Design of Shallow Foundation (Isolated Footings)

DESIGN OF ISOLATED SHALLOW FOUDATION

Data Available

Column Size = 400 mm x 400 mm

Axial Compressive Load (with Load combination of (1.5DL + 1.5LL)) = 2000 kN

Safe Bearing capacity of soil = 150kN/m2

Depth of foundation below ground level = 2m

Grade of concrete (fck) = 25 N/mm2

Grade of steel fy = 500MPa

We need to check the following for the complete design of isolated footing

• If the Gross bearing pressure under the footing is within safe bearing capacity
• Provide sufficient depth and area of reinforcement for resisiting Bending and Shear
• Sufficient anchorage for column steel into footing.
• If the bearing pressure at column footing joint is within 0.45fck (as per IS 456 cl 31.6.3.1)

Steps need to followed for the Design of isolated footing.

• Calculation of Gross pressure for area of footing
• Calculation of net pressure for internal forces(Flexural and shear design)
• Initial depth of footing
• One way shear check
• Two way shear check

For the calculation of gross pressure we should consider the Unfactored Load(Service Load)

Assume sum of surcharge and self weight of footing as 10% of unfactored load

Unfactored Load = 2000/1.5 = 1334 kN

Area of Footing required = (Unfactored load + Soil surcharge + Self weight of footing)/ Allowable soil pressure (SBC)

= ((2000/1.5) x 1.1) / 150 ( Soil surcharge + Self weight of footing = 10% unfactored load)

= 9.8 m2

Consider square footing so length and width = sqrt (9.8) = 3.12m

Take 3.2 m x 3.2 m as size of footing

Net soil pressure and initial soil depth

Net soil pressure (Pn) = Factored load/ Actual area

= 2000/(3.2x3.2) = 195.3 kN/m2

Assume initial depth of footing as 500 mm and checking it for bearing strength as per cl 34.4 IS 456

Permissible Bearing stress as per cl 34.4 IS 456 is 0.45fck

= 0.45 x 25 = 11.25 N/mm2

But as per cl 34.4 permissible bearing stresses can be incresed by a factor which is a fuction of footing and loading area which is shown below

SQRT(footing base bearing area/Loading area) ≤ 2

= SQRT((2400x2400)/(400x400)) =6 but take 2

Permissible bearing stress = 0.45fck x 2

= 0.45x25x2

= 22.5 N/mm2

actual bearing stress = Factored axial load / loaded area

= 2000x1000/(400x400)

= 12.5 N/mm2 which is less than actual bearing strength (22.5 N/mm2) so safe in bearing

If permissible bearing pressure is less than actual bearing stress, Dowel bars are need to be provided confirming to cl.34.4.1 and cl34.4.4 of IS 456 as shown below

so provided depth of 500mm is ok and used for further checks.

Calculation of flexural reinforcement

Isolated footings can be considered as balanced inverted cantilevers

Coverting net pressure into UDL = 195.3 kN/m2 x 3.2m = 625 kN

Max Bending moment at the support = wl^2/2

= 625((3.2/2)^2)/2

= 800 kNm

calculating reinforcement as per Annex G IS 456

after solving quardratic equation in turmes of ast we get

Ast = 4350 mm2

If we provide 16mm bars @ 150mm spacing area of steel will be

(3200/150) x (0.786x16x16) = 4292mm2 which is not sufficient

So provide 16 mm @ 125mm c/c both ways

Ast provided = (3200/125) (0.786*16*16) = 5146 mm2 , OK

One way shear check

Critical section for one way shear and area to be resisted by this shear as shown below . Critical section lies at a distance of d (effective depth) fron the column face for one way shear

Shear force acting on shaded area = 0.95 x 3.3 x 195.3 kN/m2

= 593.7 kN

One way shear stress = shera force / Area of resisting section (Cut sction)

= 593.7/(450x3200)

= 0.412 N/mm2

Shear stress capacity of the section (depends on area of steel provided)as per IS 456 Table 19

100 As/Bd = 100 x 5146/(3200 x 450)

= 0.35

For M25 grade concrete and value 0f 100 As/Bd , from table 19 by linear interpolation we get shear strength as

0.425 N/mm2 which is greater than actual shear stress hence safe in one wau shear.

Two way shear/ Punching shear check

Critical section for two way shear and area to be resisted by this shear as shown below . Critical section lies at a distance of d/2 (effective depth) fron the column face al around for otwo way shear

Shear force in two way shaded area = ((3200x3200)-(850x850)) x 195.3 kN/m2

= 1859 kN

Design shear stress = 1859000/(850x4)x450

= 1.21 N/mm2

Shear stress capacity as per cl 31.6.3.1 IS 456 '

Ʈ𝒸max = 0.25sqrt fck = 1.25 N/mm2

Shear capacity is Greater than actual shear stress in punching so safe

All Design checks are passed and detailing can be done and send to construction site