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

8/11/2019 Biaxially Voided Bubble Deck Slab System and Other Conventional Floor Slab Systems

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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


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C#$PTER 2

%ITER$TURE REVIEW

2.1 C%$SSI&IC$TION O& S%$BS



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SYSTEMS

ifferent 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 #onventional beam slab system

b lat slab system

c /ollow core floor slab system

d $i-axially voided bubble deck floor slab system

Slabs can be classified as follows0

2.2 BE$' S%$B

Slabs supported on beams on all sides or selected sides of each poannel are generally

termed as beam slabs. n a beam slab system, it is quite easy to visuali1e 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. 2 conventional beam slab system can

be classified as 0

3ne way slab

Two way slab

2 typical beam slab is shown in figure %.4



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2.2 &%$T S%$BS



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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. 2 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 bendingmoments in the slab panel0

1 irect design method 56

2 7quivalent frame method 576

These methods are applicable only for two way rectangular slabs.

2.2.1 DIRECT DESI(N 'ET#OD

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


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irect 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 method0

4 There must be at least three continuous spans in each direction.

% 7ach panel must be rectangular, with the long to short span ratio not exceeding%.8

3 The columns must not be offset by more than ten percent of span from either axis

between centre lines of successive columns. 2s shown in figure %.&.

4 The successive span length in each direction must not differ by more than 49&rdoflonger span.

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

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


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2.2.2. E)UIV$%ENT &R$'E 'ET#OD

The equivalent frame method 576 of design of two way beam supported slabs, flat

slabs, flat plates and waffle slab is a more general and more rigorous method than 6,

and is not sub!ected to the limitations of 6.

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 76 differs from 6 in

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


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SYSTEMS

the determination of total :negative: and :positive: design moments in the slab panel for

the condition of gravity loading. /owever, the apportioning of the moments to column strip

and middle strip is common for both methods.

2.3 #O%%OW CORE S%$B

/ollow core slabs are pre fabricated, one way spanning, concrete elements with hollow

cylinders.

&I( 2.*

ue 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 and9or

walls.

2.3.1 %I'IT$TIONS O& #O%%OW CORE S%$B

6anufactured

t requires higher capacity cranes

;revalence of post construction inflexibility

t has one way action

Department of Civil Engineering, M.S.R.I.T. Page 1$


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Two way spans can be completely without beams.

&I( 2.

2.+.2 TESTS $ND 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.+.2.1 S#E$RSTREN(T#

Tests confirm that all concrete in the slab can be taken into account when calculating any

type of forces. or safety reasons, it is recommended to use a factor of 8.) compared to

values of a solid slab of same height.



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2.+.2.2 BENDIN( STREN(T# $ND DE&%ECTION BE#$VIOUR

2 bubble deck slab has the same bending strength as a solid slab of same height. The

bending stiffness is 8.=, compared to a solid slab. $ut since the weight of the slab is only

8.)( of a solid slab, the deflection will be considerably less.

2.+.2.3 $NC#ORIN(

Tests confirm that the balls have no influence on the anchoring values. The values are

exactly the same s for a solid slab.

2.+.2.+ &IRE

2 bubble deck slab can be tailored to meet any requirements by optimi1ing 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,. 2gain, a bubble deck behaves like a solid

slab.

T$B%E 2.1

2.+.2.* SOUND



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>alues for air borne, impact sound 5vertical or hori1ontal exists. $elow are the

representative values.

T$B%E 2.2

2.+.3 $u44le 5ec6 "la4 7er"o!"

The appropriate bubble deck slab version is engineered to suit building configuration,

span length between supports, applied loadings and vertical alignment of supports.

T$B%E 2.3

2.+.+ E%E'ENT TPES

$ubble deck can be manufactured in three types of manufactured elements0



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T?;7 2 " @AR77 7@76BTS


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T?;7 #-BS/7 ;@2BFS

elivered 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.+.* POST,TENSION


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C#$PTER 3

DESI(N O& &%$T S%$BS

$N$%TIC$% P%$N

Department of Civil Engineering, M.S.R.I.T. Page 1#


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Therefore M d N %( 5clear cover

M %*( N %(

M &88 mm

3.2 %oa5 Calculato!

Self weight of slab M %( K 8.& M *.( FB 9 m%

loor finish M 4 FB 9 m%

@ive load M %.' FB 9 m%

Total M 48.= FB 9 m%

actored load M 4.( K 48.= M 4).&( FB 9 m%

3.3 E:u7ale!t &ra;e $!al


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&0 3.2

ixed end moment

6f-abM w K l% 9 4%

M - =+.4 K ).(% 9 4%

M - &'(.&= FB-m

Iue to symmetry fixed end moments are same for all spansJ

istribution factor 5 Table

Span F OF M F 9 OF2%$% 9 ).( 8.4( 4$%2%

$%#%

9 ).(

9 ).(8.&

8.(

8.(%#%

%7%

9 ).(

9 ).(8.&

8.(

8.(

7%% 9 ).( 8.& 8.(



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7%% 9 ).( 8.(

%7%

%A%

9 ).(

9 ).(8.&

8.(

8.(

A%% 9 ).( 8.4( 4 T$B%E 3.1

6oment istribution Table

Goint 2% $% #% % 7% % A%

Span 2%$% $%2% $%#% #%$% #%% %#% %7% 7%% 7%% %7% %A% A%% 4 8.( 8.( 8.( 8.( 8.( 8.( 8.( 8.( 8.( 8.( 476 -&'( &'( -&'( &'( -&'( &'( -&'( &'( -&'( &'( -&'( &'(inal6oment

-&'( &'( -&'( &'( -&'( &'( -&'( &'( -&'( &'( -&'( &'(

T$B%E 3.2

Since 2 and A are fixed ends and also due to symmetry, all moments are balanced,hence fixed end moments are equal to final moments

or edge strip



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&0 3.3

ixed end moment

6f-abM w K l%9 4%

M - '=.8( K ).(%9 4%

M - 4*%.*8 FB - m



Span F OF M F 9 OF2$ 9 ).( 8.4( 4$2

$#

9 ).(

9 ).( 8.&

8.(

8.(#

7

9 ).(

9 ).(8.&

8.(

8.(7

7

9 ).(

9 ).(8.&

8.(

8.(7

A

9 ).(

9 ).(8.&

8.(

8.(

A 9 ).( 8.4( 4 Ta4le 3.3


Goint 2 $ # 7 ASpan 2$ $2 $# #$ # # 7 7 7 A



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7 A 4 8.( 8.( 8.( 8.( 8.( 8.( 8.( 8.( 8.( 8.( 476 -4*% 4*% -4*% 4*% -4*% 4*% -4*% 4*% -4*% 4*% -4*% 4*%inal6omen

t

-4*% 4*% -4*% 4*% -4*% 4*% -4*% 4*% -4*% 4*% -4*% 4*%

T$B%E 3.+

Since 2 and A are fixed ends and also due to symmetry, all moments are balanced,hence fixed end moments are equal to final moments

2long 7-< direction

7A7 STR;

&0 3.*

ixed 7nd 6oment

6f-a4a%M w K l%9 4%

M - (&.4& K )%9 4%

M -4(&.&= FB - m





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Goint Span F =F M F 9 OF24 242% 9 ) 8.4)) 4

2%

2%24

2&2%

9 )

9 )8.&&

8.(

8.(2& 2&2% 9 ) 8.4)) 4

Ta4le 3.*


Since 24and 2&are fixed ends and also due to symmetry, all moments are balanced,hence fixed end moments are equal to final moments

Ta4le 3.

6 STR;

&0 3.

ixed 7nd 6oment

6-$4$%M < K l% 94%

M -48).%* K )%9 4%

M -&4+.+%( FB " m



Goint 24 2% 2&

Span 242% 2%24 2%2& 2&2% 4 8.( 8.( 476 -4(=.&= 4(=.&= -4(=.&= 4(=.&=inal 6oment -4(=.&= 4(=.&= -4(=.&= 4(=.&=


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Goint Span F =F M F 9 OF$4 $4$% 9 ) 8.4)) 4

$%

$%$4

$&$%

9 )

9 )8.&&

8.(

8.($& $&$% 9 ) 8.4)) 4

Ta4le 3.


Ta4le 3.>

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

&0 3.

6NiveM w K l% 9 +

M =+.4 K ).(%9 +

M (4+.8= FB " m

Department of Civil Engineering, M.S.R.I.T. Page 2!

Goint $4 $% $&Span $4$% $%$4 $%$& $&$% 4 8.( 8.( 476 -&4+.%( &4+.%( - &4+.%( &4+.%(inal 6oment -&4+.%( &4+.%( -&4+.%( &4+.%(


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3.+ 'o;e!t Calculato!

B-S direction

Begative moment calculation 5for mid strip along B-S dir

rom left support , 6-ive M 6l" 5=+.4 K 8.&(%9 %

M -&'(.&= - ).88+

M -&(4.&= FB " m

5Since all the spans are symmetrical, moment from right support will be equal to momentfrom left support

Total design moment, for span 5face to face , 6oM w K ln9 +

M =+.4 K (.+ 9 +

M *4.4%( FB - m

#alculation of 2st 5B-S direction

2dopting 6Nive for calculation of 2st, since its value is highest and reducing it by 48P inaccordance with clause &4.'.&.' of S " '().

6uM 8.=8 K (4+.8=

M ')).%+ FB " m

6u9 bd%M ')).%+ K 48 Q ) 9 5)888 K %*( %

M 4.8(

5Csing ckM &8, from table ' of S; " )

ptM 8.&8'

#onsidering 4m strip

2st M pt K b K d 9 488

M 8.&8' K 4888 K %*( 9 488

M +&) mm%

Csing 4)mm bars

Spacing M 5 K 4)% 9 ' K 4888 9 +&)

Department of Civil Engineering, M.S.R.I.T. Page 2"


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M %'8 mm

7-< irection

ue to symmetry, maximum positive moment will occur at centre

6Nive M w K l%9 +

M 48).%* K )%9 +

M '*+.%4 FB " m

6aximum negative moment,

rom left support , 6-iveM 6l" w K l%9%

M -&4+.+% " 548).%* K 8.&(%9 %

M -&%(.&& FB " m

Total design moment, 6o M w K ln 9 +

M 48).%*( K (.& 9 +

M *8.'4 FB " m

#alculation of 2st57-< direction

2dopting 6Nive for calculation of 2st, since its value is highest and reducing it by 48P in

accordance with clause &4.'.&.' of S " '().6C M .=8K'*+.%4

M '&8.&= FB-m

6C9bKd% M '&8.&=K48Q)95)(88K%*(

M 8.=85Csing ckM &8, from table ' of S; " )

ptM 8.%(=



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M 8.%(=K4888K%*(9488

M *4%.%( mm%

Csing 4)mm bars

Spacing M 5 K 4)% 9 ' K 4888 9 *4%.%(

M %+8mm

&.'.4 Shear check

v M >C 9bKd

M5 &4+.+% K % K 48& 95 )888K %*(

M 8.&+ B9mm%

or 4882st9bd M 8.&8'

Referring to table 4= of S-'()

#M 8.'8

/ence# v

Therefore S27

Detal!0 o? ?lat "la4



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&0 3.>



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C#$PTER +

DESI(N O& CONVENTION$%BE$' S%$B

+.De"0! o? Co!7e!to!al Bea; Sla4

8U"!0 l;t "tate ;etho59

+.1 De"0! o? "la4

+.1.2 Chec6 &or o!e @a< A t@o @a< t


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M 4.8+ %

Therefore slab is designed as two way slab.@x M )m@y M ).(m#onsider 4m strip2ssume slab thickness M4(8mmCsing 4(mm clear cover with 48mm bard M 4(8-4(-489% M 4&8mm

+.1.3 loa5 calculato!

ead load M 8.4(K%( M &.*( FB9m%

@ive load M %.' FB9m%

loor finish M 4 FB9m%

Total load M *.4( FB9m%

Cltimate load M 4.( K *.4( M 48.*%( FB9m%

+.1.+ Calculato! o? ;o;e!t co,e??ce!t

4.8 4.8+ 4.4 8.8)% 8.8*4) 8.8*' < 8.8)% 8.8)4% 8.8)4

Ta4le +.1

+.1.* Calculato! o? ;o;e!t"

6 M 8.8*4) K 48.*%( K )%

M %*.)' FB-m

6y M 8.8)4% K 48.*%( K )%

M %&.)& FB-m

6 max M %*.)' FB-m

+.1. Chec6 ?or 5epth



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6C M 58.&) K u-max 5 4-8.'%5u-max 9dbKd%fck 9 d

%*.)' K48) M 8.&)K8.'+5 4-8.'%K8.'+K4888Kd% K %8

d M +4.*%mm U 4(8mm

/ence safe

;rovide M4(8mmV d M4&8mm

rom S-'() V ;g *)

6 M %=.)) FB-m

Spacing M 448mm

;rovide 48mm bar W 448mm c9c along long direction and short direction

+.2 DESI(N O& BE$'S

2R72 3 4 N % M %K55).( N 8.(9%K&

M %4 m&

>olume M %4 K 8.4(

M &.4( m&


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SYSTEMS

M &%( N (8

M &*( mm

2ssuming width, b M %&8mm

+.2.2 %oa5 Calculato!

Self weight of slab M %( K 8.&*( K 8.%& M %.4) FB 9 m%

ead load due to slab M 4%.4% FB 9 m%



Total M 4*.)+ FB 9 m%

actored load M 4.( K 4*.)+ M %).(% FB 9 m%

6oment 6uM w K l%9 +

M %).(% K ).(%9 +

M 4'8.8) FB "mShear force at support, >uM w K l 9 %

M %).(% K ).( 9 %

M +).4= FB

@imiting value of moment,

6ulim M 8.&) K umaxK54- 8.'%umax9dbd% fck 9d

Referring to clause &+.4 of S " '()

or e '4(, umax9d M 8.'+

6ulim M 8.&) K 8.'+ 54-8.'%K8.'+K%&8K&%(%K%8

M )*.8&& FB " m

6u 6ulim, hence design as doubly reinforced section

umaxM 8.'+ K &%(

M 4() mm



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+.2.3 Calculato! o? area

#ompression steel

Strain M 8.88&( 5umax" d:9umax

M 8.88&( 54() " (894()

M%.&* K 48-&

rom S; 4), figure &

Stress, scM &+8 B9mm%

6u - 6ulimM scK 2sc5d-d:

48Q)54'8.8) " )*.8& M &+8 K 2sc5&%( - (8

2scM )=+.+( mm%

Csing %8mm barsBumber of bars M )=+.+( 9 5 K %8%9 '

M %.%& L & bars

/ence provide & bars of %8mm as compression steel

Tension steel

u9 d M umax9 d M5 8.+* fy2st4 958.&) fckbd

8.'+ M 58.+* K '4( K 2st4 9 58.&) K %8 K %&8 K &%(

2st4M*4(.(4mm%

2st%M 2sc K fsc 958.+* fy

M )=+.+( K &+8 9 58.+* K '4(

M *&(.(& mm%

Total area of tension steel , 2scM 2sc4N 2st%

M *4(.(4 N *&(.(&

M 4'(4.8' mm%

Csing %%mm bars

Bo of bars M 4'(4.8' 9 5 K %%% 9 '

M &.+4 L ' bars



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SYSTEMS

;rovide ' bars of %%mm as tension steel

+.2.+ De"0! ?or "hear

X>uM +).4= FB

b M %&8 mm

d M &%( mm

2ctual steel M ' K K%%% 9 '

M 4(%8.(& mm%

vM >u9 bd

M +).4= K 48 Q&95 %&8 K &%(

M 4.4* B 9 mm%

488 2st9 bd M 488 K 4(%8.(& 9 5%&8 K &%8

M %.8&

Referring to table 4= of S '(), for 6%8,

cM 8.*=

cUv

/ence provide shear reinforcement

Csing >usM 8.+* K fy K 2sv K d 9 Sv

Csing % legged , + mm stirrups,

2svM % K K +%9 '

M 488.(& mm%

>us M >u -c bd

M +).4= K 48Q&"58.*= K %&8 K &%(

M %*.4& K 48 Q&B

%*.4& K 48 Q&M 8.+* K '4( K 488.(& K &%( 9 S>

SvM '&'.)+ mm



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SYSTEMS

$ut, 2s per S '(), maximum spacing M 8.*( K d

M 8.*( K &%(

M %'&.*( mm

3R &88 mm

/ence provide % @ +mm W &88mm c9c stirrups

Detal!0 o? co!7e!to!al 4ea; "la4



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SYSTEMS

&0 +.1



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&0 +.2



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SYSTEMS

C#$PTER *

DESI(N O& BUBB%E DEC/ S%$B

*. De"0! O? Bu44le Dec6 Sla4



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SYSTEMS

*.1Sla4 thc6!e""

or deflection control

6odifying l 9 d ratio by 8.( I$S+448, product introductionJ

d H ln 9 5%)K8.=K4.( IS-'(), #lause %&.%.4J

d H )(88 9 5&(.4

d H 4+(.4= mm

2pprox. d L 4=8 mm

Therefore M d N %8 5clear cover considering

M 4=8 N %8

M %48 mm

;rovide M %&8 mm 5considering slab version $ %&8

/ence , d M %&8 " %( M %8( mm

5%(mm cover provides )8 min of fire resistance

*.2 %oa5 Calculato!

Self weight of slab M %( K 8.%& K%9 & M &.+& FB 9 m%



Total M *.%& FB 9 m%

actored load M 4.( K *.%& M 48.+'( FB 9 m%

*.3 E:u7ale!t &ra;e $!al


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SYSTEMS

2long B-S direction

or middle strip

&0 *.1

ixed end moment

6f-abM w K l% 9 4%

M - )(.8* K ).(% 9 4%

M - %%=.4 FB-m



Span F OF M F 9 OF2%$% 9 ).( 8.4( 4$%2%

$%#%

9 ).(

9 ).(8.&

8.(

8.(%#%

%7%

9 ).(

9 ).(8.&

8.(

8.(

7%%

7%%

9 ).(

9 ).(8.& 8.(

8.(%7%

%A%

9 ).(

9 ).(8.&

8.(

8.(A%% 9 ).( 8.4 4

Ta4le *.1



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SYSTEMS


Goint 2% $% #% % 7% % A%Span 2%$% $%2% $%#% #%$% #%% %#% %7% 7%% 7%% %7% %A% A%% 4 8.( 8.( 8.( 8.( 8.( 8.( 8.( 8.( 8.( 8.( 4

76 -%%= %%= -%%= %%= -%%= %%= -%%= %%= -%%= %%= -%%= %%=inal6oment

-%%= %%= -%%= %%= -%%= %%= -%%= %%= -%%= %%= -%%= %%=

Ta4le *.2

Since 2 and A are fixed ends and also due to symmetry, all moments arebalanced, hence fixed end moments are equal to final moments

or edge strip

&0 *.2

ixed end moment

6f-abM w K l%9 4%

M - &%.(&( K ).(%9 4%

M - 44'.(( FB - m





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SYSTEMS

Span F OF M F 9 OF2$ 9 ).( 8.4( 4$2

$#

9 ).(

9 ).(8.&

8.(

8.(

#

7

9 ).(

9 ).(8.&

8.(

8.(7

7

9 ).(

9 ).(8.&

8.(

8.(7

A

9 ).(

9 ).(8.&

8.(

8.(A 9 ).( 8.4( 4

Ta4le *.3


Goint 2 $ # 7 ASpan 2$ $2

$##$#

#7

77

7A

A

4 8.( 8.( 8.(8.(

8.(8.(

8.(8.(

8.(8.(

4

76 -44'.( 44'.(-44'.(

44'.(-44'.(

44'.(-44'.(

44'.(-44'.(

44'.(-44'.(

44'.(

inal6oment

-44'.( 44'.(-44'.(

44'.(-44'.(

44'.(-44'.(

44'.(-44'.(

44'.(-44'.(

44'.(

Ta4le *.+

Since 2 and A are fixed ends and also due to symmetry, all moments arebalanced, hence fixed end moments are equal to final moments

2long 7-< direction

7A7 STR;



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SYSTEMS

&0 *.3ixed 7nd 6oment

6f-a4a%M w K l%9 4%

M - &(.%' K )%9 4%

M -48(.*% FB - m



Goint Span k =F M F 9 OF

24 242% 9 ) 8.4)) 4

2%

2%24

2&2%

9 )

9 )8.&&

8.(

8.(2& 2&2% 9 ) 8.4)) 4

Ta4le *.*6oment istribution Table

Ta4le *.

Since 24and 2&are fixed ends and also due to symmetry, all moments are balanced,hence fixed end moments are equal to final moments


Goint 24 2% 2&Span 242% 2%24 2%2& 2&2% 4 8.( 8.( 4

76 -48(.*% 48(.*% -48(.*% 48(.*%inal 6oment -48(.*% 48(.*% -48(.*% 48(.*%


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SYSTEMS

6 STR;

&0 *.+

ixed 7nd 6oment

6-$4$%M < K l% 9 4%

M -*8.'=K )%9 4%

M -%44.'* FB " m



Goint Span F =F M F 9 OF$4 $4$% 9 ) 8.4)) 4

$%

$%$4

$&$%

9 )

9 )8.&&

8.(

8.($& $&$% 9 ) 8.4)) 4

Ta4le *.*




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SYSTEMS

Ta4le *.

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

&0 *.*

>a M )(.8'K).(9%

M%44.'* FB

6NiveM w K l% 9 +

M =+.4 K ).(%9 +

M (4+.8= FB " m

*.+ 'o;e!t Calculato!

B-S direction

Begative moment calculation 5for mid strip along B-S dir

rom left support , 6-ive M 6l" 5)(.8* K 8.&(%9 %

M -%%=.4- &.=+

M -%&&.8+ FB " m

5Since all the spans are symmetrical, moment from right support will be equal to moment

from left support


Goint $4 $% $&Span $4$% $%$4 $%$& $&$% 4 8.( 8.( 476 -%44.'* %44.'* - %44.'* %44.'*

inal 6oment -%44.'* %44.'* - %44.'* %44.'*


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SYSTEMS

Total design moment, for span 5face to face , 6oM w K ln9 +

M )(.8* K (.+ 9 +

M '*.4* FB - m

#alculation of 2st 5B-S direction

2dopting 6Nive for calculation of 2st, since its value is highest and reducing it by 48P inaccordance with clause &4.'.&.' of S " '().

6uM 8.=8 K &'&.8+

M &8+.**% FB " m

6u9 bd%M &8+.**% K 48 Q ) 9 5)888 K %8( %

M 4.%(

5Csing ckM &8, from table ' of S; " )

ptM 8.&)(

#onsidering 4m strip2st M pt K b K d 9 488

M 8.&)( K 4888 K %8( 9 488

M *'+.%( mm%

Csing 4)mm bars

Spacing M 5 K 4)% 9 ' K 4888 9 *'+.%(

M %*8 mm

7-< irection

ue to symmetry, maximum positive moment will occur at centre

6Nive M w K l%9 +

M *8.'= K )%9 +

M &4*.%8 FB " m

6aximum negative moment,

rom left support , 6-iveM 6l" w K l%9%



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SYSTEMS

M -%44.'* " 5*8.'=K 8.&(%9 %

M -%4(.*+ FB " m

Total design moment, 6o M w K ln 9 +

M *8.'=K (.& 9 +

M ').)= FB " m

#alculation of 2st57-< direction

2dopting 6Nive for calculation of 2st, since its value is highest and reducing it by 48P inaccordance with clause &4.'.&.' of S " '().6C M .=8K&4*.%

M %+(.'+ FB-m

6C9bKd% M %+(.'+K48Q)95)(88K%8(

M 4.8(5Csing ckM &8, from table ' of S; " )

ptM 8.&8'



M 8.&8'K4888K%8(9488

M )%&.% mm%

Csing 4)mm bars

Spacing M 5 K 4)% 9 ' K 4888 9 )%&.%

M &%( mm



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SYSTEMS

Detal!0 o? 4u44le 5ec6

&0 *.



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SYSTEMS

C#$PTER

COSTIN( $ND ESTI'$TION

.1 &%$T S%$B

).4.4 Reinforcement

2long B-S irection



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SYSTEMS

@ength of main reinforcement

@M l N % K 8.(Kd- %c

M &=*88N %K8.(K4.( " %K%(

M &=))) mm

@ength of crank bar@ M l-%c N% K8.(d N% K=d4 5d4 M -%c-d M &88-(8-4) M%&' mm

M &=*88- %K%( N%K8.(K4)N%K=K%&'

M '&+*+mm

Bumber of main reinforcement M 55span9spacingN49%

M 554%.*98.%'9%N49%

M %* bars

Bumber of cranked bars M ('-%*

M %* bars2long 7-< direction

@ength of main reinforcement,

@ M l N %K8.(d-%c

M4%*88 N %K8.(K4) -%K%(

M4%))) mm

@ength of cranked bar,

@ M l " %c N %K8.(Kd N%K=Kd4

M 4%*88 -(8 N%K8.(K4) N%K=K%&'

M 4)+*+mm


M 55&=.*98.%'N49%



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SYSTEMS

M +' bars

Bumber of cranked bars M 4)*-+'

M +& bars

Ta4le .1

Reinforcement Bumber @ength5m

olume M4%*88 K&=*88 K&88

M 4(4.& m&

Ta4le .2

;articular Yuantity Rate 5Rs 2mount 5RsSteel *.'%( 6 Ton '%888 &,44,=(8#oncrete 4(4.& m& &(88 (,%=,((8 T3T2@ M +,'4,(88

.1 BUBB%E DEC/ S%$B

).%.4 $ottom Reinforcement

2long B-S irection



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SYSTEMS


@M l N % K 8.(Kd- %c

M &=*88N %K8.(K4.( " %K%(

M &=))) mm

@ength of crank bar@ M l-%c N% K8.(d N% K=d4 5d4 M -%c-d M %&8-(8-4) M4)' mm

M &=*88- %K%( N%K8.(K4)N%K=K4)'

M '%)4+mm

2long 7-< direction

@ength of cranked bar,

@ M l " %c N %K8.(Kd N%K=Kd4

M 4%*88 -(8 N%K8.(K4) N%K=K4)'

M 4()4+mm

).%.% Top Reinforcement

2long B-S direction


@ M l N % K 8.(Kd- %c

M &=*88N %K8.(K) " %K%(

M &=)() mm

@ength of crank bar

@ M l-%c N% K8.(d N% K=d4 5d4 M -%c-d M %&8-(8-)M4*' mm

M &=*88- %K%( N%K8.(K)N%K=K4*'

M '%*++mm



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SYSTEMS

2long 7-< direction


@ M l N % K 8.(Kd- %c

M 4%*88N %K8.(K) " %K%(

M 4%)() mm

@ength of crank bar

@ M l-%c N% K8.(d N% K=d4 5d4 M -%c-d M %&8-(8-)M4*' mm

M 4%*88- %K%( N%K8.(K)N%K=K4*'

M 4(*++mm

Bumber of bars

2long B-S direction 5bottom reinforcement

Bumber of main reinforcement M 55span9spacing N 49%

M 554%.*98.%*N49%

M %' bars

Bumber of #ranked bar M '+-%'

M%' bars

2long 7-< direction 5bottom reinforcement


M 55&=.*98.%*N49%

M *' bars

Bumber of cranked bar M 4'+-*' M*' bars

2long B-S direction 5top reinforcement

Bumber of main reinforcement M55span9spacingN49%

M 55&=.*98.%N49%

M 4== bars



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SYSTEMS

2long 7-< direction 5top reinforcement


M 554%.*98.%N49%

M )' bars

etails of bottom reinforcement 5 using fe'4(steel

Ta4le .3



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SYSTEMS

T3T2@ M 4448

).%.& #oncrete

Total volume M 4%*88 K&=*88K%&8

M 44(.=) m&

Bumber of balls

2long B-S direction M 5span 9 spacing-4

M 5&=.*98.%-4

M 4=*.(

L 4=+ balls

2long 7-< direction M 5span9spacing-4

M 54%.*98.%-4

M )%.(

L )& balls

Total M 4=*K)%

M 4%,%4' balls

Reduction at columnSolid slab is to be provided for areas of high shear that is 49)th of the distance from centreto centre of column.

Therefore, area of 4 column M %888 K%4)).

Bumber of balls M 5%888K%4)).)95%888 K%888

M %8.+&

L %4 balls

Bumber of equivalent columns units M 4K(N8.(K4% N8.%(K'

M (N)N4



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SYSTEMS

M 4%

Bumber of balls to be reduced M4%K %4

M %(% balls

Therefore, total number of balls actually provided

M 4%,%4' "%(%

M 44,=)%

>olume of one ball M 5' K Kr &9&

M 5'K K8.8=8&9&

M &.8( K 48-&m&

/ence, volume of concrete M 44(.=) -&).(&

M *=.'& m&

2bstract

Ta4le .*

;articular Yuantity Rate 5Rs 2mount 5RsSteel 5fe'4( (.='% 6 Ton '%888 %,'=,()'

Steel 5fe%(8 4.44 6 Ton '%888 '),)%8#oncrete *=.'& m& &(88 %,*+,88($alls 44,=)% @ump-sump &8,888

T3T2@ M ),8',4+=

.3 BE$' S%$B

.3.1 S%$B

#oncrete>olume of concrete M ).%(K (.*( K8.4(

M (.&= m&

Total M 4% K (.&=

M )'.)+ m&

Reinforcement

2long B-S direction



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SYSTEMS

Bumber of main bars M 55span9spacingN49%

M 55(.*(98.44N49%

M %* bars

#ranked M ('-%*

M %* bars


@ M l N % K 8.(Kd- %c

M )%(8N %K8.(K4 " %K%(

M )%48 mm

@ength of crank bar @ M l-%c N% K8.(d N% K=d4

5d4 M -%c-d M 4(8-(8-48M =8 mm

M )%(8- %K%( N%K8.(K48N%K=K=8

M *+&8mm

2long 7-< direction

Bumber of main bars M 55span9spacingN49%

M 55).+(98.44N49%

M %= bars

#ranked M (=-%=

M &8 bars


@ M l N % K 8.(Kd- %c

M (*(8N %K8.(K48 " %K%(

M (*48 mm



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SYSTEMS

@ength of crank bar

@ M l-%c N% K8.(d N% K=d4

5d4 M -%c-d M 4(8-(8-48M =8 mm

M (*(8- %K%( N%K8.(K48N%K=K=8

M *&&8mm

.3.2 Bea;

2long B-S direction


@ M l N % K 8.(Kd- %c

M )(88N %K8.(K48 " %K%(

M )')8 mm

@ength of the stirrups

@ M % 5 l4 N l% N % K =Kd

M %54+8N%*( N%K=K+

M 48(' mm

Bumber of stirrups M 5).(98.&N4

M %&

2long 7-< direction


@ M l N % K 8.(Kd- %c

M )888N %K8.(K48 " %K%(

M (=)8 mm

Bumber of stirrups M 5)98.& N 4

M %4



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SYSTEMS

>olume of concrete for beam M 4+5).(K8.%&K8.&*(

M 4'5)K8.%&K8.&*(

M 4*.'& m&

etails of reinforcementTa4le .



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SYSTEMS

RESU%TS $ND DISCUSSION

*.1 IINTRODUCTION

esign and analysis of three types of slabs was done using their respective design

considerations. #osting 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.

.2 T#IC/NESS O& S%$B

$ased on the design outcome 5given in chapter &,',( comparison of thickness of slab for

the different type of floor slab systems is plotted in the figure *.4.



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SYSTEMS

&0 .1

Araph shows that bubble deck slab has considerably less thickness as compared to

conventional flat slab. Tough the conventional beam slab has least thicknessV the addition

of beam nullifies the advantage.

.3 )U$NTIT O& CONCRETE

$ased on the design outcome 5given in chapter ) comparison of quantity of concrete

used in slab for the different type of floor slab systems is plotted in the figure *.%.

&0 . 2

Araph 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. $ut

addition of beams in conventional beam slab system nullifies this advantage.

.+ )U$NTIT O& STEE%

$ased on the design outcome 5given in chapter ) comparison of quantity of steel used in

slab for the different type of floor slab systems is plotted in the figure *.&

& 0.3



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SYSTEMS

rom the graph we conclude that the bubble deck slab used least amount of steel and

usage of steel in conventional beam slab is maximum.

.* TOT$% )U$NTIT2B 7#3B36#SO& '$TERI$%S

igure *.' shows the diagrammatic comparison of quantity of steel as well as quantity of

concrete used in different type of slab systems

&0 .+

&0 .*

igure *.( shows the comparison of cost of concrete and cost of steel in slabs for different

types of floor slab systems. t 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

. ENVIRON'ENT$% CO'P$RISON

Table *.4 shows the #3% 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



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SYSTEMS

Ta4le .1

rom table *.4 we conclude that #3% emission for bubble deck slab is least and that for

conventional flat slab is most. igure *.' shows the diagrammatic comparison of quantity

of steel as well as quantity of concrete used in different type of slab systems



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SYSTEMS

C#$PTER >

CONC%USIONS $ND SCOPE &OR &UTURE WOR/



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SYSTEMS

>.1 CONC%USIONS

2 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.

esign and estimation was carried out for all the three types of slab systems. 3n the basis

of this work the following conclusions are drawn.

').& P of #oncrete was saved by using bubble deck slab instead of conventional

flat slab system

&+.* P of steel was saved in bubble deck slab system as compared to

conventional beam slab system.

2lmost %8 6.tones of #3% emission was reduced by use of bubble deck

technology

ntangibles " other intangible benefits derived from the use of bubble deck

technology are "

4 ncrease in number of floors due to less slab thickness

2 Reduction in foundation depth and si1e, which alsoi reduces the earthwork

excavation.

& Reduction in number of columns used and larger spans are possible

Department of Civil Engineering, M.S.R.I.T. Page !$


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SYSTEMS

>.2 SCOPE &OR &UTURE WOR/

The present study on bi-axially voided bubble deck slab system has the following scope

for further improvement

esign 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.

Department of Civil Engineering, M.S.R.I.T. Page !1


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SYSTEMS

ig 2.4 $all diameter

ig 2.% $ending strength design



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SYSTEMS

ig 2.& $ending stiffness

ig 2.' Shear capacity

ig 2.( Shear capacity



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SYSTEMS

ig 2.+ esign shear strength of concrete

ig 2.= 6aximum shear stress



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SYSTEMS

ig 2.48 $ending moment coefficient for slab spanning in two directions at right angles,

simply supported on four sides

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


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SYSTEMS

BIB%O(R$P#

RE&ERENCES

Department of Civil Engineering, M.S.R.I.T. Page !"


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

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