BBS of one way slab and estimation of Steel quantity,Slab bar bending schedule BBS,in this topic we know about BBS of one way slab and estimation of Steel quantity reinforcement used in slab. We know that slab is very important work which transfer load safely to beam and from to column and load acting on column safely to the bed of soil.

When we start our project and Foundation column and brickwork and Beem when it is completed than and most important work to make design of slab and how to make BBS of one way slab and estimation of Steel quantity and how much quantity of reinforcement required.

If you calculate quantity of steel required for slab you will give order and purchase Steel quantity according to requirement, as you know extra amount of Steel get corrosion when laying in open environment that is damage of reinforcement so we should buy steel according to requirement and BBS of one way slab help you to know how much quantity of steel required for your roof slab work and for your project.

in this topic we discuss about BBS of one way slab. Let us discuss about roof slab

**What are types of slab?**

● **Roof slab** :- roof slab is structural design which is put over horizontal surface on brick wall and beam and column. It is horizontal structure made of heavy concrete. In structural designing slab is of three types

1) **flat slab :**– the horizontal structure slab which is supported directly by the column or brick wall is known as flat slab. It has no wall beam, dead load and live load of flat slab directly transfer to column or brick wall and then to foundation of footing and then to bed of soil.

**.2) one way slab** :- the slab which is supported by beams on two opposite sides only in single direction that is one way slab. That’s why one way slab bend in one direction and load acting on it that is dead load and live load distributed only in two opposite sides in single direction.

If the ratio of longer span and shorter span is equal or greater than 2 then we should adopted one way slab. And it will bend in one direction that is longer direction of Span. Mostly one way slab have two parallel beam which carry load acting on it by the slab.

● **3)two way slab** :- in two way slab it is supported by beam on all four sides and load carried in both direction. And distribute the load in all four sides in equal manner.

That’s why in two way slab it’s bend in both direction and resistance is provided from all four sides of beam to withstand with gravitational load acting on it by dead load and live load. In two way slab ratio of longer span and shorter span is less than 2 then we adopted two way slab.

**Steel (reinforcement) used in flat slab**

As we know flat slab is supported directly by the column so it has no use of crank bar. Two types of bar provided for flat slab 1st straight bar having higher dimension used as main bar and provided at bottom of slab in shorter direction.

and 2nd straight bar having lesser dimension used as distribution or cross bar and provided at top of slab over main bar in longer direction.

**Steel (reinforcement) used in one way slab**

As we know there is a two types of bar used in slab one is main bar that is also known as crank bar and second one is distribution bar.

In one way slab main bar that is crank bar is provided in tension zone of slab used only in one direction of shorter span in bottom of slab and cross bar that is distribution bar is straight bar putting over main bar in longer span.

**Steel (reinforcement) used in two way slab**

In two way slab main bar that is crank bar used in bottom of slab in both direction of longer span as well as shorter span. And that is cross bar used in top of slab that is straight bar used over main bar.

**BBS ( bar bending schedule) of one way slab**

In given diagram there is cross sectional area containing x axis along the horizontal direction and the y axis along the vertical direction.

There is two types of Steel provided in one way slab work first is main bar which will be provided along Y axis and second one is distribution bar which will be provided along X axis.

**● in this topic we have to determined following**

1) number of main bar used along y axis

2) number of distribution bar used along x axis

3) cutting length and effective length of single main bar used along y axis

4) cutting length and effective length of single distribution bar used along x axis

5) estimation of reinforcement requirement in BBS of one way slab

Suppose we have given following data

Length of slab (X)= 3000 mm ( x-axis)

Length of slab (Y)= 6000 mm ( y -axis)

Ratio = longer span/shorter span

= 6000/3000 =2 there is a ratio of longer span and shorter span is equal to 2 then we should adopted one way slab.

Diameter of main bar Dm = 10mm

Diameter of distribution bar Dd=8mm

Clear cover at all side C = 25 mm

Thickness of slab H= 150 mm

Spacing between main bar Sm = 125 mm

Spacing between distribution bar Sd= 150 mm

**●1) number of main bar used parallel to x axis and shorter span**

Nm= number of main bar ?

Sm = spacing = 125 mm

Width along Y = 6000 mm

We have formula to calculate number of main bar

Nm =( width_2 cover)/spacing)+1

Nm =( Y_2C/Sm) +1

Nm = (6000mm_2×25)/125mm+1

Nm = (6000_50)/125+1

Nm = 47.6+1 = 48.6 we round of 48.6 is equal to 49

Nm = 49 nos

**●2)number of distribution bar used parallel to Y axis in longer directions**

Nd= number of distribution bar ?

Sd = spacing = 150 mm

Width along X = 3000 mm

We have formula to calculate number of distribution bar

Nd =( length_2cover)/spacing+1

Nd =( X_2C/Sd) +1

Nd = (3000mm_2×50)/150mm+1

Nd = 2950 mm/150mm +1

Nd= 19.66 +1 we round of 19.66 as 20 so

Nd = 20 +1 = 21 nos

**calculate cutting length of single piece of main bar for bbs of one way slab**

Cm1 = cutting length of single piece of main bar

Cm1 = effective length + 2(development length) + 2 (inclined bend length) _ 2 bend

**Note. Actually this in one way slab so they have use of crank bar or bent Up Bar only in one direction and distribution bar and it has provided with developmental length with the the joint section of column and slab**.

Effective length = length of slab along x axis_2 cover

Development length Ld=40d where d is diameter of main bar and if we use M20 grade of concrete and Fe415 grade of Steel.

Inclined bend length = D

Now we calculate value of D

D = thickness _2 cover _ d

Where Thickness=H = 150 mm

Cover at top and bottom =2×25mm

d = diameter of main bar=10mm

Footing all these value we get

D = 150_2×25_10 =90 mm

Bend d = 45° = d =10 mm

We have formula for calculating cutting length of single piece of main bar

Cm1= (length _ 2 cover) +2 Ld + 1×0.42×90_2×10

Where we have

Length of span along X axis = 3000 mm

Cover = 25 mm

Cm1 = (3000 _2×25) +(2× 40×10) + (1×0.42×90 mm) _2×10

Cm1 = (3000_50) + 800mm +37.80 _20

Cm1 = 2950 +800 mm +37.8_20

Cm1 = 3767.8 mm

Cm1 = 3767.8 mm = 3.768 meter

**Calculate total cutting length of main bar for bbs of one way slab**

Cm = total cutting length of main bar

Cm = No.of nos × cutting length of one main bar

Cm = Nm × Cm1

Cm = 49 × 3.768 meter = 184.632 meter

**weight calculation of main bar of bbs of one way slab**

We have formula for the weight calculation of main bar in Kilogram per metre

Weight = (D^2/162)×L

We have Dm = diameter of main bar is 10 mm and length is 110.55 meter

Weight =[ (10×10)/162]×184.632 kg/m

Weight of main bar Wm = 114 kg

**calculate cutting length of single piece of distribution bar of bbs of one way slab**

Cd1 = cutting length of single piece of distribution bar

Cd1 = effective length + 2 development length

Effective length = length of footing_2 cover

Note. Actually this is one way slab so they have use of crank bar or bent Up Bar only in single direction and distribution bar and it has provided with developmental length with the joint section of column and slab.

We have formula for calculating cutting length of single piece of distribution bar

Cd1= ( length of y axis _ 2 cover) + 2Ld

Development length Ld=40d where d is diameter of distribution bar and if we use M20 grade of concrete and Fe415 grade of Steel.

Where we have

Width of slab y = 6000 mm

Cover = 25 mm

Dd=8mm diameter of distribution bar

Cd1 = (6000 _2×25) + 2×40×8mm

Cd1 = (6000_50) + 320 mm

Cd1 = 5950 +320mm

Cd1 = 3950 +320 mm

Cd1 = 6267 mm = 6.267 m

**calculate total cutting length of distribution bar for bbs of one way slab**

Cd = total cutting length of distribution bar

Cd = No.of nos × cutting length of one piece distribution bar

Cd = Nd × Cd1

Cd = 21 × 6.267 meter = 131.67 meter

**●8) weight calculation of distribution bar**

We have formula for the weight calculation of distribution bar in Kilogram per metre

Weight = (D^2/162)×L

We have Dd = diameter of distribution bar is 8 mm and length is 131.67 meter

Weight =[ (8×8)/162]×131.67 kg/m

Weight of distribution bar Wd = 52 kg

**calculate total requirement of reinforcement for the BBS of one way slab**

Total weight = Wm + Wd

Total weight = 114 kg + 52 kg

Total weight = 166 kg

**◆You Can Follow me on Facebook and Subscribe our Youtube Channel**

**You should also visits:-**

**1) calculate self load of column**

**2) calculate self load of beam per metre**

**3) calculate slab load per square metre**

**4) calculate dead load of brick wall per metre**

**5) ultimate load carrying capacity of column**

**● now calculating Thumb Rule**

If we want to know quantity of steel required for 1 square feet

We have dimension= 3000×6000mm

Which is area = 3m ×6 m =18m2

Now converting square metre into square feet

1m2 = 10.764 sq.ft

118m2 = 18×10.764 sqft = 194 sq.ft

194 sq.ft one way slab we need about 166 kg steel

For 1sq.ft = 166/194=0.86kg

For 100 sq.ft we need = 86 kg

1) Thumb Rule for main bar

For 1sq ft = 114/194 =0.59 kg

For 100 sq ft = 59 kg

2) Thumb Rule for distribution bar

For 1sq ft = 52/194 =0.27 kg

For 100 sq ft = 27 kg

Ratio = quantity of distribution bar/ quantity of main bar

Ratio = 27/59 = 1:2.2 it means if you want to buy Main bar 10 mm dia and distribution bar of 8mm dia we keep in mind main bar is just more than double of distribution bar.