Bar bending schedule (BBS) of the one-way slab./ How to calculate the cutting length & weight of rebar's in one-way slab. ~ PARAM VISIONS

Let us calculate the cutting length & prepare the BBS of the one-way slab as shown below.

Let us redraw the above drawing with dimensions for the calculation purpose as below.

Given data:

Dimension of the slab

Lx = 6000mm

Ly = 2500mm.

Thickness = 150mm.

Main bar = 10mm@ 150 c/c

Distribution bar = 8mm @ 150 c/c

Top extra bar = 8mm @ 150 c/c

Clear cover of slab = 25mm.

Development length Ld = 40d.

Calculation steps:

1. The no. of bars:

a. Main bars:

The no. of main bars

= [Lx ÷ bar spacing ] +1

= [ 6000 ÷ 150] +1

= 40 +1

= 41 nos.

b. Distribution bars:

The no. of distribution bars

= [Ly ÷ bar spacing ] +1

= [ 2500 ÷ 150] +1

= 16.67 +1

= 17.67 nos.

By rounding off = 18 nos.

c. Top extra bars:

The no. of top extra bars on one side

= [{(Ly ÷ 4)÷ bar spacing} +1]

= [{ (2500 ÷ 4) ÷ 150} +1]

= [{ 625 ÷ 150} +1]

= [{ 4.167}] +1]

= 5.167

By rounding of = 6 nos.

The total no. of top extra bars on both sides

= [2 sides × 6nos.]

= 12 nos.

Note: The top extra bars are provided at the Ly/4 span of the main bar on both sides.

2. Cutting length:

a. Main bar:

Cutting length of the main bar

= [clear span +(2nos. × development length) + (extra crank length) - (2nos. × 45° bend)]

= [ 2500mm. + ( 2nos. × Ld) + ( 0.42D ) - ( 2nos. × 1d)]

To know why extra crank length is taken as 0.42D, Go through the article 👇

👀. What is 0.42D in a crank bar (bent-up bar) cutting length formula? - its derivation.

Here, D = [slab thickness - ( top & bottom clear cover) + bar diameter ]

= [150mm - (2nos × 25 mm.) +10mm ]

= [150mm - 60 mm]

D = 90 mm.

Cutting length of the main bar

= [ 2500mm. + ( 2nos. × 40 × 10mm. ) + ( 0.42× 90mm. ) - ( 2nos. × 1× 10mm)]

= [ 2500mm. + 800mm. + 37.8mm. - 20mm. ]

= 3317.80 mm. = 3.318m.

To know, why the bend deduction for 45° bend is taken as 1d, Go through the article 👇

👀. What are bend deductions for different angles in reinforcement bars?

b. Distribution bar:

Cutting length of the distribution bar

= [clear span + (2nos. × development length)]

= [6000mm. + ( 2nos. × 40d )]

= [ 6000mm. + ( 2nos. × 40 × 8mm. )]

= [ 6000mm. + 640mm.]

= 6640 mm. = 6.64m.

c. Top extra bar:

Cutting length of the top extra bar

= cutting length of distribution bar =6640mm.=6.64m.

Let us prepare a BBS table to calculate the weight of the rebars.

Sl. No.	Type of Bar	Dia. in mm.	Nos.	Length in m.	Total length in m.	Weight Kg/m.	Total bar wt. in kg.
1.	Main bar.	10	41	3.318	136.038	0.617	83.94
2.	Distribution bar	8	18	6.64	119.52	0.395	47.21
3.	Top extra bar.	8	12	6.64	79.68	0.395	31.47
4.	Total weight of bars =						162.62
5.	Add 3% wastage =						4.88
6.	The grand total wt. of rebar's =						167.50