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
Calculate 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 = 450 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.
Read more,
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Quick guide to design of one way slab with IS Code 456:2000
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