Let us prepare the bar bending schedule of the column drawing, as shown below.
Given data :
Longitudinal bar dia. d = 16 mm., no. of bars = 4 no.
Lateral ties bar dia. d1= 8 mm., spacing = 250 mm. cover =40 mm.
Column size x = 300mm & y = 230 mm.
Development length Ld = 50d
Length of the longitudinal bar
= [up to ground level + GL to plinth level + plinth level to slab bottom + slab cover + Ld + L- bend in footing - distance from footing bottom.]
= [1200 mm.+ 450 mm. +3000 mm.+ 20 mm + (50d) + 300 mm. - 70 mm.]
=[ 4670 +( 50 × 16mm )+ 300 mm - 70 mm.]
= [5770 mm - 70 mm]
= 5700 mm i.e. 5.70 m.
Length of the lateral ties
=[ perimeter of lateral ties + total hook length - no. of bends]
= [{2sides × ( x - 2 × cover ) + 2 sides × ( y - 2 × cover )} + ( 2nos × hook length) - (3 nos. × bend )]
( Here, we have taken hook length = 10d1 for 135°∠ & bend = 2d1 for 90°∟)
= [ {2 × (300mm - 2× 40mm.) + 2 × ( 230 mm - 2 × 40 mm.) } + ( 2 × 10 × 8mm )- (3 × 2 × 8mm )]
= [ {2 × 220 mm + 2 × 150 mm } + 160 mm - 48 mm.]
= [{440 mm + 300 mm} + 112 mm]
= 852 mm i.e. 0.852 m.
Total number of lateral ties ( stirrups )
={ [length of the longitudinal bar - (Ld + L bend over footing)] ÷ stirrup spacing } + 1
Note: Ld + L bend is deducted from the length as no stirrups are provided over that length.
={[ 5700 mm - (50 × 16 mm + 300 mm.)] ÷ 250 mm.} + 1
= {[ 5700mm - 1100mm ] ÷ 250 mm.} + 1
= {4600 mm ÷ 250 mm.} + 1
= 18.4 + 1
= 19.4 nos.
Rounding off, the number of stirrups required = 20 nos.
Now, let us prepare the BBS table for the column.
1. longitudinal 16 4 5.7 22.8 1.58 36.024
2. lateral 8 20 0.852 17.04 0.395 6.732
Total weight of bars = 42.756 kgs.
Add 5% wastage = 2.137 kgs.
Grand total of rebars = 44.893 kgs.
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