# Bar Bending Schedule For One Way Slab [BBS]

### One way Slab

Main Reinforcement bars we provide in the short span of the slabs and the distribution bars will along the longer span of the slab.

For your better understanding, let's take this example. EXAMPLE:

Suppose we have a one way slab, which having a length 5 m or width 2 m (clear span). The Main bars will be 12 mm in diameter @100 mm c/c spacing. The Distribution bars will be 8 mm in diameter @125 mm c/c spacing. The Clear cover will be 25 mm (Top or Bottom) and the thickness of the slab is 150 mm.

Calculate the quantity of steel?

Calculate the weight of steel?

GIVEN DATA:

Length = 5 m (5000 mm)

Width = 2 m (2000 mm)

Main Bar = 12 mm @ 100 mm c/c

Distribution Bar = 8 mm @ 125 mm c/c

Clear cover = 25 mm from (Top and Bottom) Thickness = 150 mm

SOLUTION:

The quantity is done into two steps.

Calculation of number of bars

Cutting length of bars

STEP 1: CALCULATION OF BARS NO'S

First calculate the number of bars required (main and distribution both).

FORMULA:

= (Total length — Clear cover)/center to center spacing + 1

Main bar:

= (5000 - [25+25])/100 + 1

= 4950 /100+1 = 51 Bars

Distribution bar:

= (2000 - + 1

= 1950/125+1

= 17 Bars

STEP 2. CUTTING LENGTHS:

MAIN BAR:

FORMULA

= (L) + (2 ×Ld)+ (1x 0.42D) - (2 x1d)

# Where

= Clear Span of the Slab

Ld = Development Length which is 40 d (where d is diameter of bar)

0.42D = Inclined length (Bend length) 1d = 45° bends (d is diameter of bar)

First calculate the length of "D".

= (Thickness) - 2 (Clear cover at Top, BOTTOM)

— Diameter of the bar

= 150 - 2(25) -12

= 88 mm

By putting Values

Cutting length:

= 2000+(2x40x12) + (1 x0.42 x 88)

- (2x1 x 12)

= 2000 + 960 + 36.96 - 24

= 2972.96 mm 2973 mm or 2.973 m

DISTRIBUTION BAR:

= Clear Span + (2 x Development Length (Ld))

= 5000 + (2 x 40 x 8)

= 5640 mm or 5.64 m Ans

## Bar Bending Schedule for Footing [BBS]

CONCLUSION:

1: Main bar: = 51 No's Length:

= 51 x 2.973 m = 151.623 m Weight:

= (D2/162) x length

= 134.776 kg

2: Distribution Bar: = 17 No's Length:

= 17 x 5.64m = 95.88 m Weight:

= (D2/162) x length

= 37.87 kg

## Format of Bar Bending Schedule as per Code IS:2502-196

Note: The weight of the bar may vary depending upon the properties of the steel.

 `Sr.No.` `Dia. of steel bar` `Weight per meter` `Round Bar (kg)` `Square Bar (kg)` `1` `6 mm` `0.22` `0.28` `2` `8 mm` `0.39` `0.50` `3` `10 mm` `0.62` `0.78` `4` `12 mm` `0.89 ` `1.13` `5` `16 mm` `1.58 ` `2.01` `6` `20 mm` `2.46 ` `3.14` `7` `25 mm` `3.85 ` `4.91` `8` `28 mm` `4.83` `6.15` `9` `32 mm` `6.31 ` `8.04` `10` `36 mm` `7.99 ` `10.17` `11` `40 mm` `9.86` `12.56` `12` `45 mm` `12.49` `15.90` `13` `50 mm` `15.41 ` `19.62`

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