International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395 -0056

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ANALYSIS, DESIGN AND ESTIMATION OF BASEMENT+G+2

RESIDENTIAL BUILDING
R.D.Deshpande1, Manoj. N. Pai2, N. Pawan2, Aashish.P.Pednekar2
1Assistant professor, Civil Department, KLS Gogte Institute of Technology, Belagavi, Karnataka, India
2Bachelore of Engineering student, KLS Gogte Institute of Technology, Belagavi, Karnataka, India
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Abstract - Structural analysis is a branch which involves building. Multi–storied buildings aim to increase the floor
determination of behavior of structures in order to predict the area of the building without increasing the area of the land,
responses of real structures such as buildings, bridges, trusses the building is built on and hence saving land and in most
etc, with economy, elegance, serviceability and durability of cases money (depending on material used and land prices in
structure. Structural engineers are facing the challenge of the area). The design process of multi storied building
striving for the most efficient and economical design with requires not only imagination and conceptual thinking but
accuracy in solution, while ensuring that the final design of a also sound knowledge of science of structural engineering [1].
building must be serviceable for its intended function over its In the present study Basement+G+2 building is considered
design lifetime. This project attempts to understand the for the analysis of building using ETABS software 15.0.0.
structural behavior of various components in the multi-
storied building. Analysis, designing and estimation of multi- The project deals with the planning and designing of
storied building has been taken up for Basement+G+2 building of reinforced concrete framed structure using IS
Building, thereby depending on the suitability of plan, layout 456:2000 code. IS 456:2000 is the basic code for general
of beams and positions of columns are fixed. Dead loads are construction in concrete structures , hence all the
calculated based on material properties and live loads are structural members are designed using limit state method
considered according to the code IS875-part 2, footings are in accordance with the IS 456:2000 code. The planning of
designed based on safe bearing capacity of soil. For the design any building in India will recognized by national building
of columns and beams frame analysis is done by limit state code (NBC) [1]. Hence the building is planned in accordance
method to know the moments they are acted upon. Slab with the NBC.
designing is done depending upon the type of slab (one way or
two way), end conditions and the loading. From the slabs the Load consideration is in accordance with IS 875-
loads are transferred to the beams, thereafter the loads from part2. Modeling of the building is done in ETABS and is
the beams are taken up by the columns and then to footing analyzed to get the deformation, bending moment, shear
finally the section is checked for the components manually and force and area of steel requirement. Structural members
using ETABS 15.0.0 software for the post analysis of like slab, beam, column and footing are designed manually
structure, maximum shear force, bending moment and based on the values of bending moment and shear force
maximum storey displacement are computed. The obtained.
quantitative estimation has been worked out.
2. LITERATURE REVIEW
Key Words: NBC, ETABS, Multi-storied Building, Isolated
Footing, Open Newel Quarter Turn Staircase, M. Mallikarjun, Dr P V Surya Prakash (2016):
Estimation. Carried study on analysis and design of a multi-storied
residential building of ung-2+G+10 by using most
1. INTRODUCTION economical column method and the dead load and live
load was applied on the various structural component like
Structural analysis means determination of the general slabs, beams and found that as the study is carried using
shape and all the specific dimensions of a particular most economical column method this was achieved by
structure so that it will perform the function for which it is reducing the size of columns at top floors as load was more
created and will safely withstand the influences which will at the bottom floor. The economizing was done by means
act on it throughout its useful life. of column orientation in longer span in longer direction as
it will reduce the amount of bending and the area of steel
Due to concentration and increase of population into was also reduced [3].
urban cities, there is a need to accommodate the influx in
urban cities. However, due to rapid increase of land cost and P.P. Chandurkar et al (2013): Had presented study
limited availability of land, constructions of Multi–storied of G+9 building having three meters height for each storey.
buildings is taking part in our daily life. A multi- storied is a The whole building design had carried out according to IS
building that has multiple floors above ground in the
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code for seismic resistant design and the building had

considered fixed at base. Structural element for design had
assumed as square or rectangular in section. They had
done modeling of building using ETAB software in that
four different models were studied with different
positioning of shear walls [5].

Ismail Sab, Prof .S.M. Hashmi (2014): Generated 3D 4. BUILDING DATA FOR ANALYSIS
analytical model of twelve storied buildings for different
buildings Models and analyzed using structural analysis tool Salient Features
ETABS. To study the effect of infill, ground soft, bare frame
and models with ground soft having concrete core wall and  Utility of Building : Residential Building
shear walls and concrete bracings at different positions
 Area of the site: 70 X 60 (ft)
during earthquake; seismic analysis using both linear static,
linear dynamic (response spectrum method) has been  Building Height: 47 ft
performed. The analytical model of the building includes all
important components that influence the mass, strength,  Number of Storey: (BASEMENT+G+2)
stiffness and deformability of the structure [7].
 Type of construction : R.C.C Framed Structure
3. METHODOLOGY
 Shape of Building :Rectangular
Collection of data
 Number of staircase: one

 Number of Lift: One

 Type of Walls : Brick Wall

Planning

Drawing AUTOCAD

Bending Moment
Analysis
Shear Force

Slabs
Designing
Beams

Columns

Footing
Estimation
Staircase

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the specified details in defining. After that we define section

size by selecting frame sections as shown below & added the
 Dimensions Of Beams B 230x300 mm required section for beams, columns etc.
1
Step - 4: Assigning of Property
B 230x380 mm After defining the property we draw the structural
2
components using command menu
 Dimensions of Column C 230x380 mm
1 Draw line for beam for beams and create columns in region
for columns by which property assigning is completed for
C 230x460 mm beams and columns.
2
Step - 5: Assigning of Supports
C 230x600 mm By keeping the selection at the base of the structure and
3
selecting all the columns we assigned supports by going to
 Thickness of Slab 4.5” (140mm) assign menu, joint\frame, Restraints (supports), fixed.
Step - 6: Defining of loads
 Thickness of External 9“ (230mm) The loads in ETABS are defined as using static load cases
Wall command in define menu
Step - 7: Assigning of Dead loads
 Thickness of Internal 5” (120mm)
Wall After defining all the loads dead loads are assigned for
(as per IS:875 part2) external walls, internal walls

 2 Step - 8: Assigning of Live loads

Live Load
 Floor Finish 1.5kN/m Live loads are assigned for the entire structure including
 All rooms and Kitchen 2 floor finishing.
 Toilet and Bathroom 2kN/m
Step - 11: Assigning of load combinations
Corridor, Staircase and 2
Balcony 2kN/m Load combinations is based on IS 875 1987 PART 5 using
load combinations command in define menu
2
3N/m Step - 12: Analysis
After the completion of all the above steps we have
performed the analysis and checked for errors.
Step - 13: Design
5. ANALYSIS AND DESIGN OF BASEMENT + G + 2 After the completion of analysis we had performed concrete
BUILDING USING ETABS design on the structure as per IS 456:2000.

Step - 1: Step by Step procedure for ETABS Analysis For this go to Design menu, concrete design, select design
combo. After this again goes to design menu, concrete
The procedure carried out for Modeling and analyzing the frame design, start design \ check of structure then ETABS
structure involves the following flow chart. performs the design for every structural element.
Step - 2: Creation of Grid points & Generation of
structure 5.1 Plan of the Building
After getting opened with ETABS we select a new model and
a window appears where we had entered the grid
dimensions and story dimensions of our building. Here itself
we had generated our 3D structure by specifying the
building details in the following window.
Step - 3: Defining of property
Here we had first defined the material property by selecting
define menu, material properties. We add new material for
our structural components (beams, columns, slabs) by giving

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Assume thickness of slab = 0.14m=140mm

(Providing clear cover of 20mm) effective depth = 0.12m

=120mm

Density of concrete = 25KN/m³

Step 2:

Effective span

Lx = 1.51m

Width = 1.0m

Step 3
Fig-1: Plan of the building
Load calculation

Dead load = 3kN/m

Live load = 2kN/m

Floor finish = 1.5kN/m

Total factored load = 1.5*(3+2+1.5) = (9.75kN/m)

Step 4

Design moment for cantilever beam

Mu =

Step 5

Main steel reinforcement

Fig-2: 3-D View of Frames and Slabs

5.2 Design of Structural components Ast(req) =

5.2.1 Slabs
Slab is plate elements forming floor and roofs of =
buildings carrying distributed loads primarily by flexure.
A concrete slab is common structural element of modern Ast(req) = 313.13mm²
buildings. Horizontal slabs of steel reinforced concrete,
typically between 4 and 20 inches (100 and 500 mm)
Ast (min) =
thick, are most often used to construct floors and ceilings,
while thinner slabs are also used for exterior paving. Consider 10mm dia bar Area = 78.53mm²
Slabs may be classified into 2 types depending on ratio of
larger side to shorter side, its recommended in IS 456 to Sv(req) =
design slabs having (ly/lx) ratio greater than 2 as one way
slab and slabs having (ly/lx) less than 2 as two way slab. Providing spacing of 230mm

Design of Slabs Ast (pro) =

Step 1: Therefore provide 10mm dia @230mmc/c

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Step 6 Therefore provide torsion reinforcement of 8mm mesh bar

@190mmc/c
Distribution steel reinforcement
5.2.2 Beams
Provide minimum reinforcement of 168.0mm²
Beams shall normally be provided under the wall or
Consider 8mm dia bar for distribution bar bellow a heavy concentrated load. Beams transfer load from
slabs to columns, beams are designed for bending. In general
we have two types of beam: single and double. Similar to
Spacing required = Sv =
columns geometry and perimeters of the beams are
Spacing provided = 280mm assigned.
There are three types of reinforced concrete beams
Step 7 1. Single reinforced beams
2. Double reinforced concrete
Check for deflection 3. Flanged beams

Pt = Design of Beam

Step 1
fs =
Beam size 230*380 Span
Kt = 2.00
=4.20m
( Breadth b =230mm

Depth D=380mm

( Effective depth =380-25-

Step 8 BM (Muy) (left) = 28.4kN-m

Check for shear SF = 66.2kN-m

fck=20N/mm²
Vu =
fy=415N/mm²
=
Step 2

Mulim = 0.138*fcb*b*d²
Since = 0.138*20*0.23*(0.349)²
Step 9 = 77.32kN-m
Torsion reinforcement
Mu

Mesh size = = 0.302m Therefore design the beam as singly reinforced beam.

A= Step 3
Area of 8mm dia = 50.4mm² Area of steel reinforcement

Sv =

Sv(pro) = 190mm Ast(req) =

= 240.44mm²
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Ast(min) = Sv =

Consider 12mm dia bar Sv = 234.19mm say 230mm

Step 6
Number of bars required =
Minimum shear reinforcement
Number of bars provided = 3 bars
Sv =
Ast provided = 3*(
Sv = 237.70mm say = 230mm
Step 4
Step 7
Check for deflection
Reinforcement

Percentage of steel (Pt) = Provide 3#12mm dia @ mid span

fs = 5#12mm dia @sides

0.58*fy*
2 legged 8mm dia @ 230mmc/c.

fs = 0.58*250* 5.2.3 Column

A column or strut is a compression member, which is used
Kt = 2.00 primary to support axial compressive loads and with a
height of at least three it is least lateral dimension. A
( reinforced concrete column is said to be subjected to axially
load when line of the resultant thrust of loads supported by
= 26*2 column is coincident with the line of C.G of the column in the
longitudinal direction.
= 52
Design of Column
(
Step 1

( Column size 230*460 Breadth

Hence safe against deflection =b=230mm

Step 5 Depth = D =460mm

Design of shear reinforcement Pu = 1316.4kN

Mux = 6.6kN-m

Muy = 1.3kN-m

Fck = 25N/mm²
Since Fy = 415N/mm²
For shear design consider 2 legged stirrups of 8mm dia d’ = 40

Vus =Vu- Step 2

= 66.2*10^3-0.42*230*349 = 32487N

Asv = 2* Mu = 1.15*sqrt*(Mux²+Muy²)

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= 1.15*sqrt*(6.6²+1.3²) Muy1 = 19.102kN-m

= 7.74kN-m Step 5

To find Puz

Puz = 0.45*fck*Ac+0.75*fy*Asc

Asc = b*D-Asc(pro)

Using sp-16 chart -45 = 230*460-678.87

= 105121mm²
=0.02
Puz = 0.45*25*105121+0.75*415*678.87
P = 0.02*fck =0.02*25
= 1393.9
= 0.5
Step 6
Asc = To find αn
Asc = 529mm²

Provide 6 bars of 12mm dia

This is exceeding value of 0.8
Asc pro = 678.67mm²
Therefore αn =2
Actual p =

Actual p = 0.641
=(
p/fck =

Step 3 To find (= 0.124

Mux1 p/fck = Hence safe

0.0257 Step 7

Pu/(fck*b*d) = Design of Ties

0.02 d’/D = 0.11 Consider diameter of Ties = 8mm

from chart 46 Maximum spacing

1) b or D = 230mm
2) 16*12=190mm
3) 300mm(therefore provide 8mm dia
Mux1 = 0.015*25*230*460² bar
@190mmc/c)
= 19.102kN-m
5.2.4 Isolated Footing
Step 4

To find Muy1 Foundations are structural elements that transfer

loads from the building or individual column to the earth.
If these loads are to be properly transmitted, foundations
Since p/fck, ,d’/D are same are also same must be designed to prevent excessive settlement or
therefore rotation, to minimize differential settlement and to
provide adequate safety against sliding and overturning.
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Design of Footing
Muy = 0.5*wu*L*(
Step 1
Muy = 0.5*257.11*3.20*(
Column size 230*460
Muy = 193.0kN-m
Factored load on footing =pu=1316.4kN
Step 5
Working load on footing =
Effective depth required
Assume self weight of footing
Dreq = sqrt(
=

Total load w = p+0.1p = sqrt (

= 877.61+87.76 = 295.7mm
= 965.37kN Increase depth 1.75 to 2 times due to shear considerations
Step 2 Depth provided = 2*dpro = 2*295.7 = 591.3mm

Area of footing required A = Effective depth provided = 600mm

Consider SBC of soil = 200KN/m² Step 6

Area of reinforcement along x-direction

Area = =4.83m²
a) Astx =
For rectangular footing

Consider length of the footing (L) = 3.20m

Astx = 1857.60mm²
Therefore the breadth (B) =
Consider 16mm dia bar
Area provided = 3.20*1.60

= 5.12m Number of bars =

Step 3 Provide 10 bars of 16mm dia along x-axis

Upward factored soil reaction (wu)

b) Asty =

Wu =

Step 4

a) Bending moment about axis x-x passing through

face of column as
=
Mux = wu*B*(
= 909.38mm²
Mux = 0.5*257.11*1.60*(
Consider 16mm dia bar
= 386.1kN-m

b) Bending moment about axis y-y passing through No. of bars =

face of the column
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Therefore provide 5 bars of 16mm dia along y- axix

Step 7 Therefore = 1.118

Check for one way shear Since The design is safe against two way shear.

Vu = w*B*(( 5.2.4 Open Newel Quarter Turn Staircase

The type of stair and its layout is governed

= 257.11*1.60(( essentially by the available size of staircase room and the
positions of the beams and columns along the boundary of
Vu = 316.8kN the staircase. The slabs, in general, are heavy compared to
floor slab-because of Greater live load on stairs than that on
floors. Therefore, longer spans for the flights are avoided as
far as possible.
Stair flight shall preferably be supported on beams or walls.
Pt = Supporting the flight on landing slab should be avoided as far
as possible. Wherever possible, landing beams may be
Pt = 0.209 provided at this end of flight to reduce the span.

Refer table no.19 IS 456-2000 for

Since

Step 8

Check for two way shear

Vu = wu*(A-((D+d)*(B+b))

= 257.1((320*1.60)-

(0.46*0.60)*(0.23*0.6)) Vu = 1090.2kN bo

= perimeter = 2*(D+d)+2*(b+d) =
Fig-3: Staircase Plan
2*(0.46+0.6)+2*(0.23+0.6) bo = 3.78m
Design of Staircase
Nominal stress
Floor height = 3m fck
= 25N/mm² fy =
500N/mm²

= 0.25*sqrt(20) Step 1

Consider flight of 1m each

= 1.118
Assume riser of 125mm Number
K=0.5+
of riser = 8 Number of treads = 7
Where
Treads = 240mm
K = 0.5+0.5 = 1.0
Total going = 7*240=1680mm
Therefore

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Width of each landing 2 and 3 = 1500mm Live load = 3.00kN/m²

Total = 9kN/m²
The landing slab is spanning longitudinally along section Factored load = 13.5kN/m²
1-1. Landing slab 2 is common to spans of 1-1 and 2-2.
Crossing at right angles clause (33.2 of IS 456). The 3) Landing slab 2 = 50% of load n landing slab 1 =
effective span of sec 1-1 shall be from centre line of edge 6.75kN/m²
beam to centre line of bridge wall. While the effective span
for section 2-2 shall be from centre line of landing slab 2 The total loads of 1, 2 and 3
to centre line of landing slab 3( CI 33.1 b of IS 456). Total loads (1) going = 16.78*1.68*1.25
= 35.23kN
Design of landing slab 1 and going Total loads (2) landing slab (1) = 13.5*1.6*1.25
= 27.00kN
Step 2
Total loads (3) landing slab (2) = 6.75*0.625*1.25
Effective span and depth of slab. = 5.27kN
Total =67.50kN
Effective span = 150+1600+1680+ = 4055mm. Step 4

Depth of waist slab is assumed as = = 202.75= D = 200mm Bending moment and shear force width = 1.250mm

The effective depth d = 175mm Vp = (27(4.055-0.875)+35.23(4.055-

2.59)+5.27(0.3125))/4.055
The landing slab is also assumed to have total depth of
200mm and effective depth of 175mm Vp =
Step 3 Vp = 34.30kN
Calculations of loads Vj = 67.5-34.30
1) loads on going (on projected plan area) Vj = 33.20kN
a) Self weight of waist slab = D*sqrt*(1+riser2/tread2) The distance x where shear force is zero is obtained
*25
34.30-27-16.78*(1.25)*(X-1.75) = 0
= 0.2* ) * 25 X = 2.098m
= 0.20*1.1275*25 = 34.30(2.098)-27(1.223)-16.78(1.25)(0.348*0.348)*0.5
= 5.63kN/m² = 37.67kN-m

b) Self weight of steps = 25* *(riser) Step 5

Check for depth

= *25*(0.125)

= 1.5625kN/m² d = sqrt ( = 104.49 mm

c) Finish load = 1.0kN/m² 104.9 175 mm

d) Live load = 35.0kN/m²
For maximum shear force N/mm²
Total load = 11.193kN/m²
For depth of slab as 200 (C.I.402.1.1 OF IS456)
Factored load = 1.5*11.193 = 16.78kN/m² 2)
K = 1.20
Landing slab 1

Self weight of slab = 250*0.20 = 5kN-m

Finish load = 1.00kN/m²

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The design is Bending moment and shear force (width =1.25m)

safe.
1) The total load is 47.88kN/m² and symmetrically
Step 6 placed to give Vg=Vh= 23.94kN.
2) The maximum bending moment at x =
Determination of areas of steel reinforcement 1.59m(centre line of span 3.18m= 23.94*1.95-
7.38(1.59-0.4375)16.78*1.25*(0.715)(0.715*(0.5)
Ast= Moment = 20.96nN-m
Shear force = 23.94kN

(Since the maximum bending moment and shear force are

= 645.97mm² less than those of other section, maximum moment = 37.67
and shear force = 34.30kN)
Use 12mm dia bars
The depth of 200mm here is ok . Accordingly the amount
reinforcing is determined. Ast = 346mm²
Sv =
Providing 12mm dia bars
Use 12mm dia bars @ 170mm c/c

Distribution steel reinforcement

Ast =
Distribution bars are of 8mm dia @200mmc/c.

Sv = = 200mm 6. ESTIMATION

Use 8mm dia @200mm c/c Estimating is the technique of calculating or

computing the various quantities and the expected
Design of landing slabs 2 and 3 expenditure to be incurred on a particular work or project.

Step 1 Following requirement are necessary for preparing an

estimate
Effective span and depth of slab a. Drawings like plan, elevation and section of
important parts.
The effective span from the centre line of landing slab 2 to b. Detailed specifications about workmanship and
the centre line of landing of 3 = 750+1680+750 properties of materials etc.
The depth of the waist slab and landing is maintained as c. Standard schedule of rates of the current year.
200mm Before taking up any work for its execution, the owner
or builder should have a thorough knowledge of work that
Step 2 can be completed within the limits of his funds or the
probable cost that may be required to complete the proposed
Calculation of loads work.
It is therefore necessary to prepare the probable cost or
(1) Load on going ( step 2(1) of A ) = 16.78kN/m² estimate for the proposed work from its plan and
(2) Loads on landing slab 2 (step 2(3)) = 6.75kN/m² specifications. Otherwise, it may so happen that the work has
(3) Loads on landing slab 3 (step2(3)) = 6.78kN/m² to be stopped before it is complete due to shortage of funds
or of materials. Besides the above, an estimate for any public
Total factored load construction work is required to be prepared and submitted
beforehand so that section of necessary funds may be
1) Going = 16.78(1.68)*(1.25) =35.23kN.
obtained from the authority concerned.
2) Landing slab A = 6.75*0.875(1.25) = 7.3kN
3) Landing slab B = 6.75(0.625)(1.25) = 5.27kN.

Total = 47.88KN.

Step 3
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Estimation is Done by Centre Line Method RUPEES ONE CRORE TWENTY NINE LAKHS NINTY
ONE THOUSAND ONLY.

Table-1: Estimation of the Building

Estimation is Done Using Microsoft Excel
7. ETABS OUTPUT

7.1 Shear Force

Basement
STOREY INR (Rs)

BASEMENT FLOOR = 28,50,000.00

GROUND FLOOR = 30,90,000.00

FIRST FLOOR = 34,40,000.00

SECOND FLOOR = 36,11,000.00

GRAND TOTAL= 129,91,000.00

Fig-4: Shear Force Values of Basement Floor
Ground Floor

Fig-8: Shear Force Values of Terrace

Fig-5: Shear Force Values of Ground Floor

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First Floor 7.2 Bending Moment

Basement

Fig-6: Shear Force Values of First Floor

Second Floor
Fig-9-: Bending Moment Values of Basement
Ground Floor

Fig-7: Shear Force Values of Second Floor

Terrace Fig-10-: Bending Moment Values of Ground Floor
First Floor

Fig-11-: Bending Moment Values of First Floor Fig-12-: Bending Moment Values of Second Floor
Terrace
Second Floor

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REFERENCES

[1] B. Pradeep Kumar, Sk. Yusuf Basha, Planning, Analysis

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[2] D. Ramya, A.V.S. Sai kumar, Comparative Study on

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Fig-13-: Bending Moment Values of Terrace [3] M. Mallikarjun, Dr. P M V Surya Prakash, Analysis and
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8. CONCLUSIONS (ung2+G+10) By Using Most Economical Column Method,
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 The Columns were designed using SP-16 and was Analysis and design of G+5 Residential Building, IOSJR
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be safe.
 The footing were designed and checked for one way [5] P.P. Chandurkar, Dr. P.S. Pajgade, Seismic Analysis of
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[6] Chaitanya Kumar. J. D, Lute Venkat, Analysis of
 In the present scenario, a structural engineer
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involves tedious procedures and complicated International Journal of civil and structural
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quiet feasible in using and offers more efficient [8] Bureau of Indian Standards: IS 456:2000, Plain and
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[9] Bureau of Indian Standards: IS-875 (part 1)-1987, Code
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 Time is saved for structural design work, so that ( second revision ), NEW DELHI.
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 We can conclude that there is a difference between [10] Bureau of Indian Standards: IS-875 (part 2)-1987,
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[12] Dr. V. L. Shah, Dr. S. R. Karve, Illustrated Design of

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BIOGRAPHIES
Prof. R. D. Deshpande,
Department of civil engineering,
KLS Gogte Institute of Technology,
Belagavi, Karnataka, India.

Mr. Manoj. N. Pai,

Department of civil engineering,
KLS Gogte Institute of Technology,
Belagavi, Karnataka, India

Mr. N. Pawan,
Department of civil engineering,
KLS Gogte Institute of Technology,
Belagavi, Karnataka, India

Mr. Aashish. P. Pednekar

Department of civil engineering,
KLS Gogte Institute of Technology,
Belagavi, Karnataka, India

© 2017, IRJET | Impact Factor value: 5.181 | ISO 9001:2008 Certified Journal | Page 2088