Tiêu đề PSAD 33 - RCD - SRCB.pdf ✅

For the depth of compression block, β1 = 0.85 for fc ′ < 28 MPa, a = β1c a = 0.85(191.176) a = 162.5 mm Equating the internal forces, the steel yields in balanced condition, C = T 0.85fc ′ab = Asfy 0.85(21)(162.5)(300) = Asb(420) Asb = 2071. 875 mm2 Computing the nominal and ultimate capacity of the beam, Mn = (0.85fc ′ab) (d − a 2 ) = 0.85(21)(162.5)(300) (325 − 162.5 2 ) = 212.108 × 106Nmm or 212. 108 kNm ∅ = 0.65 + 0.25 fs − fy 1000 − fy ( reduction factor in transition region ) = 0.65 + 0.25 ⏞ 420 steel stress is fy at balanced condition − 420 1000 − 420 = 0.65 Mu = ∅Mn = 0.65(212.108) = 137. 87 kNm Problem 3. A reinforced concrete beam has a width of 350 mm and an effective depth of 620 mm. It is to support a factored bending moment of 500 kNm. fc ′ = 35 MPa, fy = 345 MPa. Assume φ = 0.90. Use NSCP 2015. 1. Determine the theoretical steel ratio required for the beam to support the design moment. 2. What is the number of 28-mm bars required as tension reinforcement? 3. What is the maximum number of 28-mm bars allowed for singly reinforced sections, per code provisions? Solution: Mu = ∅(0.85fc ′ab) (d − a 2 ) 500 × 106 = 0.9(0.85)(35)(a)(350) (620 − a 2 ) a = 93.036 mm Equating the internal forces, the steel yields since the section is tension controlled (fs > 1000 MPa), C = T 0.85fc ′ab = Asfy 0.85(35)(93.036)(350) = As (345) As = 2807.934 mm2 ρ = As bd = 2807.934 350(620) = 0. 01294 (steel ratio)