QUANTITY OF STEEL IN ONE WAY SLAB CALCULATION (BBS)

Before calculating the quantity of steel in one way slab it is very important to clear the basic concepts. The one-way slab steel quantity calculation is done according to its supporting conditions. The one-way slab is supported on two beams and then to the columns of the building.

The bars we use in the Tension zone of the slab is called Bent up bars. Technically in the slab, we give two different names of these bars (Main and distribution bar).




Main Reinforcements bars we provide in the shorter span of the slabs and the bars will along the longer span of the slab. In the long span of the slab the steel we provide known as Distribution steel which doesn’t help to carry any types of load and the only function to distribute the load or counter the shrinkage stresses.

For your better understanding let’s take this example.

EXAMPLE:

Suppose we have a one-way slab, which has a length 5 m or width 2 m (clear span). The Main bars will be 12 mm in diameter with 100 mm c/c spacing. The Distribution bars will be 8 mm in diameter with 125 mm c/c spacing. The Clear cover will be 25 mm (Top or Bottom) and the thickness of the slab is 150 mm.

1. Calculate the quantity of steel?

2. Calculate the weight of steel?

GIVEN DATA.

Length = 5 m (5000 mm).

Width = 2 m (2000 mm).

Main Bar = 12 mm @ 100 mm c/c.

Distribution Bar = 8 mm @ 125 mm c/c.

Clear cover = 25 mm from (Top and Bottom).

Thickness = 150 mm

SOLUTION:




The quantity is done in two steps.

Step 1. (Calculation of Bars No’s)

First, calculate the number of bars required (main and distribution both).

FORMULA = (Total lengthClear cover)/center to center spacing + 1

Main bar      = (5000 – (25+25))/100 + 1

= 4950 Divided by 100 + 1

= 51 Bars.

Distribution bar = (2000 – (25+25))/125 + 1

= 1950 Divided by 125 + 1

= 17 Bars.

Step 2. (Cutting length)

MAIN BAR:

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

# Where

L = Clear Span of the Slab

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

0.42D = Inclined length (Bend length)

1d = 45° bends (d is diameter of bar)

First calculate the length of “D“.



D = (Thickness) – 2 (Clear cover at Top, BOTTOM) – Diameter of the bar.

= 150 – 2(25) -12

D = 88 mm Ans…

By putting Values.

Cutting length = 2000 + (2 x 40 x 12) + (1 x 0.42 x 88) – (2 x 1 x 12)

Cutting length = 2000 + 960 + 36.96 – 24 =2972.96 mm ~ 2973 mm or 2.973 m

DISTRIBUTION BAR:

= Clear Span + (2 x Development Length (Ld))

= 5000 + (2 x 40 x 8) = 5640 mm or 5.64 m

CONCLUSION:

Main bar:

= Numbers 51.

= Length (51 x 2.973 m) = 151.623 m.

=Weight (D^2/162) x length = 134.776 kg.

Distribution Bar:

= Numbers 17.

= Length (17 x 5.64 m) = 95.88 m.

=Weight (D^2/162) x length = 37.87 kg.

Note: The weight of the bar may vary depending upon the properties of the steel.




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