All about civil construction knowledge- PARAM VISIONS

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 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 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:

1 comment:

  1. This site is very helpful to Civil Engineers, thank for your invaluable efforts.
    Thanks,
    Donga

    ReplyDelete

Please do not enter any spam link in the comment box

Translate

Blog Archive

popular posts

Google search

Recent Posts

Pages