The Constructor

How to Design a Two-Way Continuous Slab as per Indian Standards?

Neenu S K
Design of a continuous two-way slab

Design of a continuous two-way slab

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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 ly and the shorter span is lx, then for a two-way slab, ly/lx 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:

  1. Simply supported slabs spanning in two directions
  2. 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:

  1. Adopt M20 grade concrete and  Fe500 grade steel
  2. For M20 grade concrete, fck= 20 N/mm2
  3. For Fe500 grade steel, fy= 500 N/mm2
  4. Centre-to-centre distance of longer span, ly= 4850 mm
  5. Centre-to-centre distance of longer span, lx= 3150 mm
  6. Boundary conditions is one short end discontinuous
Figure-1: Continuous two-way slab with one edge discontinuous (Slab 1)

Step 1: Check the Type of Slab

Centre-to-centre distance of longer span, ly= 4850 mm

Centre-to-centre distance of longer span, lx= 3150 mm

 = ly/ lx= 1.54 < 2

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

Step 2: Preliminary dimensioning

  1. Thickness of the slab=lx/32= 98.4 mm
  2. Provide a thickness of D=120 mm
  3. Adopt a clear cover of 20 mm and 8 mm diameter bars
  4. Effective depth=dprovided=120-20-4= 96 mm

Step 3: Load calculation

  1. Dead load of slab=25 x 0.1 = 2.5 kN/m2
  2. Floor finish=1 kN/m2
  3. Live load=3 kN/m2
  4. Total load = 6.5 kN/m2
  5. Factored load, Wu = 1.5 x 6.5 = 9.75 kN/m2

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 (lx) and long span (ly) 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.

Equation-1

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

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

ly/ lx= 1.54

Step 5: Check for Depth

Mu,limit = 0.133 x fck x b x d2 - Equation-2

drequired = 44.7mm

drequired < dprovided

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

Step 6: Reinforcement Calculation

As per IS 456:2000; Annex G

Equation-3

As per IS 456:2000 clause 26.5.2.1

Ast,minimum = 0.12% gross cross-sectional area

                = 0.0012 x 1000 x 120 = 144 mm2

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,

Ast,required = 165 mm2 > Ast,minimum

Assume 8 mm dia bars

(Ast, provided =180 mm2)

Provide 8 mm dia bars @225 mm c/c spacing 

At mid span:

Mu(+ve) = 5.03 kNm

From Equation-3,

Ast,required = 161 mm2 >  Ast,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,

Ast,required = 113 mm2 <  Ast,minimum

Hence provide Ast,minimum = 144 mm2

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,

Ast,required = 66mm2 <  Ast,minimum

Hence provide Ast,minimum = 120 mm2

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/mm2

As per IS 456:2000, Table 19

τv = (15.4 x 1000) / ( 1000 x 96)

τc = 0.343 N/mm2

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

τcmax = 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

Figure-2: Detailing of a Two-way Continuous Slab

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

Design of Two-way Slab by Coefficient Method

Understanding the Transfer of Loads from Slab to Beams

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