REINFORCED CONCRETE DESIGN 1 Design of Slab (Examples …

1 DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinREINFORCED CONCRETE DESIGN 1 DESIGN of Slab (Examples and Tutorials) Maszura Syed MohsinFaculty of Civil Engineering and Earth updated version, please click on DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 1: Simply supported One way (excludingslabself-weight) of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 1: Simply supported One way slabSLAB THICKNESSM inimum thickness for fire resistance = 80 mmEstimatedthicknessconsideringdeflectio ncontrol,h=3750/26= 150 mmDURABILITY, FIRE & BOND ,Cbond= ,Cmin,dur= ,a= ,Cmin,fire=a bar/2= (12)=14mmAllowanceindesignfordeviation, Cdev=10mm Nominalcover,Cnom=Cmin+ Cdev=20+10=Cnom=30mmUse: Cnom= 30 mmDesign of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 1: Simply supported One way slabACTIONS & ANALYSISS labself-weight= (excludingself-weight)= ,gk= ,qk= ,nd= ( )+ ( )= ,wd=ndx1m= force, V = wdL/2= kN/mBending moment,M = wdL2/8= of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 1: Simply supported One way slabMAIN REINFORCEMENTE ffectivedepth:d=h bar= (12)=114mmDesignmoment,MED= ( )/(1000x1142x25)= <Kbal= Compressionreinforcementisnotrequiredz=d [ + ]= :H12-250(452mm2/m) DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 1.

2 Simply supported One way slabMinimumandmaximumreinforcementarea,A s,min= (fctm/fyk)bd= ( )bd= (1000))(114)=160mm2/mAs,max= (1000)(150)=6000mm2/mSecondary bar: H12-450(251 mm2/m)SHEARD esign shear force, VED= kNDesign shear resistance,VRd,c= [ (100 1fck)1/3]bdk = 1 + (200/d)1/2 = 1 = As1/bd = 452/1000 x114 = ,c= [ x (100x x 25)1/3] x 1000 x 114 = kNDesign of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 1: Simply supported One way slabVmin= [ ]bd= x 23/2x 251/2x 1000 x 114 = kNThus, VRd,c= > VED = kN, OK! DEFLECTIONP ercentage of required tension reinforcement, = As,req/bd= 412 /1000 x 114= reinforcement ratio, 0= (fck)1/2x10-3= (25)1/2x 10-3= Factor for structural system, K = (L/d)basic= K[11 + 0 + ( 0 -1)3/2](L/d)basic= [11 + + ] = of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 1: Simply supported One way slabModification factor for span less than 7 m = factor for steel area provided = As,prov/As,req= , allowable span-effective depth ratio:(L/d)allowable= x 1 x = span-effective depth ratio,(L/d)actual = 3750/114 = > (L/d)allowable = , NOTOK!

3 Increase Area of steel provided to 566 mm2/m (H12-200)Modification factor for steel area provided = As,prov/As,req= , allowable span-effective depth ratio:(L/d)allowable= x 1 x = > (L/d)actual = OK! DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 1: Simply supported One way slabCRACKINGh = 175 mm < 200 mmMain bar: Smax,slab= 3h 400 = 400 mm, Max. bar spacing = 200 < Smax, slab, OKSecondary bar: Smax,slab= 450 = 450 Max. bar spacing = 450 < Smax, slab, OKDesign of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 2: Continuous one way (excludingself-weight) , ,fireexposure=REI90,exposureclass=XC1,ch aracteristicconcretestrength,fck=C25/30, highyieldsteelstrength,fyk=500N/mm2,Unit weightofconcrete=25 :Diameterofreinforcement=10mmDesign of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 2: Continuous one way slabFigure 1 DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 2: Continuous one way slabDURABILITY, FIRE & BOND ,Cbond= ,Cmin,dur= ,a= ,Cmin,fire=a bar/2= (10)=25mmAllowanceindesignfordeviation, Cdev=10mm Nominalcover,Cnom=Cmin+ Cdev=25+10=Cnom=25mmUse.

4 Cnom= 25 mmDesign of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 2: Continuous one way slabACTIONS Slabself-weight= (excludingself-weight)= ,gk= ,qk= ,nd= ( )+( )= of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 2: Continuous one way slabANALYSISC onsider 1 m width of slab, ActionMomentShearF = wl= x 4 = kNDesign of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 2: Continuous one way slabMAIN REINFORCEMENTE ffective depth:dx=150 (10)=110mmMinimumandmaximumreinforcement area,As,min= (fctm/fyk)bd= ( )bd= (1000))(110)=146mm2/mAs,max= (1000)(150)=6000mm2/mAtfirstinteriorsupp ortM= (18x106)/(1000x1102x25)= <Kbal= CompressionreinforcementisnotrequiredSec ondarybar:H10-450(175mm2/m) DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 2: Continuous one way slabz=d[ + ]= top(449 mm2/m) AtmiddleinteriorspanM= ( )/(1000x1102x25)= <Kbal= Compressionreinforcementisnotrequiredz=d [ + ]= 290 mm2/m ProvideH10-250 bottom(314 mm2/m) DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 2: Continuous one way slabNearmiddleofendspanM= ( )/(1000x1102x25)= <Kbal= Compressionreinforcementisnotrequiredz=d [ + ]= bottom(393 mm2/m) DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 2.

5 Continuous one way slabK=M/bd2fck=( )/(1000x1102x25)= <Kbal= Compressionreinforcementisnotrequiredz=d[ + ]= bottom(196 mm2/m) DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 2: Continuous one way slabSHEARM aximum DESIGN shear force, VED= = x = kNDesign shear resistance,VRd,c= [ (100 1fck)1/3]bdk = 1 + (200/110)1/2 = 1 = As1/bd = 449/1000 x110= ,c= [ x (100x x 25)1/3] x 1000 x 110 = kNVmin= [ ]bd= x 23/2x 251/2x 1000 x 110 = kNSo, VRd,c= kN> VED = kNOK! DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 2: Continuous one way slabDEFLECTIONP ercentage of required tension reinforcement, = As,req/bd= 346 /1000 x 110 = reinforcement ratio, 0= (fck)1/2x10-3= (25)1/2x 10-3= , Factor for structural system, K = (L/d)basic= K[11 + 0 + ( 0 -1)3/2](L/d)basic= [11 + + ] = factor for steel area provided = As,prov/As,req= 393/346= of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 2: Continuous one way slabTherefore, allowable span-effective depth ratio:(L/d)allowable= x 1 x = span-effective depth ratio:(L/d)actual = 4000/110 = < (L/d)allowable = , OK!

6 CRACKINGh = 150 mm < 200 mmMain bar: Smax,slab= 3h 400 = 400 Max. bar spacing = 400 Smax, slab, OK! Secondary bar: Smax,slab= 450 = 450 Max. bar spacing = 425 < 450 mm, OK! DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 3: Two way slab restrained (excludingself-weight)offinishes, ,whereas, of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 3: Two way slab restrained slabFigure 2 DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 3: Two way slab restrained slabi)Basedontheinformationgiven,calcula tetheareaofsteelrequiredfortheslabpanelC - , , )Using the information obtained in (a), calculate the DESIGN shear force in the slab and check whether the slab is safe in terms of shear.

7 Comment on your answer and give appropriate suggestionsiii)Ifthedeflectioncheckispas sed, of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 3: Two way slab restrained slabACTIONS & ANALYSISS labself-weight= (excludingself-weight)= ,gk= ,qk= ,nd= ( )+ ( )= < (Twowayslab)Case8:Threeedgesdiscontinuou s(Oneshortedgecontinuous)Shortspan:Msx1= sx1nlx2= :Msy1= sy1nlx2= sy1nlx2= of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 3: Two way slab restrained slabMAIN REINFORCEMENTE ffectivedepth:dx=150 (10)=120mmdy=150 (10)=110mmMinimumandmaximumreinforcement area,As,min= (fctm/fyk)bd= ( )bd= (1000))(120)=156mm2/mAs,max= (1000)(150)=6000mm2/mSecondarybar:H10-42 5(185mm2/m) DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 3: Two way slab restrained slabShortspan:Mid-span:Msx1= ( )/(25x1000x1202)= <Kbal= Compressionreinforcementnotrequiredz=d[ + ]= 138mm2/m ProvideH10-350 bottom(225 mm2/m) DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 3: Two way slab restrained slabLongspan:Mid-span:Msy1= ( )/(25x1000x1102)= <Kbal= Compressionreinforcementnotrequiredz=d[ + ]= 121 mm2/m ProvideH10-350 bottom(225 mm2/m) DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 3: Two way slab restrained slabLongspan:Support.

8 Msy2= ( )/(25x1000x1102)= <Kbal= Compressionreinforcementnotrequiredz=d[ + ]= 159 mm2/m ProvideH10-350 bottom(225 mm2/m) DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 3: Two way slab restrained slabSHEARS hear force, Short span;Vsx1= vs1nlx = x x 3 = span;Vsx1= vs2nlx = x x 3 = kN/mVsx2= vs2nlx = x x 3 = kN/mDesign shear force, VED= shear resistance,VRd,c= [ (100 1fck)1/3]bdk = 1 + (200/d)1/2 = of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 3: Two way slab restrained slab 1 = As1/bd = 159/1000 x110 = ,c= [ x (100x x 25)1/3] x 1000 x 110= kNVmin= [ ]bd= x 23/2x 251/2x 1000 x 110= kNSo, VRd,c= kN> VED = , OK! DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 3: Two way slab restrained slabDEFLECTIONP ercentage of required tension reinforcement, = As,req/bd= 138/1000 x 120 = reinforcement ratio, 0= (fck)1/2x10-3= (25)1/2x 10-3= , Factor for structural system, K = (L/d)basic= K[11 + 0 + ( 0 -1)3/2](L/d)basic= [11 + + ] = factor for steel area provided = As,prov/As,req= 225/138= of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinExample 3: Two way slab restrained slabTherefore, allowable span-effective depth ratio:(L/d)allowable= 1 x = span-effective depth ratio:(L/d)actual = 3000/120 = 25< (L/d)allowable OK!

9 CRACKINGh = 150 mm < 200 mmMain bar: Smax,slab= 3h 400 = 450 400 Max. bar spacing = 350 Smax, slab,OK! Secondary bar: : Smax,slab= 450 = 525 450 Max. bar spacing = 425 450, OK! DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinTutorial 1: One way continuous ,fireresistance=REI60,exposureclass=XC1, characteristicconcretestrength,fck=C30/3 7,highyieldsteelstrength,fyk=500N/mm2,un itweightofconcrete=25kN/m3,diameterofrei nforcement= of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinTutorial 1: One way continuous slabFigure 3 DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinTutorial 1: One way continuous slaba)Calculate Cnomof the slab. b)Calculate the loads on slab.

10 C)Draw the bending moment and shear force diagram. d) DESIGN the reinforcement and check shear. e)Check deflection and cracking. f)Construct the plan view detailing of the slab. DESIGN of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinTutorial 2: Two way simply supported , of Slab (Examples and Tutorials) by Sharifah Maszura Syed MohsinTutorial 2: Two way simply supported slaba)Iftheslabthicknessistakenas150mm, )Performallcheckingonshear, )Constructthedetailreinforcementobtained in(a) of Slab (Examples and Tutorials) by Sharifah Maszura Syed Mohsi