g+3 apartment building project report

DESIGN OF G+3 APARTMENT

BUILDING

Presented byVijayvenkatesh.C

Department of civil engineering

Final year

ST. JOSEPH’S COLLEGE OF ENGINEERING & TECHNOLOGY

THANJAVUR

Under guidance ofLAKSHMIKANTH.P

Assistant professor of

Civil department

Outline:

Abstract

Introduction

Literature review

Planning of apartment building

Elevation works of apartment buildings

Design of G+3 apartment building

*design of column

*design of beam

*design of foundation

*design of staircase

*design of slab

Load displacement of structural work

conclusion

ABSTRACT

Now a days, the people from villages are coming to towns for

employment and educational facilities hence with the limited land

available, individual houses are feasible.

One of the major problem facing by the INDIAN country is rapid

growth of population which restricted the availability of the land

More over, even the available houses are let out at abnormal rent

charges.

Hence an apartment building is proposed in this project.

The project consists of six houses in three floors with all basic

amenities.

The construction and the design will be based on LIMIT STATE

method and as per the schedule rates of PWD Government of

TAMILNADU.

INTRODUCTION

One of the major problem in Indian country

facing the rapidly growth of population this can be

solved to a certain extent with the construction of

multi storied apartments, which can be live many

people in available area.

The proposed site for our project is at VALLAM

town in THANJAVUR district.

The maximum numbers of people go either to

THANJAVUR town.

The project consists of six houses in three floors

with all basic amenities.

It is most economic in design and construction

and followed as per limit state design method.

GENERAL LAYOUTThe other facilities like parking, cycle

stand, toilets for either sex and bill counter have been

place with the construction for proper location.

OTHER PROVISIONApart from the space requirements and

sound treatment, for resistance has also to be

considered daily through its does not significantly

affect the layout or design.

INTRODUCTION TO LIMIT STATE

Since this method of design is

adopted widely in recent practice. We have followed

this method and all the design works are carried out

in accordance to IS 456-2000

The object of the design is to

produce a structure which is in all respects

satisfactory for which is required, having the

economy as the primary objective

As our knowledge about the characteristics to RCC

member under axial load, flexure shear and torsion

improved, the allowable stresses are periodically

adjusted to achieve greater economical design. The

adoption of limit state design concept as the latest

version of IS 456-1984 loads towards this direction.

In the limit state designs structure

shall be designed to safety withstand all loads liable to

act on throughout its life. It should be also satisfy the

serviceability requirement such as limitation on deflection

and cracking.

SPECIFICATION

EARTH WORK EXCAVATIONExcavation- foundation trenches shall be

drought to the exact width of foundation concrete and

the side should be vertical. If the soil is not good and

does not permit vertical sides. The sides should be

slopped back are protected with timber shoring

excavated earth shall not be placed within one meter of

the trench reinforcement cement concrete RCC.

BASEMENTThe basement with RR masonry in CM 1:5

using first class blue granite metal with 450mm width, 600

mm height are proposed the clear dry river sand in purpose

of filling the basement. The base concrete is also made with

CC 1:4:8 to over sand filling.

DPCThe top of the basement structure of DPC

provided with 20mm thick using crude oil 5% by weight of

cement in used.``

SUPER STRUCTUREThe super structure of the building with

brick in CM 1:5 mix using first class bricks of size 19*19*9cm

of 230mm width and height up to 3mm with parapet at the

height of 600mm is provided.

ROOFINGThe roofing of the structure will be RCC

slab of 1:1.5:3 mix using 20mm IS gauge hard broken granite

chips weathering course with jelly with lime concrete and

top finished with pressed tiles of 20*20*2cm size.

FLOORINGThe top is finished smooth with trowel floor

finish with marbles.

LITERATURE REVIEW

After referring lot of book related to

reinforced concrete structure, we have to arrive the design

and design procedures for this project as per the some

standard reference. slabs are designed by using the ratio of

longer span to shorter span; this may help to find the types

of slab. and other calculation of the slab are calculated by

using some design methods as per IS 456-2000

beams and columns are designed as per

code provision like that IS CODE AND NBC CODE. the load

from column hall is transmitted into the footing. the design

of footing maybe done by based on safe bearing capacity of

the soil. the compressive stress in concrete at the base of

column OR pedestal shall be considered during the design

of column. the live load is variable but does not exceed

three quarters of the dead load, or the nature of the live load

is the wall panels will be loaded simultaneously. moments

may be assumed to occur at all sections when full design

on design of footing(IS 456-1978) for sloped footing the

effective cross section in compression shall be limited by

the area above the general plan, and the angle of slope or

depth and location of steps shall be such that the design.

the locker room wall are constructed by RCC (1:1.5:3) with

300mm kitchen. floor may be of RCC(1:1.5:3)mm. door and

ventilation of the locker room shall be fixed as per the

manufacture specification

SPECIFICATION OF G+3 APARTMENT

BUILDINGD:

Ground floor - car parking:110.49m^2

First floor

2^nd floor

Third floor

Living room:3.81X4.03m

Kitchen:2.87X3m

Two bed room with bath room attachment:2.41X2.5 ;3X5m

Elevator lift:1.46X1m

Stair case:

rise=150mm

tread=300mm

Specified with 2-

portion:

Specification:

Components size specification use of G+3

apartment

building:

column size=0.35X0.35m

beam size=0.35X0.3m

slab size (1-panel) 4.5X3m=0.165m

Main door=1.18X2m

Window=1X1.5m

Door=1X2m

Partition external wall thickness =350mm

internal wall thickness=250 to 100mm

ventilator =0.25X0.32m

DESIGN OF G+3 APARTMENT

BUILDING

COMPONENTSGROUND FLOOR COLUMN DESIGN

C1,C2,C3,C4,C5,C6,C7,C8,C9,C10,C12,C13,C14,C15,C16,

C17,C18,C19,C20,C21

Design of column in G+3 apartment

buildingBreadth=350mm

Depth =350mm

Pu=1150KN

MuX=MuY=40KN.m

Fck=20N/MM^2

Fy=415N/MM^2

Assume d’=40mm

d’/D=0.112Equivalent moment:

The reinforcement section is design for the axial compression

load

Pu

Mu=1.15(Mux^2+Muy^2)

=1.15(40^2+40^2)

=65.0538KN.m

Non dimensional parameter:

(pu/fckbd)

1150X10^3/20X350X350

=0.46

Mu/Fckbd^2

=65.038X10^6/20X350X350^2

=0.0758

Longitudinal reinforcement:ref chart -44 of SP-16

P/fck

=0.07

P=20X0.07=1.4

asc=(pbd/100)

=1715mm^2

Provided 4 bars of 20mm diameter and 4bars of 16mm diameters

spacing =180mm c/c

p=100Xast/bd

=100x1715/350X350

=1.4

p/fck=0.07

Ref chart -44 of SP-16 and read out (Mux1/fckbd^2) corresponding to

The value of (pu/fckbd)=0.46

MuX1=(0.07 X20X350X350^2)X10^6

MuX1=Muy1=60.025KN.m

Puz=(0.45fckac+0.75fyas)

=0.45X20X((350X350)-1840)+(0.75X415X1840)

=1085940+572700

=1658KN

(pu/puz)=1150/1658

=0.693

The co-efficient αn corresponding to (pu/puz)

=0.69

αn=1.72

Check for satisfied under biaxial bendings

((Mux/Mux1)^(αn)+(Muy/Muy1)^(αn))=1

0.49+0.49=1

0.98=1

Hence the section is safe under biaxial bending.

Reinforcement detail:

provided 4-bars of 12,16,20, diameter as longitudinal reinforcement and

8MM lateral ties at 180mm c/c =(300mm c/c)

DESIGN OF = C1,C7,C13

STAAD.PRO OUTPUTS

DESIGN 0F =C2,C8,C14

`

DESIGN OF =C3,C9,C15

DESIGN OF = C16 ,C21

DESIGN OF =C5,C11,C18

DESIGN OF=C6,12,19

DESIGN OF 2ND FLOOR COLUMN

C1,C7,C13

C2,C8,C14

C2,C9,C15

C16,C21

C4,C5,C6,C10,C11,C12,C17,C18,C19

DESIGN OF 3RD FLOOR COLUMN

C1,C7,C13

C2,C8,C14

C3,C9,C15,C20

C16,C21

C4,C10,C17

C5,C6,C11,C12,C18,C19

Design of beam :

A reinforcement concrete beam is to be design

over a effective span 1.25 m

Given :

effective span =1.25m

width of beam =350mm

overall depth =300mm

service load =40KN/m

effective cover=50mm

material=M20 grade concrete

fe-415 HYSD bars

Ultimate moment and shear forces:Mu=0.125X1.5X40X1.25^2

=16.875KN/m^2

vu = 0.5X1.5X40X1.5

=45KN

Main reinforcement:Mulim=0.138fckbd^2

= 0.138X20X350X300^2

=86.94KN.M

Since Mu<Mulim

hence take singly reinforcement beam.

Mu=0.87fyastd(1-(astfy/bdfck))

ast=194.33mm^2

Provided 4 bars of 12MM diameter

Check for shear stress :vu=45KN

Ɩv=(vu/bd)

= 45X10^3/(350X250)

=0.514N/mm^2

pt=(100ast/bd)

=0.22

Ref table 19 IS:456-2000 and read out the design shear strength of concrete

Ɩc=0.173N/mm^2

Ɩc>Ɩv shear reinforcements are required

balanced shear= Vus=Vu-(Ɩcbd)

=45-(0.173X350X250)X10^-3

=29.86KN

Sv >0.75d

=0.75X250

=187.5mm

=provided 6mm diameter stirrups of at 190mm shear supports

Check for deflection control:

pt=0.22

(l/d)max=(l/d)basic XktXkcXkf

=20X1.1X1X1

=22

(l/d)actual=(1250/250)=5<22

hence safe

SPAN LENGTH : 1.25m

A reinforcement concrete beam is to be

design over a effective span =1.5m

Given:

effective span =1.5m

width of beam =350mm

D=300mm

service load =40KN/m

d’=50mm

material=M20 grade concrete

=fe415 HYSD bars

Solution:

ultimate moment and shear force :Mu=(0.125X1.5X40X1.5^2)

=16.875KN.m

vu=0.5X1.5X40X1.5

=45KN

Moment reinforcement:Mulim=0.138fckbd^2

=0.138X20X350X250^2X10^-6

=60.375KN.m

Mu<mulim

Hence single reinforced beam.

Mu=0.87fyastd(1-(astfy/bdfck))

16.875X10^6=0.87X415XastX250X(1-(ast415/350X250X20)

ast=196.069mm^2

Provided 4 bars @ 12mm diameter as tension reinforcement 2 bars of 10mm

Diameter hanger bars are compression sides

Check for shear stress:Vu=45KN

Ɩv=(vu/bd)

=0.51428N/MM^2

pt=100ast/bd

=100X197/350X250

=0.224

Table -9 of IS-456-2000

Ɩc=0.1771N/mm^2

Ɩv>Ɩc shear reinforcements are requirements

Balance shear =Vus=Vu-(Ɩcbd)

=45-(0.1771X350X250)

=15.45KN

sv>0.75d

0.75X250

=187.5mm

Provided 8mm diameter stirrups at 189mm shear support.

Check for deflection control :(l/d)max=(l/d)basicXktXkcXkf

=22

(l/d)actual<(l/d)max

hence safe.

SPAN LENGTH= 1.5m

A reinforcement concrete beam is to be

designed over a effective span =3mEffective span =3m

Width=350mm

Depth=300mm

d=250mm

d’=50mm

Service load=40KN/m

Sol:

ultimate moment and shear force :

Mu=0.125X1.5X40X3^2

=67.5KN.m

Vu=0.5X1.5X40X3

=90KN

Main reinforcement:

Mulim=0.138fckbd^2

=0.138X20X350X250^2

=60.375KN.m

Mu>Mulim design a doubly reinforced section.

(Mu-Mulim)=7.125KN.m

fsc=(0.0035(Xumax-d’)/xumax)es

=408.333N/mm^2

But fsc >0.87 , Fy=0.87X415

=361N/mm^2

asc=Mu-Mulim/fsc(d-d’)

=7.125X10^6/361X(250-50)

=98.6842mm^2

Provided 2-bars @ 16mm diameter

Ast2= ascfsc/0.87fy

=99X361/0.87fy

=98.98mm^2

Ast1=0.36fckbxulim/0.87fy

=837.557mm^2

Total tension reinforcement

ast=(ast1+ast2)

=837.557+98.98

=936.537mm^2

Provided 2bars @ 25mm diameter (ast=981.16mm^2)

Shear reinforcement:

Ɩv=vu/bd

=90X10^3/350X250

=1.02N/mm^2

pt=100ast/bd

=100X981.16/350X250

=1.12

Ɩc=0.585N/mm^2

Since Ɩv>Ɩc shear reinforcement are requared

Vus=(Vu-(Ɩcbd))

=(90X10^3-(0.585X350X250)

=38812.5X10^-3

=38.8125KN

Using 8mm diameter 2-legged stirrups

Sv=0.87fyasvd/vus

=0.87X415X2X250X50/38.8125X10^3

=232.56mm

Sv>0.75d

0.75x250

=187.5mm

Adopt spacing of 200mm near supports gradually increasing to 300mm toward

The center of span.

Check for deflection control:

(l/d)actual=3000/250

=12

(l/d)max=((l/d)basicktkckf)

pt=1.12 ; pc=100X98.64/350X250

=0.11273

Kt=0.8

Kc=0.94

Kf=0.86

(l/d)max=20X0.8x0.94X0.86

=12.9344=13

(l/d)actual<(l/d)max

Hence deflection control is satisfied.

DESIGN OF BEAM

SPAN LENGTH =3m

A reinforced concrete beam is to be designed

over a effective span =4.5mGiven:

Effective span =4.5m

b=350mm

D=300mm

d=250mm

d’=50mm

service load=40KN/m

Sol:

ultimate moment and shear force:

Mu=0.125X1.5X40X4.5^2

=151.875KN.m

vu=0.5X1.5X40X4.5

=135KN

Main reinforcement:

Mulim=0.138fckbd^2

=0.138X20X350X250^2

=60.375KN.m

Mu>Mulim

hence design a doubly reinforced beam sections

Mu-Mulim=151.875-60.375

=91.5KN.m

fsc=(0.0035(xumax-d’)/xumax)Xes

=0.0035(0.48x250)-50/0.48x250xEs

=408.33N/mm^2

But fsc=361N/mm^2

asc=Mu-Mulim/fsc(d-d’)

=91.5X10^6/361(250-50)

=1267.31mm^2

provided 6 bars @ 16mm diameters

ast2=ascXfsc/0.87fy

=1267.31X361/0.87X415

=1267.134mm^2

Ast1=0.36fckbxulim/0.87fy

=0.36X20X350X0.48X250/0.87X415

=837.557mm^2

Total tension reinforcement :

ast=ast1+ast2

ast=2104.691mm^2

Provided 4bars @ 25mm diameter

Shear reinforcement:

Ɩv=(vu/bd)

=135X10^3/350X350

=1.54N/mm^2

pt=100ast/bd

=100X2104.691/350X250

pt=2.40

Ɩc=1.255N/mm^2

Ɩv>Ɩc shear reinforcement is required

vus=(vu-(Ɩcbd))

=135-(1.255X350X250)

=109.67KN

Using 8mm diameter 2-legged stirrups

Sv= 0.87fyasvd/vus

=0.87X415X2X250/109.67X10^3

Sv=82.303mm

Sv>0.75

0.75X250

=187.5mm

Adopt a spacing of 200mm near support gradually increased to 300mm toward

the center of span.

Check for deflections:

(l/d)actual =4500/250

=18mm

pt=2.40

Kt=1.1716

Kc=1

Kf=1.8

(l/d)max=20X1.716X2.303X1.8

=61.776

(l/d)max>(l/d)actual

Hence deflection control is satisfied.

SPAN LENGTH: 4.5m

GRAPH= BEAM(3m)

RxN

GRAPHS=BEAM(1.5m)

RxN

GRAPH=(BEAM) 4.5m

RxN

GRAPH=(BEAM) 1.25m

RxN

STAAD.PRO V8I FOUNDATION DETAILS

FOR ( G+3 apartment building)

Load case = (dead load) 2.25KN

Live load :20KN/M

A square footings for a G+3 apartment

building

Given: Dead load = 507kN

Imposed load= 500KN

Breadth= depth= 350mm

Safe bearing capacity of soil (qu)= 200KN/m^2

Factored (qu)=1.5x200=300kN/m^2

Solution:

Size of footing:

Dead load + imposed load = 1007KN

Wu=1007 KN

Factored area = 1007/1.5x 200

= 3.35m^2

Side of square footing :

(3.35)^(1/2)

= 1.83m =2m

Adopt square footing of size = 1.83X1.83m

Factored soil pressure at the base:

=1007/(1.83)^2

251<300KN/m^2

Hence footing area is adequate

Factored of bending moment:

cantilever projection from the face of column

=0.5(2-0.3)

=0.85m

Bending moment of either side

=qul^2/2

=1.5(200X0.85^2)/2

=108.375KN.m

Depth of footing :

Moment

Mu =0.138fckbd^2

d=(Mu/0.138fckb)^(1/2)

=(108.375X10^6/0.138X20X1000)^(1/2)

d=198.157mm

Depth from shear consideration shear force per meter width :

=251((2000/2)-(300/2)-d)n

Ɩc = 0.36N/mm^2

For m20 grade concrete with nominal percentage of reinforcement

P=0.25

Ɩc=vu/bd

0.36=251X(850-d)/1000d

360d=213350-251d

611d=213350

d=350mm

D=350+50

=400mm

Reinforcement:

Mu=0.87XfyXastXdX(1-(astXfy/bdfck)

108.375X10^6=0.87X415XastX350X(1-5.92X10^-5ast)

Ast=906.22mm^2

Provided 16mm diameter rod and 250mmc/c

Check for shear stress:

Vu=251(850-350)X10^-3

Vu=125.5KN

p=100ast/bd

P=0.407

from tabel19 of IS456 -2000 the permissible shear

Stress

ksƖc=1X0.609=0.61N/mm^2

Nominal shear stress =Ɩv=Vu/bd

Ɩv=0.709N/mm^2

Ɩv<ksƖc shear stress within safe permissible limit.

DESIGN OF R.C SLAB FOR A G+3

APARTMENT BUILDINGS

GIVEN:Roof size=4.5X3m

LX=3m

LY=4.5m

LY/LX=1.5m

fck=25N/mm^2

fy=415N/mm^2

SOLUTION:Depth of slab:

As the ratio of long to short span 1.5 which is

<2 the slab has to be designed as a 2-way slab.

As the loading condition exceed 3KN/mm^2

Adopt a span/effective depth ratio of 25

d=4500/25

=180mm

D=180+25

=205mm

Effective span:

clear span + effective depth

=3+0.18

=3.18m

Loads:self wt=0.205X1X25=5.125KN/m^2

live load=3KN/m^2

floor finishes= 1KN/m^2

service load=9.33KN/m^2

wu=1.5X9.33=13.995KN/m^2

Ulimate design moment and shear force:αx=0.066:αy=0.048

Mux=(αxwulx^2)=9.34KN.m

Muy=(αywuly^2=6.793KN.m

Vux= 0.5wulx=22.25KN

Check for depth:consideration the maximum moment of

d=(9.34X10^6/0.138X20X10^3)^(1/2)

=58.172mm

Which is less than d=180mm

Hence the d selected is sufficient to resists the maximum design

Moment.

Reinforcement for negative and positive moment :

Short span :Mu=0.87fyastd(1-(astfy/bdfck))

9.34X10^6= 0.87X415XastX180X(1-(ast415/10^3X180X25))

ast=146mm^2

provide 10mm diameter bars

spacing=10^3X78.53/146

=537.94mm

This is more than 3d

3X180=540mm

Hence adopt maximum spacing of 540mm . Provided 10mm diameter

Bars

=78.53/540

=0.145X10^-3

ast=145.4mm

(Mux)p<(Mux)n provided the same spacing

Long span:provided 10mm diameter bar

520mm spacing on both direction

Check for shear stress:considering a short span and the width of slab .

Ɩv=vu/bd

=22.85X10^3/10^3X180

=0.1269N/mm^2

pt=100ast/bd

=0.080

Ɩc=0.135N/mm^2

kƖc=0.175N/mm^2

Which is greater than Ɩc=0.12N/mm^2

Hence the shear stress are within safe permissble limit

check for deflection control:

(l/d)basic=25

for pt=0.17;kt=1

25X1X1=25

(L/D)actual=17.67

17.65<25

Deflection control is satisfied

Check for crack control:

Spacing adopted is 360mm

Reinforcement minimum=0.12percentage

0.0012X205X1000

=246mm^2

Which is less than that provided

Diameter of reinforcement

D/8<205/8<25mm

Which is more than the diameter of rod provided as 10mm

Torsion reinforcement at corner:

Area of tension steel provided an each of the corners in 4layers

=0.75Xast

=0.75X145.4

=109.05mm^2

Length over which torsion steel is provided

=(1/5)X short span

=0.2X3000

=600mm

Provided 6mm diameter bars at 240mm center for a length of 600mm

at all 4 –corner in 4-layer

Reinforcement in edge strips:ast=0.12percentage

0.0012X10^3X205

=246mm^2/m

Provided 10mm diameter bars at 300mm centers.

PRESSURE LOAD REACTION OF SLAB

DESIGNPRESSURE LOAD=10KN/m^2

DESIGN OF STAIR CASE IN G+3

APARTMENT BUILDINGStair case (type) use in G+3 Apartment building :

dog legged stairs with waist slab , tread, risers

Given :Numbers of steps in flight =7

tread T=300mm

rise R=150mm

width of landing beams= 300mm

M20 grade concrete fck=20N/mm^2

fe415 HYSD bars fy=415N/mm^2

Effective span:L=7X300 +300

=2400mm

=2.4m

Thickness of waist slab =(span/20)

=2400/20

=120mm

Effective depth =d=95mm

Loads:Dead loads of slab on slope = ws=( 0.12X1X25)

=3KN/m

Dead load of slab on horizontal span is

W=(ws(r^2+t^2)^(1/2)/t)

W=3.35KN/m

Dead load of one step =(0.5X0.15X0.3X25)

=0.56KN/m

Load of steps per meter length =(0.56X10^3/300)

=1.86KN/m

Finishes etc =0.53KN/m

Total dead load =(3.35+1.86+0.53)

=5.74KN/m

Service live load =5KN/m

Total service load =5.74+5

=10.74KN/m

Factored load =1.5X10.74

=16.11KN/m

Bending moment:Maximum bending moment at center of span is :

M=0.125wul^2

=0.125X16.11X2.4^2

=32.22KN/m

Check for depth of waist slab :d=(mu/0.138fckb)^(1/2)

=108.5mm

94.2mm<95mm (provided)

Hence safe

Main reinforcement:Mu=0.87X415XastXd(1-(415ast/bdfck))

32.22X10^6=0.87X415astX95(1-(415ast/1000X95X20))

ast=1319.64mm^2

Provided 12mm diameter bars at 200mm centers

Distribution reinforcement :

=0.12percentX1000X120

=0.0012X1000X120

=144mm^2/m

Provided 8mm diameter bars at 200mm centers

Design using SP-16 design chart

compute the design parameter :

(Mu/bd^2)= 32.22X10^6/1000X120^2

=2.2375

Refer table 2 of SP-16 design table corresponding to fck =20N/mm^2 and

read out the percentage of reinforcement as:

ast =(pbd/100)

=0.669 X1000X120/100

=802.8mm^2/m

The reinforcement quantity is the same as that obtained by analytical

Method.

LOAD DISPLACEMENT DETAILS FOR

STRUCTURAL WORKS:

SELF WT=2.25KN

BENDING REACTIONS FOR DEADLOAD

LIVE LOAD = 20KN\m

BENDING REACTION FOR LIVE LOAD

LOAD DISPLACEMENT OF SUPER IMPOSED

LOAD

=44KN/m

Stress diagram for column

Stress diagram of beam

CONCLUSION :

The planning and designing of the AN APARTMENT

BUILDING has been completed effectively in our project. if this is

constructed in the proposed site, it will be very help full to our

people.

we all the members of our team have learned to

plan a building with referring to NATIONAL BUILDING OF

INDIA 2005. this project is a very useful to learn about the

design of the structural elements like beams , column and slabs by

using IS 456-2000.

the important think that we done were referring to

lot of books for designing, and we are very much satisfied with

exposing to field design.

THANK YOU