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Design a short axially loaded square column to carry an axial load of 2250kN. Use M20 concrete and Fe415 steel. Adopt reinforcement details.
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written 8.8 years ago by |
P=2250kN M20 & Fe415
Pu=1.5×2250=3375kN Assume 1% steel of gross c/s 99% concrete of gross c/s Asc=0.01AgAc=0.99AgNow Puc=0.4fckAc+0.67fyAsc3375×103=[0.4×20×0.99Ag+0.67×415×0.01Ag]Ag=315405.82mm2Ag=side2side=561.61≈570mm(Ag)p=570×570=324900mm2Asc=0.01×324900=3249mm2Ascp=8−25mm∅=3928mm2Acp=Agp−Ascp=324900−3928=320972mm2CheckPu=[(0.4fck(Ac)p)+(0.67fy(Asc)p)]Pu=[(0.4×20×320972)+(0.67×415×3928)]Pu=3659.95kN>3375kN Therefore Safe. Lateral Ties ∅ties≥∅main4=254=6.25mm≅8mmSpacingS1= least lateral Dimension =570mmS2=16×∅main=16×25=400mmS3=48×∅ties=48×8=384mmS4=300mm Provided 8mm ∅ ties @ 300 mm c/c