A PROJECT REPORT ON ANALYSIS AND DESIGN OF MULTI STOREY(G 6) RESIDENTIAL BUIL...
multi storey buidling
1. i
A
Project Report
on
DESIGN
OF
MULTI-STOREY
RESIDENTIAL BUILDING
Submitted for partial fulfillment of award of
BACHELOR OF TECHNOLOGY (B. Tech.)
Degree
in
CIVIL ENGINEERING
By:
ABHISEK KUMAR MAURY
AISHWARYA TIWARI
AJIT PRAJAPATI
FARHAN ALI
RAVI NANDAN SINGH
Under the Guidance of
Mr. NURUL HASSAN (Asst. Prof.)
Department of Civil Engineering
VISHVESHWARYA GROUP OF INSTITUTIONS
[20 Kms. from Ghaziabad on Ghaziabad – Bulandshahr G.T. Road]
(Near Dadri), Gautam Buddh Nagar (UP) – 203 207
Visit us at: www.viet.ac.in
(May, 2015)
2. ii
CERTIFICATE
Certified that following students have completed the project
entitled “ DESIGN OF MULTI STORY RESIDENTIAL
BUILDING ” for the award of Bachelor of Technology , in Civil
Engineering from Vishveshwarya Institute of Technology Dadri,
(G. B. Nagar) affiliated to Gautam Buddha Technical University,
Lucknow under my supervision. The project report embodies result
of original work and studies carried out by student himself and
the contents of the report do not form the basis for the award of
any other degree to the candidate or to anybody else.
GROUP MEMBER:
Sr.No. Name of Students Roll No.
1. ABHISEK KUMAR MAURY (1128600001)
2. AISHWARYA TIWARI (1128600004)
3. AJIT PRAJAPATI ( 1128600006)
4. FARHAN ALI (1128600016)
5. RAVI NANDAN SINGH ( 1128600033)
MR. NURUL HUSSAN MR. ABHAY SHANKAR RAI
ASST. PROFESSOR HOD, CIVIL ENG. DEPT.
DATE:
3. iii
ACKNOWLEDGEMENT
We would like to express our gratitude towards Mr.NURUL HUSSAN,
project supervisor for his valuable encouragement and guidance.
We would also like to thank (Mr. Abhay Shankar Rai), project
manager for his continuous support and advice throughout the entire
project of Design of Multi Storey Residential Building. We are also
thankful Mr. NURUL HUSSAN for her support and valuable guidance,
rejuvenating encouragement, positive criticism and constant supervision
all through our project session.
SUBMITTED BY:
ABHISEK KUMAR MAURY
AISHWARYA TIWARI
AJIT PRAJAPATI
FARHAN ALI
RAVI NANDAN SINGH
4. iv
OBJECTIVE
OBJECTIVE OF THE PROJECT:
• Carrying out the complete analysis and design of the main structural
elements of multi storey residential building including slab, column, shear
wall, and foundation.
• Getting real life experience with engineering practices.
5. v
ABBREVIATIONS
Unless specified otherwise, the symbols and notations used in the report shall have
the following meaning.
a0……………………………Basic Horizontal seismic coefficient
Ag…………………………...Gross area of section
Ah ………………………….. Horizontal seismic coefficient
Asc …………………………..Area of compression steel
Ast……………………………Area of tension steel
b……………………………...Width of member
C……………………………..Flexibility co-efficient
d……………………………..Effective depth of member
D……………………………..Overall depth of member
d’…………………………….Nominal cover in compression
Dia…………………………..Nominal Diameter of the bar
Fck…………………………...Characteristic compressive strength of concrete
Fy……………………………Characteristic yield strength of steel
I……………………….……..Importance factor
K……………………………..Performance factor
K1……………………………Probability factor
6. vi
K2……………………………Terrain, height and structure size factor
K3……………………………Topographical factor
Ld……………………………Development length
Lex……………………………Effective length of column about X-X axis
Ley……………………………Effective length of column about Y-Y axis
L0……………………………….Unsupported length of column
Mu………………………………Factored Moment
Mx………………………………Moment about X-X axis
My………………………………Moment about Y-Y axis
P/Pu……………………………..Axial load
Pc………………………………..Percentage compressive steel
Pt………………………………..Percentage tension steel
Pz………………………………..Design wind pressure at level z
T…………………………………Fundamental Time period
tc …………………………………Design shear strength of section
tv…………………………………Normal shear stress
V/Vu …………………………….Factored shear force
Vb ……………………………….Basic wind speed/ Base shear
Vz………………………………...Design wind speed at level z
W…………………………………Tidal weight of building
Z…………………………………..Height or level with respect to mean
ground level
7. vii
TABLE OF CONTENTS
TITLE PAGE……………………………………………………...…….i
CERTIFICATE……………………………………………...………….ii
ACKNOWLEDGEMENT……..…..………………….……………….iii
OBJECTIVE…………………………………………………………...iv
ABBREVATION…………………………….…………………....v– vi
CHAPTER 1 1 - 4
1.1 Introduction
1.1.1 Salient features of the building
1.1.2 Architectural Plan of the building
CHAPTER 2 5 - 24
2.1 Gravity Design
2.1.1 Analysis for Gravity Loads
2.2 Manual Design of Slab Panel
2.3 Manual Design of Stair case
2.4 Manual Design of Beam
2.5 Manual Design of Column
CHAPTER 3 25 – 46
3.1 Materials required by Gravity Design method
3.2 Ductility consideration
3.2.1 Requirements for Ductility
3.3 Foundation
3.3.1 Raft Foundation
3.3.2 Raft Foundation Design
3.4 Shear Wall
8. viii
CHAPTER 4 47 – 54
4.4 Design of slab
CONCLUSION 55
BIBLIOGRAPHY 56
9. 1
CHAPTER 1
INTRODUCTION
The project is to analyze and design the proposed building. The building which is to
be used as a residential building is located in the Greater Noida, U.P.The project
came under the final year project work scheme of department of Civil Engineering.
The project includes generation of floor plan in AutoCAD,design of several
component of the building viz. beams,columns,slabs,staircase,shear wall etc
manually as well as by STAAD software.this also contain the structural analysis of
the building on application of several load combination specially wind and seismic
loads.
SALIENT FEATURE OF THE BUILDING
Porpuse→Residential
No. of floors→6(G+5)
Storeys Height→3.2m
Builtup Area→859.1m2
No. of Staircase→1
No. of Lifts→2
Foundation used→Raft
10. 2
1.1.2 ARCHITECTURAL PLAN OF THE BUILDING
It covers a plan area equal to 28.13mX30.54m, consists of a ground floor plus eleven
upper floors.
The type of the building is that of a framed structure. All the floors are similar in
plan, each floor consists of four flats and each flat consists of three rooms which are
of different dimensions.
Some open area is provided in different parts of all floors in the same vertical plane
through all the floors. This open space will facilitate enough ventilation and natural
light. It is surrounded by steel railings on all the four sides.
All the rooms are provided with a wide balcony at the back face and a wide corridor
at the front face.
13. 5
CHAPTER 2
2.1 GRAVITY DESIGN
The basic analysis of the structure starts with the gravity load combinations
applied to the structure. This includes dead load due to weight of different
components of the buildings structure itself (beams, columns, Slabs stairs etc
)live load due to miscellaneous moveable components in the floors(
furniture, electrical appliances eetc. ). The presence of occupants also adds to
the live load of the structure.
Here we have analysed the structure for one load combination
• 1.5*(Dead load + Live load)
• (Dead Load+ Live load)
The beams and columns have been designed on the basis of responses obtained in
preliminary analysis for gravity loads using STAAD Pro Software. However
the slab panels have been designed manually for ine floor of the building a
model calculation for the slab panels and stair case has also been discussed.
2.1.1 ANALYSIS FOR GRAVITY LOADS
Dead Loads:
Self weight factor =1
Weight of Main Walls on Beams =14.72KN/m2
Weight of partition Walls on Beams =7.06KN/m2
Weight of parapet Walls on Beams =4.72KN/m2
Weight of Floor slabs =3KN/m2 (Discussed Later)
Weight of Floor finish =1.25 KN/m2
14. 6
Live Loads:
All floors =2KN/m2
Corridors and Staircases including fire escapes and store rooms =3 KN/m2
Roof Top =1.5 KN/m2
Based on application of this loads the structure has been designed for load
combination of 1.5(DL+LL). While the slab panels and staircases have been
designed manually or by Microsoft Excel program for the above mentioned load
conditions, the beams and columns have been designed based on the responses
obtained by STAAD pro.
2.2 MANUAL DESIGN OF FLOOR SLAB PANEL:
Slab of size (3.527×4.207)
Short span, Lx=3.527m
Long span, Ly =4.207,
Depth of slab,D=120mm,
Two adjacent age discontinuous
Load Calculation:
self wt of slab =0.120*25=3.0KN/m2
D.L due to finishing =0.05*24=1.2KN/m2
L.L onb slab =2.0KN/m2
Total load on slab(W) = 6.2KN/m2
Ultimate load on slab(Wu) =1.5×6.20=9.30KN/m2
Hence design as a two way slab
ɑx
+=0.045 ɑx
-=0.060
ɑy
+=0.035 ɑy
-=0.047
15. 7
Mux(+)=ɑx
+*Wu*Lx
2
=0.045*9.3*3.5272 = 5.206KNm
Mux(-) =0.060*9.3*3.5272 =6.94KNm
Muy(+)=0.035*9.3*3.5272 =4.094KNm
Muy(-) =0.047*9.3*3.5272 =5.437KNm
Depth oif slab required =sqrt(Mmax/(0.138*Fck*b))
= (6.94×106)/(0.138*25*1000)
=44.85mm (<100mm)
Designof Reinforcement:
Shorter span:
Area of steel required,Ast=(0.5*Fck/Fy)*(1-sqrt(1-((4.6*Mx)/(Fck*b*d2)))
=(0.5*25/415)*(1-sqrt(1-
((4.6*6.94*106)/(25*1000*1012)))
= 198.9mm2
Minimum area of steel required,Ast min =0.12%
=(0.12*b*D)/100 =(0.12*1000*120)/100
= 144mm2 (198.9mm2>144mm2) O.K
Let us provide diameter of bar 8mm
Required spacing =(1000*50.26)/198.9
=252mm
Longer span:
Area of steel required,Ast=(0.5*Fck/Fy)*(1-sqrt(1-((4.6*My)/(Fck*b*d2)))
=(0.5*25/415)*(1-sqrt(1-
((4.6*5.437*106)/(25*1000*1002)))
=163.25mm2
16. 8
Minimum area of steel required,Ast min =0.12%
=(0.12*b*D)/100 =(0.12*1000*120)/100
= 144mm2 (163.25mm2>144mm2) O.K
Let us provide diameter of bar 8mm
Required spacing =(1000*50.26)/163.25
=307.1mm
Maximum spacing for reinforcement
• Three times the effective depth ,3d=3*101=300mm
• 300mm
Provide 8mm dia bar @250m c/c on shorter span
Area of steel provided=(1000*50.26)/250=201mm2
Provide 8mm dia bar @200mmc/c on longer span
Area of steel provided=(1000*50.26)/250=201mm2
Check for deflection:
Pt=201/(103*102))*100=0.201
Fs=0.58*415*(198.9/201)
=238
Modification factor=2.15N/mm2
(l/d)max =20*2.15=43
(l/d)provided=4207/100=42.07(<43) O.K
Check for shear:
Average effective depth=(101+97)/2 =99mm
Vu=Wu*(0.5*Lx-d)
20. 12
2.4 DESIGN OF BEAM
All beams have been designed as rectangular section, of different sizes as per
optimum requirement.
The general design considerations are taken from IS: 456 -2000
Effective depth – is the distance from the centre of the tensile reinforcement to the
outermost compression fibers.
Control of deflection – the vertical deflection limit may generally assumed to be
satisfied provided that the span to depth ratios are not greater than the values
obtained as below :
a) Span to effective depth ratio for span up to 10m
Cantilever 7
Simply supported 20
Continuous 26
b) Depending upon the area and stress of steel for tension reinforcement, values in
(a) shall be modifying by multiplying with modification factor obtained as per
fig 5 (IS: 456-2000).
c) Depending upon the area of compression reinforcement, the value of span to
depth ratio is further modified by multiplying with the modification
factor obtained as per fig 5
(IS : 456-2000 ).
Development stresses in reinforcement Ld is taken directly from SP 16 (table 65),
for deform bars conforming to IS: 1786 these values shall be increased by 60% for
bars in compression, the values of bond stress for bar in tension shall be increased
by 25%.
Curtailment of tension reinforcement shall extend beyond the point at which it is no
longer required to resist flexure for distance equal to the effective depth of the
member or 12 times the bar diameter, whichever is greater except at simple support
or end of cantilever.
Positive moment reinforcement: – at least 1/3 +ve moment reinforcement in simple
member and ¼ +ve reinforcement in continuous member shall extend along the
same face of the member into the support , to length equal to Ld/3.
Spacing of reinforcement: - min. distance b/w the individual bar not be greater than
the dia. of bar if dia. are equal or dia. of larger bar if dia. are of different size and
5mm more than the nominal maximum size of course aggregate.
21. 13
Maximum distance should not be exceeded than 180mm for Fe – 415 from table –
IS: 456-2000.
Min. reinforcement should not be less than As =0.85bd/fy
Maximum reinforcement both in tension and compression shall not exceed 0.04bD.
Maximum spacing of shear reinforcement shall not exceed 0.75d for vertical stirrups
and d for inclined stirrups and in no case shall the spacing exceed 300mm and
minimum reinforcement provided as per this formula
= Asv/bsv > (0.4 /0.87fy).
The maximum spacing of shear stirrups has been kept at 200mm, subjected to
detailing consideration with respect to earthquake detailing.
At least two bars have been provided continuous over the entire span of beam.
At external joints bars with columns, top and bottom bars have been provided with
anchorage length of Ld in tension + 10 dia. of bar.
At internal joints bars have been taken continuous through the column.
The tension steel ratio on any section is not less than (0.24 fck0.5)/fy and not greater
than 0.025Mpa.
Provision for laps has been provided wherever required. Hooks shall be provided
wherever lap occurs at spacing not greater than 150mm. Further it has been taken
care not to be provided any laps in the joint within distance of 2d from any face and
within quarter length of any member. Also not more than 50% bars have been
curtailed at a section.
MANUAL DESIGN OF BEAM
LIVE LOAD ON BEAM No. 2779
DEAD LOAD ON BEAM No. 2779
1.5(DL+LL) ON BEAM No. 2779
22. 14
1.5(DL+LL) ON BEAM No. 2779
On STAAD Pro Analysis of the whole structure ,we get the follwing responses.
S.F.D
B.M.D
23. 15
Sample DesignCalculation for Beam No: 2779
Steel Reinforcement for=Tor grade 415
Concrete =M25 Grade
B=400 D=600 mm Effective L =6.12
Determination of area of steel reinforcement:
Maximum Positive Moment=206 KN-m
Maximum Negative Moment=133 KN-m
Top Reinforcement Tor 16 mm @ 150 C/C
Bottom Reinforcement Tor 10mm@ 90 mm C/C
Check for shear:
= 190 KN
= 0.47 From Is Code 456- Table-19 3.1
Since << shear reinforcement is required
=190-0.47*400*575=81.9 KN
Provide 8 mm, 2-legged stirrups@220 mm c/c
Strength of shear reinforcement
==94.8 KN > 81.9 KN OK
Development length
==825 mm
Provide (8*16mm=128mm) anchorage length and
provide a 90 degree bend in the 16 mm bars.
25. 17
B E A M N O. 2779 D E S I G N R E S U L T S
M25 Fe415 (Main) Fe415
(Sec.)
LENGTH: 6117.5 mm SIZE: 400.0 mm X 600.0 mm
COVER: 25.0 mm
SUMMARY OF REINF. AREA (Sq.mm)
----------------------------------------------------------------
SECTION 0.0 mm 1529.4 mm 3058.7 mm 4588.1 mm
6117.5 mm
----------------------------------------------------------------
TOP 718.66 0.00 0.00 0.00
1002.68
REINF. (Sq. mm) (Sq. mm) (Sq. mm) (Sq.
mm) (Sq. mm)
BOTTOM 0.00 462.89 619.86 462.89
0.00
REINF. (Sq. mm) (Sq. mm) (Sq. mm) (Sq.
mm) (Sq. mm)
----------------------------------------------------------------
SUMMARY OF PROVIDED REINF. AREA
----------------------------------------------------------------
SECTION 0.0 mm 1529.4 mm 3058.7 mm 4588.1 mm
6117.5 mm
----------------------------------------------------------------
TOP 4-16í 2-16í 2-16í 2-16í
5-16í
REINF. 1 layer(s) 1 layer(s) 1 layer(s) 1 layer(s)
1 layer(s)
BOTTOM 2-20í 3-20í 3-20í 3-20í
2-20í
REINF. 1 layer(s) 1 layer(s) 1 layer(s) 1 layer(s)
1 layer(s)
SHEAR 2 legged 8í 2 legged 8í 2 legged 8í 2 legged
8í 2 legged 8í
REINF. @ 220 mm c/c @ 220 mm c/c @ 220 mm c/c @ 220 mm
c/c @ 220 mm c/c
----------------------------------------------------------------
SHEAR DESIGN RESULTS AT DISTANCE d (EFFECTIVE DEPTH) FROM
FACE OF THE SUPPORT
SHEAR DESIGN RESULTS AT 915.0 mm AWAY FROM START SUPPORT
VY = 118.75 MX = -1.50 LD= 207
Provide 2 Legged 10í @ 220 mm c/c
SHEAR DESIGN RESULTS AT 915.0 mm AWAY FROM END SUPPORT
VY = -138.43 MX = -1.50 LD= 207
Provide 2 Legged 8í @ 220 mm c/c
26. 18
2.5 DESIGN OF COLUMNS
The columns of proposed structure have been designed as short columns with axial
load and bi axial moments. All columns have been designed using method outlined
in SP 16, (Design Aids to IS: 456-2000) using the columns interaction diagrams
with all the reinforcement distributed equally on all sides.
DESIGN APPROACH
As mentioned, all columns have been designed as short columns along both axes in
accordance with clause 25.1.1 of IS: 456-2000.
A column is said to be short when the slenderness ratio as given by the expression is
less than 12
Slenderness ratio along X-X axes
Lex /b and
Slenderness ratio along Y-Y axes
Ley/D
Where:
Lex = Effective length of column along X-X axis
Ley = Effective length of column along Y-Y axis
B=width of column along X-X axis
D=Depth of the column along Y-Y axis
27. 19
UNSUPPORTED LENGTH
The length of column ,LO was taken as the clear distance b/w the floor and the
underside of the shallower beam framing into the columns in each direction at the
next higher floor level in accordance with clause 25.1.3 of IS : 456-2000
The limit to slenderness, in accordance with clause 25.3.1 of IS: 456-2000 was also
taken into consideration.
EFFECTIVE LENGTH OF COLUMNS
The columns being restrained along both axes the effective length of columns was
taken as 0.65 Lo in accordance with table – 28 of IS: 456-2000
All columns have been designed for the following forces:-
• Axial load
• Moment about X-X axis
• Moment about Y-Y axis
• Moment due to minimum eccentricity as mentioned in clause 25.4 of
IS: 456-2000
• Shear force analysis (see article below), and
• Torsion shear due to seismic forces.
28. 20
DESIGN OF COLUMNS FOR SHEAR
As mentioned above, all columns have been designed for greater of the two.
• Factored shear force from analysis
• Shear given by the expression in IS: 13920 -1993.
In all the cases that were encountered, the factored shear force from analysis was
found greater and thus the columns designed for the same.
Design for shear was done in accordance with clause 40.1 of IS: 456-2000
by calculating the nominal shear stress given by the expression
Tv = Vu/bd
Where
Vu = Design shear force
b = Width of member
d = effective depth
Depending upon the area of tensile reinforcement and grade of the concrete used,
the design shear strength of concrete was obtained from modified given in clause
40.2.2 of IS: 456-2000
NOTE: - While calculating the design shear strength 50% area of steel was taken
into consideration by assuming that half of the steel would be in compression and
the total steel is distributed equally on all sides.
DETAILING OF REINFORCEMENT
• The cross-section of longitudinal reinforcement was kept b/w 0.8% to 4% in
accordance with clause 26.5.3.1 of IS : 456-2000
• All bars used for longitudinal reinforcement are greater than 12mm.
• Spacing of bars along periphery of column has been kept less than 300mm.
• All transverse reinforcement provided is of greater than ¼ of the largest
longitudinal bar and not exceeding the 16mm.
29. 21
• The pitch of ties should not exceed 300mm.
• All transverse reinforcement has been arranged in accordance with clause
26.5.3.2 of IS : 456-2000
Apart from these considerations, following provision of IS 13920-1993 has been
conformed to
• The least lateral dimension of the column is greater than 300mm.
• The ratio of the least lateral dimension to the perpendicular dimension is
more than 0.4.
• Lap splices wherever they occur have been proposed in the central half of
the member. Hoop with a pitch not exceeding 150mm c/c have been
provided over entire splice length.
• The transverse reinforcement consists of square hoops having 135 degree
with a 10 dia. extended at each end confined in the core.
• The parallel edges of hoops are not spaced greater than 300mm as far as
possible. A cross tie or a pair of overlapping hoops have provided engaging
all peripheral bars.
MANUAL DESIGN OF COLUMN
Unsupported Length=3200, Pu=2303KN,
Mux=11.2KNm Muy=18.67KNm ,
As the moment in X and Y direction are very small compared to axial load. We
shall design the column as axially loaded only.
Let us assume a trial section (400*700)mm2
Fck=25N/ mm2 F y=415N/mm2
30. 22
Check 1:
Effective Lenth=0.65*L=0.65*3200=2080mm,
Effective Lenth/Least Lateral dimension=2080mm/400mm=5.20(>3)
Check 2 :
Effective Length/Depth(D)=2080mm/700mm=2.97(<12)
Effective Lenth/Width(B)=2080mm/400mm=5.20(<12)
It is a short axially loaded column
Check 3:
Minimum Eccentricity
(1)Emin=Unsupported Lenth/500+lateral dimension/30
3200/500+400/30=20.73
(2) Emin=20mm => Emin=20.73
Design the member as short axially loaded column
Longitudanal Reinforcement:
We have
Pu=0.4*25*(Ag-Asc)+0.67*Fy*Asc
Pu=0.4*Fck*Ag+(0.67*Fy-0.4Fck)*Asc
Area of steel required,
Asc=(Pu-0.4*Fck*Ag)/(0.67Fy-0.4Fck)
Asc=(2303000-0.4*25*400*700)/(0.67*415-0.4*25)
Asc=1854.13mm2
Providing 12mm diameter bar
No. of bar =1844.13/(π*82)
=15.83=16 bar
31. 23
Providing 8mm dia lateral ties
The spacing of the column should not exceed
• Least dimension of the column=400mm
• Sixteen times the dia of longitudinal bar=12*16=196mm
• 300mm
Provide 8mm lateral ties at 190mm c/c spacing
33. 25
CHAPTER 3
This chapter deals with the miscellaneous topics. First of all we provide a
comparative study of the economy involved in the design with and without seismic
design. Then we move to ductile design of the building. Some theories and codal
provision have been discussed. A special mention of the reinforcement in the beam,
column and joints according to the provision of IS:13920 have been discussed. A
discussion about the type of foundation used and its design has also been given.
3.1 MATERIAL REQUIRED BY GRAVITY LOAD DESIGN
METHOD
34. 26
3.2 DUCTILITY CONSIDERATION
The basic approach of earthquake resistant design should be based on lateral
strength as well as deformability and ductile capacity of the structure with
limited damage but no collapse. The IS 13920:1993 is based on this
approach .Ductility of the structure is one of the most important factor
affecting its seismic performance. The gap between the actual and lateral
force is narrowed down by providing ductility in the structure. Ductility in
the structure will arise from inelastic material behaviour and detailing of
reinforcement in such a manner that brittle failure is avoided and ductile
behaviour is induced by allowing steel to yield.
3.2.1 REQUIREMENT FOR DUCTILITY
In order to achieve a ductile structure we must give stress on three key area
during the design process. Firstly, the overall design concept of the building
configuration must be sound. Secondly, individual member must be designed
for ductility, and finally connection and other detail need careful attention
CONSTRUCTION MATERIAL
3.1 Concrete
Concrete is a stone like hard material obtained by mixing cement, sand and
aggregate in some specific proportion and water to harden and give workability to
fill in the form of shape and dimensions desired for a structure. The chemical
interaction between cement and water binds the aggregate into a solid mass.
Concrete possesses high compressive strength but is weak in tension. This
short coming is offset by providing steel bars at appropriate location at the time of
casting the member to take up the stresses, and the compressive stresses if required.
Thus, the concrete is strengthened (i.e. reinforced) by steel and the resultant
composite mass is known as reinforced cement concrete(RCC),
3.1.1 Constituent materials:
The main constituent materials of concrete are –
35. 27
Aggregates,
Cement ,and
Water
Aggregates: The aggregates occupy approximately more than 75 percent of
the volume of concrete and, their properties have definite influence on the
strength of hardened concrete. Hence, the aggregate used for concrete should
be durable, strong, good resistance to weathering action and effects economy
in cost, of concrete.
Cement: cement is a material to having property of binding minerals
fragments into a solid mass on its chemical combination with water. Since
binding and hardening actions are due to presence of water, such cements are
called as hydraulic cement. The cement used for construction is known as
Portland cement.
Water: water plays an active role in the chemical process of hydraulic and
incurring concrete. It is, therefore, necessary that what are used for fixing
and curing should be clean and free from injurious materials like oils, acids,
alkalis, salts, sugar, organic materials or other substances that may be
deleterious to concrete and steel. Drinking water is generally considered
satisfactory for mixing concrete.
3.1.2 Concrete mix proportioning:
In reinforced concrete construction, the concrete is known by its grade and
is designated as M20, M25 in which letter M refers to mix aand the number to its
characteristic strength in axial compression at 28 days on 150mm cube, expressed in
N/mm², grades normally used in R.C construction are M20 and M25.
The structural designers specify is the required strength and properties of
concrete to achieve this, various ingredients of concrete are proportioned so that the
resulting concrete has desired strength, proper workability for placing and namely,
the cement, aggregate and water to attain the required strength is done in the
following ways:
36. 28
By designing the concrete mix: such concrete is called as ‘design mix
concrete’.
By adopting nominal mix, such concrete is called ‘nominal mix concrete’.
3.2 Reinforcement steel:
Reinforcement steel consists of bars, usually circular in cross-section. These
are at present available in different grades ways. Fe250, Fe415, Fe500, where ‘Fe’
refers to Ferrous metals and the number refers to a specified guaranteed yield stress
in N/mm².
3.2.1 Types of reinforcement:
Based on the physical and mechanical properties namely ductility, yield
strength, the following two types of steel reinforcements are mainly used in
reinforced concrete construction:
Plain round bars of mild steel.
Deformed bars of high-grade steel.
Plain round bars of mild steel: They are usually of mild steel (grade
Fe250) conforming to IS: 432-1982. It has a well-defined yield point giving
yield stress of 250N/mm² and excellent ductility.
Deformed bars of high-grade steel: These bars are usually of steel and do
not possess a well-defined yield point. The characteristics strength is given
by 0.2 percent proof stress. These bars have low ductility and low bend
ability ribs, lugs, or deformations on their surface with the result that their
bond characteristics is improved.
Detail consideration
1. GENERAL
The design and construction of reinforced concrete buildings shall be
governed by the provisions of IS 456 : 2000, except as modified by
the provisions of this code.
37. 29
For all buildings which are more than 3 storeys in height, the
minimum grade of concrete shall be M20 ( fck = 20 MPa ).
The concerned structure is G+13 storied, that’s why we have
used M25 grade of concrete.
Steel reinforcements of grade Fe 415 or less only shall be used.
2. FLEXURAL MEMBERS
2.1 General
• The factored axial stress on the member under earthquake loading shall not
exceed 0.1 fck.
• The member shall preferably have a width-to-depth ratio of more than 0.3.
• The width of the member shall not be less than 200 mm.
• The depth D of the member shall preferably be not more than 1/4 of the clear
span.
2.2 Longitudinal Reinforcement
• The top as well as bottom reinforcement shall consist of at leasttwo bars
throughout the member length.
• The tension steel ratio on any face, at any section, shall not be less than ρmin
= 0.24(fck/fy) ; where fck and fy are in MPa.
• The maximum steel ratio on any face at any section, shall not exceed ρmax =
0.025.
• The positive steel at a joint face must be at least equal to half the negative
steel at that face.
• In an external joint, both the top and the bottom bars of the beam shall be
provided with anchorage length, beyond the inner face of the column, equal
to the development length in tension plus 10 times the bar diameter minus
the allowance for 90 degree bend(s) ( see Fig. 1 ). In an internal joint, both
face bars of the beam shall be taken continuously through the column.
38. 30
• The longitudinal bars shall be spliced, only if hoops are provided over the
entire splice length, at a spacing not exceeding 150 mm.The lap length shall
not be less than the bar development length in tension. Lap splices shall not
be provided (a) within a joint, (b) within a distance of 2d from joint face, and
(c) within a quarter lengh of the member where flexural yielding may
generally occur under the effect of earthquake forces. Not more than 50
percent of the bars shall be spliced at one section.
Use of welded splices and mechanical connections may also be made, as per
25.2.5.2 of IS 456 : 1978. However, not more than half the reinforcement shall be
spliced at a section where flexural yielding may take place
39. 31
LAP SPLICES IN BEAM
2.3 Web Reinforcement
• Web reinforcement shall consist of vertical hoops. A vertical hoop is a
closed stirrup having a 135° hook with a 10 diameter extension (but not < 75
mm) at each end that is embedded In confined core.
• The minimum diameter of the bar forming a hoop shall be 6 mm. However,
inbeams with clear span exceeding 5 m, the minimum bar diameter shall be
8 mm.
• The shear force to be resisted by the vertical hoops shall be the maximum of
a) calculated factored shear force as per analysis, and
b) shear force due to formation of plastic hinges at both ends of the beam
plus the factored gravity load on the span.
40. 32
• The contribution of bent up bars and inclined hoops to shear resistance of
the section shall not be considered.
• The spacing of hoops over a length of 2d at either end of a beam shall not
exceed (a) d/4,and (b) 8 times the diameter of the smallest longitudinal bar;
however, it need not be less than 100 mm.
42. 34
3. Compression Member:
3.1General
• These requirements apply to frame members which have a factored axial
stress in excess of 0.1 fck under the effect of earthquake forces.
• The minimum dimension of the member shall not be less than 200 mm.
However, in frames which have beams with centre to centre span exceeding
5 m or columns of unsupported length exceeding 4 m, the shortest
dimension of the column shall not be less than 300 mm.
• The ratio of the shortest cross sectional dimension to the perpendicular
dimension shall preferably not be less than 0.4.
3.2Longitudinal Reinforcement
Any area of the column that extends more than 100mm beyond the confined
core due to architectural requirement shall be detailed as in diagram.
43. 35
3.3 Transverse Requirement
The detailing of the transverse reinforcement should be done in the diagram below
Transverse Reinforcement in Column
44. 36
• Special Confining reinforcements
Special confining reinforcement shall be provided over a legth lo from each
joint face, towards midspan, and on either side of anysection, where flexural
yielding may accur under the effect of earth quake forces. The length ‘lo’
shall not be less than :
• Larger dimension of the member at the section where yielding accur,
• 1/6 of clear span of member, and
• 450mm
When a column terminate into a footing or mat, special confining
reinforcement shall extend atleast 300mm into the footing or mat.
COLUMN AND JOINT DETAILING
45. 37
PROVISION OF SPECIAL CONFINING REINFORCEMENT IN FOOTING
3.3 FOUNDATION
3.3.1 RAFT OR MAT FOOTING
A raft or mat is a combined footing that covers the entire area beneath a structure
and supports all the wall and columns . When the allowable soil pressure is low ,or
the building loads are heavy, the use of spread footing Would cover more than one-
half of the area and it may prove more economical to use mat or raft foundation
.They are also used where the soil Mass contains compressible lenses or the soil is
sufficiently erratic so that the differential settlement would be difficult to control
.The raft tends to bridge over the erratic deposits and eliminates the differential
settlement. Raft foundation is also used to reduce settlement above highly
compressible soil , by making the weight of structure and raft approximately equal
to the weight of soil excavated.
Ordinarily, raft are designed as reinforced concrete flat slabs .If the C.G of loads
coincide with the centroid of the raft ,the upward load is regarded as uniform
pressure equal to the downward load divided by the area of the raft .The weight of
raft is not considered in structural design because it is assumed to be carried directly
by the subsoil .
46. 38
3.3.2 Designof RAFT Foundation
Total weight of columns =90516 KN
Assume self weight of foundation equal to 1.1 times of the total columns load
=+1.1X 90516=99567.6 KN
Area of foundation =99567/100=996
Let us provide =34.5X32=1104 ok
Net upward intensity = = 90.18 KN/
Net upward reaction/m=90.18X16=1442.88 KN/m
Maximum longitudinal bending moment =34.5X3335=116620 KNm
Factor moment= Mu =0.8X116620=93296 KNm
Equating Mu,lim to Mu
Mu,lim = 0.138fCKb
We found d=906
d = 950mm, provide 50 mm cover
D= 950+ 50 = 1000 mm
Mu/b = (93296×106)/ (34500×9502) =2.99
=0.99 % (obtained from page -49 of SP 16)
= 9405 mm2
Maximum shear force=7438.52KN
Factored shear, vu= 4867.9*0.8=3894.32KN
47. 39
τv = = 0.22 N/
But τc = 0.63 N/ (obtained from Is 456: 2000 table 19)
τv < τc . Hence OK
For =.99 %, =9405mm2
Provide 32mm dia. bars @ 80mm c/c
TRANSVERSE BENDING
Sum of all loads in outer strip =15073KN
Sum of all loads in inner strip = 7966KN
Soil pressure acting under entire width =90.18.62*32 =2885 KN/m
Maximum transverse bending moment=62046 KNm
Factored moment =0.8*62046 =49638 KNm
Equating Mu,lim to Mu
2458.6 =0.138fckb
Here we have: b=32000mm, fck = 25 N/
Hence, d is coming =599 mm
But available effective depth =1000-50-32-16=900
= =1.19
Hence = 0.343% (obtained from page -49 of SP 16)
Therefore Ast = =3037
Provide 25mm dia. bars @150mm c/c
Maximum transverse shear =0.8x3848.8 =3079 kN
Nominal shear stress (τv )= =.109 N/
and τc = 0.38 N/ (obtained from table 19 of IS 456 : 2000)
since τv < τc . ok
Using 25 mm dia. of 2 –legged vertical stirrups
A sv = 2x =981.75
Spacing =30 mm c/c
Transverse bottom steel = 0.189% b d
= x17000 x468
= 15036.8
48. 40
Provide 25 mm dia. Bars @ 80mm c/c
Longitudinal bottom steel = 0.12% of gross area
=
= 10800
Provide 25 mm dia. Bars @ 150 mm c/c
Two way punching shear force
Size of the column = 0.7mx0.4m
Depth (d) =1000mm
Critical section at d/2 = 0.5 m
Width of critical plane =0.7m
τ'c = τc
= (0.5+)
= (obtained from IS 456 : 2000 cl. 31.6.31 )
= 0.57
Hence take
Now = (0.5+0.57) (not greater than 1)
Thus, = 1.
τ'c = τc
=1x0.25 =1.25 N/> 0.34 N/.
Thus, OK.
Development of reinforcement
Development length in 25 mm dia. bars
=
= (obtained from IS 456: 2000 cl. 26.2.1.1 )
=65 x dia.
Therefore,
M.O.R = 0.87(d-0.42 x0.48xd)
= 6207.48 KNm
= 1.3 M/V + -------> here = 12 x dia.
= 694mm < Hence okay.
49. 41
3.4 SHEAR WALL
Shear walls are a type of structural system that provides lateral resistance to a
building or structure. They resist "in-plane" loads that are applied along its height.
The applied load is generally transferred to the wall by a diaphragm or collector
or drag member. They are built in wood, concrete, and CMU (masonry).
Plywood is the conventional material used in the construction of Shear Walls, but
with advances in technology and modern building methods, there are other
prefabricated options, such as Hardipanel and Simpson Strong Wall, which have
made it possible to inject shear assemblies into narrow walls that fall at either side of
an opening in a shear wall. Sheet steel and steel-backed shear panels (i.e. Sure-
Board) in the place of structural plywood in shear walls has proved to be far
stronger in seismic resistance.
50. 42
SHEAR WALL DESIGN
Detail of shear wall consideration
Lenth of the wall,lw=4950m Thickness of the wall =230mm
Height of the wall,H=39.4m Ag=113.85*103mm2
Iy=2.325*1012
51. 43
SHEAR WALL CONSIDERATION
Data collected from STADD Pro Analysis
Load case Moment(KN
-m)
Shea
r
(KN)
Axia
l
force
KN
Axial load
on boundary
element(KN
)
1.5(DL+LL) 1561.4 1010 5860 3970
1.2(DL+LL+EQZ
)
4000 2025 4420 3040
I.2(DL+LL-EQZ) 1500 410 4950 3310
1.5(DL+EQZ) 4832 2425 4890 3350
1.5(DL-EQZ) 2044 618 5550 3690
(0.9DL+1.5EQZ) 4274 2063 2800 1940
(0.9DL-1.5EQZ) 2600 980 3460 2280
Shear strength Requirement:
tv=Vu/(b*d) Vu=2425/2=1212.5KN
tw=230mm
dw0.8*lw = 0.8*4950=3960mm Effective depth of wall of the section
Now,tv=(1212.5*103)/(230*3960)=1.3312N/mm2
Table-19 IS-456 gives for pt =0.25,tc=0.36N/mm2
Table 20 IS:456 gives tc max =3.1N/mm2
Since tc<tv<tc max ,shear reinforcement is required.
Now 0.25*sqrt(fck) =0.25*sqrt(25)=1.25(tv>0.25*sqrt(25))
Also tw >200mm
SHEAR REINFORCEMENT IS REQUIRED IN TWO CURTAINS
52. 44
Area of horizontal shear reinforcement is given by:
Vus =0.87*fy*Ah*dw/Sw
Ah =Horizontal shear reinforcement area
Sv= Vertical spacing
Vus=vu-tc*tw*dw
=(tv-tc)*tw*dw
=(1.3312-0.36)*230*3960N
= 884568.9N
Spacing required for two legged 8f TOR bars
Sv=0.87*415*100.53*3960/884568.9
Sv=162.49mm
This gives the ratio As/Sv =100.53/162.49=0.64
Minimum horizontal reinforcement =0.0025*230
=0.575<0.64 (O.K)
Provide 8mm bar @150mm c-c in two curtain in horizontal reinforcement
Provide 8mm bar @150mm c-c in two curtain in vertical reinforcement
Spacing should not exceed in either direction
1.lw/s=4950/5=990mm
2. 3*tw=3*230=690mm
3. 450mm
Provide spacing 150mm . Hence O.K
Flexural strength
The moment of resistance Muv ,of the wall section shall be calculated as for column
subjected to combined axial load and uniaxial bending.The moment of resitance that
is provided by uniformly distributed vertical reinforcement in a slender rectangular
wall section,may be calculated as follows:
53. 45
Muv/(fck*tw*lw
2)=f[(1+?/f)*(0.5-0.416*xu/lw)-( xu/lw)2*(0.168+
β2/3)]
When xu/lw<=( xu
*/lw)
xu/lw=(f+?)/(2f+0.36) xu
*/lw=0.0035*Es/(0.0035Es+0.87*fy)
f=(0.87*fy*?)/fck ?=pu/( fck*tw*lw)
where ,
?=Vertical reinforcement ratio
Ast=As*lw/sv=0.64/230=0.003
f=0.87*415*0.003/25=0.044
?=2930*103/25*230*4950=0.102
pu=5860/2=2930KN
xu/lw=0.044+0.102/(2*0.044+0.36)=0.31
xu
*/lw=(0.0035*2*105)/(0.0035*2*105+0.36)=0.66
xu/lw< xu
*/lw HENCE O.K
β=0.87*415/(0.0035*2*105)=0.516
Muv/(fck*tw*lw
2)= 0.044[(1+0.102/0.044)*(0.5-0.416*0.31)-( 0.31)2*(0.168+
0.5162/3)]=0.0530
MUV=0.0530*25*230*49502
=7468KNm
The remaining moment Mu-Muv=24160-7468
=16692KNm
This much moment has to be resisted by boundary element
Pboundary element=16692/4.95=3372.2KN
fc=Pu/Ag+(Mu*lw/2)/Iy
=(2930*103)/(113.85*103)+((24160*1064950/2)/(2.325*1012)
54. 46
=25.73+25.72=51.45N/mm2 >0.2*fck
Provide boundary element
Dimension of boundary elements=(600*500)mm2
Ag =30*106mm2
Let us assume 2% longitudinal reinforcement
As=0.02*600*500=5000mm2
Axial load capacity of the boundary element
Pu =0.4*25*30*104+(0.67*415-0.4*25)6000
=4608300N
=4608.30KN >3970KN o.k
>3372.2KN O.K
Provide 12 No. of 25mm dia bar
Splicing of vertical reinforcement may be done at higher larger of
1.lw=4950mm
2.H/6=39400/6=6567mm
Splicing may be done at a height 7m above the base.
55. 47
CHAPTER – 4
DESIGN APPROACH
4.1 Working Stress Method
This has been the traditional method used for the reinforced concrete design
where it has been assumed that concrete is elastic, steel and concrete act together
elastically, and the relationship between loads and stresses is linear right up to the
collapse of the structure. The basis of the method is that the permissible stress for
steel and concrete are not exceeded anywhere in the structure when it is subjected to
the worst combination of working loads and the design is in accordance with hook’s
law.
4.2 Ultimate Load Method
In the ultimate load method, the working loads are increased by suitable
factors to obtain ultimate loads. These factors are called load factors. The structure is
then designed to resist the desired ultimate loads. This factor takes into account the
non-linear stress-strain behavior of concrete.
4.3 Limit State Method (LSM)
The discussions of the earlier two method clearly shows that the working stress
method, though ensures satisfactory performance at working loads, is unrealistic and
a rational at ultimate state and hence does not give a true margin of safety, while the
ultimate load method, though provides realistic assessment of degree of safety in
confirming with the actual behavior of the structure at or near the ultimate state, it
does not guarantee the satisfactory performance of the structure at service loads.
Undoubtedly, the ideal approach to design a structure is one which recognizes
and take into consideration all the states, like uncracked, cracked, elastic and
ultimate state through which a structure or its element and its material pass from
service loads to ultimate load, and ensures that neither the safety at the ultimate state
56. 48
nor the serviceability at the service condition is in jeopardy(danger) rendering the
structure to perform its function satisfactorily during unfit is called the limit state
philosophy of design.
4.3.1 Types and classification of limit state –
The various limit states required to be considered in structural design are
conveniently group into three major categories, namely
Limit state of collapse ,
Limit state of serviceability ,
Limit state of durability ,
4.3.1.1 Limit state of collapse:
It is the limit state on attainment of which the structure is likely to collapse
it related to stability and ultimate strength of the structure. Design to this limit state
safely of structure from collapse.
The structure failure can be any of the following types:-
Collapse of one or more members uttering as a result of force coming on the
member exceeding its strength {types (a) and (b) given below} ;
Displacement of the structure body due to lack of equilibrium between the
external forces or displacement and the resisting reactions {type (c),(d),(e)
given below}.
The various condition leading to structural failures are as follows –
(a) Failure, bright age and hence division into segment of one or more members
of the structure either due to material failure (as in case of columns) or on
account of formation of mechanism by development of plastic images at one
or more critical sections due to yielding of steel and concrete (as in case of
slabs and beams):
57. 49
(b) Elastic or plastic instability;
(c) Overturning,
(d) Sinking
This limit state is attained by providing resistance (or resisting reaction)
greater than the force coming on it and keeping a margin of safety through safety
factors.
Some of the codes consider each of the above the states as independent
limit states instead of a single limit state of collapse and prescribe different safety
factor for each of them. I S. code prescribe different safety factors for overturning
and sliding without giving any special status to sinking and buckling.
The limit state under discussion is critical in case of column and
foundations on particular and in case of a normal structure as a whole.
4.3.1.2 Limit state of serviceability:
Limit state of serviceability related to performance of behavior of structure
at working loads and based on causes effecting serviceability if the structure. They
are subdivided into following three categories:
Limit state of deflection,
Limit state of resistance to chemical and environmental actions, and
Limit state of resistance to accidental or catastrophic collapse.
58. 50
4.4 DESIGN OF SLABS
SLAB TYPE 1
Slab name S1
Condition interior panel
+αx +αy
-αx -αy
Ly, length of longer span = 4.315m
Lx, lengh of shorter span = 3.705m
Lx/Ly < 1.16 which is less than 2
Hence, design as two way slab.
Leff,whichever is less Leff = 3.705m
L/D = 26
D =3705/26 =142.5mm =150mm (say)
d = 120mm
Load calculation
Superimposed dead load =4.5KN/m2
Sumperimposed live load = 2kN
Total load = 6.5KN/m2
Edge condition
For Lx/Ly = 1.16
+αx = 0.030 +αy = 0.024
-αx = 0.040 -αy = 0.032
60. 52
SLAB TYPE S2
8mm@300mm
8mm @300mm c/c
120mm
150mm
8mm @280mm
3.705m
SLAB TYPE S3
Slab name S3
Condition one long edge discontinuous
+αx +αy
-αx -αy
Ly, length of longer span = 4.925m
Lx, lengh of shorter span = 3.705m
Lx/Ly < 1.329 which is less than 2
Hence, design as two way slab.
Leff,whichever is less Leff = 3.705m
Assumed thickness = 120mm
Effective thickness = 100mm
Load calculation
Superimposed dead load =4.5KN/m2
Sumperimposed live load = 2kN
Total load = 6.5KN/m2
Factored load = 6.5×1.5
=9.75KN/m2
61. 53
Edge condition
For Lx/Ly = 1.329
+αx = 0.0448 +αy = 0.028
-αx = 0.058 -αy = 0.037
Since (pt)reqrd/100 = (Ast)reqrd/bd = fck[1-sqrt(1- 4.59*Mu/(fckbd2)]/2fy
Grade of concrete fck 25
Grade of steel fy 415
Width b 1000
Depth of slab d 100
Mx(+) = αx*w*Lx2
Mx(+) = 5.995 KN-m
Mx(-) = 7.762 KN-m
My(+) =3.747 KN-m
My(-) = 4.952KN-m
Required depth = sqrt( Mmax/2.76b)
= 53.03mm < 100mm
Area of main reinforcement
Ast = (0.36× fck × b × 0.48 × d)/( 0.87×fy)
= ((0.36× 25 × 1000 × 0.48 × 53.03)/( 0.87×415)
=634.15mm2
Take 12mm bars
Spacing = (113.09×1000)/634.15
= 178.34mm (say 160mm)
Ast provided = (113.09 × 1000)/160
= 706.81mm2
Distribution steel
4.592×10(pow 6) = 0.87×415×Ay [43.03 –[(415 × Ay)/25×1000]]
0.0166×Ay2 - 43.03 Ay + 13715.55 =0
Ay = 372.18mm2
Use 8mm dia bars
Spacing = 135.05mm (say 130mm)
Actual area of steel = 386.615mm2
63. 55
CONCLUSION
• We have practiced real life engineering .
• We can conclude that there is difference between the theoretical and
practical work done. As the scope of understanding will be much more
when practical work is done. As we get more knowledge in such a
situation where we have great experience doing the practical work.
• At this point, we would like to thanks all the instructors,
engineers, consultant offices for their grate support.
64. 56
BIBLIOGRAPHY
• CONSTRUCTION SITE OF GREATER NOIDA U.P .
• I.S CODE 456:2000.
• I.S CODE 800:2007.
• I.S CODE 875 (PART 3) 1987
• I.S CODE 1893:2002
• FACULTY ‘S INSTRUCTIONS.
• BOOK REFERENCES:
• B.C PUNMIA
• A.K JAIN
• S. RAMAMURTHAM
![i
A
Project Report
on
DESIGN
OF
MULTI-STOREY
RESIDENTIAL BUILDING
Submitted for partial fulfillment of award of
BACHELOR OF TECHNOLOGY (B. Tech.)
Degree
in
CIVIL ENGINEERING
By:
ABHISEK KUMAR MAURY
AISHWARYA TIWARI
AJIT PRAJAPATI
FARHAN ALI
RAVI NANDAN SINGH
Under the Guidance of
Mr. NURUL HASSAN (Asst. Prof.)
Department of Civil Engineering
VISHVESHWARYA GROUP OF INSTITUTIONS
[20 Kms. from Ghaziabad on Ghaziabad – Bulandshahr G.T. Road]
(Near Dadri), Gautam Buddh Nagar (UP) – 203 207
Visit us at: www.viet.ac.in
(May, 2015)](https://image.slidesharecdn.com/067fc53d-c05d-4490-b610-d0df6fdd5ea2-160901181410/85/multi-storey-buidling-1-320.jpg)
