One Way Slab Design

 

One way Slab Design

Any slab having Longer and Shorter span ratio more than 1.5 will be considered as One way slab which means Shorter will carry the loading shorter side bars will contribute as per their specific bending moments and longer side bars will be used as distribution bars or as temperature bars.

Example

Suppose we have a slab panel sizing 20 ft x 10 ft  having imposed UDL of 150 Lbs/Ft.

Finding Longer and shorter span ratios (20/10 =2 so it is greater value of 1.5 so we will design it as One Way slab)

1) First we will find the roughly Estimated thickness of slab =span/30

= 10/30= 0.333 Ft x12 =  4 Inches

 

So self load of slab = Unit Volume of slab x density of RCC

=  1x1x4/12x150 = 50 Lbs/Sft

Total UDL =      Imposed Load+ Self Load Using estimated Thickness

150+ 50 = 200 Pounds per Sq.Ft

Formulas for Bending Moment Mx= ax.w.lx.lx

Max Bending Moment (Simply supoorted span) = w.L.L./8

 

L =Shorter span

w= UDL (Uniformly distributed Load

So Putting values in Bending Moment formulas 

Max Bending Moment = 200x10x10x1/8x 12 = 30000 Lbs-Inch

In above equation 12 is used to convert Lbs-Ft to Lbs-Inch

As per Elastic Theory Method (Using 1:2:4) concrete

Moment of resistance = Bending Moment

Moment of Resistance in Elastic theory method 

= 184.b.d1.d1

where b is unit width of slab

d1 = Effective depth of slab (depth without covers)

Putting values in above equation 

30000= 184 x 12 x d1 x d1

d1 = 3.72  inch say 3.75

So adding 1/2" concrete cover on top and bottom overall depth will be 4.75 Inches

 

Ast = Bending Moment /Fst x Lever Arm

where Ast = Area of steel in tension zone

Fst = Factored Tensile stress of steel (which is 20,000 Psi) 

Factor of safety is 3 

Lever arm = 0.857 x Effective Depth

Putting Values in Above equation 

Ast = 30000/ 20000x0.857x3.75  = 0.466 Sq.Inch

Spacing of Bars = (Area of bar x 12) /Ast

Using 1/2 inch dia bar 

Spacing = (0.196 x 12 )/0.466= 5.04 say 5 Inch c/c 

So 1/2" dia bar will be used @ 5 inch center to center on shorter side  and slab thickness will be 4.75  inch using (1:2:4) concrete.

Now Designing Distribution bar

Area of Distribution bars = 0 .2 percent of unit cross sectional area of slab

           = 0.20/100 x12x4.75 = 0.114 Sq.Inch

Spacing of Distribution bar = Area bar x 12 /Area of Distribution bars

Using 3/8 inch dia bars spacing = 0.11x12/.114= 11.57 Say 11.5 inch center to center

As per BS Code spacing should not be more than 5 times of effective depth (d1)

= 3.75 x5 = 18.75  so We can use 3/8" diameter bars @ 11.5 inch center to center