Biaxially Voided Bubble Deck Slab System and Other Conventional Floor Slab Systems

BIAXIALLY VOIDED BUBBLE DECK SLAB SYSTEM AND OTHER CONVENTIONAL FLOOR SLAB SYSTEMS

Chapter 1 INTRODUCTION

Department of Civil Engineering, M.S.R.I.T.

Page 1


1.1 OVERVIEW Reinforced concrete slabs are relatively thin, flat, structural elements, whose main function is to transmit loading acting normal to their plane. Slabs are used as floors and roofs of buildings, as walls in tanks and buildings, and as bridges to transmit relatively heavy concentrated loading. Reinforced concrete slabs are among the most common structural elements, but despite the large number of slabs design and built, the details of the elastic and plastic behavior of slabs are not always appreciated or properly taken into account. This occurs at least partially because of the mathematical complexities of dealing with elastic equations, especially for support conditions which realistically approximate those in building floor slabs. Regardless of which design method is used, the resulting slab must be serviceable at the working load level, with deflections and cracking remaining within acceptable limits .slab design methods are concerned largely with flexure, but the shear forces may also be a limiting factor .The particular problem of shear is in beamless slabs, especially when acting in combination with transfer of unbalanced moments from slabs to columns. The bi-axially voided bubble deck technology is based upon the patented integration technique - the direct way of linking air and steel. The bubble deck technology comprises of a bi axial carrying hollow slab in which plastic balls serves the purpose of eliminating concrete that has no carrying effect. In other words, it removes the non working dead load, while maintaining bi axial strength. This project report contains the details of innovative design, detailing and estimation of bi axially voided bubble deck slab and a comparative study with conventional slab systems, namely, beam slab and conventional flat slab.


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1.2 OBJECTIVES The main objective of this project is to design and compare structural, economic, and environmental panorama of the following floor slab systems – •

Conventional beam slab system

•

Conventional flat slab system

•

Bi-axially voided bubble deck system

In order to accomplish the advantages of the new bi-axially voided bubble deck floor slab systems over other conventional floor slab systems.

1.3 Project Outlines There are eight chapters in this project report. Chapter one gives introduction and objectives of this project. Chapter 2 provides literature review for this project. This chapter gives the theory behind this project. Chapter 3 gives the information related to design and detailing of flat slab system. Chapter 4 gives the information about design and detailing of conventional beam slab system. Chapter 5 gives the information about design and detailing of bi-axially voided bubble deck floor slab system. Chapter 6 contains material and economic estimation of all above mentioned floor slab systems. Chapter 7 contains results and discussions. Chapter 8 gives conclusions on the experimental outcome and the scope for future work.


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CHAPTER 2 LITERATURE REVIEW

2.1 CLASSIFICATION OF SLABS


Page 4


Different types of load patterns and support conditions require different types of floor slab systems. To accomplish this, the different types of floor slab systems used can be broadly grouped into the following four classes – a ) Conventional beam slab system b) Flat slab system c) Hollow core floor slab system d) Bi-axially voided bubble deck floor slab system Slabs can be classified as follows:

2.2 BEAM SLAB Slabs supported on beams on all sides or selected sides of each poannel are generally termed as beam slabs. In a beam slab system, it is quite easy to visualize the path from load point to columns as being from slab to beam to column and them to compute realistic moments and shears for the design of all members. A conventional beam slab system can be classified as : •

One way slab

•

Two way slab

A typical beam slab is shown in figure 2.1


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2.2 FLAT SLABS


Page 6


The term flat slab means a reinforced concrete slab with or without drops, supported generally without beams, by columns with or without flared column heads. A flat slab may be a solid slab or may have recesses formed on the soffit so that the soffit comprises a series of ribs in two direction. The following two methods are recommended by the code for determining the bending moments in the slab panel:

1) Direct design method (DDM) 2) Equivalent frame method (EFM) These methods are applicable only for two way rectangular slabs.

2.2.1 DIRECT DESIGN METHOD


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Direct design method is a simplified procedure of determining the negative and positive design moment at critical section in the slab. The code specifies that the following conditions must be satisfied by the two way slab system for the application of direct design method:

1) There must be at least three continuous spans in each direction. 2) Each panel must be rectangular, with the long to short span ratio not exceeding 2.0

3) The columns must not be offset by more than ten percent of span from either axis between centre lines of successive columns. As shown in figure 2.3.

4) The successive span length in each direction must not differ by more than 1/3rd of longer span.

5) The factored live load must not exceed three times the factored dead load.


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2.2.2. EQUIVALENT FRAME METHOD The equivalent frame method (EFM) of design of two way beam supported slabs, flat slabs, flat plates and waffle slab is a more general and more rigorous method than DDM, and is not subjected to the limitations of DDM.

The equivalent frame concept simplifies the analysis of three dimensional reinforcement concrete building by sub dividing it into a series of two dimensional frames centered on column lines in longitudinal as well as transverse direction. The EFM differs from DDM in


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the determination of total `negative` and `positive` design moments in the slab panel for the condition of gravity loading. However, the apportioning of the moments to column strip and middle strip is common for both methods.

2.3 HOLLOW CORE SLAB Hollow core slabs are pre fabricated, one way spanning, concrete elements with hollow cylinders.

FIG 2.5 Due to the pre fabrication, these are inexpensive and reduce building time, but can be used only in one way spanning construction and must be supported by beams and/or walls.

2.3.1 LIMITATIONS OF HOLLOW CORE SLAB •

Manufactured

•

It requires higher capacity cranes

•

Prevalence of post construction inflexibility

•

It has one way action


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•

It is suitable only for certain applications

2.4 BI-AXIALLY VOIDED BUBBLE DECK SLAB SYSTEM

The bi-axially voided bubble deck technology is based upon the patented integration technique - the direct way of linking air and steel. The bubble deck technology comprises of a bi axial carrying hollow slab in which plastic balls serves the purpose of eliminating concrete that has no carrying effect. In other words, it removes the non working dead load, while maintaining bi axial strength. By adopting the geometry of the ball in the mesh, an optimized concrete construction is obtained, with simultaneous maximum utility of both moment and shear zones. The construction literally creates itself as a result of the geometry of two well known components:

a) Reinforcement b) Hollow plastic balls Department of Civil Engineering, M.S.R.I.T.

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When the top and bottom reinforcement are linked in the usual way, a geometrical and statically stable bubble deck bubble-unit evolves. The reinforcement catches, distributes and locks the balls in exact position, while the balls shape the air volume, control the level of reinforcement and at the same time stabilize the spatial lattice. When the steel lattice unit is concreted, a monolithic bi-axial hollow slab is obtained.

2.4.1 GENERAL THEORY A bubble deck behaves like a solid slab, with true bi axial behavior, uniform in arbitrary direction. Tension and compression zone is not influenced by the voids. Forces can be distributed freely, with no singularities, in the three dimensional structure, hence, making all concrete effective.

Fig: 2.6


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When cutting out holes, the difference between the two deck types becomes obvious. With one way span it is necessary to place beams around the hole to transport the forces to principle beams.


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Two way spans can be completely without beams.

FIG 2.7

2.4.2 TESTS AND STUDIES The bubble deck technology has been tested thoroughly. Results confirm that a bubble deck slab behaves like a solid slab in every way. 2.4.2.1 SHEAR STRENGTH Tests confirm that all concrete in the slab can be taken into account when calculating any type of forces. For safety reasons, it is recommended to use a factor of 0.6 compared to values of a solid slab of same height.


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2.4.2.2 BENDING STRENGTH AND DEFLECTION BEHAVIOUR A bubble deck slab has the same bending strength as a solid slab of same height. The bending stiffness is 0.9, compared to a solid slab. But since the weight of the slab is only 0.65 of a solid slab, the deflection will be considerably less.

2.4.2.3 ANCHORING Tests confirm that the balls have no influence on the anchoring values. The values are exactly the same s for a solid slab. 2.4.2.4 FIRE A bubble deck slab can be tailored to meet any requirements by optimizing the actual concrete cover. The bubbles only slightly influence the patterns of heat transfer through the cover after a certain time and distance from the bottom,. Again, a bubble deck behaves like a solid slab.

TABLE 2.1

2.4.2.5 SOUND


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Values for air borne, impact sound (vertical or horizontal) exists. Below are the representative values.

TABLE 2.2

2.4.3 Bubble deck slab versions The appropriate bubble deck slab version is engineered to suit building configuration, span length between supports, applied loadings and vertical alignment of supports.

TABLE 2.3

2.4.4 ELEMENT TYPES Bubble deck can be manufactured in three types of manufactured elements:


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TYPE A – FILIGREE ELEMNTS When the bottom of the bubble reinforcement sandwich includes a 70mm thick pre cast concrete layer acting as formwork within part of the finished slab depth replacing the need for soffit shuttering. The elements are placed on temporary propping, loose joint, the shear and edge reinforcement added, perimeter and tolerance shuttered and then the remaining slab depth concreted. Most commonly specified being suitable for the majority of new-build projects. Requires fixed or mobile crane to lift into position due to weight of manufactured elements as delivered to site. TYPE B-REINFORCEMENT MODULES Comprising pre-fabricated ‘bubble reinforcement’ sandwiched elements. The modules are placed on traditional site formwork, loose joint, shear and edge reinforcement added and then concrete in two stage to the full slab depth. Suitable for suspended ground floor slabs and alteration/refurbishment projects, particularly where site access is extremely restricted. Can be manually lifted into position.


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TYPE C-FINISHED PLANKS Delivered to the building site as complete pre-cast factory made slab elements with full concrete thickness . These span in one direction only and require the inclusion of supporting beams or walls within the structure.

2.4.5 POST-TENSION When mega spans are required (above 15 meter) we can provide a post tension (PT) bubble deck solution. The above deflection limits can be increased by up-to 30 percent with post-tension bubble deck slab.

2.4.6 GREEN CREDENTIALS By virtually eliminating concrete in the middle of a slab bubble deck makes a significant contribution to reducing environmental impact. Guidance from the ODPM requires the direct environmental effects of building to be considered, including usage of natural resources and emission resulting from construction. Not only is concrete usage reduced up-to 50 percent within a building structure but knock-on benefits can be realized through reduced foundation side. Bubble deck can make a a big contribution towards achieving BREEAN targets. Every 5000 m2 of bubble deck floor slab can save:

•

1000 m2 site concrete

•

166 ready mix lorry trips

•

1798 tonnes of foundation loads –or 19 less piles

•

1745 GJ energy used in concrete production and haulage

•

278 tonnes of CO2- green house gases-emission


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CHAPTER 3 DESIGN OF FLAT SLABS

ANALYTICAL PLAN


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FIG 3.1

3. DESIGN OF FLAT SLAB (USING EQUIVALENT FRAME METHOD)

3.1 Slab thickness For deflection control d ≥ ln / 26

[IS-456, Clause 23.2.1]

Since drop is not provided d ≥ ln/(26*0.9)

[IS-456, Clause 23.2.1]

d ≥ 6500/(26*0.9) d ≥ 277.77 mm Approx. d ≈ 275 mm


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Therefore D = d + 25

(clear cover)

D = 275 + 25 D = 300 mm

3.2 Load Calculation

Self weight of slab = 25 * 0.3 = 7.5 KN / m2 1 KN / m2

Floor finish

=

Live load

= 2.4 KN / m2

Total

= 10.9 KN / m2

Factored load = 1.5 * 10.9 = 16.35 KN / m2

3.3 Equivalent Frame Analysis

Along N-S direction

Middle strip


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Fig 3.2

Fixed end moment Mf-ab = w * l2 / 12 = - 98.1 * 6.52 / 12 = - 345.39 KN-m [Due to symmetry fixed end moments are same for all spans]

Distribution factor (DF Table)

Span A2B2 B2A2

K I / 6.5 I / 6.5

∑K 0.15 I

DF = K / ∑K 1 0.5

0.3 I B2C2 D2C2

I / 6.5 I / 6.5

0.5 0.5 0.3 I

D2E2 E2F2

I / 6.5 I / 6.5

0.5 0.5 0.3 I


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E2D2

I / 6.5

F2E2

I / 6.5

0.5 0.5 0.3 I

F2G2 G2F2

I / 6.5 I / 6.5

0.15 I

0.5 1

TABLE 3.1

Moment Distribution Table

Joint Span DF FEM Final Mome nt

A2 A2B2 1 -345 -345

B2A2 0.5 345 345

B2 B2C2 0.5 -345 -345

C2B2 0.5 345 345

C2 C2D2 0.5 -345 -345

D2 D2C2 D2E2 0.5 0.5 345 -345 345 -345

E2D2 0.5 345 345

E2 E2F2 0.5 -345 -345

F2E2 0.5 345 345

F2 F2G2 0.5 -345 -345

TABLE 3.2 Since A and G are fixed ends and also due to symmetry, all moments are balanced, hence fixed end moments are equal to final moments

For edge strip


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G2 G2F2 1 345 345


Fig 3.3 Fixed end moment Mf-ab =

w * l2 / 12 = - 49.05 * 6.52 / 12 = - 172.70 KN - m

[Due to symmetry fixed end moments are same for all spans]


Span AB BA

K I / 6.5 I / 6.5

BC DC

I / 6.5 I / 6.5

DE EF

I / 6.5 I / 6.5

ED FE

I / 6.5 I / 6.5

FG GF

I / 6.5 I / 6.5

∑K 0.15 I

DF = K / ∑K 1 0.5

0.3I 0.5 0.5 0.3 I 0.5 0.5 0.3 I 0.5 0.5 0.3 I 0.5 1

0.15 I

Table 3.3


Joint Span

A AB

B BA

C BC

CB

D CD


DC

E DE

ED

F FE Page 24

G GF


DF FEM Final Momen t

1 -172 -172

0.5 172 172

0.5 -172 -172

0.5 172 172

0.5 -172 -172

0.5 172 172

0.5 -172 -172

EF 0.5 172 172

0.5 -172 -172

FG 0.5 172 172

0.5 -172 -172

TABLE 3.4 Since A and G are fixed ends and also due to symmetry, all moments are balanced, hence fixed end moments are equal to final moments

Along E-W direction

EDGE STRIP

Fig 3.5

Fixed End Moment Mf-a1a2 = w * l2 / 12 = - 53.13 * 62 / 12 = -153.39 KN - m [Due to symmetry fixed end moments are same for all spans]



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1 172 172


Joint A1 Joint A2 Span DF A3 FEM Final Moment

Span A1A2

K I/6

A2A1 A1 AA13AA22 1A3A2 -159.39 -159.39

I/6 A2A1I / 6 0.5 I / 6 159.39 159.39

∑K 0.166I A0.33I 2 0.166I

DF = K / ∑K 1 0.5 A2A3 0.5 0.5 1 -159.39 -159.39

A3 A3A2 1 159.39 159.39

Table 3.5

Moment Distribution Table Since A1 and A3 are fixed ends and also due to symmetry, all moments are balanced, hence fixed end moments are equal to final moments

Table 3.6

MID STRIP

Fig 3.6 Fixed End Moment MF-B1B2 = W * l2 / 12

= -106.27 * 62 / 12 = -318.825 KN – m [Due to symmetry fixed end moments are same for all spans]


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Joint B1

Span B1B2

K I/6

B2B1

I/6

B3B2 B3B2

I/6 I/6

B2 B3

∑K 0.166I

DF = K / ∑K 1 0.5

0.33I 0.5 1

0.166I

Table 3.7

Joint Span DF FEM Final Moment

B1 B1B2 1 -318.25 -318.25

B2 B2B1 0.5 318.25 318.25

B3 B2B3 0.5 - 318.25 -318.25

B3B2 1 318.25 318.25

Moment Distribution Table Table 3.8 Due to symmetry of span and supports, maximum positive moment will occur at centre

Fig 3.7 M+ive = w * l2 / 8 = 98.1 * 6.52 / 8 = 518.09 KN – m


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3.4 Moment Calculation N-S direction Negative moment calculation (for mid strip along N-S dir) From left support , M-ive = Ml – (98.1 * 0.352 / 2) = -345.39 - 6.008 = -351.39 KN – m (Since all the spans are symmetrical, moment from right support will be equal to moment from left support) Total design moment, for span (face to face) , Mo = w * ln / 8 = 98.1 * 5.8 / 8 = 71.125 KN - m Calculation of Ast (N-S direction) Adopting M+ive for calculation of Ast, since its value is highest and reducing it by 10% in accordance with clause 31.4.3.4 of IS – 456. Mu = 0.90 * 518.09 = 466.28 KN – m Mu / bd2 = 466.28 * 10 ^ 6 / (6000 * 275 2) = 1.05 (Using Fck = 30, from table 4 of SP – 6) pt = 0.304 Considering 1m strip Ast = pt * b * d / 100 = 0.304 * 1000 * 275 / 100 = 836 mm2 Using 16mm bars Spacing = (π * 162 / 4) * 1000 / 836


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= 240 mm

E-W Direction Due to symmetry, maximum positive moment will occur at centre M+ive = w * l2 / 8 = 106.27 * 62 / 8 = 478.21 KN – m

Maximum negative moment, From left support , M-ive = Ml – w * l2 /2 = -318.82 – (106.27 * 0.352 / 2) = -325.33 KN – m Total design moment, Mo = w * ln / 8 = 106.275 * 5.3 / 8 = 70.41 KN – m

Calculation of Ast (E-W direction) Adopting M+ive for calculation of Ast, since its value is highest and reducing it by 10% in accordance with clause 31.4.3.4 of IS – 456. MU

= .90*478.21 = 430.39 KN-m

MU/b*d2 = 430.39*10^6/(6500*275) = 0.90 (Using Fck = 30, from table 4 of SP – 6) pt = 0.259


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Considering 1m strip Ast = pt * b * d / 100 = 0.259*1000*275/100 = 712.25 mm2 Using 16mm bars Spacing = (π * 162 / 4) * 1000 / 712.25 = 280mm

3.4.1

Shear check

Ԏv = VU /b*d =( 318.82 * 2 * 103 )/( 6000* 275) = 0.38 N/mm2 For 100Ast/bd = 0.304 Referring to table 19 of IS-456 ԎC = 0.40 Hence ԎC > Ԏv Therefore SAFE

Detailing of flat slab


Page 30


Fig 3.8


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CHAPTER 4 DESIGN OF CONVENTIONAL BEAM SLAB

4. Design of Conventional Beam Slab (Using limit state method)

4.1 Design of slab

4.1.2 Check For one way / two way type of slab

Ratio of longer span to shorter span = ly / lx = 6.5/6


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= 1.08 > 2 Therefore slab is designed as two way slab. Lx = 6m Ly = 6.5m Consider 1m strip Assume slab thickness =150mm Using 15mm clear cover with 10mm bar d = 150-15-10/2 = 130mm

4.1.3

load calculation

Dead load = 0.15*25 = 3.75 KN/m2 Live load = 2.4 KN/m2 Floor finish = 1 KN/m2 Total load = 7.15 KN/m2 Ultimate load

= 1.5 * 7.15 = 10.725 KN/m2

4.1.4 Calculation of moment co-efficient

αx αy

1.0 0.062 0.062

1.08 0.0716 0.0612

1.1 0.074 0.061

Table 4.1

4.1.5 Calculation of moments MX = 0.0716 * 10.725 * 62 = 27.64 KN-m My

= 0.0612 * 10.725 * 62 = 23.63 KN-m

M max = 27.64 KN-m

4.1.6

Check for depth


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= (0.36 * Xu-max ( 1-0.42(Xu-max /d))b*d2fck )/ d

MU

27.64 *106 = 0.36*0.48( 1-0.42*0.48)*1000*d2 * 20 d

= 81.72mm < 150mm

Hence safe Provide D =150mm; d =130mm From IS-456 ; Pg 76 M

= 29.66 KN-m

Spacing

= 110mm

Provide 10mm bar @ 110mm c/c along long direction and short direction

4.2 DESIGN OF BEAMS

AREA OF 1 + 2

= =

Volume

2*((6.5 + 0.5)/2*3) 21 m3

= 21 * 0.15 =

Weight

3.15 m3

= 25 * 3.15 = 78.75 KN

Uniformly distributed load = 78.65 / 6.5 = 12.12 KN / m 4.2.1 Depth Calculation For simply supported beam (clause 23.2.2 of IS - 456) L / d = 20 d Approx. d Therefore

= 6500 / 20 ≈ 325 mm D = d + 50

(clear cover)


Page 34


D = 325 + 50 D = 375 mm Assuming width, b = 230mm 4.2.2 Load Calculation Self weight of slab = 25 * 0.375 * 0.23 = 2.16 KN / m2 Dead load due to slab

= 12.12 KN / m2

Floor finish

=

Live load

= 2.4 KN / m2

1 KN / m2

= 17.68 KN / m2

Total

Factored load = 1.5 * 17.68 = 26.52 KN / m2 Moment Mu = w * l2 / 8 = 26.52 * 6.52 / 8 = 140.06 KN –m Shear force at support,

Vu = w * l / 2 = 26.52 * 6.5 / 2 = 86.19 KN

Limiting value of moment, Mulim = 0.36 * Xumax *(1- 0.42Xumax/d)bd2 fck /d Referring to clause 38.1 of IS – 456 For Fe 415, Xumax/d = 0.48 Mulim = 0.36 * 0.48 (1-0.42*0.48)*230*3252*20 = 67.033 KN – m Mu > Mulim, hence design as doubly reinforced section Xumax = 0.48 * 325 = 156 mm


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4.2.3 Calculation of area Compression steel Strain = 0.0035 (Xumax – d`)/Xumax = 0.0035 (156 – 50)/156 =2.37 * 10-3 From SP 16, figure 3 Stress, Fsc = 380 N/mm2 Mu - Mulim = Fsc * Asc (d-d`) 10^6(140.06 – 67.03) = 380 * Asc (325 - 50) Asc= 698.85 mm2 Using 20mm bars Number of bars = 698.85 / (π * 202 / 4) = 2.23 ≈ 3 bars Hence provide 3 bars of 20mm as compression steel Tension steel Xu / d = Xumax / d =( 0.87 fyAst1) /(0.36 fck bd) 0.48 = (0.87 * 415 * Ast1) / (0.36 * 20 * 230 * 325 ) Ast1=715.51mm2 Ast2 = Asc * fsc /(0.87 fy) = 698.85 * 380 / (0.87 * 415) = 735.53 mm2 Total area of tension steel , Asc = Asc1 + Ast2 = 715.51 + 735.53 = 1451.04 mm2 Using ᴓ 22mm bars No of bars = 1451.04 / (π * 222 / 4) = 3.81

≈ 4 bars


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Provide 4 bars of 22mm as tension steel 4.2.4 Design for shear \

Vu = 86.19 KN b = 230 mm d = 325 mm Actual steel = 4 * π *222 / 4 = 1520.53 mm2 Ԏv = Vu / bd = 86.19 * 10 ^3 /( 230 * 325) = 1.17 N / mm2

100 Ast / bd = 100 * 1520.53 / (230 * 320) = 2.03 Referring to table 19 of IS 456, for M20, Ԏc = 0.79 Ԏc < Ԏv Hence provide shear reinforcement Using Vus = 0.87 * fy * Asv * d / Sv Using 2 legged , 8 mm stirrups, Asv = 2 * π * 82 / 4 = 100.53 mm2 Vus

= Vu - Ԏc bd = 86.19 * 10^3 –(0.79 * 230 * 325) = 27.13 * 10 ^3 N 27.13 * 10 ^3 = 0.87 * 415 * 100.53 * 325 / SV Sv = 434.68 mm


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But, As per IS 456, maximum spacing = 0.75 * d = 0.75 * 325 = 243.75 mm OR

300 mm

Hence provide 2 L 8mm @ 300mm c/c stirrups

Detailing of conventional beam slab


Page 38


Fig 4.1


Page 39


Fig 4.2


Page 40


CHAPTER 5 DESIGN OF BUBBLE DECK SLAB

5. Design Of Bubble Deck Slab


Page 41


5.1Slab thickness For deflection control Modifying l / d ratio by 0.5 [BS8110, product introduction] d ≥

l n

d ≥

/ (26*0.9*1.5)

[IS-456, Clause 23.2.1]

6500 / (35.1)

d ≥ 185.19 mm Approx. d ≈ 190 mm Therefore D = d + 20

(clear cover) considering

D = 190 + 20 D = 210 mm Provide D = 230 mm Hence ,

(considering slab version BD 230)

d = 230 – 25 = 205 mm

(25mm cover provides 60 min of fire resistance)

5.2 Load Calculation Self weight of slab = 25 * 0.23 *2/ 3 = 3.83 KN / m2 1 KN / m2

Floor finish

=

Live load

= 2.4 KN / m2

Total

= 7.23 KN / m2

Factored load = 1.5 * 7.23 = 10.845 KN / m2

5.3 Equivalent Frame Analysis


Page 42


Along N-S direction For middle strip

Fig 5.1 Fixed end moment Mf-ab = w * l2 / 12 = - 65.07 * 6.52 / 12 = - 229.1 KN-m [Due to symmetry fixed end moments are same for all spans]


Span A2B2 B2A2

K I / 6.5 I / 6.5

∑K 0.15 I

DF = K / ∑K 1 0.5

0.3 I B2C2 D2C2

I / 6.5 I / 6.5

D2E2 E2F2

I / 6.5 I / 6.5

E2D2 F2E2

I / 6.5 I / 6.5

F2G2 G2F2

I / 6.5 I / 6.5

0.5 0.5 0.3 I 0.5 0.5 0.3 I 0.5 0.5 0.3 I 0.1 I

0.5 1

Table 5.1


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Joint Span DF FEM Final Mome nt

A2 A2B2 1 -229 -229

B2A2 0.5 229 229

B2 B2C2 0.5 -229 -229

C2B2 0.5 229 229

C2 C2D2 0.5 -229 -229

D2 D2C2 D2E2 0.5 0.5 229 -229 229 -229

E2D2 0.5 229 229

E2 E2F2 0.5 -229 -229

F2E2 0.5 229 229

F2 F2G2 0.5 -229 -229

Table 5.2 Since A and G are fixed ends and also due to symmetry, all moments are balanced, hence fixed end moments are equal to final moments For edge strip

Fig 5.2

Fixed end moment Mf-ab =

w * l2 / 12 = - 32.535 * 6.52 / 12 = - 114.55 KN - m

[Due to symmetry fixed end moments are same for all spans]



Page 44

G2 G2F2 1 229 229


Span AB BA

K I / 6.5 I / 6.5

BC DC

I / 6.5 I / 6.5

DE EF

I / 6.5 I / 6.5

ED FE

I / 6.5 I / 6.5

FG GF

I / 6.5 I / 6.5

∑K 0.15 I

DF = K / ∑K 1 0.5

0.3 I 0.5 0.5 0.3 I 0.5 0.5 0.3 I 0.5 0.5 0.3 I 0.15 I

0.5 1

Table 5.3


Joint Span

A AB

DF

1

FEM

-114.5 114.5114.5 -114.5 114.5114.5

Final Moment

B BA BC 0.5

0.5

C CB CD 0.5 0.5 114.5114.5 114.5114.5

D DC DE 0.5 0.5 114.5114.5 114.5114.5

E ED EF 0.5 0.5 114.5114.5 114.5114.5

F FE FG 0.5 0.5 114.5114.5 114.5114.5

Table 5.4 Since A and G are fixed ends and also due to symmetry, all moments are balanced, hence fixed end moments are equal to final moments

Along E-W direction

EDGE STRIP


Page 45

G GF 1 114.5 114.5


Fig 5.3 Fixed End Moment Mf-a1a2 = w * l2 / 12 = - 35.24 * 62 / 12 = -105.72 KN - m [Due to symmetry fixed end moments are same for all spans] Distribution factor (DF Table)

Joint A1

Span A1A2

k I/6

A2A1

I/6

A3A2 A3A2

I/6 I/6

A2 A3

∑K 0.166I

DF = K / ∑K 1 0.5

0.33I 0.5 1

0.166I

Table 5.5 Moment Distribution Table


A1 A1A2 1 -105.72 -105.72

A2 A2A1 0.5 105.72 105.72

A3 A2A3 0.5 -105.72 -105.72

A3A2 1 105.72 105.72

Table 5.6

Since A1 and A3 are fixed ends and also due to symmetry, all moments are balanced, hence fixed end moments are equal to final moments


Page 46


MID STRIP

Fig 5.4 Fixed End Moment MF-B1B2 = W * l2 / 12 = -70.49* 62 / 12 = -211.47 KN – m [Due to symmetry fixed end moments are same for all spans]


Joint B1

Span B1B2

K I/6

B2B1

I/6

B2 B3

∑K 0.166I

DF = K / ∑K 1 0.5

0.33I B3B2 B3B2

I/6 I/6

0.166I

0.5 1

Table 5.5



Page 47



B1 B1B2 1 -211.47 -211.47

B2 B2B1 0.5 211.47 211.47

B2B3 0.5 - 211.47 - 211.47

B3 B3B2 1 211.47 211.47

Table 5.6

Due to symmetry of span and supports, maximum positive moment will occur at centre

Fig 5.5 Va

= 65.04*6.5/2 =211.47 KN

M+ive = w * l2 / 8 = 98.1 * 6.52 / 8 = 518.09 KN – m 5.4 Moment Calculation N-S direction Negative moment calculation (for mid strip along N-S dir) From left support , M-ive = Ml – (65.07 * 0.352 / 2) = -229.1- 3.98 = -233.08 KN – m (Since all the spans are symmetrical, moment from right support will be equal to moment from left support)


Page 48


Total design moment, for span (face to face) , Mo = w * ln / 8 = 65.07 * 5.8 / 8 = 47.17 KN - m

Calculation of Ast (N-S direction)

Adopting M+ive for calculation of Ast, since its value is highest and reducing it by 10% in accordance with clause 31.4.3.4 of IS – 456. Mu = 0.90 * 343.08 = 308.772 KN – m Mu / bd2 = 308.772 * 10 ^ 6 / (6000 * 205 2) = 1.25 (Using Fck = 30, from table 4 of SP – 6) pt = 0.365 Considering 1m strip Ast = pt * b * d / 100 = 0.365 * 1000 * 205 / 100 = 748.25 mm2 Using 16mm bars Spacing = (π * 162 / 4) * 1000 / 748.25 = 270 mm E-W Direction Due to symmetry, maximum positive moment will occur at centre M+ive = w * l2 / 8 = 70.49 * 62 / 8 = 317.20 KN – m Maximum negative moment, From left support , M-ive = Ml – w * l2 /2


Page 49


= -211.47 – (70.49* 0.352 / 2) = -215.78 KN – m Total design moment, Mo = w * ln / 8 = 70.49* 5.3 / 8 = 46.69 KN – m Calculation of Ast (E-W direction) Adopting M+ive for calculation of Ast, since its value is highest and reducing it by 10% in accordance with clause 31.4.3.4 of IS – 456. MU

= .90*317.2 = 285.48 KN-m

MU/b*d2 = 285.48*10^6/(6500*205) = 1.05 (Using Fck = 30, from table 4 of SP – 6) pt = 0.304 Considering 1m strip Ast = pt * b * d / 100 = 0.304*1000*205/100 = 623.2 mm2 Using 16mm bars Spacing = (π * 162 / 4) * 1000 / 623.2 = 325 mm


Page 50


Detailing of bubble deck

Fig 5.6


Page 51


CHAPTER 6 COSTING AND ESTIMATION

6.1 FLAT SLAB

6.1.1 Reinforcement

Along N-S Direction


Page 52


Length of main reinforcement L= l + 2 * 0.5*d- 2c = 39700+ 2*0.5*1.5 – 2*25 = 39666 mm Length of crank bar L = l-2c +2 *0.5d +2 *9d1

(d1 = D -2c-d = 300-50-16 =234 mm)

= 39700- 2*25 +2*0.5*16+2*9*234 = 43878mm Number of main reinforcement = ((span/spacing)+1)/2 = ((12.7/0.24/2)+1)/2 = 27 bars Number of cranked bars

= 54-27 = 27 bars

Along E-W direction Length of main reinforcement, L

= l + 2*0.5d-2c =12700 + 2*0.5*16 -2*25 =12666 mm

Length of cranked bar, L

= l – 2c + 2*0.5*d +2*9*d1 = 12700 -50 +2*0.5*16 +2*9*234 = 16878mm

Number of main reinforcement = ((span/spacing)+1)/2 = ((39.7/0.24)+1)/2


Page 53


= 84 bars

Number of cranked bars

= 167-84 = 83 bars

Table 6.1

Reinforcement

Number

Length (m)

Weight/meter 0.616d2 =0.616*1.62

Total weight (kilogram, kg)

N-S direction • Main

27

39.7

1.57

1683

27

44

1.57

1865

84

12.7

1.57

1675

83

16.9

• Cranked E-W direction • Main •

Cranked

1.57 TOTAL

2202 = 7425

6.1.2 CONCRETE Volume

=12700 *39700 *300 = 151.3 m3

Table 6.2

Particular Steel Concrete

Quantity 7.425 M Ton 151.3 m3

Rate (Rs) 42000 3500 TOTAL

Amount (Rs) 3,11,950 5,29,550

= 8,41,500

6.1 BUBBLE DECK SLAB

6.2.1 Bottom Reinforcement Along N-S Direction


Page 54


Length of main reinforcement L= l + 2 * 0.5*d- 2c = 39700+ 2*0.5*1.5 – 2*25 = 39666 mm Length of crank bar L = l-2c +2 *0.5d +2 *9d1

(d1 = D -2c-d = 230-50-16 =164 mm)

= 39700- 2*25 +2*0.5*16+2*9*164 = 42618mm Along E-W direction Length of cranked bar, L

= l – 2c + 2*0.5*d +2*9*d1 = 12700 -50 +2*0.5*16 +2*9*164 = 15618mm

6.2.2 Top Reinforcement Along N-S direction Length of main reinforcement, L

= l + 2 * 0.5*d- 2c = 39700+ 2*0.5*6 – 2*25 = 39656 mm

Length of crank bar L = l-2c +2 *0.5d +2 *9d1

(d1 = D -2c-d = 230-50-6=174 mm)

= 39700- 2*25 +2*0.5*6+2*9*174 = 42788mm


Page 55


Along E-W direction

Length of main reinforcement, L

= l + 2 * 0.5*d- 2c = 12700+ 2*0.5*6 – 2*25 = 12656 mm Length of crank bar

L = l-2c +2 *0.5d +2 *9d1

(d1 = D -2c-d = 230-50-6=174 mm)

= 12700- 2*25 +2*0.5*6+2*9*174 = 15788mm Number of bars Along N-S direction (bottom reinforcement) Number of main reinforcement

= ((span/spacing) + 1)/2 = ((12.7/0.27)+1)/2 = 24 bars

Number of Cranked

bar

= 48-24 =24 bars

Along E-W direction (bottom reinforcement) Number of main reinforcement

= ((span/spacing)+1)/2 = ((39.7/0.27)+1)/2 = 74 bars

Number of cranked bar

= 148-74 =74 bars

Along N-S direction (top reinforcement) Number of main reinforcement

=((span/spacing)+1)/2 = ((39.7/0.2)+1)/2 = 199 bars


Page 56


Along E-W direction (top reinforcement) Number of main reinforcement


Details of bottom reinforcement ( using fe415 steel)

Table 6.3

Reinforcement

Number

Length (m)

Weight/meter 0.616d2 =0.616*1.62



48

39.7

1.57

2992

148

12.7

1.57

2950

E-W direction • Main

TOTAL

= 5942

Details of top reinforcement (using fe250steel) Table 6.4

Reinforcement

Number

Length (m)

Weight/meter 0.616d2 =0.616*0.62



64

39.6

0.22

558

199

12.6

0.22

552

E-W direction • Main


Page 57


TOTAL

= 1110

6.2.3 Concrete Total volume = 12700 *39700*230 = 115.96 m3 Number of balls Along N-S direction = (span / spacing)-1 = (39.7/0.2)-1 = 197.5 ≈ 198 balls Along E-W direction = (span/spacing)-1 = (12.7/0.2)-1 = 62.5 ≈ 63 balls Total

= 197*62 = 12,214 balls

Reduction at column Solid slab is to be provided for areas of high shear that is 1/6th of the distance from centre to centre of column.

Therefore, area of 1 column = 2000 *2166. Number of balls

= (2000*2166.6)/(2000 *2000) = 20.83 ≈ 21 balls

Number of equivalent columns units = 1*5+0.5*12 +0.25*4 = 5+6+1


Page 58


= 12 Number of balls to be reduced

=12* 21 = 252 balls

Therefore, total number of balls actually provided = 12,214 –252 = 11,962 = (4 *π *r 3)/3

Volume of one ball

= (4* π *0.0903)/3 = 3.05 * 10-3 m3 Hence, volume of concrete

= 115.96 -36.53 = 79.43 m3

Abstract

Table 6.5

Particular Steel (fe415) Steel (fe250) Concrete Balls

Quantity 5.942 M Ton 1.11 M Ton 79.43 m3 11,962

Rate (Rs) 42000 42000 3500 Lump-sump TOTAL

Amount (Rs) 2,49,564 46,620 2,78,005 30,000

= 6,04,189

6.3 BEAM SLAB 6.3.1 SLAB Concrete Volume of concrete = 6.25* 5.75 *0.15 = 5.39 m3 Total

= 12 * 5.39 = 64.68 m3

Reinforcement Along N-S direction


Page 59


Number of main bars


Cranked

= 54-27 = 27 bars

Length of main reinforcement L

= l + 2 * 0.5*d- 2c = 6250+ 2*0.5*1 – 2*25 = 6210 mm

Length of crank bar L

= l-2c +2 *0.5d +2 *9d1

(d1 = D -2c-d = 150-50-10= 90 mm) = 6250- 2*25 +2*0.5*10+2*9*90 = 7830mm

Along E-W direction Number of main bars


Cranked

= 59-29 = 30 bars

Length of main reinforcement L

= l + 2 * 0.5*d- 2c = 5750+ 2*0.5*10 – 2*25 = 5710 mm


Page 60


Length of crank bar L

= l-2c +2 *0.5d +2 *9d1 (d1 = D -2c-d = 150-50-10= 90 mm) = 5750- 2*25 +2*0.5*10+2*9*90 = 7330mm

6.3.2 Beam

Along N-S direction Length of main reinforcement L

= l + 2 * 0.5*d- 2c = 6500+ 2*0.5*10 – 2*25 = 6460 mm

Length of the stirrups L

= 2 ( l1 + l2 ) + 2 * 9*d = 2(180+275) +2*9*8 = 1054 mm

Number of stirrups

= (6.5/0.3)+1 = 23

Along E-W direction Length of main reinforcement L

= l + 2 * 0.5*d- 2c = 6000+ 2*0.5*10 – 2*25 = 5960 mm

Number of stirrups

= (6/0.3) + 1 = 21


Page 61


Volume of concrete for beam = 18(6.5*0.23*0.375) = 14(6*0.23*0.375) = 17.43 m3

Details of reinforcement Table 6.6

Reinforcement

Number

Length (m)

Weight/meter


Slab N-S direction Main

12*27=324

6.2

0.616

1236

Cranked

12*27=324

7.8

0.616

1560

E-W direction Main

12*29=348

5.7

0.616

1222

Cranked

12*30=360

7.3

0.616

1619

18*3=54

6.5

2.46

865

Beam N-S direction Top


Page 62


Bottom

18*4=72

6.5

2.98

1395

Top

14*3=42

6

2.46

621

Bottom

14*4=56

6

2.98

1001

Stirrup

708

1.05

0.39

290

E-W direction

TOTAL

= 9809

Total quantity of concrete =17.34 +64.68 = 82.02 m3 Abstract Table 6.7

Particular Steel Concrete

Quantity 9.8 M Ton 82.02 m3

Rate (Rs) 42000 3500 TOTAL

Amount (Rs) 4,11,600 2,87,070 = 6,98,670

6.4 ABSTRACT

TYPE

VOLUME

RATE

AMOUNT( RS)

FLAT SLAB •

Steel

7.425 M-Ton

42000

3,11,850

•

Concrete

151.3 m3

3500

5,29,550

9.8

42000

4,11,600

BEAM SLAB •

Steel

M-Ton


Page 63


•

Concrete

82.02 m3

3500

2,87,070

BUBBLE DECKSLAB •

Steel (fe415 )

5.94 M-Ton

42000

2,49,480

•

Concrete

79.43 m3

3500

2,78,005

•

Recycled plastic

11,962

Lump-sump

30,000

•

Steel (fe250)

1.11 M-Ton

42000

46,620

Table 6.8

CHAPTER 7


Page 64


RESULTS AND DISCUSSION

7.1 IINTRODUCTION Design and analysis of three types of slabs was done using their respective design considerations. Costing and estimation was carried out to compute and compare the structural, economic and environmental results. The outcome of the comparison is presented in this chapter.

7.2 THICKNESS OF SLAB Based on the design outcome (given in chapter 3,4,5) comparison of thickness of slab for the different type of floor slab systems is plotted in the figure 7.1.


Page 65


Fig 7.1

Graph shows that bubble deck slab has considerably less thickness as compared to conventional flat slab. Tough the conventional beam slab has least thickness; the addition of beam nullifies the advantage.

7.3 QUANTITY OF CONCRETE Based on the design outcome (given in chapter 6) comparison of quantity of concrete used in slab for the different type of floor slab systems is plotted in the figure 7.2.

Fig 7. 2 Graph shows, the conventional flat slab system uses highest amount of concrete and conventional beam slab system and bubble deck slab uses equal amount of concrete. But addition of beams in conventional beam slab system nullifies this advantage.

7.4 QUANTITY OF STEEL Based on the design outcome (given in chapter 6) comparison of quantity of steel used in slab for the different type of floor slab systems is plotted in the figure 7.3

Fi g7.3


Page 66


From the graph we conclude that the bubble deck slab used least amount of steel and usage of steel in conventional beam slab is maximum.

7.5 TOTAL QUANTITY AND ECONOMICS OF MATERIALS Figure 7.4 shows the diagrammatic comparison of quantity of steel as well as quantity of concrete used in different type of slab systems

Fig 7.4

Fig 7.5

Figure 7.5 shows the comparison of cost of concrete and cost of steel in slabs for different types of floor slab systems. It can be seen that the bubble deck slab has least cost of both steel and concrete as compared to conventional flat slab and conventional beam slab

7.6 ENVIRONMENTAL COMPARISON

Table 7.1 shows the CO2 emissions for different types of slabs at given slab thickness The table gives relevant data with reference to designed slabs as the thickness of slabs in table are identical to the slabs designed


Page 67


Table 7.1

From table 7.1 we conclude that CO2 emission for bubble deck slab is least and that for conventional flat slab is most. Figure 7.4 shows the diagrammatic comparison of quantity of steel as well as quantity of concrete used in different type of slab systems


Page 68


CHAPTER 8 CONCLUSIONS AND SCOPE FOR FUTURE WORK


Page 69


8.1 CONCLUSIONS A floor slab was designed using three different floor slab systems, namely conventional beam slab system, conventional flat slab system, new bubble deck floor slab system. Design and estimation was carried out for all the three types of slab systems. On the basis of this work the following conclusions are drawn. •

46.3 % of Concrete was saved by using bubble deck slab instead of conventional flat slab system

•

38.7 % of steel was saved in bubble deck slab system as compared to conventional beam slab system.

•

Almost 20 M.tones of CO2 emission was reduced by use of bubble deck technology

•

Intangibles – other intangible benefits derived from the use of bubble deck technology are – 1) Increase in number of floors due to less slab thickness

2) Reduction in foundation depth and size, which alsoi reduces the earthwork excavation. 3) Reduction in number of columns used and larger spans are possible


Page 70


8.2 SCOPE FOR FUTURE WORK The present study on bi-axially voided bubble deck slab system has the following scope for further improvement

•

Design can be improved so as to provide bubbles at the areas of high punching shear

•

The technology can be extended to design of rigid pavements and design of foundation slabs.


Page 71


APPENDIX A


Page 72


Fig A.1 Ball diameter

Fig A.2 Bending strength design


Page 73


Fig A.3 Bending stiffness

Fig A.4 Shear capacity

Fig A.5 Shear capacity


Page 74


Fig A.6 Nominal cover to meet specified period of fire resistance

Fig A.7 Minimum permissible values of αc


Page 75


Fig A.8 Design shear strength of concrete

Fig A.9 Maximum shear stress


Page 76


Fig A.10 Bending moment coefficient for slab spanning in two directions at right angles, simply supported on four sides


Page 77


BIBLOGRAPHY

REFERENCES Department of Civil Engineering, M.S.R.I.T.

Page 78


[1] S UNNIKRISHNA PILLAI, DEVDAS MENON (1999) Reinforced concrete design, Tata Mcgraw Hill. [2] P.C. Varghese (2009) Design of Reinforced Concret Foundations, P H India. [3] S.S. Bhavikatti (2009) Advanced RCC Design (RCC Volume-ii), New Age International Publishers. [4] Indian Standard Plain and Reinforced Concrete – Code of Practice (FOURTH REVISION) IS 456:2000. [5] ACI code 318-02,2002. [6] Nederlands BV-QR code. [7] Bubbledeck Voided Flat slab solutions Technical Manual and Documents (June, 2008) [8] British Standard code 8110.


Page 79

Biaxially Voided Bubble Deck Slab System and Other Conventional Floor Slab Systems

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