What Is a Slab?
A slab is constructed to provide flat surfaces, typically horizontal, in building roofs, floors, bridges, and other types of structures. The slab could be supported by walls, by reinforced concrete beams normally cast monolithically with the slab, by structural steel beams, either by columns or from the ground.
A slab is a plate element having depth (D), very small as compared to its length and width. A slab is used as floor or roof in buildings, carry distribution load uniformly.
Type of Slab b
Slab May Be
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Type of slabs based on support conditions are:

 One Way Slab
 Two Way Slab
 Flat Slab Resting Direction on a Column Without Beam
 Grid Slabs or Waffle Slab
 Circular Slab and Other Shapes
What Is a OneWay Slab?
One Way Slab

 The most straightforward routine structural element for illustration of design provisions from the Code is that the oneway slab.
 A oneway slab is defined for functions of the book as a flexural member with thickness small relative to other dimensions, supporting (gravity) loads applied normal to and directly above its surface, a span in one direction between parallel supports, and fortified for flexure in this direction only.
 For purposes of analysis, oneway slabs might be restrained to some degree in the supports or possibly unrestrained. A number of Code provisions reference to “flexural members,” including one and – twoway slabs, beams, girders, footings, as well as where bending is present together with the axial walls, load, and columns.
 In general, when this code provision is intended to use to oneway slabs, the term is going to be utilized in the sense of this definition herein.
 If the is supported on two opposite sides, it is called a oneway spanning slab. In this type of slab, loads are transferred on two opposite as per the above figure. A
 If the slab is supported at four sides, and if Ly/Lx ≥ 2 one way spanning slab.
 For any slab, if Ly = Lx, the slab has a tendency to bend in both directions. With an increase of Ly, the tendency of bending along Ly is reduced and that on Lx is increased.
 When Ly/Lx ≥ 2, the slab bends only in Xdirections
 When Ly/Lx ≥ 2, the slab is called a oneway slab. In oneway slab, the main reinforcement is provided along Lx ( ShortSpan)
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What Is a TwoWay Slab?

 The design considerations of wallsupported twoway slabs are similar to those pertaining to oneway slabs.
 The thickness of the slab is generally based on deflection control criteria, and the reinforcements in the two orthogonal directions are designed to resist the calculated maximum bending moments in the respective directions at the critical sections. [Additional reinforcement may be required at the corners of twoway slabs in some cases, as explained later].
 The slab thickness should be sufficient against shear, although shear is usually not a problem in twoway slabs subjected to uniformly distributed loads.
 If the slab is supported at all four edges and if Ly/Lx < 2,
 The tendency of the slab is to bend in both directions. Such slab is called twoway slab as per above figure c
 In two way slabs, main reinforcement is provided along Lx as well as Ly direction
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What Is the Flat Slab?
Flat Slab

 When the slab is directly supported on a column, without beams, it is known as a flat slab.
 Flat slab is provided to increase the floor height and to permit a large amount of light which might be obstructed by the depth of beams.
What Is Grid Slab?
Grid Slab

 When the slab is supported on beams with column only on the periphery of the hall, the slab is called grid slab.
 Sometimes, in large halls, public places, marriage halls, auditoriums, etc. a large columnfree area is required.
 In these cases, large deep beams may be permitted, but the columns are permitted only on the periphery.
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Analysis of Slab:

 Slabs are primarily flexural members as beam and are analyzed and designed in the same manner as the beams. The analysis may be carried out as follows:

Elastic Analysis:
 A strip of 1 m width of the slab is considered, and loads are found on this strip. This strip id analyzed as a beam 1 m width.

Code Coefficients:
 This is a semiempirical method of analysis based on yield line theory. The coefficients give in code may be directly used to analyze the slabs.
 However, the redistribution of moments is not permitted in this case.

Yield Line Theory
 This is a limit state design or collapse load method developed by Johanson.

 Slabs are primarily flexural members as beam and are analyzed and designed in the same manner as the beams. The analysis may be carried out as follows:
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Difference Between One Way Slab and Two Way Slab
One Way Slab  Two Way Slab 
One way slab two opposite side support beam /wall  Two Way Slab four side mins all side supported beam /wall 
One way slab is bend only in one spanning side direction while load transfer  Two way slab is bend both spanning side direction while load transfer 
Main Reinforcement is in provide short span due to banding  Main Reinforcement is in provide short span due to banding 
Ly/Lx ≥ 2 one way slab spanning  Ly/Lx < 2 two way slab spanning 
Oneway slab near about 100mm to 150mm based on the deflection  twoway slabs is in the range of 100mm to 200mm depending upon 
one way slab economical near about 3.5 m  Twoway slab may economical for the panel sizes near about 6m x 6m. 
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Design Considerations:

One Way Slab
1. Effective Depth (d)


 For deflection control
 L/d = 20 X M.F
 M.F. Modifiction factor from— IS: 456, p.38.Fig4
 Assume % steel 0.3 to 0.6%
 Fs = 0.58 Fy X (Ast requierd / Ast Provied)
 Initially assume that Ast reqierd = Ast Provided
 Fy = 250 N/ Sq.mm —– Fs = 0.58 X 250 = 145 N/ sq.mm.
 Fy = 415 N/ Sq.mm —– Fs = 0.58 X 415 = 240 N/ sq.mm.
 Fy = 500 N/ Sq.mm —– Fs = 0.58 X 500 = 290 N/ sq.mm.

2. Effective Span:


 ClearSpan + d
 c/c of Supports

Whichever is smaller ——– as per IS 4562000 P. 34, CI 22.2.a
3. Reinforcement Requirements

 Minimum reinforcement
 For Fe250 Pt = 0.15 % of total C/s area (d x D)
 For Fe415 Pt = 0.12 % of total C/s area (d x D)
 For Fe500 Pt = 0.12 % of total C/s area (d x D) ——– as per IS 4562000 P. 48, CI 26.5.2.1
 Maximum diameter (Sp 34)
 For minbar:
 Plain bars———–10 mm Ø min dia
Deformed bars—–8 mm Ø min dia
 Plain bars———–10 mm Ø min dia
 For Distribution bars:
 Plain bars———–6 mm Ø min dia
Deformed bars—–6 mm Ø min dia
 Plain bars———–6 mm Ø min dia
 For minbar:
 Minimum reinforcement
4. Check for Cracking

 For Min Steel:
 3d ——— Where. d = Effective depth
 300mm
 For Min Steel:
Spacing should not exceed smaller these two values.

 For Distribution steel:
 5 d
 450mm
 For Distribution steel:
Spacing should not exceed smaller these two values. —— IS: 4562000, P.46
5. Check for Deflection:

 Allowable L/d = 20 X M.F.
 M.F is Obtained from IS:456200 P38 Fig 4
 Find actual, L/d
 If Actual L/d < allowable L/d ——— Ok
 Allowable L/d = 20 X M.F.
6. Check for Development Length (Ld)
IS 4562000,P.44, Cl. 26.2.3.3 C

 Ld should be ≤ 1.3 (M1/V) + L0
Where

 Ld = (Ø.σs / 4 τ bd )—————–σs = 0.87 fy As per IS 456200, P.42

 50 % of steel is bent up near support. Therefore find M.R for 50 % of steel only
 M1 = M.R. for 50% steel support
 V = Shear force at the support
 L0 = Sum of anchorage beyond the center of support

 d
 12 Ø
Take L0 as the smaller of two values.
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Two Way Slab
1. Effective Depth (d):


 For deflection control
 L/d = 35 X M.F X 0.8
 M.F. Modifiction factor from— IS: 456, p.38.Fig4
 Assume % steel 0.3 to 0.6%
 Fs = 0.58 Fy X (Ast requierd / Ast Provied)
 Initially assume that Ast reqierd = Ast Provided
 Fy = 250 N/ Sq.mm —– Fs = 0.58 X 250 = 145 N/ sq.mm.
 Fy = 415 N/ Sq.mm —– Fs = 0.58 X 415 = 240 N/ sq.mm.
 Fy = 500 N/ Sq.mm —– Fs = 0.58 X 500 = 290 N/ sq.mm.

2. Effective Span:


 ClearSpan + d
 c/c of Supports

Whichever is smaller ——– as per IS 4562000 P. 34, CI 22.2.a
3. Load Calculations:


 Total Load = D.L. + F.L. + L.L.
 Dead load of slab = (d x 25)
 Floor Finishing load = (as floor finishing near 1 kn/sq.mm)
 Live load = ( as per calculation)
 Factor Load = 1.5 x Total load
 Total Load = D.L. + F.L. + L.L.

4. Mid Span Moment:
Corners not held down conditions is given as per IS: 4562000 P90 CI D2

 Mx = ax . w . lx . lx
 My = ay . w . lx. lx
 ax and ay coefficientare obtained from IS: 456 table 26, fig 10.3 shoe nine separate possible arrangement for two way restrained slab.
5. Effective Depth of Flexure:

 Mu = 0.138 . fck . b.d.d
 Heaer find d
 Mu = Sp 16 P 10 Table C
 Fck = strength of concrete
 b = 1 m area required load
 Mu = 0.138 . fck . b.d.d
6. Reinforcement in Mild Strip :

 Along Lx
 Pt= 50 (fck/fy)x( 1√(1(4.6xMu/Fck b.d.d)))
 fck = strenth of concrete
 fy = 415 N/Sq.mm
 Mu = Sp 16 P 10 Table C =0.138 . fck . b.d.d= Mx
 Pt= 50 (fck/fy)x( 1√(1(4.6xMu/Fck b.d.d)))
 Along Lx
 Pt= 50 (fck/fy)x( 1√(1(4.6xMu/Fck b.d.d)))
 fck = strenth of concrete
 fy = 415 N/Sq.mm
 Mu = Sp 16 P 10 Table C =0.138 . fck . b.d.d= My
 Pt= 50 (fck/fy)x( 1√(1(4.6xMu/Fck b.d.d)))
 Along Lx
7. Check for Cracking:

 Along Lx:
 3d ——— Where. d = Effective depth
 300mm
 Along Lx:
Spacing should not exceed smaller these two values.

 Along Ly:
 3d ——— Where. d = Effective depth
 300mm
 Along Ly:
Spacing should not exceed smaller these two values.
8. Check for Deflection:

 Allowable L/d = 35 X M.F.X 0.8
 M.F is Obtained from IS:456200 P38 Fig 4
 Find actual, L/d
 If Actual L/d < allowable L/d ——— Ok
 Allowable L/d = 35 X M.F.X 0.8
9. Check for Development Length:
IS 4562000,P.44, Cl. 26.2.3.3 C

 Ld should be ≤ 1.3 (M1/V) + L0
Where

 Ld = (Ø.σs / 4 τ bd )—————–σs = 0.87 fy As per IS 456200, P.42

 50 % of steel is bent up near support. Therefore find M.R for 50 % of steel only
 M1 = M.R. for 50% steel support
 V = Shear force at the support
 L0 = Sum of anchorage beyond the center of support

 d
 12 Ø
Take L0 as the smaller of two values.
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