Design Steps for One 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 one way slab by limit state method(LSM).

Steps :

•  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 , 7 for cantiliver, 26 for continuous.

•Calculate the effective depth of the slab 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

•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 by using the relation.

`M_u=(w_u l^2)/8`

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

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

• Compare `M_u` and `M_(lim)`, if `M_u lt M_(lim)` then Ok 

•Now, calculate the area of steel `A_(st)` by using the relation.

`M_u = 0.87f_ytimesA_(st)d(1-(A_(st)timesf_y)/(f_(ck)timesb) )`

•Determine spacing by assuming suitable diameter bar,

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

Check for maximum spacing,

i) 3d        ii) 300mm

The value of spacing should be minimum of the above.

•Calculate for Distribution Bars

Area of distribution bars = `0.12% of  btimesD`

Check for maximum spacing,

i) 5d         ii) 450mm

The value of spacing should be minimum of above.

•Check for shear

i) Calculate shear force,

`V_u=(w_uL)/2`

ii)Calculate shear stress,

`tau_v=V_u/(bd)`

iii) Calculate the percentage of steel `(P_t)`

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

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

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/d)_(pro)

` by using the relations.

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

\[\frac{l}{d}_{pro}=\frac{Effective\ length\ of\ slab}{Effective\ depth\ of\ slab}\]

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