📖 Explanation
τuc=τuc=0.36N/mm2 for bdAst×100=0.250.48N/mm2 for bdAst×100=0.5[For M20][For M20]
Factored SF=45kN=Vu We have to calculate the dia of Fe 500 2-Legged stirrup to be used at a spacing of 325 mm c/c
τv∴%τc=bdVu=230×45045×1000=0.4348N/mm2tensile steel=230×4504×4π(12)2×100=0.437%=0.36+0.250.12×(0.437−0.25)=0.45N/mm2
since τv−τc<0 ⇒ Min shear reinforcement is required ⇒ Min shear reinforcement is given by
bSvAsvAsv=0.87fy0.4=0.87fy0.4×(Sv)(b)
since we limit fy to 415N/mm2 hence,
Asv=2×4π(ϕ)2ϕ adopt ϕ=0.87×4150.4×325×230=82.814mm2=7.26mm=8mm