Design of Shallow Foundation (Isolated Footings)

CHALLENGE 12

Aim: The aim for this challenge is to design an isolated footing.

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. Do the following checks:

• If the gross bearing pressure under the footing is within safe bearing capacity
• Provide sufficient depth and longitudinal steel for resisting bending and shear
• Sufficient anchorage for column steel into the footing
• If the bearing pressure at column-footing joint is within 0.45 times the maximum concrete compressive strength

Provide a simple sketch showing the size of the footing, maximum and minimum depth of the section, flexural reinforcement in both directions and anchorage of column steel. 

Ans:

Given Data:

Column size = 400x400

Compression axial load P = 2000 kN

Fck = M25

Fy = Fe500

Soil bearing capacity = 150 kN/m2

Design:

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)

= Pn = 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.

Permissible bearing stress 0.45 x Fck = 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

Mu = 800 kNm

Area of steel for 1m

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

Ast = 1450 mm2

Total Ast required = 4350mm2

No of bars

= Mu / Π/4 x d2.

= 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 = ((3.2-0.4)/2 – 0.45) x 3.2 x 195.3 kN

= 593.7 kN

Shear stress at one way cut face = tv = 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

=P0*(B²-b0²)

b0 = 400+450 = 850mm

= ((3.2 x 3.2) – (0.85 x 0.85)) x 195.3

= 1859 kN

Design shear stress at two way cut face

Tv = 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 16mm dia bars in both directions.
• Provide anchorage for column and footing at 750mm.

 

Vijayalakshmi S H