# How to Calculate Quantity of Steel In Two Way Slab|Two Way Slab BBS

## Two Way Slab

slab supported on four adjacent beams with a ratio of longer span to shorter span less than two that carry the load by flexure in both perpendicular directions refers to a two way slab.

If the L/B ( Length/width) ratio is less than two, a two way slab is provided. Uniformly loaded two way slabs deform into a dish or saucer-like shape, so bending moments develop in both directions.

In two-way slabs, main bars are placed in the shorter as well as longer span with the crank to prevent bending.

To learn how to calculate the steel quantity for the two way slab, let’s understand this with an example.

## Given Data

We have a Two way slab drawing with the following data

Length = 5 Meter

Width = 4 Meter

Main Bar = 12 mm, Spacing – 150 mm c/c, Alternate bent up

Distribution Bar = 10 mm, Spacing – 150 mm c/c, Alternate bent up

Slab Thickness = 150 mm

Clear Cover = 20 mm

Development Length (Ld) = 40d

## Step #1 – Bar Numbers

Nos of Bars = Total lengthspacing + 1

Nos of Main bar = 5000/150 +1

= 34 numbers

Nos of Distribution bar = 4000/150 +1

= 27 Numbers

## Step# 2 – Cutting Length

### For Main Bar

Cutting Length = Length (L) + 2 x Ld + (1 x 0.42D) – 2 x 1d

Where

L = Slab Clear Span

Ld = Development Length

0.42D = Inclined or crank length of the bar

1d = For 45° bend

Inclined Length D

D = Thickness of Slab – Top and Bottom cove – Bar Diameter

D = 150 – 2 x 20 – 12

D = 98 mm

Main Bar C L = 4000 + (2 x 40 x 12) + (0.42 x 98) – 2 x 12

= 4977.16 mm say 4977 mm or 4.977 meter

### ForDistribution Bar

Cutting Length = Length (L) + 2 x Ld + (1 x 0.42D) – 2 x 1d

Inclined Length D

D = Thickness of Slab – Top and Bottom cove – Bar Diameter

D = 150 – 2 x 20 – 10

D = 100 mm

Distribution Bar C L = 5000 + (2 x 40 x 10) + (0.42 x 100) – 2 x 10

= 5822.16 mm say 5822 mm or 5.822 meter

## Step# 3 – Top Distribution Bar

### For Shorter Span

Number of extra top bars = {(L/5) / spacing + 1} × 2(for both side)

= {(5000/5) / 150 +1} = 14 numbers

Length of the top bar = Clear Span of Slab + 2Ld

= 4000+(2 × 40 × 12)

= 4000 + 960

= 4960 mm or 4.96 m

### For Longer Span

Number of Top bars = {(L/5) / spacing + 1} × 2(for both side)

= {(4000/5) / 150 + 1} × 2 = 12 Numbers

Length of the top bar = Clear Span of Slab + 2Ld

= 5000+(2 × 40 × 10)

= 5000 + 800

= 5800 mm or 5.8 m

## Step# 4 – The calculation for Extra Top Bar

Extra top Bars are given at the top of the critical length (L/4/L/5) area to counter the bending moment.

### For Shorter Span

Numbers of Extra Top Bars = {(L – 2L/5)/spacing of bar} × 2 Side

= {5000 – 2(5000/5)}/300 × 2

= 20 Numbers

Length of Top Extra Bar = L/5 + Ld = 4000/5 + 40 × 12 = 848 mm or 0.848 meter

### For Longer Span

Numbers of Extra Top Bars = {(L – 2(L/5)/spacing of bar} × 2 Side

= {4000 – 2(4000/5)}/300 × 2

= 16 Numbers

Length of Top Extra Bar = L/5 + Ld = 5000/5 + 40 × 10 = 1040 mm or 1.04 meter

## Step# 5 – Weight of Steel

Main Bar Weight = Nos of bar x Cutting Length x  Bar Unit Weight  (12 mm = 0.888)

= 34 x 4.977 x 0.888

= 150.26 kg

Distribution Bar Weight = Nos of bar x Cutting Length x Bar Unit Weight (10 mm = 0.0.617)

= 27 x 5.822 x 0.617

= 96.98 kg

Weight of Top Distribution bars

For Shorter span = Nos of bar x Cutting Length x Bar Unit Weight (12 mm = 0.888)

= 14 × 4.96 × 0.617

= 42.84 Kg

For longer span = Nos of bar x Cutting Length x Bar Unit Weight (10 mm = 0.617)

= 12 × 5.8 × 0.617

= 42.94 Kg

Weight of Top extra bars

For Shorter span = Nos of bar x Cutting Length x Bar Unit Weight (12 mm = 0.888)

= 20 × 0.848 × 0.888

= 15.06 Kg

For longer span = Nos of bar x Cutting Length x Bar Unit Weight (12 mm = 0.888)

= 16 × 1.04 × 0.617

= 10.26 Kg

## Two Way Slab BBS

The bar bending schedule of two way slab is as follows.

I hope now you know how to calculate steel quantity for a two way slab. If you find the information valuable, please share it.

Thanks!