Design of Shallow Foundation (Isolated Footings)

AIM=Design a square footing for a column size of 400x400. The compression axial load for the load combination of (1.5 DL + 1.5 LL) is 2000 KN. The safe soil bearing capacity is 150 KN/m2 at a depth of 2 metres below E.G.L. Participants are free to go for either a tapered or stepped footing. Besides the total axial load, also account for the self-weight of the footing and soil above it. Assume M25 concrete grade.

• Steps involved in ooting design 

1. calculation of gross soil pressure for footing area 
2. calculation of Net soil pressure for flexure and shear design
3. Initial depth of footing calculation 
4. Flexural Reinforcement calculation
5. One way shear check
6. Two way shear check

• Unfactored Load with surcharge+self weight of footing /allowable soil pressure

        A  =(2000/1.5 x 1.1)/150

           A= 9.78 M^2

Here 1.5 is the factor use to calculate Factored laod  

For square footing 

B=sqrt 9.78

  B =3.12 m

Hence consider 3.20m sided islolated footing 

• Net soil pressure and intial footing depth

            =Factored load /actual area of footing 

             =  2000/(3.2x3.2)

Pn = 195.3 KN/m^2

Assume depth pf footing 500mm and check it for bearing stress as per IS 456:2000 Clause 34.4

=Sqrt(Footing base bearing area/loaded area) 

 

Permissible bearig stress 0.45 Fck =0.45x25 (M25)

                                                          = 11.25 N/mm^2

permissible bearing stress multiplied with area factor =11.25x2  =22.5 N/mm^2

Actual bearing stress at the column base 

          = (2000x1000)/(400x400)

           = 12.5 N/mm^2

12.5 N/mm^2 actual stress is less than permissible stress 22.5 N/mm^2 Hence OK

We can proceed with depth 500mm footing for further design checks

if we assume 50mm cover for footing the effective depth d shall be taken as (500-50) =450mm

• Flexural Reinforcement 

Conversion of net pressure intpo uniformly distributed load 

  = 195.3x3.2

   = 625 KN/m

Bending moment due to UDL over cantilever section

  M  =WL^2/2

M= (652x1.61.6)/2

M=800 KN-M

Reinforcement Calculation as per the 456:2000

Assume Fy=500 N/mm^2 

After calculation get Ast=5146 mm^2

if we provide 16 mm bar with 125 mm spacing 

than actual rebar provided =(3200/125x201)

Ast provided = 5146mm^2

• One way shear check 

one way shear check at the distance d from face of the column 

Shear force in teh one way shaded area

=0.95x3.2x195.3

= 593.7 KN

Design shear at one way cut face

  = 593.7x1000/(3200x450)

= 0.412 N/mm^2

Shear stress capacity of section as per IS 456:2000

 =0.425N/mm^2 

Design shear stress is less than the shear capacity Hence ok

• Two way shear ( Punching shear ) Check 

In two way shear critical section lies at the distance d/2 from face of the column 

So 450/2  =225mm 

Shear force in the two way shaded area 

= (3.2x3.2)-(0.85x.85)x 195.3

=1859 KN

Design shear stress at two way cut face 

   tau v =1859x1000/(3400x450)

             = 1.215 N/mm^2

Shear stress capacity of the section as per IS 456:2000

=0.25 Sqrt Fck

=0.25 x sqrt 25

=1.25 N/mm^2

Design shear is less than shear capacity (Hence Ok)

DETAILS OF REINFORCEMENT OF THE FOOTING