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.


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.



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.



CHAPTER 2

LITERATURE REVIEW



2.1 CLASSIFICATION OF SLABS

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


SLAB

BEAM SLAB BEAM LESS SLAB

CONVENTIONAL FLAT SLAB

HOLLOW CORE SLAB

BIAXIALLY VOIDED BUBBLE

DECK SLAB


A typical beam slab is shown in figure 2.1

2.2 FLAT SLABS



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



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.



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

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

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

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



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.



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.



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

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:



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.

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



CHAPTER 3

DESIGN OF FLAT SLABS



ANALYTICAL PLAN

FIG 3.1



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

3.1 Slab thickness

For deflection control d ≥ l

n / 26 [IS-456, Clause 23.2.1] Since drop is not provided d ≥ l

n/(26*0.9) [IS-456, Clause 23.2.1] d ≥ 6500/(26*0.9)

d ≥ 277.77 mm Approx. d ≈ 275 mm 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

Floor finish = 1 KN / m2

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

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 K ∑K DF = K / ∑KA2B2 I / 6.5 0.15 I 1B2A2

B2C2

I / 6.5

I / 6.5

0.3 I 0.5

0.5D2C2

D2E2

I / 6.5

I / 6.5

0.3 I 0.5

0.5E2F2

E2D2

I / 6.5

I / 6.50.3 I

0.5

0.5F2E2

F2G2

I / 6.5

I / 6.5

0.3 I 0.5

0.5G2F2 I / 6.5 0.15 I 1 TABLE 3.1

Moment Distribution Table

Joint A2 B2 C2 D2 E2 F2 G2

Span A2B2 B2A2 B2C2 C2B2 C2D2 D2C2 D2E2 E2D2 E2F2 F2E2 F2G2 G2F2

DF 1 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 1FEM -345 345 -345 345 -345 345 -345 345 -345 345 -345 345FinalMoment

-345 345 -345 345 -345 345 -345 345 -345 345 -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

Fig 3.3 Fixed end moment

Mf-ab = w * l2 / 12

= - 49.05 * 6.52 / 12

= - 172.70 KN - m



Span K ∑K DF = K / ∑KAB I / 6.5 0.15 I 1BA

BC

I / 6.5

I / 6.5

0.3I 0.5

0.5DC

DE

I / 6.5

I / 6.5

0.3 I 0.5

0.5EF

ED

I / 6.5

I / 6.50.3 I

0.5

0.5FE

FG

I / 6.5

I / 6.5

0.3 I 0.5

0.5GF I / 6.5 0.15 I 1 Table 3.3




Joint A B C D E F GSpan AB BA BC CB CD DC DE ED EF FE FG GFDF 1 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 1FEM -172 172 -172 172 -172 172 -172 172 -172 172 -172 172FinalMoment

-172 172 -172 172 -172 172 -172 172 -172 172 -172 172

TABLE 3.4


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





Joint Span K ∑K DF = K / ∑KA1 A1A2 I / 6 0.166I 1

A2

A2A1

A3A2

I / 6

I / 60.33I

0.5

0.5A3 A3A2 I / 6 0.166I 1

Table 3.5

Joint A1 A2 A3

Span A1A2 A2A1 A2A3 A3A2

DF 1 0.5 0.5 1FEM -159.39 159.39 -159.39 159.39Final Moment -159.39 159.39 -159.39 159.39


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



Joint Span K ∑K DF = K / ∑KB1 B1B2 I / 6 0.166I 1

B2

B2B1

B3B2

I / 6

I / 60.33I

0.5

0.5B3 B3B2 I / 6 0.166I 1

Table 3.7

Moment Distribution TableJoint B1 B2 B3

Span B1B2 B2B1 B2B3 B3B2

DF 1 0.5 0.5 1FEM -318.25 318.25 - 318.25 318.25Final Moment -318.25 318.25 -318.25 318.25

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

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 ᴓ bars

Spacing = (π * 162 / 4) * 1000 / 836

= 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


Ast = pt * b * d / 100

= 0.259*1000*275/100

= 712.25 mm2

Using 16mm ᴓ 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



Fig 3.8



CHAPTER 4

DESIGN OF CONVENTIONALBEAM 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 = 1.08 > 2

Therefore slab is designed as two way slab. Lx = 6m Ly = 6.5m Consider 1m stripAssume slab thickness =150mmUsing 15mm clear cover with 10mm ᴓ 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

1.0 1.08 1.1α x 0.062 0.0716 0.074α y 0.062 0.0612 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

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

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 ᴓ bar @ 110mm c/c along long direction and short direction

4.2 DESIGN OF BEAMS

AREA OF 1 + 2 = 2*((6.5 + 0.5)/2*3)

= 21 m3

Volume = 21 * 0.15

= 3.15 m3

Weight = 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 = 6500 / 20

Approx. d ≈ 325 mm

Therefore D = d + 50 (clear cover)

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



Total = 17.68 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

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 ᴓ bars Number of bars = 698.85 / (π * 202 / 4) = 2.23 ≈ 3 bars

Hence provide 3 bars of ᴓ 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

Provide 4 bars of ᴓ 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 , ᴓ 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

But, As per IS 456, maximum spacing = 0.75 * d

= 0.75 * 325

= 243.75 mm

OR 300 mm

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



Detailing of conventional beam slab

Fig 4.1



Fig 4.2



CHAPTER 5

DESIGN OF BUBBLE DECK SLAB



5. Design Of Bubble Deck Slab

5.1Slab thickness

For deflection control

Modifying l / d ratio by 0.5 [BS8110, product introduction]

d ≥ ln / (26*0.9*1.5) [IS-456, Clause 23.2.1]

d ≥ 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 (considering slab version BD 230)

Hence , 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



Total = 7.23 KN / m2




5.3 Equivalent Frame Analysis

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


Distribution factor (DF Table)Span K ∑K DF = K / ∑KA2B2 I / 6.5 0.15 I 1B2A2

B2C2

I / 6.5

I / 6.5

0.3 I 0.5

0.5D2C2

D2E2

I / 6.5

I / 6.5

0.3 I 0.5

0.5E2F2

E2D2

I / 6.5

I / 6.50.3 I

0.5

0.5F2E2

F2G2

I / 6.5

I / 6.5

0.3 I 0.5

0.5G2F2 I / 6.5 0.1 I 1

Table 5.1




Joint A2 B2 C2 D2 E2 F2 G2

Span A2B2 B2A2 B2C2 C2B2 C2D2 D2C2 D2E2 E2D2 E2F2 F2E2 F2G2 G2F2

DF 1 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 1FEM -229 229 -229 229 -229 229 -229 229 -229 229 -229 229FinalMoment

-229 229 -229 229 -229 229 -229 229 -229 229 -229 229

Table 5.2


For edge strip

Fig 5.2

Fixed end moment

Mf-ab = w * l2 / 12

= - 32.535 * 6.52 / 12

= - 114.55 KN - m





Span K ∑K DF = K / ∑KAB I / 6.5 0.15 I 1BA

BC

I / 6.5

I / 6.5

0.3 I 0.5

0.5DC

DE

I / 6.5

I / 6.5

0.3 I 0.5

0.5EF

ED

I / 6.5

I / 6.50.3 I

0.5

0.5FE

FG

I / 6.5

I / 6.5

0.3 I 0.5

0.5GF I / 6.5 0.15 I 1

Table 5.3


Joint A B C D E F GSpan AB BA BC CB CD DC DE ED EF FE FG GFDF 1 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 1FEM -114.5 114.5-114.5 114.5-114.5 114.5-114.5 114.5-114.5 114.5-114.5 114.5FinalMoment

-114.5 114.5-114.5 114.5-114.5 114.5-114.5 114.5-114.5 114.5-114.5 114.5

Table 5.4


Along E-W direction

EDGE STRIP



Fig 5.3Fixed End Moment

Mf-a1a2 = w * l2 / 12

= - 35.24 * 62 / 12

= -105.72 KN - m



Joint Span k ∑K DF = K / ∑KA1 A1A2 I / 6 0.166I 1

A2

A2A1

A3A2

I / 6

I / 60.33I

0.5

0.5A3 A3A2 I / 6 0.166I 1

Table 5.5 Moment Distribution Table

Joint A1 A2 A3

Span A1A2 A2A1 A2A3 A3A2

DF 1 0.5 0.5 1FEM -105.72 105.72 -105.72 105.72Final Moment -105.72 105.72 -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



MID STRIP

Fig 5.4

Fixed End Moment

MF-B1B2 = W * l2 / 12

= -70.49* 62 / 12

= -211.47 KN – m



Joint Span K ∑K DF = K / ∑KB1 B1B2 I / 6 0.166I 1

B2

B2B1

B3B2

I / 6

I / 60.33I

0.5

0.5B3 B3B2 I / 6 0.166I 1

Table 5.5




Joint B1 B2 B3

Span B1B2 B2B1 B2B3 B3B2

DF 1 0.5 0.5 1FEM -211.47 211.47 - 211.47 211.47Final Moment -211.47 211.47 - 211.47 211.47

Table 5.6

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

Fig 5.5Va = 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)

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 stripAst = pt * b * d / 100

= 0.365 * 1000 * 205 / 100

= 748.25 mm2

Using 16mm ᴓ 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

= -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


Ast = pt * b * d / 100

= 0.304*1000*205/100

= 623.2 mm2

Using 16mm ᴓ bars

Spacing = (π * 162 / 4) * 1000 / 623.2

= 325 mm



Detailing of bubble deck



Fig 5.6



CHAPTER 6

COSTING AND ESTIMATION

6.1 FLAT SLAB



6.1.1 Reinforcement

Along N-S Direction

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

= 84 bars

Number of cranked bars = 167-84

= 83 bars

Table 6.1

Reinforcement Number Length(m)

Weight/meter0.616d2

=0.616*1.62

Total weight(kilogram, kg)

N-S direction Main

Cranked

27

27

39.7

44

1.57

1.57

1683

1865

E-W direction Main

Cranked

84

83

12.7

16.9

1.57

1.57

1675

2202

TOTAL = 7425

6.1.2 CONCRETE

Volume =12700 *39700 *300

= 151.3 m3

Table 6.2Particular Quantity Rate (Rs) Amount (Rs)Steel 7.425 M Ton 42000 3,11,950Concrete 151.3 m3 3500 5,29,550



TOTAL = 8,41,500

6.2 BUBBLE DECK SLAB

6.2.1 Bottom Reinforcement

Along N-S Direction


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

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)




= ((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

Along E-W direction (top reinforcement)


= ((12.7/0.2)+1)/2

= 64 bars

Details of bottom reinforcement ( using fe415 steel)

Table 6.3Reinforcement Number Length

(m)Weight/meter

0.616d2

=0.616*1.62


N-S direction Main 48 39.7 1.57 2992

E-W direction Main 148 12.7 1.57 2950

TOTAL = 5942



Details of top reinforcement (using fe250steel)

Table 6.4Reinforcement Number Length

(m)Weight/meter

0.616d2

=0.616*0.62


N-S direction Main 64 39.6 0.22 558

E-W direction Main 199 12.6 0.22 552

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 = 12

Number of balls to be reduced =12* 21

= 252 balls

Therefore, total number of balls actually provided

= 12,214 –252

= 11,962

Volume of one ball = (4 *π *r 3)/3

= (4* π *0.0903)/3

= 3.05 * 10 -3 m3

Hence, volume of concrete = 115.96 -36.53

= 79.43 m3

Abstract

Table 6.5Particular Quantity Rate (Rs) Amount (Rs)Steel (fe415) 5.942 M Ton 42000 2,49,564Steel (fe250) 1.11 M Ton 42000 46,620Concrete 79.43 m3 3500 2,78,005Balls 11,962 Lump-sump 30,000

TOTAL = 6,04,189



6.3 BEAM SLAB

6.3.1 SLAB

ConcreteVolume of concrete = 6.25* 5.75 *0.15

= 5.39 m3

Total = 12 * 5.39 = 64.68 m3

Reinforcement

Along N-S direction

Number of main bars = ((span/spacing)+1)/2

= ((5.75/0.11)+1)/2

= 27 bars

Cranked = 54-27

= 27 bars


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 = ((span/spacing)+1)/2

= ((6.85/0.11)+1)/2

= 29 bars

Cranked = 59-29

= 30 bars


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

= 5750+ 2*0.5*10 – 2*25

= 5710 mm 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


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


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

= 6000+ 2*0.5*10 – 2*25

= 5960 mm

Number of stirrups = (6/0.3) + 1

= 21

Volume of concrete for beam = 18(6.5*0.23*0.375)

= 14(6*0.23*0.375)

= 17.43 m3



Details of reinforcementTable 6.6

Reinforcement Number Length(m)

Weight/meter Total weight(kilogram, kg)

SlabN-S direction

Main

Cranked

E-W directionMain

Cranked

12*27=324

12*27=324

12*29=348

12*30=360

6.2

7.8

5.7

7.3

0.616

0.616

0.616

0.616

1236

1560

1222

1619

BeamN-S direction

Top

Bottom

E-W direction

Top

Bottom

Stirrup

18*3=54

18*4=72

14*3=42

14*4=56

708

6.5

6.5

6

6

1.05

2.46

2.98

2.46

2.98

0.39

865

1395

621

1001

290

TOTAL = 9809

Total quantity of concrete =17.34 +64.68

= 82.02 m3

Abstract

Table 6.7Particular Quantity Rate (Rs) Amount (Rs)Steel 9.8 M Ton 42000 4,11,600Concrete 82.02 m3 3500 2,87,070

TOTAL = 6,98,670



6.4 ABSTRACT

TYPE VOLUME RATE AMOUNT( RS)

FLAT SLAB

Steel

Concrete

7.425 M-Ton

151.3 m3

42000

3500

3,11,850

5,29,550

BEAM SLAB

Steel

Concrete

9.8 M-Ton

82.02 m3

42000

3500

4,11,600

2,87,070

BUBBLE DECKSLAB

Steel (fe415 )

Concrete

Recycled plastic

Steel (fe250)

5.94 M-Ton

79.43 m3

11,962

1.11 M-Ton

42000

3500

Lump-sump

42000

2,49,480

2,78,005

30,000

46,620

Table 6.8



CHAPTER 7

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.

Beam sla

b

Flat s

lab

BUBBLEDECK

0

100

200

300

Thickness of slab in mm

Thickness of slab in mm

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.

Beam sla

b

Flat s

lab

BUBBLEDECK

020406080

100120140160

Quantity of concrete in cubic metre

Quantity of concrete in cubic metre

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

Beam sla

b

Flat s

lab

BUBBLEDECK

0

2

4

6

8

10

12

Quantity of steel in Mton(Fe415)Quantity of steel in Mton(Fe250)

Fi g7.3



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

concrete

steel

050

100150200250300350

Flat slabBeam slabBUBBLEDECK

Fig 7.4



Cost of concreteCost of steel

0

100000

200000

300000

400000

500000

600000

BUBBLEDECKBeam Slab

BUBBLEDECKFlat slabBeam Slab

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



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



CHAPTER 8

CONCLUSIONS AND SCOPE FOR FUTURE WORK



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



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.



APPENDIX A



Fig A.1 Ball diameter

Fig A.2 Bending strength design



Fig A.3 Bending stiffness

Fig A.4 Shear capacity

Fig A.5 Shear capacity



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

Fig A.7 Minimum permissible values of αc



Fig A.8 Design shear strength of concrete



Fig A.9 Maximum shear stress



Fig A.10 Bending moment coefficient for slab spanning in two directions at right angles,

simply supported on four sides



BIBLOGRAPHY



REFERENCES

[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.
