**One Way Slab**

One way slab is provided when the L/B ratio is equal to or greater than two. It is supported only on two opposite sides by rigid supports and it carries the load by flexure in the perpendicular direction to the support.

Uniformly loaded one-way slabs deform into the cylindrical surface and hence bending moment develop only in one direction.

In a one-way slab, Main bars are provided in the shorter span with the crank to counteract positive bending moment and distribution bars are provided in the longer span for distributing the load and to prevent shrinkage cracks.

For calculating the quantity of steel let’s take an example for better understanding.

**Data Given**

We have a one way slab drawing with following data

Length = 6 Meter

Width = 2 Meter

Main Bar = 12 diameter, Spacing – 100 mm c/c

Distribution Bar = 10 mm, Spacing – 125 mm c/c

Extra To Bar = 12 mm, Spacing – 250 mm c/c

Slab Thickness = 150 mm

Clear Cover = 25 mm

Development Length Ld = 40d

**Step 1- Calculation of Numbers of Bars**

Firstly we will calculate the numbers of the Main bar and Distribution bar required.

Numbers of Bars = Total length/ spacing of bar + 1

Numbers of main bar = 6000/100 +1

= 61 numbers

Numbers of Distribution bar = 2000/125 +1

= 17 Numbers

**Step 2 – Calculation of Cutting Length**

Now we will calculate the cutting length of the main bar and distribution bar.

**For Main Bar**

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

Where

L = Clear Span of Slab

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

0.42D = Crank or Inclined length of bar

1d = For 45 bend

**Calculation of inclined Length D**

D = Slab Thickness – Clear cover (top and bottom) – Diameter of Bar

D = 150 – 2 x 25 – 12

D = 88 mm

Cutting Length of Main Bar = 2000 + (2 x 40 x 12) + (0.42 x 88) – 2 x 12

= 2972.96 mm say 2972 mm or 2.972 meter

**Distribution Bar**

Cutting length of distribution Bar = Clear Span (Longer Span) + 2 Ld

= 6000 + ( 2 x 40 x 10 )

= 6800 mm or 6.8 meter

**Step 3 – Top Distribution**

Numbers of Top distribution bar for longer span = {(L/5)/spacing + 1} × 2 sides

= {(2000/4)/125 + 1} × 2

= 10 Numbers

Cutting Length of Top Distribution Bar = Clear Span + 2 Ld

= 6000 + (2 × 40 x 10)

= 6800 mm or 6.80 meter

**Step 4 –** **Extra Top Bar**

Numbers of Extra Top Bars for shorter span= {L -2(L/4)/spacing} × 2 Sides

= {6000 -2 (6000/4)}/250 × 2

= 24 Numbers

Cutting Length of Top Extra Bar = L/4 + Ld

= 2000/4 + 40 × 12

= 980 mm or 0.98 meter

**Step 5 – Calculation of Steel Weight **

Now we have to calculate the weight of steel.

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

= 61 x 2.972 x 0.888

= 160.98 kg

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

= 17 x 6.8 x 0.62

= 71.672 kg

Top Distribution Bar Weight = 10 × 6.8 × 0.617 = 41.95 Kg

Extra Top Bar Weight = 24 × 0.98 × 0.888 = 20.88 Kg

**BBS of One Way Slab**

The **bar bending schedule **of one way slab is as follows.

S.No. | Bar Description | Bar Shape | Nos | Length of Bar (m) | Total Length of Bar (M) | Bar Dia (mm) | Unit Weight (kg) | Total Weight (Kg) |
---|---|---|---|---|---|---|---|---|

1 | Main Bar | 61 | 2.972 | 181.292 | 12 | 0.888 | 160.98 | |

2 | Distribution Bar | 17 | 6.8 | 115.6 | 10 | 0.617 | 71.32 | |

3 | Top Distribution Bar | |||||||

Longer span | 10 | 6.8 | 68 | 10 | 0.617 | 41.95 | ||

4. | Extra Top Bar | |||||||

Shorter span | 24 | 0.98 | 23.52 | 12 | 0.888 | 20.88 |

**Conclusion**

10 mm diameter Steel = 113.27 Kg

12 mm diameter Steel = 181.86 Kg

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