Design of Shallow Foundation (Isolated Footings)

Answer :

Aim-

• To know the general design procedure of spread with manual calculation.
• To practice now the input s can be collaborated and arrived as data for calculation.

Steps of footing design-

1. Calculation of Gross soil pressure for footing area.
2. Calculation of net soil pressure for flexure and shear design
3. Footing Depth calculation
4. One way shear check.
5. Two way shear check.

Gross soil pressure & Footing area-

(Unfactored load with surcharge +self-weight of footing)/allowable soil pressure

Unfactored load= (2000/1.5) kN, Soil bearing capacity = 150 N/ Sq m at a depth of 2 mtr.

Assumed self weight and surcharge is 10%

So A = (1333+133)/150 = 9.78 sq m, here we have assumed square footing so

Side of footing = 3.12 m

So we are consider side footing is 3.20 mtr

Net soil pressure –

= Factored load / actual area of footing = 2000/(3.2X3.2) = 195.3 KN/ Sq m

Initial depth of footing-

Assumed depth of footing 500 mm & Checking it for bearing stress as per IS 456 clause 34.4

2400/400 = 6 is not less than 2, so take multiplied factor 2.

In limit state method permissible bearing stress 0.45 X Fck = 0.45 x 25 =11.25 ( Assumed Concrete M25)

So permissible bearing stress multiplied fit area factor =11.25x2 = 22.5 N/Sq mm,

Actual bearing stress at column = load /Area at column =2000/400X400=0.000125 KN/Sq mm =12.5 N/Sq mm

Actual stress is less than permissible stress , so this is ok.

So we can go with 500 mm footing depth for further design checks. Here we assumed 50 mm cover for footing so effective depth d= 450 mm.

Flexural Reinforcement-

Conversion of net pressure into uniformly distribution load = 195.3X3.2 = 625 KN/m

Bending moment = wL^2/2 ( here UDL over a cantilever section)

= 625x1.6X1.6/2 = 800 KN-m

R/F calculation as per IS-456 Ann-G

Assumed fy = 500 N/ Sq mm

800 = 0.87 X 500 X Ast X 450 (1-(Ast X 500)/3200X25) , Ast = 4350 Sq mm.

If we are provided 16 mm dia bar at 150mm C/C so Ast = (Area of bar X Side of Footing/ Spacing)

= (200.96 X 21.33) = 4286 Sq mm this is less than require Ast.

So we are provided 16 mm dia bar at 125 mm C/C so Ast = (200.96 X (3200/125) = 5146 Sq mm

One way shear Check –

Shear Force in the one way = 0.950 X 3.2 X 195.3 = 593.7 KN

Design shear stress at any one way cut face Tv = Vu/bXd

Tv =( 593.7x1000/3200x450)= 0.412 N/Sqmm

Shear Stress capacity of section as per IS-456 2000 Table 19

=0.425 N/ Sqmm

So design shear stress is less than shear capacity so this is ok.

Two way shear check-

Shear force in two way =( (3.2X3.2)-(0.85X0.85))X195.3 = 1859 KN

Design shear stress at two way cut face Tv = (1859 x 1000)/(3200 x 450) = 1.215 N/Sqmm

Shear Stress capacity of section as per IS-456 2000, Clause 31.6.3.1

= 0.25 Fck^1/2 = 1.25 N/sqmm

So design shear stress is less than shear capacity so this is ok.

Sketch of Reinforcement detail-

Here we are use 16 mm dia bar at spacing 125 c/c

Note- we are take here spread footing for uniform depth , If we have take tapered footing or stepped footing, depth reduce d distance and d/2 distance for checking one way shear and two way shear. That is problem with cross check of one way shear or two way shear.