1

CHAPTER 1

INTRODUCTION

1.1 GENERAL

Our project deals with the process of planning, analyzing and

designing of Restaurant. We have selected the proposed site for this
design on kottaram in kanyakumari. The project was developed inorder
to incorporate the analysis and design of Civil Engineering. Our building
is a framed structure which consists of underground floor, G+ 2 floor of
same size and shape. The structural frame would be analyzed by moment
distribution method. For the project the code book used are IS 456 – 2000
& IS 875. The drawings are drawn by using Auto Cad.

1.2 SITE SELECTION

We have selected the proposed site for the planning; analyzing &
designing of restaurant to be constructed kottaram in kanyakumari for the
purpose of satisfy the requirement of the nearby locality people.

1.3 STRUCTURAL MEMBERS

1.3.1 SLAB

Slab are the primary members of a structure which supports the

imposed loads directly on the them transit the same safely to the
supporting elements such as beams, walls, columns etc. Therefore a slab
should be safe and stable against the beams, walls, columns etc.
Therefore a slab should be safe and stable against the applied loads and
 2

should have the required strength and stiffness to satisfy the service
ability requirement.

1.3.2 CLASSIFICATION OF SLABS:

a) = 0.12/100x 1000x 190

Based on shape: Square, rectangular, circular
and polygonal in shape.
b) Based on type of support: slab supported on walls, slab supported
on beams & column, Slab supported on columns (Flat slabs).
c) Based on support or boundary condition: Simply supported,
cantilever slab, Overhanging slab, Fixed or continues slab.
d) Based on use: Roof slab, foundation slab, water tank slab, Based
on cross section or sectional configuration: Ribbed slab (or) Grid
slab, slab, Filler slab, Foundation slab, Folded Plate.
e) Based of spanning directions:
One way slab – Spanning on one direction
Two way slab – spanning in two directions

1.3.3 COLUMNS:

The column is structural member provided to carry a compressive

load and whose effective length exceeds its least lateral dimensions. In
buildings, columns are provided to support the footing and flooring
system effectively. It also transfers the foundation.

1.3.4 COLUMN FOOTING (OR) ISOLATED FOOTING:

Foundation is the bottom most important component of a strutted

which greatly lies with well below the ground level. It does not provide
any aesthetic appearance to the elevation view of building. It has to be
 3

well below the ground level and to be planned and careful designed to
ensure the safety and stability the structure.

1.3.5 BEAMS

Beams are defined as a horizontal load carrying member in a

structure. Reinforced cement concrete, pre stressed concrete and steel I-
sections are used as beam to support the slab. Thus in a structure the load
is transmitted from the slab to the beams and then to the columns. Finally
the load from column is transmitted to the foundation and thus the load is
safely transmitted to the soil. Depending upon the support condition the
beams are classified as follows
a) Simply supported beam
b) Cantilever beam
c) Over hanging beam
d) Continuous beam

1.3.6 STAIRCASE:

A staircase consist of a number of stops arranged in a series with

landing at appropriate location for the purpose of giving access to
different floors at building. The width of a staircase may depend on the
purpose for which it is provided and may generally vary between 1m for
residential building to 2m for public building.

1.4 DESIGN PHILOSOPHIES

Design is the creation of a plan or conversation of an object or a

system. Design philosophies are usually for determining design goals. A
design goal may range from solving the least significant individual
 4

problem of the smallest element, to the most holistic influential utopian

goals. Design goals may lead for guiding design, perhaps to set better
long term or ultimate goals. The design of reinforced concrete and
prestressed concrete generally contains three philosophies namely
a. Working stress method
b. Ultimate method
c. Limit state method

1.4.1 WORKING STRESS METHOD

This was the traditional method of design not only for reinforced
concrete, but also for structural steel and timber design. The method
basically assumes that the structure material behaves as a linear elastic
manner, and that adequate safety can be ensured by suitably restricting
the stress in the material induced by the expected “working loads” on the
structure. As the specified permissible stresses are kept well below the
material strength, the assumption of liner elastic behavior is considerable
justifiable. The ratio of the strength of the material to the permissible
stress is often referred to as the factor of safety.

1.4.2 ULTIMATE LOAD METHOD

With the growing realization of the short comings of working

stress method in reinforced concrete design, and with increased
understanding of the behavior of reinforced concrete at ultimate loads, the
ultimate load of design is evolved and became an alternative to working
stress method. This method is sometimes also referred to as the load
factor methods are the strength.
 5

The concept of ‘modular ratio’ and its associated problems are

avoided entirely in this method. The safety measure design is introduced
by an appropriate choice of the load factor, defined as the ratio of the
ultimate load to the working load. The ultimate load method males it
possible for different types of loads to be assigned different load factors
under combined loading conditions, thereby overcome the related
shortcoming of working stress method.

This method generally results in more slender sections, and often

economical designs of beams, and columns, particularly when high
strength reinforcing steel and concrete are used.

1.4.3 LIMIT STATE METHOD

The limit state design has originated from ultimate or plastic

design. It is a state of impending failure. Beyond which a structure ceases
to perform its intended function satisfactorily; in terms of either safety of
serviceability i.e. it either collapses or becomes unserviceable. There are
two types of state. They are

a) Ultimate limit states (limit states of collapse) – which deal with

strength, overturning, sliding, buckling, fatigue fracture etc.
b) Serviceability limit states- caused which deals with discomfort to
accuracy and or malfunction, caused by excessive deflection, crack
width, vibration leakage etc., and also loss of durability etc.

1.5 SPECIFICATION

Main door (M.D) = 2500 X 2500mm

Paneled door (D) = 2000 X 2500mm
 6

Glazed window (W) = 1500 X 1500mm

Glazed window (W1) = 1000X 1500mm
Steps:
Rise = 150mm
Tread = 250mm

1.6 OBJECTIVES

 We have selected the site for the construction of a restaurant in

kottaram for the purpose of satisfying the needs and desires of
tourists.
 The main objective of our project was to apply practically, the
various theories we have studied in the last three years.
 All the check conditions are satisfied.
 7

CHAPTER 2

METHODOLOGY

Analysis & design of Restaurant

Planning

Analysis
(Load analysis,
S frame analysis)

Structural Design

Slab Beam Column Footing Staircase

Conclusion

Fig: 2.1.1 Methodology

 8

2.1 PLANNING

We had prepared a plan by taking a reference of some plans, and by

referring books, and we finally drafted the plan by using the Auto- CAD,
and we made a section for that plan.

2.2 STRUCTURAL ANALYSIS

2.2.1 LOAD ANALYSIS

We analysed the load by using the Yield Line Theory, we used this
theory because it is conducted based on the bending moment of the
structural element at its collapsed state.

2.2.2 FRAME ANALYSIS

We made frame analysis by using the substitute frame analysis

method. It is same as that of the moment distribution method, by using two
cycles of distribution only.

2.3 STRUCTURAL DESIGN

2.3.1 SLABS

We provide the two way slab, because the ratio of the longer span
to that of the shorter span is less than two.

2.3.2 BEAMS

Beam is defined as a Horizontal (or) inclines structural members

spanning a distance between one (or) more support and carrying vertical
load. A beam is structural element that is capable of withstanding load
primarily by resisting be bending. The bending force induced into the
material of the beam as a result of the bending moment.
 9

2.3.3 STAIRCASE

A staircase provides the means of movement from one level to

another level in a structure. It consists of a number of steps arranged in a
series of flight with landing at suitable intervals, to provide comfort, ease
and safety. Various components of staircase are classified under two
categories.

2.3.3.1 Geometrical classification

2.3.3.2 Structural classification

2.3.4 COLUMN

A Column or pillar in architecture and structural engineering is a

structural element that transmits, through compression, the weight of the
structure above to other structural elements below. In other words,a column
is a compression member.

2.3.5 FOOTING

Superstructure loads are transmitted to the underlying soil strata

through a suitably designed foundation. Therefore, the foundation of a
structure is considered the most crucial structural element in a building. The
foundation may be classified into two main categories, shallow and deep
foundations.
 10

2.4 PLAN

Fig 2.1.2 Ground Floor

 11

Fig 2.1.3 First Floor

 12

Fig 2.1.4 Second Floor

 13

Fig 2.1.5 Grid Plan

 14

CHAPTER 3

STRUCTURAL ANALYSIS

3.1 LOAD ANALYSIS

(LONGITUDINAL DIRECTION)

Load acting on the beam: A1 B1 C1

Fig: 3.1.1 BeamA1B1C1

Area = (1∕2)×b×h
= 1∕2 (4×2)
= 4mm

Beam

Live load = 5×b×d

= 5×0.45×0.23
= 0.5KN∕m
Self weight of wall = (l×b×d)
= [(3.5-0.45) × 0.23×20]4
= 56.12 KN
 15

Self weight of slab:

Live load = 5×A

= 5×4
= 20 KN ∕m
Dead load = 4×0.15×25
= 15 KN
Floor finish load = 1×4
= 4 KN
Total load = 0.52+56.12+20+15+4
= 23.81 KN
Load acting on the beam:

Beam:

Area = 1∕2× (5+1)2+ 1/2×(5+1)2

= 12mm2
Self weight of the wall = (3.5-0.45)×0.23×20×5
= 70.15KN
Self weight of slab:
Live load = 5×12
= 60 KN/m
Dead load = 12×0.15×25
= 45 KN
Floor finish load = 1×12
= 12
Total load = 60+45+12+70.15+0.52
= 37.53 KN
 16

Load acting on the beam (D1 E1)

Area = 1/2 ×5×2.5

= 6.25 mm2
Beam:

Live load = 5×0.23×0.45

= 0.52 KN/m
Self weight of wall = (3.5-0.45)0.23×20×5
= 70.15 KN
Self weight of slab:

Live load = 5×6.25

= 31.25 KN
Dead load = 6.25×0.15×25
= 23.44 KN
Floor finish load = 1×6.25
= 6.25 KN
Total load = 0.52+70.15+31.25+23.44+6.25
= 26.32 KN/m
Load acting on beam (D1 D2)

Area = ½(5×2.5)+1/2(5×2.5)
= 12.5 mm2
Beam

Live load = 5×0.23×0.45

= 0.52 KN/m
Self weight of wall = (3.5-0.45)0.23×20×5
= 70.15 KN
 17

Self weight of slab:

Live load = 5×12.5

= 62.5 KN
Dead load = 0.15×25×12.5
= 46.88 KN
Floor finish load = 1×12.5
= 12.5 KN
Total load = 62.5+12.5+46.88+70.15+0.52
= 38.5 KN/m
Beam (F1G1)

Area = 1/2×3×1.5
= 2.25 mm2
Beam:

Live load = 5×0.23×0.45

= 0.52 KNm
Self weight of wall = (3.5-0.45)0.23×20×3
= 48.99 KN
Self weight of slab:
Live load = 5×2.25
= 11.25 KN
Dead load = 2.25×0.15×25
= 8.44 KN
Floor finish load = 1×2.25
= 2.25 KN
Total load = 0.52+48.99+11.25+8.44+2.25
= 23.82 KN/m
 18

Beam (F1 F2)

Area = 1/2(5×2.5) + 1/2(5+2)1.5

= 11.5 mm2
Beam

Live load = 5×0.23×0.45

= 0.52 KN/m
Self weight of wall = (3.5-0.45)0.23×20×5
= 70.15 KN
Self weight of slab:

Live load = 5×11.5

= 57.5 KN
Dead load = 11.5×0.15×25
= 43.13 KN
Floor finish = 1×11.5
= 11.5 KN
Total load = 57.5+43.13+11.5+70.15+0.52
= 36.49 KN
 19

Fig 3.1.2: Load Analysis

 20

3.2 FRAME ANALYSIS:

FRAME ANALYSIS: (X- DIRECTION)

(LONGITUDINAL DIRECTION)

Fig:3.2.1 Intermediate Frame (XX-Direction)

FIXED END MOMENT:

Mfab = -37.53×52/12 = -78.19KNm

Mfba = 37.53×52/12 = 78.19KNm
Mfbc = Mfcd = Mfde = Mfef = -78.125KNm
Mfcb= Mfdc = Mfed = Mffe = 78.125KNm

FREE BENDING MOMENT:

Mab = Mbc = Mcd = Mde = Mef = 37.53×52/8 = 117.28KNm

STIFFNESS FACTOR (K):

Ib = Ic
4EIb/5 = 4EIc/3.5
Ib = 1.425Ic
 21

JOINT A:

KAA1 = 4EIc/1.75 = 2.29Ic

KAA2 = 2.29Ic
KAB = 4E(1.425Ic)/5= 1.13

JOINT B:

KBB1=KBB2 = 2.29Ic
KBA=KBC = 1.136EIC

JOINT C:

KCC1=KCC2 = 2.29Ic
KCB=KCD = 1.136EIC

JOINT D:

KDD1=KDD2 = 2.29EIC
KDC= KDE = 1.136 EIC

JOINT E:

KEE1=KEE2 = 2.29EIC
KED=KEF = 1.136EIC

JOINT F:
KFF1=KFF2 = 2.29EIC
KFE = 1.136EIC

DISTRIBUTION FACTOR:
DfAA1 = KAA1/KAA1+KAA2+KAB
= 2.29E IC/2.29E IC+2.29E IC+1.136E IC
= 0.40
 22

DfAA2 = 0.40

DfAB = 1.136E IC/1.136E IC+2.29E IC+2.29E IC

= 0.20

JOINT B

DfBB =DfBB2 = 2.29E IC/2.29E Ic +2.29E IC+1.136E IC

= 0.33

DfAB = 1.136E IC/1.136E IC+2.29E IC+2.29E IC

= 0.17
JOINT C

DfCC1 =DfCC2 = 0.33

JOINT D

DfCD=DfDE = 0.17

JOINT E

DfEE1=DfEE2 = 0.33
DfED=DfEf = 0.17

JOINT F

DfFF1=DfFF2 = 0.33
D = 0.17
 Joint A B C D E F

Member AA1 AA2 AB BA BB1 BB2 BC CB CC1 CC2 CD DC DD1 DD2 DE ED EE1 EE2 EF FE FF1 FF2

DF 0.33 0.33 0.17 0.17 0.33 0.33 0.17 0.17 0.33 0.33 0.17 0.17 0.33 0.33 0.17 0.17 0.33 0.33 0.17 0.17 0.33 0.33

FEM 0 0 -78.19 78.19 0 0 -78.19 78.19 0 0 -78.19 78.19 0 0 -78.19 78.19 0 0 -78.19 78.19 0 0

78.19 0 0 0 0 -78.19

Balancing 0 0 13.29 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -13.29 0 0

COF 0 6.65 -6.65 0

FEM -21.36 -21.36 -64.90 85.01 2.31 2.31 -78.02 78.02 0.33 0.33 -78.02 78.02 0.33 0.33 -78.02 78.02 0 0 -84.84 78.02 0 0

64.90 -6.99 0 0 -6.82 -78.02

Table No: 3.2.1. Moment Distribution

Balancing 0 0 11.03 -1.19 1.19 0 0 0 0 -1.16 -1.16 13.26

Carry
-0.60 5.52 0 0.60 0 0 -0.58 0 -6.63 -0.58 0 0
Over

FEM -21.36 -21.36 -54.47 29.34 2.31 2.31 -76.83 78.62 0.33 0.33 -78.02 78.02 0.33 0.33 -78.60 -76.86 0.33 0.33 -92.63 12.68 0.33 0.33
23
 24

SHEAR FORCE
Span AB

Fig:3.2.2 Span AB

Taking moment about B

RAx5 -54.47-37.53x5x(5/2)+89.34 = 0
RA = 86.85KN
𝜀v = 0
RA + RB = 0
RB = 100.8KN

Span BC

Fig:3.2.3 Span BC

Taking moment about C

RB x5 -76.83 – 37.53 x5x(5/2)+78.62 = 0
 25

RB = 93.47KN
𝜀v = 0
RB+ RC = 37.53 x 5
RC = 94.18KN

Span CD

Fig:3.2.4 Span CD
Taking Moment about D
RC x 5 -78.02- 37.53x5 x(5/2)+ 78.02 = 0
RC = 93.83KN
𝜀v = 0
RC+RD = 5x37.53
RD = 93.82KN

Span DE

Fig:3.2.5 Span DE
 26

Taking Moment about E

RDx5-37.53x5x(5/2)-78.6+76.86 = 0
RD = 94.17KN
𝜀v = 0
RD+RE = 37.53x5
RE = 93.48KN

Span EF

Fig:3.2.6 Span EF

Taking Moment about F

REx5-92.63-37.53x5x(5/2)+12.68 = 0
RE = 109.82KN
𝜀v = 0
RE+RF = 37.53x5
RF = 77.83KN
 27

Fig: 3.2.7 BMD and SFD Diagram (XX-Direction)

 28

3.3. For Cross Section direction

Fig:3.3.1 Intermediate Diagram (YY-Direction)

 29

Table.No:3.3.1 Frame Analysis in Cross Section Direction

FIXED AND FINAL END

STIFFNESS DISTRIBUTION
SPAN MOMENT MOMENT
FACTOR FACTOR
(Kn/m) (Kn/m)
AA1 - 2.29 EIc 0.41 -
AA2 - 2.29 EIc 0.41 -
AB -31.75 EIc 0.18 -23.71
BA 31.75 EIc 0.15 32.24
BB1 - 2.29 EIc 0.35 -
BB2 - 2.29 EIc 0.35 -
BC -31.75 EIc 0.15 -6.42
CB 31.75 EIc 0.16 35.67
CC1 - 2.29 EIc 0.36 -
CC2 - 2.29 EIc 0.36 -
CD -54.83 0.8EIc 0.13 -49.8
DC 54.83 0.8EIc 0.13 57.4
DD1 - 2.29 EIc 0.37 -
DD2 - 2.29 EIc 0.37 -
DE -54.83 0.8EIc 0.13 -55.18
ED 54.83 0.8EIc 0.13 55
EE1 - 2.29 EIc 0.37 -
EE2 - 2.29 EIc 0.37 -
EF -54.83 0.8EIc 0.13 -58.48
FE 54.83 0.8EIc 0.12 47.28
FF1 - 2.29 EIc 0.34 -
FF2 - 2.29 EIc 0.34 -
FG -17.87 1.33EIc 0.2 -24.03
GF -17.87 1.33EIc 0.2 25.27
GG1 - 2.29 EIc 0.34 -
GG2 - 2.29 EIc 0.34 -
GH -54.83 0.8EIc 0.12 -52.71
HG 54.83 0.8EIc 0.15 46.49
HH1 - 2.29 EIc 0.43 -
HH2 - 2.29 EIc 0.43 -
 30

Fig: 3.3.2 BMD and SFD Diagram (YY-Direction)

 31

CHAPTER 4

4. STRUCTURAL DESIGN

4.1 DESIGN OF SLAB

Given data

lx = 4000mn
ly = 5000mm
L.L = 5KN/m2
F.F.L = 1KN/m2
Fck = 20N/mm2
Check
Ly / lx = 5/4 = 1.25<2
Hence it is 2 way slab

Depth of slab

Depth = Span /32

= 4000/32

= 125mm

Effective depth (d) = 125mm

Assume 10mm dia bar of
Assume clear lower as 20mm
Overall depth (D) = 125+20+10/2

Effective Span (Shorter span)

1) Effective span = clear span +one wall thickness

 32

= 4000+230
= 4230mm
2) Effective span = clear span +Effective depth
= 4000+125
= 4125mm
Lx = 4125mm

Effective Span (Longer span)

1) Effective span = clear span +one wall thickness

= 5000+230
= 5230mm

2) Effective span = clear span +Effective depth

= 5000+125
= 5125mm
Ly = 125mm

Load calculation

D.L = (l x b x dl)
= (1x1x0.125) x 25
= 3.125KN/m
L.L = 5 KN/m2
Service load (w) = self weight +L.L+D.L
= 3.125+5+1 = 9.125kn/m2
Ultimate Load (Wu) = 1.5x9.125 = 13.68kn/m2

Ultimate design moment & shear force:

Ly/lx = 1.25
 33

Condition: Two adjacent edge is discontinuous

αx(+) = 0.060
αx(-) = 0.045
αy(+) = 0.047
αy(-) = 0.035

For shorter span

Negative
Mx = αxWUL2
Mx = 0.060 x13.68x(4.125)2
= 13.37KNM
Positive
MX = 0.045X13.7X4.1252
= 10.49knm

For longer span

Negative
My = αyWul2
My = 0.047x13.7x(4.125)2
= 10.96knm
Positive

My = 0.035x13.7x4.1252
= 8.16knm
Mx = 13.37knm
My = 10.96KNM
Check for depth
Mu = 0.138fckbd2
 34

For shorter span

Negative
13.37x106 = 0.138x20x1000d2
dreq = 69.60mm
Positive
10.49x106 = 0.138x20x1000xd2
dreq = 61.65mm

For longer span

Negative
10.96x106 = 0.138x20x1000xd2
Dreq = 63.02mm
Positive
8.16x106 = 0.138x20x1000xd2
Dreq = 54.37mm

1. Ast calculation (negative)

Mu = 0.87fYAst d (1.Astfy/bdfch)
13.37x106 =0.87x415xAstx125(1.Astx415/1000x125x20)
Ast = 312.452
Min =Ast = 0.12% of bd
=0.12/100x1000x150=180mm2
Spacing Assume 10mmφbar
1. Ast x1000/Ast = 1000xπ/4x102/312.45 =251.36mm2
2. 3xd = 3x125 = 375mm
3. 300mm
Provide 10mmφbar at 300mm spacing c/c
 35

2.Ast calculation (positive)

10.49x106 =0.87x415xAst x125(1-Astx415/1000x125x20)

Ast =242.17mm2Ast

=0.12% of bd = 180mm2
Spacing Assume 10mmφ bar
1. Ast x1000/Ast = 1000xπ/4 x102/242.17 =324.31mm2
2. 3xd = 3x125 = 375mm
3. 300mm
Provide 10mmφbar at 300mm spacing c/c

1.Ast calculation (negative)

10.96x106 =0.87x415xAst x125(1-Ast 415/1000x125x20)

=253.52mm2

Ast =0.12%of bd = 180 mm2

Spacing Assume 10mm φbar

1. Ast x1000/Ast =1000xπ/4x102/253.52=309.79mm
2. 3xd2 = 375mm
3. 300mm

2.Ast calculation (positive)

8.16x106 =0.87x415 xAst x125 (1-

Ast415/1000x125x20)
Ast = 186.59mm2
Spacing Assume 10mmφbar
1.Ast x1000/Ast =1000xπ/4x102/186.59 = 420mm2
2. 3xd =3x125 = 375
 36

3.300mm
Provide 10mmφbar at 300mm spacing c/c

To find trosional ft

Ast = 3/8 x area of shorter span

= 3/8x 312.45mm2
= 117.17mm2
= 120mm2
Assume ,8mmφ torsion ft
= Area of trosion ft/are of one bar
= 120/π/4(8)2=2.39
= 3nos

Spacing = (Ast/Ast)x 1000

= π/4(8)2/120x1000
= 418.87mm = 420mm
Lx/5 = 4000/5 = 0.8
Ly/5 = 5000/5 = 1
B.M, Mu = Wu le2÷8
= 21.05(3.9)2 ÷ 8
= 40.02 KNm

Check the thickness of waist slab:

Mu = 0.138 fckbd2
40.02×106 = 0.138×20×1000d2
d = 120mm <195mm
dreq < dprovide
Hence safe.
 37

Fig: 4.1.1 Reinforcement Details for slab

 38

Table.No: 4.1.1 Design of Slab

DIMENSION OVER REINFORCEMEN REINFORCEMEN TYPE REMARK

SI
ALL T DETAILS OF T DETAILS OF OF CONDITION
NO
(mm) DEPTH SHORTER SPAN LONGER SPAN SLAB TYPE OF SLAB

Two adjacent
Two edge
S1 5000 X 4000 150 10mm@300mmc/c 10mm@300mmc/c
way
Discontinuous

Two One adjacent edge

S2 5000 X 4000 150 10mm@300mmc/c 10mm@300mmc/c
way Discontinuous

Two
S3 5000 X 4000 150 10mm@300mmc/c 10mm@300mmc/c Interior panel
way

Two One adjacent edge

S4 5000 X 5000 125 10mm@300mmc/c 10mm@300mmc/c
way
Discontinuous

Two
S5 5000 X 5000 125 10mm@300mmc/c 10mm@300mmc/c Interior panel
way

Two adjacent
Two edge
S6 5000 X 3000 125 10mm@300mmc/c 10mm@300mmc/c
way
Discontinuous

Two
S7 5000 X 3000 125 10mm@300mmc/c 10mm@300mmc/c Interior panel
way

Two adjacent
Two edge
S8 5000 X 5000 125 10mm@300mmc/c 10mm@300mmc/c
way
Discontinuous
 39

4.2 DESIGN OF BEAM

4.2.1 DESIGN OF T-BEAM
Data:
Span = 5000mm
Beam size = 230x450mm
Moment at support = 54.47x1.5
= 81.705KNm
Design shear force = 45.49 x1.5
= 68.235 kNM
Depth of flange = 150mm
Breadth of web = 230mm
Material property:
Concrete grade = M20
Steel grade = Fe415
Fck = 20N/mm2
Fy = 415N/mm2
Effective cover = 50mm
Xu,max /d = 0.48
Effective width of flange:

𝑙0
B = + bw + 6Df
6

5000
= 0.7x ( ) + 150 + (6X150)
60

= 1713.33mm
Depth of neutral axis:

Assume neutral axis lies within the flange,

𝐴𝑠𝑡 𝑋𝐹𝑦
Mu = 0.87FyAstd(1- )
𝐹𝑐𝑘𝑋𝑏𝑑
 40

𝐴𝑠𝑡𝑋415
81.705x106 = 0.87x415xAstx400(1- )
20𝑋400𝑋1713.33

Ast = 575.78mm2
𝑋𝑢 0.87𝑋𝐹𝑦𝑋𝐴𝑠𝑡
=
𝑑 0.36𝑋𝐹𝑐𝑘𝑋𝑏𝑓
0.87𝑋415𝑋575.78
=
0.36𝑋20𝑋1713.33

Xu = 16.85mm
16.85 < 150
Xu < Df
Assumption is correct
𝑋𝑢,𝑚𝑎𝑥 𝑋𝑢,𝑚𝑎𝑥
Mu,lim = 0.36x (1-0.42x )bwd2Fck
𝑑 𝑑

= 0.36x0.48(1-0.42x0.48)230x4002x20
= 101.54KN
Reinforcement at support:
Design support moment = 81.705KNm
Mu,lim > Mu
Thus under reinforcement
Moment of resistance:
𝐴𝑠𝑡𝑋𝐹𝑦
MU = 0.87FyAstd(1- )
𝐹𝑐𝑘𝑋𝑏𝑑

Ast = 575.78mm2
Provide 10mmɸ bar
575.78
N = 𝜋
𝑋102
4

= 7no’s
Provide 2 hanger bars of 8mmɸ
Provide 2 bars up at 450 at a distance = 1.414x0.9xd
= 1.414x0.9x400
= 500mm
 41

Check for shear:

𝑉𝑢
𝜏v =
𝑏𝑑
68.235𝑋103
=
1713.33𝑋400

= 0.10N/mm2
100𝑋𝐴𝑠𝑡 100𝑋575.78
=
𝑏𝑤 𝑑 1713.33𝑋400

= 0.08
𝜏c = 0.28N/mm2
𝜏c,max = 2.8N/mm2
𝜏v < 𝜏 c <𝜏c,max
Hence safe.
Assume 8mmɸ 2 legged stirrups
𝜋
Asv = 2x x82
4

= 100.53mm2
Spacing:
2.5𝑋 𝐹𝑦 𝑋𝐴𝑠𝑣
Sv =
𝑏
2.5𝑋415𝑋100.53
=
230

= 453mm
0.75d = 0.75x400
= 300m
Provide 8mmɸ 2 legged stirrups at a spacing of 300mm c/c
 42

Fig: 4.2.1 Reinforcement Details in T Beam

 43

4.3 DESIGN OF STAIRCASE

No. of steps in the flight = 12

Tread = 250mm
Rise = 150mm
Width of landing = 300mm

To fond depth of waist slab:

Eff. Span = (No of tread x width of tread) + width of landing beam

= (12x0.25) + 0.3
= 3.3mm
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑠𝑝𝑎𝑛
Thickness of waist slab =
20

= 3300/20
= 165mm
Overall depth, D = d + 25
= 165+25
= 190mm
Loads
Dead load of waist slab on horizontal span =
𝐷𝛾
√𝑅2+𝑇2
Type equation here.
250

0.19𝑥25𝑥√(150 )2=(250)2
=
250

= 5054KN/m
Dead loads of steps/m length = Area x ᵧ
= ½ xbh xᵧ
= 1/2x1010.5x25
= 1.875KN/m
 44

Floor Finishes = 1KN/m2

Live lssoad = 5KN / m2
Total Load = 5.54+1.875+1+5
= 13.415KN/m
Factored load, Wᵤ = 1.5x13.415
= 20.12KN/m
Bending moment:
Mᵤ = Wᵤle2/8
= 20.12x3.32/8
= 27.39KNm
Check for depth of waist slab:
Mᵤ = 0.133fck bd
Dreq<dpro
Hence safe.
Design of reinforcement:

Mu = 0.87fyAstd[1-fyAst / fckbd]
27.39x106 = 0.87x500xxAstx165x[Astx500/1000x165x20]
27.39x106 = 71775Ast – 10.87Ast2
Ast = 407mm2
Min Ast = 0.12 % bD
= 0.12/100x 1000x 190
= 228mm2
Check:

3d = 3x165 = 495mm

300mm
Provide 10mm dia bar@ 190mm C/C spacing
 45

Distribution reinforcement:

Ast = 0.12% bD
= 0.12/100x 1000x 190
= 228mm2
Spacing = π/4x82 / 228x1000
= 220mm
Check:
 5d = 5x165 = 825mm
 450mm
Provide 8mm dia bar@ b220mm C/C spacing

Fig: 4.3.1 Reinforcement Details in Staircase

 46

4.4 DESIGN OF COLUMN

4.4.1 DESIGN OF BIAXIAL LOADED COLUMN

Load calculation:

For column A1,

Area of load = (5/2 x 4/2 )
= 2.5x2
= 5m2
Self weight of slab = 1 x 1 x 0.12 x 25
= 3 KN / m2
Weight of floor finish = 1 KN/m2
Total dead load = 4KN/m2
Dead load = 4 x 4.2
= 16.8 KN
Total dead load = 3x(5x(3.75+1))
= 71.25 KN
Self weight of column:
Total height = 3.5+3.5+3.5+1.6
= 12.1m
Self weight = 1 x0.23 x0.23x12.1x25
= 16KN
Weight of 230mm thick wall/m = 1x1x0.23x(12.1-(3x0.45)20x3)
= 148.35 KN/m
Weight of 9” thick wall = 53.59(2+2)-0.30
= 198.28 KN
Live load:
Live load = 5x5x3
 47

= 75 KN
Load from roof = 1x5
= 5 KN
Self weight of beam/m = 1x3x0.23x0.45x25
= 7.76 KN/m
Self weight of beam = 2.58(2+2-0.3)
= 9.54 KN
Total load = 71.25+16+148.35+75+5+7.76
= 323.36 KN
Mux = 21.36 KNm
Muy = 21.36KNm
= 31.45KNm
Pu = 323.36X1.5
= 485KNm
Equivalent moment
M = 1.15√𝑀𝑢𝑥 2 + 𝑀𝑢𝑦 2
= 1.15 √21.362 + 21.362
M = 34.74KN
Non dimensional parameter:
Pu/fck bd = (485x103)/(20x230x230)
= 0.46
From sp 16 chart 46
M ux1 /fck bd2 = 0.16
M ux1 = 0.16x20x230x2302
From chart 63 from sp 16
Puz/Ag = 13.5
Puz = 13.5x Ag
 48

= 13.5x230x230
= 714.15KN
Pu/Puz = 485/714.15
= 0.7
Pu/Puz ≥ 0.8
Αn = 1.5
αn αn
(Mux /M ux1) +(Muy /M uy1) ≤ 1
= (21.36/38.93)1.5 + (21.36/38.93)1.5
= 0.18
= 0.18 < 1
Hence the section is safe against Biaxial Bending
Area of reinforcement:
As = Pbd/100
= (1.5x(230x230))/100
= 793.5 mm2
Provide 4 no’s of 16mm dia bar.
Design of lateral ties:
Diameter of ties = (1/4) x φ of main bar
= (1/4)x16
= 4mm
B = 8mm
Diameter of tie bar is 8m.
Pitch:
Least lateral dimension = 230mm
16xφ of main bar = 160mm
48xφ = 48x8 = 384
 49

Provide 8mm φ bar at a spacing of 160mm.

Fig: 4.4.1 Reinforcement Details in Column

 50

4.5 DESIGN OF FOOTING

SBC of footing = 200 kN/m2

Biaxial loading = 485 kN
Size of column = 230 x 230 mm
Material Property:
Concrete grade = M20
Steel grade = fe415
fck = 20 N/mm2
fc = 415 N/mm2
Nominal Cover = 40mm
Size of footing:
Axial load on column = 485 kN
Self weight of footing = (10/100) x 485
= 48.5 kN
Total load on column = 48.5 + 485
= 533.5 kN
Area of footing = 533.5/200
= 3 m2
Size of footing = 1.7 m x 1.7 m
= 2.89
Factored soli pressure due to column
Po = Load / Area of footing
= 48.5/(1.7 x 1.7)
= 167.87
Factored soli pressure = 167.82 x 1.9
= 251.73
= 252 kN/m2
 51

Bending Moment:
M = (Po / 8) x (B(B-b)2)
= (251.73 / 8) x (1.4(1.4 – 0.3)2)
= 54.57
Depth of footing based on moment:
Mu,lim = 0.138 fck bd2
54.57x106 = 0.138 x20 x1000 x d2
D = 140.61mm
Based on one way shear,
Critical section shall be at a distance‘d’ from face of column
Vu = PO x B x ( 0.825 – d )
= 251.73 x 1.4 x ( 0.825 – d )
= 290.75 – 352.42 d
Shear resistance by this section
Pt = 0.2 %
𝜏c = 320 KN/m2
Shear resistance by concrete
= CC x b x d
= 320 x 1.4 x d
= 448 d
Equation 1 & 2
290.75 - 352.42 d = 488 d
d = 365mm
Taking greater value of d
d = 365mm
D = 365 + 50
= 415mm
 52

Design of Reinforcement
Mu = 0.87 x 415 x fy x Ast x d x ((1) – (fyAst/fckbd))
54.57x106 = 0.87x415xAstx360x((1) –
((415xAst)/(1000x20x365))
Ast = 430 37mm2
Ast = 0.12% Bd
= 0.12 x 1000 x 365
= 438mm2
Spacing
= (1000 x (𝜋/4) xd2)/Ast
= (1000 x (𝜋/4) x122)/ 430.37
= 262.71mm
Number of bars = 438/ ( (𝜋/4) x122)
= 4 no’s
Check
1. 3d 3 x 365 = 1095mm
2. 300mm
Spacing
= (1000 x (𝜋/4) xd2)/Ast
= (1000 x (𝜋/4) x122)/ 430.37
= 265mm
Provide 12mm dia @ 265mm c/c
 53

Fig: 4.5.1 Reinforcement Details in Footing

 54

CHAPTER 5

CONCLUSION
In this project we included analysis, load calculation, design and

planning of building. The analysis of frame is done by Moment

distribution method. Limit state method is used for all designs like

slab, beam, column, footing. The reinforcement details are provided

for each design with suitable values. All designs have under correct code

provisions.

Designing of slab was based on end condition and loading. From the

slabs, the loads are transferred to the beam. After this, the designing of

column is taken up depending on end condition, moments, eccentricity,

etc. Finally, the footings are designed on the loading from the column

and also the soil bearing capacity value of particular area.

The check conditions are all satisfied. By allocating the available

space for different functions the entire structure was developed. With the

upcoming of many challenging projects, we hope that our project stands out

and become a landmark for that beautiful place.

 55

REFERENCES

1. IS-875:1987(Part1 and 2),code of practice for design loads in

buildings and structure dead load, BIS, New Delhi.

2. IS-456:2000 Indian standard code of practice for plain and reinforced

cement concrete, BIS, New Delhi.

3. SP-16,Design Aids for reinforced concrete to IS456:1978,is,New

Delhi.

4. Dr.B.C.Punmia, “Reinforced Concrete Structures”, Volume 1 Second

Edition,standards publishers,Delhi.

5. N.Krishna Raju, “Design of Reinforced Concrete Struucture”,Third

Edition, CBS Publishes,New Delhi.