Bar Bending Schedule Of Slab – BBS Of Slab

Bar bending schedule or bbs plays a significant role in estimating the quantity of steel for beams, columns, and slab. It helps to find out bar shape, size, length, weight, bending dimension, etc. In this article, I will prepare bar bending schedule of slab with examples.

Slabs are mainly two types one way slab and two way slab. One way slab is a slab that is supported on two opposite sides and carries the load in one direction. The ratio of longer span to shorter span is equal or greater than 2. i.e l/b ≥ 2.

In two way slab, the slab is supported by four sides and loads are carried along with both directions. The ratio of longer span to shorter span is less than 2. i/e l/b < 2. Go through the complete article to get an idea of how to prepare bar bending schedule of slab.

Bar Bending Schedule Of One Way Slab:

Example 1:

bar bending schedule of one way slab

Ly/Lx = Longer span/shorter span = 5000/2000 = 2.5 > 2

It is a one way slab.

Given,

Length of slab = 5000 mm

Width of slab = 2000 mm

Main bar = 12 mm @ 150 mm c/c

Distribution ber = 8 mm @ 150 mm c/c

Clear cover (Top and bottom) = 25 mm

Thickness of slab = 150 mm

Development length = 40d

Where d is dia of bar.

Bar Bending Schedule Of Slab (One Way):

Step 1:

Calculate no. of bars

First calculate number of bars required for main bars and distribution bars.

No. of bars = Length of slab/spacing + 1

No. of main bars = Lx/Spacing + 1 = 5000/150 = 34

No. of distribution bars = Ly/Spacing + 1 = 2000/150 +1 = 14

Step 2:

Calculate cutting length of main bars and distribution bars.

Cutting length of main bar = Clear span of slab + (2 x development length) + Inclined length – (Bend length)

Clear span of slab = 2000 mm

Development length = 40d

Inclined length 0.42d

1d is for every 45 bend

Where d = diameter of bar.

Now Calculate D

D = Thickness of slab – Both side clear cover (top & bottom) – dia of bar

150 – (25+25) -12 = 88 mm

Length of main bar = Ly + 2Ld+ (1 x 0.42d) -(1d x 4)

= 2000 + (2 x 40 x 12) + (2 x 0.42 x 88) – (1 x 12 x 4)

= 2838 mm = 2.838 m

Weight of main bars = d^2 x L/162 x 34 = 12^2 x 2.838/162 = 86 kg

Length Of Distribution Bar:

= Clear span of slab + 2 x development length = Lx + 2 Ld

= 5000 + 2 x 40 x 8 = 5640 mm = 5.64 m

Weight of distribution bars = d^2 x L/162 x 14

= 8^2 x 5.64/162 x 14 = 31 kg

Step 3:

Calculate Top bar (Extra bar) which are provided at the top of critical length (L/4) area.

No. of top bars = {(Ly/4)/spacing +1} x 2 = {(2000/4)150 +1} x 2 = 9

Length of extra bar = Ly – 2 x Ly/4 + 2 x 100 {Here 2 x 100 is for both side lapping of 100 mm for extra bar}

= 2000 – (2 x 2000/4) +200

= 1200 mm = 1.2 m

Weight of extra bar = d^2 x L/162 x 9 = 8^2 x 1.2/162 x 9 = 4.266 kg

Bar Bending Schedule Of Slab (Two Way)

Ly/Lx = Longer span/Shorter span = 5000/3000 = 1.66 < 2

It is a two way slab.

Given,

Length of longer span = 5000 mm

length of shorter span = 3000 mm

Main bar = 12 mm @ 150 mm c/c

Distribution ber = 8 mm @ 150 mm c/c

Clear cover (Top and bottom) = 25 mm

Thickness of slab = 200 mm

Development length = 40d

Where d is dia of bar.

Bar Bending Schedule Of Two Way Slab:

Step 1: For Section A-A

Calculate number of bars required for main bars and distribution bars.

No. of bars = Length of slab/spacing + 1

Total No. of bars = Ly/150 +1 = 4000/150 +1 = 27

No. of main bars = 14

No. of distribution bars = 13

Step 2:

Calculate cutting length of main bars and distribution bars.

Cutting length of main bar = Clear span of slab + (2 x development length) + Inclined length – (Bend length)

Clear span of slab = 5000 mm

Development length= Ld = 40d

Inclined length 0.42d

1d is or every 45° bend

Where d = diameter of bar.

Now Calculate D

D = Thickness of slab – Both side clear cover (top & bottom) – dia of bar

200 – (25+25) -12 = 138 mm

Length of main bar = Lx + (2 x Ld)+ (2 x 0.42d) -(1d x 4)

= 5000 + (2 x 40 x 12) + (2 x 0.42 x 138) – (1 x 12 x 4)

= 2838 mm = 6.02 m

For 14 bars total length = 6.02 x 14 = 84.28 m

Weight of main bars = d^2 x L/162 = 12^2 x 84.28/162 = 75 kg

Cutting Length Of Distribution Bar:

= Clear span of slab + 2 x development length = Lx + 2 Ld

= 5000 + 2 x 40 x 8 = 5640 mm = 5.64 m

Total length of distribution bar = 5.64 x 13 = 73.32

Weight of distribution bars = d^2 x L/162

= 8^2 x 73.32/162 = 29 kg

Step 3:

Calculate Top bar (Extra) which is provided at the top of critical length (L/4) area.

No. of extra bars = {(Ly/4)/spacing +1} x 2 = {(4000/4)/150 +1} x 2 = 16

Length of extra bar = Lx – 2 x Lx/4 + 2 x 100 {Here 2 x 100 is for both side lapping of 100 mm for extra bar}

= 5000 – (2 x 5000/4) + 200

= 2800 mm = 2.7 m

Total length of extra bar = 2.7 x 16 = 43.8 m

Weight of extra bar = d^L/162 = 8^2 x 43.2/162 = 17 kg

Step 1: For Section B-B

Calculate number of bars required for main bars and distribution bars.

Total no. of bars = Length of slab/spacing + 1

Total No. of bars = Lx/150 +1 = 5000/150 +1 = 35

No. of main bars = 18

No. of distribution bars = 17

Step 2:

Calculate cutting length of main bars:

Cutting length of main bar = Clear span of slab + (2 x development length) + (Inclined length) – (Bend length)

Clear span of slab = 4000 mm

Development length= Ld = 40d

Inclined length 0.42d

1d is or every 45° bend

Where d = diameter of bar.

Now Calculate D

D = Thickness of slab – Both side clear cover (top & bottom) – dia of bar

= 200 – (25+25) – 12 = 138 mm

Cutting length of main bar = Ly + (2 x Ld)+ (2 x 0.42d) – (1d x 4)

= 4000 + (2 x 40 x 12) + (2 x 0.42 x 138) – (1 x 12 x 4)

= 5028 mm = 5.02 m

For 18 bars total length = 5.02 x 18 = 90.36 m

Weight of main bars = d^2 x L/162 = 12^2 x 90.36/162 = 80 kg

Cutting Length Of Distribution Bar:

= Clear span of slab + 2 x development length = Ly + 2 x Ld

= 4000 + 2 x 40 x 8 = 4640 mm = 4.64 m

Total length of distribution bar = 4.64 x 17 = 78.88

Weight of distribution bars = d^2 x L/162

= 8^2 x 78.88/162 = 31 kg

Step 3:

Calculate no. of extra bar:

No. of extra bars = {(Lx/4)/spacing +1} x 2 = {(5000/4)150 +1} x 2 = 9

Length of extra bar = Ly – (2 x Ly/4) + (2 x 100) {Here 2 x 100 is for both side lapping of 100 mm for extra bar}

= 4000 – (2 x 4000/4) + 200

= 2200 mm = 2.2 m

Total length of extra bar = 2.2 x 9 = 19.8 m

Weight of extra bar = d^L/162 = 8^2 x 19.8/162 = 8 kg

“Happy Learning”

1 thought on “Bar Bending Schedule Of Slab – BBS Of Slab”

  1. In one way slab cutting formulae is Lx+2Ld+1*0.42d-1d*4.But you derive it as
    =2000+(2*40*12)+(2*0.42*88)-(1*12*4)
    =2000+960+73.92-48
    =2985.92 but you mentioned as 2838
    Also mentioned 1*0.42d as 2*0.42*88 how is it possible??

    Reply

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