All about civil construction knowledge- PARAM VISIONS

Bar bending schedule ( BBS ) for columns./ Calculating the cutting length & BBS of column reinforcement.

  BACK 👈PART-1   

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.


sl.    bar                    dia.     no.     length    total      weight       total   
no.                        ( mm)                 (m.)     length     kg/m      weight

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.     

BACK 👈PART-1                                                     CONTINUED 👉PART-3           

Share:

No comments:

Post a Comment

Translate

Blog Archive

popular posts

Recent Posts

Google search