Design of Multi Storied Building

CHAPTER 1

INTRODUCTION The industrial training consists of analysis and design of a Reinforced concrete multi storied residential building. The project was undertaken for a client Alliance Realtors. The proposed site is at Aluva. The building consists of ground + 13 stories+ terrace floor of moment resisting frame supported on pile foundation. During analysis, dead loads and live loads were calculated and seismic load calculated by referring IS 1893(Part1): 2002 and wind loads calculated from IS: 875(Part 3)-1987 and their combinations were applied on the space frame. The load combinations were taken to obtain the maximum design loads, moments and shear. The design is carried as per IS 456:2000 for the above load combinations. Team visits to various construction sites were conducted whose structural designs were handled by M/S Associated Structural Consultants. The site visits helped us to be aware of the latest construction methods that are adopted and being practised in the construction industry.

1

CHAPTER 2

SOFTWARES USED IN TRAINING The more frequently used softwares during my training include:1. ETABS 9 2. MS Excel 3. Auto CAD 2004 4. MS Word

2.1. ETABS 9 ETABS is a sophisticated, yet easy to use, special purpose analysis and design program developed specifically for building systems. ETABS Version 9 features an intuitive and powerful graphical interface coupled with unmatched modeling, analytical, and design procedures, all integrated using a common database. Although quick and easy for simple structures, ETABS can also handle the largest and most complex build-in models, including a wide range of nonlinear behaviors, making it the tool of choice for structural engineers in the building industry.

2.2. MS EXCEL Microsoft Excel (full name Microsoft Office Excel) is a spreadsheet program written and distributed by Microsoft for computers using the Microsoft Windows operating system and for Apple Macintosh computers. It features an intuitive interface and capable calculation and graphing tools which, along with aggressive marketing, have made Excel one of the most popular microcomputer applications to date. It is overwhelmingly the dominant spreadsheet application available for these platforms and has been so since version 5 in 1993 and its bundling as part of Microsoft Office. Microsoft Excel traditionally used a proprietary binary file format called Binary Interchange File Format (BIFF) up until Microsoft Excel 2007 that uses Office Open XML documents, similar in design but not compatible with Open Document documents. Although supporting and encouraging the use of these new formats as replacements, Excel 2007 is still backwards compatible with the traditional, binary, formats. Microsoft Excel 2002 and subsequent versions also support an XML format called "XML Spreadsheet" ("XMLSS"), although

2

this format is not able to encode VBA macros. The designs of the various structural elements were carried out in spreadsheets using MS Excel.

2.3. AUTO CAD 2007 All the drawing and detailing works were done by making use of Auto CAD 2007, released by Autodesk inc. As such, this is the pioneering software in CADD.

2.4. MS WORD MS Word is very popular word processing software issued by Microsoft Corporation. Over time, it has evolved in to a very popular package for handling all times of documentation and reporting. The software is very versatile and user friendly and has several features that makes document editing efficient. Documentation of all the work was done in MS Word.

3

CHAPTER 3

GENERAL PRINCIPLES OF DESIGN 3.1. OBJECTIVES OF STRUCTURAL DESIGN The design of a structure must satisfy three basic requirements: •

Stability:- To prevent overturning, sliding or buckling of the structure, or part of it, under the action of loads.

•

Strength:- To resist safely the stresses induced by the loads in the various structural members.

•

Serviceability:- To ensure satisfactory performance under service load conditions which implies providing adequate stiffness and reinforcement to contain deflections, crack widths and vibrations within acceptable limits, and also providing impermeability and durability.

There are other considerations that a sensible designer ought to bear in mind, viz., economy and aesthetics. One can always design a massive structure, which has more than adequate stability, strength and serviceability, but the ensuing cost of the structure may be exorbitant, and the end product, far from aesthetic.

3.2. SOIL PROFILE It was decided to take two bore holes (BH1 and BH2) at the site. In BH1, the top 7m comprise of soft lateritic clay. S.P.T value of 2 to 6 was noted in this strata. Below this clayey sand having S.P.T value of 8 to 14 was noted extending up to 12m depth. From 12m to 16.1m lateritic clay with S.P.T value of 18 to 33 was observed. This was followed by soft rock with S.P.T value greater than 100 extending up to 20m depth. From 20m depth to termination of bore hole at 21m hard rock was found. Water table was noted at a depth of 3.5m from the ground level during the time of investigation. While in BH2, the top 6m comprise of soft lateritic clay. S.P.T value of 2 to 5 was noted in this strata. Below this clayey sand having S.P.T value of 8 to 15 was noted extending up to 10m depth. From 10m to 12m lateritic clay with S.P.T value of 16 was observed. Below this up to 15m loose clayey sand having S.P.T value of 2 to 3 was noted. From 15 to 18.5m medium dense sand with S.P.T value of 20 to 34 was observed. Below this up to

4

24m hard laterite with S.P.T value greater than 50 was found. This was followed by soft rock with S.P.T value greater than 50 extending up to 24.5m depth. From 24.5m depth to termination of bore hole at 25.5m hard rock was found. Water table was noted at a depth of 3.5m from the ground level during the time of investigation. For heavy structures in this site, the ideal foundation will be bored cast in situ D.M.C pils end bearing in the hard rock strata. The safe bearing capacity of D.M.C piles end bearing on hard rock is given by Pile Diameter Pile capacity 500 mm 750 kN 600mm 1000 kN 700mm 1470 kN 800mm 1920 kN Table3.1. Pile capacities

3.3

LOADS

The various loads considered for the analysis and design are Dead load based on IS: 875(Part 1)-1987, Live load based on IS: 875(Part 2)-1987, Wind load based on IS: 875(Part 3)-1987 and Seismic load based on IS 1893(Part1): 2002. 3.3.1. Dead Loads The self weight of the structural components will be calculated by the analysis software (ETABS9). The dead load of various elements were found out as follows. Wall load= 0.22 x 2.4 x 20x clear length of wall / centre to centre length of beam Weight of partition walls on slab= (0.12 x 2.8 x 20 )/3 = 2.24 kN/m2 Weight of floor finish

= 1.5 kN/m2

3.3.2. Live Loads as per IS: 875 (Part 2): 1987 All rooms and kitchen, toilet and bath rooms = 2 kN/m2 Balconies

= 3 kN/m2

Corridors, passage, staircases including

= 5 kN/m2

fire escapes and store room

3.3.3. Wind loads as per IS: 875(Part 3)- 1987

5

Design wind speed

Vz = Vb x k1 x k2 x k3

(Clause 5.3) Basic wind speed , Vb

= 39 m/s

(From IS code figure 1) Probability factor k1

= 1.06

(From IS code table1 clause 5.3.1) Terrain, height, and structure size factor k2 (Clause 5.3.2) Category - 2 Class

-B

Topography factor

k3

=1

(Clause 5.3.3) Calculation of design wind pressure: HEIGHT

Vb

K1

K2

K3

Vz

10 15 20 25 30

39 39 39 39 39

1.06 1.06 1.06 1.06 1.06

0.98 1.02 1.05 1.075 1.1

1 1 1 1 1

35

39

1.06

1.1125

1

40.5132 42.1668 43.407 44.4405 45.474 45.9907

40

39

1.06

1

44.35

39

1.06

1.125 1.13487 5

1

5 46.5075 46.9157 3

Wind pressure, Pz 984.7916245 1066.823413 1130.500589 1184.974824 1240.730806 1269.089451 1297.768534 1320.651574

Table3.2. Design wind pressure

Calculation of wind load REGION

Pz1

Pz2

Pz AVG

6

WIDTH HEIGHT

WIND LOAD

UP TO

0

10M

984.7916

10m-15m

984.791

15m-20m

1066.82

20m-25m

1130.82 3

25m-30m

1184.97

30m-35m

1240.73

35m-40m

1269.08

40m44.8m

1297.76

492.395

1066.8234

8 1025.80

1 1130.8234

8 1098.82

1 1184.9748

3 1157.89

2 1240.7308

9 1212.85

0 1269.0894

3

5 1297.7685 3 1320.6515

1254.91 1283.42 9

33.84

10

166.6267387

33.84

5

173.5666301

33.84

5

185.9209215

33.84

5

195.9165309

33.84

5

205.2146963

33.84

5

212.3307937

33.84

5

217.1561855

4.8

145.4165791

1309.21 23.14 7 Base shear due to wind load Table 3.3. Wind load calculation

1502.149076

3.3.4. Earthquake Loads as per IS: 1893 (part 1): 2002 Dynamic forces on multi-storeyed are best computed through a detailed vibration analysis. Detailed dynamic analysis or modal analysis or pseudo static analysis should be carried out depending on the importance of problem. BIS Code 1893 (Part 1): 2002 recommends that [Ref: Cl: 7:8:1] Dynamic analysis shall be performed to obtain the design seismic force, and its distribution to different levels along the height of the building and to the various lateral load-resisting elements for the following buildings: a)

Regular buildings – those greater than 40m in height in Zone IV and Zone V, and those greater than 90m in height in Zone II and Zone III.

b)

Irregular building – all framed buildings higher than 12m in Zones IV and Zone V, and those greater than 40m in height in Zone II and III.

Since the height of the residential complex is 44.35m and it’s located in Zone III, static method of analysis was performed to find the seismic load and its distribution.

7

Static method: The base shear or total design lateral force along any principal direction shall be determined by the following expression: VB = Ah W where, VB = The design base shear Ah = Design horizontal acceleration spectrum value using the fundamental natural period T. W = Seismic weight of the building. The design horizontal seismic coefficient Ah =

Z I Sa 2Rg

Where, Z = Zone factor given in table 2, for the Maximum Considered Earthquake (MCE) and service life of structure in a zone. The factor 2 in the denominator of Z is used so as to reduce the MCE zone factor to the factor for Design Basis Earthquake (DBE) I = Importance factor, depending upon the functional use of structures, characterized by hazardous consequences of failure, post-earthquake functional needs, historical value or economic importance (Table 6 IS 1893 (Part 1):2002 R = Response reduction factor, depending on the perceived seismic damage performance of the structure, characterized by ductile or brittle deformations. However, the ratio (I/R) shall not be greater than 1.0. The values for buildings are given in Table 7 of IS 1893 (Part 1): 2002. Sa = Average response acceleration coefficient. g Distribution of Design Force The design base shear VB was distributed along the height of the buildings as per the following expressions. Qi = VB

Wi hi

2

i=n

∑W h i =1

2

i i

8

Where, Qi = Design lateral force at floor i Wi = Seismic weight of floor i hi = Height of floor i measured from base. n = Number of storeys in the building is the number of levels at which the masses are located. Seismic weight, W The seismic weight of each floor is its full dead load plus appropriate amount of imposed loads while computing the seismic weight of each floor, the weight of columns and walls in any storey shall be equally distributed to the floors above and below the storey. The seismic weight of the whole building is the sum of the seismic weights of all the floors. Any weight supported in between storeys shall be distributed to the floors above and below in inverse proportion to its distance from the floors. Imposed uniformly distributed floor

Percentage of imposed load

loads kN/m² % Upto and including 3.0 25 Above 3.0 50 Table 3.4 Percentage of imposed load to be considered in seismic weight calculation Determination Of Design Base Shear For Seismic Analysis: As per IS 1893 (Part 1):2002 = 0.075h0.75

Fundamental natural period, Ta (Clause 7.6.1) h = height of building exclude basement floor

= 43.95 m Ta = 0.075 x 44.350.75 = 1.28

Average response acceleration coefficient

Sa /g = 1.67/T = 1.67/1.28 = 1.305s

(Clause 6.4.5) Zone factor (clause 6.4.2 table 2)

Z = 0.16

Importance factor (clause6.4.2 table 6)

I=1

Response reduction factor (clause6.4.2 table7)

R=5

Design horizontal seismic coefficient (clause6.4.2), Ah = Z I Sa

9

2Rg = (0.16 x 1 x1.305)/ 2 x5 = 0.0209 Seismic weight of the building Design base shear (clause7.5.3)

W

= 117220.52 kN Vb = W X Ah =117220.52 x 0.0209 = 2449.9 kN

Distribution of base shear to different floor

= Vb ( Wi hi 2 / (Σ Wi hi 2 ))

Qi

( Clause 7.7 )

FLOOR 1

Wi 9353.84

2

7967.39

3

7967.39

4

7967.39

5

7967.39

6

7967.39

7

7967.39

8

7967.39

9

7967.39

10

7967.39

11

7967.39

12

7967.39

13

7967.39

14

7675.07

15

4543.12

Table. 3.5. Distribution of base shear hi ∑hi ( ∑hi)2 wihi2 wihi2/ ∑ wihi2 2.9 2.9 8.41 78665.7944 0.001028626 268022.999 2.9 5.8 33.64 0.00350464 6 603051.749 2.9 8.7 75.69 0.00788544 1 1072091.99 2.9 11.6 134.56 0.01401856 8 1675143.74 2.9 14.5 210.25 0.021904 8 2412206.99 2.9 17.4 302.76 0.03154176 6 3283281.74 2.9 20.3 412.09 0.042931841 5 4288367.99 2.9 23.2 538.24 0.056074241 4 5427465.74 2.9 26.1 681.21 0.070968961 2 2.9 29 841 6700574.99 0.087616001 8107695.73 2.9 31.9 1017.61 0.106015362 8 9648827.98 2.9 34.8 1211.04 0.126167042 6 11323971.7 2.9 37.7 1421.29 0.148071042 3 12651278.3 2.9 40.6 1648.36 0.165426762 9 44.3 1966.92 8935964.94 3.75 0.116845721 5 3 8 102 ∑ wihi 76476612.5 5

Qi 2.423637135 8.257597851 18.57959516 33.0303914 51.60998657 74.31838066 101.1555737 132.1215656 167.2163565 206.4399463 249.792335 297.2735226 348.8835092 389.7768824 275.3107198

3.4. LOAD COMBINATIONS The different combinations used were 1) 1.5 D.L + 1.5 LL 2) 1.5 DL + 1.5 SLX 3) 1.5 DL - 1.5 SLX 4) 1.5 DL + 1.5 SLY 5) 1.5 DL - 1.5 SLX 6) 0.9 DL + 1.5 SLX 7) 0.9 DL - 1.5 SLX 8) 0.9 DL + 1.5 SLY 9) 0.9 DL - 1.5 SLY 10) 1.2 DL + 1.2LL + 1.2 SLX 11) 1.2 DL + 1.2LL - 1.2 SLX 12) 1.2 DL + 1.2LL + 1.2 SLY 13) 1.2 DL + 1.2LL - 1.2 SLY Analysis results from the critical load combinations are used for the design of the structural members. 3.5. DETAILING Reinforcement detailing of the design is in accordance with ductile detailing code IS IS13920: 1993

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CHAPTER 4

MODELING IN ETABS 9 Modeling consisted of structure modeling, member property specification, giving support condition and loading.

4.1. STRUCTURE MODELING The first step was fixing the position of beams and columns. In the ground floor the beams are provide to get maximum column to column connectivity so that better performance during earthquake will be ensured. The beams are provided on the typical floors based on architectural aspects since these floors are used as residential apartments. The model for ground floor, typical floor and terrace floor were prepared separately using ETABS 9. Here the column points are taken as grid. The staircase is provided as an equivalent slab in the model. The slabs were meshed by giving 3 x 3 divisions. After proper checking of each floor models, they were imported to a single file and the model for the entire structure was prepared.

4.2.

MEMBER

PROPERTY

SPECIFICATIONS

AND

SUPPORT

CONDITION The dimensions of different members were fixed based on the trial design. The column dimensions provided for the modeling is as prescribed by the Architect. If necessary it will revised during the design stage. The beams are provided in such a way that torsion is released since compatibility torsion alone comes in them. The member properties assigned are as given below. 4.2.1. Slab Thickness of the slab = 120mm

12

4.2.2. Beams The dimensions of the beams are as shown below Beam Breadth, B Depth, D Position Fixed Beams 250mm 500mm Ground Floor Fixed beam 210mm 500mm First floor Fixed beam 200mm 500mm Connected to shear wall Cantilever beam 200mm 750mm Table. 4.1. Beam Dimensions 4.2.3. Column: The column dimensions are as follows: Ground floor: 400mm X 600mm, 400mm X 800mm, 350mm X 800mm, 300mm X 1300mm, 300mm X 1100mm, 300mm X 950mm, 300mm X 900mm, 300mm X 800mm First floor: 400mm X 400mm, 400mm X 600mm, 350mm X 600mm, 210mm X 1300mm, 210mm X 1100mm, 210mm X 950mm, 210mm X 900mm, 210mm X 800mm 4.2.4. Shear wall: Shear walls are 200 mm thick 4.2.5. Staircase: The staircase is provided as an equivalent slab. The thicknesses of the slab used for staircase is 250mm and 220mm. 4.2.6. Support condition Then support conditions were given to the structure. The support condition given was hinged.

13

Fig. 4.1 Beam lay out of ground floor

14

Fig.4. 2. Beam lay out of first floor

15

Fig. 4.3. Completed model

4.3. LOADING The different load cases considered are live load, dead load, wind load and seismic load. The live loads are provided based on IS:875(Part2)-1987. The self weight of the members will be taken automatically by the software. The weight of partition walls were provided to the slab based on provisions in IS:875(Part1)-1987. The wall loads were applied as a uniformly distributed load in beams. It is calculated after deducting the depth of the beam from floor height and it will be converted to equivalent centre to centre load before applying to the model. The weight of floor finish is taken as 1.5 kN/m2 . The seismic load was calculated based on IS 1893(Part1): 2002. The wind load was calculated based on IS: 875(Part 3)-1987. It was found that for the structure

16

under consideration the effect of wind load is much smaller as compared to seismic load. So for the analysis seismic load alone is considered in addition to live load and dead load.

4.4. ANALYSIS The structure was analysed as ordinary moment resisting space frames in the versatile software ETABS 9. From the analysis we got member end forces and support reactions. From these values, we design the structure.

17

CHAPTER 5

DESIGN OF RC BUILDING 5.1. DSEIGN OF SLAB 5.1.1. Slab S7 (Two way slab) Material Constants: Concrete, fck = 25 N/mm² Steel,

fy = 415 N/mm²

Loads: Using 120 mm thick slab Dead Load on Slab = 0.12 x 25 = 3 kN/m² Live Load on Slab =

2 kN/m²

Finishes

=

1.5 kN/m²

Partition load

=

2.243 kN/m²

Total

=

8.743 kN/m²

Boundary Conditions –Two Adjacent Edges Discontinuous Clear Size = 3.7 m x 4.15 m Maximum diameter of bar 1/8 of slab thickness. Assume a clear cover of 25 mm & 8 mm dia bars Eff: depth along shorter direction dx = 91 mm Eff: depth along longer direction dy = 83 mm Effective span as per IS 456: 2000 clause 22.2.b lyeff = 4.15+0.083 = 4.233 m lxeff = 3.7+0.091= 3.791 m lyeff/lxeff =1.117, Hence design as Two Way Slab. Shorter direction: -ve Bending Moment coefficient

= 0.054

Negative moment at continuous edge = 6.81 kN-m Factored bending moment, Mu Mu / (bd²)

= 10.21 kN-m = 1.233

As per IS 456:2000 Ast = 331.01 mm²

18

Minimum steel 0.12% cross sectional area =144 mm² Spacing = ast/Ast x 1000 = 50.26/331.01 x 1000 =153 mm Maximum Spacing 3d or 300mm which ever is less Maximum Spacing = 3d = 273 mm Hence, Provide reinforcement of 8 mm dia bars at 150 mm c/c +ve Bending Moment coefficient = 0.041 Positive moment at mid span

= 5.13 kN-m

Factored bending moment, Mu

= 7.70 kN-m

Mu / ( bd2 ) =0.93 As per IS 456:2000 Ast = 245.44 mm² Minimum steel 0.12% cross sectional area =144 mm² Spacing = ast/Ast x 1000 = 50.26/245.44 x 1000 =204.78 mm Maximum Spacing 3d or 300mm which ever is less Maximum Spacing = 3d = 273 mm Provide reinforcement of 8 mm dia bars at 200 mm c/c Longer Direction -ve Bending Moment coefficient

= 0.047

Negative moment at continuous edge = 5.91 kN-m Factored bending moment Mu

= 8.86 kN-m

Mu/bd²

= 1.286 Ast

= 315.68 mm²

Minimum steel 0.12% cross sectional area =144 mm² Spacing = ast/Ast x 1000 = 50.26/315.68 x 1000 =159 mm Maximum Spacing 3d or 300mm which ever is less Maximum Spacing = 3d = 249 mm Provide reinforcement of 8 mm dia bars at 150 mm c/c

19

+ve Bending moment coefficient = 0.035 Positive moment at mid span

= 4.4 kN-m

Factored bending moment, Mu

= 6.6 kN-m

Mu / ( bd² )

=0.96 Ast

= 230.9 mm²

Minimum steel 0.12% cross sectional area =144 mm² Spacing = ast/Ast x 1000 = 50.26/230.9 x 1000 =217.7 mm Maximum Spacing 3d or 300mm which ever is less. Maximum Spacing = 3d = 249 mm Provide reinforcement of 8 mm dia bars at 210 mm c/c Check for deflection: As per IS 456:2000 clause 23.2 Percentage of tension reinforcement pt = 100As/bd =0.27 Modification factor from IS 456:2000 figer-4 fs= 0.58fy Area of steel required/area of steel provided = 0.58 x 415 x 0.27/0.28 =232.1 N/ mm2 Modification factor = 1.58 Maximum span / depth ratio from code = 32 x 1.58 = 50.56 Span/ depth ratio provided = 3791/91

= 41.66

Maximum span / depth ratio from code > Span/ depth ratio provided So deflection is safe with provided depth. Check for shear: As per IS 456:2000 clause 40 page 72. Vu = WxL x l/2 = 13.1145 x 3.791x0.5 = 24.85 KN Nominal shear stress τ v = Vu/bd = 0.27 N/ mm²

20

As per IS 456:2000 maximum value of shear stress τ c = 3.1 N/ mm² As per IS 456:2000 design shear strength of concrete τ c = 0.49 N/ mm²

τ v < τ c< τ c max , so shear reinforcement is not required. 5.1.2. Slab S5 (One way slab) Material Constants: Concrete, fck = 25 N/mm² Steel,

fy = 415 N/mm²

Loads: Using 120 mm thick slab Dead Load on Slab = 0.12 x 25 = 3 kN/m² Live Load on Slab =

2 kN/m²

Finishes

=

1.5 kN/m²

Partition load

=

2.243 kN/m²

Total Dead load

=

6.74 kN/m²

Clear Size = 1.6 m x 4.35 m Maximum diameter of bar 1/8 of slab thickness. Assume a clear cover of 25 mm & 8 mm dia bars Eff: depth along shorter direction dx = 91 mm Eff: depth along longer direction dy = 83 mm Effective span as per IS 456: 2000 clause 22.2.b lyeff = 4.35+0.083 = 4.433 m lxeff = 1.6+0.091= 1.691 m lyeff/lxeff =2.62, Hence design as One Way Slab. Shorter direction: Using Table 12 of IS 456:2000 Negative moment at continuous edge = -(WDL x lxeff2/10 + WLL x lyeff2/9) = -(6.74 x 1.6912 /10+2 x 1.6912/9) = 2.2415 kNm Factored bending moment, Mu Mu / (bd²)

= 3.362 kN-m = 0.4

As per SP16 pt = 0.113

21

Minimum steel 0.12% cross sectional area =144 mm² Spacing = ast/Ast x 1000 = 50.26/144 x 1000 =349 mm Maximum Spacing 3d or 300mm which ever is less Maximum Spacing = 3d = 273 mm Hence, Provide reinforcement of 8 mm dia bars at 270 mm c/c Positive moment at mid span

= WDL x lxeff2/16 + WLL x lyeff2/12 = 6.74 x 1.6912 /16+2 x 1.6912/12 = 1.681 kNm

Factored bending moment, Mu

= 2.52 kN-m

Mu / ( bd2 ) =0.365 As per SP16 pt = 0.101 As per IS 456:2000 Minimum steel 0.12% cross sectional area =144 mm² Spacing = ast/Ast x 1000 = 50.26/144 x 1000 =349 mm Maximum Spacing 3d or 300mm which ever is less Maximum Spacing = 3d = 273 mm Provide reinforcement of 8 mm dia bars at 270 mm c/c Longer Direction: Distribution steel 0.12% crossectional area =144 mm² Provide reinforcement of 8mm dia bars at 240 mm c/c Check for deflection: As per IS 456:2000 clause 23.2 Percentage of tension reinforcement pt = 100As/bd =0.15 Modification factor from IS 456:2000 figer-4 fs= 0.58fy Area of steel required/area of steel provided = 0.58 x 415 x 0.12/0.15 =192.56 N/ mm2

22

Modification factor = 2 Maximum span / depth ratio from code = 26 x 2 = 52 Span/ depth ratio provided = 1691/91

= 18.58

Maximum span / depth ratio from code > Span/ depth ratio provided So deflection is safe with provided depth. Check for shear: As per IS 456:2000 Table13 Shear force

V =0.6 x WDL x lxeff + 0.6 x WLL x lyeff = 0.6 x 6.74 x 1.691+ 0.6 x 2 x 1.691 = 8..867 kN

Factored Shear force= 13.3 kN Nominal shear stress τ v = Vu/bd = 0.15 N/ mm² As per IS 456:2000 maximum value of shear stress τ c = 3.1 N/ mm² As per IS 456:2000 design shear strength of concrete τ c = 0.29 N/ mm²

τ v < τ c< τ c max , so shear reinforcement is not required. 5.2. DESIGN OF STAIR 5.2.1. Fire stair- First flight

23

Material Constants: Concrete, fck = 25 N/mm² Steel,

fy = 415 N/mm²

As per IS 456:2000 Clause 33.1 Effective span

= 4.55m

Thickness of slab =

leff / 20

= 227.5mm Provide an overall depth D = 230 mm Provide 20 mm dia bar Clear cover

= 25 mm.

Effective depth = 195mm Loads: Using 230 mm thick slab Landing Dead Load on Slab = 0.23 x 25 = 5.75 kN/m² Live Load on Slab =

5 kN/m²

Finishes

1.5 kN/m²

Total

= =

12.25 kN/m²

Going Self weight of waist slab

= 0.23 x 25 x 35.053/30 = 6.718 kN/m²

Self weight of step

= 0.5 x 0.1813 x 1 x 25 = 1.875 kN/m²

Finishes

=

1.5 kN/m²

Live Load on Slab

=

5 kN/m²

Total

=

15.484 kN/m²

Total reaction RA + RB

=

62.53 kN

RA

=

31.26 kN

24

RB

=

31.26 kN

Maximum moment at the centre Moment

M

Factored moment Mu

=

37.633 KN-m.

=

56.45 KN-m.

Mu / (bd²) Percentage of steel Pt

= 1.484 = 0.444

As per IS 456:2000 Ast =865.8 mm² Minimum steel 0.12% cross sectional area =192mm² Spacing = ast/Ast x 1000 = 113.1/865.8 x 1000 =130 mm Provide reinforcement of 12 mm dia bars at 130 mm c/c Maximum Spacing 3d or 300mm which ever is less Maximum Spacing = 3d = 585 mm Hence, Provide reinforcement of 12 mm dia bars at 130 mm c/c Distribution steel 0.12% cross sectional

= 276mm²

Provide 10mm dia bar. Spacing = ast/Ast x 1000 = 78.54/276 x 1000 =284.56 mm Maximum Spacing 4d or 450mm which ever is less Maximum Spacing = 4d = 780 mm Hence, Provide 10 mm dia bars at 280 mm c/c Check for deflection: As per IS 456:2000 clause 23.2 Percentage of tension reinforcement pt = 100As/bd =0.444 Modification factor from IS 456:2000 figer-4 fs= 0.58fy Area of steel required/area of steel provided = 0.58 x 415 x 865.8/869.98 =239.54 N/ mm2

25

Modification factor = 1.35 Maximum span / depth ratio from code =20 x 1.35 = 27 Span/ depth ratio provided = 4550/195

= 23.33

Maximum span / depth ratio from code > Span/ depth ratio provided So deflection is safe with provided depth. Check for shear: As per IS 456:2000 clause 40 page 72. Vu = 46.89 kN Nominal shear stress τ v = Vu/bd = 0.24 N/ mm² As per IS 456:2000 maximum value of shear stress τ c = 3.1 N/ mm² As per IS 456:2000 design shear strength of concrete τ c = 0.46 N/ mm²

τ v < τ c< τ c max , so shear reinforcement is not required 5.3. DESIGN OF BEAMS

Material Constants: Concrete, fck = 25 N/mm² Steel,

fy = 415 N/mm²

Using 200x500mm beam, Assume clear cover of 25mm and 16mm diameter main bars and 8mm diameter stirrup bars, Effective depth, d = 500 - 25 - 8 - 16/2

= 459 mm

5.3.1. BEAM B42 For beam B42, at the left end

26

Factored Bending Moment, Mu ( negative )

= 111.85 kNm

Factored Bending Moment, Mu ( positive )

= 55.56 kNm

Factored Shear Force, Vu

= 56.2 kN

Moment of resistance, Mu,lim

= 0.138 fck bd2 = 145.37 kNm

Mu < Mu,lim Hence design it as singly reinforced beam R = Mu/ (bd2)

= 2.65

From SP16, table3 Reinforcement percentage, pt

= 0.857

Ast

= 786.7 mm2

Hence provide 2 nos of Y20mm diameter bar and one Y16mm diameter bar Ast,provided

= 829.38mm2

Pt

= 0.9

Check for shear Factored Shear force, Vu

= 56.2 kN

Shear stress, τv

= Vu/ (b x d) = 0.61 N/mm2 = 0.612 N/ mm2

Shear strength of concrete, τc τc > τv

Hence safe against shear force. Provide minimum shear reinforcement From Cl.6.3.5 of IS 13920: 1993, the spacing of hoops over a length of 2d at either end of a beam shall not exceed a) d/4 and b) 8 times the diameter of the smallest longitudinal bars. Here 8xφ

d/4

=

500 / 4

=

125 mm

=

8 x 16

=

128 mm

Hence provide 8mm diameter two legged stirrups at a spacing of 120mm from both ends of beam for a distance of 1000 mm and in between this, provide spacing of 220mm for shear reinforcement

27

BEAM, B42 Bending Moment, Mu(kNm) Moment of resistance, Mulim R= Mu/ (bd2) From SP16, pt Ast required (mm2) Steel provided pt provided Factored Shear Force, Vu

b=200mm Left Negative Positive 111.85

55.56

Centre Positive 75.2

D=500 mm Right Negative Positive 104.72

56.94

0.138fck bd2 = 145.37 kNm 2.65 0.86 787.37 2,Y20mm + 1,Y16mm 0.9

1.32 0.39 358.97 2, Y16mm 0.44

1.78 2.49 0.54 0.79 499.83 728.07 2,Y16mm 2,Y20mm +

+

1,Y12mm 1,Y12mm 0.56 0.81

1.35 0.4 368.29 2, Y16mm 0.44

(kN) Shear stress, τv Shear strength, τc

107.9 1.18 0.59 Y8mm diameter two legged stirrups @100mm c/c

Shear reinforcement

for a distance of 1000mm from each end and @ 220mm c/c for the remaining distance

5.3.2. BEAM B51

28

BEAM, B51 Bending Moment, Mu(kNm) Moment of resistance, Mulim R= Mu/ (bd2) From SP16, pt Ast required (mm2) Steel provided pt provided Factored Shear Force, Vu

b=200mm Left Negative Positive 63.56

53.47

Centre Positive 48

D=500 mm Right Negative Positive 70.01

48.65

0.138fck bd2 = 145.37 kNm 1.51 0.45 414.85

1.27 0.37 344.24

1.14 0.33 306.81

1.66 0.5 461.11

1.15 0.34 311.23

2,Y20mm 2,Y16mm 2,Y16mm 2,Y20mm 2,Y16mm 0.68

0.44

0.56

0.68

0.56

(kN) Shear stress, τv Shear strength, τc

85.68 0.93 0.55 Y8mm diameter two legged stirrups @100mm c/c

Shear reinforcement

for a distance of 1000mm from each end and @ 220mm c/c for the remaining distance

5.3.3. BEAM B60 BEAM, B60 Bending Moment, Mu(kNm) Moment of resistance,

b=200mm Right Negative Positive 128.89

86.85

Centre Positive 75

D=500 mm Left Negative Positive 81.61

0.138fck bd2 = 145.37 kNm

Mulim

29

83.2

R= Mu/ (bd2) From SP16, pt Ast required (mm2) Steel provided

3.06 1.02 936.85

2.06 0.64 586.54

3,Y20mm 3,Y16mm

pt provided Factored Shear Force, Vu

1.03

0.66

1.78 0.54 497.56 2,Y16mm + 1,Y12mm 0.56

1.94 0.6 546.75

1.97 0.61 558.75

2,Y20mm 3,Y16mm 0.68

0.66

(kN) Shear stress, τv Shear strength, τc

122.27 1.33 0.55 Y8mm diameter two legged stirrups @100mm c/c

Shear reinforcement

for a distance of 1000mm from each end and @ 220mm c/c for the remaining distance

5.4. DESIGN OF COLUMN C5 5.4.1. COLUMN C5 Material Constants: Concrete M 30 fck = 30 N/mm² fy = 415 N/mm² Depth of column,

D = 400 mm

Breadth of column, b = 400 mm Support condition: - Both ends fixed Unsupported length, Lo

= 2.4 m

Multiplication factor for effective length, k [Using IS 456: 2000, Annex E] Stability index, Q

= (ΣPu x Δu)/ (Hu x hs)

Axial load on first floor columns, ΣPu= 191946.91 kN Lateral deflection, Δu

= 0.0061 m

Height of the storey, hs

= 2.9 m

Total lateral force acting within the storey, Hu = 2353.766 kN Stability index, Q

= (191946.91 x 0.0061)/ (2353.766 x 2.9) = 0.1715 > 0.04

30

Hence the column is a sway column β1

= 0.4

β2

= 0.28

Multiplication factor for effective length, k ( Lex) = 1.28 Multiplication factor for effective length, k ( Ley) = 1.28 Effective length, Lex

= 2.4 x 1.28 = 3.07

Effective length, Ley

= 2.4 x 1.28 = 3.07

Factored axial load on column load, Pu

= 2974.44 kN

Factored moment in X direction, Mux

= 58.183 kNm

Factored moment in Y, direction Muy

= 143.842 kNm

Type of Column lex / D = 3.07/0.4 = 7.68 < 12 ley / b = 3.07/0.4 = 7.68 < 12 So design as a short column with biaxial bending Calculation of eccentricity Eccentricity in the direction of longer dimension ex = lo / 500 + D / 30 = 2.4/500 + 0.4/ 30 = 0.018 m Take ex = 0.02 m Eccentricity in the direction of shorter dimension ey = lo / 500 + b / 30 = 2.4/500 + 0.4/ 30 = 0.018 m Take ey = 0.02 m Moments due to minimum eccentricity Mex = Pu x ex = 59.59 kN-m Mey = Pu x ey = 59.59 kN-m So take as actual moments are, Mux = 59.59 kN-m Muy = 143.842 kN-m

31

Assume percentage of steel

P = 3.9

( 0.8 percentage is minimum steel area of column as per IS 456: 2000 ) Assuming dia of bar

= 20 mm

Assuming clear cover

= 40 mm

d' = Clear cover + Half the bar diameter = 50 d'/D ( about X axis)

= 50/400 = 0.125

d'/D taken

= 0.125

d'/D ( about Y axis)

= 50/400 = 0.125

d'/D taken

= 0.125

The section is to be checked for biaxial bending Pu / (fck x b x D) = 0.621 p/fck = 0.13 Refer SP 16: 1980 for x axis d'/D = 0.1 Mu_____ = 0.095 fck x b x D² d'/D = 0.15 Mu_____ = 0.085 fck x b x D² Corresponding to d'/D = 0.125 Mu_____ = 0.09 fck x b x D² Mux1 = 0.09 x 30 x 400 x 400 2 = 172.8 kN-m Refer SP 16 1980 for y axis d'/D = 0.125 Mu_____ = 0.095 fck x b x D² d'/D = 0.15

32

Mu_____ = 0.085 fck x b x D² Corresponding to d'/D = 0.125 Mu_____ = 0.09 fck x b x D² Muy1 = 0.09 x 30 x 400 x 400 2 = 172.8 kN-m Puz = 0.45 x fck x Ag + 0.75 x fy x As = 0.45 x 35 x 0.952( 400 x 400) + 0.75 x 415 x 0.039 ( 400 x 400) = 4017.96 KN Pu/ Puz = 0.742 αn

= 1.903

(Mux/Mux1)αn + (Muy/Muy1) αn = 0.84 0.04

Hence the column is a sway column For Lex β1

= 0.814

β2

= 0.765

For Ley β1

= 0.232

β2

= 0.266

From Fig.27 of IS 456:2000 Multiplication factor for effective length, k ( Lex) = 2.4 Multiplication factor for effective length, k ( Ley) = 1.15 Effective length, Lex

= 2.4 x 2.4= 5.76

Effective length, Ley

= 2.4 x 1.15= 2.76

Factored axial load on column load, Pu

= 2394.67 kN

Factored moment in X direction, Mux

= 82.667 kNm

Factored moment in Y, direction Muy

= 10.017 kNm

Type of Column

34

lex / D = 5.76/0.8 = 7.20 < 12 ley / b = 2.76/0.21 = 13.14 > 12 So the column is slender about y axis e Ley =13.14 , from SP 16 clause 3.4 ; y = 0.086 b b

For

Additional moment only in Y direction. M uy = 2394.67 × 0.086 × 210 ×10 −3 = 43.43 kN-m

Assume a reinforcement percentage as p = 2.4 p 2.4 = 0.08 = f ck 30 Puz = 0.45 f ck Ac +0.75 f y As

=3468.53 kN Assuming 25 mm dia bars and 40 mm cover Pb = ( k1 + k 2

p ) f ck bD (SP: 16 Table 60) f ck

K1 =0.184 K2 =0.028 Pb = 938.65kN (About YY axis)

Ky =

Puz − Pu Puz − Pb

= 0.424 Additional moment: M ay = 43.43kNm

Minimum eccentricity consideration: ey =

l b + = 20mm 500 30

Moment due to minimum eccentricity: M uy = 47.89kNm

Total moment taken for column design is M ux = 82.67 kNm M uy = 66.33kNm

Moment capacity about YY axis

35

d' = 0.21 (About YY axis) D

Chart for

d' = 0.2 will be used. (SP 16:1980) D

Pu 2394.67 ×10 3 = f ck bD 30 × 210 × 800

=0.475 From SP 16: 1980 Chart 46

Mu = 0.07 f ck bD 2 Muy1 =74.09 kN-m Moment capacity about XX axis Assuming 25 mm dia bars and 40 mm cover d' = 0.065 D

Chart for

d' = 0.1 will be used. (SP 16:1980) D

Pu 2394.67 ×10 3 = f ck bD 30 × 210 × 800

= 0.475 From SP 16: 1980 Chart 44

Mu = 0.0875 f ck bD 2 Mux1

= 352.8 kN-m Pu = 0.069 Puz

αn = 1.817 (Mux/Mux1)αn + (Muy/Muy1) αn = 0.89 Dp/2 Hence Maximum shear force on pile cap = 1470kN Ultimate shear, Vu

= 2205 kN

Nominal shear stress, τ v

= 2.72 N/mm²

100 As/ (bd) Deign shear strength, τ c ie, τ v > τ c

= 0.55 = 0.51 N/mm²

so shear reinforcement are needed

Assume 12mm dia 6 legged stirrups Vus

= Vu - τ c bd

= 1791.9 kN

Diameter of bar

= 12 mm

Area of shear reinforcement effective in shear, Asv

= 678.58 mm²

Provide Y12 mm dia 6 legged stirrup Spacing of shear reinforcement, Sv

=

0.87 x d x fy x Asv Vus

42

= 123.05 mm c/c Provide Y12 mm dia 6 legged stirrup at 120mm c/c As per IS 456:200 Depth of pile cap is greater than 750 mm. Hence side face reinforcement is needed. Side face reinforcement

= 0.1 % of web area = 0.1 x 900 x900/100 = 810 mm²

Side face reinforcement on one face

= 405 mm²

Hence provide 4 Nos of Y12mm diameter bar on one face

5.6.2. THREE PILE GROUP Material Constants Concrete, fck = 25 N/mm² Steel,

fy = 415 N/mm²

Each pile should be connected using pile cap with a minimum of 100mm edge distance to either sides of the pile. This pile cap is designed as two simply supported beam. As per IS 2911 spacing between two pile is 2.5 x dia of pile Length of pile cap

= 2.5 x 700 + 2 x 350 + 2 x 100 =2650 mm

Depth of pile cap

= development length of column bar + cover

As per SP-16 Table 65 For 25 mm diameter bars Ldc = 806 mm Assume a 100 mm projection of pile in to the cap concrete Depth of pile cap

= 806 + 100 = 906 mm

Provide an overall depth, D = 1200mm Breadth of pilecap = diameter of pile + 100 mm overhang = 700 + 2 x 100 = 900mm Size of pile cap is as shown below

43

Effective depth, d = 900 mm b =900 mm Design of portion 1 Factored axial load on pile Pu = 2205 kN Bending moment at face of column = 1470 x 0.81 = 1190.7 kN-m Ultimate moment, Mu Mu / (bd2)

= 1786.05 kN-m = 2.205

% of tension steel, pt

= 0.69

Area of tension reinforcement, Ast = 6210mm² Provide 13 Nos of Y25 mm dia bars Area of steel provided

= 6381.36 mm²

Distance from face of the column to the centre of the pile = 0.81m > Dp/2 Hence Maximum shear force on pile cap = 1470kN Ultimate shear, Vu

= 2205 kN

Nominal shear stress, τ v

= 2.45 N/mm²

100 As/ (bd) Deign shear strength, τ c ie, τ v > τ c

= 0.7 = 0.554 N/mm²

so shear reinforcement are needed

Assume 12mm dia 6 legged stirrups

44

Vus

= Vu - τ c bd

= 1706.4 kN

Diameter of bar

= 12 mm

Area of shear reinforcement effective in shear, Asv

= 678.58 mm²

Provide Y12 mm dia 6 legged stirrup Spacing of shear reinforcement, Sv

=

0.87 x d x fy x Asv Vus

= 129.22 mm c/c Provide Y12 mm dia 6 legged stirrup at 120mm c/c As per IS 456:200 Depth of pile cap is greater than 750 mm. Hence side face reinforcement is needed. Side face reinforcement

= 0.1 % of web area = 0.1 x 1000 x900/100 = 900 mm²

Side face reinforcement on one face

= 450 mm²

Hence provide 4 Nos of Y12mm diameter bar on one face Design of portion 2 Factored axial load on pile Pu = 2205 kN Bending moment at face of column = 1470 x 0.375 = 551.25 kN-m Ultimate moment, Mu Mu / (bd2)

= 826.875 kN-m = 1.134

% of tension steel, pt

= 0.33

Area of tension reinforcement, Ast = 2694.75 mm² Provide 9 Nos of Y20 mm dia bars Area of steel provided

= 2827.43 mm²

Distance from face of the column to the centre of the pile = 0.375m > Dp/2 Hence Maximum shear force on pile cap = 1470kN Ultimate shear, Vu

= 2205 kN

Nominal shear stress, τ v

= 2.72 N/mm²

100 As/ (bd) Deign shear strength, τ c ie, τ v > τ c

= 0.35 = 0.42 N/mm²

so shear reinforcement are needed

45

Assume 12mm dia 6 legged stirrups Vus

= Vu - τ c bd

= 1864.8 kN

Diameter of bar

= 12 mm

Area of shear reinforcement effective in shear, Asv

= 678.58 mm²

Provide Y12 mm dia 6 legged stirrup Spacing of shear reinforcement, Sv

=

0.87 x d x fy x Asv Vus

= 118.22 mm c/c Provide Y12 mm dia 6 legged stirrup at 110mm c/c As per IS 456:200 Depth of pile cap is greater than 750 mm. Hence side face reinforcement is needed. Side face reinforcement

= 0.1 % of web area = 0.1 x 900 x900/100 = 810 mm²

Side face reinforcement on one face

= 405 mm²

Hence provide 4 Nos of Y12mm diameter bar on one face

5.7. DESIGN OF SHEAR WALL 5.7.1. Shear Wall P1

Axial Force 7898.48kN

Loading Moment Shear 8710.6

Force 1096.55 kN 4897.68 kN

kNm Horizontal length of wall, lw

= 6400 mm

As per IS 13920: 1993,clause 9.1.2, Thickness of the wall should not be less than 150 mm Thickness of the web provided, tw

Reaction

= 200 mm

46

Hence okay Effective depth of wall section, dw

= 0.8 x lw = 5120 mm

Shear strength Factored Shear Force, Vu

= 1096.55/2 = 548.275 kN

As per IS 13920: 1993, clause 9.1.2 Nominal Shear stress, τv

=

vu bd

= 0.535 N/ mm2 Assuming minimum reinforcement ratio for horizontal and vertical reinforcement pt

= 0.25%

From IS 456:2000, table 19 Design Shear stress, τc

= 0.36 N/ mm2 = 3.1 N/ mm2

τcmax Hence τv < τcmax and τv > τc

Therefore shear reinforcement is needed As per IS 13920: 1993, clause 9.2.5 Vus

= 0.87 x fy x Ah x dw / Sv

Vus

= Vu - τc x tw x dw = 179.635 kN

Assuming Y 8 mm diameter bars in two curtains Sv

= 1034.53 mm

Which is less than the minimum reinforcement required Hence provide minimum reinforcement, pt = 0.25 Provide Y 8mm diameter bars @ 200 mm c/c in two curtains The vertical reinforcement should not be less than the horizontal reinforcement Hence provide Y 8mm diameter bars @ 200 mm c/c in two curtains as vertical reinforcement. Flexural Strength The moment of resistance Muv of the wall section shall be calculated as for columns subjected to combined axial load and uniaxial bending

47

When xu/lw < xu*/lw, Muv / (fck x tw x lw2)

= Φ [(1+λ/ Φ)(0.5-0.416 xu /lw)- (xu /lw)2 (0.168+ β2/3)]

ρ

= 0.0025

φ=

0.87 f y xρ

λ=

f ck

= 0.0361

Pu = 0.1234 f ck twlw

λ +φ

xu/lw= 0.36 + 2φ = 0.369 xu*/lw= 0.0035/(0.0035+0.87 x fy/Es) = 0.66 β

= 0.87 fy/ (0.0035x Es) = 0.516

Hence Muv / (fck x tw x lw2) = 0.054 Muv Mu

= 11059.2 kNm = 8710.6/2 = 4355.3 kNm < Muv

Hence safe Boundary elements Area of cross section, Ag

= 6400 x 200 = 1.28x106 mm2 = tw x lw3 /12

Moment of inertia, Iy

= 4.3691 x 1012 mm4 Extreme fibre compressive stress, fc = (Pu/Ag)+(Mu x lw/ 2Iy) = 3.085 + 3.189 = 6.275 N/mm2 = 5 N/mm2

0.2 x fck fc>0.2 x fck Hence boundary element is needed

Let the boundary element be of dimension 200 x 1500 and assume 2.5% Steel Gross area of cross section

= 200 x 1500 = 300000 mm2

Area of reinforcement

= 2.5 x 300000/100

48

= 7500 mm2 Pu

= 0.4 fck Ag + (0.67 fy-0.4 fck)As = 5010.375 kN > 4897.68 kN

Hence safe The diameter of the bar should not exceed 1/10th of the wall thickness Hence use 24 nos of Y 20mm diameter bars

5.8. DESIGN OF WATER TANK (IS 3370) 5.8.1. Drinking Water Tank Water tank of dimension 3.275x2.405x1.5m having a capacity of 10000 litres. Columns are provided till top of the tank. To find the thickness of sidewall Height of wall, a

=

1.5 m

Width of wall, b

=

3.275 m

Density of liquid, w

=

10 kN/m3

Asuume M25 Concrete and Fe 415 grade steel fck = 25 N/mm² fy = 415 N/mm² b/a

=

3.275/1.5

=

2.18

Moment coefficient for individual wall panel, top and bottom hinged, vertical edges fixed from IS 3370(PartIV)-1967 My

=

0.061

Maximum horizontal moment, M

=

My ×w ×a 3

=

2.06 kNm

Permissible Tensile strength in concrete due to bending as per IS : 3370(PartII)-1967 =

1.8 N/mm2

=

82.86 mm

M / Z ≤ 1.8

Taking M / Z = 1.8 Thickness of wall, t

Using 8 mm dia bars and 25 mm clear cover Overall depth, D

=

111.86 mm

Provide 200 mm thick wall

49

Reinforcement required in vertical direction Area of steel required, Ast m

=

M/σ st jd

=

280/3σ cbc

=

11

k

=

0.384

j

=

1- k

=

0.872

=

92.64 mm2

Area of steel, Ast

3

As per IS : 3370(PartII)-1967 since t > 150 mm, reinforcement should be provided in two layers. Minimum reinforcement required, Ast min

=

0.271 % of concrete section

=

542 mm2

=

144.9 mm

Moment coefficient, Mx

=

0.046

Maximum vertical moment, M

=

Mx × w × a 3

=

1.55 kNm

=

M/σ st jd

=

280/3σ cbc

=

11

k

=

0.384

j

=

1- k

=

0.872

=

72.7 mm2

Spacing required Provide 10 mm dia bars @ 140 mm c/c Reinforcement required in vertical direction

Area of steel required, Ast m

Area of steel, Ast

3

As per IS : 3370(PartII)-1967 since t > 225 mm, reinforcement should be provided in two layers. Minimum reinforcement required, Ast mini Spacing required

=

0.271 % of concrete section

=

542 mm2

=

144.9 mm

Provide 10 mm dia bars @ 140 mm c/c

50

Design of top slab Let thickness of top slab be 100 mm Loads on the slab:-

Ly Lx

=

Live load

=

5 kN/m2

Dead load

=

2.5 kN/m2

Finish load

=

1.5 kN/m2

Total load

=

9 kN/m2

Factored load

=

13.5 kN/m2

3.338 = 1.35 2.476

Design as two way slab All edges discontinuous Area of reinforcement in shorter span:Moment coefficient

=

0.082

Moment, M

=

6.786 kNm

Provide 8 mm dia bars @ 170 mm c/c Area of reinforcement in longer span:Moment coefficient

=

0.056

Moment, M

=

4.635 kNm

Provide 8 mm dia bars @ 230 mm c/c Design of bottom slab Let thickness of bottom slab be 200 mm Loads on the slab:-

Ly Lx

=

Dead load

=

5 kN/m2

Load due to water

=

15 kN/m2

Finish load

=

1.5 kN/m2

Total load

=

21.5 kN/m2

Factored load

=

32.25 kN/m2

3.438 = 1.33 2.576

Design as two way slab All edges discontinuous

51

Area of reinforcement in shorter span:Moment coefficient

=

0.081

Moment, M

=

11.56 kNm

Area of steel required, Ast

=

M/σ st jd

=

280/3σ cbc

=

11

m k

=

0.384

j

=

1- k

=

0.872

=

519.87 mm2

Area of steel, Ast

3

Provide 10 mm dia bars @ 150 mm c/c Area of reinforcement in longer span:Moment coefficient

=

0.056

Moment, M

=

7.99 kNm

Area of steel required, Ast

=

M/σ st jd

=

280/3σ cbc

=

11

m k

=

0.384

j

=

1- k

=

0.872

=

377.07 mm2

Area of steel, Ast Provide 10 mm dia bars @ 200 mm c/c. Reinforcement details shown in figure.

52

3

CHAPTER 6

SITE VISITS 6.1. VISIT TO I.M.A SITE ON 14.01.2009 As part of the industrial training, I visited the construction site of a multistoried residential building for Indian Medical Association (I. M. A) located at Kaloor. The proposed building consist of two blocks in which the first one has G+9 stories and the later has G+5 stories. At the time of our visit, the fourth floor concreting of block two was going on. We observed the methods adopted for mixing, transporting and placing of concrete.

53

Fig. 6.1. Beam Column joint

Fig. 6.2. Beam to Beam joint

Fig. 6.3. Reinforcement details of fourth floor

6.1. VISIT TO FEDERAL CITY SITE ON 20.03.2009 I visited the construction site of Federal city, Ankamaly. It is an apartment complex promoted by Federal Bank Employees Association. I got an opportunity to observe the construction techniques adopted for boring and concreting of Drilled Mud Circulation (D.M.C) pile from this site. The hard rock available at the site was at a depth of 8m.

54

Fig. 6.4. Boring for DMC pile

Fig. 6.5. Pile Reinforcement

CHAPTER 7

CONCLUSION

55

The industrial training, taken through a period of three months allowed me to gain ample exposure to various field practice in the analysis and design of multi storeyed buildings and also in various construction techniques used in the industry. The analysis was done using ETABS9 and the detailing was done with the help of AUTOCAD 2007. All the structural components were designed manually. Though the use of the software (ETABS9) offers saving in time, the calculations are not appropriate. It takes value on safer side than manual work. Hence manual design was adopted. The analysis and design was done according to standard specifications to the possible extend.

REFERENCES 1.

S.Unnikrishna Pillai & Devadas Menon “ Reinforced Concrete Design”. Tata McGraw-Hill Publishing Company Limited, New Delhi, 2003.

56

2.

N Krishna Raju, “Advanced Reinforced Concrete Design”, C.B.S Publishers and Distributers, New Delhi, First Edition.

3.

Dr. P.C. Varghese, “Advanced Reinforced Concrete Design”, Prentice-Hall of India Private Limited, New Delhi, 2004.

4.

IS: 875 (Part I)-1987, “Indian Standard Code of Practice for Design Loads (Other than earthquake) for Building and Structures”, Bureau of Indian Standards, New Delhi.

5.

IS: 875 (Part II)-1987, “Indian Standard Code of Practice for Design Loads (Other than earthquake) for Building and Structures”, Bureau of Indian Standards, New Delhi.

6.

IS: 875 (Part III)-1987, “Indian Standard Code of Practice for Design Loads (Other than earthquake) for Building and Structures”, Bureau of Indian Standards, New Delhi.

7.

IS: 1893 (Part 1) 2002- Indian standard-“Criteria for earthquake resistant design of structures”, Bureau of Indian Standards, New Delhi.

8.

IS: 456-2000 Indian standard. ”Plain and reinforced concrete – Code of Practice” (Bureau of Indian standard, 2000, New Delhi.

9.

SP 16: 1980, “Design Aids for Reinforced Concrete to IS: 456-1978”, Bureau of Indian Standards, New Delhi.

10.

SP 34: 1987, “Hand Book on Concrete Reinforcement and Detailing”, Bureau of Indian Standards, New Delhi.

57