IE DESIGN
ASSIGNMENT
DESIGN OF
EARTHQUAKE RESISATNT BUILDING
BY
JEETU
RANA(roll no 39)
HEMANTH
KUMAR(roll no 36)
ABSTRACT
In this project work, an attempt has
been made to plan and design a G+4 storied shopping complex building. This
project work involves planning, analysis, designs, drawings and estimation of a
typical multistoried building.
The salient features of the G+4 storied building are as given below the
basement floor is 1.20m above the existing ground level. The shopping complex
consists of G+4 floors with 3.60m ceiling height. The carpet area available in
each floor is 1220sq.m.
This shopping complex having all facilities under one roof, designed
with shops, Super market, Food court, Net point, Gym, Table tennis court,
Coffee shop with ample car parking etc,
with very good water supply and sanitary arrangements.
The planned five storey commercial building frame
in modeled in STAAD Pro.Various load combinations are included in the frame analysis
and the lateral loads are calculated by seismic coefficient method for the
earthquake zone III with response reduction factor 3. The amount of concrete
and steel required along with the total cost of the building is calculated.
The structural design has been manually
done. The estimate of the building is prepared on the basis of plinth area rate.
Necessary structural drawings are enclosed at appropriate places.
LIST OF FIGURES
Fig.no. Particulars Page
no
3.2.9.1 Isometric view 22
3.2.9.2 Loading diagram 23
3.2.9.3 Shear force diagram 24
3.2.9.4 Bending moment diagram 25
4.3.2 Slab layout
35
4.3.3 Reinforcement details for Flat slab 36
4.4.2 Reinforcement details for Two way slab 47
5.3.1 Beam layout 61
5.3.1 Reinforcement details for Beam 62
6.3.1 Reinforcement details for Column 68
7.2.1 Combined footing plan 69
7.2.2 Shear force and bending moment diagram 70
7.2.3 Centre
line layout 80
7.2.3 Reinforcement details of combined footing 80
7.3.1 Isolated footing plan 81
7.3.2 Pressure diagram – 1 84
7.3.3 Pressure diagram – 2 85
7.3.4 One way shear plan 86
7.3.5 Two way shear plan 87
7.3.6 Critical section plan 88
7.3.7 Reinforcement details of Isolated footing 90
8.2.1 Reinforcement details of staircase 96
LIST OF TABLES
4.3.2 For beams 60
ABBREVIATIONS USED
The following standard letters and symbols shall have the meaning
indicated against each, where other symbols are used, they are explained at the
appropriate places.
Ast
= Area of
steel in tension.
Asc = Area of steel in compression.
Asv = Area of vertical stirrups.
b = breadth
of beam or shorter dimension of a Column.
Bw = breadth
of web or rib.
D = Over
all depth of beam or slab; dimension of a rectangular
Column in the direction
under consideration
Df = Thickness
of flange.
DL = Dead
load
d = Effective
depth of beam or slab.
d’c = Depth of compression reinforcement from
the highly
Compressed
face
Ec = Modulus
of elasticity of concrete.
fck = Characteristics compressive strength of concrete.
fy = Yield tensile strength of steel.
LL = Live
load or imposed load.
Leff = Effective span of slab.
Lx = Length of shorter side slab.
Ly = Length of longer side of slab.
L = Length of a column or beam between
adequate lateral
Restrains or the unsupported length of a
column.
M = Bending
moment.
M fd = Factored
moment.
P u = Axial
load on a compression member.
p = Percentage
of steel reinforcement.
Sv = Spacing
of stirrups
V = Shear
force.
W = Total
load.
Wd = Distributed
dead load per unit area/length.
Wi = Distributed
imposed load per unit area/length.
Z = Lever
arm.
ﺡV = Nominal shear stress.
ﺡc = design shear stress
ﺡc max = maximum Permissible shear
stress.
σcbc = Permissible
compressive stress in concrete.
σSt = Permissible
tensile stress in steel.
INTRODUCTION
1.1.
SCOPE & IMPORTANCE:
Shopping and Entertainment is
an important for each and every one. But they have short of time, so they need
a shopping complex under one roof to save the valuable time.
1.2.
LOCATION:
We have decided to choose
the site for the construction of shopping complex at Vadapalani in Chennai
city.
The site accommodates the
following special feature.
Ø Land is available in the centre of city.
Ø The site is located in main road.
Ø 24 hour transportation facilities available.
Ø Proposed site creates a pleasant environment of shopping.
1.3. OBJECTIVE OF THIS PROJECT WORK:
The objective of this project is to satisfy the
needs of people with in a single roof. In metropolitan cities like Chennai,
Mumbai etc., we have only very limited areas which are sold at high cost. So we
have to build buildings with in this limited area satisfying each of every need
of the people. This project will help us to built buildings of that type. And
this project is also designed in such a way that it would be economical.
The civil engineers have to think of construction of high raised
structures, instead of the traditional type of reinforced concrete skeletal
structure enclosed by thick walls of bricks or any other construction
materials.
A civil engineer must be
familiar with planning, analysis and design of framed structures. Hence it was
proposed to choose a problem, involving analysis and design of multistoried
framed structure as the project work.
1.4 EXISTING CONDITION OF AT SITE:
The proposed
site is approximately flat. There fore no necessary to level the site. Good soil having sufficient bearing capacity
is available at shallow depth.
1.5 OTHER FACILITIES:
The proposed
site accommodates the following important facilities.
Ø 24 hours Transport facilities are available.
Ø Drinking water facilities.
Ø Communication facilities.
Ø Electrical facilities.
Ø Police station.
Ø Nearest railway station available at Chennai Junction, which is
located at 3Km from Vadapalani
Ø Underground drainage facilities available.
2. PLANNING:
2.1.
INFRASTRUCTURE:
The proposed five-storied commercial building consist of
area of each floor is 1220 sq.m A building should be planned to make
it comfortable, economical and to meet all the requirements of the people. The
efforts of the planner should be to obtain maximum comfort with limited
available resources. Functional, utility, cost, habits, taste, requirements
etc, should also be considered in planning a building. The planning of this
multi- storied building is so planned to meet out all the above factors.
2.1.1
GROUND FLOOR:
In this floor Entrance foyer, Coffee
shop, various Shops, Escalator, Lift, Toilet blocks are provided. With entrance
foyer of 25 sq.m, coffee shop 120 sq.m, and 20 shops of 500 sq.m.
2.1.2 TYPICAL PLAN OF FIRST & SECOND FLOOR:
In this floor various Shops, Super
market, Food court, Escalator, Lift, Toilet blocks are provided. With super
market and food court of 200 sq.m and shops of 300 sq.m
2.1.3
THIRD FLOOR:
In this floor Net point, Gym, Table
tennis court, Snooker corner, various
Shops, Escalator, Lift, Toilet blocks
are provided. With Table tennis court and Snooker corner of 150 sq.m, net point
of 220 sq.m, Gym 90 sq.m and shops of 150 sq.m.
2.1.4
FOURTH FLOOR:
In this floor Office with Conference
hall and store, Escalator, Lift, Toilet blocks are provided. With Office of 300
sq.m, conference hall of 80 sq.m,
2.2. STAIR CASE:
This should be located in a place easily
accessible to all members. The minimum width of staircase should be 0.9m clear
of railing and many ranges up to 1.5m. There should be a clear head –way of
2.1m above each step and landing. The staircase should be constructed in two
flights having a landing in the middle to make it easy and comfortable to
climb. Risers and traders should be uniform throughout to keep rhythm while
climbing or descending. In our project, staircase at two corner portion to get
more access to each floors.
2.3. HEIGHT OF FLOORS:
In our project, the commercial
building each floor roof height is provided at 3.60m.
2.4. THICKNESS OF THE WALL:
For commercial buildings of one storey
one brick, 23cm thick wall is sufficient, for two storied building wall of G.F
may be one and half brick 30cm and 2nd floor wall may be one brick
23cm. in our project, the wall thickness provided as 23cm all around the
building.
2.5. SITE SELECTION:
The site for the construction was
selected at the Chennai city. As it lies in the heart of the city with good
water supply and underground drainage facilities and connected with good road
facilities and surrounded with vegetation, gives a comfort living to the
inmates and attracts demand.
2.6. SITE INVESTIGATION:
The
G+4 storied commercial building is proposed to be constructed at as site heart
of the town. The soil at the site is hard soil having a safe bearing of 250KN/m2
in Vadapalani. The size of the plot is 3000m2.
2.7. DRAINAGE AND SANITATION:
Two pipe systems had been provided to
remove and treat the sullage and human excreta, one septic tank were provided
in the commercial building for economical and efficient treatment of waste.
2.8. WATER SUPPLY:
As water is one of the basic needs
prime importance is given to planning of water supply systems. The quality of
water is calculated as per IS 1172-1963. The water is supplied for the entire
requirements form water tank under pressure. The tank is provided at the
terrace of the building with a capacity of 50000 liters. The water form
corporation main line is stored in the ground level sump and pumped to the over
head tanks.
2.9. ELECTRICAL INSTALLATIONS:
The electrical installation shall
generally be carried out in conformity with the requirements of Indian
electricity act 1910 and Indian electricity rule 1956. Electrical conduits are providing
adjacent to the lift room on either side of each floor. A generator is also
proposed as standby. It will be used in operation of the lift also in case of
power failure. Electrical installation includes electrical wiring consuming
devices accessories fittings control and protective gear and other accessories
associated with wiring situated on any premised. In this project all wiring are
concealed type.
2.10. FIRE PROTECTION:
In all buildings, sufficient automatic
fire detecting and alarm facilities shall be provided, where necessary to warn
out occupant existence of fire so that they may escape.
3. STRUCTURAL ANALYSIS AND DESIGN:
3.1. GENERAL:
STAAD Pro is structural software for
the model development, analysis, design, verification and visualization of all
aspects of structural engineering.
Following are
the main options available
STAAD-Pro
analysis and design
STAAD-PRE
graphical input generation
STAAD-POST
graphical post processing
STAAD- INTDES
Interactive design of structural components
STAAD –Pro performs analysis and
design of structure. The process and analysis are done simultaneously. The
input format con be created through CAD based input generators. The out put
generated by STAAD Pro consists of numerical results for analysis and design.
The communication with STAAD –Pro is through input file. The input file is a
text file consisting of a series of commands. These commands are executed
sequentially. The commands contain either instructions or data pertaining to
analysis and design. STAAD –Pro most general application will be the space
structure can be analyzing both frame and plate/shell elements. Any type of
structure such as plane, truss, floor structure and space structure can be
analyzed by STAAD-Pro. Most general application will be the space structure,
which is three-dimensional, framed structure with loads applied in any plane.
The input data and out put result are in engineering unit system MKS, SI and
FPS.
A structure is an assembly of
individual components such as beams, columns, slab and plate etc., in general
the term member will be such to refer to frame elements and the term element
will be used to refer to plate/shell elements. Connectivity for members will be
provided through member incidence command while connectivity for elements may
be provided with the element incidence command.
The material constants such as modulus
of Elasticity, density, and poisons ratio are provided for the analysis of
structure for the self-weight of the structure and also for calculating the
shear modulus
STAAD-Pro allows specifications for
supports such as PINNED, FIXED, or FIXED with different releases.
Loads on the structure can be
specified as joint load, member load etc, STAAD-Pro can also generate self
weight of the structure and use it as uniformly distributed member loads in
analysis. Joints loads, both forces and moments may be applied at any joint of
structure. Positive force act in the positive coordinate directions. The member
loads may be uniformly distributed load, concentrated loads and linearly
varying loads. A floor is subjected to a uniformly distributed load. Area load
command can specify the unit load per unit square area for members. The program
will calculate the tributary are for these member and provide proper member
loads.
The following
analysis facilities are available in STAAD-Pro.
Stiffness
analysis
Second order analysis
Dynamic
analysis
STAAD-Pro has the capabilities of performing concrete design base on
IS: 456(2000) Based on limit state method. The following types of cross
sections can be designed.
For
beams-prismatic: rectangular and square.
For
columns- prismatic: rectangular and square.
IS: 456(2000) these parameters can be changed to suit the particular
design. The beam design is based on flexure, shear and torsion. The columns are
designed for axial forces and biaxial bending moments at the ends.
3.2.
ANALYSIS:
3.2.1.
MATERIAL:
Grade of reinforcement : Fe415
Grade of concrete : M25
Density of concrete : 2500Kg/m3
3.2.2.
LOADING:
Dead load:
Partition
wall and other external walls, floor finish etc., as per the provisions of IS:
875-1987(part I)
Superimposed load:
As
per the provisions of IS: 875-1987(part II)
For Commercial Buildings (AL)
= 4.00 KN/m2
Seismic load
Dead load + part of live load = DL+0.5LL
3.2.3.
CODES:
Concrete design : IS: 456-2000
Steel design : IS: 800-1984
3.2.4.
PARTIAL SAFETY FACTORS:
Load
factors:
For dead load = 1.50
For live load = 1.50
The above partial safety factors are taken
from IS: 456-2000
Material
safety factor:
For reinforcement steel = 1.15
For concrete = 1.50
3.2.5.
LOAD CALCULATION:
Dead load:
At any floor level except ground floor (per
m width)
Load from slab = 0.15
x 23.5 = 3.525 KN/m2 (assuming
150mm thickness)
Partitions (G.F) = 0.23
x 4.20 x 20 =19.32 KN/m
Partitions (F.F TO F.F) = 0.23
x 3.00 x 20 =13.80 KN/m
Partitions (Terrace floor) = 0.23
x 1.00 x 20 =4.60 KN/m
Floor finishes = 1.00
KN/m2
Floor finishes (Terrace floor) = 2.00
KN/m2
B) Live load
For Commercial Buildings = 4.00
KN/m2
C) Seismic load
Dead load + part of live load = DL+0.5LL
3.2.6.
ANALYSIS ABOUT STAAD Pro:
STIFFNESS
ANALYSIS:
The stiffness analysis implemented in
STAAD is based on the matrix displacement method. In the matrix analysis of
structure is first idealized into an assembly of discrete structure components
(frame members).Each component has an assumed from of displacement in a manner
which satisfies the force equilibrium and displacement compatibility at the
joints.
Structural systems such as slabs,
which transmit loads in 2 directions, have to be discredited into a number of 3
to 4 notes connected to each other at their nodes. Loads may be applied in the
form of distributed loads on the elements as well as the plate be bending
effects are taken into consideration in the analysis.
For a complete analysis of the structure, the necessary matrices are
generated on the basis of the following assumptions.
The structure is idealized into an assembly of beam and solid type
elements jointed together at their vertices (nodes). The assemblage is loaded
and reacted by concentrated loads acting at the nodes. These loads may be both
forces and moments which may act in my specified direction.
A beam member is a longitudinal structural member having a constant,
doubly symmetric or near – doubly symmetric cross section along its length.
Beam members always carry axial forces. They may also be subjected to shear and
bending in two arbitrary perpendicular planes, and they may also be subjected
to torsion. From this point these beam members are referred to as “members” in
the manual.
A slab is a three or four nodded planar element having variable
thickness. A solid element is a 4 to 8 nodded three dimensional element. These
slab and solid elements are to as “elements” in the manual.
Internal and external loads acting on each node are in equilibrium.
If torsion or bending properties are defined for any member, six degrees of
freedom are considered at each node (i.e. three translational and three
rotational) in the generation of relevant matrices. If the member is defined as
truss member (i.e. carrying only axial forces) then only the tree degrees
(translational) of freedom are considered at each node.
Two types of coordinate systems are used in the generation of the
required matrices and are referred to as local and global systems.
Local coordinate axes are assigned to each individual element and
are oriented such that computing effort for element stiffness matrices are
generalized and minimized.
3.2.7. PROCEDURE OF ANALYSIS IN STAAD –Pro:
Concrete design is performed on some members of a space frame
structure. Design calculations consist of computation of reinforcement for
beams and columns.
In our building represents a space frame and the members are made of
concrete. The inputs will show the dimensions of the members.
3.2.8. STAAD SPACE FRAME WITH CONCRETE DESIGN:
Every input has to start with the word STAAD. The word SPACE
signifies that the structure is a space frame structure (3-D) and the geometry
is defined through all X, Y and Z coordinates.
3.2.9.1.ISOMETRIC VEIW
3.2.9.2.LOADING DIAGRAM
3.2.9.3.SHEAR FORCE DIAGRAM OF INTERIOR
PANEL IN X
DIRECTION
3.2.9.4.BENDING
MOMENT DIAGRAM OF INTERIOR PANEL IN X DIRECTION
4.
DESIGN OF SLABS
4.1.
INTRODUCTION:
When the slab supported on all four
edges, the load is transferred to all the four supports and therefore, the
bending and deflection in such slabs are considerably less then those in slabs
supported an all the four supports. The corners get lifted up when the slab is
loaded, in case the corners held down, the deflection in the central portions
are further reduced and thus the bending moment are reduced in such slabs. When
the corners are held down due to monolithic constructions of the slab with edge
beams their will be torsion reinforcement is to be provided. The bearing
moments are calculated added on the edge condition and loads. Bending moment is
obtained in our design.
If Mx and My are
the maximum bending moments per unit width in the middle strip of the slab in
the short and long span respectively then
Mx =
αxW.l x
My =
αy W.l x
Where and are coefficients depending the
ratio ly/lx
In each direction of the main bars alternate bars will be bent up at
one seventh of the span.
The slab for design is classified in
to s1, s2, s3 and s4 in floor slabs. The edge condition is considered and thickness
of the PT slab bound to be 150mm.
Live load on roof is taken as 4.0KN/m2.
When access not provided and 1KN/m2.
Dead loads of slab, weathering course and weight of walls are other load
consideration.
All slabs are designed as two way slabs based on ly/lx
ratio, 10mm dia bars are provided in both direction.
4.2. LIMIT
STATE METHOD:
Limit state of Design is a further
improvement of ultimate load design in the limit state methods a structure is
designed to with stand all loads like to act on it in the duration of its life
span and also to satisfy the service requirements like deflection and
limitation of crack width, limit means an acceptable limit, for the safety and
serviceability requirements before anything can occur.
The design provides a condition that
the structure will not become unfit for use for which it is meant or in other words
the structure will not reach a limit state.
The entire limit state that are relevant are considered in the
design to ensure an adequate degree of safety and serviceability, the structure
in general shall be designed on the basis of the most critical state and shall
also be checked for other limit states.
4.2.1.
LIMIT STATE OF COLLAPSE
The design on
limit state of collapse provides the necessary safety of the structure against
partial or total collapse of the structure.
4.2.2.
LIMIT STATE OF SERVICEABILITY
This
limit state is introduced to prevent objectionable deflection and cracking.
4.2.3.
CHARACTERISTICS STRENGTH OF CONCRETE:
|
Grade
|
M15
|
M20
|
M25
|
M30
|
M35
|
M40
|
|
Fck
N/mm2
|
15
|
20
|
25
|
30
|
35
|
40
|
4.2.4.
CHARACTERISTICS STRENGTH OF STEEL:
|
Grade
|
Fe250
|
Fe415
|
Fe500
|
|
fyN/mm2
|
250
|
415
|
500
|
4.2.5.
CHARACTERISTIC LOADS:
Characteristics load means the value of the
load, which has a 95 percent probability of not being exceeded during the life
of the structure.
Characteristics load is the weight of the structure itself.
Characteristic live load and wind load are taken as per IS875-1964
characteristic seismic loads are taken as per 1873-1975.
4.2.6.
OBJECTS OF LIMIT STATE
The
object of limit state design is the guarantee adequate safety consistent with
economy against the structure being rendered unfit for service due to cracking,
deflection, failure and such other cases. A limit sate corresponds to each of
the states in which the structure becomes unfit.
5. DESIGN
OF BEAM
5.1
GENERAL
From the STAAD Pro Analysis done we obtain the maximum positive
moment, maximum negative moment and maximum shear force from these the beams
are designed manually.
Maximum moments and shear forces
|
Beam
|
node
|
Env
|
Fx
|
Fy
|
Fz
|
Mx
|
My
|
Mz
|
|
|
|
|
kN
|
kN
|
kN
|
kNm
|
kNm
|
kNm
|
|
303
|
167
|
+ve
|
9.811
|
224.95
|
2.594
|
0.82
|
6.566
|
367.6
|
|
|
|
-ve
|
-31.04
|
-67.09
|
-2.587
|
-0.811
|
-6.585
|
-170.81
|
5.2.
DESIGN OF BEAMS BY MANUAL (Beam No: 303):
Step
- 1
Width of Beam = 300
mm
Over all depth of Beam = 600m
Thickness of slab, Df = 150mm
Breadth of web, bw = 300mm
Concrete grade = M25
Steel grade = Fe415
Step
– 2:
Bending moment and shear force
Negative moment @ interior support = 170.806
kNm
Positive moment @ centre of span =
367.60 kNm
Maximum shear force at its support, Vu =
224.952 kN
Limiting moment of Resistance
Mulimit = 0.138fck bd2
= 0.138 x25x300x5502
Mulimit = 313.088 kNm
Mu limit < Mumax
Hence the section is designed for doubly
reinforced.
Mu2 =
367.60-313.088
= 54.512kN.m
Ast
calculation:
Mu
= 0.87
fy Ast d (1- (fyAst/fckbd))
313.088x106
=
0.87 x 415 x Ast x 550
(1- (415 x
Ast / 25x300x550))
313.088
x 106 = 198577.5
Ast – 19.978 Ast2
Ast1 = 1965.19 mm2
Use 25mm dia bars
No of bars required = Ast
/ ast = 1965.19 / ((π/4) x 252)
= 4.00 Say 4 nos
Main reinforcement (excess reinforcement Positive)
Mu1 = 0.87
fyAst2(d-d’)
54.512x
106 = 0.87 x 415 x Ast2
x (550-50)
54.512
x 106 = 180525 Ast2
Ast =
301.97 mm2
No of bars required = Ast
/ ast
= 301.96 / ((π/4) x 252)
= 0.62 Say 1 nos
Provide 5 nos of
bars #25 at the Bottom tension face at centre of span section.
Asc
calculation:-
Main reinforcement (Negative)
Mu = fsc
Asc (d-d’)
d’/d = 50/550 =0.09
fsc = 353.40N/m2
170.806x
106 = 353.40 x Asc x
(550-50)
Ast = 966.64 mm2
No of bars required = Ast
/ ast = 966.64 / ((π/4) x
252)
= 1.97 Say 2 nos
Provide 2 Nos of bars #25 at the top
tension face near support
Shear
reinforcement:-
Maximum shear force V = 224.952kN
حv
= (Vu/bd)=
(224.952x 103) / (300 x 550)
= 1.36 N/ mm2
Pt =
(100 Ast/b d) = 100 x ((2945.24) / (300x550))
= 1.78%
From table no: 19 in IS 456 - 2000
حc = 0.785
N/mm2
حv > ح c (i.e.) 1.19N/mm2 > 0.785 N/mm2
Hence shear reinforcement are required.
Excess shear force,
Vus = Vu
– ح c bd
= 224.952
x 103 – (0.785x 300 x 550)
Vus = 95427
N
Assume # 10mm, 2 legged vertical stirrups
are used
Asv = 2
x (π / 4) x102 = 157.08 mm2
Spacing of stirrups (Sv)
Sv = 0.87 fy Asvd / Vus
= 0.87 x 415 x 157.08 x 550 / 95427 Sv = 326.87 mm c/c
i)
Sv min = 0.87 x fy x Asv / (0.4 x b)
Sv min = 0.87
x 415 x 157.08 / (0.4 x 300)
= 472.60 Say = 470 mm
ii)
0.75 x 550 = 412.5
mm
300mm whichever is minimum
Provide 10mm # @ 2 legged vertical stirrups at 300 mm c/c at near
support gradually increasing to 400 mm towards the centre of span
Check
for deflection:-
At centre of span
Pt = 100 Ast / (bxd) = 1.78%
Modification factor
fs =0.58 x 415 x 2460.04/ 2945.24 = 201.05
N/mm2
M.F = 0.95
Neglecting bars in compression side
Kc = 1
L/d
max = L/d basic x Kt x Kf
= 26 x 0.95 x 1.00 x1.00
= 24.70
L/d
actual = 5230 / 550 = 9.51
24.70
> 9.51
Hence defection
control is safe.
5.3.
DESIGN OF BEAMS USING STAAD-Pro:
B E A M N O.
303 D E S I G N R E S U L T S
M25 Fe415 (Main) Fe415 (Sec.)
LENGTH: 5230.0 mm
SIZE: 300.0mmX600.0mm COVER:
25.0 mm
SUMMARY
OF REINF. AREA (Sq.mm)
----------------------------------------------------------------------------SECTION 0.0 mm
1307.5 mm 2615.0 mm 3922.5 mm
5230.0 mm
----------------------------------------------------------------------------
TOP
2200.78 558.91 347.17 727.92 2286.55.94
REINF.
(Sq. mm) (Sq. mm) (Sq. mm) (Sq. mm) (Sq. mm)
BOTTOM
931.11 649.39 482.25 937.02 1255.85
REINF.
(Sq. mm) (Sq. mm) (Sq. mm) (Sq. mm)
(Sq. mm)
----------------------------------------------------------------------------
SUMMARY OF PROVIDED
REINF. AREA
--------------------------------------------------------------------------------------
SECTION
0.0 mm 1307.5 mm 2615.0 mm 3922.5
mm 5230.0 mm
--------------------------------------------------------------------------------------TOP 3-32í 3-32í 3-32í 3-32í 3-32í
REINF.
1 layer(s) 1 layer(s) 1 layer(s) 1 layer(s) 1 layer(s)
BOTTOM 3-20í 3-20í 3-20í 3-20í 4-20í
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. @190 mm c/c@190 mmc/c@190mmc/c 190mmc/c@190mmc/c
--------------------------------------------------------------------------------------
SHEAR DESIGN RESULTS AT DISTANCE d
(EFFECTIVE DEPTH) FROM FACE OF THE SUPPORT
SHEAR DESIGN RESULTS AT 790.0 mm AWAY FROM START SUPPORT
VY =
195.79 MX = 0.01 LD= 13
Provide 2 Legged 8í @ 190 mm c/c
SHEAR DESIGN RESULTS AT 790.0 mm AWAY FROM END SUPPORT
VY = -186.21 MX = 0.00 LD=
12
Provide 2 Legged 8í @ 190 mm c/c.
6. DESIGN
OF COLUMN
6.1
GENERAL
From the STAAD Pro Analysis done we obtain the maximum positive
moment, maximum negative moment and maximum shear force from these the beams
are designed manually.
Maximum moments and shear forces
Beam node Env Fx
Fy Fz Mx
My
Mz
kN kN kN kNm kNm kNm
231 86
+ve 3513.2 94.90 100.4 1.33 277.73 233.99
-ve
-63.88 -104.1 -114.8 -1.32 -248.0 -253.11
6.2. DESIGN OF COLUMNS BY MANUAL
(Beam No: 231):
Beam size
= 450 x 450mm
Concrete grade = M25
Steel grade = Fe415
Factored load Puz = 2477.56kN
Factored Moment Muz = 11.747kN.m
Muy = 262.42kN.m
Moments due to
minimum eccentricity are less than the values given above
Reinforcement is
distributed equally on four sides
As a first trail assume the reinforcement P = 3.75
P/fck
= 3.75/25 = 0.15
Uniaxial moment capacity of the section
about XX and YY axis
Effective cover d’ = cover
+dia of rod/2
= 40+25/2 = 52.5mm
Effective depth d = 450-25-(25/2)
d = 412.50mm
D = 450mm
d’/D = 52.5/450
= 0.1167
Check for d’/D 0.15 will be used
Pu/fck bd = 2477.56 x 103/ (25 x 450 x
450)
= 0.49
Referring to chart45
Mu/ fck bd2 = 0.135
Mux1 = Muy1
= 0.135
x 25 x 450 x 4502
= 307.55kN.m
Calculation
of Puz:-
Referring to chart63 corresponding to
P = 3.75, fy = 415, fck
= 25.
Puz/
Ag = 22.70Ag
= 22.70 x450 x 450/1000 =4596.75kN
Pu/Puz = 2477.56/4596.75
=0.556
Muy/Muy1 = 262.42/341.72 =0.77
Mux/Mux1 = 262.42/341.72 =0.77
Referring the
chart 64 by the permissible value of Mux/Mux1
corresponding to the above value of Muy/Muy1and Pu/Puz
is equal
As = 3.75
x 450 x 450 / 100
= 7593.75mm2
Assume 25mm # rod
16nos.
of 25mm # provided
As = 7853.98mm2
P = 7853.98/ (450 x 450) x 100
= 3.88
With the % the function may be reached as
follows
P/ fck = 3.88/25 = 0.1552
Reffering to chart 45
Mu/ fck bd2 = 0.155
Mux1
= Muy1 = 0.155 x 25 x 450 x 4502
= 353.11kN.m
Referring the chart 63
Puz/ Ag = 23 Ag
= 23 x 450 x 450/1000
= 4657.50kN
Pu/Puz = 2477.56/4657.50
=0.53
Muy/Muy1 = 262.42/353.11 =0.74
Mux/Mux1 = 262.42/353.11 =0.74
Corresponding to
the above of value of Muy/Muy1 and Pu/Puz
the permissible value of Mux/Mux1 0.68 is equal
Hence the section is ok.
Design
of lateral ties:-
Dia = 25/4 =
6.25mm
Dia 8mm which one is greater
I.e. dia = 8mm.
Pitch of lateral
ties
P = Least Lateral Dimension
= 450mm
P = 16 x 25 = 400mm
P = 300mm
Whichever is less
Provide 8mm # 300mm
c/c as lateral ties.
7. DESIGN
OF FOUNDATION:
7.1
GENERAL
The outer Column footings are designed as Isolated footings where as
the Inner column footings are designed as Combined footings. In these combined
footings the two adjacent columns are combined in the Z axis direction. From
the STAAD Pro analysis done we obtain
the Axial load for the designing of footing
7.2 Design
of combined footing: - (Node No: 89 and 91)
|
|
Axial load
Pu2 = 3938.35 say 4000 kN
To find Shear
force,
= 7.2m2
Allowable safe Bearing capacity of the soil =375+43.20=418.20kN/m2
= 203.86kN/m2
Thickness of
footing required against bending moments:-
حcu = Ks. حc
= 0.5
+ (0.45 /0.45)
Check for Effective depth required against two way
shear (or) punching shear. The critical section of two way shear is taken at a
distance of d/2 around form the pedestal
|
|
Bearing capacity of soil = 250
kN/m2
To
find length of footing
Taking moment about B,
4000 x 6.5 + 172 + 160 = 8000
X
X = 3.29m
Taking AB = 0.75m 7.2.1plan
Total length up to CG from A =3.29m+0.75m
=4.04m
Length of
footing = 2(m+n)
= 2(0.75+3.29) = 8.08m
Length of
projection CD = 0.83m
Taking 10% of total weight as self weight
of foundation
Bearing area
required = (8000+800) / 250
= 35.20 m2
Width of
foundation = 35.20 / 8.08
= 4.36 m say 4.50m
Footing Area = 8.08 x 4.50 = 36.36 m2
Net upward soil
pressure = 8000 / 36.36
= 220.02kN/m2
Net upward soil
pressure < safe bearing capacity
220.02
kN/m2 < 250 kN/m2
Hence safe.
Bearing area per meter length = 220.02
x 4.50
= 990.09kN/m2
|
|
SF @ A =0
SF @ B left =742.57kN
SF @ B right =-3257.43kN
SF @ C left =3178.15 kN
SF @ C right =-821.85 kN
SF @ D = 0
To find shear force at Zero 7.2.2.SFD
&BMD
4000 – (0.75 x
990.09) – 990.09x = 0
X = 3.29m
To
find Longitudinal bending moment:-
Maximum hogging Bending moment,
M max = 990.09 x 4.042 / 2 –
4000x3.29– 172
= -5252.08 kNm
Max
sagging Bending moment:-
At Support, B
Mx = 990.09 x 0.75 2/ 2 -172
= 106.46kN.m
At support C,
Mu = 990.09 x 0.832 / 2
= 341.04kN.m
To
find transverse reinforcement:-
We take footing in ‘X’ axis is isolated,
then finding the maximum bending moment,
Pu1 = net upward ultimate pressure under
column 1
= P1/B = 4000 / 4.5
= 888.89 kN/m
Mu1 = Pu1 (0.5B – 0.5b) 2
/ 2
= 888.89 (0.5 x 4.50 – 0.5 x 0.45)2
/ 2
= 1822.50 kN.m
Pu1 = net upward ultimate pressure under
column 2
Pu2 = P2 / B = 4500 / 4.5
= 888.89kN/m
Mu2 = Pu2
(0.5B – 0.5b) 2 / 2
= 888.89 (0.5 x 4.50 – 0.5x 0.45)2
/ 2
= 1822.50kNm
Effective
thickness calculation
Based on longitudinal moment,
b = width of the longitudinal beam
= 4500mm
d
2 = (5252.08 x 106) / (0.138 x 25
x 4500)
d = 581.63
mm say 600mm
Over all depth D = 650mm
For effective thickness based on maximum
transverse moment under column C1,
b =
width of transverse beam under
column C1
= pedestal dimension
along the width of transverse beam + effective depth of footing slab
= 550 + d mm
0.138 fck bd2 = Mu
0.138 x 25 x
(550 + d)d2 = 1822.50 x 106
3.45(550+d)d2 = 1822.50 x 106
(1897.50 +
3.45d)d2 = 1822.50 x 106
3.45d3 +1897.50d2–
1822.50 x 106 = 0
d = 660.58mm. Say 670mm
Total depth of Footing = 730 mm
For effective thickness of footing based on
maximum transverse moment under column C2
b = width of transverse beam under column C2
= pedestal dimension
along the width of transverse beam + 2 x effective depth of footing slab
= 550 + 2d
3.45(550+2d) d2 = 1822.50
x 106
(1897.50+6.90d) d2 = 1822.50
x 106
1897.50d2 + 6.90 d3 –
1822.50 x 106 = 0
d = 561.82mm Say 570mm
Total depth of footing = 630mm
Thickness
of footing based on shear:-
The effective
thickness of footing may be determined by considering that the shear is
resisted without shear reinforcement as follows,
Vumax = ﺡc b.d
d = Vumax / ﺡc b
For one way Shear:-
Vumax = Max ultimate shear at the section at distance‘d’
from the inner face of pedestal of column
C2.
=3178.15
– 990.09d
bo
= B = 4500mm
ﺡvu = shear strength of concrete in foundation slab
= Ks which may taken as
1.0, and shear strength of concrete which may be taken as its minimum value of
0.25 N/mm2
= 0.25
N/mm2
dx1000 = (3178.15
– 990.09d) x 1000 / (0.25 x 4500)
1125000d = 3178.15
x103 – 990.09x103 d
d = 1.50
m
= 1500mm. Say 1540mm
Over all depth = 1600mm
For
two way shear under column C1,
Vumax = ultimate punching shear on critical
Section at a distance of half effective
depth of foundation slab form the face of pedestal
= Pu1 – Pu
(lp1 + 0.5d) (b+d)
Where Pu = Ultimate
upward soil pressure
= 8000 / (8.08 x 4.50)
= 220.02 kN/m2
Vumax = 4000
– (220.02(0.45 + 0.5d) (0.45 + d)
= 4000
– 44.554 – 148.51d – 110.01d2
Vumax = 3955.446
– 148.51d – 110.01d2
bo
= 2(0.45 + d) + 0.5 + d
= 1.4 + 3d
حuc = 0.25
fck = 0.25
25
= 1.25 N/mm2
Where
.τc = Ks. τc, Ks = 1, τc = 0.25
fck
حv = Vu/bd
d = Vu/bحv
dx1000 = (3955.446
– 148.51d – 110.01d2) x 1000 /
(1.25 x (1.4 + 3d) x 1000)
(1.75 + 3.75d) dx106 = (3955.446
– 148.51d – 110.01d2) X 1000
d
= 0.358
m =358mm Say 360mm
Total
depth = 420mm
For
two way shear under column C2
Vumax = Pu2
– Pu (lp2 + d) (bp2 + d)
= 4000 – 220.02 (0.45+d) (0.45+d)
= 4000 – 44.554 – 198.02d – 220.02d2
Vumax = 3955.446
– 198.02d – 220.02d2
bo = 2(0.45 + d+0.45+d)
= 1.80
+ 4d
حv = Vu/bd
d = Vu/bحv
1000d = (3955.446 – 198.02d – 220.02d2)1000
1.25
X (1.80 + 4d) x 1000
(2.25 + 5d) dx106
= (3955.446
– 198.02d- 220.02d2)1000
d = 1136mm
say 1140mm
Over all depth = 1200mm
Therefore the
thickness of footing is governed by two way shear
d = 1140mm, D = 1200mm, d =
1200 – 60 = 1140mm
The base of the
footing is provided at a depth of 1300mm
DESIGN
FOR MOMENTS:-
The bottom reinforcement for transverse moments is placed below the
bottom reinforcement for longitudinal moment. Distribution reinforcement in
transverse direction.
= 0.12% of gross sectional area
= (0.12 / 100) x1000 x 1300
= 1560mm2 Provide 20mm # @ 200mmc/c
Use 20mm φ rods
Spacing
= (314.15
/ 1560) x 1000
= 201.38mm say 200mm
1)
3d = 3 x 1240 = 3720mm
2)
And mm300
Which ever is minimum
Provide 20mm #@200mmc/c
Ast
Calculation:-
Longitudinal
span moment:-
Mu = 0.87 fy
Ast d (1- (fyAst/fck bd))
5252.08 x 106 = 0.87 x 415 x Ast
x 1240 (1- (415 x Ast / 25x4500x1240))
5252.08 x 106 = 447702 Ast
– 1.33 Ast2
Ast =
12171.28mm2
Provide
min Ast
= 0.12% of gross sectional area
= (0.12 / 100) x4500 x 1240 =6696mm2
Ast per meter length = (12171.28
/ 4.5) = 2704.73mm2 /m
Use 25 mm φ rods
Spacing
= (490.57
/ 2704.43) x 1000
= 181.49 mm Say 180 mm c/c
Provide 25mm # @ 180mmc/c
Support
moment:-
Mu = 0.87
fy Ast d (1- (fyAst/fckbd))
341.04
x 106 = 447702
Ast – 1.33 Ast2
Ast = 763.49 mm2 < 6696 mm2
Ast per meter length = (6696
/ 4.5) =1488mm2/m
Use 20 mm φ rods
Spacing
= (314
/ 1488) x 1000
= 211.13 mm Say 200 mm c/c
Provide 20mm # @ 200mmc/c
Transverse
moment under column C1:-
Mu = 0.87 fyAstd(1-
(fyAst/fckbd))
b =bf
+d =450 + 1240 = 1690mm
1822.50 x 106 =
0.87 x 415 x Ast x 1240
(1- (415 x Ast / 25x1190x1240))
Ast = 4211.42mm2
Ast per meter length = (4211.42
/ 1.69)
= 2491.96mm2/m> 1488mm2
Use 25 mm φ rods
Spacing
= (490.87
/ 2491.96) x 1000
= 196.98 mm Say = 190 mm c/c
Provide 25mm # @ 190mmc/c
Transverse
moment under column C2:-
Mu = 0.87 fyAstd(1- (fyAst/fckbd))
b =bf
+2d= 450 + (2x1240) = 2930mm
7822.50 x 106 =
0.87 x 415 x Ast x 1240
(1- (415 x Ast /
25x1930x1240))
Ast = 4149.26mm2
Ast per meter length = (4149.64
/ 2.93)
= 1416.26mm2/m< 1488mm2
Use 20 mm φ rods
Spacing
= (314.15
/ 1488) x 1000
= 211.13 mm Say = 200mm c/c
Provide 20mm # @ 200mmc/c
7.3.
DESIGN OF ISOLATED FOOTING:-
DATA
FOR DESIGN:
Axial load = 1541.4kN say 1800kN
Moment, Mx = 141.271
kN
Moment, Mz = -141.19kN
Safe bearing capacity of soil = 250kN/m2
Area required for foundation = 1800
/ 250
|
|
Area required = LxB
BxB = 7.2m2
B2 = 7.2m2
B = 2.68m
L = 2.68m
Length required = 2.68 m 7.2.
Breadth required = 2.68 m 7.3.1. Plan
Length provided = 2.70 m
Breadth provided = 2.70m
Original area = 2.70m x 2.70m
= 7.29 m2
Column size:-
Length = 0.45m
Width = 0.45m
Self weight of the footing:-
Unit weight of concrete = 25.00kN/m3
Depth below Ground level = 2.40m
Depth of footing @ face of column = 1.00m
Depth of footing @ Edge of footing = 0.30m
Volume of footing:-
Volume of frusta of pyramids and concrete,
= (1.00 – 0.30) / 3 ((0.45 x 0.45) + (2.70x 2.70) +
(0.45 x 0.45 x 2.70 x 2.70)
=1.559m3
Volume of flat portion = 2.7
x 2.7 x 0.30 = 2.187m3
Total volume of concrete =
3.746m3
Weight of footing = 3.746 x 25 =93.65kN
Self
weight of soil above footing:
Self weight of soil = 18kN/m3
Volume of column portion (0.45x0.45x1.5) =0.303 m3
Volume of soil portion (2.7 x 2.7x2.40) -
(3.746 +0.303)) = 13.447m3
Weight of soil portion = 13.447
X 18 =242.05kN
Total load = Pu + self wt of footing +
self wt of soil
= 1800 + 93.65 + 242.05
p = 2135.70
kN
Soil
pressure with weight of footing and weight of soil above footing:
Zx = 2.7 x (2.7) 2 / 6 = 3.28 m3
Zz = 2.7 x (2.7) 2 / 6 = 3.28 m3
Soil Pressure = P
/A ± Mx / Zx ± Mz /Zz
Pmax = 2135.70 / (2.7x2.7) +
(141.27/3.28) + (141.19 / 3.28)
= 379.08kN/m2
Pmin = 2135.70 / (2.7x2.7) –
(141.2x3.28) - (141.19 / 3.28)
= 206.85kN/m2
Soil
pressure without weight of footing and weight of soil above footing:
Soil Pressure = P
/A ± Mx / Zx ± Mz /Zz
Pmax = 1800 / (7.29) + (141.25/3.28) +
(141.19/3.28)
= 333.00kN/m2
Pmin = 1800 / (7.27) - (141.27/3.28)
- (141.19/3.28)
= 160.84kN/m2
Net safe bearing capacity of soil =250kN/m2
Consider SBC can be increased by 50% =1.50 x 250 =375kN/m2
Can be increased by Depth of soil x density =2.40 x 18 = 43.20 kN/m2
|
|
P max is less than SBC,
Hence it is safe.
Design of Isolated footing:-
Soil pressured without weight of footing:-
Pmax = 333.00 kN/m2
Pmin = 160.824 kN/m2 7.3.2 pressure diagram A Pressure along
‘Z’ direction
Soil pressure along ‘z’ = P
/A ± Mz / Zz
Pmax = 1800
/ (2.7x2.7) + (14.146 / 3.28)
= 289.95kN/m2
Pmin = 1800
/ (2.7x2.7) - (141.19 / 3.281)
= 203.88kN/m2
Pressure along ‘X’ direction
Soil pressure along ‘x’ = P
/A ± Mx / Zx
Pmax = 1800/ (2.7x2.7) + (14.146 / 3.28)
= 289.98 kN/m2
Pmin = 1800 / (2.7x2.7) - (14.146 / 3.28)
|
|
Upward soil pressure at A
= 203.86 + (203.06 –
289.98)
2.7 x (2.7
-1.125)
= 252.56kN/m2
Over all depth of face of column=1000mm
Clear
cover = 60mm 7.3.2 pressure diagram B
Dia
of bar assumed= 12.00 mm
Effective
depth = 1000 – 60 – (12/2)
= 964.00mm
Mux = Pnu
(L-a / 2 x B) x (L-a /4)
= 252.56((2.7-.0.45) / 2) x 2.7) x (2.7 – 0.45)/
4
= 431.52kNm
Muy = Pnu
(L) x (B-b) 2 / 8
= 252.56 x 2.7 x ((2.7-0.45)2 /
8)
= 431.52kNm
Maximum Bending moment @ face of column =
431.52 kNm
Thickness of footing required against
bending moments:-
Mumax = Qubd2
431.52 x 106= 0.138 x25x 2700x d2
d = 215.23mm
Say 220mm
D = 220+
60
= 280mm
Upward soil pressure at B = 203.86
+ (203.06 – 289.98) / 2.7
X (2.7 -1.125-0.45)
= 240.102kN/m2
Mux = Muy = Pnu Lx ((B-b)
2/ 8)
= 240.102 x 2.7 x (2.7-0.45)2 /
8
= 410.24kNm
Maximum Bending moment @ face of column = 410.24kNm
|
|
Mumax = Qubd2
410.24 x 106= 0.138x25x2700 xd2
d =
(410.24 x 106)
/ 0.138x25x2700
= 209.85 mm say220mm
D = 220 + 60
= 280mm 7.3.2 one way shear plan
Check
for Effective depth required against shear:-
The critical section of shear is taken at a
distance of‘d’ from the pedestal
Vu
max =Pu B ((L-a)/2)
–d)
=252.56 x 2.70 x (2.70- 0.45) /2 –d)
=681.912(1.125
–d)
Vumax =767.15 –
681.91 d
The total shear
stress induced at critical section is resisted by the shear stress, developed
by concrete section,
|
|
Ks = 0.5
+ βc
|
|
= 1.50
>1
Ks = 1
حc = 0.25
fck
= 1.25N/mm2
7.3.2 critical section plan
حcu = 1x1.25
= 1.25N/mm2
حvu = (Vumax/bd)
حvu = (767.15 – 681.91d)/2700
x d
حvu = حcu
1.25x103 = (767.15 – 681.91d)/2.70d
d = 0.189m
d = 189 mm
D = 250mm
|
|
Vumax = Pnu ((LxB –
(a+d) (b+d))
= 252.56(2.70x2.70 –
(0.45+d)(0.45+d))
= 252.56(7.088-d2-0.9d) 7.3.3
Two way shear plan = 1719.27 – 218.30d – 252.56 d2
ﺡvu =
(Vumax)/(2(a+d)+2(b+d))d)
= 1719.27
– 218.30d – 252.56 d2 /
(2(0.45+d)
+2(0.45+d)) x d
1.25 x 103 (1.8d + 4d2) = 1719.27
– 218.30d – 252.56 d2
5252.56 d2 + 2468.30 d – 1719.21 = 0
d = 384mm say 390mm
D = 450mm Provide maximum depth d = 450mm
Reinforcement
along x direction:-
Mu = 0.87
fy Astd(1- (fyAst/fck bd))
431.52 x 106
= 0.87 x 415 x Ast x 390(1- (415
x
Ast / 25x2700x390))
Ast = 3228.94mm2
Provide 20mm dia bars.
No of bar = 3228.94 / ((π /4) x 202)
= 10.28 nos Say 11nos
Reinforcement
along y direction:-
Mu = 0.87
fy Astd(1- (fyAst/fckbd))
431.52 x 106
= 0.87 x 415 x Ast x 390(1- (415
x
Ast / 25x2700x390))
Ast = 3228.94mm2
Provide 20mm dia bars.
No of bar = 3228.94 / ((π /4) x 202)
= 10.28 nos Say 11nos
Provide 20mm dia bars 11nos in both
directions.
8. DESIGN
OF OTHER MEMBERS:
8.1.
STAIR CASE
Stairs consist of steps arranged in
series for purpose of giving access to different floors of building. Location
of stair requires good and careful consideration.
Two – dog – legged – case is arranged
for entire building i.e. one near the entrance and one in the rear face.
8.2.
DESIGN OF STAIR CASE
Assumptions:-
Imposed uniform distributed
load on staircase = 4kN/m2
The flight slab and
landing span longitudinally
Weight of brick steps = 20 kN/m3
Arrangements
of stairs:-
Vertical height between the floors = 3.60m
= 3600mm
Assuming rise of step as 150mm
No of risers = 3600/150
= 24Nos
No of flights = 2 (Dog legged stair)
Riser per flight = 24/2
= 12Nos
Providing tread as 300mm
Going in each
flight
= 11 x 300
= 3300mm
Width of
landings = 4800
– 3300
= 1500mm
Providing a clear gap of 300mm between the
two flights.
Width of flights = 2350mm
Effective span:-
Effective span of flight =
c/c distance of support
=1.5 + 3.3 + 0.3
= 5.10m
Loads:-
Assume the thickness of waist slab and
landing 180mm
Self weight of waist slab per m2
of sloped area
= 1 x 0.20 x 25 = 5.0 kN
Self weight of waist slab per m2
of plan area
= 4.5 x
(0.32 + 0.152)
/ 0.3 = 5.03kN
Weight of steps per m2 of plan
area
= 1000/300 x ½ x 0.3 x 0.15 x 1 x 20 = 1.5kN
Imposed load per m2 of plan area =
4kN
Total load on waist portion =
4 + 1.5 + 5.03 =
10.53 kN/m2
Total load on landing portion = 5.00 + 4.00 =
9.00 kN/m2
Bending moment:-
Consider 1 m width of slab throughout the
span
Reactions:-
RB x 5.26 – 10.53 x 3.76 x 3.38
– 9.0 x 1.35 x 0.75 = 0
RB = 27.17 kN
RA = 25.82kN
Max shear force = 27.17 kN
Shear force is zero at
= 27.17 / 25.82
= 1.052 m from B
Maximum bending moment at 1.091 m from B
= 27.17 x 1.052 – 10.50 x 1.052 x 0.546
= 22.55 kNm
Factored bending moment
= 1.5 x 22.55
= 33.83kNm
Maximum
shear force:-
Maximum shear force at support (Vu)
= 27.26 kN
Depth
of slab required:-
For balanced section of M25 grade concrete
and fe415 steel.
Moment of resistance = 0.138 fck bd2
=
3.45 bd2
Design bending moment = 33.83kNm
Effective depth required =
(33.83 x 106)
/ (3.45 x 1000)
= 99.02mm
Assume effective cover = 20mm
Over all depth required = 119.02mm
Over all depth provided = 119.02mm < 200mm
Hence it is safe.
Main
reinforcement:-
Mu = 0.87 fyAst180(1- (fyAst/fckbd))
33.83x 106 = 0.87
x 415 x Ast x 180(1- (415 x Ast / 25x1000x180))
Ast = 547.34mm2
No of bar = 547.34 /113.10 x 2.35
= 11.37 nos Say 12nos
Provide 12 nos of 12mm dia bar in each
flight.
Distributors:-
Area
of steel required = 0.12 % BD
= 216
mm2
Spacings = (50.26
x 1000) / 216
= 232 mm say 230mmc/c
Provide 8mm # bars @230mmc/c
Check
for shear:-
Design
shear force = 1.5 x 27.26 = 40.89 KN
Nominal
shear stress حv = Vu / bd
= (40.89 x 103) / 1000 x 160
= 0.256 N/mm2
Minimum
value of permissible shear stress = 0.35 N/mm2
Hence
safe against shear
Check
for stiffness:-
Area
of steel provided = 12 x 113.10 = 1357.20 mm2
Percentage of steel = (1357.20 / (2350 x
180)) x100
= 0.321%
Modification
factor = 1.50
Basic
value = 20
Effective depth required for Stiffness = 5100
/ (20 x 1.5)
= 170mm <180mm
Hence
it is safe.
CONCLUSION
Our project deals
with planning, analysis and design of
shopping complex using STAAD Pro at Vadapalani, Chennai.
The shopping
complex is designed with all necessary facilities such as shops, super markets,
coffee shops, Food courts, offices, Escalators, Lifts etc., as per BIS
specifications.
In this project,
the Analysis of frame is done by stiffness matrix method using STAAD Pro.
Software.
Design of footings, columns, beams & slabs
are done manually by limit state method as per IS456 – 2000, IS 1893-2002 and SP16.
BIBLIOGRAPHY
1.
Theory of structures
Dhahpat Rai & sons. -Ramamrutham
2.
Design of Reinforce d concrete
Tata McGraw Hill -
SN Sinha
3.
Indian standard code of
practice for and
Reinforced Concrete IS: 456 – 2000
4.
Design Aids for Reinforced
concrete to IS: 456 – 2000,
SP: 16-1978
5.
Design of Reinforced Structure - N. Krishnaraju
6.
Criteria for earth quake
resistant,
Design of
structures to IS: 1893 (part1):2002
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