IT MOST RESIST UNLIMITED LOADS IN APARTMENT BUILDINGS. I refer my college professor

1. DESIGN OF G+3 APARTMENT
BUILDING
Presented by
Vijayvenkatesh.C
Department of civil engineering
Final year
ST. JOSEPH’S COLLEGE OF ENGINEERING & TECHNOLOGY
THANJAVUR
Under guidance of
LAKSHMIKANTH.P
Assistant professor of
Civil department

2. Outline:
Abstract
Introduction
Literature review
Planning of apartment building
Elevation works of apartment buildings
Design of G+3 apartment building
*design of column
*design of beam
*design of foundation
*design of staircase
*design of slab
Load displacement of structural work
conclusion

3. ABSTRACT
 Now a days, the people from villages are coming to towns for
employment and educational facilities hence with the limited land
available, individual houses are feasible.
 One of the major problem facing by the INDIAN country is rapid
growth of population which restricted the availability of the land
 More over, even the available houses are let out at abnormal rent
charges.
 Hence an apartment building is proposed in this project.
 The project consists of six houses in three floors with all basic
amenities.
 The construction and the design will be based on LIMIT STATE
method and as per the schedule rates of PWD Government of
TAMILNADU.

4. INTRODUCTION
 One of the major problem in Indian country
facing the rapidly growth of population this can be
solved to a certain extent with the construction of
multi storied apartments, which can be live many
people in available area.
 The proposed site for our project is at VALLAM
town in THANJAVUR district.
The maximum numbers of people go either to
THANJAVUR town.
The project consists of six houses in three floors
with all basic amenities.
It is most economic in design and construction
and followed as per limit state design method.

5. GENERAL LAYOUT
The other facilities like parking, cycle
stand, toilets for either sex and bill counter have been
place with the construction for proper location.
OTHER PROVISION
Apart from the space requirements and
sound treatment, for resistance has also to be
considered daily through its does not significantly
affect the layout or design.

6. INTRODUCTION TO LIMIT STATE
Since this method of design is
adopted widely in recent practice. We have followed
this method and all the design works are carried out
in accordance to IS 456-2000
The object of the design is to
produce a structure which is in all respects
satisfactory for which is required, having the
economy as the primary objective

7. As our knowledge about the characteristics to RCC
member under axial load, flexure shear and torsion
improved, the allowable stresses are periodically
adjusted to achieve greater economical design. The
adoption of limit state design concept as the latest
version of IS 456-1984 loads towards this direction.
In the limit state designs structure
shall be designed to safety withstand all loads liable to
act on throughout its life. It should be also satisfy the
serviceability requirement such as limitation on deflection
and cracking.

8. SPECIFICATION
EARTH WORK EXCAVATION
Excavation- foundation trenches shall be
drought to the exact width of foundation concrete and
the side should be vertical. If the soil is not good and
does not permit vertical sides. The sides should be
slopped back are protected with timber shoring
excavated earth shall not be placed within one meter of
the trench reinforcement cement concrete RCC.

9. BASEMENT
The basement with RR masonry in CM 1:5
using first class blue granite metal with 450mm width, 600
mm height are proposed the clear dry river sand in purpose
of filling the basement. The base concrete is also made with
CC 1:4:8 to over sand filling.
DPC
The top of the basement structure of DPC
provided with 20mm thick using crude oil 5% by weight of
cement in used.
``

10. SUPER STRUCTURE
The super structure of the building with
brick in CM 1:5 mix using first class bricks of size 19*19*9cm
of 230mm width and height up to 3mm with parapet at the
height of 600mm is provided.
ROOFING
The roofing of the structure will be RCC
slab of 1:1.5:3 mix using 20mm IS gauge hard broken granite
chips weathering course with jelly with lime concrete and
top finished with pressed tiles of 20*20*2cm size.
FLOORING
The top is finished smooth with trowel floor
finish with marbles.

11. LITERATURE REVIEW
After referring lot of book related to
reinforced concrete structure, we have to arrive the design
and design procedures for this project as per the some
standard reference. slabs are designed by using the ratio of
longer span to shorter span; this may help to find the types
of slab. and other calculation of the slab are calculated by
using some design methods as per IS 456-2000
beams and columns are designed as per
code provision like that IS CODE AND NBC CODE. the load
from column hall is transmitted into the footing. the design
of footing maybe done by based on safe bearing capacity of
the soil. the compressive stress in concrete at the base of
column OR pedestal shall be considered during the design
of column. the live load is variable but does not exceed
three quarters of the dead load, or the nature of the live load
is the wall panels will be loaded simultaneously. moments

12. on design of footing(IS 456-1978) for sloped footing the
effective cross section in compression shall be limited by
the area above the general plan, and the angle of slope or
depth and location of steps shall be such that the design.
the locker room wall are constructed by RCC (1:1.5:3) with
300mm kitchen. floor may be of RCC(1:1.5:3)mm. door and
ventilation of the locker room shall be fixed as per the
manufacture specification

13. SPECIFICATION OF G+3 APARTMENT
BUILDINGD:
Ground floor - car parking:110.49m^2
First floor
2^nd floor
Third floor
Living room:3.81X4.03m
Kitchen:2.87X3m
Two bed room with bath room attachment:2.41X2.5 ;3X5m
Elevator lift:1.46X1m
Stair case:
rise=150mm
tread=300mm
Specified with 2-
portion:
Specification:

14. Components size specification use of G+3
apartment
building:
column size=0.35X0.35m
beam size=0.35X0.3m
slab size (1-panel) 4.5X3m=0.165m
Main door=1.18X2m
Window=1X1.5m
Door=1X2m
Partition external wall thickness =350mm
internal wall thickness=250 to 100mm
ventilator =0.25X0.32m

21. DESIGN OF G+3 APARTMENT
BUILDING
COMPONENTS
GROUND FLOOR COLUMN DESIGN
C1,C2,C3,C4,C5,C6,C7,C8,C9,C10,C12,C13,C14,C15,C16,
C17,C18,C19,C20,C21

22. Design of column in G+3 apartment
building
Breadth=350mm
Depth =350mm
Pu=1150KN
MuX=MuY=40KN.m
Fck=20N/MM^2
Fy=415N/MM^2
Assume d’=40mm
d’/D=0.112
Equivalent moment:
The reinforcement section is design for the axial compression
load
Pu
Mu=1.15(Mux^2+Muy^2)
=1.15(40^2+40^2)
=65.0538KN.m
Non dimensional parameter:
(pu/fckbd)

23. 1150X10^3/20X350X350
=0.46
Mu/Fckbd^2
=65.038X10^6/20X350X350^2
=0.0758
Longitudinal reinforcement:
ref chart -44 of SP-16
P/fck
=0.07
P=20X0.07=1.4
asc=(pbd/100)
=1715mm^2
Provided 4 bars of 20mm diameter and 4bars of 16mm diameters
spacing =180mm c/c
p=100Xast/bd
=100x1715/350X350
=1.4
p/fck=0.07
Ref chart -44 of SP-16 and read out (Mux1/fckbd^2) corresponding to
The value of (pu/fckbd)=0.46
MuX1=(0.07 X20X350X350^2)X10^6
MuX1=Muy1=60.025KN.m

24. Puz=(0.45fckac+0.75fyas)
=0.45X20X((350X350)-1840)+(0.75X415X1840)
=1085940+572700
=1658KN
(pu/puz)=1150/1658
=0.693
The co-efficient αn corresponding to (pu/puz)
=0.69
αn=1.72
Check for satisfied under biaxial bendings
((Mux/Mux1)^(αn)+(Muy/Muy1)^(αn))=1
0.49+0.49=1
0.98=1
Hence the section is safe under biaxial bending.
Reinforcement detail:
provided 4-bars of 12,16,20, diameter as longitudinal reinforcement and
8MM lateral ties at 180mm c/c =(300mm c/c)

25. DESIGN OF = C1,C7,C13
STAAD.PRO OUTPUTS

26. DESIGN 0F =C2,C8,C14
`

27. DESIGN OF =C3,C9,C15

28. DESIGN OF = C16 ,C21

29. DESIGN OF =C5,C11,C18

30. DESIGN OF=C6,12,19

31. DESIGN OF 2ND FLOOR COLUMN
C1,C7,C13

32. C2,C8,C14

33. C2,C9,C15

34. C16,C21

35. C4,C5,C6,C10,C11,C12,C17,C18,C19

36. DESIGN OF 3RD FLOOR COLUMN
C1,C7,C13

37. C2,C8,C14

38. C3,C9,C15,C20

39. C16,C21

40. C4,C10,C17

41. C5,C6,C11,C12,C18,C19

42. Design of beam :
A reinforcement concrete beam is to be design
over a effective span 1.25 m
Given :
effective span =1.25m
width of beam =350mm
overall depth =300mm
service load =40KN/m
effective cover=50mm
material=M20 grade concrete
fe-415 HYSD bars
Ultimate moment and shear forces:
Mu=0.125X1.5X40X1.25^2
=16.875KN/m^2
vu = 0.5X1.5X40X1.5
=45KN

43. Main reinforcement:
Mulim=0.138fckbd^2
= 0.138X20X350X300^2
=86.94KN.M
Since Mu<Mulim
hence take singly reinforcement beam.
Mu=0.87fyastd(1-(astfy/bdfck))
ast=194.33mm^2
Provided 4 bars of 12MM diameter
Check for shear stress :
vu=45KN
Ɩv=(vu/bd)
= 45X10^3/(350X250)
=0.514N/mm^2
pt=(100ast/bd)
=0.22
Ref table 19 IS:456-2000 and read out the design shear strength of concrete
Ɩc=0.173N/mm^2
Ɩc>Ɩv shear reinforcements are required
balanced shear= Vus=Vu-(Ɩcbd)

44. =45-(0.173X350X250)X10^-3
=29.86KN
Sv >0.75d
=0.75X250
=187.5mm
=provided 6mm diameter stirrups of at 190mm shear supports
Check for deflection control:
pt=0.22
(l/d)max=(l/d)basic XktXkcXkf
=20X1.1X1X1
=22
(l/d)actual=(1250/250)=5<22
hence safe

45. SPAN LENGTH : 1.25m

46. A reinforcement concrete beam is to be
design over a effective span =1.5m
Given:
effective span =1.5m
width of beam =350mm
D=300mm
service load =40KN/m
d’=50mm
material=M20 grade concrete
=fe415 HYSD bars
Solution:
ultimate moment and shear force :
Mu=(0.125X1.5X40X1.5^2)
=16.875KN.m
vu=0.5X1.5X40X1.5
=45KN

47. Moment reinforcement:
Mulim=0.138fckbd^2
=0.138X20X350X250^2X10^-6
=60.375KN.m
Mu<mulim
Hence single reinforced beam.
Mu=0.87fyastd(1-(astfy/bdfck))
16.875X10^6=0.87X415XastX250X(1-(ast415/350X250X20)
ast=196.069mm^2
Provided 4 bars @ 12mm diameter as tension reinforcement 2 bars of 10mm
Diameter hanger bars are compression sides
Check for shear stress:
Vu=45KN
Ɩv=(vu/bd)
=0.51428N/MM^2
pt=100ast/bd
=100X197/350X250
=0.224
Table -9 of IS-456-2000

48. Ɩc=0.1771N/mm^2
Ɩv>Ɩc shear reinforcements are requirements
Balance shear =Vus=Vu-(Ɩcbd)
=45-(0.1771X350X250)
=15.45KN
sv>0.75d
0.75X250
=187.5mm
Provided 8mm diameter stirrups at 189mm shear support.
Check for deflection control :
(l/d)max=(l/d)basicXktXkcXkf
=22
(l/d)actual<(l/d)max
hence safe.

49. SPAN LENGTH= 1.5m

50. A reinforcement concrete beam is to be
designed over a effective span =3m
Effective span =3m
Width=350mm
Depth=300mm
d=250mm
d’=50mm
Service load=40KN/m
Sol:
ultimate moment and shear force :
Mu=0.125X1.5X40X3^2
=67.5KN.m
Vu=0.5X1.5X40X3
=90KN
Main reinforcement:
Mulim=0.138fckbd^2
=0.138X20X350X250^2
=60.375KN.m

51. Mu>Mulim design a doubly reinforced section.
(Mu-Mulim)=7.125KN.m
fsc=(0.0035(Xumax-d’)/xumax)es
=408.333N/mm^2
But fsc >0.87 , Fy=0.87X415
=361N/mm^2
asc=Mu-Mulim/fsc(d-d’)
=7.125X10^6/361X(250-50)
=98.6842mm^2
Provided 2-bars @ 16mm diameter
Ast2= ascfsc/0.87fy
=99X361/0.87fy
=98.98mm^2
Ast1=0.36fckbxulim/0.87fy
=837.557mm^2
Total tension reinforcement
ast=(ast1+ast2)
=837.557+98.98
=936.537mm^2

52. Provided 2bars @ 25mm diameter (ast=981.16mm^2)
Shear reinforcement:
Ɩv=vu/bd
=90X10^3/350X250
=1.02N/mm^2
pt=100ast/bd
=100X981.16/350X250
=1.12
Ɩc=0.585N/mm^2
Since Ɩv>Ɩc shear reinforcement are requared
Vus=(Vu-(Ɩcbd))
=(90X10^3-(0.585X350X250)
=38812.5X10^-3
=38.8125KN
Using 8mm diameter 2-legged stirrups
Sv=0.87fyasvd/vus
=0.87X415X2X250X50/38.8125X10^3
=232.56mm

53. Sv>0.75d
0.75x250
=187.5mm
Adopt spacing of 200mm near supports gradually increasing to 300mm toward
The center of span.
Check for deflection control:
(l/d)actual=3000/250
=12
(l/d)max=((l/d)basicktkckf)
pt=1.12 ; pc=100X98.64/350X250
=0.11273
Kt=0.8
Kc=0.94
Kf=0.86
(l/d)max=20X0.8x0.94X0.86
=12.9344=13
(l/d)actual<(l/d)max
Hence deflection control is satisfied.

54. DESIGN OF BEAM
SPAN LENGTH =3m

55. A reinforced concrete beam is to be designed
over a effective span =4.5m
Given:
Effective span =4.5m
b=350mm
D=300mm
d=250mm
d’=50mm
service load=40KN/m
Sol:
ultimate moment and shear force:
Mu=0.125X1.5X40X4.5^2
=151.875KN.m
vu=0.5X1.5X40X4.5
=135KN

56. Main reinforcement:
Mulim=0.138fckbd^2
=0.138X20X350X250^2
=60.375KN.m
Mu>Mulim
hence design a doubly reinforced beam sections
Mu-Mulim=151.875-60.375
=91.5KN.m
fsc=(0.0035(xumax-d’)/xumax)Xes
=0.0035(0.48x250)-50/0.48x250xEs
=408.33N/mm^2
But fsc=361N/mm^2
asc=Mu-Mulim/fsc(d-d’)
=91.5X10^6/361(250-50)
=1267.31mm^2
provided 6 bars @ 16mm diameters
ast2=ascXfsc/0.87fy
=1267.31X361/0.87X415
=1267.134mm^2

57. Ast1=0.36fckbxulim/0.87fy
=0.36X20X350X0.48X250/0.87X415
=837.557mm^2
Total tension reinforcement :
ast=ast1+ast2
ast=2104.691mm^2
Provided 4bars @ 25mm diameter
Shear reinforcement:
Ɩv=(vu/bd)
=135X10^3/350X350
=1.54N/mm^2
pt=100ast/bd
=100X2104.691/350X250
pt=2.40
Ɩc=1.255N/mm^2
Ɩv>Ɩc shear reinforcement is required
vus=(vu-(Ɩcbd))
=135-(1.255X350X250)

58. =109.67KN
Using 8mm diameter 2-legged stirrups
Sv= 0.87fyasvd/vus
=0.87X415X2X250/109.67X10^3
Sv=82.303mm
Sv>0.75
0.75X250
=187.5mm
Adopt a spacing of 200mm near support gradually increased to 300mm toward
the center of span.
Check for deflections:
(l/d)actual =4500/250
=18mm
pt=2.40

59. Kt=1.1716
Kc=1
Kf=1.8
(l/d)max=20X1.716X2.303X1.8
=61.776
(l/d)max>(l/d)actual
Hence deflection control is satisfied.

60. SPAN LENGTH: 4.5m

61. GRAPH= BEAM(3m)
RxN

62. GRAPHS=BEAM(1.5m)
RxN

63. GRAPH=(BEAM) 4.5m
RxN

64. GRAPH=(BEAM) 1.25m
RxN

65. STAAD.PRO V8I FOUNDATION DETAILS
FOR ( G+3 apartment building)
Load case = (dead load) 2.25KN

66. Live load :20KN/M

67. A square footings for a G+3 apartment
building
Given: Dead load = 507kN
Imposed load= 500KN
Breadth= depth= 350mm
Safe bearing capacity of soil (qu)= 200KN/m^2
Factored (qu)=1.5x200=300kN/m^2
Solution:
Size of footing:
Dead load + imposed load = 1007KN

68. Wu=1007 KN
Factored area = 1007/1.5x 200
= 3.35m^2
Side of square footing :
(3.35)^(1/2)
= 1.83m =2m
Adopt square footing of size = 1.83X1.83m
Factored soil pressure at the base:
=1007/(1.83)^2
251<300KN/m^2
Hence footing area is adequate
Factored of bending moment:
cantilever projection from the face of column
=0.5(2-0.3)
=0.85m
Bending moment of either side
=qul^2/2
=1.5(200X0.85^2)/2
=108.375KN.m
Depth of footing :

69. Moment
Mu =0.138fckbd^2
d=(Mu/0.138fckb)^(1/2)
=(108.375X10^6/0.138X20X1000)^(1/2)
d=198.157mm
Depth from shear consideration shear force per meter width :
=251((2000/2)-(300/2)-d)n
Ɩc = 0.36N/mm^2
For m20 grade concrete with nominal percentage of reinforcement
P=0.25
Ɩc=vu/bd
0.36=251X(850-d)/1000d
360d=213350-251d
611d=213350
d=350mm
D=350+50
=400mm
Reinforcement:
Mu=0.87XfyXastXdX(1-(astXfy/bdfck)

70. 108.375X10^6=0.87X415XastX350X(1-5.92X10^-5ast)
Ast=906.22mm^2
Provided 16mm diameter rod and 250mmc/c
Check for shear stress:
Vu=251(850-350)X10^-3
Vu=125.5KN
p=100ast/bd
P=0.407
from tabel19 of IS456 -2000 the permissible shear
Stress
ksƖc=1X0.609=0.61N/mm^2
Nominal shear stress =Ɩv=Vu/bd
Ɩv=0.709N/mm^2
Ɩv<ksƖc shear stress within safe permissible limit.

71. DESIGN OF R.C SLAB FOR A G+3
APARTMENT BUILDINGS
GIVEN:
Roof size=4.5X3m
LX=3m
LY=4.5m
LY/LX=1.5m
fck=25N/mm^2
fy=415N/mm^2
SOLUTION:
Depth of slab:
As the ratio of long to short span 1.5 which is
<2 the slab has to be designed as a 2-way slab.
As the loading condition exceed 3KN/mm^2
Adopt a span/effective depth ratio of 25
d=4500/25
=180mm
D=180+25
=205mm

72. Effective span:
clear span + effective depth
=3+0.18
=3.18m
Loads:
self wt=0.205X1X25=5.125KN/m^2
live load=3KN/m^2
floor finishes= 1KN/m^2
service load=9.33KN/m^2
wu=1.5X9.33=13.995KN/m^2
Ulimate design moment and shear force:
αx=0.066:αy=0.048
Mux=(αxwulx^2)=9.34KN.m
Muy=(αywuly^2=6.793KN.m
Vux= 0.5wulx=22.25KN
Check for depth:
consideration the maximum moment of
d=(9.34X10^6/0.138X20X10^3)^(1/2)

73. =58.172mm
Which is less than d=180mm
Hence the d selected is sufficient to resists the maximum design
Moment.
Reinforcement for negative and positive moment :
Short span :
Mu=0.87fyastd(1-(astfy/bdfck))
9.34X10^6= 0.87X415XastX180X(1-(ast415/10^3X180X25))
ast=146mm^2
provide 10mm diameter bars
spacing=10^3X78.53/146
=537.94mm
This is more than 3d
3X180=540mm
Hence adopt maximum spacing of 540mm . Provided 10mm diameter
Bars
=78.53/540
=0.145X10^-3
ast=145.4mm

74. (Mux)p<(Mux)n provided the same spacing
Long span:
provided 10mm diameter bar
520mm spacing on both direction
Check for shear stress:
considering a short span and the width of slab .
Ɩv=vu/bd
=22.85X10^3/10^3X180
=0.1269N/mm^2
pt=100ast/bd
=0.080
Ɩc=0.135N/mm^2
kƖc=0.175N/mm^2
Which is greater than Ɩc=0.12N/mm^2
Hence the shear stress are within safe permissble limit
check for deflection control:
(l/d)basic=25
for pt=0.17;kt=1

75. 25X1X1=25
(L/D)actual=17.67
17.65<25
Deflection control is satisfied
Check for crack control:
Spacing adopted is 360mm
Reinforcement minimum=0.12percentage
0.0012X205X1000
=246mm^2
Which is less than that provided
Diameter of reinforcement
D/8<205/8<25mm
Which is more than the diameter of rod provided as 10mm
Torsion reinforcement at corner:
Area of tension steel provided an each of the corners in 4layers
=0.75Xast
=0.75X145.4
=109.05mm^2

76. Length over which torsion steel is provided
=(1/5)X short span
=0.2X3000
=600mm
Provided 6mm diameter bars at 240mm center for a length of 600mm
at all 4 –corner in 4-layer
Reinforcement in edge strips:
ast=0.12percentage
0.0012X10^3X205
=246mm^2/m
Provided 10mm diameter bars at 300mm centers.

78. PRESSURE LOAD REACTION OF SLAB
DESIGN PRESSURE LOAD=10KN/m^2

79. DESIGN OF STAIR CASE IN G+3
APARTMENT BUILDING
Stair case (type) use in G+3 Apartment building :
dog legged stairs with waist slab , tread, risers
Given :
Numbers of steps in flight =7
tread T=300mm
rise R=150mm
width of landing beams= 300mm
M20 grade concrete fck=20N/mm^2
fe415 HYSD bars fy=415N/mm^2
Effective span:
L=7X300 +300
=2400mm
=2.4m
Thickness of waist slab =(span/20)
=2400/20
=120mm
Effective depth =d=95mm

80. Loads:
Dead loads of slab on slope = ws=( 0.12X1X25)
=3KN/m
Dead load of slab on horizontal span is
W=(ws(r^2+t^2)^(1/2)/t)
W=3.35KN/m
Dead load of one step =(0.5X0.15X0.3X25)
=0.56KN/m
Load of steps per meter length =(0.56X10^3/300)
=1.86KN/m
Finishes etc =0.53KN/m
Total dead load =(3.35+1.86+0.53)
=5.74KN/m
Service live load =5KN/m
Total service load =5.74+5
=10.74KN/m
Factored load =1.5X10.74
=16.11KN/m
Bending moment:
Maximum bending moment at center of span is :

81. M=0.125wul^2
=0.125X16.11X2.4^2
=32.22KN/m
Check for depth of waist slab :
d=(mu/0.138fckb)^(1/2)
=108.5mm
94.2mm<95mm (provided)
Hence safe
Main reinforcement:
Mu=0.87X415XastXd(1-(415ast/bdfck))
32.22X10^6=0.87X415astX95(1-(415ast/1000X95X20))
ast=1319.64mm^2
Provided 12mm diameter bars at 200mm centers
Distribution reinforcement :
=0.12percentX1000X120
=0.0012X1000X120
=144mm^2/m

82. Provided 8mm diameter bars at 200mm centers
Design using SP-16 design chart
compute the design parameter :
(Mu/bd^2)= 32.22X10^6/1000X120^2
=2.2375
Refer table 2 of SP-16 design table corresponding to fck =20N/mm^2 and
read out the percentage of reinforcement as:
ast =(pbd/100)
=0.669 X1000X120/100
=802.8mm^2/m
The reinforcement quantity is the same as that obtained by analytical
Method.

83. LOAD DISPLACEMENT DETAILS FOR
STRUCTURAL WORKS:
SELF WT=2.25KN

84. BENDING REACTIONS FOR DEADLOAD

85. LIVE LOAD = 20KNm

86. BENDING REACTION FOR LIVE LOAD

88. LOAD DISPLACEMENT OF SUPER IMPOSED
LOAD
=44KN/m

89. Stress diagram for column

90. Stress diagram of beam

91. CONCLUSION :
The planning and designing of the AN APARTMENT
BUILDING has been completed effectively in our project. if this is
constructed in the proposed site, it will be very help full to our
people.
we all the members of our team have learned to
plan a building with referring to NATIONAL BUILDING OF
INDIA 2005. this project is a very useful to learn about the
design of the structural elements like beams , column and slabs by
using IS 456-2000.
the important think that we done were referring to
lot of books for designing, and we are very much satisfied with
exposing to field design.

92. THANK YOU