Presentationt design and analysis of multistorey building

DEPARTMENT OF CIVIL ENGINEERING ,
JAMIA MILLIA ISLAMIA

ACKNOWLEDGEMENT
In the name of Allah, the most gracious and the most merciful.
On the very outset of this report, we would like to extend our sincere
and heartfelt gratitude towards all the people who have helped us in
this endeavor. Without their active guidance, help, cooperation and
encouragement, this project would have not been what it turned out to
be.
We wish to express our special thanks to our mentor Dr. Mohammad
Umair sir who in spite of being extremely busy, took out his precious
time and helped us achieve our target. His supervision and support is
extremely appreciated.
We specially thank Professor Gauhar Mahmood, H.O.D Civil
Engineering, Jamia Millia Islamia, New Delhi for his support and
incessant encouragement which will always be a source of inspiration
for us.

Project Mates :
Jamil Mahmood (14-BCE-0027)
Hammad Siddiqui (14-BCE-0040)
Md Husain (14-BCE-0020)
Shivaji Mishra (14-BCE-0036)
Md Badiuzzaman(14-BCE-0042)

Noida is a Commercial city and a part of the NCR that keeps
evolving each day.Construction of Corporations and
Residential Complexes in this region has been ever-green
over the past few years. The scenario herein also
resembles akin
The aim of this Project is to design the structural features of a
Multi-storeyed Building for residential purpose and analyse its
strength perspectives , taking earthquake load into
consideration.
Multi-storeyed Buildings are very common in cities .
Construction of such tall buildings are possible only by going to a
set of rigidly interconnected beams and columns.These rigidly
inter-connected beams and columns of multi bay and multi-
storeyes are called Building Frames .
To avoid lond distance of travel, cities are growing vertically
rather than horizontally . In other words , Multi-storeyed
Apartments are preferred in cities.
INTRODUCTION

AutoCAD Modelling
The following picture holds a view of the plan of the Residential
apartment in 2D perspective :

The figure below is a 3D-model depiction of the
apartmental subdivision under consideration:

Coverage area 2.2 acres
Height of each storey 3.50 meters
Number of storeyes 12 (G+11)
Staircase Tread 250 mm
Staircase Risers 150 mm
Height of parapet wall 0.90 meter
Number of elevator-sections 1
Some of the general features of the project-apartment are :
Rest of the detailed and applied structural
features are following in the forthcoming
sections…

DESIGN OF WATER TANK
Tanks, we know, are widely used for the
storage of liquids like water,
chemicals,petroleum etc. Tanks inculcated in
Residential buildings are generally circular or
rectangular in shape .
Circular Water
Tank
Rectangular Water
Tank

In the present context we shall be
designing for circular tanks because it is
advantageous over the rectangular tank
in the following aspects :

In the present context of Water Tank design , we shall be referring
IS 3370:2009 (Part 1) & IS 3370:2009 (Part 2).
The relevent IS Code recommendations regarding detailing in
water tanks:
The minimum re-inforcement in walls , floors and roofs in each of
two directions at right angles within each surface zone should not
be less than 0.35% of the cross-section of surface-zone for HYSD
bars and 0.64% for mild steel bars.
The minimum re-inforcement can be further reduced to 0.24% for
deformed bars and 0.40% for plain bars, for tanks, not having any
dimension more than 15 meter.
In tank walls and slabs , having thickness less than 200 mm, the re-
inforcement can be placed in one face only.

For ground/base slab , having thickness less than 300 mm, the
re-inforcement should be placed on one face , as near as
possible to the upper surface consistent with the cover.
The spacing of re-inforcing bars should not exceed 300 mm or
thickness of the section , which is less.
Size of bars , distance between bars , laps and bends should be
as per IS 456:2000 .
NOTE:
For D ⩾ 500 mm , i.e thickness of the member greater than or
equal to 500 mm , re-inforcement face controls half of the total
depth (D/2) of concrete.
For D< 500 mm , i.e thickness of the member less than 500 mm ,
each re-inforcement face controls 250 mm depth of concrete ,
ignoring any central core beyond the surface depth.

The general features the Circular Tank which include
are-
Flexible base of tank
Design capacity—100000 litre
Depth of water in the tank--- 5 meter
Grade of concrete--- M25
Grade of steel--- Fe 415
(considering the ϒw = 9.8 kN/m3)
Given , volume of tank = 100000 litre = 100 m3.
Height of water in tank (H) = 5 meter
Permissible tensile stress in steel = 130 N/mm2 (for
HYSD bars , IS 3370 Part 2).

Permissible direct tensile stress = 1.30 N/mm2 ( for M25
grade of concrete, IS 3370 Part 2).
If D is the diameter of the tank , then volume of tank =
𝜋
4
∗ 𝐷2
∗ 5 = 100
⇒ D=5.05 meter
Hence , providing a diameter of 5.10 meter.
Maximum hoop Tension (T)
T = ϒH*D/2
=9.8*5*5.1/2
= 124.95 kN per meter height of the wall
Area of steel
Ast =
𝑇
𝜎𝑠𝑡
So the Area of steel =
124.95∗1000
130
=962 mm2

Using 12 mm diameter bars ,
Spacing required =
113∗1000
962
= 117 mm
Hence , provide 12 mm diameter hoops (rings)
at 110 mm c/c.
∴ Ast,provided = 1027 mm2
At a distance of 2.50 meter from top T = 62.50
kN per meter and Ast,required = 481 mm2
, hence spacing can be doubled and should not
exceed the permissible vlaues (𝜎cr).
𝜎cr >
𝑇
1000. 𝑡+ 𝑚
−
1 ∗ 𝐴𝑠𝑡
>
124.95∗1000
1000
.
𝑡
+ 11−1 ∗
1027
⇒ t> 85 mm

Hence, providing a thickness of 100 mm for tank wall
Ast,min = 0.35 % of X-section area of surface zone
=0.35/100*(1000*100/2) (t<300 mm)
= 175 mm2 , <1027 mm2.
The spacing of hoops ≯ 300 mm or the thickness of the section.
Hence, OK.
∴ Providing 12 mm diameter hoops at 110 mm c/c spacing along
the height of the wall.
The spacing increased to 220 mm c/c at a distance 2.50 meter
from the top.
Distribution re-inforcement
Distribution and temperature steel is provided at 0.35% = 175
mm2.
Providing 8 mm diameter bars at 250 mm c/c spacing , vertical
steel =
Ast = 2*50*1000/250 = 200 mm2 , > 175 mm2
Hence, OK.

Design of Base/Floor slab
Since the tank floor is resting on the ground , the load
gets directly transferred to the soil. Hence , providing a
minimium thickness of 150 mm and 0.35% minimum
steel in each direction =
0.35/100(*1000*150/2) = 263 mm2
Hence, provide 8 mm diameter bars at 180 mm c/c in
both directions at top and bottom face of the flat slab.
The details of the re-inforcement is as shown :

Presentationt design and analysis of multistorey building

Introduction to Raft Foundation
If the loads transmitted by the columns in a structure are so
heavy or the allowable soil pressure is so small that
individual footings would either overlap or cover more
than that about one –half of the area, it may be better to
provide a continuous footing under all columns and walls.
Such a footing is called raft foundation or mat foundation.

• Raft foundations are used to reduce the
settlement of structures located above highly
compressible deposits.
• If the weight of the excavated soil is equal to the
weight of the structure and that of the raft and
the centers of gravity of excavation and structure
coincide , settlement should be negligible. Owing
to the fact rafts are usually at some depth in the
ground , a large volume of excavation may be
required.
Some more additive features of raft foundation
commonly employed in field works and site-
conditions are :

Model view prelude to Raft foundation beneath the
Noida Residential Complex
Theportalstructurebelowdepictsthe
footingorientationpriortotheconstruction
oftheapartment.It’sbasicallya13X7
columndistributionconsideringittobea
supposedmodeloftheload-carriage
capability.

Plate-mesh model undelying the
building….
(pertaining to stress distribution)
The diagram on the right depicts
the distribution and variation of
stress across the mat- plane of the
building

Final model of the raft as depicted
below

DESIGN OF BEAMS AND COLUMNS
A Beam is a structural element that primarily
resists loads applied laterally to the beam's
axis. Its mode of deflection is primarily
by bending. The loads applied to the beam
result in reaction forces at the beam's support
points.
Beams are characterized by their manner of
support, profile (shape of cross-section),
length, and their material.

Singly Reinforced Beam: The beam that is
longitudinally reinforced only in tension zone,
it is known as singly reinforced beam. In
such beams, the ultimate bending moment and the
tension due to bending are
carried by the reinforcement, while the compression
is carried by the concrete.
Doubly Reinforced Beam: The beam that is reinforced
with steel both in tension and compression zone, it is
known as doubly reinforced beam. This type of beam is
mainly provided when the depth of the beam is
restricted. If a beam with limited depth is reinforced on
the tension side only it might not have sufficient
resistance to oppose the bending moment.
SOME IMPORTANT TERMINOLOGIES FOR DESIGN
PARAMETERS:

Primary Beams :
The beams that are connecting columns for transferring loads of a structure
directly to the columns are known as primary beams. Usually, primary beams
are shear connected or simply supported and they are provided in a regular
building structure. The depth of the primary beams is always greater than
secondary beams. Primary beam act as a medium between columns and
secondary beams.
Secondary Beam :
The beams that are connecting primary beams for transferring
loads of a structure to the primary beams are known as primary
beams. These beams are provided for supporting and reducing the
deflection of beams and slabs.

Specimen Design of a typical
Guest Hall Beam

The design features are –
For Beam 1 & 3 ; the load is in triangular distribution
For Beam 2 & 4, the load is in trapezoidal distribution
(because it is two way slab = 6/6 = 1, < 2).
Load on beam by triangular distribution = wLx/3
(Lx = shorter dimension of slab
w = Total load of slab per meter length)
Load on beam by Trapezoidal Distribution = wLx/2 * (1-
1
2𝛽
)
Where β = Ly/Lx
If Ly/Lx ⩾ 2 ⇒ One Way Slab
If Ly/Lx< 2 ⇒Two Way Slab

Triangular and Rectangular load
transfer distribution

So dimension and loading of beam , slab
Thickness of slab = 120 mm
Live load of slab = 3.75 kN/m2
Floor finish load = 1kN/m2
Self weight of slab = 0.12 * 25 = 3 kN/m2
∴ Total load of slab = 7.75 kN/m2
⇒Factored load = 1.5*
7.75 kN/m2 = 11.63 kN/m2
For Beam number 1
Load = Triangular load + self weight
=wLx/3 + 0.3*0.6*25*1.5
=27.76 kN/m
But slab is continuous on the other side also , so 2*(23.26
kN/m) + 4.5 kN/m
= 51 kN/m

For Beam no. 2
Load on Beam = Trapezoidal load + self weight
= wLx/2*(1-
1
2𝛽
) + 0.3*0.6*25
= 17.45 + 4.5 kN/m
= 21.95 kN/m
But other (right) also the slab is continued.
So, there will also be a load transfer from that side
= 2*17.45 + 4.5 kN/m = 40 kN/m (approx.)
Same as for beam 3, load = 21.95 kN/m
For beam 4, load = 27.76 kN/m
Step 2: Calculate design moment and design shear
Design Moment = Mu=w*l2/8 = 230 kN.m
Design shear force Vu = wl/2 = 51.02 * 6/2 = 154
kN/m

Step 3 : Calculate Limiting MOR for singly reinf. Beam section
For Fe 415 , Mu,lim = 0.138fckbd2
Let the effective cover be 50 mm,
Effective depth of beam = 600-50= 550 mm.
Mu,lim =0.138 *30*300*5502 = 375.705 kN,> 230 kN.m
Since , the design moment is 230 kN.m and Mu,lim is greater than
the design moment in singly r/f , that is why we change the size of
the beam because it is uneconomical.
So we provide Doubly r/f in high rise buildings because they
would impart greater degree of safety.
New Dimension design perspective
Beam Cross section = 250 * 500 mm2
Load on beam 1 = Imposed load + Self wt = 51.2 kN/m
Calculation
Step 2 : Calculate design moment and design shear
Design Moment Mu = wl2/8 =230 kN.m
Design shear force = Vu = wl/2 = 153.6 kN

Step 3 : Calculate limiting MOR for singly r/f beam section
Let , the effective cover be = 60 mm
Effective depth of beam = 500-60 mm = 440 mm
For Fe 415 , Mu,lim =0.138* 250* 4402*30
=200 kN.m
Mu,lim = 200 kN.m < 230 kN.m
Thus, doubly re-inforced beam section is required .
Step-4 : Calculate the amount of Compression re-inforcement
reuired
ΔMu = Mu - Mlim
= 230 -200 = 30 kN.m
fsc = (0.0035*(0.48 *500 -60)/(.48*500))*2*105=525 N/mm2
fsc ≯ 0.87fy
(= 0.87*415=361.05 N/mm2 )
Asc = (Mu – Mu,lim)/( fsc*(d-d’))= 189 mm2
Provide 2 bars of 20 mm dia , Asc (provided) = 2*π/4*202 =
628.31 mm2 , > 189 mm2 (OK).

Step 5 : Calculate the amount of Tension reinf.
Ast,2 = Area of tension steel to balance Asc
= fsc * Asc / 0.87fy = 189 mm2
Ast,1 = Ast,lim
Mu,lim = 1794.76 mm2
Ast = Ast,1 + Ast,2
=1983.77 mm2
Provide 20 mm dia bars , 1983.77/(π/4 * 202) = 7(approx)
Step-6 : Design of shear r/f
Nominal shear stress = τv =Vu/bd = 1.23 N/mm2
Percent of steel r/f = 1983.77*100/(250*500) = 1.58 %
Now, τc =0.773 N/mm2
Here, τc < τv and τc,max = 3.5 N/mm2
Shear re-inforcement = Vus = 57.5 kN
Using 2- legeged 8 mm dia stirrups Asv = 2*3.14/4*82 =100.53 mm2
Sv = 315.62 mm
Maximum spacing = 330 mm (as per IS 456)
Therefore , provide 2- legged 8 mm dia , stirrups at 30 mm c/c spacing near the
support and spacing can be increased gradually towards the mid span.
Check : Min shear =0.87 fy*Asv/(0.4*b)

Step 7:
Deflection control
(l/d)actual = 6000/500 = 12
(l/d)max =(l/d)base *kt*kc
30*0.9*1.1 =29.7 > 12 (OK)
Step 8:
Development length=0,87*fy*Ø/(4*1.5*1.6)= 752 mm
Anchorage length
For 90 degree , anchorage length (Lc) = 8*20 = 160 mm
Check : 1.3 M/V + L = 56.02 mm
So there is no need to increase the anchorage length.

So we have the sketch of the sample beam specimen below
:

DESIGN OF COLUMN
Column is basically a compression member.
If slenderness ratio less than 12, it is short column
If greater than 12 it is long column (or Slender
column)
Slenderness is the ratio between effective length to
least radius of gyration.
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.

Now let’s have a view on sample design of a typical column in a
dining Room in the Noida Residential Apartment.
Design features:
Load acting axially : 1364 kN (as calculated earlier)
Clear storey height : 3.50 m
Cross section : 700 mm* 700 mm
Since we are having framed structure , so effective length = 0.65*L
=0.65*3500 =2275 mm = 2.275 mm
Now , fck = 30 N/mm2 & fy = 415 N/mm2
Step 1 : Checking the column as short or long
Lex = Ley = 2275 mm
Dx = Dy = 700 mm
Slenderness ratio along x-direction (λr)= Lex/ Dx
Now, λx = λy =2275/700=3.25 <12
So it is short column

Step 2: Calculation of minimum eccentricities
So we have ex,min = lex/500 + Dx/30 = 30.33
& ey,min = ley/500 + Dy/30 = 30.33
Now, 0.05* Dx =0.05*700 = 35
Since the minimum eccentricity are less than 0.05 times the lateral
dimension in both the directions , the following formula given by IS
456-2000 can be used for the design of axially loaded short
columns.
Step 3: Use of column design formula
Pu = 0.4*fck*Ag + (0.67*fy -0.4*fck)*Asc
Now, axial load (P)= 1364 kN
Factored axial load = 1364 kN*1.5 =2046 kN
So we have , Pu = 0.4*fck*Ag + (0.67*fy -0.4*fck)*Asc
Imposing the respective values , we would have
1.5*1364*103 = 0.4*30*700*700+(0.67*415-0.4*30)* Asc
⇒ -451600 = (0.67*415-0.4*30)* Asc
So we see it has become uneconomical.

R/f comes out to be negative
∴We have modied the dimension to about 400*400 mm2.
Minimum eccentricity = le/500 + Dx/30 = 20.33
Again ex,min = ey,min = 20.33 ≃20
So design for axially loaded columns.
Step 3 : Evaluation of Asc
Pu =0.4*fck*Ag + (0.67*fy -0.4*fck)*Asc
Now, axial load = 1364 kN
Factored Axial load = 2046 kN (as calculated earlier)
So again imposing the requisite values in the above equation , we
would have
Asc = 473.59 mm2
Step 4: Re-inforcement design
Provide 4 bars of 20 mm dia
Asc,provided = 4*π/4*202 = 1256.63 > 473 mm2
Hence OK

Pt of r/f provided = (Asc/bd)*100 =0.8%
It seems to fulfill the minimum steel requirement in the column ,
yet again let us provide steel content of 6 bars.
Pt =
6∗
𝜋
4
∗20^2
400∗400
∗ 100%
= 1.17% ,<6%
Hence OK
Step 4 : Lateral ties
Dia of the tie = (should be greater than either of Ømain/4 or 6 mm)
=( should be greater than either of 20/4 (=5)or 6 mm )
Adopt tie bar dia = 8 mm
Spacing = (should be less than or equal to least lateral dimension
of column = 400 mm, or , 16*20 = 320 mm, or 300 mm)
So adopt spacing of 200 mm c/c

So we have a sketch of the sample column specimen below
:

DESIGN OF SLABS
Slabs are plate-elements forming floors and roofs of buildings
& carrying distributed loads primarily by flexure.
Now depending on the type of support , for
calculation purpose IS 456-2000 recommends the
following :

Let’s have a prelude to the types of slabs as per
their geometrical configuration —one-way slabs &
two-way slabs .

As sample perception pertaining to our design of slab
, let’s analyse for our Bedroom having a span of 5
m*6.114 m.
So in accordance with our IS 456-2000 design features we have
Lx= 5000 mm (5m) [length of shorter span]
Ly = 6114.4 (6.1144 m) [length of longer span]
Let’s design a slab for a simply supported case ,for a give slab
𝐿𝑜𝑛𝑔𝑒𝑟 𝑠𝑝𝑎𝑛
𝑆ℎ𝑜𝑟𝑡𝑒𝑟 𝑠𝑝𝑎𝑛
=
6114.4 𝑚𝑚
5000 𝑚𝑚
= 1.23
⇒ Ly/Lx= 1.23 , < 2⇒Two way slab
So accordingly we would have two conditions for two
way slab :
• Slab corners are helds down
• Slab corners are not held down

Except roof-slab , other slab corners are held down.
So we have simply supported , hence corner is not held down.
𝑆𝑝𝑎𝑛
𝑑
= 20
⇒ d = span /20 = 5000/ 20 =250 mm
Let us assume D = 250 mm and clear cover = 40 mm and dia of bar =
10 mm .
Now, d= (250-40 -10/2)mm = 205 mm
Length of effective span
Minimum of Leffective,x = Lex + d = 5 + 0.205 = 5.205 m…..(1)
= Lex + w/2 +w/2 = 5+ 2*0.23/2 = 5.23
m…..(2)
Minimum of (1) & (2) = 5.205 m would be taken.
Leff,y = 6.114 + 0.205 m= 6.355 m
Now,
𝐿 𝑒𝑓𝑓 𝑦
𝐿[𝑒𝑓𝑓 𝑥 ]
= 6.355/5.205 = 1.22 < 2 ⇒ Hence two way slab

Bending moment of the beam
Mx = αx *w*(lx)2
My = αy *w*(lx)2
At Ly/Lx = 1.2 , αx = 0.084, αy =0.059
At Ly/Lx = 1.3 , , αx = 0.093, αy =0.055
Via interpolation , we , αx = 0.086, αy =0.058
Calculation of load
Dead load of slab = 0.25*25*1=6.25 kN/m
Live load is as per IS 875 part 2 = 2kN/m2
Live load = 2*1 kN/m =2 kN/m
Total factored load = 1.5*(6.25 + 2) = 12.375 kN/m
Calculation of moment Mx =28.83 kN.m
My =19.44 kN.m
Effective depth of slab =
BM = 0.138 𝜎ck bd2
⇒d =
28.83∗10^6
0.138∗25∗1000
= 96.34 mm
Assume effective depth = 100 mm

D = 100 + 40 +10/2 = 145 mm
Area of steel along shorter span :
0.36∗σck
∗
b
∗
x
(
m
)
0.87∗σy
= 1196.51 mm2
Number of bars = Ast,x / (3.14/4)*d2 = 16
Spacing = 1000/16 = 60 mm(appx)
Area of tension steel along long span
BM = Force of tension * Lever arm ⇒19.44 *10^6 =0.87* 𝜎
y*Aty*(d’-
𝜎𝑦∗𝐴𝑡𝑦
𝜎𝑐𝑘∗𝑏
)
⇒Aty = 684.73 mm2
Number of bars = 14
Spacing of bars =1000/14 = 75 mm(appx)
The Code requirement for r/f , the minimum area o f steel should
be 0.12 %
= 0.12 % *B*D = 174 mm2 < Atx as well as< Aty (Hence OK).

Check for Shear
Max shear force intensity in either direction =wlx/2 =
32.205 kN
Nominal shear=τv =Vu/db =0.358 N/mm2
% of tensile r/f , = 100* Ast *0.5/bd = 0.62%
From IS 456 , τc =0.33 N/mm2 & τc,max = 0.63
For D = 145 mm
K = 1.3 , τc’ = k* τc = 0.42 N/mm2> τv
Hence OK , safe in shear .
Check of Development Length
MOR of 10 mm dia bars at spacing 75 mm c/c
M1 = 0.87* 𝜎y*Aty*(d’-
𝜎𝑦∗𝐴𝑡𝑦
𝜎𝑐𝑘∗𝑏
) = 14.5 kNm
V = SFmax = 32.205 kN

Anchorage values of bars bent at 90 degree including 60
mm starting length
L0 =60 +8Ø = 124 mm
Development length (Ld) = 0.87 𝜎y*Ø/4τbd =322.367
mm
As per IS code (Clause 26.2.3.3 )
Ld <1.3* M1/v + L0 = 725.31 mm (OK)
Check for deflection
L/d = 5000/145 = 34.48 ≮ 0.8*35
So there is increase in effective depth , d= 125 mm
& D =125 + 40 +10/2 = 170 mm
Now , L/d = 28.1 <0.8*35
Check for cracking
We know that minimum r/f = 0.12 % *bd (Cl 26.5.2.1 0f IS 456 -2000)
Provided steel = 0.62 %
Spacing of bars = 75 mm < either of (3d=375 mm) or 300 mm (OK)
Spacing of bars = 60 mm < either of (3d=375 mm) or 300 mm (OK)

Dia of bars
10 mm dia bars in tranverse direction < 1/8 * d = 21.25 mm
(OK)
8 mm dia of bars employed in the y-direction as illustrated
in the relation.
8 mm < 1/8 *d = 21.25 mm(OK)
Design is OK to control cracking.
The overal specimen design for the slab is held up below :

The 4-noded slab illustration of a typical floor (here first floor
sample example model view has only been demonstrated.) :

Seismic analysis of multistorey building
using Staad pro
Annexure 1: Code used
CODE OF EARTHQUAKE USED FOR
EXPERIMENTS AND TABLES USED
The first formal seismic code in India,
namely IS 1893, was published in 1962.
Today, the
Bureau of Indian Standards (BIS) has the
following seismic codes:
IS 1893 (Part I), 2002
IS 13920, 1993

TERMINOLOGY USED IN SEISMIC
ANALYSIS
• For the purpose of this standard, the following definitions shall apply which
are applicable generally
• to all structures.]
• {NOTE — For the definitions of terms pertaining to soil mechanics and soil
dynamics references
• may be made to IS 2809 and IS 2810}.
• Closely-Spaced Modes
• Closely-spaced modes of a structure are those of its natural modes of vibration
whose natural
• frequencies differ from each other by 10 percent or less of the lower
frequency.
• Critical Damping
• The damping beyond which the free vibration motion will not be oscillatory.
• Damping
• The effect of internal friction, imperfect elasticity of material, slipping, sliding,
etc in reducing the
• amplitude of vibration and is expressed as a percentage of critical damping..

SHEAR AND BENDING OF COLUMN
COLUMN NO 325
Bending and Shear Results
Bending about Z for Beam 325
Load Case: 1:EQX
Dist.m Fy(kN) Mz(kNm)
0.000000 0.0000 -0.0000
0.291667 0.0000 -0.0000
0.583333 0.0000 -0.0000
0.875000 0.0000 -0.0000
1.166667 0.0000 -0.0000
1.458333 0.0000 -0.0000
1.750000 0.0000 -0.0000
2.041667 0.0000 -0.0000
2.333333 0.0000 -0.0000
2.625000 0.0000 -0.0000
2.916667 0.0000 -0.0000
3.208333 0.0000 -0.0000
3.500000 0.0000 -0.0000

LOAD CASES TAKEN IN MULTI-
STOREY BUILDING
LOAD CASES
Number Name Type
1 EQX Seismic
2 EQZ Seismic
3 DEAD Dead
4 LIVE Live

Sectional property of column no 373

MATERIALS DETAILING OF BUILDING
MATERIALS
Mat Name E
(kN/mm2
)
Density
(kg/m3
) (/°C)
1 M30 0.027 0.150 2.45E+3 0.000
2 STEEL 205.000 0.300 7.83E+3 12E -6
3 STAINLESSSTEEL 197.930 0.300 7.83E+3 18E -6
4 M25 25.000 0.200 2.45E+3 0.000
5 ALUMINUM 68.948 0.330 2.71E+3 23E -6
6 CONCRETE 21.718 0.170 2.41E+3 5E -6

Deflection in column due to EQ load

Shear and bending in beam due to
dead load

Shear and bending due to earthquake
load

JOB SUMMARY
Number of Nodes 336 Highest Node 362
Number of Elements 975 Highest Beam 1034
Number of Basic Load
Cases
-2
Number of Combination
Load Cases
13
Type L/C
Primary 1 EQX
Primary 2 EQZ
Primary 3 DEAD
Primary 4 LIVE
Combination 5 1.5(DL+LL)
Combination 6 1.2(DL+LL+EQX)
Combination 7 1.2(DL+LL+EQZ)
Combination 8 1.5(DL+EQX)
Combination 9 1.5(DL+EQZ)
Combination 10 .9DL+1.5EQX
Combination 11 .9DL+1.5EQZ
Combination 12 1.2(DL+LL-EQX)
Combination 13 1.2(DL+LL-EQZ)
Combination 14 1.5(DL-EQX)
Combination 15 1.5(DL-EQZ)
Combination 16 .9DL-1.5EQX
Combination 17 .9DL-1.5EQZ
LOAD COMBINATION
FOR WHOLE
STRUCTURE

MORE ON STOREY SHEAR
REQD. STEEL AREA :
REQD. CONCRETEAREA:
6664.00 Sq.mm.
483336.00 Sq.mm.
MAIN REINFORCEMENT : Provide 60 - 12 d i a . (1.38%, 6785.84 Sq.mm.)
(Equally di st ri but ed)
TIE REINFORCEMENT : Provide 8 mmd i a . rectangular t i e s @ 190 mmc/ c
SECTION CAPACITY BASED ONREINFORCEMENT REQUIRED (KNS-MET)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Puz :
7936.53 Muz1 : 319.50 Muy1 : 319.50
INTERACTION RATIO: 0.84 (as per Cl. 39.6, IS456:2000)
SECTION CAPACITY BASED ONREINFORCEMENT PROVIDED (KNS-MET)
---------------------------------------------------------- WORST LOAD
CASE: 5
END JOINT: 2 Puz : 7980.85 Muz : 334.12 Muy : 334.12 IR: 0.77

BASE SHEAR CALCULATION
* TIME PERIOD FOR Z=0.24 AS PER
IS 1893: 2 0 0 2 i s =1.28130SEC
*SA/G PER
1893=1.061(MEDIUM TYPE
SOIL) IMPORTANCE FACTOR=
1.000
*BASE SHEAR AS PER IS 1893=
.0255*
106640.18=2719.3246KN
NOTE : THE BASE SHEAR (VB) FROM RESPONSE
SPECTRUM IS LESS THAN THE BASE SHEAR (Vb)
CALCULATED USING EMPIRICAL FORMULA FOR
FUNDAMENTAL TIME PERIOD. MULTIPLYING
FACTOR (Vb/VB) IS 16.6017

ANALYSIS TAKES PLACE IN STADD
DESIGN
350. START CONCRETE DESIGN
351. CODE INDIAN
352. FYSEC 500000 MEMB1 TO 368 371 373 TO 399 401 TO 403 406 407 409 410 413 417 -
353. 418 420 425 426 429 430 432 TO 435 437 438 442 443 446 449 450 452 TO 454 -
354. 457 459 TO 463 465 TO 469 471 TO 498 500 TO 504 506 TO 509 513 514 516 518 -
355. 519 521 523 525 527 TO 531 533 TO 535 537 539 543 548 TO 550 553 TO 555 558 -
356. 559 TO 560 562 TO 564 566 TO 568 570 TO 1034
357. FYMAIN 500000 MEMB1 TO 368 371 373 TO 399 401 TO 403 406 407 409 410 413 -
358. 417 418 420 425 426 429 430 432 TO 435 437 438 442 443 446 449 450 -
359. 452 TO 454 457 459 TO 463 465 TO 469 471 TO 498 500 TO 504 506 TO 509 513 -
360. 514 516 518 519 521 523 525 527 TO 531 533 TO 535 537 539 543 548 TO 550 -
361. 553 TO 555 558 TO 560 562 TO 564 566 TO 568 570 TO 1034
362. FC 25000 MEMB 1 TO 368 371 373 TO 399 401 TO 403 406 407 409 410 413 417 418 -
363. 420 425 426 429 430 432 TO 435 437 438 442 443 446 449 450 452 TO 454 457 -
364. 459 TO 463 465 TO 469 471 TO 498 500 TO 504 506 TO 509 513 514 516 518 519 -
365. 521 523 525 527 TO 531 533 TO 535 537 539 543 548 TO 550 553 TO 555 -
366. 558 TO 560 562 TO 564 566 TO 568 570 TO 1034
367. DESIGN BEAM1 TO 76 101 TO 138 163 TO 200 225 TO 262 287 TO 324 397 TO 399 -
368. 401 TO 403 406 407 409 410 413 417 418 420 425 426 429 430 432 TO 435 437 -
369. 438 442 443 446 449 450 452 TO 454 457 459 TO 463 465 TO 469 471 472 497 -
370. 498 500 TO 504 506 TO 509 513 514 516 518 519 521 523 525 527 TO 531 533 -
371. 534 TO 535 537 539 543 548 TO 550 553 TO 555 558 TO 560 562 TO 564 -
372. 566 TO 568 570 TO 572 597 TO 672 697 TO 772 797 TO 872 897 TO 972 -
373. 997 TO 1034

PROGRAMMING OF STADDSC:SProV8i SS5STAADPluginsStructurehammad.std 0 5 /3 0 /1 8 1 7 :3 6 :5 7 DENSITY 24
ISOTROPIC CONCRETE E
2.17184e+007 POISSON 0 . 1 7
DENSITY 23.6158 ALPHA5e-
006 DAMP0 . 0 5
TYPE CONCRETE STRENGTH FCU 27578.9
ISOTROPIC M30
E 27386.1
POISSON 0 . 1 5
DENSITY 24
ENDDEFINE MATERIAL
***************************************** MEMBER
PROPERTY AMERICAN
77 TO92 139 TO162 201 TO224 263 TO286 325 TO396 473 TO496 573 TO596 -
673 TO696 773 TO796 873 TO896 973 TO996 PRIS YD0 . 7 ZD0 . 7 MEMBERPROPERTY
AMERICAN
1 TO 76 101 TO 138 163 TO 200 225 TO 262 287 TO 324 397 TO 399 401 TO 403 -
406 407 409 410 413 417 418 420 425 426 429 430 432 TO 435 437 438 442 443 -
446 449 450 452 TO 454 457 459 TO 463 465 TO 469 471 472 497 498 500 TO 504 -
506 TO 509 513 514 516 518 519 521 523 525 527 TO 531 533 TO 535 537 539 -
543 548 TO 550 553 TO 555 558 TO 560 562 TO 564 566 TO 568 570 TO 572 597 -
598 TO672 697 TO772 797 TO872 897 TO972 997 TO1034 PRIS YD0 . 6 ZD0 . 3 MEMBERPROPERTY
AMERICAN
93 TO100 PRIS YD0 . 8 ZD0 . 8 CONSTANTS
MATERIAL M30 ALL
****************************************** SUPPORTS
170 TO174 176 TO194 FIXED
******************************************** DEFINE 1893
LOAD
ZONE0 . 2 4 RF 5 I 1 SS 2 ST 1 DM0 . 0 5
SELFWEIGHT 1 FLOOR
WEIGHT
YRANGE0 1 7 . 5 FLOAD6 MEMBERWEIGHT
1 TO 76 101 TO 138 163 TO 200 225 TO 262 287 TO 324 397 TO 399 401 TO 403 -
406 407 409 410 413 417 418 420 425 426 429 430 432 TO 435 437 438 442 443 -
446 449 450 452 TO 454 457 459 TO 463 465 TO 469 471 472 497 498 500 TO 504 -
506 TO 509 513 514 516 518 519 521 523 525 527 TO 531 533 TO 535 537 539 -
543 548 TO 550 553 TO 555 558 TO 560 562 TO 564 566 TO 568 570 TO 572 597 -
598 TO 672 697 TO 772 797 TO 872 897 TO 972 997 TO 1034 UNI 20
******************************************* CUTOFF MODE
SHAPE 30
LOAD 1 LOADTYPE Seismic TITLE EQX SELFWEIGHT X
1
SELFWEIGHT Z 1 FLOOR LOAD
YRANGE0 42 FLOAD6 GX
YRANGE0 42 FLOAD6 GZ MEMBERLOAD

MORE ON PROGRAMMING OF STADD
3 1 . 5 4 1 . 5
LOADCOMB6 1.2(DL+LL+EQX) 2 1 . 2 3 1 . 2 1 1 . 2
LOADCOMB7 1.2(DL+LL+EQZ) 3 1 . 2 2 1 . 2 4 1 . 2
LOADCOMB8 1.5(DL+EQX) 3 1 . 5 1 1 . 5
LOADCOMB9 1.5(DL+EQZ) 3 1 . 5 4 1 . 5
LOADCOMB10 .9DL+1.5EQX 3 0 . 9 1 1 . 5
LOADCOMB11 .9DL+1.5EQZ 3 0 . 9 4 1 . 5
LOADCOMB12 1.2(DL+LL-EQX) 2 1 . 2 3 1 . 2 1 1 . 2
LOADCOMB13 1.2(DL+LL-EQZ) 3 1 . 2 2 1 . 2 4 1 . 2
LOADCOMB14 1.5(DL-EQX) 3 1 . 5 1 1 . 5
LOADCOMB15 1.5(DL-EQZ) 3 1 . 5 4 1 . 5
LOADCOMB16 .9DL-1.5EQX 3 0 . 9 1 1 . 5
LOADCOMB17 .9DL-1.5EQZ 3 0 . 9 4 1 . 5
PERFORMANALYSIS START CONCRETE DESIGN CODE INDIAN
FYSEC 500000 MEMB1 TO 368 371 373 TO 399 401 TO 403 406 407 409 410 413 417 -
418 420 425 426 429 430 432 TO 435 437 438 442 443 446 449 450 452 TO 454 -
457 459 TO 463 465 TO 469 471 TO 498 500 TO 504 506 TO 509 513 514 516 518 -
519 521 523 525 527 TO 531 533 TO 535 537 539 543 548 TO 550 553 TO 555 558 -
559 TO560 562 TO564 566 TO568 570 TO1034
FYMAIN 500000 MEMB1 TO 368 371 373 TO 399 401 TO 403 406 407 409 410 413 -
417 418 420 425 426 429 430 432 TO 435 437 438 442 443 446 449 450 -
452 TO 454 457 459 TO 463 465 TO 469 471 TO 498 500 TO 504 506 TO 509 513 -
514 516 518 519 521 523 525 527 TO 531 533 TO 535 537 539 543 548 TO 550 -
553 TO 555 558 TO 560 562 TO 564 566 TO 568 570 TO 1034
FC 25000 MEMB1 TO 368 371 373 TO 399 401 TO 403 406 407 409 410 413 417 418 -
420 425 426 429 430 432 TO 435 437 438 442 443 446 449 450 452 TO 454 457 -
459 TO 463 465 TO 469 471 TO 498 500 TO 504 506 TO 509 513 514 516 518 519 -
521 523 525 527 TO 531 533 TO 535 537 539 543 548 TO 550 553 TO 555 -
558 TO560 562 TO564 566 TO568 570 TO1034
DESIGN BEAM1 TO76 101 TO138 163 TO200 225 TO262 287 TO324 397 TO399 -
401 TO 403 406 407 409 410 413 417 418 420 425 426 429 430 432 TO 435 437 -
438 442 443 446 449 450 452 TO 454 457 459 TO 463 465 TO 469 471 472 497 -
498 500 TO 504 506 TO 509 513 514 516 518 519 521 523 525 527 TO 531 533 -
534 TO 535 537 539 543 548 TO 550 553 TO 555 558 TO 560 562 TO 564 -
566 TO 568 570 TO 572 597 TO 672 697 TO 772 797 TO 872 897 TO 972 -
997 TO 1034
DESIGN COLUMN77 TO100 139 TO162 201 TO224 263 TO286 325 TO368 371 373 -
374 TO396 473 TO496 573 TO596 673 TO696 773 TO796 873 TO896 973 TO996 ENDCONCRETE DESIGN
PERFORMANALYSIS FINISH

CONCLUSION FROM SEISMIC ANALYSIS
•Seismic analysis is done in order to protect multi-
storyed building from great havoc.
•So on doing analysis we got base shear as
2719.4346kN and fundamental natural period as
1.28sec .
•As our building is RC framed building and according
to codal formula T=0.075Hpow(0.75),we get(H=42m)
as 1.24sec.
•Hence we have analysed the building’s fundamental
time period to the critical side and building can resist
the seismic destruction to some extent.

Analysis of Beam no 999
• As uppermost part of
the building is prone to
high deflection so beam
n0 999 is more critical
thing to analyze.

Assigning material to the members
Assigning concrete to beam Assigning concrete to column

Frequencies for different modes

Check to horizontal deflection
• Maxm horizontal sway of
the building can be seen
as 156.29mm .
• But IS 1893-2002 says
that maximum lateral
displacement should not
exceed 0.004 times of
storey height .Here it
comes out to be
140mm.So this is unsafe
hence lateral bracing is
required.

Maximum & minimum reactions at
different nodes

Analysis of structure as per dead load

Analysis of structure as per Earthquake load

Graphical analysis of storey shear
• Here we can say clearly
that due to Earthquake
load in x direction the
base shear increases
but horizontal sway
decreases
• Static analysis gives the
same result .

Comparison of nodes
Node 170 (Lowest node) Node 339 (Highest node)

Check of vertical deflection in beam 999
• Maximum Vertical
deflection in beam is
L/250 for continious
beam .It comes out to
be 259.007 mm under
the dead load which is
more than 20 mm.
• The higher deflection
may be controlled by
prestresing and camber
provision

CONCLUSION (OVERALL PROJECT)
From the above Project analysis , we have come to know about the
pragmatic application of structural designs in field conditions and
their impact on strength, economy , durability and sustainability of a
Multi-storeyed Residential Apartment.
It is important to inculcate that though the design implementations
are being ameliorated day after day ,the current designs employed
herein in this Project have been carried as per the latest IS Code
provisions . Besides, the incorporation of Seismic design and analysis
in modern Residential buildings has always been a matter of great
envisage and prudence .
The Major Project—Design of Multi-storeyed Building & its Strength
Analysis-- 2018-- was a great platform to correlate the theoretically
gained ideas of Structural Designs to that of on-site constructional
works keeping in view the satisfaction of both the Client & the
Contractor in toto.
Eventually, the project-analysis helped serve a great deal of
knowledge intake and outplay to pour out our cognizance of B Tech
Civil engineering Graduates.

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