# Quick guide to design of one way slab – with IS Code 456:2000

Slabs are those structure which transfers dead load and lives load over large span or area to the beam and column also on the wall sometimes. In this article, I am going to explain to you the design of one way slab step by span. so, read it carefully to get full knowledge about it.

The slab is the member without which, there is no meaning of constructing any structure like a building, water tank, bridge, culvert, etc. Its thickness is considerably smaller than another member of the structure. It may be of different materials like metal or concrete. The slab is frequently used for the construction of different structures like building, bridge, culvert, at top and bottom of the water tank, stair and so on.

### Different types of slab

Depending upon the support condition, the slab may be simply supported, continuous, or cantilever slab. but also according to shape slab may be divided into rectangular, square, trapezoidal, circular, or maybe triangular. in most cases, the slab is rectangular in shape.

By the way additionally there are two types of slab. they are :-

1. One way slab
2. Two way slab

Now, lets us discuss different between one way slab and two way slab.

Now, let us move toward the design of one way slab. Keep in mind, while analyzing or designing slab, we always take the width of the slab as one meter.

### Procedure to design of one way slab

1) At first check weather the slab is one way or two way slab by ly and lx ratio. where ly is length of longer side and ly is length of shorter side.

If lx/ly = >2 It is one way slab.

If lx/ly < 2It is one two slab.

2) Now, assume the depth of the slab so that, it satisfies deflection control criteria.

The basic value of a simply supported slab is 20. ( From clause 23.2.1 of IS code 456:2000, page 37)

So, to satisfy deflection control criteria span to depth ratio should be greater than or equal to 25. for the simply supported slab. It should be taken according to IS Code I will explain later in numerical below the post.

Hence, assume the depth of slab = (Span/25) for simply supported slab and (10 for Cantilever slab)

3) Calculate effective span of slab according to ( clause 22.2 of code IS 456:2000)

a) clear span + effective depth

b) center to center distance of support.

You should choose the effective span of slab according to the above condition whichever is less among (a) and (b).

A clear span is a distance between support without the width of support. and center to center distance is the distance between mid point of support on the opposite side. you can see the picture below.

4) Calculate the total load and hence factored load on the slab.

The factored load is calculated by multiplying the total load on the slab by 1.5 for the safety factor.

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5) Calculate total shear force and bending moment for simply supported slab or cantilever slab if it is.

6) Now check whether the depth of the slab you assumed is adequate or not.

To check this you have to assume the slab is at a balanced section, and you have to apply the formula given below to calculate the required depth of the slab. (From annex G of IS 456:2000)

Where, Mu,lim is calculated moment

fck is strength of concrete.

and Xu,max should be calculated from clause, 38.1 of code IS 456:2000

7) if the assumed depth is more than calculated depth then the depth is adequate otherwise reassume depth and start the process again.

8) If the assumed depth is sufficient then calculate the area of reinforcement required for slab from Annex G of code IS 456:2000.

a) For deign of main bars or longitudinal bars.

where, fy is grade of steel.

Ast is area of steel to be calculated.

b is width of slab.

d is depth of slab.

Assume diameter of bars and calculate spacing of bars.

b) For design of distribution bar/Transverse bar

From clause 26.5.2.1 of code IS 456:2000

Now assume diameter of bars and calculate spacing between bars.

9) Check for shear ( using Table 19 of code IS 456:2000).

10) Check for deflection ( Using clause 23.2.1 of code IS 456:2000 and from figure of IS 456:2000)

#### Now let us have a numerical on it to get a full concept on the procedure to design of one way slab.

##### Q) Design a floor slab for an interior room, with clear dimensions of 3.5 m x 9 m. The slab is resting on 230 mm thick masonry walls. Assume live load as 4 kN/m sq. and dead load to finish, partition, etc…as 1.5 kN/m sq. Use M25 concrete and Fe 415 steel.

Solution:-

Given, Length of Slab (ly) = 9 m

Breadth of slab (lx) = 3.5 m

Width of wall or bearig (b’) = 230 mm

Live load (LL) = 4 kN/ m sq.

fck = 25 N/mm sq.

fy = 415 N/mm sq.

Step-1:- Checking one way or two way slab

Since, (ly/lx) = (9/3.5) = 2.571 > 2, ( This slab is a one way slab), prefere procedure (1)

Step-2:- Assuming depth of slab

Prefere procedure-(2)

For simply supported one way slab,

Assuming, diameter of main bars is 10 mm, and total nominal cover 30 mm,

Then total depth (D) = d + cover + (diameter of bars/2)

= 140 + 30 + (10/2)

= 175 mm

Step-3:- Calculating effective span.

1. clear span + effective depth = lc + d = 3.5 + 0.14 + 3.64 m
2. c/c of support = lc + b’/2 + b’/2 = 3.5 + 0.230 = 3.73 m

Prefering procedure (3),

Effective span = (le) = 4.64 m

Self load (Dead load) of slab = (depth of slab * unit width * unit weight of RCC)

= D x B x yRCC

= 0.175 x 1 x 25 kN/m sq.

= 4.375 kN/m sq.

Dead due to finish = 1.50 kN/m sq.

= 5.875 kN/m sq.

and Live load (LL) = 4 kN/m sq.

Hence, total load = DL + LL

= 5.875 + 5

= 9.875 kN/m sq.

= 1.5 x 9.875

= 14.8125 kN/m sq.

(Prefer procedure (4) )

Step-5:- Calculating shear force and bending moment

Step-6:- Checking depth of slab

Prefering Procedure (6)

Mu, lim = 0.36 x fck x b x (Xu,max) x (d – 0.42 x Xu,max)

But, from clause 38.1 of code IS 456:2000

For, Fe 415 steel,

Xu,max/d = 0.48

Then,

Step-7:- Design of main bars

The depth is greater than required for bending, hence the section is under reinforced.

(Prefering procedure 8.a)

We have,

Hence, we provide

Ast = 517.037 mm sq. (Providing 10 mm diameter, then)

Step-8:Design of distribution bars.

Preferring, procedure (8.b),

(Ast)d = 0.12 % of (b x D)

= (0.12/100) x 1000 x 175

= 210 mm sq.

Providing, 8 mm bars, then

Step-9:- Checking for shear

Nominal shear stress = (Vu/bd)

= (26.9588 x 1000)/(1000 x 140)

= 0.193 N/mm sq.

Percentage of steel (Pt) = ( 524/(1000 x 140) x 100 )

= 0.374 %

Assuming tension steel at support, = 0.374/2 = 0.19%

preferring procedure (9)

Provided shear, 0.318 N/mm sq. > 0.193 N/mm sq.

Hence, the slab is safe in shear.

Step-10:- Checking for deflection

Preffereng procedure (10).

Basic span to depth ratio = 20,

fs = 0.58 x fy x (Area of cross section of steel required/Area of cross section of steel provided)

fs = 0.58 x 415 x (517/524) {( calculate the area of steel provided above = 524)}

fs = 237.48 N/mm sq.

fs = 237.50/mm sq.

From figure 4 of IS 456:2000,

modification facctor for Pt = 0.374 and fs = 237.50 is 1.384

Hence, allowable (L/d) = 1.384 x 20 = 27.68 > 25 assumed.

Hence, slab is safe against deflection.

##### Details Design of Slab

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### 2 thoughts on “Quick guide to design of one way slab – with IS Code 456:2000”

1. Informative post thanks for sharing it.

2. How shear comes out to be 0.318 in shear check?