group 1212b
DESCRIPTION
tall building designTRANSCRIPT
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Analysis and Design of S Shaped
Commercial Office Tower
Under the Guidance:
Dr. Nirjhar Dhang
Group S (M.tech)
1. Somraj Biswas (14CE65R01) 2. Saudip Sahu (14CE65R13) 3. Zankar Sanket D. (14CE65R29) 4. Sachin Y. Rangari (14CE65R31)
High Rise Structures
Department of Civil Engineering Indian Institute of Technology, Kharagpur
Analysis and Design of S Shaped Commercial Office Tower
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Contents
Chapter 1. Drawings
Chapter 2. Area Statement
Chapter 3. Calculation of Column and beam cross
section Chapter 4. Earthquake Load Calculation
Chapter 5. Analysis of the Frame and checks
Chapter 6. Design of Beams and Columns
Analysis and Design of S Shaped Commercial Office Tower
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Chapter 1. Drawings
General Layout
Analysis and Design of S Shaped Commercial Office Tower
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Column Spacing
Floor Plan
Analysis and Design of S Shaped Commercial Office Tower
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Chapter 2. Area Statement
Area Calculation
Foot Print Area =90*60-2*0.5*20*10-0.5*20*20-2*10*20=4600 sq.
Carpet Area at Each Floor =footprint area 829.5m2=3770.5m2 Total carpet area req. = 150000 sq.m
Total Area for Lifts No. of lifts provided=20 Total Area required for lifts=20*(2*5)=200Sq.m
Total column area Total No. of columns provided=62 Area occupied by columns=62*2*2=248sq.m
Area of staircases
No. of staircases provided=10
Total area of staircases=10*2*5=100Sq.m
Washroom area No of washrooms=5
Total area of all washrooms=5*4*5=100Sq.m
Wall Area
Thickness of outer wall=0.25m
Thickness of inner wall=0.2m
Total area occupied by walls= 181.5Sq.m
Specifications
Floor Carpet Area = 3770.5m2
Carpet Area= 150000 m2
Structural form adopted - Rigid Frame Footprint Area= 4600sq.m (
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Chapter 3. Calculation of Column and beam cross section
Load Calculations for typical floors
Dead Load Self wt. of slab=10.00 KN/m Floor Finish=4KN/m
Self Weight of beam=5.25KN/m
Load From Walls=19 KN/m
Live Load Let us assume, L.L=10KN/m Total Factored UDL=1.5*(DL + LL) = 1.5*(38.25 +10) =72.375 KN
Column positions & Load Calculation
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Load Acting on Critical Frame
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Load On Columns
Assuming the columns to be square columns of size 2m
Column load per floor=22254
=400KN
Load on C1(outer column)=(3(wl/2) +400) no of storeys
=(3289.5+400) 40
=50740KN
Load on C2(inner column)=(4(wl/2)+400) no of storeys
=(4289.5+400) 40
=62320KN
Dimensions of Columns OUTER COLUMN
Pu=0.4fck0.96Ag+0.67fy0.04Ag
Here the reinforcement considered is 4% of the gross area
5074010^3=0.4350.96Ag +0.674150.04Ag
Ag=2065792.69mm2
Since we have considered the column to be square
Size of the column=1.431.43m
INNER COLUMN
Pu=0.4fck0.96Ag+0.67fy0.04Ag
Here the reinforcement considered is 4% of the gross area
6232010^3=0.4350.96Ag +0.674150.04Ag
Ag=2537252.67mm2
Since we have considered the column to be square
Size of the column=1.61.6m
Column size considered in the design is taken to be 2m which is safe
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REINFORCEMENT
OUTER COLUMN
Ast1 = 0.042065793=82631.68 mm2
No of 32mm reinforcement =102.74=103
INNER COLUMN
Ast2 = 0.04 2537252 = 101490 mm2
No of 32mm reinforcement =126.23=127
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Chapter 4. Earthquake Load Calculation
Earthquake Load Calculation Given : Building Located in Kolkata Zone III Z=0.16 Table 2 of IS: 1893
I=1.5 Table 6 of IS:1893
R=5.0 Table 7 of IS: 1893 Part 1
Floor Area =4600 sq.m Live Load=10KN/m
Wt. of individual Floors W1=W2W39=4600*(9.69*5*0.5)
=56093.2 KN
Wt. of top Floor W40=4600*(5.521)=25397.46 KN
Total Weight =56093.2*39+25397=2213024KN
Fundamental period
T = 0.09h / sqrt(d) Clause 7.6.2. of IS: 1893 Part 1
Along X direction=0.09*207/sqrt(48)=2.689 seconds Sa/g=0.505
Along Y direction=0.09*207/sqrt(120)=1.7 seconds Sa/g=0.8
Design horizontal Seismic coefficient Ah Along X direction=0.01216 Along Y direction=0.0192
IN X DIRECTION
T(in X direction)=0.09h/d
=0.09160/60
=1.85 sec
Assuming the building to be constructed in zone 2
Z=0.1 I=1.5 R=5
From the graph we can find Sa/g =0.8
Ah=(ZI/2R) Sa/g=0.012
So design base shear in X direction =AhW
=26556.3 KN
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IN Y DIRECTION
T(in Y direction)=0.09h/d
=0.09160/90
=1.518 sec
Assuming the building to be constructed in zone 2
Z=0.1 I=1.5 R=5
From the graph we can find Sa/g =0.7
Ah=(ZI/2R) Sa/g=0.0105
So design base shear in Y direction =AhW
=23237 KN
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Chapter 5. Analysis of the Frame and checks G+40 storey
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Chapter 6. Design of Beams and Columns
Design of Composite Beams:
Given Data: The Bending Moment and Shear force was obtained from OpenSees:
Maximum Bending Moment ( M ) = 1123.23 KN-m Maximum Shear Force ( V ) = 578.101 KN
Span = 8 m
beff = span/4 = 2.00 m
Assumed Data: The Composite Beam Properties are:
Grade of Concrete ( fck ) = M 30 Grade of Steel ( Fy ) = Fe 250
Steel Section Used = ISMB 500
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Position of Plastic Neutral Axis: Ds=150mm
Aa=11100mm2
Dc=325mm
= 20.1 Aa=223110 beffds =2000*150=300000
Plastic neutral axis lies in slab
Plastic Moment Capacity:
Xu= Aa/ beffds Mp=0.87*fy*Aa(250 +150 -0.42Xu) = 1275.08KN-m >1123 Hence section is safe.
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Design of Column 1:
Given Data:
INTERNAL Columns The Axial force and moments was obtained from Open Sees:
Maximum Axial Load ( P ) Maximum Bending Moment (Mz1) Maximum Bending Moment(Mz2)
=46315 KN-m =1755.5KN-m =-1132 KN-m
Aa=15600 sq.mm As=1256 sq.mm Ac=3983144 sq.mm Plastic resistance of concrete= Aafy + Ac*(fck)cu + Asfsk 62510KN > 62320 KN So safe
EXTERNAL COLUMNS
Maximum Axial Load ( P ) Maximum Bending Moment (Mz1) Maximum Bending Moment(Mz2)
=46315 KN-m =1755.5KN-m =-1132 KN-m
Aa=15600 sq.mm As=1256 sq.mm Ac=3983144 sq.mm Plastic resistance of concrete= Aafy + Ac*(fck)cu + Asfsk 62510KN > 50740 KN So safe