## Reinforcement in one way slab

Reinforcement is nothing but the steel rod. So, here we will discuss the quantity and weight of Reinforcement i.e steel rood in the one-way slab.

**Q) Calculate the quantity and weight of steel rod in a one-way slab having length 6 m, width 2.5 m. The diameter of the main bar is 12 mm in diameter with 100 mm c/c spacing and the diameter of the distribution bar is 8 mm with 130mm c/c spacing. The clear cover is 30 mm ( top to bottom ) and the thickness of the slab is 150 mm.**

Solution,

Given that;

Length = 6 m

Width = 2.5 m

Main bar = 12 mm @ 100 mm c/c

Distribution bar = 8 mm @ 130 mm c/c

Clear cover = 30 mm From top to bottom

Thickness = 150 mm

**Step -1 : Calculation of no. of bars**

** ****C**alculate the number of bar required in both main and distribution bar.

Formula = ( Total length – clear cover)/c/c spacing +1

No. of main bar = ( 6000-(30+30))/100 +1

= 5940/100 +1

= 60.4 61 bars

No. of distribution bar = (2500 – (30+30)/130 +1

= 2440/100 +1

= 25.4 26 bars

**Step -2 : Cutting length**

** For main bar:**

** ** Formula: L+(2* Ld) +(1*0.42D)- (2*1d)

Where,

L = Clear span of the slab

Ld = Development length which is equal to 40d ( d is diameter of bar)

0.42D = Inclined length

1d = 45^{0} bends

First we should calculate the length of ‘D’.

D = Thickness – 2* clear cover at top,bottom – dia of bar

= 150- 2*30 -12

D = 78 mm

Putting the values in above equation

Cutting length = 2500 + (2*40*12) +(1*0.42*78) – (2*1*12)

Cutting length = 2500+960 +32.76 – 24

= 3468.76 mm 3469 mm (or3.469 m)

**For distribution bar:**

= clear span + (2* development length)

**= **6000 + ( 2*40*8)

= 6640 mm ( or 6.64 m)

**Conclusion:**

**Main bar:**

Number of bar = 61

Length = 61 * 3.469 m = 211.609 m

**Distribution bar:**

Number of bar = 26

Length = 26 * 6.64 m = 172.64 m

I hope you understood the procedure to calculate reinforcement in one way slab.

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