HI-TECH shopping mall provides shopping along with driving.This will be an evolution in indian shopping centre

1. PLANING,ANALYSIS AND DESIGN OF HI-
TECH SHOPPING MALL
Batch members:
 R. Arshana
 A.S.Aswathi
 A. Lekshmi
 P.Pemila
Reg No:
960515103008
960515103012
960515103025
960515103030
Guided By,
Dr . B.Thanukumari,ME,PhD
HOD,Department of civil .
Cape Institute Of Technology

2. INTRODUCTION
• In this project, planned to construct a shopping mall in
Anjugramam which is a main town ,situated near the National
Highway .
• In high tech shopping mall we are introducing a new era (shop
with drive ).
• The plinth area of each shop is 54 square meter.
• We are sure about that high tech shopping mall will be useful for
all sectors of people.

3. ABSTRACT
• Our project deals with the planning,analysis and design of Hi Tech
shopping mall.
• The proposed site of the project is in Anjugramam.
• The structural elements such as slab,beam,column and foundation will
be designed by using IS456-2000 and the materials used for the
design are M20 grade concrete and Fe415 grade steel.
• The plan,section and elevation will be drawn by using Autocad
software and the analysis is done by using STADD.PRO

4. SCOPE
• Hi Tech shopping mall plays an anchor role in the Indian
shopping centre.
• It can be established in crowded area without any parking
facilities.
• The shopping mall is used as a best entertainment place for
the people.
• Shopping can be done within a fraction of minute.

5. OBJECTIVES
• In high tech shopping mall digital screen system help to
guide customers in product information and availability
of product.
• Friendly users interface and log in systems makes more
secured.
• ‘Shop and drive facilities with digital collection method
attracts more customers.
• Walk and shop makes the shopping more realistic.

6. SITE PLAN

13. ANALYSIS
• Analysis can be done by using stadd.pro
BENDING MOMENT DIAGRAM
20
20
20
20
40
40
40
40
60
60
60
60
10 20 3030.7
9 12
48.4 48.4
-23.7
15.4
Mz(kNm)
20
20
20
20
40
40
40
40
10 20 3030.7
9 12
30.8
-30.8
Fy(kN)
2
2
2
2
4
4
4
4
10 20 3030.7
9 12
-3.15 -3.15
Fx(kN)

14. FOOTING

15. COLUMN

16. DESIGN OF TWO WAY SLAB
(ONE LONG EDGE DISCONTINUOUS)
Datas:
Room size= 6m×9m
Live load= 5kN/m^2
Floor finish load=1kN/m^2
Fck=20 N/mm2
Fy=415N/mm2
Check for span ratio:
Ly/Lx = 9/6 =1.5m<2m
Hence it is a two way slab.

17. Depth calculation:
D = span / 25 = 6000/25 = 240mm
Say D = 200mm
Cover = 20mm
Effective depth = 180mm
Effective span :
Lx= 6.18 m
Ly = 9000 + 180 = 9180mm
Load calculation:
Dead load = 1m×1m×0.25×25 = 5kN/m2
Live load = 5 kN/m2
Floor finish load = 1 kN/m2
Total load = 11kN/m2
Factored load = 1.5 ×11 = 16.5 kN/m2

18. Moment calculation :
αx=0.067 ; αx=0.051
αy=0.037 ; αy=0.028
MX = 0.067 × 16.5 × 62 = 39.79 kN m
MX = 0.051 × 16.5 × 62 = 30.29 kN m
MY = 0.037 ×16.5 × 62 = 21.98 kN m
MY = 0.028 × 16.5 × 62 = 16.63 kN m
To find effective depth :
Mu= Mulim
Mu = 2.75 bd2
d = 120.28mm
say d = 120mm<180mm
Hence safe

19. Ast Calculation:
1) Main Reinforcement (short span ) :
39.79×106 = 0.87×415×Ast×180(1-415Ast/1000×20×180)
Ast=662.92mm2
Provide 16 mm dia bars
Ast prov = π×102/4
=78.53mm2
i)Spacing=1000Astprov /Ast
=1000×78.53/662.92
=118.46mm
Say 110 mm
ii) 3×d = 3× 180 = 540mm
iii) 450 mm
least value to be taken
provide 16mm dia bars @ 110 mm c/c

20. 2) Distribution Reinforcement (shorter span ):
30.29×106= 0.87×415×Ast×180(1-415×Ast/20×1000×180)
Ast=494.23mm2
Provide 10 mm dia rods
Ast (p)=π/4×102= 78.53mm2
Spacing=1000×78053/494.23= 158.89mm
1)say spacing=150mm
2)3d=3×180=540mm
3)450mm
Least value is to be taken
Provide 10mm dia bars at 150mm c/c

21. 3)Main reinforcement(longer span):
29.18×106=0.87×415×Ast×180(1-415×Ast/1000×20×180)
Ast=475mm2
Provide 16 mm dia bars
Ast(p)=π/4×102=78.53mm2
Spacing=1000×78.53/475=165.32mm
1)Say spacing=160mm
2)3d=3×180=540mm
3)450mm
The least value to be taken
Provide 16mm dia bars at 160 mm c/c

22. 4) Distribution reinforcement(longer span):
16.63×106=0.87×415×Ast×180(1-415×Ast/1000×20×180)
Ast=263.92mm2
Provide 10 mm dia bars
Ast(p)=π/4×102=78.53mm2
Spacing=1000×78.53/263.92=297.5mm
1)say spacing=290mm
2)3d=3×180=540mm
3)450mm
Least value to be taken
Provide 10mm dia bars at 290mm c/c

23. EDGE STRIP
Ast=Astmin=263.92mm2
Spacing=(1000×π/4×82)/263.92=190mm
Provide 8mm dia bars at 190mm c/c where the edge strips
are both longer and shorter span
CHECK FOR SHEAR
τv=Vu/bd
Vu=16.5×6.18/2=50.985KN
τv =50.99×103/1000×180=0.28N/mm2
pt=100×Ast/bd=100×662.92/1000×180=0.37N/mm2
τc=0.417N/mm2

24. CHECK FOR DEPTH
D=200mm
K=1.20
k ×τc=1.20×0.417=0.5
k ×τc > τv
0.5>0.28N/mm2 Hence ok

25. DESIGN OF T- BEAM
Datas:
Length of span = 9m
Spacing of beam = 6m
Fy = 415 N/mm2
Fck = 20 N/mm2
Depth of the flange = 200mm
Width of the flange = 9.27m
Cross sectional dimensions
d = span/12 = 9000/12 = 750mm
overall depth = 750 + 50 =800mm
Load calculation
Self weight of beam = 0.2 ×0.75×25=3.75kN/m2
Load from slab = 1×6×0.2×25= 30kN/m2
Floor finish = 1 kN/m2
Live load = 5 kN/m2
Total load = 38.75 kN/m2
Factored load = 1.5 ×38.75 = 58.13 kN/m2

26. Ultimate load:
Mu = Wu × l2/ 8
=58.13×62 /8
=261.56 kN/m
Shear force :
Vu= Wu ×L /2
=58.13×6/2
=174.39kN
Effective depth of flange:
Bf = L0/6 + bw +6 Df
=9/6 + 0.23 +(6×0.2)
=2.93m
Centre to centre of beam= 6-0.23 = 5.77mm
Take least value bf = 2.93m
Moment capacity of flange:
Mu= 0.36 × fck × bf × df (d-0.42 Df)
=0.36 × 20 × 2930×200×(750-0.42×200)
Mu=1755.18×106 kN m> 261.56 kNm
Hence ok

27. Reinforcement calculation:
261.56×106= 0.87×415×Ast×750(1-415×Ast/20×1000×750)
Ast=993.22mm2
Provide 16 mm dia rods
Ast (p)=π/4×162= 201.06mm2
No of bars = Ast/ Ast (p) = 4.9
Provide 5 numbers of 16 mm dia bars.
Shear reinforcement:
τv=Vu/bd
=174.39×103/230×750 = 1.01 N/mm2
pt=100×Ast/bd=100×993.22/230×750=0.56N/mm2
τc=0.49N/mm2
0.49<1.01 N/mm2
Vus = Vu- τc×bd
=174.39×103-0.49×230×750
=89.865 kN
Use 8mm dia of 2 legged stirrups

28. Spacing:
1) Sv = 0.87 × fy ×Asv ×d / Vus
= 150 mm
2) 0.75d = 0.75×750 = 562.5
3) 300mm
Provide 8 mm dia rods @ 150 mm c/c

29. DESIGN OF L- BEAM
Datas:
Length of span = 9m
Spacing of beam = 6m
Fy = 415 N/mm2
Fck = 20 N/mm2
Depth of the flange = 200mm
Cross sectional dimensions :
d = span/12 = 9000/12 = 750mm
overall depth =780mm
effective span=9+0.3=9.3m
Load calculation:
Self weight of beam = 0.2 ×0.75×25=3.75kN/m2
Load from slab = 1×6×0.2×25= 30kN/m2
Floor finish = 1 kN/m2

30. Live load = 5 kN/m2
Total load = 39.75 kN/m2
Factored load = 1.5 ×39.75 = 59.63 kN/m2
Effective width of flange:
Bf=lo+bw+3Df=9.3/12+0.23+0.6=1.605m
Say Bf=1.6m
ultimate bending moment :
Mu = Wu × l2/ 12
=195.35 kN/m
Shear force :
Vu= Wu ×L /2
=31.16kN
Torsional moment at the support=20KN/m
Equivalent bending moment and shear force:
Me1=Mu+M

31. Mu=50.12kNm
Me1=50.12+195.35=245.48kNm
Mu lim=0.36×fck×bd2=0.36×20×230×7502=931.5kNm
Me1 <Mu lim
Ast=930.49mm2
Provide 16 mm dia rods
Ast (p)=π/4×162= 201.06mm2
Area of steel required at the centre with
Mu=50.12kNm
Ast min=0.85bd/Fy=0.85×230×750/415=353.31mm2
Provide 2 rods of 8mm dia bars
Ast (p)=π/4×82= 100.53mm2
Shear reinforcement:
τv=Ve/bd

32. =170.29×103/230×750 = 0.99N/mm2
p t=100×Ast/bd=100×930.49/230×750=0.54N/mm2
τc =0.49N/mm2
0.49<0.99 N/mm2
Hence safe

33. DESIGN OF COLUMN
Given data:
Size of column=600×480mm
Mux=48.36kNm
Muy=22.174mm
Fck=20N/mm2
Fy=415N/mm2
Load calculation:
Self weight of column=7.2kNm
Load due to T beam=38.75kNm
Load due to L beam=38.75kNm
Total load=85 kNm
Design load=128.55kN/m
Load Pu=128.55×9/2+128.55×6/2=710Kn/m
Mu=61.18KNm

34. Provide 4 bars of 25mm dia and 4 bars of 16mm distributed equally on all
faces with 4 bars on each face
P=100×2766/480×600=0.96
p/fck=0.96/20=0.048
Because of symmetry
Mux1=Muy1=221.18kNm
Puz=0.45fckac+0.75fyasc=4076kN
Pu/puz=710/4076=0.17
Check for safety against biaxial bending
(mux/mux1)αn+(muy/muy1)αn greater than or egual to 1.
(48.367/221.18)+(22.174/221018) greater than or equal to 1
=1.32
Hence the section is safe against biaxial bending

36. DESIGN OF FOOTING
Given data:
Column size=600×480mm
Total load=710Kn
SBC of soil=350Kn/m
Fck =20N/mm2
Fy=415N/mm2
Assume 10 %of column load as self weight of footing
= 107×10/100
=71KN
Total load on footing=710+71=781Kn
Area of footing=781/350=1.5m
Since it’s a rectangular footing
A=B×L
A=1.5B2
B2=2.23/1.5=1.48m
B=1.5m
h=2.5m
size of footing =1.5×2.5m

37. Fu=710/(1.5×2.5)=189.33KN/m2
Ultimate SBC of soil=1.5×189.33=284KN/m2
Fu<ultimate SBC=284KN/m2
Hence the footing area is adequate
Factored moment:
1)cantilever projection=B-b=1.5-0.25=1.25m
2)bending moment=Ful2/2=189.33×1.252 /2=147.91KNm
Depth of footing:
(a)from moment consideration
Mu=0.138fckbd2
D=230mm
(b)from shear consideration
For Fe 415&M20combination
Pt=0.96%

38. K=1
k.τc =1×0.61=0.61N/mm2
τv= Vu/bd
=620mm
Overall depth D=620+30=650mm
Reinforcement in footing
Mu=0.87×fy×Ast×d(1-fy×Ast/bd×fck)
147.91×106=0.87×415×Ast×650(1-415Ast/1000×650×20)
Ast=643mm2
Use 16 mm dia bars
Ast(p)=π/4×162 =201.06mm2
Spacing=Ast(p)/Astr×1000
=(π/4×162 ) /643×1000
=313mm
Say310mm
Provide 16mm dia bars at 310mm c/c
Check for shear stress:
Actual shear stress τv=Vu/bd

39. =0.394N /mm2
Allowable shear stress =ks×τc
τc =0.25√fck
=0.25√20
=1.12N /mm2 Ks=1
Allowable shear stress =ks ×τc
=1.12> τv
Safe against shear

40. DESIGN OF STAIRCASE
Datas :
Height between the floors = 4m
Rise = 150mm
Tread = 250mm
Landing width = 1.83m
Live load = 5kN/m2
Floor finish load = 1kN/m2
Width of support = 230mm
fck = 20N/mm2
fy = 415N/mm2
Height of each floor = 4/2 = 2m
Number of rise = 2/0.15 = 13 nos
Number of steps = 13-1 = 12 nos
Effective length = 0.23/2 + 1.83 +(12*0.25)+ 1.8+(.23/2)= = 6.89m

41. Load calculation:
Self weight of stairslab = ws(R^2+T^2)^(1/2)/T
Thickness of waist slab = Leff/20
= 6.89/20
say 350mm
Ws = (1*1*0.35)25 = 8.75kN/m2
Self weight of stairslab = 8.75(0.15^2+0.25^2)^(1/2)/0.25
= 10.204kN/m2
Self weight of steps = R/2*25
= 0.15/2*25 = 1.875 KN/m2
Live load = 5kN/m2
Floor finish = 1kN/m2
Factored load = 1.5*18.579 = 27.86kN/m2

42. Load on landings:
Assume thickness of landing slab = 200mm
Self weight of landing slab = 0.2*25 = 5kN/m2
Live load = 5kN/m2
Floor finish = 1kN/m2
Total load = 11.5kN/m2
Factored load = 1.5*11.5 =17.25kN/m
Moment @ centre
Taking moment about C
Mc=2.45RB-17.25*1.945(1.5+1.945/2)–27.86*1.5*1.5/2
Mc = 70.28 kNm

43. Check for depth:
Mulim = 0.138 fckbd^2
D = 159.5mm
say D = 160mm
Assume 12mm dia bars
To find Ast :
Ast = 2577.94mm2 Mu = 0.87fyAstd(1-fyAst/fckbd)
Ast = 2580mm2
Provide 12mm dia bars
Astp = 3.14*12^2/4 = 113.097mm2
Spacing = Astp/ Astr *1000
= 113.097/2580*1000
= 43.83mm ~40mm
Provide 12mm dia bar @40mm c/c

44. Ast = 0.12% of bD
= 0.12/100*1000*160
= 192mm2
Provide 8mm dia bars
Ast p = 3.14*8^2/4 = 50.26mm2
Spacing = 50.26/192*1000
= 261.7 ~260mm
Provide 8mm dia bar @ 260mm c/c

45. 3D MODEL OF GROUND FLOOR

46. CONCLUSION
In this project work we have carried out load calculations, analysis and
design of the hi-tech shopping mall. Designs are done by using relevant
codes. We have adopted Limit State Method for the design of slabs, beams,
columns, stairs, etc.
Main purpose for doing this project is to gain a clear idea about
structural design of elements of a framed building. It is to be noted that a
theory cannot be brought to practice completely, a number of amendments
have to be make a design during execution of the work.During this project
we gathered more knowledge about the structural design.
This project is to be submitted to the management of Cape Institute of
Technology for implementation.