Calculating the quantity of materials in the RCC footings of a compound wall. / Estimating & costing of a compound wall ( part 3 ).

 PART 2    👈 BACK

Let us make a footing of size 1.5 ft × 1.5 ft. having 10" ( 0.833 ft. ) thickness as shown in the drawing.

 

Given data :

Footing length = 1.5 ft., 

Width = 1.5 ft. ,

 Thickness = 0.833 ft.

Rebar diameter = 10 mm.(0.0328 ft.), 

 Spacing =5" (0.416 ft. ) c/c,  

Cover = 2"(0.166 ft ) on all the sides.

From part 1, the number of footings = 14nos. 

4a. The volume of footing for the compound wall

  = [total nos. × length × breadth × thickness]

  = [14 nos. × 1.5 ft. × 1.5 ft. × 0.833 ft.]

  =26.24 cu ft.i.e.0.743 cum. 

4b. BBS for the footings:

No. of bars along the x-axis

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

  = [ {( 1.5 ft.) - ( 2 ×0.166 ft.)} ÷ 0.416 ft.] +1

  = [ { 1.168 ft. } ÷ 0.416 ft. ] +1

  = 2.807 +1 

  = 4 nos.

( By rounding off )

No. of bars along the y-axis

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

   = [ {( 1.5 ft.) - ( 2 ×0.166 ft.)} ÷ 0.416 ft.] +1

  = [ { 1.168 ft. } ÷ 0.416 ft. ] +1

  = 2.807 +1 

  = 4 nos.

( By rounding off )

Cutting length of the bar along the x-axis

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

      ( we have deducted 2 times bar dia i.e. 2d for the  90° bend of the bar. )

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

= [ { 1.5 ft. - 2 × 0.166 ft. } + 2 × { 0.833 ft. - 2 × 0.166 ft. } ] - 2 × ( 2 × 0.0328 ft.)

= [ 1.168 ft. + 1.002 ft. ] - 0.131ft.

= 2.17 ft. - 0.131 ft.

= 2.039 ft. i.e. 0.6214 m.

Cutting length of the bar along the y -axis

=  [ {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. )

= [ { 1.5 ft. - 2 × 0.166 ft. } + 2 × { 0.833 ft. - 2 × 0.166 ft. } ] - 2 × ( 2 × 0.0328 ft.)

= [ 1.168 ft. + 1.002 ft. ] - 0.131ft.

= 2.17 ft. - 0.131 ft.

= 2.039 ft. i.e. 0.6214 m.

Note: The cutting length & number of bars in both ( x & y ) directions will be the same, in the case of square footing having a similar bar diameter. 

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    10    4    0.6214     2.4856     0.62         1.54   

2.   y - axis   10     4     0.6214     2.4856      0.62         1.54

                                            Total weight of the bars = 3.08kgs               

                                             Add 2 % wastage            = 0.0616kgs

                  A grand total of rebar for a footing    =  3.1416 kgs.

Note : Weight of 10mm dia bar /meter is 0.62 kg.

The total weight of the 10mm bar for all the footings 

   = [14 nos. × 3.1416 kgs.]

  = 43.98 kgs.

PART 2    👈 BACK              CONTINUED 👉   PART 4

Share: