weight , brick work load, floor finish load, etc.)

Live load

3.1.2 LOAD COMBINATIONS

(D.L + L.L + Floor finish)

where,

D.L = Dead load

L.L = Live load

3.2 CODES

3.2.1 INDIAN STANDARDS

IS 456-2000 Code of practice of plain and reinforced concrete.

IS 875-1987 code of practice for design loads for building.

3.2.2 SP-16

The charts and tables are used for design purpose

3.3 SOFTWARE USED

1.Plan and centre line sketch -AUTOCADD

2.Analyis - STAAD PRO

3.4 ANLYSIS OF STRUCTURAL MEMBERS

The analysis of structural members were done with the help of STAAD Pro
software. Different types of loads and load combinations were given as input
wherever necessary in the software. The building frame as shown in figure
3.1

18
Fig 3.1 skeleton of building -STAAD Pro model

Fig 3.2 Rendered view

19
 Fig 3.4 Shear force Diagram

Fig 3.5 Bending Moment Diagram

20
 BEAM REINFORCEMENT DETAILS

Fig 3.6Marking of designed beam

Fig 3.7 Reinforcement details for marked beam

21
 COLUMN REINFORCEMENT DETAILS

Fig 3.8 Marking of designed Column

Fig 3.9 Reinforcement details for marked column

22
 DESIGN PRICIPLES

Limit State Method:

The aim of design is to probabilities that the structure will not

become unfit for the use which it is intended.

A structure or a part of a structure must be safe against collapse and

also serviceable in use, the acceptable limits for the safety and serviceability
requirement before failure occur is called “LIMIT STATE”.

Partial safety factor 1.5 for concrete is used in this method to

determine the design loads and their design strength of materials from their
characteristic values.

The main factor has to be considered in the design of slab. They are
follows:

• Strength of slab against flexure, shear and twist.

• Stiffness against deflection.

23
 CHAPTER 4

DESIGN OF STRUCTURAL MEMBERS

4.1DESIGN OF SLAB

1. ONE WAY CONTINIOUS SLAB (Day care centre)

Data:
Dimension = 3×6m
Live load = 3kN/m
fck = 20N/mm²
fy = 415N/mm²
Step 1: Type of slab (Ly/ Lx)
Lx = 3m
Ly = 6m
𝐿𝑌 6
= =2≤2
𝐿𝑋 3
The slab is designed as one way continuous slab
Step 2: Effective depth of slab
𝐿𝑒𝑓𝑓
= 28
𝑑
3000
d = =107mm ~ 110mm
28
D = d+Clear cover +(D/2)
D = 110+15+5
D = 130mm
Step 3: Load calculation
Dead load
Self weight of slab = 1×1×0.13×25 = 3.25kN/m²
Floor finish(0.6-1kN/m²) = 1kN/m²
Dead load (Wd) = 4.25kN/m²
Factored dead load(Wdu) =4.25×1.5 =6.375 kN/m²
Live load (Wl) = 3kN/m²
Factored dead load(Wlu) =3×1.5 =4.5 kN/m²
Step4: Span Moments
𝐿𝑥2 𝐿𝑥2
M1 =Wdu +Wlu
12 10
32 32
= (6.375× ) + (4.5× )
12 10
= 7.975kNm

24
 𝐿𝑥2 𝐿𝑥2
M2 = Wdu +Wlu
10 9
32 32
= (6.375× ) + (4.5× )
10 9
=10.24kNm
M2> M1
Interior panel:
𝐿𝑥2 𝐿𝑥2
M3 = Wdu + Wlu
16 12
32 32
=(6.375× ) + (4.5× )
16 12
= 6.97kNm
𝐿𝑥2 𝐿𝑥2
M4 = Wdu + Wlu
12 9
32 32
= (6.375× ) + (4.5× )
12 9
= 9.23kNm
M4> M3
Step5:Astand spacing for End span
M2 = 0.87 fyAst (d -0.42 Xu)
0.42×0.87×415×𝐴𝑠𝑡
10.24×106 = 0.87×415×Ast [110 –( )]
0.36×20×1000
10.24×106 = 36105Ast – 7.604Ast²
Ast = 271mm2
𝜋
×82 ×1000
4
Spacing, S = =185~180mm
271
S =180mm c/c
Provide 8mm dia @ 180mmc/c as main reinforcement in end span.
Ast and spacing for Interior span
M4 = 0.87 fyAst(d -0.42 Xu)
0.42×0.87×415×𝐴𝑠𝑡
9.23×106 = 0.87×415×Ast [110 – ( )]
0.36×20×1000
9.23×106 = 36105Ast – 7.604Ast²
Ast = 244mm²
𝜋
×82 ×1000
4
Spacing, S = =206~200mm
244
S=200mm c/c
Provide 8mm dia bar @ 200mmc/c as main reinforcement in interior span.

25
Distribution Reinforcement
0.12
Ast(min) = ×b×D
100
0.12
= ×1000×13 = 156mm²
100
𝜋
×62 ×1000
Spacing, S = 4 =170
156
S =170mm c/c
Provide 6mm dia bar @170mmc/c as distribution reinforcement

(Fig.No-4.1.1.1)

26
2.TWO WAY SLAB (Room 15)

Data:
Dimension =3×3.655m
Live load = 3kN/m
fck = 20N/mm²
fy = 415N/mm²
Step 1: Type of slab (Ly/Lx)
𝐿𝑦 3.655
= = 1.22<2
𝐿𝑥 3
Hence slab is designed as two way slab.
Step 2: Effective depth of slab
Leff/d = 28
d= 3000/28 =107mm ~ 110mm
D= d+Clear cover +(D/2)
D= 110+15+5 = 130mm
Step 3: Load calculation
Self weight of slab = 1×1×0.13×25 = 3.25KN/m2
Floor finish (0.6-1kN/m²) = 1kN/m²
Live load (WL) = 3kN/m²
Total load (WT) = 7.25kN/m²
Factored load (Wu) = 7.25×1.5 = 10.88kN/m²
Step4: Moment Calculation Mux & Muy
Mux = αx Wu Leffx²
Muy = αy Wu Leffx²
From table 26 of IS456 – 2000, pg.no 91( one short edge discontinuous)
αx = 0.052 αy = 0.035
Mux = 7.56kNm
Muy = 3.43kNm
Step5: Astx & Asty
Astx:
Mux = 0.87 fy Ast (d -0.42 Xu)
0.42×0.87×415×𝐴𝑠𝑡
7.56 ×106 = 0.87×415×Ast× [110 – ( )]
0.36×20×1000
7.56×106 = 39715Ast – 7.604Ast²
Astx = 100mm2

27
 𝜋
×82 ×1000
4
Spacing , S =
100
S=300mm c/c.
Provide 8mm dia @ 300mmc/c along X-direction.
Asty:
Muy = 0.87 fy Ast (d -0.42 Xu)
0.42×0.87×415×𝐴𝑠𝑡
5.38 ×106 = 0.87×415×Ast× [110 – ( )]
0.36×20×1000
5.38×106 = 39715Ast – 7.604Ast²
Asty = 100mm2

𝜋
×82 ×1000
4
Spacing , S =
100
S=300mm c/c.
Provide 8mm dia @ 300mmc/c along Y-direction.

(Fig.No-4.1.2.2)

28
3. CANTILEVER SLAB (Sun shade)
Data:
Dimension = 715×1600mm
Live load = 3kN/m
fck = 20N/mm²
fy = 415N/mm²
Step 1:Effective depth of slab
𝐿𝑒𝑓𝑓
= 10
𝑑
715
d= =71.5mm ~ 75mm
10
D = d+Clear cover +(D/2)
D = 75+15+5 = 95mm
Step 2: Load calculation
Self weight of slab = 1×1×0.08×25 = 2kN/m2
Floor finish (0.6-1kN/m²) = 1kN/m²
Live load (Wl) = 2kN/m²
Total load (WT) = 5kN/m²
Factored load (Wu) = 5×1.5 = 7.5KN/m²

Step3: Moment Calculation

𝐿𝑒𝑓𝑓 2
Mu = Wu
2
7152
= 7.5×
2
= 1.91kNm
Step4: To find Ast
Mu(max) = 0.87 fy Ast (d -0.42 Xu)
0.42×0.87×415×𝐴𝑠𝑡
1.91×106 = 0.87×415×Ast [75 – ( )]
0.36×20×1000
1.91×106 = 21663Ast – 7.604Ast²
Ast = 191mm²

𝜋
×82 ×1000
4
Spacing, S =
191
=230mm c/c
Provide 8mm dia @ 230mmc/c as main reinforcement.

29
Step5: Distribution Reinforcement
0.12
Ast (min) = ×b×D
100
0.12
= ×1000×95
100
= 114mm²
𝜋
×62 ×1000
4
Spacing, S =
114
=200mm c/c
Provide 6mm dia @ 200mmc/c as distribution reinforcement.

(Fig.No-4.1.3.3)

30
 4.2.DESIGN OF BEAM

1. BEAM(ROOM15)

DATA:

b = 230mm

d = 400mm

𝑓𝑐𝑘 = 20 𝑁⁄𝑚𝑚2

𝑓𝑦 = 415𝑁⁄𝑚𝑚2

L = 3m (Fig.No-4.2.1.1)

Step 1: Load Calculation

𝟏 𝟏
Area =( ×1.885×3)+( ×1.615×3)=5.25m2 (Fig.No-4.2.1.2)
𝟐 𝟐

Self-weight of beam = 0.23× 0.4× 25×3 = 6.9kN

Brick work = 0.23× 3.2× 19.81x3 = 43.74kN

Live load = 3×5.25 =15.75kN

Self-weight of slab = 5.25× 0.15×25 =19.68kN

Floor finish = 1×5.25 =5.25kN

WT =91.32kN

Wu = WT× 1.5 = 91.32 × 1.5 =136.98kN

136.98
Wu = = 45.664kN/m
3

Step 2: Moment Calculation

Support moment:
𝑙𝑒𝑓𝑓2 45.664×32
𝑀𝑢 = 𝑊𝑢 × ⁄ = =34.25kN-m
12 12

Mid span moment:

31
 𝑙𝑒𝑓𝑓2 2
𝑀𝑢 = 𝑊𝑢 × ⁄ = 45.664×3 = 17.124kN-m
24 24

Step 3: To find 𝑨𝒔𝒕

Support:
0.87×415×𝐴𝑠𝑡
𝑀𝑢 = 0.87×415× 𝐴𝑠𝑡 [d-(0.42× )]
0.36×20×230

34.25x106 = 0.87× 415 × 𝐴𝑠𝑡 [ 370−(0.42× 0.218 𝐴𝑠𝑡 )]

𝐴𝑠𝑡 = 266.47𝑚𝑚2
𝐴𝑠𝑡 266.476
No of bar = =п = 2.35 nos.~ 3 nos
𝑎𝑠𝑡 ⁄4×122

Provide 3nos of 12mm dia bar as Tension Reinforcement at support

Mid span:

𝑀𝑢 = 0.87×415× 𝐴𝑠𝑡 [370-(0.42𝑋𝑢 )]

17.124× 106 = 0.87×415× 𝐴𝑠𝑡 [370 − 0.42 × 0.218𝐴𝑠𝑡 ]

𝐴𝑠𝑡 = 130.61𝑚𝑚2
𝐴𝑠𝑡 130.61
No of bar = =п = 2.3 nos. ~ 3 nos
𝑎𝑠𝑡 ⁄4×102

Provide 3nos of 10mm dia bar as Compression Reinforcement at Mid span.

Step 4: Design of shear reinforcement

wu×leff 45.66×3
vu = = =68.49kN
2 2

𝑉𝑢 68.49×103 2
𝜏𝑣 = = = 0.8 𝑁⁄𝑚𝑚 (refer IS 456 :2000 clauses 40.1 pg.no:72)
𝑏×𝑑 230×370

100× 𝐴𝑠 𝑡 100×266.47 2
𝜏𝑐 = = =0.31 =0.39 𝑁⁄𝑚𝑚
𝑏×𝑑 230×370

(refer IS 456 :2000 Table 19 pg.no:73 using M20 grade)

2
𝜏𝑐 𝑚𝑎𝑥 = 2.8 𝑁⁄𝑚𝑚 (refer IS 456 :2000 Table 20 pg.no:73 using M20 grade)

𝜏𝑣 > 𝜏𝑐 <𝜏𝑐 𝑚𝑎𝑥 0.8 >0.39< 2.8 (Design shear reinforcement)

Spacing:

32
 0.87×𝑓𝑦 ×𝐴𝑠𝑣 × 𝑑
𝑠𝑣 = (refer IS 456 :2000 clauses 40.4 pg.no:72
𝑉𝑢𝑠

𝑉𝑢𝑠 = 𝑣𝑢 − 𝜏𝑐 × 𝑏 × 𝑑

= 68.49× 103 − 0.39× 230 ×370 = 35 kN

𝜋
0.87×𝑓𝑦 ×𝐴𝑠𝑣 × 𝑑 0.87 × 415 ×2× ×82 ×370
4
𝑠𝑣 = = = 250 mm
𝑉𝑢𝑠 35×103

(i) 0.75× d = 0.75 × 370 = 270 mm

(ii) 300 mm

(iii) 𝑠𝑣 = 200 𝑚𝑚

Provide 8 mm 𝑑𝑖𝑎 2 legged stirrups @ 250 mm c/c as shear reinforcement

(Fig.No4.2.1.3)

2. BEAM (ROOM14)
33
DATA:

L = 3.77 m

b = 230mm

D = 400mm

d = 370 mm

𝑓𝑐𝑘 = 20 𝑁⁄𝑚𝑚2

𝑓𝑦 = 415𝑁⁄𝑚𝑚2 (Fig.No-4.2.2.1) (Fig.No-4.2.2.2)

Step 1:Load calculation

𝟏
Area =( ×(3.77+0.77)) ×1.5×2=6.81m2
𝟐

Self-weight of beam = 0.23× 0.4× 25× 3.77 = 8.7kN

Brick work = 0.23× 19.81× 3.2× 3.77 = 54.97kN

Live load = 3×6.81 = 20.43kN

Floor finish = 1× 6.81 = 6.81 kN

Self-weight of slab = 6.81× 0.15×25 =25.54kN

W T= 116.45 kN

Wu = WT× 1.5 =116.45× 1.5 = 174.68kN

174.68
Wu = = 46.33 kN/m
3.77

Step 2: Moment calculation

Support Moment:
𝑙𝑒𝑓𝑓2 46.33×3.772
𝑀𝑢 = 𝑊𝑢 × ⁄ = = 54.88 kN-m
12 12

Mid span moment:

34
 𝑙𝑒𝑓𝑓2 2
𝑀𝑢 = 𝑊𝑢 × ⁄ = 46.33×3.77 = 27.44kN-m
24 24

Step 3: To find 𝑨𝒔𝒕

Support :
0.87×415×𝐴𝑠𝑡
𝑀𝑢 = 0.87×415× 𝐴𝑠𝑡 [d-(0.42× )]
0.36×20×230

54.88x106 = 0.87× 415 × 𝐴𝑠𝑡 [ 370−(0.42× 0.218 𝐴𝑠𝑡 )]

𝐴𝑠𝑡 = 438.13𝑚𝑚2
𝐴𝑠𝑡 438
No of bar = =п = 2.17 nos ~3 nos
𝑎𝑠𝑡 ⁄4×162

Provide 3nos of 16mm dia bar as Tension Reinforcement at Support

Mid span:

𝑀𝑢 = 0.87×415× 𝐴𝑠𝑡 [370-(0.42 𝑋𝑢 )]

27.44× 106 = 0.87 ×415× 𝐴𝑠𝑡 [370 − 0.42 × 0.218𝐴𝑠𝑡 ]

𝐴𝑠𝑡 =211 𝑚𝑚2

𝐴𝑠𝑡 211
No of bar = =п = 3 nos.
𝑎𝑠𝑡 ⁄4×122

Provide 3nos of 12mm dia bar as Compression Reinforcement at Mid span.

Step 4: Design of shear reinforcement

wu×leff 46.33× 3.77
vu = = = 87.33 kN
2 2

𝑉𝑢 87.33×103 2
𝜏𝑣 = = = 1.026 𝑁⁄𝑚𝑚 (refer IS 456 :2000 clauses 40.1 pg.no:72)
𝑏×𝑑 230×370

100× 𝐴𝑠𝑡 100×438 2

𝜏𝑐 = = =0.5 = 0.48 𝑁⁄𝑚𝑚
𝑏×𝑑 230×370

(refer IS 456 :2000 Table 19 pg.no:73 using M20 grade)

2
𝜏𝑐 𝑚𝑎𝑥 = 2.8 𝑁⁄𝑚𝑚 (refer IS 456 :2000 Table 20 pg.no:73 using M20 grade)

𝜏𝑣 > 𝜏𝑐 <𝜏𝑐 𝑚𝑎𝑥 1.026> 0.48 < 2.8 (Design shear reinforcement)

Spacing:

35
 0.87×𝑓𝑦 ×𝐴𝑠𝑣 × 𝑑
𝑠𝑣 = (refer IS 456 :2000 clauses 40.4 pg.no:72)
𝑉𝑢𝑠

𝑉𝑢𝑠 = 𝑣𝑢 − 𝜏𝑐 × 𝑏 × 𝑑

= 87.33 × 103 − 0.48 × 230 ×370

= 46.48 kN
𝜋
0.87×𝑓𝑦 ×𝐴𝑠𝑣 × 𝑑 0.87 × 415 ×2× ×82 ×370
4
𝑠𝑣 = = =288.79 mm
𝑉𝑢𝑠 46.48×103

(i) 0.75× d = 0.75 × 370 = 250 mm

(ii) 300 mm

(iii) 𝑠𝑣 = 288.79 𝑚𝑚

Provide 8 mm 𝑑𝑖𝑎 2 legged stirrups @ 250 mm c/c as shear reinforcement

(Fig.No-4.2.2.3)

3.

36
BEAM(CORRIDOR)

DATA:

L = 3.23 m

b = 230mm

D = 400mm

d = 370 mm

𝑓𝑐𝑘 = 20 𝑁⁄𝑚𝑚2 (Fig.No-4.2.3.1) (Fig.No-4.2.3.2)

𝑓𝑦 = 415𝑁⁄𝑚𝑚2

Step 1:Load calculation

𝟏
Area =( ×(3.23+0.23)) ×1.5×2=7.785m2
𝟐

Self-weight of beam = 0.23× 0.4× 25× 3.23 = 7.429kN

Brick work = 0.23× 19.81× 3.2× 3.23 =47.09 kN

Live load = 3× 7.785 = 23.355kN

Floor finish = 1× 7.785 = 7.785 kN

Self-weight of slab = 7.785× 0.15×25 =29.19 kN

WT =114.857 kN

Wu = WT×1.5=114.857 × 1.5 = 172.28 kN

172.28
Wu = = 53.34 kN/m
3.23

Step 2: Moment calculation

Supporrt moment:
𝑙𝑒𝑓𝑓2 53.34×3.232
𝑀𝑢 = 𝑊𝑢 × ⁄ = =46.373 kN-m
12 12

37
Mid span moment:
𝑙𝑒𝑓𝑓2 2
𝑀𝑢 = 𝑊𝑢 × ⁄ = 53.34×3.23 =23.18 kN-m
24 24

Step 3: To find 𝑨𝒔𝒕

Support :
0.87×415×𝐴𝑠𝑡
𝑀𝑢 = 0.87×415× 𝐴𝑠𝑡 [d-(0.42× )]
0.36×20×230

46.37x106 = 0.87× 415 × 𝐴𝑠𝑡 [ 370−(0.42× 0.218 𝐴𝑠𝑡 )]

𝐴𝑠𝑡 = 366𝑚𝑚2
𝐴𝑠𝑡 366
No of bar = =п = 3.2 nos ~4nos
𝑎𝑠𝑡 ⁄4×122

Provide 4nos of 12mm dia bar as Tension Reinforcement at Support

Mid span:

𝑀𝑢 = 0.87×415× 𝐴𝑠𝑡 [d-(0.42 𝑋𝑢 )]

23.187× 106 = 0.87 ×415× 𝐴𝑠𝑡 [370 − 0.42 × 0.218𝐴𝑠𝑡 ]

𝐴𝑠𝑡 =178𝑚𝑚2
𝐴𝑠𝑡 178
No of bar = =п = 2.26 nos ~3 nos
𝑎𝑠𝑡 ⁄4×102

Provide 3nos of 10mm dia bar as Compression Reinforcement at Mid span.

Step 4: Design of shear reinforcement

wu×leff 53.33× 3.23
vu = = = 86.13 kN
2 2

𝑉𝑢 86.13×103 2
𝜏𝑣 = = = 1.01𝑁⁄𝑚𝑚 (refer IS 456 :2000 clauses 40.1 pg.no:72)
𝑏×𝑑 230×370

100× 𝐴𝑠𝑡 100×366 2

𝜏𝑐 = = = 0.43 =0.45𝑁⁄𝑚𝑚
𝑏×𝑑 230×370

(refer IS 456 :2000 Table 19 pg.no:73 using M20 grade)

2
𝜏𝑐 𝑚𝑎𝑥 = 2.8 𝑁⁄𝑚𝑚 (refer IS 456 :2000 Table 20 pg.no:73 using M20 grade)

𝜏𝑣 > 𝜏𝑐 <𝜏𝑐 𝑚𝑎𝑥 1.01> 0.45< 2.8 (Design shear reinforcement)

38
Spacing:
0.87×𝑓𝑦 ×𝐴𝑠𝑣 × 𝑑
𝑠𝑣 = (refer IS 456 :2000 clauses 40.4 pg.no:72)
𝑉𝑢𝑠

𝑉𝑢𝑠 = 𝑣𝑢 − 𝜏𝑐 × 𝑏 × 𝑑

= 86.13× 103 – 0.45× 230 ×370 =44kN

𝜋
0.87×𝑓𝑦 ×𝐴𝑠𝑣 × 𝑑 0.87 × 415 ×2× ×82 ×370
4
𝑠𝑣 = = = 200 mm
𝑉𝑢𝑠 44×103

(i) 0.75× d = 0.75 × 370 = 270 mm

(ii) 300 mm

(iii) 𝑠𝑣 = 200 𝑚𝑚

Provide 8 mm 𝑑𝑖𝑎 2 legged stirrups @ 200 mm c/c as shear reinforcement

(Fig.No-4.2.3.3)

39
 4.3DESIGN OF COLUMN

1. COLUMN DESIGN (UNI–AXIALLY

LOADED)

Step 1: Load Calculation

3 3.77
Beam Load =(45.66× ) × 2 + (46.33 × ) =224.31×3
2 2

= 448.6kN

Self wt of Column=0.23×0.23×3.2×25 =4.324kN

Total Load, P =452.8kN

Pu =1.5×452.8 =679.26kN

Step2: Parameters
Pu 679.2×103
= =0.64
𝑓𝑐𝑘 𝑏 𝐷 20×230×230

𝐿 𝐷 3.2 230
e= + = + =7.67mm<20mm
500 30 500 30

Mu 679.2×103 ×7.67
= =0.021
𝑓𝑐𝑘 𝑏 𝐷ˆ2 20×230×230ˆ2

40
d’/D = =0.17 (Fig.No-4.3.1.1)
230

Step 3:Select Chart 34 of SP-16 pg.no.119:

𝑃
=0.06
𝑓𝑐𝑘

P = 0.06×20

P = 1.2 %
1.2
Asc = ×b×D
100

Asc = 634.8 mm2

Step 4: No of Bars
Asc 𝟔𝟑𝟒
= =𝝅 = 6nos
𝒂𝒔𝒄 ×𝟏𝟔𝟐
𝟒

40
Step 5: Design of lateral ties
𝜙𝐿 16
𝜙𝑡 < 6mm (or) =
4 4

𝜙𝑡 = 6mm (or) 4mm

𝜙𝑡 = 6mmφ

Pitch of lateral ties

i) b = 230mm
ii) 16𝜙𝑙 = 16x16 = 256mm
iii) 48𝜙𝑡 = 48x6 = 288mm
iv) 300mm

Pitch = 230 𝑚𝑚 𝑐⁄𝑐

Provide 6 nos of 16mm dia bars as longitudinal reinforcement.

Provide 6mm dia lateral ties @ 230 𝑚𝑚 𝑐⁄𝑐 pitch.

(Fig.No-4.3.2.2)

41
2. DESIGN COLUMN ( AXIALLY LOADED)

Step 1: Load Calculation

3.77 3.77 3.23
Beam Load =(45.66× ) × 2 + (46.33 × ) + (53.33 × )
2 2 2

=310.44×3 = 931.32kN

Self wt of Column=0.23×0.23×3.2×25 =4.324kN

Total Load, P=935.6kN

Pu =1.5×935.6=1410kN

Step2: To find Ag

Assume

Asc = 2% (0.8%-4%) axially loaded

Asc = 2%

Asc = 0.02 Ag

Pu = 0.4 fck Ac + 0.67 fy Asc

1410 = 0.4 ×20× (Ag - Asc) +0.67×415× Asc

Ag = 105.22× 103 mm2 (Fig.No-4.3.2.1)

Ac = Ag - Asc

Ac = 103.11× 103 mm2

Asc =0.02× 105.22 =2.1× 103 mm2

Size of the Column = 230 ×230 mm

Asc required.

Pu = 0.4 fck Ac + 0.67 fy Asc

1410 ×103 = 423200+270.05 Asc

Asc = 3672 mm2

42
No of Bars:
3672
= (Asc/asc) =(𝜋 ) = 7.48 nos~ 8 nos
×252
4

Step 3: Design of lateral ties

i) 6mm
𝜙𝑙 25
ii) = = 8mm.
4 4

Pitch of lateral ties

i) b = 230mm
ii) 16𝜙𝑙 = 16x25 =400mm
iii) 48𝜙𝑡 = 48x8 = 384mm
iv) 300mm

Pitch = 230 𝑚𝑚 𝑐⁄𝑐

Provide 8 nos of 25mm dia bars as longitudinal reinforcement.

Provide 8mm dia lateral ties @ 230 𝑚𝑚 𝑐⁄𝑐 pitch.

(Fig.No-4.3.2.2)
43
3. DESIGN COLUMN ( BI-AXIALLY LOADED)

Data:

Factored Load,Pu = 525kN

Size of column =230×230 mm

Moment along y-axis = 11.875 kNm

Moment along z-axis = 3.69 kNm

Solution:
Pu 525×103
= =0.5
𝑓𝑐𝑘 𝑏 𝐷 20×230×230

40
d’/D = =0.17
230

Step 1:Select Chart 46 of SP-16 :

𝑀𝑢𝑦1
=0.09
𝑓𝑐𝑘 𝑏 𝐷2

𝑀𝑢𝑦1 = 0.09×20×230×2302 =21.9 kNm (Fig.No-4.3.3.1)

40
d’/b = =0.17
230

Muz1 = 21.9 kNm

Refer Chart 63 of SP-16 pg.no:148

( Puz/Ag) =13

Puz = 13×230×230 =687 kN

(Pu/ Puz) =0.76>0.2 ∝𝑛 =2

Step 2:To Check the Safety of Column

𝑀𝑢𝑦 𝑀𝑢𝑧
( )∝𝑛 + ( )∝𝑛 ≤1.0
𝑀𝑢𝑦1 𝑀𝑢𝑧1

17.8 2 5.85 2
( ) +( ) =0.72 ≤1.0
21.9 21.9

44
 The Column is Safe.

Step 3: To find Asc

Assume 2% of Asc
2
Asc = ×b×D = 1058 mm2
100

Step 4: No of Bars
Asc 𝟏𝟎𝟓𝟖
= =𝝅 = 12nos
𝒂𝒔𝒄 ×𝟏𝟐𝟐
𝟒

Step 5: Design of lateral ties

𝜙𝑙 12
𝜙𝑡 < 6mm (or) = = 3mm
4 4

Pitch of lateral ties

i) b = 230mm
ii) 16𝜙𝑙 = 16x12 = 192mm
48𝜙𝑡 = 48x6 = 288mm
iii) 300mm

Pitch = 190 𝑚𝑚 𝑐⁄𝑐

Provide 12 nos of 12mm dia bars as longitudinal reinforcement.

Provide 6mm dia lateral ties at 190 𝑚𝑚 𝑐⁄𝑐 pitch.

(Fig.No-4.3.3.2)
45
 4.4DESIGN OF FOOTING
1.Footing(Bi-axially loaded Column)
Data:
Factored load,Pu =525kN
Size of column =230×230mm
SBC =200kN/m2
𝛾 =20kN/m3
fck = 20N/mm²
fy = 415N/mm²
Solution:
Step 1:Depth of Foundation
Assume φ=300
𝑆𝐵𝐶 1−sin φ 2
D = [ ]
𝛾 1+sin φ
200 1−sin 30 2
D = [ ]
20 1+sin 30
D =1.1m>0.5m
Step 2: Area of Footing
1.1×𝑃𝑈
Area of footing =
𝑆𝐵𝐶
1.1 ×525
=
200
Area of footing =2.89m2
A =B2
B=1.78m B~1.8m
Area of footing (provided) =1.8×1.8 =3.24m2
1.1×525
Net upward pressure = =178.24kN/m2<SBC
3.24
Step 3:Depth of Footing
Bending moment:
1.5×𝑊×L2 1.5×178.24×0.7852
Mu= = = 82.38kNm
2 2
Mu =82.38kNm
Depth of footing:
Mu(max) =0.138fckbd2
82.38×106 =0.138×20×1000×d2
d =172mm~180mm

46
NOTE:
The depth of footing has to be increased 1.5 to 2 times considering the shear
requirement.
Effective depth of footing =2×180=360mm.
Assume clear cover =60mm.
D=360+60=420mm
D=420mm

Step4:To find Ast

Mu =0.87fyAst (d-0.42Xu)
82.38×106=129978Ast-7.604 Ast2
Ast=659mm2
Spacing:
𝜋
×122 ×1000
4
S= =158mm~150mm.
659
Provide 12mm 𝑑𝑖𝑎 at 150mmc/c along Both ways.
Check for one way shear:
𝑣
𝜏 v= 𝑢
𝑏𝑑
Vu=1.5[𝑊 × (𝐿 − 𝑑)]
=1.5[178.24 × 0.425]
vu=112.63kN
𝑣 112.63×103
𝜏 v= 𝑢 = =0.31N/mm2
𝑏𝑑 1000×380
𝝉 c:
Refer table 19 of IS456-2000 pg.no.73
100 𝐴𝑠𝑡 100×659
%Ast= =
𝐵×𝐷 1000×420
%Ast=0.17
P=0.17% fck=20N/mm2 𝜏 c=0.32N/mm2
τc > 𝜏 v
Check for two way shear (or) Punching shear:
𝜏 v > 𝜏𝑐̀
𝜏𝑐̀ = τc K
K= (0.5+𝛽) >1.0
𝜏𝑐 =0.25√𝑓𝑐𝑘

47
 𝜏𝑐 =1.12N/mm2
𝜏𝑐̀ =1.12×1=1.12 N/mm2
Critical section for two way shear (or) punching shear @ d/2 distance all-
round the face of column.
Shear force along punching area,
𝑉𝑢
𝜏𝑣 =
𝑏.𝑑
Vu=W[𝐴 − 𝑃𝑢𝑛𝑐ℎ𝑖𝑛𝑔 𝑠ℎ𝑒𝑎𝑟]
=178.24[(1.8 × 1.8) − (0.59 × 0.59)]
Vu=514.45kN
514.45×103
𝜏𝑣 =
4×590×360
𝜏𝑣 =0.6N/mm2
𝜏𝑐̀ > 𝜏 v
Hence safe in two way shear.

(Fig.No-4.4.1.1)
48
2.Footing(Axially loaded Column)
Data:
Factored load,Pu =1482kN
Size of column =230×230mm
SBC =200KN/m2
𝛾 =20KN/m3
fck = 20N/mm²
fy = 415N/mm²
Solution:
Step 1:Depth of Foundation
Assume φ=300
𝑆𝐵𝐶 1−sin φ 2
D = [ ]
𝛾 1+sin φ
200 1−sin 30 2
D = [ ]
20 1+sin 30
D =1.1m>0.5m
Step 2: Area of Footing
1.1×𝑃𝑈
Area of footing =
𝑆𝐵𝐶
1.1 ×1482
=
200
Area of footing =8.15m2
A =B2
B=2.85m B~2.9m
Area of footing (provided) =2.9×2.9 =8.41m2
1.1×1482
Net upward pressure = =193kN/m2<SBC
8.41
Step 3:Depth of Footing
Bending moment:
1.5×𝑊×L2 1.5×193×1.3352
Mu= = = 257kNm
2 2
Mu =257kNm
Depth of footing:
Mu(max) =0.138fckbd2
257×106 =0.138×20×1000×d2
d =305mm~310mm

49
NOTE:
The depth of footing has to be increased 1.5 to 2 times considering the shear
requirement.

Effective depth of footing =2×310=620mm.

Assume clear cover =60mm.
D=620+60=680mm
D=680mm
Step 4:To find Ast
Mu =0.87fyAst (d-0.42Xu)
0.42×0.87×415×𝐴𝑠𝑡
257× 106=0.87×415×Ast [620 –( )]
0.36×20×1000
257×106=223851Ast-7.604 Ast2
Ast=1196mm2
Spacing:
𝜋
×122 ×1000
4
S= =102mm~100mm.
1196
Provide 12mm 𝑑𝑖𝑎 at 100mmc/c along Both ways.
Check for one way shear:
𝝉 v:
𝑣
𝜏 v= 𝑢
𝑏𝑑
vu=1.5[𝑊 × (𝐿 − 𝑑)]
=1.5[193 × 0.715]
vu=206kN
𝑣 206×103
𝜏 v= 𝑢 = =0.32N/mm2
𝑏𝑑 1000×620
𝝉 c:
Refer table 19 of IS456-2000 pg.no.73
100 𝐴𝑠𝑡 100×1196
%Ast= =
𝐵×𝐷 1000×680
%Ast=0.19%
P=0.19% fck=20N/mm2 𝜏 c=0.33N/mm2
τc > 𝜏 v
Check for two way shear (or) Punching shear:
𝜏 v > 𝜏𝑐̀
𝜏𝑐̀ = τc K
K= (0.5+𝛽) >1.0

50
 𝜏𝑐 =0.25√𝑓𝑐𝑘
𝜏𝑐 =1.12N/mm2 𝜏𝑐̀ =1.12×1=1.12 N/mm2

Critical section for two way shear (or) punching shear @ d/2 distance all-
round the face of column.
Shear force along punching area,
𝑉𝑢
𝜏𝑣 =
𝑏.𝑑
Vu=W[𝐴 − 𝑃𝑢𝑛𝑐ℎ𝑖𝑛𝑔 𝑠ℎ𝑒𝑎𝑟]
=193×[(2.9 × 2.9) − (0.85 × 0.85)]
Vu=1483kN
1483×103
𝜏𝑣 =
4×850×620
𝜏𝑣 =0.7N/mm2
𝜏𝑐̀ > 𝜏 v
Hence safe in two way shear.

(Fig.No-4.4.2.1)
3.Footing(Uni-Axially loaded Column)
51
Data:
Factored load,Pu =733.47kN
Size of column =230×230mm
SBC =200kN/m2
𝛾 =20kN/m3
fck = 20N/mm²
fy = 415N/mm²
Solution:
Step 1:Depth of Foundation
Assume φ=300
𝑆𝐵𝐶 1−sin φ 2
D = [ ]
𝛾 1+sin φ
200 1−sin 30 2
D = [ ]
20 1+sin 30
D =1.1m>0.5m
Step 2: Area of Footing
1.1×𝑃𝑈
Area of footing =
𝑆𝐵𝐶
1.1 ×733.47
=
200
Area of footing =4.03m2
A =B2
B=2.08m B~2.1m
Area of footing (provided) =2.1×2.1 =4.41m2
1.1×733.47
Net upward pressure = =182.95kN/m2<SBC
4.41
Step 3:Depth of Footing
Bending moment:
1.5×𝑊×L2 1.5×183×0.9352
Mu= = = 119.98kNm
2 2
Mu =120kNm
Depth of footing:
Mu(max) =0.138fckbd2
120×106 =0.138×20×1000×d2
d =208.5mm~210mm

NOTE:

52
 The depth of footing has to be increased 1.5 to 2 times considering the shear
requirement.

Effective depth of footing =2×210=420mm.

Assume clear cover =60mm.
D=420+60=480mm
D=480mm
Step 4:To find Ast
Mu =0.87fyAst (d-0.42Xu)
0.42×0.87×415×𝐴𝑠𝑡
120× 106=0.87×415×Ast [420 –( )]
0.36×20×1000
120×106=151641Ast-7.604 Ast2
Ast=825mm2
Spacing:
𝜋
×122 ×1000
4
S= =130mm.
825
Provide 12mm 𝑑𝑖𝑎 at 130mmc/c along Both ways.
Check for one way shear:
𝝉 v:
𝑣
𝜏 v= 𝑢
𝑏𝑑
vu=1.5[𝑊 × (𝐿 − 𝑑)]
=1.5[183 × 0.515]
vu=141.4kN
𝑣 141.4×103
𝜏 v= 𝑢 = =0.32N/mm2
𝑏𝑑 1000×420
𝝉 c:
Refer table 19 of IS456-2000 pg.no.73
100 𝐴𝑠𝑡 100×825
%Ast= =
𝐵×𝐷 1000×480
%Ast=0.19%
P=0.19% fck=20N/mm2 𝜏 c=0.335N/mm2
τc > 𝜏 v
Check for two way shear (or) Punching shear:
𝜏 v > 𝜏𝑐̀
𝜏𝑐̀ = τc K
K= (0.5+𝛽) >1.0
𝜏𝑐 =0.25√𝑓𝑐𝑘
53
 𝜏𝑐 =1.12N/mm2 𝜏𝑐̀ =1.12×1=1.12 N/mm2
Critical section for two way shear (or) punching shear @ d/2 distance all-
round the face of column.
Shear force along punching area,
𝑉𝑢
𝜏𝑣 =
𝑏.𝑑
Vu=W[𝐴 − 𝑃𝑢𝑛𝑐ℎ𝑖𝑛𝑔 𝑠ℎ𝑒𝑎𝑟]
=183×[(2.1 × 2.1) − (0.65 × 0.65)]
Vu=729kN
729×103
𝜏𝑣 =
4×650×420
𝜏𝑣 =0.67N/mm2
𝜏𝑐̀ > 𝜏 v
Hence safe in two way shear.

(Fig.No-4.4.3.1)
4.5 STAIRCASE DESIGN

54
Given:
Height of floor = 3.6m
Riser = 150mm
Tread = 300mm
Width of landing = 1.45m
Live load = 3kN/m2
Floor finish = 0.6kN/m2
Solution:
No of steps = 3600/150 = 24nos (Fig.No-4.5.1)
No of tread = 12nos
No of riser = 11nos
Length of going = 11×300 = 3300mm

Step 1: Effective depth

3300/d = 20
d = 165mm
D = 165+20 = 185mm
Length of inclination, x = √3002 + 150² =335.4mm
Step 2: Loading on going
Self weight of slab = 5.164kN/m2
(25×0.185×(0.335÷0.3))
Self weight of steps = 0.56kN/m2
(25×(1/2×0.15×0.3))
Thread finish(0.6×0.3) = 0.18kN/m2
Live load = 3kN/m2
Total = 8.91kN/m2
Factored load (1.5×8.91) = 13.36kN/m2
Step 3:Loading on landing
Self weight of slab = 4.625kN/m2
(25×0.185)
Floor finish = 0.6kN/m2
Live load = 3kN/m2
Total = 8.225kN/m2
Factored load(1.5×8.225) = 12.34kN/m2

Step 4: Reaction RA & RB

55
Moment about A
RB×4.9-12.34×0.8×(4.9-
(0.8/2))-13.36×3.3×(4.9/2)-
12.34×(0.8/2)2
RB= 31.92kN
(Fig.No-4.5.2)
RA =RB

Moment at center
MC =RB×(4.9/2)-(12.34×0.8×2.45-(0.8/2))-13.36×(3.3/2)×(3.3/4)
MC= 39.77kNm
Step 5: To find Ast
Mu=0.87fyAst (d-0.42Xu)
0.42×0.87×415×𝐴𝑠𝑡
39.77×106 = 0.87×415×Ast [185 –( )]
0.36×20×1000
Ast=642mm2
Spacing:
𝜋
×122 ×1000
4
S=
642
=160mmc/c
Provide 12mm dia at 160mmc/c as main reinforcement
Distribution Reinforcement:
0.12
Ast(min)= ×b×D
100
0.12
= ×1000×185
100
=186mm2
Spacing:
𝜋
×82 ×1000
4
S=
220
=220mmc/c
Provide 8mm dia at 220mmc/c as distribution reinforcement

56
(Fig.No-4.5.3)

57
 ESTIMATION

S.NO DETAILS QUANTITY UNIT RATE PER AMOUNT

(Rs) (Rs)

1. Area of Ground floor 821 m² 22,000 1,80,62,000

2. Area of First floor 764.6 m² 19,000 1,45,27,400

3. Area of Second floor 764.6 m² 20,000 1,52,92,000

Total construction cost

[CIVIL WORK] 4,78,58,000

4. Water supply& Sanitary

installation @ 5% of cost of
construction 5 % 23,92,900

5. Electrical work @ 2% of 9,57,160

construction cost 2 %

6. Services(EB, Drainage, Water 28,71,480

supply) @ 6% of construction
cost 6 %

Total 62,21,540

7. Contingencies(incidental and 14,35,740

miscellaneous expenses) at 3%
of construction cost 3 %

8. supervision charges at 8 % of 38,28,640

total cost of construction 8 %

Total 52,64,380

Grand Total 5,93,43,920

58
 PROGRAM OUTCOMES

PO1 Engineering Knowledge

Apply the knowledge of mathematics,science,engineering fundamentals and an

engineering specialization to the solution of complex engineering problems.

PO2 Problem Analysis

Identify, formulate ,review research literature,and analyze complex engineering

problems reaching substantiated conclusions using first principles of
mathematics, natural sciences and engineering sciences.

PO3 Design and Development of Solutions

Design solutions for complex engineering problems and design system

components or processes that meet the specified needs with appropriate
consideration for the public health and safety ,and the cultural, societal and
environmental considerations.

PO4 Conduct investigations of Complex problems

Use research-based knowledge and research methods including design of

experiments, analysis and interpretation of data and synthesis of the information
to provide valid conclusions.

PO5 Modern Tool usage

Create, select and apply appropriate techniques resources and modern

engineering and IT tools including prediction and modeling to complex
engineering activities with an understanding of the limitations.

PO6 The Engineer & Society

Apply reasoning informed by the contextual knowledge to assess societal,health,

safety,legal and cultural issues and the consequent responsibilities relevant to the
professional engineering practice.

59
PO7 Environment & Sustainability

Understand the impact of the professional engineering solutions in societal and

environmental contexts and demonstrate the knowledge of and need for
sustainable development.

PO8 Ethics

Apply ethical principles and commit to professional ethics and responsibilities

and norms of the engineering practice.

PO9 Individual and Team Work

Function effectively as an individual and as a member or leader in diverse teams

and in multidisciplinary settings.

PO10 Communication

Communicate effectively on complex engineering activities with the engineering

community and with society at large such as being able to comprehend and write
effective reports and design documentation, make effective presentation and give
and receive clear instructions.

PO11 Project Management and Finance

Demonstrate knowledge and understanding of the engineering and management

principle and apply these to one’s own work as a member and leader in team to
manage project and in multidisciplinary environments.

PO12 Lifelong Learning

Recognize the need for and have the preparation and ability to engage in
independent and life-long learning in the broadest context of technological
change.

60
 JUSTIFICATION OF PROGRAM OUTCOMES

PO1: ENGINEERING KNOWLEDGE

Justification: We applied the knowledge of engineering fundamentals,

engineering specification. Hence it is highly correlated with this outcome.

PO2: PROBLEM ANALYSIS

Justification: We have identified the problem, formulated, analysed and designed

the structure. Hence it is highly correlated with this outcome.

PO3: DESIGN AND DEVELOPMENT OF SOLUTIONS

We have given a design solution for complex engineering problem like design of
hospital building which we hope it will meet the specified needs with appropriate
consideration for the public health and safety, and the cultural, societal and
environmental considerations. Hence our project is highly correlated with this
outcome.

PO4: CONDUCT INVESTIGATIONS OF COMPLEX PROBLEMS

We used research-based knowledge and research methods including data

collection, review of past reports, arriving at aim and objectives, analysis and
design of structure, interpretation of data in the form of reports/presentations to
provide valid conclusions. This justifies for a high correlation with this outcome.

PO5: MODERN TOOL USAGE

Justification: We have used modern software’s like Staad Pro and AutoCAD in
our project. Hence it is highly correlated.

PO6: THE ENGINEER & SOCIETY

Justification: This project involves the contextual knowledge to assess societal,

health, safety, legal and cultural issues like treating the patients with multi-
diseases.

PO7: ENVIRONMENT & SUSTAINABILITY

Justification: This project comprises the impact of professional engineering

solutions in society and environmental context and need for sustainable
development. For example, we studied about design of hospital building in this
project which may have impact in societal sustainability for future generation.

61
PO8: ETHICS

Justification: We acquired the knowledge and understood the responsibility and

norms of engineering practice liking writing reports, communicate and interpret
the report to the engineering society, non-plagriasm. Hence it is highly
correlated.

PO9: INDIVIDUAL AND TEAM WORK

We did individual as well as team work related to various topics of this project.
Hence our project is highly correlated with this outcome.

PO10: COMMUNICATION

We had a chance to communicate effectively on our project activities with the

engineering community and non-engineering community and with society at
large, such as being able to comprehend and write effective reports and design
documentation which ultimately resulted to learn how to make effective
presentations. Hence our project has high correlation with this PO.

PO11: PROJECT MANAGEMENT AND FINANCE

Our project has low correlation with this outcome since our project do not
involve any project management and financial aspects, except the approximate
estimation part.

PO12: LIFELONG LEARNING

Justification: The knowledge we gained from this project work/activities will

definitely be helpful to learn more and more - both in academic as well as in
non-academic. So, a high level of correlation exists between our project and the
last outcome of the Civil Engineering Program.

62
 Project – PO Mapping

PO10

PO11

PO12
PO1

PO2

PO3

PO4

PO5

PO6

PO7

PO8

PO9
CE6712 3 3 3 2 2 2 2 2 3 2 1 3
Design
Project

3 – Highly Correlated, 2 – Moderately Correlated, 1- Low Correlation. – No

Correlation.

63
 CHAPTER 5

CONCLUSION

In this project the work is done in such a manner to cover all the major
aspects of a marriage hall to be provided. The design work is carried out by
taking into the consideration of loads. The manual designs of various structural
elements were carried out s per IS codes. The marriage hall is planned in such a
way as to provide the necessary needs required for the people around its locality.

The fundamental principle of analysis and design of framed building was

applied during project work. During this design process, we have to learn various
design methodology and basic concept of project. This concept will help us to
develop our knowledge in the process of analysis and design.

Hence an overall approach has been made to put forward for the project to
be economical, energy efficient and application oriented. Thus our aim of
proposing a marriage hall is being satisfied through the project.

64
 REFERENCE

1. Punmia, B.C., “Reinforced Concrete Design”, 8th Edition, Laxmi

Publications, New Delhi 110 051.
2. Shah, H.J., “Reinforced Concrete Design – Vol.II”, Charotar Publishers,
New Delhi.
3. Krishnaraju, N, “Structural Engineering Design and Drawing – RCC and
Steel”, Oxford Publishers, New Delhi.
4. IS:456-2000, “Code of Practice for Plain and Reinforced Concrete”,
Bureau of Indian Standards, New Delhi.
5. SP: 16, “Design Aids for Reinforced Concrete for IS: 456-1978”, Bureau
of Indian Standards, New Delhi.

65