Design Steps of Two Way Slab by Limit State Method (LSM)

 Slab:

Slab is defined as two dimensional structural element typically made up of reinforced concrete. It serves as floor, roof and ceiling in a building.

There are two types of slab.

i) One way slab

If the ratio of longer span to the shorter span is greater than 2 then the slab is known as one way slab.

ie. `l_y/l_xgt2`

ii) Two way slab

If the ratio of longer span to the shorter span is lesser than or equal to 2 then the slab is known as two way slab.

ie `l_y/l_xle2`

Here we talk about how we can design a two way slab by limit state method(LSM). There are two conditions for designing two way slab.

1) When the corners are not held down.(Corners are Not Restrained/ Simply supported)

• First check the type of slab by using the formula

`l_(y)/(l_(x)`, if `l_(y)/l_(x) gt 2` then the slab is one way slab.

•Take Basic Value 20 for simply supported slab, 7 for cantiliver, 26 for continuous.

•Calculate the effective depth of the beam by using the relation. `d=l_x/(Basic value times 1.3)`

•Assume effective cover ranging from 20mm to 25mm.

•Find the total depth, `D = d ` + effective cover. Take Always `b=1000`

•Find the effective span of the slab. The value of the effective span should be minimum of below.

-Center to center distance of the support (shorter span)

-Clear Span + effective depth 

•Calculate Loads

-Dead Load : `D times 1 times 25`

-Imposed Load : Given imposed `times 1`

-Load of floor finishes : `0.8 times 1`

•Calculate total load

Total load (`w`) = Dead load + Imposed Load + Load of floor finishes 

•Calculate Factored load

Factored load (`w_u`) = Total load `times` 1.5

•Calculate Bending Moment in both span by using the relation.

`M_(ux)=alpha_xw_u l_x^2`

`M_(uy)=alpha_yw_u l_x^2`

Here the value of `alpha_x` and `alpha_y` is taken from IS 456 Table No. 27 (Page No. 91)

•Calculate `M_(lim)` by using the relation.

`M_(lim)=0.36x_m/d(1-0.42x_m/d)bd^2f_(ck)`

• Compare `M_(ux)` and `M_(uy)` with `M_(lim)`, if `M_(ux)` and `M_(uy) lt M_(lim)` then Ok 

•Now, calculate the area of steel in both span `A_(stx)` and `A_(sty)` by using the relation.

`M_(ux) = 0.87f_ytimesA_(stx)d(1-(A_(stx)timesf_y)/(f_(ck)timesb) )`

`M_(uy) = 0.87f_ytimesA_(sty)d(1-(A_(sty)timesf_y)/(f_(ck)timesb) )`

•Determine spacing by assuming suitable diameter bar,

Spacing `=(1000timesA_(ba r))/A_(stx)`

Spacing ` =(1000timesA_(b ar))/A_(sty)`

Check for maximum spacing,

i) 3d      ii) 300mm

The value of spacing should be minimum of the above.

•Check for shear

i) Calculate shear force,

`V_u=(w_uL)/2`

(Take clear span of shorter length for L)

ii)Calculate Nominal shear stress,

`tau_v=V_u/(bd)`

iii) Calculate the percentage of steel in shorter span `(P_(tx))`

`P_(tx)=(((A_(b ar)times1000)/(Spaci ng))/(bd))times100`

iv) Determine `tau_c` from IS 456 Table No. 19 (Page No. 73) using `P_(tx)`.

If `tau_c gt tau_v` then no shear reinforcement required.

If `tau_c lt tau_v` then provide shear reinforcement and follow the steps.

i) Calculate shear taken by stirrups`V_(us)`,

`V_(us) = V_u - tau_cbd`

ii)Assume a suitable diameter bar (2 legged) and calculate the area,

Area(`A_(sv)`) = `(2piphi^2)/4`

ii)Calculate spacing of stirrups

1)Spacing`(S_v)=(0.87f_yA_(sv)d)/V_us`

2)`A_(sv)/(bS_v)ge(0.4)/(0.87f_y)`

3) `300mm`

4) `0.75d`

Spacing should be minimum of the above.

•Check for deflection

i) Calculate `f_s` by using the relation

`f_s=(A_(st)(required))/(A_(st)(provided))`

ii) Determine the Modification factor`M_f` by using the values of `f_s` (IS 456 Figure No.4, Page No. 38).

iii)Calculate `(l/d)_(max)` and `(l_x/d)_(pro)` by using the relations.

`(l/d)_(max)=20timesMF`

`(l_x/d)_(pro)=(Effective \ l eng th\ of\ slab\ i n\ sho rter\ span)/(Effective\ depth\ of\ slab)`

if `(l/d)_(max)gt(l/d)_(pro)`, then the slab is safe from deflection.

2) When the corners of the slab are held down (Corners are restrained)

The steps for designing the slab when the corners are held down are same as above condition. 

But the value of `alpha_x` and `alpha_y` is taken from IS 456 Table No. 26 (Page No. 91)

In this case, after checking for deflection, corner reinforcement should be provided.

For this,

i)Calculate area of corner reinforcement 

Area of corner reinforcement `A_(cr)=0.75ofA_(stx)`

ii)Calculate spacing for corner reinforcement by sutable dia bar.

Spacing`=(1000timesA_(bar))/A_(cr)`

Notations

`l_(y)` is effective length of shorter span

`l_(x)` is effective length of longer span 

`L` is clear dimension of shorter span

`D ` is Total depth of slab

 `d ` is Effective depth of slab 

`b` is width of slab (Taken 1000mm)

`w` is Total load on slab

`w_u` is factored load 

`M_(ux)` is moment on slab along shorter span

`M_(uy)`is moment on slab along longer span

`M_(lim)` is limiting moment

`A_(stx)` is area of steel in shorter span

`A_(sty)`is area of steel in longer span

`f_(ck)`is stress on concrete

`f_y`is stress on steel

`P_(tx)` is percentage of steel in shorter span

`V_u` is shear force