Car Parking Design

7/30/2019 Car Parking Design

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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.


(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.


(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


Mnm/meter =

= 65.52 kNm/m

(ii) Positive moment

(a) Column strips : the width of column strip of the interior panel is

2.18m


Mnc/meter =

= 85.25kNm/m

(b) Middle strips : the width of middle strip of the interior panel is4.39m


Mnm/meter =

= 56.45 kNm/m

(A)

Strip of span BC


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(i) Negative moment

(a) column strip : width = 2.18m


Mnm/meter =

= 168.36 kNm/m

Middle strip : width = 5.64m


Mnm/meter =

= 43.12 kNm/m

(ii) Positive moment

(a) Column strips : Width = 2.18m


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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25/109


26/109

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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32/109

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


Provide 10mm dia

S =

x 1000

ast =

x d2

=

x 10

2

= 78.53mm2

=

x 1000

= 436.27mm400mm


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) =


36/109

=

= 34.1

34.1


37/109

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



38.82 x 106

= 0.138 x 20 x 1000 x d2

d = 118.59mm < 150mm

Hence safe


39/109


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


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


a) 3xd = 3 x 150= 450mm

b) 450mmProvide 10mm @ 90mm spacing

EDGE STRIP

Ast(min) = 0.12% of bd

=

x 1000 x 150

= 180mm


Provide 10mm dia

S =

x 1000

ast =

x d

2

=

x 102

= 78.53mm2

=

x 1000


41/109

= 436.27mm400mm


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


42/109

= 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


43/109


44/109

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)


45/109

Mu (x) =0.0551x 24 x 8.982

= 106.63kNm

Mu (y) =0.037 x 24x8.982

= 71.6kNm



106.63 x 106

= 0.138 x 20 x 1000 x d2

d = 196.55mm < 275mm

Hence safe


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


46/109


47/109


48/109


49/109

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


50/109


51/109

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


52/109

Carry Over 0 -32.29 32.29 0

Total 174.41 -206.95 206.95 -174.41

BENDING MOMENT DIAGRAM:


53/109

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


54/109

Since RA + RB = 0

RA + 53 = 0

Therefore RA = -53kN

SHEAR FORCE DIAGRAM:


55/109


56/109

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


57/109


58/109

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:


59/109

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


60/109

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


61/109

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


62/109

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


63/109


64/109

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


65/109


66/109


67/109

=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


68/109


69/109

=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.


70/109

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


71/109


72/109


73/109

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)/


74/109


75/109

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


76/109

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


77/109


78/109

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


79/109

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


80/109


81/109

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


82/109


83/109


84/109

= 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


85/109

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


= 7.5kN/m2


Total dead load =8.5 kN/m2


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


86/109

= 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


87/109


88/109

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


89/109

= 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


90/109

6.4 DESIGN OF COLUMN C3

LOAD CALCULATION


= 5kN/m2


Total dead load =6 kN/m2



= 30m2

Dead load from floor = 6 x 30

=180kN


= 11m length of beam


= 4.6kN/m2


= 3.01875kN/m

Total load = 4.6 + 3.01875

= 7.61875kN/m

Dead load of wall & self weight of beam=7.61875 x11


91/109

=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


92/109


93/109

6.3.4 DESIGN OF COLUMN

LOAD CALCULATION


= 5kN/m2


Total dead load =6 kN/m2



= 30m2

Dead load from floor = 6 x 30

=180kN


= 11m length of beam


= 4.6kN/m2


= 3.01875kN/m

Total load = 4.6 + 3.01875

= 7.61875kN/m

Dead load of wall & self weight of beam=7.61875 x11


94/109


95/109


96/109


97/109


98/109


99/109


100/109

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


101/109

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


102/109


103/109

=36

Actual bearing stress =(5390.625 x 103)/(1200x1200)

=3.74

3.74


104/109


105/109

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


106/109


107/109

BUILDING

DRAWINGS


108/109


109/109

Documents

Car Parking Design