Design of Shallow Foundation (Isolated Footings)
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
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 = 9.77m^2
We assume a square footing.
Therefore,
Square root of 9.77m^2= 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/m^2
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/mm^2
Permissible bearing stress multiplied by the area factor of 2 = 11.25 x 2 = 22.5 N/mm^2
Actual bearing stress at the column base of the footing.
2000 x 1000 / 400x 400 =12.5 N/mm^2
Actual bearing stress < permissible bearing stress.
12.5 N/mm2<22.5N/mm^2
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 mm^2
Total Ast required = 4350mm^2
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/mm^2
Shear capacity check as per IS 456-2000 table 19.
= 0.425 N/mm^2
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/mm^2
Check for shear capacity as per IS 456-2000 cl 31.6.3.1
= 0.25x √fck
= 0.25x √25
= 1.25 N/mm^2
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.