Design of Shallow Foundation (Isolated Footings)

Aim- designing a footing for a column size of 400x400.

Given Data-

Column size = 400x400

Compression axial load P = 2000 KN.

Fck = M25

Fy = Fe500

Soil bearing capacity = 150 kn/m²

Procedure –

Gross Soil Pressure And Footing area

Unfactored load with surcharge + self weight of the footing / allowable soil pressure

10% of the load is taken for self weight.

= (2000/1.5 x 1.1)/150 = 9.77m²

We assume a square footing.

Therefore,

Square root of 9.77m² = 3.12m each side.

Net soil pressure and depth of the footing.

Factored load / actual area of footing.

2000 / (3.2 x3.2)

= P_n = 195.3 kN/m²

Assume the depth of the footing 500 mm.

Check it for bearing stress as per IS 456-2000 clause 34.4.

Therefore,

Permissible bearing stress 0.45 x F_ck = 0.45 x 25 = 11.25 N/mm²

Permissible bearing stress multiplied by the area factor of 2 = 11.25 x 2 = 22.5 N/mm²

Actual bearing stress at the column base of the footing.

2000 x 1000 / 400x 400 =12.5 N/mm²

Actual bearing stress < permissible bearing stress.

12.5 N/mm²<22.5N/mm²

Hence 500 mm depth is okay and provide 50 mm for cover.

Effective depth = 450mm

Bending Moment and flexural reinforcement.

Bending moment due to UDL = wL²/ 2

Where,

w= net pressure into UDL.

w = 195.3 x 3.2 = 652 kN/m

= wL² / 2

= 625 x 1.6² / 2

M_u = 800 kNm

Area of steel for 1m

A_st = 0.5x 25/ 500 x (1- √1- (4.6x 800 x 10⁶)/ 25 x 1000 x 500²) x 1000 x500

A^st = 1450 mm²

Total A_st required = 4350mm²

No of bars

= M_u / Π/4 x d^2.

= 4350 / Π/4 x 16²

= 22 nos

Spacing,

S = L / Nos of bars

S = 3200 / 22

S = 145.45 mm

Therefore,

Provide 22 bars of 16mm dia at 140mm spacing.

Check for One way shear stress

Shear force in the one way area = 0.95 x 3.2 x 195.3 kN

= 593.7 kN

Shear stress at one way cut face = t_v = 593.7 x 1000 / (3200 x 450)

= 0.412 N/mm²

Shear capacity check as per IS 456-2000 table 19.

= 0.425 N/mm2

Design shear stress is less than capacity shear stress.

Hence okay.

Check for two way shear or punching shear.

Shear force in two way

= (3.2 x 3.2) – (0.85 x 0.85) x 195.3

= 1859 kN

Design shear stress at two way cut face

T_{v =} 1859 x 1000 / 3450 x 450

= 1.25 N/mm2

Check for shear capacity as per IS 456-2000 cl 31.6.3.1

= 0.25x √fck

= 0.25x √25

= 1.25 N/mm2

Design shear stress is less than shear capacity hence okay.

RESULT-

Provide a 3.2x3.2m footing with a depth of 500 mm.

Provide reinforcement of 20mm dia bars in both directions.

Provide anchorage for column and footing at 750mm.

AutoCAD design.