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箱涵支架设计计算

2023-05-21 来源:爱go旅游网
Formwork & Falsework Design for Single cell Box CulvertReferenceCalculationsStructures 5025C at 50+423, 5204C at 52+137, 5212C at 52+767 , 5812C at 58+994 & 5903C at 59+092Refer attched table No. 01OutputConsider RCBC 5812C at 58+994 as a sigle cell typical structure for the calculationData Top Slab Thickness Bottum Slab Thickness Wall Thickness Clear Span Clear Height Wall Height== ====750700600993054655200mmmmmmmmmmmmCodesGenerally the designs are undertaken to the following standards and modified wherer shown(a). BS 5975:1996 Code of practice for Falsework(b). BS 5950 : Part 1 , 1990 Structural Use of Steel Work in Building(c). BS EN 10210 Circular Hollow Hot Form Section(d). CIRIA Report 108 Concrete Pressure on FormworkDesign of Top SlabIntroductionLoadsfromwetconcrete,formworkandworkingloadsaredirectlyreston15mmplywood.Theplywoodissupportedby50mmdiameterand2mmthickG.I.pipes,whicharespacedat150mmcenters.ThentheGIpipesaresupportedby2/50G.I.pipesspacedat775mmcenters.TheG.I.supportsto50G.I.pipesat775mmand785mmcenters.thetotalloadsfromconcrete,workingandformwork will be transformed to these scafoliding frams and then to bottom slabLoad EvaluationBS 5975 1996cl. 8.3.1cl. 8.3.2Concrete DensityWeight of formworkLive load on formworkDead Load=====2521.5KN/m3KN/m2KN/m2slab thickness25* 0.750 + 220.75KN/m21

ReferenceCalculationsFormwork/falsework are designed for nominal loads ( unfactored )Unfactored Load Intensity (n1)===1.0gk +1.0 qk1.0 * 20.75 + 1.0 * 1.522.25KN/m2OutputDesign of 15 mm PlywoodPlywood is supported by 50 mm GI pipes15 mm Flywood 150 mmConsidering 1m wide stripUDL ( w )=Check for BendingMaximum Bending Moment (M)=== Maximum Bending Stress (s)= 0.06322.25KN/m50 mm G.I. pipesWL28KNmKNmMZ0.03710310310001526 1.67N/mm2=NmmPlywood=< 11.37N/mm2(Allow. Bending Stress)15mm ok.Therefore Bending Stress is Satisfactory.Check for Shear Maximum Shear on plywood==WL212.980.1502=1.669Shear StressKN0.974103=100015=0.111N/mm2N/mm22.58N/mm2(Allow. Shear Stress)OK2

ReferenceCheck for DeflectionCalculationsMaximum Deflection=OutputW =22.250L =150E =70001WL4384EIKN/mmmN/mm2=13.82150413384100015700012=Span/Deflection=0.0149150mm0.0149Span/Deflection=Hence 15 mm plywood is satisfactory10067> 200OKDesign of GI PipesTheplywoodissupportedby50mmdiameterand2mmthickGIpipes,whicharespacedat150mmcenters. Then the GI pipes are supported by 2/50 GI pipes spaced at 775 mm centers.Check for Bending0.1502/50 G.I. PipesG.I. Pipe0.1500.775Maximum Bending Moment (M)=WL28=12.980.1500.7752KNm8 0.251= Bending Stress (s)Check for G.I PipesG. I. Pipe 48.3Hand Book ofstructural steelwork A.D.weller 1997KNm=MZ thickness 2 mmId41d426433.142848.3444.34ReferenceId41d42OutputCalculations64I=EN 102103.142848.3444.34641047.81cm4 3.238cm3 .Z= Therefore Bending Stress (s)=0.1463.238106KN/m2= Bending Stress (s)BS 5975Ann-B77385KN/m2 200OKHence GI pipe 150 c/c and supporting at 775 mm are satisfactory4

ReferenceCalculationsDesign of 2/50 G.I pipes just under the GI pipeOutput1502/50 GI Pipes785 mm2/50 GI Pipes 0.775 m0.775 m Load from GI pipe is a UDLW=W=Maximum Bending Moment (M)=12.980.775KN/m10.06KN/mWL2810.060.78528=KNm= Bending Stress (s)Check for G.I PipesG. I. Pipe 48.3Hand Book ofstructural steelwork A.D.weller 1997 0.775KNm=MZ thickness 2 mmId41d42643.142848.3444.34641047.81cm4 3.238cm3 .I=EN 10210Z=5

Reference Therefore Bending Stress (s)Calculations=Output0.77522(3.238106)KN/m= Bending Stress (s)BS 5975Ann-B119652KN/m2 200OKHence 2/50 GI pipe 775 c/c and supporting at 785mm are satisfactory6

ReferenceCalculationsDesign of GI Pipe Supports.0.785 mOutput 0.775 m0.775 m 0.785 mLoad on a Leg==Assume - Effective length of GI pipe Radius of GyrationSlenderness ratio==12.98 x 0.775 x 0.785 7.901500KNKNScaffold legmm1.62cmLer==(150 / 1.62)93BS 5950CL:7.4Capacity of GI pipePcAgpc==290.9 x 12436.07 KN 36Capacity of a Leg( GI Pipe Scaffolding leg)=KNcapacity ok.Therefore the condition is Satisfactory.Design of WallsLateralpressurefromwetconcretearedirectlyappliedto15mmplywood.Theplywoodissupportedby50mmdiameterand2mmthickGIpipes,whicharespacedat150mmcenters.ThentheGIpipesare supported by 2(50) GI pipes, spaced at 600 mm centers.Maximum concrete pressure on side formworkCIRIAReport 108PmaxDC1RC2KHC1R.............1Where :-D=C1=C2=R=24 kN/m3 ( Unit weight of concrete )1 ( as a wall )0.45 ( retarded concrete )1.5 m/hr. ( concrete pouring rate )7

ReferenceCalculationsH=T=K=1.730 m ( Wall height )32 0C ( concrete temparature at pouring ) 36Tc +16K= 3632 +16K=0.562522OutputPmax2411.50.450.56251.73011.5Pmax.=33.71KN/m2Design of Side PlywoodPmaxD50mm G.I. pipes1.410.320 150 150 150 15mm Flywood 50 mm GI PipeConsidering 1m wide stripCheck for BendingUDL for 1m strip ( w )=Maximum Bending Moment (M)== 33.712(50) mm Gi pipe33.71KN/m2KN/mWL28=42.670.1502KN.m80.09KNm Maximum Bending Stress (s)=MZ0.0910610001526=N/mm28

ReferenceCalculations= Maximum Bending Stress (s)2.40OutputN/mm2 200OKDesign of GI PipesTheplywoodissupportedby50mmdiameterand2mmthickGIpipes,whicharespacedat150mmcenters.9

ReferenceCheck for BendingMaximum Bending Moment (M)CalculationsOutput=WL2833.710.1500.60028 0.228== Bending Stress (s)Check for G.I PipesG. I. Pipe 48.3Hand Book ofstructural steelwork A.D.weller 1997KNmKNm=MZ thickness 2 mmId41d42643.142848.3444.3447.81cm64104I=EN 10210Z= Therefore Bending Stress (s)== Bending Stress (s) 3.238cm3 .0.2283.23810670413.84KN/m2KN/m2ReferenceCheck for DeflectionCalculationsOutputMaximum Deflection=1WL4384EIW =5.057L =150mmKN/mN/mm2EI = 2898 x 107=415.0576003842898107=Span/Deflection=0.059mm6000.059Span/Deflection=10188> 200OKHence GI pipe 150 c/c and supporting at 600 mm are satisfactoryCheck for M12 Tie bar 600 mm460 mmForce on M12 Tie bar==Direct Tensile Stress on M12 Thread bars = 33.71 x 0.460 x 0.6009.30KN9.3103113.098N/mm2=Braking Load of Thread bars=Permissible Stress on Thread bars =Direct Tensile Stress on M12 tie bar ReferenceCalculationsDesign of GI Pipes Support for wallsOutputBS 5975Cl: 6.3.1.3Consider 1m strip of the wall and assume all structure sway at one side0.7 (100As)1/3(fcu)1/3Vertical force of the wall== =Vertical force of the top slab==Horizontal force of the wall==Force on GI pipes/Adjutable jack=R Cos 45=Reaction on GI PipeConcrete weight + F/W weight + Live Load{(1.7300.255124)2(1.73012)(1.73011.5)}240.21KN(3.5600.395124)(3.56012)(3.56011.5)46.21KN86.422.5%2.16KNR Cos 452.163.05R=KNBS 5950CL:7.4Capacity of GI pipePcAgpc( Effective length 1.8 m, Radious of Giration 1.62 = 111)Pc290.9133PcCapacity of Ajestable jack==38.6938.69KNKNReaction on GI pipes ReferenceCalculationsOutput13

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