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# Reinforced concrete design

## by Fahim Sediqi, Civil Engineering Student at Dawat University on May 12, 2013

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reinforced concrete design

reinforced concrete design
Civil Engineering

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## Reinforced concrete designPresentation Transcript

•  fcr is the average cylinder strength f’c compressive strength for design f’c ~2500 psi - 18,000 psi, typically 3000 - 6000 psi Ec estimated as:where w = weight of concrete, lb/ft3f’c in psiE in psifor normal weight concrete ~145 lb/ft3E w fc c33 15.E fc c57 000, www.facebook.com/civilengineeringmaterials
• Strain in concrete will be caused by loading, creep,shrinkage, and temperature change.For scale, consider a 20’ section of concrete,f’c = 4000 psi, under a stress, fc = 1800 psi.Determine the change in length.www.facebook.com/civilengineeringmaterials
•  Tensile strength of concrete is about ~300 – 600 psi Tensile strength of concrete is ignored in design Steel reinforcement is placed where tensile stressesoccurWhere do tensile stresses occur?f c10www.facebook.com/civilengineeringmaterials
• Restrained shrinkageslab on gradeshrinkage strain, ε = 0.0006σ = εE = 0.0006 x 3600 ksi = 2.16 ksiFlexural membercompressiontensionwww.facebook.com/civilengineeringmaterials
•  Deformed steel reinforcing bars Welded wire fabric 7-strand wire (for pre-stressing)www.facebook.com/civilengineeringmaterials
•  Grade 60 (most common in US) Sizes #3 → #18 (number indicatesdiameter in ⅛ inch)www.facebook.com/civilengineeringmaterials
• Readily available fabricsDesignation:longitudinal wire spacing x transverse wire spacing –cross-sectional areas of longitudinal wire x transverse wires in hundredths ofin2www.facebook.com/civilengineeringmaterials
• Two codes for reinforced concrete design: ACI 318 Building Code Requirements for StructuralConcrete AASHTO Specifications for Highway BridgesWe will design according to ACI 318 which is an ‘LRFD’design. Load and resistance factors for ACI 318 aregiven on page 7, notes.www.facebook.com/civilengineeringmaterials
•  Short - slenderness does not need to beconsidered – column will not buckle Only axial loadLCross-sectional Areas:As = Area of steelAc = Area of concreteAg = Total areaFs = stress in steelFc = stress in concreteFrom Equilibrium:P = Acfc + AsfsPLPIf bond is maintained εs = εcwww.facebook.com/civilengineeringmaterials
• For ductile failure – must assure that steelreinforcement will yield before concrete crushes. Strain in steel at yield ~0.002 ε = 0.002 corresponds to max. stress in concrete. Concrete crushes at a strain ~ 0.003Equilibrium at failure: P = AsFy +Acf’cwww.facebook.com/civilengineeringmaterials
•  ρ = As/Ag ACI 318 limits on ρ for columns:0.01≤ρ≤0.08 (practical ρmax = 0.06) Substitute ρ=As/Ag andAg=As+Ac intoequilibrium equation:P = Ag[ρfy +f’c(1- ρ)]www.facebook.com/civilengineeringmaterials
• P = Ag[ρfy +f’c(1- ρ)]Safety Factors Resistance factor, Ф = 0.65 (tied), Ф = 0.70 (spiral) When fc>0.85f’c, over time, concrete will collapse Stray moment factor for columns, K1 K1=0.80 for tied reinforcement K1=0.85 for spiral reinforcementФPn = Ф K1 Ag[ρfy +0.85f’c(1- ρ)]www.facebook.com/civilengineeringmaterials
• ФPn = Ф K1 Ag[ρfy +0.85f’c(1- ρ)]for design, Pu ≤ ФPncgucyfAKPff85.0)85.0(11)1(85.01 cyugffKPAwww.facebook.com/civilengineeringmaterials
• Used to resist bulge of concrete and buckling of steelwww.facebook.com/civilengineeringmaterials
• Used to protect steel reinforcement andprovide bond between steel and concretewww.facebook.com/civilengineeringmaterials
• Design a short, interior, column for a service deadload of 220 kips and a service live load of 243 kips.Consider both a circular and a square cross section.Assume that this column will be the prototype for anumber of columns of the same size to takeadvantage of the economy to be achieved throughrepetition of formwork. Also assume that this columnwill be the most heavily loaded (“worst first”).Available materials are concrete with f’c = 4 ksi andgrade 60 steel.www.facebook.com/civilengineeringmaterials