HS-250 ( HOLLOW CORE SLAB )

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Deirdre Powell
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1 Type of slab : Design Data : Section : Slab thickness (d) ( mm ) 250 Effective width ( be ) ( mm ) 1180 No. of Core ( nc ) ( Ppcs. ) 6 Area of Core ( mm2 ) Spacing between core ( mm ) 40 Weight of slab ( Kg/m2 ) 346 Strands : Diameter ( Ds ) ( mm ) 12.5 Nominal Area ( As ) ( mm2 ) 93 No. of strand ( Ns ) ( pcs. ) 9 Tensile Strength ( fpu ) ( N/mm2 ) 1860 Modulus of elasticity (Es ) ( N/mm2 ) Concrete Strength for Pre-cast : 28 days ( Mpa ) 38 transfer ( Mpa ) 26 Concrete density ( Kg/m3 ) 2450 Concrete Strength for Topping : Topping thickness ( dt ) ( mm ) 28 days ( Mpa ) 40 Loads : HS-250 ( HOLLOW CORE SLAB ) Thickness 250mm, Width 1200mm. Topping 100mm, 4 Strands dia. 9.3 mm. In Bottom Weight of slab ( KN/m2 ) Weight of topping ( KN/m2) Super imposed D. load ( KN/m2) 5 Live load ( KN/m2) 5 Length : Length of clear span ( m ) 6.8 Length of slab span ( L ) ( m ) 7 1

2 2 Section Properties : Area ( A) = mm2 Center of gravity, bot. ( Yb ) = mm Center of gravity, top. ( Yt ) = mm Moment of inertia ( I ) = mm4 Section modulus, bot ( Zb ) = mm3 Section modulus, top ( Zt ) = mm3 Modulus of elasticity ( E ) = 4700 fc' 28 days = Mpa transfer = Mpa 28 days for topping = Mpa Section Properties with 60 mm Topping : Modular Ratio ( MR ) = Et / Ec = % Effective Width of topping ( bet ) = MR x be = mm Area of Topping ( At ) = bet x dt = mm Slab area with Topping ( Ac ) = A + At = mm2 { A x Yb } + { At ( d dt) Center of gravity, bot. ( Ybc ) = Ac = mm Center of gravity, top. ( Ytc ) = d - Ybc + dt = mm Moment of inertia ( Ic ) = { I + A ( Yb-Ybc) 2 } + { bet (dt) 3 / 12 } + { At ((d + 0.5dt)-Ybc) 2 } = mm4 Section modulus, bot ( Zbc ) = Ic / Ybc = mm3 Section modulus, top ( Ztc ) = Ic / Ytc = mm3

3 3 Loads Calculations : Self Weight of slab ( Ws ) = 4.13 KN/m Weight of topping ( Wt ) = KN/m Super imposed load ( Wsi ) = 6KN/m Total Dead load ( D.L ) = KN/m Live Load ( L.L ) = 6KN/m Total service load ( W ) = KN/m Total Ultimate load ( Wu ) = 1.4 D.L L.L = KN/m Moments : Moments for Simply Supported Span = W x L 2 8 Self weight moment ( Ms ) = KN.m Topping moment ( Mt ) = KN.m Super imposed moment ( Msi ) = KN.m D.L Moment ( MD.L ) = KN.m L.L Moment ( ML.L ) = KN.m Service Moment ( M ) = KN.m Ultimate Moment ( Mu ) = 1.4 MD.L ML.L = KN.m Prestressing Forces : No. of strands ( Ns ) = 9 Nominal Area of strand ( As ) = 93 mm2 Ultimate Tensile Strength ( fpu ) = 1860 N/mm2 Total area of steel ( Ast ) = Ns x As = 837 mm2 Initial Tensile Strength ( fpi ) = 0.7 x fpu = 1302 N/mm2 Tensile Strength Capacity ( Fpu) = Ast x Fpu = N Initial prestessing force ( Fi ) = 70 % of Ultimate tensile strenght of strand = 0.7 Fpu = N = KN Stressing Eccentricity ( e ) = Yb-( 40+Ds/2) = mm

4 Losses calculation of Prestress : I ) Elastic Shortening ( ESL ) : fcir = Kcir { Fi /A + ( Fi x e 2 /I ) } - ( Ms x e/i ) Kcir = 0.9 ESL = = 8.65 N/mm2 Kcs x Es x fcir Kcs = 1 Ei ESL = N/mm2 II ) Creep of Concrete ( CR ) : Mt+si = Mt + Msi = KN.m = N.mm fcde = Mt+si x e I = 2.16 N/mm2 CR =Kcr x ( fcir - fcde) Es/Ec Kcr = 2 CR = N/mm2 III ) Shrinkage Concete ( SC ) : V = A/ = 263 inches S = 2( be + d ) / 25.4 (Perimeter) = inches RH = 75 SC = x Kah x Es ( V/S) (100 - RH) Kah = 1 SC = N/mm2 IIII ) Relaxation of strands ( RE ) : Ast = 837 mm2 Rfp = fpi / fbu = 0.7 From PCI table & : Kre = 34.48, J = 0.04, C = 0.75 RE =C { Kre - J ( SC+ CR + ESL ) } RE = N/mm2 Total Losses ( TL ) = ( ESL + CR + SC + RE ) = N/mm2 Percentage of Losses ( PL ) = TL / fpi x 100 = % 4

5 5 Prestressing Force after Losses : Net Stress force after, PL % Fn = N ( Fn) = ( PL/100) Fi Stress Transfer, Assume 10 % Losses ( Ft ) = (100-10%/100 ) Fi Ft = N Check of Stresses : The following calculation is to confirm that the actual stress is less than allowable stress. I ) Stress at Sopport : Stbs = (Ft/A) + (Ft x e/zb) < 0.6 fci' = N/mm2 (Bottom) = N/mm2 Safe Stts = (Ft/A) - (Ft x e/zt) < 0.5 fci' = N/mm2 (Top) = N/mm2 Safe II ) Stress at Midspan : Stbm = (Ft/A) + (Ft x e/zb) - (Ms/Zb) < 0.6 fci' = N/mm2 (Bottom) = N/mm2 Safe Sttm = (Ft/A) - (Ft x e/zt) + (Ms/Zt) < 0.6 fci' = N/mm2 (Top) = 0.91 N/mm2 Safe III ) Stress after full load : Fn/A = 5.39 N/mm2 Fn/A = 5.39 N/mm2 Fn x e/zb = 6.53 N/mm2 Fn x e/zt = N/mm2 Ms/Zb = N/mm2 Ms/Zt = 2.54 N/mm2 Mt/Zb = N/mm2 Mt/Zt = N/mm2 Msi/Zbc = N/mm2 Msi/Ztc = 3.69 N/mm2 MLL/Zbc = N/mm2 MLL/Ztc = 3.69 N/mm2 Snb = (Fn/A) + (Fn x e/zb) - (Ms/Zb) - (Mt/Zb) - (Msi/Zbc) - (ML.L/Zbc) (Bottom) = 2.52 N/mm2 < 0.5 fc' = N/mm2 Safe Snt = (Fn/A) - (Fn x e/zt) + (Ms/Zt) + (Mt/Zt) + (Msi/Ztc) + (ML.L/Ztc) (Top) = 8.41 N/mm2 < 0.45 fc' = 17.1 N/mm2 Safe

6 6 Check of Ulltimate Moment : de = d + dt - ( 40 + Ds/2 ) = mm (effective depth) ƒ = Ast / be x de = C = T C = 0.85 x fc'topping x a x be T =Ast x fps = KN f ps = fpu { 1 - ( 0.5 x ƒ x fpu / fc' ) } = N/mm2 a = T / 0.85 x fc'topping x be = mm Mu = Ø Mn Ø Mn = 0.9 x Ast x fps x (de - a/2) / KN.m Actual Mu = KN.m < Ø Mn Safe Check of Shear : Maximum shear at ( 2 x de) distance from Support, ( Vu ) Vu = Wu ( L - 2 de ) / 2 = KN nc = No. of Core = 6pcs. wc = Width of Core = 140 mm bw = be - ( nc x wc ) = 340 mm V (low) = 0.85 fc' x bw x de / 6 = 60.5 KN V (up) = 0.4 x 0.85 fc' x bw x de = KN V(trans.) = 0.3 x 0.85 fi' x bw x de = KN V (low) < Vu < V (up) Vu < V (transfer) Safe

7 7 Check of deflection : In Short term : Prestressing Camber Δc = Fn x e x L 2 8 x Ei x I = mm Due self weight Δs = 5 x Ws x L x Ei x I = 4.21 mm Due Topping Δt = 5 x Wt x L 4 = 384 x Ec x I mm Due Super impose Δsi = 5 x Wsi x L 4 = x Ec x Ic mm Due Live load ΔL = 5 x L.L x L 4 = 5.06 mm 384 x Ec x Ic Deflection ( final ) Δf = 2.2Δc + 2.4Δs + 2.3Δt + 3Δsi = mm Deflection ( Erection ) Δe = 1.8Δc Δs + Δt + Δsi = mm Total Deflection TΔ = [ Δe - Δf ] + ΔL.L = mm Allowable deflection = L / 360 = mm > TΔ = mm Safe The Results : Reference above calculation 1 ) Stressing Safe 2 ) Moment Safe 3 ) Shear Safe 4 ) Deflection Safe The slab HS-250 Thickness 250mm, Width 1200mm. Topping 100mm, 4 Strands dia. 9.3 mm. In Bottom Span (m) = 7 OK

8 8 SPAN (M)LOAD (KN/M2) 3 STR. 4 STR. 5 STR. 6 STR. 7 STR. 8 STR. 9 STR

9 Yb-Ybc

10 Fi /A Fi x e2 /I Ms x e/i

11 3.36E E E E E E E E E+16 Safe Unsafe OK

12 6 Hollow Core Load Span 250 mm Load (Kn/(m2) STR. 4 STR. 5 STR. 6 STR. 7 STR. 8 STR. 9 STR Spam (m)

Dr. Hazim Dwairi. Example: Continuous beam deflection

Dr. Hazim Dwairi. Example: Continuous beam deflection Example: Continuous beam deflection Analyze the short-term and ultimate long-term deflections of end-span of multi-span beam shown below. Ignore comp steel Beam spacing = 3000 mm b eff = 9000/4 = 2250