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In this blog, you will learn step by step about the manual design of a two way slab for one floor of the G + 5 building procedure using I.S. 456:2000. For daily blogs, subscribe to our blog page and learn complete information about the structural engineering industry Given :- 1) Grade of Concrete = M202) Grade of Steel = Fe5003) Clear Cover to Reinforcement (c) = 20 mm4) Dimension of Slab =  (3050mm X 4180 mm)5) Assume Beam Support (Bearing) = 230 mm6) Unit Weight of Concrete = 25 kN/m2 Step 1 :- Aspect Ratio (Refer Cl. No. D-1.11, Page No. 90 , I.S. 456:2000) Lx = 3050 mm & Ly = 4180 mm  …(Given)Therefore, Ly/Lx = 4180/3050 = 1.37 < 2  …(Two Way Slab) Step 2 :- Depth of Slab (D) Basic Value (B.V.) = L / D = 26Assume % of Steel = 0.25 %Therefore, Modification Factor (M.F.) = 1.35 …(Refer Fig. 4 , Page No. 38 , I.S.456:2000)Therefore, Effective Depth (d) = SPAN/B.V.X.M.F =  [3050 / (26 X 1.35)] = 86.894 mm Assume Bars to be Provided of Diameter = 8 mm & Provide Clear Cover to Reinforcement = 20 mmTherefore, Overall Depth of Slab (D) = d + c + (8/2) = 86.894 + 20 + 4 = 110.894 ≈ 125 mmTherefore, Effective Depth Provided (dprovided) = D – c – (8/2) = 125 – 20 – 4 = 101 mm Step 3 :- Effective Span (le)                Minimum of,       i) Le = Lx + Bearing = 3050 + 230 = 3280 mm                                          ii) Le = Lx + d = 3050 + 101 = 3151 mm               Therefore,  Effective Span (Le) = 3151 mm Step 4 :- Load Calculations Consider 1 m Strip of Slab,i) Self Weight of Slab (DL) = D X Unit Weight of Concrete                                             = 0.125 X 25                                             = 3.125 kN/m2ii) Live Load (LL)               = 2 kN/m2  …[Refer Table No. 1 , Page No. 7 , I.S. 875  (Part 2) : 1987]iii) Super Imposed Dead Load (SIDL) =      Floor Finish = Wt. of Screeding (50 mm Thk.) + Flooring (10 mm Thk.)                          = (0.05 X 24) + (0.01 X 22) …[Refer I.S. 875 (Part 1) : 1987]                          = 1.42 kN/m2                          ≈ 1.5 kN/m2 Total Load (W) = 3.125 + 2 + 1.5                          = 6.625 kN/m2                          ≈ 6.63 kN/m2Total Ultimate Load (Wu) = 6.63 X 1.5                                           =  9.945 kN/m2                                            ≈ 9.95 kN/m2 Step 5 :- Maximum Design Bending Moment (Mumax) Ly/Lx 1.3 1.37 1.4 -αx 0.065 ? 0.071 +αx 0.049 ? 0.053 Refer Table No. 26, Page No. 90 , I.S. 456:2000 Shorter Span Moments :- 1) -ve Moment at Continuous Edge :- By Interpolation, -αx = 0.065+[{(0.071-0.065) / (1.4-1.3)} X (1.37 – 1.3)]      = 0.0692 Mu = -αx . Wu . le2     = 0.0692 X 9.95 X  3.1512        =  6.836 kN.m 2) +ve Moment at Mid-Span :-By Interpolation,+αx = 0.049+[{(0.053-0.049) / (1.4-1.3)} X (1.37 – 1.3)]      = 0.0518 Mu = +αx . Wu . le2     = 0.0518 X 9.95 X  3.1512       =  5.117 kN.m Longer Span Moments :- 3) -ve Moment at Continuous Edge :-Mu = -αy . Wu . le2     = 0.047 X 9.95 X  3.1512        = 4.64 kN.m 4) +ve Moment at Mid-Span :-Mu = +αy . Wu . le2     = 0.035 X 9.95 X  3.1512       = 3.46 kN.m Therefore, Maximum Design Bending Moment (Mumax) =  6.836 kN.m Step 6 :- Check For Depth               Equate   Mumax & Mulim,              Mumax =  Mulim             6.836 X 106 = 0.133 X fck X b X dreq2              6.836 X 106 = 0.133 X 20 X 1000 X dreq2dreq= 50.7 mm  < 101 mm   …(dreq< dprovided)              Therefore,  Safe Step 7 :- Area of Steel Calculation :- Spacing of Bars :- As per Cl. No. 26.3.3 (b) , Page No. 46 , I.S. 456 : 2000Maximum SpacingBetween Bars          < Minimum of   i) 3d                                                          ii) 300 mmTherefore,Maximum Spacing Between Bars = i) 3 X 101 = 303 mm                                                         ii) 300 mm Note:- Generally We Provide 200 mm Maximum Spacing Between Bars. Spacing of Bars = Area of one bar/ Area of steel * 1000 Along Shorter Span (Lx)  :- 1) At Support :-          = 162.18 mm2Spacing of Bars = (Area of One Bar / Area of Steel) . 1000Consider Bars to Provide are of Diameter 8mm.Therefore, Spacing of Bars = [(π/4) X 82 / 162.18] X 1000 = 3709.98 mmNote:- Generally We Provide 200 mm Maximum Spacing Between Bars. 2) At Mid-Span :-          = 120.1 mm2Spacing of Bars = (Area of One Bar / Area of Steel) . 1000Consider Bars to be Provide are of Diameter 8mm.Therefore, Spacing of Bars = [(π/4) X 82 / 120.1 ] X 1000 = 418.58 mmNote:- Generally We Provide 200 mm Maximum Spacing Between Bars. Along Longer Span (Ly)  :-   d =  dprovided – diameter of bar provided along shorter span = 101 – 8 = 93 mm 1) At Support :-          = 118.53 mm2Spacing of Bars = (Area of One Bar / Area of Steel) . 1000Consider Bars to be Provide are of Diameter 8mm.Therefore, Spacing of Bars = [(π/4) X 82 / 118.53] X 1000 = 424.13mmNote:- Generally We Provide 200 mm Maximum Spacing Between Bars. 2) At Mid-Span :-          = 87.74 mm2Spacing of Bars = (Area of One Bar / Area of Steel) . 1000Consider Bars to be Provide are of Diameter 8mm.Therefore, Spacing of Bars = [(π/4) X 82 / 87.74] X 1000 = 572.97 mmNote:- Generally We Provide 200 mm Maximum Spacing Between Bars. Note:- Calculation of Distribution Steel is Eliminated, Since it is a Two Way Slab. Step 8 :- Check For Deflection :- % of Steel = Pt = 100. Astreq / b.dTherfore, Pt = (100 X 120.1) / (1000 X 101) = 0.11 %Therefore,  Fs = 0.58 X 500 X (120.1 / 251.36) = 138.56  N/mm2% of Steel Provided = Pt = 100. Astprovided  / b.dTherefore , Pt = (100 X 251.36) / (1000 X 101) = 0.248 %M.F. = 2.0Now,