G + 9 residential building design

7/26/2019 G + 9 residential building design

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A PROJECT ON

G+9 RESIDENTIAL BUILDING

ByJENERIUS MINJ

University Roll No : 10301313132University Registration No : 131030120098

Submitted in partial fulfillment of

The requirements for award of B. Tech Degree in Civil EngineeringOf theMaulana Abul Kalam Azad University of Technology, West Bengal

Guider by Assistants Prof. KAUSIK BERA, Assistants Prof. GOUTAM DUTTA.

Department Of Civil Engineering

Haldia Institute Of Technology, Haldia

Haldia Institute Of Technology


Icare Complex, Hit Campus

Purba Medinipur

Haldia721657 (West Bengal) India


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FORWARDIt is my pleasure to forward this project on the planning and design of G+9

Residential Building to our respective Prof. and project adviser Assistant Prof.

It helps to establish a good concept on paining, design, construction of a malty-storied residential building.

Assistants Prof. KAUSIK BERA, Assistants Prof. GOUTAM DUTTA



..

(N. K. Yadav)

H.O.D




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ACKNOWLEDGEMENT

I express my deep sense of gratitude to my guide ass department of civilengineering haldia institute of technology, for the valuable guidance.

I express my sincere thanks to Mr, N. K. Yadav, head of the department,department of civil engineering.

Above all I thank God, the almighty for his grace without which it would nothave been possible to complete this work in time.


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Sr. No. Topic

1. Introductioni). Effective Spanii). Stiffness

iii). Loadsiv). Analysis

2. Typical Floor Plan

3. Moment Calculation by Moment DistributionMethod

4. Design of One way Slab

5. Design of Two way Slab

6. Design of T-Beam

7. Design of Column

8. Design of Staircase

9. Design of Flat Footing

10. Conclusion

11. References

Table of Contents


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Multistory Building

1.1. INTRODUCTION

The aim of this project is to design a Multistory Building (G+9) for residential purpose,

taking earthquake load into consideration.

Multistory buildings are very commonly seen in cities. Construction of such tall buildings are

possible only by going to a set of rigidly interconnected beams and column. These rigidly

interconnected beams and columns of multi bay and multistoried are called Buildings frames.

To avoid long distance of travel, cities are growing vertically rather than horizontally. Inother words multistory buildings are preferred in cities.

Building laws of many cities permits construction of ground plus three storey buildings

without lifts.

The loads from walls and beams are transformed to beams, rotation of beams take place.

Since, beams are rigidly connected to column, the rotation of column also take place. Thus

any load applied any where on beam is shared by entire network of beam and columns.


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1.2. EFFECTIVE SPAN

As per IS 456-2000, in the analysis of frames, the effective length of members shall be center

to centre distance (clause 22.2 d)

1.3. STIFFNESS

For the analysis of frame, the relative stiffness values of various members are required. IS

456-2000 clause suggests the relative stiffness of the members may be based on the moment

of inertia of the section.

The made shall be consistent for all the members of the structure throughout analysis. It

needs arriving at member sizes before designing. The sizes are selected on the basis of

architectural, economic and structural considerations.

For Beamsspan to depth ratio preferred is 12 to 15. Width is kept (1/3) to (1/2) of depth, but

sometimes they are fixed on architectural consideration.

Columnsizes are to be selected on the basis of experience.

It is to be noted that in Multistory frames, columns of upper stories carry less axial force but

more moments, while columns of lower story carry more axial loads and less moments.

Design can roughly estimate the axial load on lower story column and arrive at sizes of the

column.

Next two to three stories can have same size. Beyond that, sizes may be reduced. Stiffness of

member is given by (I / L).

1.4. LOADS

For Multistory frames Dead load, imposed load (live load), wind load and earthquake loads

are important for designing.

The IS code suggests following load combination to get designed loads:

1. 1.5DL + 1.5IL

2. 1.5DL + 1.5WL

3. 1.5DL + 1.5EL

4. 1.2DL + IL + 1.2WL

5. 1.2DL + IL + 1.2EL


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1.5 ANALYSIS

It may be analyzed as a set of intersecting frames taking care of loads from triangular

pattern of loads from floors. However, IS 456-2000 (Clause 22.42) permits the analysis of

frames by approximate methods like:

Portal method, cantilever method, Substitute frame method for Dead loads, factor method for

wind loads; to arrive at design moments, shear and other forces.

We have adopted moment distribution method for frame analysis.


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TYPICAL FLOOR PLAN

LIFT

3X2.5


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MOMENT CALCULATION:-

D.F Calculation:-

JOINT MEMBER STIFFNESS (K) D.F.=

A

AB

+

0.32

AE

0.34

AF

0.34

B

BC

+2 +

0.19

BA 0.26

BG

0.27

BH

0.27

C

CD

+

+

0.28

CB 0.19

CI

0.27

CJ

0.27

D

DC

+

0.34

DK

0.33

DL

0.33


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ANALYSIS

Case 1:-Putting design dead load and imposed load in all spans AB, BC &CD.

A B C D

D.F.

F.E.M. -38.86 +38.86 -101.15 +101.15 -29.5 +29.5

Balance +12.44 16.2 11.84 -13.61 -20.1 -10.03

C.O.M. 8.1 6.22 -6.81 5.92 -5.02 -10.05

Balance -2.6 0.15 0.11 -0.17 -0.25 3.42

Final -20.92 61.43 -96.01 93.29 -54.87 12.84

Case 2:- Design dead load in all the span & imposed load on the longest span

SPAN TOTALLOAD

TOTALMOMENT

AB 35.98 38.86

BC 56.68 101.15

CD 32.51 29.5

SPAN TOTALLOAD

TOTALMOMENT

AB 31.48 34

BC 56.68 108.83

CD 28.01 25.42

0.32 0.26 0.190.19 0.28 0.34


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A B C D

D.F.

F.E.M. -34+34

-108.83+108.83

-25.42+25.42

Balance +10.8819.46

14.22 -15.85

-23.35 -8.64

C.O.M. 9.735.44

-7.937.11

-4.32 -11.68

Balance -3.110.65

0.47 -0.53

-0.783.97

Final -16.559.55

-102.0799.56

-53.879.07

FREEMOMENT ATCENTRE OF

SPAN[(wl2)/8]

51 163.24 38.13

MID SPANMOMENT

13 62.43 6.66

ELASTICSHEAR

11.96 -0.52 -13.58

FREE SPANSHEAR

56.66 136.03 46.22

TOTAL 68.62 135.51 32.64


SPAN[(wl2)/8]

51 163.24 38.13

MID SPANMOMENT

13 62.43 6.66

ELASTICSHEAR

11.96 -0.52 -13.58

FREE SPANSHEAR

56.66 136.03 46.22

TOTAL 68.62 135.51 32.64

0.32 0.26 0.190.19 0.28 0.34


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Design moment in the column:

Colum moment =Distribution factor of the column (-1)[F.E.M.S+C.O.M.S]

This calculation is shown in the tabular form:

Case I:-Live load on all the span.

A B C D

D.F.

F.E.M. -38.86 +38.86 -101.15 +101.15 -29.5 +29.5

C.O.M. 8.1 6.22 -6.81 5.92 -5.02 -10.05

C.O.M.+F.E.M.

-30.76 -62.88 72.55 19.45

AT TOP 10.46 16.98 -19.6 -6.42

ATBOTTOM

10.46 16.98 -19.6 -6.42

Case II :- Live load on longest span

A B C D

D.F.

F.E.M. -34 +34 -108.83 +108.83 -25.42 +25.42

C.O.M. 9.73 5.44 -7.93 7.11 -4.32 -11.68

C.O.M.+

F.E.M.

-24.27 -77.32 86.2 13.74

AT TOP 8.25 20.88 -23.27 -4.53

ATBOTTOM

8.25 20.88 -23.27 -4.53

0.34 0.27 0.27 0.34

0.34 0.27 0.27 0.34


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BEAM ALONG X- DIRECTION :

Distribution factor Calculation:-

JOINT MEMBER STIFFNESS (K) D.F.=

P

PQ

+

0.28

PA

0.36

PG

0.36

Q

QP

+

2

0.15

QR

0.45

QB

0.2

QH

0.2

R

RQ +

+

0.43

RS

0.2

RC

0.185

RI

0.185

S

SR

+ +

0.2

ST

0.43

SD

0.185

SJ

0.185


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T

TS

+ +

0.45

TU

0.15

TE

0.2

TK

0.2

U UT

+

0.28

UF

0.36

UL

0.36

Analysis:-

Case 1:- Putting design dead load and imposed load in all spans.

P Q R S T U

D.F.

F.E.M. -94.42 94.42 -2.8 2.8 -27.11 27.11 -2.8 2.8 -94.42 94.42

Balance 26.44 -13.74 -41.2 10.45 4.86 -4.86 -10.4 41.23 13.74 -26.44

C.O.M. -6.87 13.22 5.22 -20.62 -2.43 2.43 20.62 -5.22 -13.22 6.87

Balance 1.92 -2.77 -8.3 9.91 4.61 -4.61 -9.91 8.3 2.77 -1.92

Final -72.9 91.13 -47.11 2.54 -20.07 20.07 -2.54 47.11 -91.13 72.93

SPAN TOTALLOAD

TOTALMOMENT

PQ 55.95 94.42

QR 15.2 2.8

RS 31.77 27.11ST 15.2 2.8

TU 55.95 94.42

0.28 0.15 0.430.45 0.2 0.2 0.43 0.45 0.15 0.28


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SPAN[(wl2)/8]

141.62 4.28 40.66 4.28 141.62

MID SPANMOMENT

59.59 -20.54 20.59 -20.54 59.59

ELASTICSHEAR

4.044 -29.71 0 -29.71 4.044

FREE SPANSHEAR

129.375 12.34 52.83 12.34 129.375

TOTAL 133.42 -17.37 52083 -17.37 133.42

Case 2:- Design dead load in all the span & imposed load on PQ,RS &TU

P Q R S T U

D.F.

F.E.M. -94.42 94.42 -2.01 2.01 -27.11 27.11 -2.01 2.01 -94.42 94.42

Balance 26.44 -13.86 -41.58 10.79 5.02 -5.02 -10.79 41.58 13.86 -26.44

C.O.M. -6.93 13.22 5.4 -20.79 -2.51 2.51 20.79 -5.2 -13.22 6.93

Balance 1.94 -2.79 -8.38 10.02 4.66 -4.66 -10.02 8.38 2.79 -1.94

Final -72.97 90.99 -46.57 2.03 -19.94 19.94 -2.03 46.57 -90.99 72.97


SPAN[(wl2)/8]

141.62 3.01 40.66 3.01 141.62

MID SPAN

MOMENT

59.64 -21.29 20.72 -21.29 59.64

SPAN TOTAL

LOAD

TOTAL

MOMENT

PQ 55.95 94.42

QR 10.7 2.01

RS 31.77 27.11

ST 10.7 2.01

TU 55.95 94.42

0.28 0.15 0.430.45 0.2 0.2 0.43 0.45 0.15 0.28


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Design of One way Slab:

Datagiven (m)

Clear span (or Room size ) = 7mX3mL.L = 1.5 KN/m , support thickness = 200mmSurface finishing = 1 KN/m

Using M20 & Fe 415

Step 1 :- Design constant for M20 concrete & Fe415 steel

Fck =20 N/mm , Fy = 415 N/mm

Mulimit = 0.138 fck bd

Xu = 0.479 d

Step 2:- Type of Slab- ly/lx = 7/3 = 2.33 > 2

therefore design One way slab,

considering shorter span.

Step 3:- Effective depth of span

For continuous slab one way

Deff. = l/ (26 X M.F)

Assume Modification factor

M.F =1.3 (IS456:2000 Page - 38)

= 3000/(26 X 1.3)

Provide depth = 88.75 90 mm ,

TakeDeff .=125mmOverall depth D = d +(c.c +/2) assume dia. of bar 10mm

= 125 +(20+10/2) c.c= 20mm

= 125+25 =150 mm


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Fig. Diagrammatic Representation

Step 4:-Effective Span (l eff)-

(1) L+ b = 3000 + 200 = 3200 mm

(2) L+ b = 3000 + 125 = 3125 mm (whichever is less)

thus leff = 3.125 m

Step 5 :- Load Calculation-

(1) Dead load of slab = 1x 1x(d/1000) rcc

= (150/1000)x 25 =3.75KN/m

(2) Live load = 1.5 KN/m

(3) Finishing load = 1 KN/m

Working load w = 6.25 KN/m

Factored load wu = 1.5w = 1.5x6.25

= 9.375 KN/m


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Step 6:- Factored Bending Moment (Mu)-

Mu=coeff. x wu x leff

From : IS 456:2000 Page 36 Tabel no.12

[BM coefficients of Continuous slab at the mid of interior

span for dead load & imposed load (fixed) + 1/16 ]

Mu = (9.375x3.125)/16

Mu = 5.722 KNm per meter width of slab

Step 7 :- Check for depth (dreq.)-

Effective depth required dreq. =(Mu/0.138fck b

=(5.722x10)/(0.138x20x1000)dreq. = 45.53 mm

dreq. dprovided

OK-SAFE.

Step 8 :-Main Steel

Ast = 0.5 fck/fy [ 1-1-(4.6 Mu/fck bd) ] bd

Ast = 0.5x20/415[1-1-(4.6 x 5.722 x 10/ 20 x 1000 x 125)] 1000 x 125Ast = 129.638 130 mmandAstmin = 0.0012 bD

= 0.0012 x 1000 x 150= 180 mm

here, Astmin > Asttherefore use Astmin i.e. 180 mm

Step 9:-Spacing Of Main Bar

(1) (1000 x Ast) / Astmin = (1000 x /4 x 10 ) / 180= 437 mm

(2) 3d = 3x125 = 375 mm(3) 300 mm = 300 mm

(which ever is less )provide ( = 300 mm )

= 10 mm @ 300 mm c/c spacing along shorter span.Length of rod = 3000 (2 x clear cover )

= 3000 (2 x 20 ) = 2960 mm

Provide 10 @ 200 mm c/c & extra at top upto l/4 i.e. 0.8 m both supports


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Step 10 :-Spacing Of Distribution steel

hereAstmin = 180 mm( assuming dia. Of bar 8 mm )

(1) (1000x /4 x 8 )/180 = 279.25 280 mm(2) 5d = 5x125 = 625 mm(3) 450 mm

(whichever is less )provide 8 mm dia. Of distribution bar @ 280 mm c/c spacing across main bar

Fig. Reinforcement Details in One way Slab.

Design of Two way Slab:

Given Data-Size of slab (m) = 7 x 4.75Live load = 2 KN/msupport thickness = 200 mmFinishing = 1 KN/m

Use M20 & Fe415


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Step 1):-Design constant-fck = 20 MPa, fy = 415 MPaMu lim = 0.138 fck bdXu = 0.479 d

Step 2):-Type of Slab-ly/lx = 7/4.75 = 1.5 < 2

(Two way slab)Step 3):-Deffx = lx/26 x 1.5 = 4750/26 x 1.5 = 121.7 mm = dx

125 mm = dxAssume 10 , clear cover 20 mmdy = 125-10 = 115 mmOverall depth of slab D = d+ (c.c. + /2)

D = 125 + 20 + 5 = 150 mm

Step 4):-Effective length of Slabhere support thickness = 200 mm

Shorter Span Longer Span

i). Clear span + dx4750 + 125=4875mm

i). Clear span + dy7000 + 115=7115mm

ii). Clr span + support width

4750 + 200=4950mm

ii). Clear span + b

7000 + 200=7200mm

(whichever is less)lx = 4.875 m ly = 7.115 m

Step 5):-Load-i). D.L. = 1x1x150/1000x25 = 3.75KN/mii). Live load = 2KN/miii). Finishing = 1KN/m

Working load = 6.75KN/m

Wu = 1.5 x 6.75 = 10.125KN/m

Step 6):-Moments-ly/lx = 7.115/4.875 = 1.46


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Moment coefficients:

ly/lx x y

1.4 0.099 0.0511.46

1.5 0.104 0.046

x = 0.099 + (0.104-0.099)/(1.5-1.4) x (1.46-1.4)= 0.102

y = 0.051 + (0.046-0.05)/(1.5-1.4) x (1.46-1.4)= 0.048

Mx = x Wu lx = 0.102x10.125x4.875 = 24.54 KNmMy = y Wu ly = 0.048x10.125x4.875 = 11.55 KNm

Step 7:-Check for depth-

drequired= (Mx/0.138x20x1000)= [(24.54x10)/(0.138x20x1000)]= 94.29 mm 95 mm

dreq < dprovidedOK SAFE.

Step 8:-Area of Main Steel-

Astx = 0.5(fck/fy) [1-1-{(4.6XMx)/(fckbdx)}]b dx= 0.5(20/415) [1-1-{(4.6 X 24.54X10) /(20X1000 X125)}]1000X125

= 604.72 mm

Asty= 0.5(20/415) [1-1-{(4.6 X11.55X10) /(20X1000X115)}]1000X11.5= 293.89 mm

Astmin = (0.0012 X bD) = (0.0012 X 1000 X 150)= 180 mm

Astx &Asty>AstminHence, use Astx &Asty .

Step 10:-Spacing of main bar

Assume dia. of main bar = 10 mm


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Shorter span Long span

(1) 1000 X /4 X 10/ Astx = 129.88 120mm

(1) 1000 X /4 X 10/ 293.89 =267.24 260 mm

(2) 3dx = 3 X 12 = 375 (2) 3dy = 3 X 115 = 345

(3) 300 mm (3) 300 mm

(which ever is less)provide 10 @ 120 c/cprovide 10 @ 260 c/c

(3/4 l ) span middle strip

Step 11:-Distribution Steel

Astmin = 180 mmspacing assume = 8 mm

(1) 1000 X /4 X 8/180 =279.25 mm

(2) 5dx = 5X125 = 625

5dy = 5X 115 = 575

(3) 450 mm

provide 8 @ 270 c/c edge strip (span/ 8)

Step 12:-Check for deflection

dprovided = l/(26 X MF)Astprovided= (1000 X /4 X 10)/120

= 654.5 mm

Astrequired= 604.72 mm% of steel =Astprovided/(b X d X 1000)

= 0.37 %

F5 = 0.58 X fy Astrequired/AstprovidedF5 = 222.4 ( IS 456 : 2000)

MF = 1.5

drequired 121.8 mm

dprovided 125 mmdrequireddprovided OK-SAFE


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Step-4 Ultimate BM and Shear force

Step-5 Effective width of flange(bf):

=[(4.95/6)+0.2+(6X0.15)= 1925 mm

Step-6 Moment capacity of Flange section(Muf):-

Muf = bf Df 0.36fck (d0.416Df)

=1925 X 150 X 0.36 X 20 X (3200.416X150)

= 535.55 KN-m

Since, Mu < Muf i.e. Neutral axis is within the Flange,

Hence, the section is treated as Rectangular with b=bf for designing reinforcement.

Step-7 Tension Reinforcements:-


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514.55X10= Ast X0.87X415X320{1-(AstX415)/(1925X320X20)}Ast = 545.651 mm

Ast = 545.651

Provide 3 nos. 14 at bottom,

2 nos. 10 at top, & provide (l/4) extra at top

total Ast = 618.89

Step-8 Shear Reinforcement:-

v = (Vu / bw d) = 415.8X10/(200X320)

= 6.49 N/mm

Pt = 100 Ast /bwd = 100X545.651/(200X320)

= 0.853 m from IS 456:2000, page no.73,table-19,

Design shear strength of concrete (M20)

c = 0.28 N/mm

Balance Shear => Vus = [Vus(c bd)]

Vus = [415.8(0.28X200X320 )10]

= 397.88 KN

Using 8 mm dia, 2 legged stirupps,

Spacing is given by,

SV = (0.87fy Asv d/Vus)

SV = (0.87X415 X(4)X8/397.88X10)

Sv = 220 mm 200 mm

provide spacing of 100 mm and gradually increase to 200 mm at centre of span.

Step 9:- Check for deflection Control


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Pt = 100 Ast/(bf d)

= (100 X 5378 )/( 2025 X 320 ) = 0.83

bw/bf = 200/2025 = 0.099

(L/d)provided = L/d x Kt x Kc x Kf

4950/320 = 20x1.05x1x0.94

15.46 19.74

hence, check for deflection is satisfactory.

Fig. Reinforcement Details in T-beam.

Design of Column:


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Data-

Axial load on column = 400 KN3737

Length (L) = 3.3 KN

Column size = 200X300

Adopt M20 and Fe415

Fck = 20 N/mm Fy = 415N/mm

Step 1:- Effective length of column-

both end fixed l = 0.6 L

= 0.65 X 3.3 = 2.145 m

factored load Pu = 1.5 X 400 = 600 KN

Step 2:-Slenderness ratio-

unsupported length/least lateral dimension

{Leff/D = 2145/200 =10.725 12

hence column is designed as short column

Step 3:-Minimum Eccentricity-

emin = [(l/500)+(D/30)] or 20 mm

= 10.96 mm or 20 mm

emin = 20 mm

Check,

10.96/200 = 0.05 0.05

OK .

Hence, codal formula for short column is applicable.

Step 4:- Main steel ( Longitudinal reinforcement )-


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Pu = [(0.4XfckAc) + (0.67FyAsc)]

Ac = area of concrete

Asc = area of steel

Ag = gross area (200x300 = 60000 mm)

600X10 = 0.4X20X0.99Ag + 0.67X415X0.01Ag

Ag = 56072.15 mm

Asc = 0.01 Ag = 561 mm

Ascmin = 0.08 Ag = 448.57 mm 449 mm

provide 12 - 6Nos( Total Area of steel = 678.58 mm)

Step 5:- Design of Lateral Ties-

(1) Dia. of ties tie = tie / 4 =12/4 = 3 mm

tie = 8 mm (for Fe 415)

Spacing-

a) least lateral dimension = 200 mm

b) 16 X main = 16X12= 192 mm

c) 300 mm

which ever is less

provide 8 @ 200c/c


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Design of Stair case Dog legged)

Data,

ht. Of storey = 3.3 m

size of stair hall =4.5mX3m

L.L = 2 KN/m

supported width = 200 mm

Step 1 :- Design constants

using M20 and fe415

Fck = 20 Mpa

Fy = 415 Mpa

Mulimit = 0.138 Fckbd

Step 2 :- Arrangement of stair-

Ht. Of storey = 3.3 m

Ht. Of flight = 3.3/2 = 1.65 m

assume R = 150 mm , T = 300 mm

No. Or riser = 1650/150 = 11


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No. Of tread = 11-1 = 10

Going G = no. Of tread X T

= 10 X 300 = 3000 mm

Fig. Arrangement of Steps in Staircase.

Step 3 :- Effective length-

leff = c/c dist. b/w support

= 3000 + 1500 +200/2 = 4600 mm

Step 4 :- Effective depth of waist slab

d l/25 = 4600/25 = 184 180

assume 10 and clear cover 15 mm

D = 180 + ( 15+10/2) = 200 mm

but we adopted D = 150 mm

Step 5 :- Load calculation (unit area )

(1) Self wt. Of waist slab in horizontal area

= w s X (R+T)/T


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= (1X1XD/1000)rcc X (150+300)/300

= 4.19 KN/m

(2) Self wt. Of step per meter length

= (R/2)pcc = (150/2)24 = 1.8 KN/m

(3) Finishing load minimum = 0.75 KN/m

(4) L.L = 2 KN/m

w = 8.74

wu =1.5 w = 13.11 KN/m

Step 6:- Bending moment

Mu = wl/8 = (13.11 X 4.6)/8 =34.67 KN/m

Step 7:- Check for effective depth

drequired = (Mu/0.138fckb)

= (34.67X10/0.138X20X1000)

drequired = 112.078 mm

drequired dprovided (i.e.= 150 )

OK SAFE.

Step 8:- Main steel

Ast = 0.5X20/415[1-1-{(4.6X34.67X10)/(20X1000X150)}]

711 mm

Astmin = 0.0012X1000X150 = 180 mm

use Ast = 711 mm


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Step 9:- Spacing of Main bar-

(1) (1000X/4X10)/711 assume 10

=110.46 mm

(2) 3X150

(3) 300mm

which ever is less

Main bar provide 10 @ 100 c/c

Step 10:- Distribution bar-

use Astmin = 180 assume = 8 mm

(1) (1000X /4X8)/180 = 279.15 mm

(2) 5D = 5X150 =750 mm

(3) 450 mm

distribution bar provide 8 @ 250 c/c spacing

Fig. Reinforcement Details in Stairs


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Design of Flat Footing

Data:

Assume SBC of soil = 200 KN/m

Reinforcement concrete column size = 200 X 300

Axial service load P = 400 KN

Adopt M20 & Fe415

Step 1: Calculation of Load-

a) Load on column = 400KNb) Self wt. of footing = 10% of column

= 400 X (10/100) = 40 KN

Total load = 440 KN

Factored load Wu = 1.5 X 440 = 660 KN

Step 2: Area of footing-

=Load/SBC = 440/200 = 2.2 m2

Assuming square footing,

Size of footing=2.2 = 1.45m

Adopt size of footing = 1.5m X 1.5m

Step 3: Net upward pressure-

Pu=660/(1.5 x 1.5)=293.33 KN/m2

Step 4: Bending Moment calculation-

Maximum bending moment


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will be on the face of column,

M = F X Distance of C.G.

= (area X stress) x (0.65/2)m = 92.95 KNm

Step 5: Depth of Footing

Drequired=M/0.138fckbx 106/0.138 x 20 x 20=410.35Taken 420 mm.

Fig. Depth of Footing


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Assume cover = 60mm

Thus, Overall Depth = 420+60 = 480mm

Step 6: Main Steel calculation-

Ast=0.5fck/fy(1-1-(0.46 Mu/ fckBd2))Bd =623.18 mm2

Ast=0.0012 x BD = 864 mm2

Provide 10 @ 100 c/c in each direction at bottom of footing i.e. 12 nos .

Step 7: Check for Shear-

The critical; section will be at a distance (d/2) from column face.

Shear Force = Stress X Area; here, Ar ea =[B2-(b+d)

2]

= 293.33X{ 1.5-[(0.200+0.420) X (0.300+0.420)] }

= 529.05 KN

Shear stress

v=V/b0 d = 1260KN/m2= 0.00126 N/mm2

Here, b0 = 2(l+b)=2(0.2 + 0.3)=1m

Here b0 = 2(l+b) = 2(0.2 + 0.3) = 1 m.

Permisible shear stress =0.25fck = 0.2520 = 1.11>c

OK SAFE


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Fig. Sectional View

Fig. Plan


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CONCLUSION

In this report, a design of Multistory building for residential purpose is presented. We havesuccessfully completed the planning and designing of a multistory (G+9) structure.

The main key features of project are as follows:

Plot size = 20m X 20m Total construction area = 65% of plot size. Total no. of 1BHK Flats = 48


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References

A.K. Jain, Advanced R.C.C. Design. N. Krishna Raju, Reinforced Concrete Design. S.S. Bhavikatti, Advanced R.C.C. Design.

IS 456-2000 IS 1893(Part 1) 2002 IS 800-2007

G + 9 residential building design

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