Question

For a braced rectangular tied column with bars equally distributed along four sides, using the following design data: Axial dead load Column size 270x270 mm Pu = 981 kN Mu = 42.4 kN.m fc' = 20.7 N/mm², fy = = 414 N/mm2, y =0.6 Bar diameter 16mm with cross-sectional area 200.96 mm2

Answer 1

Given Column size = 270x270mm Pu = 98IKN Mu = 4204 KN-m to = 2017 N/mm² fy = 414 N/mm2 y = 0.6 Bardiameter = 16mm ( 200.96 mm?) Step I : calculation of longitudinal steel We know that Com ) = (42.4x106 = 2.15 270x2702 N bh (981x102 :) = 13.45 270x270 Diameter of longitudinal bars (4) = 16mm Assume diameter of links ($)=8m cover = 30mm d-h-cover-$-$ Assume =(27 270- 30.- 8-16 2 d = 224 mm 6) = (224 0.82 2.70

then 100 Asc)=2 2 from code bh Asc =xbxh) = (2x270x270 = 14.58 mm² Too Number of bars = 7.25 8 1458 206.96 Provide 8H16 as longitudinal reinforcement ( Asc=1607668 mm2) Step I - Calculation of links The diameter of links. is one-Quarter times the diameter of largest longitudinal bar ².e *16) = 4mm but not less than 8mm The spacing of the links is lesser of: (a) 12 times diameter of the smallest longitudinal bar zoe (2x46) = 192 ~200mm (b) the smallest cross-sectional dimension of the column ie 270 mm .. Provide H8 links at 200mm centres. K 270mm 4H16 H8-200 uwolz 4H16 fig:-Reinforcement detailing