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Bar Bending Schedule Of Footing

Bar Bending Schedule Of Footing

Let us prepare the bar bending schedule of the footing for the below-given drawing.

BBS

Given data :

footing length = 2000 mm., width = 1500 mm. , depth = 300 mm.

rebar diameter = 12 mm.  spacing =150 mm, cover = 50 mm on all the sides.

No. of bars along x-direction

= [ {( footing length ) – ( 2 × cover )} ÷ spacing ] + 1

= [ {( 2000 mm.) – ( 2 × 50 mm.)} ÷ 150 mm.] +1

= [ { 1900 mm } ÷ 150 mm. ] +1

= 12.67 +1

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  = 14 nos.

( by rounding off )

No. of bars along y-direction

= [ {( footing width ) – ( 2 × cover )} ÷ spacing ] + 1

= [ {( 1500 mm.) – ( 2 × 50 mm.)} ÷ 150 mm.] +1

= [ { 1400 mm } ÷ 150 mm. ] +1

= 9.33 +1

 = 10 nos.

Cutting length of the bar along the x-direction

=  [ {bar length in x – axis } +{ 2 nos. ×( L- bend length)}] – 2nos.× ( 2 times bar dia. for 90° bend.)

=[ { footing length – 2 × cover } + 2nos.×{ footing height – 2 × cover}] – 2× ( 2 × bar dia. )

= [ { 2000 mm. – 2 × 50 mm. } + 2 × { 300 mm. – 2 × 50 mm. } ] – 2 × ( 2 × 12 mm.)

= [ 1900 mm. + 400 mm ] – 48 mm.

= 2300 mm – 48 mm

= 2252 mm. i.e. 2.252 m.

 

Cutting length of the bar along the y -direction

=  [ {bar length in y – axis } +{ 2 nos. ×( L- bend length)}] – 2nos.× ( 2 times bar dia. for 90° bend.)

=[ { footing width – 2 × cover } + 2nos.×{ footing height – 2 × cover}] – 2× ( 2 × bar dia. )

= [ { 1500 mm. – 2 × 50 mm. } + 2 × { 300 mm. – 2 × 50 mm. } ] – 2 × ( 2 × 12 mm.)

= [ 1400 mm. + 400 mm ] – 48 mm.

= 1800 mm – 48 mm

= 1752 mm. i.e. 1.752 m.

Now, we will prepare BBS of the footing, from calculated data.

sl       bar      dia.   no.   length      total     weight         total     

no.    type    mm.            in m.    length    in kg/m       weight       

 1.    x- axis    12     10    2.252      22.52       0.89          20.043    

2.   y – axis   12      14     1.752      24.528      0.89          21.829

                                            Total weight of the bars = 41.872kgs               

                                             Add 2 % wastage            = 0.837kgs

                  The grand total of rebar for footing    =  42.709 kgs.

 

Note : Weight of 12mm dia bar /meter is 0.89 kg.

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Bar Bending Schedule of the column drawing.

bbs of colmn

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.

 

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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. 


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About Raja Junaid Iqbal

Hi My name is Raja Junaid Iqbal and i am Land Surveyor by Profession but I've specialized in the field of Q,S and land development. As Professional Surveyor I engaged my self in all over the Gulf since 14 years.

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