Comparative Study of Waffle Slabs with Flat Slabs and Conventional RCC Slabs

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 456-2000 and shown with comparison of various points.

Keywords Waffle Slabs, flat slabs, joists, flexural tensile stresses.

1. 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 two-directional 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 solid-slab 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.

2. 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:-

   1. Savings on weights and material

   2. Vertical penetration between ribs is easy.

   3. Attractive soffit appearance if exposed.

   4. Can be used for long span also.

   5. Economical when reusable formwork used.

   6. Reduction in dead load of slab.

   7. Economical for structure having repeated works.

3. 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.

4. 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 456-2000. 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.

5. DESIGN OF RCC SLAB SIZE = 7.5 × 7.5 m

   Given:-

   Ly = 7.5m Lx = 7.5m

   Live Load, LL = 4 KN/m² Floor Finish, F.F = 1 KN/m² 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 × M.F

   d

   d reqd. = 7500

   26 ×1.2

   dc = 20 mm

   = 240 mm

   D = d + dc = 240 + 20 D = 260 mm

   Load Analysis:-

   D.L of Slab, DL = 0.24 × 25 × 1 = 6 KN/m² Floor Finish, FF = 1 KN/m²

   Live Load, LL = 4 KN/m²

   Total Load = DL + FF + LL = W = 11 KN/m²

   Total Factored Load = Wu = 1.5 × W = 1.5 × 11 = 16.5 KN/m²

   Analysis of Slab:-

   = 7.5 = 1 < 2 ( It is a two way slab.)

   7.5

   Refer Tb 26, Case 4, IS 456-2000.

   x = 0.047 y = 0.047

   x = 0.035 y = 0.035

   Wu.Lx² = 16.5 × 7.5² = 928.13 KN-m

   Mux = x × Wu × Lx² Mux = x × Wu × Lx²

   = 0.035 × 928.13 = 0.047 × 928.13

   = 32.48 KN = 43.62 KN

   Muy = y × Wu × Ly² Muy = y × Wu × Ly²

   = 0.035 × 928.13 = 0.047 × 928.13

   = 32.48 KN = 43.62 KN

   Check for d:- Mu = Ru.b.d ²

   43.62 × 106 = 2.76 × 1000 × d ²

   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 mm²

   AT = 3 × 388 = 291 mm²

   4

   8 3 nos.

   Check for Shear.:-

   4

   4

   3

   3

   Reaction = Wu.Lx + Mx = 16.5×7.5 + 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 ×103

   = 0.26 < c max

   bd 1000 ×240

   100 ×Ast

   bd

   = 0.16

   Refer Tb. 19, Pg. – 73, IS 456-2000,

   c = 0.29 N/mm ²

   Refer Pg. 72, IS 456-2000,

   k = 1.10

   k. c = 1.10 × 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 456-2000,

   M.F = 2

   Basic Value × M.F = 26 × 2 = 52

   31.25 52

   Safe.

   5 (Wu×L4) = 5 × (16.5 ×103×75004) = 26.44 mm

   384 EI 384 22.36 ×1.15 ×109

6. DESIGN OF FLAT SLAB SIZE = 7.5 × 7.5 m.

Given:-

Ly = 7.5 m Lx = 7.5 m

Live Load, LL = 4 KN/m² Floor Finish, F.F = 1 KN/m²

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 × L2 = 0.25 × 7.5 = 1.875 m

But it should not be greater than 0.25L1 = 0.25 × 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 × 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 456-2000)

It is safe to provide a depth = 315 mm.

Load Calculation:-

Dead Load, DL = 0.315 × 25 = 7.875 KN/m²

Live Load, LL = 6 KN/m² Floor Finish, FF = 2 KN/m²

Total DL = DL + FF = 7.875 + 2 = 9.875 KN/m²

Total LL = 6 KN/m²

Total Load = DL + LL + FF = 7.875 + 6 + 2 = w = 15.875 KN/m²

Total Factored Load, Wu = 1.5 × w = 1.5 × 15.875 = 23.81 KN/m²

Check:- LL = 6

= 0.61 < 3 OK

DL 9.875

Design Moment for span:- Mo = W.Ln

8

Wu = 23.81 KN/m²

a = () × ² = () × 1.8²

4 4

a = 2.54 m²

Ln = 7.5 ((1) × 2.54) ((1) × 2.54)

2 2

Ln = 4.96 m > ( 0.65.L1 = 0.65 × 7.5 = 4.875 m ) W = Wu × L1 × Ln = 23.81 × 7.5 × 4.96

W = 885.73 KN

Mo = (W.Ln) = 885.73 × 4.96

8 8

Mo = 549.15 KN-m 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 × 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:-

1. Column Strip

   1+(1)

   1+(1)

   Negative BM at exterior support = (0.65Mo) × 1.0

   1+( 1 )

   1+( 1 )

   = (0.65 ×550) × 1.0 = 218.93 KN-m

   1.58

   1+(1)

   1+(1)

   Positive Span BM = ( 0.63 ( 0.28 ) × Mo × 0.60 )

   1+( )

   1+( )

   = ( 0.63 ( 0.28 ) × 550 × 0.60 ) = 151.31 KN-m

   1 1.58

   1+( 1 )

   1+( 1 )

   Negative Span BM at interior supports = ( 0.75 ( 0.10 ) × Mo × 0.75 )

   c

   1+( )

   1+( )

   = ( 0.75 ( 0.10 ) × 550 × 0.75 ) = 284.11 KN-m

   1 1.58

2. MIDDLE STRIP

1+(1)

1+(1)

Negative BM at exterior support = (0.65 Mo) × 0 = 0 KN-m

1+(1)

1+(1)

Positive span BM = (0.63 ( 0.28 ) × Mo × 0.40 )

1+( )

1+( )

= ( 0.63 ( 0.28 ) × 550 × 0.40 ) = 100.87 KN-m

1 1.58

1+(1)

1+(1)

Negative BM at interior support = (0.75 ( 0.10 )) × Mo × 0.75

1+( )

1+( )

= (0.75 ( 0.10 )) × 550 × 0.75 = – 94.70 KN-m

For effective depth of slab,

1 1.58

Maximum positive BM occurs in the column strip = 151.31 KN-m

Mu = 0.138.fck.b.d²

151.31 × 106 = 0.138 × 25 × 3750 × d²

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 KN-m

Mu = 0.138.fck.b.d²

284.11 × 106 = 0.138 × 25 × 1875 × d²

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 × 4 = 10640 mm Vo = Wu ( L1 × L2 (Side) ² )

= 23.81 ( 7.5 × 7.5 2.66 ² ) Vo = 1170.84 KN

Nominal Shear Stress = v = = 1170.84 ×103

10640 ×160

v = 0.69 N/mm ²

c = 0.2525 = 1.25 N/mm ² v < c ( Safe.)

Check for shear in drop:-

bo = ( D + do ) = ( 1.8 + 0.29 ) = 6.57 m

V = 23.81 ( 7.5 × 7.5

4

(1.8 + 0.29)² ) = 1257.63 KN

v = 1257.63 ×103 = 0.66 N/mm ²

6570 ×290

c = 1.25 N/mm ²

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 = L× = 7.5 m

Negative Exterior Reinforcement:- Mu = 0.87.Fy.Ast ( d 0.42xu )

218.93 × 106 = 0.87 × 415 × Ast. × ( 139 ( 0.42 × 0.48 × 139 ))

Ast. = 5463.9 mm ² Use 12 mm bars

Number of bars = 5463.9 = 48.35 49 bars

113

Spacing = 1.875 ×1000 = 39 mm c/c

48

Positive Reinforcement:-

Mu = 0.87.Fy.Ast ( d 0.42xu )

151.31 × 106 = 0.87 × 415 × Ast. × ( 148 ( 148 ( 0.42 × 0.48 × 148 ))

Ast. = 3546.6 mm ² Use 12 mm bars

Number of bars = 3546.6 31 bars

113

Spacing = ( 3.75 ×1000 ) = 120 mm c/c

31

Deflection Check:-

E = 2.1 × 105 N/mm²

I = 3 = 1000 ×2903 = 2032.41 × 106 mm4

12 12

( 5 ) (Wu .L4) = ( 5 ) × ( 23.81 ×75004

) = 22.9 mm

384 EI

384

(2.1 ×105) ×( 2032.41 ×106)

Span = 7500 = 30 mm

250 250

22.9 mm < 30 mm

Safe.

Given:-

1. DESIGN OF FLAT SLAB SIZE = 15 × 7.5 m.

   Ly = 15 m Lx = 7.5 m

   Live Load, LL = 4 KN/m² Floor Finish, F.F = 1 KN/m²

   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

   1. 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

      1. 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 × 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 × 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 re-used 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.

2. DESIGN OF WAFFLE SLAB SIZE = 7.5 × 7.5 m

Given:-

Size of Grid = 7.5 × 7.5 m

Spacing of ribs = 1.5 m a1 = 1500 mm ; b1 = 1500 mm M20, fck = 20 N/mm²

Fe415, Fy = 415 N/mm² Live Load, LL = 4 KN/m² Floor Finish, FF = 1 KN/m²

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 × 1000

= 5000 × 20 × 1000

E = 22.36 × 106

400

100

300

Load Calculations:- Total weight of slab

= 25 × Df × Lx × Ly

= 25 × 0.1 × 7.5 × 7.5 = 140.63 KN

Total weight of beams in X Direction

= 25 × bw × D × Nx × L×

= 25 × 0.175 × 0.40 × 6 × 7.5 = 78.75 KN

Total weight of beams in Y Direction

= 25 × bw × D × Ny × ( Ly ( bw × Nx ) )

= 25 × 0.175 × 0.40 × 6 × ( 7.5 ( 0.175 × 6 ) ) = 67.73 KN Total Live Load, LL = LL × Ly × Lx = 4 × 7.5 × 7.5 = 225 KN

Total Floor Finish, FF = FF × Ly × Lx = 1 × 7.5 × 7.5 = 56.25 KN Total Load = 140.63 + 78.75 + 67.73 + 225 + 56.25 = 568.38 KN

175

Total Load / m² = q = 568.75

7.5 ×7.5

= 10.10 KN/m²

Total Factored Load = Q = 1.5 × q = 1.5 × 10.10 = 15.15 KN/m²

Design Parameters:-

Df = 100 = 0.25

D 400

bf = 1500 = 8.57

bw 175

Moment of Inertia:-

I = Kx × bw × D3 = 3 ×0.175 ×0.43 ( Kx = 3 SP 16, Tb. 18 )

12 12

I = 2.8 × 103 m4 2.8 × 109 mm4

Flexural Rigidity of Ribs:-

Dx = × = 22.36 ×106 ×2.8 ×109

1

Dx = 4.7 × 104

1500

Dy = × = 22.36 ×106 ×2.8 ×109

1

Dy = 4.7 × 104

Modulus of Shear:-

1500

G = 2 ( 1+ )

= 22.36 ×106 ( = 0.15 assume )

2 ( 1+0.15 )

G = 9.72 × 106 KN/m2

Torsional Constants ( Polar Sectional Modulus ):-

C1 = 1 (0.63 × (bw)) (bw3 × (D)) = 1 (0.63 × (0.175)) (0.1753 × (0.4)) = 5.18 × 104 m3

D 3 4 3

C2 = 1 (0.63 × (bw)) (D3 × (bw)) = 1 (0.63 × (0.175)) (0.43 × (0.175)) = 2.7 × 103 m3

D 3 0.4 3

Torsional Rigidity:-

Cx = ( G ×C1 ) = 9.72 ×106×5.18 ×104

= 3.35 × 103

b1 1.5

Cy = G×C2 = 9.72 ×106 ×2.7 ×103

= 1.75 × 104

a1 1.5

2H = Cx + Cy = ( 3.35 × 10³) + ( 1.75 × 104 ) = 2.09 × 104

Dx Lx4 =

Dy Ly4 =

4.75 × 104

7.54 = 13.9

4.75 × 104

7.54 = 13.9

2H 2.09 × 104

Lx2 × Ly2 = 7.52 × 7.52 = 6.60

Deflection Check:-

16 ×(Q) 2H

Dy 16 ×(15.15)

=

=

Dx + ( 2

2) + (

4) = + 6.60 + 13.19 = 7.65 mm

L×4

Lx ×Ly Ly

13.19

Long Term Deflections:-

Lt defl. = 3 × = 3 × 7.65 = 22.94 mm

Span = 7500 = 30 mm ( Lt defl. < Span

Safe.)

250 250 250

Maximum Moment and Shear Values:-

Mx = Dx × ( )2 × = 4.74 × 104 × (

2

) × 7.65 = 56 KN-m

2

2

L× 7500

My = Dy × ( ) × = 4.7 × 104 × (

2

) × 7.65 = 56 KN-m

7500

Maximum Torsional Moment:-

7500

Mxy = Cx × 2 × = 3.35 ×103 ×2 ×7.65 = 5 KN-m

Ly ×Lx 7500 ×7500

Shear Force:-

Qx = (( Dx × ( )3 ) + ( Cy × ( 3

))) ×

Lx

= ((4.74 × 104 × (

a1 ×b12

3 3

) ) + ( 1.75 × 104 × (

p>2))) × 7.65 = 25 KN

Qy = ((Dy × (

7500

3 3

) ) + (Cx × (

))) ×

1500 ×1500

7500

= ((4.74 × 104 × (

a12 ×b1

3 3

) + (3.35 × 104 × (

))) × 7.65 = 25 KN

)

7500

15002

×1500

Reinforcement Details:-

Astx = Asty = 0.5 ×fck × (1 1 4.6Mu ) × b × d ( as Ly = Lx = 7.5 m )

fy fck.b.d2

= 0.5 ×20 × (1 1 4.6 ×56 ×106

) × 175 × 370

415

Astx = Asty = 499.29 mm2 Use 12 mm bars

Provide 5 bars of 12 mm .

20 ×175 ×3702

Provide minimum steel in slab portion i.e. flange of waffle slab. Ast min. = 0.12 × b × d = 0.12 × 1000 × 100

100

Ast min. = 120 mm2

Given:-

Size of Grid = 15 × 7.5 m

100

IX. DESIGN OF WAFFLE SLAB SIZE = 15 × 7.5 m

Spacing of ribs = 1.5 m a1 = 1500 mm ; b1 = 1500 mm M20, fck = 20 N/mm²

Fe415, Fy = 415 N/mm² Live Load, LL = 4 KN/m² Floor Finish, FF = 1 KN/m²

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 × 1000 = 5000 × 20 × 1000 E = 22.36 × 106

Load Calculation:-

Total weight of slab = 25 × Df × Lx × Ly

= 25 × 0.10 × 7.5 × 15 = 281.25 KN

150

480

Total weight of beams in X Direction = 25 × bw × D × Nx × Lx = 25 × 0.150 × 0.580 × 6 × 7.5 = 97.88 KN Total weight of beams in Y Direction = 25 × bw × D × Ny × ( Ly ( bw × Nx ) )

= 25 × 0.15 × 0.25 × 11 × ( 15 ( 0.15 × 6 ) ) = 337.34 KN Total Live Load, LL = LL × Lx × Ly = 4 × 7.5 × 15 = 450 KN

Total Floor Finish, FF = FF × Lx × Ly = 1 × 7.5 × 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 ×15

Total Factored load / m2 = Q = 1.5 × q = 1.5 × 11.37 =17.05 KN/m2

Design Parameters:-

Df = 100 = 0.17

D 580

bf = 1500 = 10

bw 150

Moment of Inertia:-

I = Kx ×bw ×D3

12

= 2.3 ×150 ×5803 ( Kx = 2.3, SP – 16, Tb. 88 )

12

I = 5.60 × 109 mm4 = 5.61 × 103 m4

Flexural Rigidity of Ribs:-

Dx = E×I = 22.36 ×106 ×5.60 ×109 = 8.35 × 1013

a1 1500

Dy = E ×I = 22.36 ×106 ×5.60 ×109 = 8.35 × 1013

b1 1500

Modulus of Shear:-

G = E 2(1+)

= 22.36 ×106

2(1+0.15)

( = 0.15 assume ) = 9.72 × 106

Torsional Constants ( Polar Sectional Modulus ):-

C1 = 1 0.63 (bw) (bw3 × (D)) = 1 0.63 (150) (1503 × (580)) = 5.46 × 104 m3

d 3 580 3

C2 = 1 0.63 (bw) (D3 × (bw)) = 1 0.63 (150) (5803 × (150)) = 8.17 × 103 m3

D 3 580 3

Torsional Rigidity:-

Cx = G ×C1 = 9.72 ×106 ×5.46 ×104 = 3.54 × 103

b1 1.5

Cy = G ×C2 = 9.72 ×106 ×8.17 ×103 = 5.29 × 104

a1 1.5

2H = Cx + Cy = ( 3.54 × 103 ) + ( 5.29 × 104) = 5.65 × 104

Dx Lx4 =

Dy Ly4 =

8.36 × 104

7.54 = 26.43

8.36 × 104

154 = 1.65

2H 5.65 × 104

Lx2 × Ly2 = 7.52 × 152 = 4.46

Deflection Check:-

16 ×(Q) 2H

Dy 16 ×(17.05)

=

=

Dx + (

2) + ( 4) = ( ) + 4.46 + 1.65 = 8.72 mm

Lx4

Lx ×Ly Ly

26.43

Long Term Deflection:-

Lt defl. = 3 × = 3 × 8.72 = 26.16 mm

Span = 7500 = 30 mm

250 250

Lt defl. < Span

250

Safe.

Maximum Moment and Shear Values:-

Mx = D × × ( )2 × = 8.64 × 104 × ( )2 × 8.72 = 128 KN-m

L× 7.5

My = Dy × ( )2 × = 8.36 × 104 × ( )2 × 8.72 = 32 KN-m

Ly 15

Maximum Torsional Moment:-

Mxy = C× × 2 ×

Lx × Ly

= 3.54 ×103 × 2 ×8.72 = 3 KN-m

7.5 ×15

Shear Force:-

Qx = ((D × × ( )3) + (Cy × ( 3 ))) ×

Lx a1 × b12

= ((8.36 × 104 × ( )3) + (5.29 × 104 × ( 3 ))) × 8.72 = 54 KN

7.5 1.5 ×1.52

Qy = ((Dy × ( )3) + (C × × ( 3 ))) ×

Ly a12 ×b1

= ((8.36 × 104 × ( )3) + ( 3.54 × 103 × ( 3 ))) × 8.72 = 8 KN

15 1.52 ×1.5

Reinforcement Details:- For X Direction,

Mu = 128 KN-m, b= 150 mm, d = 550mm

Astx = 0.5 ×fck × (1 1 ( 4.6Mu )) × b × d

Fy fck.b.d2

= 0.5×20 × ( 1 1 ( 4.6 ×128 ×106 )) × 150 × 550

415 20 ×150 ×5502

Astx = 810 mm2 Use 16 mm bars.

Provide 4 bars of 16 mm . For Y Direction,

Mu = 32 KN-m, b = 150 mm, d = 550 mm

Asty = 0.5 ×fck × ( 1 1 ( 4.6Mu )) × b × d

Fy fck.b.d2

= 0.5 ×20 × ( 1 1 ( 4.6 ×32 ×106 )) × 150 × 550

415 20 ×150 ×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 × × ) = (0.12 × 1000 × 100 )

100

Ast min. = 120 mm2

100

1. DESIGN SUMMARY

   	Size (m)	Slab Thickness ( mm )	Deflection ( mm )	Total Ast (mm2)	Max. Moment (KN-m)	Max. Shear (KN)	Factored Load () 2	Economy & Construction	Uses
   Waffle Slab	7.5 × 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 × 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 × 7.5	315	22.9	9010.5	284.11	1170.84	23.81	Suitable for medium spans, beam-less construction, difficulty while construction pods and post-tensioning, complicated designs	Beam-less construction, aesthetics, less complication while construction
   Flat Slab	15 × 7.5	600	–	–	–	–	–	Not for large spans, Suitable for medium spans, beam-less construction, difficulty while construction pods and post-tensioning, complicated designs	Beam-less construction, aesthetics, less complication while construction & suitable for medium spans
   Conventional RCC Slab	7.5 × 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.

2. 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

3. 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.

4. REFERENCE

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

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

3. Design of Waffle Slab – Ariel Mayer.

4. Analysis and Design of Waffle Slabs-2011-Jose Pablo Rosales Sanchez.

5. Waffle Slab Khairul Salleh Baharudin.

6. Structural Analysis , by S.S Bhavikatti.

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