G + 9 residential building design

A PROJECT ON G+9 RESIDENTIAL BUILDING

By JENERIUS MINJ University Roll No : 10301313132 University Registration No : 131030120098 Submitted in partial fulfillment of The requirements for award of B. Tech Degree in Civil Engineering Of the Maulana 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 Department Of Civil Engineering Icare Complex, Hit Campus Purba Medinipur Haldia – 721657 (West Bengal) India 1

FORWARD It 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 maltystoried residential building.

Assistants Prof. KAUSIK BERA, Assistants Prof. GOUTAM DUTTA Department Of Civil Engineering Haldia Institute Of Technology

…………………………………………….. (N. K. Yadav) H.O.D Department Of Civil Engineering Haldia Institute Of Technology

2

ACKNOWLEDGEMENT I express my deep sense of gratitude to my guide ass department of civil engineering 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 not have been possible to complete this work in time.

3

Table of Contents Sr. No.

Topic

1.

Introduction i). Effective Span ii). Stiffness iii). Loads iv). Analysis

2.

Typical Floor Plan

3.

Moment Calculation by Moment Distribution Method

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

4

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. In other 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.

5

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 Beams span 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. Column sizes 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

6

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.

7

LIFT 3X2.5

TYPICAL FLOOR PLAN 8

MOMENT CALCULATION:D.F Calculation:JOINT

MEMBER

STIFFNESS (K)

∑

D.F.=∑ 0.32

AB

+ AE

0.34

AF

0.34

BC

0.19

A

+2×

BA

B

+

0.26

BG

0.27

BH

0.27

CD

0.28

+

CB

0.19

+ C

D

CI

0.27

CJ

0.27

DC

0.34

DK

+

0.33 0.33

DL

9

SPAN

TOTAL LOAD

TOTAL MOMENT

AB

35.98

38.86

BC

56.68

101.15

CD

32.51

29.5

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

B

D.F.

0.32

0.26

0.19

F.E.M.

-38.86

+38.86

Balance

+12.44

C.O.M.

C

D

0.19

0.28

0.34

-101.15

+101.15

-29.5

+29.5

16.2

11.84

-13.61

-20.1

-10.03

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

-96.01

93.29

-54.87

12.84

61.43

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

TOTAL LOAD

TOTAL MOMENT

AB

31.48

34

BC

56.68

108.83

CD

28.01

25.42

10

A

B

D.F.

0.32

F.E.M.

-34 +34 +10.88 19.46

Balance

0.26

C 0.19

D

0.19

-108.83 +108.83

0.28 -25.42 +25.42

9.73 5.44 -3.11 0.65 -16.5 59.55

14.22 15.85 -7.93 7.11 0.47 0.53 -102.07 99.56

FREE MOMENT AT CENTRE OF SPAN[(w×l2)/8] MID SPAN MOMENT ELASTIC SHEAR FREE SPAN SHEAR TOTAL

51

163.24

38.13

13

62.43

6.66

11.96

-0.52

-13.58

56.66

136.03

46.22

68.62

135.51

32.64

FREE MOMENT AT CENTRE OF SPAN[(w×l2)/8] MID SPAN MOMENT ELASTIC SHEAR FREE SPAN SHEAR TOTAL

51

163.24

38.13

13

62.43

6.66

11.96

-0.52

-13.58

56.66

136.03

46.22

68.62

135.51

32.64

C.O.M. Balance Final

0.34

-

-

11

-23.35 8.64 -4.32 11.68 -0.78 3.97 -53.87 9.07

-

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

0.27

0.27

D

D.F.

0.34

F.E.M.

-38.86

C.O.M.

8.1

C.O.M.+ F.E.M. AT TOP

-30.76

-62.88

72.55

19.45

10.46

16.98

-19.6

-6.42

AT BOTTOM

10.46

16.98

-19.6

-6.42

+38.86

-101.15

6.22 -6.81

+101.15

5.92

0.34

-29.5

+29.5

-5.02

-10.05

Case II :- Live load on longest span

A

B

C

0.27

0.27

D

D.F.

0.34

F.E.M.

-34

C.O.M.

9.73

C.O.M.+ F.E.M. AT TOP

-24.27

-77.32

86.2

13.74

8.25

20.88

-23.27

-4.53

AT BOTTOM

8.25

20.88

-23.27

-4.53

+34

-108.83

5.44 -7.93

0.34

+108.83

-25.42

+25.42

7.11

-4.32

-11.68

12

BEAM ALONG X- DIRECTION : Distribution factor Calculation:JOINT

MEMBER

STIFFNESS (K)

∑

D.F.=∑ 0.28

PQ

+ PA

0.36

PG

0.36

QP

0.15

P

+

QR Q

2×

QB

0.2

QH

0.2

RQ

0.43

+

RS R

0.45

0.2

0.185

RC

+ RI

0.185

SR

0.2

+

ST

+

0.43

S SD

0.185

SJ

0.185

13

TS

0.45

+

TU

+

0.15

T

U

TE

0.2

TK

0.2

UT

+

UF

0.28 0.36 0.36

UL SPAN

TOTAL LOAD

TOTAL MOMENT

PQ

55.95

94.42

QR

15.2

2.8

RS

31.77

27.11

ST

15.2

2.8

TU

55.95

94.42

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

Q

R

D.F.

0.28

F.E.M.

-94.42

94.42

-2.8

Balance

26.44

-13.74

-41.2

C.O.M.

-6.87

13.22

Balance

1.92

Final

-72.9

0.15

0.45

0.43

S

T

0.2

0.2

0.43

0.15

0.28

-27.11

27.11

-2.8

2.8

-94.42

94.42

10.45

4.86

-4.86 -10.4

41.23

13.74

-26.44

5.22

-20.62

-2.43

2.43 20.62

-5.22

-13.22

6.87

-2.77

-8.3

9.91

4.61

-4.61 -9.91

8.3

91.13

-47.11

2.54

-20.07

20.07 -2.54

47.11

2.8

14

0.45

U

2.77

-1.92

-91.13

72.93

FREE MOMENT AT CENTRE OF SPAN[(w×l2)/8] MID SPAN MOMENT ELASTIC SHEAR FREE SPAN SHEAR TOTAL

141.62

4.28

40.66

4.28

141.62

59.59

-20.54

20.59

-20.54

59.59

4.044

-29.71

0

-29.71

4.044

129.375

12.34

52.83

12.34

129.375

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 D.F.

0.28

F.E.M.

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

Q

R

0.15

0.45

-94.42

94.42

-2.01

Balance

26.44

-13.86

-41.58

C.O.M.

-6.93

13.22

5.4

Balance

1.94

-2.79

-8.38

10.02

Final

-72.97 90.99

-46.57

2.03

FREE MOMENT AT CENTRE OF SPAN[(w×l2)/8] MID SPAN MOMENT

0.43 2.01

10.79 -20.79

S

T

0.2

0.2

0.43

-27.11

27.11

-2.01

5.02

0.15

0.28

2.01

-94.42

94.42

-5.02 -10.79

41.58

13.86

-26.44

-2.51

2.51 20.79

-5.2

-13.22

6.93

4.66

-4.66 -10.02

8.38

2.79

-1.94

-90.99

72.97

-19.94

19.94

0.45

U

-2.03

46.57

141.62

3.01

40.66

3.01

141.62

59.64

-21.29

20.72

-21.29

59.64

15

Design of One way Slab: Data given‚ (m) Clear span (or Room size ) = 7mX3m L.L = 1.5 KN/m² , support thickness = 200mm Surface finishing = 1 KN/m² Using M20 & Fe 415

Step 1 :- Design constant for M20 concrete & Fe415 steel Fck =20 N/mm² , Fy = 415 N/mm² Mulimit = 0.138 fck bd² 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 ,

Take Deff .=125mmOverall depth D = d +(c.c +Ø/2) = 125 +(20+10/2) = 125+25 =150 mm

16

assume dia. of bar 10mm c.c= 20mm

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) ρrcc = (150/1000)x 25 =3.75KN/m² (2) Live load

= 1.5 KN/m²

(3) Finishing load = 1 KN/m² Working load w = 6.25 KN/m² Factored load wu = 1.5w

= 1.5x6.25 = 9.375 KN/m² 17

Step 6:- Factored Bending Moment (Mu)Mu=coeff. x wu x leff² 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 = (9.375x3.125²)/16 Mu = 5.722 KNm

per meter width of slab

Step 7 :- Check for depth (dreq.)Effective depth required dreq. =√(Mu/0.138fck b =√(5.722x10⁶)/(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 bd²) ] bd Ast = 0.5x20/415[1- √1-(4.6 x 5.722 x 10⁶ / 20 x 1000 x 125²)] 1000 x 125 Ast = 129.638 ≈ 130 mm² and Astmin = = = here, Astmin >

0.0012 bD 0.0012 x 1000 x 150 180 mm² Ast therefore use Astmin i.e. 180 mm²

Step 9:- Spacing Of Main Bar – (1) (1000 x Ast) / Astmin = (1000 x π/4 x 10² ) / 180 = 437 mm (2) 3d = 3x125 = 375 mm (3) 300 mm = 300 mm (which ever is less ) provide (δ = 300 mm ) Ø = 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 Ø @ 200 mm c/c & extra at top upto l/4 i.e. 0.8 m both supports 18

Step 10 :- Spacing Of Distribution steel – here Astmin = 180 mm² ( assuming dia. Of bar 8 mm ) (1) (1000x π/4 x 8² )/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 DataSize of slab (m) = 7 x 4.75 Live load = 2 KN/m² support thickness = 200 mm Finishing = 1 KN/m² Use M20 & Fe415

19

Step 1):- Design constantfck = 20 MPa, fy = 415 MPa Mu lim = 0.138 fck bd² Xu = 0.479 d

Step 2):- Type of Slably/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 = dx Assume 10 Ø, clear cover 20 mm dy = 125-10 = 115 mm Overall depth of slab D = d+ (c.c. + Ø/2) D = 125 + 20 + 5 = 150 mm

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

Shorter Span

Longer Span

i). Clear span + dx 4750 + 125=4875mm

i). Clear span + dy 7000 + 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):- Loadi). D.L. = 1x1x150/1000x25 = 3.75KN/m² ii). Live load = 2KN/m² iii). Finishing = 1KN/m² Working load = 6.75KN/m² Wu = 1.5 x 6.75 = 10.125KN/m²

Step 6):- Momentsly/lx = 7.115/4.875 = 1.46

20

Moment coefficients: ly/lx

αx

αy

1.4

0.099

0.051

0.104

0.046

1.46 1.5

αx = 0.099 + (0.104-0.099)/(1.5-1.4) x (1.46-1.4) = 0.102 αy = 0.051 + (0.046-0.05)/(1.5-1.4) x (1.46-1.4) = 0.048 Mx = αx Wu lx² = 0.102x10.125x4.875² = 24.54 KNm My = αy Wu ly² = 0.048x10.125x4.875² = 11.55 KNm

Step 7:-Check for depth-

drequired

= √(Mx/0.138x20x1000) = √[(24.54x10⁶)/(0.138x20x1000)] = 94.29 mm ≈ 95 mm

dreq

<

dprovided OK SAFE.

Step 8:- Area of Main Steel-

Astx

/

= 0.5(fck fy) [1- √1-{(4.6 X

Mx

) /(fck

b d²x)}]b dx

= 0.5(20/415) [1- √1-{(4.6 X 24.54X10⁶ ) /(20X1000 X125²)}]1000X125 = 604.72 mm²

Asty = 0.5(20/415) [1- √1-{(4.6 X11.55X10⁶ ) /(20X1000X115²)}]1000X11.5 = 293.89 mm²

Astmin = (0.0012 X bD) = (0.0012 X 1000 X 150) = 180 mm²

Astx & Asty > Astmin Hence, use Astx & Asty . Step 10:- Spacing of main bar – Assume dia. of main bar Ø = 10 mm

21

Shorter span

Long span

(1) 1000 X π/4 X 10²/ Astx = 129.88 ≈ 120 (1) 1000 X π/4 X 10²/ 293.89 = mm 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 Ø @ 120 c/c provide 10 @ 260 c/c (3/4 l ) span middle strip

Step 11:- Distribution Steel –

Astmin = 180 mm² spacing assume Ø = 8 mm (1) 1000 X π/4 X 8²/180 =279.25 mm (2) 5dx = 5X125 = 625 5dy = 5X 115 = 575 (3) 450 mm provide 8 Ø @ 270 c/c edge strip (span/ 8)

Step 12:- Check for deflection –

dprovided = l/(26 X MF) Astprovided = (1000 X π/4 X 10²)/120 = 654.5 mm²

Astrequired = 604.72 mm² % of steel = Astprovided /(b X d X 1000) = 0.37 % F5 = 0.58 X fy Astrequired/

Astprovided

F5 = 222.4

( IS 456 : 2000)

MF = 1.5

drequired 121.8 mm dprovided 125 mm drequired ˂ dprovided

22

OK-SAFE

Fig.Reinforcement Details in Two way Slab.

Design of T- BEAM Data : Clear span(L) = 4.75 m, Depth of flange (Df) = 150 mm, Depth of web (bw) = 200 mm Imposed Load = 112 KN/m,

Step-1

fck = 20 fy = 415

Effective Depth (d):

Adopt D = 320 + 20 + 25 = 360 mm

Step-2

Effective Span (leff):

The least of (i) Centre to centre of support = 4.75+0.2 = 4.95 m (ii) Clear span + effective depth = 4.75 + 0.32 = 5.1 m Effective span = 4.95 m

Step-3

Loads

Imposed load = 112 KN/m Ultimate load = 1.5 X 112 = 168 KN/m

23

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 (d – 0.416Df) =1925 X 150 X 0.36 X 20 X (320 – 0.416X150) = 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:24

514.55X10⁶ = Ast X0.87X415X320{1-(AstX415)/(1925X320X20)} Ast = 545.651 mm² Ast = 545.651 Provide 3 nos. 14Ø at bottom, 2 nos. 10Ø at top, & provide (l/4) extra at top total Ast = 618.89

Step-8

Shear Reinforcement:-

τv = (Vu / bw d) = 415.8X10³/(200X320) = 6.49 N/mm² 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) τc = 0.28 N/mm² Balance Shear => Vus = [Vus – (τc bd)] Vus = [415.8 – (0.28X200X320 )10¯³] = 397.88 KN Using 8 mm dia, 2 legged stirupps, Spacing is given by, SV = (0.87fy Asv d/Vus) SV = (0.87X415 X(π÷4)X8²/397.88X10³) 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 – 25

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: 26

DataAxial load on column = 400 KN3737 Length (L)

= 3.3 KN

Column size = 200X300 Adopt M20 and Fe415 Fck = 20 N/mm²

Fy = 415N/mm²

Step 1:- Effective length of columnboth 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 ratiounsupported length/least lateral dimension {Leff/D = 2145/200 =10.725 ˂ 12 hence column is designed as short column

Step 3:-Minimum Eccentricityemin = [(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 )-

27

Pu = [(0.4XfckAc) + (0.67FyAsc)] Ac = area of concrete Asc = area of steel Ag = gross area (200x300 = 60000 mm²) 600X10³ = 0.4X20X0.99Ag + 0.67X415X0.01Ag Ag = 56072.15 mm² Asc = 0.01 Ag = 561 mm² Ascmin = 0.08 Ag = 448.57 mm² ≈ 449 mm² provide 12Ø - 6Nos( Total Area of steel = 678.58 mm²)

Step 5:- Design of Lateral Ties(1) Dia. of ties Øtie = Øtie / 4 =12/4 = 3 mm Øtie = 8 mm (for Fe 415) Spacinga) least lateral dimension = 200 mm b) 16 X Ømain = 16X12= 192 mm c) 300 mm which ever is less provide 8 Ø @ 200c/c

28

Design of Stair case (Dog legged) Data, ht. Of storey = 3.3 m size of stair hall =4.5mX3m L.L = 2 KN/m² supported width = 200 mm

Step 1 :- Design constants – using M20 and fe415 Fck = 20 Mpa Fy = 415 Mpa Mulimit = 0.138 Fckbd²

Step 2 :- Arrangement of stairHt. 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 29

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 lengthleff = 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 Ø 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 √(R²+T²)/T 30

= (1X1XD/1000)ρrcc X √(150²+300²)/300 = 4.19 KN/m² (2) Self wt. Of step per meter length = (R/2)ρpcc = (150/2)24 = 1.8 KN/m² (3) Finishing load minimum = 0.75 KN/m² (4) L.L = 2 KN/m² w

= 8.74

wu =1.5 w = 13.11 KN/m²

Step 6:- Bending moment – Mu = wl²/8 = (13.11 X 4.6²)/8 =34.67 KN/m

Step 7:- Check for effective depth –

drequired = √(Mu/0.138fckb) = √(34.67X10⁶/0.138X20X1000)

drequired =

112.078 mm

drequired ˂ dprovided (i.e.= 150 ) OK SAFE.

Step 8:- Main steel –

Ast = 0.5X20/415[1-√1-{(4.6X34.67X10⁶)/(20X1000X150²)}] ≈ 711 mm²

Astmin = 0.0012X1000X150 = 180 mm² use

Ast

= 711 mm²

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Step 9:- Spacing of Main bar(1) (1000Xπ/4X10²)/711

assume 10Ø

=110.46 mm (2) 3X150 (3) 300mm which ever is less Main bar provide 10Ø @ 100 c/c

Step 10:- Distribution baruse Astmin = 180

assume Ø = 8 mm

(1) (1000X π/4X8²)/180 = 279.15 mm (2) 5D = 5X150 =750 mm (3) 450 mm distribution bar provide 8Ø @ 250 c/c spacing

Fig. Reinforcement Details in Stairs 32

Design of Flat Footing Data: Assume SBC of soil = 200 KN/m² Reinforcement concrete column size = 200 X 300 Axial service load P = 400 KN Adopt M20 & Fe415

Step 1: Calculation of Loada) Load on column = 400KN b) 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 33

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/0.138fckb

x 106/0.138 x 20 x 20 =410.35

Taken 420 mm.

Fig. Depth of Footing 34

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 10Ø @ 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, Area =[B2-(b+d)2] = 293.33X{ 1.5²-[(0.200+0.420) X (0.300+0.420)] } = 529.05 KN Shear stress Τv=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 =0.25√fck = 0.25√20 = 1.11>τc 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 have successfully 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

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