Design of Shallow Foundation (Isolated Footings)

Aim: To design a square footing with the provided data

Introduction:

The foundation of a building is constructed in order to transfer the superstructure's load to the ground. An isolated footing is one of the most used types of foundation to support single-columns when they are arranged at a long distance.

An isolated footing must be designed to avoid exceeding its bearing capacity and provide safety against overturning or sliding while preventing the ground from settling. Isolated footing must be applied when the mechanical properties of the ground do not drastically change in the foundation area. It consists the financially most beneficial kind of foundation.

Procedure:

Given:

Size of footing = 400 x 400 mm

Soil bearing capacity (SBC) = 150 kN/m^2

Compression axial load = 2000 kN

Concrete = M25

Step 1: Size of footing

Assume self weight of footing (Ww)= 10% of superimposed load = .01 x 2000 = 200 kN

Total load (W)= 2000+200 = 2200 kN

Area of footing (A) = W / SBC = 2200 / 150 = 14.66 kN/m^2

Lets consider L=B, so L= `sqrt 14.66` = 3.82 m

Let's make it 4. So take 4x4 m size of square footing with area of 16m^2.

Step 2:

Net upward pressure (Po) = W*1.5/(Area of footing) = 2200*1.5/16 = 206.25 kN/m^2

Step 3: Depth of footing on the basis of B.M

The max. B.M at the face of column is given by M= {Po* B*(B-b)^2}/{8}= {206.25 * 4*(4-0.4)^2 }/8

= 1336.5 kN/m

Ultiate B.M = 1.5 M =1.5 x 1336.5 = 2004.75 kN/m

Depth of the footing =1336.5 = 0.138 Fck Bd^2

From here , we get d= 381.14

By thumb rule we take 2.5 time higher value for shear check

Depth of footing = 952.85

D= 952.85 + 50 = 1003mm

Step 4: Area of reinforcement

Ast = (0.5*Fck/Fy)*[1- `sqrt{1-(4.6*Mu)/Fck*b*d^2}]*Bd` = 5985.3 mm^2

% of reinforcement = Ast / Bd*100 = 0.157%

Assume 20 mm dia bar as reinforcement

Then Spacing of bar = 200 mm c/c

Step5: One way shear

Critical section located at 'd' distance from the face of column

Shear force = PoB{1/2*(B-b) - d} = 699.6 kN

Ultimate shear force(Vu) = 1.5*shear force = 1.5*699.6 = 1049.4 =1050 kN

As per IS 456:2000

Nominal shear stress = Vu/Bd = 0.27 N/mm^2

Design shear strength of concrete = 0.28 n/mm^2 according to code

Nominal shear is less than design shear so OK

Vu /Bd = k.`tau`

From here we got d = 937 mm < d provided

Step 6 : Two way shear check

Section lies at distance d/2 from column face

Width = 400 + 953 = 1353 mm

Net Shear force acting on perimeter = Po(B^2-b^2) = 449.10 kN

Ultimate shear force =1.5*449.10 = 673.63 kN

Now from clause 31.6.2 IS456:2000

`tau` =0.23 kN/mm^2

Permissible `tau`0.25`sqrt Fck` =1.25

which is safe.

Result: With this our challenge is complete