DESIGN OF SLAB INTERIOR PANEL SLAB: Step 1: Room size 4.23mx5.23m Ly/Lx= 5.23/4.23 =1.24<2. Therefore two way slab used. Cover for mild exposure= 25mm As the span is large and loading heavy , assume a span depth ratio of 25 instead of usual 35. d = 4230/25 =169.2mm=170mm h
= 170+5+15=190mm.
Step 2: Load calculation Dead load = 0.19x25x1= 4.75 kN /m^2 Water proof = 2kN/m^2 Floor finish = 1kN/m^2 Live load = 1.5kN/m^2 Total load = 9.25kN/m^2 Factored load = 1.5x9.25 = 13.875kN/m^2. Step 3: Checking for depth: Ly/Lx = 5.23/4.23 = 1.24. αx = 0.045. αy = 0.032. 1
Mx = αx.w.lx^2 = 0.045x13.875x4.23^2 =11.17kN.m My = αy.w.lx^2 =0.032x13.875x4.23^2 =7.94 kN.m M max = 0.138.f ck.b.d^2 d^2 = M max/ 0.138.f ck.b = 11.17x10^6/(0.138x25x1000) d = 56.9< 170 mm hence safe. Step 4: Check for shear, Ly/Lx = 1.24 Co-efficient for shear = γx in ly direction= 0.33 =γx in lx direction = 0.393 Max .design shear = γx.w.lx = 0.398x13.875x4.23 =23.4kN. V/bd = 23.4x10^3/(1000x170) = 0.138 Safe minimum: shear for M25 concrete is equal to 0.29 and hence slab is safe and shear.
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Step5: Area of steel, d = 170mm.
steel for short direction. M/bd^2 = (11.17x10^6)/(1000x170^2). = 0.39 {From SP 16, table3:} Pt = 0.11% As=(0.11x1000x170)/100 = 187.34mm^2. spacing = (π/4 x 10^2)/187.34 = 419mm. Spacing = 400mm. Provide 10mm dia @ 400mm c/c.
Steel for longer direction, d = 170-10=160mm. M/(bd^2) = 7.94x10^6/(1000x160^2) = 0.087. Ast = (0.087x1000x160)/100 = 139.2 mm^2. Spacing = ((π/4 x10^2)x1000)/139 = 565 mm. Spacing = 550mm. Provide 10 mm dia @ 300 mm c/c. 3
Step 6: Check deflection , Span depth ratio = 20. Pt = 0.11%. F1 = 2. Allowable l/d ratio = 20x2 = 40. Actual span/depth ratio = 4.23/0.170 = 24.9. There fore assumed span depth ratio is enough to control deflection.
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INTERIOR PANEL SLAB Step 1: Thickness of slab and durability consideration Lx = 4230mm. Ly = 5230mm. Cover for mild exposure= 15mm As the span is large and loading heavy , assume a span depth ratio of 25 instead of usual 35. d= 4230/25 = 169.2mm = 170mm. h = 170+6+15=191mm. step 2: Design of loads.
Dead load= 0.19x25x1= 4.75 kN /m^2 Floor finish = 1kN/m^2 Live load = 4 kN/m^2 Total load = 9.775kN/m^2 Factored load = 1.5x9.775 = 14.66kN/m^2. Step 3: Checking for depth: Ly/Lx = 5.23/4.23 = 1.24. From the table , αx = 0.045. αy = 0.032. Mx = αx.w.lx^2 5
= 0.045x14.66x4.23^2 =11.8kN.m My = αy.w.lx^2 =0.032x14.66x4.23^2 =8.394 kN.m Check for depth from maximum. bending moment M max = 0.138.f ck.b.d^2 d2 = M max/ 0.138.f ck.b = 11.8x10^6/(0.138x25x1000) d = 58.5< 170 mm hence safe.
Step 4: Check for shear, Ly/Lx = 1.24 Co-efficient for shear γx in ly direction= 0.33 γx in lx direction = 0.393 Max .design shear = γx.w.lx = 0.398x14.66x4.23 =24.687kN. V/bd = 24.68x10^3/(1000x170) = 0.145 Safe min: shear for M25 concrete is equal to 0.29 and hence slab is safe and shear.
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Step5: Area of steel, steel for short direction. d = 170mm. m/bd2 = (11.8x10^6)/(1000x170^2). = 0.408. {From sp 16, table3:} Pt = 0.11% As=(0.11x1000x170)/100 = 187.34mm^2. spacing =1000x( (π/4 x 10^2)/187.34) = 419mm. Spacing = 400mm. Provide 10mm dia @ 400mm c/c. Steel for longer direction, d = 170-10=160mm. M/(bd^2) = 8.394x10^6/(1000x160^2) = 0.328. Pt = 0.092. Ast = (0.092x1000x160)/100 = 147.2mm^2. Spacing = ((π/4 x8^2)x1000)/147 = 340 mm. Spacing = 340mm. Provide 8 mm dia @ 340 mm c/c.
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Step 6: Check deflection : Span depth ratio = 20. Pt = 0.11%. F1 = 2.0 Allowable l/d ratio = 20x2 = 40. Actual span/depth ratio = 4.23/0.170 =24.88. There fore assumed span to depth ratio is enough to control deflection.
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ONE LONG EDGE DISCONTINUOUS SLAB Step 1: Thickness of slab and durability consideration Lx = 4230mm. Ly = 5230mm. {From IS 456-2000}
Cover for mild exposure= 15mm
As the span is large and loading heavy , assume a span to depth ratio of 25 instead of usual 35. d= 4230/25 = 169.2mm = 170mm. h = 170+6+15=191mm. step 2: Design of loads :
Dead load = 0.19x25x1= 4.75 kN /m^2 Floor finish = 1kN/m^2 Live load = 4 kN/m^2 Total load = 9.775kN/m^2 Factored load = 1.5x9.775 = 14.66kN/m^2. Step 3: Checking for depth: Ly/Lx = 5.23/4.23 = 1.24. From the table 26 {IS 456-2000} αx = 0.054. αy = 0.037. Mx = αx.w.lx^2 9
= 0.054x14.66x4.23^2 =14.16N.m My = αy.w.l-+^2 =0.037x14.66x4.23^2 =9.7 kN.m Check for depth from max. bending moment M max = 0.138.f ck.b.d^2 d^2 = M max/ 0.138.f ck.b = 14.16x10^6/(0.138x25x1000) d= 64.1< 170 mm hence safe.
Step 4: Check for shear, Ly/Lx = 1.24 Co-efficient for shear
{from IS 456-2000}
γx in ly direction= 0.33 γx in lx direction = 0.393 Max .design shear = γx.w.lx = 0.398x14.66x4.23 = 24.687kN. V/bd = 24.68x10^3/(1000x170) = 0.145 Safe min: shear for M25 concrete is equal to 0.29 and hence slab is safe and shear.
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Step5: Area of steel, steel for short direction. d = 170mm. M/bd2 = (14.16x10^6)/(1000x170^2). = 0.49. {From sp 16, table3:} Pt = 0.138% As=(0.138x1000x170)/100 = 234.6 mm^2. spacing =1000x( (π/4 x 10^2)/234.6) Spacing = 300mm. Provide 10mm dia @ 300mm c/c.
Steel for longer direction, d = 170-10=160mm. M/(bd2) = 9.7x10^6/(1000x160^2) = 0.34. {from IS 456-2000}
Pt = 0.096. Ast = (0.096x1000x160)/100 = 163.2mm^2. Spacing = ((π/4 x8^2)x1000)/163 = 308 mm. Spacing = 300mm. Provide 8 mm dia @ 300 mm c/c.
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Step 6: Check deflection : Span depth ratio = 20. {from IS 456-2000}
Pt = 0.138%. F1 = 2.0
Allowable l/d ratio = 20x2 = 40. Actual span/depth ratio = 4.23/0.170 =24.88. There fore assumed span depth ratio is enough to control deflection.
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DESIGN OF BEAMS DESIGN OF T- BEAM:
Step 1: Datas L = 5.23m. Spacing =4.23m. Df = 190mm. Fck = 20 N/mm^2 Fy = 415 N/mm^2 Step 2: Cross sectional dimensions, Basic span/depth ratio for simply supported beam is 20 For tee beams, assuming the width of rib = 230mm and flange width = 4.23m. the ratio of web width to flange width is equal to 230/4230 = 0.054. Reduction factor = 0.8. Hence, basic span/depth = 20x0.8 = 16. d = span/16 5230/16 = 326.88mm. Adopt overall depth = 400mm Cover = 50mm. Hence tee beam parameters are, d = 350mm D = 400mm 13
[from fig 6 IS 456]
bw = 230mm Df = 190mm Step 3: Loads Self weight of slab = 0.19x25x4.23 = 20.09kN/m Floor finish = 1x4.23 = 4.23 kN/m Self weight of rib = 0.23x0.21x25 = 1.21kN/m Plaster finish = 0.45 kN/m Water proofing = 2x4.23=8.46 kN/m Weight of parapet wall = 0.6x0.23x19 = 2.62 kN/m Total dead load = 37.06 kN/m Live load = 1.5 kN/m Design ultimate load = 1.5x(37.06+1.5) =57.84 kN/m Step 4: Ultimate moment and shear force Mu = (wl^2)/8 = (57.84x5.23^2)/8 = 197.76kN.m Vu = wl/2 = (57.84x5.23)/2 = 151.25kN
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Step 5 : Effective width of flange 1) bf = [Lo/6+bw+6Df] = [5.23/6+0.23+(6x0.19)] = 2.24m 2) c/c of rib = (4.23-0.23) = 4m Hence the least value of 1 & 2is bf = 2.42m. step 6: moment capacity of flange Muf = 0.36*fck.bf.df.(d-0.42df) = 0.36x20x2242x190x(350-0.42x190) = 828.72kN. Since Mu < Muf, Xu < Df Hence the section is consider as rectangular width b =bf Step 7: Reinforcements Mu = 0.87.fy.Ast.d(1-[fy.Ast/bdfck]) 197.76x10^6= 0.87x415xAstx350x(1-415xAst/2242x350x20) Ast = 1635.67mm^2. Provide 4 numbers of 25mm dia (Ast= 1963mm^2) and 2 hanger bars of 12mm dia on compresion force
Step 8: Shear reinforcement Ʈv = Vu/bw.d = 151.25x10^3/(230x350) 15
= 1.88 N/mm^2. Pt = 100 Ast/bw.d = (100x1963)/230x350 =2.44 τc = 0.818 balance shear = Vu – Ʈ c.bw.d = 151.25x10^3-0.818x230x350 = 85.40kN. Using 8mm dia two legged strirup the spacing is Sv = (0.87x415x2x50x350)/85.4x10^3 = 147.97mm Sv >or= 0.75d 0.75x350 = 262.5mm. Sv = 300mm. Whichever is less Provide 8mm dia two legged strirup at 145mm c/c length of beam.
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DESIGN OF L BEAM Step 1: Datas L = 5m. Spacing=4.23m. Df = 190mm. fck = 20 N/mm^2 fy = 415 N/mm^2 Step 2: Cross sectional dimensions, since L beam is subjected to bending torsion and shear force assume a trial section having span / depth ratio = 12. d = span/12 5000/12 = 416.67mm. d = 450mm. Adopt overall depth D = 500mm bw = 230mm. step 3 : Effective span Effective span is the least of 1) c/c of supports = 5.23m 2) Clear span + eff.depth = 5+0.45 = 5.45m.
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Step 4: Loads Self weight of slab = 0.19x25x0.5x4.23 = 10.05kN/m Floor finish = 1x0.5x4.23 = 2.13 kN/m Self weight of rib = 0.23x0.31x25 = 1.78kN/m Water proofing = 2x0.5x4.23=4.23 kN/m Weight of parapet wall = 0.6x0.23x19 = 2.62 kN/m Total dead load = 23.97 kN/m Live load = 1.5 x0.5x4.23= 3.17kN/m Step 5: Effective flange width 1) bf = [Lo/12+bw+3Df] =[ 5230/12+230+(3x190)] = 1235.8mm. 2) bf = bw +0.5 time spacing b/n ribs = 230+(0.5x4000) = 2230mm bf = 1235mm. step 6 : ultimate bending moment and shear force at support s/c. Mu = 1.5(23.97x5.23^2)/12 = 81.93kN.m 18
Vu = 1.5(0.5x23.97x5.23) = 94 kN. At centre span s/c Mu = 1.5(23.97x5.23^2)/24 = 40.98 kN.m Step 7 : Torsional moments at support s/c. (Working load /m – rip s.w)= (23.97-1.78) = 22.19 Kn/m. Total ultimate load on slab = 1.5(22.19x5.3) = 176.4 kN. Total ultimate shear force = 0.5x176.4 = 88.21 kN. Distance of centroid of shear force from c/c = (0.5x1235-190) = 427.5mm. Ultimate torsional moment Tu = (88.21x0.427) = 37.67 kN.m Step 8: Equivalent bending moment and shear force : {Is 456:2000 clause 41.42} Mel = (Mu+Mt) Mt = Tu[1+(D/b)/1.7] = 37.67[1+(500/230)/1.7] = 70.3 kN.m Mel = (81.93+70.3) = 152.3 kN.m
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Ve = Vu +1.6(Tu/B) = 94+1.6(37.67/0.23) = 356.1 kN. Step 9: Main longitudinal reinforcement Mel = 152.3 kN.m Mu,lim = 0.138.f ck.b.d^2 = 0.138x20x230x450^2 = 128.5 kN.m Mel < Mu,lim Mel = 0.87.fy.Ast.d[1-Ast.fy/bd.fck] 152.3x10^6 = 0.87x415xAstx450[1-415xAst/230x450x20] Ast = 1251mm^2. Provide 3nos of 25mm dia (Ast = 1472mm^2) Ast min = (0.85.bw.d/fy) = (0.85x230x450/415) = 212mm^2. = 212/(π/4x12^2) = 1.85 = 2nos. Provide 2 nos of 12mm dia (Ast = 226mm^2) Step 10: Shear reinforcement : τve = (ve/bw.d) =(356.1x10^3/(230x450)) = 3.4N/mm^2. Pt = (100Ast/bw.d) 20
= (100x1472/230x450) =1.4 τc = 0.7N/mm^2 < τve . hence shear reinforcement are required. Using 10mm dia and 25mm side covers and bottom cover 50mm.b1 = 180mm, d1 = 400mm,Asv = 157mm^2. 1).Sv = 92mm. 2).Sv = 91.3mm.
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DESIGN OF COLUMN
Step 1: Slenderness ratio consideration : Le/D = (0.65x3600)/230 = 10.17<12. Step 2: Eccentricity consideration Le/b = (0.65x3600)/230 = 10.17<12. X = Lo/500 + D/300 = 3600/500 + 230/300 = 14.87<20. Y = Lo/500 + D/300 = 3600/500 + 230/300 = 14.87<20. Hence it is short column. Step 3: Load on column. Load from 5.23m span = 57.84x5.23/2 = 151.25kN. Load from 4.23m span = 65.04x4.23/2 =137.6 kN. Total load = (2x151.23)+137.6 = 440.1 kN. 22
Step 4: Pu = 440.1 kN. b = 230mm. D = 230mm. Fck = 20N/mm^2. Fy = 415N/mm^2. Step 5: Assume Pt = 3%. Pt = 100xAst/b.d Ast = (3x230x190)/100 = 1311mm^2. Provide 5 nos of 20mm dia bars (Ast=1570mm^2) Ag = 230x230 = 52,900mm^2. Ac = Ag- Ast = 52,900-1570 = 51,330mm^2. Pu = 0.4 fck. Ac + 0.67.fy.Ast = 0.4x20x51,330 + (0.67x415x1570) = 859.7 kN > 440.1Kn. Step 6: Detail of longitudinal steel. Use cover = 40mm Steel spacing = 230-40-40-20 = 130mm. = 130/2 = 65mm. 23
Clear spacing between bars = 65-20 = 45 < 230. Step 7: Transerverse steel : Dia of link not less than ,20/4 or 6mm. Use 10mm, spacing not less than 1) Dimension of column = 230mm. 2)16 times dia of long steel = 16x20 =320mm. 3) 300mm. Adopt 300mm spacing between each link.
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DESIGN OF COLUMN Step 1: Slenderness ratio consideration : Le/D = (0.65x3600)/230 = 10.17<12. Step 2: Eccentricity consideration Le/b = (0.65x3600)/230 = 10.17<12. X = Lo/500 + D/300 = 3600/500 + 230/300 = 14.87<20. Y = Lo/500 + D/300 = 3600/500 + 230/300 = 14.87<20. Hence it is short column. Step 3: Load on column. Load from 5.23m span = 57.84x5.23/2 = 151.25kN.
Load from 4.23m span = 65.04x4.23/2 =137.6 kN. Total load = 151.23+137.6= 288.83 kN. 25
Step 4: Pu = 288.83 kN. b = 230mm. D = 230mm. Fck = 20N/mm^2. Fy = 415N/mm^2. Step 5: From IS 456-2000
Assume Pt = 3%. Pt = 100xAst/b.d Ast = (3x230x190)/100 = 1311mm^2.
Provide 5 nos of 20mm dia bars (Ast=1570mm^2) Ag = 230x230 = 52,900mm^2. Ac = Ag- Ast = 52,900-1570 = 51,330mm^2. Pu = 0.4 fck. Ac + 0.67.fy.Ast = 0.4x20x51330 + (0.67x415x1570) =785.08 kN > 288.3Kn.
Step 6: Detail of longitudinal steel. Use cover = 40mm Steel spacing = 230-40-40-20 = 130mm. 26
= 130/2 = 65mm. Clear spacing between bars = 65-20 = 45 < 230. Step 7: Transerverse steel Dia of link not less than ,20/4 or 6mm. Use 10mm, spacing not less than 1) Dimension of column = 230mm. 2)16 times dia of long steel = 16x20 =320mm. 3) 300mm. Adopt 300mm spacing between each link.
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Design of column on ground floor Step 1: Slenderness ratio consideration : Le/D = (0.65x3600)/230 = 10.17<12. Step 2: Eccentricity consideration Le/b = (0.65x3600)/230 = 10.17<12. X = Lo/500 + D/300 = 3600/500 + 230/300 = 14.87<20. Y = Lo/500 + D/300 = 3600/500 + 230/300 = 14.87<20. Hence it is short column. Step 3: Load on column. Load from 5.23m span = 57.84x5.23/2 = 151.25kN. Load from 4.23m span = 65.04x4.23/2 =137.6 kN. 28
Total load = (2x151.23)+137.6+440.1 = 880.2 kN. Step 4: Pu = 880.2 kN. b = 230mm. D = 230mm. Fck = 20N/mm^2. Fy = 415N/mm^2. Step 5: Assume Pt = 4%. Pt = 100xAst/b.d Ast = (4x230x190)/100 = 1748 mm^2. Provide 6 nos of 20mm dia bars (Ast=1885mm^2) Ag = 230x230 = 52,900mm^2. Ac = Ag- Ast = 52,900-1885 = 51,015mm^2. Pu = 0.4 fck. Ac + 0.67.fy.Ast = 0.4x20x51,015 + (0.67x415x1885) = 932.24kN > 880.2. kN
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Step 6: Detail of longitudinal steel. Use cover = 40mm Steel spacing = 230-40-40-20 = 130mm. = 130/2 = 65mm. Clear spacing between bars = 65-20 = 45 < 230. Step 7: Transerverse steel Dia of link not less than ,20/4 or 6mm. Use 10mm, spacing not less than 1). Dimension of column = 230mm. 2).16 times dia of long steel = 16x20 =320mm. 3). 300mm. Adopt 300mm spacing between each link.
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DESIGN OF FOOTING
Step 1: Pu = 932 kN. b = 230mm. D = 230mm. P = 200kN/m^2. Fck = 20N/mm^2. Fy = 415N/mm^2. Step 2: Load on column = 932 kN. Assume s.w of footing (10%) = 93.2 kN. Total factored load Wu = 1025.2 kN. Footing area = [1025.25/(1.5x200)] Area = 3.42 m^2. L x B = 3.42 m^2. (1.5BxB) = 3.42 m^2. 1.5B^2 = 3.42 m^2. B = 1.5m. L = 1.5 x B = 1.5x1.5 L = 2.25m. Factored soil pressure at base is computed as Pu = [932.2/(1.5x2.25)] = 276.2 kN/m^2. 31
1.5x200 = 300 > 276.2 kN/m^2.
Hence the footing area is adequate since the soil pressure developed at base is less than the factored bearing of soil. Step 3: Factored B.M : Cantilever projection from the short side face on column} = 0.5(2.25-0.23) = 1.01m.
Cantilever projection from the long side face on column} = 0.5(1.5-0.23) = 0.635m. B.m at short side face of column is (0.5.Pu.L^2) = 0.5x276.2x1.01^2 = 140.8 kN.m B.m at short side face of column is = (0.5x276.2x0.635^2) = 55.69 kN.m Step 4: Depth of footing : a) From moment consideration we have Mu = 0.138.fck.b.d^2. d = 225.9mm. b) From shear stress consideration use have the critical s/c for one way shear is located at a distance d from the face of column. Vu =276.2[1000.23-d]. Assuming the shear strength of τc = 0.36 N/mm^2 For M25 grade concrete with Pt = 0.25. 32
ƮC = (Vul/b.d) 0.36 = 276.2[1000.23-d] / (1000xd) 0.36x1000xd = 276.2[1000.23-d]. = 434mm. d = 450mm. D = 500mm. Step 5 : Reinfocement in footing : 1). Longer direction Mu = 0.87.fy.Ast.d[1-Ast.fy/b.d.fck] 140.8x10^6 = 0.87x415xAstx450 . [1-415xAst/ 1000x450x20] Ast = 896mm^2. Adopt 16mm dia bars of 5 nos [Ast = 1005mm^2] Spacing = 200mm c/c. 2).shorter direction: 55.69x10^6 = 162472.5 Ast- 5.99 Ast^2. Ast = 347 mm^2. Ratio of long to short side Bc = [2.25/1.5] = 1.5. Reinfocement in cement bond width 2m is [2/B+1] Ast = [2/1.5+1]2x347 = 555.2mm^2. Min . reinforcement = 0.12/100 x 1000x500 = 600mm^2. Provide 12mm dia bar 6 nos (Ast = 1206.4mm^2) Spacing = 150mm.
Step 6 : Check for shear stress : The critical s/c for one way shear is located at distance d from the face of column . ultimate shear force per width in the longer direction . Vu = (276.2x0.635) = 175.4kN. 100xAst/b.d = 100x1005.5/ (1000x230) = 0.44. {Table 19 from IS 456:2000} 33
= ks . τc = 2x0.34 = 0.68. τc = Vu/bd = 175.4x10^3/(1000x450) = 0.39. τv < Ks. τc. Hence safe.
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DESIGN OF FOOTING Step 1: Pu = 859.7 kN. b = 230mm. D = 230mm. P = 200kN/m^2. fck = 25N/mm^2. fy = 415N/mm^2. Step 2: Load on column = 859.7 kN. Assume s.w of footing (10%) = 85.97 kN. Total factored load Wu = 945.67kN. Footing area = [945.67/(1.5x200)] Area = 3.15 m^2. L x B = 3.15 m^2. (1.5BxB) = 3.15 m^2. 1.5B^2 = 3.15m^2. B = 1.45m. L = 1.5 x B = 1.5x1.45 L = 2.175m. Provide L = 2.5m, B = 2m. Factored soil pressure at base is computed as Pu = [945.6/(2.5x2)] = 189.12 kN/m^2. 1.5x200 = 300 > 189.12 kN/m^2. 35
Hence the footing area is adequate since the soil pressure developed at base is less than the factored bearing of soil. Step 3: Factored B.M : Cantilever projection from the short side face on column} = 0.5(2.5-0.45) = 1.025m.
Cantilever projection from the long side face on column} = 0.5(2-0.30) = 0.85m. B.m at short side face of column is (0.5.Pu.L^2) = 0.5x189.12x1.025^2 = 99.35 kN.m B.m at short side face of column is = (0.5x189.12x0.85^2) = 80.4 kN.m Step 4: Depth of footing : a) From moment consideration we have Mu = 0.138.fck.b.d^2. d = 170 mm. b) From shear stress consideration use have the critical s/c for one way shear is located at a distance d from the face of column. Vu =189.12[689.7 -d]. Assuming the shear strength of τc = 0.36 N/mm^2 For M25 grade concrete with Pt = 0.25. Ʈc = (Vul/b.d) 0.36 = 189.12[689.7-d] / (1000xd) 36
0.36x1000xd = 189.12[1000.23-d]. = 237.5mm. d = 300 mm. D = 350 mm.
Step 5 : Reinfocement in footing. 1). Longer direction Mu = 0.87.fy.Ast.d[1-Ast.fy/b.d.fck] 99.35x10^6 = 0.87x415xAstx300 . [1-415xAst/ 1000x300x20] Ast = 970 mm^2. Adopt 16mm dia bars of 5 nos [Ast = 1008mm^2] Spacing = 200mm c/c. 2).shorter direction: 80.4x10^6 = 108315 Ast - 5.99 Ast ^2. Ast = 775 mm^2. Ratio of long to short side Bc = [2.5/2] = 1.25. Reinfocement in cement bond width 2m is [2/B+1] Ast = [2/1.5+1]2x775 = 1240 mm^2. Min . reinforcement = 0.12/100 x 1000x230 = 276 mm^2. Provide 10mm dia bar 4 nos (Ast = 314.16mm^2) Spacing = 285mm. Provide 10mm dia @ 250mm c/c. Step 6 : Check for shear stress : The critical s/c for one way shear is located at distance d from the face of column . ultimate shear force per width in the longer direction . Vu = (189.12x0.885) = 167.4kN. 100xAst/b.d = 100x1008/ (1000x300) 37
= 0.336. {Table 19 from IS 456:2000} = ks . τc = 2x0.41 = 0.82. τc = Vu/bd = 167.4x10^3/(1000x300) = 0.56 τv < Ks. τc.
Hence safe.
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DESIGN OF STAIR CASE Data: Types of stair case Dog-legged Tread T =250mm Rise R = 150mm Fck=20N/mm2 Fy=415 N/mm2 Width of each flight = 1.6m Height of each flight=5/2=2.5m Number of rises required =2.5/0.15=17 Number of tread flight
=17-1=16
Space occupied by tread =16×250=4000mm=4.0m Keep the landing with=2m Passage=1.5m
Effective span: The loading on each flight The loading slab is assumed to span in the some direction as the stairs and is considered to act together to form a single slab Bearing of the slab =300mm The effective span =4+1.5+0.300/2=5.65m
Loads: Assume the waist slab thickness =200mm Weight of slab w on slope =0.2×1×25000 =5000N/m2 Dead weight of horizontal span W=Ws√
/T = 39
=5000√
/0.250
=3830.95N/m Dead weight of step is given by W= (R /2×1000)1×25000 =(150/2000)1×25000 W =1875N/m2 Weight of finishing =1000N/m Live load =5000N/m Total load= 5830.95+1875+1000+5000 =13706N/m Factored load =1.5×13706 =10559N/m
Bending moment: M =Wul2/8=20559×5.652/8 =82036.55N-m
Check depth of waist slap: D=√
/0.138×Fckb
=√
3/0.138×20×1000
Main reinforcement : Mu=0.87Fy Astd [1-Ast Fy/bd Fck] 93.505×106 = 0.87×415×Ast×180[1- Ast×415/1000×230×20] =1042.75mm2 Numbers of bars needed in 1.6m width 40
=1.6×1042.75/π/4×122 = 15 No's Adopt 15 No's of 12mmØ rod Spacing =1600/15=107mm≈100mm Provide 15 No's of 12mmØ bars at 100mm c/c in main reinforcement
Distribution reinforcement: =0.12 % cross area =0.12/100×1000×200=240mm2 Adopt 10mm Ø bars Spacing =1000ast/Ast=1000×π/4×102/240=327mm Provide 10mmØ bars @300mm c/c
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CONCLUSION
This project deals with the design considering thevertical loads alone. The extent of the project can be done by considering the earthquake load. More forces will be considered based on the strength criteria in our codal provisions. The HOSPITAL BUILDING is framed structure and designed by LIMIT STATE METHOD and the structure provided to be safe. The drawings were prepared using AUTO-CAD. Thus the object of the project has been successfully achieved and evaluated in all aspects. The project has enlightened us in the field of analysis and design.
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BIBLIOGRAPHY
IS 456-2000 “Indian Standard Plain and Reinforcement Concrete of Practice”. SP-16 “Design Aids for Reinforcement Concrete to IS 456-2000”.
References: Varghes.P.C, (2002) “Limit State Design of Reinforced Concrete. ” Krishna Raju.N, (2003) “Design of Reinforced Concrete Structures.”
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