# 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 slabs. 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 to 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.

Contents

## Bar Bending Schedule Of One Way Slab

### Example 1

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 the 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”

### 3 thoughts 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??