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CHAPTER-I
INTRODUCTION
1.1GeneralThe unprecedented growth of personalized vehicles and the unplanned road
infrastructure have made the provision for parking an important aspect of
transportation planning. Since most of the places are frequented by public and
busy with floating population, the demand for parking is very high.
On-Street parking is observed on all roads. This has reduced the capacity of
the carriageway and endangering pedestrians and motorists alike. The frontage
of almost all the roads has been convertedinto commercial land use without
taking into account the demand for parking of vehicles. There is no planned
parking space available.
1.2Planning and design concept
The unprecedented design of multi-storey car parking is based on the limit
sate design method and involves RCC. The plan of the building is done based
on the guidelines given by the Bureau of Indian Standards. The loads are taken
from IS 875 and for the design purpose the following codes are used:
1. IS 456-2000 for RCC Design2. SP 16 Design aids for RCC Design
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CHAPTER-II
INTRODUCTION
2.1 General:
The multi-storey car parking is planned for parking 256 cars. A place which
is near airport having good transport facilities is chosen. The basic provisions
for parking area, fire safety, security room, workshop, electric room, toilet and
other common areas are provided. The proposed site for construction is near
Coimbatore airport. This s 1km from national highway. The site is clear and
ready for construction.
2.2 General principles of site selection:
Site selection has a major role in the planning and design of buildings.
Natural defects already existing in a site will involve considerable expenditure
on construction and maintenance of the building. The following points shouldbe borne in mind while selecting a site for the proposed building:
(i)The site should be selected where there is a need for good parkingfacilities.
(ii)The site should be situated in a locality which is already fullydeveloped or which is fast developing.
(iii) The site should be constructed where there is a good transportationfacility.
(iv) The site should be permit unobstructed natural light and air.(v)The site should be clear and away from dumped wastages.
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2.3 Site plan:
The site plan is included so as to locate the area within the limits of the
building. This plan should be prepared before the construction of the proposedbuilding and should have:
(i)Boundary of the plot and shape of building.(ii)Name of existing roads and nearby sites.(iii)North direction.
2.4 Planning:
The multi-storey car parking is planned for parking 256 cars. The
provision for parking area, workshop, electric room, fire safety, security room
and other required areas necessary for standard construction are provided. We
have planned to provide rooms for drivers working here. In this project we have
planned to provide Hydraulic Jack for lifting of cars to different floors.
The project is planned to construct in an area for about 1.63 acres, with
plot area 2 acres and location is near Coimbatore airport.
Here we have planned to provide parking lot of size 5.5m x 2.5m, 2 lift of
size 1.83m x 1.83m per each lift, 4 hydraulic jack of size 6m x 5.5m per each
jack, a workshop room of sie 6m x 12m, a fire safety room of size 6m x 3m, a
electric room of size 6m x 10m, 2 toilet rooms of size 4m x 7m per each room, 2
security rooms of size 3m x 3m per each room, a canteen of size 6m x 6m, a rest
room of size 6m x 8m a office room of size 6m x 4m and a store room of size
6m x 7m.
The parking area is alone planned to elevate for 3 storeys (ie) G + 3.
Others structures are planned to built only in ground floor.
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2.5 Detailed Plan:
1. Ground floor plan
2. First floor plan
3.Second floor plan
4.Third floor plan
5. Terrace Plan
6. Elevation
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CHAPTER-III
DESIGN OF SLAB
3.1 General
Reinforced concrete slabs constitute the most common type of structural
elements used to cover floor and roofs of buildings. A reinforcement concrete
slab is a broad, flat plate usually horizontal, with top and bottom surfaces
parallel or nearly so. It may support by reinforced concrete beams, by masonry
or reinforced concrete walls, by structural steel members, directly by columns,
or continuously by the ground.
3.2 Types of slab
Slabs are classified according to the system of supports used as under
1. One way spinning slab2. Two way spinning slab3. Flat slab supported directly columns without beams4. Gird slab or wattle slabs5. Circular and other shapes6. One way continuous slab7. Two way continuous slab8. Cantilever slab
3.2.1 One way slab
One way slabs are supported opposite sides and the loads are transmitted
in one direction. Reinforced concrete slabs supported on two opposite sides or
on all four sides with the ratio of long to short span exceeding 2 are referred to
as one way slabs.
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3.2.2 Two way slab
Reinforced concrete slabs supported on all the four edges with the ratio of
long to short span not exceeding 2 are referred to as two way slab. In this type,the loads are transmitted to the supporting both directions with main
reinforcements provided in mutually perpendicular directions.
3.2.3 Flat slab
The term flat means a reinforced concrete slabs with or without drops,
supported generally without beams by columns with or without drops are called
flat slabs. The flat slabs may be solid or may have recesses formed on the soffit.
So that the soffit comprises of a series of a series of ribs in tow directions. The
recesses may be formed by removable or permanent filler blocks.
3.2.3.1 Thickness of flat slab
The thickness of flat slab shall be generally controlled by span to
effective depth ratio similar to beams. However the minimum thickness of flat
slab shall be taken as 125mm.
3.2.3.2 Drop
The drop in each direction shall have a length of not less than one third of
panel length in that direction.
3.2.3.3 Method of analysis
1.Direct Design methods.
2.Equivalent frame method
We have adopted direct design method because direct method we should
have minimum of three continuous spans in each direction.
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3.3 DESIGN OF FLAT SLAB
3.3.2 DIRECT DESIGN METHOD
Design: M20 grade of concrete & HYSD steel bars of grade fe415 will be
used for that flat slabs.
Design constants: the ratio of limiting value of the depth of neutral axis to
the effective depth of the slabf
Limiting moment of resistance factor for singly reinforced resistance
rectangular section
Max percentage of tensile reinforcement
STEP 1 : CLEAR SPAN
The flat slab shall be designed without drop panels and the supporting
column shall have column head of 400mm height with 45 inclination and
additional vertical height of 400mm. As per IS 456-1978, the column of circular
section shall be treated as square section having the same cross sectional area.
Therefore let a be the size of the square column.
=1.06m
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Clear span along AB (long span)
ll-n =[10-
=8.94m
Clear span along BC (short span)
ll-n =[8.75-
=7.69m
STEP 2 : ESTIMATED THICKNESS OF THE FLAT SLAB
The flat slab remains continuous over minimum of three span (direct
design method)
For slabs without drops:
Where is the modification factor
For HYSD steel of Fe415 grade
=0.96%
=1.4
Estimated effective depth
Let the overall estimated thickness be 200mm (more than 125mm
minimum thickness specified
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STEP 3 : ESTIMATED LOADS
(A) Characteristic Dead LoadIt consists of the weight of finishing surface plus the self weight of the
slab.
Weight of finishing surface =1k N/
Self weight of slab =
=5k N/
Total Wd =6k N/
(B)Characteristic live LoadLive load =7.5k N/
Total Load=wd + w1 =6 + 7.5
=13.5k N/
(C)Factored (design) load in the panelWfd =1.5 x (wd + w1)
=1.5 x 13.5=20.25k N/
STEP 4 : WIDTH OF THE STRIPS FOR INTERIOR PANEL
(A) Width of strip along AB (long span)Span AB, l1=10m
Span BC, l2 =8.75m
Width of column strip
0.25 l2 = 0.25 x 8.75
=2.18m (adopted)
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Width of middle strip
(0.75 - 2.18 - 2.18) = 4.39m
Width of strip
0.25 l2 = 0.25 x 10
=2.5m
>0.25 l1 = 0.25 x 8.75
=2.18m (adopted)
Width of middle strip
(102.18 2.18) = 5.64m
STEP 5 : FACTRED (DESIGN) MOMENT FOR SPAN
(A) Width of strip along AB (long span)l1=10m, ll-n = 8.94m, l2 = 8.75m
Total design load over an area, l2 x ll-n
WFd = 1.5 (Wd + W1) x l2 x ll-n
WFd = 20.25 x 8.75 x 8.94
= 1584 kN
Total factored design moment along the direction AB
Mo.AB =
=
=1770.12 kNm
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(B)Total design moment, Mo along BC (short span)l1=8.75, ll-n = 7.69, l2 = 10m
Total design load over an area, l2 x l1-n
WFd = 1.5 (Wd + W1) x l2 x ll-n
WFd = 20.25 x 10 x 7.69
= 1557.22 kN
Total factored design moment along the direction BC
Mo.AB =
=
=1496.87 kNm
STEP 6 : NEGATIVE AND POSITIVE DESIGN MOMENTS
In an interior span, the total design moment
Mo is distributed in the following proportions in each direction
(A)Distribution of total design moment MO.ABNegative design moment,
Mn.AB = 0.65 MO.AB
Mn.AB = 0.65 x 1770.12
= 1150.57 kNm
Positive design moment,
MP.AB = 0.35 MO.AB
Mp.AB = 0.35 x 1770.12
= 619.54 kNm
(B)Distribution of total design moment MO.BC
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Negative design moment,
Mn.BC = 0.65 MO.BC
Mn.BC = 0.65 x 1496.87
= 972.96 kNm
Positive design moment,
MP. BC = 0.35 MO.BC
Mp.BC = 0.35 x 1496.87
= 523.90 kNm
STEP 7 : EFFECTS OF PATTERN OF LOADING
The effects of pattern of loading for negative and positive BM maybe
seen before distribution of these
Moments across the panel width.
Ratio of live load to dead load
The ratio of
exceeds 0.5
(A)For the distribution along ABThe ratio of flexural stiffness of the columns above and below the slab
to the flexural stiffness of the
Slabs at a joint in the direction along AB.
(I)Stiffness coefficient for slabSize of equivalent square column (diameter of column is 1200mm)
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c/c distance between the columns
c1 = [
= 1063mm
Along AB, l1=10m
Along BC, l2=8.75m
C1/l1 =1.06/10
=0.106
C2/l2 =1.06/8.75
=0.121
As per SP241983, stiffness factor for slab without drop panel
K=4.18
For the slab of uniform thickness
Ks =
x1
=
x
Ks = 2.438 Ec x 10-3
kNm
(II) Stiffness coefficient for column
Length of column section a = 100mm = 0.1m
Depth of 45 degree capital (below the soffit of the slab)
b1 = 440mm
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= 0.44m
lc = 2.8 + 0.1 + 0.1
= 3m
a/lc = 0.1/3
b1/lc =0.44/3
=0.146
Column stiffness coeffients for columns with 45 degree tapered capital
for above ratios
For upper (capital) end,
Kc = 5.474
For lower (base) end,
Kc = 4.964
Stiffness of column for upper (capital) end
Kc =
=
x
x (1.2)4
Kc = 0.315 Ec kNm
Stiffness of column for lower (base) end
Kc =
x
x (1.2)
4
Kc = 0.285 Ec kNm
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c = 123.05
= 0.875
=1.25
From table 9.1 (as per IS 4561978) by interpolation
c min =0.7
c min is lessthan c = 123.05
Therefore, the correction for pattern of loading in the direction AB is not
necessary.
(A) For the direction along BCThe ratio of flexural stiffness of the columns above and below the slab to the
flexural stiffness of the slab at a joint in the direction along is:
1. Stiffness coefficient for slab.Size of equivalent square column (diameter of column is 1200mm)
c1 =[
=1063mm
c/c distance between columns
Along BC, l1 = 8.75m
Along AB, l2 = 10m
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= 0.121
= 0.106
From table 9.2, stiffness factor for the slab without drop panel (as per SP 24 :
1983)
K = 4.26
Ks =
x
x (0.2)3
Ks =3.245 x 10-3 kNm
(ii) Stiffness coefficient for column
Length of column section (considered rigid)
A = 100mm
=0.1m
Depth of 45 capital (below the soffit of slab)
b1 = 440mm
lc = 2.8 + 0.1 + 0.1
= 3m
= 0.033 ~ 0.35
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= 0.146
From 9.5 table, column stiffness coefficient for colums with 45 tapered capital
For upper (capital) end for above ratios
Kc = 5.474
For lower (base) end.
Kc = 4.964
Stiffness for column for upper (capital) end
Kc =
x
x (1.2)4
Kc = 0.315 Ec kNm
Stiffness for column for lower (base) end
Kc =
x
x (1.2)4
Kc = 0.285 Ec kNm
= 92.44
= 0.875
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=1.25
From table 9.1 (as per IS 4561978) by interp
c min = 0.9
c min =0.9 less than c = 89.98
The correction for pattern of loading in the direction BC is not needed
STEP 8: DISTRIBUTION OF BM ACROSS THE PANEL WIDTH
(A) Across the strip of span AB(i) Negative moment at an interior support(a)Column strip : 75% of the total ve moment in the panel at this
support is resisted by the column strips.
Therefore, the moment resisted by each column strip
(Since MnAB=1150.57 kNm)
1/2 Mnc =
x 0.75 MnAB
=
x 0.75 x 1150.57
= 431.46 kNm
(b) Middle strip : Theve moment resisted by the middle strip
Mnm = 0.25 Mnm
= 0.25 x 1150.57
= 287.64 kNm
Check : ( 431.46 + 287.64 + 431.46 ) = 1150.56 kNm
(ii)Positive moment at an interior support
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(a)Column Strips : 60% of total +ve moment in the panel is resisted by
the column strips.
Therefore, the moment resisted by each column strips
Mp.AB = 619.54 kNm
x Mp.c =
x 0.60 Mp.AB
= (
x 0.60 x 619.54 )
= 185.86 kNm
(b)Middle strips : The +ve moment resisted by the middle strips
Mpm =0.40 Mp.AB
= 0.40 x 619.54
= 247.816 kNm
Check : 185.86 + 247.81 + 185.86 = 619.53 kNm
(B)Across the strips of span BC(i) ve moment at an interior supporColumn strips : 75% of total -ve moment in the panel at this support is
resisted by the column strips.
Therefore, the moment resisted by each column strip
(since Mn.BC = 972.96 kNm)
1/2 Mnc =
x 0.75 MnBC
=
x 0.75 x 972.96
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= 364.86 kNm
(c)Middle strips : The -ve moment resisted by the middle stripsMnm =0.25 Mn. BC
= 0.25 x 972.96
= 247.24 kNm
Check : 364.86+ 243.24 + 364.86 = 972.96 kNm
(ii)Positive moment at an interior support.(a)Column strips : 60% of total positive moment in the panel is by the
column strips.
Therefore, the moment resisted by each column strip
(since Mn.BC = 972.96 kNm)
MpBC = 523.90 kNm
1/2 Mpc =
x 0.60 x MnBC
=
x 0.60 x 523.90
= 157.57 kNm
(b)Middle strips : The +ve moment resisted by the middle strips.Mp.m =0.40 Mp. BC
= 0.40 x 523.90
= 209.56 kNm
Check : 157.17+ 209.56 + 157.17 = 523.90 kNm
STEP 9: STRIP OF SPAN AB
A.Strip of span AB(i) Negative moment
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(a) Column strips : the width of column strip of the interior panel is
2.18m
Therefore, the moment per meter width
Mnc/meter =
= 197.91kNm/m
(b) Middle strips : the width of middle strip of the interior panel is
4.39m
Therefore, the moment per meter width
Mnm/meter =
= 65.52 kNm/m
(ii) Positive moment
(a) Column strips : the width of column strip of the interior panel is
2.18m
Therefore, the moment per meter width
Mnc/meter =
= 85.25kNm/m
(b) Middle strips : the width of middle strip of the interior panel is4.39m
Therefore, the moment per meter width
Mnm/meter =
= 56.45 kNm/m
(A)
Strip of span BC
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(i) Negative moment
(a) column strip : width = 2.18m
Therefore, the moment per meter width
Mnm/meter =
= 168.36 kNm/m
Middle strip : width = 5.64m
Therefore, the moment per meter width
Mnm/meter =
= 43.12 kNm/m
(ii) Positive moment
(a) Column strips : Width = 2.18m
Therefore, the moment per meter width
Mnm/meter =
= 37.15 kNm/m
STEP 10: EFFECTIVE DEPTH OF FLAT SLAB
The absolute maximum factored B.M out of eight value of the step 9
is 197.91kNm/m
The effective depth of the flat slab
0.36 fckx
x (1-0.42 x
) bd2 = MFD
[0.36 x 20 x 0.48 x (1-0.42 x 0.48 ) x
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x1000 x d2] = 197.91 x 10
3456 (1-0.2016)d2 = 197.91 x 10
3456d2
696.72d2
=197.91 x 106
2759.28d2
=197.91 x 106
d =267.81mm
Assume d = 275mm
Let the effective cover be 25mm
Therefore, overall thickness of slab D=300mm
STEP 11 : SLAB REINFORCEMENT
The spacing of bars in flat slab shall not exceed 2 times the slab
thickness
2x300 = 600mm2
Therefore, provide 12mm diameter HYSD steel bars shall be
provided
Ast =
= 113.04mm2
(A)Reinforcement along the span AB(i) Negative moment reinforcement(a)Column strips
0.87 fy Ast d (1-
) = MFD
0.87 x 415 x Ast x 275 (1-
) =197.91 x 10
6
99288.75 Ast - Ast2
=197.91 x 106
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0.87 fy Ast d (1-
) = MFD
0.87 x 415 x Ast x 275 (1-
) =43.12 x 106
Ast2
13253.01 Ast +5756240.82 = 0
Ast =449.58mm2
S =
x 1000
=
x 1000
= 251.43mm
Therefore,provide bars 12 nos @ 250mm spacing at bottom
(a)Column strips0.87 fy Ast d (1-
) = MFD
0.87 x 415 x Ast x 275 (1-
) =72.09 x 10
6
Ast213253.01 Ast +9623548.25 = 0
Ast =770.99mm2
S =
x 1000
=
x 1000
= 146.61mm
Therefore,provide bars 12 nos @ 145mm spacing at bottom
(b) Middle strips
0.87 fy Ast d (1-
) = MFD
0.87 x 415 x Ast x 275 (1-
) =37.15 x 106
Ast213253.01 Ast +5756240.82 = 0
Ast =385.40mm2
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(i) Column strips:
(a) 50% at top
Projection = half the size of square column support + 0.31l-n
=(
x 1.06 + (0.22 x 7.69))
=2.83m
(b)50% at top
Projection =(
x 1.06 + (0.2 x 7.69))
=2.06m
(ii) Middle Strips : (100% at top)
(d)50% at topProjection = half the size of square column support + 0.221l-n
=(
x 1.06 + (0.22 x 7.69))
=2.22m
STEP 12 : SHEAR IN FLAT SLABS
The diameter of column head is 2000mm and the effective depth of flat slab, d is
275mm.
The diameter of critical section for two way round the column head
(DE+
= (2000+
)
= 2275mm
Peripheral distance
bo = x 2.275
=7.143m
factored share force at critical section
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V = 1.5 (Wd + W1) (l . l2
(DE + d)
2)
V = 1.5 (6 + 7.5) (10 x 8.75
(2.275)2)
V = 1689.60KN
Normal shear stress
=
v =
= 0.8601 N/mm2
Or permissible shear stress
The grade of concrete is M20. The permissible shear stress in flat slab in concrete
(limit state method of design)
v =0.25 (fck)1/2
v =0.25 (20)1/2
= 1.118 N/mm2
When the shear reinforcement is not provided the nominal shear stress calculated
at the critical section shall not exceed KsTc
Ks = (0.5 + c) and not greater than 1.0
c = (1.00 for circular column
Ks = 1.0
KsTc = 1 x 1.118
=1.118 N/mm2
v = 0.8601 N/mm2 not greater than (KsTc=1.118 N/mm2)
Therefore Stisfiled
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3.4 DESIGN OF SLAB
3.4.1 DESIGN OF TWO WAY SLAB : S2
Clear size =5m x 10m
Wall thickness =230mm
SIDE RATIO:
Side ratio,
=
= 2 2m
Hence the slab is to be designed as two way slab
value = 32
Basic value =
Depth (d) =
= 156.25mm
Adopt depth =150mm
Effective cover (d) = 20mm
Assume 10 mm dia bars
Effective cover = 20+
x10=25mm
Overall depth (D) = d + d
= 150 + 25
= 175mm
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Two adjacent edges contines
=
= 1.97
x = 0.09101
y =0.047 (IS 456 : 2000, pg
no: 91)
Mu (x) =0.09101 x 19.312 x 5.152
= 46.18kNm
Mu (y) =0.047 x 19.312 x 5.152
= 24.07kNm
CHECK FOR THE DEPTH:
Mu(lim) = 0.138 x fckx b x d2
= 0.138 x 20 x 1000 x d2
D = 129.35mm < 150mm
Hence safe
AREA OF THE REINFORCEMENT:
Mu (x) =0.87 x fy x Ast (x) x d x d
[1-
]
46.18 x 106
= 0.87 x 415 x Ast (x)
X 150 ( 1-
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Ast (x) = 987.74mm2
Mu (y) =0.87 x fy x Ast (y) x d
x [1-
]
24.07 x 106 = 0.87 x 415 x Ast (y)
X 140 ( 1-
Ast (y) = 426.98mm2
Spacing of reinforcement:
Provide 10mm dia:
S =
x 1000
ast =
x d
2
=
x 10
2
= 78.53mm2
=
x 1000
= 79.5mm75mm
Sy =
x 1000
= 183.91mm180mm
Maximum permitted spacing:
a) 3xd = 3 x 150
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= 450mm
b) 450mmProvide 10mm @ 75mm spacing
EDGE STRIP
Ast(min) = 0.12% of bd
=
x 1000 x 150
= 180mm
Spacing of reinforcement:
Provide 10mm dia
S =
x 1000
ast =
x d2
=
x 10
2
= 78.53mm2
=
x 1000
= 436.27mm400mm
Maximum permitted spacing:
c) 5xd = 5 x 150= 750mm
d) 450mm
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Provide 10mm @ 400mm c/c spacing
CHECK FOR SHEAR:
V =
u =
V =
= 0.33 N/mm2
K=1.25 [ for 175mm depth) (IS 456:2000, page no : 72]
% of =
x 100
=
x100
= 0.66
= 0.5312 (table no : 19, page
no:85, IS 456 : 2000]
xk = 0.5312 x 1.25
= 0.664 N/mm2
c(max) = 2.8N/mm2
[table 20]
CHECK FOR DEFLECTION:
L/d (max) = 32 [IS 456 : 2000, page no:38]
L/d (max) = 32 x 1.2 = 38.4
L/d (actual) =
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=
= 34.1
34.1
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Assume 10 mm dia bars
Effective cover = 20+
x10=25mm
Overall depth (D) = d + d
= 150 + 25
= 175mm
EFFECTIVE LENTH @ SUPPORT:
1. Clear span + effective depth = 5+0.15= 5.15m
2. Clear span + c/c distance of the support =5+
+
= 5.23m
Effective length = 5.15m
LOAD CALCULATION:
Self weight = L x B x density of
concrete
= 0.175 x 1 x 25 x 1
= 4.375 kN/m2
Live Load = 7.5 kN/m2
Floor finish = 1 kN/m2
Total load (W) = 4.375 + 7.5 + 1
= 12.875 kN/m
2
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Wu = 12.875 x 1.5
= 19.31kN/m2
DESIGN BENDING MOMENT:
Mu = x x Wu x lx2
Mu = x x Wu x lx2 (IS 456 : 2000, pg
no: 90)
One adjacent edges discontinues
=
= 1.72
x = 0.0758
y =0.037(IS 456 : 2000, pg no:
91)
Mu (x) =0.0758 x 19.312 x 5.152
= 38.82kNm
Mu (y) =0.037 x 19.312 x 5.152
= 18.95kNm
CHECK FOR THE DEPTH:
Mu(lim) = 0.138 x fckx b x d2
38.82 x 106
= 0.138 x 20 x 1000 x d2
d = 118.59mm < 150mm
Hence safe
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AREA OF THE REINFORCEMENT:
Mu (x) =0.87 x fy x Ast (x) x d x [1-
]
38.82 x 106 = 0.87 x 415 x Ast (x)
X 150 ( 1-
Ast (x) = 806.97mm2
Mu (y) =0.87 x fy x Ast (y) x d x [1-
]
18.95 x 106
= 0.87 x 415 x Ast (y)
X 140 ( 1-
Ast (y) = 398.42mm
2
Spacing of reinforcement:
Provide 10mm dia:
S =
x 1000
ast =
x d
2
=
x 102
= 78.53mm2
=
x 1000
Sx = 97.31mm90mm
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Sy =
x 1000
= 197.1mm190mm
Maximum permitted spacing:
a) 3xd = 3 x 150= 450mm
b) 450mmProvide 10mm @ 90mm spacing
EDGE STRIP
Ast(min) = 0.12% of bd
=
x 1000 x 150
= 180mm
Spacing of reinforcement:
Provide 10mm dia
S =
x 1000
ast =
x d
2
=
x 102
= 78.53mm2
=
x 1000
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= 436.27mm400mm
Maximum permitted spacing:
c) 5xd = 5 x 150= 750mm
d) 450mmProvide 10mm @ 400mm c/c spacing
CHECK FOR SHEAR:
V =
Vu =
V =
V =
= 0.33N/mm2
K=1.25 [ for 175mm depth) (IS 456:2000, page no : 72]
% of steel =
x 100
=
x100
= 0.53
= 0.4896 (table no : 19,
Page no:85, IS 456 : 2000]
cxk = 0.4896 x 1.25
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= 0.612 N/mm2
c(max) = 2.8N/mm2 [table 20]
Hence shear is safe and design is ok
CHECK FOR DEFLECTION:
L/d (max) = 32 [IS 456 : 2000, page no:38]
L/d (max) = 32 x 1.2 = 38.4
L/d (actual) =
=
= 34.1
34.1
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LOAD CALCULATION:
Self weight = L x B x density of
concrete
= 0.35 x 1 x 25 x 1
= 7.5 kN/m2
Live Load = 7.5 kN/m2
Floor finish = 1 kN/m2
Total load (W) = 7.5 + 7.5 + 1
= 16 kN/m2
Wu = 16 x 1.5
= 24 kN/m2
DESIGN BENDING MOMENT:
Mu = x x Wu x lx2
Mu = x x Wu x lx2 (IS 456 : 2000, pg
no: 90)
One adjacent edges discontinues
=
= 1.13
x = 0.0551
y =0.047(IS 456 : 2000, pg no:
91)
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Mu (x) =0.0551x 24 x 8.982
= 106.63kNm
Mu (y) =0.037 x 24x8.982
= 71.6kNm
CHECK FOR THE DEPTH:
Mu(lim) = 0.138 x fckx b x d2
106.63 x 106
= 0.138 x 20 x 1000 x d2
d = 196.55mm < 275mm
Hence safe
AREA OF THE REINFORCEMENT:
Mu (x) =0.87 x fy x Ast (x) x d x [1-
]
38.82 x 106
= 0.87 x 415 x Ast (x)
X 275 ( 1-
Ast (x) = 1178.76mm2
Mu (y) =0.87 x fy x Ast (y) x d x [1-
]
18.95 x 106 = 0.87 x 415 x Ast (y)
X 265 ( 1-
Ast (y) = 798.29mm2
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CHAPTER-IV
INTRODUCTION
4.1 General
Multi-storied buildings are usually constructed for offices, residential
flats, hotels, hospitals,social centers etc. The framing of multi-stored building
consists of colums and beams which supports roof and floor load.
A multi-stored, Multi-planned frame is a complicated statically
indeterminate structure. It consists of a number of beams and columns built
monolithically, framing network.
The doors and walls are supported on beams which transmit the load to
the colums.A building frames is objected to both verticals as well as horizontal
loads. The vertical loads consist of the dead weight of the structural components
such as beams, slabs, columns, etc and live load.
Structural behavior of the multi-storied buildings subjected to the lateral
forces complex and highly indeterminate. There are three are three recognized
types of joints between beams and columns, simper, semi rigid and rigid joints.
Frames with flexible joints have no internal resistance against horizontal loads.
4.2 Analysis Methods
The bending moments is the beam and columns of a substitute frame may
be computed for the following methods:
1. Slope deflection method2. Moment distribution method3. Building frame formulae4. Kanis method
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MF BA = 0
MF BC = -Wl2/12
=-(37.5 x 8.752
)/12
=-239.25kN.m
MF CB = Wl2/12
=-(37.5 x 8.752)/12
=-239.25kN.m
MFCD = 0
MFDC = 0
Table 4.3.1.1 Load and FEM:
SL No MEMBER LOAD (kN/m FEM(kNm)
1 BC 37.5 -239.25
2 CD 37.5 239.25
Table 4.3.1.2 DISTRIBUTION FACTOR
Joint Member K K D.F
B BA 0.308 E1 0.422 E1 0.729
BC 0.114 E1 0.270
C CB 0.114 E1 0.422 E1 0.270CD 0.308 E1 0.729
Table 4.3.1.3 DISTRIBUTION TABLE:
Joing B C
BA BC CB CD
D.F 0.729 0.270 0.270 0.729
F.E.M 0 -239.25 239.25 0
1 distribution 174.41 64.59 -64.59 -174.41
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Carry Over 0 -32.29 32.29 0
Total 174.41 -206.95 206.95 -174.41
BENDING MOMENT DIAGRAM:
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RB + RC =37.5 x 8.75
= 328.12 kN
MB=0
-Rc x 8.75 + 206.95 + 37.5 x 8.75 x 4.375206.95 = 0
-8.75 Rc = -1435.57
Rc = 164.0kN
RB + 164.0 = 328.12kN
RB
MA=0
-RB x 3.24 + 174.4 = 0
-RB = -53kN
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Since RA + RB = 0
RA + 53 = 0
Therefore RA = -53kN
SHEAR FORCE DIAGRAM:
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Table 4.3.1.1 Load and FEM:
SL No MEMBER LOAD (kN/m FEM(kNm)
1 BC 24.4 -73.2
2 CD 24.4 -73.2
Table 4.3.2.2 DISTRIBUTION FACTOR
Joint Member K K D.F
B BA 0.34 E1 0.51 E1 0.67
BC 0.17 E1 0.34
C CB 0.17 E1 0.51 E1 0.34
CD 0.34 E1 0.67
Table 4.3.1.3 DISTRIBUTION TABLE:
Joing B C
BA BC CB CD
D.F 0.67 0.34 0.34 0.67
F.E.M 0 -73.2 73.2 0
1ST
distribution
49.044 24.88 -24.88 -49.044
Carry Over 0 -12.44 12.44 0
Total 49.044 -60.76 60.79 -49.044
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RB = 73.2kN
MA=0
-RB x 3 + 49.04 = 0
-RB = -16.3kN
Since RA + RB = 0
RA + 16.3 = 0
Therefore RA = -16.3kN
SHEAR FORCE DIAGRAM:
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CHAPTER-V
DESIGN OF BEAMS
4.1 General
Structural concrete beam elements are designed to support a given
system of external loads such as walls and slabs of roof and floor systems. The
cross sectional dimensions are generally assumed based on serviceability
requirements. The width is fixed based on thickness of walls and housing of
reinforcement ad the depth is selected to control deflections within safe
permissible limits.
5.2 Classification of beams
According to support condition:
1. Simply supported beam2. Cantilever beam3. Fixed beam4. Over hanging beam5. Continuous beamAccording to structural steel configuration:
1. Simple beam2. Compound beam3. Plate beam4. Turn beam
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5.3 DESIGN OF BEAMS
5.3.1 DESIGN OF BEAMS B1
CONTINOUS BEAM
GIVEN DATA
L = 10m
Live load =7.5kN/m2
fy = 415N/mm2
fck = 20 N/mm2
CROSS SECTIONAL DIMENTIONS
Depth = spm / 12
= 10000 / 12
Adopt d = 800mm
D = 850 mm
b = 300 mm
cover = 50mm
LOAD CALCULATION
Self weight of the slab = (7.5 x 2.5 x 0.3 x 25)/10
= 14.1 kN/m
Self weight of the beam =(0.3 x 0.85 x 25)
= 6.4 kN/M2
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Self weight of the wall = (0.23 x 1 x 20)
= 4.6 kN/m2
Finishes = 1 kN/m2
Total dead load = 26.1 kN/m2
Live load = 7.5 kN/m2
Total load = 33.6 kN/m2
BENDING MOMENTS AND SHEAR FORCE
Negative moment
Mu(-ve) = 1.5[dead load / 10 + live load / 9]
=1.5[26.1 x 10 x 10/10 + 7.5
x 10 x 10/9]
=516.5 kN.m
Positive moment
Mu(+ve) = 1.5[dead load / 12 +
live load / 10]
=1.5[26.1 x 10 x 10/12 + 7.5 x 10
x 10/10]
=438.75 kN.m
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MAXIMUM SHEAR FORCE
Vu = 1.5x0.6 x span[dead load +
live load ]
=1.5 x 0.4 x 10 x (26.1 + 7.5)
= 201.6 kN
LIMITING MOMENT OF RESISTANCE
Mulimt = 0.138 fckbd
2
=0.138 x 20 x 300 x 800
2
= 529.92 kN.m
Since Mu
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CHECK FOR DEFLECTION CONTROL
(L/d)max =(L/d)basic x Kt x Kc x Kf
(L/d)actual =10000/800
=12.5
[from IS 456:2000]
Kt = 1.4
Kc =1.50
Kf =1
(L/d)max = 20 x 1.4 x 1.50 x 1
= 54.6mm
12.5 < 54.6mm
(L/d)actual < (L/d)max
Hence deflection is safe
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=1.5 x 0.5 x 6 x (20.6 + 7.5)
= 126.45 kN
LIMITING MOMENT OF RESISTANCE
Mulimt = 0.138 fckbd2
=0.138 x 20 x 300 x 5002
= 207 kN.m
Since Mu
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=12
[from IS 456:2000]
Kt = 1.2
Kc =0.75
Kf =1
(L/d)max = 26 x 1.2 x 0.75 x 1
= 23.4mm
12 < 23.4mm
(L/d)actual < (L/d)max
Hence deflection is safe.
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5.3.3 DESIGN OF BEAMS B3
SIMPLY SUPPORTED BEAM
DATA
Breadth of beam = width of wall
ie; breadth of beam, b =230mm
effective cover, d =25mm
depth of beam = span/12
=6000/12
=500mm
overall depth, D = 500 + 25 = 525mm
span of beam, L =6m
LOAD CALCULATION
Load on beam due to two way slab= x h(a+b) x D x density
Where D is the depth of slab = x 2 x ( 2 + 6) x 0.15 x 25
=30kN
Uniformly distributed load (udl) =30/6
Total load from slab to beam = 5kN/m
Brickwork load = 1 x 0.23 x 3 x 20
Live load = 2 kN/m2
Floor finishes = 1 kN/m2
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Ast =45.47 + 1100.78
= 1146.25mm2
Provide 4 bars of 20mm dia (Ast = 1256 mm2
)
SHEAR REINORCEMENT
V =Vu/bd
=(111 x 103)/(230 x 500)
=0.965 N/mm2
Pt =(100xAst)/bd
=(100 x256) / (230 x 500)
=1.09N/mm2
[from IS456:2000,table 19]
c =0.68N/mm2
Since Tv>Tc
Hence Shear reinforcement is required
Vus =(VuTc bd)
=(111-(0.68 x 230 x 500) x 103)
=32.80kN
Using 8mm dia 2 legged stirrups
Sv =(0.87 x fyAscd)/Vus
=(0.87x415x2x25x500)/
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5.3.4 DESIGN OF BEAM
SIMPLY SUPPORTED BEAM
DATA
Breadth of beam = width of wall
ie; breadth of beam, b =230mm
effective cover, d =25mm
depth of beam = span/12
=6000/12
=500mm
overall depth, D = 500 + 25
= 525mm
span of beam, L =6m
LOAD CALCULATION
Load on beam due to two way slab = x h(a+b) x D x density +
(1/2 x b x h x D x density)
Where D is the depth of slab = x 2 x ( 2 + 6) x 0.15 x 25
(0.5 x 6 x 3 x 0.15 x 25)
=63.75kN
Uniformly distributed load (udl) =63.75kN + 10
= 10.6 kN
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Total load from slab to beam = 10.6kN/m
Brickwork load = 1 x 0.23 x 3 x 20
= 13.8
Live load = 2 kN/m2
Floor finishes = 1 kN/m2
Self weight of beam = 1 x 0.23 x 0.50 x 25
Total load = 3kN/m
= 10.6 + 13.8 + 2 + 1 + 3
= 30.4 kN
Design load, Wu =1.5 x 30.4
= 45.6kN
CALCULATION
Mu = 0.125 x WuL2
= 0.125 x 37 x 62
= 205.2 kNm
Vu = 0.5 x WuL = 0.5 x 45.6 x 6
= 136.8 kN
MAIN REINFORCEMENT
Mu(limit) = 0.138 x fckx bd2
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Total tension reinforcement Ast = (Ast 1 + Ast2)
Ast =271 + 1100.8
= 11371.8mm2
Provide 4 bars of 22mm dia (Ast = 1519.76 mm2)
SHEAR REINORCEMENT
V =Vu/bd
=(136.8 x 103)/(230 x 500)
=1.19 N/mm2
Pt =(100xAst)/bd
=(100 x1519.76) / (230 x 500)
=1.32N/mm2
[from IS456:2000,table 19]
c =0.68N/mm2
Since Tv>Tc
Hence Shear reinforcement is required
Vus =(VuTc bd)
=(136.8-(0.68 x 230 x 500) x 103)
=136.72kN
Using 8mm dia 2 legged stirrups
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Sv =(0.87 x fyAscd)/Vus
=(0.87x415x2x25x500)/(136.72 x
103)
=66mm
Sv>0.75d =(0.75 x 500)
=375mm
Adopt a spacing of 65mm near supports gradually increasing to
300mm towards c/c of span
CHECK FOR DEFLECTION CONTROL
(L/d)Max =(L/d)basic x Kt x Kc x Kf
(L/d)Actual =6000/500
=12
[from IS 456:2000], Kt = 1.4, Kc = 2, Kf= 1
(L/d)Max =20 x 1.4 x 2 x 1
= 56mm
21 < 56mm
(L/d)Actual < (L/d)Max
Hence deflection is safe
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3. Classification based on slenderness ratio:a. Short column
b. Log columns6.4 DESIGN OF COLUMN
6.4.1 DESIGN OF COLUMN C1
DESIGN LOAD
No.of floors % reduction
1 0
2 10
3 20
4 30
LOAD CALCULATION
Dead la load of RC slab = 1x1x0.3x25
= 7.5kN/m2
Weight of floor finish = 1 kN/m2
Total dead load =8.5 kN/m2
Assume each interior column take
Load from an area = 10 x 8.75
= 87.5m2
Dead load from floor = 8.5 x 87.5
=743.75kN
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= 498.73x103mm2
Asc =0.01(503.77 x 103)
5037.7mm2
[ Provide 11 numbers of 25mm dia bars with a nominal cover of 40
mm (5399.60mm2)
Or
{ Provide 14nos of 22mm dia bars with a nominal cover of 40mm
(5321.85mm2)
LATERAL TIES
of the dia of largest longitudinal bar
= x 25
=6.25mm
= x 22
= 5.5mm
= 6mm
Provide 6mm dia of lateral ties
PITCH
Least lateral dimension of column =1200mm
16x25 =400mm
16x22 = 352mm
300 mm
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Provide 350mm c/c of lateral ties
6.4 DESIGN OF COLUMN
6.4.1 DESIGN OF COLUMN C1
DESIGN LOAD
No.of floors % reduction
1 0
2 10
3 20
4 30
LOAD CALCULATION
Dead la load of RC slab = 1x1x0.3x25
= 7.5kN/m2
Weight of floor finish = 1 kN/m2
Total dead load =8.5 kN/m2
Assume each interior column take
Load from an area = 10 x 5
= 50m2
Dead load from floor = 8.5 x 50
=425kN
Assume each column takes load from = 10+10
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= 20m long beam
Dead load f wall = 0.23 x 1 x 20
= 4.6kN/m2
Dead load of beam = 0.3 x 0.85x25
= 6.375kN/m2
Total load = 4.6 + 6.375
= 10.975kN/m2
Dead load of wall & self weight of beam=10.975 x 20
=219.5kN
Total dead load from each floor = 425 + 219.5
=644.5kN
Including self weight of column,
load from each flor = 700kN
Dead load on ground floor column of
4 storey building =700 x 4
= 2800kN
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5390.625 x 103 =0.4x20 (Ag -0.01 Ag) + 0.67 x 415
x 0.01 Ag
Ag =424.04x103mm2
Therefore the side of the square column=(Ag)1/2
= 651.18mm
We adopt 1200 x 1200mm as side of the column
Ac =(Ag-0.01Ag)
= 0.99Ag
= 0.99(424.04 x 103)
= 419x103mm
2
Asc =0.01(424.04 x 103)
=4240.4mm2
[ Provide 14 numbers of 20mm dia bars with a nominal cover of 40
mm (4561.59mm2)
Or
{ Provide 14nos of 22mm dia bars with a nominal cover of 40mm
(5321.85mm2)
LATERAL TIES
of the dia of largest longitudinal bar
= x 20=5mm
= x 22= 5.5mm
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= 6mm
Provide 6mm dia of lateral ties
PITCH
Least lateral dimension of column =1000mm
16x25 =352mm
16x22 = 320mm
300 mm
Provide 300mm c/c of lateral ties
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6.4 DESIGN OF COLUMN C3
LOAD CALCULATION
Dead la load of RC slab = 1x1x0.2x25
= 5kN/m2
Weight of floor finish = 1 kN/m2
Total dead load =6 kN/m2
Assume each interior column take
Load from an area = 6 x 5
= 30m2
Dead load from floor = 6 x 30
=180kN
Assume each column takes load from = 6+5
= 11m length of beam
Dead load f wall = 0.23 x 1 x 20
= 4.6kN/m2
Dead load of beam = 0.23 x 0.525x25
= 3.01875kN/m
Total load = 4.6 + 3.01875
= 7.61875kN/m
Dead load of wall & self weight of beam=7.61875 x11
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=83.806kN
Total dead load from each floor = 180 + 83.806
=263.806kN
Including self weight of column,
load from each flor = 275kN
Live load from each floor = 2x6x5
= 60kN
Total dead load = 275kN
Total live load = 60kN
Design load = 1.5(d.1+1.1)
` = 1.5x(275 + 60)
= 502.kN
SIZE OF THE COLUMN
Assume 1% as the longitudinal steel in the column
Asc = 0.01Ag
Ac =(Ag - Asc)
Ac =0.99Ag
Pu 0.4fckAc + 0.67fy Asc
502.5 x 103
=0.4x20 (Ag -0.01 Ag) + 0.67 x 415
x 0.01 Ag
Ag =46960.42mm2
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6.3.4 DESIGN OF COLUMN
LOAD CALCULATION
Dead la load of RC slab = 1x1x0.2x25
= 5kN/m2
Weight of floor finish = 1 kN/m2
Total dead load =6 kN/m2
Assume each interior column take
Load from an area = 6 x 5
= 30m2
Dead load from floor = 6 x 30
=180kN
Assume each column takes load from = 6+5
= 11m length of beam
Dead load f wall = 0.23 x 1 x 20
= 4.6kN/m2
Dead load of beam = 0.23 x 0.525x25
= 3.01875kN/m
Total load = 4.6 + 3.01875
= 7.61875kN/m
Dead load of wall & self weight of beam=7.61875 x11
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Assuming 32mm dia bars
Ast =(3.14 x 322/4)
=803.84mm2
Number of bars =Ast/ast
=10483.7/490.62
=21.3
=21numbers
Provide 21nos of 25mm dia barss
CHECK FOR DEVELOPMENT LENGTH
=0.7fyd/4xTbd
=(0.87 x 415 x 25) / ( 4x1.2 )
=1880.46mm
Development length available =1400-(25/2)
=1387.50mm
D =(16/2)-(a/2)-clear cover
=5.95 m
SBC of soil
Column load =5390.625kN
Self weight =4x4x1400
=560kN
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Total load =5918.35kN
SBC of soil =total load/area
=5918.35/(4x4)
=369.8Tvy
Hence safe
CHECK FOR TRANSVERSE SHEAR
=Wx[AXB-(a+b)x(b+d)/(2a+2b+4d)d]
={336.9x[(16x4)-(x+1.2)(1.2+1.4)]/[
=)-(x+1.2)(1.2+1.4)]/
=[(2x4)+(2x1.2)+(4x1.4))]1.4}
=14020.92/22.4
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=36
Actual bearing stress =(5390.625 x 103)/(1200x1200)
=3.74
3.74
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Maximum bending moment at section XX,
Mx =125.6x2x0.75x0.75/2
=70.65kNm
=70.65x106Nmm
DEPTH OF FOOTING
MR = 2.98bd2
Effective depth required to resist the
Bending moment,
D =[(70.65x106)/(2.98x2000)]1/2
=108mm
=110mm
TENSION REINFROCEMENT
Maximum bending moment = 70.65x106mm
=0.87x415xastx216
[1-(4.8x10-5
)Ast2]
Ast2-20833Ast+18872916.6 =0
Ast =1898.3mm2
Provide 10 numbers of 16mm dia bars in each direction
DEVELOPMENT LENGTH
Development length required for the
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BUILDING
DRAWINGS
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