DESIGN OF DECKSLABClear Span = 7.450 m Thickness of slab = 65 cm (assumed) Thickness of w.c = 75 mm Clear cover = 4.00 cm (as per clause 304.3.1 and table 10 of IRC bridge code 21-2000 section III ) Main reinforcement = 20 mm dia. HYSD bars conforming to IS-1786 (Deformed bars) Concrete Mix is M 20 grade Bearing(assumed) = 49 cm Effective depth = = 65 - 4.00 - 1.00 60 cm Effective Span As per clause 305.1.2 of IRC code 21-2000 Effective span shall be the least of the following i). Effective span where = l1 + d l1 = clear span = 7.45 m d = effective depth = 0.6 m Effective span = 7.45 + 0.6000 = 8.050 m ii) l = distance from centre of supports = 7.45 + 2x 0.490 2 = 7.940 m Effective span shall be least of (i) and (ii) Effective span = 7.940 m Carriage way width = 4.25 m (clause 113.1 of IRC 5-1985) Kerb width Width of slab = 0.225 m = 4.25 + 0.225 x 2 = 4.700 m Loading IRC class 'A' As per clause 303.1 and Table 6 of IRC code 21-2000, for M 20 Permissible flexural compressive stresses 66.67 c allowable = 6.667 M.Pa = 66.67 kg/cm2 Modular ratio = Es = 10.00 Ec grade RCC 00 + = 0.917 = 2000 Q = = 1/2*c*n*j 7.1 of IRC code 21-2000 Permissible tensile stress in steel for combined bending For steel S 415 t = 200 M.67 66.00 n = mc / (mc+t) n = 10.As per clause 303.00 x 66.640 .Pa m = 10.67 x 10.250 j = 1 n/3 = 1 0.2.25 /3 0. 940 = 0.5 0.72 1.2 of IRC code 6 Impact factor fraction = 4. = The effective span as indicated in clause 305.5 6 + 7.70 m lo = Effective span = 7.6 For b/lo x + 0.1 of IRC code 21-2000 Ratio b / lo where b = width of slab = 4.72 7.941 As per clause 305.70 7.94 0.65 0.940 m Bending Moment due to dead load Ratio = = 4. the effective width may be calculated in accordance with the following equation where bef bef lo a b1 = a (1 a/lo) +b1 = The effective width of slab on which the load acts.940 8 8 = 14224.323 Dead Load Bending Moment Weight of slab Weight of w.075 * * 2500 2400 = = 1625 kg/m 180 kg/m 1805 kg/m wl2 = 1805 x 7.2. i.592 for simply supported slab 1.As per clause 211.94 = b/lo = = 1.24 * 0.c Total dead load = = 0.21225 kg-m = 1422421.5 6+L where L is Effective span= 7.13.940 m Impact factor fraction = 4.1 = The distance of the centre of gravity of the concentrated load from the nearer support = The breadth of concentration area of the load.1.96 0.e the dimension .2(1) IRC 21-2000 Solid slabs spanning in one direction For a single concentrated load.1 0.092 1.3.225 kg-cm Live Load Bending Moment As per clause 305. 3 3 Class 'A' Train For maximum bending moment.269 1.4 0.of the tyre or track contact area over the road surface of the slab in a direction at right angles to the span plus twice the thickness of the wearing coat or surface finish above the structural slab.2 4.2 D 7.4) = 86.70 Left Dispersion calculated (bef/2) = 0.4) 25.7 11.8 4.099 0.4 + 11.7 t load bef = a( 1 a/lo) + b1 = 1.225 0.1 0.940 11.970 3.4 1.8 m Dispersion Widths do not overlap 0.4 x 3.941 a from A = a = 0.2 0.2 0. Position of Loads 2.40 t should be kept such that the resultant of the load system and the load under consideration should be equidistant from the centre of span.5 CG from D = 0.7 A C 11.8 0. the two loads of 11.8 6.970 7.2)+(11.398 .4 x 4.7 11.4 3.002 0.64 = (2.543 m < 1.198 2.869 A C D E B 3.671 a from B = lo = 7.4 3. = a constant depending upon the ratio b/lo where b is the width of the slab 2.4 1.2 6.4 R 11.7713 left dispersion means the possible dispersion on the left side from the centre 3.940 Dispersion Width Under 'C' 2.671 3.2 E B CG of the load sytem from C = (11.2 2.2 1.099 1.7 + 11.671 7.94 b1 = 0.2 11.35 m bef = 1. 65 m bef = 4.2/2 0.35 (1.00 + 0.of the left wheel of the vehicle.871 lo = 7.433 t 11.941 a = 3.543 0.246 m 1.225 + 0.94 b1 = 0.7713 here left is going beyond the slab edge.475 + /2 + Intensity of load under 'C' = Dispersion Width Under D bef = a (1 = 1.15 + 0.48 = Dispersion width under one wheel 1.323 ) 1.500 m Dispersion Widths overlap = 1.871 a from B = 4.4 t load a/lo) +b1 a from A = 3. hence the left dispersion is limited to 0.069 > 0 1.8 m .475 < 0.= kerb width + kerb to wheel gap + tyre width / 2 = = 0.246 = 1. 00 + 0.225 + 0.8 m Dispersion Widths overlap 0.869 lo = 7.099 1.4 ( 1.330 0.5 0.323 ) 4.941 a from A = 5.433 3.25 1.97 3.2 2.70 Left Dispersion calculated (bef/2) = 2.225 0.323 ) 4.675 m Intensity of load under 'D' = 11.2 0.206 /2 + 1.4 ( 1. hence the left dispersion is limited 0.625 < 2.00 + 0. hence the left dispersion is limited to 0.2 0. = kerb width + kerb to wheel gap + tyre width / 2 = 0.103 left dispersion means the possible dispersion on the left side from the centre of the left wheel of the vehicle.25 1.675 = 3.206 m > 1.250 here left dispersion means the possible dispersion on the left side from the centre of the left wheel of the vehicle.15 + 0.250 here left is going beyond the slab edge.8 0.6250 Combined dispersion width = 0.103 here left dispersion is going beyond the slab edge.8 = 4.528 m Intensity of load under 'E' = 11.5 4.869 C D E 3.071 a = 2.528 = 3.940 b1 = 0.5/2 0.94 B .225 0.671 3.65 m bef = 4.625 + 4.6250 Combined dispersion width = 0.225 + 0.226 t Dispersion Width Under 'E' 11.970 7.869 a from B = 2.625 < 2.8 0.330 t A 1.0.5 0.70 Left Dispersion calculated (bef/2) = 2.15 + 0.099 0.6250 + 4.500 /2 + 1.8 = 4.226 R 3. = kerb width + kerb to wheel gap + tyre width / 2 = = 0.002 0.2 1.5/2 0.5 4.4 t load bef = a (1 a/lo) +b1 = 1. 219 Ra = 4.433 + Rb = 3.869 + 7.805 t 2.184 t Rb = 1.9 = 3.226 + 3.330 - * 4.269 3.330 * 1.226 3.433 * 7.184 .9 = 33.Ra x Ra x Taking moments about 'B' 7.07 4. 917 x 60 2 = 23.L B.184 = = = Total B.L B.23 = 632807. shrinkage.910 cm As per clause 305.50 tjd 2000 x 0. Of distribution 12 mm HYSD Effective depth = 58.4 2 = 5.M = = = Effective depth required = = Steel steel at bottom Main steel reinforcement required = * 3.40 cm Ast required = M = 632807.M + Live Load B.2*D.1 of IRC code 21-2000 Resisting moment = 0.15 cm Distribution Steel As per clause 305.4993 kg-cm Assuming the dia.5 + 2583499 kg-cm M Q*b = 58.486 cm As per clause 305.871 ) .19 of IRC code 21-2000 Minimum area of tension reinforcement = 0. + 0.640 x100 < OK = ### 60 cm 2583499 2000 x 0. temperature etc.486 provide 20 mm dia.5 + 0.40 100 2 2 = 7.3 * L.80 cm < 23.917 x 58. HYSD bars at 130 mm c/c 2 OK Ast provided = 24.12 x 100 x 58.18.61078 t-m 1161078 kg-cm Dead load B.2 times the moment due to other loads such as dead load.008 cm > 5.2 1161077.12 * 100 * 65 * 100 2 cm2 = 7.153 cm M tjd ) 2583499 7.3x 1161077.196 4.910 cm . Mom.( 1.3 times the moment due to concentrated live loads plus 0. Mom.12 % of total cross sectional area On each face = 0.585 11.433 * 16.19 of IRC code 21-2000 Minimum distribution reinforcem = 0. = 0.12 % of total cross section area = 0.M 1422421. = 0.Maximum live load Bending Moment = ( 4.2x 1422421. 008 cm provide 12 mm dia. HYSD bars at 130 mm c/c 2 OK Ast provided = 8.2 Hence provide minimun distribution reinforceme 7.695 cm . 5 0.5/2 = 0.625 < 1.941 a from A = 0.4 t get max.850 a from B = 7.Top Reinforcement 2 Min.8 0. HYSD bars at 130 mm c/c Check for shear As per clause 305.008 cm Provide 12 mm dia. shear bef = a (1 a/lo) +b1 = 1.225 + 0.70 /2 = 0.3 D 3. Reinforcement = 7.25 + 2 ( 0.590 E 3.25 1.3 of IRC code 21-2000 Dispersion of loads along the span Longitudinal dispersion = The effect of contact of wheel or track load in the direction of span length shall be taken as equal to the dimension of the tyre contact area over the wearing surface of the slab in the direction of the span plus twice the overall depth of the slab inclusive of the thickness of the wearing surface.85 m from the support to get max.940 b1 = 0.225 0. Longitudinal dispersion = 0.40 t may be kept at 1.970 7.090 lo = 7.2 C 4.15 + 0. shear 11.4 6.075 ) = 1.65 m bef = 2.4 11.13.85 1.65 + 0.85 m from the support to Dispersion Width Under 'C' 11.8 m Dispersion Widths overlap 0.15 0.0616 left dispersion means the possible dispersion on the left side from the centre of the left wheel of the vehicle. hence the left dispersion is limited to 0.70 m so 11 t load may be k 0. = kerb width + kerb to wheel gap + tyre width / 2 = = 0.940 B The load of 11.625 .5 4.970 A 1. HYSD bars at 130 mm c/c Also provide distribution steel of 12 mm dia.123 m > 1.70 Left Dispersion (bef/2) = 1.8 t 0.0616 here left is going beyond the slab edge.850 a = 0. 625 + 2.487 m 1.Combined dispersion widt= = 0.8 .123 /2 + 3. 6250 + 3.226 m Intensity of load under 'D' = 11.864 m Intensity of load under 'E' = 6.0 + 0.2 0.941 a from A = 6.8009 left dispersion means the possible dispersion on the left side from the centre of the left wheel of the vehicle.00 + 0. = kerb width + kerb to wheel gap + tyre width / 2 = 0.323 ) 3.05 a = 2.565 < 1.8 t bef = a (1 a/lo) +b1 = 1.80 = 4.590 lo = 7.53 m bef = 2.38 0.998 /2 + 1.4 ( 1.15 + 0.94 b1 = 0.890 lo = 7. hence the left dispersion is limited to 0.00 + 0.602 m > 1.4 ( 1.8 = 3. = kerb width + kerb to wheel gap + tyre width / 2 = = 0.8 m Dispersion Widths overlap Left calculated Dispersion (bef/2) = 1.487 4.8 0.565 Combined dispersion width = 0.8 m Dispersion Widths overlap 0.4991 here left is going beyond the slab edge.940 b1 = 0.565 + 2. hence the left dispersion is limited to 0.94 a from A = 2.590 a from B = 1.38/2 = 0.19 1.050 a from B = 5.65 m bef = 3.998 m > 1.4991 left dispersion means the possible dispersion on the left side from the centre of the left wheel of the vehicle.325 t Dispersion Width Under 'D' 11.4 t bef = a (1 a/lo) +b1 = 1.864 .6250 Combined dispersion width 0.8009 here left is going beyond the slab edge.323 ) 3.Intensity of load under 'C' = = 11.350 a = 1.5 4.226 = 3.225 0.15 + 0.602 /2 + 1.5/2 0.8 ( 1.323 ) 4.569 t Dispersion Width Under 'E' 6.225 + 0.625 < 1.225 + 0.70 Left Dispersion calculated (bef/2) = 1. 2 D 3.328 4.3 1.59 E 3.970 2.940 B .569 0.328 t 4.= 2.85 C A 1.325 3.970 7. 328 Ra x 7.975 t 3.1.325 + 2.09 1.Ra Taking moments about 'B' x 7.9 kgs 6975.385 6.1 of IRC 21-2000 Max.9 9. or the breadth of the rib in the case of flanged beam 14141 kgs 100 cms 60 cms = 14141 100x 60 = 2.36 kg/cm2 As per clause 304.1 of IRC 21-2000 τc = Design shear stress where V d b = = = here V b d = = = 7165. permissible shear s 20 τcmax = 1.690 Kg/cm > 2 2.94 = 4.154 cm 100 Ast / b d from Table 12B of IRC:21-2000 τc = 0.94 Ra Rb Shear due to dead load = = = = = * * 7.0 0.403 grade concrete 0.690 = 2 2.940 + 3.569 * 5.8 N/mm2 for solid slabs the permissible shear stress shall not exceed half the value of τcmax given in Table 12 A τcmax Hence = 0.7.59 55.5 x 1.850 kgs V bd The design shear across the section The depth of the section The breadth of the rectangular beam or slab.7.00 (Table 12C of IRC:21-2000) k* τc= 1. permissible stress = = Ast provided = N/mm2 kg/cm2 2 24.301 7165.269 N/mm2 2 = 2.7.2 of IRC:21-2000) k for solid slab of 65 cm thick = 1.1.8 Hence max.0 x = 2.89 2 Total shear due to dead load & live loa = = As per clause 304.3.690 kg/cm τc For solid slabs the permissible shear in concrete k * (vide cl.246 t wl 2 1805 x 7.45 + 14141. 304.36 kg/cm . .Hence safe against shear.
Report "SLRB Design of 7.45m clear span for IRC Class A loading"