 Open Access
 Total Downloads : 778
 Authors : Sarita R. Khot, Himanshu V. Mahajan, Purval D. Shiram, Vishwajit V. Jadhav, Siddharth V. Tupe, Kumar T. Bharekar
 Paper ID : IJERTV5IS041092
 Volume & Issue : Volume 05, Issue 04 (April 2016)
 DOI : http://dx.doi.org/10.17577/IJERTV5IS041092
 Published (First Online): 30042016
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Comparative Study of Waffle Slabs with Flat Slabs and Conventional RCC Slabs
Mrs. Sarita R. Khot
Asst. Professor Department of Civil Engineering
RMD Sinhgad School of Engineering, Warje Pune 411058 (MS), India
Mr. Vishwajit V. Jadhav
BE Student Department of Civil Engineering
RMD Sinhgad School of Engineering, Warje Pune 411058 (MS), India
Mr. Purval D. Shiram
BE Student Department of Civil Engineering
RMD Sinhgad School of Engineering, Warje Pune 411058 (MS), India
Mr. Kumar T. Bharekar
BE Student Department of Civil Engineering
RMD Sinhgad School of Engineering, Warje Pune 411058 (MS), India
Mr. Himanshu V. Mahajan
BE Student Department of Civil Engineering
RMD Sinhgad School of Engineering, Warje Pune 411058 (MS), India
Mr. Siddharth V. Tupe
BE Student Department of Civil Engineering
RMD Sinhgad School of Engineering, Warje Pune 411058 (MS), India
Abstract Waffle slab construction consists of concrete joists at right angles to each other with solid heads at the column which is needed for shear requirements or with solid wide beam sections on the column centerlines for uniform depth construction. Waffle slab construction allows considerable reduction in the dead load of the overall structure as compared to flat slabs and conventional RCC slabs. The thickness of waffle slabs can be minimized to great extent as compared to flat slabs and RCC slabs. The bottom portion of waffle slab has many square projections with ribs spanning in both directions. The ribs are reinforced with steel to resist flexural tensile stresses. The design of waffle slab is done in such manner so as to achieve better load distribution. This paper deals with the comparative study of Waffle slabs with flat slabs and conventional RCC slabs and highlights the advantage waffle slabs have over flat slabs and RCC slabs. This comparison is shown with the help of a case study by designing of waffle slabs along with flat slabs and RCC slabs with the help of IS 4562000 and shown with comparison of various points.
Keywords Waffle Slabs, flat slabs, joists, flexural tensile stresses.

INTRODUCTION
Waffle Slabs can be defined as A reinforced concrete slab with equally spaced ribs parallel to the sides, having a waffle appearance from below. A Waffle Slab is a type of building material that has twodirectional reinforcement on the outside of the material, giving it the shape of the pockets on a waffle. The top of a waffle slab is generally smooth, like a traditional building surface, but underneath has a shape reminiscent of a waffle. Straight lines run the entire width & length of the slab, generally raised several inches from the surface. These ridges form the namesake square pockets of the entire length and width of slab. It helps insulate the floor since hot air gets trapped in the pockets. Waffle slabs have a thick solidslab floor from which the bottom layer concrete in tension is partially replaced by their ribs along orthogonal directions. The ribs are reinforced with steel. This type of reinforcement is common on concrete, wood and metal construction. A waffle slab gives a substance significantly more structural stability without using a lot of additional material. This makes a waffle slab perfect for large flat areas like foundations or floors. Reinforced concrete floors and roof construction employing a square grid of deep ribs with coffers in the intertices.

OBJECTIVE OF STUDY
The basic aim of this study is to design and show the comparative benefits of Waffle slabs over flat slabs and RCC slabs. Waffle slabs have some advantages over Flat slabs and RCC Slabs which helps us to use the Waffle Slabs.
The objectives are:

Savings on weights and material

Vertical penetration between ribs is easy.

Attractive soffit appearance if exposed.

Can be used for long span also.

Economical when reusable formwork used.

Reduction in dead load of slab.

Economical for structure having repeated works.


WHAT ARE FLAT SLABS?
Flat slabs system of construction is one in which the beams used in t he conventional methods of constructions are done away with. The slab directly rests on the column and load from the slab is directly transferred to the columns and then to the foundation. To support heavy loads the thickness of slab near the support with the column is increased and these are called drops, or columns are generally provided with enlarged heads called column heads or capitals. Absence of beam gives a plain ceiling, thus giving better architectural appearance and also less vulnerability in case of fire than in usual cases where beams are used. Plain ceiling diffuses light better, easier to construct and requires cheaper form work.

CASE STUDY
In the following case study a typical plan of shopping centre is taken into consideration. The design of various slabs in the entire building is done by standard design steps from the IS 4562000. The comparison of various slabs i.e.
1. Waffle Slabs; 2. Flat Slabs; 3. Conventional RCC Slabs are done with respect to its design and construction aspects.

DESIGN OF RCC SLAB SIZE = 7.5 Ã— 7.5 m
Given:
Ly = 7.5m Lx = 7.5m
Live Load, LL = 4 KN/mÂ² Floor Finish, F.F = 1 KN/mÂ² M20, fck = 20 MPa Fe415, Fy = 415 MPa
Design Constants:
fck = 20 MPa Fy = 415 MPa
Pt max. = 0.95 Ru = 2.76
Ld. = 47
Depth of Slab from Deflection Criteria: Let pt = 0.5
Refer Pg. 38, IS 456 2000, M.F = 1.2
Short Span = 26 Ã— M.F
d
d reqd. = 7500
26 Ã—1.2
dc = 20 mm
= 240 mm
D = d + dc = 240 + 20 D = 260 mm
Load Analysis:
D.L of Slab, DL = 0.24 Ã— 25 Ã— 1 = 6 KN/mÂ² Floor Finish, FF = 1 KN/mÂ²
Live Load, LL = 4 KN/mÂ²
Total Load = DL + FF + LL = W = 11 KN/mÂ²
Total Factored Load = Wu = 1.5 Ã— W = 1.5 Ã— 11 = 16.5 KN/mÂ²
Analysis of Slab:
= 7.5 = 1 < 2 ( It is a two way slab.)
7.5
Refer Tb 26, Case 4, IS 4562000.
x = 0.047 y = 0.047
x = 0.035 y = 0.035
Wu.LxÂ² = 16.5 Ã— 7.5Â² = 928.13 KNm
Mux = x Ã— Wu Ã— LxÂ² Mux = x Ã— Wu Ã— LxÂ²
= 0.035 Ã— 928.13 = 0.047 Ã— 928.13
= 32.48 KN = 43.62 KN
Muy = y Ã— Wu Ã— LyÂ² Muy = y Ã— Wu Ã— LyÂ²
= 0.035 Ã— 928.13 = 0.047 Ã— 928.13
= 32.48 KN = 43.62 KN
Check for d: Mu = Ru.b.d Â²
43.62 Ã— 106 = 2.76 Ã— 1000 Ã— d Â²
Span
Position
Mu
d prov.
Ast. reqd.
Ast. min.
Spacing
Ast. prov.
Short
Midspan
32.48
240
388
288
10 200
390
Continous
43.62
240
528
288
10 140
557
Long
Midspan
32.48
240
388
288
10 200
390
Continous
43.62
240
528
288
10 140
557
Span
Position
Mu
d prov.
Ast. reqd.
Ast. min.
/td>
Spacing
Ast. prov.
Short
Midspan
32.48
240
388
288
10 200
390
Continous
43.62
240
528
288
10 140
557
Long
Midspan
32.48
240
388
288
10 200
390
Continous
43.62
240
528
288
10 140
557
d = 125.71 < 240 mm ( Safe) Design Table:
1
1
2
2
Provision of Torsion Steel at corners.: In this case,
Astx. = 388 mmÂ²
AT = 3 Ã— 388 = 291 mmÂ²
4
8 3 nos.
Check for Shear.:
4
4
3
3
Reaction = Wu.Lx + Mx = 16.5Ã—7.5 + 43.62
2 Lx 2
= 67.69 KN
7.5
Vu = 67.9 16.5 ( 0.3 + 0.24 ) = 61.26 KN
2
v = Vu = 61.26 Ã—103
= 0.26 < c max
bd 1000 Ã—240
100 Ã—Ast
bd
= 0.16
Refer Tb. 19, Pg. – 73, IS 4562000,
c = 0.29 N/mm Â²
Refer Pg. 72, IS 4562000,
k = 1.10
k. c = 1.10 Ã— 0.29 = 0.319 MPa
v < k.c ( Safe)
Check for Deflection:
( Short Span ) = 7500
= 31.25
d 240
fs = 240 MPa , pt = 0.16
Refer Pg. 38, Fig. 4, IS 4562000,
M.F = 2
Basic Value Ã— M.F = 26 Ã— 2 = 52
31.25 52
Safe.
5 (WuÃ—L4) = 5 Ã— (16.5 Ã—103Ã—75004) = 26.44 mm
384 EI 384 22.36 Ã—1.15 Ã—109

DESIGN OF FLAT SLAB SIZE = 7.5 Ã— 7.5 m.
Given:
Ly = 7.5 m Lx = 7.5 m
Live Load, LL = 4 KN/mÂ² Floor Finish, F.F = 1 KN/mÂ²
M20, fck = 25 MPa Fe415, Fy = 415 MPa
Division of slab into Column Strip and Middle Strip:
As the length of slab in both directions is same i.e. Ly = Lx = 7.5 m ( L1 = L2 ), no separate calculations needed for long span and short as the values would be identical.
L1 = 7.5 m ; L2 = 7.5 m
Column Strip:
= 0.25 Ã— L2 = 0.25 Ã— 7.5 = 1.875 m
But it should not be greater than 0.25L1 = 0.25 Ã— 7.5 = 1.875 m Middle Strip = 7.5 ( 1.875 + 1.875 ) = 3.75 m
Column Strip = 1.875 m 
Middle Strip = 3.75 m 
Column Strip = 1.875 m 
Column Strip = 1.875 m 
Middle Strip = 3.75 m 
Column Strip = 1.875 
Column Strip = 1.875 m 
Middle Strip = 3.75 m 
Column Strip = 1.875 m 
Column Strip = 1.875 m 
Middle Strip = 3.75 m 
Column Strip = 1.875 
Along X Direction Along Y Direction
Since, the span is large, it is desirable to provide drop.
L1 = 7.5 m, L2 = 7.5 m
It should not be less than, L1 = 7.5 = 2.5 m
3 3
Hence provide a drop of size 2.5 Ã— 2.5 m i.e. in column strip width.
Column Head:
The diameter of column head should not be greater than L1 = 7.5 = 1.875 m
4 4
Adopting the diameter of Column Head = 1.875 m
Depth of Flat Slab:
By considering the flat slab as a continuous slab over a span not exceeding 10 m .
L = 26 d = L
d 26
d = 7500 = 288.46 mm
26
d = 290 mm
Taking effective cover = 25 mm
D = 290 + 25 = 315 mm > 125 mm … ( 125 mm = min. slab thck. as per IS 4562000)
It is safe to provide a depth = 315 mm.
Load Calculation:
Dead Load, DL = 0.315 Ã— 25 = 7.875 KN/mÂ²
Live Load, LL = 6 KN/mÂ² Floor Finish, FF = 2 KN/mÂ²
Total DL = DL + FF = 7.875 + 2 = 9.875 KN/mÂ²
Total LL = 6 KN/mÂ²
Total Load = DL + LL + FF = 7.875 + 6 + 2 = w = 15.875 KN/mÂ²
Total Factored Load, Wu = 1.5 Ã— w = 1.5 Ã— 15.875 = 23.81 KN/mÂ²
Check: LL = 6
= 0.61 < 3 OK
DL 9.875
Design Moment for span: Mo = W.Ln
8
Wu = 23.81 KN/mÂ²
a = () Ã— Â² = () Ã— 1.8Â²
4 4
a = 2.54 mÂ²
Ln = 7.5 ((1) Ã— 2.54) ((1) Ã— 2.54)
2 2
Ln = 4.96 m > ( 0.65.L1 = 0.65 Ã— 7.5 = 4.875 m ) W = Wu Ã— L1 Ã— Ln = 23.81 Ã— 7.5 Ã— 4.96
W = 885.73 KN
Mo = (W.Ln) = 885.73 Ã— 4.96
8 8
Mo = 549.15 KNm 550 KN
Stiffness Calculation: Height of floor = 4.0 m
Clear ht. of column = ht. of floor depth of drop thickness of slab thickness of head
= 4.0 0.16 0.315 0.415 = 3.11 m 3110 mm
Effective Ht. of Column = 0.8 Ã— 3.11 = 2.49 m ( assuming one end hinged and other end fixed )
c = Kc
Ks
Kc = (4EI) bottom + (4EI) top
L L
= 1.58
L2 = 1 ( from Tb. 17 )
L1
Wl = 1
Wd
c min. = 0.7
c = 1.58 > c min.
Safe.
Distribution of BM across panel width:

Column Strip
1+(1)
1+(1)
Negative BM at exterior support = (0.65Mo) Ã— 1.0
1+( 1 )
1+( 1 )
= (0.65 Ã—550) Ã— 1.0 = 218.93 KNm
1.58
1+(1)
1+(1)
Positive Span BM = ( 0.63 ( 0.28 ) Ã— Mo Ã— 0.60 )
1+( )
1+( )
= ( 0.63 ( 0.28 ) Ã— 550 Ã— 0.60 ) = 151.31 KNm
1 1.58
1+( 1 )
1+( 1 )
Negative Span BM at interior supports = ( 0.75 ( 0.10 ) Ã— Mo Ã— 0.75 )
c
1+( )
1+( )
= ( 0.75 ( 0.10 ) Ã— 550 Ã— 0.75 ) = 284.11 KNm
1 1.58

MIDDLE STRIP
1+(1)
1+(1)
Negative BM at exterior support = (0.65 Mo) Ã— 0 = 0 KNm
1+(1)
1+(1)
Positive span BM = (0.63 ( 0.28 ) Ã— Mo Ã— 0.40 )
1+( )
1+( )
= ( 0.63 ( 0.28 ) Ã— 550 Ã— 0.40 ) = 100.87 KNm
1 1.58
1+(1)
1+(1)
Negative BM at interior support = (0.75 ( 0.10 )) Ã— Mo Ã— 0.75
1+( )
1+( )
= (0.75 ( 0.10 )) Ã— 550 Ã— 0.75 = – 94.70 KNm
For effective depth of slab,
1 1.58
Maximum positive BM occurs in the column strip = 151.31 KNm
Mu = 0.138.fck.b.dÂ²
151.31 Ã— 106 = 0.138 Ã— 25 Ã— 3750 Ã— dÂ²
d = 108.15 120 mm Using 12 mm main bars
D = 120 + 25 = 145 mm 160 mm
Depth ( along longitudinal direction) i. = 160 15 12 = 139 mm
2
ii. = 160 12 = 148 mm
Thickness of drop from maximum negative moment anywhere in the panel. Maximum negative BM occurs in the column strip = 248.11 KNm
Mu = 0.138.fck.b.dÂ²
284.11 Ã— 106 = 0.138 Ã— 25 Ã— 1875 Ã— dÂ²
d = 209.57 mm 220 mm Using 12 mm main bars
Overall thickness of Flat Slab, D = 220 + 15 + 12 = 241 mm 250 mm
2
Shear in Flat Slab:
Check for shear stress developed in slab.
The critical section for shear for the slab will be at a distance from face of the drop.
2
Side of critical section = 2500 + 160 + 160
2 2
= 2660 mm
Perimeter of critical section = 2660 Ã— 4 = 10640 mm Vo = Wu ( L1 Ã— L2 (Side) Â² )
= 23.81 ( 7.5 Ã— 7.5 2.66 Â² ) Vo = 1170.84 KN
Nominal Shear Stress = v = = 1170.84 Ã—103
10640 Ã—160
v = 0.69 N/mm Â²
c = 0.2525 = 1.25 N/mm Â² v < c ( Safe.)
Check for shear in drop:
bo = ( D + do ) = ( 1.8 + 0.29 ) = 6.57 m
V = 23.81 ( 7.5 Ã— 7.5
4
(1.8 + 0.29)Â² ) = 1257.63 KN
v = 1257.63 Ã—103 = 0.66 N/mm Â²
6570 Ã—290
c = 1.25 N/mm Â²
v < c
Safe.
Reinforcement Details:
As the lengths of both the spans are equal i.e. Ly = Lx = 7.5 m, the reinforcement in both the directions wold be same. Hence, calculations would be same.
Ly = LÃ— = 7.5 m
Negative Exterior Reinforcement: Mu = 0.87.Fy.Ast ( d 0.42xu )
218.93 Ã— 106 = 0.87 Ã— 415 Ã— Ast. Ã— ( 139 ( 0.42 Ã— 0.48 Ã— 139 ))
Ast. = 5463.9 mm Â² Use 12 mm bars
Number of bars = 5463.9 = 48.35 49 bars
113
Spacing = 1.875 Ã—1000 = 39 mm c/c
48
Positive Reinforcement:
Mu = 0.87.Fy.Ast ( d 0.42xu )
151.31 Ã— 106 = 0.87 Ã— 415 Ã— Ast. Ã— ( 148 ( 148 ( 0.42 Ã— 0.48 Ã— 148 ))
Ast. = 3546.6 mm Â² Use 12 mm bars
Number of bars = 3546.6 31 bars
113
Spacing = ( 3.75 Ã—1000 ) = 120 mm c/c
31
Deflection Check:
E = 2.1 Ã— 105 N/mmÂ²
I = 3 = 1000 Ã—2903 = 2032.41 Ã— 106 mm4
12 12
( 5 ) (Wu .L4) = ( 5 ) Ã— ( 23.81 Ã—75004
) = 22.9 mm
384 EI
384
(2.1 Ã—105) Ã—( 2032.41 Ã—106)
Span = 7500 = 30 mm
250 250
22.9 mm < 30 mm
Safe.
Given:

DESIGN OF FLAT SLAB SIZE = 15 Ã— 7.5 m.
Ly = 15 m Lx = 7.5 m
Live Load, LL = 4 KN/mÂ² Floor Finish, F.F = 1 KN/mÂ²
M20, fck = 25 MPa Fe415, Fy = 415 MPa
Division of slab into Column Strip and Middle Strip: Long Span
L1 = 15 m , L2 = 7.5 m

Column Strip = 0.25.L2 = 1.875 m (But not greater than (0.25L1 = 3.75 m ) ii. Middle Strip = 7.5 ( 1.875 + 1.875 ) = 3.75 m
Short Span
L1 = 7.5 m , L2 = 15 m

Column Strip= 0.25.L2 = 3.75 m (But not greater than (0.25L1 = 1.875m)

ii. Middle Strip = 15 (3.75 + 3.75) = 7.5 m
Column Strip = 3.75
Middle Strip = 7.5
Column Strip = 3.75
Column Strip = 1.875
Middle Strip = 3.75
Column Strip = 1.875
Column Strip = 3.75
Middle Strip = 7.5
Column Strip = 3.75
Column Strip = 1.875
Middle Strip = 3.75
Column Strip = 1.875
Along X Direction = 7.5 m Along Y Direction = 15 m Since the span is large, it is desirable to provide drop.
Long Span
L1 = 15 m , L2 = 7.5 m
Not less than L1 = 15 = 5 m
3 3
Short Span
L1 = 7.5 m , L2 = 15 m
Not less than 1 = 7.5 = 2.5 m
3 3
Hence, provide a drop of size 5 Ã— 5 m in column strip width.
Column Head.: Long Span
L1 = 15 m , L2 = 7.5 m
Not greater than 1 = 15 = 3.75 m
4 4
Short Span
L1 = 7.5 m , L2 = 15 m
Not greater than 1 = 7.5
= 1.875 m
4 4
Adopting the diameter of column head = 1.75 m 1750 mm
Depth of Flat Slab:
L = 26 d = L
d 26
Long Span
L1 = 15 m , L2 = 7.5 m
d = L = 15000
26 26
d = 573.92 575 mm
Short Span
L1 = 7.5 m , L2 = 15 m
d = L = 7500
26 26
d = 288.46 mm d 290 mm
Taking effective cover of 25 mm.
Overall depth of Flat Slab, D = 575 + 25 = 600 mm.
As the depth of Flat Slab for the overall dimensions of 15 Ã— 7.5 m is 600 mm, it is not practically feasible to provide a slab of such great depth. Also it is would not be economical to provide slab of such depth. Besides economical and practical problems, this may also have problems during execution stage as the concrete required would be great so the concrete batches would be more & also the heat of hydration would be very large.
So, for providing slab of such greater spans Waffle Slabs are the alternative for flat slabs. The slab depth in waffle slabs is very much less as compared to Flat Slabs and any other slabs such conventional RCC slabs, etc. Due to decrease in the depth of slab it is practically more feasible to provide Waffle Slabs instead of Flat Slabs. The execution would also be easier in Waffle Slabs in comparison to Flat Slabs. Though the formwork required for Waffle Slabs is more it can be reused many times as well as the construction is fast.
Hence, in this case study in comparison for Conventional RCC Slabs and Flat Slabs, we are comparing Waffle Slabs to lighten the benefits of using Waffle Slabs for large spans of slabs in replacement of other slabs. The construction, cost and practical benefits of Waffle Slabs in comparison with other slabs can be clearly elaborated from these case studies.


DESIGN OF WAFFLE SLAB SIZE = 7.5 Ã— 7.5 m
Given:
Size of Grid = 7.5 Ã— 7.5 m
Spacing of ribs = 1.5 m a1 = 1500 mm ; b1 = 1500 mm M20, fck = 20 N/mmÂ²
Fe415, Fy = 415 N/mmÂ² Live Load, LL = 4 KN/mÂ² Floor Finish, FF = 1 KN/mÂ²
Dimensions of Slab & Beam: Adopt thickness of slab = 100 mm
Depth of Rib = Span = 7500 = 288.46 mm 400 mm
1500
26 26
Width of Rib = 175 mm
Number of beams in X Direction, Nx = 6 Number of beams in Y Direction, Ny = 6 E = 5000fck Ã— 1000
= 5000 Ã— 20 Ã— 1000
E = 22.36 Ã— 106
400
100
300
Load Calculations: Total weight of slab
= 25 Ã— Df Ã— Lx Ã— Ly
= 25 Ã— 0.1 Ã— 7.5 Ã— 7.5 = 140.63 KN
Total weight of beams in X Direction
= 25 Ã— bw Ã— D Ã— Nx Ã— LÃ—
= 25 Ã— 0.175 Ã— 0.40 Ã— 6 Ã— 7.5 = 78.75 KN
Total weight of beams in Y Direction
= 25 Ã— bw Ã— D Ã— Ny Ã— ( Ly ( bw Ã— Nx ) )
= 25 Ã— 0.175 Ã— 0.40 Ã— 6 Ã— ( 7.5 ( 0.175 Ã— 6 ) ) = 67.73 KN Total Live Load, LL = LL Ã— Ly Ã— Lx = 4 Ã— 7.5 Ã— 7.5 = 225 KN
Total Floor Finish, FF = FF Ã— Ly Ã— Lx = 1 Ã— 7.5 Ã— 7.5 = 56.25 KN Total Load = 140.63 + 78.75 + 67.73 + 225 + 56.25 = 568.38 KN
175
Total Load / mÂ² = q = 568.75
7.5 Ã—7.5
= 10.10 KN/mÂ²
Total Factored Load = Q = 1.5 Ã— q = 1.5 Ã— 10.10 = 15.15 KN/mÂ²
Design Parameters:
Df = 100 = 0.25
D 400
bf = 1500 = 8.57
bw 175
Moment of Inertia:
I = Kx Ã— bw Ã— D3 = 3 Ã—0.175 Ã—0.43 ( Kx = 3 SP 16, Tb. 18 )
12 12
I = 2.8 Ã— 103 m4 2.8 Ã— 109 mm4
Flexural Rigidity of Ribs:
Dx = Ã— = 22.36 Ã—106 Ã—2.8 Ã—109
1
Dx = 4.7 Ã— 104
1500
Dy = Ã— = 22.36 Ã—106 Ã—2.8 Ã—109
1
Dy = 4.7 Ã— 104
Modulus of Shear:
1500
G = 2 ( 1+ )
= 22.36 Ã—106 ( = 0.15 assume )
2 ( 1+0.15 )
G = 9.72 Ã— 106 KN/m2
Torsional Constants ( Polar Sectional Modulus ):
C1 = 1 (0.63 Ã— (bw)) (bw3 Ã— (D)) = 1 (0.63 Ã— (0.175)) (0.1753 Ã— (0.4)) = 5.18 Ã— 104 m3
D 3 4 3
C2 = 1 (0.63 Ã— (bw)) (D3 Ã— (bw)) = 1 (0.63 Ã— (0.175)) (0.43 Ã— (0.175)) = 2.7 Ã— 103 m3
D 3 0.4 3
Torsional Rigidity:
Cx = ( G Ã—C1 ) = 9.72 Ã—106Ã—5.18 Ã—104
= 3.35 Ã— 103
b1 1.5
Cy = GÃ—C2 = 9.72 Ã—106 Ã—2.7 Ã—103
= 1.75 Ã— 104
a1 1.5
2H = Cx + Cy = ( 3.35 Ã— 10Â³) + ( 1.75 Ã— 104 ) = 2.09 Ã— 104
Dx Lx4 =
Dy Ly4 =
4.75 Ã— 104
7.54 = 13.9
4.75 Ã— 104
7.54 = 13.9
2H 2.09 Ã— 104
Lx2 Ã— Ly2 = 7.52 Ã— 7.52 = 6.60
Deflection Check:
16 Ã—(Q) 2H
Dy 16 Ã—(15.15)
=
=
Dx + ( 2
2) + (
4) = + 6.60 + 13.19 = 7.65 mm
LÃ—4
Lx Ã—Ly Ly
13.19
Long Term Deflections:
Lt defl. = 3 Ã— = 3 Ã— 7.65 = 22.94 mm
Span = 7500 = 30 mm ( Lt defl. < Span
Safe.)
250 250 250
Maximum Moment and Shear Values:
Mx = Dx Ã— ( )2 Ã— = 4.74 Ã— 104 Ã— (
2
) Ã— 7.65 = 56 KNm
2
2
LÃ— 7500
My = Dy Ã— ( ) Ã— = 4.7 Ã— 104 Ã— (
2
) Ã— 7.65 = 56 KNm
7500
Maximum Torsional Moment:
7500
Mxy = Cx Ã— 2 Ã— = 3.35 Ã—103 Ã—2 Ã—7.65 = 5 KNm
Ly Ã—Lx 7500 Ã—7500
Shear Force:
Qx = (( Dx Ã— ( )3 ) + ( Cy Ã— ( 3
))) Ã—
Lx
= ((4.74 Ã— 104 Ã— (
a1 Ã—b12
3 3
) ) + ( 1.75 Ã— 104 Ã— (
p>2))) Ã— 7.65 = 25 KN
Qy = ((Dy Ã— (
7500
3 3
) ) + (Cx Ã— (
))) Ã—
1500 Ã—1500
7500
= ((4.74 Ã— 104 Ã— (
a12 Ã—b1
3 3
) + (3.35 Ã— 104 Ã— (
))) Ã— 7.65 = 25 KN
)
7500
15002
Ã—1500
Reinforcement Details:
Astx = Asty = 0.5 Ã—fck Ã— (1 1 4.6Mu ) Ã— b Ã— d ( as Ly = Lx = 7.5 m )
fy fck.b.d2
= 0.5 Ã—20 Ã— (1 1 4.6 Ã—56 Ã—106
) Ã— 175 Ã— 370
415
Astx = Asty = 499.29 mm2 Use 12 mm bars
Provide 5 bars of 12 mm .
20 Ã—175 Ã—3702
Provide minimum steel in slab portion i.e. flange of waffle slab. Ast min. = 0.12 Ã— b Ã— d = 0.12 Ã— 1000 Ã— 100
100
Ast min. = 120 mm2
Given:
Size of Grid = 15 Ã— 7.5 m
100
IX. DESIGN OF WAFFLE SLAB SIZE = 15 Ã— 7.5 m
Spacing of ribs = 1.5 m a1 = 1500 mm ; b1 = 1500 mm M20, fck = 20 N/mmÂ²
Fe415, Fy = 415 N/mmÂ² Live Load, LL = 4 KN/mÂ² Floor Finish, FF = 1 KN/mÂ²
1500
Dimensions of Slab & Beam: Adopt thickness of slab = 100 mm
Depth of Rib = Span = 15000 = 580 mm
580
100
26 26
Width of Rib = 150 mm
Number of beams in X Direction, Nx = 6 Number of beams in Y Direction, Ny = 11 E = 500020 Ã— 1000 = 5000 Ã— 20 Ã— 1000 E = 22.36 Ã— 106
Load Calculation:
Total weight of slab = 25 Ã— Df Ã— Lx Ã— Ly
= 25 Ã— 0.10 Ã— 7.5 Ã— 15 = 281.25 KN
150
480
Total weight of beams in X Direction = 25 Ã— bw Ã— D Ã— Nx Ã— Lx = 25 Ã— 0.150 Ã— 0.580 Ã— 6 Ã— 7.5 = 97.88 KN Total weight of beams in Y Direction = 25 Ã— bw Ã— D Ã— Ny Ã— ( Ly ( bw Ã— Nx ) )
= 25 Ã— 0.15 Ã— 0.25 Ã— 11 Ã— ( 15 ( 0.15 Ã— 6 ) ) = 337.34 KN Total Live Load, LL = LL Ã— Lx Ã— Ly = 4 Ã— 7.5 Ã— 15 = 450 KN
Total Floor Finish, FF = FF Ã— Lx Ã— Ly = 1 Ã— 7.5 Ã— 15 = 112.5 KN Total Load = 281.25 + 97.88 + 337.34 + 450 + 112.5 = 1278.97 KN
Total Load / m2 = q = 1278.97 = 11.37 KN/m2
7.5 Ã—15
Total Factored load / m2 = Q = 1.5 Ã— q = 1.5 Ã— 11.37 =17.05 KN/m2
Design Parameters:
Df = 100 = 0.17
D 580
bf = 1500 = 10
bw 150
Moment of Inertia:
I = Kx Ã—bw Ã—D3
12
= 2.3 Ã—150 Ã—5803 ( Kx = 2.3, SP – 16, Tb. 88 )
12
I = 5.60 Ã— 109 mm4 = 5.61 Ã— 103 m4
Flexural Rigidity of Ribs:
Dx = EÃ—I = 22.36 Ã—106 Ã—5.60 Ã—109 = 8.35 Ã— 1013
a1 1500
Dy = E Ã—I = 22.36 Ã—106 Ã—5.60 Ã—109 = 8.35 Ã— 1013
b1 1500
Modulus of Shear:
G = E 2(1+)
= 22.36 Ã—106
2(1+0.15)
( = 0.15 assume ) = 9.72 Ã— 106
Torsional Constants ( Polar Sectional Modulus ):
C1 = 1 0.63 (bw) (bw3 Ã— (D)) = 1 0.63 (150) (1503 Ã— (580)) = 5.46 Ã— 104 m3
d 3 580 3
C2 = 1 0.63 (bw) (D3 Ã— (bw)) = 1 0.63 (150) (5803 Ã— (150)) = 8.17 Ã— 103 m3
D 3 580 3
Torsional Rigidity:
Cx = G Ã—C1 = 9.72 Ã—106 Ã—5.46 Ã—104 = 3.54 Ã— 103
b1 1.5
Cy = G Ã—C2 = 9.72 Ã—106 Ã—8.17 Ã—103 = 5.29 Ã— 104
a1 1.5
2H = Cx + Cy = ( 3.54 Ã— 103 ) + ( 5.29 Ã— 104) = 5.65 Ã— 104
Dx Lx4 =
Dy Ly4 =
8.36 Ã— 104
7.54 = 26.43
8.36 Ã— 104
154 = 1.65
2H 5.65 Ã— 104
Lx2 Ã— Ly2 = 7.52 Ã— 152 = 4.46
Deflection Check:
16 Ã—(Q) 2H
Dy 16 Ã—(17.05)
=
=
Dx + (
2) + ( 4) = ( ) + 4.46 + 1.65 = 8.72 mm
Lx4
Lx Ã—Ly Ly
26.43
Long Term Deflection:
Lt defl. = 3 Ã— = 3 Ã— 8.72 = 26.16 mm
Span = 7500 = 30 mm
250 250
Lt defl. < Span
250
Safe.
Maximum Moment and Shear Values:
Mx = D Ã— Ã— ( )2 Ã— = 8.64 Ã— 104 Ã— ( )2 Ã— 8.72 = 128 KNm
LÃ— 7.5
My = Dy Ã— ( )2 Ã— = 8.36 Ã— 104 Ã— ( )2 Ã— 8.72 = 32 KNm
Ly 15
Maximum Torsional Moment:
Mxy = CÃ— Ã— 2 Ã—
Lx Ã— Ly
= 3.54 Ã—103 Ã— 2 Ã—8.72 = 3 KNm
7.5 Ã—15
Shear Force:
Qx = ((D Ã— Ã— ( )3) + (Cy Ã— ( 3 ))) Ã—
Lx a1 Ã— b12
= ((8.36 Ã— 104 Ã— ( )3) + (5.29 Ã— 104 Ã— ( 3 ))) Ã— 8.72 = 54 KN
7.5 1.5 Ã—1.52
Qy = ((Dy Ã— ( )3) + (C Ã— Ã— ( 3 ))) Ã—
Ly a12 Ã—b1
= ((8.36 Ã— 104 Ã— ( )3) + ( 3.54 Ã— 103 Ã— ( 3 ))) Ã— 8.72 = 8 KN
15 1.52 Ã—1.5
Reinforcement Details: For X Direction,
Mu = 128 KNm, b= 150 mm, d = 550mm
Astx = 0.5 Ã—fck Ã— (1 1 ( 4.6Mu )) Ã— b Ã— d
Fy fck.b.d2
= 0.5Ã—20 Ã— ( 1 1 ( 4.6 Ã—128 Ã—106 )) Ã— 150 Ã— 550
415 20 Ã—150 Ã—5502
Astx = 810 mm2 Use 16 mm bars.
Provide 4 bars of 16 mm . For Y Direction,
Mu = 32 KNm, b = 150 mm, d = 550 mm
Asty = 0.5 Ã—fck Ã— ( 1 1 ( 4.6Mu )) Ã— b Ã— d
Fy fck.b.d2
= 0.5 Ã—20 Ã— ( 1 1 ( 4.6 Ã—32 Ã—106 )) Ã— 150 Ã— 550
415 20 Ã—150 Ã—5502
Asty = 168.36 mm2 Use 10 mm bars.
Provide 3 bars of 10 mm .
Provide minimum steel in slab portion i.e. flange of waffle slab.
Ast min. = (0.12 Ã— Ã— ) = (0.12 Ã— 1000 Ã— 100 )
100
Ast min. = 120 mm2
100

DESIGN SUMMARY
Size (m)
Slab Thickness ( mm )
Deflection ( mm )
Total Ast (mm2)
Max. Moment (KNm)
Max. Shear (KN)
Factored Load ()
2
Economy & Construction
Uses
Waffle Slab
7.5 Ã— 7.5
100
7.65
1118.58
56
25
15.15
Economical for repetitive works, requires more time for construction as compared to RCC slabs.
Suitable for large loads, large spans, repetitive works, aesthetic appearance, etc.
Waffle Slab
15 Ã—
7.5
100
8.72
1160
128
54
17.05
Economical for large spans and repetitive works, skilled labors required, aesthetically more useful structures, no need of finishing in various cases
. Suitable for large loads, large spans, repetitive works, aesthetic appearance, etc.
Flat Slab
7.5 Ã— 7.5
315
22.9
9010.5
284.11
1170.84
23.81
Suitable for medium spans, beamless construction, difficulty while construction pods and posttensioning, complicated designs
Beamless construction, aesthetics, less complication while construction
Flat Slab
15 Ã—
7.5
600
–
–
–
–
–
Not for large spans, Suitable for medium spans, beamless construction, difficulty while construction pods and posttensioning, complicated designs
Beamless construction, aesthetics, less complication while construction & suitable for medium spans
Conventional RCC Slab
7.5 Ã— 7.5
260
31.25
1894
43.62
61.26
16.5
Suitable for short spans, most easiest way of construction, skilled labors not required, can be constructed in rural areas very easily.
Easiest way of construction, less complicated designs, residential buildings, etc.

ACKNOWLEDGEMENT
Sincere thanks to Dr. C.B. Bangal, Principal RMD Sinhgad School of Engineering, Prof. Sarita R. Khot, our guide and Mrs. P.S Shete Head of Department of Civil Engineering

CONCLUSION
From the following paper with the help of the case study, we conclude that waffle slabs are more advantageous as compared to other slabs such as flat slabs and RCC slabs, in terms of loading, large spans, aesthetic appearance, etc.

REFERENCE

Research Paper – Optimum design of reinforced conrete waffle slabs, by – Alaa C. Galeb, Zainab F. Atiyah Civil Engineering Department, University of Basrah, Iraq

Research Paper – Waffle Slab Analysis By Different Methods, by Naziya Ghanchi, Chitra V.

Design of Waffle Slab – Ariel Mayer.

Analysis and Design of Waffle Slabs2011Jose Pablo Rosales Sanchez.

Waffle Slab Khairul Salleh Baharudin.

Structural Analysis , by S.S Bhavikatti.

Reinforced Concrete Design Theory & Examples , By T.J. Macginley & B.S. Choo.