U18CEP7703 PROJECT PHASE-1
PLANNING ANALYSIS AND DESIGNING OF
RESIDENTIAL APARTMENT
DESIGN PROJECT REPORT
Submitted by
KISHORE BABU A - 18BCE085
LINGESHWARAN A - 18BCE091
MADHAN DEV M - 18BCE109
in partial fulfilment for the award of the degree of
BACHELOR OF ENGINEERING
IN
CIVIL ENGINEERING
KUMARAGURU COLLEGE OF TECHNOLOGY
COIMBATORE - 641049
JANUARY 2022
U18CEP7703 PROJECT PHASE-1
PLANNING ANALYSIS AND DESIGNING OF
RESIDENTIAL APARTMENT
DESIGN PROJECT REPORT
Submitted by
KISHORE BABU A - 18BCE085
LINGESHWARAN A - 18BCE091
MADHAN DEV M - 18BCE109
in partial fulfilment for the award of the degree of
BACHELOR OF ENGINEERING
IN
CIVIL ENGINEERING
KUMARAGURU COLLEGE OF TECHNOLOGY
COIMBATORE - 641049
JANUARY 2022
KUMARAGURU COLLEGE OF TECHNOLOGY
COIMBATORE - 641049
BONAFIDE CERTIFICATE
Certified that this report titled “PLANNING ANALYSIS AND DESIGNING OF
RESIDENTIAL APARTMENT” is a bonafide work of KISHORE BABU A
(18BCE085), LINGESHWARAN A (18BCE091), MADHAN DEV M (18BCE109)
who carried out the work under my supervision. Certified further that to the best of my
knowledge the work reported herein does not form part of any other thesis or
dissertation on the basis of which a degree or award was conferred on an earlier
occasion on this or any other candidate.
Dr.V.Selvan Mr.J.Viswanath
Associate Professor and Head Associate Professor
Department of Civil Engineering Department of Civil Engineering
Kumaraguru Collegeof Technology Kumaraguru College ofTechnology
Coimbatore - 641049 Coimbatore - 641049
Submitted for the project viva voice examination held on …………
INTERNAL EXAMINER EXTERNAL EXAMINER
iii
ACKNOWLEDGEMENT
First of all, we express our profound gratefulness to Dr.D.Saravanan,
Principal, Kumaraguru College of Technology for providing all the necessary lab
facilities that helped us to complete the project effectively.
We deem great pride in expressing heartfelt gratitude to Dr.V.Selvan, Associate
Professor and Head, Department of Civil Engineering, Kumaraguru College of
Technology for being concerned about the student’s progress always.
We heartfully express our gratitude and regards to our project guide
Mr.J.Viswanath, Associate Professor, Department of Civil Engineering, Kumaraguru
College of Technology for her moral support, valuable guidance and constant
encouragement in carrying out our design project which played a vital role in
motivating us to learn throughout this venture.
Finally, we would like to acknowledge our teachers, parents and friends who have
helped us in gathering many realistic information and completing our project with ease.
KISHORE BABU A - 18BCE085
LINGESHWARAN A - 18BCE091
MADHAN DEV M - 18BCE109
iv
ABSTRACT
Our project location is in Fairlands is a rapidly growing town in Salem district in
the Indian state of Tamil Nadu. Structural planning and design is an art and science of
designing with economy and durable structure.The entire process of structural planning
and designing requires not only imagination and conceptual thinking,such as relevant
design codes.
For this purpose a site is selected in which the building has four floors including
a ground floor, It consists of all the rooms required for a residential house like bed
room,dining,toilets,hall and kitchen. This report is prepared and presented with the
details of structural features and design calculations. This particular layout has been
prepared to suit the requirements of the place while keeping the economy of
construction in proper perspective. The concrete mix used for slabs, beams, footings
and columns is M20 Concrete. The steel used for all the members is a high yield
strength deformed bars of grade Fe415.The entire structure is modelled using STAAD
Pro Software and sketched in Auto CAD software. The design output is available in
the STAAD Pro software as well as the manual design is also performed for beams,
columns, footings and slabs.
In this project work,an attempt is made according to building bye laws and
design of RESIDENTIAL APARTMENT as per IS:456-2000,and SP-34 specification.
Keywords: RESIDENTIAL APARTMENT, STAAD Pro, Auto CAD, Manual design
v
TABLE OF CONTENTS
CHAPTER TITLE PAGE NO
ABSTRACT IV
LIST OF FIGURES 8
LIST OF SYMBOLS AND ABBREVIATIONS 9
1. INTRODUCTION
1.1 GENERAl 1
1.1.1 GENERAL SPECIFICATIONS OF 2
RESIDENTIAL BUILDING
1.2 NEED FOR THE STUDY 3
2 SURVEY WORK/DATA COLLECTION
2.1 SITE LOCATION 4
2.1.1 CRITERIA FOR SITE LOCATION 4
2.2 CODAL PROVISION REFERRED 5
2.3 PROJECT DETAILS 6
vi
2.3.1 GENERAL SPECIFICATIONS 6
2.3.2 SPECIAL FEATURES 6
3 DRAWINGS (DRAFTING)
3.1PLAN 8
3.1.1 FIRST FLOOR 8
3.1.2 SECOND FLOOR 9
3.1.3 THIRD FLOOR 10
3.2 SECTIONAL VIEW 11
3.3 STRUCTURAL DRAWINGS 12
3.3.1 COLUMN LOCATION AND ORIENTATION 12
4 STRUCTURAL ANALYSIS AND DESIGN
4.1 STRUCTURAL ANALYSIS 13
4.1.1 ANALYSIS PROCEDURE 14
4.1.2 ANALYSIS OF FORCES 15
4.1.3 OUTPUT FILE 16
4.2 DESIGN OF STRUCTURAL ELEMENTS 18
4.2.1 DESIGN OF SLAB 19
4.2.2 DESIGN OF BEAM 27
4.2.3 DESIGN OF COLUMN 31
4.2.4 DESIGN OF FOOTING 34
vii
5 SUMMARY AND CONCLUSIONS 39
REFERENCES 40
6 ANNEXURE 41
viii
viii
LIST OF FIGURES
FIGURE TITLE PAGE NO.
3.1 FIRST FLOOR PLAN 8
3.2 SECOND FLOOR PLAN 9
3.3 THIRD FLOOR PLAN 10
3.4 SECTIONAL VIEW OF THE BUILDING 11
3.5 COLUMN LOCATION AND ORIENTATION 12
4.1 STRUCTURAL ANALYSIS 13
4.2 GRAPHICAL REPRESENTATION OF THE SUPPORTS 14
4.3 SECTIONAL VIEW OF THE BUILDING 15
4.4 GRAPHICAL REPRESENTATION OF THE BEAMS 15
4.5 LOAD DISTRIBUTION DUE TO SELF WEIGHT 16
4.6 RENDERED VIEW OF THE BUILDING 16
4.7 BEAM DESIGN OUTPUT 17
4.8 COLUMN DESIGN OUTPUT 18
4.9 DESIGN OF SLAB 19
4.8 DESIGN OF BEAM 27
4.9 DESIGN OF COLUMN 31
4.10 DESIGN OF FOOTING 34
ix
LIST OF SYMBOLS AND ABBREVIATIONS
fck = Characteristic compressive strength
Fy = Characteristic strength of steel
Ast = Area of steel reinforcement
Asc = Area of compression steel
Ag = Gross area of section
Asv = Cross section area of vertical stirrups
Mu = Design moment
Pu = Factored axial load
Pt = Percentage of steel
Sv = Spacing of stirrups
BM = Bending Moment
dl = Effective cover of beam or slab
B = Breadth of beam and slab
D = Overall depth of beam or slab
B = Breadth of column
D = Effective depth of beam or slab
MR = Moment of Resistance
P = Axial load
V = Shear force
W = Total load
SBC = Soil Bearing Capacity
M = Modular ratio
L = Length of column
c/c = Centre to Centre
Dia = Diameter of rods
1
CHAPTER-1
INTRODUCTION
Every human being has an inherent liking for a peaceful environment
needed for his pleasant living, this object is achieved by having a place of living
situated at the safe and convenient living requires considered and kept in view. Our
project is based on the design and analysis of the multi storied buildings. Analysis is
done through using the STAAD PRO. Notation adopted throughout the project is
same as in IS 456-2000.Multistory buildings are very commonly seen in cities.
Construction of such tall buildings are possible only by going to a set of rigidly
interconnected beams and columns of multi bay and multi storied are called building
frames.
The load from the walls and beams are transformed to beams, rotation of
beams take place. since, beams are rigidly connected to column, the rotation of
column also take place. Thus, any load applied anywhere on beam is shared by entire
network of beam and columns.
The engineer has to kept in mind the municipal conditions, building bye laws,
environment, financial capacity, water supply, sewage arrangement, provision of future,
aeration, ventilation etc.
In such building sleeping accommodation is provided. It includes the living room,
bathroom, kitchen, hall, toilet and master bedroom.
2
1.RESIDENTIAL APARTMENT
A RESIDENTIAL APARTMENT is the building which provides more than half
of its floor area for dwelling purposes. In other words, RESIDENTIAL APARTMENT
provides sleeping accommodation with or without cooking or dining or both facilities.
RESIDENTIAL APARTMENTS are divided into following types are individual houses
or private dwellings, Lodging or rooming houses, Dormitories, Apartments and Hotel.
Apartments or Flats are big buildings which consists separate dwellings for
different families. Apartment will resides minimum three or more families living
independently of each other.
1.SPECIFICATIONS
⚫ SUPERSTRUCTURE AND ROOFING
Ground floor is taken to the height of 3.2 m roofing consists of RCC slab roofing 0.2
m thick. First floor wall is taken to the height of 3 m. Roof slab in the terrace is 0.12 m thick
and it is provided with weather proof course of 0.015 m .
⚫ FLOORING
It consists of 0.02 m thick CM plaster over a bed of cement concrete (1:4:8) 0.1 m
thick. The remaining part of basement is filled and compacted with sand.
⚫ LINTELS
RCC lintels on all openings like doors, windows are 0.15 m thick having a bearing of
0.15 m on either side
⚫ DOOR AND WINDOWS
Door,D1 - 1.50 × 1.40 m
Door,D2 - 0.78 × 0.78 m
Window,W - 1.37 × 0.22 m
Window,W1 - 1.23 × 0.19 m
Window,W2 - 1.48 × 0.22 m
Ventilator,V - 0.60 × 0.11 m
3
1.1 NEED FOR STUDY
• To study the Stakeholder necessities and define the project scope.
• To draw the plan, section and elevation of the proposed three star hotel
building .
• To draft the building using Auto CADD software as per NBC norms.
• To design the structural elements manually and to analyse the building
using Stadd pro software.
• To know about the design process clearly.
4
CHAPTER-2
SURVEY WORK/DATA COLLECTION
2.1 SITE LOCATION
Our Site is located at Fairlands Salem which is located at
2 km from Salem bus stand
5km From Salem railway station
4km from SONA college of technology.
2.1.1 CRITERIA FOR SITE SELECTION
Site should be developed or developing area.
Make sure that electric lines, pipelines, drainage, water lines areproperly
provided.
Site should be easy to access the road and railways.
Site should not be near to industries and factories.
The site should have easy access to schools and hospitals.
The soil condition should be good
ur site satisfies all the above criteria
5
2.2 CODAL PROVISIONS REFERRED
IS 456:2000 - Code of practice for plain and reinforced concrete
IS 800:2007- Code of practice for general construction in steel
National Building Code of India
Tamil Nadu Building bye laws code book
Design aids
6
2.3 PROJECT DETAILS
2.3.1 GENERAL SPECIFICATIONS
⚫ Our site Area: 8400 sq.ft
⚫ Build up Area: 6863 sq. ft
⚫ Height of the Building: 10 meters
⚫ Soil Type: Red Soil
⚫ SBC: 350N/𝑚𝑚2
⚫ Parking Area: groundfloor Parking.
⚫ North Facing, G+3 floor
⚫ Setbacks are provided as per NBC
⚫ All provisions are based on NBC norms.
2.3.2 SPECIAL FEATURES
⚫ Ventilation
⚫ Provision of sufficient Lift and staircase
⚫ Provision of sufficient Car Parking area
⚫ Solar Panel
⚫ Rainwater harvesting system
⚫ All provisions are based on NBC norms
7
3.DRAWING (DRAFTING)
3.1SITE LAYOUT
8
3.2 PLAN
3.2.1 FIRST FLOOR PLAN
Fig 3.5 FIRST FLOOR OF
RESIDENTIAL APARTMENT
9
3.2.2 SECOND FLOOR PLAN
Fig 3.5 SECOND FLOOR OF
RESIDENTIAL APARTMENT
10
3.2.3 THIRD FLOOR
Fig 3.7 THIRD FLOOR
RESIDENTIAL APARTMENT
11
3.3 SECTIONAL VIEW:
Fig 3.8 SECTIONL VIEW OF RESIDENTIAL
APARTMENT
12
3.4 STRUCTURAL DRAWING
3.4.1 COLUMN LOCATION AND ORIENTATION
Fig 3.10 COLUMN LOCATION AND ORIENTATION
13
CHAPTER 4
STRUCTURAL ANALYSIS AND DESIGN
4.1 STRUCTURAL ANALYSIS
The model is generated using STAAD Pro Software and the analysis of frames is
done using the stiffness method in the software. STAAD Pro is a general purpose tool for
performing the analysis and design of a wide variety of type of structures. The 3 basic
activities which are carried out by the software is,
1. Model Generation
2. The calculations to obtain the analytical results
3. Reports in post processing and verification of data
After the load calculations, the analysis part is the most important to check the
stability of the structure. Perform analysis is selected once and in the output file, the
results are displayed. Then the postprocessing tab helps to gather the reports of Shear
Forces, Bending Moments, Displacements and the Reaction forces can be determined.
This can be viewed as a Summary of the entire project or even for a single
node/beam/column. The loads given as an input are also considered along with the
combination of all maximum loads generated if given. This report ensures the stability
check for a structure.
14
4.1.1 ANALYSIS PROCEDURE
The procedure for analysis using the Staad Pro Software
1. The skeleton of the structure is formed using the node (coordinates) method.
2. Fixed support is provided at all nodes at the base.
3. Load definition are given.
4. Perform analysis command is selected.
5. The required design parameters are added and the concrete design for the beams and
columns is taken off.
6. The analysis is done using Perform Analysis option once and Run Analysis
option after several modifications.
7. The postprocessing tab is used for further report making purposes
Some of the Pictures showing the analysis process of the project are as follows:
FIG 4.1 GRAPHICAL REPRESENTATION OF SUPPORTS
15
FIG 4.2 SECTIONAL VIEW OF THE BUILDING
FIG 4.3 GRAPHICAL REPRESENTATION OF BEAMS
16
FIG 4.4 LOAD DISTRIBUTION DUE TO SELF WEIGHT
RENDERING VIEW
FIG 4.5 RENDERED VIEW OF THE BUILDING
17
4.1.2 OUTPUT FILE
The final output contains error free design results. The reinforcement details of
beams and columns are displayed in the output file. It also gives the quantity estimate of
the materials. Some results are as follows:
FIG 4.12 BEAM DESIGN OUTPUT
18
FIG 4.13 COLUMN DESIGN OUTPUT
19
4.2 DESIGN OF STRUCTUAL MEMBERS
4.2.1 DESIGN OF SLAB
Lx = 3.46
Ly = 3.48
Grade of concrete = M20
Grade of steel = Fe415
Types of slab
𝑙𝑦
𝑙𝑥
=
𝑙𝑜𝑛𝑔𝑒𝑟 𝑠𝑝𝑎𝑛
𝑠ℎ𝑜𝑟𝑡𝑒𝑟 𝑠𝑝𝑎𝑛
=
3.48
3.46
= 1 < 2
Hence it is a two-way slab.
Dimensions of slab
l/b = 40 x 0.8
D = 3460 / 40 x 0.8
= 110 mm
dx = D – cover
= 110 – 20
= 90 mm
dy = 90 -10
= 80 mm
20
B = 1000mm
Effective span
1. Centre to Centre supports = 3.46 + 0.26
= 3.72 m
2. Clear span + effective depth = 3.46 + 0.09
Effective span = 3.55
Load calculation
Self-weight of slab = b x D x unit weight of RCC
= 1 x 0.11 x 25
= 2.75 KN / m
Total load = self-weight + L.L + floor finish
= 2.75 + 4 + 1.5
= 8.25 KN / m
Load factor (Wu) = 1.5 x 8.25
= 12.37 KN/ m
Bending moment and shear force
Mx = αx x w x lx2
My = αy x w x lx2
Ly/ lx = 3.48 / 3.46
= 1
21
Short span coefficient
(-ve) moment coefficient = 0.032
(+ve) moment coefficient =0.024
Positive moment
Mx = αx x w x lx2
Mx = 0.024 x 12.37 x (3.55)2
Mx = 3.741 KN /m
Negative span
Mx = αx x w x lx2
Mx = 0.032 x 12.37 x (3.55)2
Mx = 4.988KN /m
Long span coefficient
(-ve) moment coefficient = 0.032
(+ve) moment coefficient =0.024
Positive moment
My = αx x w x lx2
My = 0.024 x 12.37 x (3.55)2
My = 3.741 KN /m
Negative span
My = αx x w x lx2
My = 0.032 x 12.37 x (3.55)2
22
My = 4.988KN /m
Shear force
V = W x L /2
= 12.37 x 3.55 / 2
= 21.96 KN
Reinforcement
For shorter span (- ve) moment
Mu limit = 0.138 fck b d2
= 0.138 x 20 x 1000 x (90)2
= 22.3 KNm
Mu < Mu limit Under reinforcement
Mu = 0.87 x fy x Ast x d [1-
𝐴𝑠𝑡 𝑋 𝑓𝑦
𝑏 𝑋 𝑑 𝑋 𝐹𝑐𝑘
]
4.988 x 10⁶ = 0.87 x 415 x Ast x 90 [1-
𝐴𝑠𝑡 𝑋 415
1000 𝑋 90 𝑋 20
]
Ast = 159.35mm2
Spacing of bar =
𝜋
4
𝑋 102
159.35
x 1000
= 492.6
Provide 10 mm dia bar at 300 mm c/c giving
For shorter span (+ve) moment
Mu = 0.87 x fy x Ast x d [1-
𝐴𝑠𝑡 𝑋 𝑓𝑦
𝑏 𝑋 𝑑 𝑋 𝐹𝑐𝑘
]
3.741 x 10⁶ = 0.87 x 415 x Ast x 90 [1-
𝐴𝑠𝑡 𝑋 415
1000 𝑋 90 𝑋 20
]
23
Ast = 118.3mm2
Spacing of bar =
𝜋
4
𝑋 102
118.3
x 1000
= 663.5
Provide 10 mm dia bar at 300 mm c/c giving
For longer span (-ve) moment
Mu limit = 0.138 x fck x b x d2
= 0.138 x 20 x 1000 x (80)2
= 17.664 KNm
Mu < mu limit Under reinforcement
Mu = 0.87 x fy x Ast x d [1-
𝐴𝑠𝑡 𝑋 𝑓𝑦
𝑏 𝑋 𝑑 𝑋 𝐹𝑐𝑘
]
4.988 x 10⁶ = 0.87 x 415 x Ast x 80 [1-
𝐴𝑠𝑡 𝑋 415
1000 𝑋 80 𝑋 20
]
Ast = 181.20mm2
Spacing of bar =
𝜋
4
𝑋 102
181.20
x 1000
= 433.2
Provide 10 mm dia bar at 300 mm c/c giving
For long span (+ve) moment
Mu = 0.87 x fy x Ast x d [1-
𝐴𝑠𝑡 𝑋 𝑓𝑦
𝑏 𝑋 𝑑 𝑋 𝐹𝑐𝑘
]
3.741 x 10⁶ = 0.87 x 415 x Ast x 80 [1-
𝐴𝑠𝑡 𝑋 415
1000 𝑋 80 𝑋 20
]
Ast = 134.17mm2
24
Spacing of bar =
𝜋
4
𝑋 102
134.17
x 1000
= 585.07
Provide 10 mm dia bar at 300 mm c/c giving
Check for shear
𝜏v < k x 𝜏c
𝜏v =
𝑣
𝑏 𝑋 𝑑
=
21.956 𝑋 103
1000 𝑋 90
𝜏v = 0.243N /mm2
Permissible shear
K = 1.3
100 𝑋 𝐴𝑠𝑡
𝑏 𝑑
=
100 𝑋 118.3
1000 𝑋 90
= 0.131
𝜏c = 0.79
𝜏c x k = 0.79 x 1.3
= 1.027N/mm2
𝜏v < 𝜏c x k
Slab is safe in shear
Check for deflection
(
𝑳
𝒅
)max = (
𝑳
𝒅
)basic x kt x kc x k f
25
(
𝑳
𝒅
) max = 40 x 1.68 x 1
= 67.2
(
𝑳
𝒅
) actual =
3.550
90
= 39.4
Hence it is safe
Check for crack
1. Steel provide is more the min percentage of 0.12%
2. Spacing of main steel < 3d or 300mm which is small.
3. Dia of reinforcement <
𝐷
8
𝐷
8
=
110
8
= 13.75mm
cracker will be within permissible limit.
26
RESULT
27
4.1.2 DESIGN OF BEAM:
Clear span = 3.4m
Wide of support = 0.267m
Stress and load factor
fck = 20 N/mm²
fy = 415
load factor = 1.5 for (D.L & L.L).
Dimension of beam
Effective depth = span / 15
= 340 /15
= 260
D = effective depth + clear cover
= 260 + 40
D = 300 ( assume b = 260 mm)
Effective span
1) Centre to Centre = 3.4 + 0.26
= 3.66
2) Clear span + effective depth = 3.4 + 0.26
= 3.66
Effective span is 3.66m
Load calculation
Self-weight of beam = 0.26 x 0.3 x 25
= 1.95KN/m
Parapet wall load = 1 x 0.26 x 20
= 5.2KN/m
Slab load =
𝑊 𝑋 𝑙𝑥
3
= 12.37 x 3.4 / 3
28
= 14.01KN/m
Total load = 1.5 (1.95 + 5.2) +14.01
= 24.73KN/m
Bending moment
Mu =
𝑤 𝑋 𝐿²
8
=
24.73 𝑋 (3.66)²
8
=41.40KN/m
Shear force
V = (w x L) / 2
= 43.82KN
Check for Mu and Mu limit
Mu limit = 0.138 x 20 x 260 x (260) ²
= 48.509KNm
Mu < Mu limit under reinforcement
Area of reinforcement
Mu = 0.87 x fy x Ast x d [1-
𝐴𝑠𝑡 𝑋 𝑓𝑦
𝑏 𝑋 𝑑 𝑋 𝐹𝑐𝑘
]
41.40 x 10⁶ = 0.87x415x Ast x260[1-
𝐴𝑠𝑡 𝑋 415
260 𝑋 260 𝑋 20
]
Ast = 525.9
No of bars =
525.9
𝜋
4
𝑋 16²
= 2.61 (3 bars)
Provide 3 bars of 16 mm dia.
Check for shear
Nominal shear
τv = v / b x d
= 45.25 x 103
/ 260 x 260
= 0.669N/mm²
Permissible shear
29
100 𝑋 𝐴𝑠𝑡
𝑏 𝑋 𝑑
=
100 𝑋 602.8
260 𝑋 260
= 0.89
τc = 0.59
τv > τc
provide permissible shear reinforcement using 2legged 8mm bar
Vus =
0.87 𝑋 𝑓𝑦 𝑋 𝐴𝑠𝑡 𝑋 𝑑
𝑆𝑣
Sv =
0.87 𝑋 415 𝑋 (2 𝑋
𝜋
4
𝑋 82𝑋 260)
𝑆𝑣
Sv = 1367.80mm
Check ͢ 0.75 d = 0.75 X 260
= 195 (300mm)
Provide 8mm dia bars of 2 legged stirrups with 200mm c/c spacing
Check for deflection
(
𝐿
𝑑
)max = 15 x 1.38 x 1
= 20.7
(
𝐿
𝑑
)actual = 3660/260
= 14.076
Hence it safe
30
RESULT
31
4.1.3 DESIGN OF COLOUMN
Pu = 2000KN
Dimension = 260 x 450
Effective length
Effective length = 0.65 x L
= 0.65 x 3
= 1.95 m
Slenderness ratio
L eff / b = 1950 / 260
= 7.5 < 12
Short column
Strength of short column
Pu = 0.4 fck x Ac + 0.67 fy Asc
2000 x 10³ = 0.4 x 20 x [(260 x 450)- Asc] + 0.67 x 415 Asc
1064 x 10³ = 270.05 Asc
Asc = 3933.45
Ac = As – Asc
Ac = (260 x 450) – 3933.45
= 113066.55
Minimum eccentricity
E min =
𝐿
500
+
𝑏
30
=
3000
500
+
450
30
= 6 + 15
= 21 mm
0.05 D = 0.05 x 450
= 22.5 mm
E < 0.050
32
Hence design is axially loaded short column
Longitudinal reinforcement
Asc = 3933.45
No of bars =
3933.45
𝜋
4
𝑋 25²
=
3933.45
490.625
= 8.01 (8 bars)
Provide 8 bars of 25mm dia
Asc provide = 8 x π / 4 x 25²
= 3925mm²
Lateral ties
1. Diameter of ties = 1 / 4 x 25
= 6.25 (8mm)
Provide 8mm dia ties
2. Spacing
• 16 ɸ = 16 x 25
= 400
• 48 d = 48 x 8
= 384
• 300mm
33
RESULT
34
4.1.2 DESIGN OF FOOTING
Beam 1
(Min spacing between individual rods for no.of. bars)
Slab: 3.05m x 2.3m
3.05
2.3
= 1.32 < 2
2.3
2
= 1.15
Area of trapezoidal
= 1/2 (a + b) h
=1/2(3.05 + 0.76) 1.15
= 2.2m²
Load = 2.2 x 7.125
=15.6KN
Beam 2:
Area of the triangle = 1 / 2 x 23 x 1.15
=1.32²
Load = 1.32 x 7.125
= 9.40KN
Load on column:
Self wt. of beam = 0.26 x 0.46 x 3.05 x 25
=8.92
Self wt. of beam 2 = 0.26 x 0.3 x 2.3 x 25
= 4.48
Wall load on beam 1 = 3.05 x 0.26 x 3.05 x 20
= 48.3
Wall load on beam 2 = 3.05 x 0.26 x 2.3 x 20
= 36.47
Load from beam 1:
35
Load 1 =
15.6+8.92+48.3
2
= 36.41KN
Load from beam 2:
Load 2 =
9.40+4.48+36.47
2
= 25 .17KN
Self wt. of column
=0.26 x 0.46 x 3.048 x 25
=9.11
Total on footing
=4 x (36.41 + 25.17 + 9.11)
= 282.76
=300KN
Design data
Column size = 260 x 450 mm²
SBC = 200 KN/m ²
Fe 415
M20
Load 300KN
Assume 10 % of load as self-weight
Self wt = 300 x 10/100
= 30 KN
Total load = 330 KN
Area =
𝑡𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑
𝑆𝐵𝐶
=
330
200
=1.65m²
L = 450 / 260
= 1.73
36
L x B = 1.65
2B² = 1.65
B = 0.90m
Provide
L = 2m
B = 1m
Size of footing = 2m x 1m
Net upward pressure (po) = 300 / 2 x 1
= 150KN/m²
= 150 X 1.5
Po = 225 kN/m²
Depth
Based on BM
Mx = po B (L-l) ² / 8
=225 x 1 x (2-0.46) ² /8
Mx = 66.70 KNm
My = po L (B-b) ² / 8
= 225 x 2 x (1- 0.46) ² / 8
My = 31.5 KNm
Taking max moment to calculate depth
Mu = 2.7 Bd²
66.70 x 10⁶ = 2.7 x 1000 x d²
D = 156.59
provide (d = 160mm) and (D= 210mm)
based on one-way slab:
critical section for shear is at a distance from face to column.
Vs =po B[
𝐿−𝑙
𝛼
-d]
=225×1[
2−0.46
2
− 0.16]
37
=225× [
1.5
2
− 0.16]
=132.75 KN
Vy =PoL [
𝐵−𝑏
𝑏
-d]
=225×2[
1−0.26
2
-0.16]
=94.5 KN
P = 0.21
Ꚍc = [
0.36−0.28
0.25−0.15
][0.21-0.15]
= 0.328 N/mm^2
Based on punching shear
=Po [LB-(l+d) (b+d)]
=225[2-(0.46+0.16) (0.26+0.16)]
=391.41 KN
Surface area
=2[(l+d) +(b+d)] d
=2[(0.46+0.16) (0.26+0.16)]0.16
=0.729 m^2
𝜏v =
391.41×10^3
0.729×10^6
=
391410
729000
=0.536
Ks =0.5+(
0.26
0.46
)
Ks =1.06>1
𝜏c =0.26√𝑓𝑐𝑘
𝜏c =1.16 N/mm^2
𝜏v < 𝜏c
Hence it is safe punching shear
Area of reinforcement (along longer direction)
38
Mx =0.87 fy Ast d[1-
𝐴𝑠𝑡 𝑓𝑦
𝑏𝑑 𝑓𝑐𝑘
]
66.70×10^6 =0.87×415×Ast×100[1-
𝐴𝑠𝑡×415
1000×500×20
]
Ast =175.10 mm^2
No of bars(along longer direction)
Assume 12mm dia bars
No of bars =
𝐴𝑠𝑡
𝐴𝑟𝑒𝑎 𝑜𝑓 1 𝑏𝑎𝑟
=
372.25
(𝜋÷4)12^2
=3.29
= 4 rods
Provide 4 no.s of 12mm dia bars along longer direction
As per IS 456,Pg:65,
Reinforcement in central band =
2
𝐵+1
B =2
Reinforcement in central band =
2
3
(175.10)
=116.73 mm^2
Reinforcement in edge strip =175.10-116.73
=58.37 mm^2
Assume 12mm dia bars
No of bars =
175.10
(𝜋÷4)10^2
=1.59
=2 rods
Provide 2 no of 12mm dia bars along shorter direction.
39
CHAPTER-5
SUMMARY AND CONCLUSIONS
CONCLUSION
• In this project, an attempt has been made to plan, analyse and design G+3
apartment based on the provisions of IS 456:2000.
• The structure composed of beams and columns is modeled in the Staad Pro
v8i Software and are analysed with various loading combinations.
• For detailing codebooks are referred and the detailed drawings of the
various components are presented.
• The design of components such as Beams, Columns and Slab are carried out
completely using Staad Pro and Manual Calculations.
• To plot structural drawings, the drafting software AutoCAD is used.
• We are now familiarized with Staad Pro v8i design software which is
majorly used for analysis of structures and modeling of structures.
• The design concepts studied for the various RCC elements used inthis
project helped us to reinforce our confidence in structural design area
40
REFERENCES
1) “S.Ramamrutham”,”Reinforced Concrete Structures”,New Delhi
2) “B.C.Punmia”,”Reinforced Concrete Structures”,New Delhi
3) IS 875 (Part-1) 1987, Code of practice for Design of loads for buildings
4) IS 456-2000, Plain and Reinforced Concrete Code of Practice
5) SP 16, Designs aids for Reinforced Concrete
41
CHAPTER 6
ANNEXURE
The lists of drawings attached in the reports are
1. Site layout
2. Ground floor plan
3. First floor plan
4. Second floor plan
5. Third floor plan
6. Section
7. Column orientation
8. Beam orientation

Final report 85,91,109

  • 1.
    U18CEP7703 PROJECT PHASE-1 PLANNINGANALYSIS AND DESIGNING OF RESIDENTIAL APARTMENT DESIGN PROJECT REPORT Submitted by KISHORE BABU A - 18BCE085 LINGESHWARAN A - 18BCE091 MADHAN DEV M - 18BCE109 in partial fulfilment for the award of the degree of BACHELOR OF ENGINEERING IN CIVIL ENGINEERING KUMARAGURU COLLEGE OF TECHNOLOGY COIMBATORE - 641049 JANUARY 2022
  • 2.
    U18CEP7703 PROJECT PHASE-1 PLANNINGANALYSIS AND DESIGNING OF RESIDENTIAL APARTMENT DESIGN PROJECT REPORT Submitted by KISHORE BABU A - 18BCE085 LINGESHWARAN A - 18BCE091 MADHAN DEV M - 18BCE109 in partial fulfilment for the award of the degree of BACHELOR OF ENGINEERING IN CIVIL ENGINEERING KUMARAGURU COLLEGE OF TECHNOLOGY COIMBATORE - 641049 JANUARY 2022
  • 3.
    KUMARAGURU COLLEGE OFTECHNOLOGY COIMBATORE - 641049 BONAFIDE CERTIFICATE Certified that this report titled “PLANNING ANALYSIS AND DESIGNING OF RESIDENTIAL APARTMENT” is a bonafide work of KISHORE BABU A (18BCE085), LINGESHWARAN A (18BCE091), MADHAN DEV M (18BCE109) who carried out the work under my supervision. Certified further that to the best of my knowledge the work reported herein does not form part of any other thesis or dissertation on the basis of which a degree or award was conferred on an earlier occasion on this or any other candidate. Dr.V.Selvan Mr.J.Viswanath Associate Professor and Head Associate Professor Department of Civil Engineering Department of Civil Engineering Kumaraguru Collegeof Technology Kumaraguru College ofTechnology Coimbatore - 641049 Coimbatore - 641049 Submitted for the project viva voice examination held on ………… INTERNAL EXAMINER EXTERNAL EXAMINER
  • 4.
    iii ACKNOWLEDGEMENT First of all,we express our profound gratefulness to Dr.D.Saravanan, Principal, Kumaraguru College of Technology for providing all the necessary lab facilities that helped us to complete the project effectively. We deem great pride in expressing heartfelt gratitude to Dr.V.Selvan, Associate Professor and Head, Department of Civil Engineering, Kumaraguru College of Technology for being concerned about the student’s progress always. We heartfully express our gratitude and regards to our project guide Mr.J.Viswanath, Associate Professor, Department of Civil Engineering, Kumaraguru College of Technology for her moral support, valuable guidance and constant encouragement in carrying out our design project which played a vital role in motivating us to learn throughout this venture. Finally, we would like to acknowledge our teachers, parents and friends who have helped us in gathering many realistic information and completing our project with ease. KISHORE BABU A - 18BCE085 LINGESHWARAN A - 18BCE091 MADHAN DEV M - 18BCE109
  • 5.
    iv ABSTRACT Our project locationis in Fairlands is a rapidly growing town in Salem district in the Indian state of Tamil Nadu. Structural planning and design is an art and science of designing with economy and durable structure.The entire process of structural planning and designing requires not only imagination and conceptual thinking,such as relevant design codes. For this purpose a site is selected in which the building has four floors including a ground floor, It consists of all the rooms required for a residential house like bed room,dining,toilets,hall and kitchen. This report is prepared and presented with the details of structural features and design calculations. This particular layout has been prepared to suit the requirements of the place while keeping the economy of construction in proper perspective. The concrete mix used for slabs, beams, footings and columns is M20 Concrete. The steel used for all the members is a high yield strength deformed bars of grade Fe415.The entire structure is modelled using STAAD Pro Software and sketched in Auto CAD software. The design output is available in the STAAD Pro software as well as the manual design is also performed for beams, columns, footings and slabs. In this project work,an attempt is made according to building bye laws and design of RESIDENTIAL APARTMENT as per IS:456-2000,and SP-34 specification. Keywords: RESIDENTIAL APARTMENT, STAAD Pro, Auto CAD, Manual design
  • 6.
    v TABLE OF CONTENTS CHAPTERTITLE PAGE NO ABSTRACT IV LIST OF FIGURES 8 LIST OF SYMBOLS AND ABBREVIATIONS 9 1. INTRODUCTION 1.1 GENERAl 1 1.1.1 GENERAL SPECIFICATIONS OF 2 RESIDENTIAL BUILDING 1.2 NEED FOR THE STUDY 3 2 SURVEY WORK/DATA COLLECTION 2.1 SITE LOCATION 4 2.1.1 CRITERIA FOR SITE LOCATION 4 2.2 CODAL PROVISION REFERRED 5 2.3 PROJECT DETAILS 6
  • 7.
    vi 2.3.1 GENERAL SPECIFICATIONS6 2.3.2 SPECIAL FEATURES 6 3 DRAWINGS (DRAFTING) 3.1PLAN 8 3.1.1 FIRST FLOOR 8 3.1.2 SECOND FLOOR 9 3.1.3 THIRD FLOOR 10 3.2 SECTIONAL VIEW 11 3.3 STRUCTURAL DRAWINGS 12 3.3.1 COLUMN LOCATION AND ORIENTATION 12 4 STRUCTURAL ANALYSIS AND DESIGN 4.1 STRUCTURAL ANALYSIS 13 4.1.1 ANALYSIS PROCEDURE 14 4.1.2 ANALYSIS OF FORCES 15 4.1.3 OUTPUT FILE 16 4.2 DESIGN OF STRUCTURAL ELEMENTS 18 4.2.1 DESIGN OF SLAB 19 4.2.2 DESIGN OF BEAM 27 4.2.3 DESIGN OF COLUMN 31 4.2.4 DESIGN OF FOOTING 34
  • 8.
    vii 5 SUMMARY ANDCONCLUSIONS 39 REFERENCES 40 6 ANNEXURE 41
  • 9.
  • 10.
    viii LIST OF FIGURES FIGURETITLE PAGE NO. 3.1 FIRST FLOOR PLAN 8 3.2 SECOND FLOOR PLAN 9 3.3 THIRD FLOOR PLAN 10 3.4 SECTIONAL VIEW OF THE BUILDING 11 3.5 COLUMN LOCATION AND ORIENTATION 12 4.1 STRUCTURAL ANALYSIS 13 4.2 GRAPHICAL REPRESENTATION OF THE SUPPORTS 14 4.3 SECTIONAL VIEW OF THE BUILDING 15 4.4 GRAPHICAL REPRESENTATION OF THE BEAMS 15 4.5 LOAD DISTRIBUTION DUE TO SELF WEIGHT 16 4.6 RENDERED VIEW OF THE BUILDING 16 4.7 BEAM DESIGN OUTPUT 17 4.8 COLUMN DESIGN OUTPUT 18 4.9 DESIGN OF SLAB 19 4.8 DESIGN OF BEAM 27 4.9 DESIGN OF COLUMN 31 4.10 DESIGN OF FOOTING 34
  • 11.
    ix LIST OF SYMBOLSAND ABBREVIATIONS fck = Characteristic compressive strength Fy = Characteristic strength of steel Ast = Area of steel reinforcement Asc = Area of compression steel Ag = Gross area of section Asv = Cross section area of vertical stirrups Mu = Design moment Pu = Factored axial load Pt = Percentage of steel Sv = Spacing of stirrups BM = Bending Moment dl = Effective cover of beam or slab B = Breadth of beam and slab D = Overall depth of beam or slab B = Breadth of column D = Effective depth of beam or slab MR = Moment of Resistance P = Axial load V = Shear force W = Total load SBC = Soil Bearing Capacity M = Modular ratio L = Length of column c/c = Centre to Centre Dia = Diameter of rods
  • 12.
    1 CHAPTER-1 INTRODUCTION Every human beinghas an inherent liking for a peaceful environment needed for his pleasant living, this object is achieved by having a place of living situated at the safe and convenient living requires considered and kept in view. Our project is based on the design and analysis of the multi storied buildings. Analysis is done through using the STAAD PRO. Notation adopted throughout the project is same as in IS 456-2000.Multistory buildings are very commonly seen in cities. Construction of such tall buildings are possible only by going to a set of rigidly interconnected beams and columns of multi bay and multi storied are called building frames. The load from the walls and beams are transformed to beams, rotation of beams take place. since, beams are rigidly connected to column, the rotation of column also take place. Thus, any load applied anywhere on beam is shared by entire network of beam and columns. The engineer has to kept in mind the municipal conditions, building bye laws, environment, financial capacity, water supply, sewage arrangement, provision of future, aeration, ventilation etc. In such building sleeping accommodation is provided. It includes the living room, bathroom, kitchen, hall, toilet and master bedroom.
  • 13.
    2 1.RESIDENTIAL APARTMENT A RESIDENTIALAPARTMENT is the building which provides more than half of its floor area for dwelling purposes. In other words, RESIDENTIAL APARTMENT provides sleeping accommodation with or without cooking or dining or both facilities. RESIDENTIAL APARTMENTS are divided into following types are individual houses or private dwellings, Lodging or rooming houses, Dormitories, Apartments and Hotel. Apartments or Flats are big buildings which consists separate dwellings for different families. Apartment will resides minimum three or more families living independently of each other. 1.SPECIFICATIONS ⚫ SUPERSTRUCTURE AND ROOFING Ground floor is taken to the height of 3.2 m roofing consists of RCC slab roofing 0.2 m thick. First floor wall is taken to the height of 3 m. Roof slab in the terrace is 0.12 m thick and it is provided with weather proof course of 0.015 m . ⚫ FLOORING It consists of 0.02 m thick CM plaster over a bed of cement concrete (1:4:8) 0.1 m thick. The remaining part of basement is filled and compacted with sand. ⚫ LINTELS RCC lintels on all openings like doors, windows are 0.15 m thick having a bearing of 0.15 m on either side ⚫ DOOR AND WINDOWS Door,D1 - 1.50 × 1.40 m Door,D2 - 0.78 × 0.78 m Window,W - 1.37 × 0.22 m Window,W1 - 1.23 × 0.19 m Window,W2 - 1.48 × 0.22 m Ventilator,V - 0.60 × 0.11 m
  • 14.
    3 1.1 NEED FORSTUDY • To study the Stakeholder necessities and define the project scope. • To draw the plan, section and elevation of the proposed three star hotel building . • To draft the building using Auto CADD software as per NBC norms. • To design the structural elements manually and to analyse the building using Stadd pro software. • To know about the design process clearly.
  • 15.
    4 CHAPTER-2 SURVEY WORK/DATA COLLECTION 2.1SITE LOCATION Our Site is located at Fairlands Salem which is located at 2 km from Salem bus stand 5km From Salem railway station 4km from SONA college of technology. 2.1.1 CRITERIA FOR SITE SELECTION Site should be developed or developing area. Make sure that electric lines, pipelines, drainage, water lines areproperly provided. Site should be easy to access the road and railways. Site should not be near to industries and factories. The site should have easy access to schools and hospitals. The soil condition should be good ur site satisfies all the above criteria
  • 16.
    5 2.2 CODAL PROVISIONSREFERRED IS 456:2000 - Code of practice for plain and reinforced concrete IS 800:2007- Code of practice for general construction in steel National Building Code of India Tamil Nadu Building bye laws code book Design aids
  • 17.
    6 2.3 PROJECT DETAILS 2.3.1GENERAL SPECIFICATIONS ⚫ Our site Area: 8400 sq.ft ⚫ Build up Area: 6863 sq. ft ⚫ Height of the Building: 10 meters ⚫ Soil Type: Red Soil ⚫ SBC: 350N/𝑚𝑚2 ⚫ Parking Area: groundfloor Parking. ⚫ North Facing, G+3 floor ⚫ Setbacks are provided as per NBC ⚫ All provisions are based on NBC norms. 2.3.2 SPECIAL FEATURES ⚫ Ventilation ⚫ Provision of sufficient Lift and staircase ⚫ Provision of sufficient Car Parking area ⚫ Solar Panel ⚫ Rainwater harvesting system ⚫ All provisions are based on NBC norms
  • 18.
  • 19.
    8 3.2 PLAN 3.2.1 FIRSTFLOOR PLAN Fig 3.5 FIRST FLOOR OF RESIDENTIAL APARTMENT
  • 20.
    9 3.2.2 SECOND FLOORPLAN Fig 3.5 SECOND FLOOR OF RESIDENTIAL APARTMENT
  • 21.
    10 3.2.3 THIRD FLOOR Fig3.7 THIRD FLOOR RESIDENTIAL APARTMENT
  • 22.
    11 3.3 SECTIONAL VIEW: Fig3.8 SECTIONL VIEW OF RESIDENTIAL APARTMENT
  • 23.
    12 3.4 STRUCTURAL DRAWING 3.4.1COLUMN LOCATION AND ORIENTATION Fig 3.10 COLUMN LOCATION AND ORIENTATION
  • 24.
    13 CHAPTER 4 STRUCTURAL ANALYSISAND DESIGN 4.1 STRUCTURAL ANALYSIS The model is generated using STAAD Pro Software and the analysis of frames is done using the stiffness method in the software. STAAD Pro is a general purpose tool for performing the analysis and design of a wide variety of type of structures. The 3 basic activities which are carried out by the software is, 1. Model Generation 2. The calculations to obtain the analytical results 3. Reports in post processing and verification of data After the load calculations, the analysis part is the most important to check the stability of the structure. Perform analysis is selected once and in the output file, the results are displayed. Then the postprocessing tab helps to gather the reports of Shear Forces, Bending Moments, Displacements and the Reaction forces can be determined. This can be viewed as a Summary of the entire project or even for a single node/beam/column. The loads given as an input are also considered along with the combination of all maximum loads generated if given. This report ensures the stability check for a structure.
  • 25.
    14 4.1.1 ANALYSIS PROCEDURE Theprocedure for analysis using the Staad Pro Software 1. The skeleton of the structure is formed using the node (coordinates) method. 2. Fixed support is provided at all nodes at the base. 3. Load definition are given. 4. Perform analysis command is selected. 5. The required design parameters are added and the concrete design for the beams and columns is taken off. 6. The analysis is done using Perform Analysis option once and Run Analysis option after several modifications. 7. The postprocessing tab is used for further report making purposes Some of the Pictures showing the analysis process of the project are as follows: FIG 4.1 GRAPHICAL REPRESENTATION OF SUPPORTS
  • 26.
    15 FIG 4.2 SECTIONALVIEW OF THE BUILDING FIG 4.3 GRAPHICAL REPRESENTATION OF BEAMS
  • 27.
    16 FIG 4.4 LOADDISTRIBUTION DUE TO SELF WEIGHT RENDERING VIEW FIG 4.5 RENDERED VIEW OF THE BUILDING
  • 28.
    17 4.1.2 OUTPUT FILE Thefinal output contains error free design results. The reinforcement details of beams and columns are displayed in the output file. It also gives the quantity estimate of the materials. Some results are as follows: FIG 4.12 BEAM DESIGN OUTPUT
  • 29.
    18 FIG 4.13 COLUMNDESIGN OUTPUT
  • 30.
    19 4.2 DESIGN OFSTRUCTUAL MEMBERS 4.2.1 DESIGN OF SLAB Lx = 3.46 Ly = 3.48 Grade of concrete = M20 Grade of steel = Fe415 Types of slab 𝑙𝑦 𝑙𝑥 = 𝑙𝑜𝑛𝑔𝑒𝑟 𝑠𝑝𝑎𝑛 𝑠ℎ𝑜𝑟𝑡𝑒𝑟 𝑠𝑝𝑎𝑛 = 3.48 3.46 = 1 < 2 Hence it is a two-way slab. Dimensions of slab l/b = 40 x 0.8 D = 3460 / 40 x 0.8 = 110 mm dx = D – cover = 110 – 20 = 90 mm dy = 90 -10 = 80 mm
  • 31.
    20 B = 1000mm Effectivespan 1. Centre to Centre supports = 3.46 + 0.26 = 3.72 m 2. Clear span + effective depth = 3.46 + 0.09 Effective span = 3.55 Load calculation Self-weight of slab = b x D x unit weight of RCC = 1 x 0.11 x 25 = 2.75 KN / m Total load = self-weight + L.L + floor finish = 2.75 + 4 + 1.5 = 8.25 KN / m Load factor (Wu) = 1.5 x 8.25 = 12.37 KN/ m Bending moment and shear force Mx = αx x w x lx2 My = αy x w x lx2 Ly/ lx = 3.48 / 3.46 = 1
  • 32.
    21 Short span coefficient (-ve)moment coefficient = 0.032 (+ve) moment coefficient =0.024 Positive moment Mx = αx x w x lx2 Mx = 0.024 x 12.37 x (3.55)2 Mx = 3.741 KN /m Negative span Mx = αx x w x lx2 Mx = 0.032 x 12.37 x (3.55)2 Mx = 4.988KN /m Long span coefficient (-ve) moment coefficient = 0.032 (+ve) moment coefficient =0.024 Positive moment My = αx x w x lx2 My = 0.024 x 12.37 x (3.55)2 My = 3.741 KN /m Negative span My = αx x w x lx2 My = 0.032 x 12.37 x (3.55)2
  • 33.
    22 My = 4.988KN/m Shear force V = W x L /2 = 12.37 x 3.55 / 2 = 21.96 KN Reinforcement For shorter span (- ve) moment Mu limit = 0.138 fck b d2 = 0.138 x 20 x 1000 x (90)2 = 22.3 KNm Mu < Mu limit Under reinforcement Mu = 0.87 x fy x Ast x d [1- 𝐴𝑠𝑡 𝑋 𝑓𝑦 𝑏 𝑋 𝑑 𝑋 𝐹𝑐𝑘 ] 4.988 x 10⁶ = 0.87 x 415 x Ast x 90 [1- 𝐴𝑠𝑡 𝑋 415 1000 𝑋 90 𝑋 20 ] Ast = 159.35mm2 Spacing of bar = 𝜋 4 𝑋 102 159.35 x 1000 = 492.6 Provide 10 mm dia bar at 300 mm c/c giving For shorter span (+ve) moment Mu = 0.87 x fy x Ast x d [1- 𝐴𝑠𝑡 𝑋 𝑓𝑦 𝑏 𝑋 𝑑 𝑋 𝐹𝑐𝑘 ] 3.741 x 10⁶ = 0.87 x 415 x Ast x 90 [1- 𝐴𝑠𝑡 𝑋 415 1000 𝑋 90 𝑋 20 ]
  • 34.
    23 Ast = 118.3mm2 Spacingof bar = 𝜋 4 𝑋 102 118.3 x 1000 = 663.5 Provide 10 mm dia bar at 300 mm c/c giving For longer span (-ve) moment Mu limit = 0.138 x fck x b x d2 = 0.138 x 20 x 1000 x (80)2 = 17.664 KNm Mu < mu limit Under reinforcement Mu = 0.87 x fy x Ast x d [1- 𝐴𝑠𝑡 𝑋 𝑓𝑦 𝑏 𝑋 𝑑 𝑋 𝐹𝑐𝑘 ] 4.988 x 10⁶ = 0.87 x 415 x Ast x 80 [1- 𝐴𝑠𝑡 𝑋 415 1000 𝑋 80 𝑋 20 ] Ast = 181.20mm2 Spacing of bar = 𝜋 4 𝑋 102 181.20 x 1000 = 433.2 Provide 10 mm dia bar at 300 mm c/c giving For long span (+ve) moment Mu = 0.87 x fy x Ast x d [1- 𝐴𝑠𝑡 𝑋 𝑓𝑦 𝑏 𝑋 𝑑 𝑋 𝐹𝑐𝑘 ] 3.741 x 10⁶ = 0.87 x 415 x Ast x 80 [1- 𝐴𝑠𝑡 𝑋 415 1000 𝑋 80 𝑋 20 ] Ast = 134.17mm2
  • 35.
    24 Spacing of bar= 𝜋 4 𝑋 102 134.17 x 1000 = 585.07 Provide 10 mm dia bar at 300 mm c/c giving Check for shear 𝜏v < k x 𝜏c 𝜏v = 𝑣 𝑏 𝑋 𝑑 = 21.956 𝑋 103 1000 𝑋 90 𝜏v = 0.243N /mm2 Permissible shear K = 1.3 100 𝑋 𝐴𝑠𝑡 𝑏 𝑑 = 100 𝑋 118.3 1000 𝑋 90 = 0.131 𝜏c = 0.79 𝜏c x k = 0.79 x 1.3 = 1.027N/mm2 𝜏v < 𝜏c x k Slab is safe in shear Check for deflection ( 𝑳 𝒅 )max = ( 𝑳 𝒅 )basic x kt x kc x k f
  • 36.
    25 ( 𝑳 𝒅 ) max =40 x 1.68 x 1 = 67.2 ( 𝑳 𝒅 ) actual = 3.550 90 = 39.4 Hence it is safe Check for crack 1. Steel provide is more the min percentage of 0.12% 2. Spacing of main steel < 3d or 300mm which is small. 3. Dia of reinforcement < 𝐷 8 𝐷 8 = 110 8 = 13.75mm cracker will be within permissible limit.
  • 37.
  • 38.
    27 4.1.2 DESIGN OFBEAM: Clear span = 3.4m Wide of support = 0.267m Stress and load factor fck = 20 N/mm² fy = 415 load factor = 1.5 for (D.L & L.L). Dimension of beam Effective depth = span / 15 = 340 /15 = 260 D = effective depth + clear cover = 260 + 40 D = 300 ( assume b = 260 mm) Effective span 1) Centre to Centre = 3.4 + 0.26 = 3.66 2) Clear span + effective depth = 3.4 + 0.26 = 3.66 Effective span is 3.66m Load calculation Self-weight of beam = 0.26 x 0.3 x 25 = 1.95KN/m Parapet wall load = 1 x 0.26 x 20 = 5.2KN/m Slab load = 𝑊 𝑋 𝑙𝑥 3 = 12.37 x 3.4 / 3
  • 39.
    28 = 14.01KN/m Total load= 1.5 (1.95 + 5.2) +14.01 = 24.73KN/m Bending moment Mu = 𝑤 𝑋 𝐿² 8 = 24.73 𝑋 (3.66)² 8 =41.40KN/m Shear force V = (w x L) / 2 = 43.82KN Check for Mu and Mu limit Mu limit = 0.138 x 20 x 260 x (260) ² = 48.509KNm Mu < Mu limit under reinforcement Area of reinforcement Mu = 0.87 x fy x Ast x d [1- 𝐴𝑠𝑡 𝑋 𝑓𝑦 𝑏 𝑋 𝑑 𝑋 𝐹𝑐𝑘 ] 41.40 x 10⁶ = 0.87x415x Ast x260[1- 𝐴𝑠𝑡 𝑋 415 260 𝑋 260 𝑋 20 ] Ast = 525.9 No of bars = 525.9 𝜋 4 𝑋 16² = 2.61 (3 bars) Provide 3 bars of 16 mm dia. Check for shear Nominal shear τv = v / b x d = 45.25 x 103 / 260 x 260 = 0.669N/mm² Permissible shear
  • 40.
    29 100 𝑋 𝐴𝑠𝑡 𝑏𝑋 𝑑 = 100 𝑋 602.8 260 𝑋 260 = 0.89 τc = 0.59 τv > τc provide permissible shear reinforcement using 2legged 8mm bar Vus = 0.87 𝑋 𝑓𝑦 𝑋 𝐴𝑠𝑡 𝑋 𝑑 𝑆𝑣 Sv = 0.87 𝑋 415 𝑋 (2 𝑋 𝜋 4 𝑋 82𝑋 260) 𝑆𝑣 Sv = 1367.80mm Check ͢ 0.75 d = 0.75 X 260 = 195 (300mm) Provide 8mm dia bars of 2 legged stirrups with 200mm c/c spacing Check for deflection ( 𝐿 𝑑 )max = 15 x 1.38 x 1 = 20.7 ( 𝐿 𝑑 )actual = 3660/260 = 14.076 Hence it safe
  • 41.
  • 42.
    31 4.1.3 DESIGN OFCOLOUMN Pu = 2000KN Dimension = 260 x 450 Effective length Effective length = 0.65 x L = 0.65 x 3 = 1.95 m Slenderness ratio L eff / b = 1950 / 260 = 7.5 < 12 Short column Strength of short column Pu = 0.4 fck x Ac + 0.67 fy Asc 2000 x 10³ = 0.4 x 20 x [(260 x 450)- Asc] + 0.67 x 415 Asc 1064 x 10³ = 270.05 Asc Asc = 3933.45 Ac = As – Asc Ac = (260 x 450) – 3933.45 = 113066.55 Minimum eccentricity E min = 𝐿 500 + 𝑏 30 = 3000 500 + 450 30 = 6 + 15 = 21 mm 0.05 D = 0.05 x 450 = 22.5 mm E < 0.050
  • 43.
    32 Hence design isaxially loaded short column Longitudinal reinforcement Asc = 3933.45 No of bars = 3933.45 𝜋 4 𝑋 25² = 3933.45 490.625 = 8.01 (8 bars) Provide 8 bars of 25mm dia Asc provide = 8 x π / 4 x 25² = 3925mm² Lateral ties 1. Diameter of ties = 1 / 4 x 25 = 6.25 (8mm) Provide 8mm dia ties 2. Spacing • 16 ɸ = 16 x 25 = 400 • 48 d = 48 x 8 = 384 • 300mm
  • 44.
  • 45.
    34 4.1.2 DESIGN OFFOOTING Beam 1 (Min spacing between individual rods for no.of. bars) Slab: 3.05m x 2.3m 3.05 2.3 = 1.32 < 2 2.3 2 = 1.15 Area of trapezoidal = 1/2 (a + b) h =1/2(3.05 + 0.76) 1.15 = 2.2m² Load = 2.2 x 7.125 =15.6KN Beam 2: Area of the triangle = 1 / 2 x 23 x 1.15 =1.32² Load = 1.32 x 7.125 = 9.40KN Load on column: Self wt. of beam = 0.26 x 0.46 x 3.05 x 25 =8.92 Self wt. of beam 2 = 0.26 x 0.3 x 2.3 x 25 = 4.48 Wall load on beam 1 = 3.05 x 0.26 x 3.05 x 20 = 48.3 Wall load on beam 2 = 3.05 x 0.26 x 2.3 x 20 = 36.47 Load from beam 1:
  • 46.
    35 Load 1 = 15.6+8.92+48.3 2 =36.41KN Load from beam 2: Load 2 = 9.40+4.48+36.47 2 = 25 .17KN Self wt. of column =0.26 x 0.46 x 3.048 x 25 =9.11 Total on footing =4 x (36.41 + 25.17 + 9.11) = 282.76 =300KN Design data Column size = 260 x 450 mm² SBC = 200 KN/m ² Fe 415 M20 Load 300KN Assume 10 % of load as self-weight Self wt = 300 x 10/100 = 30 KN Total load = 330 KN Area = 𝑡𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 𝑆𝐵𝐶 = 330 200 =1.65m² L = 450 / 260 = 1.73
  • 47.
    36 L x B= 1.65 2B² = 1.65 B = 0.90m Provide L = 2m B = 1m Size of footing = 2m x 1m Net upward pressure (po) = 300 / 2 x 1 = 150KN/m² = 150 X 1.5 Po = 225 kN/m² Depth Based on BM Mx = po B (L-l) ² / 8 =225 x 1 x (2-0.46) ² /8 Mx = 66.70 KNm My = po L (B-b) ² / 8 = 225 x 2 x (1- 0.46) ² / 8 My = 31.5 KNm Taking max moment to calculate depth Mu = 2.7 Bd² 66.70 x 10⁶ = 2.7 x 1000 x d² D = 156.59 provide (d = 160mm) and (D= 210mm) based on one-way slab: critical section for shear is at a distance from face to column. Vs =po B[ 𝐿−𝑙 𝛼 -d] =225×1[ 2−0.46 2 − 0.16]
  • 48.
    37 =225× [ 1.5 2 − 0.16] =132.75KN Vy =PoL [ 𝐵−𝑏 𝑏 -d] =225×2[ 1−0.26 2 -0.16] =94.5 KN P = 0.21 Ꚍc = [ 0.36−0.28 0.25−0.15 ][0.21-0.15] = 0.328 N/mm^2 Based on punching shear =Po [LB-(l+d) (b+d)] =225[2-(0.46+0.16) (0.26+0.16)] =391.41 KN Surface area =2[(l+d) +(b+d)] d =2[(0.46+0.16) (0.26+0.16)]0.16 =0.729 m^2 𝜏v = 391.41×10^3 0.729×10^6 = 391410 729000 =0.536 Ks =0.5+( 0.26 0.46 ) Ks =1.06>1 𝜏c =0.26√𝑓𝑐𝑘 𝜏c =1.16 N/mm^2 𝜏v < 𝜏c Hence it is safe punching shear Area of reinforcement (along longer direction)
  • 49.
    38 Mx =0.87 fyAst d[1- 𝐴𝑠𝑡 𝑓𝑦 𝑏𝑑 𝑓𝑐𝑘 ] 66.70×10^6 =0.87×415×Ast×100[1- 𝐴𝑠𝑡×415 1000×500×20 ] Ast =175.10 mm^2 No of bars(along longer direction) Assume 12mm dia bars No of bars = 𝐴𝑠𝑡 𝐴𝑟𝑒𝑎 𝑜𝑓 1 𝑏𝑎𝑟 = 372.25 (𝜋÷4)12^2 =3.29 = 4 rods Provide 4 no.s of 12mm dia bars along longer direction As per IS 456,Pg:65, Reinforcement in central band = 2 𝐵+1 B =2 Reinforcement in central band = 2 3 (175.10) =116.73 mm^2 Reinforcement in edge strip =175.10-116.73 =58.37 mm^2 Assume 12mm dia bars No of bars = 175.10 (𝜋÷4)10^2 =1.59 =2 rods Provide 2 no of 12mm dia bars along shorter direction.
  • 50.
    39 CHAPTER-5 SUMMARY AND CONCLUSIONS CONCLUSION •In this project, an attempt has been made to plan, analyse and design G+3 apartment based on the provisions of IS 456:2000. • The structure composed of beams and columns is modeled in the Staad Pro v8i Software and are analysed with various loading combinations. • For detailing codebooks are referred and the detailed drawings of the various components are presented. • The design of components such as Beams, Columns and Slab are carried out completely using Staad Pro and Manual Calculations. • To plot structural drawings, the drafting software AutoCAD is used. • We are now familiarized with Staad Pro v8i design software which is majorly used for analysis of structures and modeling of structures. • The design concepts studied for the various RCC elements used inthis project helped us to reinforce our confidence in structural design area
  • 51.
    40 REFERENCES 1) “S.Ramamrutham”,”Reinforced ConcreteStructures”,New Delhi 2) “B.C.Punmia”,”Reinforced Concrete Structures”,New Delhi 3) IS 875 (Part-1) 1987, Code of practice for Design of loads for buildings 4) IS 456-2000, Plain and Reinforced Concrete Code of Practice 5) SP 16, Designs aids for Reinforced Concrete
  • 52.
    41 CHAPTER 6 ANNEXURE The listsof drawings attached in the reports are 1. Site layout 2. Ground floor plan 3. First floor plan 4. Second floor plan 5. Third floor plan 6. Section 7. Column orientation 8. Beam orientation