# Solving the Riddle of Concrete Shear Strength Shear ... - SEAoT

Solving the Riddle of Concrete Shear Strength Shear ... - SEAoT

11/5/2009Solving the Riddle of ConcreteShear StrengthShear ResistanceRobert J. Frosch, Ph.D., P.E.Professor of Civil Engineeringd• Beam width, b w• Effective depth, d• Tensile strength,'f cA sb wShear StrengthFRP Reinforcement7619635Sand CoatedwVbdtestf'c432Wrapped and Sand Coated100 1 2 3 4 5 6 7Effective Reinforcement Ratio, Ratio, ρ (%) ρ eff (%)Fabric TextureStress (ksi)FRP Bars140120Aramid (6,850 ksi)100 Steel Glass 1 (5,900 ksi)8060Glass 2 (5,450 ksi)402000 5 10 15 20 25 30 35 40Strain, ε (0.001)Test SetupP aa / d = 3.44’-0” 4’-0”8’-0”d1

11/5/2009V'= 5 f b ccalc c wVcalc= 2f'cbwdV testV calc3.02.52.01.51.00.50.00 1 2 3 4 5 6 7Effective Reinforcement Ratio, ρ eff (%)V testV calc3.02.52.01.51.00.50.00 1 2 3 4 5 6 7Effective Reinforcement Ratio, ρ eff (%)Factors Influencing Strength• Flexural Reinforcement Ratio• Concrete Strength• Member SizeV testV calc3.02.52.01.51.00.5Concrete StrengthV= 5f b c'calc c w0.00 2 4 6 8 10 12 14 16Concrete Compressive Strength (ksi)3.02.5Concrete StrengthV= 2f b d'calc c wMember Depth(Equal A s )V'= 2 f b dc c wV testV calc2.01.51.00.5Beam AA sdBeam BA s2dVV c (Beam B) = 2 V c (Beam A)'= 5 f b cc c wV c (Beam B) = 1.5 V c (Beam A)0.00 2 4 6 8 10 12 14 16Concrete Compressive Strength (ksi)b wb w4

11/5/2009Member Depth(Equal ρ)14Influence of Beam DepthV'= 2 f b dc c w1210Beam AdBeam C2dVV c (Beam C) = 2 V c (Beam A)'= 5 f b cc c wA s2A sV c (Beam C) = 2 V c (Beam A)0Vtestbc fwc'864Lower-bound TrendAverage Trend2b wb w0 5 10 15 20 25 30 35 40 45 50Effective Depth, d (in.)Influence of Beam DepthFlanged SectionsVtestbd fwc'7654345oNAA effNA2100 5 10 15 20 25 30 35 40 45 50Vcalc= 5f'cAeffNAEffective Depth, d (in.)VV testcalc4.54.03.53.02.52.01.51.00.50.0Flanged SectionsVcalc= 2f 'c bwdVcalc0.0 2.0 4.0 6.0 8.0 10.0Reinforcement Ratio, ρ w (%)'c= 5 f AeffFindings• Shear Strength influenced byLongitudinal Stiffness'• V = 5 f b c = 5kccw'f b d• Improved Insight of Shear Resistancecw5

11/5/2009LL =1.5 k/ftV u (k)M u (k-ft)150100500-50-100-1500400800120016002000Design Method ExampleDL = 0.5 k/ft70’d = 35.2”95.1 kM cr =1215 k-ftM ult =1815 k-ft14.9’Proposed Design Equation• Calculate the Neutral Axis Depth– Strain Compatibility•V ci DeterminedV= 5ci c eff•V cw Determinedf′AShear (kips)Shear Strength160Uncracked Cracked140V120n100 V cw = 105 k8060V ci40V ci, ACI20M0cr0 5 10 15 20 25 30 35Distance from Support (ft)Conclusions• Simple Method for Design• Unified Method– Reinforced– Prestressed– Partially Prestressed– Varying Reinforcement (FRP, Steel)V= 5f b c'calc c w8

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