1Cantilever Type Stairs2RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Stairs Stairs are needed to transfer people vertically in the building between levels. Stairs width 0.9 1.5m and stair well is 0.3-0.6m23RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Stairs Components Landings Flights: contains 10-14 steps; each hasRisers: 150 200mmGoing (treads): 250 300mmwidthrisergoingFlightLanding4RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Structural System for Cantilever Stairs35RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Forces & Reinforcement LdLclearWuPu6RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Minimum DimensionsLclear+ tavg Leff= min. edge to C.L.2cos**risert tttavgs+ ==tstavgrisert*47RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Loading Dead LoadsSelf weight = tavg* concCovering= 0.81.0 kN/m2 Live Loads3 4 kN/m2 Parapet LoadConcentrated L.L. @ edge = 1.5kN/mtstavgriserExample59RC Design 2 -Dr. Hany NessimMonday, December 10, 2012ExampleDesign a cantilever type stair case for a floor height of 3.0mLoads:Covering material 0.8kN/m2Live Load 3.0 kN/m2Materials:fcu= 35N/mm2fy= 400 N/mm210RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution1] Design of the FlightAssume the riser height is 150mm and the going width is 300mm. = tan-1(riser/going) = tan-1(150/300) = 26.561.1) Load CalculationsAssume ts=140mmtstavgrisert*mmrisert tmmttavgs5 . 23121505 . 15625 . 156) 56 . 26 cos(140cos**= + = + == = =611RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution1.1) Load Calculations (contd)Dead Load = conctavg+ covering materials= 25*0.2315 + 0.8 = 6.5875 kN/m2Wu= 1.4DL + 1.6LL = 1.4*6.5875 + 1.6*3 = 14.023 kN/m2Pu= 1.6*1.5 = 2.4 kN/m1.2) Effective Lengthtstavgrisert*)`= += +=)` +=mmedget LLavg cleareffective725 . 1 125 . 0 6 . 18315 . 1 2315 . 0 6 . 1. minCL to. min12RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution1.3) Bending MomentsMu= Wul2/2 + Pul= 14.02*1.7252/2 + 2.4*1.725 = 25kN.m/m1.4) Design of ReinforcementAssume 20mm cover d= tavg 20mm = 211.5mmfrom curves = 0.019Wu= 14kN/mPu= 2.4kN016 . 05 . 211 * 1000 * 3510 25262 1= = =xbd fMRcuu2346 5 . 211 * 100040035019 . 0 mm bdffAycus= = = 713RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution1.5) Check for Minimum ReinforcementAs/step = As x step width (going)= 347 x 0.30 = 104mm2choose 2 10 (157mm2)/stepsysA mm bdfA < = = =2min ,317 5 . 211 * 10004006 . 0 6 . 0 ok14RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution2] Design of The Landingtake thickness = ts= 140mmWu= 1.4DL + 1.6LL= 1.4(25*0.14+0.8) + 1.6*3 = 10.82kN/m2As an approximation take half of the load in each direction; Wul= 10.82/2 = 5.41kN/mMu= WulxL2effective/2 = 5.41*1.7252/2 = 8.05 kN.mTake d = ts 20mm = 120mm815RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution2] Design of The Landing (contd)From curves = 0.019Choose 510mm (393mm2)>As,min ok0159 . 0120 * 1000 * 3510 05 . 8262 1= = =xbd fMRcuu2196 120 * 100040035019 . 0 mm bdffAycus= = = 16RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution3] Design of the Supporting Beam3.1) Calculation of LoadsAssume the beam cross section is 250 x 800mm.To obtain weight in the halprojection use t*bSelf weight = 1.4*c*b*t*b= 1.4*25*0.25*0.8944= 7.83 kN/mhw=3-floor beam-t*b-1.5/2 = 3-0.6-0.89-0.75=0.755mmmttb4 . 894) 56 . 26 cos(800cos*= = =917RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution3.1) Calculation of Loads (contd)Wall load = 1.4(w*tw+0.9)*hw= 1.4(16*0.25+0.9) * 0.755 = 5.18kN/mWu= self wt. + Wuxflight width + wall load + edge live loadWubf= 7.83 + 14.02x1.6 + 5.18 + 2.4 = 37.84 kN/m18RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution3.1) Calculation of Loads (contd)The load transmitted to the beam @ the landing levelWubl= self wt. + Wulxlanding width + wall load Wubl= 7.83 + 5.41x1.6 + 5.18 = 21.67 kN/m1019RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution3.2) Design for FlexureLocation of maximum moment x3.0m 1.6mWubf= 37.84kN/mWubl= 21.67kN/mR1R2Mmax= X = kNx x x xRkNx x x xR66 . 656 . 4) 8 . 0 3 ( 6 . 1 67 . 21 5 . 1 3 84 . 37253 . 826 . 48 . 0 6 . 1 67 . 21 ) 5 . 1 6 . 1 ( 3 84 . 371=+ +==+ +=mWRxubf18 . 284 . 3753 . 82 1' = = =22.18* 37.84 - 2.18 * 82.53 2'' 122max= =xW x R Mubf=90 kN.m3.0m 1.6mWubf= 37.84kN/mWubl= 21.67kN/mR1R220RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution3.2) Design for Flexure (contd)From Curve =0.02196Check for minimum reinforcementAs< As,min use As,min= 468.4mm2(Choose 316, 603mm2)0183 . 0750 * 250 * 3510 90262 1= = =xbd fMRcuu23 . 360 750 * 2504003502196 . 0 mm bdffAycus= = = )`= =< = ==22min ,4 . 468 3 . 360 * 3 . 1 3 . 1623 750 * 25040035 225 . 0 225 . 0ofmm AA mm bdffsmaller Assycus1121RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution3.3) Design for Shear & Torsiondue to inclination the reaction or any vertical force on the beam will result in:shear; Q = V.cosNormal; N = V.sin3.3.1) Shear StressesThe cretical section is @ d/2 from face of support;Assuming the column width is 600mm critical section is at distance = (0.6/2 + 0.75/2)V = R1-Wubf(0.6/2+0.75/2) = 82.53 37.84 (0.3+0.375)= 57 kNQu= V.cos = 57xcos(26.56)= 51 kN3.0m 1.6mWubf= 37.84kN/mWubl= 21.67kN/mR1R2NQ22RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution3.3.1) Shear Stresses (contd)qu= Qu/bd = 51x103/(250x750) = 0.272 N/mm23.3.2) Shear Reinforcementqu< qcu shear reinforcement is not needed2umax2umax/ 38 . 35 . 1357 . 0 q&N/mm 1.165 . 13524 . 07 . 0 q& 24 . 0mm N qf fqcuccuccucu= = = == = 1223RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution3.3.3) Torsion Stresses x1= 250-2x40 = 170mmy1= 800-2x40 = 720mmph= 2x(x1+ y1) = 2x(170 + 720) = 1780mmAoh= x1.y1= 170x720 = 122400mm2Ao= 0.85Aoh= 0.85x122400 = 104040mm2te= Aoh/ph= 122400/1780 = 68.76mmX1Y124RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution3.3.3) Torsion Stresses (contd)Critical section for torsion is @ d/2 from face of supportTu= 52.8 25 (0.6/2+0.75/2)= 35.9 kN.mMTu= Tu.cos = 35.9xcos26.56= 32.1 kN.m3.0m 1.6mm kNx x x xRT. 8 . 526 . 48 . 0 6 . 1 05 . 8 ) 5 . 1 6 . 1 ( 3 251=+ +=3.0m 1.6mRT1RT2MTuf= 25kN.mMTul= 8.05kN.m26/ 24 . 276 . 68 104040 210 1 . 322mm Nx xxt AMqe otutu= = =1325RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution3.3.4) Check for adequacy of concrete dimensions 3.3.5) Torsional ReinforcementQtu> qtu,min torsional reinforcement is neededAssume stirrups spacing is 100mm38 . 3 26 . 2 24 . 2 272 . 02 2max2 2 = + +u tu uq q q ok2min ,/ 29 . 05 . 13506 . 0 06 . 0 mm Nfqccutu= = =26.97 . 7315 . 1240104040 2100 10 1 . 322mmx xx xfAS MAsystrotustr=||

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\|=26RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution3.3.5) Torsional Reinforcement (contd)Check for minimum longitudinal reinforcementChoose 812 (904mm2)2 ..7904002401001780 97 . 73mmxffSP AAyystrh strsl= ||

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\| = ok1427RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution3.3.6) Reinforcement for combined shear and torsionAstr+ Ast/2 = 73.97+0 = 73.97mm2Choose 10mm @ 100mm (78.5mm2)Check for minimum stirrups reinforcementAst,min= (0.4/fy)b.s = (0.4/240)250x100 = 41.67mm2Total area of stirrups = 2Astr+ Ast= 2x73.97 + 0 = 147.94mm2> Ast,min ok28RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution1529RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution30RC Design 2 -Dr. Hany NessimMonday, December 10, 2012Solution