# Design of two way slab

The design of the RCC slab classified by the spanning as one way and two-way slab.

We have calculated the one-way slab as per IS: 456-2000. now, we will calculate and the design of a two-way slab.

To calculate the RCC slab manually is a lengthy process, I suggest to use tools, which will save time.

In this article, we will calculate all the necessary steps manually and provide a final reinforcement detail for a two-way slab.

## Two-way slab

As we discussed in previous articles, the ratio of longer to shorter span is less than two and the slab is supported in all the four edges, then the slab is called a two-way slab.

The slab will bend in both the direction.

## Design steps for two-way slab

The steps for designing the two-way slab is similar to the one-way slab design.

(1) Calculate the Effective depth and effective span.

(2) Calculate the total factored load on the slab.

(3) Calculate the mid span moment using the formula given in the Indian standard 456-2000 and Pages 90,91. and, check the effective depth for flexure.

(4) Calculate the steel percentage, area of required steel, and spacing of the bars along the shorter and the longer span and provide the reinforcement detail.

(5) Perform the checks for cracking, deflection, development length, and shear and draw a sketch.

## Two-way slab design calculation

Slab layout |

You can see the S-7 has a longer direction of 13.125 feet and a shorter direction of 12.75 feet in the given plan.

The use of the slab will be a bedroom. so, 2 kilonewtons per square meter will be taken as a live load.

In the design calculation, we will consider a 1-meter strip. which means b= 1000 m.

#### Data

f_{ck = 20}

fy = 415

shorter span, lx = 3.89 m

longer span, ly = 4 m

live load = 2 kN/ mm^{2}

^{}

floor finish = 1 kN/ mm

^{2}### Effective depth

Ix = 3.89 m > 3.5

live load = 2 kN/ mm

^{2}If the shorter span is less than or equal to 3.5 and the live load is less than 3 then the following formula must be multiplied by 0.8

l/d = 20 X M.F ( IS: 456-2000, P.- 38,39)

Assume 0.3 % steel,

fy = 415 N/mm

^{2}M.F = 1.5

3890/d = 20 X 1.5

d = 129.66 mm

Provide,

**d = 135 mm**Provide, d = 135- 10 =

**125 mm**(for longer span)( Assume, 10 mm dia. bars )

Overall depth, D = 135 + 5 + 20

**D = 160 mm**

### Effective span

( IS: 456-2000, P.- 34, CL. 22.2.a)

For lx,

(1) clear span + d = 3890 + 135 = 4025 mm

(2) c/c of support = 3890 + 230 = 4120 mm

Whichever is smaller, 4025

For ly,

(1) clear span + d = 4000 + 125 = 4125 mm

(2) c/c of support = 4000 + 230 = 4230 mm

Whichever is smaller, 4125

Effective span for,

**lx = 4025 mm**

**ly = 4125 mm**

**Load calculation**

(1) Dead load of slab = 0.160 X 25 = 4 kN/ mm

^{2}(2) live load = 2 kN/ mm

^{2}(3) floor finish = 1 kN/ mm

^{2}total load = 4 + 2 + 1 = 7 kN/ mm

^{2}Factored load = 1.5 X 7 = 10.5 kN/ mm

^{2}^{}

^{Mid span moment}

^{}

^{Corners are not held down.}

^{}

^{From, IS: 456-2000, P.- 90, CL.D -2}

^{}

$$M_{x} = \alpha_{x} w l_x^2$$

$$M_{y} = \alpha_{y} w l_x^2$$

ly/lx = 4125/4025 = 1.02

From, IS: 456 - 2000, P.- 91, table- 27

`\alpha_{x}` = 0.064

`\alpha_{y}` = 0.0618

`M_{x} = \alpha_{x}\cdot w l_x^2`

`= 0.064 X 10.5 X 4.025^{2}`

`M_{x}` = 10.88 kN⋅m

`M_{y} = \alpha_{y}\cdot w l_x^2`

`= 0.0618 X 10.5 X 4.025^{2}`

`M_{y}` = 10.51 kN⋅m

### Effective depth for flexure

From, SP. -16, P.- 10, table- c

`M_{u} = 0.138 f_{ck} b d^{2}`

`10.88 X 10^{6} = 0.138 X 20 X 1000 X d^{2}`

d = 62.79 mm (required ) < 135 mm (provided) .....O.K.

### Reinforcement

#### Along lx,

`p_{t} = 50 \frac{f_{ck}}{f_{y}} [ 1 - \sqrt{1- \frac{4.6 M_{x}}{f_{ck} b d^{2}}}]`

`p_{t} = 50 X \frac{20}{415} [ 1 - \sqrt{1- \frac{4.6 X 10.88 X 10^{6}}{20 X 1000 X 135^{2}}}]`

`p_{t}` = 0.172 %

`A_{st} = \frac{p_{t}}{100} b d`

Ast = 0.172/100 X 1000 X 135

`A_{st}` = 232.2 mm

^{2}For 10 mm bars = `a_{st} = \frac{\pi}{4} 10^{2}` = 78.53 mm

spacing = 78.53/232.2 X 1000 = 338.20 mm

Provide, 330 mm spacing.

Ast = 78.53/330 X 1000

Ast = 237.97 mm

^{2}^{}

^{Provide, 10 mm Î¦ 330 mm c/c along the shorter span.}

^{}

^{Along ly,}

^{}

`p_{t} = 50 X \frac{20}{415} [ 1 - \sqrt{1- \frac{4.6 X 10.51 X 10^{6}}{20 X 1000 X 125^{2}}}]`

`p_{t}` = 0.194 %

A_{st} = 0.194/100 X 1000 X 135

`A_{st}` = 242.8 mm

^{2}For 10 mm bars = `a_{st}` = 78.53 mm

spacing = `\frac{78.53}{242.8} X 1000` = 323.43 mm

Provide, 320 mm spacing.

Ast = 78.53/320 X 1000

Ast = 245.4 mm

^{2}^{}

^{Provide, 10 mm Î¦ 320 mm c/c along the longer span.}

^{}

^{Check for cracking}

^{}

^{IS: 456-2000, P.- 46, CL. 26.3.3 (b)}

^{}

^{lx,}

^{}

^{(1) 3d = 3 X 135 = 405 mm}

^{}

^{(2) 300}

^{}

^{135 mm < 300 mm ..... O.K.}

^{}

^{ly,}

^{}

^{(1) 3d = 3 X 125 = 375}

^{}

^{(2) 300 mm }

^{}

^{125 < 300 mm ..... O.K.}

^{}

^{Check for deflection}

^{}

^{Check for deflection can be done along lx,}

^{}

^{Allowable, l/d = 20 X M.F = 20 X 1.5 = 30}

^{}

^{Actual, l/d = 4025/135 = 29.81}

^{}

^{29.81 < 30 ..... O.K.}

^{}

^{Check for development length}

^{}

^{We will consider the longer span because the bond is critical in the longer direction.}

^{}

^{From IS: 456-2000, P. - 44}

^{}

^{`L_{d} \leq 1.3 \frac{M_{1}}{V} + L_{0}`}

^{}

`L_{0}`

(1) d = 125 mm

(2) 12 Î¦ = 12 X 10 = 120 mm

Greater of these two values,

`L_{0}` = 125 mm

50 % bars are bent up near the support,

Ast = 245.4 / 2 = 122.7 mm

^{2}^{}

^{`M_{1} = 0.87 f_{y} A_{st} d (1 - \frac{f_{y} A_{st}}{f_{ck} b d})`}

`= 0.87 X 415 X 122.7 X 125 (1 - \frac{415 X 122.7}{20 X 1000 X 125})`

`M_{1} = 5.42 X 10^{6} N\cdot mm `

S.F at support, `V_{u} = \frac{w l_{x}}{2}`

`= \frac{10.5 X4.025}{2}` =

**21.13 kN**

`1.3 \frac{M_{1}}{V} + L_{0} = 1.3 X \frac{5.42 X 10^{6}}{21.13 X 10^{3}} + 125` = 458.46 mm

From SP. - 16 P.- 184, table - 64,

`L_{d}` =

**453 mm**453 < 458.46 mm ..... O.K.

### Check for shear

This check is also done along the longer span because, the shear critical along ly.

Maximum S.F. = `V_{u}` =

**21.13 kN**

`\tau_{v} = \frac{V_{u}}{bd} = \frac{21.13 X 10^{3}}{1000 X 125} = 0.17` N/mm

^{2}`p_{t} = \frac{A_{st} 100}{bd}`

`= \frac{245.4 X 100}{1000 X 135} = 0.182` %

from, IS: 456-2000, P.- 73, table- 19,

`\tau_{c}` = 0.31 N/mm

^{2}k = 1.27

`\tau_{c}' = k \tau_{c}`

= 1.27 X 0.31

`\tau_{c}'` = 0.393 N/mm

^{2}0.17 < 0.393 ..... O.K.

### Reinforcement

For two way slab, the plan view will look like following,

And the section view,

After the completion of the slab design, we will further move towards the design of columns.

If you have any doubts, you can ask me.

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