Complete slab design as per IS 456 code | structural design | Two way slab design |

In this article I will explain you complete slab design as per IS 456 code by using manual method. Basically slab is the horizontal member which is rests on the beam section. The manual design process included the reinforcement calculation for the main bars and distribution bars as per the considerations and as per the problem statement.

For the present slab design am assuming slab section which is having 6.5m length and 5m width as the dimensions. The beams which are exists below the slab are considered as 0.23mX0.23m dimensions. The super imposed load is taken as 4.5kN/m³, grade of the concrete is taken as M20 and grade of steel is considered as Fe500 grade steel under consideration.

• Beam size =0.23mX0.23m
• Super Imposed load = 4.5kN/m3
• M20 and Fe500

Before design for the slab design the following are the basic steps which is taken as per the IS code provisions. If we follow the 6 steps then it is easy for us to make reinforcement details.

1. Type of slab (one way or two way)
2. Trial depth and effective span
3. Load on the slab
4. Design of BM and check for depth
5. Area of reinforcement
6. Detailing

1. Type of slab 

In this step initially we will determine the type of slab which is related to the one way slab or two way slab system.

If the ratio of L/D less than 2 we will consider slab as two way slab system and if the slab having L/D ratio greater than 2 we will take it as one way slab system. Where L is the length of the slab and D is the width of the slab.

For the present slab section which is having 6.5mX5m dimensions

The L/D =6.5/5 = 1.3<2

So the slab is considered as two way slab which supported with beams of 0.23mX0.23m dimensions

2. Trial depth and effective span in complete slab design 

The trial depth and effective depth is the second step in the slab design as per IS code provisions. Initially approximate depth is calculated for the slab by using d/300 formula. And for calculating the effective span we need to consider equations which are presented in IS code. The effective spans are calculated in X direction and Y direction.

The complete calculation of trial depth and effective span is shown in below

By assuming approximate depth        =       d/300

=       5000/300

=       166mm

And assume D =180 and cover =15mm for mild exposure condition

d        =      180-15-10/2

=       160mm

So take trial depth as 160mm = 0.16m

Now effective span is taken as lesser of two values calculated below

Ly=6.5+0.23 = 6.73m                                  Lx =5+0.23 =5.23.

And Ly=6.5+0.16 = 6.66m                                  Lx =5+0.16 =5.16.

So take ly =6.66 and lx =5.16

Now ly/lx =6.66/5.16 =1.3

So a = 1.3

3. Loads on the slab 

The third step in the slab design is calculation of loads in this case the loads related to the gravity are taken which are mainly related to the dead load and live load condition systems. The sum of the self weight and super imposed loads are taken initially and ultimate loads are calculated by using 1.5 factor of safety.

Self weight of slab             =      0.18*25 = 4.50kN/m²

Super imposed load         =       4.5 kN/m²

Total load                          =       4.5+4.5

=       9kN/m²

Ultimate load                  =        1.5(total load)

=        1.5*9

=       13.5kN/m²

4. Design of BM and check for depth in Complete slab design 

The fourth step in the slab design is calculating the bending moment and check for the depth. The bending moment values calculation part is taken as per Clauses 24.4 and 37.1.2 in page number 90 in IS 456-2000 code.

From pg no 91 IS 456-2000

Mx   =  α_xwlx²

My   =  α_ywly²

For ly/lx =1.3

As per table 26 IS 456-2000

αx=0.079 and αy=0.056

Mx                      =       0.079*13.5*5.16²

=       28.40kNm

And My              =      0.056X13.5X6.66²

=       20.13kNm

So the minimum depth for max BM is calculated as

Mu              =       0.138 bd²

=       0.138X20X1000Xd²

So d = 103mm

However provide d =160mm

5. Area of reinforcement required in Complete slab design 

Area of the reinforcement is the fifth step in the slab design as per this provisions the amount of the steel required Ast is determined in the span along X direction and Y direction for the two way slab system. The complete calculation of reinforcement values are shown in the below steps for the 6.5mX5m slab section.

Steel along shorter span (Mx)

Mux=0.87fyAstd(1-(fyAst/fckbd))

28.40X10⁶=0.87X500XAstX160(1-(500Ast/20X1000X160))

                                                Ast = 438mm²

By using 10mm diameter bars

S                         =       (ast/Ast)X1000

=      ((Π/4)10² /438)/1000

=       179.30mm

Use 175mm

So provide 10mm @ 175C/C distance.

Steel along longer span (My)

Since longer span bars are placed above the shorter span

Then d = 160-10 =150mm

Muy=0.87fyAstd(1-(fyAst/fckbd))

20.13X10⁶=0.87X500XAstX150(1-(500Ast/20X1000X150))

                                                Ast = 327mm²

By using 10mm diameter bars

S       =      (ast/Ast)X1000

=       ((Π/4)10² /327)X1000

=       241mm

Use 240mm

So provide 10mm @ 240C/C distance.

Detailing

The detailing is the last step in the slab reinforcement calculation in this case the area of torsion steel, size of mesh are determined by taking the Ast vales for both X and Y direction condition. Along with this reinforcement values the steel in edge strip is determined by taking 0.12% times the gross area for the steel in slab section. The complete final detailing part calculation for two way slab system is shown below.

Torsion steel

1. Area of torsion steel  =       75XAst(x)

=       0.75X438

=       328mm²

2. By using 8mm diameter bars

S       =      (ast/Ast)X1000

=       ((Π/4)8² /328)X1000

=       153 mm

3. Size of mesh                =       lx/5

=       5160/5

=       1032mm

So provide 8mm @ 150 C/C in both X and Y direction for 1035mm length mesh at top and bottom and the steel in shorter and longer direction is to be provided only in the idle strips.

Since steel in edge strip contain 0.12% gross area

So, steel in edge strip  is (0.12/100)1000X180

= 216mm²

By using 8mm diameter bars

S       =      (ast/Ast)X1000

=       ((Π/4)8² /216)X1000

=       232.70 mm

So provide 8mm dia bars with 230 C/C distance.

The below diagram indicates the full reinforcement detailing of two way slab system as per Is code standards and as per the calculation. It is consisting of 10mm dia bars with 175C/C horizontal reinforcement, 10mm dia bars with 240C/C vertical reinforcement and 8mm dia bars with 230C/C mesh reinforcement.

The complete concepts of slab design by manual method is shown in my YouTube Channel please click here to read.

Conclusions for Slab reinforcement details

Well now the above explained concepts are related to the complete slab reinforcement details calculations for 6.5mx5m slab as per IS 456-2000 code. The reinforcement details are calculated for slab are classified in to six stages type of slab (one way or two way), trial depth and effective span, load on the slab, design of BM and check for depth and Area of reinforcement.

As per the calculation the reinforcement values obtained as 10mm dia bars with 175C/C horizontal reinforcement, 10mm dia bars with 240C/C vertical reinforcement and 8mm dia bars with 230C/C mesh reinforcement.

Please follow our previous posts here

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Volume of concrete per cubic meter click here to read

What is slab? Types of slabs? Difference between one way slab and two way slab ? click here to read

 

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