A two-way slab is supported on all its four sides by beams. The supports carry the loads along both directions of the slab, therefore it is called a two-way slab. Whereas, a continuous slab extends over three or more support beams in a given direction. These slabs are constructed as a single unit with the beam supports.

If the longer span of the slab is l_{y} and the shorter span is l_{x}, then for a two-way slab, l_{y}/l_{x} is less than 2. If the ratio is greater than 2, it is identified as a one-way slab.

The two-way slabs distribute the loads in both directions. Hence, the reinforcement is provided along shorter and longer sides of the slabs.

A two-way slab can be provided in the following ways:

- Simply supported slabs spanning in two directions
- Continuous two-way slabs

This article explains the design of a continuous two-way slab whose one short edge is discontinuous.

## Example: Design of a Two-Way Continuous Slab

Design a slab with the following details:

- Adopt M20 grade concrete and Fe500 grade steel
- For M20 grade concrete, f
_{ck}= 20 N/mm^{2} - For Fe500 grade steel, f
_{y}= 500 N/mm^{2} - Centre-to-centre distance of longer span, l
_{y}= 4850 mm - Centre-to-centre distance of longer span, l
_{x}= 3150 mm - Boundary conditions is one short end discontinuous

### Step 1: Check the Type of Slab

Centre-to-centre distance of longer span, l_{y}= 4850 mm

Centre-to-centre distance of longer span, l_{x}= 3150 mm

= l_{y}/ l_{x}= 1.54 < 2

Hence, the slab is designed as a two-way slab.

### Step 2: **Preliminary dimensioning**

- Thickness of the slab=l
_{x}/32= 98.4 mm - Provide a thickness of D=120 mm
- Adopt a clear cover of 20 mm and 8 mm diameter bars
- Effective depth=d
_{provided}=120-20-4= 96 mm

### Step 3: **Load calculation**

- Dead load of slab=25 x 0.1 = 2.5 kN/m
^{2} - Floor finish=1 kN/m
^{2} - Live load=3 kN/m
^{2} - Total load = 6.5 kN/m
^{2} - Factored load, Wu = 1.5 x 6.5 = 9.75 kN/m
^{2}

### Step 4: Moment Calculation

For continuous slabs, the support ends possess negative moments and the middle strip is subjected to a positive moment. Hence, negative and positive moments are determined along the short span (l_{x}) and long span (l_{y}) of the slab.

The moments are calculated using the bending moment coefficients given by IS:456-2000. Based on the edge conditions, the coefficients vary. Once the coefficients are determined, the moments are calculated by the formula below, as per IS:456-2000, Annex D-1.1.

The edge condition given for this slab is one short end discontinuous.

**As per Annex D, Table 26 of IS 456: 2000**

l_{y}/ l_{x}= 1.54

### Step 5: Check for Depth

M_{u,limit} = 0.133 x f_{ck} x b x d^{2} - Equation-2

d_{required }= 44.7mm

d_{required }< d_{provided}

Hence, the effective depth selected is sufficient to resist the design ultimate moment.

### Step 6: Reinforcement Calculation

**As per IS 456:2000; Annex G**

**As per IS 456:2000 clause 26.5.2.1**

A_{st,minimum} = 0.12% gross cross-sectional area

= 0.0012 x 1000 x 120 = 144 mm^{2}

**As per IS 456:2000 clause 26.3.3 (b)**

Maximum spacing = (3d or 300 mm) whichever is less

= (3 x 96=288 mm or 300 mm)

= 280 mm

**6.1. Reinforcement Along Shorter direction**

__At supports:__

Mu(-ve) = 6.67 kNm

From Equation-3,

A_{st,required} = 165 mm^{2 }>_{ }A_{st,minimum}

Assume 8 mm dia bars

(Ast, provided =180 mm^{2})

__Provide 8 mm dia bars @225 mm c/c spacing __

__At mid span:__

Mu(+ve) = 5.03 kNm

From Equation-3,

A_{st,required} = 161 mm^{2 }_{> }A_{st,minimum}

Assume 8 mm dia bars

Spacing = 312 mm

__Provide 8 mm dia bars @225 mm c/c spacing__

**6.2. Longer direction**

__At supports:__

Mu(-ve) = 3.58 kNm

From Equation-3,

A_{st,required} = 113 mm^{2 }_{< }A_{st,minimum}

Hence provide A_{st,minimum} = 144 mm^{2}

Assume 8 mm dia bars

Spacing = 349 mm > maximum spacing, hence

__Provide 8mm dia bars @225 mm c/c spacing__

__At mid span:__

Mu (+ve)= 2.71 kNm

From Equation-3,

A_{st,required} = 66mm^{2 }_{< }A_{st,minimum}

Hence provide A_{st,minimum} = 120 mm^{2}

Assume 8 mm dia bars

Spacing = 418mm > maximum spacing, hence

__Provide 8mm dia bars @225 mmc/c spacing__

### Step 7: **Check for shear stress **

**According to IS 456:2000 Cl 40.1, 40.2.3 and 40.2.3.1**

Considering unit width of the slab

V = wl/2 = (9.75 x 3.150 )/2

V=15.4 kN

= 0.160 N/mm^{2}

**As per IS 456:2000, Table 19**

τ_{v} = (15.4 x 1000) / ( 1000 x 96)

τc = 0.343 N/mm^{2}

**As per IS 456:2000, cl. 40.2.1.1**,

k=1.3

k τc = 1.3 x 0.343 = 0.446 > τv

**As per IS 456:2000, Table 20**

τ_{cmax} = 3.1 N/mm2

τv < kτc < τcmax

Hence, the slab is safe against shear.

**Step 8: Check for deflection**

Assuming 1 m width

**According to IS 456:2000, Cl 23.2.1**

**As per IS 456:2000, fig 4**

Modification factor = 1.5

**As per IS 456:2000, clause 23.2.1**

**Hence safe for deflection**

### Step 9: Detailing of Two-Way Slab Design

## FAQs

**What is a two-way slab?**

A two-way slab is a slab that is supported on all its four sides by beams. The supports carry the loads along both directions of the slab. Hence, it is called a two-way slab. If the longer span of the slab is ly and the shorter span is lx, then if it is a two-way slab, ly/lx is less than 2.

**How to provide reinforcement for two-way slab?**

A two-way slab distributes loads along both the direction. Hence, the steel reinforcement is provided in both directions.

**How are two way slab constructed?**

A two-way slab can be constructed either as:

1. Simply supported two-way slabs

2. Continuous two-way slabs

**Read More**